solutions manual foundations of mathematical economics

Solutions Manual

Foundations of Mathematical Economics

Michael Carter

November 15, 2002

Solutions for Foundations of Mathematical Economicsc⃝ 2001 Michael Carter

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Chapter 1: Sets and Spaces

1.1

{ 1, 3, 5, 7 . . .} or {𝑛 ∈ 𝑁 : 𝑛 is odd }1.2 Every 𝑥 ∈ 𝐴 also belongs to 𝐵. Every 𝑥 ∈ 𝐵 also belongs to 𝐴. Hence 𝐴,𝐵 haveprecisely the same elements.

1.3 Examples of finite sets are

∙ the letters of the alphabet {A, B, C, . . . , Z }∙ the set of consumers in an economy

∙ the set of goods in an economy

∙ the set of players in a game.

Examples of infinite sets are

∙ the real numbers ℜ∙ the natural numbers 𝔑

∙ the set of all possible colors

∙ the set of possible prices of copper on the world market

∙ the set of possible temperatures of liquid water.

1.4 𝑆 = { 1, 2, 3, 4, 5, 6 }, 𝐸 = { 2, 4, 6 }.1.5 The player set is 𝑁 = { Jenny,Chris }. Their action spaces are

𝐴𝑖 = {Rock, Scissors,Paper } 𝑖 = Jenny,Chris

1.6 The set of players is 𝑁 = {1, 2, . . . , 𝑛 }. The strategy space of each player is the setof feasible outputs

𝐴𝑖 = { 𝑞𝑖 ∈ ℜ+ : 𝑞𝑖 ≤ 𝑄𝑖 }where 𝑞𝑖 is the output of dam 𝑖.

1.7 The player set is 𝑁 = {1, 2, 3}. There are 23 = 8 coalitions, namely

𝒫(𝑁) = {∅, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}}There are 210 coalitions in a ten player game.

1.8 Assume that 𝑥 ∈ (𝑆 ∪ 𝑇 )𝑐. That is 𝑥 /∈ 𝑆 ∪ 𝑇 . This implies 𝑥 /∈ 𝑆 and 𝑥 /∈ 𝑇 ,or 𝑥 ∈ 𝑆𝑐 and 𝑥 ∈ 𝑇 𝑐. Consequently, 𝑥 ∈ 𝑆𝑐 ∩ 𝑇 𝑐. Conversely, assume 𝑥 ∈ 𝑆𝑐 ∩ 𝑇 𝑐.This implies that 𝑥 ∈ 𝑆𝑐 and 𝑥 ∈ 𝑇 𝑐. Consequently 𝑥 /∈ 𝑆 and 𝑥 /∈ 𝑇 and therefore𝑥 /∈ 𝑆 ∪ 𝑇 . This implies that 𝑥 ∈ (𝑆 ∪ 𝑇 )𝑐. The other identity is proved similarly.

1.9 ∪𝑆∈𝒞𝑆 = 𝑁

∩𝑆∈𝒞𝑆 = ∅

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0-1 1𝑥1

-1

1𝑥2

Figure 1.1: The relation { (𝑥, 𝑦) : 𝑥2 + 𝑦2 = 1 }

1.10 The sample space of a single coin toss is {𝐻,𝑇 }. The set of possible outcomes inthree tosses is the product

{𝐻,𝑇 } × {𝐻,𝑇 } × {𝐻,𝑇 } ={

(𝐻,𝐻,𝐻), (𝐻,𝐻, 𝑇 ), (𝐻,𝑇,𝐻),

(𝐻,𝑇, 𝑇 ), (𝑇,𝐻,𝐻), (𝑇,𝐻, 𝑇 ), (𝑇, 𝑇,𝐻), (𝑇, 𝑇, 𝑇 )}

A typical outcome is the sequence (𝐻,𝐻, 𝑇 ) of two heads followed by a tail.

1.11

𝑌 ∩ ℜ𝑛+ = {0}

where 0 = (0, 0, . . . , 0) is the production plan using no inputs and producing no outputs.To see this, first note that 0 is a feasible production plan. Therefore, 0 ∈ 𝑌 . Also,0 ∈ ℜ𝑛+ and therefore 0 ∈ 𝑌 ∩ ℜ𝑛+.

To show that there is no other feasible production plan in ℜ𝑛+, we assume the contrary.That is, we assume there is some feasible production plan y ∈ ℜ𝑛+ ∖ {0}. This impliesthe existence of a plan producing a positive output with no inputs. This technologicalinfeasible, so that 𝑦 /∈ 𝑌 .

1.12 1. Let x ∈ 𝑉 (𝑦). This implies that (𝑦,−x) ∈ 𝑌 . Let x′ ≥ x. Then (𝑦,−x′) ≤(𝑦,−x) and free disposability implies that (𝑦,−x′) ∈ 𝑌 . Therefore x′ ∈ 𝑉 (𝑦).

2. Again assume x ∈ 𝑉 (𝑦). This implies that (𝑦,−x) ∈ 𝑌 . By free disposal,(𝑦′,−x) ∈ 𝑌 for every 𝑦′ ≤ 𝑦, which implies that x ∈ 𝑉 (𝑦′). 𝑉 (𝑦′) ⊇ 𝑉 (𝑦).

1.13 The domain of “<” is {1, 2} = 𝑋 and the range is {2, 3} ⫋ 𝑌 .

1.14 Figure 1.1.

1.15 The relation “is strictly higher than” is transitive, antisymmetric and asymmetric.It is not complete, reflexive or symmetric.

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1.16 The following table lists their respective properties.

< ≤ =reflexive × √ √transitive

√ √ √symmetric × √ √asymmetric

√ × ×anti-symmetric

√ √ √complete

√ √ ×Note that the properties of symmetry and anti-symmetry are not mutually exclusive.

1.17 Let ∼ be an equivalence relation of a set 𝑋 ∕= ∅. That is, the relation ∼ is reflexive,symmetric and transitive. We first show that every 𝑥 ∈ 𝑋 belongs to some equivalenceclass. Let 𝑎 be any element in 𝑋 and let ∼ (𝑎) be the class of elements equivalent to𝑎, that is

∼(𝑎) ≡ { 𝑥 ∈ 𝑋 : 𝑥 ∼ 𝑎 }Since ∼ is reflexive, 𝑎 ∼ 𝑎 and so 𝑎 ∈ ∼(𝑎). Every 𝑎 ∈ 𝑋 belongs to some equivalenceclass and therefore

𝑋 =∪𝑎∈𝑋∼(𝑎)

Next, we show that the equivalence classes are either disjoint or identical, that is∼(𝑎) ∕= ∼(𝑏) if and only if f∼(𝑎) ∩ ∼(𝑏) = ∅.First, assume ∼(𝑎) ∩ ∼(𝑏) = ∅. Then 𝑎 ∈ ∼(𝑎) but 𝑎 /∈ ∼(𝑏). Therefore ∼(𝑎) ∕= ∼(𝑏).

Conversely, assume ∼(𝑎) ∩ ∼(𝑏) ∕= ∅ and let 𝑥 ∈ ∼(𝑎) ∩ ∼(𝑏). Then 𝑥 ∼ 𝑎 and bysymmetry 𝑎 ∼ 𝑥. Also 𝑥 ∼ 𝑏 and so by transitivity 𝑎 ∼ 𝑏. Let 𝑦 be any elementin ∼(𝑎) so that 𝑦 ∼ 𝑎. Again by transitivity 𝑦 ∼ 𝑏 and therefore 𝑦 ∈ ∼(𝑏). Hence∼(𝑎) ⊆ ∼(𝑏). Similar reasoning implies that ∼(𝑏) ⊆ ∼(𝑎). Therefore ∼(𝑎) = ∼(𝑏).

We conclude that the equivalence classes partition 𝑋 .

1.18 The set of proper coalitions is not a partition of the set of players, since any playercan belong to more than one coalition. For example, player 1 belongs to the coalitions{1}, {1, 2} and so on.

1.19

𝑥 ≻ 𝑦 =⇒ 𝑥 ≿ 𝑦 and 𝑦 ∕≿ 𝑥𝑦 ∼ 𝑧 =⇒ 𝑦 ≿ 𝑧 and 𝑧 ≿ 𝑦

Transitivity of ≿ implies 𝑥 ≿ 𝑧. We need to show that 𝑧 ∕≿ 𝑥. Assume otherwise, thatis assume 𝑧 ≿ 𝑥 This implies 𝑧 ∼ 𝑥 and by transitivity 𝑦 ∼ 𝑥. But this implies that𝑦 ≿ 𝑥 which contradicts the assumption that 𝑥 ≻ 𝑦. Therefore we conclude that 𝑧 ∕≿ 𝑥and therefore 𝑥 ≻ 𝑧. The other result is proved in similar fashion.

1.20 asymmetric Assume 𝑥 ≻ 𝑦.𝑥 ≻ 𝑦 =⇒ 𝑦 ∕≿ 𝑥

while

𝑦 ≻ 𝑥 =⇒ 𝑦 ≿ 𝑥

Therefore

𝑥 ≻ 𝑦 =⇒ 𝑦 ∕≻ 𝑥

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transitive Assume 𝑥 ≻ 𝑦 and 𝑦 ≻ 𝑧.𝑥 ≻ 𝑦 =⇒ 𝑥 ≿ 𝑦 and 𝑦 ∕≿ 𝑥𝑦 ≻ 𝑧 =⇒ 𝑦 ≿ 𝑧 and 𝑧 ∕≿ 𝑦

Since ≿ is transitive, we conclude that 𝑥 ≿ 𝑧.It remains to show that 𝑧 ∕≿ 𝑥. Assume otherwise, that is assume 𝑧 ≿ 𝑥. Weknow that 𝑥 ≿ 𝑦 and transitivity implies that 𝑧 ≿ 𝑦, contrary to the assumptionthat 𝑦 ≻ 𝑧. We conclude that 𝑧 ∕≿ 𝑥 and

𝑥 ≿ 𝑧 and 𝑧 ∕≿ 𝑥 =⇒ 𝑥 ≻ 𝑧This shows that ≻ is transitive.

1.21 reflexive Since ≿ is reflexive, 𝑥 ≿ 𝑥 which implies 𝑥 ∼ 𝑥.transitive Assume 𝑥 ∼ 𝑦 and 𝑦 ∼ 𝑧. Now

𝑥 ∼ 𝑦 ⇐⇒ 𝑥 ≿ 𝑦 and 𝑦 ≿ 𝑥

𝑦 ∼ 𝑧 ⇐⇒ 𝑦 ≿ 𝑧 and 𝑧 ≿ 𝑦

Transitivity of ≿ implies

𝑥 ≿ 𝑦 and 𝑦 ≿ 𝑧 =⇒ 𝑥 ≿ 𝑧

𝑧 ≿ 𝑦 and 𝑦 ≿ 𝑥 =⇒ 𝑧 ≿ 𝑥

Combining

𝑥 ≿ 𝑧 and 𝑧 ≿ 𝑥 =⇒ 𝑥 ∼ 𝑧

symmetric

𝑥 ∼ 𝑦 ⇐⇒ 𝑥 ≿ 𝑦 and 𝑦 ≿ 𝑥⇐⇒ 𝑦 ≿ 𝑥 and 𝑥 ≿ 𝑦⇐⇒ 𝑦 ∼ 𝑥

1.22 reflexive Every integer is a multiple of itself, that is 𝑚 = 1𝑚.

transitive Assume 𝑚 = 𝑘𝑛 and 𝑛 = 𝑙𝑝 where 𝑘, 𝑙 ∈ 𝑁 . Then 𝑚 = 𝑘𝑙𝑝 so that 𝑚 is amultiple of 𝑝.

not symmetric If 𝑚 = 𝑘𝑛, 𝑘 ∈ 𝑁 , then 𝑛 = 1𝑘𝑚 and 𝑘 /∈ 𝑁 . For example, 4 is a

multiple of 2 but 2 is not a multiple of 4.

1.23

[𝑎, 𝑏] = { 𝑎, 𝑦, 𝑏, 𝑧 }(𝑎, 𝑏) = { 𝑦 }

1.24

≿ (𝑦) = {𝑏, 𝑦, 𝑧 }≻ (𝑦) = {𝑏, 𝑧 }≾ (𝑦) = {𝑎, 𝑥, 𝑦 }≺ (𝑦) = {𝑎, 𝑥 }

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1.25 Let 𝑋 be ordered by ≿. 𝑥 ∈ 𝑋 is a minimal element there is no element whichstrictly precedes it, that is there is no element 𝑦 ∈ 𝑋 such that 𝑦 ≺ 𝑥. 𝑥 ∈ 𝑋 is thefirst element if it precedes every other element, that is 𝑥 ≾ 𝑦 for all 𝑦 ∈ 𝑋 .

1.26 The maximal elements of 𝑋 are 𝑏 and 𝑧. The minimal element of 𝑋 is 𝑥. Theseare also best and worst elements respectively.

1.27 Assume that 𝑥 is a best element in 𝑋 ordered by ≿. That is, 𝑥 ≿ 𝑦 for all 𝑦 ∈ 𝑋 .This implies that there is no 𝑦 ∈ 𝑋 which strictly dominates 𝑥. Therefore, 𝑥 is maximalin 𝑋 . In Example 1.23, the numbers 5, 6, 7, 8, 9 are all maximal elements, but none ofthem is a best element.

1.28 Assume that the elements are denoted 𝑥1, 𝑥2, . . . , 𝑥𝑛. We can identify the maximalelement by constructing another list using the following recursive algorithm

𝑎1 = 𝑥1

𝑎𝑖 =

{𝑥𝑖 if 𝑥𝑖 ≻ 𝑎𝑖−1𝑎𝑖−1 otherwise

By construction, there is no 𝑥𝑖 which strictly succedes 𝑎𝑛. 𝑎𝑛 is a maximal element.

1.29

𝑥∗ is maximal ⇐⇒ there does not exist 𝑥 ≻ 𝑥∗

that is

≻(𝑥∗) = { 𝑥 : 𝑥 ≻ 𝑥∗ } = ∅

𝑥∗ is best ⇐⇒ 𝑥∗ ≿ 𝑥 for every 𝑥 ∈ 𝑋⇐⇒ 𝑥 ≾ 𝑥∗ for every 𝑥 ∈ 𝑋

That is, every 𝑥 ∈ 𝑋 belongs to ≾(𝑥∗) or ≾(𝑥∗) = 𝑋 .

1.30 Let 𝐴 be a nonempty set of a set 𝑋 ordered by ≿. 𝑥 ∈ 𝑋 is a lower bound for𝐴 if it precedes every element in 𝐴, that is 𝑥 ≾ 𝑎 for all 𝑎 ∈ 𝐴. It is a greatest lowerbound if it dominates every lower bound, that is 𝑥 ≿ 𝑦 for every lower bound 𝑦 of 𝐴.

1.31 Any multiple of 60 is an upper bound for 𝐴. Thus, the set of upper bounds of 𝐴is {60, 120, 240, . . .}. The least upper bound of 𝐴 is 60. The only lower bound is 1,hence it is the greatest lower bound.

1.32 The least upper bounds of interval [𝑎, 𝑏] are 𝑏 and 𝑧. The least upper bound of(𝑎, 𝑏) is 𝑦.

1.33

𝑥 is an upper bound of 𝐴 ⇐⇒ 𝑥 ≿ 𝑎 for every 𝑎 ∈ 𝐴⇐⇒ 𝑎 ≾ 𝑥 for every 𝑎 ∈ 𝐴⇐⇒ 𝐴 ⊆ ≾(𝑥)

Similarly

𝑥 is a lower bound of 𝐴 ⇐⇒ 𝑥 ≾ 𝑎 for every 𝑎 ∈ 𝐴⇐⇒ 𝑎 ≿ 𝑥 for every 𝑎 ∈ 𝐴⇐⇒ 𝐴 ⊆ ≿(𝑥)

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1.34 For every 𝑥 ∈ ℜ2,

𝑥 ≻ 𝑦 if 𝑥1 > 𝑦1 or 𝑥1 = 𝑦1 and 𝑥2 > 𝑦2

Since all elements 𝑥 ∈ ℜ2 are comparable, ≻ is complete; it is a total order.

1.35 Assume ≿𝑖 is complete for every 𝑖. Then for every 𝑥, 𝑦 ∈ 𝑋 and for all 𝑖 =1, 2, . . . , 𝑛, either 𝑥𝑖 ≿𝑖 𝑦𝑖 or 𝑦𝑖 ≿𝑖 𝑥𝑖 or both. Either

𝑥𝑖 ∼𝑖 𝑦𝑖 for all 𝑖 Then define 𝑥 ∼ 𝑦.𝑥𝑖 ∕∼𝑖 𝑦𝑖 for some 𝑖 Let 𝑘 be the first individual with a strict preference, that is 𝑘 =

min𝑖(𝑥𝑖 ∕∼ 𝑦𝑖). (Completeness of ≿𝑖 ensures that 𝑘 is defined). Then define

𝑥 ≻ 𝑦 if 𝑥𝑘 ≻𝑖 𝑦𝑘

𝑦 ≻ 𝑥 otherwise

1.36 Let 𝑆, 𝑇 and 𝑈 be subsets of a finite set 𝑋 . Set inclusion ⊆ is

reflexive since 𝑆 ⊆ 𝑆.

transitive since 𝑆 ⊆ 𝑇 and 𝑇 ⊆ 𝑈 implies 𝑆 ⊆ 𝑈 .

anti-symmetric since 𝑆 ⊆ 𝑇 and 𝑇 ⊆ 𝑆 implies 𝑆 = 𝑇

Therefore ⊆ is a partial order.

1.37 Assume 𝑥 and 𝑦 are both least upper bounds of 𝐴. That is 𝑥 ≿ 𝑎 for all 𝑎 ∈ 𝐴and 𝑦 ≿ 𝑎 for all 𝑎 ∈ 𝐴. Further, if 𝑥 is a least upper bound, 𝑦 ≿ 𝑥. If 𝑦 is a leastupper bound, 𝑥 ≿ 𝑦. By anti-symmetry, 𝑥 = 𝑦.

1.38

𝑥 ∼ 𝑦 =⇒ 𝑥 ≿ 𝑦 and 𝑦 ≿ 𝑥

which implies that 𝑥 = 𝑦 by antisymmetry. Each equivalence class

∼ (𝑥) = { 𝑦 ∈ 𝑋 : 𝑦 ∼ 𝑥 }

comprises just a single element 𝑥.

1.39 max𝒫(𝑋) = 𝑋 and min𝒫(𝑋) = ∅.1.40 The subset {2, 4, 8} forms a chain. More generally, the set of integer powers of agiven number {𝑛, 𝑛2, 𝑛3, . . . } forms a chain.

1.41 Assume 𝑥 and 𝑦 are maximal elements of the chain 𝐴. Then 𝑥 ≿ 𝑎 for all 𝑎 ∈ 𝐴and in particular 𝑥 ≿ 𝑦. Similarly, 𝑦 ≿ 𝑎 for all 𝑎 ∈ 𝐴 and in particular 𝑦 ≿ 𝑥. Since≿ is anti-symmetric, 𝑥 = 𝑦.

1.42 1. By assumption, for every 𝑡 ∈ 𝑇 ∖𝑊 , ≺(𝑡) is a nonempty finite chain. Hence,it has a unique maximal element, 𝑝(𝑡).

2. Let 𝑡 be any node. Either 𝑡 is an initial node or 𝑡 has a unique predecessor 𝑝(𝑡).Either 𝑝(𝑡) is an initial node, or it has a unique predecessor 𝑝(𝑝(𝑡)). Continuingin this way, we trace out a unique path from 𝑡 back to an initial node. We canbe sure of eventually reaching an initial node since 𝑇 is finite.

1.43

(1, 2) ∨ (3, 1) = (3, 2) and (1, 2) ∧ (3, 2) = (1, 2)

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1.44 1. 𝑥∨𝑦 is an upper bound for { 𝑥, 𝑦 }, that is x∨y ≿ 𝑥 and x∨y ≿ 𝑦. Similarly,𝑥 ∨ 𝑦 is a lower bound for { 𝑥, 𝑦 }.

2. Assume 𝑥 ≿ 𝑦. Then 𝑥 is an upper bound for { 𝑥, 𝑦 }, that is 𝑥 ≿ 𝑥 ∨ 𝑦. If 𝑏 isany upper bound for { 𝑥, 𝑦 }, then 𝑏 ≿ 𝑥. Therefore, 𝑥 is the least upper boundfor { 𝑥, 𝑦 }. Similarly, 𝑦 is a lower bound for { 𝑥, 𝑦 }, and is greater than anyother lower bound. Conversely, assume 𝑥∨ 𝑦 = 𝑥. Then 𝑥 is an upper bound for{ 𝑥, 𝑦 }, that is 𝑥 ≿ 𝑦.

3. Using the preceding equivalence

𝑥 ≿ 𝑥 ∧ 𝑦 =⇒ 𝑥 ∨ (𝑥 ∧ 𝑦) = 𝑥

𝑥 ∨ 𝑦 ≿ 𝑥 =⇒ (𝑥 ∨ 𝑦) ∧ 𝑥 = 𝑥

1.45 A chain 𝑋 is a complete partially ordered set. For every 𝑥, 𝑦 ∈ 𝑋 with 𝑥 ∕= 𝑦,either 𝑥 ≻ 𝑦 or 𝑦 ≻ 𝑥. Therefore, define the meet and join by

𝑥 ∧ 𝑦 =

{𝑦 if 𝑥 ≻ 𝑦𝑥 if 𝑦 ≻ 𝑥

𝑥 ∨ 𝑦 =

{𝑥 if 𝑥 ≻ 𝑦𝑦 if 𝑦 ≻ 𝑥

𝑋 is a lattice with these operations.

1.46 Assume 𝑋1 and 𝑋2 are lattices, and let 𝑋 = 𝑋1×𝑋2. Consider any two elementsx = (𝑥1, 𝑥2) and y = (𝑦1, 𝑦2) in 𝑋 . Since 𝑋1 and 𝑋2 are lattices, 𝑏1 = 𝑥1 ∨ 𝑦1 ∈ 𝑋1and 𝑏2 = 𝑥2 ∨ 𝑦2 ∈ 𝑋2, so that b = (𝑏1, 𝑏2) = (𝑥1 ∨ 𝑦1, 𝑥2 ∨ 𝑦2) ∈ 𝑋 . Furthermoreb ≿ x and b ≿ y in the natural product order, so that b is an upper bound for the{x,y}. Every upper bound b = (��1, ��2) of {x,y} must have 𝑏𝑖 ≿𝑖 𝑥𝑖 and 𝑏𝑖 ≿𝑖 𝑦𝑖,

so that b ≿ b. Therefore, b is the least upper bound of {x,y}, that is b = x ∨ y.Similarly, x ∧ y = (𝑥1 ∧ 𝑦1, 𝑥2 ∧ 𝑦2).1.47 Let 𝑆 be a subset of 𝑋 and let

𝑆∗ = { 𝑥 ∈ 𝑋 : 𝑥 ≿ 𝑠 for every 𝑠 ∈ 𝑆 }

be the set of upper bounds of 𝑆. Then 𝑥∗ ∈ 𝑆∗ ∕= ∅. By assumption, 𝑆∗ has a greatestlower bound 𝑏. Since every 𝑠 ∈ 𝑆 is a lower bound of 𝑆∗, 𝑏 ≿ 𝑠 for every 𝑠 ∈ 𝑆.Therefore 𝑏 is an upper bound of 𝑆. Furthermore, 𝑏 is the least upper bound of 𝑆,since 𝑏 ≾ 𝑥 for every 𝑥 ∈ 𝑆∗. This establishes that every subset of 𝑋 also has a leastupper bound. In particular, every pair of elements has a least upper and a greatestlower bound. Consequently 𝑋 is a complete lattice.

1.48 Without loss of generality, we will prove the closed interval case. Let [𝑎, 𝑏] be aninterval in a lattice 𝐿. Recall that 𝑎 = inf[𝑎, 𝑏] and 𝑏 = sup[𝑎, 𝑏]. Choose any 𝑥, 𝑦 in[𝑎, 𝑏] ⊆ 𝐿. Since 𝐿 is a lattice, 𝑥 ∨ 𝑦 ∈ 𝐿 and

𝑥 ∨ 𝑦 = sup{ 𝑥, 𝑦 } ≾ 𝑏

Therefore 𝑥 ∨ 𝑦 ∈ [𝑎, 𝑏]. Similarly, 𝑥 ∧ 𝑦 ∈ [𝑎, 𝑏]. [𝑎, 𝑏] is a lattice. Similarly, for anysubset 𝑆 ⊆ [𝑎, 𝑏] ⊆ 𝐿, sup𝑆 ∈ 𝐿 if 𝐿 is complete. Also, sup𝑆 ≾ 𝑏 = sup[𝑎, 𝑏]. Thereforesup𝑆 ∈ [𝑎, 𝑏]. Similarly inf 𝑆 ∈ [𝑎, 𝑏] so that [𝑎, 𝑏] is complete.

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1.49 1. The strong set order ≿𝑆 is

antisymmetric Let 𝑆1, 𝑆2 ⊆ 𝑋 with 𝑆1 ≿𝑆 𝑆2 and 𝑆2 ≿𝑆 𝑆1. Choose 𝑥1 ∈ 𝑆1and 𝑥2 ∈ 𝑆2. Since 𝑆1 ≿𝑆 𝑆2, 𝑥1 ∨ 𝑥2 ∈ 𝑆1 and 𝑥1 ∧ 𝑥2 ∈ 𝑆2. On the otherhand, since 𝑆2 ≿ 𝑆1, 𝑥1 = (𝑥1 ∨ (𝑥1 ∧ 𝑥2) ∈ 𝑆2 and 𝑥2 = 𝑥2 ∧ (𝑥1 ∨ 𝑥2) ∈ 𝑆1(Exercise 1.44. Therefore 𝑆1 = 𝑆2 and ≿𝑆 is antisymmetric.

transitive Let 𝑆1, 𝑆2, 𝑆3 ⊆ 𝑋 with 𝑆1 ≿𝑆 𝑆2 and 𝑆2 ≿𝑆 𝑆3. Choose 𝑥1 ∈ 𝑆1,𝑥2 ∈ 𝑆2 and 𝑥3 ∈ 𝑆3. Since 𝑆1 ≿𝑆 𝑆2 and 𝑆2 ≿𝑆 𝑆3, 𝑥1 ∨ 𝑥2 and 𝑥2 ∧ 𝑥3are in 𝑆2. Therefore 𝑦2 = 𝑥1 ∨ (𝑥2 ∧ 𝑥3) ∈ 𝑆2 which implies

𝑥1 ∨ 𝑥3 = 𝑥1 ∨((𝑥2 ∧ 𝑥3) ∨ 𝑥3

)=

(𝑥1 ∨ (𝑥2 ∧ 𝑥3)

) ∨ 𝑥3= 𝑦2 ∨ 𝑥3 ∈ 𝑆3

since 𝑆2 ≿𝑆 𝑆3. Similarly 𝑧2 = (𝑥1 ∨ 𝑥2) ∧ 𝑥3 ∈ 𝑆2 and

𝑥1 ∧ 𝑥3 =(𝑥1 ∧ (𝑥1 ∨ 𝑥2)

) ∧ 𝑥3= 𝑥1 ∧

((𝑥1 ∨ 𝑥2) ∧ 𝑥3

)= 𝑥1 ∧ 𝑧2 ∈ 𝑆1

Therefore, 𝑆1 ≿𝑆 𝑆3.

2. 𝑆 ≿𝑆 𝑆 if and only if, for every 𝑥1, 𝑥2 ∈ 𝑆, 𝑥1 ∨ 𝑥2 ∈ 𝑆 and 𝑥1 ∧ 𝑥2 ∈ 𝑆, whichis the case if and only if 𝑆 is a sublattice.

3. Let 𝐿(𝑋) denote the set of all sublattices of 𝑋 . We have shown that ≿𝑆 isreflexive, transitive and antisymmetric on 𝐿(𝑋). Hence, it is a partial order on𝐿(𝑋).

1.50 Assume 𝑆1 ≿𝑆 𝑆2. For any 𝑥1 ∈ 𝑆1 and 𝑥2 ∈ 𝑆2, 𝑥1 ∨ 𝑥2 ∈ 𝑆1 and 𝑥1 ∧ 𝑥2 ∈ 𝑆2.Therefore

sup𝑆1 ≿ 𝑥1 ∨ 𝑥2 ≿ 𝑥2 for every 𝑥2 ∈ 𝑆2which implies that sup𝑆1 ≿ sup𝑆2. Similarly

inf 𝑆2 ≾ 𝑥1 ∧ 𝑥2 ≾ 𝑥1 for every 𝑥1 ∈ 𝑆1which implies that inf 𝑆2 ≾ inf 𝑆1. Note that completeness ensures the existence ofsup𝑆 and inf 𝑆 respectively.

1.51 An argument analogous to the preceding exercise establishes =⇒ . (Complete-ness is not required, since for any interval 𝑎 = inf[𝑎, 𝑏] and 𝑏 = sup[𝑎, 𝑏]).

To establish the converse, assume that 𝑆1 = [𝑎1, 𝑏1] and 𝑆2 = [𝑎2, 𝑏2]. Consider any𝑥1 ∈ 𝑆1 and 𝑥2 ∈ 𝑆2. There are two cases.

Case 1. 𝑥1 ≿ 𝑥2 Since 𝑋 is a chain, 𝑥1 ∨ 𝑥2 = 𝑥1 ∈ 𝑆1. 𝑥1 ∧ 𝑥2 = 𝑥2 ∈ 𝑆2.Case 2. 𝑥1 ≺ 𝑥2 Since 𝑋 is a chain, 𝑥1 ∨ 𝑥2 = 𝑥2. Now 𝑎1 ≾ 𝑥1 ≺ 𝑥2 ≾ 𝑏2 ≾ 𝑏2.

Therefore, 𝑥2 = 𝑥1 ∨ 𝑥2 ∈ 𝑆1. Similarly 𝑎2 ≾ 𝑎1 ≾ 𝑥1 ≺ 𝑥2 ≾ 𝑏2. Therefore𝑥1 ∧ 𝑥2 = 𝑥1 ∈ 𝑆2.

We have shown that 𝑆1 ≿𝑆 𝑆2 in both cases.

1.52 Assume that ≿ is a complete relation on 𝑋 . This means that for every 𝑥, 𝑦 ∈ 𝑋 ,either 𝑥 ≿ 𝑦 or 𝑦 ≿ 𝑥. In particular, letting 𝑥 = 𝑦, 𝑥 ≿ 𝑥 for 𝑥 ∈ 𝑋 . ≿ is reflexive.

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1.53 Anti-symmetry implies that each indifference class contains a single element. If theconsumer’s preference relation was anti-symmetric, there would be no baskets of goodsbetween which the consumer was indifferent. Each indifference curve which consist asingle point.

1.54 We previously showed (Exercise 1.27) that every best element is maximal. Toprove the converse, assume that 𝑥 is maximal in the weakly ordered set 𝑋 . We have toshow that 𝑥 ≿ 𝑦 for all 𝑦 ∈ 𝑋 . Assume otherwise, that is assume there is some 𝑦 ∈ 𝑋for which 𝑥 ∕≿ 𝑦. Since ≿ is complete, this implies that 𝑦 ≻ 𝑥 which contradicts theassumption that 𝑥 is maximal. Hence we conclude that 𝑥 ≿ 𝑦 for 𝑦 ∈ 𝑋 and 𝑥 is abest element.

1.55 False. A chain has at most one maximal element (Exercise 1.41). Here, uniquenessis ensured by anti-symmetry. A weakly ordered set in which the order is not anti-symmetric may have multiple maximal and best elements. For example, 𝑎 and 𝑏 areboth best elements in the weakly ordered set {𝑎 ∼ 𝑏 ≻ 𝑐}.1.56 1. For every 𝑥 ∈ 𝑋 , either 𝑥 ≿ 𝑦 =⇒ 𝑥 ∈ ≿(𝑦) or 𝑦 ≿ 𝑥 =⇒ 𝑥 ∈ ≾(𝑦) since

≿ is complete. Consequently, ≿(𝑦) ∪ ≺(𝑦) = 𝑋 If 𝑥 ∈ ≿(𝑦) ∩ ≾(𝑦), then 𝑥 ≿ 𝑦and 𝑦 ≿ 𝑥 so that 𝑥 ∼ 𝑦 and 𝑥 ∈ 𝐼𝑦 .

2. For every 𝑥 ∈ 𝑋 , either 𝑥 ≿ 𝑦 =⇒ 𝑥 ∈ ≿(𝑦) or 𝑦 ≻ 𝑥 =⇒ 𝑥 ∈ ≺(𝑦) since ≿ iscomplete. Consequently, ≿(𝑦) ∪≺(𝑦) = 𝑋 and ≿(𝑦) ∩≺(𝑦) = ∅.

3. For every 𝑦 ∈ 𝑋 , ≻(𝑦) and 𝐼𝑦 partition ≿(𝑦) and therefore ≻(𝑦), 𝐼𝑦 and ≺(𝑦)partition 𝑋 .

1.57 Assume 𝑥 ≿ 𝑦 and 𝑧 ∈ ≿(𝑥). Then 𝑧 ≿ 𝑥 ≿ 𝑦 by transitivity. Therefore 𝑧 ∈ ≿(𝑦).This shows that ≿(𝑥) ⊆ ≿(𝑦).

Similarly, assume 𝑥 ≻ 𝑦 and 𝑧 ∈ ≻(𝑥). Then 𝑧 ≻ 𝑥 ≻ 𝑦 by transitivity. Therefore𝑧 ∈ ≻(𝑦). This shows that ≿(𝑥) ⊆ ≿(𝑦). To show that ≿(𝑥) ∕= ≿(𝑦), observe that𝑥 ∈ ≻(𝑦) but that 𝑥 /∈ ≻(𝑥)

1.58 Every finite ordered set has a least one maximal element (Exercise 1.28).

1.59 Kreps (1990, p.323), Luenberger (1995, p.170) and Mas-Colell et al. (1995, p.313)adopt the weak Pareto order, whereas Varian (1992, p.323) distinguishes the two or-ders. Osborne and Rubinstein (1994, p.7) also distinguish the two orders, utilizing theweak order in defining the core (Chapter 13) but the strong Pareto order in the Nashbargaining solution (Chapter 15).

1.60 Assume that a group 𝑆 is decisive over 𝑥, 𝑦 ∈ 𝑋 . Let 𝑎, 𝑏 ∈ 𝑋 be two other states.We have to show that 𝑆 is decisive over 𝑎 and 𝑏. Without loss of generality, assumefor all individuals 𝑎 ≿𝑖 𝑥 and 𝑦 ≿𝑖 𝑏. Then, the Pareto order implies that 𝑎 ≻ 𝑥 and𝑦 ≻ 𝑏.Assume that for every 𝑖 ∈ 𝑆, 𝑥 ≿𝑖 𝑦. Since 𝑆 is decisive over 𝑥 and 𝑦, the socialorder ranks 𝑥 ≿ 𝑦. By transitivity, 𝑎 ≿ 𝑏. By IIA, this holds irrespective of individualpreferences on other alternatives. Hence, 𝑆 is decisive over 𝑎 and 𝑏.

1.61 Assume that 𝑆 is decisive. Let 𝑥, 𝑦 and 𝑧 be any three alternatives and assume𝑥 ≿ 𝑦 for every 𝑖 ∈ 𝑆. Partition 𝑆 into two subgroups 𝑆1 and 𝑆2 so that

𝑥 ≿𝑖 𝑧 for every 𝑖 ∈ 𝑆1 and 𝑧 ≿𝑖 𝑦 for every 𝑖 ∈ 𝑆2Since 𝑆 is decisive, 𝑥 ≿ 𝑦. By completeness, either

𝑥 ≿ 𝑧 in which case 𝑆1 is decisive over 𝑥 and 𝑧. By the field expansion lemma (Exercise1.60), 𝑆1 is decisive.

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𝑧 ≻ 𝑥 which implies 𝑧 ≿ 𝑦. In this case, 𝑆2 is decisive over 𝑦 and 𝑧, and therefore(Exercise 1.60) decisive.

1.62 Assume ≻ is a social order which is Pareto and satisfies Independence of IrrelevantAlternatives. By the Pareto principle, the whole group is decisive over any pair ofalternatives. By the previous exercise, some proper subgroup is decisive. Continuingin this way, we eventually arrive at a decisive subgroup of one individual. By theField Expansion Lemma (Exercise 1.60), that individual is decisive over every pair ofalternatives. That is, the individual is a dictator.

1.63 Assume 𝐴 is decisive over 𝑥 and 𝑦 and 𝐵 is decisive over 𝑤 and 𝑧. That is, assume

𝑥 ≻𝐴 𝑦 =⇒ 𝑥 ≻ 𝑦𝑤 ≻𝐵 𝑧 =⇒ 𝑤 ≻ 𝑧

Also assume

𝑦 ≿𝑖 𝑤 for every 𝑖

𝑧 ≿𝑖 𝑥 for every 𝑖

This implies that 𝑦 ≿ 𝑤 and 𝑧 ≿ 𝑥 (Pareto principle). Combining these preferences,transitivity implies that

𝑥 ≻ 𝑦 ≿ 𝑤 ≻ 𝑧which contradicts the assumption that 𝑧 ≿ 𝑥. Therefore, the implied social ordering isintransitive.

1.64 Assume 𝑥 ∈ core. In particular this implies that there does not exist any 𝑦 ∈ 𝑊 (𝑁)such that 𝑦 ≻ 𝑥. Therefore 𝑥 ∈ Pareto.

1.65 No state will accept a cost share which exceeds what it can achieve on its own, sothat if 𝑥 ∈ core then

𝑥𝐴𝑃 ≤ 1870

𝑥𝑇𝑁 ≤ 5330

𝑥𝐴𝑃 ≤ 860

Similarly, the combined share of the two states AP and TN should not exceed 6990,which they could achieve by proceeding without KM, that is

𝑥𝐴𝑃 + 𝑥𝑇𝑁 ≤ 6990

Similarly

𝑥𝐴𝑃 + 𝑥𝐾𝑀 ≤ 1960

𝑥𝑇𝑁 + 𝑥𝐾𝑀 ≤ 5020

Finally, the sum of the shares should equal the total cost

𝑥𝐴𝑃 + 𝑥𝑇𝑁 + 𝑥𝐾𝑀 = 6530

The core is the set of all allocations of the total cost which satisfy the precedinginequalities.

For example, the allocation (𝑥𝐴𝑃 = 1500, 𝑥𝑇𝑁 = 5000, 𝑥𝐾𝑀 = 30) does not belongto the core, since TN and KM will object to their combined share of 5030; since theycan meet their needs jointly at a total cost of 5020. One the other hand, no groupcan object to the allocation (𝑥𝐴𝑃 = 1510, 𝑥𝑇𝑁 = 5000, 𝑥𝐾𝑀 = 20), which thereforebelongs to the core.

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1.66 The usual way to model a cost allocation problem as a TP-coalitional game isto regard the potential cost savings from cooperation as the sum to be allocated. Inthis example, the total joint cost of 6530 represents a potential saving of 1530 overthe aggregate cost of 8060 if each region goes its own way. This potential saving of1530 measures 𝑤(𝑁). Similarly, undertaking a joint development, AP and TN couldsatisfy their combined requirements at a total cost of 6890. This compares with thestandalone costs of 7100 (= 1870 (AP) + 5330 (TN)). Hence, the potential cost savingsfrom their collaboration are 210 (= 7100 - 6890), which measures 𝑤(𝐴𝑃, 𝑇𝑁). Bysimilar calculations, we can compute the worth of each coalition, namely

𝑤(𝐴𝑃 ) = 0 𝑤(𝐴𝑃, 𝑇𝑁) = 210

𝑤(𝑇𝑁) = 0 𝑤(𝐴𝑃,𝐾𝑀) = 770 𝑤(𝑁) = 1530

𝑤(𝐾𝑀) = 0 𝑤(𝐾𝑀,𝑇𝑁) = 1170

An outcome in this game is an allocation of the total cost savings 𝑤(𝑁) = 1530 amongstthe three players. This can be translated into final cost shares by subtracting eachplayers share of the cost savings from their standalone cost. For example, a specificoutcome in this game is (𝑥𝐴𝑃 = 370, 𝑥𝑇𝑁 = 330, 𝑥𝐾𝑀 = 830), which corresponds tofinal cost shares of 1500 for AP, 5000 for TN and 30 for KM.

1.67 Let

𝐶 = {x ∈ 𝑋 :∑𝑖∈𝑆𝑥𝑖 ≥ 𝑤(𝑆) for every 𝑆 ⊆ 𝑁}

1. 𝐶 ⊆ core Assume that x ∈ 𝐶. Suppose x /∈ core. This implies there exists some

coalition 𝑆 and outcome y ∈ 𝑤(𝑆) such that y ≻𝑖 x for every 𝑖 ∈ 𝑆.

∙ y ∈ 𝑤(𝑆) implies∑

𝑖∈𝑆 𝑦𝑖 ≤ 𝑤(𝑆) while

∙ y ≻𝑖 x for every 𝑖 ∈ 𝑆 implies 𝑦𝑖 > 𝑥𝑖 for every 𝑖 ∈ 𝑆. Summing, thisimplies ∑

𝑖∈𝑆𝑦𝑖 >

∑𝑖∈𝑆𝑥𝑖 ≥ 𝑤(𝑆)

This contradiction establishes that x ∈ core.

2. core ⊆ 𝐶 Assume that x ∈ core. Suppose x /∈ 𝐶. This implies there exists some

coalition 𝑆 such that∑

𝑖∈𝑆 𝑥𝑖 < 𝑤(𝑆). Let 𝑑 = 𝑤(𝑆)−∑𝑖∈𝑆 𝑥𝑖 and consider the

allocation y obtained by reallocating 𝑑 from 𝑆𝑐 to 𝑆, that is

𝑦𝑖 =

{𝑥𝑖 + 𝑑/𝑠 𝑖 ∈ 𝑆𝑥𝑖 − 𝑑/(𝑛− 𝑠) 𝑖 /∈ 𝑆

where 𝑠 = ∣𝑆∣ is the number of players in 𝑆 and 𝑛 = ∣𝑁 ∣ is the number in 𝑁 .Then 𝑦𝑖 > 𝑥𝑖 for every 𝑖 ∈ 𝑆 so that y ≻𝑖 x for every 𝑖 ∈ 𝑆. Further, y ∈ 𝑤(𝑆)since

∑𝑖∈𝑆 𝑦𝑖 =

∑𝑖∈𝑆 𝑥𝑖 + 𝑑 = 𝑤(𝑆) and y ∈ 𝑋 since

∑𝑖∈𝑁𝑦𝑖 =

∑𝑖∈𝑆

(𝑥𝑖 + 𝑑/𝑠) +∑𝑖/∈𝑆

(𝑥𝑖 − 𝑑/(𝑛− 𝑠)) =∑𝑖∈𝑁𝑥𝑖 = 𝑤(𝑁)

This contradicts our assumption that x /∈ core, establishing that x ∈ 𝐶.

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1.68 The 7 unanimity games for the player set 𝑁 = {1, 2, 3} are

𝑢{1}(𝑆) =

{1 S = {1}, {1,2}, {1,3}, N

0 otherwise

𝑢{2}(𝑆) =

{1 S = {2}, {1,2}, {2,3}, N

0 otherwise

𝑢{3}(𝑆) =

{1 S = {3}, {1,3}, {2,3}, N

0 otherwise

𝑢{1,2}(𝑆) =

{1 S = {1,2}, N

0 otherwise

𝑢{1,3}(𝑆) =

{1 S = {1,3}, N

0 otherwise

𝑢{2,3}(𝑆) =

{1 S = {2,3}, N

0 otherwise

𝑢𝑁(𝑆) =

{1 S = N

0 otherwise

1.69 Firstly, consider a simple game which is a unanimity game with essential coalition𝑇 and let 𝑥 be an outcome in which

𝑥𝑖 ≥ 0 for every 𝑖 ∈ 𝑇𝑥𝑖 = 0 for every 𝑖 /∈ 𝑇

and

∑𝑖∈𝑁𝑥𝑖 = 1

We claim that 𝑥 ∈ core.

Winning coalitions If 𝑆 is winning coalition, then 𝑤(𝑆) = 1. Furthermore, if it is awinning coalition, it must contain 𝑇 , that is 𝑇 ⊆ 𝑆 and∑

𝑖∈𝑆𝑥𝑖 ≥

∑𝑖∈𝑇𝑥𝑖 = 1 = 𝑤(𝑆)

Losing coalitions If 𝑆 is a losing coalition, 𝑤(𝑆) = 0 and∑𝑖∈𝑆𝑥𝑖 ≥ 0 = 𝑤(𝑆)

Therefore 𝑥 ∈ core and so core ∕= ∅.Conversely, consider a simple game which is not a unanimity game. Suppose thereexists an outcome 𝑥 ∈ core. Then ∑

𝑖∈𝑁𝑥𝑖𝑤(𝑁) = 1 (1.15)

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Since there are no veto players (𝑇 = ∅), 𝑤(𝑁 ∖ {𝑖}) = 1 for every player 𝑖 ∈ 𝑁 and∑𝑗 ∕=𝑖𝑥𝑗 ≥ 𝑤(𝑁 ∖ {𝑖}) = 1

which implies that 𝑥𝑖 = 0 for every 𝑖 ∈ 𝑁 contradicting (1.15). Thus we conclude thatcore = ∅.1.70 The excesses of the proper coalitions at x1 and x2 are

x1 x2

{AP} -180 -200{KM} -955 -950{TN} -395 -380{AP, KM} -365 -380{AP, TN} -365 -370{KM, TN} -180 -160

Therefore

𝑑(x1) = (−180,−180,−365,−365,−395,−955)

and

𝑑(x2) = (−160,−200,−370,−380,−380,−950)

d(x1) ≺𝐿 d(x2) which implies x1 ≻𝑑 x2.

1.71 It is a weak order on 𝑋 , that is ≿ is reflexive, transitive and complete. Reflexivityand transitivity flow from the corresponding properties of ≿𝐿 on ℜ2𝑛 . Similarly, forany x,y ∈ 𝑋 , either d(x) ≾𝐿 d(y) or d(y) ≾𝐿 d(x) since ≿𝐿 is complete on ℜ2𝑛 .Consequently either x ≿ y or y ≿ x (or both).

≿ is not a partial order since it is not antisymmetric

d(x) ≾𝐿 d(y) and d(y) ≾𝐿 d(x) does not imply x = y

1.72

𝑑(𝑆,x) = 𝑤(𝑆) −∑𝑖∈𝑆𝑥𝑖

so that

𝑑(𝑆,x) ≤ 0 ⇐⇒∑𝑖∈𝑆𝑥𝑖 ≥ 𝑤(𝑆)

1.73 Assume to the contrary that x ∈ Nu but that x /∈ core. Then, there exists acoalition 𝑇 with a positive deficit 𝑑(𝑇,x) > 0. Since core ∕= ∅, there exists some y ∈ 𝑋such that 𝑑(𝑆,y) ≤ 0 for every 𝑆 ⊆ Nu. Consequently, d(y) ≺ d(x) and y ≻ x, sothat x /∈ Nu. This contradiction establishes that Nu ⊆ core.

1.74 For player 1, 𝐴1 = {𝐶,𝑁} and

(𝐶,𝐶) ≿1 (𝐶,𝐶)

(𝐶,𝐶) ≿1 (𝑁,𝐶)

Similarly for player 2

(𝐶,𝐶) ≿2 (𝐶,𝐶)

(𝐶,𝐶) ≿2 (𝐶,𝑁)

Therefore, (𝐶,𝐶) satisfies the requirements of the definition of a Nash equilibrium(Example 1.51).

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1.75 If a∗𝑖 is the best element in (𝐴𝑖,≿′𝑖) for every player 𝑖, then

(𝑎∗𝑖 , a−𝑖) ≻𝑖 (𝑎𝑖, a−𝑖) for every 𝑎𝑖 ∈ 𝐴𝑖 and a−𝑖 ∈ 𝐴−𝑖

for every 𝑖 ∈ 𝑁 . Therefore, a∗ is a Nash equilibrium.

To show that it is unique, assume that a is another Nash equilibrium. Then for everyplayer 𝑖 ∈ 𝑁

(��𝑖, a−𝑖) ≿𝑖 (𝑎𝑖, a−𝑖) for every 𝑎𝑖 ∈ 𝐴𝑖which implies that a is a maximal element of ≿′

𝑖. To see this, assume not. That is,assume that there exists some 𝑎𝑖 ∈ 𝐴𝑖 such that 𝑎𝑖 ≻′

𝑖 ��𝑖 which implies

(��𝑖, a−𝑖) ≻𝑖 (��𝑖, a−𝑖) for every a−𝑖 ∈ 𝐴−𝑖

In particular

(��𝑖, a−𝑖) ≻𝑖 (𝑎∗𝑖 , a∗−𝑖)

which contradicts the assumption that a∗ is a Nash equilibrium. Therefore, a is an-other Nash equilibrium, then ��𝑖 is maximal in ≿′

𝑖 and hence also a best element of ≿′𝑖

(Exercise 1.54), which contradicts the assumption that 𝑎∗𝑖 is the unique best element.Consequently, we conclude that a∗ is the unique Nash equilibrium of the game.

1.76 We show that 𝜌(𝑥, 𝑦) = ∣𝑥− 𝑦∣ satisfies the requirements of a metric, namely

1. ∣𝑥− 𝑦∣ ≥ 0.

2. ∣𝑥− 𝑦∣ = 0 if and only if 𝑥 = 𝑦.

3. ∣𝑥− 𝑦∣ = ∣𝑦 − 𝑥∣.To establish the triangle inequality, we can consider various cases. For example, if𝑥 ≤ 𝑦 ≤ 𝑧

∣𝑥− 𝑧∣+ ∣𝑧 − 𝑦∣ ≥ ∣𝑥− 𝑧∣ = 𝑧 − 𝑥 ≥ 𝑦 − 𝑥 = ∣𝑥− 𝑦∣If 𝑥 ≤ 𝑧 ≤ 𝑦

∣𝑥− 𝑧∣+ ∣𝑧 − 𝑦∣ = 𝑧 − 𝑥+ 𝑦 − 𝑧 = 𝑦 − 𝑥 = ∣𝑥− 𝑦∣and so on.

1.77 We show that 𝜌∞𝑥, 𝑦 = max𝑛𝑖=1 ∣𝑥𝑖 − 𝑦𝑖∣ satisfies the requirements of a metric,namely

1. max𝑛𝑖=1 ∣𝑥𝑖 − 𝑦𝑖∣ ≥ 0

2. max𝑛𝑖=1 ∣𝑥𝑖 − 𝑦𝑖∣ = 0 if and only if 𝑥𝑖 = 𝑦𝑖 for all 𝑖.

3. max𝑛𝑖=1 ∣𝑥𝑖 − 𝑦𝑖∣ = max𝑛𝑖=1 ∣𝑦𝑖 − 𝑥𝑖∣4. For every 𝑖, ∣𝑥𝑖 − 𝑦𝑖∣ ≤ ∣𝑥𝑖 − 𝑧𝑖∣+ ∣𝑧𝑖 − 𝑦𝑖∣ from previous exercise. Therefore

max ∣𝑥𝑖 − 𝑦𝑖∣ ≤ max (∣𝑥𝑖 − 𝑧𝑖∣+ ∣𝑧𝑖 − 𝑦𝑖∣)≤ max ∣𝑥𝑖 − 𝑧𝑖∣+ max ∣𝑧𝑖 − 𝑦𝑖∣

1.78 For any 𝑛, any neighborhood of 1/𝑛 contains points of 𝑆 (namely 1/𝑛) and pointsnot in 𝑆 (1/𝑛+ 𝜖). Hence every point in 𝑆 is a boundary point. Also, 0 is a boundarypoint. Therefore b(𝑆) = 𝑆 ∪ {0}. Note that 𝑆 ⊂ b(𝑆). Therefore, 𝑆 has no interiorpoints.

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1.79 1. Let 𝑥 ∈ int 𝑆. Thus 𝑆 is a neighborhood of 𝑥. Therefore, 𝑇 ⊇ 𝑆 is aneighborhood of 𝑥, so that 𝑥 is an interior point of 𝑇 .

2. Clearly, if 𝑥 ∈ 𝑆, then 𝑥 ∈ 𝑇 ⊆ 𝑇 . Therefore, assume 𝑥 ∈ 𝑆 ∖ 𝑆 which impliesthat 𝑥 is a boundary point of 𝑆. Every neighborhood of 𝑥 contains other pointsof 𝑆 ⊆ 𝑇 . Hence 𝑥 ∈ 𝑇 .

1.80 Assume that 𝑆 is open. Every 𝑥 ∈ 𝑆 has a neighborhood which is disjoint from𝑆𝑐. Hence no 𝑥 ∈ 𝑆 is a closure point of 𝑆𝑐. 𝑆𝑐 contains all its closure points and istherefore closed.

Conversely, assume that 𝑆 is closed. Let 𝑥 be a point its complement 𝑆𝑐. Since 𝑆is closed and 𝑥 /∈ 𝑆, 𝑥 is not a boundary point of 𝑆. This implies that 𝑥 has aneighborhood 𝑁 which is disjoint from 𝑆, that is 𝑁 ⊆ 𝑆𝑐. Hence, 𝑥 is an interior pointof 𝑆𝑐. This implies that 𝑆𝑐 contains only interior points, and hence is open.

1.81 Clearly 𝑥 is a neighborhood of every point 𝑥 ∈ 𝑋 , since 𝐵𝑟(𝑥) ⊆ 𝑋 for every𝑟 > 0. Hence, every point 𝑥 ∈ 𝑋 is an interior point of 𝑥. Similarly, every point 𝑥 ∈ ∅is an interior point (there are none). Since 𝑥 and ∅ are open, there complements ∅ and𝑥 are closed.

Alternatively, ∅ has no boundary points, and is therefore is open. Trivialy, on the otherhand, ∅ contains all its boundary points, and is therefore closed.

1.82 Let 𝑋 be a metric space. Assume 𝑋 is the union of two disjoint closed sets 𝐴 and𝐵, that is

𝑋 = 𝐴 ∪𝐵 𝐴 ∩𝐵 = ∅

Then 𝐴 = 𝐵𝑐 is open as is 𝐵 = 𝐴𝑐. Therefore 𝑋 is not connected.

Conversely, assume that 𝑋 is not connected. Then there exist disjoint open sets 𝐴 and𝐵 such that 𝑋 = 𝐴 ∪𝐵. But 𝐴 = 𝐵𝑐 is also closed as is 𝐵 = 𝐴𝑐. Therefore 𝑋 is theunion of two disjoint closed sets.

1.83 Assume 𝑆 is both open and closed, ∅ ⊂ 𝑆 ⊂ 𝑋 . We show that we can represent𝑋 as the union of two disjoint open sets, 𝑆 and 𝑆𝑐. For any 𝑆 ⊂ 𝑋 , 𝑋 = 𝑆 ∪ 𝑆𝑐 and𝑆 ∩ 𝑆𝑐 = ∅. 𝑆 is open by assumption. It complement 𝑆𝑐 is open since 𝑆 is closed.Therefore, 𝑋 is not connected.

Conversely, assume that 𝑆 is not connected. That is, there exists two disjoint opensets 𝑆 and 𝑇 such that 𝑋 = 𝑆 ∪ 𝑇 . Now 𝑆 = 𝑇 𝑐, which implies that 𝑆 is closed since𝑇 is open. Therefore 𝑆 is both open and closed.

1.84 Assume that 𝑆 is both open and closed. Then so is 𝑆𝑐 and 𝑋 is the disjoint unionof two closed sets

𝑥 = 𝑆 ∪ 𝑆𝑐

so that

b(𝑆) = 𝑆 ∩ 𝑆𝑐 = 𝑆 ∩ 𝑆𝑐 = ∅

Conversely, assume that b(𝑆) = 𝑆 ∩ 𝑆𝑐 = ∅. This implies that Consider any 𝑥 ∈ 𝑆.Since 𝑆 ∩ 𝑆𝑐 = ∅, 𝑥 /∈ 𝑆𝑐. A fortiori, x /∈ 𝑆𝑐 which implies that 𝑥 ∈ 𝑆 and therefore𝑆 ⊆ 𝑆. 𝑆 is closed. Similarly we can show that 𝑆𝑐 ⊆ 𝑆𝑐 so that 𝑆𝑐 is closed andtherefore 𝑆 is open. 𝑆 is both open and closed.

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1.85 1. Let {𝐺𝑖} be a (possibly infinite) collection of open sets. Let 𝐺 = ∪𝑖𝐺𝑖. Let𝑥 be a point in 𝐺. Then there exists some particular 𝐺𝑗 which contains 𝑥. Since𝐺𝑗 is open, 𝐺𝑗 is a neighborhood of 𝑥. Since 𝐺𝑗 ⊆ 𝐺, 𝑥 is an interior point of 𝐺.Since 𝑥 is an arbitrary point in 𝐺, we have shown that every 𝑥 ∈ 𝐺 is an interiorpoint. Hence, 𝐺 is open.

What happens if every 𝐺𝑖 is empty? In this case, 𝐺 = ∅ and is open (Exercise1.81). The other possibility is that the collection {𝐺𝑖} is empty. Again 𝐺 = ∅which is open.

Suppose {𝐺1, 𝐺2, . . . , 𝐺𝑛 } is a finite collection of open sets. Let 𝐺 = ∩𝑖𝐺𝑖. If𝐺 = ∅, then it is trivially open. Otherwise, let 𝑥 be a point in 𝐺. Then 𝑥 ∈ 𝐺𝑖

for all 𝑖 = 1, 2, . . . , 𝑛. Since the sets 𝐺𝑖 are open, for every 𝑖, there exists an openball 𝐵(𝑥, 𝑟𝑖) ⊆ 𝐺𝑖 about 𝑥. Let 𝑟 be the smallest radius of these open balls, thatis 𝑟 = min{ 𝑟1, 𝑟2, . . . , 𝑟𝑛 }. Then 𝐵𝑟(𝑥) ⊆ 𝐵(𝑥, 𝑟𝑖), so that 𝐵𝑟(𝑥) ⊆ 𝐺𝑖 for all i.Hence 𝐵𝑟(𝑥) ⊆ 𝐺. 𝑥 is an interior point of 𝐺 and 𝐺 is open.

To complete the proof, we need to deal with the trivial case in which the collectionis empty. In that case, 𝐺 = ∩𝑖𝐺𝑖 = 𝑋 and hence is open.

2. The corresponding properties of closed sets are established analogously.

1.86 1. Let 𝑥0 be an interior point of 𝑆. This implies there exists an open ball 𝐵 ⊆ 𝑆about 𝑥0. Every 𝑥 ∈ 𝐵 is an interior point of 𝑆. Hence 𝐵 ⊆ int 𝑆. 𝑥0 is aninterior point of int 𝑆 which is therefore open.

Let 𝐺 be any open subset of 𝑆 and 𝑥 be a point in 𝐺. 𝐺 is neighborhood of 𝑥,which implies that 𝑆 ⊇ 𝐺 is also neighborhood of 𝑥. Therefore 𝑥 is an interiorpoint of 𝑆. Therefore int 𝑆 contains every open subset 𝐺 ⊆ 𝑆, and hence is thelargest open set in 𝑆.

2. Let 𝑆 denote the closure of the set 𝑆. Clearly, 𝑆 ⊆ 𝑆. To show the converse, let𝑥 be a closure point of 𝑆 and let 𝑁 be a neighborhood of 𝑥. Then 𝑁 containssome other point 𝑥′ ∕= 𝑋 which is a closure point of 𝑆. 𝑁 is a neighborhood of𝑥′ which intersects 𝑆. Hence 𝑥 is a closure point of 𝑆.

Consequently 𝑆 = 𝑆 which implies that 𝑆 is closed.

Assume 𝐹 is a closed subset of containing 𝑆. Then

𝑆 ⊆ 𝐹 = 𝐹

since 𝐹 is closed. Hence, 𝑆 is a subset of every closed set containing 𝑆.

1.87 Every 𝑥 ∈ 𝑆 is either an interior point or a boundary point. Consequently, theinterior of 𝑆 is the set of all 𝑥 ∈ 𝑆 which are not boundary points

int 𝑆 = 𝑆 ∖ b(𝑆)

1.88 Assume that 𝑆 is closed, that is

𝑆 = 𝑆 ∪ b(𝑆) = 𝑆

This implies that b(𝑆) ⊆ 𝑆. 𝑆 contains its boundary.

Assume that 𝑆 contains its boundary, that is 𝑆 ⊇ b(𝑆). Then

𝑆 = 𝑆 ∪ b(𝑆) = 𝑆

𝑆 is closed.

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1.89 Assume 𝑆 is bounded, and let 𝑑 = 𝑑(𝑆). Choose any 𝑥 ∈ 𝑆. For all 𝑦 ∈ 𝑆,𝜌(𝑥, 𝑦) ≤ 𝑑 < 𝑑 + 1. Therefore, 𝑦 ∈ 𝐵(𝑥, 𝑑 + 1). 𝑆 is contained in the open ball𝐵(𝑥, 𝑑+ 1).

Conversely, assume 𝑆 is contained in the open ball 𝐵𝑟(𝑥). Then for any 𝑦, 𝑧 ∈ 𝑆

𝜌(𝑦, 𝑧) ≤ 𝜌(𝑦, 𝑥) + 𝜌(𝑥, 𝑧) < 2𝑟

by the triangle inequality. Therefore 𝑑(𝑆) < 2𝑟 and the set is bounded.

1.90 Let 𝑦 ∈ 𝑆 ∩𝐵𝑟(𝑥0). For every 𝑥 ∈ 𝑆, 𝜌(𝑥, 𝑦) < 𝑟 and therefore

𝜌(𝑥, 𝑥0) ≤ 𝜌(𝑥, 𝑦) + 𝜌(𝑦, 𝑥0) < 𝑟 + 𝑟 = 2𝑟

so that 𝑥 ∈ 𝐵2𝑟(𝑥0).1.91 Let y0 ∈ 𝑌 . For any 𝑟 > 0, let y′ = y− 𝑟 be the production plan which is 𝑟 unitsless in every commodity. Then, for any y ∈ 𝐵𝑟(y

′)

𝑦𝑖 − 𝑦′𝑖 ≤ 𝜌∞(y,y′) < 𝑟 for every 𝑖

and therefore y < y0. Thus 𝐵𝑟(y′) ⊂ 𝑌 and so y′ ∈ int 𝑌 ∕= ∅.

1.92 For any 𝑥 ∈ 𝑆1𝜌𝑥 = 𝜌(𝑥, 𝑆2) > 0

Similarly, for every 𝑦 ∈ 𝑆2𝜌𝑦 = 𝜌(𝑦, 𝑆1) > 0

Let

𝑇1 =∪𝑥∈𝑆1

𝐵𝜌𝑥/2(𝑥)

𝑇2 =∪𝑦∈𝑆2

𝐵𝜌𝑦/2(𝑥)

Then 𝑇1 and 𝑇2 are open sets containing 𝑆1 and 𝑆2 respectively.

To show that 𝑇1 and 𝑇2 are disjoint, suppose to the contrary that 𝑧 ∈ 𝑇1 ∩ 𝑇2. Then,there exist points 𝑥 ∈ 𝑆1 and 𝑦 ∈ 𝑆2 such that

𝜌(𝑥, 𝑧) < 𝜌𝑥/2, 𝜌(𝑦, 𝑧) < 𝜌𝑦/2

Without loss of generality, suppose that 𝜌𝑥 ≤ 𝜌𝑦 and therefore

𝜌(𝑥, 𝑦) ≤ 𝜌(𝑥, 𝑧) + 𝜌(𝑦, 𝑧) < 𝜌𝑥/2 + 𝜌𝑦/2 ≤ 𝜌𝑦which contradicts the definition of 𝜌𝑦 and shows that 𝑇1 ∩ 𝑇2 = ∅.1.93 By Exercise 1.92, there exist disjoint open sets 𝑇1 and 𝑇2 such that 𝑆1 ⊆ 𝑇1 and𝑆2 ⊆ 𝑇2. Since 𝑆2 ⊆ 𝑇2, 𝑆2 ∩ 𝑇 𝑐2 = ∅. 𝑇 𝑐2 is a closed set which contains 𝑇1, andtherefore 𝑆2 ∩ 𝑇1 = ∅. 𝑇 = 𝑇1 is the desired set.

1.94 See Figure 1.2.

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1 2

1𝑆

𝐵1/2((2, 0))

Figure 1.2: Open ball about (2, 0) relative to 𝑋

1.95 Assume 𝑆 is connected. Suppose 𝑆 is not an interval. This implies that thereexists numbers 𝑥, 𝑦, 𝑧 such that 𝑥 < 𝑦 < 𝑧 and 𝑥, 𝑧 ∈ 𝑆 while 𝑦 /∈ 𝑆. Then

𝑆 = (𝑆 ∩ (−∞, 𝑦)) ∪ (𝑆 ∩ (𝑦,∞))

represents 𝑆 as the union of two disjoint open sets (relative to 𝑆), contradicting theassumption that 𝑆 is connected.

Conversely, assume that 𝑆 is an interval. Suppose that 𝑆 is not connected. That is,𝑆 = 𝐴∪𝐵 where 𝐴 and 𝐵 are nonempty disjoint closed sets. Choose 𝑥 ∈ 𝐴 and 𝑧 ∈ 𝐵.Since 𝐴 and 𝐵 are disjoint, 𝑥 ∕= 𝑧. Without loss of generality, we may assume 𝑥 < 𝑧.Since 𝑆 is an interval, [𝑥, 𝑧] ⊆ 𝑆 = 𝐴 ∪𝐵. Let

𝑦 = sup{ [𝑥, 𝑧] ∩ 𝑆 }Clearly 𝑥 ≤ 𝑦 ≤ 𝑧 so that 𝑦 ∈ 𝑆. Now 𝑦 belongs to either 𝐴 or 𝐵. Since 𝐴 is closed in 𝑆,[𝑥, 𝑧]∩𝐴 is closed and 𝑦 = sup{ [𝑥, 𝑧]∩𝑆 } ∈ 𝐴. This implies the 𝑦 < 𝑧. Consequently,𝑦 + 𝜖 ∈ 𝐵 for every 𝜖 > 0 such that 𝑦 + 𝜖 ≤ 𝑧. Since 𝐵 is closed, 𝑦 ∈ 𝐵. This impliesthat 𝑦 belongs to both 𝐴 and 𝐵 contradicting the assumption that 𝐴 ∩ 𝐵 = ∅. Weconclude that 𝑆 must be connected.

1.96 Assume 𝑥𝑛 → 𝑥 and also 𝑥𝑛 → 𝑦. We have to show that 𝑥 = 𝑦. Suppose not,that is suppose 𝑥 ∕= 𝑦 (see Figure 1.3). Then 𝜌(𝑥, 𝑦) = 𝑅 > 0. Let 𝑟 = 𝑅/3 > 0. Since𝑥𝑛 → 𝑥, there exists some 𝑁𝑥 such that 𝑥𝑛 ∈ 𝐵𝑟(𝑥) for all 𝑛 ≥ 𝑁𝑥. Since 𝑥𝑛 → 𝑦,there exists some 𝑁𝑦 such that 𝑥𝑛 ∈ 𝐵𝑟(𝑦) for all 𝑛 ≥ 𝑁𝑦. But these statements arecontradictory since 𝐵𝑟(𝑥) ∩ 𝐵(𝑦, 𝑟) = ∅. We conclude that the successive terms of aconvergent sequence cannot get arbitrarily close to two distinct points, so that the limita convergent sequence is unique.

1.97 Let (𝑥𝑛) be a sequence which converges to 𝑥. There exists some 𝑁 such that

𝜌(𝑥𝑛 − 𝑥) < 1

for all 𝑛 ≥ 𝑁 . Let

𝑅 = max{ 𝜌(𝑥1 − 𝑥), 𝜌(𝑥2 − 𝑥), . . . , 𝜌(𝑥𝑁−1 − 𝑥), 1 }Then for all 𝑛, 𝜌(𝑥𝑛−𝑥) ≤ 𝑅. That is every element 𝑥𝑛 in the sequence (𝑥𝑛) belongs to𝐵(𝑥,𝑅+ 1), the open ball about 𝑥 of radius 𝑅+ 1. Therefore the sequence is bounded.

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𝐵(𝑦, 𝑟)

𝑟𝑟

𝑅𝐵(𝑥, 𝑟)

𝑥 𝑦

Figure 1.3: A convergent sequence cannot have two distinct limits

1.98 The share 𝑠𝑛 of the 𝑛th guest is

𝑠𝑛 =1

2

𝑛

lim 𝑠𝑛 = 0

However, 𝑠𝑛 > 0 for all 𝑛. There is no limit to the number of guests who will get ashare of the cake, although the shares will get vanishingly small for large parties.

1.99 Suppose 𝑥𝑛 → 𝑥. That is, there exists some 𝑁 such that 𝜌(𝑥𝑛, 𝑥) < 𝜖/2 for all𝑛 ≥ 𝑁 . Then, for all 𝑚,𝑛 ≥ 𝑁

𝜌(𝑥𝑚, 𝑥𝑛) ≤ 𝜌(𝑥𝑚, 𝑥) + 𝜌(𝑥, 𝑥𝑛)

< 𝜖/2 + 𝜖/2 = 𝜖

1.100 Let (𝑥𝑛) be a Cauchy sequence. There exists some 𝑁 such that

𝜌(𝑥𝑛 − 𝑥𝑁 ) < 1

for all 𝑛 ≥ 𝑁 . Let

𝑅 = max{ 𝜌(𝑥1 − 𝑥𝑁 ), 𝜌(𝑥2 − 𝑥𝑁 ), . . . , 𝜌(𝑥𝑁−1 − 𝑥𝑁 ), 1 }Every 𝑥𝑛 belongs to 𝐵(𝑥𝑁 , 𝑅+ 1), the ball of radius 𝑅+ 1 centered on 𝑥𝑁 .

1.101 Let (𝑥𝑛) be a bounded increasing sequence in ℜ and let 𝑆 = { 𝑥𝑛 } be the set ofelements of (𝑥𝑛). Let 𝑏 be the least upper bound of 𝑆. We show that 𝑥𝑛 → 𝑏.First observe that 𝑥𝑛 ≤ 𝑏 for every 𝑛 (since 𝑏 is an upper bound). Since 𝑏 is the leastupper bound, for every 𝜖 > 0 there exists some element 𝑥𝑁 such that 𝑥𝑁 > 𝑏− 𝜖. Since(𝑥𝑛) is increasing, we must have

𝑏− 𝜖 < 𝑥𝑛 ≤ 𝑏 for every 𝑛 ≥ 𝑁That is, for every 𝜖 > 0 there exists an 𝑁 such that

𝜌(𝑥𝑛, 𝑥) < 𝜖 for every 𝑛 ≥ 𝑁𝑥𝑛 → 𝑏.1.102 If 𝛽 > 1, the sequence 𝛽, 𝛽2, 𝛽3, . . . is unbounded.

Otherwise, if 𝛽 ≤ 1, 𝛽𝑛 ≤ 𝛽𝑛−1 and the sequence is decreasing and bounded by 𝛽 ≤ 1.Therefore the sequence converges (Exercise 1.101). Let 𝑥 = lim𝑛→∞. Then

𝛽𝑛+1 = 𝛽𝛽𝑛

and therefore

𝑥 = lim𝑛→∞𝛽

𝑛+1 = 𝛽 lim𝑛→∞ 𝛽

𝑛 = 𝛽𝑥

which can be satisfied if and only if

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∙ 𝛽 = 1, in which case 𝑥 = lim 1𝑛 = 1

∙ 𝑥 = 0 when 0 ≤ 𝛽 < 1

Therefore

𝛽𝑛 → 0 ⇐⇒ 𝛽 < 1

1.103 1. For every 𝑥 ∈ ℜ(𝑥−

√2)2 ≥ 0

Expanding

𝑥2 − 2√

2𝑥+ 2 ≥ 0

𝑥2 + 2 ≥ 2√

2𝑥

Dividing by 𝑥

𝑥+2

𝑥≥ 2√

2

for every 𝑥 > 0. Therefore

1

2

(𝑥+

2

𝑥

)≥ √2

2. Let (𝑥𝑛) be the sequence defined in Example 1.64. That is

𝑥𝑛 =1

2

(𝑥𝑛−1 +

2

𝑥𝑛−1

)

Starting from 𝑥0 = 2, it is clear that 𝑥𝑛 ≥ 0 for all 𝑛. Substituting in

1

2

(𝑥+

2

𝑥

)≥√

2

𝑥𝑛 =1

2

(𝑥𝑛−1 +

2

𝑥𝑛−1

)≥√

2

That is 𝑥𝑛 ≥ √2 for every 𝑛. Therefore for every 𝑛

𝑥𝑛 − 𝑥𝑛+1 = 𝑥𝑛 − 1

2

(𝑥𝑛 +

2

𝑥𝑛

)

=1

2

(𝑥𝑛 − 2

𝑥𝑛

)

≥ 1

2

(𝑥𝑛 − 2√

2

)= 𝑥𝑛 −

√2

≥ 0

This implies that 𝑥𝑛+1 ≤ 𝑥𝑛. Consequently√

2 ≤ 𝑥𝑛 ≤ 2 for every 𝑛. (𝑥𝑛) is abounded monotone sequence. By Exercise 1.101, 𝑥𝑛 → 𝑥. The limit 𝑥 satisfiesthe equation

𝑥 =1

2

(𝑥+

2

𝑥

)

Solving, this implies 𝑥2 = 2 or 𝑥 =√

2 as required.

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1.104 The following sequence approximates the square root of any positive number 𝑎

𝑥1 = 𝑎

𝑥𝑛+1 =1

2

(𝑥𝑛 +

𝑎

𝑥𝑛)

1.105 Let 𝑥 ∈ 𝑆. If 𝑥 ∈ 𝑆, then 𝑥 is the limit of the sequence (𝑥, 𝑥, 𝑥, . . . ). If 𝑥 /∈ 𝑆,then 𝑥 is a boundary point of 𝑆. For every 𝑛, the ball 𝐵(𝑥, 1/𝑛) contains a point𝑥𝑛 ∈ 𝑆. From the sequence of open balls 𝐵(𝑥, 1/𝑛) for 𝑛 = 1, 2, 3, . . . , we can generateof a sequence of points 𝑥𝑛 which converges to 𝑥.

Conversely, assume that 𝑥 is the limit of a sequence (𝑥𝑛) of points in 𝑆. Either 𝑥 ∈ 𝑆and therefore 𝑥 ∈ 𝑆. Or 𝑥 /∈ 𝑆. Since 𝑥𝑛 → 𝑥, every neighborhood of 𝑥 contains points𝑥𝑛 of the sequence. Hence, 𝑥 is a boundary point of 𝑆 and 𝑥 ∈ 𝑆.

1.106 𝑆 is closed if and only if 𝑆 = 𝑆. The result follows from Exercise 1.105.

1.107 Let 𝑆 be a closed subset of a complete metric space 𝑋 . Let (𝑥𝑛) be a Cauchysequence in 𝑆. Since 𝑋 is complete, 𝑥𝑛 → 𝑥 ∈ 𝑋 . Since 𝑆 is closed, 𝑥 ∈ 𝑆 (Exercise1.106).

1.108 Since 𝑑(𝑆𝑛) → 0, 𝑆 cannot contain more than one point. Therefore, it sufficesto show that 𝑆 is nonempty. Choose some 𝑥𝑛 from each 𝑆𝑛. Since 𝑑(𝑆𝑛)→ 0, (𝑥𝑛) isa Cauchy sequence. Since 𝑋 is complete, there exists some 𝑥 ∈ 𝑋 such that 𝑥𝑛 → 𝑥.Choose some𝑚. Since the sets are nested, the subsequence { 𝑥𝑛 : 𝑛 ≥ 𝑚 } ⊆ 𝑆𝑚. Since𝑆𝑚 is closed, 𝑥 ∈ 𝑆𝑚 (Exercise 1.106). Since 𝑥 ∈ 𝑆𝑚 for every 𝑚

𝑥 ∈∞∩

𝑚=1

𝑆𝑚

1.109 If player 1 picks closed balls whose radius decreases by at least half after eachpair of moves, then {𝑆1, 𝑆3, 𝑆5, . . . } is a nested sequence of closed sets which has anonempty intersection (Exercise 1.108).

1.110 Let (𝑥𝑛) be a sequence in 𝑆 ⊆ 𝑇 with 𝑆 closed and 𝑇 compact. Since 𝑇 iscompact, there exists a convergent subsequence 𝑥𝑚 → 𝑥 ∈ 𝑇 . Since 𝑆 is closed,we must have 𝑥 ∈ 𝑆 (Exercise 1.106). Therefore (𝑥𝑛) contains a subsequence whichconverges in 𝑆, so that 𝑆 is compact.

1.111 Let (𝑥𝑛) be a Cauchy sequence in a metric space. For every 𝜖 > 0, there exists𝑁 such that

𝜌(𝑥𝑚, 𝑥𝑛) < 𝜖/2 for all 𝑚,𝑛 ≥ 𝑁

Trivially, if (𝑥𝑛) converges, it has a convergent subsequence (the whole sequence).

Conversely, assume that (𝑥𝑛) has a subsequence (𝑥𝑚) which converges to 𝑥. That is,there exists some 𝑀 such that

𝜌(𝑥𝑚, 𝑥) < 𝜖/2 for all 𝑚 ≥𝑀

Therefore, by the triangle inequality

𝜌(𝑥𝑛, 𝑥) ≤ 𝜌(𝑥𝑛, 𝑥𝑀 ) + 𝜌(𝑥𝑀 , 𝑥) < 𝜖/2 + 𝜖/2 = 𝜖

for all 𝑛 ≥ max𝑀,𝑁

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1.112 We proceed sequentially as follows. Choose any 𝑥1 in 𝑋 . If the open ball 𝐵(𝑥1, 𝑟)contains 𝑋 , we are done. Otherwise, choose some 𝑥2 /∈ 𝐵(𝑥1, 𝑟) and consider the set∪2

𝑖=1 𝐵(𝑥𝑖, 𝑟). If this set contains 𝑋 , we are done. Otherwise, choose some 𝑥3 /∈∪2𝑖=1 𝐵(𝑥𝑖, 𝑟) and consider

∪3𝑖=1 𝐵(𝑥𝑖, 𝑟)

The process must terminate with a finite number of open balls. Otherwise, if the processcould be continued indefinitely, we could construct an infinite sequence (𝑥1, 𝑥2, 𝑥3, . . . )which had no convergent subsequence. The would contradict the compactness of 𝑋 .

1.113 Assume 𝑋 is compact. The previous exercise showed that 𝑋 is totally bounded.Further, since every sequence has a convergent subsequence, every Cauchy sequenceconverges (Exercise 1.111). Therefore 𝑋 is complete.

Conversely, assume that𝑋 is complete and totally bounded and let 𝑆1 = { 𝑥11, 𝑥21, 𝑥31, . . . }be an infinite sequence of points in 𝑋 . Since 𝑋 is totally bounded, it is covered by afinite collection of open balls of radius 1/2. 𝑆1 has a subsequence 𝑆2 = { 𝑥12, 𝑥22, 𝑥32, . . . }all of whose points lie in one of the open balls. Similarly, 𝑆2 has a subsequence𝑆3 = { 𝑥13, 𝑥23, 𝑥33, . . . } all of whose points lie in an open ball of radius 1/3. Contin-uing in this fashion, we construct a sequence of subsequences, each of which lies in aball of smaller and smaller radius. Consequently, successive terms of the “diagonal”subsequence { 𝑥11, 𝑥22, 𝑥33, . . . } get closer and closer together. That is, 𝑆 is a Cauchysequence. Since 𝑋 is complete, 𝑆 converges in 𝑋 and 𝑆1 has a convergent subsequence𝑆. Hence, 𝑋 is compact.

1.114 1. Every big set 𝑇 ∈ ℬ has a least two distinct points. Hence 𝑑(𝑇 ) >0 for every 𝑇 ∈ ℬ.

2. Otherwise, there exists 𝑛 such that 𝑑(𝑇 ) ≥ 1/𝑛 for every 𝑇 ∈ ℬ and therefore𝛿 = inf𝑇∈ℬ 𝑑(𝑇 ) ≥ 1/𝑛 > 0.

3. Choose a point 𝑥𝑛 in each 𝑇𝑛. Since 𝑋 is compact, the sequence (𝑥𝑛) has aconvergent subsequence (𝑥𝑚) which converges to some point 𝑥0 ∈ 𝑋 .

4. The point 𝑥0 belongs to at least one 𝑆0 in the open cover 𝒞. Since 𝑆0 is open,there exists some open ball 𝐵𝑟(𝑥0) ⊆ 𝑆0.

5. Consider the concentric ball 𝐵𝑟/2(𝑥0). Since (𝑥𝑚) is a convergent subsequence,there exists some 𝑀 such that 𝑥𝑚 ∈ 𝐵𝑟/2(𝑥) for every 𝑚 ≥𝑀 .

6. Choose some 𝑛0 ≥ min{𝑀, 2/𝑟 }. Then 1/𝑛0 < 𝑟/2 and 𝑑(𝑇𝑛0) < 1/𝑛0 < 𝑟/2.𝑥𝑛0 ∈ 𝑇𝑛0 ∩𝐵𝑟/2(𝑥) and therefore (Exercise 1.90) 𝑇𝑛0 ⊆ 𝐵𝑟(𝑥) ⊆ 𝑆0.

This contradicts the assumption that 𝑇𝑛 is a big set. Therefore, we conclude that𝛿 > 0.

1.115 1. 𝑋 is totally bounded (Exercise 1.112). Therefore, for every 𝑟 > 0, thereexists a finite number of open balls 𝐵𝑟(𝑥𝑛) such that

𝑋 =

𝑛∪𝑖=1

𝐵𝑟(𝑥𝑖)

2. 𝑑(𝐵𝑟(𝑥𝑖)) = 2𝑟 < 𝛿. By definition of the Lebesgue number, every 𝐵𝑟(𝑥𝑖) iscontained in some 𝑆𝑖 ∈ 𝒞.

3. The collection of open balls {𝐵𝑟(𝑥𝑖)} covers 𝑋 . Therefore, fore every 𝑥 ∈ 𝑋 ,there exists 𝑖 such that

𝑥 ∈ 𝐵𝑟(𝑥𝑖) ⊆ 𝑆𝑖

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Therefore, the finite collection 𝑆1, 𝑆2, . . . , 𝑆𝑛 covers 𝑋 .

1.116 For any family of subsets 𝒞∩𝑆∈𝒞𝑆 = ∅ ⇐⇒

∪𝑆∈𝒞𝑆𝑐 = 𝑋

Suppose to the contrary that 𝒞 is a collection of closed sets with the finite intersectionproperty, but that

∩𝑆∈𝒞 𝑆 = ∅. Then {𝑆𝑐 : 𝑆 ∈ 𝒞 } is a open cover of 𝑋 which does

not have a finite subcover. Consequently 𝑋 cannot be compact.

Conversely, assume every collection of closed sets with the finite intersection propertyhas a nonempty intersection. Let ℬ be an open cover of 𝑋 . Let

𝒞 = {𝑆 ⊆ 𝑋 : 𝑆𝑐 ∈ ℬ }That is ∪

𝑆∈𝒞𝑆𝑐 = 𝑋 which implies

∩𝑆∈𝒞𝑆 = ∅

Consequently, 𝒞 does not have the finite intersection property. There exists a finitesubcollection {𝑆1, 𝑆2, . . . , 𝑆𝑛 } such that

𝑛∩𝑖=1

𝑆𝑖 = ∅

which implies that

𝑛∪𝑖=1

𝑆𝑐𝑖 = 𝑋

{𝑆𝑐1, 𝑆𝑐2, . . . , 𝑆𝑐𝑛 } is a finite subcover of 𝑋 . Thus, 𝑋 is compact.

1.117 Every finite collection of nested (nonempty) sets has the finite intersection prop-erty. By Exercise 1.116, the sequence has a non-empty intersection. (Note: every set𝑆𝑖 is a subset of the compact set 𝑆1.)

1.118 (1) =⇒ (2) Exercises 1.114 and 1.115.

(2) =⇒ (3) Exercise 1.116

(3) =⇒ (1) Let 𝑋 be a metric space in which every collection of closed subsets withthe finite intersection property has a finite intersection. Let (𝑥𝑛) be a sequencein 𝑋 . For any 𝑛, let 𝑆𝑛 be the tail of the sequence minus the first 𝑛 terms, thatis

𝑆𝑛 = { 𝑥𝑚 : 𝑚 = 𝑛+ 1, 𝑛+ 2, . . . }The collection (𝑆𝑛) has the finite intersection property since, for any finite set ofintegers {𝑛1, 𝑛2, . . . , 𝑛𝑘 }

𝑘∩𝑗=1

𝑆𝑛𝑗 ⊆ 𝑆𝐾 ∕= ∅

where 𝐾 = max{𝑛1, 𝑛2, . . . , 𝑛𝑘 }. Therefore

∞∩𝑛=1

𝑆𝑛 ∕= ∅

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Choose any 𝑥 ∈ ∪∞𝑛=1 𝑆𝑛. That is, 𝑥 ∈ 𝑆𝑛 for each 𝑛 = 1, 2, . . . . Thus, for every

𝑟 > 0 and 𝑛 = 1, 2, . . . , there exists some 𝑥𝑛 ∈ 𝐵𝑟(𝑥) ∩ 𝑆𝑛We construct a subsequence as follows. For 𝑘 = 1, 2, . . . , let 𝑥𝑘 be the firstterm in 𝑆𝑘 which belongs to 𝐵1/𝑘(𝑥). Then, (𝑥𝑘) is a subsequence of (𝑥𝑛) whichconverges to 𝑥. We conclude that every sequence has a convergent subsequence.

1.119 Assume (𝑥𝑛) is a bounded sequence in ℜ. Without loss of generality, we canassume that { 𝑥𝑛 } ⊂ [0, 1]. Divide 𝐼0 = [0, 1] into two sub-intervals [0, 1/2] and[1/2, 1]. At least one of the sub-intervals must contain an infinite number of terms ofthe sequence. Call this interval 𝐼1. Continuing this process of subdivision, we obtaina nested sequence of intervals

𝐼0 ⊃ 𝐼1 ⊃ 𝐼2 ⊃ . . .

each of which contains an infinite number of terms of the sequence. Consequently,we can construct a subsequence (𝑥𝑚) with 𝑥𝑚 ∈ 𝐼𝑚. Furthermore, the intervals getsmaller and smaller with 𝑑(𝐼𝑛) → 0, so that (𝑥𝑚) is a Cauchy sequence. Since ℜ iscomplete, the subsequence (𝑥𝑚) converges to 𝑥 ∈ ℜ.

Note how we implicitly called on the Axiom of Choice (Remark 1.5) in choosing asubsequence from the nested sequence of intervals.

1.120 Let (𝑥𝑛) be a Cauchy sequence in ℜ. That is, for every 𝜖 > 0, there exists 𝑁 suchthat ∣𝑥𝑛 − 𝑥𝑚∣ < 𝜖 for all 𝑚,𝑛 ≥ 𝑁 . (𝑥𝑛) is bounded (Exercise 1.100) and hence by theBolzano-Weierstrass theorem, it has a convergent subsequence (𝑥𝑚) with 𝑥𝑚 → 𝑥 ∈ ℜ.Choose 𝑥𝑟 from the convergent subsequence such that 𝑟 ≥ 𝑁 and ∣𝑥𝑟 − 𝑥∣ < 𝜖/2. Bythe triangle inequality

∣𝑥𝑛 − 𝑥∣ ≤ ∣𝑥𝑛 − 𝑥𝑟∣+ ∣𝑥𝑟 − 𝑥∣ < 𝜖/2 + 𝜖/2 = 𝜖

Hence the sequence (𝑥𝑛) converges to 𝑥 ∈ ℜ.

1.121 Since 𝑋1 and 𝑋2 are linear spaces, x1 + y1 ∈ 𝑋1 and x2 + y2 ∈ 𝑋2, so that(x1 + y1,x2 + y2) ∈ 𝑋1 × 𝑋2. Similarly (𝛼x1, 𝛼x2) ∈ 𝑋1 × 𝑋2 for every (x1,x2) ∈𝑋1 ×𝑋2. Hence, 𝑋 = 𝑋1 ×𝑋2 is closed under addition and scalar multiplication.

With addition and scalar multiplication defined component-wise, 𝑋 inherits the arith-metic properties (like associativity) of its constituent spaces. Verifying this wouldproceed identically as for ℜ𝑛. It is straightforward though tedious. The zero elementin 𝑋 is 0 = (01, 02) where 01 is the zero element in 𝑋1 and 02 is the zero element in𝑋2. Similarly, the inverse of x = (x1,x2) is −x = (−x1,−x2).1.122 1.

x+ y = x+ z

−x+ (x+ y) = −x+ (x+ z)

(−x+ x) + y = (−x+ x) + z

0+ y = 0+ z

y = z

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2.

𝛼x = 𝛼y

1

𝛼(𝛼x) =

1

𝛼(𝛼y)(

1

𝛼𝛼

)x =

(1

𝛼𝛼

)y

x = y

3. 𝛼x = 𝛽x implies

(𝛼− 𝛽)x = 𝛼x− 𝛽x = 0

Provided x = 0, we must have

(𝛼− 𝛽)x = 0x

That is 𝛼− 𝛽 = 0 which implies 𝛼 = 𝛽.

4.

(𝛼− 𝛽)x = (𝛼+ (−𝛽))x

= 𝛼x+ (−𝛽)x

= 𝛼x− 𝛽x

5.

𝛼(x− y) = 𝛼(x+ (−1)y)

= 𝛼x+ 𝛼(−1)y

= 𝛼x− 𝛼y

6.

𝛼0 = 𝛼(x + (−x))= 𝛼x+ 𝛼(−x)= 𝛼x− 𝛼x= 0

1.123 The linear hull of the vectors {(1, 0), (0, 2)} is

lin {(1, 0), (0, 2)} =

{𝛼1

(10

)+ 𝛼2

(02

)}

={ (

𝛼1𝛼2

)}= ℜ2

The linear hull of the vectors {(1, 0), (0, 2)} is the whole plane ℜ2. Figure 1.4 illustrateshow any vector in ℜ2 can be obtained as a linear combination of {(1, 0), (0, 2)}.1.124 1. From the definition of 𝛼,

𝛼𝑆 = 𝑤(𝑆) −∑𝑇⊊𝑆

𝛼𝑇

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(−2, 3)

2

3

1

1-1-2 0

Figure 1.4: Illustrating the span of { (1, 0), (0, 2) }.

for every 𝑆 ⊆ 𝑁 . Rearranging

𝑤(𝑆) = 𝛼𝑆 +∑𝑇⊊𝑆

𝛼𝑇

=∑𝑇=𝑆

𝛼𝑇 +∑𝑇⊊𝑆

𝛼𝑇

=∑𝑇⊆𝑆

𝛼𝑇

2. ∑𝑇⊆𝑁

𝛼𝑇𝑤𝑇 (𝑆) =∑𝑇⊆𝑆

𝛼𝑇𝑤𝑇 (𝑆) +∑𝑇 ∕⊆𝑆

𝛼𝑇𝑤𝑇 (𝑆)

=∑𝑇⊆𝑆

𝛼𝑇 1 +∑𝑇 ∕⊆𝑆

𝛼𝑇 0

=∑𝑇⊆𝑆

𝛼𝑇 1

= 𝑤(𝑆)

1.125 1. Choose any x ∈ 𝑆. By homogeneity 0x = 𝜃 ∈ 𝑆.

2. For every x ∈ 𝑆, −x = (−1)x ∈ 𝑆 by homogeneity.

1.126 Examples of subspaces in ℜ𝑛 include:

1. The set containing just the null vector {0} is subspace.

2. Let x be any element in ℜ𝑛 and let 𝑇 be the set of all scalar multiples of x

𝑇 = {𝛼x : 𝛼 ∈ ℜ}𝑇 is a line through the origin in ℜ𝑛 and is a subspace.

3. Let 𝑆 be the set of all 𝑛-tuples with zero first coordinate, that is

𝑆 = { (𝑥1, 𝑥2, . . . , 𝑥𝑛) : 𝑥1 = 0, 𝑥𝑗 ∈ ℜ, 𝑗 ∕= 1 }For any x,y ∈ 𝑆

x+ y = (0, 𝑥2, 𝑥3, . . . , 𝑥𝑛) + (0, 𝑦2, 𝑦3, . . . , 𝑦𝑛)

= (0, 𝑥2 + 𝑦2, 𝑥3 + 𝑦3, . . . , 𝑥𝑛 + 𝑦𝑛) ∈ 𝑆

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Similarly

𝛼x = 𝛼(0, 𝑥2, 𝑥3, . . . , 𝑥𝑛)

= (0, 𝛼𝑥2, 𝛼𝑥3, . . . , 𝛼𝑥𝑛) ∈ 𝑆

Therefore 𝑆 is a subspace of ℜ𝑛. Generalizing, any set of vectors with one ormore coordinates identically zero is a subspace of ℜ𝑛.

4. We will meet some more complicated subspaces in Chapter 2.

1.127 No, −x /∈ ℜ𝑛+ if x ∈ ℜ𝑛+ unless x = 0. ℜ𝑛+ is an example of a cone (Section 1.4.5).

1.128 lin 𝑆 is a subspace Let x,y be two elements in lin 𝑆. x is a linear combinationof elements of 𝑆, that is

x = 𝛼1𝑥1 + 𝛼2𝑥2 + . . . 𝛼𝑛𝑥𝑛

Similarly

y = 𝛽1𝑥1 + 𝛽2𝑥2 + . . . 𝛽𝑛𝑥𝑛

and

x+ y = (𝛼1 + 𝛽1)𝑥1 + (𝛼2 + 𝛽2)𝑥2 + ⋅ ⋅ ⋅+ (𝛼𝑛 + 𝛽𝑛)𝑥𝑛 ∈ lin 𝑆

and

𝛼x = 𝛼𝛼1𝑥1 + 𝛼𝛼2𝑥2 + ⋅ ⋅ ⋅+ 𝛼𝛼𝑛𝑥𝑛 ∈ lin 𝑆

This shows that lin 𝑆 is closed under addition and scalar multiplication and henceis a subspace.

lin 𝑆 is the smallest subspace containing 𝑆 Let 𝑇 be any subspace containing 𝑆.Then 𝑇 contains all linear combinations of elements in 𝑆, so that lin 𝑆 ⊂ 𝑇 .Hence lin 𝑆 is the smallest subspace containing S.

1.129 The previous exercise showed that lin 𝑆 is a subspace. Therefore, if 𝑆 = lin 𝑆,𝑆 is a subspace.

Conversely, assume that 𝑆 is a subspace. Then 𝑆 is the smallest subspace containing𝑆, and therefore 𝑆 = lin 𝑆 (again by the previous exercise).

1.130 Let x,y ∈ 𝑆 = 𝑆1 ∩ 𝑆2. Hence x,y ∈ 𝑆1 and for any 𝛼, 𝛽 ∈ ℜ, 𝛼x + 𝛽y ∈ 𝑆1.Similarly 𝛼x+ 𝛽y ∈ 𝑆2 and therefore 𝛼x+ 𝛽y ∈ 𝑆. 𝑆 is a subspace.

1.131 Let 𝑆 = 𝑆1+𝑆2. First note that 0 = 0 + 0 ∈ 𝑆. Suppose x,y belong to 𝑆. Thenthere exist s1, t1 ∈ 𝑆1 and s2, t2 ∈ 𝑆2 such that x = s1 + s2 and y = t1 + t2. For any𝛼, 𝛽 ∈ ℜ,

𝛼x+ 𝛽y = 𝛼(s1 + s2) + 𝛽(t1 + t2)

= (𝛼s1 + 𝛽t1) + (𝛼s2 + 𝛽t2) ∈ 𝑆

since 𝛼s1 + 𝛽t1 ∈ 𝑆1 and 𝛼s2 + 𝛽t2 ∈ 𝑆2.1.132 Let

𝑆1 = {𝛼(1, 0) : 𝛼 ∈ ℜ}𝑆2 = {𝛼(0, 1) : 𝛼 ∈ ℜ}

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𝑆1 and 𝑆2 are respectively the horizontal and vertical axes in ℜ2. Their union is not asubspace, since for example(

11

)=

(10

)+

(01

)/∈ 𝑆1 ∪ 𝑆2

However, any vector in ℜ2 can be written as the sum of an element of 𝑆1 and an elementof 𝑆2. Therefore, their sum is the whole space ℜ2, that is

𝑆1 + 𝑆2 = ℜ2

1.133 Assume that 𝑆 is linearly dependent, that is there exists x1, . . . ,x𝑛 ∈ 𝑆 and𝛼2, . . . , 𝛼𝑛 ∈ 𝑅 such that

x1 = 𝛼2x2 + 𝛼3x3 + . . . , 𝛼𝑛x𝑛

Rearranging, this implies

1x1 − 𝛼2x2 − 𝛼3x3 − . . . 𝛼𝑛x𝑛 = 0

Conversely, assume there exist x1,x2, . . . ,x𝑛 ∈ x and 𝛼1, 𝛼2, . . . , 𝛼𝑛 ∈ ℜ such that

𝛼1x1 + 𝛼2x2 . . .+ 𝛼𝑛x𝑛 = 0

Assume without loss of generality that 𝛼1 ∕= 0. Then

x1 = −𝛼2𝛼1x2 − 𝛼3

𝛼1x3 − . . .− 𝛼𝑛

𝛼1x𝑛

which shows that

x1 ∈ lin 𝑆 ∖ {x1}1.134 Assume {(1, 1, 1), (0, 1, 1), (0, 0, 1)} are linearly dependent. Then there exists𝛼1, 𝛼2, 𝛼3 such that

𝛼1

⎛⎝1

11

⎞⎠ + 𝛼2

⎛⎝0

11

⎞⎠ + 𝛼3

⎛⎝0

01

⎞⎠ =

⎛⎝0

00

⎞⎠

or equivalently

𝛼1 = 0

𝛼1 + 𝛼2 = 0

𝛼1 + 𝛼2 + 𝛼3 = 0

which imply that

𝛼1 = 𝛼2 = 𝛼3 = 0

Therefore {(1, 1, 1), (0, 1, 1), (0, 0, 1)} are linearly independent.

1.135 Suppose on the contrary that 𝑈 is linearly dependent. That is, there exists aset of games { 𝑢𝑇1, 𝑢𝑇2 , . . . , 𝑢𝑇𝑚 } and nonzero coefficients (𝛼1, 𝛼2, . . . , 𝛼𝑚) such that(Exercise 1.133)

𝛼1𝑢𝑇1 + 𝛼2𝑢𝑇2 + . . .+ 𝛼𝑚𝑢𝑇𝑚 = 0 (1.16)

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Assume that the coalitions are ordered so that 𝑇1 has the smallest number of playersof any of the coalitions 𝑇1, 𝑇2, . . . , 𝑇𝑚. This implies that no coalition 𝑇2, 𝑇3, . . . , 𝑇𝑚 isa subset of 𝑇1 and

𝑢𝑇𝑗 (𝑇1) = 0 for every 𝑗 = 2, 3, . . . , 𝑛 (1.17)

Using (1.39), 𝑢𝑇1 can be expressed as a linear combination of the other games,

𝑢𝑇1 = −1/𝛼1

𝑚∑𝑗=2

𝛼𝑗𝑢𝑇𝑗 (1.18)

Substituting (1.40) this implies that

𝑢𝑇1(𝑇1) = 0

whereas

𝑢𝑇 (𝑇 ) = 1 for every 𝑇

by definition. This contradiction establishes that the set 𝑈 is linearly independent.

1.136 If 𝑆 is a subspace, then 0 ∈ 𝑆 and

𝛼x1 = 0

with 𝛼 ∕= 0 and x1 = 0 (Exercise 1.122). Therefore 𝑆 is linearly dependent (Exercise1.133).

1.137 Suppose x has two representations, that is

x = 𝛼1x1 + 𝛼2x2 + . . .+ 𝛼𝑛x𝑛

x = 𝛽1x1 + 𝛽2x2 + . . .+ 𝛽𝑛x𝑛

Subtracting

0 = (𝛼1 − 𝛽1)x1 + (𝛼2 − 𝛽2)x2 + . . .+ (𝛼𝑛 − 𝛽𝑛)x𝑛 (1.19)

Since {x1,x2, . . . , ,x𝑛} is linearly independent, (1.19) implies that 𝛼𝑖 = 𝛽𝑖 = 0 for all𝑖 (Exercise 1.133)

1.138 Let 𝑃 be the set of all linearly independent subsets of a linear space 𝑋 . 𝑃 ispartially ordered by inclusion. Every chain 𝐶 = {𝑆𝛼} ⊆ 𝑃 has an upper bound, namely∪

𝑆∈𝐶 𝑆. By Zorn’s lemma, 𝑃 has a maximal element 𝐵. We show that 𝐵 is a basisfor 𝑋 .

𝐵 is linearly independent since 𝐵 ∈ 𝑃 . Suppose that 𝐵 does not span 𝑋 so thatlin 𝐵 ⊂ 𝑋 . Then there exists some x ∈ 𝑋 ∖ lin 𝐵. The set 𝐵 ∪ {x} is a linearlyindependent and contains 𝐵, which contradicts the assumption that 𝐵 is the maximalelement of 𝑃 . Consequently, we conclude that 𝐵 spans 𝑋 and hence is a basis.

1.139 Exercise 1.134 established that the set 𝐵 = { (1, 1, 1), (0, 1, 1), (0, 0, 1)} is linearlyindependent. Since dim𝑅3 = 3, any other vectors must be linearly dependent on 𝐵.That is lin 𝐵 = ℜ3. 𝐵 is a basis.

By a similar argument to exercise 1.134, it is readily seen that {(1, 0, 0), (0, 1, 0), (0, 0, 1)}is linearly independent and hence constitutes a basis.

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1.140 Let 𝐴 = {a1, a2, . . . , a𝑛} and 𝐵 = {b1,b2, . . . ,b𝑚} be two bases for a linearspace 𝑋 . Let

𝑆1 = {𝑏1} ∪ 𝐴 = {b1, a1, a2, . . . , a𝑛 }𝑆 is linearly dependent (since 𝑏1 ∈ lin 𝐴) and spans 𝑋 . Therefore, there exists𝛼1, 𝛼2, . . . , 𝛼𝑛 and 𝛽1 such that

𝛽1b1 + 𝛼1a1 + 𝛼2a2 + . . .+ 𝛼𝑛a𝑛 = 0

At least one 𝛼𝑖 ∕= 0. Deleting the corresponding element a𝑖, we obtain another set 𝑆′1

of 𝑛 elements

𝑆′1 = {b1, a1, a2, . . . , a𝑖−1, a𝑖+1, . . . , a𝑛}

which is also spans 𝑋 . Adding the second element from 𝐵, we obtain the 𝑛+1 elementset

𝑆2 = {b1,b2, a1, a2, . . . , a𝑖−1, a𝑖+1, . . . , a𝑛}which again is linearly dependent and spans 𝑋 .

Continuing in this way, we can replace 𝑚 vectors in 𝐴 with the 𝑚 vectors from 𝐵 whilemaintaining a spanning set. This process cannot eliminate all the vectors in 𝐴, becausethis would imply that 𝐵 was linearly dependent. (Otherwise, the remaining b𝑖 would belinear combinations of preceding elements of 𝐵.) We conclude that necessarily 𝑚 ≤ 𝑛.Reversing the process and replacing elements of 𝐵 with elements of 𝐴 establishes that𝑛 ≤ 𝑚. Together these inequalities imply that 𝑛 = 𝑚 and 𝐴 and 𝐵 have the samenumber of elements.

1.141 Suppose that the coalitions are ordered in some way, so that

𝒫(𝑁) = {𝑆0, 𝑆1, 𝑆2, . . . , 𝑆2𝑛−1}with 𝑆0 = ∅. There are 2𝑛 coalitions. Each game 𝐺 ∈ 𝒢𝑁 corresponds to a unique listof length 2𝑛 of coalitional worths

v = (𝑣0, 𝑣1, 𝑣2, . . . , 𝑣2𝑛−1)

with 𝑣0 = 0. That is, each game defines a vector 𝑣 = (0, 𝑣1, . . . , 𝑣2𝑛−1) ∈ ℜ2𝑛 andconversely each vector 𝑣 ∈ ℜ2𝑛 (with 𝑣0 = 0) defines a game. Therefore, the space ofall games 𝒢𝑁 is formally identical to the subspace of ℜ2𝑛 in which the first componentis identically zero, which in turn is equivalent to the space ℜ2𝑛−1. Thus, 𝒢𝑁 is a2𝑛 − 1-dimensional linear space.

1.142 For illustrative purposes, we present two proofs, depending upon whether thelinear space is assumed to be finite dimensional or not. In the finite dimensional case,a constructive proof is possible, which forms the basis for practical algorithms forconstructing a basis.

Let 𝑆 be a linearly independent set in a linear space 𝑋 .

𝑋 is finite dimensional Let 𝑛 = dim𝑋 . Assume 𝑆 has 𝑚 elements and denote it𝑆𝑚.

If lin 𝑆𝑚 = 𝑋 , then 𝑆𝑚 is a basis and we are done. Otherwise, there exists somex𝑚+1 ∈ 𝑋 ∖ lin 𝑆𝑚. Adding x𝑚+1 to 𝑆𝑚 gives a new set of 𝑚+ 1 elements

𝑆𝑚+1 = 𝑆𝑚 ∪ {x𝑚+1 }

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which is also linearly independent ( since x𝑚+1 /∈ lin 𝑆𝑚).

If lin 𝑆𝑚+1 = 𝑋 , then 𝑆𝑚+1 is a basis and we are done. Otherwise, there existssome x𝑚+2 ∈ 𝑋 ∖ lin 𝑆𝑚+1. Adding x𝑚+2 to 𝑆𝑚+1 gives a new set of 𝑚 + 2elements

𝑆𝑚+2 = 𝑆𝑚+1 ∪ {x𝑚+2 }which is also linearly independent ( since x𝑚+2 /∈ lin 𝑆𝑚+2).

Repeating this process, we can construct a sequence of linearly independent sets𝑆𝑚, 𝑆𝑚+1, 𝑆𝑚+2 . . . such that lin 𝑆𝑚 ⫋ lin 𝑆𝑚+1 ⫋ lin 𝑆𝑚+2 ⋅ ⋅ ⋅ ⊆ 𝑋 . Eventu-ally, we will reach a set which spans 𝑋 and hence is a basis.

𝑋 is possibly infinite dimensional For the general case, we can adapt the proofof the existence of a basis (Exercise 1.138), restricting 𝑃 to be the class of alllinearly independent subsets of 𝑋 containing 𝑆.

1.143 Otherwise (if a set of 𝑛 + 1 elements was linearly independent), it could beextended to basis at least 𝑛 + 1 elements (exercise 1.142). This would contradict thefundamental result that all bases have the same number of elements (Exercise 1.140).

1.144 Every basis is linearly independent. Conversely, let 𝐵 = {x1,x2, . . . ,x𝑛 } be aset of linearly independent elements in an 𝑛-dimensional linear space 𝑋 . We have toshow that lin 𝐵 = 𝑋 .

Take any x ∈ 𝑋 . The set

𝐵 ∪ {x} = {x1,x2, . . . ,x𝑛,x }must be linearly dependent (Exercise 1.143). That is there exists numbers 𝛼1, 𝛼2, . . . , 𝛼𝑛, 𝛼,not all zero, such that

𝛼1x1 + 𝛼2x2 + ⋅ ⋅ ⋅+ 𝛼𝑛x𝑛 + 𝛼x = 0 (1.20)

Furthermore, it must be the case that 𝛼 ∕= 0 since otherwise

𝛼1x1 + 𝛼2x2 + ⋅ ⋅ ⋅+ 𝛼𝑛x𝑛 = 0

which contradicts the linear independence of 𝐴. Solving (1.20) for x, we obtain

x =1

𝛼𝛼1x1 + 𝛼2x2 + ⋅ ⋅ ⋅+ 𝛼𝑛x𝑛

Since x was an arbitrary element of 𝑋 , we conclude that 𝐵 spans 𝑋 and hence 𝐵 is abasis.

1.145 A basis spans 𝑋 . To establish the converse, assume that 𝐵 = {x1,x2, . . . ,x𝑛 }is a set of 𝑛 elements which span 𝑋 . If 𝑆 is linearly dependent, then one elementis linearly dependent on the other elements. Without loss of generality, assume thatx1 ∈ lin 𝐵 ∖ {x1}. Deleting x1 the set

𝐵 ∖ {x1} = {x2,x3, . . . ,x𝑛}also spans 𝑋 . Continuing in this fashion by eliminating dependent elements, we finishwith a linearly independent set of 𝑚 < 𝑛 elements which spans 𝑋 . That is, we canfind a basis of 𝑚 < 𝑛 elements, which contradicts the assumption that the dimensionof 𝑋 is 𝑛 (Exercise 1.140). Thus any set of 𝑛 vectors which spans 𝑋 must be linearlyindependent and hence a basis.

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1.146 We have previously shown

∙ that the set 𝑈 is linearly independent (Exercise 1.135).

∙ the space 𝒢𝑁 has dimension 2𝑛−1 (Exercise 1.141).

There are 2𝑛−1 distinct T-unanimity games 𝑢𝑇 in 𝑈 . Hence 𝑈 spans the 2𝑛−1 space𝒢𝑁 . Alternatively, note that any game 𝑤 ∈ 𝒢𝑁 can be written as a linear combinationof T-unanimity games (Exercise 1.75).

1.147 Let 𝐵 = {x1,x2, . . . ,x𝑚 } be a basis for 𝑆. Since 𝐵 is linearly independent,𝑚 ≤ 𝑛 (Exercise 1.143). There are two possibilities.

Case 1: 𝑚 = 𝑛. 𝐵 is a set of 𝑛 linearly independent elements in an 𝑛-dimensionalspace 𝑋 . Hence 𝐵 is a basis for 𝑋 and 𝑆 = lin 𝐵 = 𝑋 .

Case 2: 𝑚 < 𝑛. Since 𝐵 is linearly independent but cannot be a basis for the 𝑛-dimensional space 𝑋 , we must have 𝑆 = lin 𝐵 ⊂ 𝑋 .

Therefore, we conclude that if 𝑆 ⊂ 𝑋 is a proper subspace, it has a lower dimensionthan 𝑋 .

1.148 Let 𝛼1, 𝛼2, 𝛼3 be the coordinates of (1, 1, 1) for the basis {(1, 1, 1), (0, 1, 1), (0, 0, 1)}.That is ⎛

⎝111

⎞⎠ = 𝛼1

⎛⎝1

11

⎞⎠ + 𝛼2

⎛⎝0

11

⎞⎠ + 𝛼3

⎛⎝0

01

⎞⎠

which implies that 𝛼1 = 1, 𝛼2 = 𝛼3 = 0. Therefore (1, 0, 0) are the required coordinatesof the (1, 1, 1) with respect to the basis {(1, 1, 1), (0, 1, 1), (0, 0, 1)}.(1, 1, 1) are the coordinates of the vector (1, 1, 1) with respect to the standard basis.

1.149 A subset 𝑆 of a linear space 𝑋 is a subspace of 𝑋 if

𝛼x + 𝛽y ∈ 𝑆 for every x,y ∈ 𝑆 and for every 𝛼, 𝛽 ∈ ℜLetting 𝛽 = 1− 𝛼, this implies that

𝛼x+ (1− 𝛼)y ∈ 𝑆 for every x,y ∈ 𝑆 and 𝛼 ∈ ℜ𝑆 is an affine set.

Conversely, suppose that 𝑆 is an affine set containing 0, that is

𝛼x+ (1− 𝛼)y ∈ 𝑆 for every x,y ∈ 𝑆 and 𝛼 ∈ ℜLetting y = 0, this implies that

𝛼x ∈ 𝑆 for every x ∈ 𝑆 and 𝛼 ∈ ℜso that 𝑆 is homogeneous. Now letting 𝛼 = 1

2 , for every x and y in 𝑆,

1

2x+

1

2y ∈ 𝑆

and homogeneity implies

x+ y = 2

(1

2x+

1

2y

)∈ 𝑆

𝑆 is also additive. Hence 𝑆 is subspace.

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1.150 For any x ∈ 𝑆, let

𝑉 = 𝑆 − x = {v ∈ 𝑋 : v + x ∈ 𝑆 }

𝑉 is an affine set For any v1,v2 ∈ 𝑉 , there exist corresponding s1, s2 ∈ 𝑆 such thatv1 = s1 − x and v2 = s2 − x and therefore

𝛼v1 + (1 − 𝛼)v2 = 𝛼(s1 − x) + (1 − 𝛼)(s1 − x)= 𝛼s1 + (1− 𝛼)s2 − 𝛼x + (1− 𝛼)x

= s− x

where s = 𝛼𝑠1 + (1− 𝛼)𝑠2 ∈ 𝑆. There 𝑉 is an affine set.

𝑉 is a subspace Since x ∈ 𝑆, 0 = x − x ∈ 𝑉 . Therefore 𝑉 is a subspace (Exercise1.149).

𝑉 is unique Suppose that there are two subspaces 𝑉 1 and 𝑉 2 such that 𝑆 = 𝑉 1+x1

and 𝑆 = 𝑉 2 + x2. Then

𝑉1 + x1 = 𝑉2 + x2

𝑉1 = 𝑉2 + (x2 − x1)= 𝑉2 + x

where x = x2 − x1 ∈ 𝑋 . Therefore 𝑉1 is parallel to 𝑉2.

Since 𝑉1 is a subspace, 0 ∈ 𝑉1 which implies that −x ∈ 𝑉2. Since 𝑉2 is a subspace,this implies that x ∈ 𝑉2 and 𝑉2+x ⊆ 𝑉2. Therefore 𝑉1 = 𝑉2+x ⊆ 𝑉2. Similarly,𝑉2 ⊆ 𝑉1 and hence 𝑉1 = 𝑉2. Therefore the subspace 𝑉 is unique.

1.151 Let 𝑆 ∥ 𝑇 denote the relation 𝑆 is parallel to 𝑇 , that is

𝑆 ∥ 𝑇 ⇐⇒ 𝑆 = 𝑇 + x for some x ∈ 𝑋

The relation ∥ is

reflexive 𝑆 ∥ 𝑆 since 𝑆 = 𝑆 + 0

transitive Assume 𝑆 = 𝑇 + x and 𝑇 = 𝑈 + y. Then 𝑆 = 𝑈 + (x+ y)

symmetric 𝑆 = 𝑇 + x =⇒ 𝑇 = 𝑆 + (−x)Therefore ∥ is an equivalence relation.

1.152 See exercises 1.130 and 1.162.

1.153 1. Exercise 1.150

2. Assume x0 ∈ 𝑉 . For every x ∈ 𝐻

x = x0 + v = w ∈ 𝑉

which implies that 𝐻 ⊆ 𝑉 . Conversely, assume 𝐻 = 𝑉 . Then x0 = 0 ∈ 𝑉 since𝑉 is a subspace.

3. By definition, 𝐻 ⊂ 𝑋 . Therefore 𝑉 = 𝐻 − x ⊂ 𝑋 .

4. Let x1 /∈ 𝑉 . Suppose to the contrary

lin {x1, 𝑉 } = 𝑉 ′ ⊂ 𝑋

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Then

𝐻 ′ = x0 + 𝑉 ′

is an affine set (Exercise 1.150) which strictly contains 𝐻 . This contradicts thedefinition of 𝐻 as a maximal proper affine set.

5. Let x1 /∈ 𝑉 . By the previous part, x ∈ lin {x1, 𝑉 }. That is, there exists 𝛼 ∈ ℜsuch that

x = 𝛼x1 + v for some v ∈ 𝑉To see that 𝛼 is unique, suppose that there exists 𝛽 ∈ ℜ such that

x = 𝛽x1 + v′ for some v′ ∈ 𝑉Subtracting

0 = (𝛼− 𝛽)x1 + (v − v′)which implies that 𝛼 = 𝛽 since x1 /∈ 𝑉 .

1.154 Assume x,y ∈ 𝑋 . That is, x,y ∈ ℜ𝑛 and∑𝑖∈𝑁𝑥𝑖 =

∑𝑖∈𝑁𝑦𝑖 = 𝑤(𝑁)

For any 𝛼 ∈ ℜ, 𝛼x + (1− 𝛼)y ∈ ℜ𝑛 and∑𝑖∈𝑁𝛼𝑥𝑖 + (1− 𝛼)𝑦𝑖 = 𝛼

∑𝑖∈𝑁𝑥𝑖 + (1− 𝛼)

∑𝑖∈𝑁𝑦𝑖

= 𝛼𝑤(𝑁) + (1− 𝛼)𝑤(𝑁)

= 𝑤(𝑁)

Hence 𝑋 is an affine subset of ℜ𝑛.

1.155 See Exercise 1.129.

1.156 No. A straight line through any two points in ℜ𝑛+ extends outside ℜ𝑛+. Putdifferently, the affine hull of ℜ𝑛+ is the whole space ℜ𝑛.

1.157 Let

𝑉 = aff 𝑆 − x1= aff {0,x2 − x1,x3 − x1, . . . ,x𝑛 − x1 }

𝑉 is a subspace (0 ∈ 𝑉 ) and

aff 𝑆 = 𝑉 + x1

and

dim aff 𝑆 = dim𝑉

Note that the choice of x1 is arbitrary.

𝑆 is affinely dependent if and only if there exists some x𝑘 ∈ 𝑆 such that x𝑘 ∈aff (𝑆 ∖ {x𝑘}). Since the choice of x1 is arbitrary, we assume that x𝑘 ∕= x1.

x𝑘 ∈ aff (𝑆 ∖ {x𝑘}) ⇐⇒ x𝑘 ∈ (𝑉 + x1) ∖ {x𝑘}⇐⇒ x𝑘 − x1 ∈ 𝑉 ∖ {x𝑘 − x1}⇐⇒ x𝑘 − x1 ∈ lin {x2 − x1,x3 − x1, . . . ,x𝑘−1 − x1,

. . . ,x𝑘+1 − x1, . . . ,x𝑛 − x1 }

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Therefore, 𝑆 is affinely dependent if and only if {x2 − x1,x3 − x1, . . . ,x𝑛 − x1 } islinearly independent.

1.158 By the previous exercise, the set 𝑆 = {x1,x2, . . . ,x𝑛 } is affinely dependent ifand only if the set {x2 − x1,x3 − x1, . . . ,x𝑛 − x1 } is linearly dependent, so that thereexist numbers 𝛼2, 𝛼3, . . . , 𝛼𝑛, not all zero, such that

𝛼2(x2 − x1) + 𝛼3(x3 − x1) + ⋅ ⋅ ⋅+ 𝛼𝑛(x𝑛 − x1) = 0

or

𝛼2x2 + 𝛼3x3 + ⋅ ⋅ ⋅+ 𝛼𝑛x𝑛 −𝑛∑𝑖=2

𝛼𝑖x1 = 0

Let 𝛼1 = −∑𝑛𝑖=2 𝛼𝑖. Then

𝛼1x1 + 𝛼2x2 + . . .+ 𝛼𝑛x𝑛 = 0

and

𝛼1 + 𝛼2 + . . .+ 𝛼𝑛 = 0

as required.

1.159 Let

𝑉 = aff 𝑆 − x1 = aff { 0,x2 − x1,x3 − x1, . . . ,x𝑛 − x1 }

Then

aff 𝑆 = x1 + 𝑉

If 𝑆 is affinely independent, every x ∈ aff 𝑆 has a unique representation as

x = x1 + v, v ∈ 𝑉

with

v = 𝛼2(x2 − x1) + 𝛼3(x3 − x1) + ⋅ ⋅ ⋅+ 𝛼𝑛(x𝑛 − x1)

so that

x = x1 + 𝛼2(x2 − x1) + 𝛼3(x3 − x1) + ⋅ ⋅ ⋅+ 𝛼𝑛(x𝑛 − x1)

Define 𝛼1 = 1−∑𝑛𝑖=2 𝛼𝑖. Then

x = 𝛼1x1 + 𝛼2x2 + ⋅ ⋅ ⋅+ 𝛼𝑛x𝑛with

𝛼1 + 𝛼2 + ⋅ ⋅ ⋅+ 𝛼𝑛 = 1

x is a unique affine combination of the elements of 𝑆.

1.160 Assume that 𝑥, 𝑦 ∈ (𝑎, 𝑏) ⊆ ℜ. This means that 𝑎 < 𝑥 < 𝑏 and 𝑎 < 𝑦 < 𝑏. Forevery 0 ≤ 𝛼 ≤ 1

𝛼𝑥 + (1− 𝛼)𝑦 > 𝛼𝑎+ (1− 𝛼)𝑎

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and

𝛼𝑥 + (1− 𝛼)𝑦 < 𝛼𝑏 + (1− 𝛼)𝑏

Therefore 𝑎 < 𝛼𝑥+(1−𝛼)𝑦 < 𝑏 and 𝛼𝑥+(1−𝛼)𝑦 ∈ (𝑎, 𝑏). (𝑎, 𝑏) is convex. Substituting≤ for < demonstrates that [𝑎, 𝑏] is convex.

Let 𝑆 be an arbitrary convex set in ℜ. Assume that 𝑆 is not an interval. This impliesthat there exist numbers 𝑥, 𝑦, 𝑧 such that 𝑥 < 𝑦 < 𝑧 and 𝑥, 𝑧 ∈ 𝑆 while 𝑦 /∈ 𝑆. Define

𝛼 =𝑧 − 𝑦𝑧 − 𝑥

so that

1− 𝛼 =𝑦 − 𝑥𝑧 − 𝑥

Note that 0 ≤ 𝛼 ≤ 1 and that

𝛼𝑥+ (1− 𝛼)𝑧 =𝑧 − 𝑦𝑧 − 𝑥𝑥+

𝑦 − 𝑥𝑧 − 𝑥𝑧 = 𝑦 /∈ 𝑆

which contradicts the assumption that 𝑆 is convex. We conclude that every convex setin ℜ is an interval. Note that 𝑆 may be a hybrid interval such (𝑎, 𝑏] or [𝑎, 𝑏) as well asan open (𝑎, 𝑏) or closed [𝑎, 𝑏] interval.

1.161 Let (𝑁,𝑤) be a TP-coalitional game. If core(𝑁,𝑤) = ∅ then it is trivially convex.Otherwise, assume core(𝑁,𝑤) is nonempty and let x1 and x2 belong to core(𝑁,𝑤).That is ∑

𝑖∈𝑆𝑥1𝑖 ≥ 𝑤(𝑆) for every 𝑆 ⊆ 𝑁

∑𝑖∈𝑁𝑥1𝑖 = 𝑤(𝑁)

and therefore for any 0 ≤ 𝛼 ≤ 1∑𝑖∈𝑆𝛼𝑥1𝑖 ≥ 𝛼𝑤(𝑆) for every 𝑆 ⊆ 𝑁

∑𝑖∈𝑁𝛼𝑥1𝑖 = 𝛼𝑤(𝑁)

Similarly ∑𝑖∈𝑆

(1− 𝛼)𝑥2𝑖 ≥ (1 − 𝛼)𝑤(𝑆) for every 𝑆 ⊆ 𝑁∑𝑖∈𝑁

(1− 𝛼)𝑥2𝑖 = (1 − 𝛼)𝑤(𝑁)

Summing these two systems∑𝑖∈𝑆𝛼𝑥1𝑖 + (1− 𝛼)𝑥2𝑖 ≥ 𝛼𝑤(𝑆) + (1− 𝛼)𝑤(𝑆) = 𝑤(𝑆) for every 𝑆 ⊆ 𝑁

∑𝑖∈𝑁𝛼𝑥1𝑖 + (1− 𝛼)𝑥2𝑖 = 𝛼𝑤(𝑁) + (1− 𝛼)𝑤(𝑁) = 𝑤(𝑁)

That is, 𝛼𝑥1𝑖 + (1− 𝛼)𝑥2𝑖 belongs to core(𝑁,𝑤).

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1.162 Let ℭ be a collection of convex sets and let x, y belong to∩

𝑆∈ℭ 𝑆. for every𝑆 ∈ ℭ, x,y ∈ 𝑆 and therefore 𝛼x+ (1− 𝛼)y ∈ 𝑆 for all 0 ≤ 𝛼 ≤ 1 (since 𝑆 is convex).Therefore 𝛼x + (1− 𝛼)y ∈ ∩

𝑆∈ℭ 𝑆.

1.163 Fix some output 𝑦. Assume that x1,x2 ∈ 𝑉 (𝑦). This implies that both (𝑦,−x1)and (𝑦,−x2) belong to the production possibility set 𝑌 . If 𝑌 is convex

𝛼(𝑦,−x1) + (1− 𝛼)(𝑦,−x2) = (𝛼𝑦 + (1− 𝛼)𝑦, 𝛼x1 + (1− 𝛼)x2)

= (𝑦, 𝛼x1 + (1− 𝛼)x2) ∈ 𝑌

for every 𝛼 ∈ [0, 1]. This implies that 𝛼x1 + (1 − 𝛼)x2 ∈ 𝑉 (𝑦). Since the choice of 𝑦was arbitrary, this implies that 𝑉 (𝑦) is convex for every 𝑦.

1.164 Assume 𝑆1 and 𝑆2 are convex sets. Let 𝑆 = 𝑆1 + 𝑆2. Suppose x,y belong to 𝑆.Then there exist s1, t1 ∈ 𝑆1 and s2, t2 ∈ 𝑆2 such that x = s1 + s2 and y = t1 + t2. Forany 𝛼 ∈ [0, 1]

𝛼x+ (1 − 𝛼)y = 𝛼s1 + s2 + (1− 𝛼)t1 + t2

= 𝛼s1 + (1− 𝛼)t1 + 𝛼s2 + (1 − 𝛼)t2 ∈ 𝑆

since 𝛼s1 + (1−𝛼)t1 ∈ 𝑆1 and 𝛼s2 + (1−𝛼)t2 ∈ 𝑆2. The argument readily extends toany finite number of sets.

1.165 Without loss of generality, assume that 𝑛 = 2. Let 𝑆 = 𝑆1×𝑆2 ⊆ 𝑋 = 𝑋1×𝑋2.Suppose x = (𝑥1, 𝑥2) and y = (𝑦1, 𝑦2) belong to 𝑆. Then

𝛼x+ (1− 𝛼)y = 𝛼(𝑥1, 𝑥2) + (1 − 𝛼)(𝑦1, 𝑦2)

= (𝛼𝑥1, 𝛼𝑥2) + ((1− 𝛼)𝑦1, (1− 𝛼)𝑦2)

= (𝛼𝑥1 + (1 − 𝛼)𝑦1, 𝛼𝑥2 + (1− 𝛼)𝑦2) ∈ 𝑆

1.166 Let 𝛼x, 𝛼y be points in 𝛼𝑆 so that x,y ∈ 𝑆. Since 𝑆 is convex, 𝛽x+(1−𝛽)y ∈ 𝑆for every 0 ≤ 𝛽 ≤ 1. Multiplying by 𝛼

𝛼(𝛽x+ (1 − 𝛽)y) = 𝛽(𝛼x) + (1− 𝛽)(𝛼y) ∈ 𝛼𝑆

Therefore, 𝛼𝑆 is convex.

1.167 Combine Exercises 1.164 and 1.166.

1.168 The inclusion 𝑆 ⊆ 𝛼𝑆 + (1 − 𝛼)𝑆 is true for any set (whether convex or not),since for every x ∈ 𝑆

x = 𝛼x+ (1− 𝛼)x ∈ 𝛼𝑆 + (1− 𝛼)𝑆

The reverse inclusion 𝛼𝑆+(1−𝛼)𝑆 ⊆ 𝑆 follows directly from the definition of convexity.

1.169 Given any two convex sets 𝑆 and 𝑇 in a linear space, the largest convex setcontained in both is 𝑆 ∩ 𝑇 ; the smallest convex set containing both is conv 𝑆 ∪ 𝑇 .Therefore, the set of all convex sets is a lattice with

𝑆 ∧ 𝑇 = 𝑆 ∩ 𝑇𝑆 ∨ 𝑇 = conv 𝑆 ∪ 𝑇

The lattice is complete since every collection {𝑆𝑖} has a least upper bound conv ∪ 𝑆𝑖and a greatest lower bound ∩𝑆𝑖.

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1.170 If a set contains all convex combinations of its elements, it contains all convexcombinations of any two points, and hence is convex.

Conversely, assume that 𝑆 is convex. Let x be a convex combination of elements in 𝑆,that is let

x = 𝛼1x1 + 𝛼2x2 + . . .+ 𝛼𝑛x𝑛

where x1,x2, . . . ,x𝑛 ∈ 𝑆 and 𝛼1, 𝛼2, . . . , 𝛼𝑛 ∈ ℜ+ with 𝛼1 + 𝛼2 + . . . + 𝛼𝑛 = 1. Weneed to show that x ∈ 𝑆.

We proceed by induction of the number of points 𝑛. Clearly, x ∈ 𝑆 if 𝑛 = 1 or 𝑛 = 2.To show that it is true for 𝑛 = 3, let

x = 𝛼1x1 + 𝛼2x2 + 𝛼3x3

where x1,x2,x3 ∈ 𝑆 and 𝛼1, 𝛼2, 𝛼3 ∈ ℜ+ with 𝛼1 + 𝛼2 + 𝛼3 = 1. Assume that 𝛼𝑖 > 0for all 𝑖 (otherwise 𝑛 = 1 or 𝑛 = 2) so that 𝛼1 < 1. Rewriting

x = 𝛼1x1 + 𝛼2x2 + 𝛼3x3

= 𝛼1x1 + (1 − 𝛼1)(𝛼2

1− 𝛼1x2 +𝛼2

1− 𝛼1x3)

= 𝛼1x1 + (1 − 𝛼1)y

where

y =

(𝛼2

1− 𝛼1x2 +𝛼2

1− 𝛼1x3)

y is a convex combination of two elements x2 and x3 since

𝛼21− 𝛼1 +

𝛼21− 𝛼1 =

𝛼2 + 𝛼31− 𝛼1 = 1

and 𝛼2 + 𝛼3 = 1 − 𝛼1. Hence y ∈ 𝑆. Therefore x is a convex combination of twoelements x1 and 𝑦 and is also in 𝑆. Proceeding in this fashion, we can show that everyconvex combination belongs to 𝑆, that is conv 𝑆 ⊆ 𝑆.

1.171 This is precisely analogous to Exercise 1.128. We observe that

1. conv 𝑆 is a convex set.

2. if 𝐶 is any convex set containing 𝑆, then conv 𝑆 ⊆ 𝐶.

Therefore, conv 𝑆 is the smallest convex set containing S.

1.172 Note first that 𝑆 ⊆ conv 𝑆 for any set 𝑆. The converse for convex sets followsfrom Exercise 1.170.

1.173 Assume x ∈ conv (𝑆1 + 𝑆2). Then, there exist numbers 𝛼1, 𝛼2, . . . , 𝛼𝑛 and vec-tors x1,x2, . . . ,x𝑛 in 𝑆1 + 𝑆2 such that

x = 𝛼1x1 + 𝛼2x2 + ⋅ ⋅ ⋅+ 𝛼𝑛x𝑛For every x𝑖, there exists x1𝑖 ∈ 𝑆1 and x2𝑖 ∈ 𝑆2 such that

x𝑖 = x1𝑖 + x2𝑖

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and therefore

x =

𝑛∑𝑖=1

𝛼𝑖x1𝑖 +

𝑛∑𝑖=1

𝛼𝑖x2𝑖

= x1 + x2

where x1 =∑𝑛

𝑖=1 𝛼𝑖x1𝑖 ∈ 𝑆1 and x2 =

∑𝑛𝑖=1 𝛼𝑖x

2𝑖 ∈ 𝑆2. Therefore x ∈ conv 𝑆1 +

conv 𝑆2.

Conversely, assume that x ∈ conv 𝑆1 + conv 𝑆2. Then x = x1 + x2, where

x1 =

𝑛∑𝑖=1

𝛼𝑖𝑥1𝑖 , x1𝑖 ∈ 𝑆1

x2 =

𝑚∑𝑗=1

𝛽𝑗𝑥2𝑗 , x2𝑖 ∈ 𝑆2

and

x = x1 + x2 =

𝑛∑𝑖=1

𝛼𝑖𝑥1𝑖 +

𝑚∑𝑗=1

𝛽𝑗𝑥2𝑗 ∈ conv (𝑆1 + 𝑆2)

since x1𝑖 ,x2𝑗 ∈ 𝑆1 + 𝑆2 for every 𝑖 and 𝑗.

1.174 The dimension of the input requirement set 𝑉 (𝑦) is 𝑛. Its affine hull is ℜ𝑛.

1.175 1. Let

x = 𝛼1x1 + 𝛼2x2 + . . .+ 𝛼𝑛x𝑛 (1.21)

If 𝑛 > dim𝑆+1, the elements x1,x2, . . . ,x𝑛 ∈ 𝑆 are affinely dependent (Exercise1.157 and therefore there exist numbers 𝛽1, 𝛽2, . . . , 𝛽𝑛, not all zero, such that(Exercise 1.158)

𝛽1x1 + 𝛽2x2 + . . .+ 𝛽𝑛x𝑛 = 0 (1.22)

and

𝛽1 + 𝛽2 + . . .+ 𝛽𝑛 = 0

2. Combining (1.21) and (1.22)

x = x− 𝑡0

=

𝑛∑𝑖=1

𝛼𝑖x𝑖 − 𝑡𝑛∑𝑖=1

𝛽𝑖x𝑖

=𝑛∑𝑖=1

(𝛼𝑖 − 𝑡𝛽𝑖)x𝑖 (1.23)

for any 𝑡 ∈ ℜ.

3. Let 𝑡 = min𝑖{

𝛼𝑖

𝛽𝑖: 𝛽𝑖 > 0

}=

𝛼𝑗

𝛽𝑗

We note that

∙ 𝑡 > 0 since 𝛼𝑖 > 0 for every 𝑖.

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∙ If 𝛽𝑖 > 0, then 𝛼𝑖/𝛽𝑖 ≥ 𝛼𝑗/𝛽𝑗 ≥ 𝑡 and therefore 𝛼𝑖 − 𝑡𝛽𝑖 ≥ 0

∙ If 𝛽𝑖 ≤ 0 then 𝛼𝑖 − 𝑡𝛽𝑖 > 0 for every 𝑡 > 0.

∙ Therefore 𝛼𝑖 − 𝑡𝛽𝑖 ≥ 0 for every 𝑡 and

∙ 𝛼𝑖 − 𝑡𝛽𝑖 = 0 for 𝑖 = 𝑗.

Therefore, (1.23) represents x as a convex combination of only 𝑛− 1 points.

4. This process can be repeated until x is represented as a convex combination ofat most dim𝑆 + 1 elements.

1.176 Assume x is not an extreme point of 𝑆. Then there exists distinct x1 and x2 inS such that

x = 𝛼x1 + (1 − 𝛼)x2

Without loss of generality, assume 𝛼 ≤ 1/2 and let y = x2 − x. Then x+ y = x2 ∈ 𝑆.Furthermore

x− y = x− x2 + x

= 2x− x2= 2(𝛼x1 + (1 − 𝛼)x2)− x2= 2𝛼x1 + (1− 2𝛼)x2 ∈ 𝑆

since 𝛼 ≤ 1/2.

1.177 1. For any x = (𝑥1, 𝑥2) ∈ 𝐶2, there exists some 𝛼1 ∈ [0, 1] such that

𝑥1 = 𝛼1𝑐+ (1− 𝛼1)(−𝑐) = (2𝛼1 − 1)𝑐

In fact, 𝛼1 is defined by

𝛼1 =𝑥1 + 𝑐

2𝑐

Therefore (see Figure 1.5)(𝑥1𝑐

)= 𝛼1

(𝑐𝑐

)+ (1 − 𝛼1)

( −𝑐𝑐

)(𝑥1−𝑐

)= 𝛼1

(𝑐−𝑐

)+ (1− 𝛼1)

( −𝑐−𝑐

)

Similarly 𝑥2 = 𝛼2𝑐+ (1− 𝛼2)(−𝑐) where

𝛼2 =𝑥2 + 𝑐

2𝑐

Therefore, for any x ∈ 𝐶2,

x =

(𝑥1𝑥2

)= 𝛼2

(𝑥1𝑐

)+ (1 − 𝛼2)

(𝑥1−𝑐

)

= 𝛼1𝛼2

(𝑐𝑐

)+ (1− 𝛼1)𝛼2

( −𝑐𝑐

)

+ 𝛼1(1− 𝛼2)(𝑐−𝑐

)+ (1− 𝛼1)(1− 𝛼2)

( −𝑐−𝑐

)

= 𝛽1

(𝑐𝑐

)+ 𝛽2

( −𝑐𝑐

)+ 𝛽3

(𝑐−𝑐

)+ 𝛽4

( −𝑐−𝑐

)

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(𝑥1, c)

(𝑥1, -c)

𝑥1

x

Figure 1.5: A cube in ℜ2

where 0 ≤ 𝛽𝑖 ≤ 1 and

𝛽1 + 𝛽2 + 𝛽3 + 𝛽4 = 𝛼1𝛼2 + (1− 𝛼1)𝛼2 + 𝛼1(1− 𝛼2) + (1− 𝛼1)(1 − 𝛼2)= 𝛼1𝛼2 + 𝛼2 − 𝛼1𝛼2 + 𝛼1 − 𝛼1𝛼2 + 1− 𝛼1 − 𝛼2 + 𝛼1𝛼2

= 1

That is

𝑥 ∈ conv

{(𝑐𝑐

),

( −𝑐𝑐

),

(𝑐−𝑐

),

( −𝑐−𝑐

)}

2. (a) For any point (𝑥1, 𝑥2, . . . , 𝑥𝑛−1, 𝑐) which lies on face of the cube 𝐶𝑛, (𝑥1, 𝑥2, . . . , 𝑥𝑛−1) ∈𝐶𝑛−1 and therefore

(𝑥1, 𝑥2, . . . , 𝑥𝑛−1) ∈ conv {±𝑐,±𝑐, . . . ,±𝑐) } ⊆ ℜ𝑛−1

so that

x ∈ conv { (±𝑐,±𝑐, . . . ,±𝑐, 𝑐) } ⊆ ℜ𝑛

Similarly, any point (𝑥1, 𝑥2, . . . , 𝑥𝑛−1,−𝑐) on the opposite face lies in theconvex hull of the points { (±𝑐,±𝑐, . . . ,±𝑐,−𝑐) }.

(b) For any other point x = (𝑥1, 𝑥2, . . . , 𝑥𝑛) ∈ 𝐶𝑛, let

𝛼𝑛 =𝑥𝑛 + 𝑐

2𝑐

so that

𝑥𝑛 = 𝛼𝑛𝑐+ (1− 𝛼𝑛)(−𝑐)

Then

x =

⎛⎜⎜⎜⎜⎝𝑥1𝑥2. . .𝑥𝑛−1𝑥𝑛

⎞⎟⎟⎟⎟⎠ = 𝛼𝑛

⎛⎜⎜⎜⎜⎝𝑥1𝑥2. . .𝑥𝑛−1𝑐

⎞⎟⎟⎟⎟⎠ + (1 − 𝛼𝑛)

⎛⎜⎜⎜⎜⎝𝑥1𝑥2. . .𝑥𝑛−1−𝑐

⎞⎟⎟⎟⎟⎠

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Hence

x ∈ conv { (±𝑐,±𝑐, . . . ,±𝑐) } ⊂ ℜ𝑛

In other words

𝐶𝑛 ⊆ conv { (±𝑐,±𝑐, . . . ,±𝑐) } ⊂ ℜ𝑛

3. Let 𝐸 denote the set of points of the form { (±𝑐,±𝑐, . . . ,±𝑐) } ⊆ ℜ𝑛. Clearly,every point in 𝐸 is an extreme point of 𝐶𝑛. Conversely, we have shown that𝐶𝑛 ⊆ conv 𝐸. Therefore, no point x ∈ 𝐶𝑛 ∖𝐸 can be an extreme point of 𝐶𝑛. 𝐸is the set of extreme points of 𝐶𝑛.

4. Since 𝐶𝑛 is convex, and 𝐸 ⊂ 𝐶𝑛, conv 𝐸 ⊆ 𝐶𝑛. Consequently, 𝐶𝑛 = conv 𝐸.

1.178 Let x, y belong to 𝑆 ∖ 𝐹 is convex. For any 𝛼 ∈ [0, 1]

∙ 𝛼x+ (1− 𝛼)y ∈ 𝑆 since 𝑆 convex

∙ 𝛼x+ (1− 𝛼)y /∈ 𝐹 since 𝐹 is a face

Thus 𝛼x + (1− 𝛼)y ∈ 𝑆 ∖ 𝐹 which is convex.

1.179 1. Trivial.

2. Let {𝐹𝑖} be a collection of faces of 𝑆 and let 𝐹 =∪𝐹𝑖. Choose any x,y ∈ 𝑆.

If the line segment between x and y intersects 𝐹 , then it intersects some face 𝐹𝑖which implies that x,y ∈ 𝐹𝑖. Therefore, x,y ∈ 𝐹 =

∪𝐹𝑖.

3. Let {𝐹𝑖} be a collection of faces of 𝑆 and let 𝐹 =∩𝐹𝑖. Choose any x,y ∈ 𝑆. if

the line segment between x and y intersects 𝐹 , then it intersects every face 𝐹𝑖which implies that x,y ∈ 𝐹𝑖 for every 𝑖. Therefore, x,y ∈ 𝐹 =

∪𝐹𝑖.

4. Let 𝔉 be the collection of all faces of 𝑆. This is partially ordered by inclusion.By the previous result, every nonempty subcollection 𝔊 has a least upper bound(∩

𝐹∈𝔊 𝐹 ). Hence 𝔉 is a complete lattice (Exercise 1.47).

1.180 Let 𝑆 be a polytope. Then 𝑆 = conv {x1,x2, . . . ,x𝑛 }. Note that every extremepoint belongs to {x1,x2, . . . ,x𝑛 }. Now choose the smallest subset whose convex hullis still 𝑆, that is delete elements which can be written as convex combinations of otherelements. Suppose the minimal subset is {x1,x2, . . . ,x𝑚 }. We claim that each ofthese elements is an extreme point of 𝑆, that is {x1,x2, . . . ,x𝑚 } = 𝐸.

Assume not, that is assume that x𝑚 is not an extreme point so that there exists x,y ∈ 𝑆with

x𝑚 = 𝛼x+ (1− 𝛼)y with 0 < 𝛼 < 1 (1.24)

Since x,y ∈ conv {x1,x2, . . . ,x𝑚}

x =𝑚∑𝑖=1

𝛼𝑖x𝑖 y =𝑚∑𝑖=1

𝛽x𝑖

Substituting in (1.24), we can write x𝑚 as a convex combination of {x1,x2, . . . ,x𝑚}.

x𝑚 =

𝑚∑𝑖=1

(𝛼𝛼𝑖 + (1 − 𝛼)𝛽𝑖

)x𝑖 =

𝑚∑𝑖=1

𝛾𝑖x𝑖

where

𝛾𝑖 = 𝛼𝛼𝑖 + (1− 𝛼)𝛽𝑖

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Note that 0 ≤ 𝛾𝑖 ≤ 1, so that either 𝛾𝑚 < 1 or 𝛾𝑚 = 1. We show that both cases leadto a contradiction.

∙ 𝛾𝑚 < 1. Then

x𝑚 =1

1− 𝛾𝑚𝑚−1∑𝑖=1

(𝛼𝛼𝑖 + (1− 𝛼)𝛽𝑖

)x𝑖

which contradicts the minimality of the set {x1,x2, . . . ,x𝑚}.∙ 𝛾𝑚 = 1. Then 𝛾𝑖 = 0 for every 𝑖 ∕= 𝑚. That is

𝛼𝛼𝑖 + (1− 𝛼)𝛽𝑖 = 0 for every 𝑖 ∕= 𝑚which implies that 𝛼𝑖 = 𝛽𝑖 for every 𝑖 ∕= 𝑚 and therefore x = y.

Therefore, if {x1,x2, . . . ,x𝑚} is a minimal spanning set, every point must be an extremepoint.

1.181 Assume to the contrary that one of the vertices is not an extreme point of thesimplex. Without loss of generality, assume this is x1. Then, there exist distincty, z ∈ 𝑆 and 0 < 𝛼 < 1 such that

x1 = 𝛼y + (1− 𝛼)z (1.25)

Now, since y ∈ 𝑆, there exist 𝛽1, 𝛽2, . . . , 𝛽𝑛 such that

y =

𝑛∑𝑖=1

𝛽𝑖x𝑖,

𝑛∑𝑖=1

𝛽𝑖 = 1

Similarly, there exist 𝛿1, 𝛿2, . . . , 𝛿𝑛 such that

z =𝑛∑𝑖=1

𝛿𝑖x𝑖,𝑛∑𝑖=1

𝛿𝑖 = 1

Substituting in (1.25)

x1 = 𝛼

𝑛∑𝑖=1

𝛽𝑖x𝑖 + (1− 𝛼)

𝑛∑𝑖=1

𝛿𝑖x𝑖

=

𝑛∑𝑖=1

(𝛼𝛽𝑖 + (1− 𝛼)𝛿𝑖

)x𝑖

Since∑𝑛

𝑖=1

(𝛼𝛽𝑖 + (1 − 𝛼)𝛿𝑖

)= 𝛼

∑𝑛𝑖=1 𝛽𝑖 + (1− 𝛼)

∑𝑖=1 𝛿𝑖 = 1

x1 =

𝑛∑𝑖=1


)x𝑖

Subtracting, this implies

0 =

𝑛∑𝑖=2


)(x𝑖 − x1)

This establishes that the set {x2−x1,x3−x1, . . . ,x𝑛−x1 } is linearly dependent andtherefore 𝐸 = {x1,x2, . . . ,x𝑛 } is affinely dependent (Exercise 1.157). This contradictsthe assumption that 𝑆 is a simplex.

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1.182 Let 𝑛 be the dimension of a convex set 𝑆 in a linear space 𝑋 . Then 𝑛 = dim aff 𝑆and there exists a set {x1,x2, . . . ,x𝑛+1 } of affinely independent points in 𝑆. Define

𝑆′ = conv {x1,x2, . . . ,x𝑛+1 }Then 𝑆′ is an 𝑛-dimensional simplex contained in 𝑆.

1.183 Let w = (𝑤({1}), 𝑤({2}), . . . , 𝑤({𝑛})) denote the vector of individual worths andlet 𝑠 denote the surplus to be distributed, that is

𝑠 = 𝑤(𝑁)−∑𝑖∈𝑁𝑤({𝑖})

𝑠 > 0 if the game is essential. For each player 𝑖 = 1, 2, . . . , 𝑛, let

y𝑖 = w + 𝑠e𝑖

be the outcome in which player 𝑖 receives the entire surplus. (e𝑖 is the 𝑖th unit vector.)Note that

𝑦𝑖𝑗 =

{𝑤({𝑖}) + 𝑠 𝑗 = 𝑖

𝑤({𝑖}) 𝑗 ∕= 𝑖

Each y𝑖 is an imputation since 𝑦𝑖𝑗 ≥ 𝑤({𝑗}) and∑𝑗∈𝑁𝑦𝑖𝑗 =

∑𝑗∈𝑁𝑤({𝑗}) + 𝑠 = 𝑤(𝑁)

Therefore {y1,y2, . . . ,y𝑛 } ⊆ 𝐼. Since 𝐼 is convex (why ?), 𝑆 = conv {y1,y2, . . . ,y𝑛 } ⊆𝐼. Further, for every 𝑖, 𝑗 ∈ 𝑁 the vectors

y𝑖 − y𝑗 = 𝑠(e𝑖 − e𝑗)are linearly independent. Therefore 𝑆 is an 𝑛− 1-dimensional simplex in ℜ𝑛.

For any x ∈ 𝐼 define

𝛼𝑖 =𝑥𝑖 − 𝑤({𝑖})

𝑠

so that

𝑥𝑗 = 𝑤({𝑗}) + 𝛼𝑗𝑠

Since x is an imputation

∙ 𝛼𝑖 ≥ 0

∙ ∑𝑖∈𝑁 𝛼𝑖 =

(∑𝑖∈𝑁 𝑥𝑖 −

∑𝑖∈𝑁 𝑤({𝑖})) /𝑠 = 1

We claim that x =∑

𝑖∈𝑁 𝛼𝑖y𝑖 since for each 𝑗 = 1, 2, . . . , 𝑛∑

𝑖∈𝑁𝛼𝑖𝑦

𝑖𝑗 =

∑𝑖∈𝑁𝛼𝑖𝑤({𝑗}) + 𝛼𝑗𝑠

= 𝑤({𝑗}) + 𝛼𝑗𝑠

= 𝑥𝑗

Therefore x ∈ conv {y1,y2, . . . ,y𝑛 } = 𝑆, that is 𝐼 ⊆ 𝑆. Since we previously showedthat 𝑆 ⊆ 𝐼, we have established that 𝐼 = 𝑆, which is an 𝑛− 1 dimensional simplex inℜ𝑛.

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𝑥1

𝑥2

𝑥1

𝑥2

𝑥1

𝑥2

3. A convex cone1. A non-convex cone 2. A convex set

Figure 1.6: A cone which is not convex, a convex set and a convex cone


1.185 Let x = (𝑥1, 𝑥2, . . . , 𝑥𝑛) belong to ℜ𝑛+, which means that 𝑥𝑖 ≥ 0 for every 𝑖. Forevery 𝛼 > 0

𝛼x = (𝛼x1, 𝛼x2, . . . , 𝛼x𝑛)

and 𝛼𝑥𝑖 ≥ 0 for every 𝑖. Therefore 𝛼x ∈ ℜ𝑛+. ℜ𝑛+ is a cone in ℜ𝑛.

1.186 Assume

𝛼x+ 𝛽y ∈ 𝑆 for every x,y ∈ 𝑆 and 𝛼, 𝛽 ∈ ℜ+ (1.26)

Letting 𝛽 = 0, this implies that

𝛼x ∈ 𝑆 for every x ∈ 𝑆 and 𝛼 ∈ ℜ+so that 𝑆 is a cone. To show that 𝑆 is convex, let x and y be any two elements in 𝑆.For any 𝛼 ∈ [0, 1], (1.26) implies that

𝛼x + (1− 𝛼)y ∈ 𝑆

Therefore 𝑆 is convex.

Conversely, assume that 𝑆 is a convex cone. For any 𝛼, 𝛽 ∈ ℜ+ and x,y ∈ 𝑆𝛼

𝛼+ 𝛽x+

𝛽

𝛼+ 𝛽y ∈ 𝑆

and therefore

𝛼x + 𝛽y ∈ 𝑆

1.187 Assume 𝑆 satisfies

1. 𝛼𝑆 ⊆ 𝑆 for every 𝛼 ≥ 0

2. 𝑆 + 𝑆 ⊆ 𝑆

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By (1), 𝑆 is a cone. To show that it is convex, let x and y belong to 𝑆. By (1), 𝛼x and(1− 𝛼)y belong to 𝑆, and therefore 𝛼x+ (1− 𝛼)y belongs to 𝑆 by (2). 𝑆 is convex.

Conversely, assume that 𝑆 is a convex cone. Then

𝛼𝑆 ⊆ 𝑆 for every 𝛼 ≥ 0

Let x and y be any two elements in 𝑆. Since 𝑆 is convex, 𝑧 = 𝛼12x + (1 − 𝛼)12y ∈ 𝑆and since it is a cone, 2𝑧 = x+ y ∈ 𝑆. Therefore

𝑆 + 𝑆 ⊆ 𝑆

1.188 We have to show that 𝑌 is convex cone. By assumption, 𝑌 is convex. To showthat 𝑌 is a cone, let y be any production plan in 𝑌 . By convexity

𝛼y = 𝛼y + (1− 𝛼)0 ∈ 𝑌 for every 0 ≤ 𝛼 ≤ 1

Repeated use of additivity ensures that

𝛼y ∈ 𝑌 for every 𝛼 = 1, 2, . . .

Combining these two conclusions implies that

𝛼y ∈ 𝑌 for every 𝛼 ≥ 0

1.189 Let 𝒮 ⊂ 𝒢𝑁 denote the set of all superadditive games. Let 𝑤1, 𝑤2 ∈ 𝑆 be twosuperadditive games. Then, for all distinct coalitions 𝑆, 𝑇 ⊂ 𝑁 with 𝑆 ∩ 𝑇 = ∅

𝑤1(𝑆 ∪ 𝑇 ) ≥ 𝑤1(𝑆) + 𝑤1(𝑇 )

𝑤2(𝑆 ∪ 𝑇 ) ≥ 𝑤2(𝑆) + 𝑤2(𝑇 )

Adding

(𝑤1 + 𝑤2)(𝑆 ∪ 𝑇 ) = 𝑤1(𝑆 ∪ 𝑇 ) + 𝑤2(𝑆 ∪ 𝑇 )

≥ 𝑤1(𝑆) + 𝑤2(𝑆) + 𝑤1(𝑇 ) + 𝑤2(𝑇 )

= (𝑤1 + 𝑤2)(𝑆) + (𝑤1 + 𝑤2)(𝑇 )

so that 𝑤1 +𝑤2 is superadditive. Similarly, we can show that 𝛼𝑤1 is superadditive forall 𝛼 ∈ ℜ+. Hence 𝒮 is a convex cone in 𝒢𝑁 .

1.190 Let x belong to∩𝑛

𝑖=1 𝑆𝑖. Then x ∈ 𝑆𝑖 for every 𝑖. Since each 𝑆𝑖 is a cone,𝛼x ∈ 𝑆𝑖 for every 𝛼 ≥ 0 and therefore 𝛼x ∈ ∩𝑛

𝑖=1 𝑆𝑖.

Let 𝑆 = 𝑆1 + 𝑆2 + ⋅ ⋅ ⋅ + 𝑆𝑛 and assume x belongs to 𝑆. Then there exist x𝑖 ∈ 𝑆𝑖,𝑖 = 1, 2, . . . , 𝑛 such that

x = x1 + x2 + ⋅ ⋅ ⋅+ x𝑛For any 𝛼 ≥ 0

𝛼x = 𝛼(x1 + x2 + ⋅ ⋅ ⋅+ x𝑛)

= 𝛼x1 + 𝛼x2 + ⋅ ⋅ ⋅+ 𝛼x𝑛 ∈ 𝑆

since 𝛼x𝑖 ∈ 𝑆𝑖 for every 𝑖.

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1.191 1. Suppose that y ∈ 𝑌 . Then, there exist 𝛼,𝛼2, . . . , 𝛼8 ≥ 0 such that

y =

8∑𝑖=1

𝛼𝑖y𝑖

and for the first commodity

y1 =

8∑𝑖=1

𝛼𝑖𝑦𝑖1

If y ∕= 0, at least one of the 𝛼𝑖 > 0 and hence y1 < 0 since 𝑦𝑖1 < 0 for 𝑖 =1, 2, . . . , 8.

2. Free disposal requires that y ∈ 𝑌,y′ ≤ y =⇒ y′ ∈ 𝑌 . Consider the productionplan y′ = (−2,−2,−2,−2). Note that y′ ≤ y3 and y′ ≤ y6. Suppose thaty′ ∈ 𝑌 . Then there exist 𝛼1, 𝛼2, . . . , 𝛼8 ≥ 0 such that

y =

8∑𝑖=1

𝛼𝑖y𝑖

For the third commodity (component), we have

4𝛼1 + 3𝛼2 + 3𝛼3 + 3𝛼4 + 12𝛼5 − 2𝛼6 + 5𝛼8 = −2 (1.27)

and for the fourth commodity

2𝛼2 − 1𝛼3 + 1𝛼4 + 5𝛼6 + 10𝛼7 − 2𝛼8 = −2 (1.28)

Adding to (1.28) to (1.27) gives

4𝛼1 + 5𝛼2 + 2𝛼3 + 4𝛼4 + 12𝛼5 + 3𝛼6 + 10𝛼7 + 3𝛼8 = −4

which is impossible given that 𝛼𝑖 ≥ 0. Therefore, we conclude that y′ /∈ 𝑌 .

3.

y2 = (−7,−9, 3, 2) ≥ (−8,−13, 3, 1) = y4

3y1 = (−9,−18, 12, 0) ≥ (−11,−19, 12, 0) = y5

y7 = (−8,−5, 0, 10) ≥ (−8,−6,−4, 10) = 2y6

2y3 = (−2,−4, 6,−2) ≥ (−2,−4, 5,−2) = y8

4.

2y3 + y7 = (−2,−4, 6,−2) + (−8,−5, 0, 10)

= (−10,−9, 6, 8)

≥ (−14,−18, 6, 4) = 2y2

20y3 + 2y7 = 20(−1,−2, 3,−1) + 2(−8,−5, 0, 10)

= (−20,−40, 60,−20) + (−16,−10, 0, 20)

= (−36,−50, 60, 0)

≥ (−45,−90, 60, 0) = 15y1

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5. Eff(𝑌 ) = cone {y3,y7 }.1.192 This is precisely analogous to Exercise 1.128. We observe that

1. cone 𝑆 is a convex cone.

2. if 𝐶 is any convex cone containing 𝑆, then conv 𝑆 ⊆ 𝐶.

Therefore, cone 𝑆 is the smallest convex cone containing S.

1.193 For any set 𝑆, 𝑆 ⊆ cone 𝑆. If 𝑆 is a convex cone, Exercise 1.186 implies thatcone 𝑆 ⊆ 𝑆.

1.194 1. If 𝑛 > 𝑚 = dim cone 𝑆 = dim lin 𝑆, the elements x1,x2, . . . ,x𝑛 ∈ 𝑆 arelinearly dependent and therefore there exist numbers 𝛽1, 𝛽2, . . . , 𝛽𝑛, not all zero,such that (Exercise 1.134)

𝛽1x1 + 𝛽2x2 + . . .+ 𝛽𝑛x𝑛 = 0 (1.29)

2. Combining (1.14) and (1.29)

x = x− 𝑡0

=

𝑛∑𝑖=1

𝛼𝑖x𝑖 − 𝑡𝑛∑𝑖=1

𝛽𝑖x𝑖

=

𝑛∑𝑖=1

(𝛼𝑖 − 𝑡𝛽𝑖)x𝑖 (1.30)

for any 𝑡 ∈ ℜ.

3. Let 𝑡 = min𝑖{

𝛼𝑖

𝛽𝑖: 𝛽𝑖 > 0

}=

𝛼𝑗

𝛽𝑗

We note that

∙ 𝑡 > 0 since 𝛼𝑖 > 0 for every 𝑖.

∙ If 𝛽𝑖 > 0, then 𝛼𝑖/𝛽𝑖 ≥ 𝛼𝑗/𝛽𝑗 ≥ 𝑡 and therefore 𝛼𝑖 − 𝑡𝛽𝑖 ≥ 0.

∙ If 𝛽𝑖 ≤ 0 then 𝛼𝑖 − 𝑡𝛽𝑖 > 0 for every 𝑡 > 0.

∙ Therefore 𝛼𝑖 − 𝑡𝛽𝑖 ≥ 0 for every 𝑡 and

∙ 𝛼𝑖 − 𝑡𝛽𝑖 = 0 for 𝑖 = 𝑗.

Therefore, (1.30) represents x as a nonnegative combination of only 𝑛− 1 points.

4. This process can be repeated until x is represented as a convex combination ofat most 𝑚 points.

1.195 1. The affine hull of 𝑆 is parallel to the affine hull of 𝑆. Therefore

dim𝑆 = dim aff 𝑆 = dim aff 𝑆

Since

(00

)/∈ aff 𝑆,

dim cone 𝑆 = dim aff 𝑆 + 1 = dim𝑆 + 1

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2. For every x ∈ conv 𝑆,

(x1

)∈ conv 𝑆 and there exist (Exercise 1.194) 𝑚 + 1

points

(x𝑖1

)∈ 𝑆 such that

(x1

)∈ conv {

(x11

),

(x21

), . . .

(x𝑚+1

1

)}

This implies that

x ∈ conv {x1,x2, . . . ,x𝑚+1 }

with x1,x2, . . . ,x𝑚+1 ∈ 𝑆.

1.196 A subsimplex with precisely one distinguished face is completely labeled. Supposea subsimplex has more than one distinguished face. This means that it has verticeslabeled 1, 2, . . . , 𝑛. Since it has 𝑛 + 1 vertices, one of these labels must be repeated(twice). The distinguished faces lie opposite the repeated vertices. There are preciselytwo distinguished faces.

1.197 1. 𝜌(x,y) = ∥x− y∥ ≥ 0.

2. 𝜌(x,y) = ∥x− y∥ = 0 if and only if x− y = 0, that is x = y.

3. Property (3) ensures that ∥−x∥ = ∥x∥ and therefore ∥x− y∥ = ∥y − x∥ so that

𝜌(x,y) = ∥x− y∥ = ∥y − x∥ = 𝜌(y,x)

4. For any z ∈ 𝑋

𝜌(x,y) = ∥x− y∥= ∥x− z+ z− y∥≤ ∥x− z∥ + ∥z− y∥= 𝜌(x, z) + 𝜌(z,y)

Therefore 𝜌(x,y) = ∥x− y∥ satisfies the properties required of a metric.

1.198 Clearly ∥x∥∞ ≥ 0 and ∥x∥∞ = 0 if and only if x = 0. Thirdly

∥𝛼x∥ =𝑛

max𝑖=1∣𝛼𝑥𝑖∣ = ∣𝛼∣ 𝑛

max𝑖=1∣𝑥𝑖∣ = ∣𝛼∣ ∥x∥

To prove the triangle inequality, we note that for any 𝑥𝑖, 𝑦𝑖 ∈ ℜ

max(𝑥𝑖 + 𝑦𝑖) ≤ max𝑥𝑖 + max 𝑦𝑖

Therefore

∥x∥ =𝑛

max𝑖=1

(𝑥𝑖 + 𝑦𝑖) ≤ 𝑛max𝑖=1𝑥𝑖 +

𝑛max𝑖=1𝑦𝑖 = ∥x∥ + ∥y∥

1.199 Suppose that producing one unit of good 1 requires two units of good 2 and threeunits of good 3. The production plan is (1,−2,−3) and the average net output, −2,is negative. A norm is required to be nonnegative. Moreover, the quantities of inputsand outputs may balance out yielding a zero average. That is, (

∑𝑛𝑖=1 𝑦𝑖)/𝑛 = 0 does

not imply that 𝑦𝑖 = 0 for all 𝑖.

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1.200

∥x∥ − ∥y∥ = ∥x− y + y∥ − ∥y∥≤ ∥x− y∥ + ∥y∥ − ∥y∥= ∥x− y∥

1.201 Using the previous exercise

∥x𝑛∥ − ∥x∥ ≤ ∥x𝑛 − x∥ → 0

1.202 First note that each term x𝑛 + y𝑛 ∈ 𝑋 by linearity. Similarly, x + y ∈ 𝑋 . Fixsome 𝜖 > 0. There exists some 𝑁x such that ∥x𝑛 − x∥ < 𝜖 for all 𝑛 ≥ 𝑁x. Similarly,there exists some𝑁y such that ∥y𝑛 − y∥ < 𝜖 for all 𝑛 ≥ 𝑁y. For all 𝑛 ≥ max{𝑁x, 𝑁y },

∥(x𝑛 + y𝑛)− (x+ y)∥ = ∥(x𝑛 − x) + (y𝑛 − y)∥≤ ∥x𝑛 − x∥+ ∥y𝑛 − y∥< 𝜖

Similarly, for every 𝑛 ≥ 𝑁x

∥𝛼x𝑛 − 𝛼x∥ = ∣𝛼∣ ∥x𝑛 − x∥≤ ∣𝛼∣ 𝜖/2→ 0 as 𝜖→ 0

1.203 Let x𝑛 be a sequence in 𝑆 + 𝑇 converging to x. For every 𝑛, there exists y𝑛 ∈ 𝑆and z𝑛 ∈ 𝑇 such that x𝑛 = y𝑛 +z𝑛. Since 𝑇 is compact, there exists a subsequence z𝑚converging to z ∈ 𝑇 . Let y = lim 𝑦𝑚. Then y ∈ 𝑆 since 𝑆 is closed. By the previousexercise, y𝑚 + z𝑚 → y+ z. By assumption, y𝑚 + z𝑚 → x so that x = y+ z ∈ 𝑆 + 𝑇 .𝑆 + 𝑇 is closed.

1.204 Yes. Apply Exercise 1.202.

1.205 The 𝑛th partial sum of the series is

s𝑛 = x+ 𝛽x+ 𝛽2x+ ⋅ ⋅ ⋅+ 𝛽𝑛−1xMultiplying this equation by 𝛽 gives

𝛽s𝑛 = 𝛽x+ 𝛽2x+ 𝛽3x+ ⋅ ⋅ ⋅+ 𝛽𝑛xSubtracting this equation from the previous one and canceling common terms gives

(1− 𝛽)s𝑛 = x− 𝛽𝑛x = (1− 𝛽𝑛)x

Provided that 𝛽 ∕= 1

s𝑛 =x− 𝛽𝑛x

1− 𝛽 =x

1− 𝛽 −𝛽𝑛x

1− 𝛽 (1.31)

If 𝛽 < 1, then 𝛽𝑛 → 0 (Exercise 1.102) and therefore 𝑠𝑛 converges to x/(1− 𝛽).

1.206

1 +1

2+

1

4+

1

8+

1

16+ . . .

is a geometric series 1 +𝛽+𝛽2+𝛽3+ . . . with 𝛽 = 1/2. The series converges (Exercise1.205) to

1 +1

2+

1

4+

1

8+

1

16+ ⋅ ⋅ ⋅ = 1

1− 𝛽 =1

1− 12

= 2

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1.207 The present value of the 𝑛 payments is the 𝑛th partial sum of the geometricseries 𝑥+ 𝛽𝑥 + 𝛽2𝑥+ 𝛽3𝑥+ . . . which (using (1.31)) is given by

Present value = 𝑠𝑛 =𝑥− 𝛽𝑛𝑥

1− 𝛽1.208 By Exercise 1.93, there exists an open set 𝑇 ⊇ 𝑆1 such that 𝑇 ∩𝑆2 = ∅. For everyx ∈ 𝑆1, there exists an open ball 𝐵(x) such that 𝐵(x) ⊆ 𝑇 and therefore 𝐵(x)∩𝑆2 = ∅.The collection {𝐵(x) } of open balls is an open cover for 𝑆1. Since 𝑆1 is compact thereexists a finite subcover, that is there exists points x1,x2, . . . ,x𝑛 in 𝑆1 such that

𝑆1 ⊆𝑛∪𝑖=1

𝐵(x𝑖)

Furthermore, for every x𝑖, there exists 𝑟𝑛 such that

𝐵(x𝑖) = x𝑖 + 𝑟𝑛𝐵

where 𝐵 is the unit ball. Let 𝑟 = min 𝑟𝑛. 𝑈 = 𝑟𝐵 is the required neighborhood.

1.209 Clearly 𝑋 × 𝑌 is a normed linear space. To show that it is complete, let (z𝑛) bea Cauchy sequence in 𝑋 × 𝑌 where z𝑛 = (x𝑛,y𝑛). For every 𝜖 > 0, there exists some𝑁 such that

∥z𝑛 − z𝑚∥ = max{ ∥x𝑛 − x𝑚∥ , ∥y𝑛 − y𝑚∥ } < 𝜖

for every 𝑛,𝑚 ≥ 𝑁 . This implies that (x𝑛) and (y𝑛) are Cauchy sequences in 𝑋 and𝑌 respectively. Since 𝑋 and 𝑌 are complete, both sequences converge. That is, thereexists x ∈ 𝑋 and y ∈ 𝑌 such that ∥x𝑛 − x∥ → 0 and ∥y𝑛 − y∥ → 0. In other words,given 𝜖 > 0 there exists 𝑁 such that ∥x𝑛 − x∥ < 𝜖 and ∥y𝑛 − y∥ < 𝜖 for every 𝑛 ≥ 𝑁 .Let z = (x,y). Then, for every 𝑛 ≥ 𝑁

∥z𝑛 − z∥ = max{ ∥x𝑛 − x∥ , ∥y𝑛 − y∥ } < 𝜖

z𝑛 → z.1.210 1. By assumption, for every 𝑚 = 1, 2, . . . , there exists a point y𝑚 such that

∥y∥ < 1

𝑚

(𝑛∑𝑖=1

∣𝛼𝑖∣)

where

y = 𝛼1x1 + 𝛼2x2 + ⋅ ⋅ ⋅+ 𝛼𝑛x𝑛Let 𝑠𝑚 =

∑𝑛𝑖=1 ∣𝛼𝑖∣. By assumption 𝑠𝑚 > 𝑚 ∥y𝑚∥ ≥ 0. Define

x𝑚 =1

𝑠𝑚y𝑚

Then

x𝑚 = 𝛽𝑚1 x1 + 𝛽𝑚2 x2 + ⋅ ⋅ ⋅+ 𝛽𝑚𝑛 x𝑛where 𝛽𝑚𝑖 = 𝛼𝑚𝑖 /𝑠

𝑚,∑𝑛

𝑖=1 ∣𝛽𝑚𝑖 ∣ = 1 and ∥x𝑚∥ < 1𝑚 for every 𝑛 = 1, 2, . . . .

Consequently ∥x𝑚∥ → 0.

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2. Since∑𝑛

𝑖=1 ∣𝛽𝑚𝑖 ∣ = 1, ∣𝛽𝑚𝑖 ∣ ≤ 1 for every 𝑖. Consequently, for every coordinate𝑖, the sequence (𝛽𝑚𝑖 ) is bounded. By the Bolzano-Weierstrass theorem (Exercise1.119), the sequence (𝛽𝑚1 ) has a convergent subsequence with 𝛽𝑚1 → 𝛽1. Let x𝑚,1

denote the corresponding subsequence of x𝑚.

Similarly, 𝛽𝑚,12 has a subsequence converging to 𝛽2. Let (x𝑚,2) denote the corre-

sponding subsequence of (x𝑚). Proceeding coordinate by coordinate, we obtaina subsequence (x𝑚,𝑛) where each term is

x𝑚,𝑛 = 𝛽𝑚,𝑛x1 + 𝛽𝑚,𝑛x2 + ⋅ ⋅ ⋅+ 𝛽𝑚,𝑛x𝑛

and each coefficient converges 𝛽𝑚,𝑛𝑖 → 𝛽𝑖. Let

x = 𝛽1x1 + 𝛽2x2 + ⋅ ⋅ ⋅+ 𝛽2x𝑛Then x𝑚,𝑛 → x (Exercise 1.202).

3. Since∑𝑛

𝑖=1 ∣𝛽𝑚𝑖 ∣ = 1 for every 𝑚,∑𝑛

𝑖=1 ∣𝛽𝑖∣ = 1. Consequently, at least oneof the coefficients 𝛽𝑖 ∕= 0. Since x1,x2, . . . ,x𝑛 are linearly independent, x ∕= 0(Exercise 1.133) and therefore ∥x∥ ∕= 0. But (x𝑚,𝑛) is a subsequence of (x𝑚).This contradicts the earlier conclusion (part 1) that ∥x𝑚∥ → 0.

1.211 1. Let 𝑋 be a normed linear space 𝑋 of dimension 𝑛 and let {x1,x2, . . . ,x𝑛 }be a basis for 𝑋 . Let (x𝑚) be a Cauchy sequence in 𝑋 . Each term x𝑚 has aunique representation

x𝑚 = 𝛼𝑚1 x1 + 𝛼𝑚2 x2 + ⋅ ⋅ ⋅+ 𝛼𝑚𝑛 x𝑛We show that each of the sequences 𝛼𝑚𝑖 is a Cauchy sequence in ℜ.

Since x𝑚 is a Cauchy sequence, for every 𝜖 > 0 there exists an 𝑁 such that∥x𝑚 − x𝑟∥ < 𝜖 for all 𝑚, 𝑟 ≥ 𝑁 . Using Lemma 1.1, there exists 𝑐 > 0 such that

𝑐

𝑛∑𝑖=1

∣𝛼𝑚𝑖 − 𝛼𝑟𝑖 ∣ ≤∥∥∥∥∥

𝑛∑𝑖=1

(𝛼𝑚𝑖 − 𝛼𝑟𝑖 )x𝑖∥∥∥∥∥ = ∥x𝑚 − x𝑟∥ < 𝜖

for all 𝑚, 𝑟 ≥ 𝑁 . Dividing by 𝑐 > 0 we get for every 𝑖

∣𝛼𝑚𝑖 − 𝛼𝑟𝑖 ∣ ≤𝑛∑𝑖=1

∣𝛼𝑚𝑖 − 𝛼𝑟𝑖 ∣ <𝜖

𝑐

Thus each sequence 𝛼𝑚𝑖 is a Cauchy sequence in ℜ. Since ℜ is complete, eachsequence converges to some limit 𝛼𝑖 ∈ ℜ.

2. Let

x = 𝛼1x1 + 𝛼2x2 + ⋅ ⋅ ⋅+ 𝛼𝑛x𝑛Then x ∈ 𝑋 and

∥x𝑚 − x∥ =

∥∥∥∥∥𝑛∑𝑖=1

(𝛼𝑚𝑖 − 𝛼𝑖)x𝑖∥∥∥∥∥ ≤

𝑛∑𝑖=1

∣𝛼𝑚𝑖 − 𝛼𝑖∣ ∥x𝑖∥

Since 𝛼𝑚𝑖 → 𝛼𝑖 for every 𝑖, ∥x𝑚 − x∥ → 0 which implies that x𝑚 → x.3. Since (x𝑚) was an arbitrary Cauchy sequence, we have shown that every Cauchy

sequence in 𝑋 converges. Hence 𝑋 is complete.

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1.212 Let 𝑆 be an open set according to the ∥⋅∥𝑎 and let x0 be a point in 𝑆. Since 𝑆 isopen, it contains an open ball in the ∥⋅∥𝑎 topology about x0, namely 𝐵𝑎(x0, 𝑟) = {x ∈𝑋 : ∥x− x0∥𝑎 < 𝑟 } ⊆ 𝑆 Let

𝐵𝑏(x0, 𝑟) = {x ∈ 𝑋 : ∥x− x0∥𝑏 < 𝑟 }

be the open ball about x0 in the ∥⋅∥𝑏 topology. The condition (1.15) implies that𝐵𝑏(x0, 𝑟) ⊆ 𝐵𝑎(x0, 𝑟) ⊆ 𝑆 and therefore

x0 ∈ 𝐵𝑏(x0, 𝑟) ⊂ 𝑆

𝑆 is open in the ∥⋅∥𝑏 topology. Similarly, any 𝑆 open in the ∥⋅∥𝑏 topology is open inthe ∥⋅∥𝑎 topology.

1.213 Let 𝑋 be a normed linear space of dimension 𝑛. and let {x1,x2, . . . ,x𝑛 } bea basis for 𝑋 . Let ∥⋅∥𝑎 and ∥⋅∥𝑏 be two norms on 𝑋 . Every x ∈ 𝑋 has a uniquerepresentation

x = 𝛼1x1 + 𝛼2x2 + ⋅ ⋅ ⋅+ 𝛼𝑛x𝑛Repeated application of the triangle inequality gives

∥x∥𝑎 = ∥𝛼1x1 + 𝛼2x2 + ⋅ ⋅ ⋅+ 𝛼𝑛x𝑛∥𝑎≤

𝑛∑𝑖=1

∥𝛼𝑖x𝑖∥𝑎

=

𝑛∑𝑖=1

∣𝛼𝑖∣ ∥x𝑖∥𝑎

≤ 𝑘𝑛∑𝑖=1

∣𝛼𝑖∣

where 𝑘 = max𝑖 ∥x𝑖∥.By Lemma 1.1, there is a positive constant 𝑐 such that

𝑛∑𝑖=1

∣𝛼𝑖∣ ≤ ∥x∥𝑏 /𝑐

Combining these two inequalities, we have

∥x∥𝑎 ≤ 𝑘 ∥x∥𝑏 /𝑐

Setting 𝐴 = 𝑐/𝑘 > 0, we have shown

𝐴 ∥x∥𝑎 ≤ ∥x∥𝑏The other inequality in (1.15) is obtained by interchanging the roles of ∥⋅∥𝑎 and ∥⋅∥𝑏.1.214 Assume x𝑛 → x = (𝑥1, 𝑥2, . . . , 𝑥𝑛). Then, for every 𝜖 > 0, there exists some 𝑁such that ∥x𝑛 − x∥∞ < 𝜖. Therefore, for 𝑖 = 1, 2, . . . , 𝑛

∣𝑥𝑛𝑖 − 𝑥𝑖∣ ≤ max𝑖∣𝑥𝑛𝑖 − 𝑥𝑖∣ = ∥x𝑛 − x∥∞ < 𝜖

Therefore 𝑥𝑛𝑖 → x𝑖.

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Conversely, assume that (x𝑛) is a sequence in ℜ𝑛 with 𝑥𝑛𝑖 → 𝑥𝑖 for every coordinate 𝑖.Choose some 𝜖 > 0. For every 𝑖, there exists some integer 𝑁𝑖 such that

∣𝑥𝑛𝑖 − 𝑥𝑖∣ < 𝜖 for every 𝑛 ≥ 𝑁𝑖

Let 𝑁 = max𝑖{𝑁1, 𝑁2, . . . , 𝑁𝑛 }. Then

∣𝑥𝑛𝑖 − 𝑥𝑖∣ < 𝜖 for every 𝑛 ≥ 𝑁and

∥x𝑛 − x∥∞ = max𝑖∣𝑥𝑛𝑖 − 𝑥𝑖∣ < 𝜖 for every 𝑛 ≥ 𝑁

That is, x𝑛 → x.A similar proof can be given using the Euclidean norm ∥⋅∥2, but it is slightly morecomplicated. This illustrates an instance where the sup norm is more tractable.

1.215 1. Let 𝑆 ⊆ 𝑋 be closed and bounded and let x𝑚 be a sequence in 𝑆. Everyterm x𝑚 has a representation

x𝑚 =𝑛∑𝑖=1

𝛼𝑚𝑖 x𝑖

Since 𝑆 is bounded, so is x𝑚. That is, there exists 𝑘 such that ∥x𝑚∥ ≤ 𝑘 for all𝑚. Applying Lemma 1.1, there is a positive constant 𝑐 such that

𝑐

𝑛∑𝑖=1

∣𝛼𝑖∣ ≤ ∥x𝑚∥ ≤ 𝑘

Hence, for every 𝑖, the sequence of scalars 𝛼𝑛𝑖 is bounded.

2. By the Bolzano-Weierstrass theorem (Exercise 1.119), the sequence 𝛼𝑚1 has aconvergent subsequence with limit 𝛼1. Let 𝑥𝑚(1) be the corresponding subsequenceof x𝑚.

3. Similarly, 𝑥𝑚(1) has a subsequence for which the corresponding scalars 𝛼𝑚2 con-

verge to 𝛼2. Repeating this process 𝑛 times (this is were finiteness is impor-tant), we deduce the existence of a subsequence 𝑥𝑚(𝑛) whose scalars converge to

(𝛼1, 𝛼2, . . . , 𝛼𝑛).

4. Let

x =

𝑛∑𝑖=1

𝛼𝑖x𝑖

Since 𝛼𝑚𝑖 → 𝛼𝑖 for every 𝑖, ∥x𝑚 − x∥ → 0 which implies that x𝑚 → x.5. Since 𝑆 is closed, x ∈ 𝑆.

6. We have shown that every sequence in 𝑆 has a subsequence which converges in𝑆. 𝑆 is compact.

1.216 Let x and y belong to 𝐵 = {x : ∥x∥ < 1 }, the unit ball in the normed linearspace 𝑋 . Then ∥x∥ , ∥y∥ < 1. By the triangle inequality

∥𝛼x + (1− 𝛼)y∥ ≤ 𝛼 ∥x∥ + (1− 𝛼) ∥y∥ ≤ 𝛼+ (1− 𝛼) = 1

Hence 𝛼x+ (1− 𝛼)y ∈ 𝐵.

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1.217 If int 𝑆 is empty, it is trivially convex. Therefore, assume int 𝑆 ∕= ∅ and letx,y ∈ int 𝑆. We must show that z = 𝛼x + (1− 𝛼)y ∈ int 𝑆.

Since x,y ∈ int 𝑆, there exists some 𝑟 > 0 such that the open balls 𝐵(x, 𝑟) and 𝐵(y, 𝑟)are both contained in int 𝑆. Let w be any vector with ∥w∥ < 𝑟. The point

z+w = 𝛼(x+w) + (1− 𝛼)(y +w) ∈ 𝑆since x +w ∈ 𝐵(x, 𝑟) ⊂ 𝑆 and y +w ∈ 𝐵(y, 𝑟) ⊂ 𝑆 and 𝑆 is convex. Hence z is aninterior point of 𝑆.

Similarly, if 𝑆 is empty, it is trivially convex. Therefore, assume 𝑆 ∕= ∅ and let x,y ∈ 𝑆.Choose some 𝛼. We must show that 𝑧 = 𝛼x + (1− 𝛼)y ∈ 𝑆.

There exist sequences (x𝑛) and (y𝑛) in 𝑆 which converge to x and y respectively(Exercise 1.105). Since 𝑆 is convex, the sequence (𝛼x𝑛 + (1 − 𝛼)y𝑛) lies in 𝑆 andmoreover converges to 𝛼x + (1 − 𝛼)y = z (Exercise 1.202). Therefore 𝑧 is the limit ofa sequence in 𝑆 and hence 𝑧 ∈ 𝑆. Therefore, 𝑆 is convex.

1.218 Let x = 𝛼x1 + (1− 𝛼)x2 for some 𝛼 ∈ (0, 1). Since x1 ∈ 𝑆,

x1 ∈ 𝑆 + 𝑟𝐵

𝛼x1 ∈ 𝛼(𝑆 + 𝑟𝐵)

The open ball about x of radius 𝑟 is

𝐵(x, 𝑟) = x+ 𝑟𝐵

= 𝛼x1 + (1 − 𝛼)x2 + 𝑟𝐵

⊆ 𝛼(𝑆 + 𝑟𝐵) + (1 − 𝛼)x2 + 𝑟𝐵

= 𝛼𝑆 + (1 − 𝛼)x2 + (1 + 𝛼)𝑟𝐵

= 𝛼𝑆 + (1 − 𝛼)

(x2 +

1 + 𝛼

1− 𝛼𝑟𝐵)

Since x2 ∈ int 𝑆

x2 +1 + 𝛼

1− 𝛼𝑟𝐵 = 𝐵

(x2,

1 + 𝛼

1− 𝛼𝑟)⊆ 𝑆

for sufficiently small 𝑟. For such 𝑟

𝐵(x, 𝑟) ⊆ 𝛼𝑆 + (1− 𝛼)𝑆

= 𝑆

by Exercise 1.168. Therefore x ∈ int 𝑆.

1.219 It is easy to show that

𝑆 ⊆∩𝑖∈𝐼𝑆𝑖

To show the converse, choose any x ∈ 𝑆 and let x0 ∈ 𝑆𝑖 for every 𝑖 ∈ 𝐼. By Exercise1.218, 𝛼x + (1 − 𝛼)x0 ∈ 𝑆𝑖 for all 0 < 𝛼 < 1. This implies that 𝛼x + (1 − 𝛼)x0 ∈∩𝑖∈𝐼𝑆𝑖 = 𝑆 for all 0 < 𝛼 < 1, and therefore that x0 = lim𝛼→0 𝛼x + (1− 𝛼)x0 ∈ 𝑆.

1.220 Assume that x ∈ int 𝑆. Then, there exists some 𝑟 such that

𝐵(x, 𝑟) = x+ 𝑟𝐵 ⊆ 𝑆

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Let y be any element in the unit ball 𝐵. Then −y ∈ 𝐵 and

x1 = x+ 𝑟y ∈ 𝑆x2 = x− 𝑟y ∈ 𝑆

so that

x =1

2x1 +

1

2x2

x is not an extreme point. We have shown that no interior point is an extreme point;hence every extreme point must be a boundary point.

1.221 We showed in Exercise 1.220 that ext(𝑆) ⊆ b(𝑆). To show the converse, assumethat x is a boundary point which is not an extreme point. That is, there exist x1,x2 ∈ 𝑆such that

x = 𝛼x1 + (1− 𝛼)x2 0 < 𝛼 < 1

This contradicts the assumption that 𝑆 is strictly convex.

1.222 If 𝑆 is open, int 𝑆 = 𝑆. Since 𝑆 is convex

𝛼x+ (1 − 𝛼)y ∈ 𝑆 = int 𝑆 for every 0 ≤ 𝛼 ≤ 1

A fortiori for every x ∕= y

𝛼x+ (1 − 𝛼)y ∈ 𝑆 = int 𝑆 for every 0 < 𝛼 < 1

𝑆 is strictly convex.

1.223 Let 𝑆 be open and x ∈ conv 𝑆. That is

x =

𝑛∑𝑖=1

𝛼𝑖x𝑖

with x𝑖 ∈ 𝑆, 𝛼𝑖 ∈ [0, 1] and∑

𝑖 𝛼𝑖 = 1. The open ball about x

𝐵(x, 𝑟) = x+ 𝑟𝐵

=

(𝑛∑𝑖=1

𝛼𝑖x𝑖

)+ 𝑟𝐵

=

(𝑛∑𝑖=1

𝛼𝑖x𝑖

)+

(𝑛∑𝑖=1

𝛼𝑖𝑟𝐵

)

=

𝑛∑𝑖=1

(𝛼𝑖x𝑖 + 𝑟𝐵)

Since 𝑆 is open, there exists some 𝑟 such that x𝑖 + 𝑟𝐵 ∈ 𝑆 for all 𝑖. For this 𝑟

𝐵(x, 𝑟) ⊆ conv 𝑆

Therefore conv 𝑆 is open.

1.224

conv 𝑆 = { (𝑥1, 𝑥2) ∈ ℜ2 : 𝑥2 > 0 }

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1.225 𝑆 is closed and bounded (Proposition 1.1).

1. 𝑆 is bounded, that is there exists some 𝐾 such that ∥x∥ < 𝐾 for every x ∈ 𝑆.Let x ∈ conv 𝑆. x is a convex combination of a finite number of points in 𝑆, thatis

x =𝑚∑𝑖=1

𝛼𝑖x𝑖

with 𝑥𝑖 ∈ 𝑆, 𝛼𝑖 ≥ 0 and∑𝑚

𝑖=1 𝛼𝑖 = 1. By the triangle inequality

∥x∥ ≤𝑚∑𝑖=1

𝛼𝑖 ∥x𝑖∥ < 𝐾

Therefore conv 𝑆 is bounded.

2. Let x belong to conv 𝑆. Then, there exists a sequence (x𝑘) in conv 𝑆 which con-verges to x. By Caratheodory’s theorem, each term x𝑘 is a convex combinationof at most 𝑛+ 1 points, that is

x𝑘 =

𝑛+1∑𝑖=1

𝛼𝑘𝑖 x𝑘𝑖

where x𝑘𝑖 ∈ 𝑆.

For each 𝑖, the sequence (x𝑘𝑖 ) lies in a compact set 𝑆 and hence contains a conver-gent subsequence. Similarly, the sequence of coefficients (𝛼𝑘𝑖 ) ∈ [0, 1] is boundedand contains a convergent subsequence (Bolzano-Weierstrass theorem, Exercise1.119). Proceeding coordinate by coordinate as in Exercise 1.215, we can con-struct convergent subsequences 𝛼𝑘𝑖 → 𝛼𝑖 and x𝑘𝑖 − x𝑖.

3. Let

x =𝑛+1∑𝑖=1

𝛼𝑖x𝑖

Since

∥∥x𝑘 − x∥∥ =

∥∥∥∥∥𝑛+1∑𝑖=1

(𝛼𝑘𝑖 x𝑘𝑖 − 𝛼𝑖x𝑖)

∥∥∥∥∥≤

𝑛+1∑𝑖=1

∥∥𝛼𝑘𝑖 x𝑘𝑖 − 𝛼𝑖x𝑖∥∥=

𝑛+1∑𝑖=1

∥∥𝛼𝑘𝑖 x𝑘𝑖 − 𝛼𝑖x𝑘𝑖 + 𝛼𝑖x𝑘𝑖 − 𝛼𝑖x𝑖

∥∥=

𝑛+1∑𝑖=1

∣∣𝛼𝑘𝑖 − 𝛼𝑖∣∣ ∥∥x𝑘𝑖 ∥∥ +

𝑛+1∑𝑖=1

∣𝛼𝑖∣∥∥x𝑘𝑖 − x𝑖∥∥

→ 0

as 𝛼𝑘𝑖 → 𝛼𝑖, x𝑘𝑖 − x𝑖, 𝛼𝑖 and x𝑘 are bounded. Therefore x𝑘 → x.4. Since 𝛼𝑘𝑖 ≥ 0 and

∑𝑛+1𝑖=1 𝛼

𝑘𝑖 = 1 for every 𝑘, we conclude that 𝛼𝑖 = lim𝛼𝑘𝑖 ≥ 0 and∑𝑛+1

𝑖=1 𝛼𝑖 = 1. Furthermore, since 𝑆 is closed, x𝑖 ∈ 𝑆 for every 𝑖 and therefore

x =∑𝑛+1

𝑖=1 𝛼𝑖x𝑖 ∈ conv 𝑆.

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5. We have shown that conv 𝑆 ⊆ conv 𝑆, that is conv 𝑆 is closed.

6. conv 𝑆 is a closed and bounded subset of a finite dimensional space, and henceconv 𝑆 is compact (Proposition 1.4 and Exercise 1.215).

1.226 1. 𝑆 is bounded. Therefore, there exists some 𝑐 such that ∥x∥∞ = max𝑖 ∣𝑥𝑖∣ <𝑐 for every x ∈ 𝑆. That is −𝑐 ≤ 𝑥𝑖 ≤ 𝑐 so that

x ∈ 𝐶 = {x = (𝑥1, 𝑥2, . . . , 𝑥𝑛) ∈ ℜ𝑛 : −𝑐 ≤ 𝑥𝑖 ≤ 𝑐 for every 𝑖 }Therefore 𝑆 ⊂ 𝐾.

2. Exercise 1.177.

3. 𝐶 is the convex hull of a finite set and hence is compact (Exercise 1.225)

4. 𝑆 is a closed subset of a compact set and hence is compact (Exercise 1.110).

1.227 A polytope is the convex hull of a finite set. Any finite set is compact.

1.228 The unit simplex Δ𝑛−1 in ℜ𝑛 is the convex hull of the unit vectors e1, e2, . . . , e𝑛,that is

Δ𝑛−1 = conv { e1, e2, . . . , e𝑛 }=

{(𝑥1, 𝑥2, . . . , 𝑥𝑛) ∈ ℜ𝑛 : 𝑥𝑖 ≥ 0 and

∑𝑥𝑖 = 1

}This simplex has a nonempty relative interior, namely

ri 𝑆 ={

(𝑥1, 𝑥2, . . . , 𝑥𝑛) ∈ ℜ𝑛 : 𝑥𝑖 > 0 and∑𝑥𝑖 < 1

}1.229 Let 𝑛 = dim𝑆. By Exercise 1.182, 𝑆 contains a simplex 𝑆𝑛 of the same dimen-sion. That is, there exist 𝑛 vertices v1,v2, . . . ,v𝑛 such that

𝑆𝑛 = conv {v1,v2, . . . ,v𝑛 }=

{𝛼1v1 + 𝛼2v2 + ⋅ ⋅ ⋅+ 𝛼𝑛v𝑛 :

𝛼1, 𝛼2, . . . , 𝛼𝑛 ≥ 0,

𝛼1 + 𝛼2 + . . .+ 𝛼𝑛 = 1}

Analogous to the previous part, the relative interior of 𝑆𝑛 is

ri 𝑆𝑛 = conv {v1,v2, . . . ,v𝑛 }=

{𝛼1v1 + 𝛼2v2 + ⋅ ⋅ ⋅+ 𝛼𝑛v𝑛 :

𝛼1, 𝛼2, . . . , 𝛼𝑛 > 0,

𝛼1 + 𝛼2 + . . .+ 𝛼𝑛 = 1}

which is nonempty.

Note, the proposition is trivially true for a set containing a single point (𝑛 = 0), sincethis point is the whole affine space.

1.230 If int 𝑆 ∕= ∅, then aff 𝑆 = 𝑋 and ri 𝑆 = int 𝑆. The converse follows from Exercise1.229.

1.231 Since

𝑚 > infx∈𝑋

𝑛∑𝑖=1

𝑝𝑖𝑥𝑖

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there exists some x ∈ 𝑋 such that𝑛∑𝑖=1

𝑝𝑖𝑥𝑖 ≤ 𝑚

Therefore x ∈ 𝑋(p,𝑚) which is nonempty.

Let 𝑝 = min𝑖 𝑝𝑖 be the lowest price of the 𝑛 goods. Then 𝑋(p,𝑚) ⊆ 𝐵(0,𝑚/𝑝) andso is bounded. (That is, no component of an affordable bundle can contain more than𝑚/𝑝 units.)

To show that 𝑋(p,𝑚) is closed, let (x𝑛) be a sequence of consumption bundles in𝑋(p,𝑚). Since 𝑋(p,𝑚) is bounded, x𝑛 → x ∈ 𝑋 . Furthermore

𝑝1𝑥𝑛1 + 𝑝2𝑥

𝑛2 + ⋅ ⋅ ⋅+ 𝑝𝑛𝑥𝑛𝑛 ≤ 𝑚 for every 𝑛

This implies that

𝑝1𝑥1 + 𝑝2𝑥2 + ⋅ ⋅ ⋅+ 𝑝𝑛𝑥𝑛 ≤ 𝑚so that x𝑛 → x ∈ 𝑋(p,𝑚). Therefore 𝑋(p,𝑚) is closed.

We have shown that 𝑋(p,𝑚) is a closed and bounded subset of ℜ𝑛. Hence it is compact(Proposition 1.4).

1.232 Let x,y ∈ 𝑋(p,𝑚). That is

𝑛∑𝑖=1

𝑝𝑖𝑥𝑖 ≤ 𝑚𝑛∑𝑖=1

𝑝𝑖𝑦𝑖 ≤ 𝑚

For any 𝛼 ∈ [0, 1], the cost of the weighted average bundle z = 𝛼x + (1 − 𝛼)y (whereeach component 𝑧𝑖 = 𝛼𝑥𝑖 + (1− 𝛼)𝑦𝑖) is

𝑛∑𝑖=1

𝑝𝑖𝑧𝑖 =

𝑛∑𝑖=1

𝑝𝑖(𝛼𝑥𝑖 + (1− 𝛼)𝑦𝑖

= 𝛼

𝑛∑𝑖=1

𝑝𝑖𝑥𝑖 + (1 − 𝛼)

𝑛∑𝑖=1

𝑝𝑖𝑦𝑖

≤ 𝛼𝑚+ (1− 𝛼)𝑚

= 𝑚

Therefore z ∈ 𝑋(p,𝑚). The budget set 𝑋(p,𝑚) is convex.

1.233 1. Assume that ≻ is strongly monotone. Let x,y ∈ 𝑋 with x ≥ y.Either x ≩ y so that x ≻ y by strong monotonicity

or x = y so that x ≿ y by reflexivity.

In either case, x ≿ y so that ≿ is weakly monotonic.

2. Again, assume that ≿ is strongly monotonic and let y ∈ 𝑋 . 𝑋 is open (relativeto itself). Therefore, there exists some 𝑟 such that

𝐵(y, 𝑟) = y + 𝑟𝐵 ⊆ 𝑋Let x = y + 𝑟e1 be the consumption bundle containing 𝑟 more units of good 1.Then e1 ∈ 𝐵, x ∈ 𝐵(y, 𝑟) and therefore ∥x− y∥ < 𝑟. Furthermore, x ≩ y andtherefore x ≻ y.

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3. Assume ≿ is locally nonsatiated. Then, for every x ∈ 𝑋 , there exists some y ∈ 𝑋such that y ≻ x. Therefore, there is no best element.

1.234 Assume otherwise, that is assume that x∗ ≿ x for every x ∈ 𝐵(p,𝑚) but that∑𝑛𝑖=1 𝑝𝑖𝑥𝑖 < 𝑚. Let 𝑟 = 𝑚−∑𝑛

𝑖=1 𝑝𝑖𝑥𝑖 be the unspent income. Spending the residualon good 1, the commodity bundle x = x∗ + 𝑟

𝑝1e1 is affordable

𝑛∑𝑖=1

𝑝𝑖𝑥𝑖 =

𝑛∑𝑖=1

𝑝𝑖𝑥∗𝑖 + 𝑝1

𝑟

𝑝1= 𝑚

Moreover, since x ≩ x∗, x ≻ x∗, which contradicts the assumption that x∗ is the bestelement in 𝑋(p,𝑚).

1.235 Assume otherwise, that is assume that x∗ ≿ x for every x ∈ 𝐵(p,𝑚) but that∑𝑛𝑖=1 𝑝𝑖𝑥

∗𝑖 < 𝑚. This implies that x∗ ∈ int 𝑋(p,𝑚). Therefore, there exists a neigh-

borhood 𝑁 of x∗ with 𝑁 ⊆ 𝑋(p,𝑚). Within this neighborhood, there exists somex ∈ 𝑁 ⊆ 𝑋(p,𝑚) with x ≻ x∗, which contradicts the assumption that x∗ is the bestelement in 𝑋(p,𝑚).

1.236 1. Assume ≿ is continuous. Choose some y ∈ 𝑋 . For any x0 in ≻(y), x0 ≻ yand (since ≿ is continuous) there exists some neighborhood 𝑆(x0) such that x ≻ yfor every x ∈ 𝑆(x0). That is, 𝑆(x0) ⊆ ≻(y) and ≻(y) is open.

Similarly, for any x0 ∈ ≺(y), x0 ≺ y and there exists some neighborhood 𝑆(x0)such that x ≺ y for every x ∈ 𝑆(x0). Thus 𝑆(x0) ⊆ ≺(y) and ≺(y) is open.

2. Conversely, assume that the sets ≻(y) = {x : x ≻ y } and ≺(y) = {x : x ≺ y }are open in x. Assume x0 ≻ y0.(a) Suppose there exists some y such that x0 ≻ y ≻ z0. Then x0 ∈ ≻(y), which

is open by assumption. That is, ≻(y) is an open neighborhood of x0 andx ≻ y for every x ∈ ≻(y). Similarly, ≺(y) is an open neighborhood of z0 forwhich y ≻ z for every z ∈ ≺(y). Therefore 𝑆(x0) = ≻(y) and 𝑆(z0) = ≺(y)are the required neighborhoods of x0 and z0 respectively such that

x ≻ y ≻ z for every x ∈ 𝑆(x0) and y ∈ 𝑆(z0)

(b) Suppose there is no y such that x0 ≻ y ≻ z0.i. By assumption

∙ ≻(z0) is open

∙ x0 ≻ z0 which implies x0 ∈ ≻(z0),

Therefore ≻(z0) is an open neighborhood of x0.

ii. Since ≿ is complete, either y ≺ x0 or y ≿ x0 for every y ∈ 𝑋 (Exercise1.56. Since there is no y such that x0 ≻ y ≻ z0

y ≻ z0 =⇒ y ∕≺ x0 =⇒ y ≿ x0Therefore ≻(z0) = ≿(x0).

iii. Since x ≿ x0 ≻ z0 for every x ∈ ≿(x0) = ≻(z0)

x ≻ z0 for every x ∈ ≻(z0)

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iv. Therefore 𝑆(x0) = ≻(z0) is an open neighborhood of x0 such that

x ≻ z0 for every x ∈ 𝑆(x0)

Similarly, 𝑆(z0) = ≺(x0) is an open neighborhood of z0 such that z ≺ x0for every z ∈ 𝑆(z0). Consequently

x ≻ z for every x ∈ 𝑆(x0) and z ∈ 𝑆(z0)

3. ≿(y) =(≺(y)

)𝑐(Exercise 1.56). Therefore, ≿(y) is closed if and only if ≺(y) is

open (Exercise 1.80). Similarly, ≾(y) is closed if and only if ≻(y) is open.

1.237 1. Let 𝐹 = { (x,y) ∈ 𝑋×𝑋 : x ≿ y }. Let ((x𝑛,y𝑛)) be a sequence in 𝐹 whichconverges to (x,y). Since (x𝑛,y𝑛) ∈ 𝐹 , x𝑛 ≿ y𝑛 for every 𝑛. By assumption,x ≿ y. Therefore, (x,y) ∈ 𝐹 which establishes that 𝐹 is closed (Exercise 1.106)

Conversely, assume that 𝐹 is closed and let ((x𝑛,y𝑛)) be a sequence convergingto (x,y) with x𝑛 ≿ y𝑛 for every 𝑛. Then ((x𝑛,y𝑛)) ∈ 𝐹 which implies that(x,y) ∈ 𝐹 . Therefore x ≿ y.

2. Yes. Setting y𝑛 = y for every 𝑛, their definition implies that for every sequence(x𝑛) in 𝑋 with x𝑛 ≿ y, x = limx𝑛 ≿ y. That is, the upper contour set≿(y) = {x : x ≿ y } is closed. Similarly, the lower contour set ≾(y) is closed.

Conversely, assume that the preference relation is continuous (in our definition).We show that the set 𝐺 = { (x,y) : x ≺ y } is open. Let (x0,y0) ∈ 𝐺. Thenx0 ≺ y0. By continuity, there exists neighborhoods 𝑆(x0) and 𝑆(y0) of x0 andy0 such that x ≺ y for every x ∈ 𝑆(x0) and y ∈ 𝑆(y0). Hence, for every(x,y) ∈ 𝑁 = 𝑆(x0) × 𝑆(y0), x ≺ y. Therefore 𝑁 ⊆ 𝐺 which implies that 𝐺 isopen. Consequently 𝐺𝑐 = { (x,y) : x ≿ y } is closed.

1.238 Assume the contrary. That is, assume there is no y with x ≻ y ≻ z. Since ≿ iscomplete, either y ≺ x0 or y ≿ x0 for every y ∈ 𝑋 (Exercise 1.56). Since there is noy such that x0 ≻ y ≻ z0

y ≻ z0 =⇒ y ∕≺ x0 =⇒ y ≿ x0Therefore ≻(z0) = ≿(x0). By continuity, ≻(z0) is open and ≿(x0) is closed. Hence≻(z0) = ≿(x0) is both open and closed (Exercise 1.83).

Alternatively, ≿(x0) and ≾(z0) are both open sets which partition 𝑋 . This contradictsthe assumption that 𝑋 is connected.

1.239 Let 𝑋∗ denote the set of best elements. As demonstrated in the preceding proof

𝑋∗ =∩y∈𝑋

≿(y𝑖)

Therefore 𝑋∗ is closed (Exercise 1.85) and hence compact (Exercise 1.110).

1.240 Assume for simplicity that 𝑝1 = 𝑝2 = 1 and that 𝑚 = 1. Then, the budget set is

𝐵(1, 1) = {x ∈ ℜ2++ : 𝑥1 + 𝑥2 ≤ 1 }

The consumer would like to spend as much as possible of her income on good 1.However, the point (1, 0) is not feasible, since (1, 0) /∈ 𝑋 .

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1.241 Essentially, consumer theory (in economics) is concerned with predicting the wayin which consumer purchases vary with changes in observable parameters such as pricesand incomes. Predictions are deduced by assuming that the consumer will consistentlychoose the best affordable alternative in her budget set. The theory would be emptyif there was no such optimal choice.

1.242 1. Let 𝑋0 = 𝑋 ∩ ℜ𝑛+. Then 𝑋0 is compact and 𝑋1 ⊆ 𝑋0. Define the orderx ≿1 y if and only if 𝑑1(x) ≤ 𝑑1(y). Then ≿1 is continuous on 𝑋 and

𝑋1 = {x ∈ 𝑋 : 𝑑1(x) ≤ 𝑑1(y) for every y ∈ 𝑋 }is the set of best elements in 𝑋 with respect to the order ≿1. By Exercise 1.239,𝑋1 is nonempty and compact.

2. Assume 𝑋𝑘−1 is compact. Define the order x ≿𝑘 y if and only if 𝑑𝑘(x) ≤ 𝑑𝑘(y).Then ≿𝑘 is continuous on 𝑋𝑘−1 and

𝑋𝑘 = {x ∈ 𝑋𝑘−1 : 𝑑𝑘(x) ≤ 𝑑𝑘(y) for every y ∈ 𝑋𝑘−1 }is the set of best elements in 𝑋𝑘−1 with respect to the order ≿𝑘. By Exercise1.239, 𝑋𝑘 is nonempty and compact.

3. Assume x ∈ Nu. Then

x ≿ y for every y ∈ 𝑋d(x) ≾𝐿 d(y) for every y ∈ 𝑋

For every 𝑘 = 1, 2, . . . , 2𝑛

d𝑘(x) ≤ d𝑘(y) for every y ∈ 𝑋which implies x ∈ 𝑋𝑘. In particular x ∈ 𝑋2𝑛 . Therefore Nu ⊆ 𝑋2𝑛 .

Suppose Nu ⊂ 𝑋2𝑛 . Then there exists some x,y ∈ 𝑋 , x /∈ 𝑋2𝑛 and y ∈ 𝑋2𝑛 suchthat x ≿𝑑 y. Let 𝑘 be the smallest integer such that x /∈ 𝑋𝑘. Then 𝑑𝑘(x) > 𝑑𝑘(y).But x ∈ 𝑋 𝑙 for every 𝑙 < 𝑘 and therefore 𝑑𝑙(x) = 𝑑𝑙(y) for 𝑙 = 1, 2, . . . , 𝑘 − 1.This means that d(y) ≺𝐿 d(x) so that x ≺𝑑 y. This contradiction establishesthat Nu = 𝑋2

𝑛

.

1.243 Assume ≿ is convex. Choose any y ∈ 𝑋 and let x ∈ ≿(y). Then x ≿ y. Since≿ is convex, this implies that

𝛼x+ (1 − 𝛼)y ≿ y for every 0 ≤ 𝛼 ≤ 1

and therefore

𝛼x+ (1 − 𝛼)y ∈ ≿(y) for every 0 ≤ 𝛼 ≤ 1

Therefore ≿(y) is convex.

To show the converse, assume that ≿(y) is convex for every y ∈ 𝑋 . Choose x,y ∈ 𝑋 .Interchanging x and y if necessary, we can assume that x ≿ y so that x ∈ ≿(y). Ofcourse, y ∈ ≿(y). Since ≿(y) is convex

𝛼x+ (1 − 𝛼)y ∈ ≿(y) for every 0 ≤ 𝛼 ≤ 1

which implies

𝛼x+ (1 − 𝛼)y ≿ y for every 0 ≤ 𝛼 ≤ 1

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1.244 𝑋∗ may be empty, in which case it is trivially convex. Otherwise, let x∗ ∈ 𝑋∗.For every x ∈ 𝑋∗

x ≿ x∗ which implies x ∈ ≿(x∗)

Therefore 𝑋∗ ⊆ ≿(x∗). Conversely, by transitivity

x ≿ x∗ ≿ y for every y ∈ 𝑋for every x ∈ ≿(x∗) which implies ≿(x∗) ⊆ 𝑋∗. Therefore, 𝑋∗ = ≿(x∗) which isconvex.

1.245 To show that ≿𝑑 is strictly convex, assume that x,y ∈ 𝑋 are such d(x) = d(y)with x ∕= y. Suppose

d(x) =(𝑑(𝑆1,x), 𝑑(𝑆2,x) . . . , 𝑑(𝑆2𝑛 ,x)

)In the order 𝑆1, 𝑆2, . . . , 𝑆2𝑛 , let 𝑆𝑘 be the first coalition for which 𝑑(𝑆𝑘,x) ∕= 𝑑(𝑆𝑘,y).That is

𝑑(𝑆𝑗 ,x) = 𝑑(𝑆𝑗 ,y) for every 𝑗 < 𝑘 (1.32)

Since 𝑑(𝑆𝑘,x) ∕= 𝑑(𝑆𝑘,y) and d(x) is listed in descending order, we must have

𝑑(𝑆𝑘,x) > 𝑑(𝑆𝑘,y) (1.33)

and

𝑑(𝑆𝑘,x) ≥ 𝑑(𝑆𝑗 ,y) for every 𝑗 > 𝑘 (1.34)

Choose 0 < 𝛼 < 1 and let z = 𝛼x+ (1− 𝛼)y. For any coalition 𝑆

𝑑(𝑆, z) = 𝑤(𝑆) −∑𝑖∈𝑆𝑧𝑖

= 𝑤(𝑆) −∑𝑖∈𝑆

(𝛼𝑥𝑖 + (1− 𝛼)𝑦𝑖

)= 𝑤(𝑆) − 𝛼

∑𝑖∈𝑆𝑥𝑖 − (1 − 𝛼)

∑𝑖∈𝑆𝑦𝑖

= 𝛼

(𝑤(𝑆) −

∑𝑖∈𝑆𝑥𝑖

)+ (1− 𝛼)

(𝑤(𝑆) −

∑𝑖∈𝑆𝑦𝑖

)

= 𝛼𝑑(𝑆,x) + (1 − 𝛼)𝑑(𝑆,y)

Using (1.55) to (1.57), this implies that

𝑑(𝑆𝑗 , z) = 𝑑(𝑆𝑗 ,x), 𝑗 < 𝑘

𝑑(𝑆𝑘, z) < 𝑑(𝑆𝑘,x)

𝑑(𝑆𝑘, z) ≤ 𝑑(𝑆𝑗 ,x), 𝑗 > 𝑘

for every 0 < 𝛼 < 1, Therefore d(z) ≺𝐿 d(x). Thus z ≻𝑑 x, which establishes that ≿is strictly convex.

The set of feasible outcomes is convex. Assume x,y ∈ Nu ⊆ 𝑋 , x ∕= y. Thend(x) = d(y) and

z = 𝛼x+ (1− 𝛼)y ≻𝑑 x

for every 0 < 𝛼 < 1 which contradicts the assumption that x ∈ Nu. We conclude thatthe nucleolus contains only one element.

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1.246 1. (a) Clearly ≺(x0) ⊆ ≾(x0) and ≻(y0) ⊆ ≿(y0). Consequently ≺(x0) ∪≻(y0) ⊆ ≾(x0) ∪≿(y0). We claim that these sets are in fact equal.

Let z ∈ ≾(x0) ∪ ≿(y0). Suppose that z ∈ ≾(x0) but z /∈ ≺(x0). Thenz ≿ x0. By transitivity, z ≿ x0 ≻ y0 which implies that z ∈ ≻(y0).Similarly z ∈ ≿(y0) ∖ ≻(y0) implies z ∈ ≺(x0). Therefore

≺(x0) ∪ ≻(y0) = ≾(x0) ∪≿(y0)

(b) By continuity, ≺(x0)∪≻(y0) is open and ≾(x0)∪≿(y0) = ≺(x0)∪≻(y0) isclosed. Further x0 ≻ y0 implies that x0 ∈ ≻(y0) so that ≺(x0)∪≻(y0) ∕= ∅.We have established that ≺(x0) ∪ ≻(y0) is a nonempty subset of 𝑋 whichis both open and closed. Since 𝑋 is connected, this implies (Exercise 1.83)that

≺(x0) ∪ ≻(y0) = 𝑋

2. (a) By definition, x /∈ ≺(x). So ≺(x) ∩ ≺(y) = 𝑋 implies x ∈ ≻(y), that isx ≿ y contradicting the noncomparability of x and y. Therefore

≺(x) ∩ ≺(y) ∕= 𝑋

(b) By assumption, there exists at least one pair x0,y0 such that x0 ≻ y0. Bythe previous part

≺(x0) ∪ ≻(y0) = 𝑋

This implies either x ≺ x0 or x ≻ y0. Without loss of generality, assumex ≻ y0. Again using the previous part, we have

≺(x) ∪ ≻(y0) = 𝑋

Since x and y are not comparable, y /∈ ≺(x) which implies that y ∈ ≻(y0).Therefore x ≻ y0 and y ≻ y0 or alternatively

y0 ∈ ≺(x0) ∩ ≻(y0) ∕= ∅

(c) Clearly ≺(x) ⊆ ≾(x) and ≻(y) ⊆ ≿(y). Consequently

≺(x) ∩ ≺(y) ⊆ ≾(x) ∩≾(y)

Let z ∈ ≾(x)∩≾(y). That is, z ≾ x. If x ≾ z, then transitivity implies x ≾z ≾ y, which contradicts the noncomparability of x and y. Consequentlyx ∕≾ z which implies z ≺ x and z ∈ ≺(x). Similarly z ∈ ≺(y) and therefore

≺(x) ∩ ≺(y) = ≾(x) ∩≾(y)

3. If x and y are noncomparable, ≺(x)∩≺(y) is a nonempty proper subset of 𝑋 . Bycontinuity ≺(x)∩≺(y) = ≾(x)∩≾(y) is both open and closed which contradictsthe assumption that 𝑋 is connected (Exercise 1.83). We conclude that ≿ mustbe complete.

1.247 Assume x ≻ y. Then x ∈ ≻(y). Since ≻(y) is open, x ∈ int ≻(y). Alsoy ∈ ≻(y). By Exercise 1.218, 𝛼x + (1 − 𝛼)y ∈ int ≻(y) for every 0 < 𝛼 < 1, whichimplies

𝛼x+ (1 − 𝛼)y ≻ y for every 0 < 𝛼 < 1

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1.248 For every x ∈ 𝑋 , there exists some z such that z ≻ x (Nonsatiation). For any 𝑟,choose some 𝛼 ∈ (0, 𝑟/ ∥x− z∥) and let y = 𝛼z+ (1 − 𝛼)x. Then

∥x− y∥ = 𝛼 ∥x− z∥ < 𝑟

Moreover, since ≿ is strictly convex,

y = 𝛼z + (1− 𝛼)x ≻ x

Thus, ≿ is locally nonsatiated.

We have previously shown that local nonsatiation implies nonsatiation (Exercise 1.233).Consequently, these two properties are equivalent for strictly convex preferences.

1.249 Assume that x is not strongly Pareto efficient. That is, there exist allocation ysuch that y ≿𝑖 x for all 𝑖 and some individual 𝑗 for which y ≻𝑗 x. Take 1− 𝑡 percentof 𝑗’s consumption and distribute it equally to the other participants. By continuity,there exists some 𝑡 such that 𝑡y ≻𝑗 x. The other agents receive y𝑖 + 1−𝑡

𝑛−1y𝑗 which, bymonotonicity, they strictly prefer to x𝑖.

1.250 Assume that (p∗,x∗) is a competitive equilibrium of an exchange economy, butthat x∗ does not belong to the core of the corresponding market game. Then thereexists some coalition 𝑆 and allocation y ∈ 𝑊 (𝑆) such that y𝑖 ≻𝑖 x

∗𝑖 for every 𝑖 ∈ 𝑆.

Since y ∈𝑊 (𝑆), we must have∑

𝑖∈𝑆 y𝑖 =∑

𝑖∈𝑆 w𝑖.

Since x∗ is a competitive equilibrium and y𝑖 ≻𝑖 x∗𝑖 for every 𝑖 ∈ 𝑆, y must be unaf-

fordable, that is

𝑙∑𝑗=1

𝑝𝑗𝑦𝑖𝑗 >

𝑙∑𝑗=1

𝑝𝑗w𝑖𝑗 for every 𝑖 ∈ 𝑆

and therefore

∑𝑖∈𝑆

𝑙∑𝑗=1

𝑝𝑗𝑦𝑖𝑗 >∑𝑖∈𝑆

𝑙∑𝑗=1

𝑝𝑗w𝑖𝑗

which contradicts the assumption that 𝑦 ∈𝑊 (𝑆).

1.251 Combining the previous exercise with Exercise 1.64

x∗ ∈ core ⊆ Pareto

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Chapter 2: Functions

2.1 In general, the birthday mapping is not one-to-one since two individuals may havethe same birthday. It is not onto since some days may be no one’s birthday.

2.2 The origin 0 is fixed point for every 𝜃. Furthermore, when 𝜃 = 0, 𝑓 is an identityfunction and every point is a fixed point.

2.3 For every 𝑥 ∈ 𝑋 , there exists some 𝑦 ∈ 𝑌 such that 𝑓(𝑥) = 𝑦, whence 𝑥 ∈ 𝑓−1(𝑦).Therefore, every 𝑥 belongs to some contour. To show that distinct contours are disjoint,assume 𝑥 ∈ 𝑓−1(𝑦1) ∩ 𝑓−1(𝑦2). Then 𝑓(𝑥) = 𝑦1 and also 𝑓(𝑥) = 𝑦2. Since 𝑓 is afunction, this implies that 𝑦1 = 𝑦2.

2.4 Assume 𝑓 is one-to-one and onto. Then, for every 𝑦 ∈ 𝑌 , there exists 𝑥 ∈ 𝑋 suchthat 𝑓(𝑥) = 𝑦. That is, 𝑓−1(𝑦) ∕= ∅ for every 𝑦 ∈ 𝑌 . If 𝑓 is one to one, 𝑓(𝑥) = 𝑦 = 𝑓(𝑥′)implies 𝑥 = 𝑥′. Therefore, 𝑓−1(𝑦) consists of a single element. Therefore, the inversefunction 𝑓−1 exists.

Conversely, assume that 𝑓 : 𝑋 → 𝑌 has an inverse 𝑓−1. As 𝑓−1 is a function mapping𝑌 to 𝑋 , it must be defined for every 𝑦 ∈ 𝑌 . Therefore 𝑓 is onto. Assume thereexists 𝑥, 𝑥′ ∈ 𝑋 and 𝑦 ∈ 𝑌 such that 𝑓(𝑥) = 𝑦 = 𝑓(𝑥′). Then 𝑓−1(𝑦) = 𝑥 andalso 𝑓−1(𝑦) = 𝑥′. Since 𝑓−1 is a function, this implies that 𝑥 = 𝑥′. Therefore 𝑓 isone-to-one.

2.5 Choose any 𝑥 ∈ 𝑋 and let 𝑦 = 𝑓(𝑥). Since 𝑓 is one-to-one, 𝑥 = 𝑓−1(𝑦) = 𝑓−1(𝑓(𝑥)).The second identity is proved similarly.

2.6 (2.2) implies for every 𝑥 ∈ ℜ

𝑒𝑥𝑒−𝑥 = 𝑒0 = 1

and therefore

𝑒−𝑥 =1

𝑒𝑥(2.28)

For every 𝑥 ≥ 0

𝑒𝑥 = 1 +𝑥

1+𝑥2

2+𝑥3

6+ ⋅ ⋅ ⋅ > 0

and therefore by (2.28) 𝑒𝑥 > 0 for every 𝑥 ∈ ℜ. For every 𝑥 ≥ 1

𝑒𝑥 = 1 +𝑥

1+𝑥2

2+𝑥3

6+ ⋅ ⋅ ⋅ ≥ 1 + 𝑥→∞ as 𝑥→∞

and therefore 𝑒𝑥 →∞ as 𝑥→∞. By (2.28) 𝑒𝑥 → 0 as 𝑥→ −∞.

2.7

𝑒𝑥

𝑥=𝑒𝑥/2𝑒𝑥/2

2𝑥2

=1

2

(𝑒𝑥/2

𝑥/2

)𝑒𝑥/2 →∞ as 𝑥→∞

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since the term in brackets is strictly greater than 1 for any 𝑥 > 0. Similarly

𝑒𝑥

𝑥=

(𝑒𝑥/(𝑛+1))𝑛𝑒𝑥/(𝑛+1)

(𝑛+ 1)𝑛( 𝑥𝑛+1 )𝑛

=1

(𝑛+ 1)𝑛

(𝑒𝑥/(𝑛+1)

𝑥/(𝑛+ 1)

)𝑛

𝑒𝑥/(𝑛+1) →∞

2.8 Assume that 𝑆 ⊆ ℜ is compact. Then 𝑆 is bounded (Proposition 1.1), and thereexists 𝑀 such that ∣𝑥∣ ≤𝑀 for every 𝑥 ∈ 𝑆. For all 𝑛 ≥ 𝑚 ≥ 2𝑀

∣𝑓𝑛(𝑥)− 𝑓𝑚(𝑥)∣ =∣∣∣∣∣

𝑛∑𝑘=𝑚+1

𝑥𝑘

𝑘!

∣∣∣∣∣ ≤∣∣∣∣∣ 𝑥

𝑚+1

(𝑚+ 1)!

𝑛−𝑚∑𝑘=0

( 𝑥𝑚

)𝑘∣∣∣∣∣≤

∣∣∣∣∣ 𝑀𝑚+1

(𝑚+ 1)!

𝑛−𝑚∑𝑘=0

(𝑀

𝑚

)𝑘∣∣∣∣∣

≤ 𝑀𝑚+1

(𝑚+ 1)!

(1 +

1

2+

1

4+ ⋅ ⋅ ⋅+

(1

2

)𝑛−𝑚)

≤ 2𝑀𝑚+1

(𝑚+ 1)!≤ 2

(𝑀

𝑚+ 1

)𝑚+1

≤(

1

2

)𝑚

by Exercise 1.206. Therefore 𝑓𝑛 converges to 𝑓 for all 𝑥 ∈ 𝑆.

2.9 This is a special case of Example 2.8. For any 𝑓, 𝑔 ∈ 𝐹 (𝑋), define

(𝑓 + 𝑔) = 𝑓(𝑥) + 𝑔(𝑥)

(𝛼𝑓)(𝑥) = 𝛼𝑓(𝑥)

With these definitions 𝑓 + 𝑔 and 𝛼𝑓 also map 𝑋 to ℜ. Hence 𝐹 (𝑋) is closed underaddition and scalar multiplication. It is straightforward but tedious to verify that 𝐹 (𝑋)satisfies the other requirements of a linear space.

2.10 The zero element in 𝐹 (𝑋) is the constant function 𝑓(𝑥) = 0 for every 𝑥 ∈ 𝑋 .

2.11 1. From the definition of ∥𝑓∥ it is clear that

∙ ∥𝑓∥ ≥ 0.

∙ ∥𝑓∥ = 0 of and only 𝑓 is the zero functional.

∙ ∥𝛼𝑓∥ = ∣𝛼∣ ∥𝑓∥ since sup𝑥∈𝑋 ∣𝛼𝑓(𝑥)∣ = ∣𝛼∣ sup𝑥∈𝑋 ∣𝑓(𝑥)∣It remains to verify the triangle inequality, namely

∥𝑓 + 𝑔∥ = sup𝑥∈𝑋∣(𝑓 + 𝑔)(𝑥)∣

= sup𝑥∈𝑋∣𝑓(𝑥) + 𝑔(𝑥)∣

≤ sup𝑥∈𝑋

{ ∣𝑓(𝑥)∣+ ∣𝑔(𝑥)∣ }≤ sup

𝑥∈𝑋∣(𝑓(𝑥)∣ + sup

𝑥∈𝑋∣𝑔(𝑥)∣

= ∥𝑓∥+ ∥𝑔∥

2. Consequently, for any 𝑓 ∈ 𝐵(𝑋), 𝛼𝑓(𝑥) ≤ ∣𝛼∣ ∥𝑓∥ for every 𝑥 ∈ 𝑋 and therefore𝛼𝑓 ∈ 𝐵(𝑋). Similarly, for any 𝑓, 𝑔 ∈ 𝐵(𝑋), (𝑓 + 𝑔)(𝑥) ≤ ∥𝑓∥ + ∥𝑔∥ for every

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𝑥 ∈ 𝑋 and therefore 𝑓 + 𝑔 ∈ 𝐵(𝑋). Hence, 𝐵(𝑋) is closed under addition andscalar multiplication; it is a subspace of the linear space 𝐹 (𝑋). We conclude that𝐵(𝑋) is a normed linear space.

3. To show that 𝐵(𝑋) is complete, assume that (𝑓𝑛) is a Cauchy sequence in 𝐵(𝑋).For every 𝑥 ∈ 𝑋

∣𝑓𝑛(𝑥) − 𝑓𝑚(𝑥)∣ ≤ ∥𝑓𝑛 − 𝑓𝑚∥ → 0

Therefore, for 𝑥 ∈ 𝑋 , 𝑓𝑛(𝑥) is a Cauchy sequence of real numbers. Since ℜ iscomplete, this sequence converges. Define the function

𝑓(𝑥) = lim𝑛→∞ 𝑓

𝑛(𝑥)

We need to show

∙ ∥𝑓𝑛 − 𝑓∥ → 0 and

∙ 𝑓 ∈ 𝐵(𝑋)

(𝑓𝑛) is a Cauchy sequence. For given 𝜖 > 0, choose 𝑁 such that ∥𝑓𝑛 − 𝑓𝑚∥ < 𝜖/2for very 𝑚,𝑛 ≥ 𝑁 . For any 𝑥 ∈ 𝑋 and 𝑛 ≥ 𝑁 ,

∣𝑓𝑛(𝑥) − 𝑓(𝑥)∣ ≤ ∣𝑓𝑛(𝑥)− 𝑓𝑚(𝑥)∣ + ∣𝑓𝑚(𝑥) − 𝑓(𝑥)∣≤ ∥𝑓𝑛 − 𝑓𝑚∥+ ∣𝑓𝑚(𝑥) − 𝑓(𝑥)∣

By suitable choice of 𝑚 (which may depend upon 𝑥), each term on the right canbe made smaller than 𝜖/2 and therefore

∣𝑓𝑛(𝑥) − 𝑓(𝑥)∣ < 𝜖for every 𝑥 ∈ 𝑋 and 𝑛 ≥ 𝑁 .

∥𝑓𝑛 − 𝑓∥ = sup𝑥∈𝑋∣𝑓𝑛(𝑥)− 𝑓(𝑥)∣ ≤ 𝜖

Finally, this implies ∥𝑓∥ = lim𝑛→∞ ∥𝑓𝑛∥. Therefore 𝑓 ∈ 𝐵(𝑋).

2.12 If the die is fair, the probability of the elementary outcomes is

𝑃 ({1}) = 𝑃 ({2}) = 𝑃 ({3}) = 𝑃 ({4}) = 𝑃 ({5}) = 𝑃 ({6}) = 1/6

By Condition 3

𝑃 ({2, 4, 6}) = 𝑃 ({2}) + 𝑃 ({4}) + 𝑃 ({6}) = 1/2

2.13 The profit maximization problem of a competitive single-output firm is to choosethe combination of inputs x ∈ ℜ𝑛+ and scale of production 𝑦 to maximize net profit.This is summarized in the constrained maximization problem

maxx,𝑦𝑝𝑦 −

𝑛∑𝑖=1

𝑤𝑖𝑥𝑖

subject to x ∈ 𝑉 (𝑦)

where 𝑝𝑦 is total revenue and∑𝑛

𝑖=1 𝑤𝑖𝑥𝑖 total cost. The profit function, which dependsupon both 𝑝 and w, is defined by

Π(𝑝,w) = max𝑦,x∈𝑉 (𝑦)

𝑝𝑦 −𝑛∑𝑖=1

𝑤𝑖𝑥𝑖

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For analysis, it is convenient to represent the technology 𝑉 (𝑦) by a production function(Example 2.24). The firm’s optimization can then be expressed as

maxx∈ℜ𝑛

+

𝑝𝑓(x)−𝑛∑𝑖=1

𝑤𝑖𝑥𝑖

and the profit function as

Π(𝑝,w) = maxx∈ℜ𝑛

+

𝑝𝑓(x)−𝑛∑𝑖=1

𝑤𝑖𝑥𝑖

2.14 1. Assume that production is profitable at p. That is, there exists some y ∈ 𝑌such that 𝑓(y,p) > 0. Since the technology exhibits constant returns to scale, 𝑌is a cone (Example 1.101). Therefore 𝛼y ∈ 𝑌 for every 𝛼 > 0 and

𝑓(𝛼y,p) =∑𝑖

𝑝𝑖(𝛼𝑦𝑖) = 𝛼∑𝑖

𝑝𝑖𝑦𝑖 = 𝛼𝑓(y,p)

Therefore { 𝑓(𝛼y,p) : 𝛼 > 0 } is unbounded and

Π(p) = supy∈𝑌𝑓(y,p) ≥ sup

𝛼>0𝑓(𝛼y,p) = +∞

2. Assume to the contrary that there exists p ∈ ℜ𝑛+ with Π(p) = 𝜋 /∈ { 0,+∞,−∞}.There are two possible cases.

(a) 0 < 𝜋 < +∞. Since 𝜋 = sup𝑦∈𝑌 𝑓(y,p) > 0, there exists y ∈ 𝑌 such that𝑓(y,p) > 0 The previous part implies Π(p) = +∞.

(b) −∞ < 𝜋 < 0. Then there exists y ∈ 𝑌 such that 𝑓(y,p) < 0 By a similarargument to the previous part, this implies Π(p) = −∞.

2.15 Assume x∗ is a solution to (2.4).

𝑓(x∗, 𝜽) ≥ 𝑓(x, 𝜽) for every x ∈ 𝐺(𝜽)

and therefore

𝑓(x∗, 𝜽) ≥ supx∈𝐺(𝜽)

𝑓(x, 𝜽) = 𝑣(𝜽)

On the other hand x∗ ∈ 𝐺(𝜽) and therefore

𝑣(𝜽) = supx∈𝐺(𝜽)

𝑓(x, 𝜽) ≥ 𝑓(x∗, 𝜽)

Therefore, x∗ satisfies (2.5). Conversely, assume x∗ ∈ 𝐺(𝜽) satisfies (2.5). Then

𝑓(x∗, 𝜽) = 𝑣(𝜽) = supx∈𝐺(𝜽)

𝑓(x, 𝜽) ≥ 𝑓(x, 𝜽) for every x ∈ 𝐺(𝜽)

x∗ solve (2.4).

2.16 The assumption that 𝐺(𝑥) ∕= ∅ for every 𝑥 ∈ 𝑋 implies Γ(𝑥0) ∕= ∅ for every𝑥0 ∈ 𝑋 . There always exist feasible plans from any starting point. Since 𝑢 is bounded,there exists 𝑀 such that ∣𝑓(𝑥𝑡, 𝑥𝑡+1)∣ ≤ 𝑀 for every x ∈ Γ(𝑥0). Consequently, forevery x ∈ Γ(𝑥0), 𝑈(x) ∈ ℜ and

∣𝑈(x)∣ =∣∣∣∣∣∞∑𝑡=0

𝛽𝑡𝑓(𝑥𝑡, 𝑥𝑡+1)

∣∣∣∣∣ ≤∞∑𝑡=0

𝛽𝑡 ∣𝑓(𝑥𝑡, 𝑥𝑡+1)∣ ≤∞∑𝑡=0

𝛽𝑡𝑀 =𝑀

1− 𝛽

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using the formula for a geometric series (Exercise 1.108). Therefore

𝑣(𝑥0) = supx∈Γ(𝑥0)

𝑈(x) ≤ 𝑀

1− 𝛽

and 𝑣 ∈ 𝐵(𝑋). Next, we note that for every feasible plan x ∈ Γ(𝑥0)

𝑈(x) =∞∑𝑡=0


= 𝑓(𝑥0, 𝑥1) + 𝛽

∞∑𝑡=0

𝛽𝑡𝑓(𝑥𝑡+1, 𝑥𝑡+2)

= 𝑓(𝑥0, 𝑥1) + 𝛽𝑈(x′) (2.29)

where x′ = (𝑥1, 𝑥2, . . . ) is the continuation of the plan x starting at 𝑥1. For any 𝑥0 ∈ 𝑋and 𝜖 > 0, there exists a feasible plan x ∈ Γ(𝑥0) such that

𝑈(x) ≥ 𝑣(𝑥0)− 𝜖

Let x′ = (𝑥1, 𝑥2, . . . ) be the continuation of the plan x starting at 𝑥1. Using (2.29)and the fact that 𝑈(x′) ≤ 𝑣(𝑥1), we conclude that

𝑣(𝑥0)− 𝜖 ≤ 𝑈(x)

= 𝑓(𝑥0, 𝑥1) + 𝛽𝑈(x′)≤ 𝑓(𝑥0, 𝑥1) + 𝛽𝑣(𝑥1)

≤ sup𝑦∈𝐺(𝑥)

{ 𝑓(𝑥0, 𝑦) + 𝛽𝑣(𝑦) }

Since this is true for every 𝜖 > 0, we must have

𝑣(𝑥0) ≤ sup𝑦∈𝐺(𝑥)

{ 𝑓(𝑥0, 𝑦) + 𝛽𝑣(𝑦) } (2.30)

On the other hand, choose any 𝑥1 ∈ 𝐺(𝑥0) ⊆ 𝑋 . Since


𝑈(x)

there exists a feasible plan x′ = (𝑥1, 𝑥2, . . . ) starting at 𝑥1 such that

𝑈(x′) ≥ 𝑣(𝑥1)− 𝜖

Moreover, the plan x = (𝑥0, 𝑥1, 𝑥2, . . . ) is feasible from 𝑥0 and

𝑣(𝑥0) ≥ 𝑈(x) = 𝑓(𝑥0, 𝑥1) + 𝛽𝑈(x′) ≥ 𝑓(𝑥0, 𝑥1) + 𝛽𝑣(𝑥1)− 𝛽𝜖

Since this is true for every 𝜖 > 0 and 𝑥1 ∈ 𝐺(𝑥0), we conclude that

𝑣(𝑥0) ≥ sup𝑦∈𝐺(𝑥)

{ 𝑓(𝑥0, 𝑦) + 𝛽𝑣(𝑦) }

Together with (2.30) this establishes the required result, namely

𝑣(𝑥0) = sup𝑦∈𝐺(𝑥)

{ 𝑓(𝑥0, 𝑦) + 𝛽𝑣(𝑦) }

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2.17 Assume x is optimal, so that

𝑈(x∗) ≥ 𝑈(x) for every x ∈ Γ(𝑥0)

This implies (using (2.39))

𝑓(𝑥0, 𝑥∗1) + 𝛽𝑈(x∗′) ≥ 𝑓(𝑥0, 𝑥1) + 𝛽𝑈(x′)

where x′ = (𝑥1, 𝑥2, . . . ) is the continuation of the plan x starting at 𝑥1 and x∗′ =(𝑥∗1, 𝑥

∗2, . . . ) is the continuation of the plan x∗. In particular, this is true for every plan

x ∈ Γ(𝑥0) with 𝑥1 = 𝑥∗1 and therefore

𝑓(𝑥0, 𝑥∗1) + 𝛽𝑈(x∗′) ≥ 𝑓(𝑥0, 𝑥

∗1) + 𝛽𝑈(x′) for every x′ ∈ Γ(𝑥∗1)

which implies that

𝑈(x∗′) ≥ 𝑈(x′) for every x′ ∈ Γ(𝑥∗1)

That is, x∗′ is optimal starting at 𝑥∗1 and therefore 𝑈(x∗′) = 𝑣(𝑥∗1) (Exercise 2.15).Consequently

𝑣(𝑥0) = 𝑈(x∗) = 𝑓(𝑥0, 𝑥∗1) + 𝛽𝑈(x∗′) = 𝑓(𝑥0, 𝑥

∗1) + 𝛽𝑣(𝑥∗1)

This verifies (2.13) for 𝑡 = 0. A similar argument verifies (2.13) for any period 𝑡.

To show the converse, assume that x∗ = (𝑥0, 𝑥∗1, 𝑥

∗2, . . . ) ∈ Γ(𝑥0) satisfies (2.13). Suc-

cessively using (2.13)

𝑣(𝑥0) = 𝑓(𝑥0, 𝑥∗1) + 𝛽𝑣(𝑥∗1)

= 𝑓(𝑥0, 𝑥∗1) + 𝛽𝑓(𝑥∗1, 𝑥

∗2) + 𝛽2𝑣(𝑥∗1)

=

1∑𝑡=0

𝛽𝑡𝑓(𝑥∗𝑡 , 𝑥∗𝑡+1) + 𝛽2𝑣(𝑥∗2)

=

2∑𝑡=0

𝛽𝑡𝑓(𝑥∗𝑡 , 𝑥∗𝑡+1) + 𝛽3𝑣(𝑥∗3)

=

𝑇−1∑𝑡=0

𝛽𝑡𝑓(𝑥𝑡, 𝑥𝑡+ 1) + 𝛽𝑇 𝑣(𝑥∗𝑇 )

for any 𝑇 = 1, 2, . . . . Since 𝑣 is bounded (Exercise 2.16), 𝛽𝑇 𝑣(𝑥∗𝑇 )→ 0 as 𝑇 →∞ andtherefore

𝑣(𝑥0) =∞∑𝑡=0

𝛽𝑡𝑓(𝑥𝑡, 𝑥𝑡+1) = 𝑈(x∗)

Again using Exercise 2.15, x∗ is optimal.

2.18 We have to show that

∙ for any 𝑣 ∈ 𝐵(𝑋), 𝑇𝑣 is a functional on 𝑋 .

∙ 𝑇𝑣 is bounded.

Since 𝐹 ∈ 𝐵(𝑋 ×𝑋), there exists 𝑀1 <∞ such that ∣𝑓(𝑥, 𝑦)∣ ≤𝑀1 for every (𝑥, 𝑦) ∈𝑋 ×𝑋 . Similarly, for any 𝑣 ∈ 𝐵(𝑋), there exists 𝑀2 < ∞ such that ∣𝑣(𝑥)∣ ≤ 𝑀2 forevery 𝑥 ∈ 𝑋 . Consequently for every (𝑥, 𝑦) ∈ 𝑋 ×𝑋 and 𝑣 ∈ 𝐵(𝑋)

∣𝑓(𝑥, 𝑦) + 𝛽𝑣(𝑦)∣ ≤ ∣𝑓(𝑥, 𝑦)∣+ 𝛽 ∣𝑣(𝑦)∣ ≤𝑀1 + 𝛽𝑀2 <∞ (2.31)

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For each 𝑥 ∈ 𝑋 , the set

𝑆𝑥 ={𝑓(𝑥, 𝑦) + 𝛽𝑣(𝑦) : 𝑦 ∈ 𝐺(𝑥)

}is a nonempty bounded subset of ℜ, which has least upper bound. Therefore

(𝑇𝑣)(𝑥) = sup𝑆𝑥 = sup𝑦∈𝐺(𝑥)

𝑓(𝑥, 𝑦) + 𝛽𝑣(𝑦)

defines a functional on 𝑋 . Moreover by (2.31)

∣(𝑇𝑣)(𝑘)∣ ≤𝑀1 + 𝛽𝑀2 <∞

Therefore 𝑇𝑣 ∈ 𝐵(𝑋).

2.19 Let 𝑁 = {1, 2, 3}. Any individual is powerless so that

𝑤({𝑖}) = 0 𝑖 = 1, 2, 3

Any two players can allocate the $1 to between themselves, leaving the other playerout. Therefore

𝑤({𝑖, 𝑗}) = 1 𝑖, 𝑗 ∈ 𝑁, 𝑖 ∕= 𝑗

The best that the three players can achieve is to allocate the $1 amongst themselves,so that

𝑤(𝑁) = 1

2.20 If the consumers preferences are continuous and strictly convex, she has a uniqueoptimal choice x∗ for every set of prices p and income 𝑚 in 𝑃 (Example 1.116). There-fore, the demand correspondence is single valued.

2.21 Assume 𝑠∗𝑖 ∈ 𝐵(s∗) for every 𝑖 ∈ 𝑁 . Then for every player 𝑖 ∈ 𝑁

(𝑠𝑖, s−𝑖) ≿𝑖 (𝑠′𝑖, s−𝑖) for every 𝑠′𝑖 ∈ 𝑆𝑖s∗ = (𝑠∗1, 𝑠

∗2, . . . , 𝑠

∗𝑛) is a Nash equilibrium. Conversely, assume s∗ = (𝑠∗1, 𝑠

∗2, . . . , 𝑠

∗𝑛) is

a Nash equilibrium. Then for every player 𝑖 ∈ 𝑁

(𝑠𝑖, s−𝑖) ≿𝑖 (𝑠′𝑖, s−𝑖) for every 𝑠′𝑖 ∈ 𝑆𝑖which implies that

𝑠∗𝑖 ∈ 𝐵(s∗) for every 𝑖 ∈ 𝑁

2.22 For any nonempty compact set 𝑇 ⊆ 𝑆, 𝐵(𝑇 ) is nonempty and compact provided≿𝑖 is continuous (Proposition 1.5) and 𝐵(𝑇 ) ⊆ 𝑇 . Therefore

𝐵1𝑖 ⊇ 𝐵2𝑖 ⊇ 𝐵3𝑖 . . .

is a nested sequence of nonempty compact sets. By the nested intersection theorem(Exercise 1.117), 𝑅𝑖 =

∩∞𝑛=0𝐵

𝑛𝑖 ∕= ∅.

2.23 If s∗ is a Nash equilibrium, 𝑠𝑖 ∈ 𝐵𝑛𝑖 for every 𝑛.

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2.24 For any 𝜽, let x∗ ∈ 𝜑(𝜽). Then

𝑓(x∗, 𝜽) ≥ 𝑓(x, 𝜽) for every x ∈ 𝐺(𝜽)

Therefore

𝑓(x∗, 𝜽) ≥ 𝑣(𝜽) = supx∈𝐺(𝜽)

𝑓(x, 𝜽)

Conversely

𝑣(𝜽) = supx∈𝐺(𝜽)

𝑓(x, 𝜽) ≥ supx∈𝐺(𝜽)

𝑓(x, 𝜽) ≥ 𝑓(x∗, 𝜽) for every x∗ ∈ 𝜑(𝜽)

Consequently

𝑣(𝜽) = 𝑓(x∗, 𝜽) for any x∗ ∈ 𝜑(𝜽)

2.25 The graph of 𝑉 is

graph(𝑉 ) = { (𝑦,x) ∈ ℜ+ ×ℜ𝑛+ : x ∈ 𝑉 (𝑦) }

while the production possibility set 𝑌 is

𝑌 = { (𝑦,−x) ∈ ℜ+ ×ℜ𝑛+ : 𝑥 ∈ 𝑉 (𝑦) }

Assume that 𝑌 is convex and let (𝑦𝑖,x𝑖) ∈ graph(𝑉 ), 𝑖 = 1, 2. This means that

(𝑦1,−x1) ∈ 𝑌 and (𝑦2,−x2) ∈ 𝑌

Let

𝑦 = 𝛼𝑦1 + (1− 𝛼)𝑦2 and x = 𝛼x1 + (1− 𝛼)x2

for some 0 ≤ 𝛼 ≤ 1. Since 𝑌 is convex

(𝑦,−x) = 𝛼(𝑦1,−x1) + (1 − 𝛼)(𝑦2,−x2) ∈ 𝑌

and therefore x ∈ 𝑉 (𝑦) so that (𝑦, x) ∈ graph(𝑉 ). That is graph(𝑉 ) is convex.

Conversely, assuming graph(𝑉 ) is convex, if (𝑦𝑖,−x𝑖) ∈ 𝑌 , 𝑖 = 1, 2, then (𝑦𝑖,x𝑖) ∈graph(𝑉 ) and therefore

(𝑦, x) ∈ graph(𝑉 ) =⇒ x ∈ 𝑉 (𝑦) =⇒ (𝑦,−x) ∈ 𝑌

so that 𝑌 is convex.

2.26 The graph of 𝜑 is

graph(𝐺) = { (𝜽,x) ∈ Θ×𝑋 : x ∈ 𝐺(𝜽) }

Assume that (𝜽𝑖,x𝑖) ∈ graph(𝐺), 𝑖 = 1, 2. This means that x𝑖 ∈ 𝐺(𝜽) and therefore𝑔𝑗(x, 𝜽) ≤ 𝑐𝑗 for every 𝑗 and 𝑖 = 1, 2. Since 𝑔𝑗 is convex

𝑔(𝛼x1 + (1− 𝛼)x2, 𝛼𝜽1 + (1− 𝛼)𝜽2) ≥ 𝛼𝑔(x1, 𝜽1) + (1− 𝛼)𝑔(x2, 𝜽2) ≥ 𝑐𝑗Therefore 𝛼x1+(1−𝛼)x2 ∈ 𝐺(𝛼𝜽1+(1−𝛼)𝜽2) and (𝛼𝜽1+(1−𝛼)𝜽2, 𝛼x1+(1−𝛼)x2) ∈graph(𝐺). 𝐺 is convex.

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2.27 The identity function 𝐼𝑋 : 𝑋 → 𝑋 is defined by 𝐼𝑋(𝑥) = 𝑥 for every 𝑥 ∈ 𝑋 .Therefore

𝑥2 ≻𝑋 𝑥1 =⇒ 𝐼𝑋(𝑥2) = 𝑥2 ≻𝑋 𝑥1 = 𝐼𝑋(𝑥1)

2.28 Assume that 𝑓 and 𝑔 are increasing. Then, for every 𝑥1, 𝑥2 ∈ 𝑋 with 𝑥2 ≿𝑋 𝑥1

𝑓(𝑥2) ≿𝑌 𝑓(𝑥1) =⇒ 𝑔(𝑓(𝑥2)) ≿𝑍 𝑔(𝑓(𝑥1))

𝑔 ∘ 𝑓 is also increasing. Similarly, if 𝑓 and 𝑔 are strictly increasing,

𝑥2 ≻𝑋 𝑥1 =⇒ 𝑓(𝑥2) ≻𝑌 𝑓(𝑥1) =⇒ 𝑔(𝑓(𝑥2)) ≻𝑍 𝑔(𝑓(𝑥1))

2.29 For every 𝑦 ∈ 𝑓(𝑋), there exists a unique 𝑥 ∈ 𝑋 such that 𝑓(𝑥) = y. (For if 𝑥1, 𝑥2are such that 𝑓(𝑥1) = 𝑓(𝑥2), then 𝑥1 = 𝑥2.) Therefore, 𝑓 is one-to-one and onto 𝑓(𝑋),and so has an inverse (Exercise 2.4). Further

𝑥2 > 𝑥2 ⇐⇒ 𝑓(𝑥2) > 𝑓(𝑥1)

Therefore 𝑓−1 is strictly increasing.

2.30 Assume 𝑓 : 𝑋 → ℜ is increasing. Then, for every 𝑥2 ≿ 𝑥1, 𝑓(𝑥2) ≥ 𝑓(𝑥1) whichimplies that −𝑓(𝑥2) ≤ −𝑓(𝑥1). −𝑓 is decreasing.

2.31 For every 𝑥2 ≿ 𝑥1.

𝑓(𝑥2) ≥ 𝑓(𝑥1)

𝑔(𝑥2) ≥ 𝑔(𝑥1)Adding

(𝑓 + 𝑔)(𝑥2) = 𝑓(𝑥2) + 𝑔(𝑥2) ≥ 𝑓(𝑥1) + 𝑓(𝑥1) = (𝑓 + 𝑔)(𝑥1)

That is, 𝑓 + 𝑔 is increasing. Similarly for every 𝛼 ≥ 0

𝛼𝑓(𝑥2) ≥ 𝛼𝑓(𝑥1)

and therefore 𝛼𝑓 is increasing. By Exercise 1.186, the set of all increasing functionalsis a convex cone in 𝐹 (𝑋).

If 𝑓 is strictly increasing, then for every 𝑥2 ≻ 𝑥1,𝑓(𝑥2) > 𝑓(𝑥1)

𝑔(𝑥2) ≥ 𝑔(𝑥1)Adding

(𝑓 + 𝑔)(𝑥2) = 𝑓(𝑥2) + 𝑔(𝑥2) > 𝑓(𝑥1) + 𝑔(𝑥1) = (𝑓 + 𝑔)(𝑥1)

𝑓 + 𝑔 is strictly increasing. Similarly for every 𝛼 > 0

𝛼𝑓(𝑥2) > 𝛼𝑓(𝑥1)

𝛼𝑓 is strictly increasing.

2.32 For every 𝑥2 ≻ 𝑥1.𝑓(𝑥2) > 𝑓(𝑥1) > 0

𝑔(𝑥2) > 𝑔(𝑥1) > 0

and therefore

(𝑓𝑔)(𝑥2) = 𝑓(𝑥2)𝑔(𝑥2) > 𝑓(𝑥2)𝑔(𝑥1) > 𝑓(𝑥1)𝑔(𝑥1) = (𝑓𝑔)(𝑥1)

using Exercise 2.31.

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2.33 By Exercise 2.31 and Example 2.53, each 𝑔𝑛 is strictly increasing on ℜ+. That is

𝑥1 < 𝑥2 =⇒ 𝑔𝑛(𝑥1) < 𝑔𝑛(𝑥2) for every 𝑛 (2.32)

and therefore

lim𝑛→∞ 𝑔𝑛(𝑥1) ≤ lim

𝑛→∞ 𝑔𝑛(𝑥2)

This suffices to show that 𝑔(𝑥) = lim𝑛→∞ 𝑔𝑛(𝑥) is increasing (not strictly increasing).However, 1 + 𝑥 is strictly increasing, and therefore by Exercise 2.31

𝑒𝑥 = 1 + 𝑥+ 𝑔(𝑥)

is strictly increasing on ℜ+. While it is the case that 𝑔 = lim 𝑔𝑛 is strictly increasingon ℜ+, (2.32) does not suffice to show this.

2.34 For every 𝑎 > 0, 𝑎 log 𝑥 is strictly increasing (Exercise 2.32) and therefore 𝑒𝑎 log 𝑥

is strictly increasing (Exercise 2.28). For every 𝑎 < 0, −𝑎 log𝑥 is strictly increasingand therefore (Exercise 2.30 𝑎 log 𝑥 is strictly decreasing. Therefore 𝑒𝑎 log 𝑥 is strictlydecreasing (Exercise 2.28).

2.35 Apply Exercises 2.31 and 2.28 to Example 2.56.

2.36 𝑢 is (strictly) increasing so that

𝑥2 ≿ 𝑥1 =⇒ 𝑢(𝑥2) ≥ 𝑢(𝑥1)To show the converse, assume that 𝑥1, 𝑥2 ∈ 𝑋 with 𝑢(𝑥2) ≥ 𝑢(𝑥1). Since ≿ is complete,either 𝑥2 ≿ 𝑥1 or 𝑥1 ≻ 𝑥2. However, the second possibility cannot occur since 𝑢 isstrictly increasing and therefore

𝑥1 ≻ 𝑥2 =⇒ 𝑢(𝑥1) > 𝑢(𝑥2)

contradicting the hypothesis that 𝑢(𝑥2) ≥ 𝑢(𝑥1). We conclude that

𝑢(𝑥2) ≥ 𝑢(𝑥1) =⇒ 𝑥2 ≿ 𝑥1

2.37 Assume 𝑢 represents the preference ordering ≿ on 𝑋 and let 𝑔 : ℜ → ℜ be strictlyincreasing. Then composition 𝑔 ∘ 𝑢 : 𝑋 → ℜ is strictly increasing (Exercise 2.28).Therefore 𝑔 ∘ 𝑢 is a utility function (Example 2.58). Since 𝑔 is strictly increasing

𝑔(𝑢(𝑥2)) ≥ 𝑔(𝑢(𝑥1)) ⇐⇒ 𝑢(𝑥2) ≥ 𝑢(𝑥1) ⇐⇒ 𝑥2 ≿ 𝑥1

for every 𝑥1, 𝑥2 ∈ 𝑋 Therefore, 𝑔 ∘ 𝑢 also represents ≿.

2.38 1. (a) Let 𝑧 = max𝑛𝑖=1 𝑥𝑖. Then z = 𝑧1 ≿ x and therefore z ∈ 𝑍+x . Similarly,let 𝑧 = min𝑛𝑖=1 𝑥𝑖. Then z = 𝑧1 ∈ 𝑍−

x . Therefore, 𝑍+x and 𝑍−x are both

nonempty. By continuity, the upper and lower contour sets ≿(x) and ≾(x)are closed. 𝑍 is a closed cone. Since

𝑍+x = ≿(x) ∩ 𝑍 and 𝑍−x = ≾(x) ∩ 𝑍

𝑍+x and 𝑍−x are closed.

(b) By completeness, 𝑍+x ∪ 𝑍−x = 𝑍. Since 𝑍 is connected, 𝑍+x ∩ 𝑍−

x ∕= ∅.(Otherwise, 𝑍 is the union of two disjoint closed sets and hence the unionof two disjoint open sets.)

(c) Let zx ∈ 𝑍+x ∩ 𝑍−x . Then z ≿ x and also z ≾ x. That is, z ∼ x.

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(d) Suppose x ∼ z1x and x ∼ z2x with z1x ∕= z2x. Then either z1x > z2x or

z1x < z2x. Without loss of generality, assume z2x > z

1x. Then monotonicity

and transitivity imply

x ∼ z2x ≻ z1x ∼ xwhich is a contradiction. Therefore zx is unique.

Let 𝑧x denote the scale of zx, that is zx = 𝑧x1. For every x ∈ ℜ𝑛+, there is aunique zx ∼ x and the function 𝑢 : ℜ𝑛+ → ℜ given by 𝑢(x) = 𝑧x is well-defined.Moreover

x2 ≿ x1 ⇐⇒ zx2 ≿ zx1

⇐⇒ 𝑧x2 ≥ 𝑧x1

⇐⇒ 𝑢(x2) ≥ 𝑢(x1)𝑢 represents the preference order ≿.

2.39 1. For every 𝑥1 ∈ ℜ, (𝑥1, 2) ≻𝐿 (𝑥1, 1) in the lexicographic order. If 𝑢 represents≿𝐿, 𝑢 is strictly increasing and therefore 𝑢(𝑥1, 2) > 𝑢(𝑥1, 1). There exists arational number 𝑟(𝑥1) such that 𝑢(𝑥1, 2) > 𝑟(𝑥1) > 𝑢(𝑥1, 1).

2. The preceding construction associates a rational number with every real number𝑥1 ∈ ℜ. Hence 𝑟 is a function from ℜ to the set of rational numbers 𝑄. For any𝑥11, 𝑥

21 ∈ ℜ with 𝑥21 > 𝑥

11

𝑟(𝑥21) > 𝑢(𝑥21, 1) > 𝑢(𝑥11, 2) > 𝑟(𝑥11)

Therefore

𝑥21 > 𝑥11 =⇒ 𝑟(𝑥21) > 𝑟(𝑥

11)

𝑟 is strictly increasing.

3. By Exercise 2.29, 𝑟 has an inverse. This implies that 𝑟 is one-to-one and onto,which is impossible since 𝑄 is countable and ℜ is uncountable (Example 2.16).This contradiction establishes that ≿𝐿 has no such representation 𝑢.

2.40 Let a1, a2 ∈ 𝐴 with a1 ≿2 a2. Since the game is strictly competitive, a2 ≿1 a1.Since 𝑢1 represents ≿1, 𝑢1(a2) ≥ 𝑢1(a1) which implies that −𝑢1(a2) ≤ −𝑢1(a1), thatis 𝑢2(a

1) ≥ 𝑢2(a2) where 𝑢2 = −𝑢1. Similarly

𝑢2(a1) ≥ 𝑢2(a2) =⇒ 𝑢1(a

1) ≤ 𝑢1(a2) ⇐⇒ a1 ≾1 a2 =⇒ a1 ≿2 a2

Therefore 𝑢2 = −𝑢1 represents ≿2 and

𝑢1(a) + 𝑢2(a) = 0 for every a ∈ 𝐴2.41 Assume 𝑆 ⫋ 𝑇 . By superadditivity

𝑤(𝑇 ) ≥ 𝑤(𝑆) + 𝑤(𝑇 ∖ 𝑆) ≥ 𝑤(𝑆)

2.42 Assume 𝑣, 𝑤 ∈ 𝐵(𝑋) with 𝑤(𝑦) ≥ 𝑣(𝑦) for every 𝑦 ∈ 𝑋 . Then for any 𝑥 ∈ 𝑋𝑓(𝑥, 𝑦) + 𝛽𝑤(𝑦) ≥ 𝑓(𝑥, 𝑦) + 𝛽𝑣(𝑦) for every 𝑦 ∈ 𝑋

and therefore

(𝑇𝑤)(𝑥) = sup𝑦∈𝐺(𝑥)

{𝑓(𝑥, 𝑦) + 𝛽𝑤(𝑦)} ≥ sup𝑦∈𝐺(𝑥)

{𝑓(𝑥, 𝑦) + 𝛽𝑣(𝑦)} = (𝑇𝑣)(𝑥)

T is increasing.

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2.43 For every 𝜃2 ≥ 𝜃1 ∈ Θ, if 𝑥1 ∈ 𝐺(𝜃1) and 𝑥2 ∈ 𝐺(𝜃2), then 𝑥1 ∧ 𝑥2 ≤ 𝑥1 andtherefore 𝑥1 ∧ 𝑥2 ∈ 𝐺(𝜃1). If 𝑥1 ≥ 𝑥2, then 𝑥1 ∨ 𝑥2 = 𝑥1 ≤ 𝑔(𝜃1) ≤ 𝑔(𝜃2) and therefore𝑥1 ∨ 𝑥2 ∈ 𝐺(𝜃2). On the other hand, if 𝑥1 ≤ 𝑥2, then 𝑥1 ∨ 𝑥2 = 𝑥2 ∈ 𝐺(𝜃2).

2.44 Assume 𝜑 is increasing, and let 𝑥1, 𝑥2 ∈ 𝑋 with 𝑥2 ≿ 𝑥1. Let 𝑦1 ∈ 𝜑(𝑥1). Chooseany 𝑦′ ∈ 𝜑(𝑥2). Since 𝜑 is increasing, 𝜑(𝑥2) ≿𝑆 𝜑(𝑥1) and therefore 𝑦2 = 𝑦1 ∨ 𝑦′ ∈𝜑(𝑥2). 𝑦2 ≿ 𝑦1 as required. Similarly, for every 𝑦2 ∈ 𝜑(𝑥2), there exists some 𝑦′ ∈ 𝜑(𝑥2)such that 𝑦1 = 𝑦′ ∧ 𝑦2 ∈ 𝜑(𝑥1) with 𝑦2 ≿ 𝑦1.2.45 Since 𝜑(𝑥) is a sublattice, sup𝜑(𝑥) ∈ 𝜑(𝑥) for every 𝑥. Therefore, the function

𝑓(𝑥) = sup𝜑(𝑥)

is a selection. Similarly

𝑔(𝑥) = inf 𝜑(𝑥)

is a selection. Both 𝑓 and 𝑔 are increasing (Exercise 1.50).

2.46 Let 𝑥1, 𝑥2 belong to 𝑋 with 𝑥2 ≿ 𝑥1. Choose y1 = (𝑦11 , 𝑦12, . . . , 𝑦

1𝑛) ∈ ∏

𝑖 𝜑𝑖(𝑥1)

and y2 = (𝑦21 , 𝑦22 , . . . , 𝑦

2𝑛) ∈ ∏

𝑖 𝜑𝑖(𝑥2). Then, for each 𝑖 = 1, 2, . . . , 𝑛, 𝑦1𝑖 ∈ 𝜑𝑖(𝑥1) and

𝑦2𝑖 ∈ 𝜑𝑖(𝑥2). Since each 𝜑𝑖 is increasing, 𝑦1𝑖 ∧ 𝑦2𝑖 ∈ 𝜑𝑖(𝑥1) and 𝑦1𝑖 ∨ 𝑦2𝑖 ∈ 𝜑𝑖(𝑥2) foreach 𝑖. Therefore y1 ∧ y2 ∈ ∏

𝑖 𝜑𝑖(𝑥1) and y1 ∨ y2 ∈ ∏

𝑖 𝜑𝑖(𝑥2). 𝜑(𝑥) =

∏𝑖 𝜑𝑖(𝑥) is

increasing.

2.47 Let 𝑥1, 𝑥2 belong to 𝑋 with 𝑥2 ≿ 𝑥1. Choose 𝑦1 ∈ ∩𝑖 𝜑𝑖(𝑥

1) and 𝑦2 ∈ ∩𝑖 𝜑𝑖(𝑥

2).Then 𝑦1 ∈ 𝜑𝑖(𝑥1) and 𝑦2 ∈ 𝜑𝑖(𝑥2) for each 𝑖 = 1, 2, . . . , 𝑛. Since each 𝜑𝑖 is increasing,𝑦1 ∧ 𝑦2 ∈ 𝜑𝑖(𝑥1) and 𝑦1 ∨ 𝑦2 ∈ 𝜑𝑖(𝑥2) for each 𝑖. Therefore 𝑦1 ∧ 𝑦2 ∈ ∩

𝑖 𝜑𝑖(𝑥1) and

𝑦1 ∨ 𝑦2 ∈ ∩𝑖 𝜑𝑖(𝑥

2). 𝜑 =∩

𝑖 𝜑 is increasing.

2.48 Let 𝑓 be a selection from an always increasing correspondence 𝜑 : 𝑋 ⇉ 𝑌 . Forevery 𝑥1, 𝑥2 ∈ 𝑋 , 𝑓(𝑥1) ∈ 𝜑(𝑥1) and 𝑓(𝑥2) ∈ 𝜑(𝑥2). Since 𝜑 is always increasing

𝑥1 ≿𝑋 𝑥2 =⇒ 𝑓(𝑥1) ≿𝑌 𝑓(𝑥2)

𝑓 is increasing. Conversely, assume every selection 𝑓 ∈ 𝜑 is increasing. Choose any𝑥1, 𝑥2 ∈ 𝑋 with 𝑥1 ≿ 𝑥2. For every 𝑦1 ∈ 𝜑(𝑥1) and 𝑦2 ∈ 𝜑(𝑥2), there exists a selection𝑓 with 𝑦𝑖 = 𝜑(𝑥𝑖), 𝑖 = 1, 2. Since 𝑓 is increasing,

𝑥1 ≿𝑋 𝑥2 =⇒ 𝑦1 ≿𝑌 𝑦2

𝜑 is increasing.

2.49 Let 𝑥1, 𝑥2 ∈ 𝑋 . If 𝑋 is a chain, either 𝑥1 ≿ 𝑥2 or 𝑥2 ≿ 𝑥1. Without loss ofgenerality , assume 𝑥2 ≿ 𝑥1. Then 𝑥1∨𝑥2 = 𝑥2 and 𝑥1∧𝑥2 = 𝑥1 and (2.17) is satisfiedas an identity.

2.50

(𝑓 + 𝑔)(𝑥1 ∨ 𝑥2) + (𝑓 + 𝑔)(𝑥1 ∧ 𝑥2) = 𝑓(𝑥1 ∨ 𝑥2) + 𝑔(𝑥1 ∨ 𝑥2) + 𝑓(𝑥1 ∧ 𝑥2) + 𝑔(𝑥1 ∧ 𝑥2)= 𝑓(𝑥1 ∨ 𝑥2) + 𝑓(𝑥1 ∧ 𝑥2) + 𝑔(𝑥1 ∨ 𝑥2) + 𝑔(𝑥1 ∧ 𝑥2)≥ 𝑓(𝑥1) + 𝑓(𝑥2) + 𝑔(𝑥1) + 𝑔(𝑥2)

= (𝑓 + 𝑔)(𝑥1) + (𝑓 + 𝑔)(𝑥2)

Similarly

𝑓(𝑥1 ∨ 𝑥2) + 𝑓(𝑥1 ∧ 𝑥2) ≥ 𝑓(𝑥1) + 𝑓(𝑥2)

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implies

𝛼𝑓(𝑥1 ∨ 𝑥2) + 𝛼𝑓(𝑥1 ∧ 𝑥2) ≥ 𝛼𝑓(𝑥1) + 𝛼𝑓(𝑥2)

for all 𝛼 ≥ 0. By Exercise 1.186, the set of all supermodular functions is a convex conein 𝐹 (𝑋).

2.51 Since 𝑓 is supermodular and 𝑔 is nonnegative definite,

𝑓(𝑥1 ∨ 𝑥2)𝑔(𝑥1 ∨ 𝑥2) ≥(𝑓(𝑥1) + 𝑓(𝑥2)− 𝑓(𝑥1 ∧ 𝑥2)

)𝑔(𝑥1 ∨ 𝑥2)

= 𝑓(𝑥2)𝑔(𝑥1 ∨ 𝑥2) +(𝑓(𝑥1)− 𝑓(𝑥1 ∧ 𝑥2)

)𝑔(𝑥1 ∨ 𝑥2)

for any 𝑥1, 𝑥2 ∈ 𝑋 . Since 𝑓 and 𝑔 are increasing, this implies

𝑓(𝑥1 ∨ 𝑥2)𝑔(𝑥1 ∨ 𝑥2) ≥ 𝑓(𝑥2)𝑔(𝑥1 ∨ 𝑥2) +(𝑓(𝑥1)− 𝑓(𝑥1 ∧ 𝑥2)

)𝑔(𝑥1) (2.33)

Similarly, since 𝑓 is nonnegative definite, 𝑔 supermodular, and 𝑓 and 𝑔 increasing

𝑓(𝑥2)𝑔(𝑥1 ∨ 𝑥2) ≥ 𝑓(𝑥2)(𝑔(𝑥1) + 𝑔(𝑥2)− 𝑔(𝑥1 ∧ 𝑥2)

)= 𝑓(𝑥2)𝑔(𝑥2) + 𝑓(𝑥2)

(𝑔(𝑥1)− 𝑔(𝑥1 ∧ 𝑥2)

)≥ 𝑓(𝑥2)𝑔(𝑥2) + 𝑓(𝑥1 ∧ 𝑥2)

(𝑔(𝑥1)− 𝑔(𝑥1 ∧ 𝑥2)

)Combining this inequality with (2.33) gives

𝑓(𝑥1 ∨ 𝑥2)𝑔(𝑥1 ∨ 𝑥2) ≥ 𝑓(𝑥2)𝑔(𝑥2) + 𝑓(𝑥1 ∧ 𝑥2)(𝑔(𝑥1)− 𝑔(𝑥1 ∧ 𝑥2)

)+

(𝑓(𝑥1)− 𝑓(𝑥1 ∧ 𝑥2)

)𝑔(𝑥1)

= 𝑓(𝑥2)𝑔(𝑥2) + 𝑓(𝑥1 ∧ 𝑥2)𝑔(𝑥1)− 𝑓(𝑥1 ∧ 𝑥2)𝑔(𝑥1 ∧ 𝑥2)+ 𝑓(𝑥1)𝑔(𝑥1)− 𝑓(𝑥1 ∧ 𝑥2)𝑔(𝑥1)

= 𝑓(𝑥2)𝑔(𝑥2)− 𝑓(𝑥1 ∧ 𝑥2)𝑔(𝑥1 ∧ 𝑥2) + 𝑓(𝑥1)𝑔(𝑥1)

or

𝑓𝑔(𝑥1 ∨ 𝑥2) + 𝑓𝑔(𝑥1 ∧ 𝑥2) ≥ 𝑓𝑔(𝑥1) + 𝑓𝑔(𝑥2)

𝑓𝑔 is supermodular. (I acknowledge the help of Don Topkis in formulating this proof.)

2.52 Exercises 2.49 and 2.50.

2.53 For simplicity, assume that the firm produces two products. For every productionplan y = (𝑦1, 𝑦2),

y = (𝑦1, 0) ∨ (0, 𝑦2)

0 = (𝑦1, 0) ∧ (0, 𝑦2)

If 𝑐 is strictly submodular

𝑐(w,y) + 𝑐(w,0) < 𝑐(w, (𝑦1, 0)) + 𝑐(w, (0, 𝑦2))

Since 𝑐(w,0) = 0

𝑐(w,y) < 𝑐(w, (𝑦1, 0)) + 𝑐(w, (0, 𝑦2))

The technology displays economies of scope.

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2.54 Assume (𝑁,𝑤) is convex, that is

𝑤(𝑆 ∪ 𝑇 ) + 𝑤(𝑆 ∩ 𝑇 ) ≥ 𝑤(𝑆) + 𝑤(𝑇 ) for every 𝑆, 𝑇 ⊆ 𝑁

For all disjoint coalitions 𝑆 ∩ 𝑇 = ∅

𝑤(𝑆 ∪ 𝑇 ) ≥ 𝑤(𝑆) + 𝑤(𝑇 )

𝑤 is superadditive.

2.55 Rewriting (2.18), this implies

𝑤(𝑆 ∪ 𝑇 )− 𝑤(𝑇 ) ≥ 𝑤(𝑆) − 𝑤(𝑆 ∩ 𝑇 ) for every 𝑆, 𝑇 ⊆ 𝑁 (2.34)

Let 𝑆 ⊂ 𝑇 ⊂ 𝑁 ∖ {𝑖} and let 𝑆′ = 𝑆 ∪ {𝑖}. Substituting in (2.34)

𝑤(𝑆′ ∪ 𝑇 )− 𝑤(𝑇 ) ≥ 𝑤(𝑆′)− 𝑤(𝑆′ ∩ 𝑇 )

Since 𝑆 ⊂ 𝑇

𝑆′ ∪ 𝑇 = (𝑆 ∪ {𝑖}) ∪ 𝑇 = 𝑇 ∪ {𝑖}𝑆′ ∩ 𝑇 = (𝑆 ∪ {𝑖}) ∩ 𝑇 = 𝑆

Substituting in the previous equation gives the required result, namely

𝑤(𝑇 ∪ {𝑖})− 𝑤(𝑇 ) ≥ 𝑤(𝑆 ∪ {𝑖})− 𝑤(𝑆)

Conversely, assume that

𝑤(𝑇 ∪ {𝑖})− 𝑤(𝑇 ) ≥ 𝑤(𝑆 ∪ {𝑖})− 𝑤(𝑆) (2.35)

for every 𝑆 ⊂ 𝑇 ⊂ 𝑁 ∖ {𝑖}. Let 𝑆 and 𝑇 be arbitrary coalitions. Assume 𝑆 ∩ 𝑇 ⊂ 𝑆and 𝑆 ∩ 𝑇 ⊂ 𝑇 (otherwise (2.18) is trivially satisfied). This implies that 𝑇 ∖ 𝑆 ∕= ∅.Assume these players are labelled 1, 2, . . . ,𝑚, that is 𝑇 ∖ 𝑆 = {1, 2, . . . ,𝑚}. By (2.35)

𝑤(𝑆 ∪ {1})− 𝑤(𝑆) ≥ 𝑤((𝑆 ∩ 𝑇 ) ∪ {1})− 𝑤(𝑆 ∩ 𝑇 ) (2.36)

Successively adding the remaining players in 𝑇 ∖ 𝑆

𝑤(𝑆 ∪ {1, 2})− 𝑤(𝑆 ∪ {1}) ≥ 𝑤((𝑆 ∩ 𝑇 ) ∪ {1, 2})− 𝑤((𝑆 ∩ 𝑇 ) ∪ {1})...

𝑤(𝑆 ∪ (𝑇 ∖ 𝑆))− 𝑤(𝑆 ∪ {1, 2, . . . ,𝑚− 1}) ≥ 𝑤(𝑆 ∩ 𝑇 ) ∪ (𝑇 ∖ 𝑆))

− 𝑤((𝑆 ∩ 𝑇 ) ∪ {1, 2, . . . ,𝑚− 1})

Adding these inequalities to (2.36), we get

𝑤(𝑆 ∪ (𝑇 ∖ 𝑆))− 𝑤(𝑆) ≥ 𝑤(𝑆 ∩ 𝑇 ) ∪ (𝑇 ∖ 𝑆))− 𝑤(𝑆 ∩ 𝑇 )

This simplifies to

𝑤(𝑆 ∪ 𝑇 )− 𝑤(𝑆) ≥ 𝑆(𝑇 )− 𝑤(𝑆 ∩ 𝑇 )

which can be arranged to give (2.18).

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2.56 The cost allocation game is not convex. Let 𝑆 = {𝐴𝑃,𝐾𝑀}, 𝑇 = {𝐾𝑀,𝑇𝑁}.Then 𝑆 ∪ 𝑇 = {𝐴𝑃,𝐾𝑀,𝑇𝑁} = 𝑁 and 𝑆 ∩ 𝑇 = {𝐾𝑀} and

𝑤(𝑆 ∪ 𝑇 ) + 𝑤(𝑆 ∩ 𝑇 ) = 1530 < 1940 = 770 + 1170 = 𝑤(𝑆) + 𝑤(𝑇 )

Alternatively, observe that TN’s marginal contribution to coalition {𝐾𝑀,𝑇𝑁} is 1170,which is greater than its marginal contribution to the grand coalition {𝐴𝑃,𝐾𝑀,𝑇𝑁}(1530− 770 = 760).

2.57 𝑓 is supermodular if

𝑓(𝑥1 ∨ 𝑥2) + 𝑓(𝑥1 ∧ 𝑥2) ≥ 𝑓(𝑥1) + 𝑓(𝑥2)

which can be rearranged to give

𝑓(𝑥1 ∨ 𝑥2)− 𝑓(𝑥2) ≥ 𝑓(𝑥1)− 𝑓(𝑥1 ∧ 𝑥2)If the right hand side of this inequality is nonnegative, then so a fortiori is the lefthand side, that is

𝑓(𝑥1) ≥ 𝑓(𝑥1 ∧ 𝑥2) =⇒ 𝑓(𝑥1 ∨ 𝑥2) ≥ 𝑓(𝑥2)

If the right hand side is strictly positive, so must be the left hand side

𝑓(𝑥1) > 𝑓(𝑥1 ∧ 𝑥2) =⇒ 𝑓(𝑥1 ∨ 𝑥2) > 𝑓(𝑥2)

2.58 Assume 𝑥2 ≿ 𝑥1 ∈ 𝑋 and 𝑦2 ≿𝑌 𝑦2 ∈ 𝑌 . Assume that 𝑓 displays increasingdifferences in (𝑥, 𝑦), that is

𝑓(𝑥2, 𝑦2)− 𝑓(𝑥1, 𝑦2) ≥ 𝑓(𝑥2, 𝑦1)− 𝑓(𝑥1, 𝑦1) (2.37)

Rearranging

𝑓(𝑥2, 𝑦2)− 𝑓(𝑥2, 𝑦1) ≥ 𝑓(𝑥1, 𝑦2)− 𝑓(𝑥1, 𝑦1) (2.38)

Conversely, (2.38) implies (2.37) .

2.59 We showed in the text that supermodularity implies increasing differences. Toshow that reverse, assume that 𝑓 : 𝑋 × 𝑌 → ℜ displays increasing differences in (𝑥, 𝑦).Choose any (𝑥1, 𝑦1), (𝑥2, 𝑦2) ∈ 𝑋×𝑌 . If (𝑥1, 𝑦1), (𝑥2, 𝑦2) are comparable, so that either(𝑥1, 𝑦1) ≿ (𝑥2, 𝑦2) or (𝑥1, 𝑦1) ≾ (𝑥2, 𝑦2), then (2.17) holds has an equality. Thereforeassume that (𝑥1, 𝑦1), (𝑥2, 𝑦2) are incomparable. Without loss of generality, assume that𝑥1 ≾ 𝑥2 while 𝑦1 ≿ 𝑦2. (This is where we require that 𝑋 and 𝑌 be chains). This implies

(𝑥1, 𝑦1) ∧ (𝑥2, 𝑦2) = (𝑥1, 𝑦2) and (𝑥1, 𝑦1) ∨ (𝑥2, 𝑦2) = (𝑥2, 𝑦1) (2.39)

Increasing differences implies that

𝑓(𝑥2, 𝑦1)− 𝑓(𝑥1, 𝑦1) ≥ 𝑓(𝑥2, 𝑦2)− 𝑓(𝑥1, 𝑦2)

which can be rewritten as

𝑓(𝑥2, 𝑦1) + 𝑓(𝑥1, 𝑦2) ≥ 𝑓(𝑥1, 𝑦1) + 𝑓(𝑥2, 𝑦2)

Substituting (2.39)

𝑓((𝑥1, 𝑦1) ∨ (𝑥2, 𝑦2)

)+ 𝑓

((𝑥1, 𝑦1) ∧ (𝑥2, 𝑦2)

) ≥ 𝑓(𝑥1, 𝑦1) + 𝑓(𝑥2, 𝑦2)

which establishes the supermodularity of 𝑓 on 𝑋 × 𝑌 (2.17).

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2.60 In the standard Bertrand model of oligopoly

∙ the strategy space of each firm is ℜ+, a lattice.

∙ 𝑢𝑖(𝑝𝑖,p−𝑖) is supermodular in 𝑝𝑖 (Exercise 2.51).

∙ If the other firm’s increase their prices from p1−𝑖 to p2−𝑖, the effect on the demandfor firm 𝑖’s product is

𝑓(𝑝𝑖,p2−𝑖)− 𝑓(𝑝𝑖,p

1−𝑖) =

∑𝑖∕=𝑗𝑑𝑖𝑗(𝑝

2𝑗 − 𝑝1𝑗)

If the goods are gross substitutes, demand for firm 𝑖 increases and the amountof the increase is independent of 𝑝𝑖. Consequently, the effect on profit will be in-creasing in 𝑝𝑖. That is the payoff function (net revenue) has increasing differencesin (𝑝𝑖,p−𝑖). Specifically,

𝑢(𝑝𝑖,p2−𝑖)− 𝑢(𝑝𝑖,p1−𝑖) =

∑𝑖∕=𝑗𝑑𝑖𝑗(𝑝𝑖 − 𝑐𝑖)(𝑝2𝑗 − 𝑝1𝑗)

For any price increase p2−𝑖 ≩ p1−𝑖, the change in profit 𝑢(𝑝𝑖,p

2−𝑖) − 𝑢(𝑝𝑖,p1−𝑖) is

increasing in 𝑝𝑖.

Hence, the Bertrand oligopoly model is a supermodular game.

2.61 Suppose 𝑓 displays increasing differences so that for all 𝑥2 ≿ 𝑥1 and 𝑦2 ≿ 𝑦1

𝑓(𝑥2, 𝑦2)− 𝑓(𝑥1, 𝑦2) ≥ 𝑓(𝑥2, 𝑦1)− 𝑓(𝑥1, 𝑦1)

Then

𝑓(𝑥2, 𝑦1)− 𝑓(𝑥1, 𝑦1) ≥ 0 =⇒ 𝑓(𝑥2, 𝑦2)− 𝑓(𝑥1, 𝑦2) ≥ 0

and

𝑓(𝑥2, 𝑦1)− 𝑓(𝑥1, 𝑦1) > 0 =⇒ 𝑓(𝑥2, 𝑦2)− 𝑓(𝑥1, 𝑦2) > 0

2.62 For any 𝜽 ∈ Θ∗, let x1,x2 ∈ 𝜑(𝜽). Supermodularity implies

𝑓(x1 ∨ x2, 𝜽) + 𝑓(x1 ∧ x2, 𝜽) ≥ 𝑓(x1, 𝜽) + 𝑓(x2, 𝜽)


𝑓(x1 ∨ x2, 𝜽)− 𝑓(x2, 𝜽) ≥ 𝑓(x1, 𝜽)− 𝑓(x1 ∧ x2, 𝜽) (2.40)

However x1 and x2 are both maximal in 𝐺(𝜽).

𝑓(x2, 𝜽) ≥ 𝑓(x1 ∨ x2, 𝜽) =⇒ 𝑓(x1 ∨ x2, 𝜽)− 𝑓(x2, 𝜽) ≤ 0

𝑓(x1, 𝜽) ≥ 𝑓(x1 ∧ x2, 𝜽) =⇒ 𝑓(x1, 𝜽)− 𝑓(x1 ∧ x2, 𝜽) ≥ 0

Substituting in (2.40), we conclude

0 ≥ 𝑓(x1 ∨ x2, 𝜽)− 𝑓(x2, 𝜽) ≥ 𝑓(x1, 𝜽)− 𝑓(x1 ∧ x2, 𝜽) ≥ 0

This inequality must be satisfied as an equality with

𝑓(x1 ∨ x2, 𝜽) = 𝑓(x2, 𝜽)

𝑓(x1 ∧ x2, 𝜽) = 𝑓(x1, 𝜽)

That is x1 ∨ x2 ∈ 𝜑(𝜽) and x1 ∧ x2 ∈ 𝜑(𝜽). By Exercise 2.45, 𝜑 has an increasingselection.

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2.63 As in the proof of the theorem, let 𝜽1, 𝜽2 belong to Θ with 𝜽2 ≿ 𝜽1. Choose anyoptimal solutions x1 ∈ 𝜑(𝜽1) and x2 ∈ 𝜑(𝜽2). We claim that x2 ≿𝑋 x1. Assumeotherwise, that is assume x2 ∕≿𝑋 x1. This implies (Exercise 1.44) that x1 ∧ x2 ∕= x1.Since x1 ≿ x1 ∧ x2, we must have x1 ≻ x1 ∧ x2. Strictly increasing differences implies

𝑓(x1, 𝜽2)− 𝑓(x1, 𝜽1) > 𝑓(x1 ∧ x2, 𝜽2)− 𝑓(x1 ∧ x2, 𝜽1)which can be rearranged to give

𝑓(x1, 𝜽2)− 𝑓(x1 ∧ x2, 𝜽2) > 𝑓(x1, 𝜽1)− 𝑓(x1 ∧ x2, 𝜽1) (2.41)

Supermodularity implies

𝑓(x1 ∨ x2, 𝜽2) + 𝑓(x1 ∧ x2, 𝜽2) ≥ 𝑓(x1, 𝜽2) + 𝑓(x2, 𝜽2)


𝑓(x1 ∨ x2, 𝜽2)− 𝑓(x2, 𝜽2) ≥ 𝑓(x1, 𝜽2)− 𝑓(x1 ∧ x2, 𝜽2)Combining this inequality with (2.41) gives

𝑓(x1 ∨ x2, 𝜽2)− 𝑓(x2, 𝜽2) > 𝑓(x1, 𝜽1)− 𝑓(x1 ∧ x2, 𝜽1) (2.42)

However x1 and x2 are optimal for their respective parameter values, that is

𝑓(x2, 𝜽2) ≥ 𝑓(x1 ∨ x2, 𝜽2) =⇒ 𝑓(x1 ∨ x2, 𝜽2)− 𝑓(x2, 𝜽2) ≤ 0

𝑓(x1, 𝜽1) ≥ 𝑓(x1 ∧ x2, 𝜽1) =⇒ 𝑓(x1, 𝜽1)− 𝑓(x1 ∧ x2, 𝜽1) ≥ 0

Substituting in (2.42), we conclude

0 ≥ 𝑓(x1 ∨ x2, 𝜽2)− 𝑓(x2, 𝜽2) > 𝑓(x1, 𝜽1)− 𝑓(x1 ∧ x2, 𝜽1) ≥ 0

This contradiction implies that our assumption that x2 ∕≿𝑋 x1 is false. x2 ≿𝑋 x1 asrequired. 𝜑 is always increasing.

2.64 The budget correspondence is descending in p and therefore ascending in −p.Consequently, the indirect utility function

𝑣(p,𝑚) = supx∈𝑋(p,𝑚)

𝑢(x)

is increasing in −p, that is decreasing in p.

2.65 ⇐= Let 𝜽2 ≿ 𝜽1 and 𝐺2 ≿𝑆 𝐺1. Select x1 ∈ 𝜑(𝜽1, 𝐺1) and x2 ∈ 𝜑(𝜽2, 𝐺2).Since 𝐺2 ≿𝑆 𝐺1, x1 ∧ x2 ∈ 𝐺1. Since x1 is optimal (x1 ∈ 𝜑(𝜽1, 𝐺1)), 𝑓(x1, 𝜽1) ≥𝑓(x1 ∧x2, 𝜽1). Quasisupermodularity implies 𝑓(x1 ∨x2, 𝜽1) ≥ 𝑓(x2, 𝜽1). By the singlecrossing condition 𝑓(x1 ∨ x2, 𝜽2) ≥ 𝑓(x2, 𝜽2). Therefore x1 ∨ x2 ∈ 𝜑(𝜽2, 𝐺2).

Similarly, since 𝐺2 ≿𝑆 𝐺1, x1 ∨ x2 ∈ 𝐺(𝜽2). But x2 is optimal, which implies that𝑓(x2, 𝜽2) ≥ 𝑓(x1 ∨ x2, 𝜽2) or 𝑓(x1 ∨ x2, 𝜽2) ≤ 𝑓(x2, 𝜽2). The single crossing conditionimplies that a similar inequality holds at 𝜽1, that is 𝑓(x1 ∨ x2, 𝜽1) ≤ 𝑓(x2, 𝜽1). Quasi-supermodularity implies that 𝑓(x1, 𝜽1) ≤ 𝑓(x1∧x2, 𝜽1). Therefore x1∧x2 ∈ 𝜑(𝜽1, 𝐺1).Since x1 ∨ x2 ∈ 𝜑(𝜽2, 𝐺2) and x1 ∧ x2 ∈ 𝜑(𝜽1, 𝐺1), 𝜑 is increasing in (𝜽, 𝐺).

=⇒ To show that 𝑓 is quasisupermodular, suppose that 𝜽 is fixed. Choose anyx1,x2 ∈ 𝑋 . Let 𝐺1 = {x1,x1 ∧x2} and 𝐺2 = {x2,x1 ∨x2}. Then 𝐺2 ≿𝑆 𝐺1. Assumethat 𝑓(x1, 𝜽) ≥ 𝑓(x1∧x2, 𝜽). Then x1 ∈ 𝜑(𝜽, 𝐺1) which implies that x1∨x2 ∈ 𝜑(𝜽, 𝐺2).(If x2 ∈ 𝜑(𝜽, 𝐺2), then also x1 ∨ x2 ∈ 𝜑(𝜽, 𝐺2) since 𝜑 is increasing in (𝜽, 𝐺)). Butthis implies that 𝑓(x1 ∨ x2, 𝜽) ≥ 𝑓(x2, 𝜽). 𝑓 is quasisupermodular in 𝑋 .

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To show that 𝑓 satisfies the single crossing condition, choose any x2 ≿ x1 and let𝐺 = {x1,x2}. Assume that 𝑓(x2, 𝜽1) ≥ 𝑓(x1, 𝜽1). Then x2 ∈ 𝜑(𝜽1, 𝐺) which impliesthat x2 ∈ 𝜑(𝜽2, 𝐺) for any 𝜽2 ≿ 𝜽1. (If x1 ∈ 𝜑(𝜽2, 𝐺), then also x1∨x2 = x2 ∈ 𝜑(𝜽2, 𝐺)since 𝜑 is increasing in (𝜽, 𝐺).) But this implies that 𝑓(x2, 𝜽2) ≥ 𝑓(x1, 𝜽2). 𝑓 satisfiesthe single crossing condition.

2.66 First, assume that 𝑓 is continuous. Let 𝑇 be an open subset in 𝑌 and 𝑆 = 𝑓−1(𝑇 ).If 𝑆 = ∅, it is open. Otherwise, choose 𝑥0 ∈ 𝑆 and let 𝑦0 = 𝑓(𝑥0) ∈ 𝑇 . Since 𝑇 isopen, there exists a neighborhood 𝑁(𝑦0) ⊆ 𝑇 . Since 𝑓 is continuous, there exists acorresponding neighborhood 𝑁(𝑥0) with 𝑓(𝑁(𝑥0)) ⊆ 𝑁(𝑓(𝑥0)). Since 𝑁(𝑓(𝑥0)) ⊆ 𝑇 ,𝑁(𝑥0) ⊆ 𝑆. This establishes that for every 𝑥0 ∈ 𝑆 there exist a neighborhood 𝑁(𝑥0)contained in 𝑆. That is, 𝑆 is open in 𝑋 .

Conversely, assume that the inverse image of every open set in 𝑌 is open in 𝑋 . Choosesome 𝑥0 ∈ 𝑋 and let 𝑦0 = 𝑓(𝑥0). Let 𝑇 ⊂ 𝑌 be a neighborhood of 𝑦0. 𝑇 contains anopen ball 𝐵𝑟(𝑦0) about 𝑦0. By hypothesis, the inverse image 𝑆 = 𝑓−1(𝐵𝑟(𝑦0)) is openin 𝑋 . Therefore, there exists a neighborhood 𝑁(𝑥0) ⊆ 𝑓−1(𝐵𝑟(𝑦0)). Since 𝐵𝑟(𝑦0) ⊆ 𝑇 ,𝑓(𝑁(𝑥0)) ⊆ 𝑇 . Since the choice of 𝑥0 was arbitrary, we conclude that 𝑓 is continuous.

2.67 Assume 𝑓 is continuous. Let 𝑇 be a closed set in 𝑌 and let 𝑆 = 𝑓−1(𝑇 ). Then,𝑇 𝑐 is open. By the previous exercise, 𝑓−1(𝑇 𝑐) = 𝑆𝑐 is open and therefore 𝑆 is closed.Conversely, for every open set 𝑇 ⊆ 𝑌 , 𝑇 𝑐 is closed. By hypothesis, 𝑆𝑐 = 𝑓−1(𝑇 𝑐) isclosed and therefore 𝑆 = 𝑓−1(𝑇 ) is open. 𝑓 is continuous by the previous exercise.

2.68 Assume 𝑓 is continuous. Let 𝑥𝑛 be a sequence converging to 𝑥 Let 𝑇 be a neigh-borhood of 𝑓(𝑥). Since 𝑓 is continuous, there exists a neighborhood 𝑆 ∋ 𝑥 such that𝑓(𝑆) ⊆ 𝑇 . Since 𝑥𝑛 converges to 𝑥, there exists some 𝑁 such that 𝑥𝑛 ∈ 𝑆 for all𝑛 ≥ 𝑁 . Consequently 𝑓(𝑥𝑛) ∈ 𝑇 for every 𝑛 ≥ 𝑁 . This establishes that 𝑓(𝑥𝑛)→ 𝑓(𝑥).

Conversely, assume that for every sequence 𝑥𝑛 → 𝑥, 𝑓(𝑥𝑛)→ 𝑓(𝑥). We show that if 𝑓were not continuous, it would be possible to construct a sequence which violates thishypothesis. Suppose then that 𝑓 is not continuous. Then there exists a neighborhood𝑇 of 𝑓(𝑥) such that for every neighborhood 𝑆 of 𝑥, there is 𝑥′ ∈ 𝑆 with 𝑓(𝑥′) /∈ 𝑇 . Inparticular, consider the sequence of open balls 𝐵1/𝑛(𝑥). For every 𝑛, choose a point𝑥𝑛 ∈ 𝐵1/𝑛(𝑥) with 𝑓(𝑥𝑛) /∈ 𝑇 . Then 𝑥𝑛 → 𝑥 but 𝑓(𝑥𝑛) does not converge to 𝑓(𝑥).This contradicts the assumption. We conclude that 𝑓 must be continuous.

2.69 Since 𝑓 is one-to-one and onto, it has an inverse 𝑔 = 𝑓−1 which maps 𝑌 onto𝑋 . Let 𝑆 be an open set in 𝑋 . Since 𝑓 is open, 𝑇 = 𝑔−1(𝑆) = 𝑓(𝑆) is open in 𝑌 .Therefore 𝑔 = 𝑓−1 is continuous.

2.70 Assume 𝑓 is continuous. Let (𝑥𝑛, 𝑦𝑛) be a sequence of points in graph(𝑓) con-verging to (𝑥, 𝑦). Then 𝑦𝑛 = 𝑓(𝑥𝑛) and 𝑥𝑛 → 𝑥. Since 𝑓 is continuous, 𝑦 = 𝑓(𝑥) =lim𝑛→∞ 𝑓(𝑥𝑛) = lim𝑛→∞ 𝑦𝑛. Therefore (𝑥, 𝑦) ∈ graph(𝑓) which is therefore closed.

2.71 By the previous exercise, 𝑓 continuous implies graph(𝑓) closed. Conversely, sup-pose graph(𝑓) is closed and let 𝑥𝑛 be a sequence converging to 𝑥. Then (𝑥𝑛, 𝑓(𝑥𝑛)))is a sequence in graph(𝑓). Since 𝑌 is compact, 𝑓(𝑥𝑛) contains a subsequence whichconverges 𝑦. Since graph(𝑓) is closed, (𝑥, 𝑦) ∈ graph(𝑓) and therefore 𝑦 = 𝑓(𝑥) and𝑓(𝑥𝑛)→ 𝑓(𝑥).

2.72 Let 𝑇 be an open set in 𝑍. Since 𝑓 and 𝑔 are continuous, 𝑔−1(𝑇 ) is open in 𝑌and 𝑓−1(𝑔−1(𝑇 )) is open in 𝑋 . But 𝑓−1(𝑔−1(𝑇 )) = (𝑓 ∘ 𝑔)−1(𝑇 ). Therefore 𝑓 ∘ 𝑔 iscontinuous.

2.73 Exercises 1.201 and 2.68.

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2.74 Let 𝑢 be defined as in Exercise 2.38. Let (x𝑛) be a sequence converging to x. Let𝑧𝑛 = 𝑢(x𝑛) and 𝑧 = 𝑢(x). We need to show that 𝑧𝑛 → 𝑧.(𝑧𝑛) has a convergent subsequence. Let 𝑧 = max𝑖 𝑥𝑖 and 𝑧 = min𝑖 𝑥𝑖. Then 𝑧 ∈

[𝑧, 𝑧]. Fix some 𝜖 > 0. Since x𝑛 → x, there exists some 𝑁 such that ∥x𝑛 − x∥∞ <𝜖 for every 𝑛 ≥ 𝑁 . Consequently, for all 𝑛 ≥ 𝑁 , the terms of the sequence (𝑧𝑛)lie in the compact set [𝑧 − 𝜖, 𝑧 + 𝜖]. Hence, (𝑧𝑛) has a convergent subsequence(𝑧𝑚).

Every convergent subsequence (𝑧𝑚) converges to 𝑧. Suppose not. That is, sup-pose there exists a convergent subsequence which converges to 𝑧′. Without lossof generality, assume 𝑧′ > 𝑧. Let 𝑧 = 1

2 (𝑧 + 𝑧′) and let z = 𝑧1, z′ = 𝑧′1, z = 𝑧1be the corresponding commodity bundles (see Exercise 2.38). Since 𝑧𝑚 → 𝑧′ > 𝑧,there exists some 𝑀 such that 𝑧𝑚 > 𝑧 for every 𝑚 ≥𝑀 . This implies that

x𝑚 ∼ z𝑚 ≻ z for every 𝑚 ≥𝑀

by monotonicity. Now x𝑚 → x and continuity of preferences implies that x ≿ z.However x ∼ z which implies that z ≿ z which contradicts monotonicity, sincez > z. Consequently, every convergent subsequence (𝑧𝑚) converges to 𝑧.

2.75 Assume 𝑋 is compact. Let 𝑦𝑛 be a sequence in 𝑓(𝑋). There exists a sequence𝑥𝑛 in 𝑋 with 𝑦𝑛 = 𝑓(𝑥𝑛). Since 𝑋 is compact, it contains a convergent subsequence𝑥𝑚 → 𝑥. If 𝑓 is continuous, the subsequence 𝑦𝑚 = 𝑓(𝑥𝑚) converges in 𝑓(𝑋) (Exercise2.68). Therefore 𝑓(𝑋) is compact.

Assume 𝑋 is connected but 𝑓(𝑋) is not. This means there exists open subsets 𝐺 and𝐻 in 𝑌 such that 𝑓(𝑋) ⊂ 𝐺∪

𝐻 and (𝐺 ∩ 𝑓(𝑋))∩

(𝐻 ∩ 𝑓(𝑋)) = ∅. This implies that𝑋 = 𝑓−1(𝐺)∪ 𝑓−1(𝐻) is a disconnection of 𝑋 , which contradicts the connectedness of𝑋 .

2.76 Let 𝑆 be any open set in 𝑋 . Its complement 𝑆𝑐 is closed and therefore compact.Consequently, 𝑓(𝑆𝑐) is compact (Exercise 2.3) and hence closed. Since 𝑓 is one-to-oneand onto, 𝑓(𝑆) is the complement of 𝑓(𝑆𝑐), and thus open in 𝑌 . Therefore, 𝑓 is anopen mapping. By Exercise 2.69, 𝑓−1 is continuous and 𝑓 is a homeomorphism.

2.77 Assume 𝑓 continuous. The sets {𝑓(𝑥) ≥ 𝑎} and {𝑓(𝑥) ≤ 𝑎} are closed subsets ofthe ℜ and hence ≿(𝑎) = 𝑓−1{𝑓(𝑥) ≥ 𝑎} and ≾(𝑎) = 𝑓−1{𝑓(𝑥) ≤ 𝑎} are closed subsetsof 𝑋 (Exercise 2.67).

Conversely, assume that all upper ≿(𝑎) and lower ≾(𝑎) contour sets are closed. Thisimplies that the sets ≻(𝑎) and ≺(𝑎) are open.

Let 𝐴 be an open set in ℜ. Then for every 𝑎 ∈ 𝐴, there exists an open ball 𝐵𝑟𝑎(𝑎) ⊆ 𝐴

𝐴 =∪𝑎∈𝐴𝐵𝑟𝑎(𝑎)

For every 𝑎 ∈ 𝐴, 𝐵𝑟𝑎(𝑎) = (𝑎− 𝑟𝑎, 𝑎+ 𝑟𝑎) and

𝑓−1 (𝐵𝑟𝑎(𝑎)) = ≻(𝑎− 𝑟𝑎) ∩≺(𝑎+ 𝑟𝑎)

which is open. Consequently

𝑓−1(𝐴) =∪𝑎∈𝐴𝑓−1 (𝐵𝑟𝑎(𝑎)) =

∪𝑎∈𝐴

(≻(𝑎− 𝑟𝑎) ∩ ≺(𝑎+ 𝑟𝑎))

is open. 𝑓 is continuous by Exercise 2.66.

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2.78 Choose any 𝑥0 ∈ 𝑋 and 𝜖 > 0. Since 𝑓 is continuous, there exists 𝛿1 such that

𝜌(𝑥, 𝑥0) < 𝛿1 =⇒ ∣𝑓(𝑥)− 𝑓(𝑥0)∣ < 𝜖/2Similarly, there exists 𝛿2 such that

𝜌(𝑥, 𝑥0) < 𝛿2 =⇒ ∣𝑔(𝑥)− 𝑔(𝑥0)∣ < 𝜖/2Let 𝛿 = min{𝛿1, 𝛿2}. Then, provided 𝜌(𝑥, 𝑥0) < 𝛿

∣(𝑓 + 𝑔)(𝑥)− (𝑓 + 𝑔)(𝑥0)∣ = ∣𝑓(𝑥) + 𝑔(𝑥)− 𝑓(𝑥0)− 𝑔(𝑥0)∣≤ ∣𝑓(𝑥)− 𝑓(𝑥0)∣+ ∣𝑔(𝑥)− 𝑔(𝑥0)∣< 𝜖

This establishes 𝑓 + 𝑔 is continuous at 𝑥0. Since 𝑥0 was arbitrary, 𝑓 + 𝑔 is continuousfor every 𝑥0 ∈ 𝑋 . The continuity of 𝛼𝑓 is shown similarly.

2.79 Choose any 𝑥0 ∈ 𝑋 . Given 0 < 𝜂 ≤ 1, there exists 𝛿 > 0 such that

∣𝑓(𝑥)− 𝑓(𝑥0)∣ < 𝜂 and ∣𝑔(𝑥)− 𝑔(𝑥0)∣ < 𝜂whenever 𝜌(𝑥, 𝑥0) < 𝛿. Consequently, while 𝜌(𝑥, 𝑥0) < 𝛿

∣𝑓(𝑥)∣ ≤ ∣𝑓(𝑥)− 𝑓(𝑥0)∣+ ∣𝑓(𝑥0)∣< 𝜂 + ∣𝑓(𝑥0)∣≤ 1 + ∣𝑓(𝑥0)∣

and

∣(𝑓𝑔)(𝑥)− (𝑓𝑔)(𝑥0)∣ = ∣𝑓(𝑥)𝑔(𝑥) − 𝑓(𝑥0)𝑔(𝑥0)∣= ∣𝑓(𝑥)(𝑔(𝑥) − 𝑔(𝑥0)) + 𝑔(𝑥0)(𝑓(𝑥) − 𝑓(𝑥0))∣≤ ∣𝑓(𝑥)∣ ∣𝑔(𝑥)− 𝑔(𝑥0)∣+ ∣𝑔(𝑥0)∣ ∣𝑓(𝑥)− 𝑓(𝑥0)∣< 𝜂(1 + ∣𝑓(𝑥0)∣+ ∣𝑔(𝑥0)∣)

Given 𝜖 > 0, let 𝜂 = min{1, 𝜖/(1 + ∣𝑓(𝑥0)∣+ ∣𝑔(𝑥0)∣)}. Then, we have shown that thereexists 𝛿 > 0 such that

𝜌(𝑥, 𝑥0) < 𝛿 =⇒ ∣(𝑓𝑔)(𝑥) − (𝑓𝑔)(𝑥0)∣ < 𝜖Therefore, 𝑓𝑔 is continuous at 𝑥0.

2.80 Apply Exercises 2.78 and 2.72.

2.81 For any 𝑎 ∈ ℜ, the upper and lower contour sets of 𝑓 ∨ 𝑔, namely

{ 𝑥 : max{𝑓(𝑥), 𝑔(𝑥)} ≥ 𝑎} = {𝑥 : 𝑓(𝑥) ≥ 𝑎 } ∪ { 𝑥 : 𝑔(𝑥) ≥ 𝑎 }

{ 𝑥 : max{𝑓(𝑥), 𝑔(𝑥)} ≤ 𝑎} = {𝑥 : 𝑓(𝑥) ≤ 𝑎 } ∩ { 𝑥 : 𝑔(𝑥) ≤ 𝑎 }are closed. Therefore 𝑓 ∨ 𝑔 is continuous (Exercise 2.77). Similarly for 𝑓 ∧ 𝑔.2.82 The set 𝑇 = 𝑓(𝑋) is compact (Proposition 2.3). We want to show that 𝑇 hasboth largest and smallest elements. Assume otherwise, that is assume that 𝑇 hasno largest element. Then, the set of intervals {(−∞, 𝑡) : 𝑡 ∈ 𝑇 } forms an opencovering of 𝑇 . Since 𝑇 is compact, there exists a finite subcollection of intervals{(−∞, 𝑡1), (−∞, 𝑡2), . . . , (−∞, 𝑡𝑛)} which covers 𝑇 . Let 𝑡∗ be the largest of these 𝑡𝑖.Then 𝑡∗ does not belong to any of the intervals {(−∞, 𝑡1), (−∞, 𝑡2), . . . , (−∞, 𝑡𝑛)},contrary to the fact that they cover 𝑇 . This contradiction shows that, contrary to ourassumption, there must exist a largest element 𝑡∗ ∈ 𝑇 , that is 𝑡∗ ≥ 𝑡 for all 𝑡 ∈ 𝑇 .Let 𝑥∗ ∈ 𝑓−1(𝑡∗). Then 𝑡∗ = 𝑓(𝑥∗) ≥ 𝑓(𝑥) for all 𝑥 ∈ 𝑋 . The existence of a smallestelement is proved analogously.

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2.83 By Proposition 2.3, 𝑓(𝑋) is connected and hence an interval (Exercise 1.95).

2.84 The range 𝑓(𝑋) is a compact subset of ℜ (Proposition 2.3). Therefore 𝑓 is bounded(Proposition 1.1).

2.85 Let 𝐶(𝑋) denote the set of all continuous (not necessarily bounded) functionalson 𝑋 . Then

𝐶(𝑋) = 𝐵(𝑋) ∩ 𝐶(𝑋)

𝐵(𝑋), 𝐶(𝑋) are a linear subspaces of the set of all functionals 𝐹 (𝑋) (Exercises 2.11,2.78 respectively). Therefore 𝐶(𝑋) = 𝐵(𝑋) ∩ 𝐶(𝑋) is a subspace of 𝐹 (𝑋) (Exercise1.130). Clearly 𝐶(𝑋) ⊆ 𝐵(𝑋). Therefore 𝐶(𝑋) is a linear subspace of 𝐵(𝑋).

Let 𝑓 be a bounded function in the closure of 𝐶(𝑋), that is 𝑓 ∈ 𝐶(𝑋). We show that𝑓 is continuous. For any 𝜖 > 0, there exists 𝑓0 ∈ 𝐶(𝑋) such that ∥𝑓 − 𝑓0∥ < 𝜖/3.Therefore ∣𝑓(𝑥)− 𝑓0(𝑥)∣ < 𝜖/3 for every 𝑥 ∈ 𝑋 . Choose some 𝑥0 ∈ 𝑋 . Since 𝑓0 iscontinuous, there exists 𝛿 > 0 such that

𝜌(𝑥, 𝑥0) < 𝛿 =⇒ ∣𝑓0(𝑥)− 𝑓0(𝑥0)∣ < 𝜖/3Therefore, for every 𝑥 ∈ 𝑋 such that 𝜌(𝑥, 𝑥0) < 𝛿

∣𝑓(𝑥)− 𝑓(𝑥0)∣ = ∣𝑓(𝑥)− 𝑓0(𝑥) + 𝑓0(𝑥)− 𝑓0(𝑥0) + 𝑓0(𝑥0)− 𝑓(𝑥0)∣≤ ∣𝑓(𝑥)− 𝑓0(𝑥)∣ + ∣𝑓0(𝑥)− 𝑓0(𝑥0)∣+ ∣𝑓0(𝑥0)− 𝑓(𝑥0)∣< 𝜖/3 + 𝜖/3 + 𝜖/3 = 𝜖

Therefore 𝑓 is continuous at 𝑥0. Since 𝑥0 was arbitrary, we conclude that is continuouseverywhere, that is 𝑓 ∈ 𝐶(𝑋). Therefore 𝐶(𝑋) = 𝐶(𝑋) and 𝐶(𝑋) is closed in 𝐵(𝑋).

Since 𝐵(𝑋) is complete (Exercise 2.11), we conclude that 𝐶(𝑋) is complete (Exercise1.107). Therefore 𝐶(𝑋) is a Banach space.

2.86 For every 𝛼 ∈ ℜ,

{ 𝑥 : 𝑓(𝑥) ≥ 𝛼 } = {𝑥 : −𝑓(𝑥) ≤ −𝛼 }and therefore

{ 𝑥 : 𝑓(𝑥) ≥ 𝛼 } is closed ⇐⇒ {𝑥 : −𝑓(𝑥) ≤ −𝛼 } is closed

2.87 Exercise 2.77.

2.88 1 implies 2 Suppose 𝑓 is upper semi-continuous. Let 𝑥𝑛 be a sequence converg-ing to 𝑥0. Assume 𝑓(𝑥𝑛) → 𝜇. For every 𝛼 < 𝜇, there exists some 𝑁 such that𝑓(𝑥𝑛) > 𝛼 for every 𝑛 ≥ 𝑁 . Hence

𝑥0 ∈ { 𝑥 : 𝑓(𝑥) ≥ 𝛼 } = { 𝑥 : 𝑓(𝑥) ≥ 𝛼 }since 𝑓 is upper semi-continuous. That is, 𝑓(𝑥0) ≥ 𝛼 for every 𝛼 < 𝜇. Hence𝑓(𝑥0) ≥ 𝜇 = lim𝑛→∞ 𝑓(𝑥𝑛).

2 implies 3 Let (𝑥𝑛, 𝑦𝑛) be a sequence in hypo 𝑓 which converges to (𝑥, 𝑦). That is,𝑥𝑛 → 𝑥, 𝑦𝑛 → 𝑦 and 𝑦𝑛 ≤ 𝑓(𝑥𝑛). Condition 2 implies that 𝑓(𝑥) ≥ 𝑦. Hence,(𝑥, 𝑦) ∈ hypo 𝑓 . Therefore hypo 𝑓 is closed.

3 implies 1 For fixed 𝛼 ∈ ℜ, let 𝑥𝑛 be a sequence in { 𝑥 : 𝑓(𝑥) ≥ 𝛼 }. Suppose𝑥𝑛 → 𝑥0. Then, the sequence (𝑥𝑛, 𝛼) converges to (𝑥0, 𝛼) ∈ hypo 𝑓 . Hence𝑓(𝑥0) ≥ 𝛼 and 𝑥0 ∈ { 𝑥 : 𝑓(𝑥) ≥ 𝛼 }, which is therefore closed (Exercise 1.106).

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2.89 Let 𝑀 = sup𝑥∈𝑋 𝑓(𝑥), so that

𝑓(𝑥) ≤𝑀 for every 𝑥 ∈ 𝑋 (2.43)

There exists a sequence 𝑥𝑛 in 𝑋 with 𝑓(𝑥𝑛) → 𝑀 . Since 𝑋 is compact, there existsa convergent subsequence 𝑥𝑚 → 𝑥∗ and 𝑓(𝑥𝑚) → 𝑀 . However, since 𝑓 is uppersemi-continuous, 𝑓(𝑥∗) ≥ lim 𝑓(𝑥𝑚) = 𝑀 . Combined with (2.43), we conclude that𝑓(𝑥∗) =𝑀 .

2.90 Choose some 𝜖 > 0. Since 𝑓 is uniformly continuous, there exists some 𝛿 > 0 suchthat 𝜌(𝑓(𝑥𝑚), 𝑓(𝑥𝑛)) < 𝜖 for every 𝑥𝑚, 𝑥𝑛 ∈ 𝑋 such that 𝜌(𝑥𝑚, 𝑥𝑛) < 𝛿. Let (𝑥𝑛)be a Cauchy sequence in 𝑋 . There exists some 𝑁 such that 𝜌(𝑥𝑚, 𝑥𝑛) < 𝛿 for every𝑚,𝑛 ≥ 𝑁 . Uniform continuity implies that 𝜌(𝑓(𝑥𝑚), 𝑓(𝑥𝑛)) < 𝜖 for every 𝑚,𝑛 ≥ 𝑁 .(𝑓(𝑥𝑛)) is a Cauchy sequence.

2.91 Suppose not. That is, suppose 𝑓 is continuous but not uniformly continuous. Thenthere exists some 𝜖 > 0 such that for 𝑛 = 1, 2, . . . , there exist points 𝑥1𝑛, 𝑥

2𝑛 such that

𝜌(𝑥1𝑛, 𝑥2𝑛) < 1/𝑛 but 𝜌(𝑓(𝑥1𝑛), 𝑓(𝑥2𝑛)) ≥ 𝜖 (2.44)

Since 𝑋 is compact, (𝑥1𝑛) has a subsequence (𝑥1𝑚) converging to some 𝑥 ∈ 𝑋 . Byconstruction (𝜌(𝑥1𝑛, 𝑥

2𝑛) < 1/𝑛), the sequence (𝑥2𝑚) also converges to 𝑥 and by continuity

lim𝑚→∞ 𝑓(𝑥1𝑚) = lim

𝑚→∞ 𝑓(𝑥2𝑚)

which contradicts (2.44).

2.92 Assume 𝑓 is Lipschitz with constant 𝛽. For any 𝜖 > 0, let 𝛿 = 𝜖/2𝛽. Then,provided 𝜌(𝑥, 𝑥0) ≤ 𝛿

𝜌(𝑓(𝑥), 𝑓(𝑥0)) ≤ 𝛽𝜌(𝑥, 𝑥0) = 𝛽𝛿 = 𝛽𝜖

2𝛽=𝜖

2< 𝜖

𝑓 is uniformly continuous.

2.93 Let 𝑓, 𝑔 ∈ 𝐵(𝑋).F Since 𝐵(𝑋) is a normed linear space, for every 𝑥 ∈ 𝑋𝑓(𝑥)− 𝑔(𝑥) = (𝑓 − 𝑔)(𝑥) ≤ ∥𝑓 − 𝑔∥

which implies that

𝑓(𝑥) ≤ 𝑔(𝑥) + ∥𝑓 − 𝑔∥Since 𝑇 is increasing and satisfies (2.21)

𝑇 (𝑓) ≤ 𝑇 (𝑔 + ∥𝑓 − 𝑔∥) = 𝑇 (𝑔) + 𝛽 ∥𝑓 − 𝑔∥or

𝑇 (𝑓)− 𝑇 (𝑔) ≤ 𝛽 ∥𝑓 − 𝑔∥That is, for every 𝑥 ∈ 𝑋

(𝑇𝑓 − 𝑇𝑔)(𝑥) ≤ 𝛽 ∥𝑓 − 𝑔∥and consequently

∥𝑇𝑓 − 𝑇𝑔∥ ≤ 𝛽 ∥𝑓 − 𝑔∥𝑇 is a contraction with modulus 𝛽.

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2.94 We have previously shown that 𝑇 is increasing (Exercise 2.42). By direct calcula-tion, for any constant 𝑐 ∈ ℜ,

𝑇 (𝑣 + 𝑐)(𝑥) = sup𝑦∈𝐺(𝑥)

{𝑓(𝑥, 𝑦) + 𝛽

(𝑣(𝑦) + 𝑐

)}

= sup𝑦∈𝐺(𝑥)

{𝑓(𝑥, 𝑦) + 𝛽𝑣(𝑦)

}+ 𝛽𝑐

= 𝑇 (𝑣)(𝑥) + 𝛽𝑐

2.95 Assume that 𝐹 is a compact subset of 𝐶(𝑋). Then 𝐹 is bounded (Proposition1.1). To show that 𝐹 is equicontinuous, choose 𝜖 > 0. 𝐹 is totally bounded (Exercise1.113), so that there exist finite set of functions {𝑓1, 𝑓2, . . . , 𝑓𝑛} in F such that

𝑛min𝑘=1∥𝑓 − 𝑓𝑘∥ ≤ 𝜖/3

Each 𝑓𝑘 is uniformly continuous (Exercise 2.91), so that there exists 𝛿𝑘 > 0 such that

𝜌(𝑥, 𝑥0) ≤ 𝛿 =⇒ 𝜌(𝑓𝑘(𝑥), 𝑓𝑘(𝑥0) < 𝜖/3

Let 𝛿 = min{𝛿1, 𝛿2, . . . , 𝛿𝑘}. Given any 𝑓 ∈ 𝐹 , let 𝑘 be such that ∥𝑓 − 𝑓𝑘∥ < 𝜖/3. Thenfor any 𝑥, 𝑥0 ∈ 𝑋 , 𝜌(𝑥, 𝑥0) ≤ 𝛿 implies

𝜌(𝑓(𝑥), 𝑓(𝑥0) ≤ 𝜌(𝑓(𝑥), 𝑓𝑘(𝑥)) + 𝜌(𝑓𝑘(𝑥), 𝑓𝑘(𝑥0)) + 𝜌(𝑓𝑘(𝑥0), 𝑓(𝑥0)) <𝜖

3+𝜖

3+𝜖

3= 𝜖

for every 𝑓 ∈ 𝐹 . Therefore, 𝐹 is equicontinuous.

Conversely, assume that 𝐹 ⊆ 𝐶(𝑋) is closed, bounded and equicontinuous. Let (𝑓𝑛)be a bounded equicontinuous sequence of functions in 𝐹 . We show that (𝑓𝑛) has aconvergent subsequence.

1. First, we show that for any 𝜖 > 0, there is exists a subsequence (𝑓𝑚) such that∥𝑓𝑚 − 𝑓𝑚′∥ < 𝜖 for every 𝑓𝑚, 𝑓𝑚′ in the subsequence. Since the functions areequicontinuous, there exists 𝛿 > 0 such that

𝜌(𝑓𝑛(𝑥)− 𝑓𝑛(𝑥0) <𝜖

3

for every 𝑥, 𝑥0 in 𝑋 with 𝜌(𝑥, 𝑥0) ≤ 𝛿. Since 𝑋 is compact, it is totally bounded(Exercise 1.113). That is, there exist a finite number of open balls 𝐵𝛿(𝑥𝑖),𝑖 = 1, 2 . . . , 𝑘 which cover 𝑋 . The sequence (𝑓𝑛(𝑥1), 𝑓𝑛(𝑥2, . . . , 𝑓𝑛(𝑥𝑘)) is abounded sequence in ℜ𝑛. By the Bolzano-Weierstrass theorem (Exercise 1.119),this sequence has a convergent subsequence (𝑓𝑚(𝑥1), 𝑓𝑚(𝑥2), . . . , 𝑓𝑚(𝑥𝑘)) suchthat 𝑓𝑚(𝑥𝑖)− 𝑓𝑚′(𝑥𝑖) < 𝜖/3 for 𝑖 and every 𝑓𝑚, 𝑓𝑚′ in the subsequence. Conse-quently, for any 𝑥 ∈ 𝑋 , there exists 𝑖 such that

𝜌(𝑓𝑚(𝑥), 𝑓𝑚′(𝑥) ≤ 𝜌(𝑓𝑚(𝑥), 𝑓𝑚(𝑥𝑖)) + 𝜌(𝑓𝑚(𝑥𝑖), 𝑓𝑚′(𝑥𝑖)) + 𝜌(𝑓𝑚′(𝑥𝑖), 𝑓𝑚′(𝑥))

<𝜖

3+𝜖

3+𝜖

3= 𝜖

That is, ∥𝑓𝑚 − 𝑓𝑚′∥ < 𝜖 for every 𝑓𝑚, 𝑓𝑚′ in the subsequence.

2. Choose a ball 𝐵1 of radius 1 in 𝐶(𝑋) which contains infinitely many elements of(𝑓𝑛). Applying step 1, there exists a ball 𝐵2 of radius 1/2 containing infinitelymany elements of (𝑓𝑛). Proceeding in this fashion, we obtain a nested sequence𝐵1 ⊇ 𝐵2 ⊇ . . . of balls in 𝐶(𝑋) such that (a) 𝑑(𝐵𝑖)→ 0 and (b) each 𝐵𝑖 containsinfinitely many terms of (𝑓𝑛). Choosing 𝑓𝑛𝑖 ∈ 𝐵𝑖 gives a convergent subsequence.

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2.96 Let 𝑔 ∈ 𝐹 . Then for every 𝜖 > 0 there exists 𝛿 > 0 and 𝑓 ∈ 𝐹 such that∥𝑓 − 𝑔∥ < 𝜖/3 and

𝜌(𝑥, 𝑥0) ≤ 𝛿 =⇒ 𝜌(𝑓(𝑥), 𝑓(𝑥0) < 𝜖/3

so that if 𝜌(𝑥, 𝑥0) ≤ 𝛿∥𝑔(𝑥)− 𝑔(𝑥0)∥ ≤ ∥𝑓(𝑥)− 𝑔(𝑥)∥ + ∥𝑓(𝑥)− 𝑓(𝑥0)∥+ ∥𝑓(𝑥0)− 𝑔(𝑥0)∥ < 𝜖

3+𝜖

3+𝜖

3= 𝜖

2.97 For every 𝑇 ⊆ 𝑌𝜑−(𝑇 𝑐) = { 𝑥 ∈ 𝑋 : 𝜑(𝑥) ∩ 𝑇 𝑐 ∕= ∅ }𝜑+(𝑇 ) = { 𝑥 ∈ 𝑋 : 𝜑(𝑥) ⊆ 𝑇 }

For every x ∈ 𝑋 either 𝜑(𝑥) ⊆ 𝑇 or 𝜑(𝑥) ∩ 𝑇 𝑐 ∕= ∅ but not both. Therefore

𝜑+(𝑇 ) ∪ 𝜑−(𝑇 𝑐) = 𝑋

𝜑+(𝑇 ) ∩ 𝜑−(𝑇 𝑐) = ∅That is

𝜑+(𝑇 ) =(𝜑−(𝑇 𝑐)

)𝑐2.98 Assume 𝑥 ∈ 𝜑(𝑇 )−1. Then 𝜑(𝑥) = 𝑇 , 𝜑(𝑥) ⊆ 𝑇 and 𝑥 ∈ 𝜑+(𝑇 ). Now assume𝑥 ∈ 𝜑+(𝑇 ) so that 𝜑(𝑥) ⊆ 𝑇 . Consequently, 𝜑(𝑥) ∩ 𝑇 = 𝜑(𝑥) ∕= ∅ and 𝑥 ∈ 𝜑−(𝑇 ).

2.99 The respective inverses are:

𝜑−12 𝜑+2 𝜑−2{𝑡1} ∅ ∅ {𝑠1}{𝑡2} ∅ ∅ {𝑠1, 𝑠2}{𝑡1, 𝑡2} {𝑠1} {𝑠1} {𝑠1, 𝑠2}{𝑡2, 𝑡3} {𝑠2} {𝑠2} {𝑠1, 𝑠2}{𝑡1, 𝑡2, 𝑡3} ∅ {𝑠1, 𝑠2} {𝑠1, 𝑠2}

2.100 Let 𝑇 be an open interval meeting 𝜑(1), that is 𝜑(1) ∩ 𝑇 ∕= ∅. Since 𝜑(1) = {1},we must have 1 ∈ 𝑇 and therefore 𝜑(𝑥) ∩ 𝑇 ∕= ∅ for every 𝑥 ∈ 𝑋 . Therefore 𝜑 is lhc at𝑥 = 1. On the other hand, the open interval 𝑇 = (1/2, 3/2) contains 𝜑(1) but it doesnot contain 𝜑(𝑥) for any 𝑥 > 1. Therefore, 𝜑 is not uhc at 𝑥 = 1.

2.101 Choose any open set 𝑇 ⊆ 𝑌 and 𝑥 ∈ 𝑋 . Since 𝜑(𝑥) = 𝐾 = 𝜑(𝑥′) for every𝑥, 𝑥′ ∈ 𝑋∙ 𝜑(𝑥) ⊆ 𝑇 if and only if 𝜑(𝑥′) ⊆ 𝑇 for every 𝑥, 𝑥′ ∈ 𝑋∙ 𝜑(𝑥) ∩ 𝑇 ∕= ∅ if and only if 𝜑(𝑥′) ∩ 𝑇 ∕= ∅ for every 𝑥, 𝑥′ ∈ 𝑋 .

Consequently, 𝜑 is both uhc and lhc at all 𝑥 ∈ 𝑋 .

2.102 First assume that the 𝜑 is uhc. Let 𝑇 be any open subset in 𝑌 and 𝑆 = 𝜑+(𝑇 ).If 𝑆 = ∅, it is open. Otherwise, choose 𝑥0 ∈ 𝑆 so that 𝜑(𝑥0) ⊆ 𝑇 . Since 𝜑 is uhc,there exists a neighborhood 𝑆(𝑥0) such that 𝜑(𝑥) ⊆ 𝑇 for every 𝑥 ∈ 𝑆(𝑥0). That is,𝑆(𝑥0) ⊆ 𝜑+(𝑇 ) = 𝑆. This establishes that for every 𝑥0 ∈ 𝑆 there exist a neighborhood𝑆(𝑥0) contained in 𝑆. That is, 𝑆 is open in 𝑋 .

Conversely, assume that the upper inverse of every open set in 𝑌 is open in 𝑋 . Choosesome 𝑥0 ∈ 𝑋 and let 𝑇 be an open set containing 𝜑(𝑥0). Let 𝑆 = 𝜑+(𝑇 ). 𝑆 is an openset containing 𝑥0. That is, 𝑆 is a neighborhood of 𝑥0 with 𝜑(𝑥) ⊆ 𝑇 for every 𝑥 ∈ 𝑆.Since the choice of 𝑥0 was arbitrary, we conclude that 𝜑 is uhc.

The lhc case is analogous.

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2.103 Assume 𝜑 is uhc and 𝑇 be any closed set in 𝑌 . By Exercise 2.97

𝜑−(𝑇 ) =[𝜑+(𝑇 𝑐)

]𝑇 𝑐 is open. By the previous exercise, 𝜑+(𝑇 𝑐) is open which implies that 𝜑−(𝑇 ) isclosed.

Conversely, assume 𝜑−(𝑇 ) is closed for every closed set 𝑇 . Let 𝑇 be an open subset of𝑌 so that 𝑇 𝑐 is closed. Again by Exercise 2.97,

𝜑+(𝑇 ) =[𝜑−(𝑇 𝑐)

]By assumption 𝜑−(𝑇 𝑐) is closed and therefore 𝜑+(𝑇 ) is open. By the previous exercise,𝜑 is uhc.

The lhc case is analogous.

2.104 Assume that 𝜑 is uhc at 𝑥0. We first show that (𝑦𝑛) is bounded and hence hasa convergent subsequence. Since 𝜑(𝑥0) is compact, there exists a bounded open set 𝑇containing 𝜑(𝑥0). Since 𝜑 is uhc, there exists a neighborhood 𝑆 of 𝑥0 such that 𝜑(𝑥) ⊆𝑇 for 𝑥 ∈ 𝑆. Since 𝑥𝑛 → 𝑥0, there exists some 𝑁 such that 𝑥𝑛 ∈ 𝑆 for every 𝑛 ≥ 𝑁 .Consequently, 𝜑(𝑥𝑛) ⊆ 𝑇 for every 𝑛 ≥ 𝑁 and therefore 𝑦𝑛 ∈ 𝑇 for every 𝑛 ≥ 𝑁 .The sequence 𝑦𝑛 is bounded and hence has a convergent subsequence 𝑦𝑚 → 𝑦0.To complete the proof, we have to show that 𝑦0 ∈ 𝜑(𝑥0). Assume not, assume that𝑦0 /∈ 𝜑(𝑥0). Then, there exists an open set 𝑇 containing 𝜑(𝑥0) such that 𝑦0 /∈ 𝑇(Exercise 1.93). Since 𝜑 is uhc, there exists 𝑁 such that 𝜑(𝑥𝑛) ⊆ 𝑇 for every 𝑛 ≥ 𝑁 .This implies that 𝑦𝑚 ∈ 𝑇 for every 𝑚 ≥ 𝑁 . Since 𝑦𝑚 → 𝑦0, we conclude that 𝑦0 ∈ 𝑇 ,contradicting the specification of 𝑇 .

Conversely, suppose that for every sequence 𝑥𝑛 → 𝑥0, 𝑦𝑛 ∈ 𝜑(𝑥𝑛), there is a subse-quence of 𝑦𝑚 → 𝑦0 ∈ 𝜑(𝑥0). Suppose that 𝜑 is not uhc at 𝑥0. That is, there existsan open set 𝑇 ⊇ 𝜑(𝑥0) such that every neighborhood contains some 𝑥 with 𝜑(𝑥) ∕⊆ 𝑇 .From the sequence of neighborhoods 𝐵1/𝑛(𝑥0), we can construct a sequence 𝑥𝑛 → 𝑥and 𝑦𝑛 ∈ 𝜑(𝑥𝑛) but 𝑦𝑛 /∈ 𝑇 . Such a sequence cannot have a subsequence which con-verges to 𝑦0 ∈ 𝜑(𝑥), contradicting the hypothesis. We conclude that 𝜑 must be uhc at𝑥0.

2.105 Assume that 𝜑 is lhc. Let 𝑥𝑛 be a sequence converging to 𝑥0 and 𝑦0 ∈ 𝜑(𝑥0).Consider the sequence of open balls 𝐵1/𝑚(𝑦0),𝑚 = 1, 2, . . . . Note that every 𝐵1/𝑚(𝑦0)meets 𝜑(𝑥0). Since 𝜑 is lhc, there exists a sequence (𝑆𝑚) of neighborhoods of 𝑥0 suchthat 𝜑(𝑥) ∩𝐵1/𝑚 ∕= ∅ for every 𝑥 ∈ 𝑆𝑚. Since 𝑥𝑛 → 𝑥, for every 𝑚, there exists some𝑁𝑚 such that 𝑥𝑛 ∈ 𝑆𝑚 for every 𝑛 ≥ 𝑁𝑚. Without loss of generality, we can assumethat 𝑁1 < 𝑁2 < 𝑁3 . . . . We can now construct the desired sequence 𝑦𝑛. For each𝑛 = 1, 2, . . . , choose 𝑦𝑛 in the set 𝜑(𝑥𝑛) ∩𝐵1/m where 𝑁𝑚 ≤ 𝑛 ≤ 𝑁𝑚+1 since

𝑛 ≥ 𝑁𝑚 =⇒ 𝑥𝑛 ∈ 𝑆𝑚 =⇒ 𝜑(𝑥𝑛) ∩𝐵1/𝑚 ∕= ∅

Since 𝑦𝑛 ∈ 𝐵1/m(𝑦0), the sequence (𝑦𝑛) converges to 𝑦0 and 𝑛→∞.

Conversely, assume that 𝜑 is not lhc at 𝑥0, that is there exists an open set 𝑇 with𝑇 ∩𝜑(𝑥0) ∕= ∅ such that every neighborhood 𝑆 ∋ 𝑥0 contains some 𝑥 with 𝜑(𝑥)∩𝑇 = ∅.Therefore, there exists a sequence 𝑥𝑛 → 𝑥 with 𝜑(𝑥)∩𝑇 = ∅. Choose any 𝑦0 ∈ 𝜑(𝑥0)∩𝑇 .By assumption, there exists a sequence 𝑦𝑛 → 𝑦 with 𝑦𝑛 ∈ 𝜑(𝑥𝑛). Since 𝑇 is open and𝑦0 ∈ 𝑇 , there exists some 𝑁 such that 𝑦𝑛 ∈ 𝑇 for all 𝑛 ≥ 𝑁 , for which 𝜑(𝑦𝑛) ∩ 𝑇 ∕= ∅.This contradiction establishes that 𝜑 is lhc at 𝑥0.

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2.106 1. Assume 𝜑 is closed. For any 𝑥 ∈ 𝑋 , let (𝑦𝑛) be a sequence in 𝜑(𝑥). Since𝜑 is closed, 𝑦𝑛 → 𝑦 ∈ 𝜑(𝑥). Therefore 𝜑(𝑥) is closed.

2. Assume 𝜑 is closed-valued and uhc. Choose any (𝑥, 𝑦) /∈ graph(𝜑). Since 𝜑(𝑥) isclosed, there exist disjoint open sets 𝑇1 and 𝑇2 in 𝑌 such that 𝑦 ∈ 𝑇1 and 𝜑(𝑥) ⊆𝑇2 (Exercise 1.93). Since 𝜑 is uhc, 𝜑+(𝑇2) is a neighborhood of 𝑥. Therefore𝜑+(𝑇2) × 𝑇1 is a neighborhood of (𝑥, 𝑦) disjoint from graph(𝜑). Therefore thecomplement of graph(𝜑) is open, which implies that graph(𝜑) is closed.

3. Since 𝜑 is closed and 𝑌 compact, 𝜑 is compact-valued. Let (𝑥𝑛) → 𝑥 be asequence in 𝑋 and (𝑦𝑛) a sequence in 𝑌 with 𝑦𝑛 ∈ 𝜑(𝑥𝑛). Since 𝑌 is compact,there exists a subsequence 𝑦𝑚 → 𝑦. Since 𝜑 is closed, 𝑦 ∈ 𝜑(𝑥). Therefore, byExercise 2.104, 𝜑 is uhc.

2.107 Assume 𝜑 is closed-valued and uhc. Then 𝜑 is closed (Exercise 2.106). Con-versely, if 𝜑 is closed, then 𝜑(𝑥) is closed for every 𝑥 (Exercise 2.106). If 𝑌 is compact,then 𝜑 is compact-valued (Exercise 1.110). By Exercise 2.104, 𝜑 is uhc.

2.108 𝜑1 is closed-valued (Exercise 2.106). Similarly, 𝜑2 is closed-valued (Proposition1.1). Therefore, for every 𝑥 ∈ 𝑋 , 𝜑(𝑥) = 𝜑1(𝑥) ∩ 𝜑2(𝑥) is closed (Exercise 1.85) andhence compact (Exercise 1.110). Hence 𝜑 is compact-valued.

Now, for any 𝑥0 ∈ 𝑋 , let 𝑇 be an open neighborhood of 𝜑(𝑥0). We need to show thatthere is a neighborhood 𝑆 of 𝑥0 such that 𝜑(𝑆) ⊆ 𝑇 .

Case 1 𝑇 ⊇ 𝜑2(𝑥0): Since 𝜑2 is uhc, there exists a neighborhood such that 𝑆 ∋ 𝑥0such that 𝜑2(𝑆) ⊆ 𝑇 which implies that 𝜑(𝑆) ⊆ 𝜑2(𝑆) ⊆ 𝑇

Case 2 𝑇 ∕⊇ 𝜑2(𝑥0): Let 𝐾 = 𝜑2(𝑥0) ∖ 𝑇 ∕= ∅. For every 𝑦 ∈ 𝐾, there exist neighbor-hoods 𝑆𝑦(𝑥0) and 𝑇 (𝑦) such that 𝜑1(𝑆𝑦(𝑥0))∩𝑇 (𝑦) = ∅ (Exercise 1.93). The sets𝑇 (𝑦) constitute an open covering of 𝐾. Since 𝐾 is compact, there exists a finitesubcover, that is there exists a finite number of elements 𝑦1, 𝑦2, . . . 𝑦𝑛 such that

𝐾 ⊆𝑛∪𝑖=1

𝑇 (𝑦𝑖)

Let 𝑇 (𝐾) denote∪𝑛

𝑖=1 𝑇 (𝑦𝑖). Note that 𝑇∪𝑇 (𝐾) is an open set containing 𝜑2(𝑥0).Since 𝜑2 is uhc, there exists a neighborhood 𝑆′(𝑥0) such that 𝜑2(𝑆

′(𝑥0)) ⊆ 𝑇 ∪𝑇 (𝐾). Let

𝑆(𝑥0) =

𝑛∩𝑖=1

𝑆𝑦𝑖(𝑥0) ∩ 𝑆′(𝑥0)

𝑆(𝑥0) is an open neighborhood of 𝑥0 for which

𝜑1(𝑆(𝑥0)) ∩ 𝑇 (𝐾) = ∅ and 𝜑2(𝑆(𝑥0)) ⊆ 𝑇 ∪ 𝑇 (𝐾)

from which we conclude that

𝜑(𝑆(𝑥0)) = 𝜑1(𝑆(𝑥0)) ∩ 𝜑2(𝑆(𝑥0)) ⊆ 𝑇

2.109 1. Let x ∈ 𝑋(p,𝑚) ∩ 𝑇 . Then x ∈ 𝑋(p,𝑚) and∑𝑛

𝑖=1 𝑝𝑖𝑥𝑖 ≤ 𝑚. Since 𝑇 isopen, there exists 𝛼 < 1 such that x = 𝛼x ∈ 𝑇 and

𝑛∑𝑖=1

𝑝𝑖𝑥𝑖 = 𝛼

𝑛∑𝑖=1

𝑝𝑖𝑥𝑖 <

𝑛∑𝑖=1

𝑝𝑖𝑥𝑖 ≤ 𝑚

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2. (a) Suppose that 𝑋(p,𝑚) is not lhc. Then for every neighborhood 𝑆 of (p,𝑚),there exists (p′,𝑚′) ∈ 𝑆 such that 𝑋(p′,𝑚′) ∩ 𝑇 = ∅. In particular, forevery open ball 𝐵𝑛(p,𝑚), there exists a point (p𝑛,𝑚𝑛) ∈ 𝐵𝑛(p,𝑚) suchthat 𝑋(p𝑛,𝑚𝑛) ∩ 𝑇 = ∅. ((p𝑛,𝑚𝑛)) is the required sequence.

(b) By construction, ∥p𝑛 − p∥ < 1/𝑛→ 0 which implies that 𝑝𝑛𝑖 → 𝑝𝑖 for every𝑖. Therefore (Exercise 1.202)∑

𝑝𝑛𝑖 ��𝑖 →∑𝑝𝑖𝑥𝑖 < 𝑚 and 𝑚𝑛 → 𝑚

and therefore there exists 𝑁 such that∑𝑝𝑁𝑖 ��𝑖 < 𝑚

𝑁

which implies that

x ∈ 𝑋(p𝑁 ,𝑚𝑁 )

(c) Also by construction 𝑋(p𝑁 ,𝑚𝑁 ) ∩ 𝑇 = ∅ which implies 𝑋(p𝑁 ,𝑚𝑁 ) ⊆ 𝑇 𝑐and therefore

x ∈ 𝑋(p𝑛,𝑚𝑛) =⇒ x /∈ 𝑇

The assumption that 𝑋(p,𝑚) is not lhc at (p,𝑚) implies that x /∈ 𝑇 , contra-dicting the conclusion in part 1 that x ∈ 𝑇 .

3. This contradiction establishes that (p,𝑚) is lhc at (p,𝑚). Since the choice of(p,𝑚) was arbitrary, we conclude that the budget correspondence 𝑋(p,𝑚) is lhcfor all (p,𝑚) ∈ 𝑃 (assuming 𝑋 = ℜ𝑛+).

4. In the previous example (Example 2.89), we have shown that 𝑋(p,𝑚) is uhc.Hence, the budget correspondence is continuous for all (p,𝑚) such that 𝑚 >infx∈𝑋

∑𝑚𝑖=1 𝑝𝑖𝑥𝑖.

2.110 We give two alternative proofs.

Proof 1 Let 𝒞 = {𝑆} be an open cover of 𝜑(𝐾). For every 𝑥 ∈ 𝐾, 𝜑(𝑥) ⊆ 𝜑(𝐾) iscompact and hence can be covered by a finite number of the sets 𝑆 ∈ 𝒞. Let𝑆𝑥 denote the union of the finite cover of 𝜑(𝑥). Since 𝜑 is uhc, every 𝜑+(𝑆𝑥)is open in 𝑋 . Therefore {𝜑+(𝑆𝑥) : 𝑥 ∈ 𝐾 } is an open covering of 𝐾. If 𝐾 iscompact, it contains an finite covering {𝜑+(𝑆𝑥1), 𝜑+(𝑆𝑥2), . . . , 𝜑+(𝑆𝑥𝑛) }. Thesets 𝑆𝑥1 , 𝑆𝑥2 , . . . , 𝑆𝑥𝑛 are a finite subcovering of 𝜑(𝐾).

Proof 2 Let (𝑦𝑛) be a sequence in 𝜑(𝐾). We have to show that (𝑦𝑛) has a convergentsubsequence with a limit in 𝜑(𝐾). For every 𝑦𝑛, there is an 𝑥𝑛 with 𝑦𝑛 ∈𝜑(𝑥𝑛). Since 𝐾 is compact, the sequence (𝑥𝑛) has a convergent subsequence𝑥𝑚 → 𝑥 ∈ 𝐾. Since 𝜑 is uhc, the sequence (𝑦𝑚) has a subsequence (𝑦𝑝) whichconverges to 𝑦 ∈ 𝜑(𝑥) ⊆ 𝜑(𝐾). Hence the original sequence (𝑦𝑛) has a convergentsubsequence.

2.111 The sets 𝑋,𝜑(𝑋), 𝜑2(𝑋), . . . form a sequence of nonempty compact sets. Since𝜑(𝑋) ⊆ 𝑋 , 𝜑2(𝑋) ⊆ 𝜑(𝑋) and so on, the sequence of sets 𝜑𝑛𝑋 is decreasing. Let

𝐾 =

∞∩𝑛=1

𝜑𝑛(𝑋)

By the nested intersection theorem (Exercise 1.117), 𝐾 ∕= ∅. Since 𝐾 ⊆ 𝜑𝑛−1(𝑋),𝜑(𝐾) ⊆ 𝜑𝑛(𝑋) for every 𝑛, which implies that 𝜑(𝐾) ⊆ 𝐾.

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To show that 𝐾 ⊆ 𝜑(𝐾), let 𝑦 ∈ 𝐾. For every 𝑛 there exists an 𝑥𝑛 ∈ 𝜑𝑛(𝑋) suchthat 𝑦 ∈ 𝜑(𝑥𝑛). Since 𝑋 is compact, there exists a subsequence 𝑥𝑚 → 𝑥0. Since𝑥𝑚 ∈ 𝜑𝑚(𝑋) for every 𝑚, 𝑥0 ∈ 𝐾. The sequence (𝑥𝑚, 𝑦) → (𝑥0, 𝑦). Since 𝜑 is closed(Exercise 2.107), 𝑦 ∈ 𝜑(𝑥0). Therefore 𝑦 ∈ 𝜑(𝐾) which implies that 𝐾 ⊆ 𝜑(𝐾).

2.112 𝜑(𝑥) is compact for every 𝑥 ∈ 𝑋 by Tychonoff’s theorem (Proposition 1.2).Let 𝑥𝑘 → 𝑥 be a sequence in 𝑋 and let 𝑦𝑘 = (𝑦𝑘1 , 𝑦

𝑘2 , . . . , 𝑦

𝑘𝑛) with 𝑦𝑘𝑖 ∈ 𝜑(𝑥𝑘) be

a corresponding sequence of points in 𝑌 . For each 𝑦𝑘𝑖 , 𝑖 = 1, 2, . . . , 𝑛, there exists a

subsequence 𝑦𝑘′

𝑖 → 𝑦𝑖 with 𝑦𝑖 ∈ 𝜑𝑖(𝑥) (Exercise 2.104). Therefore 𝑦 = (𝑦1, 𝑦2, . . . , 𝑦𝑛) ∈𝜑(𝑥) which implies that 𝜑 is uhc.

2.113 Let 𝑣 ∈ 𝐶(𝑋). For every x ∈ 𝑋 , the maximand 𝑓(𝑥, 𝑦) + 𝛽𝑣(𝑦) is a continuousfunction on a compact set 𝐺(𝑥). Therefore the supremum is attained, and max canreplace sup in the definition of the operator 𝑇 (Theorem 2.2). 𝑇𝑣 is the value functionfor the constrained optimization problem

max𝑦∈𝐺(𝑥)

{ 𝑓(𝑥, 𝑦) + 𝛽𝑣(𝑦) }

satisfying the requirements of the continuous maximum theorem (Theorem 2.3), whichensures that 𝑇𝑣 is continuous on 𝑋 . We have previously shown that 𝑇𝑣 is bounded(Exercise 2.18). Therefore 𝑇𝑣 ∈ 𝐶(𝑋).

2.114 1. 𝑆 has a least upper bound since 𝑋 is a complete lattice. Let 𝑠∗ = sup𝑆.Then 𝑆∗ = ≿(𝑠∗) is a complete sublattice of 𝑋 (Exercise 1.48).

2. For every 𝑠 ∈ 𝑆, 𝑠 ≾ 𝑠∗ and since 𝑓 is increasing and 𝑠 is a fixed point

𝑠 = 𝑓(𝑠) ≾ 𝑓(𝑠∗)

Therefore 𝑓(𝑠∗) ∈ 𝑆∗. (𝑓(𝑠∗) is an upper bound of 𝑆). Again, since 𝑓 is increas-ing, this implies that 𝑓(𝑥) ≿ 𝑓(𝑠∗) for every 𝑥 ∈ 𝑆∗. Therefore 𝑓(𝑆∗) ⊆ 𝑆∗.

3. Let 𝑔 be the restriction of 𝑓 to the sublattice 𝑆∗. Since 𝑓(𝑆∗) ⊆ 𝑆∗, 𝑔 is anincreasing function on a complete lattice. Applying Theorem 2.4, 𝑔 has a smallestfixed point ��.

4. �� is a fixed point of 𝑓 , that is �� ∈ 𝐸. Furthermore, 𝑥 ∈ 𝑆∗. Therefore 𝑥 isan upper bound for 𝑆 in 𝐸. Moreover, �� is the smallest fixed point of 𝑓 in 𝑆∗.Therefore, �� is the least upper bound of 𝑆 in 𝐸.

5. By Exercise 1.47, this implies that 𝐸 is a complete lattice.

In Example 2.91, if 𝑆 = {(2, 1), (1, 2)}, 𝑆∗ = {(2, 2), (3, 2), (2, 3), (3, 3)} and 𝑥 = (3, 3).

2.115 1. For every 𝑥 ∈𝑀 , there exists some 𝑦𝑥 ∈ 𝜑(𝑥) such that 𝑦𝑥 ≾ 𝑥. Moreover,since 𝜑 is increasing and �� ≾ 𝑥, there exists some 𝑧𝑥 ∈ 𝜑(��) such that

𝑧𝑥 ≾ 𝑦𝑥 ≾ 𝑥 for every 𝑥 ∈𝑀

2. Let 𝑧 = inf{𝑧𝑥}(a) Since 𝑧𝑥 ≾ 𝑥 for every 𝑥 ∈𝑀 , 𝑧 = inf{𝑧𝑥} ≾ inf{𝑥} = ��.

(b) Since 𝜑(��) is a complete sublattice of 𝑋 , 𝑧 = inf{𝑧𝑥} ∈ 𝜑(��).

3. Therefore, �� ∈𝑀 .

4. Since 𝑧 ≾ 𝑥 and 𝜑 is increasing, there exists some 𝑦 ∈ 𝜑(𝑧) such that

𝑦 ≾ 𝑧 ∈ 𝜑(��)

Hence 𝑧 ∈𝑀 .

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5. This implies that �� ≾ 𝑧. Therefore

�� = 𝑧 ∈ 𝜑(��)

�� is a fixed point of 𝜑.

6. Since 𝐸 ⊆𝑀 , 𝑥 = inf𝑀 is the least fixed point of 𝜑.

2.116 1. Let 𝑆 ⊆ 𝐸 and 𝑠∗ = sup𝑆. For every 𝑥 ∈ 𝑆, 𝑥 ∈ 𝜑(𝑥). Since 𝜑 isincreasing, there exists some 𝑧𝑥 ∈ 𝜑(𝑠∗) such that 𝑧𝑥 ≿ 𝑥.

2. Let 𝑧∗ = sup 𝑧𝑥. Then

(a) Since 𝑧𝑥 ≿ 𝑥 for every 𝑥 ∈ 𝑆, 𝑧∗ = sup 𝑧𝑥 ≿ sup𝑥 = 𝑠∗

(b) 𝑧∗ ∈ 𝜑(𝑠∗) since 𝜑(𝑠∗) is a complete sublattice.

3. Define

𝑆∗ = { 𝑥 ∈ 𝑋 : 𝑥 ≿ 𝑠 for every 𝑠 ∈ 𝑆 }

𝑆∗ is the set of all upper bounds of 𝑆 in 𝑋 . Then 𝑆∗ is a complete lattice, since

𝑆∗ = ≿(𝑠∗)

4. Let 𝜇 : 𝑆∗ ⇉ 𝑆∗ be the correspondence

𝜇(𝑥) = 𝜑(𝑥) ∩ 𝜓(𝑥)

where 𝜓 : 𝑆∗ ⇉ 𝑆∗ is the constant correspondence defined by 𝜓(𝑥) = 𝑆∗ for every 𝑥 ∈𝑆∗. Then

(a) Since 𝜑 is increasing, for every 𝑥 ≿ 𝑠∗, there exists some 𝑦𝑥 ∈ 𝜑(𝑥) suchthat 𝑦𝑥 ≿ 𝑠∗. Therefore 𝜇(𝑥) ∕= ∅ for every 𝑥 ∈ 𝑆∗.

(b) Both 𝜑(𝑥) and 𝜓(𝑥) are complete sublattices for every 𝑥 ∈ 𝑆∗. Therefore𝜇(𝑥) is a complete sublattice for every 𝑥 ∈ 𝑆∗.

(c) Since both 𝜑 and 𝜓 are increasing on 𝑆∗, 𝜇 is increasing on 𝑆∗ (Exercise2.47).

5. By the previous exercise, 𝜇 has a least fixed point 𝑥.

6. �� ∈ 𝑆∗ is an upper bound of 𝑆. Therefore �� is the least upper bound of 𝑆 in 𝐸.

7. By the previous exercise, 𝐸 has a least element. Since we have shown everysubset 𝑆 ⊆ 𝐸 has a least upper bound, this establishes that 𝐸 is complete lattice(Exercise 1.47).

2.117 For any 𝑖, let a1−𝑖, a2−𝑖 ∈ 𝐴−𝑖 with a2−𝑖 ≿ a1−𝑖. Let ��1𝑖 = 𝑓(a1−𝑖) and ��2𝑖 = 𝑓(a2−𝑖).

We want to show that ��2𝑖 ≿ ��1𝑖 . Since ��1𝑖 ∈ 𝐵(a1−𝑖) and 𝐵(a−𝑖) is increasing, thereexists some 𝑎𝑖 ∈ 𝐵(a2−𝑖) such that 𝑎𝑖 ≿ ��1𝑖 . (Exercise 2.44). Therefore

sup𝐵(a−𝑖) = ��2𝑖 ≿ 𝑎𝑖 ≿ ��1𝑖

𝑓𝑖 is increasing.

2.118 For any player 𝑖, their best response correspondence 𝐵𝑖(a−𝑖) is

1. increasing by the monotone maximum theorem (Theorem 2.1).

2. a complete sublattice of 𝐴𝑖 for every a−𝑖 ∈ 𝐴−𝑖 (Corollary 2.1.1).

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The joint best response correspondence

𝐵(a) = 𝐵1(a−1)×𝐵2(a−2)× ⋅ ⋅ ⋅ ×𝐵𝑛(a−𝑛)

is also

1. increasing (Exercise 2.46)

2. a complete sublattice of 𝐴 for every a ∈ 𝐴Therefore, the best response correspondence 𝐵(a) satisfies the conditions of Zhou’stheorem, which implies that the set 𝐸 of fixed points of 𝐵 is a nonempty completelattice. 𝐸 is precisely the set of Nash equilibria of the game.

2.119 In proving the theorem, we showed that

𝜌(𝑥𝑛, 𝑥𝑛+𝑚) ≤ 𝛽𝑛

1− 𝛽 𝜌(𝑥0, 𝑥1)

for every 𝑚,𝑛 ≥ 0. Letting 𝑚→∞, 𝑥𝑛+𝑚 → 𝑥 and therefore

𝜌(𝑥𝑛, 𝑥) ≤ 𝛽𝑛

1− 𝛽 𝜌(𝑥0, 𝑥1)

Similarly, for every 𝑛,𝑚 ≥ 0

𝜌(𝑥𝑛, 𝑥𝑛+𝑚) ≤ 𝜌(𝑥𝑛, 𝑥𝑛+1) + 𝜌(𝑥𝑛+1, 𝑥𝑛+2) + ⋅ ⋅ ⋅+ 𝜌(𝑥𝑛+𝑚−1, 𝑥𝑛+𝑚)

≤ (𝛽 + 𝛽2 + ⋅ ⋅ ⋅+ 𝛽𝑚)𝜌(𝑥𝑛−1, 𝑥𝑛)

≤ 𝛽(1 − 𝛽𝑚)

1− 𝛽 𝜌(𝑥𝑛−1, 𝑥𝑛)

Letting 𝑚→∞, 𝑥𝑛+𝑚 → 𝑥 and 𝛽𝑚 → 0 so that

𝜌(𝑥𝑛, 𝑥) ≤ 𝛽

1− 𝛽 𝜌(𝑥𝑛−1, 𝑥𝑛)

2.120 First observe that 𝑓(𝑥) ≥ 1 for every 𝑥 ≥ 1. Therefore 𝑓 : 𝑋 → 𝑋 . For any𝑥, 𝑧 ∈ 𝑋

𝑓(𝑥)− 𝑓(𝑦)

𝑥− 𝑦 =𝑥− 𝑦 + 2

𝑥 − 2𝑦

2(𝑥− 𝑦) =1

2− 1

𝑥𝑦

Since 1𝑥𝑦 ≤ 1 for all 𝑥, 𝑦 ∈ 𝑋

−1

2≤ 𝑓(𝑥)− 𝑓(𝑦)

𝑥− 𝑦 ≤ 1

2

so that ∣∣∣∣𝑓(𝑥)− 𝑓(𝑦)

𝑥− 𝑦∣∣∣∣ =∣𝑓(𝑥)− 𝑓(𝑦)∣∣𝑥− 𝑦∣ ≤ 1

2

or

∣𝑓(𝑥)− 𝑓(𝑦)∣ ≤ 1

2∣𝑥− 𝑦∣

𝑓 is a contraction on 𝑋 with modulus 1/2.

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𝑋 is closed and hence complete (Exercise 1.107). Therefore, 𝑓 has a fixed point. Thatis, there exists 𝑥0 ∈ 𝑋 such that

𝑥0 = 𝑓(𝑥0) =1

2(𝑥0 +

2

𝑥0)

Rearranging

2𝑥20 = 𝑥20 + 2 =⇒ 𝑥20 = 2

so that 𝑥0 =√

2.

Letting 𝑥0 = 2

𝑥1 =1

2(2 + 1) =

3

2

Using the error bounds in Corollary 2.5.1,

𝜌(𝑥𝑛,√

2) ≤ 𝛽𝑛

1− 𝛽 𝜌(𝑥0, 𝑥1)

=(1/2)𝑛

1/21/2

=1

2𝑛

=1

1024< 0.001

when 𝑛 = 10. Therefore, we conclude that 10 iterations are ample to reduce theerror below 0.001. Actually, with experience, we can refine this a priori estimate. InExample 1.64, we calculated the first five terms of the sequence to be

(2, 1.5, 1.416666666666667, 1.41421568627451, 1.41421356237469)

We observe that

𝜌(𝑥3, 𝑥4) = 1.41421568627451− 1.41421356237469) = 0.0000212389982

so that using the second inequality of Corollary 2.5.1

𝜌(𝑥4,√

2) ≤ 1/2

1/20.0000212389982< 0.001

𝑥4 = 1.41421356237469 is the desired approximation.

2.121 Choose any 𝑥0 ∈ 𝑆. Define the sequence 𝑥𝑛 = 𝑓(𝑥𝑛) = 𝑓𝑛(𝑥0). Then (𝑥𝑛) is aCauchy sequence in 𝑆 converging to 𝑥. Since 𝑆 is closed, 𝑥 ∈ 𝑆.

2.122 By the Banach fixed point theorem, 𝑓𝑁 has a unique fixed point 𝑥. Let 𝛽 be theLipschitz constant of 𝑓𝑁 . We have to show

𝑥 is a fixed point of 𝑓

𝜌(𝑓(𝑥), 𝑥) = 𝜌(𝑓(𝑓𝑁 (𝑥), 𝑓𝑁 (𝑥)) = 𝜌(𝑓𝑁 (𝑓(𝑥), 𝑓𝑁 (𝑥)) ≤ 𝛽𝜌(𝑓(𝑥), 𝑥)

Since 𝛽 < 1, this implies that 𝜌(𝑓(𝑥), 𝑥) = 0 or 𝑓(𝑥) = 𝑥.

𝑥 is the only fixed point of 𝑓 Suppose 𝑧 = 𝑓(𝑧) is another fixed point of 𝑓 . Then𝑧 is a fixed point of 𝑓𝑁 and

𝜌(𝑥, 𝑧) = 𝜌(𝑓𝑁 (𝑥), 𝑓𝑁 (𝑧)) ≤ 𝛽𝜌(𝑥, 𝑧)which implies that 𝑥 = 𝑧.

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2.123 By the Banach fixed point theorem, for every 𝜃 ∈ Θ, there exists 𝑥𝜃 ∈ 𝑋 suchthat 𝑓𝜃(𝑥𝜃) = 𝑥𝜃. Choose any 𝜃0 ∈ Θ.

𝜌(𝑥𝜃, 𝑥𝜃0) = 𝜌(𝑓𝜃(𝑥𝜃), 𝑓𝜃0(𝑥𝜃0))

≤ 𝜌(𝑓𝜃(𝑥𝜃), 𝑓𝜃(𝑥𝜃0)) + 𝜌(𝑓𝜃(𝑥𝜃0), 𝑓𝜃0(𝑥𝜃0))

≤ 𝛽𝜌(𝑥𝜃 , 𝑥𝜃0) + 𝜌(𝑓𝜃(𝑥𝜃0), 𝑓𝜃0(𝑥𝜃0))

(1− 𝛽)𝜌(𝑥𝜃 , 𝑥𝜃0) ≤ 𝜌(𝑓𝜃(𝑥𝜃0), 𝑓𝜃0(𝑥𝜃0))

𝜌(𝑥𝜃, 𝑥𝜃0) ≤ 𝜌(𝑓𝜃(𝑥𝜃0), 𝑓𝜃0(𝑥𝜃0))

(1− 𝛽)→ 0

as 𝜃 → 𝜃0. Therefore 𝑥𝜃 → 𝑥𝜃0 .

2.124 1. Let x be a fixed point of 𝑓 . Then x satisfies

x = (𝐼 −𝐴)x+ c = x−𝐴x+ 𝑐

which implies that 𝐴x = 𝑐.

2. For any x1,x2 ∈ 𝑋 ∥∥𝑓(x1)− 𝑓(x2)∥∥ =

∥∥(𝐼 −𝐴)(x1 − x2)∥∥≤ ∥𝐼 −𝐴∥ ∥∥x1 − x2∥∥

Since 𝑎𝑖𝑖 = 1, the norm of 𝐼 −𝐴 is

∥𝐼 −𝐴∥ = max𝑖

∑𝑗 ∕=𝑖∣𝑎𝑖𝑗 ∣ = 𝑘

and ∥∥𝑓(x1)− 𝑓(x2)∥∥ ≤ 𝑘 ∥∥x1 − x2∥∥

By the assumption of strict diagonal dominance, 𝑘 < 1. Therefore 𝑓 is a contrac-tion and has a unique fixed point x.

2.125 1.

𝜑(𝑥) = { 𝑦∗ ∈ 𝐺(𝑥) : 𝑓(𝑥, 𝑦∗) + 𝛽𝑣(𝑦∗) = 𝑣(𝑥) }= {𝑦∗ ∈ 𝐺(𝑥) : 𝑓(𝑥, 𝑦∗) + 𝛽𝑣(𝑦∗) = sup

𝑦∈𝐺(𝑥){𝑓(𝑥, 𝑦) + 𝛽𝑣(𝑦)}}

= {𝑦∗ ∈ 𝐺(𝑥) : 𝑓(𝑥, 𝑦∗) + 𝛽𝑣(𝑦∗) ≥ 𝑓(𝑥, 𝑦) + 𝛽𝑣(𝑦) for every 𝑦 ∈ 𝐺(𝑥)}= arg max

𝑦∈𝐺(𝑥){𝑓(𝑥, 𝑦) + 𝛽𝑣(𝑦)}

2. 𝜑(𝑥) is the solution correspondence of a standard constrained maximization prob-lem, with 𝑥 as parameter and 𝑦 the decision variable. By assumption the maxi-mand 𝑓(𝑥, 𝑦) = 𝑓(𝑥, 𝑦) + 𝛽𝑣(𝑦) is continuous and the constraint correspondence𝐺(𝑥) is continuous and compact-valued. Applying the continuous maximum the-orem (Theorem 2.3), 𝜑 is nonempty, compact-valued and uhc.

3. We have just shown that 𝜑(𝑥) is nonempty for every 𝑥 ∈ 𝑋 . Starting at 𝑥0,choose some 𝑥∗1 ∈ 𝜑(𝑥0). Then choose 𝑥∗2 ∈ 𝜑(𝑥∗1). Proceeding in this way,we can construct a plan x∗ = 𝑥0, 𝑥

∗1, 𝑥

∗2, . . . such that 𝑥∗𝑡+1 ∈ 𝜑(𝑥∗𝑡 ) for every

𝑡 = 0, 1, 2, . . . .

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4. Since 𝑥∗𝑡+1 ∈ 𝜑(𝑥∗𝑡 ) for every 𝑡, x satisfies Bellman’s equation, that is

𝑣(𝑥∗𝑡 ) = 𝑓(𝑥∗𝑡 , 𝑥∗𝑡+1) + 𝛽𝑣(𝑥∗𝑡+1), 𝑡 = 0, 1, 2, . . .

Therefore x is optimal (Exercise 2.17).

2.126 1. In the previous exercise (Exercise 2.125) we showed that the set 𝜑 of solu-tions to Bellman’s equation (Exercise 2.17) is the solution correspondence of theconstrained maximization problem

𝜑(𝑥) = arg max𝑦∈𝐺(𝑥)

{ 𝑓(𝑥, 𝑦) + 𝛽𝑣(𝑦) }

This problem satisfies the requirements of the monotone maximum theorem (The-orem 2.1), since the objective function 𝑓(𝑥, 𝑦) + 𝛽𝑣(𝑦)

∙ supermodular in 𝑦

∙ displays strictly increasing differences in (𝑥, 𝑦) since for every 𝑥2 ≥ 𝑥1

𝑓(𝑥2, 𝑦) + 𝛽𝑣(𝑦)− 𝑓(𝑥1, 𝑦) + 𝛽𝑣(𝑦) = 𝑓(𝑥2, 𝑦)− 𝑓(𝑥1, 𝑦)

∙ 𝐺(𝑥) is increasing.

By Corollary 2.1.2, 𝜑(𝑥) is always increasing.

2. Let x∗ = (𝑥0, 𝑥∗1, 𝑥

∗2, . . . ) be an optimal plan. Then (Exercise 2.17)

𝑥∗𝑡+1 ∈ 𝜑(𝑥∗𝑡 ), 𝑡 = 0, 1, 2, . . .

Since 𝜑 is always increasing

𝑥∗𝑡 ≥ 𝑥∗𝑡−1 =⇒ 𝑥∗𝑡+1 ≥ 𝑥∗𝑡for every 𝑡 = 1, 2, . . . . Similarly

𝑥∗𝑡 ≤ 𝑥∗𝑡−1 =⇒ 𝑥∗𝑡+1 ≤ 𝑥∗𝑡x∗ = (𝑥0, 𝑥

∗1, 𝑥

∗2, . . . ) is a monotone sequence.

2.127 Let 𝑔(𝑥) = 𝑓(𝑥)− 𝑥. 𝑔 is continuous (Exercise 2.78) with

𝑔(0) ≥ 0 and 𝑔(1) ≤ 0

By the intermediate value theorem (Exercise 2.83), there exists some point 𝑥 ∈ [0, 1]with 𝑔(𝑥) = 0 which implies that 𝑓(𝑥) = 𝑥.

2.128 1. To show that a label min{ 𝑖 : 𝛽𝑖 ≤ 𝛼𝑖 ∕= 0 } exists for every x ∈ 𝑆, assumeto the contrary that, for some x ∈ 𝑆, 𝛽𝑖 > 𝛼𝑖 for every 𝑖 = 0, 1, . . . , 𝑛. Thisimplies

𝑛∑𝑖=0

𝛽𝑖 >

𝑛∑𝑖=0

𝛼𝑖 = 1

contradicting the requirement that

𝑛∑𝑖=0

𝛽𝑖 = 1 for every 𝑓(x) ∈ 𝑆

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2. The barycentric coordinates of vertex x𝑖 are 𝛼𝑖 = 1 with 𝛼𝑗 = 0 for every 𝑗 ∕= 𝑖.Therefore the rule assigns vertex x𝑖 the label 𝑖.

3. Similarly, if x belongs to a proper face of 𝑆, it coordinates relative to the verticesnot in that face are 0, and it cannot be assigned a label corresponding to a vertexnot in the face. To be concrete, suppose that x ∈ conv {x1,x2,x4}. Then

x = 𝛼1x1 + 𝛼2x2 + 𝛼4x4, 𝛼1 + 𝛼2 + 𝛼4 = 1

and 𝛼𝑖 = 0 for 𝑖 /∈ {1, 2, 4}. Therefore

x +−→ min{ 𝑖 : 𝛽𝑖 ≤ 𝛼𝑖 ∕= 0 } ∈ {1, 2, 4}2.129 1. Since 𝑆 is compact, it is bounded (Proposition 1.1) and therefore it is

contained in a sufficiently large simplex 𝑇 .

2. By Exercise 3.74, there exists a continuous retraction 𝑟 : 𝑇 → 𝑆. The composition𝑓 ∘ 𝑟 : 𝑇 → 𝑆 ⊆ 𝑇 . Furthermore as the composition of continuous functions, 𝑓 ∘ 𝑟is continuous (Exercise 2.72). Therefore 𝑓 ∘ 𝑟 has a fixed point x∗ ∈ 𝑇 , that is𝑓 ∘ 𝑟(x∗) = x∗.

3. Since 𝑓 ∘ 𝑟(x) ∈ 𝑆 for every x ∈ 𝑇 , we must have 𝑓 ∘ 𝑟(x∗) = x∗ ∈ 𝑆. Therefore,𝑟(x∗) = x∗ which implies that 𝑓(x∗) = x∗. That is, x∗ is a fixed point of 𝑓 .

2.130 Convexity of 𝑆 is required to ensure that there is a continuous retraction of thesimplex onto 𝑆.

2.131 1. 𝑓(𝑥) = 𝑥2 on 𝑆 = (0, 1) or 𝑓(𝑥) = 𝑥+ 1 on 𝑆 = ℜ+.

2. 𝑓(𝑥) = 1− 𝑥 on 𝑆 = [0, 1/3]∪ [2/3, 1].

3. Let 𝑆 = [0, 1] and define

𝑓(𝑥) =

{1 0 ≤ 𝑥 < 1/2

0 otherwise

2.132 Suppose such a function exists. Define 𝑓(x) = −𝑟(x). Then 𝑓 : 𝐵 → 𝐵 conti-nously, and has no fixed point since for

∙ x ∈ 𝑆, 𝑓(x) = −𝑟(x) = −x ∕= x∙ x ∈ 𝐵 ∖ 𝑆, 𝑓(x) /∈ 𝐵 ∖ 𝑆 and therefore𝑓(x) ∕= x

Therefore 𝑓 has no fixed point contradicting Brouwer’s theorem.

2.133 Suppose to the contrary that 𝑓 has no fixed point. For every x ∈ 𝐵, let 𝑟(z)denote the point where the line segment from 𝑓(x) through x intersects the boundary𝑆 of 𝐵. Since 𝑓 is continuous and 𝑓(x) ∕= x, 𝑟 is a continuous function from 𝐵 to itsboundary, that is a retraction, contradicting Exercise 2.132. We conclude that 𝑓 musthave a fixed point.

2.134 No-retraction =⇒ Brouwer Note first that the no-retraction theorem (Ex-ercise 2.132) generalizes immediately to a closed ball about 0 of arbitrary radius.Assume that 𝑓 is a continuous operator on a compact, convex set 𝑆 in a fi-nite dimensional normed linear space. There exists a closed ball 𝐵 containing 𝑆(Proposition 1.1). Define 𝑔 : 𝐵 → 𝑆 by

𝑔(y) = {x ∈ 𝑆 : x is closest to y }As in Exercise 2.129, 𝑔 is well-defined, continuous and 𝑔(x) = x for every x ∈ 𝑆.𝑓 ∘ 𝑔 : 𝐵 → 𝑆 ⊆ 𝐵 and has a fixed point x∗ = 𝑓(𝑔(x∗)) by Exercise 2.133. Since

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𝑓 ∘ 𝑔(x) ∈ 𝑆 for every x ∈ 𝐵, we must have 𝑓 ∘ 𝑔(x∗) = x∗ ∈ 𝑆. Therefore,𝑔(x∗) = x∗ which implies that 𝑓(x∗) = x∗. That is, x∗ is a fixed point of 𝑓 .

Brouwer =⇒ no-retraction Exercise 2.132.

2.135 Let Λ𝑘, 𝑘 = 1, 2, . . . be a sequence of simplicial partitions of 𝑆 in which themaximum diameter of the subsimplices tend to zero as 𝑘 → ∞. By Sperner’s lemma(Proposition 1.3), every partition Λ𝑘 has a completely labeled subsimplex with verticesx𝑘0 ,x

𝑘1 , . . . ,x

𝑘𝑛. By construction of an admissible labeling, each x𝑘𝑖 belongs to a face

containing x𝑖, that is

x𝑘𝑖 ∈ conv {x𝑖, . . . }and therefore

x𝑘𝑖 ∈ 𝐴𝑖, 𝑖 = 0, 1, . . . , 𝑛

Since 𝑆 is compact, each sequence x𝑘𝑖 has a convergent subsequence x𝑘′

𝑖 . Moreover,since the diameters of the subsimplices converge to zero, these subsequences mustconverge to the same point, say x∗. That is,

lim𝑘′→∞

x𝑘′

𝑖 = x∗, 𝑖 = 0, 1, . . . , 𝑛

Since the sets 𝐴𝑖 are closed, x∗ ∈ 𝐴𝑖 for every 𝑖 and therefore

x∗ ∈𝑛∩𝑖=0

𝐴𝑖 ∕= ∅

2.136 =⇒ Let 𝑓 : 𝑆 → 𝑆 be a continuous operator on an 𝑛-dimensional simplex 𝑆with vertices x0,x1, . . . ,x𝑛. For 𝑖 = 0, 1, . . . , 𝑛, let

𝐴𝑖 = {x ∈ 𝑆 : 𝛽𝑖 ≤ 𝛼𝑖 }where 𝛼0, 𝛼1, . . . , 𝛼𝑛 and 𝛽0, 𝛽1, . . . , 𝛽𝑛 are the barycentric coordinates of x and𝑓(x) respectively. Then

∙ 𝑓 continuous =⇒ 𝐴𝑖 closed for every 𝑖 = 0, 1, . . . , 𝑛 (Exercise 1.106)

∙ Let x ∈ conv {x𝑖 : 𝑖 ∈ 𝐼 } for some 𝐼 ⊆ { 0, 1, . . . , 𝑛 }. Then

∑𝑖∈𝐼𝛼𝑖 = 1 =

𝑛∑𝑖=0

𝛽𝑖

which implies that 𝛽𝑖 ≤ 𝛼𝑖 for some 𝑖 ∈ 𝐼, so that x ∈ 𝐴𝑖. Therefore

conv {x𝑖 : 𝑖 ∈ 𝐼 } ⊆∪𝑖∈𝐼𝐴𝑖

Therefore the collection 𝐴0, 𝐴1, . . . , 𝐴𝑛 satisfies the hypotheses of the K-K-Mtheorem and their intersection is nonempty. That is, there exists

x∗ ∈𝑛∩𝑖=0

𝐴𝑖 ∕= ∅ with 𝛽∗𝑖 ≤ 𝛼∗𝑖 , 𝑖 = 0, 1, . . . , 𝑛

where 𝛼∗ and 𝛽∗ are the barycentric coordinates of x∗ and 𝑓(x∗) respectively.Since

∑𝛽∗𝑖 =

∑𝛼∗𝑖 = 1, this implies that

𝛽∗𝑖 = 𝛼∗𝑖 𝑖 = 0, 1, . . . , 𝑛

In other words, 𝑓(x∗) = x∗.

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⇐= Let 𝐴0, 𝐴1, . . . , 𝐴𝑛 be closed subsets of an 𝑛 dimensional simplex 𝑆 with verticesx0,x1, . . . ,x𝑛 such that

conv {x𝑖 : 𝑖 ∈ 𝐼 } ⊆∪𝑖∈𝐼𝐴𝑖

for every 𝐼 ⊆ { 0, 1, . . . , 𝑛 }. For 𝑖 = 0, 1, . . . , 𝑛, let

𝑔𝑖(x) = 𝜌(x, 𝐴𝑖)

For any x ∈ 𝑆 with barycentric coordinates 𝛼0, 𝛼1, . . . , 𝛼𝑛, define

𝑓(x) = 𝛽0x0 + 𝛽1x1 + ⋅ ⋅ ⋅+ 𝛽𝑛x𝑛where

𝛽𝑖 =𝛼𝑖 + 𝑔𝑖(x)

1 +∑𝑛

𝑗=0 𝑔𝑗(x)𝑖 = 0, 1, . . . , 𝑛 (2.45)

By construction 𝛽𝑖 ≥ 0 and∑𝑛

𝑖=0 𝛽𝑖 = 1. Therefore 𝑓(x) ∈ 𝑆. That is, 𝑓 : 𝑆 → 𝑆.Furthermore 𝑓 is continuous. By Brouwer’s theorem, there exists a fixed point𝑥∗ with 𝑓(x∗) = x∗. That is 𝛽∗𝑖 = 𝛼∗𝑖 for 𝑖 = 0, 1, . . . , 𝑛.

Now, since the collection 𝐴0, 𝐴1, . . . , 𝐴𝑛 covers 𝑆, there exists some 𝑖 for which𝜌(x∗, 𝐴𝑖) = 0. Substituting 𝛽∗𝑖 = 𝛼∗𝑖 in (2.45) we have

𝛼∗𝑖 =𝛼∗𝑖

1 +∑𝑛

𝑗=0 𝑔𝑗(x∗)

which implies that 𝑔𝑗(x∗) = 0 for every 𝑗. Since the 𝐴𝑖 are closed, x∗ ∈ 𝐴𝑖 for

every 𝑖 and therefore

x∗ ∈𝑛∩𝑖=0

𝐴𝑖 ∕= ∅

2.137 To simplify the notation, let 𝑧+𝑘 (p) = max(0, z𝑖(p)

). Assume p∗ is a fixed point

of 𝑔. Then for every 𝑘 = 1, 2, . . . , 𝑛

𝑝∗𝑘 =𝑝𝑘 + 𝑧+𝑘 (p∗)

1 +∑𝑛

𝑗=1 𝑧+𝑗 (p∗)

Cross-multiplying

𝑝∗𝑘 + 𝑝∗𝑘𝑛∑

𝑗=1

𝑧+𝑗 (p) = 𝑝∗𝑘 + 𝑧+𝑘 (p∗)

or

𝑝∗𝑘𝑛∑

𝑗=1

𝑧+𝑗 (p) = 𝑧+𝑘 (p∗) 𝑘 = 1, 2, . . . 𝑛

Multiplying each equation by 𝑧𝑘(p) we get

𝑝∗𝑘𝑧𝑘(p∗)

𝑛∑𝑗=1

𝑧+𝑖 (p) = 𝑧𝑘(p∗)𝑧+𝑘 (p∗) 𝑘 = 1, 2, . . . 𝑛

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Summing over 𝑘

𝑛∑𝑘=1

𝑝∗𝑘𝑧𝑘(p∗)

𝑛∑𝑗=1

𝑧+𝑖 (p) =

𝑛∑𝑘=1

𝑧𝑘(p∗)𝑧+𝑘 (p∗)

Since∑𝑛

𝑘=1 𝑝∗𝑘𝑧𝑘(p∗) = 0 this implies that

𝑛∑𝑘=1

𝑧𝑘(p∗)𝑧+𝑘 (p∗) = 0

Each term of this sum is nonnegative, since it is either 0 or(𝑧𝑘(p∗)

)2. Consequently,

every term must be zero which implies that 𝑧𝑘(p∗) ≤ 0 for every 𝑘 = 1, 2, . . . , 𝑙. Inother words, z(p∗) ≤ 0.2.138 Every individual demand function x𝑖(p,𝑚) is continuous (Example 2.90) in pand 𝑚. For given endowment 𝝎𝑖

𝑚𝑖 =𝑙∑

𝑗=1

𝑝𝑗𝝎𝑖𝑗

is continuous in p (Exercise 2.78). Therefore the excess demand function

z𝑖(p) = x𝑖(p,𝑚)− 𝝎𝑖

is continuous for every consumer 𝑖 and hence the aggregate excess demand function iscontinuous.

Similarly, the consumer’s demand function x𝑖(p,𝑚) is homogeneous of degree 0 in pand 𝑚. For given endowment 𝝎𝑖, the consumer’s wealth is homogeneous of degree 1 inp and therefore the consumer’s excess demand function z𝑖(p) is homogeneous of degree0. So therefore is the aggregate excess demand function z(p).

2.139

z(p) =

𝑛∑𝑖=1

z𝑖(p)

=

𝑛∑𝑖=1

(x𝑖(p,𝑚)− 𝝎𝑖

)

and therefore

p𝑇 z(p) =

𝑛∑𝑖=1

p𝑇x𝑖(p,𝑚)−𝑛∑𝑖=1

p𝑇𝝎𝑖

Since preferences are nonsatiated and strictly convex, they are locally nonsatiated(Exercise 1.248) which implies (Exercise 1.235) that every consumer must satisfy hisbudget constraint

p𝑇x𝑖(p,𝑚) = p𝑇𝝎𝑖 for every 𝑖 = 1, 2, . . . , 𝑛

Therefore in aggregate

p𝑇 z(p) =

𝑛∑𝑖=1

p𝑇x𝑖(p,𝑚)−𝑛∑𝑖=1

p𝑇𝝎𝑖 = 0

for every p.

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2.140 Assume there exists p∗ such that z(p∗) ≤ 0. That is

z(p∗) =𝑛∑𝑖=1

z𝑖(p) =𝑛∑𝑖=1

(x𝑖(p,𝑚)− 𝝎𝑖

)=

𝑛∑𝑖=1

x𝑖(p,𝑚)−𝑛∑𝑖=1

𝝎𝑖 ≤ 0

or ∑𝑖∈𝑁x𝑖 ≤

∑𝑖∈𝑁

𝝎𝑖

Aggregate demand is less or equal to available supply.

Let 𝑚∗𝑖 =

∑𝑙𝑗=1 𝑝

∗𝑗𝝎𝑖𝑗 denote the wealth of consumer 𝑖 when the price system is p∗

and let x∗𝑖 = x(p∗,𝑚∗) be his chosen consumption bundle. Then

x∗𝑖 ≿ x𝑖 for every x𝑖 ∈ 𝑋(p∗,𝑚𝑖)

Let x∗ = (x∗1,x∗2, . . . ,x

∗𝑛) be the allocation comprising these optimal bundles. The

pair (p∗,x∗) is a competitive equilibrium.

2.141 For each x𝑘, let 𝑆𝑘 denote the subsimplex of Λ𝑘 which contains x𝑘 and letx𝑘0 ,x

𝑘1 , . . . ,x

𝑘𝑛 denote the vertices of 𝑆𝑘. Let 𝛼𝑘0 , 𝛼

𝑘1 , . . . , 𝛼

𝑘𝑛 denote the barycentric

coordinates (Exercise 1.159) of x with respect to the vertices of 𝑆𝑘 and let y𝑘𝑖 = 𝑓𝑘(x𝑘𝑖 ),𝑖 = 0, 1, . . . , 𝑛, denote the images of the vertices. Since 𝑆 is compact, there existssubsequences x𝑘

′𝑖 , y𝑘

′𝑖 and 𝛼𝑘

′such that

x𝑘𝑖 → x∗𝑖 y𝑘𝑖 → y∗𝑖 and 𝛼𝑘𝑖 → 𝛼∗𝑖 𝑖 = 0, 1, . . . , 𝑛

Furthermore, 𝛼∗𝑖 ≥ 0 and 𝛼∗0+𝛼∗1+⋅ ⋅ ⋅+𝛼∗𝑛 = 1. Since the diameters of the subsimplicesconverge to zero, their vertices must converge to the same point. That is, we must have

x∗0 = x∗1 = ⋅ ⋅ ⋅ = x∗𝑛 = x∗

By definition of 𝑓𝑘

𝑓𝑘(x𝑘) = 𝛼𝑘0𝑓(x𝑘0) + 𝛼𝑘1𝑓(x𝑘1) + ⋅ ⋅ ⋅+ 𝛼𝑘𝑛𝑓(x𝑘𝑛)

Substituting y𝑘𝑖 = 𝑓𝑘(x𝑘𝑖 ), 𝑖 = 0, 1, . . . , 𝑛 and recognizing that x𝑘 is a fixed point of𝑓𝑘, we have

𝑥𝑘 = 𝑓𝑘(x𝑘) = 𝛼𝑘0y𝑘0 + 𝛼𝑘1y

𝑘1 + ⋅ ⋅ ⋅+ 𝛼𝑘𝑛y𝑘𝑛

Taking limits

x∗ = 𝛼∗0y∗0 + 𝛼∗1y

∗1 + ⋅ ⋅ ⋅+ 𝛼∗𝑛y∗𝑛 (2.46)

For each coordinate 𝑖, (x𝑘𝑖 ,y𝑘𝑖 ) ∈ graph(𝜑) for every 𝑘 = 0, 1, . . . . Since 𝜑 is closed,

(x∗𝑖 ,y∗𝑖 ) ∈ graph(𝜑). That is, y∗𝑖 ∈ 𝜑(x∗𝑖 ) = 𝜑(x∗) for every 𝑖 = 0, 1, . . . , 𝑛. Therefore,

(2.46) implies

x∗ ∈ conv 𝜑(x∗)

Since 𝜑 is convex valued,

x∗ ∈ 𝜑(x∗)

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2.142 Analogous to Exercise 2.129, there exists a simplex 𝑇 containing 𝑆 and a retrac-tion of 𝑇 onto 𝑆, that is a continuous function 𝑔 : 𝑇 → 𝑆 with 𝑔(x) = x for every x ∈ 𝑆.Then 𝜑 ∘ 𝑔 : 𝑇 ⇉ 𝑆 ⊂ 𝑇 is closed-valued (Exercise 2.106) and uhc (Exercise 2.103).By the argument in the proof, there exists a point x∗ ∈ 𝑇 such that x∗ ∈ 𝜑 ∘ 𝑔(x∗).However, since 𝜑 ∘ 𝑔(x∗) ⊆ 𝑆, we must have x∗ ∈ 𝑆 and therefore 𝑔(x∗) = x∗. Thisimplies x∗ ∈ 𝜑(x∗). That is, x∗ is a fixed point of 𝜑.

2.143 𝐵 = 𝐵1 ×𝐵2 × . . .×𝐵𝑛 is the Cartesian product of uhc, compact- and convex-valued correspondences. Therefore 𝐵 is also compact-valued and uhc (Exercise 2.112and also convex-valued (Exercise 1.165). By Exercise 2.106, 𝐵 is closed.

2.144 Strict quasiconcavity ensures that the best response correspondence is in fact afunction 𝐵 : 𝑆 → 𝑆. Since the hypotheses of Example 2.96 apply, there exists at leastone equilibrium. Suppose that there are two Nash equilibria s and s′. Since 𝐵 is acontraction,

𝜌(𝐵(s), 𝐵(s′) ≤ 𝛽𝜌(s, s′)for some 𝛽 < 1. However

𝐵(s) = s and 𝐵(s′) = s′

and (2.46) implies that

𝜌(s, s′) ≤ 𝛽𝜌(s, s′)which is possible if and only if s = s′. This implies that the equilibrium must be unique.

2.145 Since 𝐾 is compact, it is totally bounded (Exercise 1.112). There exists a finiteset of points x1,x2, . . . ,x𝑛 such that

𝐾 ⊆𝑛∩𝑖=1

𝐵𝜖(x𝑖)

Let 𝑆 = conv {x1,x2, . . . ,x𝑛}. For 𝑖 = 1, 2, . . . , 𝑛 and x ∈ 𝑋 , define

𝛼𝑖(x) = max{0, 𝜖− ∥x− x𝑖∥}Then for every x ∈ 𝐾,

0 ≤ 𝛼𝑖(x) ≤ 𝜖, 𝑖 = 1, 2, . . . , 𝑛

and

𝛼𝑖(x) > 0 ⇐⇒ x ∈ 𝐵𝜖(x𝑖)

Note that 𝛼𝑖(x) > 0 for some 𝑖. Define

ℎ(x) =

∑𝛼𝑖(x)x𝑖∑𝛼𝑖(x)

Then ℎ(x) ∈ 𝑆 and therefore ℎ : 𝐾 → 𝑆. Furthermore, ℎ is continuous and

∥ℎ(x) − x∥ =

∥∥∥∥∑𝛼𝑖(x)x𝑖∑𝛼𝑖(x)

− x∥∥∥∥

=

∥∥∥∥∑𝛼𝑖(x)(x𝑖 − x)∑𝛼𝑖(x)

∥∥∥∥=

∑𝛼𝑖(x) ∥x𝑖 − x∥∑

𝛼𝑖(x)

≤∑𝛼𝑖(x)𝜖∑𝛼𝑖(x)

= 𝜖

since 𝛼𝑖(x) > 0 ⇐⇒ ∥x𝑖 − x∥ ≤ 𝜖.

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2.146 1. For every x ∈ 𝑆𝑘, 𝑓(x) ∈ 𝑆 and therefore 𝑔𝑘(x) = ℎ𝑘(𝑓(x)

) ∈ 𝑆𝑘.

2. For any x ∈ 𝑆𝑘, let y = 𝑓(x) ∈ 𝑓(𝑆) and therefore

∥∥ℎ𝑘(y) − y∥∥ < 1

𝑘

which implies

∥∥𝑔𝑘(x)− 𝑓(x)∥∥ ≤ 1

𝑘for every x ∈ 𝑆𝑘

2.147 By the Triangle inequality∥∥x𝑘 − 𝑓(x)∥∥ ≤ ∥∥𝑔𝑘(x𝑘)− 𝑓(x𝑘)

∥∥ +∥∥𝑓(x𝑘)− 𝑓(x)

∥∥As shown in the previous exercise

∥∥𝑔𝑘(x𝑘)− 𝑓(x𝑘)∥∥ ≤ 1

𝑘→ 0

as 𝑘 →∞. Also since 𝑓 is continuous∥∥𝑓(x𝑘)− 𝑓(x)∥∥→ 0

Therefore ∥∥x𝑘 − 𝑓(x)∥∥→ 0 =⇒ x = 𝑓(x)

x is a fixed point of 𝑓 .

2.148 𝑇 (𝐹 ) is bounded and equicontinuous and so therefore is 𝑇 (𝐹 ) (Exercise 2.96). ByAscoli’s theorem (Exercise 2.95), 𝑇 (𝐹 ) is compact. Therefore 𝑇 is a compact operator.Applying Corollary 2.8.1, 𝑇 has a fixed point.

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Chapter 3: Linear Functions

3.1 Let x1,x2 ∈ 𝑋 and 𝛼1, 𝛼2 ∈ ℜ. Homogeneity implies that

𝑓(𝛼1x1) = 𝛼1𝑓(𝑥1)

𝑓(𝛼2x2) = 𝛼2𝑓(𝑥2)

and additivity implies

𝑓(𝛼1x1 + 𝛼2x2) = 𝛼1𝑓(x1) + 𝛼2𝑓(x2)

Conversely, assume

𝑓(𝛼1x1 + 𝛼2x2) = 𝛼1𝑓(x1) + 𝛼2𝑓(x2)

for all x1,x2 ∈ 𝑋 and 𝛼1, 𝛼2 ∈ ℜ. Letting 𝛼1 = 𝛼2 = 1 implies

𝑓(x1 + x2) = 𝑓(x1) + 𝑓(x2)

while setting x2 = 0 implies

𝑓(𝛼1x1) = 𝛼1𝑓(x1)

3.2 Assume 𝑓1, 𝑓2 ∈ 𝐿(𝑋,𝑌 ). Define the mapping 𝑓1 + 𝑓2 : 𝑋 → 𝑌 by

(𝑓1 + 𝑓2)(x) = 𝑓1(x) + 𝑓2(x)

We have to confirm that 𝑓1 + 𝑓2 is linear, that is

(𝑓1 + 𝑓2)(x1 + x2) = 𝑓1(x1 + x2) + 𝑓2(x1 + x2)

= 𝑓1(x1) + 𝑓1(x2) + 𝑓2(x1) + 𝑓2(x2)

= 𝑓1(x1) + 𝑓2(x1) + 𝑓1(x1) + 𝑓2(x2)

= (𝑓1 + 𝑓2)(x1) + (𝑓1 + 𝑓2)(x2)

and

(𝑓1 + 𝑓2)(𝛼x) = 𝑓1(𝛼x) + 𝑓2(𝛼x)

= 𝛼(𝑓1(x) + 𝑓2(x))

= 𝛼(𝑓1 + 𝑓2)(x)

Similarly let 𝑓 ∈ 𝐿(𝑋,𝑌 ) and define 𝛼𝑓 : 𝑋 → 𝑌 by

(𝛼𝑓)(x) = 𝛼𝑓(x)

𝛼𝑓 is also linear, since

(𝛼𝑓)(𝛽x) = 𝛼𝑓(𝛽x)

= 𝛼𝛽𝑓(x)

= 𝛽𝛼𝑓(x)

= 𝛽(𝛼𝑓)(x)

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3.3 Let x,x1,x2 ∈ ℜ2. Then

𝑓(x1 + x2) = 𝑓(𝑥11 + 𝑥21, 𝑥12 + 𝑥22)

=((𝑥11 + 𝑥21) cos 𝜃 − (𝑥12 + 𝑥22) sin 𝜃, (𝑥11 + 𝑥21) sin 𝜃 − (𝑥12 + 𝑥22) cos 𝜃

)=

((𝑥11 cos 𝜃 − 𝑥12 sin 𝜃) + (𝑥21 cos 𝜃 − 𝑥22 sin 𝜃), (𝑥11 sin 𝜃 + 𝑥12 cos 𝜃) + (𝑥21 sin 𝜃 − 𝑥22 cos 𝜃)

)=

((𝑥11 cos 𝜃 − 𝑥12 sin 𝜃, 𝑥11 sin 𝜃 + 𝑥12 cos 𝜃) + (𝑥21 cos 𝜃 − 𝑥22 sin 𝜃, 𝑥21 sin 𝜃 − 𝑥22 cos 𝜃

)= 𝑓(𝑥11, 𝑥

12) + 𝑓(𝑥21, 𝑥

22)

= 𝑓(x1) + 𝑓(x2)

and

𝑓(𝛼x) = 𝑓(𝛼𝑥1, 𝛼𝑥2)

= (𝛼𝑥1 cos 𝜃 − 𝛼𝑥2 sin 𝜃, 𝛼𝑥1 sin 𝜃 + 𝛼𝑥2 cos 𝜃)

= 𝛼 (𝑥1 cos 𝜃 − 𝑥1 sin 𝜃, 𝑥1 sin 𝜃 + 𝑥2 cos 𝜃)

= 𝛼𝑓(𝑥1, 𝑥2)

= 𝛼𝑓(x)

3.4 Let x,x1,x2 ∈ ℜ3.𝑓(x1 + x2) = 𝑓(𝑥11 + 𝑥2, 𝑥12 + 𝑥22, 𝑥

13 + 𝑥23)

= (𝑥11 + 𝑥21, 𝑥12 + 𝑥22, 0)

= (𝑥11, 𝑥12, 0) + (𝑥21, 𝑥

22, 0)

= 𝑓(𝑥11, 𝑥12, 𝑥

13) + 𝑓(𝑥21, 𝑥

22, 𝑥

23)

= 𝑓(x1) + 𝑓(x2)

and

𝑓(𝛼x) = 𝑓(𝛼𝑥1, 𝛼𝑥2, 𝛼𝑥3)

= (𝛼𝑥1, 𝛼𝑥2, 0)

= 𝛼(𝑥1, 𝑥2, 0)

= 𝛼𝑓(𝑥1, 𝑥2, 𝑥3)

= 𝛼𝑓(x)

This mapping is the projection of 3-dimensional space onto the (2-dimensional) plane.

3.5 Applying the definition

𝑓(𝑥1, 𝑥2) =

(0 11 0

)(𝑥1𝑥2

)= (𝑥2, 𝑥1)

This function interchanges the two coordinates of any point in the plane ℜ2. Its actioncorresponds to reflection about the line 𝑥1 = 𝑥2 ( 45 degree diagonal).

3.6 Assume (𝑁,𝑤) and (𝑁,𝑤′) are two games in 𝒢𝑁 . For any coalition 𝑆 ⊆ 𝑁(𝑤 + 𝑤′)(𝑆)− (𝑤 + 𝑤′)(𝑆 ∖ {𝑖}) = 𝑤(𝑆) + 𝑤(𝑆′)− 𝑤(𝑆 ∖ {𝑖})− 𝑤′(𝑆 ∖ {𝑖})

= (𝑤(𝑆) − 𝑤(𝑆 ∖ {𝑖})) + (𝑤′(𝑆)− 𝑤′(𝑆 ∖ {𝑖}))= 𝜑𝑖(𝑤) + 𝜑𝑖(𝑤

′)

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3.7 The characteristic function of cost allocation game is

𝑤(𝐴𝑃 ) = 0 𝑤(𝐴𝑃, 𝑇𝑁) = 210

𝑤(𝑇𝑁) = 0 𝑤(𝐴𝑃,𝐾𝑀) = 770 𝑤(𝑁) = 1530

𝑤(𝐾𝑀) = 0 𝑤(𝐾𝑀,𝑇𝑁) = 1170

The following table details the computation of the Shapley value for player 𝐴𝑃 .

𝑆 𝛾𝑆 𝑤(𝑆) 𝑤(𝑆 ∖ {𝑖}) 𝛾𝑆(𝑤(𝑆) − 𝑤(𝑆 ∖ {𝑖}))𝐴𝑃 1/3 0 0 0

𝐴𝑃, 𝑇𝑁 1/6 210 0 35𝐴𝑃,𝐾𝑀 1/6 770 0 1281/3

𝐴𝑃, 𝑇𝑁,𝐾𝑀 1/3 1530 1170 120𝜑𝑓 (𝑤) 2831/3

Thus 𝜑𝐴𝑃𝑤 = 2831/3. Similarly, we can calculate that 𝜑𝑇𝑁𝑤 = 4831/3 and 𝜑𝐾𝑀𝑤 =7631/3.

3.8

∑𝑖∈𝑁𝜑𝑖𝑤 =

∑𝑖∈𝑁

(∑𝑆∋𝑖𝛾𝑆(𝑤(𝑆) − 𝑤(𝑆 ∖ {𝑖}))

)

=∑𝑆∋𝑖

(∑𝑖∈𝑁𝛾𝑆(𝑤(𝑆) − 𝑤(𝑆 ∖ {𝑖}))

)

=∑𝑆⊆𝑁

∑𝑖∈𝑆𝛾𝑆𝑤(𝑆)−

∑𝑆⊆𝑁

∑𝑖∈𝑆𝛾𝑆𝑤(𝑆 ∖ {𝑖})

=∑𝑆⊆𝑁

𝑠× 𝛾𝑆𝑤(𝑆)−∑𝑆⊆𝑁

𝛾𝑆

(∑𝑖∈𝑆𝑤(𝑆 ∖ {𝑖})

)

=∑𝑆⊆𝑁

𝑠× 𝛾𝑆𝑤(𝑆)−∑𝑆⊂𝑁

𝑠× 𝛾𝑆𝑤(𝑆)

= 𝑛× 𝛾𝑁𝑤(𝑁)

= 𝑤(𝑁)

3.9 If 𝑖, 𝑗 ∈ 𝑆

𝑤(𝑆 ∖ {𝑖}) = 𝑤(𝑆 ∖ {𝑖, 𝑗} ∪ {𝑗}) = 𝑤(𝑆 ∖ {𝑖, 𝑗} ∪ {𝑖}) = 𝑤(𝑆 ∖ {𝑖})

𝜑𝑖(𝑤) =∑𝑆∋𝑖𝛾𝑆(𝑤(𝑆) − 𝑤(𝑆 ∖ {𝑖}))

=∑𝑆∋𝑖,𝑗

𝛾𝑆(𝑤(𝑆)− 𝑤(𝑆 ∖ {𝑖})) +∑

𝑆∋𝑖,𝑆 ∕∋𝑗𝛾𝑆(𝑤(𝑆) − 𝑤(𝑆 ∖ {𝑖}))


𝛾𝑆(𝑤(𝑆)− 𝑤(𝑆 ∖ {𝑗})) +∑𝑆 ∕∋𝑖,𝑗

𝛾𝑆(𝑤(𝑆 ∪ {𝑖})− 𝑤(𝑆))


𝛾𝑆(𝑤(𝑆)− 𝑤(𝑆 ∖ {𝑗})) +∑𝑆′ ∕∋𝑖,𝑗

𝛾𝑆′(𝑤(𝑆′ ∪ {𝑗})− 𝑤(𝑆′))


𝛾𝑆(𝑤(𝑆)− 𝑤(𝑆 ∖ {𝑗})) +∑

𝑆 ∕∋𝑖,𝑆∋𝑗𝛾𝑆(𝑤(𝑆) − 𝑤(𝑆 ∖ {𝑗}))

= 𝜑𝑗(𝑤)

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3.10 For any null player

𝑤(𝑆) − 𝑤(𝑆 ∖ {𝑖}) = 0

for every 𝑆 ⊆ 𝑁 . Consequently

𝜑𝑖(𝑤) =∑𝑆⊆𝑁

𝛾𝑆(𝑤(𝑆) − 𝑤(𝑆 ∖ {𝑖})) = 0

3.11 Every 𝑖 /∈ 𝑇 is a null player, so that

𝜑𝑖(𝑢𝑇 ) = 0 for every 𝑖 /∈ 𝑇Feasibility requires that ∑

𝑖∈𝑇𝜑𝑖(𝑢𝑇 ) =

∑𝑖∈𝑁𝜑𝑖(𝑢𝑇 ) = 1

Further, any two players in 𝑇 are substitutes, so that symmetry requires that

𝜑𝑖(𝑢𝑇 ) = 𝜑𝑗(𝑢𝑇 ) for every 𝑖, 𝑗 ∈ 𝑇Together, these conditions require that

𝜑𝑖(𝑢𝑇 ) =1

𝑡for every 𝑖 ∈ 𝑇

The Shapley value of the a T-unanimity game is

𝜑𝑖(𝑢𝑇 ) =

{1𝑡 𝑖 ∈ 𝑇0 𝑖 /∈ 𝑇

where 𝑡 = ∣𝑇 ∣.3.12 Any coalitional game can be represented as a linear combination of unanimitygames 𝑢𝑇 (Example 1.75)

𝑤 =∑𝑇

𝛼𝑇𝑢𝑇

By linearity, the Shapley value is

𝜑𝑤 = 𝜑

⎛⎝∑

𝑇⊆𝑁

𝛼𝑇𝑢𝑇

⎞⎠

=∑𝑇⊆𝑁

𝛼𝑇𝜑𝑢𝑇

and therefore for player 𝑖

𝜑𝑖𝑤 =∑𝑇⊆𝑁

𝛼𝑇𝜑𝑖𝑢𝑇

=∑𝑇⊆𝑁𝑇∋𝑖

1

𝑡𝛼𝑇

=∑𝑇⊆𝑁

1

𝑡𝛼𝑇 −

∑𝑇⊆𝑁𝑖/∈𝑇

1

𝑡𝛼𝑇

= 𝑃 (𝑁,𝑤) − 𝑃 (𝑁 ∖ {𝑖}, 𝑤)

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Using Exercise 3.8

𝑤(𝑁) =∑𝑖∈𝑁𝜑𝑖𝑤

=∑𝑖∈𝑁

(𝑃 (𝑁,𝑤) − 𝑃 (𝑁 ∖ {𝑖}, 𝑣)

)

= 𝑛𝑃 (𝑁,𝑤)−∑𝑖∈𝑁𝑃 (𝑁 ∖ {𝑖}, 𝑣)

which implies that

𝑃 (𝑁,𝑤) =1

𝑛

(𝑤(𝑁)−

∑𝑖∈𝑁𝑃 (𝑁 ∖ {𝑖}, 𝑣)

)

3.13 Choose any x ∕= 0 ∈ 𝑋 .

0𝑋 = x− x

and by additivity

𝑓(0𝑋) = 𝑓(x− x)= 𝑓(x)− 𝑓(x)

= 0𝑌

3.14 Let x1, x2 belong to 𝑋 . Then

𝑔 ∘ 𝑓(x1 + x2) = 𝑔 ∘ 𝑓(x1 + x2)

= 𝑔(𝑓(x1) + 𝑓(x2)

)= 𝑔

(𝑓(x1)

)+ 𝑔

(𝑓(x2)

)= 𝑔 ∘ 𝑓(x1) + 𝑔 ∘ 𝑓(x2)

and

𝑔 ∘ 𝑓(𝛼x) = 𝑔 (𝑓(𝛼x))

= 𝑔 (𝛼𝑓(x))

= 𝛼𝑔 (𝑓(x))

= 𝛼𝑔 ∘ 𝑓(x)

Therefore 𝑔 ∘ 𝑓 is linear.

3.15 Let 𝑆 be a subspace of 𝑋 and let y1, y2 belong to 𝑓(𝑆). Choose any x1 ∈ 𝑓−1(y1)and x2 ∈ 𝑓−1(y2). Then for 𝛼1, 𝛼2 ∈ ℜ

𝛼1x1 + 𝛼2x2 ∈ 𝑆Since 𝑓 is linear (Exercise 3.1)

𝛼1y1 + 𝛼2y2 = 𝛼1𝑓(x1) + 𝛼2𝑓(x2) = 𝑓(𝛼1x1 + 𝛼2x2) ∈ 𝑓(𝑆)

𝑓(𝑆) is a subspace.

Let 𝑇 be a subspace of 𝑌 and let x1, x2 belong to 𝑓−1(𝑇 ). Let y1 = 𝑓(x1) andy2 = 𝑓(x2). Then y1,y2 ∈ 𝑇 . For every 𝛼1, 𝛼2 ∈ ℜ

𝛼1y1 + 𝛼2y2 = 𝛼1𝑓(x1) + 𝛼2𝑓(x2) ∈ 𝑇

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Since 𝑓 is linear, this implies that

𝑓(𝛼1x1 + 𝛼2x2) = 𝛼1𝑓(x1) + 𝛼2𝑓(x2) ∈ 𝑇

Therefore

𝛼1x1 + 𝛼2x2 ∈ 𝑓−1(𝑇 )

We conclude that 𝑓−1(𝑇 ) is a subspace.

3.16 𝑓(𝑋) is a subspace of 𝑌 . rank 𝑓(𝑋) = rank 𝑌 implies that 𝑓(𝑋) = 𝑌 . 𝑓 is onto.

3.17 This is a special case of the previous exercise, since {0𝑌 } is a subspace of 𝑌 .

3.18 Assume not. That is, assume that there exist two distinct elements x1 and x2with 𝑓(x1) = 𝑓(x2). Then x1 − x2 ∕= 0𝑋 but

𝑓(x1 − x2) = 𝑓(x1)− 𝑓(x2) = 0𝑌

so that x1 − x2 ∈ kernel 𝑓 which contradicts the assumption that kernel 𝑓 = {0}.3.19 If 𝑓 has an inverse, then it is one-to-one and onto (Exercise 2.4), that is 𝑓−1(0) = 0and 𝑓(𝑋) = 𝑌 . Conversely, if kernel 𝑓 = {0} then 𝑓 is one-to-one by the previousexercise. If furthermore 𝑓(𝑋) = 𝑌 , then 𝑓 is one-to-one and onto, and therefore hasan inverse (Exercise 2.4).

3.20 Let 𝑓 be a nonsingular linear function from 𝑋 to 𝑌 with inverse 𝑓−1. Choosey1,y2 ∈ 𝑌 and let

x1 = 𝑓−1(y1)

x2 = 𝑓−1(y2)

so that

y1 = 𝑓(x1)

y2 = 𝑓(x2)

Since 𝑓 is linear

𝑓(x1 + x2) = 𝑓(x1) + 𝑓(x2) = y1 + y2

which implies that

𝑓−1(y1 + y2) = x1 + x2 = 𝑓−1(y1) + 𝑓−1(y2)

The homogeneity of 𝑓−1 can be demonstrated similarly.

3.21 Assume that 𝑓 : 𝑋 → 𝑌 and 𝑔 : 𝑌 → 𝑍 are nonsingular. Then (Exercise 3.19)

∙ 𝑓(𝑋) = 𝑌 and 𝑔(𝑌 ) = 𝑍

∙ kernel 𝑓 = {0𝑋} and kernel 𝑔 = {0𝑌 }We have previously shown (Exercise 3.14) that ℎ = 𝑔 ∘ 𝑓 : 𝑋 → 𝑍 is linear. To showthat ℎ is nonsingular, we note that

∙ ℎ(𝑋) = 𝑔 ∘ 𝑓(𝑋) = 𝑔(𝑌 ) = 𝑍

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∙ If x ∈ kernel (ℎ) then

ℎ(x) = 𝑔 (𝑓(x)) = 0

and 𝑓(x) ∈ kernel 𝑔 = {0𝑌 }. Therefore 𝑓(x) = 0𝑌 which implies that x = 0𝑋 .Thus kernel ℎ = {0𝑋}.

We conclude that ℎ is nonsingular.

Finally, let z be any point in 𝑍 and let

x1 = ℎ−1(z) = (𝑔 ∘ 𝑓)−1(z)

y = 𝑔−1(z)

x2 = 𝑓−1(y)

Then

z = ℎ(x1) = 𝑔 ∘ 𝑓(x1)

z = 𝑔(y) = 𝑔 ∘ 𝑓(x2)

which implies that x1 = x2.

3.22 Suppose 𝑓 were one-to-one. Then kernel 𝑓 = {0} ⊆ kernel ℎ and 𝑔 = ℎ ∘ 𝑓−1 is awell-defined linear function mapping 𝑓(𝑋) to 𝑌 with

𝑔 ∘ 𝑓 =(ℎ ∘ 𝑓−1) ∘ 𝑓 = ℎ

We need to show that this still holds if 𝑓 is not one-to-one. In this case, for arbitraryy ∈ 𝑓(𝑋), 𝑓−1(y) may contain more than one element. Suppose x1 and x2 are distinctelements in 𝑓−1(y). Then

𝑓(x1 − x2) = 𝑓(x1)− 𝑓(x2) = y − y = 0

so that x1 − x2 ∈ kernel 𝑓 ⊆ kernel ℎ (by assumption). Therefore

ℎ(x1)− ℎ(x2) = ℎ(x1 − x2) = 0

which implies that ℎ(x1) = ℎ(x2) for all x1,x2 ∈ 𝑓−1(y). Thus 𝑔 = ℎ∘𝑓−1 : 𝑓(𝑋)→ 𝑍is well defined even if 𝑓 is many-to-one.

To show that 𝑔 is linear, choose y1, y2 in 𝑓(𝑋) and let

x1 ∈ 𝑓−1(y1)x2 ∈ 𝑓−1(y2)

Since 𝑓(x1 + x2) = 𝑓(x1) + 𝑓(x2) = y1 + y2

x1 + x2 ∈ 𝑓−1(y1 + y2)

and

𝑔(y1 + y2) = ℎ(x1 + x2)

Therefore

𝑔(y1) + 𝑔(y2) = ℎ(x1) + ℎ(x2)

= ℎ(x1 + x2)

= 𝑔(y1 + y2)

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Similarly 𝛼x1 ∈ 𝑓−1(𝛼y1) and

𝑔(𝛼y1) = ℎ(𝛼x1)

= 𝛼ℎ(x1)

= 𝛼𝑔(y1)

We conclude that 𝑔 = ℎ ∘ 𝑓−1 is a linear function mapping 𝑓(𝑋) to 𝑍 with ℎ = 𝑔 ∘ 𝑓 .

3.23 Let y be an arbitrary element of 𝑓(𝑋) with x ∈ 𝑓−1(y). Since B is a basis for𝑋 , x can be represented as a linear combination of elements of 𝐵, that is there existsx1,x2, ..,x𝑚 ∈ 𝐵 and 𝛼1, ..., 𝛼𝑚 ∈ 𝑅 such that

x =𝑚∑𝑖=1

𝛼𝑖x𝑖

y = 𝑓(x)

= 𝑓

(∑𝑖

𝛼𝑖x𝑖

)

=∑𝑖

𝛼𝑖𝑓(x𝑖)

Since 𝑓(x𝑖) ∈ 𝑓(𝐵), we have shown that y can be written as a linear combination ofelements of 𝑓(𝐵), that is

y ∈ lin 𝐵

Since the choice of y was arbitrary, 𝑓(𝐵) spans 𝑓(𝑋), that is

lin 𝐵 = 𝑓(𝑋)

3.24 Let 𝑛 = dim𝑋 and 𝑘 = dim kernel 𝑓 . Let x1, . . . ,x𝑘 be a basis for the kernel of𝑓 . This can be extended (Exercise 1.142) to a basis 𝐵 for 𝑋 . Exercise 3.23 showed

lin 𝐵 = 𝑓(𝑋)

Since x1,x2, . . . ,x𝑘 ∈ kernel 𝑓 , 𝑓(x𝑖) = 0 for 𝑖 = 1, 2, . . . , 𝑘. This implies that{𝑓(x𝑘+1), . . . , 𝑓(x𝑛)} spans 𝑓(𝑋), that is

lin {(x𝑘+1), ..., 𝑓(x𝑛)} = 𝑓(𝑋)

To show that dim 𝑓(𝑋) = 𝑛−𝑘, we have to show that {𝑓(𝑥𝑘+1), 𝑓(𝑥𝑘+2), . . . , 𝑓(𝑥𝑛)} islinearly independent. Assume not. That is, assume there exist 𝛼𝑘+1, 𝛼𝑘+2, ..., 𝛼𝑛 ∈ 𝑅such that

𝑛∑𝑖=𝑘+1

𝛼𝑖𝑓(x𝑖) = 0

This implies that

𝑓

(𝑛∑

𝑖=𝑘+1

𝛼𝑖x𝑖

)= 0

or

x =

𝑛∑𝑖=𝑘+1

𝛼𝑖𝑥𝑖 ∈ kernel 𝑓

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This implies that x can also be expressed as a linear combination of elements in{x1, 𝑥2, ...,x𝑘}, that is there exist scalars 𝛼1, 𝛼2, . . . , 𝛼𝑘 such that

x =

𝑘∑𝑖=1

𝛼𝑖x𝑖

or

x =

𝑘∑𝑖=1

𝛼𝑖x𝑖 =

𝑛∑𝑖=𝑘+1

𝛼𝑖x𝑖

which contradicts the assumption that 𝐵 is a basis for𝑋 . Therefore {𝑓(x𝑘+1), . . . , 𝑓(x𝑛)}is a basis for 𝑓(𝑋) and therefore dim 𝑓(𝑥) = 𝑛− 𝑘. We conclude that

dim kernel 𝑓 + dim 𝑓(𝑋) = 𝑛 = dim𝑋

3.25 Equation (3.2) implies that nullity 𝑓 = 0, and therefore 𝑓 is one-to-one (Exercise3.18).

3.26 Choose some x = (𝑥1, 𝑥2, . . . , 𝑥𝑛) ∈ 𝑋 . x has a unique representation in terms ofthe standard basis (Example 1.79)

x =𝑛∑

𝑗=1

𝑥𝑗e𝑗

Let y = 𝑓(x). Since 𝑓 is linear

y = 𝑓(x) = 𝑓

⎛⎝ 𝑛∑

𝑗=1

𝑥𝑗e𝑗

⎞⎠ =

𝑛∑𝑗=1

x𝑗𝑓(e𝑗)

Each 𝑓(e𝑗) has a unique representation of the form

𝑓(e𝑗) =𝑚∑𝑖=1

𝑎𝑖𝑗e𝑖

so that

y = 𝑓(x)

=

𝑛∑𝑗=1

𝑥𝑗

(𝑚∑𝑖=1

𝑎𝑖𝑗e𝑖

)

=

𝑚∑𝑖=1

⎛⎝ 𝑛∑

𝑗=1

𝑎𝑖𝑗𝑥𝑗

⎞⎠ e𝑖

=

⎛⎜⎜⎜⎝

∑𝑛𝑗=1 𝑎1𝑗𝑥𝑗∑𝑛𝑗=1 𝑎2𝑗𝑥𝑗

...∑𝑛𝑗=1 𝑎𝑚𝑗𝑥𝑗

⎞⎟⎟⎟⎠

= 𝐴x

where

𝐴 =

⎛⎜⎜⎝𝑎11 𝑎12 . . . 𝑎1𝑛𝑎21 𝑎22 . . . 𝑎2𝑛. . . . . . . . . . . . . . . . . . . . .𝑎𝑚1 𝑎𝑚2 . . . 𝑎𝑚𝑛

⎞⎟⎟⎠

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3.27 (1 0 00 1 0

)

3.28 We must specify bases for each space. The most convenient basis for 𝐺𝑁 is theT-unanimity games. We adopt the standard basis for ℜ𝑛. With respect to these bases,the Shapley value 𝜑 is represented by the 2𝑛−1×𝑛matrix where each row is the Shapleyvalue of the corresponding T-unanimity game.

For three player games (𝑛 = 3), the matrix is⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

1 0 00 1 00 0 112

12 0

12 0 1

20 1

212

13

13

13

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

3.29 Clearly, if 𝑓 is continuous, 𝑓 is continuous at 0.

To show the converse, assume that 𝑓 : 𝑋 → 𝑌 is continuous at 0. Let (x𝑛) be a sequencewhich converges to x ∈ 𝑋 . Then the sequence (x𝑛 − x) converges to 0𝑋 and therefore𝑓(x𝑛−x)→ 0𝑌 by continuity (Exercise 2.68). By linearity, 𝑓(x𝑛)−𝑓(x) = 𝑓(x𝑛−x)→0𝑌 and therefore 𝑓(x𝑛) converges to 𝑓(x). We conclude that 𝑓 is continuous at x.

3.30 Assume that 𝑓 is bounded, that is

∥𝑓(x)∥ ≤𝑀 ∥x∥ for every x ∈ 𝑋

Then 𝑓 is Lipschitz at 0 (with Lipschitz constant 𝑀) and hence continuous (by theprevious exercise).

Conversely, assume 𝑓 is continuous but not bounded. Then, for every positive integer𝑛, there exists some x𝑛 ∈ 𝑋 such that ∥𝑓(x𝑛)∥ > 𝑛 ∥x𝑛∥ which implies that∥∥∥∥𝑓

(x𝑛

𝑛 ∥x𝑛∥)∥∥∥∥ > 1

Define

y𝑛 =x𝑛

𝑛 ∥x𝑛∥Then y𝑛 → 0 but 𝑓(y𝑛) ∕→ 0. This implies that 𝑓 is not continuous at the origin,contradicting our hypothesis.

3.31 Let {x1,x2, . . . ,x𝑛} be a basis for 𝑋 . For every x ∈ 𝑋 , there exists numbers𝛼1, 𝛼2, . . . , 𝛼𝑛 such that

x =

𝑛∑𝑖=1

𝛼𝑖x𝑖

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and

𝑓(x) =

𝑛∑𝑖=1

𝛼𝑖𝑓(x𝑖)

∥𝑓(x)∥ =

∥∥∥∥∥𝑛∑𝑖=1

𝛼𝑖𝑓(x𝑖)

∥∥∥∥∥≤

𝑛∑𝑖=1

∣𝛼𝑖∣ ∥𝑓(x𝑖)∥

≤(

𝑛max𝑖=1∥𝑓(x𝑖)∥

) 𝑛∑𝑖=1

∣𝛼𝑖∣

By Lemma 1.1, there exists a constant 𝑐 such that

𝑛∑𝑖=1

∣𝛼𝑖∣ ≤ 1

𝑐

∥∥∥∥∥𝑛∑𝑖=1

𝛼𝑖x𝑖

∥∥∥∥∥ =1

𝑐∥x∥

Combining these two inequalities

∥𝑓(x)∥ ≤𝑀 ∥x∥where 𝑀 = max𝑛𝑖=1 ∥𝑓(x𝑖)∥ /𝑐.3.32 For any x ∈ 𝑋 , let 𝑎 = ∥x∥ and define y = x/𝑎. Linearity implies that

∥𝑓∥ = supx ∕=0

∥𝑓(x)∥𝑎

= supx ∕=0∥𝑓(x/𝑎)∥ = sup

∥y∥=1∥𝑓(y)∥

3.33 ∥𝑓∥ is a norm Let 𝑓 ∈ 𝐵𝐿(𝑋,𝑌 ). Clearly

∥𝑓∥ = sup∥x∥=1

∥𝑓(x)∥ ≥ 0

Further, for every 𝛼 ∈ ℜ,

∥𝛼𝑓∥ = sup∥x∥=1

∥𝛼𝑓(x)∥ = ∣𝛼∣ ∥𝑓∥

Finally, for every 𝑔 ∈ 𝐵𝐿(𝑋,𝑌 ),

∥𝑓 + 𝑔∥ = sup∥x∥=1

∥𝑓(x) + 𝑔(x)∥ ≤ sup∥x∥=1

∥𝑓(x)∥ + sup∥x∥=1

∥𝑔(x)∥ ≤ ∥𝑓∥+ ∥𝑔∥

verifying the triangle inequality. There ∥𝑓∥ is a norm.

𝐵𝐿(𝑋,𝑌 ) is a linear space Let 𝑓, 𝑔 ∈ 𝐵𝐿(𝑋,𝑌 ). Since 𝐵𝐿(𝑋,𝑌 ) ⊆ 𝐿(𝑋,𝑌 ), 𝑓 + 𝑔is linear, that is 𝑓+𝑔 ∈ 𝐿(𝑋,𝑌 ) (Exercise 3.2). Similarly, 𝛼𝑓 ∈ 𝐿(𝑋,𝑌 ) for every𝛼 ∈ ℜ. Further, by the triangle inequality ∥𝑓 + 𝑔∥ ≤ ∥𝑓∥+ ∥𝑔∥ and therefore forevery x ∈ 𝑋

∥(𝑓 + 𝑔)(x)∥ ≤ ∥𝑓 + 𝑔∥ ∥x∥ ≤ (∥𝑓∥+ ∥𝑔∥) ∥x∥Therefore 𝑓 + 𝑔 ∈ 𝐵𝐿(𝑋,𝑌 ). Similarly

∥(𝛼𝑓)(x)∥ ≤ (∣𝛼∣ ∥𝑓∥) ∥x∥so that 𝛼𝑓 ∈ 𝐵𝐿(𝑋,𝑌 ).

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𝐵𝐿(𝑋,𝑌 ) is complete with this norm Let (𝑓𝑛) be a Cauchy sequence in 𝐵𝐿(𝑋,𝑌 ).

For every x ∈ 𝑋∥𝑓𝑛(x) − 𝑓𝑚(x)∥ ≤ ∥𝑓𝑛 − 𝑓𝑚∥ ∥x∥

Therefore (𝑓𝑛(x)) is a Cauchy sequence in 𝑌 , which converges since 𝑌 is complete.Define the function 𝑓 : 𝑋 → 𝑌 by 𝑓(x) = lim𝑛→∞ 𝑓𝑛(x).

𝑓 is linear since

𝑓(x1 + x2) = lim 𝑓𝑛(x1 + x2) = lim 𝑓𝑛(x1) + lim 𝑓𝑛(x2) = 𝑓(x1) + 𝑓(x2)

and

𝑓(𝛼x) = lim 𝑓𝑛(𝛼x) = 𝛼 lim 𝑓𝑛(x) = 𝛼𝑓(x)

To show that 𝑓 is bounded, we observe that

∥𝑓(x)∥ =∥∥∥lim

𝑛𝑓𝑛(x)

∥∥∥ = lim𝑛∥𝑓𝑛(x)∥ ≤ sup

𝑛∥𝑓𝑛(x)∥ ≤ sup

𝑛∥𝑓𝑛∥ ∥x∥

Since (𝑓𝑛) is a Cauchy sequence, (𝑓𝑛) is bounded (Exercise 1.100), that is thereexists 𝑀 such that ∥𝑓𝑛∥ ≤𝑀 . This implies

∥𝑓(x)∥ ≤ sup𝑛∥𝑓𝑛∥ ∥x∥ ≤𝑀 ∥x∥

Thus, 𝑓 is bounded.

To complete the proof, we must show 𝑓𝑛 → 𝑓 , that is ∥𝑓𝑛 − 𝑓∥ → 0. Since (𝑓𝑛)is a Cauchy sequence, for every 𝜖 > 0, there exists 𝑁 such that ∥𝑓𝑛 − 𝑓𝑚∥ ≤ 𝜖for every 𝑛,𝑚 ≥ 𝑁 and consequently

∥𝑓𝑛(x) − 𝑓𝑚(x)∥ = ∥(𝑓𝑛 − 𝑓𝑚)(x)∥ ≤ 𝜖 ∥x∥

Letting 𝑚 go to infinity,

∥𝑓𝑛(x) − 𝑓(x)∥ = ∥(𝑓𝑛 − 𝑓)(x)∥ ≤ 𝜖 ∥x∥

for every x ∈ 𝑋 and 𝑛 ≥ 𝑁 and therefore

∥𝑓𝑛 − 𝑓∥ = sup∥x∥=1

{𝑓𝑛 − 𝑓)(x)} ≤ 𝜖

for every 𝑛 ≥ 𝑁 .

3.34 1. Since 𝑋 is finite-dimensional, 𝑆 is compact (Proposition 1.4). Since 𝑓 iscontinuous, 𝑓(𝑆) is a compact set in 𝑌 (Exercise 2.3). Since 0𝑋 /∈ 𝑆, 0𝑌 =𝑓(0𝑋) /∈ 𝑓(𝑆).

2. Consequently,(𝑓(𝑆)

)𝑐is an open set containing 0𝑌 . It contains an open ball

𝑇 ⊆ (𝑓(𝑆)

)𝑐around 0𝑌 .

3. Let y ∈ 𝑇 and choose any x ∈ 𝑓−1(y) and consider y/ ∥x∥. Since 𝑓 is linear,

y

∥x∥ =𝑓(x)

∥x∥ = 𝑓

(x

∥x∥)∈ 𝑓(𝑆)

and therefore y/ ∥x∥ /∈ 𝑇 since 𝑇 ∩ 𝑓(𝑆) = ∅.

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Suppose that y /∈ 𝑓(𝐵). Then ∥x∥ ≥ 1 and therefore

y ∈ 𝑇 =⇒ y

∥x∥ ∈ 𝑇

since 𝑇 is convex. This contradiction establishes that y ∈ 𝑓(𝐵) and therefore𝑇 ⊆ 𝑓(𝐵). We conclude that 𝑓(𝐵) contains an open ball around 0𝑌 .

4. Let 𝑆 be any open set in 𝑋 . We need to show that 𝑓(𝑆) is open in 𝑌 . Chooseany y ∈ 𝑓(𝑆) and x ∈ 𝑓−1(y). Then x ∈ 𝑆 and, since 𝑆 is open, there existssome 𝑟 > 0 such that 𝐵𝑟(x) ⊆ 𝑆. Now 𝐵𝑟(x) = x+ 𝑟𝐵 and

𝑓(𝐵𝑟(x)) = y + 𝑟𝑓(𝐵) ⊆ 𝑓(𝑆)

by linearity. As we have just shown, there exists an open ball T about 0𝑌 suchthat 𝑇 ⊆ 𝑓(𝐵). Let 𝑇 (x) = y + 𝑟𝑇 . 𝑇 (x) is an open ball about y. Since𝑇 ⊆ 𝑓(𝐵), 𝑇 (x) = y + 𝑟𝑇 ⊆ 𝑓(𝐵𝑟(x)) ⊆ 𝑓(𝑆). This implies that 𝑓(𝑆) is open.Since 𝑆 was an arbitrary open set, 𝑓 is an open map.

5. Exercise 2.69.

3.35 𝑓 is linear

𝑓(𝛼+ 𝛽) =

𝑛∑𝑖=1

(𝛼𝑖 + 𝛽𝑖)x𝑖 =

𝑛∑𝑖=1

𝛼𝑖x𝑖 +

𝑛∑𝑖=1

𝛽𝑖x𝑖 = 𝑓(𝛼) + 𝑓(𝛽)

Similarly for every 𝑡 ∈ ℜ

𝑓(𝑡𝛼) = 𝑡𝑛∑𝑖=1

𝛼𝑖x𝑖 = 𝑡𝑓(𝑡𝛼)

𝑓 is one-to-one Exercise 1.137.

𝑓 is onto By definition of a basis lin {x1,x2, . . . ,x𝑛} = 𝑋

𝑓 is continuous Exercise 3.31

𝑓 is an open map Proposition 3.2

3.36 𝑓 is bounded and therefore there exists 𝑀 such that ∥𝑓(x)∥ ≤ 𝑀 ∥x∥. Similarly,𝑓−1 is bounded and therefore there exists 𝑚 such that for every x

𝑓−1(y) ≤ 1

𝑚∥y∥

where y = 𝑓(x). This implies

𝑚 ∥x∥ ≤ ∥𝑓(x)∥

and therefore for every x ∈ 𝑋 .

𝑚 ∥x∥ ≤ ∥𝑓(x)∥ ≤𝑀 ∥x∥

By the linearity of 𝑓 ,

𝑚 ∥x1 − x2∥ ≤ ∥𝑓(x1 − x2)∥ = ∥𝑓(x1)− 𝑓(x2)∥ ≤𝑀 ∥x1 − x2∥

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3.37 For any function, continuity implies closed graph (Exercise 2.70). To show the con-verse, assume that 𝐺 = graph(𝑓) is closed. 𝑋×𝑌 with norm ∥(x,y)∥ = max{∥x∥ , ∥y∥}is a Banach space (Exercise 1.209). Since 𝐺 is closed, 𝐺 is complete. Also, 𝐺 is a sub-space of 𝑋 × 𝑌 . Consequently, 𝐺 is a Banach space in its own right.

Consider the projection ℎ : 𝐺 → 𝑋 defined by ℎ(x, 𝑓(x)) = x. Clearly ℎ is linear,one-to-one and onto with

ℎ−1(x) = (x, 𝑓(x))

It is also bounded since

∥ℎ(x, 𝑓(x))∥ = ∥x∥ ≤ ∥(x, 𝑓(x)∥By the open mapping theorem, ℎ−1 is bounded. For every x ∈ 𝑋

∥𝑓(x)∥ ≤ ∥(x, 𝑓(x))∥ =∥∥ℎ−1(x)∥∥ ≤ ∥∥ℎ−1∥∥ ∥x∥

We conclude that 𝑓 is bounded and hence continuous.

3.38 𝑓(1) = 5, 𝑓(2) = 7 but

𝑓(1 + 2) = 𝑓(3) = 9 ∕= 𝑓(1) + 𝑓(2)

Similarly

𝑓(3× 2) = 𝑓(6) = 15 ∕= 3× 𝑓(2)

3.39 Assume 𝑓 is affine. Let y = 𝑓(0) and define

𝑔(x) = 𝑓(x)− y

𝑔 is homogeneous since for every 𝛼 ∈ ℜ𝑔(𝛼x) = 𝑔(𝛼x+ (1− 𝛼)0)

= 𝑓(𝛼x+ (1− 𝛼)0) − y= 𝛼𝑓(x) + (1− 𝛼)𝑓(0)− y= 𝛼𝑓(x) + (1− 𝛼)y − y= 𝛼𝑓(x)− 𝛼y= 𝛼(𝑓(𝑥) − y)= 𝛼𝑔(x)

Similarly for any x1,x2 ∈ 𝑋𝑔(𝛼x1 + (1− 𝛼)x2) = 𝑓(𝛼x1 + (1− 𝛼)x2)− 𝑦

= 𝛼𝑓(x1) + (1 − 𝛼)𝑓(x2)− 𝑦

Therefore, for 𝛼 = 1/2

𝑔(1

2x1 +

1

2x2) =

1

2𝑓(x1) +

1

2𝑓(x2)− 𝑦

=1

2(𝑓(x1)− 𝑦) +

1

2(𝑓(x2)− 𝑦)

=1

2𝑔(x1) +

1

2𝑔(x2)

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Since 𝑔 is homogeneous

𝑔(x1 + x2) = 𝑔(x1) + 𝑔(x2)

which shows that 𝑔 is additive and hence linear.

Conversely if

𝑓(x) = 𝑔(x) + y

with 𝑔 linear

𝑓(𝛼x1 + (1− 𝛼)x2) = 𝛼𝑔(x1) + (1− 𝛼)𝑔(x2) + 𝑦

= 𝛼𝑔(x1) + 𝑦 + (1− 𝛼)𝑔(x2) + 𝑦

= 𝛼𝑓(x1) + (1− 𝛼)𝑓(x2)

3.40 Let 𝑆 be an affine subset of 𝑋 and let y1, y2 belong to 𝑓(𝑆). Choose any x1 ∈𝑓−1(y1) and x2 ∈ 𝑓−1(y2). Then for any 𝛼 ∈ ℜ

𝛼x1 + (1 − 𝛼)x2 ∈ 𝑆

Since 𝑓 is affine

𝛼y1 + (1− 𝛼)y2 = 𝛼𝑓(x1) + (1− 𝛼)𝑓(x2) = 𝑓(𝛼x1 + (1− 𝛼)x2) ∈ 𝑓(𝑆)

𝑓(𝑆) is an affine set.

Let 𝑇 be an affine subset of 𝑌 and let x1, x2 belong to 𝑓−1(𝑇 ). Let y1 = 𝑓(x1) andy2 = 𝑓(x2). Then y1,y2 ∈ 𝑇 . For every 𝛼 ∈ ℜ

𝛼y1 + (1− 𝛼)y2 = 𝛼𝑓(x1) + (1− 𝛼)𝑓(x2) ∈ 𝑇

Since 𝑓 is affine, this implies that

𝑓(𝛼x1 + (1− 𝛼)x2) = 𝛼𝑓(x1) + (1− 𝛼)𝑓(x2) ∈ 𝑇

Therefore

𝛼x1 + (1− 𝛼)x2 ∈ 𝑓−1(𝑇 )

We conclude that 𝑓−1(𝑇 ) is an affine set.

3.41 For any y1,y2 ∈ 𝑓(𝑆), choose x1,x2 ∈ 𝑆 such that y𝑖 = 𝑓(x𝑖). Since 𝑆 is convex,𝛼x1 + (1− 𝛼)x2 ∈ 𝑆 and therefore

𝑓(𝛼x1 + (1− 𝛼)x2) = 𝛼𝑓(x1) + (1 − 𝛼)𝑓(x2)

= 𝛼y1 + (1 − 𝛼)y2 ∈ 𝑓(𝑆)

Therefore 𝑓(𝑆) is convex.

3.42 Suppose otherwise that y is not efficient. Then there exists another productionplan y′ ∈ 𝑌 such that y′ ≥ y. Since p > 0, this implies that py′ > py, contradictingthe assumption that y maximizes profit.

3.43 The random variable 𝑋 can be represented as the sum

𝑋 =∑𝑠∈𝑆𝑋(𝑠)𝜒{𝑠}

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where 𝜒{𝑠} is the indicator function of the set {𝑠}. Since 𝐸 is linear

𝐸(𝑋) =∑𝑠∈𝑆𝑋(𝑠)𝐸(𝜒{𝑠})

=∑𝑠∈𝑆𝑝𝑆𝑋(𝑠)

since𝐸(𝜒{𝑠} = 𝑃 ({𝑠}) = 𝑝𝑠 ≥ 0. For the random variable𝑋 = 1,𝑋(𝑠) = 1 for every 𝑠 ∈𝑆 and

𝐸(1) =∑𝑠∈𝑆𝑝𝑆 = 1

3.44 Let 𝑥1, 𝑥2 ∈ 𝐶[0, 1]. Recall that addition in C[0,1] is defined by

(𝑥1 + 𝑥2)(𝑡) = 𝑥1(𝑡) + 𝑥2(𝑡)

Therefore

𝑓(𝑥1 + 𝑥2) = (𝑥1 + 𝑥2)(1/2) = 𝑥1(1/2) + 𝑥2(1/2) = 𝑓(𝑥1) + 𝑓(𝑥2)

Similarly

𝑓(𝛼𝑥1) = (𝛼𝑥1)(1/2) = 𝛼𝑥1(1/2) = 𝛼𝑓(𝑥1)

3.45 Assume that x∗ = x∗1+x∗2+ ⋅ ⋅ ⋅+x∗𝑛 maximizes 𝑓 over 𝑆. Suppose to the contrarythat there exists y𝑗 ∈ 𝑆𝑗 such that 𝑓(y𝑗) > 𝑓(x∗𝑗 ). Then y = x∗1 +x∗2+ ⋅ ⋅ ⋅+y𝑗 + ⋅ ⋅ ⋅+x∗𝑛 ∈ 𝑆 and

𝑓(y) =∑𝑖∕=𝑗𝑓(x∗𝑖 ) + 𝑓(y𝑖) >

∑𝑖

𝑓(x∗𝑖 ) = 𝑓(x∗)

contradicting the assumption at 𝑓 is maximized at x∗.

Conversely, assume

𝑓(x∗𝑖 ) ≥ 𝑓(x𝑖) for every x𝑖 ∈ 𝑆𝑖for every 𝑖 = 1, 2, . . . , 𝑛. Summing

𝑓(x∗) = 𝑓(∑x∗𝑖 ) =

∑𝑓(𝑥∗𝑖 ) ≥

∑𝑓(x𝑖) = 𝑓(

∑x𝑖) = 𝑓(x) for every x ∈ 𝑆

x∗ = x∗1 + x∗2 + ⋅ ⋅ ⋅+ x∗𝑛 maximizes 𝑓 over 𝑆.

3.46 1. Assume (𝑥𝑡) is a sequence in 𝑙1 with 𝑠 =∑∞

𝑡=1 ∣𝑥𝑗 ∣ < ∞. Let (𝑠𝑡) denotethe sequence of partial sums

𝑠𝑡 =𝑡∑

𝑗=1

∣𝑥𝑗 ∣

Then (𝑠𝑡) is a bounded monotone sequence in ℜ𝑛 which converges to 𝑠. Con-sequently, (𝑠𝑡) is a Cauchy sequence. For every 𝜖 > 0 there exists an 𝑁 suchthat

𝑚+𝑘∑𝑛=𝑚

∣𝑥𝑡∣ < 𝜖

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for every 𝑚 ≥ 𝑁 and 𝑘 ≥ 0. Letting 𝑘 = 0

∣𝑥𝑡∣ < 𝜖 for every 𝑛 ≥ 𝑁We conclude that 𝑥𝑡 → 0 so that (𝑥𝑡) ∈ 𝑐0. This establishes 𝑙1 ⊆ 𝑐0.To see that the inclusion is strict, that is 𝑙1 ⊂ 𝑐0, we observe that the sequence(1/𝑛) = (1, 1/2, 1/3, . . . ) converges to zero but that since

∞∑𝑛=1

∣∣∣∣ 1

𝑛

∣∣∣∣ = 1 +1

2+

1

3+ =∞

(1/𝑛) /∈ 𝑙1.Every convergent sequence is bounded (Exercise 1.97). Therefore 𝑐0 ⊂ 𝑙∞.

2. Clearly, every sequence (𝑝𝑡) ∈ 𝑙1 defines a linear functional 𝑓 ∈ 𝑐′0 given by

𝑓(x) =

∞∑𝑛=1

𝑝𝑡𝑥𝑡

for every x = (𝑥𝑡) ∈ 𝑐0. To show that 𝑓 is bounded we observe that every(𝑥𝑡) ∈ 𝑐0 is bounded and consequently

∣𝑓(x)∣ ≤∞∑𝑛=1

∣𝑝𝑡∣ ∣𝑥𝑡∣ ≤ ∥(𝑥𝑡)∥∞∞∑𝑛=1

∣𝑝𝑡∣ = ∥(𝑝𝑡)∥1 ∥(𝑥𝑡)∥∞

Therefore 𝑓 ∈ 𝑐∗0.To show the converse, let e𝑡 denote the unit sequences

e1 = (1, 0, 0, . . . )

e2 = (0, 1, 0, . . . )

e3 = (0, 0, 1, . . . )

{e1, e2, e3, . . . , } form a basis for 𝑐0. Then every sequence (𝑥𝑡) ∈ 𝑐0 has a uniquerepresentation

(𝑥𝑡) =∞∑𝑛=1

𝑥𝑡e𝑡

Let 𝑓 ∈ 𝑐∗0 be a continuous linear functional on 𝑐0. By continuity and linearity

𝑓(x) =

∞∑𝑛=1

𝑥𝑡𝑓(e𝑡)

Let

𝑝𝑡 = 𝑓(e𝑡)

so that

𝑓(x) =

∞∑𝑛=1

𝑝𝑡𝑥𝑡

Every linear function is determined by its action on a basis (Exercise 3.23).

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We need to show that the sequence (𝑝𝑡) ∈ 𝑙1. For any 𝑁 , consider the sequencex𝑡 = (𝑥1, 𝑥2, . . . , 𝑥𝑡, 0, 0, . . . ) where

𝑥𝑡 =

⎧⎨⎩

0 𝑝𝑡 = 0 or 𝑛 ≥ 𝑁∣𝑝𝑡∣𝑝𝑡

otherwise

Then (x𝑡) ∈ 𝑐0, ∥x𝑡∥∞ = 1 and

𝑓(x𝑡) =

𝑡∑𝑛=1

𝑝𝑡𝑥𝑡 =

𝑡∑𝑛=1

∣𝑝𝑡∣

Since 𝑓 ∈ 𝑐∗0, 𝑓 is bounded and therefore

𝑓(x𝑡) ≤ ∥𝑓∥ ∥x𝑡∥ = ∥𝑓∥ <∞and therefore

𝑡∑𝑛=1

∣𝑝𝑡∣ <∞ for every 𝑁 = 1, 2, . . .

Consequently

∞∑𝑛=1

∣𝑝𝑡∣ = sup𝑁

𝑡∑𝑛=1

∣𝑝𝑡∣ ≤ ∥𝑓∥ <∞

We conclude that (𝑝𝑡) ∈ 𝑙1 and therefore 𝑐∗0 = 𝑙1

3. Similarly, every sequence (𝑝𝑡) ∈ 𝑙∞ defines a linear functional 𝑓 on 𝑙1 given by

𝑓(x) =∞∑𝑛=1

𝑝𝑡𝑥𝑡

for every x = (𝑥𝑡) ∈ 𝑙1. Moreover 𝑓 is bounded since

∣𝑓(x)∣ ≤∞∑𝑛=1

∣𝑝𝑡∣ ∣𝑥𝑡∣ ≤ ∥(𝑝𝑡)∥∞∑𝑛=1

∣𝑥𝑡∣ <∞

for every x = (𝑥𝑡) ∈ 𝑙1 Again, given any linear functional 𝑓 ∈ 𝑙∗1 , let 𝑝𝑡 = 𝑓(e𝑡)where e𝑡 is the 𝑛 unit sequence. Then 𝑓 has the representation

𝑓(x) =

∞∑𝑛=1

𝑝𝑡𝑥𝑡

To show that (𝑝𝑡) ∈ 𝑙∞, for𝑁 = 1, 2, . . . , consider the sequence x𝑡 = (0, 0, . . . , 𝑥𝑡, 0, 0, . . . )where

𝑥𝑡 =

⎧⎨⎩∣𝑝𝑡∣𝑝𝑡

𝑛 = 𝑁 and 𝑝𝑡 ∕= 0

0 otherwise

Then x𝑡 ∈ 𝑙1, ∥x𝑡∥1 = 1 and

𝑓(x𝑡) = ∣𝑝𝑡∣Since 𝑓 ∈ 𝑙∗1 , 𝑓 is bounded and therefore∣∣𝑝𝑁 ∣∣ = 𝑓(x𝑁 ) ≤ ∥𝑓∥ ∥x𝑛∥ = ∥𝑓∥for every 𝑁 . Consequently (𝑝𝑁 ) ∈ 𝑙∞. We conclude that 𝑙∗1 = 𝑙∞

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3.47 By linearity

𝜑(𝑥, 𝑡) = 𝜑(𝑥, 0) + 𝜑(0, 𝑡)

= 𝜑(𝑥, 0) + 𝜑(0, 1)𝑡

Considered as a function of 𝑥, 𝜑(𝑥, 0) is a linear functional on 𝑋 . Define

𝑔(𝑥) = 𝜑(𝑥, 0)

𝛼 = 𝜑(0, 1)

Then

𝜑(𝑥, 𝑡) = 𝑔(𝑥) + 𝛼𝑡

3.48 Suppose

𝑚∩𝑗=1

kernel 𝑔𝑗 ⊆ kernel 𝑓

Define the function 𝐺 : 𝑋 → ℜ𝑛 by

𝐺(x) = (𝑔1(x), 𝑔2(x), . . . , 𝑔𝑚(x))

Then

kernel 𝐺 = {x ∈ 𝑋 : 𝑔𝑗(x) = 0, 𝑗 = 1, 2, . . .𝑚 }

=

𝑚∩𝑗=1

kernel 𝑔𝑗

⊆ kernel 𝑓

𝑓 : 𝑋 → ℜ and𝐺 : 𝑋 → ℜ𝑛. By Exercise 3.22, there exists a linear function𝐻 : ℜ𝑛 → ℜsuch that 𝑓 = 𝐻 ∘𝐺. That is, for every 𝑥 ∈ 𝑋

𝑓(x) = 𝐻 ∘𝐺(x) = 𝐻(𝑔1(x), 𝑔2(x), . . . , 𝑔𝑚(x))

Let 𝛼𝑗 = 𝐻(e𝑗) where e𝑗 is the 𝑗-th unit vector in ℜ𝑚. Since every linear mapping isdetermined by its action on a basis, we must have

𝑓(x) = 𝛼1𝑔1(x) + 𝛼2𝑔2(x) + ⋅ ⋅ ⋅+ 𝛼𝑚𝑔𝑚(x) for every 𝑥 ∈ 𝑋That is

𝑓 ∈ lin 𝑔1, 𝑔2, . . . , 𝑔𝑚

Conversely, suppose

𝑓 ∈ lin 𝑔1, 𝑔2, . . . , 𝑔𝑚

That is

𝑓(x) = 𝛼1𝑔1(x) + 𝛼2𝑔2(x) + ⋅ ⋅ ⋅+ 𝛼𝑚𝑔𝑚(x) for every 𝑥 ∈ 𝑋For every x ∈ ∩𝑚

𝑗=1 kernel 𝑔𝑗 , 𝑔𝑗(x) = 0, 𝑗 = 1, 2, . . . ,𝑚 and therefore 𝑓(x) = 0.Therefore 𝑥 ∈ kernel 𝑓 . That is

𝑚∩𝑗=1

kernel 𝑔𝑗 ⊆ kernel 𝑓

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3.49 Let 𝐻 be a hyperplane in 𝑋 . Then there exists a unique subspace 𝑉 such that𝐻 = x0 + 𝑉 for some x0 ∈ 𝐻 (Exercise 1.153). There are two cases to consider.

Case 1: x0 /∈ 𝑉 . For every x ∈ 𝑋 , there exists unique 𝛼x ∈ ℜ such

x = 𝛼xx0 + 𝑣 for some 𝑣 ∈ 𝑉

Define 𝑓(x) = 𝛼x. Then 𝑓 : 𝑋 → ℜ. It is straightforward to show that 𝑓 is linear.

Since 𝐻 = x0 + 𝑉 , 𝛼x = 1 if and only if x ∈ 𝐻 . Therefore

𝐻 = {x ∈ 𝑋 : 𝑓(x) = 1 }

Case 2: x0 ∈ 𝑉 . In this case, choose some x1 /∈ 𝑉 . Again, for every x ∈ 𝑋 , thereexists a unique 𝛼x ∈ ℜ such

x = 𝛼xx1 + 𝑣 for some 𝑣 ∈ 𝑉

and 𝑓(x) = 𝛼x is a linear functional on 𝑋 . Furthermore x0 ∈ 𝑉 implies 𝐻 = 𝑉(Exercise 1.153) and therefore 𝑓(x) = 0 if and only if x ∈ 𝐻 . Therefore

𝐻 = {x ∈ 𝑋 : 𝑓(x) = 0 }

Conversely, let 𝑓 be a nonzero linear functional in 𝑋 ′. Let 𝑉 = kernel 𝑓 and choosex0 ∈ 𝑓−1(1). (This is why we require 𝑓 ∕= 0). For any x ∈ 𝑋

𝑓 (x− 𝑓(x)x0) = 𝑓(x)− 𝑓(x)× 1 = 0

so that x − 𝑓(x)x0 ∈ 𝑉 . That is, x = 𝑓(x)x0 + 𝑣 for some 𝑣 ∈ 𝑉 . Therefore,𝑋 = lin (x0, 𝑉 ) so that 𝑉 is a maximal proper subspace.

For any 𝑐 ∈ ℜ, let x1 ∈ 𝑓−1(𝑐). Then, for every x ∈ 𝑓−1(𝑐), 𝑓(x− x1) = 0 and

{x : 𝑓(x) = 𝑐} = {x : 𝑓(x− x1) = 0 } = x1 + 𝑉

which is a hyperplane.

3.50 By the previous exercise, there exists a linear functional 𝑔 such that

𝐻 = { 𝑥 ∈ 𝑋 : 𝑓(𝑥) = 𝑐 }

for some 𝑐 ∈ ℜ. Since 0 /∈ 𝐻 , 𝑐 ∕= 0. Without loss of generality, we can assume that𝑐 = 1. (Otherwise, take the linear functional 1𝑐𝑓).

To show that 𝑓 is unique, assume that 𝑔 is another linear functional with

𝐻 = {x : 𝑓(𝑥) = 1} = {x : 𝑔(𝑥) = 1 }

Then

𝐻 ⊆ {x : 𝑓(𝑥)− 𝑔(𝑥) = 0 }

Since 𝐻 is a maximal subset, 𝑋 is the smallest subspace containing 𝐻 . Therefore𝑓(𝑥) = 𝑔(𝑥) for every 𝑥 ∈ 𝑋 .

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3.51 By Exercise 3.49, there exists a linear functional 𝑓 such that

𝐻 = { 𝑥 ∈ 𝑋 : 𝑓(𝑥) = 0 }

Since x0 /∈ 𝐻 , 𝑓(x0) ∕= 0. Without loss of generality, we can normalize so that 𝑓(x0) =1. (If 𝑓(x0) = 𝑐 ∕= 1, then the linear functional 𝑓 ′ = 1/c𝑓 has 𝑓 ′(x0) = 1 and kernel 𝑓 ′ =𝐻 .)

To show that 𝑓 is unique, suppose that 𝑔 is another linear functional with kernel 𝑔 = 𝐻and 𝑔(x0) = 1. For any x ∈ 𝑋 , there exists 𝛼 ∈ ℜ such that

x = 𝛼x0 + v

with 𝑣 ∈ 𝐻 (Exercise 1.153). Since 𝑓(v) = 𝑔(v) = 0 and 𝑓(x0) = 𝑔(x0) = 1

𝑔(x) = 𝑔(𝛼x0 + v) = 𝛼𝑔(𝑥0) = 𝛼𝑓(x0) = 𝑓(𝛼x0 + v) = 𝑓(x)

3.52 Assume 𝑓 = 𝜆𝑔, 𝜆 ∕= 0. Then

𝑓(𝑥) = 0 ⇐⇒ 𝑔(𝑥) = 0

Conversely, let 𝐻 = 𝑓−1(0) = 𝑔−1(0). If 𝐻 = 𝑋 , then 𝑓 = 𝑔 = 0. Otherwise, 𝐻is a hyperplane containing 0. Choose some x0 /∈ 𝐻 . Every x ∈ 𝑋 has a uniquerepresentation x = 𝛼x0 + v with v ∈ 𝐻 (Exercise 1.153) and

𝑓(x) = 𝛼𝑓(x0)

𝑔(x) = 𝛼𝑔(x0)

Let 𝜆 = 𝑓(x0)/𝑔(x0) so that 𝑓(x0) = 𝜆𝑔(x0). Substituting

𝑓(x) = 𝛼𝑓(x0) = 𝛼𝜆𝑔(x0) = 𝜆𝑔(x)

3.53 𝑓 continuous implies that the set { 𝑥 ∈ 𝑋 : 𝑓(𝑥) = 𝑐 } = 𝑓−1(𝑐) is closed for every𝑐 ∈ ℜ (Exercise 2.67). Conversely, let 𝑐 = 0 and assume that 𝐻 = { 𝑥 ∈ 𝑋 : 𝑓(𝑥) = 0 }is closed. There exists x0 ∕= 0 such that 𝑋 = lin {𝑥0, 𝐻} (Exercise 1.153). Let x𝑛 → xbe a convergent sequence in 𝑋 . Then there exist 𝛼𝑛, 𝛼 ∈ ℜ and v𝑛, 𝑣 ∈ 𝐻 such thatx𝑛 = 𝛼𝑛x0 + v𝑛, x = 𝛼x0 + v and

∥x𝑛 − x∥ = ∥𝛼𝑛x0 + v𝑛 − 𝛼x0 + v∥= ∥𝛼𝑛x0 − 𝛼x0 + v𝑛 − v∥≤ ∣𝛼𝑛 − 𝛼∣ ∥x0∥+ ∥v𝑛 − v∥→ 0

which implies that 𝛼𝑛 → 𝛼. By linearity

𝑓(x𝑛) = 𝛼𝑛𝑓(x0) + 𝑓(v𝑛) = 𝛼𝑛𝑓(x0)

since v𝑛 ∈ 𝐻 and therefore

𝑓(x𝑛) = 𝛼𝑛𝑓(x0)→ 𝛼𝑓(x0) = 𝑓(x)

𝑓 is continuous.

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3.54

𝑓(x+ x′,y) =

𝑚∑𝑖=1

𝑛∑𝑗=1

𝑎𝑖𝑗(𝑥𝑖 + 𝑥′𝑖)𝑦𝑗

=𝑚∑𝑖=1

𝑛∑𝑗=1

𝑎𝑖𝑗𝑥𝑖𝑦𝑗 +𝑚∑𝑖=1

𝑛∑𝑗=1

𝑎𝑖𝑗𝑥′𝑖𝑦𝑗

= 𝑓(x,y) + 𝑓(x′,y)

Similarly, we can show that

𝑓(x,y + y′) = 𝑓(x,y) + 𝑓(x,y′)

and

𝑓(𝛼x,y) = 𝛼𝑓(x,y) = 𝑓(x, 𝛼y) for every 𝛼 ∈ ℜ3.55 Let x1,x2, . . . ,x𝑚 be a basis for 𝑋 and y1,y2, . . . ,y𝑛 be a basis for 𝑌 . Let thenumbers 𝑎𝑖𝑗 represent the action of 𝑓 on these bases, that is

𝑎𝑖𝑗 = 𝑓(x𝑖,y𝑗) 𝑖 = 1, 2, . . . ,𝑚, 𝑗 = 1, 2, . . . , 𝑛

and let 𝐴 be the 𝑚× 𝑛 matrix of numbers 𝑎𝑖𝑗 .

Choose any x ∈ 𝑋 and y ∈ 𝑌 and let their representations in terms of the bases be

x =

𝑚∑𝑖=1

𝛼𝑖x𝑖 and y =

𝑛∑𝑗=1

𝛽𝑖y𝑗

respectively. By the bilinearity of 𝑓

𝑓(x,y) = 𝑓(∑𝑖

𝛼𝑖x𝑖,∑𝑗

𝛼𝑗y𝑗)

=∑𝑖

𝛼𝑖𝑓(x𝑖,∑𝑗

𝛼𝑗y𝑗)

=∑𝑖

𝛼𝑖∑𝑗

𝛼𝑗𝑓(x𝑖,y𝑗)

=∑𝑖

𝛼𝑖∑𝑗

𝛼𝑗𝑎𝑖𝑗

=∑𝑖

𝛼𝑖𝐴y

= x′𝐴y

3.56 Every y ∈ 𝑋 ′ is a linear functional on 𝑋 . Hence

y(x + x′) = y(x) + y(x′)y(𝛼x) = 𝛼y(x)

and therefore

𝑓(x+ x′,y) = y(x + x′) = y(x) + y(x′) = 𝑓(x,y) + 𝑓(x′,y)𝑓(𝛼x,y) = y(𝛼x) = 𝛼y(x) = 𝛼𝑓(x,y)

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In the dual space 𝑋 ′

(y + y′)(x) ≡ y(x) + y′(x)(𝛼y)(x) ≡ 𝛼y(x)

and therefore

𝑓(x,y + y′) = (y + y′)(x) = y(x) + y′(x) = 𝑓(x,y) + 𝑓(x,y′)𝑓(x, 𝛼y) = (𝛼y)(x) = 𝛼y(x) = 𝛼𝑓(x,y)

3.57 Assume 𝑓1, 𝑓2 ∈ 𝐵𝑖𝐿(𝑋 × 𝑌, 𝑍). Define the mapping 𝑓1 + 𝑓2 : 𝑋 × 𝑌 → 𝑍 by

(𝑓1 + 𝑓2)(x,y) = 𝑓1(x,y) + 𝑓2(x,y)

We have to confirm that 𝑓1 + 𝑓2 is bilinear, that is

(𝑓1 + 𝑓2)(x1 + x2,y) = 𝑓1(x1 + x2,y) + 𝑓2(x1 + x2,y)

= 𝑓1(x1,y) + 𝑓1(x2,y) + 𝑓2(x1,y) + 𝑓2(x2,y)

= 𝑓1(x1,y) + 𝑓2(x1,y) + 𝑓1(x1,y) + 𝑓2(x2,y)

= (𝑓1 + 𝑓2)(x1,y) + (𝑓1 + 𝑓2)(x2,y)

Similarly, we can show that

(𝑓1 + 𝑓2)(x,y1 + y2) = (𝑓1 + 𝑓2)(x,y1) + (𝑓1 + 𝑓2)(x,y2)

and

(𝑓1 + 𝑓2)(𝛼x,y) = 𝛼(𝑓1 + 𝑓2)(x,y) = (𝑓1 + 𝑓2)(x, 𝛼y)

For every 𝑓 ∈ 𝐵𝑖𝐿(𝑋 × 𝑌, 𝑍) define the function 𝛼𝑓 : 𝑋 × 𝑌 → 𝑍 by

(𝛼𝑓)(x,y) = 𝛼𝑓(x,y)

𝛼𝑓 is also bilinear, since

(𝛼𝑓)(x1 + x2,y) = 𝛼𝑓(x1 + x2,y)

= 𝛼𝑓(x1,y) + 𝛼𝑓(x2,y)

= (𝛼𝑓)(x1,y) + (𝛼𝑓(x2,y)

Similarly

(𝛼𝑓)(x,y1 + y2) = (𝛼𝑓)(x,y1) + (𝛼𝑓)(x,y2)

(𝛼𝑓)(𝛽x,y) = 𝛽(𝛼𝑓)(x,y) = (𝛼𝑓)(x, 𝛽y)

Analogous to (Exercise 2.78), 𝑓1 + 𝑓2 and 𝛼𝑓 are also continuous

3.58 1. 𝐵𝐿(𝑌, 𝑍) is a linear space and therefore so is 𝐵𝐿(𝑋,𝐵𝐿(𝑌, 𝑍)) (Exercise3.33).

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2. 𝜑x is linear and therefore

𝑓(x,y1 + y2) = 𝜑(x)(y1 + y2) = 𝜑(x)(y1) + 𝜑(x)(y2) = 𝑓(x,y1) + 𝑓(x,y2)

and

𝑓(x, 𝛼y) = 𝜑(x)(𝛼y) = 𝛼𝜑(x)(y) = 𝛼𝑓(x,y)

Similarly, 𝜑 is linear and therefore

𝑓(x1 + x2,y) = 𝜑x1+x2(y) = 𝜑x1(y) + 𝜑x2(y) = 𝑓(x1,y) + 𝑓(x2,y)

and

𝑓(𝛼x,y) = 𝜑𝛼x(y) = 𝛼𝜑x(y) = 𝛼𝑓(x,y)

𝑓 is bilinear

3. Let 𝑓 ∈ 𝐵𝑖𝐿(𝑋 × 𝑌, 𝑍). For every x ∈ 𝑋 , the partial function 𝑓x : 𝑌 → 𝑍 islinear. Therefore 𝑓x ∈ 𝐵𝐿(𝑌, 𝑍) and 𝜑 ∈ 𝐵𝐿(𝑋,𝐵𝐿(𝑌, 𝑍)).

3.59 Bilinearity and symmetry imply

𝑓(x− 𝛼y,x− 𝛼y) = 𝑓(x,x− 𝛼y)− 𝛼𝑓(y,x− 𝛼y)= 𝑓(x,x)− 𝛼𝑓(x,y) − 𝛼𝑓(y,x) + 𝛼2𝑓(y,y)

= 𝑓(x,x)− 2𝛼𝑓(x,y) + 𝛼2𝑓(y,y)

Nonnegativity implies

𝑓(x− 𝛼y,x − 𝛼y) = 𝑓(x,x)− 2𝛼𝑓(x,y) + 𝛼2𝑓(y,y) ≥ 0 (3.38)

for every x,y ∈ 𝑋 and 𝛼 ∈ ℜCase 1 𝑓(x,x) = 𝑓(y,y) = 0 Then (3.38) becomes

−2𝛼𝑓(x,y) ≥ 0

Setting 𝛼 = 𝑓(x,y) generates

−2(𝑓(x,y)

)2 ≥ 0

which implies that

𝑓(x,y) = 0

Case 2 Either 𝑓(x,x) > 0 or 𝑓(y,y) > 0. Without loss of generality, assume 𝑓(y,y) >0 and set 𝛼 = 𝑓(x,y)/𝑓(y,y) in (3.38). That is

𝑓(x,x)− 2

(𝑓(x,y)

𝑓(y,y)

)𝑓(x,y) +

(𝑓(x,y)

𝑓(y,y)

)2𝑓(y,y) ≥ 0

or

𝑓(x,x)− 𝑓(x,y)2

𝑓(y,y)≥ 0

which implies (𝑓(x,y)

)2 ≤ 𝑓(x,x)𝑓(y,y) for every x,y ∈ 𝑋

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3.60 A Euclidean space is a finite-dimensional normed space, which is complete (Propo-sition 1.4).

3.61 𝑓(x,y) = x𝑇y satisfies the requirements of Exercise 3.59 and therefore

(x𝑇y)2 ≤ (x𝑇x)(y𝑇y)

Taking square roots ∣∣x𝑇y∣∣ ≤ ∥x∥ ∥y∥3.62 By definition, the inner product is a bilinear functional. To show that it is contin-uous, let 𝑋 be an inner product space with inner product denote by x𝑇y. Let x𝑛 → xand y𝑛 → y be sequences in 𝑋 .∣∣(x𝑛)𝑇y𝑛 − x𝑇y∣∣ =

∣∣(x𝑛)𝑇y𝑛 − (x𝑛)𝑇y + (x𝑛)𝑇y − x𝑇y∣∣≤ ∣∣(x𝑛)𝑇y𝑛 − (x𝑛)𝑇y

∣∣ +∣∣(x𝑛)𝑇y − x𝑇y∣∣

≤ ∣∣(x𝑛)𝑇 (y𝑛 − y)∣∣ +∣∣(x𝑛 − x)𝑇y∣∣

Applying the Cauchy-Schwartz inequality

∣∣(x𝑛)𝑇y𝑛 − x𝑇y∣∣ ≤ ∥x𝑛∥ ∥y𝑛 − y∥ + ∥x𝑛 − x∥ ∥y∥

Since the sequence x𝑛 converges, it is bounded, that is there exists𝑀 such that ∥x𝑛∥ ≤𝑀 for every 𝑛. Therefore∣∣(x𝑛)𝑇y𝑛 − x𝑇y∣∣ ≤ ∥x𝑛∥ ∥y𝑛 − y∥+ ∥x𝑛 − x∥ ∥y∥ ≤𝑀 ∥y𝑛 − y∥ + ∥x𝑛 − x∥ ∥y∥ → 0

3.63 Applying the properties of the inner product

∙ ∥x∥ =√x𝑇x ≥ 0

∙ ∥x∥ =√x𝑇x = 0 if and only if x = 0

∙ ∥𝛼x∥ =√

(𝛼x)𝑇 (𝛼x) =√𝛼2x𝑇x = ∣𝛼∣ ∥x∥

To prove the triangle inequality, observe that bilinearity and symmetry imply

∥x+ y∥2 = (x+ y)𝑇 (x+ y)

= x𝑇x+ x𝑇y + y𝑇x+ z𝑇 z

= x𝑇x+ 2x𝑇y + y𝑇y

= ∥x∥2 + 2x𝑇y + ∥y∥2

≤ ∥x∥2 + 2∣∣x𝑇y∣∣ + ∥y∥2

Applying the Cauchy-Schwartz inequality

∥x+ y∥2 ≤ ∥x∥2 + 2 ∥x∥ ∥y∥+ ∥y∥2= (∥x∥ + ∥y∥)2

3.64 For every y ∈ 𝑋 , the partial function 𝑓y(x) = x𝑇y is a linear functional on 𝑋(since x𝑇y is bilinear). Continuity follows from the Cauchy-Schwartz inequality, sincefor every x ∈ 𝑋

∣𝑓y(x)∣ = ∣∣x𝑇y∣∣ ≤ ∥y∥ ∥x∥130


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which shows that ∥𝑓y∥ ≤ ∥y∥. In fact, ∥𝑓y∥ = ∥y∥ since

∥𝑓y∥ = sup∥x∥=1

∣𝑓y(x)∣

≥∣∣∣∣𝑓y

(y

∥y∥)∣∣∣∣

=

∣∣∣∣∣(y

∥y∥)𝑇

y

∣∣∣∣∣=

1

∥y∥y𝑇y = ∥y∥

3.65 By the Weierstrass Theorem (Theorem 2.2), the continuous function 𝑔(x) = ∥x∥attains a maximum on the compact set 𝑆 at some point x0.

We claim that x0 is an extreme point. Suppose not. Then, there exist x1,x2 ∈ 𝑆 suchthat

x0 = 𝛼x1 + (1− 𝛼)x2 = x2 + 𝛼(x1 − x2)Since x0 maximizes ∥x∥ on 𝑆

∥x2∥2 ≤ ∥x0∥2 =(x2 + 𝛼(x1 − x2)

)𝑇 (x2 + 𝛼(x1 − x2)

)= ∥x2∥2 + 2𝛼x𝑇2 (x1 − x2) + 𝛼2 ∥x1 − x2∥2

or

2x𝑇2 (x1 − x2) + 𝛼 ∥x1 − x2∥2 ≥ 0 (3.39)

Similarly, interchanging the role of x1 and x2

2x𝑇1 (x2 − x1) + 𝛼 ∥x2 − x1∥2 ≥ 0

or

−2x𝑇1 (x1 − x2) + 𝛼 ∥x1 − x2∥2 ≥ 0 (3.40)

Adding the inequalities (3.39) and (3.40) yields

2(x2 − x1)𝑇 (x1 − x2) + 2𝛼 ∥x1 − x2∥2 ≥ 0

or

2(x2 − x1)𝑇 (x2 − x1) = −2(x2 − x1)𝑇 (x1 − x2) ≤ 2𝛼 ∥x1 − x2∥2

and therefore

∥x2 − x1∥ ≤ 𝛼 ∥x2 − x1∥Since 0 < 𝛼 < 1, this implies that ∥x1 − x2∥ = 0 or x1 = x2 which contradicts ourpremise that x0 is not an extreme point.

3.66 Using bilinearity and symmetry of the inner product

∥x+ y∥2 + ∥x− y∥2 = (x+ y)𝑇 (x+ y) + (x− y)𝑇 (x− y)= x𝑇x+ x𝑇y + y𝑇x+ y𝑇y +

x𝑇x− x𝑇y − y𝑇x+ y𝑇y

= 2x𝑇x+ 2y𝑇y

= 2 ∥x∥2 + 2 ∥y∥2

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3.67 Note that ∥𝑥∥ = ∥𝑦∥ = 1 and

∥𝑥+ 𝑦∥ = sup0≤𝑡≤1

(𝑥(𝑡) + 𝑦(𝑡)

)= sup0≤𝑡≤1

(1 + 𝑡) = 2

∥𝑥− 𝑦∥ = sup0≤𝑡≤1

(𝑥(𝑡)− 𝑦(𝑡)) = sup

0≤𝑡≤1(1− 𝑡) = 1

so that

∥𝑥+ 𝑦∥2 + ∥𝑥− 𝑦∥2 = 5 ∕= 2 ∥𝑥∥2 + 2 ∥𝑥∥2

Since 𝑥 and 𝑦 do not satisfy the parallelogram law (Exercise 3.66), 𝐶(𝑋) cannot be aninner product space.

3.68 Let {x1,x2, . . . ,x𝑛} be a set of pairwise orthogonal vectors. Assume

0 = 𝛼x1 + 𝛼2x2 + ⋅ ⋅ ⋅+ 𝛼𝑛x𝑛Using bilinearity, this implies

0 = 0𝑇x𝑗 =𝑛∑𝑖=1

𝛼𝑖x𝑇𝑖 x𝑗 = 𝛼𝑗 ∥x𝑗∥

for every 𝑗 = 1, 2, . . . , 𝑛. Since x𝑗 ∕= 0, this implies 𝛼𝑗 = 0 for every 𝑗 = 1, 2, . . . , 𝑛.We conclude that the set {x1,x2, . . . ,x𝑛} is linearly independent (Exercise 1.133).

3.69 Let x1,x2, . . . ,x𝑛 be a orthonormal basis for 𝑋 . Since 𝐴 represents 𝑓

𝑓(x𝑗) =

𝑛∑𝑖=1

𝑎𝑖𝑗x𝑖

for 𝑗 = 1, 2, . . . , 𝑛. Taking the inner product with x𝑖,

x𝑇𝑖 𝑓(x𝑗) = x𝑇𝑖

(𝑛∑𝑖=1

𝑎𝑖𝑗x𝑖

)=

𝑛∑𝑖=1

𝑎𝑖𝑗x𝑇𝑖 x𝑗

Since {x1,x2, . . . ,x𝑛} is orthonormal

x𝑇𝑘 x𝑗 =

{1 if 𝑖 = 𝑗

0 otherwise

so that the last sum simplifies to

x𝑇𝑖 𝑓(x𝑗) = 𝑎𝑖𝑗 for every 𝑖, 𝑗

3.70 1. By the Cauchy-Schwartz inequality∣∣x𝑇y∣∣ ≤ ∥x∥ ∥x∥for every x and y, so that

∣cos 𝜃∣ =∣∣∣∣ x𝑇y∥x∥ ∥y∥

∣∣∣∣ ≤ 1

which implies

−1 ≤ cos 𝜃 ≤ 1

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2. Since cos 90 = 0, 𝜃 = 90 implies that x𝑇y = 0 or x ⊥ y. Conversely, if x ⊥ y,x𝑇y = 0 and cos 𝜃 = 0 which implies 𝜃 = 90 degrees.

3.71 By bilinearity

∥x+ y∥2 = (x+ y)𝑇 (x+ y) = ∥x∥2 + x𝑇y + y𝑇x+ ∥y∥2

If x ⊥ y, x𝑇y = y𝑇x = 0 and

∥x+ y∥2 = ∥x∥2 + ∥y∥2

3.72 1. Choose some x ∈ 𝑆 and let 𝑆 be the set of all x ∈ 𝑆 which are closer to ythan x, that is

𝑆 = {x ∈ 𝑆 : ∥x− y∥ ≤ ∥x− y∥ }

𝑆 is compact (Proposition 1.4).

By the Weierstrass theorem (Theorem 2.2), the continuous function 𝑔(x) = ∥xy∥attains a minimum on 𝑆 at some point x0 ∈ 𝑆. That is

∥x0 − y∥ ≤ ∥x− y∥ for every x ∈ 𝑆

A fortiori


2. Suppose there exists some x1 ∈ 𝑆 such that

∥x1 − y∥ = ∥x0 − y∥ = 𝛿

By the parallelogram law (Exercise 3.66)

∥x0 − x1∥2 = ∥x0 − y + y − x1∥2

= 2 ∥x0 − y∥2 + 2 ∥x1 − y∥2 − ∥(x0 − y) − (y − x1)∥2

= 2 ∥x0 − y∥2 + 2 ∥x1 − y∥2 − 22∥∥∥∥1

2(x0 + x1)− y

∥∥∥∥2

= 2𝛿2 + 2𝛿2 − 22∥∥∥∥1

2(x0 + x1)− y

∥∥∥∥2

since 12 (x0 + x1) ∈ 𝑆 and therefore

∥∥ 12 (x0 + x1)− y

∥∥ ≥ 𝛿 so that

∥x0 − x1∥2 ≤ 2𝛿2 + 2𝛿2 − 4𝛿2 = 0

which implies that x1 = x0.

3. Let x ∈ 𝑆. Since 𝑆 is convex, the line segment 𝛼x+(1−𝛼)x0 = x0+𝛼(x−x0) ∈ 𝑆and therefore (since x0 is the closest point)

∥x0 − y∥2 ≤∥∥(x0 + 𝛼(x − x0)

)− y∥∥2= ∥(x0 − y) + 𝛼(x− x0)∥2

=((x0 − y) + 𝛼(x − x0)

)𝑇 ((x0 − y) + 𝛼(x− x0)

)= ∥x0 − y∥2 + 2𝛼(x0 − y)𝑇 (x− x0) + 𝛼2 ∥x− x0∥2

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which implies that

2𝛼(x0 − y)𝑇 (x− x0) + 𝛼2 ∥x− x0∥2 ≥ 0

Dividing through by 𝛼

2(x0 − y)𝑇 (x− x0) + 𝛼 ∥x− x0∥2 ≥ 0

which inequality must hold for every 0 < 𝛼 < 1. Letting 𝛼→ 0, we must have

(x0 − y)𝑇 (x − x0) ≥ 0

as required.

3.73 1. Using the parallelogram law (Exercise 3.66),

∥x𝑚 − x𝑛∥2 = ∥(x𝑚 − y) + (y − x𝑛)∥2

= 2 ∥x𝑚 − y∥2 + 2 ∥y − x𝑛∥2 − 2 ∥x𝑚 + x𝑛∥2

for every𝑚,𝑛. Since 𝑆 is convex, (x𝑚+x𝑛)/2 ∈ 𝑆 and therefore ∥x𝑚 + x𝑛∥ ≥ 2𝑑.Therefore

∥x𝑚 − x𝑛∥2 = 2 ∥x𝑚 − y∥2 + 2 ∥y − x𝑛∥2 − 4𝑑2

Since ∥x𝑚 − y∥ → 𝑑 and ∥x𝑛 − y∥ → 𝑑 as 𝑚,𝑛 → ∞, we conclude that

∥x𝑚 − x𝑛∥2 → 0. That is, (x𝑛) is a Cauchy sequence.

2. Since 𝑆 is a closed subspace of complete space, there exists x0 ∈ 𝑆 such thatx𝑛 → x0. By continuity of the norm

∥x0 − y∥ = lim𝑛→∞ ∥x

𝑛 − y∥ = 𝑑

Therefore


Uniqueness follows in the same manner as the finite-dimensional case.

3.74 Define 𝑔 : 𝑇 → 𝑆 by

𝑔(y) = {x ∈ 𝑆 : x is closest to y }

The function 𝑔 is well-defined since for every y ∈ 𝑇 there exists a unique point x ∈ 𝑆which is closest to y (Exercise 3.72). Clearly, for every x ∈ 𝑆, x is the closest point tox. Therefore 𝑔(x) = x for every x ∈ 𝑆.

To show that 𝑔 is continuous, choose any y1 and y2 in 𝑇

x1 = 𝑔(y1) and x2 = 𝑔(y2)

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be the corresponding closest points in 𝑆. Then

∥(y1 − y2)− (x1 − x2)∥2 =((y1 − y2)− (x1 − x2)

)𝑇 ((y1 − y1)− (x1 − x2)

)= (y1 − y2)𝑇 (y1 − y2) + (x1 − x2)𝑇 (x1 − x2)− 2(y1 − y2)𝑇 (x1 − x2)

= ∥y1 − y2∥2 + ∥x1 − x2∥2 − 2(y1 − y2)𝑇 (x1 − x2)= ∥y1 − y2∥2 + ∥x1 − x2∥2 − 2(y1 − y2)𝑇 (x1 − x2)− 2 ∥x1 − x2∥2 + 2(x1 − x2)𝑇 (x1 − x2)

= ∥y1 − y2∥2 − ∥x1 − x2∥2

+ 2((x1 − x2)− (y1 − y2)

)𝑇(x1 − x2)

= ∥y1 − y2∥2 − ∥x1 − x2∥2+ 2(x1 − y1)𝑇 (x1 − x2)− 2(x2 − y2)𝑇 (x1 − x2)

= ∥y1 − y2∥2 − ∥x1 − x2∥2− 2(x1 − y1)𝑇 (x2 − x1)− 2(x2 − y2)𝑇 (x1 − x2)

so that

∥y1 − y2∥2 − ∥x1 − x2∥2 = ∥(y1 − y2)− (x1 − x2)∥2+ 2(x1 − y1)𝑇 (x2 − x1) + 2(x2 − y2)𝑇 (x1 − 𝑥2)

Using Exercise 3.72

(x1 − y1)𝑇 (x2 − x1) ≥ 0 and (x2 − y2)𝑇 (x1 − x2) ≥ 0

which implies that the left-hand side

∥y1 − y2∥2 − ∥x1 − x2∥2 ≥ 0

or

∥x1 − x2∥ = ∥𝑔(y1)− 𝑔(y2)∥ ≤ ∥y1 − y2∥𝑔 is Lipschitz continuous.

3.75 Let 𝑆 = kernel 𝑓 . Then 𝑆 is a closed subspace of 𝑋 . If 𝑆 = 𝑋 , then 𝑓 is the zerofunctional and y = 0 is the required element. Otherwise chose any y /∈ 𝑆 and let x0be the closest point in 𝑆 (Exercise 3.72). Define z = x0 − y. Then z ∕= 0 and

z𝑇x ≥ 0 for every x ∈ 𝑆Since 𝑆 is subspace, this implies that

z𝑇x = 0 for every x ∈ 𝑆that is z is orthogonal to 𝑆.

Let 𝑆 be the subset of 𝑋 defined by

𝑆 = { 𝑓(x)z− 𝑓(z)x : x ∈ 𝑋 }For every x ∈ 𝑆

𝑓(x) = 𝑓(𝑓(x)z − 𝑓(z)x

)= 𝑓(x)𝑓(z)− 𝑓(z)𝑓(x) = 0

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Therefore 𝑆 ⊆ 𝑆. For every x ∈ 𝑋(𝑓(x)z− 𝑓(z)x

)𝑇z = 𝑓(x)z𝑇 z− 𝑓(z)x𝑇 z = 0

since z ∈ 𝑆⊥. Therefore

𝑓(x) =𝑓(z)

∥z∥2x𝑇 z = x𝑇

(z𝑓(z)

∥z∥2)

= x𝑇y

where

y =z𝑓(z)

∥z∥2

3.76 𝑋∗ is always complete (Proposition 3.3). To show that it is a Hilbert space, wehave to that it has an inner product. For this purpose, it will be clearer if we use analternative notation < x,y > to denote the inner product of x and y. Assume 𝑋 is aHilbert space. By the Riesz representation theorem (Exercise 3.75), for every 𝑓 ∈ 𝑋∗

there exists y𝑓 ∈ 𝑋 such that

𝑓(x) =< x,y𝑓 > for every x ∈ 𝑋Furthermore, if y𝑓 represents 𝑓 and y𝑔 represents 𝑔 ∈ 𝑋∗, then y𝑓 + y𝑔 represents𝑓 + 𝑔 and 𝛼y𝑓 represents 𝛼𝑓 since

(𝑓 + 𝑔)(x) = 𝑓(x) + 𝑔(x) =< x,y𝑓 > + < x,y𝑔 >=< x,y𝑓 + y𝑔 >

(𝛼𝑓)(x) = 𝛼𝑓(x) = 𝛼 < x,y𝑓 >=< x, 𝛼y𝑓 >

Define an inner product on 𝑋∗ by

< 𝑓, 𝑔 >=< y𝑔,y𝑓 >

We show that it satisfies the properties of an inner product, namely

symmetry < 𝑓, 𝑔 >=< y𝑔,y𝑓 >=< y𝑓 ,y𝑔 >=< 𝑔, 𝑓 >

additivity < 𝑓1 + 𝑓2, 𝑔 >=< y𝑔,y𝑓1+𝑓2 >=< y𝑔,y𝑓1 + y𝑓2 >=< 𝑓1, 𝑔 > + < 𝑓2, 𝑔 >

homogeneity < 𝛼𝑓, 𝑔 >=< y𝑔, 𝛼y𝑓 >= 𝛼 < y𝑔,y𝑓 >= 𝛼 < 𝑓, 𝑔 >

positive definiteness < 𝑓, 𝑔 >=< y𝑔,y𝑓 >≥ 0 and < 𝑓, 𝑔 >=< y𝑔,y𝑓 >= 0 if andonly if 𝑓 = 𝑔.

Therefore, 𝑋∗ is a complete inner product space, that is a Hilbert space.

3.77 Let𝑋 be a Hilbert space. Applying the previous exercise a second time, 𝑋∗∗ is alsoa Hilbert space. Let 𝐹 be an arbitrary functional in 𝑋∗∗. By the Riesz representationtheorem, there exists 𝑔 ∈ 𝑋∗ such that

𝐹 (𝑓) =< 𝑓, 𝑔 > for every 𝑓 ∈ 𝑋∗

Again by the Riesz representation theorem, there exists x𝑓 (representing 𝑓) and x𝐹(representing 𝑔) in 𝑋 such that

𝐹 (𝑓) =< 𝑓, 𝑔 >=< x𝐹 ,x𝑓 >

and

𝑓(x) =< x,x𝑓 >

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In particular,

𝑓(x𝐹 ) =< x𝐹 ,x𝑓 >= 𝐹 (𝑓)

That is, for every 𝐹 ∈ 𝑋∗∗, there exists an element x𝐹 ∈ 𝑋 such that

𝐹 (𝑓) = 𝑓(x𝐹 )

𝑋 is reflexive.

3.78 1. Adapt Exercise 3.64.

2. By Exercise 3.75, there exists unique x∗ ∈ 𝑋 such that

𝑓y(x) = x𝑇x∗

3. Substituting

𝑓(x)𝑇y = 𝑓y(x) = x𝑇x∗ = 𝑥𝑇 𝑓∗(y)

4. For every y1,y2 ∈ 𝑌x𝑇

(𝑓∗(y1 + y2)

)= 𝑓(x)𝑇

(y1 + y2

)= 𝑓(x)𝑇y1 + 𝑓(x)𝑇y1 = x𝑇 𝑓∗(y1) + x𝑇 𝑓∗(y1)

and for every y ∈ 𝑌x𝑇 𝑓∗(𝛼y) = 𝑓(x)𝑇𝛼y = 𝛼𝑓(x)𝑇y = 𝛼x𝑇 𝑓∗(y) = x𝑇𝛼𝑓∗(y)

3.79 The zero element 0𝑋 is a fixed point of every linear operator (Exercise 3.13).

3.80 𝐴𝐴−1 = 𝐼 so that

det(𝐴) det(𝐴−1) = det(𝐼) = 1

3.81 Expanding along the 𝑖th row using (3.8)

det(𝐶) =𝑛∑

𝑗=1

(−1)𝑖+𝑗(𝛼𝑎𝑖𝑗 + 𝛽𝑏𝑖𝑗) det(𝐶𝑖𝑗)

= 𝛼

𝑛∑𝑗=1

(−1)𝑖+𝑗𝑎𝑖𝑗 det(𝐶𝑖𝑗) + 𝛽

𝑛∑𝑗=1

(−1)𝑖+𝑗𝑏𝑖𝑗 det(𝐶𝑖𝑗)

But the matrices differ only in the 𝑖th row and therefore

𝐴𝑖𝑗 = 𝐵𝑖𝑗 = 𝐶𝑖𝑗 , 𝑗 = 1, 2, . . . 𝑛

so that

det(𝐶) = 𝛼

𝑛∑𝑗=1

(−1)𝑖+𝑗𝑎𝑖𝑗 det(𝐴𝑖𝑗) + 𝛽

𝑛∑𝑗=1

(−1)𝑖+𝑗𝑏𝑖𝑗 det(𝐵𝑖𝑗)

= 𝛼 det(𝐴) + 𝛽 det(𝐵)

3.82 Suppose that x1 and x2 are eigenvectors corresponding to the eigenvalue 𝜆. Bylinearity

𝑓(x1 + x2) = 𝑓(x1) + 𝑓(x2) = 𝜆x1 + 𝜆x2 = 𝜆(x1 + x2)

and

𝑓(𝛼x1) = 𝛼𝑓(x1) = 𝛼𝜆x

Therefore x1 + x2 and 𝛼x1 are also eigenvectors.

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3.83 Suppose 𝑓 is singular. Then there exists x ∕= 0 such that 𝑓(x) = 0. Therefore xis an eigenvector with eigenvalue 0. Conversely, if 0 is an eigenvalue

𝑓(x) = 0x = 0

for any x ∕= 0. Therefore 𝑓 is singular.

3.84 Since 𝑓(x) = 𝜆x

𝑓(x)𝑇x = 𝜆x𝑇x = 𝜆x𝑇x

3.85 By Exercise 3.69

𝑎𝑖𝑗 = x𝑇𝑖 𝑓(x𝑗)

𝑎𝑗𝑖 = x𝑇𝑗 𝑓(x𝑖) = 𝑓(x𝑖)𝑇x𝑗

and therefore

𝑎𝑖𝑗 = 𝑎𝑗𝑖 ⇐⇒ x𝑇𝑖 𝑓(x𝑗) = 𝑓(x𝑖)𝑇x𝑗

3.86 By bilinearity

x𝑇1 𝑓(x2) = x𝑇1 𝜆2x2 = 𝜆2x𝑇1 x2

𝑓(x1)𝑇x2 = 𝜆1x

𝑇1 x2 = 𝜆1x

𝑇1 x2

Since 𝑓 is symmetric, this implies

(𝜆1 − 𝜆2)x𝑇1 x2 = 0

and 𝜆1 ∕= 𝜆2 implies x𝑇1 x2 = 0.

3.87 1. Since 𝑆 compact and 𝑓 is continuous (Exercises 3.31, 3.62), the maximum isattained at some x0 ∈ 𝑆 (Theorem 2.2), that is

𝜆 = 𝑓(x0)𝑇x0 ≥ 𝑓(x)𝑇x for every x ∈ 𝑆

Hence

𝑔(x,y) =(𝜆x− 𝑓(x)

)𝑇y

is well-defined.

2. For any x ∈ 𝑋

𝑔(x,x) =(𝜆x − 𝑓(x)

)𝑇x

= 𝜆x𝑇x− 𝑓(x)𝑇x

= 𝜆 ∥x∥2 − 𝑓(x)𝑇x

= 𝜆 ∥x∥2 − ∥x∥2 𝑓(x

∥x∥2)𝑇 (

x

∥x∥2)

= ∥x∥2 (𝜆− 𝑓(z)𝑇 z) ≥ 0

since z = x/ ∥x∥ ∈ 𝑆.

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3. Since 𝑓 is symmetric

𝑔(y,x) =(𝜆y − 𝑓(y)

)𝑇x

= 𝜆y𝑇x− 𝑓(y)𝑇x

= 𝜆x𝑇y − 𝑓(x)𝑇 𝑦

=(𝜆x − 𝑓(x)

)𝑇y = 𝑔(x,y)

4. 𝑔 satisfies the conditions of Exercise 3.59 and therefore

(𝑔(x,y))2 ≤ 𝑔(x,x)𝑔(y,y) for every x,y ∈ 𝑋 (3.41)

By definition 𝑔(x0,x0) = 0 and (3.41) implies that

𝑔(x0,y) = 0 for every y ∈ 𝑋

That is

𝑔(x0,y) =(𝜆x0 − 𝑓(x0)

)𝑇y = 0 for every 𝑦 ∈ 𝑋

and therefore

𝜆x0 − 𝑓(x0) = 0

or

𝑓(x0) = 𝜆x0

In other words, x0 is an eigenvector. By construction, ∥x0∥ = 1.

3.88 1. Suppose x2,x3 ∈ 𝑆. Then

(𝛼x2 + 𝛽x3

)𝑇x1 = 𝛼x𝑇2 x1 + 𝛽x𝑇3 x1 = 0

so that 𝛼x2 + 𝛽x3 ∈ 𝑆. 𝑆 is a subspace.

Let {x1,x2, . . . ,x𝑛} be a basis for 𝑋 (Exercise 1.142). For x ∈ 𝑋 , there exists(Exercise 1.137) unique 𝛼1, 𝛼2, . . . , 𝛼𝑛 such that

x = 𝛼1x1 + 𝛼2x2 + ⋅ ⋅ ⋅+ 𝛼𝑛x𝑛If x ∈ 𝑆

x𝑇x1 = 𝛼1x𝑇1 x1 = 0

which implies that 𝛼1 = 0. Therefore, x2,x3, . . . ,x𝑛 span 𝑆 and thereforedim𝑆 = 𝑛− 1.

2. For every x ∈ 𝑆,

𝑓(x)𝑇x0 = x𝑇 𝑓(x0) = x𝑇𝜆x0 = 𝜆x𝑇x0 = 0

since 𝑓 is symmetric. Therefore 𝑓(x) ∈ {x0}⊥ = 𝑆.

3.89 Let 𝑓 be a symmetric operator. By the spectral theorem (Proposition 3.6), thereexists a diagonal matrix 𝐴 which represents 𝑓 . The elements of 𝐴 are the eigenvalues of𝑓 . By Proposition 3.5, the determinant of 𝐴 is the product of these diagonal elements.

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3.90 By linearity

𝑓(x) =∑𝑗

𝑥𝑗𝑓(x𝑗)

𝑄 defines a quadratic form since

𝑄(x) = x𝑇 𝑓(x) =

(∑𝑖

𝑥𝑖x𝑖

)𝑇⎛⎝∑

𝑗

𝑥𝑗𝑓(x𝑗)

⎞⎠ =

∑𝑖

∑𝑗

𝑥𝑖𝑥𝑗x𝑇𝑖 𝑓(x𝑗) =

∑𝑖

∑𝑗

𝑎𝑖𝑗𝑥𝑖𝑥𝑗

by Exercise 3.69.

3.91 Let 𝑓 be the symmetric linear operator defining 𝑄

𝑄(x) = x𝑇 𝑓(x)

By the spectral theorem (Proposition 3.6), there exists an orthonormal basis x1,x2, . . . ,x𝑛comprising the eigenvectors of 𝑓 . Let 𝜆1, 𝜆2, . . . , 𝜆𝑛 be the corresponding eigenvalues,that is

𝑓(x𝑖) = 𝜆𝑖x𝑖 𝑖 = 1, 2 . . . , 𝑛

Then for x = 𝑥1x1 + 𝑥2x2 + ⋅ ⋅ ⋅+ 𝑥𝑛x𝑛𝑄(x) = x𝑇 𝑓(x)

= (𝑥1x1 + 𝑥2x2 + ⋅ ⋅ ⋅+ 𝑥𝑛x𝑛)𝑇 𝑓(𝑥1x1 + 𝑥2x2 + ⋅ ⋅ ⋅+ 𝑥𝑛x𝑛)

= (𝑥1x1 + 𝑥2x2 + ⋅ ⋅ ⋅+ 𝑥𝑛x𝑛)𝑇 (𝑥1𝑓(x1) + 𝑥2𝑓(x2) + ⋅ ⋅ ⋅+ 𝑥𝑛𝑓(x𝑛))

= (𝑥1x1 + 𝑥2x2 + ⋅ ⋅ ⋅+ 𝑥𝑛x𝑛)𝑇 (𝑥1𝜆1x1 + 𝑥2𝜆2x2 + ⋅ ⋅ ⋅+ 𝑥𝑛𝜆𝑛x𝑛)

= 𝑥1𝜆1𝑥1 + 𝑥2𝜆2𝑥2 + ⋅ ⋅ ⋅+ 𝑥𝑛𝜆𝑛𝑥𝑛= 𝜆1𝑥

21 + 𝜆2𝑥

22 + ⋅ ⋅ ⋅+ 𝜆𝑛𝑥2𝑛

3.92 1. Assuming that 𝑎11 ∕= 0, the quadratic form can be rewritten as follows

𝑄(𝑥1, 𝑥2) = 𝑎11𝑥21 + 2𝑎12𝑥1𝑥2 + 𝑎22𝑥

22

= 𝑎11𝑥21 + 2𝑎12𝑥1𝑥2 +

𝑎212𝑎11𝑥22 −

𝑎212𝑎11𝑥22 + 𝑎22𝑥

22

= 𝑎11

(𝑥21 + 2

𝑎12𝑎11𝑥1𝑥2 +

(𝑎12𝑎11𝑥2

)2)+

(𝑎22 − 𝑎

212

𝑎11

)𝑥22

= 𝑎11

(𝑥1 +

𝑎12𝑎11𝑥2

)2+

(𝑎11𝑎22 − 𝑎212

𝑎11

)𝑥22

2. We observe that 𝑞 must be positive for every 𝑥1 and 𝑥2 provided 𝑎11 > 0 and𝑎11𝑎22− 𝑎212 > 0. Similarly 𝑞 must be negative for every 𝑥1 and 𝑥2 if 𝑎11 > 0 and𝑎11𝑎22 − 𝑎212 > 0. Otherwise, we can choose values for 𝑥1 and 𝑥2 which make 𝑞both positive and negative.

Note that the condition 𝑎11𝑎22 > 𝑎212 > 0 implies that 𝑎11 and 𝑎12 must have the

same sign.

3. If 𝑎11 = 𝑎22 = 0, then 𝑞 is indefinite. Otherwise, if 𝑎11 = 0 but 𝑎22 ∕= 0, then the𝑞 can we can “complete the square” using 𝑎22 and deduce

𝑞 is

{nonnegativenonpositive

}definite if and only if

{𝑎11, 𝑎22 ≥ 0𝑎11, 𝑎22 ≤ 0

}and 𝑎11𝑎22 ≥ 𝑎212

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3.93 Let 𝑄 : 𝑋 → ℜ be a quadratic form on 𝑋 . Then there exists a linear operator 𝑓such that

𝑄(x) = x𝑇 𝑓(x)

and (Exercise 3.13)

𝑄(0) = 0𝑇 𝑓(0) = 0

3.94 Suppose to the contrary that the positive (negative) definite matrix 𝐴 is singular.Then there exists x ∕= 0 such that 𝐴x = 0 and therefore x′𝐴x = 0 contradicting thedefiniteness of 𝐴.

3.95 Let e1, e2, . . . , e𝑛 be the standard basis for ℜ𝑛 (Example 1.79). Then for every 𝑖

e′𝑖𝐴e𝑖 = 𝑎𝑖𝑖 > 0

3.96 Let 𝑄 be the quadratic form defined by 𝐴. By Exercise 3.91, there exists anorthonormal basis such that

𝑄(x) = 𝜆1x21 + 𝜆2x

22 + ⋅ ⋅ ⋅+ 𝜆𝑛x2𝑛

where 𝜆1, 𝜆2, . . . , 𝜆𝑛 are the eigenvalues of 𝐴. This implies⎧⎨⎩𝑄(x) > 0𝑄(x) ≥ 0𝑄(x) < 0𝑄(x) ≤ 0

⎫⎬⎭ ⇐⇒

⎧⎨⎩𝜆𝑖 > 0𝜆𝑖 ≥ 0𝜆𝑖 < 0𝜆𝑖 ≤ 0

⎫⎬⎭ 𝑖 = 1, 2, . . . , 𝑛

3.97 Let 𝜆1, 𝜆2, . . . , 𝜆𝑛 be the eigenvalues of 𝐴. By Exercise 3.89

det(𝐴) = 𝜆1𝜆2 . . . 𝜆𝑛

By Exercise 3.96, 𝜆𝑖 ≥ 0 for every 𝑖 and therefore det(𝐴) ≥ 0. We conclude that

det(𝐴) > 0 ⇐⇒ 𝜆𝑖 > 0 for every 𝑖 ⇐⇒ 𝐴 is positive definite

by Exercise 3.96.

3.98 1. 𝐴0 = 0. Therefore, 0 is always a solution.

2. Assume x1 and x2 are solutions, that is

𝐴x1 = 0 and 𝐴x2 = 0

Then

𝐴(x1 + x2) = 𝐴x1 +𝐴x2 = 0

x1 + x2 is also a solution.

3. Let 𝑓 be the linear function defined by

𝑓(x) = 𝐴x

The system of equations 𝐴x = 0 has a nontrivial solution if and only if

kernel 𝑓 ∕= {0} ⇐⇒ nullity 𝑓 > 0

By the rank theorem (Exercise 3.24)

rank𝑓 + nullity𝑓 = dim𝑋

so that

nullity 𝑓 > 0 ⇐⇒ rank𝑓 < dim𝑋 = 𝑛

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3.99 1. Assume x1 and x2 are solutions of (3.16). That is

𝐴x1 = c and 𝐴x2 = c

Subtracting

𝐴x1 −𝐴x2 = 𝐴(x1 − x2) = 0

2. Assume x𝑝 solves (3.16) while x is any solution to (3.17). That is

𝐴x𝑝 = c and 𝐴x = 0

Adding

𝐴x𝑝 +𝐴x = 𝐴(x𝑝 + x) = c

We conclude that x𝑝 + x solves (3.16) for every x ∈ 𝐾.

3. If 0 is the only solution of (3.17), 𝐾 = {0}. Assume x1 and x2 are solutions of(3.16). Then x1 − x2 ∈ 𝐾 = {0} which implies x1 = x2.

3.100 Let 𝑆 = {x : 𝐴x = 𝑐 }. For every x,y ∈ 𝑆 and 𝛼 ∈ ℜ

𝐴𝛼x + (1− 𝛼)y = 𝛼𝐴x + (1− 𝛼)𝐴y = 𝛼c + (1− 𝛼)c = 𝑐

Therefore, z = 𝛼x+ (1− 𝛼)y ∈ 𝑆. 𝑆 is affine.

3.101 Let 𝑆 ∕= ∅ be an affine set ℜ𝑛. Then there exists a unique subspace 𝑉 such that

𝑆 = x0 + 𝑉

for some x0 ∈ 𝑆 (Exercise 1.150). The orthogonal complement of 𝑉 is

𝑉 ⊥ = { a ∈ 𝑋 : ax = 0 for every x ∈ 𝑉 }

Let (a1, a2, . . . , a𝑚) be a basis for 𝑉 ⊥. Then

𝑉 = (𝑉 ⊥)⊥ = {x : a𝑖x = 0, 𝑖 = 1, 2, . . .𝑚}

Let 𝐴 be the 𝑚×𝑛 matrix whose rows are a1, a2, . . . , a𝑚. Then 𝑉 is the set of solutionsto the homogeneous linear system 𝐴x = 0, that is

𝑉 = { 𝑥 : 𝐴x = 0 }

Therefore

𝑆 = x0 + 𝑉

= x0 + {x : 𝐴x = 0 }= {x : 𝐴(x− x0) = 0 }= {x : 𝐴x = c }

where c = 𝐴x0.

3.102 Consider corresponding homogeneous system

𝑥1 + 3𝑥2 = 0

𝑥1 − 𝑥2 = 0

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Multiplying the second equation by 3

𝑥1 + 6𝑥2 = 0

3𝑥1 − 3𝑥2 = 0

and adding yields

4𝑥1 = 0

for which the only solution is 𝑥1 = 0. Substituting in the first equation implies 𝑥2 = 0.The kernel of 𝑓 = 𝐴x is {0}. Therefore dim 𝑓(ℜ2) = 2, and the system 𝐴x = 𝑐 has aunique solution for every 𝑐1, 𝑐2.

3.103 We can write the system 𝐴x = c in the form

𝑥1

⎛⎜⎝𝑎11

...𝑎𝑛1

⎞⎟⎠ + ⋅ ⋅ ⋅+ 𝑥𝑗

⎛⎜⎝𝑎1𝑗...𝑎𝑛𝑗

⎞⎟⎠ + ⋅ ⋅ ⋅+ 𝑥𝑛

⎛⎜⎝𝑎1𝑛

...𝑎𝑛𝑛

⎞⎟⎠ =

⎛⎜⎝𝑐1...𝑐𝑛

⎞⎟⎠

Subtracting c from the 𝑗th column gives

𝑥1

⎛⎜⎝𝑎11

...𝑎𝑛1

⎞⎟⎠ + ⋅ ⋅ ⋅+

⎛⎜⎝𝑥𝑗𝑎1𝑗 − 𝑐1

...𝑥𝑗𝑎𝑛𝑗 − 𝑐𝑛

⎞⎟⎠ + ⋅ ⋅ ⋅+ 𝑥𝑛

⎛⎜⎝𝑎1𝑛

...𝑎𝑛𝑛

⎞⎟⎠ = 0

so that the columns of the matrix

𝐶 =

⎛⎜⎝𝑎11 . . . (𝑥𝑗𝑎1𝑗 − 𝑐1) . . . 𝑎1𝑛

......

...𝑎𝑛1 . . . (𝑥𝑗𝑎𝑛𝑗 − 𝑐𝑛) . . . 𝑎𝑛𝑛

⎞⎟⎠

are linearly dependent (Exercise 1.133). Therefore det(𝐶) = 0. Let 𝐵𝑗 denote thematrix obtained from 𝐴 by replacing the 𝑗th column with c. Then 𝐴, 𝐵𝑗 and 𝐶 differonly in the 𝑗th column, with the 𝑗th column of 𝐶 being a linear combination of the𝑗th columns of 𝐴 and 𝐵𝑗 . ⎛

⎜⎝𝑐1𝑗...𝑐𝑛𝑗

⎞⎟⎠ = 𝑥𝑗

⎛⎜⎝𝑎1𝑗...𝑎𝑛𝑗

⎞⎟⎠−

⎛⎜⎝𝑐1𝑗...𝑐𝑛𝑗

⎞⎟⎠

By Exercise 3.81

det(𝐶) = x𝑗 det(𝐴)− det(𝐵𝑗) = 0

and therefore

𝑥𝑗 =det(𝐵𝑗)

det(𝐴)

as required.

3.104 Let (𝑎 𝑏𝑐 𝑑

)−1=

(𝐴 𝐵𝐶 𝐷

)

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The inverse satisfies the equation(𝑎 𝑏𝑐 𝑑

)(𝐴 𝐵𝐶 𝐷

)=

(1 00 1

)

In particular, this means that 𝐴 and 𝐶 satisfy the equation(𝑎 𝑏𝑐 𝑑

)(𝐴𝐶

)=

(10

)

By Cramer’s rule (Exercise 3.103)

𝐴 =

∣∣∣∣1 𝑏0 𝑑

∣∣∣∣∣∣∣∣𝑎 𝑏𝑐 𝑑

∣∣∣∣=𝑑

Δ

where Δ = 𝑎𝑑− 𝑏𝑐. Similarly

𝐶 =

∣∣∣∣𝑎 1𝑐 0

∣∣∣∣∣∣∣∣𝑎 𝑏𝑐 𝑑

∣∣∣∣=−𝑐Δ

𝐵 and 𝐷 are determined analogously.

3.105 A portfolio is duplicable if and only if there is a different portfolio y ∕= x suchthat

𝑅x = 𝑅y

or

𝑅(x− y) = 0

There exists a duplicable portfolio if and only if this homogeneous system has a non-trivial solution, that is if rank 𝑅 < 𝐴.

3.106 State 𝑠 is insurable if there is a solution to the linear system

𝑅x = e𝑠 (3.42)

where e𝑠 is the 𝑠-th unit vector (the 𝑠 Arrow-Debreu security). (3.42) has a solutionfor every state 𝑠 if and only if 𝑓(ℜ𝐴) = ℜ𝑆 , that is rank 𝑅 = 𝑆.

3.107 Figure 3.1.

3.108 Let 𝑆 be an affine subset of ℜ𝑛. Then there exists (Exercise 3.101) a system oflinear equations 𝐴x = c such that

𝑆 = {x : 𝐴x = c }Let a𝑖 denote the 𝑖-th row of 𝐴. Then

𝑆 = { 𝑥 : a𝑖x = 𝑐𝑖, 𝑖 = 1, 2, . . . , 𝑛 } =

𝑛∩𝑖=1

{x : a𝑖x = 𝑐𝑖 }

where each {x : a𝑖x = 𝑐𝑖 } is a hyperplane in ℜ𝑛 (Example 3.21).

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Figure 3.1: The solutions of three equations in two unknowns

3.109 Let 𝑆 = {x : 𝐴x ≤ c }. For every x,y ∈ 𝑆 and 0 ≤ 𝛼 ≤ 1

𝐴x ≤ c𝐴y ≤ c

and therefore

𝐴𝛼x + (1− 𝛼)y = 𝛼𝐴x + (1− 𝛼)𝐴y ≤ 𝛼c + (1− 𝛼)c = 𝑐

Therefore, z = 𝛼x+ (1− 𝛼)y ∈ 𝑆. 𝑆 is a convex set.

3.110 We have already seen that 𝑆 = {x : 𝐴x ≤ 0 } is convex. To show that it is acone, let x ∈ 𝑆. Then

𝐴x ≤ 0𝐴𝛼x ≤ 0

so that 𝛼x ∈ 𝑆. 𝑆 is a convex cone.

3.111 1. Each column 𝐴𝑗 is a vector in ℜ𝑚. If the set {𝐴1, 𝐴2, . . . , 𝐴𝑘} is linearlyindependent, it has at most 𝑚 elements, that is 𝑘 ≤ 𝑚 and x is a basic feasiblesolution.

2. (a) Assume {𝐴1, 𝐴2, . . . , 𝐴𝑘} are linearly dependent. Then (Exercise 1.133)there exist numbers 𝑦1, 𝑦2, . . . , 𝑦𝑘, not all zero, such that

𝑦1𝐴1 + 𝑦2𝐴2 + ⋅ ⋅ ⋅+ 𝑦𝑘𝐴𝑘 = 0

y = (𝑦1, 𝑦2, . . . , 𝑦𝑘) is a nontrivial solution to the homogeneous system.

(b) For every 𝑡 ∈ ℜ, −𝑡y ∈ kernel 𝑓 = 𝐴x and x′ = x − 𝑡y is a solution ofthe corresponding nonhomogeneous system 𝐴x = c. To see this directly,subtract

𝐴𝑡y = 0

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from

𝐴x = c

to give

𝐴x′ = 𝐴(x− 𝑡y) = c

(c) Note that x > 0 and therefore 𝑡 > 0 which implies that ��𝑗 > 0 for every𝑦𝑗 ≤ 0. For every 𝑦𝑗 > 0, 𝑥𝑗/𝑦𝑗 ≥ 𝑡, which implies that 𝑥𝑗 ≥ 𝑡𝑦𝑗 , so that

𝑥𝑗 ≥ 𝑥𝑗 − 𝑡𝑦𝑗 ≥ 0

Therefore, x is a feasible solution.

(d) There exists some coordinate ℎ such that 𝑡 = 𝑥ℎ/𝑦ℎ so that

��ℎ = 𝑥ℎ − 𝑡𝑦ℎ = 0

so that

c =

𝑘∑𝑗

𝑗 ∕=ℎ=1

𝑥𝑗𝐴𝐽

�� is a feasible solution with one less positive component.

3. Starting with any nonbasic feasible solution, this elimination technique can berepeated until the remaining vectors are linearly independent and a basic feasiblesolution is obtained.

3.112 1. Exercise 1.173.

2. For each 𝑖, there exists 𝑙𝑖 elements x𝑖𝑗 and coefficients 𝑎𝑖𝑗 > 0 such that

x𝑖 =

𝑙𝑖∑𝑖=1

𝑎𝑖𝑗x𝑖𝑗

and∑𝑙𝑖

𝑗=1 𝑎𝑖𝑗 = 1. Hence

x =

𝑛∑𝑖=1

x𝑖 =

𝑛∑𝑖=1

𝑙𝑖∑𝑗=1

𝑎𝑖𝑗x𝑖𝑗

3. Direct computation.

4. Regarding the 𝑎𝑖𝑗 as “variables” and the points 𝑧𝑖𝑗 as coefficents,

z =

𝑛∑𝑖=1

𝑙𝑖∑𝑗=1

𝑎𝑖𝑗z𝑖𝑗

is a linear equation system in which variables are restricted to be nonnegative.By the fundamental theorem of linear programming (Exercise 3.111), there exists

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a basic feasible solution. That is, there exists coefficients 𝑏𝑖𝑗 ≥ 0 and 𝑏𝑖𝑗 > 0 forat most (𝑚+ 𝑛) components such that

z =

𝑛∑𝑖=1

𝑙𝑖∑𝑗=1

𝑏𝑖𝑗z𝑖𝑗 (3.43)

Decomposing, (3.43) implies

x =

𝑛∑𝑖=1

𝑙𝑖∑𝑗=1

𝑏𝑖𝑗x𝑖𝑗

and

𝑙𝑖∑𝑗=1

𝑏𝑖𝑗 = 1 for every 𝑖

5. (3.43) implies that at least one 𝑏𝑖𝑗 > 0 for every 𝑖. This accounts for at least𝑛 of the positive 𝑏𝑖𝑗 . Since there are at most (𝑚+ 𝑛) coefficients 𝑏𝑖𝑗 which arestrictly positive, there are at most 𝑚 indices 𝑖 which have more than one positivecoefficient 𝑏𝑖𝑗 . For the remaining 𝑚− 𝑛 indices, x𝑖 = x𝑖𝑗 for some 𝑗; that isx𝑖 ∈ 𝑆𝑖.

3.113 1. Since 𝐴 is productive, there exists x ≥ 0 such that 𝐴x > 0. Consider anyz for which 𝐴z ≥ 0. For every 𝛼 > 0

𝐴(x+ 𝛼z) = 𝐴x+ 𝛼𝐴z > 0 (3.44)

Suppose to the contrary that z ∕≥ 0. That is, there exists some component 𝑧𝑖 < 0.Let

𝛼 = max{− 𝑧𝑖𝑥𝑖}

Without loss of generality, 𝑧1 attains this maximum, that is assume 𝛼 = 𝑧1/𝑥1.Then

𝑥1 + 𝛼𝑧1 = 0

and

𝑥𝑖 + 𝛼𝑧𝑖 ≥ 0

for every 𝑖.

Now consider the matrix 𝐵 = 𝐼 − 𝐴. By the assumptions of the Leontief model(Example 3.35), the matrix 𝐴 has 1 along the diagonal and negative off-diagonalelements. That is

𝑎𝑖𝑖 = 1 𝑖 = 1, 2, . . . , 𝑛

𝑎𝑖𝑗 ≤ 0 𝑖, 𝑗 = 1, 2, . . . , 𝑛, 𝑗 ∕= 𝑗

Therefore

𝐵 = 𝐼 −𝐴 ≥ 0

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That is, every element of 𝐵 is nonnegative. Consequently since x+ 𝛼z ≥ 0𝐵(x+ 𝛼z) ≥ 0 (3.45)

On the other hand, substituting 𝐴 = 𝐼 −𝐵 in (3.45)

(𝐼 −𝐵)(x + 𝛼z) > 0

x+ 𝛼z > 𝐵(x+ 𝛼z)

which implies that the first component of 𝐵(x+ 𝛼z) is negative, contradicting (3.45).This contradiction establishes that z ≥ 0.Suppose 𝐴x = 0. A fortiori 𝐴x ≥ 0. By the previous part this implies x ≥ 0. On theother hand, it also implies that −𝐴x = 𝐴(−x) = 0 so that −x ≥ 0. We conclude thatx = 0 is the only solution to 𝐴x = 0. 𝐴 is nonsingular.

Since 𝐴 is nonsingular, the system 𝐴x = y has a unique solution x for any y ≥ 0. Bythe first part, x ≥ 0.3.114 Suppose 𝐴 is productive. By the previous exercise, 𝐴 is nonsingular with inverse𝐴−1. Let e𝑖 be the 𝑖th unit vector. Since e𝑖 ≥ 0, there exists x𝑖 ≥ 0 such that

𝐴x𝑖 = e𝑖

Multiplying by 𝐴−1

x𝑖 = 𝐴−1𝐴x𝑖 = 𝐴−1e𝑖 = 𝐴−1𝑖

where 𝐴−1𝑖 is the 𝑖 column of 𝐴−1. Since x𝑖 ≥ 0 for every 𝑖, we conclude that 𝐴−1 ≥ 0.

Conversely, assume that 𝐴−1 ≥ 0. Let 1 = (1, 1, . . . , 1) denote a net output of 1 foreach commodity. Then

x = 𝐴−11 ≥ 0and

𝐴x = 1 > 0

𝐴 is productive.

3.115 Takayama 1985, p.383, Theorem 4.C.4.

3.116 Let a0 = (𝑎01, 𝑎02, . . . , 𝑎0𝑛) be the vector of labour requirements and 𝑤 the wagerate. The unit profit of industry 𝑖 is

𝜋𝑖 = 𝑝𝑖 +∑𝑗 ∕=𝑖𝑎𝑖𝑗𝑝𝑗 − 𝑤𝑎0

Recall that 𝑎𝑖𝑗 ≤ 0 for 𝑗 ∕= 𝑖. The vector of unit profits for all industries is

Π = 𝐴p− 𝑤𝑎0Profits will be zero in all industries if there exists a price system p such that

Π = 𝐴p− 𝑤𝑎0 = 0

or

𝐴p = 𝑤𝑎0 (3.46)

By the previous results, (3.46) has a unique nonnegative solution p = 𝐴−1𝑤𝑎0 if thetechnology 𝐴 is productive. Furthermore, 𝐴−1 is nonnegative. Since 𝑎0 > 0, so isp > 0.

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3.117 Let 𝑢𝐵 denote the steady state unemployment rate for blacks. Then 𝑢𝐵 satisfiesthe equation

𝑢𝐵 = 0.0038(1− 𝑢𝐵) + 0.8975𝑢𝐵

which implies that 𝑢𝐵 = 0.036. That is, the data implies an unemployment rate of 3.6percent for blacks. Similarly, the unemployment rate for white males 𝑢𝑊 satisfies theequation

𝑢𝑊 = 0.0022(1− 𝑢𝑊 ) + 0.8614𝑢𝑊

which implies that 𝑢𝑊 = 0.016 or 1.6 percent.

3.118 The transition matrix is

𝑇 =

(.6 .25.4 .75

)

If the current state vector is x0 = (.4, .6), the state vector after a single mailing will be

x1 = 𝑇x0

=

(.6 .25.4 .75

)(.4.6

)

=

(0.39.61

)

Following a single mailing, the number of subscribers will drop to 30 percent of themailing list, comprising 24 percent from renewals and 15 percent new subscriptions.

3.119 Let 𝑓(𝑥) = 𝑥2. For every 𝑥1, 𝑥2 ∈ ℜ and 0 ≤ 𝛼 ≤ 1

𝑓(𝛼𝑥1 + (1 − 𝛼)𝑥2) = (𝛼𝑥1 + (1− 𝛼)𝑥2)2

= (𝛼𝑥1 + (1− 𝛼)𝑥2)(𝛼𝑥1 + (1− 𝛼)𝑥2)

= 𝛼2𝑥21 + 2𝛼(1− 𝛼)𝑥1𝑥2 + (1− 𝛼)2𝑥22

= 𝛼𝑥21 + (1− 𝛼)𝑥22 − 𝛼𝑥21 − (1 − 𝛼)𝑥22 + 𝛼2𝑥21 + 2𝛼(1 − 𝛼)𝑥1𝑥2 + (1− 𝛼)2𝑥22

= 𝛼𝑥21 + (1− 𝛼)𝑥22 −(𝛼(1 − 𝛼)𝑥21 − 2𝛼(1− 𝛼)𝑥1𝑥2 + 𝛼(1− 𝛼)𝑥22

)= 𝛼𝑥21 + (1− 𝛼)𝑥22 − 𝛼(1− 𝛼)(𝑥1 − 𝑥2)2≤ 𝛼𝑥21 + (1− 𝛼)𝑥22

= 𝛼𝑓(𝑥1) + (1− 𝛼)(𝑥2)

3.120 𝑓(𝑥) = 𝑥 is linear and therefore convex. In the previous exercise we showed that𝑥2 is convex. Therefore 𝑓(𝑥) = 𝑥𝑛 is convex for 𝑛 = 1, 2. Assume that 𝑓 is convex for

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𝑛− 1. Then

𝑓(𝛼𝑥1 + (1 − 𝛼)𝑥2) = (𝛼𝑥1 + (1− 𝛼)𝑥2)𝑛

= (𝛼𝑥1 + (1− 𝛼)𝑥2)(𝛼𝑥1 + (1− 𝛼)𝑥2)𝑛−1

≤ (𝛼𝑥1 + (1− 𝛼)𝑥2)(𝛼𝑥𝑛−11 + (1− 𝛼)𝑥𝑛−12 ) (since 𝑥𝑛−1 is convex)

= 𝛼2𝑥𝑛1 + 𝛼(1 − 𝛼)𝑥𝑛−11 𝑥2 + 𝛼(1 − 𝛼)𝑥1𝑥𝑛−12 + (1− 𝛼)2𝑥𝑛2

= 𝛼𝑥𝑛1 + (1− 𝛼)𝑥𝑛2 − 𝛼𝑥𝑛1 − (1− 𝛼)𝑥𝑛2

+ 𝛼2𝑥𝑛1 + 𝛼(1− 𝛼)𝑥𝑛−11 𝑥2 + 𝛼(1 − 𝛼)𝑥1𝑥𝑛−12 + (1− 𝛼)2𝑥𝑛2

= 𝛼𝑥𝑛1 + (1− 𝛼)𝑥𝑛2 − 𝛼(1− 𝛼)(𝑥𝑛1 − 𝑥1𝑥𝑛−12 − 𝑥𝑛−11 𝑥2 + 𝑥𝑛2

)= 𝛼𝑥𝑛1 + (1− 𝛼)𝑥𝑛2 − 𝛼(1− 𝛼)

(𝑥𝑛−11 (𝑥1 − 𝑥2)− 𝑥𝑛−12 (𝑥1 − 𝑥2)

)= 𝛼𝑥𝑛1 + (1− 𝛼)𝑥𝑛2 − 𝛼(1− 𝛼)

((𝑥1 − 𝑥2)(𝑥𝑛−11 − 𝑥𝑛−12 )

)Since 𝑥𝑚 is monotonic (Example 2.53)

𝑥𝑛−11 − 𝑥𝑛−12 ≥ 0 ⇐⇒ 𝑥1 − 𝑥2 ≥ 0

and therefore

(𝑥1 − 𝑥2)(𝑥𝑛−11 − 𝑥𝑛−12 ) ≥ 0

We conclude that

𝑓(𝛼𝑥1 + (1 − 𝛼)𝑥2) ≤ 𝛼𝑥𝑛1 + (1− 𝛼)𝑥𝑛2 = 𝛼𝑓(𝑥1) + (1− 𝛼)(𝑥2)

𝑓 is convex for all 𝑛 = 1, 2, . . . .

3.121 For given x1,x2 ∈ 𝑆, define 𝑔 : [0, 1]→ 𝑆 by

𝑔(𝑡) = (1− 𝑡)x1 + 𝑡x2

Then 𝑔(0) = x1, 𝑔(1) = x2 and ℎ = 𝑔 ∘ 𝑓 .

Assume 𝑓 is convex. For any 𝑡1, 𝑡2 ∈ [0, 1], let

𝑔(𝑡1) = x1 and 𝑔(𝑡2) = x2

For any 𝛼 ∈ [0, 1]

𝑔(𝛼𝑡1 + (1− 𝛼)𝑡2

)= 𝛼x1 + (1 − 𝛼)x2

ℎ(𝛼𝑡1 + (1− 𝛼)𝑡2

)= 𝑓

(𝛼x1 + (1− 𝛼)x2

)≤ 𝛼𝑓(x1) + (1 − 𝛼)𝑓(x2)

≤ 𝛼ℎ(𝑡1) + (1 − 𝛼)𝑡2)

ℎ is convex.

Conversely, assume ℎ is convex for any x1,x2 ∈ 𝑆. For any 𝛼 ∈ [0, 1]

𝑔(𝛼) = 𝛼x1 + (1− 𝛼)x2

and

𝑓(𝛼x1 + (1− 𝛼)x2

)= ℎ(𝛼)

≤ 𝛼ℎ(0) + (1− 𝛼)ℎ(1)

= 𝛼𝑓(x1) + (1− 𝛼)𝑓(x2)

Since this is true for any x1,x2 ∈ 𝑆, we conclude that 𝑓 is convex.

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3.122 Assume 𝑓 is convex which implies epi 𝑓 is convex. The points (x𝑖, 𝑓(x𝑖)) ∈ epi 𝑓 .Since epi 𝑓 is convex

𝛼1(x1, 𝑓(x1)) + 𝛼2(x1, 𝑓(x1)) + ⋅ ⋅ ⋅+ (x𝑛, 𝑓(x𝑛)) ∈ epi 𝑓

that is

𝑓(𝛼1x1 + 𝛼2x2 + ⋅ ⋅ ⋅+ 𝛼𝑛x𝑛) ≤ 𝛼1𝑓(x1) + 𝛼2𝑓(x1) + ⋅ ⋅ ⋅+ 𝛼𝑛𝑓(x𝑛))

Conversely, letting 𝑛 = 2 and 𝛼 = 𝛼1, (3.25) implies that

𝑓(𝛼x1 + (1− 𝛼)x2) ≤ 𝛼𝑓(x1) + (1− 𝛼)𝑓(x2)

Jensen’s inequality can also be proved by induction from the definition of a convexfunction (see for example Sydsaeter + Hammond 1995; p.624).

3.123 For each 𝑖, let 𝑦𝑖 = log 𝑥𝑖 so that

𝑥𝑖 = 𝑒𝑦𝑖

𝑥𝛼𝑖

𝑖 = 𝑒𝛼𝑖𝑦𝑖

Since 𝑒𝑥 is convex (Example 3.41)

𝑥𝑎11 𝑥

𝑎22 . . . 𝑥

𝑎𝑛𝑛 𝑎𝑖 > 0 =

∏exp(𝛼𝑖𝑦𝑖) = exp

(∑𝛼𝑖𝑦𝑖

)≤

∑𝛼𝑖𝑒

𝑦𝑖 =∑𝛼𝑖𝑥𝑖

by Jensen’s inequality. Setting 𝛼𝑖 = 1/𝑛, we have

(𝑥1𝑥2 . . . 𝑥𝑛)1/𝑛 ≤ 1

𝑛

𝑛∑𝑖=1

𝑥𝑖

as required.

3.124 Assume 𝑓 is concave. That is for every x1,x2 ∈ 𝑆 and 0 ≤ 𝛼 ≤ 1

𝑓(𝛼x1 + (1− 𝛼)x2) ≥ 𝛼𝑓(x1) + (1− 𝛼)𝑓(x2)

Multiplying through by −1 reverses the inequality so that

−𝑓(𝛼x1 + (1− 𝛼)x2) ≤ −𝛼𝑓(x1) + (1− 𝛼)𝑓(x2) = 𝛼− 𝑓(x1) + (1− 𝛼)− 𝑓(x2)

which shows that −𝑓 is concave. The converse follows analogously.

3.125 Assume that 𝑓 is concave. Then −𝑓 is convex and by Theorem 3.7

epi − 𝑓 = { (𝑥, 𝑦) ∈ 𝑋 ×ℜ : 𝑦 ≥ −𝑓(𝑥), 𝑥 ∈ 𝑋 }

is convex. But

epi − 𝑓 = { (𝑥, 𝑦) ∈ 𝑋 ×ℜ : 𝑦 ≥ −𝑓(𝑥), 𝑥 ∈ 𝑋 } = { (𝑥, 𝑦) ∈ 𝑋 ×ℜ : 𝑦 ≤ 𝑓(𝑥), 𝑥 ∈ 𝑋 } = hypo 𝑓

Therefore hypo 𝑓 is convex.

Conversely, if hypo 𝑓 is convex, epi − 𝑓 is convex which implies that −𝑓 is convex andhence 𝑓 is concave.

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3.126 Suppose that x1 minimizes the cost of producing 𝑦 at input prices w1 while x2

minimizes cost at w2. For some 𝛼 ∈ [0, 1], let w be the weighted average price, that is

w = 𝛼w1 + (1 − 𝛼)w2

and suppose that x minimizes cost at w. Then

𝑐(w, 𝑦) = wx

= (𝛼w1 + (1 − 𝛼)w2)x

= 𝛼w1x+ (1 − 𝛼)w2x

But since x1 and x2 minimize cost at w1 and w2 respectively

𝛼w1x ≥ 𝛼w1x1 = 𝛼𝑐(w1, 𝑦)

(1− 𝛼)w2x ≥ (1− 𝛼)w2x2 = (1 − 𝛼)𝑐(w2, 𝑦)

so that

𝑐(w, 𝑦) = 𝑐(𝛼w1 + (1− 𝛼)w2, 𝑦) = 𝛼w1x+ (1− 𝛼)w2x ≥ 𝛼𝑐(w1, 𝑦) + (1− 𝛼)𝑐w2, 𝑦)

This establishes that the cost function 𝑐 is concave in w.

3.127 Since 𝑢 is concave, Jensen’s inequality implies

𝑢

(𝑇∑𝑡=1

1

𝑇𝑐𝑡

)≥

𝑇∑𝑡=1

1

𝑇𝑢(𝑐𝑡) =

1

𝑇

𝑇∑𝑡=1

𝑢(𝑐𝑡)

for any consumption stream 𝑐1, 𝑐2, . . . , 𝑐𝑇 so that

𝑈 =

𝑇∑𝑡=1

𝑢(𝑐𝑡) ≤ 𝑇𝑢(

𝑇∑𝑡=1

1

𝑇𝑐𝑡

)= 𝑇𝑢(𝑐)

It is impossible to do better than consume a constant fraction 𝑐 = 𝑤/𝑇 of wealth ineach period.

3.128 If 𝑥1 = 𝑥3, the inequality is trivially satisfied. Now assume 𝑥1 ∕= 𝑥3. Since𝑥2 ∈ [𝑥1, 𝑥3], there exists 𝛼 ∈ [0, 1] such that

𝑥2 = 𝛼𝑥1 + (1− 𝛼)𝑥2

Let �� = 𝑥1 − 𝑥2 + 𝑥3. Then �� ∈ [𝑥1, 𝑥3] and there exists 𝛽 ∈ [0, 1] such that

𝑥 = 𝛽𝑥1 + (1− 𝛽)𝑥2

Adding

𝑥+ 𝑥2 = (𝛼+ 𝛽)𝑥1 +((1− 𝛼) + (1− 𝛽)

)𝑥3

or

𝑥1 − 𝑥3 = (𝛼+ 𝛽)(𝑥3 − 𝑥1)which implies that 𝛼+ 𝛽 = 1 and therefore 𝛽 = 1− 𝛼. Since 𝑓 is convex

𝑓(𝑥2) ≤ 𝛼𝑓(𝑥1) + (1− 𝛼)𝑓(𝑥2

𝑓(��) ≤ 𝛽𝑓(𝑥1) + (1− 𝛽)𝑓(𝑥2)

= (1− 𝛼)𝑓(𝑥1) + 𝛼𝑓(𝑥3)

Adding

𝑓(��) + 𝑓(𝑥2) ≤ 𝑓(𝑥1) + 𝑓(𝑥3)

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3.129 Let 𝑥1, 𝑥2, 𝑦1, 𝑦2 ∈ ℜ with 𝑥1 < 𝑥2 and 𝑦1 < 𝑦2. Note that 𝑥1 − 𝑦2 ≤ 𝑥2 − 𝑦2 ≤𝑥2 − 𝑦1 and therefore (Exercise 3.128)

𝑓(𝑥1 − 𝑦2)− (𝑥2 − 𝑦2) + (𝑥2 − 𝑦1)

)> 𝑓(𝑥1 − 𝑦2)− 𝑓(𝑥2 − 𝑦2) + 𝑓(𝑥2 − 𝑦1)

That is

𝑓(𝑥1 − 𝑦1) > 𝑓(𝑥1 − 𝑦2)− 𝑓(𝑥2 − 𝑦2) + 𝑓(𝑥2 − 𝑦1)Rearranging

𝑓(𝑥2 − 𝑦2)− 𝑓(𝑥1 − 𝑦2) > 𝑓(𝑥2 − 𝑦1)− 𝑓(𝑥1 − 𝑦1)as required.

3.130 A functional is affine if and only if inequalities (3.24) and (3.26) are satisfied asequalities.

3.131 Since 𝑓 and 𝑔 are convex on 𝑆

𝑓(𝛽x1 + (1− 𝛽)x2) ≤ 𝛽𝑓(x1) + (1− 𝛽)𝑓(x2) (3.47)

𝑔(𝛽x1 + (1− 𝛽)x2) ≤ 𝛽𝑔(x1) + (1− 𝛽)𝑔(x2) (3.48)

for every x1,x2 ∈ 𝑆 and 𝛽 ∈ [0, 1]. Adding

(𝑓 + 𝑔)(𝛽x1 + (1− 𝛽)x2) ≤ 𝛽(𝑓 + 𝑔)(x1) + (1 − 𝛽)𝑓(x2)

𝑓 + 𝑔 is convex. Multiplying (3.47) by 𝛼 ≥ 0

𝛼𝑓(𝛽x1 + (1− 𝛽)x2) ≤ 𝛼(𝛽𝑓(x1) + (1− 𝛽)𝑓(x2))

= (𝛽𝛼𝑓(x1) + (1− 𝛽)𝛼𝑓(x2))

𝛼𝑓 is convex.

Moreover, if 𝑓 is strictly convex,

𝑓(𝛽x1 + (1− 𝛽)x2) < 𝛽𝑓(x1) + (1− 𝛽)𝑓(x2) (3.49)

for every x1,x2 ∈ 𝑆, x1 ∕= x2 and 𝛽 ∈ (0, 1). Adding this to (3.48)

(𝑓 + 𝑔)(𝛽x1 + (1− 𝛽)x2) < 𝛽(𝑓 + 𝑔)(x1) + (1 − 𝛽)𝑓(x2)

so that 𝑓 + 𝑔 is strictly convex. Multiplying (3.49) by 𝛼 > 0

𝛼𝑓(𝛽x1 + (1− 𝛽)x2) < 𝛼(𝛽𝑓(x1) + (1− 𝛽)𝑓(x2))

= (𝛽𝛼𝑓(x1) + (1− 𝛽)𝛼𝑓(x2))

𝛼𝑓 is strictly convex.

3.132

x ∈ epi (𝑓 ∨ 𝑔) ⇐⇒ x ∈ epi 𝑓 and x ∈ epi 𝑔

That is

epi (𝑓 ∨ 𝑔) = epi 𝑓 ∩ epi 𝑔

Therefore epi 𝑓 ∨ 𝑔 is convex (Exercise 1.162) and therefore 𝑓 is convex (Proposition3.7).

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3.133 If 𝑓 is convex

𝑓(𝛼x1 + (1− 𝛼)x2) ≤ 𝛼𝑓(x1) + (1− 𝛼)𝑓(x2)

Since 𝑔 is increasing

𝑔(𝑓(𝛼x1 + (1 − 𝛼)x2)

) ≤ 𝑔(𝛼𝑓(x1) + (1− 𝛼)𝑓(x2))

≤ 𝛼𝑔(𝑓(x1))

+ (1 − 𝛼)𝑔(𝑓(x2)

)since 𝑔 is also convex. The concave case is proved similarly.

3.134 Let 𝐹 = log 𝑓 . If 𝐹 is convex, 𝑓(x) = 𝑒𝐹 (x) is an increasing convex function of aconvex function and is therefore convex (Exercise 3.133).

3.135 If 𝑓 is positive and concave, then log 𝑓 is concave (Exercise 3.51). Therefore

log1

𝑓= log 1− log 𝑓 = − log 𝑓

is convex. By the previous exercise (Exercise 3.134), this implies that 1/𝑓 is convex.

If 𝑓 is negative and convex, then −𝑓 is positive and concave, 1/ − 𝑓 is convex, andtherefore 1/𝑓 is concave.

3.136 Consider the identity

𝑔(𝑓(𝑥1 ∨ 𝑥2)

)+ 𝑔

(𝑓(𝑥1 ∧ 𝑥2)

)− 𝑔(𝑓(𝑥1))− 𝑔(𝑓(𝑥2)

)= 𝑔

(𝑓(𝑥1 ∨ 𝑥2)

)+ 𝑔

(𝑓(𝑥1 ∧ 𝑥2)

)− 𝑔(𝑓(𝑥1))− 𝑔(𝑓(𝑥1 ∨ 𝑥2)

)+ 𝑓(𝑥1 ∧ 𝑥2)

)− 𝑓(𝑥1))

+ 𝑔(𝑓(𝑥1 ∨ 𝑥2) + 𝑓(𝑥1 ∧ 𝑥2)− 𝑓(𝑥1)

)− 𝑔(𝑓(𝑥2))

(3.50)

Define

𝜑(𝑥1, 𝑥2) = 𝑔(𝑓(𝑥1 ∨ 𝑥2)

)+ 𝑔

(𝑓(𝑥1 ∧ 𝑥2)

)− 𝑔(𝑓(𝑥1))− 𝑔(𝑓(𝑥2)

)Then 𝑔 ∘ 𝑓 is supermodular if 𝜑 is nonnegative definite and submodular if 𝜑 is nonpos-itive definite. Using the identity (3.50), 𝜑 can be decomposed into two components

𝜑(𝑥1, 𝑥2) = 𝜑1(𝑥1, 𝑥2) + 𝜑2(𝑥1, 𝑥2)

𝜑1(𝑥1, 𝑥2) = 𝑔(𝑓(𝑥1 ∨ 𝑥2)

)+ 𝑔

(𝑓(𝑥1 ∧ 𝑥2)

)− 𝑔(𝑓(𝑥1))

− 𝑔(𝑓(𝑥1 ∨ 𝑥2) + 𝑓(𝑥1 ∧ 𝑥2)− 𝑓(𝑥1))

(3.51)

𝜑2(𝑥1, 𝑥2) = 𝑔(𝑓(𝑥1 ∨ 𝑥2) + 𝑓(𝑥1 ∧ 𝑥2)− 𝑓(𝑥1)

)− 𝑔(𝑓(𝑥2))

𝜑 will definite if both components are definite.

For any 𝑥1, 𝑥2 ∈ 𝑥1, let 𝑎 = 𝑓(𝑥1 ∧ 𝑥2), 𝑏 = 𝑓(𝑥1) and 𝑐 = 𝑓(𝑥1 ∨ 𝑥2). Provided 𝑓 ismonotone, 𝑏 lies between 𝑎 and 𝑐. Substituting in (3.51)

𝜑1(𝑥1, 𝑥2) = 𝑔(𝑐) + 𝑔(𝑎)− 𝑔(𝑏)− 𝑔(𝑐+ 𝑎− 𝑏)and Exercise 3.128 implies

𝜑1(𝑥1, 𝑥2) = 𝑔(𝑐) + 𝑔(𝑎)− 𝑔(𝑏)− 𝑔(𝑐+ 𝑎− 𝑏){≥ 𝑂≤ 0

}if 𝑔 is

{convex

concave

}(3.52)

Now consider 𝜑2.

𝑓(𝑥1 ∨ 𝑥2) + 𝑓(𝑥1 ∧ 𝑥2)− 𝑓(𝑥1) is

{≥≤

}𝑓(𝑥2) if 𝑓 is

{supermodular

submodular

}

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and therefore since 𝑔 is increasing

𝜑2(𝑥1, 𝑥2) =

{≥ 0 if 𝑓 is supermodular

≤ 0 if 𝑓 is submodular(3.53)

Together (3.52) and (3.53) gives the desired result.

3.137 1. Assume that 𝑓 is bounded above in a neighborhood of x0. Then thereexists a ball 𝐵(𝑥0) and constant 𝑀 such that

𝑓(x) ≤𝑀 for every x ∈ 𝐵(𝑥0)

Since 𝑓 is convex

𝑓(𝛼x+ (1 − 𝛼)x0) ≤ 𝛼𝑓(x) + (1 − 𝛼)𝑓(x0) ≤ 𝛼𝑀 + (1 − 𝛼)𝑓(x0) (3.54)

2. Given x ∈ 𝐵(𝑥0) and 𝛼 ∈ [0, 1] let

z = 𝛼x + (1− 𝛼)x0 (3.55)

Subtracting 𝑓(x0) from (3.54) gives

𝑓(z)− 𝑓(x0) ≤ 𝛼(𝑀 − 𝑓(x0)) (3.56)

Rewriting (3.55)

(1− 𝛼)x0 = z− 𝛼x(1 + 𝛼)x0 = z+ 𝛼(2x0 − x)

x0 =1

1 + 𝛼z+

𝛼

1 + 𝛼(2x0 − x)

3. Note that

(2x0 − x) = x0 − (x− x0) ∈ 𝐵(x0)

so that

𝑓(2x0 − x) ≤𝑀and therefore

𝑓(x0) ≤ 1

1 + 𝛼𝑓(z) +

𝛼

1 + 𝛼𝑓(2x0 − x) ≤ 1

1 + 𝛼𝑓(z) +

𝛼

1 + 𝛼𝑀

which implies

(1 + 𝛼)𝑓(x0) ≤ 𝑓(z) + 𝛼𝑀

𝛼(𝑓(x0)−𝑀) ≤ 𝑓(z)− 𝑓(x0)

4. Combined with (3.56) we have

𝛼(𝑓(x0)−𝑀) ≤ 𝑓(z)− 𝑓(x0) ≤ 𝛼(𝑀 − 𝑓(x0))

or

∣𝑓(z)− 𝑓(x0)∣ ≤ 𝛼(𝑀 − 𝑓(x0))

and therefore 𝑓(z)→ 𝑓(x0) as z→ x0. 𝑓 is continuous.

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3.138 1. Since 𝑆 is open, there exists a ball 𝐵𝑟(x1) ⊆ 𝑆. Let 𝑡 = 1 + 𝑟2 . Then

x0 + 𝑡(x1 − x0) ∈ 𝐵𝑟(𝑥1) ⊆ 𝑆.

2. Let 𝑠 = 𝑡−1𝑡 𝑟. The open ball 𝐵𝑠(x1) of radius 𝑠 centered on x1 is contained in 𝑇 .

Therefore 𝑇 is a neighborhood of x1.

3. Since 𝑓 is convex, for every y ∈ 𝑇

𝑓(y) ≤ (1− 𝛼)𝑓(x) + 𝛼𝑓(z) ≤ (1− 𝛼)𝑀 + 𝛼𝑓(z) ≤𝑀 + 𝑓(z)

Therefore 𝑓 is bounded on 𝑇 .

3.139 The previous exercise showed that 𝑓 is locally bounded from above for everyx ∈ 𝑆. To show that it is also locally bounded from below, choose some x0 ∈ 𝑆. Thereexists some 𝐵(x0 and 𝑀 such that

𝑓(x) ≤𝑀 for every x ∈ 𝐵(x0)

Choose some 𝑥1 ∈ 𝐵(x0) and let x2 = 2x0 − x1. Then

x2 = 2x0 − x1 = x0 − (x1 − x0) ∈ 𝐵(x0)

and 𝑓(x2) ≤𝑀 . Since 𝐹 is convex

𝑓(x) ≤ 1

2𝑓(x1) +

1

2𝑓(x2)

and therefore

𝑓(x1) ≥ 2𝑓(x)− 𝑓(x2)

Since 𝑓(x2) ≤𝑀 , −𝑓(x2) ≥ −𝑀 and therefore

𝑓(x1) ≥ 2𝑓(x)−𝑀

so that 𝑓 is bounded from below.

3.140 Let 𝑓 be a convex function defined on an open convex set 𝑆 in a normed linearspace, which is bounded from above in a neighborhood of a single point x0 ∈ 𝑆. ByExercise 3.138, 𝑓 is bounded above at every x ∈ 𝑆. This implies (Exercise 3.137) that𝑓 is continuous at every x ∈ 𝑆.

3.141 Without loss of generality, assume 0 ∈ 𝑆. Assume 𝑆 has dimension 𝑛 and letx1,x2, . . . ,x𝑛 be a basis for the subspace containing 𝑆. Choose some 𝜆 > 0 smallenough so that

𝑈 = conv {0, 𝜆x1, 𝜆x2, . . . , 𝜆𝑥𝑛} ⊆ 𝑆

Any x ∈ 𝑈 is a convex combination of the points 0,x1,x2, . . . ,x𝑛 and so there exists𝛼0, 𝛼1, 𝛼2, . . . , 𝛼𝑛 ≥ 0,

∑𝛼𝑖 = 1 such that x = 𝛼00+ 𝛼1x1 + ⋅ ⋅ ⋅+ 𝛼𝑛x𝑛. By Jensen’s

inequality

𝑓(x) = 𝑓(𝛼00+ 𝛼1x1 + ⋅ ⋅ ⋅+ 𝛼𝑛x𝑛) ≤ 𝛼0𝑓(0) + 𝛼1𝑓(x1) + ⋅ ⋅ ⋅+ 𝛼𝑛𝑓(x𝑛)

≤ max{ 𝑓(0), 𝑓(x1), . . . , 𝑓(x𝑛) }

Therefore, 𝑓 is bounded above on a neighbourhood of some x0 ∈ int 𝑈 (which isnonempty by Exercise 1.229). By Proposition 3.8, 𝑓 is continuous on 𝑆.

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3.142 Clearly, if 𝑓 is convex, it is locally convex at every x ∈ 𝑆, where 𝑆 is the requiredneighborhood. To prove the converse, assume to the contrary that 𝑓 is locally convexat every x ∈ 𝑆 but it is not globally convex. That is, there exists x1,x2 ∈ 𝑆 such that

𝑓(𝛼x1 + (1− 𝛼)x2) > 𝛼𝑓(x1) + (1− 𝛼)𝑓(x2)

Let

ℎ(𝑡) = 𝑓(𝑡x1 + (1 − 𝑡)x2

)Local convexity implies that 𝑓 is continuous at every x ∈ 𝑆 (Corollary 3.8.1), andtherefore continuous on 𝑆. Therefore, ℎ is continuous on [0, 1]. By the continuousmaximum theorem (Theorem 2.3),

𝑇 = arg maxx∈[x1,x2]

ℎ(𝑡)

is nonempty and compact. Let 𝑡0 = max𝑇 . For every 𝜖 > 0,

ℎ(𝑡0 − 𝜖) ≤ ℎ(𝑡0) and ℎ(𝑡0 + 𝜖) < ℎ(𝑡0)

Let

x0 = 𝑡0x1 + (1 − 𝑡0)x2 and x𝜖 = (𝑡0 + 𝜖)x1 + (1− 𝑡0 − 𝜖)x2Every neighborhood 𝑉 of x0 contains x−𝜖,x𝜖 ∈ [x1,x2] with

1

2𝑓(x−𝜖) +

1

2𝑓(x𝜖) =

1

2ℎ(𝑡0 − 𝜖) +

1

2ℎ(𝑡0 + 𝜖) < ℎ(𝑡0) = 𝑓(x0) = 𝑓

(1

2x−𝜖 +

1

2x𝜖

)

contradicting the local convexity of 𝑓 at x0.

3.143 Assume 𝑓 is quasiconcave. That is for every x1,x2 ∈ 𝑆 and 0 ≤ 𝛼 ≤ 1

𝑓(𝛼x1 + (1 − 𝛼)x2) ≥ min{𝑓(x1), (x2)}

Multiplying through by −1 reverses the inequality so that

−𝑓(𝛼x1 + (1− 𝛼)x2) ≤ −min{𝑓(x1), 𝑓(x2)} = max{−𝑓(x1),−𝑓(x2)}

which shows that −𝑓 is quasiconvex. The converse follows analogously.

3.144 Assume 𝑓 is concave, that is

𝑓(𝛼x1 + (1− 𝛼)x2) ≥ 𝛼𝑓(x1) + (1− 𝛼)𝑓(x2) for every x1,x2 ∈ 𝑆 and 0 ≤ 𝛼 ≤ 1

Without loss of generality assume that 𝑓(x1) ≤ 𝑓(x2). Then

𝑓(𝛼x1 + (1 − 𝛼)x2) ≥ 𝛼𝑓(x1) + (1− 𝛼)𝑓(x2) ≥ 𝛼𝑓(x1) + (1− 𝛼)𝑓(x1) = 𝑓(x1) = min{𝑓(x1), 𝑓(x2)}

𝑓 is quasiconcave.

3.145 Let 𝑓 : ℜ → ℜ. Choose any 𝑥1, 𝑥2 in ℜ with 𝑥1 < 𝑥2. If 𝑓 is increasing, then

𝑓(𝑥1) ≤ 𝑓(𝛼𝑥1 + (1− 𝛼)𝑥2) ≤ 𝑓(𝑥2)

for every 0 ≤ 𝛼 ≤ 1. The first inequality implies that

𝑓(𝑥1) = min{𝑓(𝑥1), 𝑓(𝑥2)} ≤ 𝑓(𝛼𝑥1 + (1− 𝛼)𝑥2)

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so that 𝑓 is quasiconcave. The second inequality implies that

𝑓(𝛼𝑥1 + (1− 𝛼)𝑥2) ≤ max{𝑓(𝑥1), 𝑓(𝑥2)} = 𝑓(𝑥2)

so that 𝑓 is also quasiconvex.

Conversely, if 𝑓 is decreasing

𝑓(𝑥1) ≥ 𝑓(𝛼𝑥1 + (1− 𝛼)𝑥2) ≥ 𝑓(𝑥2)

for every 0 ≤ 𝛼 ≤ 1. The first inequality implies that

𝑓(𝑥1) = max{𝑓(𝑥1), 𝑓(𝑥2)} ≥ 𝑓(𝛼𝑥1 + (1− 𝛼)𝑥2)

so that 𝑓 is quasiconvex. The second inequality implies that

𝑓(𝛼𝑥1 + (1− 𝛼)𝑥2) ≤ max{𝑓(𝑥1), 𝑓(𝑥2)} = 𝑓(𝑥2)

so that 𝑓 is also quasiconcave.

3.146

≾𝑓(𝑐) = {x ∈ 𝑋 : 𝑓(x) ≤ 𝑎 } = {x ∈ 𝑋 : −𝑓(x) ≥ −𝑐} = ≿−𝑓(−𝑐)3.147 For given 𝑐 and 𝑚, choose any p1 and p2 in ≾𝑣(𝑐). For any 0 ≤ 𝛼 ≤ 1, letp = 𝛼p1 + (1 − 𝛼)p2. The key step is to show that any commodity bundle x which isaffordable at p is also affordable at either p1 or p2. Assume that x is affordable at p,that is x is in the budget set

x ∈ 𝑋(p,𝑚) = {x : px ≤ 𝑚 }To show that x is affordable at either p1 or p2, that is

x ∈ 𝑋(p1,𝑚) or x ∈ 𝑋(p2,𝑚)

assume to the contrary that

x /∈ 𝑋(p1,𝑚) and x /∈ 𝑋(p2,𝑚)

This implies that

p1x > 𝑚 and p2x > 𝑚

so that

𝛼p1x > 𝛼𝑚 and (1− 𝛼)p2 > (1− 𝛼)𝑚

Summing these two inequalities

px = (𝛼p1 + (1− 𝛼)p2)x > 𝑚

contradicting the assumption that x ∈ 𝑋(p,𝑚). We conclude that

𝑋(p,𝑚) ⊆ 𝑋(p1,𝑚) ∪𝑋(p2,𝑚)

Now

𝑣(𝑝,𝑚) = sup{ 𝑢(x) : x ∈ 𝑋(p,𝑚) }≤ sup{ 𝑢(x) : x ∈ 𝑋(p1,𝑚) ∪𝑋(p2,𝑚) }≤ 𝑐

Therefore p ∈ ≾𝑣(𝑐) for every 0 ≤ 𝛼 ≤ 1. Thus, ≾𝑣(𝑐) is convex and so 𝑣 is quasiconvex(Exercise 3.146).

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3.148 Since 𝑓 is quasiconcave

𝑓(𝛼x1 + (1− 𝛼)x2) ≥ min{𝑓(x1), 𝑓(x2)} for every x1,x2 ∈ 𝑆 and 0 ≤ 𝛼 ≤ 1

Since 𝑔 is increasing

𝑔(𝑓(𝛼x1 + (1− 𝛼)x2)

) ≥ 𝑔((min{𝑓(x1), 𝑓(x2)})) ≥ min{𝑔(𝑓(x1)), 𝑔

(𝑓(x2)

)}𝑔 ∘ 𝑓 is quasiconcave.

3.149 When 𝜌 ≥ 1, the function

ℎ(x) = 𝛼1𝑥𝜌1 + 𝛼2𝑥

𝜌2 + . . . 𝛼𝑛𝑥

𝜌𝑛

is convex (Example 3.58) as is 𝑦1/𝜌. Therefore

𝑓(x) = (ℎ(x))1/𝜌

is an increasing convex function of a convex function and is therefore convex (Exercise3.133).

3.150 𝑓 is a monotonic transformation of the concave function ℎ(x) = x.

3.151 By Exercise 3.39, there exist linear functionals 𝑓 and 𝑔 and scalars 𝑏 and 𝑐 suchthat

𝑓(x) = 𝑓(x) + 𝑏 and 𝑔(x) = 𝑔(x) + 𝑐

The upper contour set

≿ℎ(𝑎) = { 𝑥 ∈ 𝑆 : ℎ(x) ≥ 𝑎 }

= { 𝑥 ∈ 𝑆 :𝑓(𝑥) + 𝑏

𝑔(𝑥) + 𝑐≥ 𝑎 }

= { 𝑥 ∈ ℜ𝑛+ : 𝑓(x) + 𝑏 ≥ 𝑎𝑔(x) + 𝑎𝑐 }= { 𝑥 ∈ ℜ𝑛+ : 𝑓(x)− 𝑎𝑔(x) ≥ 𝑏− 𝑎𝑐 }

which is a halfspace in 𝑋 and therefore convex. Similarly, the lower contour set

≾ℎ(𝑎) = { 𝑥 ∈ 𝑆 : ℎ(x) ≥ 𝑎 }is also a halfspace and hence convex. Therefore ℎ is both quasiconcave and quasiconvex.

3.152 For 𝑎 ≤ 0

≿(𝑎) = { 𝑥 ∈ 𝑆 : ℎ(x) ≥ 0 } = 𝑆

which is convex. For 𝑎 > 0

≿ℎ(𝑎) = { 𝑥 ∈ 𝑆 : ℎ(x) ≥ 𝑎 }

= { 𝑥 ∈ 𝑆 :𝑓(x)

𝑔(x)≥ 𝑎 }

= { 𝑥 ∈ 𝑆 : 𝑓(x) ≥ 𝑎𝑔(x) }= { 𝑥 ∈ 𝑆 : 𝑓(x)− 𝑎𝑔(x) ≥ 0 }

is convex since 𝑓 −𝑎𝑔 = 𝑓 +𝑎(−𝑔) is concave (Exercises 3.124 and 3.131). Since ≿ℎ(𝑎)is convex for every 𝑎, ℎ is quasiconcave.

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3.153

ℎ(x) =𝑓(x)

𝑔(x)

where 𝑔 = 1/𝑔 is positive and convex by Exercise 3.135. By the previous exercise, ℎ isquasiconcave.

3.154 Let 𝐹 = log 𝑓 . If 𝐹 is concave, 𝑓(x) = 𝑒𝐹 (x) is an increasing function of(quasi)concave function, and hence is quasiconcave (Exercise 3.148).

3.155 Let

𝐹 (x) = log 𝑓(x) =𝑛∑𝑖=1

𝛼𝑖 log 𝑓𝑖(x)

As the sum of concave functions, 𝐹 is concave (Exercise 3.131). By the previousexercise, 𝑓 is quasiconcave.

3.156 Assume x1, x2 and x are optimal solutions for 𝜽1, 𝜽2 and �� = 𝛼𝜽1 + (1 − 𝛼)𝜽2respectively. That is

𝑓(x1, 𝜽1) = 𝑣(𝜽1)

𝑓(x2, 𝜽2) = 𝑣(𝜽2)

𝑓(x, ��) = 𝑣(��)

Since 𝑓 is convex in 𝜽

𝑣(��) = 𝑓(x, ��)

= 𝑓(x, 𝛼𝜽1 + (1 − 𝛼)𝜽2)

≤ 𝛼𝑓(x, 𝜽1) + (1− 𝛼)𝑓(x∗, 𝜽2)≤ 𝛼𝑓(x1, 𝜽1) + (1− 𝛼)𝑓(x2, 𝜽2)

= 𝛼𝑣(𝜽1) + (1− 𝛼)𝑣(𝜽2)

𝑣 is convex.

3.157 Assume to the contrary that x1 and x2 are distinct optimal solutions, that isx1,x2 ∈ 𝜑(𝜽), x1 ∕= x2, for some 𝜽 ∈ Θ∗, so that

𝑓(x1, 𝜽) = 𝑓(x2, 𝜽) = 𝑣(𝜽) ≥ 𝑓(x, 𝜽) for every x ∈ 𝐺(𝜽)

Let x = 𝛼x1 + (1 − 𝛼)x2 for 𝛼 ∈ (0, 1). Since 𝐺(𝜽) is convex, x is feasible. Since 𝑓 isstrictly quasiconcave

𝑓(x, 𝜽) > min{ 𝑓(x1, 𝜽), 𝑓(x2, 𝜽) } = 𝑣(𝜽)

contradicting the optimality of x1 and x2. We conclude that 𝜑(𝜽) is single-valued forevery 𝜽 ∈ Θ∗. In other words, 𝜑 is a function.

3.158 1. The value function is


𝑈(x)

where

𝑈(x) =

∞∑𝑡=0


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and

Γ(𝑥0) = {x ∈ 𝑋∞ : 𝑥𝑡+1 ∈ 𝐺(𝑥𝑡), 𝑡 = 0, 1, 2, . . .}Since an optimal policy exists (Exercise 2.125), the maximum is attained and

𝑣(𝑥0) = maxx∈Γ(𝑥0)

𝑈(x) (3.57)

It is straightforward to show that

∙ 𝑈(x) is strictly concave and

∙ Γ(𝑥0) is convex

Applying the Concave Maximum Theorem (Theorem 3.1) to (3.57), we concludethat the value function 𝑣 is strictly concave.

2. Assume to the contrary that x′ and x′′ are distinct optimal plans, so that

𝑣(𝑥0) = 𝑈(x′) = 𝑈(x′′)

Let x = 𝛼x′ + (1− 𝛼)x′′. Since Γ(𝑥0) is convex, x is feasible and

𝑈(x) > 𝛼𝑈(x′) + (1− 𝛼)𝑈(x′′) = 𝑈(x′)

which contradicts the optimality of x′. We conclude that the optimal plan isunique.

3.159 We observe that

∙ 𝑢(𝐹 (𝑘)− 𝑦) is supermodular in 𝑦 (Exercise 2.51)

∙ 𝑢(𝐹 (𝑘)− 𝑦) displays strictly increasing differences in (𝑘, 𝑦) (Exercise 3.129)

∙ 𝐺(𝑘) = [0, 𝐹 (𝑘)] is increasing.

Applying Exercise 2.126, we can conclude that the optimal policy (𝑘0, 𝑘∗1 , 𝑘

∗2 , . . . ) is a

monotone sequence. Since 𝑋 is compact, k∗ is a bounded monotone sequence, whichconverges monotonically to some steady state 𝑘∗ (Exercise 1.101).

3.160 Suppose there exists (x∗,y∗) ∈ 𝑋 × 𝑌 such that

𝑓(x,y∗) ≤ 𝑓(x∗,y∗) ≤ 𝑓(x∗,y) for every x ∈ 𝑋 and y ∈ 𝑌Let 𝑣 = 𝑓(x∗,y∗). Since

𝑓(x,y∗) ≤ 𝑣 for every x ∈ 𝑋maxx∈𝑋

𝑓(𝑥,y∗) ≤ 𝑣

and therefore

miny∈𝑌

maxx∈𝑋

𝑓(x,y) ≤ maxx∈𝑋

𝑓(x,y∗) ≤ 𝑣

Similarly

maxx∈𝑋

miny∈𝑦 𝑓(x,y) ≥ 𝑣

Combining the last two inequalities, we have

maxx∈𝑋

miny∈𝑦 𝑓(x,y) ≥ 𝑣 ≥ min

y∈𝑌maxx∈𝑋

𝑓(𝑥,y)

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Together with (3.28), this implies equality

maxx∈𝑋

miny∈𝑌𝑓(x,y) = min

y∈𝑌maxx∈𝑋

𝑓(x,y)

Conversely, suppose that

maxx∈𝑋

miny∈𝑌𝑓(x,y) = 𝑣 = min

y∈𝑌maxx∈𝑋

𝑓(x,y)

The function

𝑔(x) = miny∈𝑌𝑓(x,y)

is a continuous function (Theorem 2.3) on a compact set𝑋 . By the Weierstrass theorem(Theorem 2.2), there exists x∗ which maximizes 𝑔 on 𝑋 , that is

𝑔(x∗) = miny∈𝑌𝑓(x∗,y) = max

x∈𝑋𝑔(x) = max

x∈𝑋miny∈𝑌𝑓(x,y) = 𝑣

which implies that

𝑓(x∗,y) ≥ 𝑣 for every y ∈ 𝑌

Similarly, there exists y ∈ 𝑌 such that

𝑓(x,y∗) ≤ 𝑣 for every x ∈ 𝑋

Combining these inequalities, we have

𝑓(x,y∗) ≤ 𝑣 ≤ 𝑓(x∗,y) for every x ∈ 𝑋 and y ∈ 𝑌

In particular, we have

𝑓(x∗,y∗) ≤ 𝑣 ≤ 𝑓(x∗,y∗)

so that 𝑣 = 𝑓(x∗,y∗) as required.

3.161 For any 𝑥 ∈ 𝑋 and 𝑦 ∈ 𝑌 , let

𝑔(𝑥) = min𝑦∈𝑌𝑓(𝑥, 𝑦) and ℎ(𝑦) = max

𝑥∈𝑋𝑓(𝑥, 𝑦)

Then

𝑔(𝑥) = min𝑦∈𝑌𝑓(𝑥, 𝑦) ≤ max

𝑥∈𝑋𝑓(𝑥, 𝑦) = ℎ(𝑦)

and therefore

max𝑥∈𝑋

𝑔(𝑥) ≤ max𝑦∈𝑌

ℎ(𝑦)

That is

max𝑥

min𝑦𝑓(𝑥, 𝑦) ≤ 𝑓𝑦 max

𝑥𝑓(𝑥, 𝑦)

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3.162 Clearly 𝑓(𝑥) = 𝑥𝑎 his homogeneous of degree 𝑎. Conversely assume 𝑓 is homo-geneous of degree 𝑎, that is

𝑓(𝑡𝑥) = 𝑡𝑎𝑓(𝑥)

Letting 𝑥 = 1

𝑓(𝑡) = 𝑡𝑎𝑓(1)

Setting 𝑓(1) = 𝐴 ∈ ℜ and interchanging 𝑥 and 𝑡 yields the result.

3.163

𝑓(𝑡x) = (𝑎1(𝑡𝑥1)𝜌 + 𝑎2(𝑡𝑥2)

𝜌 + . . . 𝑎𝑛(𝑡𝑥𝑛)𝜌)1/𝜌

= 𝑡 (𝑎1𝑥𝜌1 + 𝑎2𝑥

𝜌2 + . . . 𝑎𝑛𝑥

𝜌𝑛)1/𝜌

= 𝑡𝑓(x)

3.164 For 𝛽 ∈ ℜ++ℎ(𝛽𝑡) = 𝑓(𝛽𝑡x0) = 𝛽𝑘𝑓(𝑡x0) = 𝛽𝑘ℎ(𝑡)

3.165 Suppose that x∗ minimizes the cost of producing output 𝑦 at prices w. That is

w𝑇x∗ ≤ w𝑇x for every x ∈ 𝑉 (𝑦)

It follows that

𝑡w𝑇x∗ ≤ 𝑡w𝑇x for every x ∈ 𝑉 (𝑦)

for every 𝑡 > 0, verifying that x∗ minimizes the cost of producing 𝑦 at prices 𝑡w.Therefore

𝑐(𝑡w, 𝑦) = (𝑡w)x∗ = 𝑡(w𝑇x∗) = 𝑡𝑐(w, 𝑦)

𝑐(w, 𝑦) homogeneous of degree one in input prices w.

3.166 For given prices w, let x∗ minimize the cost of producing one unit of output, sothat 𝑐(w, 1) = w𝑇x∗. Clearly 𝑓(x∗) = 1 where 𝑓 is the production function.

Now consider any output 𝑦. Since 𝑓 is homogeneous

𝑓(𝑦x∗) = 𝑦𝑓(x∗) = 𝑦

Therefore 𝑦x∗ is sufficient to produce 𝑦, so that

𝑐(w, 𝑦) ≤ w𝑇 (𝑦x∗) = 𝑦w𝑇x∗ = 𝑦𝑐(w, 1)

Suppose that

𝑐(w, 𝑦) < w𝑇 (𝑦x∗) = 𝑦𝑐(w, 1)

Then there exists x′ such that 𝑓(x′) = 𝑦 and

w𝑇x′ < w𝑇 (𝑦x∗)

which implies that

w𝑇

(x′

𝑦

)< w𝑇x∗ = 𝑐(w, 1)

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Since 𝑓 is homogeneous

𝑓

(x′

𝑦

)=

1

𝑦𝑓(x′) = 1

Therefore, x′ is a lower cost method of producing one unit of output, contradicting thedefinition of x∗. We conclude that

𝑐(w, 𝑦) = 𝑦𝑐(w, 1)

𝑐(w, 𝑦) is homogeneous of degree one in 𝑦.

3.167 If the consumer’s demand is invariant to proportionate changes in all prices andincome, so also will the derived utility. More formally, suppose that x∗ maximizesutility at prices p and income 𝑚, that is

x∗ ≿ x for every x ∈ 𝑋(p,𝑚)

Then

𝑣(p,𝑚) = 𝑢(x∗)

Since 𝑋(𝑡p, 𝑡𝑚) = 𝑋(p,𝑚)

x∗ ≿ x for every x ∈ 𝑋(𝑡p, 𝑡𝑚)

and

𝑣(𝑡p, 𝑡𝑚) = 𝑢(x∗) = 𝑣(p,𝑚)

3.168 Assume 𝑓 is homogeneous of degree one, so that

𝑓(𝑡x) = 𝑡𝑓(x) for every 𝑡 > 0

Let (x, 𝑦) ∈ epi 𝑓 , so that

𝑓(x) ≤ 𝑦For any 𝑡 > 0

𝑓(𝑡x) = 𝑡𝑓(x) ≤ 𝑡𝑦which implies that (𝑡x, 𝑡𝑦) ∈ epi 𝑓 . Therefore epi 𝑓 is a cone.

Conversely assume epi 𝑓 is a cone. Let x ∈ 𝑆 and define 𝑦 = 𝑓(x). Then (x, 𝑦) ∈ epi 𝑓and therefore (𝑡x, 𝑡𝑦) ∈ epi 𝑓 so

𝑓(𝑡x) ≤ 𝑡𝑦Now suppose to the contrary that

𝑓(𝑡x) = 𝑧 < 𝑡𝑦 = 𝑡𝑓(x) (3.58)

Then (𝑡x, 𝑧) ∈ epi 𝑓 . Since epi 𝑓 is a cone, we must have (x, 𝑧/𝑡) ∈ epi 𝑓 so that

𝑓(x) ≤ 𝑧𝑡

and

𝑡𝑓(x) ≤ 𝑧 = 𝑓(𝑡x)

contradicting (3.58). We conclude that

𝑓(𝑡x) = 𝑡𝑓(x) for every 𝑡 > 0

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3.169 Take any x1 and x2 in 𝑆 and let

𝑦1 = 𝑓(x1) > 0 and 𝑦2 = 𝑓(x2) > 0

Since 𝑓 is homogeneous of degree one,

𝑓

(x1𝑦1

)= 𝑓

(x2𝑦2

)= 1

Since 𝑓 is quasiconcave

𝑓

(𝛼x1𝑦1

+ (1− 𝛼)x2𝑦2

)≥ 1

for every 0 ≤ 𝛼 ≤ 1. Choose 𝛼 = 𝑦1/(𝑦1 + 𝑦2) so that (1− 𝛼) = 𝑦2/(𝑦1 + 𝑦2). Then

𝑓

(x1

𝑦1 + 𝑦2+x2

𝑦1 + 𝑦2

)≥ 1

Again using the homogeneity of 𝑓 , this implies

𝑓 (x1 + x2) ≥ 𝑦1 + 𝑦2 = 𝑓(x1) + 𝑓(x2)

3.170 Let 𝑓 ∈ 𝐹 (𝑆) be a strictly positive definite, quasiconcave functional which ishomogeneous of degree one. For any x1, x2 in 𝑆 and 0 ≤ 𝛼 ≤ 1 𝛼x1, (1 − 𝛼)x2 in 𝑆and therefore

𝑓(𝛼x1 + (1− 𝛼)x2) ≥ 𝑓(𝛼x1) + 𝑓((1− 𝛼)x2)

since 𝑓 is superadditive (Exercise 3.169). But

𝑓(𝛼x1) = 𝛼𝑓(x1)

𝑓((1− 𝛼)x2) = (1 − 𝛼)𝑓(x2)

by homogeneity. Substituting in (3.58), we conclude that

𝑓(𝛼x1 + (1− 𝛼)x2) ≥ 𝛼𝑓(x1) + (1 − 𝛼)𝑓((1 − 𝛼)x2)

𝑓 is concave.

3.171 Assume that 𝑓 is strictly positive definite, quasiconcave and homogeneous ofdegree 𝑘, 0 < 𝑘 < 1. Define

ℎ(x) = (𝑓(x))1/𝑘

Then ℎ is quasiconcave (Exercise 3.148. Further, for every 𝑡 > 0

ℎ(𝑡x) = (𝑓(𝑡x))1/𝑘

=(𝑡𝑘𝑓(x)

)1/𝑘= 𝑡 (𝑓(x))

1/𝑘

= 𝑡ℎ(x)

so that ℎ is homogeneous of degree 1. By Exercise 3.170, ℎ is concave.

𝑓(x) = (ℎ(x))𝑘

That is 𝑓 = 𝑔 ∘ ℎ where

𝑔(𝑦) = 𝑦𝑘

is monotone and concave provided 𝑘 ≤ 1. By Exercise 3.133, 𝑓 = 𝑔 ∘ ℎ is concave.

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3.172 Continuity is a necessary and sufficient condition for the existence of a utilityfunction representing ≿ (Remark 2.9).

Suppose 𝑢 represents the homothetic preference relation ≿. For any x1,x2 ∈ 𝑆

𝑢(x1) = 𝑢(x2) =⇒ x1 ∼ x2 =⇒ 𝑡x1 ∼ 𝑡x2 =⇒ 𝑢(𝑡x1) = 𝑢(𝑡x2) for every 𝑡 > 0

Conversely, if 𝑢 is a homothetic functional,

x1 ∼ x2 =⇒ 𝑢(x1) = 𝑢(x2) =⇒ 𝑢(𝑡x1) = 𝑢(𝑡x2) =⇒ 𝑡x1 ∼ 𝑡x2 for every 𝑡 > 0

3.173 Suppose that 𝑓 = 𝑔 ∘ ℎ where 𝑔 is strictly increasing and ℎ is homogeneous ofdegree 𝑘. Then

ℎ(x) =(ℎ(x)

)1/𝑘is homogeneous of degree one and 𝑓 = 𝑔 ∘ ℎ where

𝑔(𝑦) = 𝑔(𝑦𝑘)

)is increasing.

3.174 Assume x1,x2 ∈ 𝑆 with

𝑓(x1) = 𝑔(ℎ(x1)) = 𝑔(ℎx2)) = 𝑓(x2)

Since 𝑔 is strictly increasing, this implies that

ℎ(x1) = ℎ(x2)

Since ℎ is homogeneous

ℎ(𝑡x1) = 𝑡𝑘ℎ(x1) = 𝑡𝑘ℎ(x2) = ℎ(𝑡x2)

for some 𝑘. Therefore

𝑓(𝑡x1) = 𝑔(ℎ(𝑡x1)) = 𝑔(ℎ(𝑡x2)) = 𝑓(𝑡x2)

3.175 Let x0 ∕= 0 be any point in 𝑆, and define 𝑔 : ℜ → ℜ by

𝑔(𝛼) = 𝑓(𝛼x0)

Since 𝑓 is strictly increasing, so is 𝑔 and therefore 𝑔 has a strictly increasing inverse𝑔−1. Let ℎ = 𝑔−1 ∘ 𝑓 so that 𝑓 = 𝑔 ∘ ℎ.We need to show that ℎ is homogeneous. For any x ∈ 𝑆, there exists 𝛼 such that

𝑔(𝛼) = 𝑓(𝛼x0) = 𝑓(x)

that is 𝛼 = ℎ(x) = 𝑔−1(𝑓(x)). Since 𝑓 is homothetic

𝑔(𝑡𝛼) = 𝑓(𝑡𝛼x0)𝑓(𝑡x) for every 𝑡 > 0

and therefore

ℎ(𝑡x) = 𝑔−1(𝑓(𝑡x)) = 𝑔−1(𝑓(𝑡𝛼x0)) = 𝑔−1𝑔(𝑡𝛼) = 𝑡𝛼 = 𝑡ℎ(x)

ℎ is homogeneous of degree one.

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3.176 Let 𝑓 be the production function. If 𝑓 is homothetic, there exists (Exercise 3.175)a linearly homogeneous function ℎ and strictly increasing function 𝑔 such that 𝑓 = 𝑔∘ℎ.

𝑐(w, 𝑦) = minx{w𝑇x : 𝑓(x) ≥ 𝑦 }

= minx{w𝑇x : 𝑔(ℎ(x)) ≥ 𝑦 }

= minx{w𝑇x : ℎ(x) ≥ 𝑔−1(𝑦) }

= 𝑔−1(𝑦)𝑐(w, 1)

by Exercise 3.166.

3.177 Let 𝑓 : 𝑆 → ℜ be positive, strictly increasing, homothetic and quasiconcave. ByExercise 3.175, there exists a linearly homogeneous function ℎ : 𝑆 → ℜ and strictlyincreasing function 𝑔 ∈ 𝐹 (𝑅) such that 𝑓 = 𝑔 ∘ℎ. ℎ = 𝑔−1 ∘ 𝑓 is positive, quasiconcave(Exercise 3.148) and homogeneous of degree one. By Proposition 3.12, ℎ is concaveand therefore 𝑓 = 𝑔 ∘ ℎ is concavifiable.

3.178 Since 𝐻𝑓 (𝑐) is a supporting hyperplane to 𝑆 at x0, then

𝑓(x0) = 𝑐

and either

𝑓(x) ≥ 𝑐 = 𝑓(x0) for every x ∈ 𝑆

or

𝑓(x) ≤ 𝑐 = 𝑓(x0) for every x ∈ 𝑆3.179 Suppose to the contrary that y = (ℎ, 𝑞) ∈ int 𝐴 ∩ 𝐵. Then y ≿ y∗. By strictconvexity

y𝛼 = 𝛼y + (1− 𝛼)y∗ ≻ y∗ for every 𝛼 ∈ (0, 1)

Since y ∈ int 𝐴, y𝛼 ∈ 𝐴 for 𝛼 sufficiently small. That is, there exists some 𝛼 such thaty𝛼 is feasible and y𝛼 ≻ y∗, contradicting the optimality of y∗.

3.180 For notational simplicity, let 𝑓 be the linear functional which separates 𝐴 and 𝐵in Example 3.77. 𝑓(y) measure the cost of the plan y = (ℎ, 𝑞), that is 𝑓(y) = 𝑤ℎ+ 𝑝𝑞.

Assume to the contrary there exists a preferred lifestyle in 𝑋 , that is there exists somey = (ℎ, 𝑞) ∈ 𝑋 such that y ≻ y∗ = (ℎ∗, 𝑞∗). Since y ∈ 𝐵, 𝑓(y) ≥ 𝑓(y∗) by (3.29). Onthe other hand, y ∈ 𝑋 which implies that 𝑓(y) ≤ 𝑓(y∗). Consequently, 𝑓(y) = 𝑓(y∗).

By continuity, there exists some 𝛼 < 1 such that 𝛼y ≻ y∗ which implies that 𝛼y ∈ 𝐵.By linearity

𝑓(𝛼y) = 𝛼𝑓(y) < 𝑓(y) = 𝑓(y∗) = 𝛼

contrary to (3.29). This contradiction establishes that y∗ is the best choice in budgetset 𝑋 .

3.181 By Proposition 3.7, epi 𝑓 is a convex set in 𝑋 × ℜ with (x0, 𝑓(x0)) a point onits boundary. By Corollary 3.2.2 of the Separating Hyperplane Theorem, there existslinear a functional 𝜑 ∈ (𝑋 ×ℜ)′ such that

𝜑(x, 𝑦) ≥ 𝜑(x0, 𝑓(x0)) for every (x, 𝑦) ∈ epi 𝑓 (3.59)

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𝜑 can be decomposed into two components (Exercise 3.47)

𝜑(x, 𝑦) = −𝑔(x) + 𝛼𝑦

The assumption that x0 ∈ int 𝑆 ensures that 𝛼 > 0 and we can normalize so that𝛼 = 1. Substituting in (3.59)

−𝑔(x) + 𝑓(x) ≥ −𝑔(x0) + 𝑓(x0)

𝑓(x) ≥ 𝑓(x0) + 𝑔(x− x0)for every x ∈ 𝑆.

3.182 By Exercise 3.72, there exists a unique point x0 ∈ 𝑆 such that

(x0 − y)𝑇 (x− x0) ≥ 0 for every x ∈ 𝑆Define the linear functional (Exercise 3.64)

𝑓(x) = (x0 − y)𝑇xand let 𝑐 = 𝑓(x0). For all x ∈ 𝑆

𝑓(x)− 𝑓(x0) = 𝑓(x− x0) = (x0 − y)𝑇 (x− x0) ≥ 0

and therefore

𝑓(x) ≥ 𝑓(x0) = 𝑐 for every x ∈ 𝑆Furthermore

𝑓(x0)− 𝑓(y) = 𝑓(x0 − y) = (x0 − y)𝑇 (x0 − y) = ∥x0 − y∥2 > 0

since y ∕= x0. Therefore 𝑓(x0) > 𝑓(y) and

𝑓(y) < 𝑐 ≤ 𝑓(x) for every x ∈ 𝑆3.183 If y ∈ b(𝑆), y ∈ 𝑆𝑐 and there exists a sequence of points {y𝑛} ∈ 𝑆𝑐 convergingto y (Exercise 1.105). That is, there exists a sequence of nonboundary points {y𝑛} /∈ 𝑆converging to y. For every point y𝑛, there is a linear functional 𝑔𝑛 ∈ 𝑋∗ and 𝑐𝑛 suchthat

𝑔𝑛(y𝑛) < 𝑐𝑛 ≤ 𝑔𝑛(x) for every x ∈ 𝑆Define 𝑓𝑛 = 𝑔𝑛/ ∥𝑔𝑛∥. By construction, the sequence of linear functionals 𝑓𝑛 belongto the unit ball in 𝑋∗ (since ∥𝑓∥ = 1). Since 𝑋∗ is finite dimensional, the unit ball iscompact as so 𝑓𝑛 has a convergent subsequence with limit 𝑓 such that

𝑓(y) ≤ 𝑓(x) for every 𝑥 ∈ 𝑆A fortiori

𝑓(y) ≤ 𝑓(x) for every 𝑥 ∈ 𝑆3.184 There are two possible cases.

y /∈ 𝑆 By Exercise 3.182, there exists a hyperplane which separates y and 𝑆 which afortiori separates y and 𝑆, that is

𝑓(y) ≤ 𝑓(x) for every x ∈ 𝑆

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y ∈ 𝑆 Since y /∈ 𝑆, y must be a boundary point of 𝑆. By the previous exercise,there exists a supporting hyperplane at y, that is there exists a continuous linearfunctional 𝑓 ∈ 𝑋∗ such that

𝑓(y) ≤ 𝑓(x) for every x ∈ 𝑆

3.185 1. 𝑓(𝑆) ⊆ ℜ.

2. 𝑓(𝑆) is convex and hence an interval (Exercise 1.160.

3. 𝑓(𝑆) is open in ℜ (Proposition 3.2).

3.186 𝑆 is nonempty and convex and 0 /∈ 𝑆. (Otherwise, there exists x ∈ 𝐴 and y ∈ 𝐵such that 0 = y + (−x) which implies that x = y contradicting the assumption that𝐴 ∩𝐵 = ∅.) Thus there exists a continuous linear functional 𝑓 ∈ 𝑋∗ such that

𝑓(y − x) ≥ 𝑓(0) = 0 for every x ∈ 𝐴,y ∈ 𝐵

so that

𝑓(x) ≤ 𝑓(y) for every x ∈ 𝐴,y ∈ 𝐵

Let 𝑐 = supx∈𝐴 𝑓(x). Then

𝑓(x) ≤ 𝑐 ≤ 𝑓(y) for every x ∈ 𝐴,y ∈ 𝐵

By Exercise 3.185, 𝑓(int 𝐴) is an open interval in (−∞, 𝑐], hence 𝑓(int 𝐴) ⊆ (−∞, 𝑐),so that 𝑓(x) < 𝑐 for every x ∈ int 𝐴. Similarly, 𝑓(int 𝐵) > 𝑐 and

𝑓(x) < 𝑐 < 𝑓(y) for every x ∈ int 𝐴,y ∈ int 𝐵

3.187 Since int 𝐴 ∩ 𝐵 = ∅, int 𝐴 and 𝐵 can be separated. That is, there exists acontinuous linear functional 𝑓 ∈ 𝑋∗ and a number 𝑐 such that

𝑓(x) ≤ 𝑐 ≤ 𝑓(y) for every x ∈ 𝐴,y ∈ int 𝐵

which implies that


since

𝑐 ≤ infy∈int 𝐵

𝑓(y) = infy∈𝐵𝑓(y)

Conversely, suppose that 𝐴 and 𝐵 can be separated. That is, there exists 𝑓 ∈ 𝑋∗ suchthat


Then 𝑓(int 𝐴) is an open interval in [𝑐,∞), which is disjoint from the interval 𝑓(𝐵) ⊆(−∞, 𝑐]. This implies that int 𝐴 ∩𝐵 = ∅.3.188 Since x0 ∈ b(𝑆), {x0} ∩ int 𝑆 = ∅ and int 𝑆 ∕= ∅. By Corollary 3.2.1, {x0} and𝑆 can be separated, that is there exist 𝑓 ∈ 𝑋∗ such that

𝑓(x0) ≤ 𝑓(x) for every x ∈ 𝑆

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3.189 Let x ∈ 𝐶. Since 𝐶 is a cone, 𝜆x ∈ 𝐶 for every 𝜆 ≥ 0 and therefore

𝑓(𝜆x) ≥ 𝑐

or

𝑓(x) ≥ 𝑐/𝜆 for every 𝜆 ≥ 0

Taking the limit as 𝜆→∞ implies that

𝑓(x) ≥ 0 for every x ∈ 𝐶

3.190 First note that 0 ∈ 𝑍 and therefore 𝑓(0) = 0 ≤ 𝑐 so that 𝑐 ≥ 0. Suppose thatthere exists some z ∈ 𝑍 for which 𝑓(z) = 𝜖 ∕= 0. By linearity, this implies

𝑓(2𝑐

𝜖z) =

2𝑐

𝜖𝑓(z) = 2𝑐 > 𝑐

which contradicts the requirement

𝑓(z) ≤ 𝑐 for every z ∈ 𝑍

3.191 By Corollary 3.2.1, there exists 𝑓 ∈ 𝑋∗ such that

𝑓(z) ≤ 𝑐 ≤ 𝑓(x) for every x ∈ 𝑆, z ∈ 𝑍

By Exercise 3.190

𝑓(z) = 0 for every z ∈ 𝑍

and therefore

𝑓(x) ≥ 0 for every x ∈ 𝑆

Therefore 𝑍 is contained in the hyperplane 𝐻𝑓 (0) which separates 𝑆 from 𝑍.

3.192 Combining Theorem 3.2 and Corollary 3.2.1, there exists a hyperplane 𝐻𝑓 (𝑐)such that


and such that

𝑓(x) < 𝑐 ≤ 𝑓(y) for every x ∈ int 𝐴,y ∈ 𝐵

Since int 𝐴 ∕= ∅, there exists some x ∈ int 𝐴 with 𝑓(x) < 𝑐. Hence 𝐴 ⊈ 𝑓−1(𝑐) = 𝐻𝑓 (𝑐).

3.193 Follows directly from the basic separation theorem, since 𝐴 = int 𝐴 and 𝐵 =int 𝐵.

3.194 Let 𝑆 = 𝐵 −𝐴. Then

1. 𝑆 is a nonempty, closed, convex set (Exercise 1.203).

2. 0 /∈ 𝑆.

There exists a continuous linear functional 𝑓 ∈ 𝑋∗ such that

𝑓(x) ≥ 𝑐 > 𝑓(0) = 0

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𝐴

𝐵

Figure 3.2: 𝐴 and 𝐵 cannot be strongly separated.

for every z ∈ 𝑆 (Exercise 3.182). For every x ∈ 𝐴,y ∈ 𝐵, z = y − x ∈ 𝑆 and

𝑓(z) = 𝑓(y) − 𝑓(x) ≥ 𝑐 > 0

or

𝑓(x) + 𝑐 ≤ 𝑓(y)

which implies that

supx∈𝐴𝑓(x) + 𝑐 ≤ inf

y∈𝐵𝑓(y)

and

supx∈𝐴𝑓(x) < inf

y∈𝐵𝑓(y)

3.195 No. See Figure 3.2.

3.196 1. Assume that there exists a convex neighborhood 𝑈 ∋ 0 such that

(𝐴+ 𝑈) ∩𝐵 = ∅Then (𝐴+ 𝑈) is convex and 𝐴 ⊂ int (𝐴+ 𝑈) ∕= ∅ and int (𝐴+ 𝑈) ∩ 𝐵 = ∅. ByCorollary 3.2.1, there exists continuous linear functional such that

𝑓(x+ u) ≤ 𝑓(y) for every x ∈ 𝐴,u ∈ 𝑈,y ∈ 𝐵Since 𝑓(𝑈) is an open interval containing 0, there exists some u0 with 𝑓(u0) =𝜖 > 0.

𝑓(x) + 𝜖 ≤ 𝑓(y) for every x ∈ 𝐴,y ∈ 𝐵which implies that

supx∈𝐴𝑓(x) < inf

y∈𝐵𝑓(y)

Conversely, assume that 𝐴 and 𝐵 can be strongly separated. That is, there existsa continuous linear functional 𝑓 ∈ 𝑋∗ and number 𝜖 > 0 such that

𝑓(x) ≤ 𝑐− 𝜖 < 𝑐+ 𝜖 ≤ 𝑓(y) for every x ∈ 𝐴,y ∈ 𝐵Let 𝑈 = { 𝑥 ∈ 𝑋 : ∣𝑓(𝑥)∣ < 𝜖 }. 𝑈 is a convex neighborhood of 0 such that(𝐴+ 𝑈) ∩𝐵 = ∅.

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2. Let 𝐴 and 𝐵 be nonempty, disjoint, convex subsets in a normed linear space 𝑋with 𝐴 compact and 𝐵 closed. By Exercise 1.208, there exists a convex neigh-borhood 𝑈 ∋ 0 such that (𝐴 + 𝑈) ∩ 𝐵 = ∅. By the previous part, 𝐴 and 𝐵 canbe strongly separated.

3.197 Assume 𝜌(𝐴,𝐵) = inf{ ∥x− y∥ : x ∈ 𝐴,y ∈ 𝐵 } = 2𝜖 > 0. Let 𝑈 = 𝐵𝜖(0) bethe open ball around 0 of radius 𝜖. For every x ∈ 𝐴,u ∈ 𝑈,y ∈ 𝐵

∥x+ (−u)− y∥ = ∥x− y − u∥ ≥ ∥x− y∥ − ∥u∥

so that

𝜌(𝐴+ 𝑈,𝐵) = infx,u,y

∥x+ (−u)− y∥ ≥ infx,u,y

(∥x− y∥ − ∥u∥)≥ inf

x,y∥x− y∥ − sup

u∥u∥)

= 2𝜖− 𝜖= 𝜖 > 0

Therefore (𝐴+ 𝑈) ∩𝐵 = ∅ and so 𝐴 and 𝐵 can be strongly separated.

Conversely, assume that 𝐴 and 𝐵 can be strongly separated, so that there exists aconvex neighborhood 𝑈 of 0 such that (𝐴+ 𝑈) ∩𝐵 = ∅. Therefore, there exists 𝜖 > 0such that 𝐵𝜖(0) ⊆ 𝑈 and

𝐴+𝐵𝜖 ∩𝐵 = ∅This implies that

𝜌(𝐴,𝐵) = inf{ ∥x− y∥ : x ∈ 𝐴,y ∈ 𝐵 } > 𝜖 > 0

3.198 Take 𝐴 = {y} and 𝐵 =𝑀 in Proposition 3.14. There exists 𝑓 ∈ 𝑋∗ such that

𝑓(y) < 𝑐 ≤ 𝑓(x) for every x ∈𝑀By Corollary 3.2.3, 𝑐 = 0.

3.199 1. Consider the set

𝑍 = { 𝑓(𝑥),−𝑔1(𝑥),−𝑔2(𝑥), . . . ,−𝑔𝑚(𝑥) : 𝑥 ∈ 𝑋 }𝑍 is the image of a linear mapping from 𝑋 to 𝑌 = ℜ𝑚+1 and hence is a subspaceof ℜ𝑚+1.

2. By hypothesis, the point e0 = (1, 0, 0, . . . , 0) ∈ ℜ𝑚+1 does not belong to 𝑍.Otherwise, we have an 𝑥 ∈ 𝑋 such that 𝑔𝑖(𝑥) = 0 for every 𝑖 but 𝑓(𝑥) = 1.

3. By the previous exercise, there exists a linear functional 𝜑 ∈ 𝑌 ∗ such that

𝜑(e0) > 0

𝜑(z) = 0 for every z ∈ 𝑍

4. In other words, there exists a vector 𝜆 = (𝜆0, 𝜆1, . . . , 𝜆𝑚) ∈ 𝑌 = ℜ(𝑚+1)∗ suchthat

𝜆e0 > 0 (3.60)

𝜆z = 0 for every z ∈ 𝑍 (3.61)

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Equation (3.61) states that

𝜆z = 𝜆0z0 + 𝜆1z1 + ⋅ ⋅ ⋅+ 𝜆𝑚z𝑚 = 0 for every z ∈ 𝑍That is, for every 𝑥 ∈ 𝑋 ,

𝜆0𝑓(x)− 𝜆1𝑔1(x) − 𝜆2𝑔2(x) − . . .− 𝜆𝑚𝑔𝑚(x) = 0

5. Inequality (3.60) establishes that 𝜆0 > 0. Without loss of generality we cannormalize so that 𝜆0 = 1.

6. Therefore

𝑓(𝑥) =

𝑚∑𝑖=1

𝜆𝑖𝑔𝑖(𝑥)

3.200 For every x ∈ 𝑆, 𝑔𝑗(x) = 0, 𝑗 = 1, 2 . . .𝑚 and therefore

𝑓(x) =

𝑚∑𝑖=1

𝜆𝑖𝑔𝑖(x) = 0

3.201 The set

𝑍 = { 𝑔1(𝑥), 𝑔2(𝑥), . . . , 𝑔𝑚(𝑥) : 𝑥 ∈ 𝑋 }is a closed subspace in ℜ𝑚. If the system is inconsistent, c = (𝑐1, 𝑐2, . . . , 𝑐𝑚) /∈ 𝑍. ByExercise 3.198, there exists a linear functional 𝜑 on ℜ𝑚 such

𝜑(z) = 0 for every z ∈ 𝑍𝜑(c) > 0

That is, there exist numbers 𝜆1, 𝜆2, . . . , 𝜆𝑚 such that

𝑚∑𝑗=1

𝜆𝑗𝑔𝑗(x) = 0

and𝑚∑𝑗=1

𝜆𝑗𝑐𝑗 > 0

which contradicts the hypothesis

𝑚∑𝑗=1

𝜆𝑗𝑔𝑗 = 0 =⇒𝑚∑𝑗=1

𝜆𝑗𝑐𝑗 = 0

Conversely, if for some x ∈ 𝑋𝑔𝑗(x) = 𝑐𝑗 𝑗 = 1, 2, . . . ,𝑚

then𝑚∑𝑗=1

𝜆𝑗𝑔𝑗(x) =

𝑚∑𝑗=1

𝜆𝑗𝑐𝑗

and𝑚∑𝑗=1

𝜆𝑗𝑔𝑗 = 0 =⇒𝑚∑𝑗=1

𝜆𝑗𝑐𝑗 = 0

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3.202 The set �� = {x ∈ 𝐾 : ∥x∥1 = 1 } is

∙ compact (the unit ball is compact if and only if 𝑋 is finite-dimensional)

∙ convex (which is why we need the 1 norm)

By Proposition 3.14, there exists a linear functional 𝑓 ∈ 𝑋∗ such that

𝑓(x) > 0 for every x ∈ ��𝑓(x) = 0 for every x ∈𝑀

For any x ∈ 𝐾,x ∕= 0, define x = x/ ∥x∥1 ∈ ��. Then

𝑓(x) = 𝑓(∥x∥1 x) = ∥x∥1 𝑓(x) > 0

3.203 1. Let

𝐴 = { (x, 𝑦) : 𝑦 ≥ 𝑔(x),x ∈ 𝑋 }𝐵 = { (x, 𝑦) : 𝑦 = 𝑓0(x),x ∈ 𝑍 }

𝐴 is the epigraph of a convex functional and hence convex. 𝐵 is a subspace of𝑌 = 𝑋 ×ℜ and also convex.

2. Since 𝑔 is convex, int 𝐴 ∕= ∅. Furthermore

𝑓0(x) ≤ 𝑔(x) =⇒ int 𝐴 ∩𝐵 = ∅

3. By Exercise 3.2.3, there exists linear functional 𝜑 ∈ 𝑌 ∗ such that

𝜑(x, 𝑦) ≥ 0 for every (x, 𝑦) ∈ 𝐴𝜑(x, 𝑦) = 0 for every (x, 𝑦) ∈ 𝐵

There exists 𝑦 such that 𝑦 > 𝑔(0) and therefore (0, 𝑦) ∈ int 𝐴 and 𝜑(0, 𝑦) > 0.Therefore

𝜑(0, 1) =1

𝑦𝜑(0, 1) > 0

4. Let 𝑓 ∈ 𝑋∗ be defined by

𝑓(x) = −1

𝑐𝜑(x, 0)

where 𝑐 = 𝜑(0, 1). Since

𝜑(x, 0) = 𝜑(x, 𝑦)− 𝜑(0, 𝑦)

= 𝜑(x, 𝑦)− 𝑐𝑦

𝑓(x) = −1

𝑐(𝜑(x, 𝑦)− 𝑐𝑦) = −1

𝑐𝜑(x, 𝑦) + 𝑦

for every 𝑦 ∈ ℜ5. For every x ∈ 𝑍

𝑓(x) = −1

𝑐𝜑(x, 𝑓0(x)) + 𝑓0(x)

= 𝑓0(x)

since 𝜑(x, 𝑓0(x)) = 0 for every x ∈ 𝑍. Thus 𝑓 is an extension of 𝑓0.

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6. For any x ∈ 𝑋 , let 𝑦 = 𝑔(x). Then (x, 𝑦) ∈ 𝐴 and 𝜑(x, 𝑦) ≥ 0. Therefore

𝑓(x) = −1

𝑐𝜑(x, 𝑦) + 𝑦

= −1

𝑐𝜑(x, 𝑦) + 𝑔(x)

≤ 𝑔(x)Therefore 𝑓 is bounded by 𝑔 as required.

3.204 Let 𝑔 ∈ 𝑋∗ be defined by

𝑔(x) = ∥𝑓0∥𝑍 ∥x∥Then 𝑓0(x) ≤ 𝑔(x) for all x ∈ 𝑍. By the Hahn-Banach theorem (Exercise 3.15), thereexists an extension 𝑓 ∈ 𝑋∗ such that

𝑓(x) ≤ 𝑔(x) = ∥𝑓0∥𝑍 ∥x∥Therefore

∥𝑓∥𝑋 = sup∥x∥=1

∥𝑓(x)∥ = ∥𝑓0∥𝑍

3.205 If x0 = 0, any bounded linear functional will do. Therefore, assume x0 ∕= 0. Onthe subspace lin {x0} = {𝛼x0 : 𝛼 ∈ ℜ}, define the function

𝑓0(𝛼x0) = 𝛼 ∥x0∥𝑓0 is a bounded linear functional on lin {x0} with norm 1. By the previous part, 𝑓0can be extended to a bounded linear functional 𝑓 ∈ 𝑋∗ with the same norm, that is∥𝑓∥ = 1 and 𝑓(x0) = ∥x0∥.3.206 Since x1 ∕= x2, x1 − x2 ∕= 0. There exists a bounded linear functional such that

𝑓(x1 − x2) = ∥x1 − x2∥ ∕= 0

so that

𝑓(x1) ∕= 𝑓(x2)

3.207 1. ∙ 𝔉 is a complete lattice (Exercise 1.179).

∙ The intersection of any chain is

– nonempty (since 𝑆 is compact)

– a face (Exercise 1.179)

Hence every chain has a minimal element.

∙ By Zorn’s lemma (Remark 1.5), 𝔉 has a minimal element 𝐹0.

2. Assume to the contrary that 𝐹0 contains two distinct elements x1,x2. Then(Exercise 3.206) there exists a continuous linear functional 𝑓 ∈ 𝑋∗ such that

𝑓(x1) ∕= 𝑓(x2)

Let 𝑐 be in the minimum value of 𝑓(x) on 𝐹0 and let 𝐹1 be the set on which itattains this minimum. (Since 𝐹0 is compact, 𝑐 is well-defined and 𝐹1 is nonempty.That is

𝑐 = min{ 𝑓(𝑥) : x ∈ 𝐹0 }𝐹1 = {x ∈ 𝑆 : 𝑓(x) = 𝑐 }

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Now 𝐹1 ⊂ 𝐹0 since 𝑓(x1) ∕= 𝑓(x2).

To show that 𝐹1 is a face of 𝐹0, assume that 𝛼x+(1−𝛼)y ∈ 𝐹1 for some x,y ∈ 𝐹0.Then 𝑐 = 𝑓(𝛼x + (1 − 𝛼)y) = 𝛼𝑓(x) + (1 − 𝛼)𝑓(y) = 𝑐. Since x,y ∈ 𝐹0, thisimplies that 𝑓(x) = 𝑓(y) = 𝑐 so that x,y ∈ 𝐹1. Therefore 𝐹1 is a face.

We have shown that, if 𝐹0 contains two distinct elements, there exists a smallerface 𝐹1 ⊂ 𝐹0, contradicting the minimality of 𝐹0. We conclude that 𝐹0 comprisesa single element x0.

3. 𝐹0 = {x0} which is an extreme point of 𝑆.

3.208 Let 𝐻 = 𝐻𝑓 (𝑐) be a supporting hyperplane to 𝑆. Without loss of generalityassume

𝑓(x) ≤ 𝑐 for every x ∈ 𝑆 (3.62)

and there exists some x∗ ∈ 𝑆 such that

𝑓(x∗) = 𝑐

That is 𝑓 is maximized at x∗.

Version 1 By the previous exercise, 𝑓 achieves its maximum at an extreme point.That is, there exists an extreme point x0 ∈ 𝑆 such that

𝑓(x0) ≥ 𝑓(x) for every x ∈ 𝑆

In particular, 𝑓(x0) ≥ 𝑓(x∗) = 𝑐. But (3.62) implies 𝑓(x0) ≤ 𝑐. Therefore, weconclude that 𝑓(x0) = 𝑐 and therefore x0 ∈ 𝐻 .

Version 2 The set 𝐻 ∩ 𝑆 is a nonempty, compact, convex subset of a linear space.Hence, by Exercise 3.207, 𝐻 ∩ 𝑆 contains an extreme point, say x0. We showthat x0 is an extreme point of 𝑆.

Assume not, that is assume that there exists x1,x2 ∈ 𝑆 such that x0 = 𝛼x1 +(1 − 𝛼)x2 for some 𝛼 ∈ (0, 1). Since x0 is an extreme point of 𝐻 ∩ 𝑆, at leastone of the points x1,x2 must lie outside 𝐻 . Assume x1 /∈ 𝐻 which implies that𝑓(x1) < 𝑐. Since 𝑓(x2) ≤ 𝑐

𝑓(x0) = 𝛼𝑓(x1) + (1− 𝛼)𝑓(x2) < 𝑐 (3.63)

However, since x0 ∈ 𝐻 ∩ 𝑆, we must have

𝑓(x0) = 𝑐

which contradicts (3.63).

Therefore x0 is an extreme point of 𝑆. In fact, we have shown that every extremepoint of 𝐻 ∩ 𝑆 must be an extreme point of 𝑆.

3.209 Let 𝑆 denote the closed, convex hull of the extreme points of 𝑆. (The closed,convex hull of a set is simply the closure of the convex hull.) Clearly 𝑆 ⊂ 𝑆 and itremains to show that 𝑆 contains all of 𝑆.

Assume not. That is, assume 𝑆 ⊊ 𝑆 and let x0 ∈ 𝑆 ∖ 𝑆. By the Strong SeparationTheorem, there exists a linear functional 𝑓 ∈ 𝑋∗ such that

𝑓(x0) > 𝑓(x) for every x ∈ 𝑆 (3.64)

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On the other hand, by Exercise 3.16, 𝑓 attains its maximum at an extreme point of 𝑆.That is, there exists x1 ∈ 𝑆 such that

𝑓(x1) ≥ 𝑓(x) for every x ∈ 𝑆In particular

𝑓(x1) ≥ 𝑓(x0)

since x0 ∈ 𝑆 ⊂ 𝑆. This contradicts (3.64) since x1 ∈ 𝑆.

Thus our assumption that 𝑆 ⊊ 𝑆 yields a contradiction. We conclude that

𝑆 = 𝑆

3.210 1. (a) 𝑃 is compact and convex, since it is the product of compact, convexsets (Proposition 1.2, Exercise 1.165).

(b) Since x ∈ ∑𝑛𝑖=1 conv 𝑆𝑖, there exist x𝑖 ∈ conv 𝑆𝑖 such that x =

∑𝑛𝑖=1 x𝑖.

(x1,x2, . . . ,x𝑛) ∈ 𝑃 (x) so that 𝑃 (x) ∕= ∅.(c) By the Krein-Millman theorem (or Exercise 3.207), 𝑃 (x) has an extreme

point z = (z1, z2, . . . , z𝑛) such that

∙ z𝑖 ∈ conv 𝑆𝑖 for every 𝑖

∙ ∑𝑛𝑖=1 z𝑖 = x.

since z ∈ 𝑃 (x).

2. (a) Exercise 1.176

(b) Since 𝑙 > 𝑚 = dim𝑋 , the vectors y1,y2, . . . ,y𝑙 are linearly dependent(Exercise 1.143). Consequently, there exists numbers 𝛼′1, 𝛼

′2, . . . , 𝛼

′𝑙, not all

zero, such that

𝛼′1y1 + 𝛼′2y2 + ⋅ ⋅ ⋅+ 𝛼′𝑙y𝑙 = 0

(Exercise 1.133). Let

𝛼𝑖 =𝛼′𝑖

max𝑖 ∣𝛼𝑖∣Then ∣𝛼𝑖∣ ≤ 1 for every 𝑖 and

𝛼1y1 + 𝛼2y2 + ⋅ ⋅ ⋅+ 𝛼𝑙y𝑙 = 0

(c) Since ∣𝛼𝑖∣ ≤ 1, z𝑖 + 𝛼𝑖y𝑖 ∈ conv 𝑆𝑖 for every 𝑖 = 1, 2, . . . , 𝑙. Furthermore

𝑛∑𝑖=1

z+𝑖 =

𝑛∑𝑖=1

z𝑖 +

𝑙∑𝑖=1

𝛼𝑖y𝑖 =

𝑛∑𝑖=1

z𝑖 = x

Therefore, z+ ∈ 𝑃 (x). Similarly, z− ∈ 𝑃 (x).

(d) By direct computation

z =1

2z+ +

1

2z−

which implies that z is not an extreme point of 𝑃 (x), contrary to our as-sumption. This establishes that at least 𝑛−𝑚 z𝑖 are extreme points of thecorresponding conv 𝑆𝑖.

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0 1 2 3 4

1

2

3

4

conv 𝑆2

P(x)

conv 𝑆1

(0, 2.5)

(.5, 2)

Figure 3.3: Illustrating the proof of the Shapley Folkman theorem.

3. Every extreme point of conv 𝑆𝑖 is an element of 𝑆𝑖.


3.212 Let {𝑆1, 𝑆2, . . . , 𝑆𝑛} be a collection of nonempty subsets of an 𝑚-dimensionallinear space and let x ∈ conv

∑𝑛𝑖=1 𝑆𝑖 =

∑𝑛𝑖=1 conv 𝑆𝑖. That is, there exists x𝑖 ∈

conv 𝑆𝑖 such that x =∑𝑛

𝑖=1 x𝑖. By Caratheodory’s theorem, there exists for every x𝑖a finite number of points x𝑖1,x𝑖2, . . . ,x𝑖𝑙𝑖 such that x𝑖 ∈ conv {x𝑖1,x𝑖2, . . . ,x𝑖𝑙𝑖}.For every 𝑖 = 1, 2, . . . , 𝑛, let

𝑆𝑖 = {x𝑖𝑗 : 𝑗 = 1, 2, . . . , 𝑙𝑖 }

Then

x =

𝑛∑𝑖=1

x𝑖, x𝑖 ∈ conv 𝑆𝑖

That is, x ∈ ∑conv 𝑆𝑖 = conv

∑𝑆𝑖. Moreover, the sets 𝑆𝑖 are compact (in fact

finite). By the previous exercise, there exists 𝑛 points z𝑖 ∈ 𝑆𝑖 such that

x =

𝑛∑𝑖=1

z𝑖, z𝑖 ∈ conv 𝑆𝑖

and moreover z𝑖 ∈ 𝑆𝑖 ⊆ 𝑆𝑖 for at least 𝑛−𝑚 indices 𝑖.

3.213 Let 𝑆 be a closed convex set in a normed linear space. Clearly, 𝑆 is contained inthe intersection of all the closed halfspaces which contain 𝑆.

For any y /∈ 𝑆, there exists a hyperplane which strongly separates {y} and 𝑆. Oneof its closed halfspaces contains 𝑆 but not y. Consequently, y does not belong to theintersection of all the closed halfspaces containing 𝑆.

3.214 1. Since 𝑉 ∗(𝑦) is the intersection of closed, convex sets, it is closed and convex.Assume x is feasible, that is x ∈ 𝑉 (𝑦). Then w𝑇x ≤ (𝑐w, 𝑦) and x ∈ 𝑉 ∗(𝑦).That is, 𝑉 (𝑦) ⊆ 𝑉 ∗(𝑦).

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2. Assume 𝑉 (𝑦) is convex. For any x0 /∈ 𝑉 (𝑦) there exists w such that

w𝑇x0 < infx∈𝑉 (𝑦)

w𝑇x = 𝑐(w, 𝑦)

by the Strong Separation Theorem. Monotonicity ensures that w ≥ 0 and hencex0 /∈ 𝑉 ∗(𝑦).

3.215 Assume x ∈ 𝑉 (𝑦) = 𝑉 ∗(𝑦). That is

w𝑇x ≥ 𝑦𝑐(w) for every x

Therefore, for any 𝑡 ∈ ℜ+𝑡w𝑇x ≥ 𝑡𝑦𝑐(w) for every x

which implies that 𝑡x ∈ 𝑉 ∗(𝑦) = 𝑉 (𝑦).

3.216 A polyhedron

𝑆 = { 𝑥 ∈ 𝑋 : 𝑔𝑖(𝑥) ≤ 𝑐𝑖, 𝑖 = 1, 2, . . . ,𝑚 }

=

𝑚∩𝑖=1

{x ∈ 𝑋 : 𝑔𝑖(x) ≤ 𝑐𝑖 }

is the intersection of a finite number of closed convex sets.

3.217 Each row a𝑖 = (𝑎𝑖1, 𝑎𝑖2, . . . 𝑎𝑖𝑛) of 𝐴 defines a linear functional 𝑔𝑖(x) = 𝑎𝑖1𝑥1 +𝑎𝑖2𝑥2 + ⋅ ⋅ ⋅+ 𝑎𝑖𝑛𝑥𝑛 on ℜ𝑛. The set 𝑆 of solutions to 𝐴x ≤ c is

𝑆 = { 𝑥 ∈ 𝑋 : 𝑔𝑖(x) ≤ 𝑐𝑖, 𝑖 = 1, 2, . . . ,𝑚 }is a polyhedron.

3.218 For simplicity, we assume that the game is superadditive, so that 𝑤(𝑖) ≥ 0 forevery 𝑖. Consequently, in every core allocation x, 0 ≤ 𝑥𝑖 ≤ 𝑤(𝑁) and

core ⊆ [0, 𝑤(𝑁)]× [0, 𝑤(𝑁)]× ⋅ ⋅ ⋅ × [0, 𝑤(𝑁)] ⊂ ℜ𝑛

Thus, the core is bounded. Since it is the intersection of closed halfspaces, the core isalso closed. By Proposition 1.1, the core is compact.

3.219 polytope =⇒ polyhedron Assume that 𝑃 is a polytope generated by thepoints {x1,x2, . . . ,x𝑚 } and let 𝐹1, 𝐹2, . . . , 𝐹𝑘 denote the proper faces of 𝑃 . Foreach 𝑖 = 1, 2, . . . , 𝑘, let 𝐻𝑖 denote the hyperplane containing 𝐹𝑖 so that 𝐹𝑖 =𝑃 ∩ 𝐻𝑖. For every such hyperplane, there exists a nonzero linear functional 𝑔𝑖and constant 𝑐𝑖 such that 𝑔𝑖(x) = 𝑐𝑖 for every x ∈ 𝐻𝑖. Furthermore, every suchhyperplane is a bounding hyperplane of 𝑃 . Without loss of generality, we canassume that 𝑔𝑖(x) ≤ 𝑐 for every x ∈ 𝑃 . Let

𝑆 = {x ∈ 𝑋 : 𝑔𝑖(x) ≤ 𝑐𝑖, 𝑖 = 1, 2, . . . ,𝑚 }

Clearly 𝑃 ⊆ 𝑆. To show that 𝑆 ⊆ 𝑃 , assume not. That is, assume that thereexists y ∈ 𝑆∖𝑃 and let x ∈ ri 𝑃 . (ri 𝑃 is nonempty by exercise 1.229). Since 𝑃 isclosed (Exercise 1.227), there exists a some 𝛼 such that x = 𝛼x+(1−𝛼)y belongsto the relative boundary of 𝑃 , and there exists some 𝑖 such that x ∈ 𝐹𝑖 ⊆ 𝐻𝑖.

Let 𝐻+𝑖 = {x ∈ 𝑋 : 𝑔𝑖(x) ≤ 𝑐𝑖 } denote the closed half-space bounded by 𝐻𝑖 andcontaining 𝑃 . 𝐻𝑖 is a face of𝐻+𝑖 containing x = 𝛼x+(1−𝛼)y, which implies thatx,y ∈ 𝐻𝑖. This in turn implies that x ∈ 𝐹𝑖, which contradicts the assumptionthat x ∈ ri 𝑃 . We conclude that 𝑆 = 𝑃 .

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polyhedron =⇒ polytope Conversely, assume 𝑆 is a nonempty compact polyhedralset in a normed linear space. Then, there exist linear functionals 𝑔1, 𝑔2, . . . , 𝑔𝑚in 𝑋∗ and numbers 𝑐1, 𝑐2, . . . , 𝑐𝑚 such that 𝑆 = {x ∈ 𝑋 : 𝑔𝑖(x) ≤ 𝑐𝑖, 𝑖 =1, 2, . . . ,𝑚 }. We show that 𝑆 has a finite number of extreme points. Let 𝑛 denotethe dimension of 𝑆. If 𝑛 = 1, 𝑆 is either a single point or closed line segment(since 𝑆 is compact), and therefore has a finite number of extreme points (thatis, 1 or 2).

Now assume that every compact polyhedral set of dimension 𝑛 − 1 has a finitenumber of extreme points. Let 𝐻𝑖, 𝑖 = 𝑖 = 1, 2, . . . ,𝑚 denote the hyperplanesassociated with the linear functionals 𝑔𝑖 defining 𝑆 (Exercise 3.49). Let x bean extreme point of 𝑆. Then 𝑆 is a boundary point of 𝑆 (Exercise 1.220) andtherefore belongs to some 𝐻𝑗 . We claim that x is also an extreme point of the set𝑆 ∩𝐻𝑗 . To see this, assume otherwise. That is, assume that x is not an extremepoint of 𝑆∩𝐻𝑗 . Then, there exists x1,x2 ∈ 𝑆∩𝐻𝑗 such that x = 𝛼x1+(1−𝛼)x2.But then x1,x2 ∈ 𝑆 and x is not an extreme point of 𝑆. Therefore, every extremepoint of 𝑆 is an extreme point of some 𝑆 ∩𝐻𝑖, which is a compact polyhedral setof dimension 𝑛 − 1. By hypothesis, each 𝑆 ∩𝐻𝑖 has a finite number of extremepoints. Since there are only 𝑚 such hyperplanes 𝐻𝑖, 𝑆 has a finite number ofextreme points.

By the Krein-Milman theorem (Exercise 3.209), 𝑆 is the closed convex hull of itsextreme points. Since there are only finite extreme points, 𝑆 is a polytope.

3.220 1. Let 𝑓, 𝑔 ∈ 𝑆∗ so that 𝑓(x) ≤ 0 and 𝑔(x) ≤ 0 for every x ∈ 𝑆. For every𝛼, 𝛽 ≥ 0

𝛼𝑓(x) + 𝛽𝑓(x) ≤ 0

for every 𝑥 ∈ 𝑆. This shows that 𝛼𝑓 + 𝛽𝑔 ∈ 𝑆∗. 𝑆∗ is a convex cone.

To show that 𝑆∗ is closed, let 𝑓 be the limit of a sequence (𝑓𝑛) of functionals in𝑆∗. Then, for every x ∈ 𝑆,

𝑓𝑛(x) ≤ 0

so that

𝑓(x) = lim 𝑓𝑛(x) ≤ 0

2. Let x,y ∈ 𝑆∗∗. Then, for every 𝑓 ∈ 𝑆∗

𝑓(x) ≤ 0 and 𝑓(y) ≤ 0

and therefore

𝑓(𝛼x+ 𝛽y) = 𝛼𝑓(x) + 𝛽𝑓(y) ≤ 0

for every 𝛼, 𝛽 ≥ 0. There 𝛼x+ 𝛽y ∈ 𝑆∗∗. 𝑆∗∗ is a convex cone.

To show that 𝑆∗∗ is closed, let x𝑛 be a sequence of points in 𝑆∗∗ converging to𝑥. For every 𝑛 = 1, 2, . . .

𝑓(x𝑛) ≤ 0 for every 𝑓 ∈ 𝑆∗

By continuity

𝑓(x) = lim 𝑓(x𝑛) ≤ 0 for every 𝑓 ∈ 𝑆∗

Consequently x ∈ 𝑆∗∗ which is therefore closed.

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3. Let x ∈ 𝑆. Then 𝑓(x) ≤ 0 for every 𝑓 ∈ 𝑆∗ so that x ∈ 𝑆∗∗.

4. Exercise 1.79.

3.221 Let 𝑓 ∈ 𝑆∗2 . Then 𝑓(x) ≤ 0 for every x ∈ 𝑆2. A fortiori, since 𝑆1 ⊆ 𝑆2,

𝑓(x) ≤ 0 for every x ∈ 𝑆1. Therefore 𝑓 ∈ 𝑆∗1 .

3.222 Exercise 3.220 showed that 𝑆 ⊆ 𝑆∗∗. To show the converse, let y /∈ 𝑆. ByProposition 3.14, there exists some 𝑓 ∈ 𝑋∗ and 𝑐 such that

𝑓(y) > 𝑐

𝑓(x) < 𝑐 for every x ∈ 𝑆Since 𝑆 is a cone, 0 ∈ 𝑆 and 𝑓(0) = 0 < 𝑐. Since 𝛼𝑆 = 𝑆 for every 𝛼 > 0 then

𝑓(x) < 0 for every x ∈ 𝑆so that 𝑓 ∈ 𝑆∗. 𝑓(y) > 0, y /∈ 𝑆∗∗. That is

y /∈ 𝑆 =⇒ y /∈ 𝑆∗∗

from which we conclude that 𝑆∗∗ ⊆ 𝑆.

3.223 Let

𝐾 = cone {𝑔1, 𝑔2, . . . , 𝑔𝑚}

= { 𝑔 ∈ 𝑋∗ : 𝑔 =

𝑚∑𝑗=1

𝜆𝑗𝑔𝑗, 𝜆𝑗 ≥ 0 }

be the set of all nonnegative linear combinations of the linear functionals 𝑔𝑗 . 𝐾 is aclosed convex cone.

Suppose that 𝑓 /∈ cone {𝑔1, 𝑔2, . . . , 𝑔𝑚}, that is assume that 𝑓 /∈ 𝐾. Then {𝑓} is acompact convex set disjoint from 𝐾. By Proposition 3.14, there exists a continuouslinear functional 𝜑 and number 𝑐 such that

sup𝑔∈𝐾𝜑(𝑔) < 𝑐 < 𝜑(𝑓)

Since 0 ∈ 𝐾, 𝑐 ≥ 0 and so 𝜑(𝑓) > 0. Further, for every 𝑔 ∈ 𝐺

𝜑(𝑔) = 𝜑(

𝑚∑𝑗=1

𝜆𝑗𝑔𝑗)

=

𝑚∑𝑗=1

𝜆𝑗𝜑(𝑔𝑗) < 𝑐 for every 𝜆𝑗 ≥ 0

Since 𝜆𝑗 can be made arbitrarily large, this last inequality implies that

𝜑(𝑔𝑗) ≤ 0 𝑗 = 1, 2, . . . ,𝑚

By the Riesz representation theorem (Exercise 3.75), there exists x ∈ 𝑋𝜑(𝑔𝑗) = 𝑔𝑗(x) and 𝜑(𝑓) = 𝑓(x)

Since

𝜑(𝑔𝑗) = 𝑔𝑗(x) ≤ 0

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x ∈ 𝑆. By hypothesis

𝑓(x) = 𝜑(𝑓) ≤ 0

contradicting the conclusion that 𝜑(𝑓) > 0. This contradiction establishes that 𝑓 ∈ 𝐾,that is

𝑓(𝑥) =

𝑚∑𝑗=1

𝜆𝑗𝑔𝑗(𝑥), 𝜆𝑗 ≥ 0

3.224 Let a1, a2, . . . , a𝑚 denote the rows of𝐴 and define the linear functional 𝑓, 𝑔1, 𝑔2, . . . , 𝑔𝑚by

𝑓(x) = cx

𝑔𝑗(x) = a𝑗x 𝑗 = 1, 2, . . . ,𝑚

Assume cx ≤ 0 for every x satisfying 𝐴x ≤ 0, that is 𝑓(x) ≤ 0 for every x ∈ 𝑆 where

𝑆 = {x ∈ 𝑋 : 𝑔𝑗(x) ≤ 0, 𝑗 = 1, 2, . . . ,𝑚 }By Proposition 3.18, there exists y ∈ ℜ𝑚+ such that

𝑓(x) =

𝑚∑𝑗=1

𝑦𝑗𝑔𝑗(x)

or

c =

𝑚∑𝑗=1

𝑦𝑗a𝑗 = 𝐴𝑇y


c = 𝐴𝑇y =

𝑚∑𝑗=1

𝑦𝑗a𝑗

Then

𝐴x ≤ 0 =⇒ a𝑗x ≤ 0 for every 𝑗 =⇒ cx ≤ 0

3.225 Let 𝑁 = ℜ𝑛+ denote the positive orthant of ℜ𝑛. 𝑁 is a convex set (indeed cone)with a nonempty interior. By Corollary 3.2.1, there exists a hyperplane 𝐻p(𝑐) suchthat

p𝑇x ≤ 𝑐 ≤ py for every x ∈ 𝑆,y ∈ 𝑁Since 0 ∈ 𝑁

p0 = 0 ≥ 𝑐which implies that 𝑐 ≤ 0 and

p𝑇x ≤ 𝑐 ≤ 0 for every 𝑥 ∈ 𝑆To show that p is nonnegative, let e1, e2, . . . , e𝑛 denote the standard basis for ℜ𝑛.Each e𝑖 belongs to 𝑁 so that

pe𝑖 = 𝑝𝑖 ≥ 0 for every 𝑖

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3.226 Assume y∗ is an efficient production plan in 𝑌 and let 𝑆 = 𝑌 − 𝑦∗. 𝑆 is convex.We claim that 𝑆∩ℜ𝑛++ = ∅. Otherwise, if there exists some z ∈ 𝑆∩ℜ𝑛++, let y′ = y∗+z

∙ z ∈ 𝑆 implies y′ ∈ 𝑌 while

∙ z ∈ ℜ𝑛++ implies y′ > y∗

contradicting the efficiency of y∗. Therefore, 𝑆 is a convex set which contains nointerior points of the nonnegative orthant ℜ𝑛+. By Exercise 3.225, there exists a pricesystem p such that

p𝑇x ≤ 0 for every x ∈ 𝑆

Since 𝑆 = 𝑌 − 𝑦∗, this implies

p(y − y∗) ≤ 0 for every y ∈ 𝑌

or

py∗ ≥ py for every y ∈ 𝑌

𝑦∗ maximizes the producer’s profit at prices p.

3.227 Consider the set 𝑆− = {x ∈ ℜ𝑛 : −x ∈ 𝑆 }.

𝑆 ∩ int ℜ𝑛− = ∅ =⇒ 𝑆− ∩ int ℜ𝑛+ = ∅

From the previous exercise, there exists a hyperplane with nonnegative normal p ≩ 0such that

p𝑇x ≤ 0 for every x ∈ 𝑆−

Since p ≩ 0, this implies

p𝑇x ≥ 0 for every x ∈ 𝑆

3.228 1. Suppose x ∈ ≿(x∗). Then, there exists an allocation (x1,x2, . . . ,x𝑛) suchthat

x =

𝑛∑𝑖=1

x𝑖

where x𝑖 ∈ ≿(x∗𝑖 ) for every 𝑖 = 1, 2, . . . , 𝑛. Conversely, if (x1,x2, . . . ,x𝑛) is anallocation with x𝑖 ∈ ≿(x∗𝑖 ) for every 𝑖 = 1, 2, . . . , 𝑛, then x =

∑𝑛𝑖=1 x𝑖 ∈ ≿(x∗).

2. For every agent 𝑖, x∗𝑖 ∈ ≿(x∗𝑖 ), which implies that

x∗ =

𝑛∑𝑖=1

x∗𝑖 ∈ ≿(x∗)

and therefore

0 ∈ 𝑆 = ≿(x∗)− x∗ ∕= ∅

Since individual preferences are convex, ≿(x∗𝑖 ) is convex for each 𝑖 and therefore𝑆 = ≿(x∗)− x∗ =

∑𝑖≿(x∗𝑖 )− x∗ is convex (Exercise 1.164).

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Assume to the contrary that 𝑆 ∩ int ℜ𝑙− ∕= ∅. That is, there exists some z ∈ 𝑆with z < 0. This implies that there exists some allocation (x1,x2, . . . ,x𝑛) suchthat

z =∑𝑖

x𝑖 − x∗ < 0

and x𝑖 ≿ x∗𝑖 for every 𝑖 ∈ 𝑁 . Distribute z equally to all the consumers. That is,consider the allocation

y𝑖 = x𝑖 + z/𝑛

By strict monotonicity, y𝑖 ≻ x𝑖 ≿ x∗𝑖 for every 𝑖 ∈ 𝑁 . Since∑𝑖

y𝑖 =∑𝑖

x𝑖 + z = x∗ =∑𝑖

x∗𝑖

(y1,y2, . . . ,y𝑛) is a reallocation of the original allocation x∗ which is strictlypreferred by all consumers. This contradicts the assumed Pareto efficiency of x∗.We conclude that

𝑆 ∩ int ℜ𝑙− ∕= ∅

3. Applying Exercise 3.227, there exists a hyperplane with nonnegative normal p∗ ≩0 such that

p∗z ≥ 0 for every z ∈ 𝑆That is

p∗(x− x∗) ≥ 0 or p∗x ≥ p∗x∗ for every x ∈ ≿(x∗) (3.65)

4. Consider any allocation which is strictly preferred to x∗ by consumer 𝑗, that isx𝑗 ∈ ≻𝑗(x

∗𝑗 ). Construct another allocation y by taking 𝜖 > 0 of each commodity

away from agent 𝑗 and distributing amongst the other agents to give

y𝑗 = (1− 𝜖)x𝑗y𝑖 = x∗𝑖 +

𝜖

𝑛− 1x𝑗 , 𝑖 ∕= 𝑗

By continuity, there exists some 𝜖 > 0 such that y𝑗 = (1 − 𝜖)x𝑗 ≻𝑗 x∗𝑗 . By

monotonicity, y𝑖 ≻𝑖 x∗𝑖 for every 𝑖 ∕= 𝑗. We have constructed an allocation y

which is strictly preferred to x∗ by all the agents, so that y =∑

𝑖 y𝑖 ∈ ≿(x∗).(3.65) implies that

py ≥ px∗

That is

p

⎛⎝(1 − 𝜖)x𝑗 +

∑𝑖∕=𝑗

(x∗𝑖 +

𝜖

𝑛− 1x𝑗

)⎞⎠ = p

⎛⎝x𝑗 +

∑𝑖∕=𝑗x∗𝑖

⎞⎠ ≥ p

⎛⎝x∗𝑗 +

∑𝑖∕=𝑗x∗𝑖

⎞⎠

which implies that

px𝑗 ≥ px∗𝑗 for every x𝑗 ∈ ≻(x∗𝑗 ) (3.66)

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5. Trivially, x∗ is a feasible allocation with endowments w𝑖 = x∗𝑖 and 𝑚𝑖 = p∗w𝑖 =p∗x∗𝑖 . To show that (p∗,x∗) is a competitive equilibrium, we have to show thatx∗𝑖 is the best allocation in the budget set𝑋𝑖(p,𝑚𝑖) for each consumer 𝑖. Supposeto the contrary there exists some consumer 𝑗 and allocation y𝑗 such that y𝑗 ≻ x𝑗and py𝑗 ≤ 𝑚𝑗 = px∗𝑗 . By continuity, there exists some 𝛼 ∈ (0, 1) such that𝛼y𝑗 ≻𝑖 x

∗𝑗 and

p(𝛼y𝑗) = 𝛼py𝑗 < py𝑗 ≤ px∗

contradicting (3.66). We conclude that

x∗𝑖 ≿𝑖 x𝑖 for every x ∈ 𝑋(p∗,𝑚𝑖)

for every consumer 𝑖. (p∗,x∗) is a competitive equilibrium.

3.229 By the previous exercise, there exists a price system p∗ such that x∗𝑖 is optimalfor each consumer 𝑖 in the budget set 𝑋(p∗,p∗x∗𝑖 ), that is

x∗𝑖 ≿𝑖 x𝑖 for every x𝑖 ∈ 𝑋(p∗,p∗x∗𝑖 ) (3.67)

For each consumer, let 𝑡𝑖 be the difference between her endowed wealth p∗w𝑖 and herrequired wealth p∗x∗𝑖 . That is, define

𝑡𝑖 = p∗x∗𝑖 − p∗w𝑖 = p∗(x∗𝑖 −w𝑖)

Then

p∗x∗𝑖 = p∗ +w𝑖 (3.68)

By assumption x∗ is feasible, so that∑𝑖

x∗𝑖 −∑𝑖

w𝑖 =∑𝑖

(x∗𝑖 −w𝑖) = 0

so that ∑𝑖

𝑡𝑖 = p∗∑𝑖

(x∗𝑖 −w𝑖) = 0

Furthermore, for 𝑚𝑖 = 𝑝∗w𝑖 + 𝑡𝑖, (3.68) implies

𝑋(p∗,𝑚𝑖) = {x𝑖 : p∗x𝑖 ≤ p∗w𝑖 + 𝑡𝑖 } = {x𝑖 : p∗x𝑖 ≤ p∗x∗𝑖 } = 𝑋(p∗,p∗x∗𝑖 )

for each consumer 𝑖. Using (3.67) we conclude that

x∗𝑖 ≿𝑖 x𝑖 for every x𝑖 ∈ 𝑋(p∗,𝑚𝑖)

for every agent 𝑖. (p∗,x∗) is a competitive equilibrium where each consumer’s after-taxwealth is

𝑚𝑖 = pw𝑖 + 𝑡𝑖

3.230 Apply Exercise 3.202 with 𝐾 = ℜ𝑛+.

3.231

𝐾∗ = {p : p𝑇x ≤ 0 for every x ∈ 𝐾 }No such hyperplane exists if and only if 𝐾∗ ∩ ℜ𝑛++ = ∅. Assume this is the case. ByExercise 3.225, there exists x ≩ 0 such that

xp = p𝑇x ≤ 0 for every p ∈ 𝐾∗

In other words, x ∈ 𝐾∗∗. By the duality theorem 𝐾∗∗ = 𝐾 which implies that x ∈ 𝐾as well as ℜ𝑛+, contrary to the hypothesis that 𝐾 ∩ ℜ𝑛+ = {0}. This contradictionestablishes that 𝐾∗ ∩ ℜ𝑛++ ∕= ∅.

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3.232 Given a set of financial assets with prices p and payoff matrix 𝑅, let

𝑍 = { (−px, 𝑅𝑥) : x ∈ ℜ𝑛 }𝑍 is the set of all possible (cost, payoff) pairs. It is a subspace of ℜ𝑆+1. Let 𝑁 be thenonnegative orthant in ℜ𝑆+1. The no arbitrage condition

𝑅x ≥ 0 =⇒ p𝑇x ≥ 0

implies that 𝑍 ∩𝑁 = {0}. By Exercise 3.230, there exists a hyperplane with positivenormal 𝜆 = 𝜆0, 𝜆1, . . . , 𝜆𝑆 such that

𝜆z = 0 for every z ∈ 𝑍𝜆z > 0 for every z ∈ ℜ𝑆+1+ ∖ {0}

That is

−𝜆0px+ 𝜆𝑅x = 0 for every x ∈ ℜ𝑛

or

p𝑇x = 𝜆/𝜆0𝑅x for every x ∈ ℜ𝑛

𝜆/𝜆0 is required state price vector.

Conversely, if a state price vector exists

𝑝𝑎 =

𝑆∑𝑎=1

𝑅𝑎𝑠𝜋𝑠

then clearly

𝑅x ≥ 0 =⇒ p𝑇x ≥ 0

No arbitrage portfolios exist.

3.233 Apply the Farkas lemma to the system

−𝐴x ≤ 0

−c𝑇x > 0

3.234 The inequality system 𝐴𝑇y ≥ c has a nonnegative solution if and only if thecorresponding system of equations

𝐴𝑇y − z = c

has a nonnegative solution y ∈ ℜ𝑚+ , z ∈ ℜ𝑛+. This is equivalent to the system

𝐵′(yz

)= c (3.69)

where 𝐵′ = (𝐴𝑇 ,−𝐼𝑛) and 𝐼𝑛 is the 𝑛 × 𝑛 identity matrix. By the Farkas lemma,system (3.69) has no solution if and only if the system

𝐵x ≤ 0 and c𝑇x > 0

has a solution x ∈ ℜ𝑛. Since 𝐵 =

(𝐴−𝐼

), 𝐵x ≤ 0 implies

𝐴x ≤ 0 and − 𝐼x ≤ 0

and the latter inequality implies x ∈ ℜ𝑛+. Thus we have established that the system𝐴𝑇y ≥ c has no nonnegative solution if and only if

𝐴x ≤ 0 and c𝑇x > 0 for some x ∈ ℜ𝑛+

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3.235 Assume system I has a solution, that is there exists x ∈ ℜ𝑛+ such that

𝐴x = 0, cx > 0, x ≥ 0

Then x = x/cx satisfies the system

𝐴x = 0, cx = 1, x ≥ 0 (3.70)

which is equivalent to

x′𝐴𝑇 = 0, xc = 1, x ≥ 0 (3.71)

Suppose y ∈ ℜ𝑚 satisfies

𝐴y ≥ c

Multiplying by x ≥ 0 gives

x′𝐴𝑇y ≥ xcSubstituting (3.71), this implies the contradiction

0 ≥ 1

We conclude that system II cannot have a solution if I has a solution.

Now, assume system I has no solution. System I is equivalent to (3.70) which in turnis equivalent to the system (

𝐴c

)x =

(01

)

or

𝐵x = b (3.72)

where 𝐵 =

( −𝐴c

)is (𝑚+ 1)×𝑛 and b =

(01

)∈ ℜ𝑚+1. If (3.72) has no solution,

there exists (by the Farkas alternative) some z ∈ ℜ𝑚+1 such that

𝐵′z ≤ 0 and bz > 0

Decompose z into z = (y, 𝑧) with y ∈ ℜ𝑚 and 𝑧 ∈ ℜ. The second inequality impliesthat

(0, 1)′(y, 𝑧) = 0y + 𝑧 = 𝑧 > 0

Without loss of generality, we can normalize so that 𝑧 = 1 and z = (y, 1).

Now 𝐵′ = (−𝐴𝑇 , c) and so the first inequality implies that

(−𝐴𝑇 , c)(y1

)= −𝐴𝑇y + c ≤ 0

or

𝐴𝑇y ≥ c

We conclude that II has a solution.

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3.236 For every linear functional 𝑔𝑗 , there exists a vector a ∈ ℜ𝑛 such that

𝑔𝑗(x) = a𝑗 x

(Proposition 3.11). Let 𝐴𝑇 be the matrix whose rows are a𝑗 , that is

𝐴 =

⎛⎜⎜⎝a1

a2

. . .a𝑚

⎞⎟⎟⎠

Then, the system of inequalities (3.31) is

𝐴𝑇x ≥ cwhere c = (𝑐1, 𝑐2, . . . , 𝑐𝑚). By the preceding exercise, this system is consistent if andonly there is no solution to the system

𝐴𝜆 = 0 c𝜆 > 0 𝜆 ≥ 0Now

𝐴𝜆 = 0 ⇐⇒𝑚∑𝑗=1

𝜆𝑗𝑔𝑖 = 0 𝑖 = 1, 2, . . . ,𝑚

Therefore, the inequalities (3.31) is consistent if an only if

𝑚∑𝑗=1

𝜆𝑗𝑔𝑖 = 0 =⇒𝑚∑𝑗=1

𝜆𝑗𝑐𝑗 ≤ 0

for every set of nonnegative numbers 𝜆1, 𝜆2, . . . , 𝜆𝑚.

3.237 Let 𝐵 be the 2𝑚× 𝑛 matrix comprising 𝐴 and −𝐴 as follows

𝐵 =

(𝐴−𝐴

)

Then the Fredholm alternative I

𝐴x = 0 c𝑇x = 1

is equivalent to the system

𝐵x ≤ 0 cx > 0 (3.73)

By the Farkas alternative theorem, either (3.73) has a solution or there exists 𝜆 ∈ 𝑅2𝑚+such that

𝐵′𝜆 = c (3.74)

Decompose 𝜆 into two 𝑚-vectors

𝜆 = (𝜇, 𝛿), 𝜇, 𝛿 ∈ 𝑅𝑚+

so that (3.74) can be rewritten as

𝐵′𝜆 = 𝐴𝑇𝜇−𝐴𝑇 𝛿 = 𝐴𝑇 (𝜇− 𝛿) = c

Define y = 𝜇− 𝛿 ∈ ℜ𝑚 We have established that either (3.73) has a solution or thereexists a vector y ∈ ℜ𝑚 such that

𝐴𝑇y = c

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3.238 Let a𝑗 , 𝑗 = 1, 2, . . . ,𝑚 denote the rows of 𝐴. Each a𝑖 defines linear functional𝑔𝑗(𝑥) = a𝑗𝑥 on ℜ𝑛, and c defines another linear functional 𝑓(𝑥) = c𝑇x. Assume that𝑓(𝑥) = c𝑇x = 0 for every x ∈ 𝑆 where

𝑆 = {x : 𝑔𝑗(x) = a𝑖x = 0, 𝑗 = 1, 2, . . . ,𝑚 }Then the system

𝐴𝑥 = 0

has no solution satisfying the constraint c𝑇x > 0. By Exercise 3.20, there exists scalars𝑦1, 𝑦2, . . . , 𝑦𝑚 such that

𝑓(x)=

𝑚∑𝑗=1

𝑦𝑗𝑔𝑗(x)

or

c =

𝑚∑𝑗=1

𝑦𝑗𝑎𝑗 = 𝐴𝑇y

That is y = (𝑦1, 𝑦2, . . . , 𝑦𝑚) solves the related nonhomogeneous system

𝐴𝑇y = c

Conversely, assume that 𝐴𝑇y = c for some 𝑦 ∈ ℜ𝑚. Then

c𝑇x = 𝑦𝐴𝑥 = 0

for all 𝑥 such that 𝐴𝑥 = 0 and therefore there is no solution satisfying the constraintc𝑇x = 1.

3.239 Let

𝑆 = { z : z = 𝐴x, x ∈ ℜ}the image of 𝑆. 𝑆 is a subspace. Assume that system I has no solution, that is

𝑆 ∩ ℜ𝑚++ = ∅By Exercise 3.225, there exists y ∈ ℜ𝑚+ ∖ {0} such that

yz = 0 for every z ∈ 𝑆That is

y𝐴x = 0 for every x ∈ ℜ𝑛

Letting x = 𝐴𝑇y, we have y𝐴𝐴𝑇y = 0 which implies that

𝐴𝑇y = 0

System II has a solution y.

Conversely, assume that x is a solution to I. Suppose to the contrary there also existsa solution y to II. Then, since 𝐴x > 0 and y ≩ 0, we must have y𝐴x = x𝐴𝑇 y > 0.On the other hand, 𝐴𝑇 y = 0 which implies x𝐴𝑇 y = 0, a contradiction. Hence, weconclude that II cannot have a solution if I has a solution.

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3.240 We have already shown (Exercise 3.239) that the alternatives I and II are mutu-ally incompatible. If Gordan’s system II

𝐴𝑇y = 0

has a semipositive solution y ≩ 0, then we can normalize y such that 1y = 1 and thesystem

𝐴𝑇y = 0

1y = 1

has a nonnegative solution.

Conversely, if Gordan’s system II has no solution, the system

𝐵′y = c

where 𝐵′ =

(𝐴𝑇

1

)and c = (0, 1) = (0, 0, . . . , 0, 1), 0 ∈ ℜ𝑚, is the (𝑚 + 1)st unit

vector has no solution y ≥ 0. By the Farkas lemma, there exists z ∈ ℜ𝑛+1 such that

𝐵z ≥ 0

cz < 0

Decompose z into z = (x, 𝑥) with x ∈ ℜ𝑛. The second inequality implies that 𝑥 < 0since

cz = (0, 1)′(x, 𝑥) = 𝑥 < 0

Since 𝐵 = (𝐴,1), the first inequality implies that

𝐵z = (𝐴,1)(x, 𝑥) = 𝐴x+ 1𝑥 ≥ 0

or

𝐴x ≥ −1𝑥 > 0

x solves Gordan’s system I.

3.241 Let a1, a2, . . . , a𝑚 be a basis for 𝑆. Let

𝐴 = (a1, a2, . . . , a𝑚)

be the matrix whose columns are a𝑗 . To say that 𝑆 contains no positive vector meansthat the system

𝐴x > 0

has no solution. By Gordan’s theorem, there exists some y ≩ 0 such that

𝐴𝑇y = 0

that is

a𝑗y = ya𝑗 = 0, 𝑗 = 1, 2, . . . ,𝑚

so that y ∈ 𝑆⊥.

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3.242 Let 𝑍 be the subspace 𝑍 = { z : 𝐴x : x ∈ ℜ𝑛 }. System I has no solution 𝐴x ≩ 0if and only if 𝑍 has no nonnegative vector z ≩ 0. By the previous exercise, 𝑍⊥ containsa positive vector y > 0 such that

yz = 0 for every z ∈ 𝑍Letting x = 𝐴𝑇y, we have y𝐴𝐴𝑇y = 0 which implies that

𝐴𝑇y = 0


3.243 Let

𝑆 = { z : z = 𝐴x, x ∈ ℜ}the image of 𝑆. 𝑆 is a subspace. Assume that system I has no solution, that is

𝑆 ∩ ℜ𝑚+ = {0}By Exercise 3.230, there exists y ∈ ℜ𝑚++ such that

yz = 0 for every z ∈ 𝑆That is

y𝐴x = 0 for every x ∈ ℜ𝑛

Letting x = 𝐴𝑇y, we have y𝐴𝐴𝑇y = 0 which implies that

𝐴𝑇y = 0


Conversely, assume that x is a solution to I. Suppose to the contrary there also existsa solution y to II. Then, since 𝐴x ≩ 0 and y > 0, we must have y𝐴x = x𝐴𝑇 y > 0.On the other hand, 𝐴𝑇 y = 0 which implies x𝐴𝑇 y = 0, a contradiction. Hence, weconclude that II cannot have a solution if I has a solution.

3.244 The inequality system 𝐴𝑇y ≤ 0 has a nonnegative solution if and only if thecorresponding system of equations

𝐴𝑇y + z = 0

has a nonnegative solution y ∈ ℜ𝑚+ , z ∈ ℜ𝑛+. This is equivalent to the system

𝐵′(yz

)= 0 (3.75)

where 𝐵′ = (𝐴𝑇 , 𝐼𝑛) and 𝐼𝑛 is the 𝑛×𝑛 identity matrix. By Gordan’s theorem, system(3.75) has no solution if and only if the system

𝐵x > 0

has a solution x ∈ ℜ𝑛. Since 𝐵 =

(𝐴𝐼

), 𝐵x > 0 implies

𝐴x > 0 and 𝐼x > 0

and the latter inequality implies x ∈ ℜ𝑛++. Thus we have established that the system𝐴𝑇y ≤ 0 has no nonnegative solution if and only if

𝐴x > 0 for some x ∈ ℜ𝑛++

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3.245 Assume system II has no solution, that is there is no y ∈ ℜ𝑛 such that

𝐴y ≤ 0, y ≩ 0

This implies that the system

−𝐴y ≥ 01y ≥ 1

has no solution y ∈ ℜ𝑚+ . Defining 𝐵′ =

( −𝐴1′

), the latter can be written as

𝐵′y ≥ −e𝑚+1 (3.76)

where −e𝑚+1 = (0, 1), 0 ∈ ℜ𝑚.

By the Gale alternative (Exercise 3.234), if system (3.76) has no solution, the alterna-tive system

𝐵z ≤ 0, −e𝑚+1z > 0

has a nonnegative solution z ∈ ℜ𝑛+1+ . Decompose z into z = (x, 𝑧) where x ∈ ℜ𝑛+ and𝑧 ∈ ℜ+. The second inequality implies 𝑧 > 0 since e𝑚+1z = 𝑧.

𝐵 = (−𝐴𝑇 ,1) and the first inequality implies

𝐵z = (−𝐴𝑇 ,1)(x𝑧

)= −𝐴𝑇x+ 1𝑧 ≤ 0

or

𝐴𝑇x ≥ 1𝑧 > 0

Thus system I has a solution x ∈ ℜ𝑛+. Since x = 0 implies 𝐴x = 0, we conclude thatx ≩ 0.

Conversely, assume that II has a solution y ≩ 0 such that 𝐴y ≤ 0. Then, for everyx ∈ ℜ𝑛+

x𝐴𝑇y = y′𝐴𝑇x ≤ 0

Since y ≩ 0, this implies

𝐴𝑇x ≤ 0

for every x ∈ ℜ𝑛+ which contradicts I.

3.246 We give a constructive proof, by proposing an algorithm which will generatethe desired decomposition. Assume that x satisfies 𝐴x ≩ 0. Arrange the rows of 𝐴such that the positive elements of 𝐴x are listed first. That is, decompose 𝐴 into twosubmatrices such that

𝐵1x > 0

𝐶1x = 0

Either

Case 1 𝐶1x ≩ 0 has no solution and the result is proved or

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Case 2 𝐶1x ≩ 0 has a solution x′.

Let x be a linear combination of x and x′. Specifically, define

x = 𝛼x+ x′

where

𝛼 > max−b𝑗xb𝑗x′

where b𝑗 is the 𝑗th row of 𝐵1. 𝛼 is chosen so that

𝛼𝐵1x > 𝐵1x′

By direct computation

𝐵1x = 𝛼𝐵1x+𝐵1x′ > 0

𝐶1x = 𝛼𝐶1x+ 𝐶1x′ ≩ 0

since 𝐶1x = 0 and 𝐶1x′ ≩ 0. By construction, x is another solution to 𝐴x ≩ 0such that 𝐴x has more positive components than 𝐴x. Again, collect all the positivecomponents together, decomposing 𝐴 into two submatrices such that

𝐵2x > 0

𝐶2x = 0

Either

Case 1 𝐶2x ≩ 0 has no solution and the result is proved or

Case 2 𝐶2x ≩ 0 has a solution x′′.

In the second case, we can repeat the previous procedure, generating another decom-position 𝐵3, 𝐶3 and so on. At each stage 𝑘, the matrix 𝐵𝑘 get larger and 𝐶𝑘 smaller.The algorithm must terminate before 𝐵𝑘 equals 𝐴, since we began with the assumptionthat 𝐴x > 0 has no solution.

3.247 There are three possible cases to consider.

Case 1: y = 0 is the only solution of 𝐴𝑇y = 0. Then 𝐴x > 0 has a solution x′ byGordan’s theorem and

𝐴x′ + 0 > 0

Case 2: 𝐴𝑇y = 0 has a positive solution y > 0 Then 0 is the only solution 𝐴x ≥ 0by Stiemke’s theorem and

𝐴0+ y > 0

Case 3 𝐴𝑇y = 0 has a solution y ≩ 0 but y ∕> 0. By Gordan’s theorem 𝐴x > 0 hasno solution. By the previous exercise, 𝐴 can be decomposed into two consistentsubsystems

𝐵x > 0

𝐶x = 0

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such that 𝐶x ≩ 0 has no solution. Assume that 𝐵 is 𝑘 × 𝑛 and 𝐶 is 𝑙× 𝑛 where𝑙 = 𝑚− 𝑘. Applying Stiemke’s theorem to 𝐶, there exists z > 0, z ∈ ℜ𝑙. Definey ∈ ℜ𝑚+ by

𝑦𝑗 =

{0 𝑗 = 1, 2, . . . , 𝑘

𝑦𝑗 = 𝑧𝑗−𝑘 𝑗 = 𝑘 + 1, 𝑘 + 2, . . . ,𝑚

Then x, y is the desired solution since for every 𝑗, 𝑗 = 1, 2, . . . ,𝑚 either 𝑦𝑗 > 0or (𝐴x)𝑗 = (𝐵x)𝑗 > 0.

3.248 Consider the dual pair(𝐴𝐼

)x ≥ 0 and (𝐴𝑇 , 𝐼)

(yz

)= 0, y ≥ 0, z ≥ 0

By Tucker’s theorem, this has a solution x∗,y∗, z∗ such that

𝐴x∗ ≥ 0, x∗ ≥ 0, 𝐴𝑇y∗ + z∗ = 0, y∗ ≥ 0, z∗ ≥ 0𝐴x+ y > 0

𝐼x∗ + 𝐼z > 0

Substituting z∗ = −𝐴𝑇y∗ implies

𝐴𝑇y ≤ 0

and

x−𝐴𝑇y∗ > 0

3.249 Consider the dual pair

𝐴x ≥ 0 and 𝐴𝑇y = 0, y ≥ 0

where 𝐴 is an 𝑚 × 𝑛 matrix. By Tucker’s theorem, there exists a pair of solutionsx∗ ∈ ℜ𝑛 and y∗ ∈ ℜ𝑚 such that

𝐴x∗ + y∗ > 0 (3.77)

Assume that 𝐴x > 0 has no solution (Gordan I). Then there exists some 𝑗 such that(𝐴x∗)𝑗 = 0 and (3.77) implies that 𝑦∗𝑗 > 0. Therefore y∗ ≩ 0 and solves Gordan II.

Conversely, assume that 𝐴𝑇y = 0 has no solution y > 0 (Stiemke II). Then, thereexists some 𝑗 such that 𝑦∗𝑗 = 0 and (3.77) implies that (𝐴x∗)𝑗 > 0). Therefore x∗

solves 𝐴x ≩ 0 (Stiemke I).

3.250 We have already shown that Farkas I and II are mutually inconsistent. Assumethat Farkas system I

𝐴x ≥ 0, c𝑇x < 0

has no solution. Define the (𝑚+ 1)×𝑛 matrix 𝐵 =

(𝐴−c′

). Our assumption is that

the system

𝐵x ≥ 0

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has no solution with (𝐵x)𝑚+1 = −cx > 0. By Tucker’s theorem, the dual system

𝐵′z = 0

has a solution z ∈ ℜ𝑚+1+ with z𝑚+1 > 0. Without loss of generality, we can normalizeso that z𝑚+1 = 1. Decompose z into z = (y, 1) with y ∈ ℜ𝑚+ . Since 𝐵′ = (𝐴𝑇 ,−c),𝐵′z = 0 implies

𝐵′z = (𝐴𝑇 ,−c)(y, 1) = 𝐴𝑇y − c = 0

or

𝐴𝑇y = c

y ∈ ℜ𝑚+ solves Farkas II.

3.251 If x ≥ 0 solves I, then

x′(𝐴𝑇y1 +𝐵′y2 + 𝐶′y3) = x′𝐴𝑇y1 + x′𝐵′y2 + x′𝐶′y3) > 0

since x′𝐴𝑇y1 = y1𝐴x > 0, x′𝐵′y2 = y2𝐵x ≥ 0 and x′𝐶′y3 = y3𝐶x = 0 whichcontradicts II.

The equation 𝐶x = 0 is equivalent to the pair of inequalities 𝐶x ≥ 0,−𝐶x ≥ 0. ByTucker’s theorem the dual pair

𝐴x ≥ 0 𝐴𝑇y1 +𝐵′y2 + 𝐶′y3 − 𝐶′y4 = 0

𝐵x ≥ 0𝐶x ≥ 0−𝐶x ≥ 0

has solutions 𝑥 ∈ ℜ𝑛, y1 ∈ ℜ𝑚1 , y2 ∈ ℜ𝑚2 , u3,v3 ∈ ℜ𝑚3 such that

y1 ≥ 0 𝐴x+ y1 > 0

y2 ≥ 0 𝐵x+ y2 > 0

u3 ≥ 0 𝐶x + u3 > 0

v3 ≥ 0 −𝐶x+ v3 > 0

Assume Motzkin I has no solution. That is, there is y1 ≩ 0. Define y3 = u3 − v3.Then y1,y2,y3 satisfies Motzkin II.

3.252 1. For every a ∈ 𝑆, let 𝑆∗a be the polar set

𝑆∗a = {x ∈ ℜ𝑛 : ∥x∥ = 1,xa ≥ 0 }

𝑆∗a is nonempty since 0 ∈ 𝑆∗

a. Let x be the limit of a sequence x𝑛 of points in 𝑆∗a.

Since x𝑛a ≥ 0 for every 𝑛, xa ≥ 0 so that x ∈ 𝑆∗a. Hence 𝑆∗

a is a closed subset of𝐵 = {x ∈ ℜ𝑛 : ∥x∥ = 1 }.

2. Let {a1, a2, . . . , a𝑚} be any finite set of points in 𝑆. Since 0 /∈ 𝑆, the system

𝑚∑𝑖=1

𝑦𝑖a𝑖 = 0,

𝑚∑𝑖=1

𝑦𝑖 = 1, 𝑦𝑖 ≥ 0

has no solution. A fortiori, the system

𝑚∑𝑖=1

𝑦𝑖a𝑖 = 0

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has no solution 𝑦 ∈ ℜ𝑚+ . If 𝐴 is the 𝑚×n matrix whose rows are a𝑖, the lattersystem can be written as

𝐴𝑇y = 0

3. By Gordan’s theorem, the system

𝐴x > 0 (3.78)

has a solution x ∕= 0.4. Without loss of generality, we can take ∥x∥ = 1. (3.78) implies that

a𝑖x = xa𝑖 > 0

for every 𝑖 = 1, 2 . . . ,𝑚 so that x ∈ 𝑆∗a𝑖

. Hence

x ∈𝑚∩𝑖=1

𝑆∗a𝑖

5. We have shown that for every finite set {a1, a2, . . . , a𝑚} ⊆ 𝑆,∩𝑚

𝑖=1 𝑆∗a𝑖

is non-empty closed subset of the compact set 𝐵 = {𝑥 ∈ ℜ𝑛 : ∥x∥ = 1}. By the Finiteintersection property (Exercise 1.116)∩

a∈𝑆𝑆∗a ∕= ∅

6. For every p ∈ ∩a∈𝑆 𝑆

∗a

pa ≥ 0 for every a ∈ 𝑆p defines a hyperplane 𝑓(a) = pa which separates 𝑆 from 0.

3.253 The expected outcome if player 1 adopts the mixed strategy p = (𝑝1, 𝑝2, . . . , 𝑝𝑚)and player 2 plays her 𝑗 pure strategy is

𝑢(p, 𝑗) =

𝑚∑𝑖=1

𝑝𝑖𝑎𝑖𝑗 = pa𝑗

where a𝑗 is the 𝑗th column of 𝐴. The expected payoff to 1 for all possible responsesof player 2 is the vector (p𝐴)′ = 𝐴𝑇p. The mixed strategy p ensures player 1 anonnegative security level provided 𝐴𝑇p ≥ 0.Similarly, if 2 adopts the mixed strategy q = (𝑞1, 𝑞2, . . . , 𝑞𝑛), the expected payoff to 2if 1 plays his 𝑖 strategy is a𝑖q where a𝑖 is the 𝑖th row of 𝐴. The expected outcome forall the possible responses of player 1 is the vector 𝐴q. The mixed strategy q ensuresplayer 2 a nonpositive security level provided 𝐴q ≤ 0.By the von Neumann alternative theorem (Exercise 3.245), at least one of these alter-natives must be true. That is, either

Either I 𝐴𝑇p > 0, p ≩ 0 for some p ∈ ℜ𝑚or II 𝐴q ≤ 0, q ≩ 0 for some q ∈ ℜ𝑛Since p ≩ 0 and q ≩ 0, we can normalize so that p ∈ Δ𝑚−1 and q ∈ Δ𝑛−1. At leastone of the players has a strategy which guarantees she cannot lose.

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3.254 1. For any 𝑐 ∈ ℜ, define the game

𝑢(a1, a2) = 𝑢(a1, a2)− 𝑐with

𝑣1 = maxp

min𝑗��(p, 𝑗) = max

pmin𝑗𝑢(p, 𝑗)− 𝑐 = 𝑣1 − 𝑐

𝑣2 = minq

max𝑖��(𝑖,q) = min

qmax𝑖𝑢(𝑖,q)− 𝑐 = 𝑣2 − 𝑐

By the previous exercise,

Either 𝑣1 ≥ 0 or 𝑣𝑦 ≤ 0

That is

Either 𝑣1 ≥ 𝑐 or 𝑣2 ≤ 𝑐

2. Since this applies for arbitrary 𝑐 ∈ ℜ, it implies that while

𝑣1 ≤ 𝑣2and there is no 𝑐 such that

𝑣1 < 𝑐 < 𝑣2

Therefore, we conclude that 𝑣1 = 𝑣2 as required.

3.255 1. The mixed strategies p of player 1 are elements of the simplex Δ𝑚−1,which is compact (Example 1.110). Since 𝑣1(p) = min𝑛𝑗=1 𝑢(p, 𝑗) is continuous(Maximum theorem 2.3), 𝑣1(p) achieves its maximum on Δ𝑚−1 (Weierstrasstheorem 2.2). That is, there exists p∗ ∈ Δ𝑚−1 such that

𝑣1 = 𝑣1(p∗) = max

p𝑣1(p)

Similarly, there exists q∗ ∈ Δ𝑛−1 such that

𝑣2 = 𝑣2(q∗) = min

q𝑣2(q)

2. Let 𝑢(p,q) denote the expected outcome when player 1 adopts mixed strategy pand player 2 plays q. That is

𝑢(p,q) =𝑚∑𝑖=1

𝑛∑𝑗=1

𝑝𝑖𝑞𝑖𝑎𝑖𝑗

Then

𝑣 = 𝑢(p∗,q∗) = max𝑖𝑢(𝑖,q∗) ≥

∑𝑖

𝑝𝑖𝑢(𝑖,q∗) = 𝑢(p,q∗) for every p ∈ Δ𝑚−1

Similarly

𝑣 = 𝑢(p∗,q∗) = min𝑗𝑢(p∗, 𝑗) ≤

∑𝑗

𝑞𝑗𝑢(p∗, 𝑗) = 𝑢(p∗,q) for every q ∈ Δ𝑛−1

(p∗,q∗) is a Nash equilibrium.

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3.256 By the Minimax theorem, every finite two person zero-sum game has a value.The previous result shows that this is attained at a Nash equilibrium.

3.257 If player 2 adopts the strategy 𝑡1

𝑓p(𝑡1) = −𝑝1 + 2𝑝2 < 0 if 𝑝1 > 2𝑝2

If player 2 adopts the strategy 𝑡5

𝑓p(𝑡5) = 𝑝1 − 2𝑝2 < 0 if 𝑝1 < 2𝑝2

Therefore

𝑣1(p) = min𝑧∈𝑍𝑓p(z) ≤ min{𝑓p(𝑡1), 𝑓p(𝑡5)} < 0

for every p such that 𝑝1 ∕= 𝑝2. Since 𝑝1 + 𝑝2 = 1, we conclude that

𝑣1(p)

{= 0 p = p∗ = (2/3, 1/3)

< 0 otherwise

We conclude that

𝑣1 = maxp𝑣1(p) = 0

which is attained at p∗ = (2/3, 1/3).

3.258 1.

𝑣2 = minz∈𝑍

𝑚max𝑖=1𝑧𝑖

Since 𝑍 is compact, 𝑣2 = 0 implies there exists z ∈ 𝑍 such that

𝑚max𝑖=1𝑧𝑖 = 0

which implies that z ≤ 0. Consequently 𝑍 ∩ ℜ𝑛− ∕= ∅.2. Assume to the contrary that there exists

z ∈ 𝑍 ∩ int ℜ𝑛−That is, there exists some strategy q ∈ Δ𝑛−1 such that 𝐴q < 0 and therefore𝑣2 < 0, contrary to the hypothesis.

3. There exists a hyperplane with nonnegative normal separating 𝑍 from ℜ𝑛− (Ex-ercise 3.227). That is, there exists p∗ ∈ ℜ𝑛+, p∗ ∕= 0 such that

𝑓p∗(z) ≥ 0 for every z ∈ 𝑍

and therefore

𝑣1(p∗) = min

z∈𝑍𝑓p∗(z) ≥ 0

Without loss of generality, we can normalize so that∑𝑛

𝑖=1 𝑝∗𝑖 = 1 and therefore

p∗ ∈ Δ𝑚−1.

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4. Consequently

𝑣1 = maxp𝑣1(p) ≥ 𝑣1(p∗) ≥ 0

On the other hand, we know that 𝑍 contains a point z ≤ 0. For every p ≥ 0𝑓p(z) ≤ 0

and therefore

𝑣1(p) = minz∈𝑍𝑓p(z) ≤ 𝑓p(𝑧) ≤ 0

so that

𝑣1 = maxp𝑣1(p) ≤ 0

We conclude that

𝑣1 = 0 = 𝑣2

3.259 Consider the game with the same strategies and the payoff function

��(a1, a2) = 𝑢(a1, a2)− 𝑐The expected value to player 2 is

𝑣2 = minq

max𝑖��(𝑖,q) = min

qmax𝑖𝑢(𝑖,q)− 𝑐 = 𝑣2 − 𝑐 = 0

By the previous exercise 𝑣1 = 𝑣2 = 0 and

𝑣1 = maxp

min𝑗𝑢(p, 𝑗) = max

qmin𝑗��(p, 𝑗) + 𝑐 = 𝑣1 + 𝑐 = 𝑐 = 𝑣2

3.260 Assume that p1 and p2 are both optimal strategies for player 1. Then

𝑢(p1,q) ≥ 𝑣 for every q ∈ Δ𝑛−1

𝑢(p2,q) ≥ 𝑣 for every q ∈ Δ𝑛−1

Let p = 𝛼p1,p2 + (1− 𝛼). Since 𝑢 is bilinear

𝑢(p,q) = 𝛼𝑢(p1,q) + (1− 𝛼)𝑢(p2,q) ≥ 𝑣 for every q ∈ Δ𝑛−1

Consequently, p is also an optimal strategy for player 1.

3.261 𝑓 is the payoff function of some 2 person zero-sum game in which the playershave 𝑚 + 1 and 𝑛 + 1 strategies respectively. The result follows from the MinimaxTheorem.

3.262 1. The possible partitions of 𝑁 = {1, 2, 3} are:

{1}, {2}, {3}{𝑖, 𝑗}, {𝑘}, 𝑖, 𝑗, 𝑘 =∈ 𝑁, 𝑖 ∕= 𝑗 ∕= 𝑘{1, 2, 3}

In any partition, at most one coalition can have two or more players, and therefore

𝐾∑𝑘=1

𝑤(𝑆𝑘) ≤ 1

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2. Assume x = (𝑥1, 𝑥2, 𝑥3) ∈ core. Then x must satisfy the following system ofinequalities

𝑥1 + 𝑥2 ≥ 1 = 𝑤({1, 2})𝑥1 + 𝑥3 ≥ 1 = 𝑤({1, 3})𝑥2 + 𝑥3 ≥ 1 = 𝑤({2, 3})

which can be summed to yield

2(𝑥1 + 𝑥2 + 𝑥3) ≥ 3

or

𝑥1 + 𝑥2 + 𝑥3 ≥ 3/2

which implies that x exceeds the sum available. This contradiction establishesthat the core is empty.

Alternatively, observe that the three person majority game is a simple game withno veto players. By Exercise 1.69, its core is empty.

3.263 Assume that the game (𝑁,𝑤) is not cohesive. Then there exists a partition{𝑆1, 𝑆2, . . . , 𝑆𝐾} of 𝑁 such that

𝑤(𝑁) <𝐾∑𝑘=1

𝑤(𝑆𝑘)

Assume x ∈ core. Then ∑𝑖∈𝑆𝑘

𝑥𝑖 ≥ 𝑤(𝑆𝑘) 𝑘 = 1, 2, . . . ,𝐾

Since {𝑆1, 𝑆2, . . . , 𝑆𝐾} is a partition

∑𝑖∈𝑁𝑥𝑖 =

𝐾∑𝑘=1

∑𝑖∈𝑆𝑘

𝑥𝑖 ≥𝑁∑𝑘=1

𝑤(𝑆𝑘) > 𝑤(𝑁)

which contradicts the assumption that x ∈ core. This establishes that cohesivity isnecessary for the existence of the core.

To show that cohesivity is not sufficient, we observe that the three person majoritygame is cohesive, but its core is empty.

3.264 The other balanced families of coalitions in a three player game are

1. ℬ = {𝑁} with weights

𝑤(𝑆) =

{1 𝑆 = 𝑁

0 otherwise

2. ℬ = {{1}, {2}, {3}} with weights 𝑤(𝑆) = 1 for every 𝑆 ∈ ℬ3. ℬ = {{𝑖}, {𝑗, 𝑘}}, 𝑖, 𝑗, 𝑘 ∈ ℬ, 𝑖 ∕= 𝑗 ∕= 𝑘 with weights 𝑤(𝑆) = 1 for every 𝑆 ∈ ℬ

3.265 The following table lists some nontrivial balanced families of coalitions for a fourplayer game. Other balanced families can be obtained by permutation of the players.

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Weights{123}, {124}, {34} 1/2, 1/2, 1/2{12}, {13}, {23}, {4} 1/2, 1/2, 1/2, 1{123}, {14}, {24}, {3} 1/2, 1/2, 1/2, 1/2{123}, {14}, {24}, {34} 2/3, 1/3, 1/3, 1/3{123}, {124}, {134}, {234} 1/3, 1/3, 1/3, 1/3

3.266 Both sides of the expression

e𝑁 =∑𝑆∈ℬ𝜆𝑆e𝑆

are vectors, with each component corresponding to a particular player. For player 𝑖,the 𝑖𝑡ℎ component of e𝑁 is 1 and the 𝑖𝑡ℎ component of e𝑆 is 1 if 𝑖 ∈ 𝑆 and 0 otherwise.Therefore, for each player 𝑖, the preceding expression can be written∑

𝑆∈ℬ∣𝑆∋𝑖𝜆𝑆 = 1

For each coalition 𝑆, the share of the coalition 𝑆 at the allocation x is

𝑔𝑆(x) =∑𝑖 ∈ 𝑆𝑥𝑖 = e𝑆x (3.79)

The condition

𝑔𝑁 =∑𝑆∈ℬ𝜆𝑆𝑔𝑆

means that for every x ∈ 𝑋

𝑔𝑁 (x) =∑𝑆∈ℬ𝜆𝑆𝑔𝑆(x)

Substituting (3.79)

e𝑁 x =∑𝑆∈ℬ𝜆𝑆𝑒𝑆x

which is equivalent to the condition∑𝑆∈ℬ𝜆𝑆e𝑆 = e𝑁

3.267 By construction, 𝜇 ≥ 0. If 𝜇 = 0,∑𝑆⊆𝑁

𝜆𝑆𝑔𝑆 − 𝜇𝑔𝑁 = 0

implies that 𝜆𝑆 = 0 for all 𝑆 and consequently∑𝑆⊆𝑁

𝜆𝑆𝑤(𝑆)− 𝜇𝑤(𝑁) ≤ 0

is trivially satisfied. On the other hand, if 𝜇 > 0, we can divide both conditions by 𝜇.)

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3.268 Let (𝑁,𝑤1) and (𝑁,𝑤2) be balanced games. By the Bondareva-Shapley theorem,they have nonempty cores. Let x1 ∈ core(𝑁,𝑤1) and x2 ∈ core(𝑁,𝑤2). That is,

𝑔𝑆(x1) ≥ 𝑤1(𝑆) for every 𝑆 ⊆ 𝑁𝑔𝑆(x2) ≥ 𝑤2(𝑆) for every 𝑆 ⊆ 𝑁

Adding, we have

𝑔𝑆(x1) + 𝑔𝑆(x2) = 𝑔𝑆(x1 + x2) ≥ 𝑤1(𝑆) + 𝑤2(𝑆) for every 𝑆 ⊆ 𝑁which implies that x1 + x2 belongs to core(𝑁,𝑤1 + 𝑤2). Therefore (𝑁,𝑤1 + 𝑤2) isbalanced. Similarly, if x ∈ core(𝑁,𝑤), then 𝛼x belongs to core(𝑁,𝛼𝑤) for every𝛼 ∈ ℜ+. That is (𝑁,𝛼𝑤) is balanced for every 𝛼 ∈ ℜ+.

3.269 1. Assume otherwise. That is assume there exists some y ∈ 𝐴 ∩ 𝐵. Takingthe first 𝑛 components, this implies that

e𝑁 =∑𝑆⊆𝑁

𝜆𝑠e𝑆

for some (𝜆𝑆 ≥ 0 : 𝑆 ⊆ 𝑁). Let ℬ = {𝑆 ⊂ 𝑁 ∣ 𝜆𝑆 > 0} be the set of coalitionswith strictly positive weights. Then ℬ is a balanced family of coalitions withweights 𝜆𝑆 (Exercise 3.266).

However, looking at the last coordinate, y ∈ 𝐴 ∩𝐵 implies∑𝑆∈ℬ𝜆𝑠𝑤(𝑆) = 𝑤(𝑁) + 𝜖 > 𝑤(𝑁)

which contradicts the assumption that the game is balanced. We conclude that𝐴 and 𝐵 are disjoint if the game is balanced.

2. (a) Substituting y = (e∅, 0) in (3.36) gives

(z, 𝑧0)′(0, 0) = 0 ≥ 𝑐

which implies that 𝑐 ≤ 0.

NOTE We still have to show that 𝑐 ≥ 0.

(b) Substituting (e𝑁 , 𝑤𝑦(𝑁)) in (3.36) gives

𝑧e𝑁 + 𝑧0𝑤(𝑁) > 𝑧e𝑁 + 𝑧0𝑤(𝑁) + 𝑧0𝜖

for all 𝜖 > 0, which implies that 𝑧0 < 0.

3. Without loss of generality, we can normalize so that 𝑧0 = −1. Then the separatinghyperplane conditions become

(z,−1)′y ≥ 0 for every y ∈ 𝐴 (3.80)

(z,−1)′(e𝑁 , 𝑤(𝑁) + 𝜖) < 0 for every 𝜖 > 0 (3.81)

For any 𝑆 ⊆ 𝑁 , (e𝑆 , 𝑤(𝑆)) ∈ 𝐴. Substituting y = (e𝑆 , 𝑤(𝑆)) in (3.80) gives

e′𝑆z− 𝑤(𝑆) ≥ 0

that is

𝑔𝑆(z) = e′𝑆z =≥ 𝑤(𝑆)

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while (3.81) implies

𝑔𝑁 (z) = e′𝑁z > 𝑤(𝑁) + 𝜖 for every 𝜖 > 0

This establishes that z belongs to the core. Hence the core is nonempty.

3.270 1. Let 𝛼 = 𝑤(𝑁) −∑𝑖∈𝑁 𝑤𝑖 > 0 since (𝑁,𝑤) is essential. For every 𝑆 ⊆ 𝑁 ,

define

𝑤0(𝑆) =1

𝛼

(𝑤(𝑆) −

∑𝑖∈𝑆𝑤𝑖

)

Then

𝑤0({𝑖}) = 0 for every 𝑖 ∈ 𝑁𝑤0(𝑁) = 1

𝑤0 is 0–1 normalized.

2. Let y ∈ core(𝑁,𝑤0). Then for every 𝑆 ⊆ 𝑁∑𝑖∈𝑆𝑦𝑖 ≥ 𝑤0(𝑆) (3.82)

∑𝑖∈𝑁𝑦𝑖 = 1 (3.83)

Let w = (𝑤1, 𝑤2, . . . , 𝑤𝑛) where 𝑤𝑖 = 𝑤({𝑖}). Let x = 𝛼y+w. Using (3.82) and(3.83) ∑

𝑖∈𝑆𝑥𝑖 =

∑𝑖∈𝑆

(𝛼𝑦𝑖 + 𝑤𝑖)

= 𝛼∑𝑖∈𝑆𝑦𝑖 +


≥ 𝛼𝑤0(𝑆) +∑𝑖∈𝑆𝑤𝑖

= 𝛼1

𝛼

(𝑤(𝑆) −


)+


= 𝑤(𝑆)∑𝑖∈𝑁𝑥𝑖 =

∑𝑖∈𝑁

(𝛼𝑦𝑖 + 𝑤𝑖)

= 𝛼+∑𝑖∈𝑁𝑤𝑖

= 𝑤(𝑁)

Therefore, x = 𝛼y +w ∈ core(𝑁,𝑤). Similarly, we can show that

x ∈ core(𝑁,𝑤) =⇒ y =1

𝛼(x−w) ∈ core(𝑁,𝑤0)

and therefore

core(𝑁,𝑤) = 𝛼core(𝑁,𝑤0) +w

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3. This immediately implies

core(𝑁,𝑤) = ∅ ⇐⇒ core(𝑁,𝑤0) = ∅3.271 (𝑁,𝑤) is 0–1 normalized, that is

𝑤({𝑖} = 0 for every 𝑖 ∈ 𝑁𝑤(𝑁) = 1

Consequently, x belongs to the core of (𝑁,𝑤) if and only if

𝑥𝑖 ≥ 𝑤𝑖 = 0 (3.84)∑𝑖∈𝑁𝑥𝑖 = 𝑤(𝑁) = 1 (3.85)

∑𝑖∈𝑆𝑥𝑖 ≥ 𝑤(𝑆) for every 𝑆 ∈ 𝒜 (3.86)

(3.84) and (3.85) ensure that x = (𝑥1, 𝑥2, . . . , 𝑥𝑛) is a mixed strategy for player 1 in thetwo-person zero-sum game. Using this mixed strategy, the expected payoff to player Ifor any strategy 𝑆 of player II is

𝑢(x, 𝑆) =∑𝑖∈𝑁𝑥𝑖𝑢(𝑖, 𝑆) =

∑𝑖∈𝑆𝑥𝑖

1

𝑤(𝑆)

(3.86) implies

𝑢(x, 𝑆) =∑𝑖∈𝑆𝑥𝑖

1

𝑤(𝑆)≥ 1 for every 𝑆 ∈ 𝒜

That is any x ∈ core(𝑁,𝑤) provides a mixed strategy for player I which ensures apayoff at least 1. That is

core(𝑁,𝑤) ∕= ∅ =⇒ 𝛿 ≥ 1

Conversely, if the 𝛿 < 1, there is no mixed strategy for player I which satisfies (3.86) andconsequently no x which satisfies (3.84), (3.85) and (3.86). In other words, core(𝑁,𝑤) =∅.3.272 If 𝛿 is the value of 𝐺, there exists a mixed strategy which will guarantee that IIpays no more than 𝛿. That is, there exists numbers 𝑦𝑆 ≥ 0 for every coalition 𝑆 ∈ 𝒜such that ∑

𝑆∈𝒜𝑦𝑆 = 1

and ∑𝑆∈𝒜𝑦𝑆𝑢(𝑖, 𝑆) ≤ 𝛿 for every 𝑖 ∈ 𝑁

that is

∑𝑆∈𝒜𝑆∋𝑖

𝑦𝑆1

𝑤(𝑆)≤ 𝛿 for every 𝑖 ∈ 𝑁

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or

∑𝑆∈𝒜𝑆∋𝑖

𝑦𝑆𝛿𝑤(𝑆)

≤ 1 for every 𝑖 ∈ 𝑁 (3.87)

For each coalition 𝑆 ∈ 𝒜 let

𝜆𝑆 =𝑦𝑆𝛿𝑤(𝑆)

in (3.87) ∑𝑆∈𝒜𝑆∋𝑖

𝜆𝑆 ≤ 1

Augment the collection 𝒜 with the single-player coalitions to form the collection

ℬ = 𝒜 ∪ { {𝑖} : 𝑖 ∈ 𝑁 }

and with weights {𝜆𝑆 : 𝑆 ∈ 𝒜} and

𝜆{𝑖} = 1−∑𝑆∈𝒜𝜆𝑆

Then ℬ is a balanced collection.

Since the game (𝑁,𝑤) is balanced

1 = 𝑤(𝑁) ≥∑𝑆∈ℬ𝜆𝑆𝑤(𝑆)

=∑𝑆∈𝒜𝜆𝑆𝑤(𝑆)

=∑𝑆∈ℬ

𝑦𝑆𝛿𝑤(𝑆)

𝑤(𝑆)

=1

𝛿

∑𝑆∈ℬ𝑦𝑆

=1

𝛿

that is

1 ≥ 1

𝛿(3.88)

If I plays the mixed strategy x = (1/𝑛, 1/𝑛, . . . , 1/𝑛), the payoff is

𝑢(x, 𝑆) =∑𝑖∈𝑁

1

𝑛𝑤(𝑆)=

1

𝑤(𝑆)> 0 for every 𝑆 ⊆ 𝒜

Therefore 𝛿 > 0 and (3.88) implies that

𝛿 ≥ 1

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3.273 Assume core(𝑁,𝑤) ∕= ∅ and let 𝑥 ∈ core(𝑁,𝑤). Then

𝑔𝑆(x) ≥ 𝑤(𝑆) for every 𝑆 ⊆ 𝑁 (3.89)

where 𝑔𝑆 =∑

𝑖∈𝑆 𝑥𝑖 measures the share coalition 𝑆 at the allocation x.

Let ℬ be a balanced family of coalitions with weights 𝜆𝑆 . For every 𝑆 ∈ ℬ, (3.89)implies

𝜆𝑆𝑔𝑆(x) ≥ 𝜆𝑆𝑤(𝑆)

Summing over all 𝑆 ∈ ℬ ∑𝑆∈ℬ𝜆𝑆𝑔𝑆(x) ≥

∑𝑆∈ℬ𝜆𝑆𝑤(𝑆) (3.90)

Evaluating the left hand side of this inequality∑𝑆∈ℬ𝜆𝑆𝑔𝑆(x) =

∑𝑆∈ℬ𝜆∑𝑖∈𝑆𝑥𝑖

=∑𝑖∈𝑁

∑𝑆∈ℬ𝑆∋𝑖

𝜆𝑥𝑖

=∑𝑖∈𝑁𝑥𝑖

∑𝑆∈ℬ𝑆∋𝑖

𝜆

=∑𝑖∈𝑁𝑥𝑖

= 𝑤(𝑁)

Substituting this in (3.90) gives

𝑤(𝑁) ≥∑𝑆∈ℬ𝜆𝑆𝑤(𝑆)

The game is balanced.

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Chapter 4: Smooth Functions

4.1 Along the demand curve, price and quantity are related according to the equation

𝑝 = 10− 𝑥

This is called the inverse demand function. Total revenue 𝑅(𝑥) (price times quantity)is given by

𝑅(𝑥) = 𝑝𝑥

= (10− 𝑥)𝑥= 10𝑥− 𝑥2= 𝑓(𝑥)

𝑔(𝑥) can be rewritten as

𝑔(𝑥) = 21 + 4(𝑥− 3)

At 𝑥 = 3, the price is 7 but the marginal revenue of an additional unit is only 4. Thefunction 𝑔 decomposes (approximately) the total revenue into two components — therevenue from the sale of 3 units (21 = 3 × 7) plus the marginal revenue from the saleof additional units (4(𝑥− 3)).

4.2 If your answer is 5 per cent, obtained by subtracting the inflation rate from thegrowth rate of nominal GDP, you are implicitly using a linear approximation. To seethis, let

𝑝 = price level at the beginning of the year𝑞 = real GDP at the beginning of the year𝑑𝑝 = change in prices during year𝑑𝑞 = change in output during year

We are told that nominal GDP at the end of the year, (𝑝 + 𝑑𝑝)(𝑞 + 𝑑𝑞), equals 1.10times nominal GDP at the beginning of the year, 𝑝𝑞. That is

(𝑝+ 𝑑𝑝)(𝑞 + 𝑑𝑞) = 1.10𝑝𝑞 (4.42)

Furthermore, the price level at the end of the year, 𝑝+ 𝑑𝑝 equals 1.05 times the pricelevel of the start of year, 𝑝:

𝑝+ 𝑑𝑝 = 1.05𝑝

Substituting this in equation (4.38) yields

1.05𝑝(𝑞 + 𝑑𝑞) = 1.10𝑝𝑞

which can be solved to give

𝑑𝑞 = (1.10

1.05− 1)𝑞 = 0.0476

The growth rate of real GDP (𝑑𝑞/𝑞) is equal to 4.76 per cent.

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To show how the estimate of 5 per cent involves a linear approximation, we expand theexpression for real GDP at the end of the year.

(𝑝+ 𝑑𝑝)(𝑞 + 𝑑𝑞) = 𝑝𝑞 + 𝑝𝑑𝑞 + 𝑞𝑑𝑝+ 𝑑𝑝𝑑𝑞

Dividing by 𝑝𝑞

(𝑝+ 𝑑𝑝)(𝑞 + 𝑑𝑞)

𝑝𝑞= 1 +

𝑑𝑞

𝑞+𝑑𝑝

𝑝+𝑑𝑝𝑑𝑞

𝑝𝑞

The growth rate of nominal GDP is

(𝑝+ 𝑑𝑝)(𝑞 + 𝑑𝑞)− 𝑝𝑞𝑝𝑞

=(𝑝+ 𝑑𝑝)(𝑞 + 𝑑𝑞)

𝑝𝑞− 1

=𝑑𝑞

𝑞+𝑑𝑝

𝑑𝑝+𝑑𝑝𝑑𝑞

𝑝𝑞

= Growth rate of output

+ Inflation rate

+ Error term

For small changes, the error term 𝑑𝑝𝑑𝑞/𝑝𝑞 is insignificant, and we can approximate thegrowth rate of output according to the sum

Growth rate of nominal GDP = Growth rate of output + Inflation rate

This is a linear approximation since it approximates the function (𝑝 + 𝑑𝑝)(𝑞 + 𝑑𝑞) bythe linear function 𝑝𝑞 + 𝑝𝑑𝑞 + 𝑞𝑑𝑝. In effect, we are evaluating the change output atthe old prices, and the change in prices at the old output, and ignoring in interactionbetween changes in prices and changes in quantities. The use of linear approximationin growth rates is extremely common in practice.

4.3 From (4.2)

∥x∥ 𝜂(x) = 𝑓(x0 + x)− 𝑓(x0)− 𝑔(x)

and therefore

𝜂(x) =𝑓(x0 + x)− 𝑓(x0)− 𝑔(x)

∥x∥

𝜂(x)→ 0𝑌 as x→ 0𝑋can be expressed as

limx→0𝑋

𝜂(x) = 0𝑌

4.4 Suppose not. That is, there exist two linear maps such that

𝑓(x0 + x) = 𝑓(x0) + 𝑔1(x) + ∥x∥ 𝜂1(x)𝑓(x0 + x) = 𝑓(x0) + 𝑔2(x) + ∥x∥ 𝜂2(x)

with

limx→0𝜂𝑖(x) = 0, 𝑖 = 1, 2

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Subtracting we have

𝐿1(x)− 𝐿2(x) = ∥x∥ (𝜂1(x)− 𝜂2(x))

and

limx→0

𝑔1(x)− 𝑔2(x)∥x∥ = 0

Since 𝑔1 − 𝑔2 is linear, (4) implies that 𝑔1(x) = 𝑔2(x) for all x ∈ 𝑋 .

To see this, we proceed by contradiction. Again, suppose not. That is, suppose thereexists some x ∈ 𝑋 such that

𝑔1(x) ∕= 𝑔2(x)

For this x, let

𝜂 =𝑔1(x)− 𝑔2(x)

∥x∥By linearity,

𝑔1(𝑡x)− 𝑔2(𝑡x)∥𝑡x∥ = 𝜂 for every ∀𝑡 > 0

and therefore

lim𝑡→0𝑔1(𝑡x)− 𝑔2(𝑡x)

∥𝑡x∥ = 𝜂 ∕= 0

which contradicts (4). Therefore 𝑔1(x) = 𝑔2(x) for all x ∈ 𝑋 .

4.5 If 𝑓 : 𝑋 → 𝑌 is differentiable at x0, then

𝑓(x0 + x) = 𝑓(x0) + 𝑔(x) + 𝜂(x) ∥x∥

where 𝜂(x) → 0𝑌 as x → 0𝑋 . Since 𝑔 is a continuous linear function, 𝑔(x) → 0𝑌 asx→ 0𝑋 . Therefore

limx→0𝑓(x0 + x) = lim

x→0𝑓(x0) + lim

x→0𝑔(x) + lim

x→0𝜂(x) ∥x∥

= 𝑓(x0)

𝑓 is continuous.

4.6

4.7

4.8 The approximation error at the point (2, 16) is

𝑓(2, 16) =8.0000𝑔(2, 16) =11.3333

Absolute error =-3.3333Percentage error =-41.6667

Relative error =-4.1667

By contrast, ℎ(2, 16) = 8 = 𝑓(2, 16). Table 4.1 shows that ℎ gives a good approximationto 𝑓 in the neighborhood of (2, 16).

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Table 4.1: Approximating the Cobb-Douglas function at (2, 16)Approximation Error

x x0 + x 𝑓(x0 + x) ℎ(x0 + x) Percentage Relative

At their intersection:(0.0, 0.0) (2.0, 16.0) 8.0000 8.0000 0.0000 NIL

Around the unit circle:(1.0, 0.0) (3.0, 16.0) 9.1577 9.3333 -1.9177 -0.1756(0.7, 0.7) (2.7, 16.7) 9.1083 9.1785 -0.7712 -0.0702(0.0, 1.0) (2.0, 17.0) 8.3300 8.3333 -0.0406 -0.0034

(-0.7, 0.7) (1.3, 16.7) 7.1196 7.2929 -2.4342 -0.1733(-1.0, 0.0) (1.0, 16.0) 6.3496 6.6667 -4.9934 -0.3171

(-0.7, -0.7) (1.3, 15.3) 6.7119 6.8215 -1.6323 -0.1096(0.0, -1.0) (2.0, 15.0) 7.6631 7.6667 -0.0466 -0.0036(0.7, -0.7) (2.7, 15.3) 8.5867 8.7071 -1.4018 -0.1204

Around a smaller circle:(0.10, 0.00) (2.1, 16.0) 8.1312 8.1333 -0.0266 -0.0216(0.07, 0.07) (2.1, 16.1) 8.1170 8.1179 -0.0103 -0.0083(0.00, 0.10) (2.0, 16.1) 8.0333 8.0333 -0.0004 -0.0003

(-0.07, 0.07) (1.9, 16.1) 7.9279 7.9293 -0.0181 -0.0143(-0.10, 0.00) (1.9, 16.0) 7.8644 7.8667 -0.0291 -0.0229

(-0.07, -0.07) (1.9, 15.9) 7.8813 7.8821 -0.0110 -0.0087(0.00, -0.10) (2.0, 15.9) 7.9666 7.9667 -0.0004 -0.0003(0.07, -0.07) (2.1, 15.9) 8.0693 8.0707 -0.0171 -0.0138

Parallel to the x1 axis:(-2.0, 0.0) (0.0, 16.0) 0.0000 5.3333 NIL -2.6667(-1.0, 0.0) (1.0, 16.0) 6.3496 6.6667 -4.9934 -0.3171(-0.5, 0.0) (1.5, 16.0) 7.2685 7.3333 -0.8922 -0.1297(-0.1, 0.0) (1.9, 16.0) 7.8644 7.8667 -0.0291 -0.0229(0.0, 0.0) (2.0, 16.0) 8.0000 8.0000 0.0000 NIL(0.1, 0.0) (2.1, 16.0) 8.1312 8.1333 -0.0266 -0.0216(0.5, 0.0) (2.5, 16.0) 8.6177 8.6667 -0.5678 -0.0979(1.0, 0.0) (3.0, 16.0) 9.1577 9.3333 -1.9177 -0.1756(2.0, 0.0) (4.0, 16.0) 10.0794 10.6667 -5.8267 -0.2936(4.0, 0.0) (6.0, 16.0) 11.5380 13.3333 -15.5602 -0.4488

Parallel to the x2 axis:(0.0, -4.0) (2.0, 12.0) 6.6039 6.6667 -0.9511 -0.0157(0.0, -2.0) (2.0, 14.0) 7.3186 7.3333 -0.2012 -0.0074(0.0, -1.0) (2.0, 15.0) 7.6631 7.6667 -0.0466 -0.0036(0.0, -0.5) (2.0, 15.5) 7.8325 7.8333 -0.0112 -0.0018(0.0, -0.1) (2.0, 15.9) 7.9666 7.9667 -0.0004 -0.0003(0.0, 0.0) (2.0, 16.0) 8.0000 8.0000 0.0000 NIL(0.0, 0.1) (2.0, 16.1) 8.0333 8.0333 -0.0004 -0.0003(0.0, 0.5) (2.0, 16.5) 8.1658 8.1667 -0.0105 -0.0017(0.0, 1.0) (2.0, 17.0) 8.3300 8.3333 -0.0406 -0.0034(0.0, 2.0) (2.0, 18.0) 8.6535 8.6667 -0.1522 -0.0066(0.0, 4.0) (2.0, 20.0) 9.2832 9.3333 -0.5403 -0.0125

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4.9 To show that 𝑟 is nonlinear, consider

𝑟((1, 2, 3, 4, 5) + (66, 55, 75, 81, 63)) = 𝑟(67, 57, 78, 85, 68)

= (85, 78, 68, 67, 58)

∕= (5, 4, 3, 2, 1) + (81, 75, 67, 63, 55)

To show that 𝑟 is differentiable, consider a particular point, say (66, 55, 75, 81, 63).Consider the permutation 𝑔 : ℜ𝑛 → ℜ𝑛 defined by

𝑔(𝑥1, 𝑥2, . . . , 𝑥5) = (𝑥4, 𝑥3, 𝑥1, 𝑥5, 𝑥2)

𝑔 is linear and

𝑔(66, 55, 75, 81, 63) = (81, 75, 67, 63, 55) = 𝑟(66, 55, 75, 81, 63)

Furthermore, 𝑔(x) = 𝑟(x) for all x close to (66, 55, 75, 81, 63). Hence, 𝑔(x) approxi-mates 𝑟(x) in a neighborhood of (66, 55, 75, 81, 63) and so 𝑟 is differentiable at (66, 55, 75, 81, 63).The choice of (66, 55, 75, 81, 63) was arbitrary, and the argument applies at every x suchthat x𝑖 ∕= x𝑗 .In summary, each application of 𝑟 involves a permutation, although the particularpermutation depends upon the argument, x. However, for any given x0 with x0𝑖 ∕=x0𝑗 , the same permutation applies to all x in the neighborhood of x0, so that the

permutation (which is a linear function) is the derivative of 𝑟 at x0.

4.10 Using (4.3), we have for any x

lim𝑡x→0

𝑓(x0 + 𝑡x)− 𝑓(x0)−𝐷𝑓 [x0](𝑡x)

∥𝑡x∥ = 0

or

lim𝑡→0𝑓(x0 + 𝑡x)− 𝑓(x0)− 𝑡𝐷𝑓 [x0](x)

𝑡 ∥x∥ = 0

For ∥x∥ = 1, this implies

lim𝑡→0𝑓(x0 + 𝑡x)− 𝑓(x0)

𝑡=𝑡𝐷𝑓 [x0](x)

𝑡

that is

��x𝑓 [x0] = lim𝑡→0𝑓(x0 + 𝑡x)− 𝑓(x0)

𝑡= 𝐷𝑓 [x0](x)

4.11 By direct calculation

𝐷𝑥𝑖𝑓 [x0] = lim𝑡→0ℎ(x0𝑖 + 𝑡)− ℎ(x0𝑖 )

𝑡

= lim𝑡→0𝑓(x01,x

02, . . . ,x

0𝑖 + 𝑡, . . . ,x0𝑛)− 𝑓(x01,x

02, . . . ,x

0𝑖 , . . . ,x

0𝑛)

𝑡

= lim𝑡→0𝑓(x0 + 𝑡e𝑖)− 𝑓(x0)

𝑡

= ��e𝑖𝑓 [x0]

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4.12 Define the function

ℎ(𝑡) = 𝑓((8, 8) + 𝑡(1, 1)

)= (8 + 𝑡)1/3(8 + 𝑡)2/3

= 8 + 𝑡

The directional derivative of 𝑓 in the direction (1, 1) is

��(1,1)𝑓(8, 8) = lim𝑡→0ℎ(𝑡)− ℎ(0)

𝑡

= 1

Generalization of this example reveals that the directional derivative of 𝑓 along anyray through the origin equals 1, that is ��x0𝑓 [x0] = 1 for every x0. Economically,this means that increasing inputs in the same proportions leads to a proportionateincrease in output, which is the property of constant returns to scale. We will studythis property of homogeneity is some depth in Section 4.6.

4.13 Let p = ∇𝑓(x0). Each component of p represents the action of the derivative onan element of the standard basis {e1, e2, . . . , e𝑛}(see proof of Theorem 3.4)

𝑝𝑖 = 𝐷𝑓 [x0](e𝑖)

Since ∥e𝑖∥ = 1, 𝐷𝑓 [x0](e𝑖) is the directional derivative at x0 in the direction e𝑖 (Exer-cise 4.10)

𝑝𝑖 = 𝐷𝑓 [x0](e𝑖) = ��e𝑖(x0)

But this is simply the 𝑖 partial derivative of 𝑓 (Exercise 4.11)

𝑝𝑖 = 𝐷𝑓 [x0](e𝑖) = ��e𝑖(x0) = 𝐷𝑥𝑖𝑓(x0)

4.14 Using the standard inner product on ℜ𝑛 (Example 3.26) and Exercise 4.13

< ∇𝑓(x0),x >=𝑛∑𝑖=1

𝐷𝑥𝑖𝑓 [x0]x𝑖 = 𝐷𝑓 [x0](x)

4.15 Since 𝑓 is differentiable

𝑓(x1 + 𝑡x) = 𝑓(x1) +∇𝑓(x0)𝑇 𝑡x+ 𝜂(𝑡x) ∥𝑡x∥

with 𝜂(𝑡x) → 0 as 𝑡x→ 0. If 𝑓 is increasing, 𝑓(x1 + 𝑡x) ≥ 𝑓(x1) for every x ≥ 0 and𝑡 > 0. Therefore

∇𝑓(x0)𝑇 𝑡x+ 𝜂(𝑡x) ∥𝑡x∥ = 𝑡∇𝑓(x0)𝑇x+ 𝑡𝜂(𝑡x) ∥x∥ ≥ 0

Dividing by 𝑡 and letting 𝑡→ 0

∇𝑓(x0)𝑇x ≥ 0 for every x ≥ 0

In particular, this applies for unit vectors e𝑖. Therefore

𝐷𝑥𝑖𝑓(x1) ≥ 0, 𝑖 = 1, 2, . . . , 𝑛

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4.16 The directional derivative ��x𝑓(x0) measures the rate of increase of 𝑓 in the di-rection x. Using Exercises 4.10, 4.14 and 3.61, assuming x has unit norm,

��x𝑓(x0) = 𝐷𝑓 [x0](x) =< ∇𝑓(x0),x >≤ ∥∥∇𝑓(x0)∥∥

This bound is attained when x = ∇𝑓(x0)/∥∥∇𝑓(x0)

∥∥ since

��x𝑓(x0) =< ∇𝑓(x0),∇𝑓(x0)

∥∇𝑓(x0)∥ >=

∥∥∇𝑓(x0)∥∥2

∥∇𝑓(x0)∥ =∥∥∇𝑓(x0)

∥∥The directional derivative is maximized when ∇𝑓(x0) and x are aligned.

4.17 Using Exercise 4.14

𝐻 = { x ∈ 𝑋 :< ∇𝑓 [x0],x >= 0 }4.18 Assume each 𝑓𝑗 is differentiable at x0 and let

𝐷𝑓 [x0] = (𝐷𝑓1[x0], 𝐷𝑓2[x0], . . . , 𝐷𝑓𝑚[x0])

Then

f(x0 + x)− f [x0]−𝐷f [x0]x =

⎛⎜⎜⎜⎝

𝑓1(x0 + x)− 𝑓1[x0]−𝐷𝑓1[x0]x𝑓2(x0 + x)− 𝑓2[x0]−𝐷𝑓2[x0]x

...𝑓𝑚(x0 + x)− 𝑓𝑚(x0)−𝐷𝑓𝑚[x0]x

⎞⎟⎟⎟⎠

and

𝑓𝑗(x0 + x)− 𝑓𝑗(x0)−𝐷𝑓𝑗[x0]x∥x∥ → 0 as ∥x∥ → 0

for every 𝑗 implies

f(x0 + x)− f(x0)−𝐷f [x0](x)∥x∥ → 0 as ∥x∥ → 0 (4.43)

Therefore f is differentiable with derivative

𝐷f [x0] = 𝐿 = (𝐷𝑓1(x0), 𝐷𝑓2[x0], . . . , 𝐷𝑓𝑚[x0])

Each 𝐷𝑓𝑗 [x0] is represented by the gradient ∇𝑓𝑗 [x0] (Exercise 4.13) and therefore𝐷𝑓 [x0] is represented by the matrix

𝐽 =

⎛⎜⎜⎜⎝∇𝑓1[x0]∇𝑓2[x0]

...∇𝑓𝑚[x0]

⎞⎟⎟⎟⎠ =

⎛⎜⎜⎜⎝𝐷𝑥1𝑓1[x0] 𝐷𝑥2𝑓1[x0] . . . 𝐷𝑥𝑛𝑓1[x0]𝐷𝑥1𝑓2[x0] 𝐷𝑥2𝑓2[x0] . . . 𝐷𝑥𝑛𝑓2[x0]

......

. . ....

𝐷𝑥1𝑓𝑚[x0] 𝐷𝑥2𝑓𝑚[x0] . . . 𝐷𝑥𝑛𝑓𝑚[x0]

⎞⎟⎟⎟⎠

Conversely, if f is differentiable, its derivative 𝐷f [x0] : ℜ𝑛 → ℜ𝑚 be decomposed into𝑚 component 𝐷𝑓1[x0], 𝐷𝑓2[x0], . . . , 𝐷𝑓𝑚[x0] functionals such that

f(x0 + x)− f(x0)−𝐷f [x0]x =

⎛⎜⎜⎜⎝𝑓1(x0 + x)− 𝑓1(x0)−𝐷𝑓1[x0]x𝑓2(x0 + x)− 𝑓2(x0)−𝐷𝑓2[x0]x

...𝑓𝑚(x0 + x)− 𝑓𝑚(x0)−𝐷𝑓𝑚[x0]x

⎞⎟⎟⎟⎠

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(4.43) implies that

𝑓𝑗(x0 + x)− 𝑓𝑗(x0)−𝐷𝑓𝑗[x0]x∥x∥ → 0 as ∥x∥ → 0

for every 𝑗.

4.19 If 𝐷𝑓 [x0] has full rank, then it is one-to-one (Exercise 3.25) and onto (Exercise3.16). Therefore 𝐷𝑓 [x0] is nonsingular. The Jacobian 𝐽𝑓 (x0) represents 𝐷𝑓 [x0], whichis therefore nonsingular if and only if det 𝐽𝑓 (x0) ∕= 0.

4.20 When 𝑓 is a functional, rank 𝑋 ≥ 𝑟𝑎𝑛𝑘𝑌 = 1. If 𝐷𝑓 [x0] has full rank (1), then𝐷𝑓 [x0] maps 𝑋 onto ℜ (Exercise 3.16), which requires that ∇𝑓(x0) ∕= 0.4.21

4.23 If 𝑓 : 𝑋 × 𝑌 → 𝑍 is bilinear

𝑓(x0 + x,y0 + y) = 𝑓(x0,y0) + 𝑓(x0,y) + 𝑓(x,y0) + 𝑓(x,y)

Defining

𝐷𝑓 [x0,y0](x,y) = 𝑓(x0,y) + 𝑓(x,y0)

𝑓(x0 + x,y0 + y) = 𝑓(x0,y0) +𝐷𝑓 [x0,y0](x,y) + 𝑓(x,y)

Since 𝑓 is continuous, there exists 𝑀 such that

𝑓(x,y) ≤𝑀 ∥x∥ ∥y∥ for every x ∈ 𝑋 and y ∈ 𝑌

and therefore

NOTE This is not quite right. See Spivak p. 23. Avez (Tilburg) has

∥𝑓(x,y)∥ ≤𝑀 ∥x∥ ∥y∥ ≤𝑀( ∥x∥ + ∥y∥ )2 ≤𝑀 ∥(x,y)∥2which implies that

∥𝑓(x,y)∥∥(x,y)∥ → 0 as (x,y)→ 0

limx1,x2→0

𝑓(x1,x2)

∥x1∥ ∥x2∥ = 0

Therefore 𝑓 is differentiable with derivative

𝐷𝑓 [x0,y0] = 𝑓(x0,y) + 𝑓(x,y0)

4.24 Define 𝑚 : ℜ2 → ℜ by

𝑚(𝑧1, 𝑧2) = 𝑧1𝑧2

Then 𝑚 is bilinear (Example 3.23) and continuous (Exercise 2.79) and therefore differ-entiable (Exercise 4.23) with derivative

𝐷𝑚[𝑧1, z2] = 𝑚(z1, ⋅) +𝑚(⋅, z2)

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The function 𝑓𝑔 is the composition of 𝑚 with 𝑓 and 𝑔,

𝑓𝑔(x,y) = 𝑚(𝑓(x), 𝑔(y))

By the chain rule, the derivative of 𝑓𝑔 is

𝐷𝑓𝑔[x,y] = 𝐷𝑚[𝑧1, z2](𝐷𝑓 [x], 𝐷𝑔[x]

)= 𝑚(z1, 𝐷𝑔[y]) +𝑚(𝐷𝑓 [x], z2)

= 𝑓 [x]𝐷𝑔[y]) + 𝑔(y)𝐷𝑓 [x]

where z1 = 𝑓(x) and z2 = 𝑔(y).

4.25 For 𝑛 = 1, 𝑓(𝑥) = 𝑥 is linear and therefore (Exercise 4.6) 𝐷𝑓 [𝑥] = 1 (𝐷𝑓 [𝑥](𝑥) =𝑥). For 𝑛 = 2, let 𝑔(𝑥) = 𝑥 so that 𝑓(𝑥) = 𝑥2 = 𝑔(𝑥)𝑔(𝑥). Using the product rule

𝐷𝑓 [x] = 𝑔(𝑥)𝐷𝑔(𝑥) + 𝑔(𝑥)𝐷𝑔(𝑥) = 2𝑥

Now assume it is true for 𝑛 − 1 and let 𝑔(𝑥) = 𝑥𝑛−1, so that 𝑓(x) = 𝑥𝑔(𝑥). By theproduct rule

𝐷𝑓 [x] = 𝑥𝐷𝑔[𝑥] + 𝑔(𝑥)1

By assumption 𝐷𝑔[𝑥] = (𝑛− 1)𝑥𝑛−2 and therefore

𝐷𝑓 [x] = 𝑥𝐷𝑔[𝑥] + 𝑔(𝑥)1 = 𝑥(𝑛− 1)𝑥𝑛−2 + 𝑥𝑛−1 = 𝑛𝑥𝑛−1

4.26 Using the product rule (Exercise 4.24)

𝐷𝑥𝑅(𝑥0) = 𝑓(𝑥0)𝐷𝑥𝑥+ 𝑥0𝐷𝑥𝑓(𝑥0)

= 𝑝0 + 𝑥0𝐷𝑥𝑓(𝑥0)

where 𝑝0 = 𝑓(𝑥0). Marginal revenue equals one unit at the current price minus thereduction in revenue caused by reducing the price on existing sales.

4.27 Fix some x0 and let 𝑔 =(𝐷𝑓 [x0]

)−1. Let y0 = 𝑓(x0). For any y, let x =

𝑓−1(y0 + y) − 𝑓−1(y0) so that 𝑔(y) = 𝑓(x0 + x)− 𝑓(x) and∥∥𝑓−1(y0 + y) − 𝑓−1(y0)− 𝑔(y)∥∥ =∥∥(x− 𝑔(𝑓(x0 + x)− 𝑓(x0))

)∥∥Since 𝑓 is differentiable at x0 with 𝐷𝑓 [x0] = 𝑔−1

𝑓(x0 + x)− 𝑓(x0) = 𝑔−1(x) + 𝜂(x) ∥x∥

Substituting

∥∥𝑓−1(y0 + y) − 𝑓−1(y0)− 𝑔(y)∥∥ =∥∥∥x− 𝑔(𝑔−1(x) + 𝜂(x) ∥x∥

)∥∥∥=

∥∥∥𝑔(𝜂(x) ∥x∥)∥∥∥= ∥x∥

∥∥∥𝑔(𝜂(x))∥∥∥with 𝜂(x)→ 0𝑌 as x→ 0𝑋 . Since 𝑓−1 and 𝑔 are continuous, 𝑔

(𝜂(x)

)→ 0𝑋 as y→ 0.

We conclude that 𝑓−1 is differentiable with derivative 𝑔 =(𝐷𝑓 [x0]

)−1.

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4.28

log 𝑓(𝑥) = 𝑥 log 𝑎

and therefore

𝑓(𝑥) = exp(

log 𝑓(𝑥))

= 𝑒𝑥 log 𝑎

By the Chain Rule, 𝑓 is differentiable with derivative

𝐷𝑥𝑓(𝑥) = 𝑒𝑥 log 𝑎 log 𝑎 = 𝑎𝑥 log 𝑎

4.29 By Exercise 4.15, the function 𝑔 : ℜ → ℜ defined by 𝑔(𝑦) = 1𝑦 = 𝑦−1 is differen-

tiable with derivative

𝐷𝑦𝑔[𝑦] = −𝑦−2 = − 1

𝑦2

Applying the Chain Rule, 1/𝑓 = 𝑔 ∘ 𝑓 is differentiable with derivative

𝐷1

𝑓[x] = 𝐷𝑔[𝑓(x)]𝐷𝑓 [x] = − 𝐷𝑓 [x](

𝑓(x))2

4.30 Applying the Product Rule to 𝑓 × (1/𝑔)

𝐷𝑓

𝑔[x,y] = 𝑓(x)𝐷

1

𝑔[y] +

1

𝑔(y)𝐷𝑓 [x]

= −𝑓(x)𝐷𝑔[y](𝑔(y)

)2 +1

𝑔(y)𝐷𝑓 [x]

=𝑔(y)𝐷𝑓 [x]− 𝑓(x)𝐷𝑔[y](

𝑔(y))2

4.31 In the particular case where

𝑓(x1,x2) = x1/31 x

2/32

the partial derivatives at the point (8, 8) are

𝐷𝑥1𝑓 [(8, 8)] =2

3and 𝐷𝑥2𝐹 [(8, 8)] =

1

3

4.32 The partial derivatives of 𝑓(x) are from Table 4.4

𝐷𝑥𝑖𝑓 [x] = 𝑥𝑎11 𝑥

𝑎22 . . . 𝑎𝑖𝑥

𝑎𝑖−1𝑖 . . . 𝑥𝑎𝑛

𝑛

= 𝑎𝑖𝑓(x)

𝑥𝑖

so that the gradient is

∇𝑓(x) =

(𝑎1𝑥1,𝑎2𝑥2, . . . ,

𝑎𝑛𝑥𝑛

)𝑓(x)

4.33 Applying the chain rule (Exercise 4.22) to general power function (Example 4.15),the partial derivatives of the CES function are

𝐷𝑥𝑖𝑓 [x] =1

𝜌(𝑎1𝑥

𝜌1 + 𝑎2𝑥

𝜌2 + ⋅ ⋅ ⋅+ 𝑎𝑛𝑥𝜌𝑛)

1𝜌−1 𝑎𝑖𝜌𝑥

𝜌−1𝑖

= 𝑎𝑖𝑥𝜌−1𝑖 (𝑎1𝑥

𝜌1 + 𝑎2𝑥

𝜌2 + ⋅ ⋅ ⋅+ 𝑎𝑛𝑥𝜌𝑛)

1−𝜌𝜌

= 𝑎𝑖

(𝑓(x)

𝑥𝑖

)1−𝜌

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4.34 Define

ℎ(𝑥) = 𝑓(𝑥)− 𝑓(𝑏)− 𝑓(𝑎)

𝑏− 𝑎 (𝑥− 𝑎)

Then ℎ is continuous on [𝑎, 𝑏] and differentiable on (𝑎, 𝑏) with

ℎ(𝑏) = 𝑓(𝑏)− 𝑓(𝑏)− 𝑓(𝑎)

𝑏− 𝑎 (𝑏 − 𝑎)𝑓(𝑎) = ℎ(𝑎)

By Rolle’s theorem (Exercise 5.8), there exists 𝑥 ∈ (𝑎, 𝑏) such that

ℎ′(𝑥) = 𝑓 ′(𝑥)− 𝑓(𝑏)− 𝑓(𝑎)

𝑏− 𝑎 = 0

4.35 Assume ∇𝑓(x) ≥ 0 for every x ∈ 𝑋 . By the mean value theorem, for any x2 ≥ x1in 𝑋 , there exists x ∈ (x1,x2) such that

𝑓(x2) = 𝑓(x1) +𝐷𝑓 [x](x2 − x1)

Using (4.6)

𝑓(x2) = 𝑓(x1) +

𝑛∑𝑖=1

𝐷𝑥𝑖𝑓(x)(𝑥2𝑖 − 𝑥1𝑖 ) (4.44)

∇𝑓(x) ≥ 0 and x2 ≥ x1 implies that

𝑛∑𝑖=1

𝐷𝑥𝑖𝑓(x)(𝑥2𝑖 − 𝑥1𝑖 ) ≥ 0

and therefore 𝑓(x2) ≥ 𝑓(x1). 𝑓 is increasing. The converse was established in Exercise4.15

4.36 ∇𝑓(x) > 0 and x2 ≥ x1 implies that

𝑛∑𝑖=1

𝐷𝑥𝑖𝑓(x)(𝑥2𝑖 − 𝑥1𝑖 ) > 0


𝑓(x2) = 𝑓(x1) +𝑛∑𝑖=1

𝐷𝑥𝑖𝑓(x)(𝑥2𝑖 − 𝑥1𝑖 ) > 𝑓(x1)

𝑓 is strictly increasing.

4.37 Differentiability implies the existence of the gradient and hence the partial deriv-atives of 𝑓 (Exercise 4.13). Continuity of 𝐷𝑓 [x] implies the continuity of the partialderivatives.

To prove the converse, choose some x0 ∈ 𝑆 and define for the partial functions

ℎ𝑖(𝑡) = 𝑓(𝑥01, 𝑥02, . . . , 𝑥

0𝑖−1, 𝑡, 𝑥

0𝑖+1 + 𝑥𝑖+1, . . . , 𝑥

0𝑛 + 𝑥𝑛) 𝑖 = 1, 2, . . . , 𝑛

so that ℎ′𝑖(𝑡) = 𝐷𝑥𝑖𝑓(x𝑖) where x𝑖 = (𝑥01, 𝑥02, . . . , 𝑥

0𝑖 , 𝑡, 𝑥

0𝑖+1 + 𝑥𝑖+1, . . . , 𝑥

0𝑛 + 𝑥𝑛). Fur-

ther, ℎ1(𝑥01 + 𝑥1) = 𝑓(x0 + x), ℎ𝑛(𝑥0𝑛) = 𝑓(x0), and ℎ𝑖(𝑥

0𝑖 + 𝑥𝑖) = ℎ𝑖−1(𝑥0𝑖 ) so that

𝑓(x0 + x) − 𝑓(x0) =

𝑛∑𝑖=1

(ℎ𝑖(𝑥

0𝑖 + 𝑥𝑖)− ℎ𝑖(𝑥0𝑖 )

)

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By the mean value theorem, there exists, for each 𝑖, 𝑡𝑖 between 𝑥0𝑖 + 𝑥𝑖 and 𝑥𝑖 suchthat

ℎ𝑖(𝑥0𝑖 + 𝑥𝑖)− ℎ𝑖(𝑥𝑖) = 𝐷𝑥𝑖𝑓(x𝑖)𝑥𝑖

where x𝑖 = (𝑥01, 𝑥02, . . . , 𝑥

0𝑖 , 𝑡, 𝑥

0𝑖+1 + 𝑥𝑖+1, . . . , 𝑥

0𝑛 + 𝑥𝑛). Therefore

𝑓(x0 + x)− 𝑓(x0) =

𝑛∑𝑖=1

𝐷𝑥𝑖𝑓(x𝑖)𝑥𝑖

Define the linear functional

𝑔(x) =𝑛∑𝑖=1

𝐷𝑥𝑖𝑓(x0)𝑥𝑖

Then

𝑓(x0 + x)− 𝑓(x0)− 𝑔(x) =

𝑛∑𝑖=1

(𝐷𝑥𝑖𝑓(x𝑖)−𝐷𝑥𝑖𝑓(x0)

)𝑥𝑖

and

∥∥𝑓(x0 + x) − 𝑓(x0)− 𝑔(x)∥∥ ≤ 𝑛∑𝑖=1

∥∥(𝐷𝑥𝑖𝑓(x𝑖)−𝐷𝑥𝑖𝑓(x0)∥∥ ∣𝑥𝑖∣

so that

limx→0

∥∥𝑓(x0 + x)− 𝑓(x0)− 𝑔(x)∥∥∥x∥ ≤

𝑛∑𝑖=1

∥∥(𝐷𝑥𝑖𝑓(x𝑖)−𝐷𝑥𝑖𝑓(x0)∥∥ ∣𝑥𝑖∣∥x∥

≤𝑛∑𝑖=1

∥∥(𝐷𝑥𝑖𝑓(x𝑖)−𝐷𝑥𝑖𝑓(x0)∥∥

= 0

since the partial derivatives 𝐷𝑥𝑖𝑓(x) are continuous. Therefore 𝑓 is differentiable withderivative

𝑔(x) =

𝑛∑𝑖=1

𝐷𝑥𝑖𝑓 [x0]𝑥𝑖

4.38 For every x1,x2 ∈ 𝑆∥𝑓(x1)− 𝑓(x2)∥ ≤ sup

x∈[x1,x2]

∥𝐷𝑓(x)∥ ∥x1 − x2∥

by Corollary 4.1.1. If 𝐷𝑓 [x] = 0 for every x ∈ 𝑋 , then

∥𝑓(x1)− 𝑓(x2)∥ = 0

which implies that 𝑓(x1) = 𝑓(x2). We conclude that 𝑓 is constant on 𝑆. The conversewas established in Exercise 4.7.

4.39 For any x0 ∈ 𝑆, let 𝐵 ⊆ 𝑆 be an open ball of radius of radius 𝑟 centered on x0.Applying the mean value inequality (Corollary 4.1.1) to 𝑓𝑛 − 𝑓𝑚 we have∥∥𝑓𝑛(x) − 𝑓𝑚(x) − (

𝑓𝑛(x0)− 𝑓𝑚(x0))∥∥ ≤ sup

x∈𝐵∥𝐷𝑓𝑛[x]−𝐷𝑓𝑚[x]∥ ∥x− x0∥

≤ 𝑟 supx∈𝐵∥𝐷𝑓𝑛[x]−𝐷𝑓𝑚[x]∥

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for every x ∈ 𝐵. Given 𝜖 > 0, there exists 𝑁 such that for every 𝑚,𝑛 > 𝑁

∥𝐷𝑓𝑛 −𝐷𝑓𝑚∥ < 𝜖/𝑟 and ∥𝐷𝑓𝑛 − 𝑔∥ < 𝜖Letting 𝑚→∞ ∥∥𝑓𝑛(x)− 𝑓(x)− (

𝑓𝑛(x0)− 𝑓(x0))∥∥ ≤ 𝜖 ∥x− x0∥ (4.45)

for 𝑛 ≥ 𝑁 and x ∈ 𝐵. Applying the mean value inequality to 𝑓𝑛, there exists 𝛿 suchthat

∥𝑓𝑛(x) − 𝑓𝑛(x0)∥ ≤ 𝜖 ∥x− x0∥ (4.46)

Using (4.45) and (4.46) and the fact that ∥𝐷𝑓𝑛 − 𝑔∥ < 𝜖 we deduce that

∥𝑓(x)− 𝑓(x0)− 𝑔(x0)∥ ≤ 3𝜖 ∥x− x0∥𝑓 is differentiable with derivative 𝑔.

4.40 Define

𝑓(𝑥) =𝑒𝑥+𝑦

𝑒𝑦

By the chain rule (Exercise 4.22)

𝑓 ′(𝑥) =𝑒𝑥+𝑦

𝑒𝑦= 𝑓(𝑥)

which implies (Example 4.21) that


𝑒𝑦= 𝐴𝑒𝑥 for some 𝐴 ∈ ℜ

Evaluating at 𝑥 = 0 using 𝑒0 = 1 gives

𝑓(0) =𝑒𝑦

𝑒𝑦= 𝐴 for some 𝐴 ∈ ℜ

so that


𝑒𝑦=𝑒𝑦

𝑒𝑦𝑒𝑥

which implies that

𝑒𝑥+𝑦 = 𝑒𝑥𝑒𝑦

4.41 If 𝑓 = 𝐴𝑥𝑎, 𝑓 ′(𝑥) = 𝑎𝐴𝑥𝑎−1 and

𝐸(𝑥) = 𝑥𝑎𝐴𝑥𝑎−1

𝐴𝑥𝑎= 𝑎

To show that this is the only function with constant elasticity, define

𝑔(𝑥) =𝑓(𝑥)

𝑥𝑎

𝑔 is differentiable (Exercise 4.30) with derivative

𝑔′(𝑥) =𝑥𝑎𝑓 ′(𝑥)− 𝑓(𝑥)𝑎𝑥𝑎−1

𝑥2𝑎=𝑥𝑓 ′(𝑥) − 𝑎𝑓(𝑥)

𝑥𝑎+1(4.47)

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If

𝐸(𝑥) = 𝑥𝑓 ′(𝑥)𝑓(𝑥)

= 𝑎

then

𝑥𝑓 ′(𝑥) = 𝑎𝑓(𝑥)


𝑔′(𝑥) =𝑥𝑓 ′(𝑥)− 𝑎𝑓(𝑥)

𝑥𝑎+1= 0 for every 𝑥 ∈ ℜ

Therefore, 𝑔 is a constant function (Exercise 4.38). That is, there exists 𝐴 ∈ ℜ suchthat

𝑔(𝑥) =𝑓(𝑥)

𝑥𝑎= 𝐴 or 𝑓(𝑥) = 𝐴𝑥𝑎

4.42 Define 𝑔 : 𝑆 → 𝑌 by

𝑔(x) = 𝑓(x)−𝐷𝑓 [x0](x)

𝑔 is differentiable with

𝐷𝑔[x] = 𝐷𝑓 [x]−𝐷𝑓 [x0]

Applying Corollary 4.1.1 to 𝑔,

∥𝑔(x1)− 𝑔(x2)∥ ≤ supx∈[x1,x2]

∥𝐷𝑔[x]∥ ∥x1 − x2∥

for every x1,x2 ∈ 𝑆. Substituting for 𝑔 and 𝐷𝑔

∥𝑓(x1)−𝐷𝑓 [x0](x1)− 𝑓(x2) +𝐷𝑓 [x0](x2)∥ = ∥𝑓(x1)− 𝑓(x2)−𝐷𝑓 [x0](x1 − x2)∥≤ sup

x∈[x1,x2]

∥𝐷𝑓 [x]−𝐷𝑓 [x0]∥ ∥x1 − x2∥

4.43 Since 𝐷𝑓 is continuous, there exists a neighborhood 𝑆 of x0 such that

∥𝐷𝑓 [x]−𝐷𝑓 [x0]∥ < 𝜖 for every x ∈ 𝑆and therefore for every x1,x2 ∈ 𝑆

supx∈[x1,x2]

∥𝐷𝑓 [x]−𝐷𝑓 [x0]∥ < 𝜖

By the previous exercise (Exercise 4.42)

∥𝑓(x1)− 𝑓(x2)−𝐷𝑓 [x0](x1 − x2)∥ ≤ 𝜖 ∥x1 − x2∥4.44 By the previous exercise (Exercise 4.43), there exists a neighborhood such that

∥𝑓(x1)− 𝑓(x2)−𝐷𝑓 [x0](x1 − x2)∥ ≤ 𝜖 ∥x1 − x2∥The Triangle Inequality (Exercise 1.200) implies

∥𝑓(x1)− 𝑓(x2)∥ − ∥𝐷𝑓 [x0](x1 − x2)∥ ≤ ∥𝑓(x1)− 𝑓(x2)−𝐷𝑓 [x0](x1 − x2)∥ ≤ 𝜖 ∥x1 − x2∥and therefore

∥𝑓(x1)− 𝑓(x2)∥ ≤ ∥𝐷𝑓 [x0](x1 − x2)∥+ 𝜖 ∥x1 − x2∥ ≤ ∥𝐷𝑓 [x0] + 𝜖∥ ∥x1 − x2∥

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4.45 Assume not. That is, assume that

y = 𝑓(x1)− 𝑓(x2) ∕∈ conv 𝐴

Then by the (strong) separating hyperplane theorem (Proposition 3.14) there exists alinear functional 𝜑 on 𝑌 such that

𝜑(y) > 𝜑(a) for every a ∈ 𝐴 (4.48)

where

𝜑(𝑦) = 𝜑(𝑓(x1)− 𝑓(x2)) = 𝜑(𝑓(x1))− 𝜑(𝑓(x2))

𝜑𝑓 is a functional on 𝑆. By the mean value theorem (Theorem 4.1), there exists somex ∈ [x1,x2] such that

𝜑 ∘ 𝑓(x1)− 𝜑 ∘ 𝑓(x2)) = 𝐷(𝜑 ∘ 𝑓)[x](x1 − x2) = 𝜑 ∘𝐷𝑓 [x](x− x2) = 𝜑(𝑎)

for some 𝑎 ∈ 𝐴 contradicting (4.44).

4.46 Define ℎ : [𝑎, 𝑏]→ ℜ by

ℎ(𝑥) =(𝑓(𝑏)− 𝑓(𝑎)

)𝑔(𝑥)− (

𝑔(𝑏)− 𝑔(𝑎))𝑓(𝑥)

ℎ ∈ 𝐶[𝑎, 𝑏] and is differentiable on 𝑎, 𝑏) with

ℎ(𝑎) =(𝑓(𝑏)− 𝑓(𝑎)

)𝑔(𝑎)− (

𝑔(𝑏)− 𝑔(𝑎))𝑓(𝑎) = 𝑓(𝑏)𝑔(𝑎)− 𝑓(𝑎)𝑔(𝑏) = ℎ(𝑏)

By Rolle’s theorem (Exercise 5.8), there exists 𝑥 ∈ (𝑎, 𝑏) such that

ℎ′(𝑥) =(𝑓(𝑏)− 𝑓(𝑎)

)𝑔′(𝑥)− (

𝑔(𝑏)− 𝑔(𝑎))𝑓 ′(𝑥) = 0

4.47 The hypothesis that lim𝑥→𝑎𝐷𝑓(𝑥)/𝐷𝑔(𝑥) exists contains two implicit assump-tions, namely

∙ 𝑓 and 𝑔 are differentiable on a neighborhood 𝑆 of 𝑎 (except perhaps at 𝑎)

∙ 𝑔′(𝑥) ∕= 0 in this neighborhood (except perhaps at 𝑎).

Applying the Cauchy mean value theorem, for every 𝑥 ∈ 𝑆, there exists some 𝑦𝑥 ∈ (𝑎, 𝑥)such that

𝑓 ′(𝑦𝑥)

𝑔′(𝑦𝑥)=𝑓(𝑥)− 𝑓(𝑎)

𝑔(𝑥)− 𝑔(𝑎) =𝑓(𝑥)

𝑔(𝑥)

and therefore

lim𝑥→𝑎

𝑓(𝑥)

𝑔(𝑥)= lim

𝑥→𝑎

𝑓 ′(𝑦𝑥)

𝑔′(𝑦𝑥)= lim

𝑥→𝑎

𝑓 ′(𝑥)𝑔′(𝑥)

4.48 Let 𝐴 = 𝑎1 + 𝑎2 + ⋅ ⋅ ⋅+ 𝑎𝑛 ∕= 1. Then from (4.12)

lim𝜌→0𝑔(𝜌) =

𝑎1 log 𝑥1 + 𝑎2 log 𝑥2 + . . . 𝑎𝑛 log 𝑥𝑛𝐴

=𝑎1𝐴

log 𝑥1 +𝑎2𝐴

log 𝑥2 + . . .𝑎𝑛𝐴

log 𝑥𝑛

and therefore

lim𝜌→0

log 𝑓(𝜌,x) =𝑎1𝐴

log 𝑥1 +𝑎2𝐴

log 𝑥2 + . . .𝑎𝑛𝐴

log 𝑥𝑛

so that

lim𝜌→0𝑓(𝜌,x) = 𝑥

𝑎1𝐴1 𝑥

𝑎1𝐴2 . . . 𝑥

𝑎1𝐴𝑛

which is homogeneous of degree one.

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4.49 Average cost is given by 𝑐(𝑦)/𝑦 which is undefined at 𝑦 = 0. We seek lim𝑦→0 𝑐(𝑦)/𝑦.By L’Hopital’s rule

lim𝑦→0

𝑐(𝑦)

𝑦= lim

𝑦→0𝑐′(𝑦)

1

= 𝑐′(0)

which is marginal cost at zero output.

4.50 1. Since lim𝑥→∞ 𝑓 ′(𝑥)/𝑔′(𝑥) = k, for every 𝜖 > 0 there exists 𝑎 such that∣∣∣∣𝑓 ′(��)𝑔′(��)− 𝑘

∣∣∣∣ < 𝜖/2 for every �� > 𝑎 (4.49)

For every 𝑥 > 𝑎, there exists (Exercise 4.46) �� ∈ (𝑎, 𝑥) such that

𝑓(𝑥) − 𝑓(𝑎)

𝑔(𝑥) − 𝑔(𝑎) =𝑓 ′(��)𝑔′(��)

and therefore by (4.49)∣∣∣∣𝑓(𝑥)− 𝑓(𝑎)

𝑔(𝑥)− 𝑔(𝑎) − 𝑘∣∣∣∣ < 𝜖/2 for every 𝑥 > 𝑎

2.

𝑓(𝑥)

𝑔(𝑥)=𝑓(𝑥)− 𝑓(𝑎)

𝑔(𝑥)− 𝑔(𝑎) ×𝑓(𝑥)

𝑓(𝑥)− 𝑓(𝑎)× 𝑔(𝑥)− 𝑔(𝑎)

𝑔(𝑥)

=𝑓(𝑥)− 𝑓(𝑎)

𝑔(𝑥)− 𝑔(𝑎) ×1− 𝑔(𝑎)

𝑔(𝑥)

1− 𝑓(𝑎)𝑓(𝑥)

For fixed 𝑎

lim𝑥→∞

1− 𝑔(𝑎)𝑔(𝑥)

1− 𝑓(𝑎)𝑓(𝑥)

= 1

and therefore there exists 𝑎2 such that

1− 𝑔(𝑎)𝑔(𝑥)

1− 𝑓(𝑎)𝑓(𝑥)

< 2 for every 𝑥 > 𝑎2

which implies that∣∣∣∣𝑓(𝑥)

𝑔(𝑥)− 𝑘

∣∣∣∣ < 𝜖2 × 2 for every 𝑥 > 𝑎 = max{𝑎1, 𝑎2}

4.51 We know that the result holds for 𝑛 = 1 (Exercise 4.22). Assume that the resultholds for 𝑛− 1. By the chain rule

𝐷(𝑔 ∘ 𝑓)[x] = 𝐷𝑔[𝑓(x)] ∘𝐷𝑓 [x]

If 𝑓, 𝑔 ∈ 𝐶𝑛, the 𝐷𝑓,𝐷𝑔 ∈ 𝐶𝑛−1 and therefore (by assumption) 𝐷(𝑔 ∘ 𝑓) ∈ 𝐶𝑛−1,which implies that 𝑔 ∘ 𝑓 ∈ 𝐶𝑛.

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4.52 The partial derivatives of the quadratic function are

𝐷1𝑓 = 2𝑎𝑥1 + 2𝑏𝑥2

𝐷2𝑓 = 2𝑏𝑥1 + 2𝑐𝑥2

The second-order partial derivatives are

𝐷11𝑓 = 2𝑎 𝐷21𝑓 = 2𝑏

𝐷12𝑓 = 2𝑏 𝐷22𝑓 = 2𝑐

4.53 Apply Exercise 4.37 to each partial derivative 𝐷𝑖𝑓 [x].

4.54

𝐻(x0) =

(𝐷11𝑓𝑓 [x0] 𝐷12𝑓𝑓 [x0]𝐷21𝑓𝑓 [x0] 𝐷22𝑓𝑓 [x0]

)= 2

(𝑎 𝑏𝑐 𝑑

)

4.55

4.56 For any 𝑥1 ∈ 𝑆, define 𝑔 : 𝑆 → ℜ by

𝑔(𝑡) = 𝑓(𝑡) + 𝑓 ′[𝑡](𝑥1 − 𝑡) + 𝑎2(𝑥1 − 𝑡)2

𝑔 is differentiable on 𝑆 with

𝑝′(𝑡) = 𝑓 ′[𝑡]− 𝑓 ′[𝑡] + 𝑓 ′′[𝑡](𝑥1 − 𝑡)− 2𝑎2(𝑥1 − 𝑡) = 𝑓 ′′[𝑡](𝑥1 − 𝑡)− 2𝑎2(𝑥1 − 𝑡)Note that 𝑔(𝑥1) = 𝑓(𝑥1) and

𝑔(𝑥0) = 𝑓(𝑥0) + 𝑓 ′(𝑥0)(𝑥1 − 𝑥0) + 𝑎2(𝑥1 − 𝑥0)2 (4.50)

is a quadratic approximation for 𝑓 near 𝑥0. If we require that this be exact at 𝑥1 ∕= 𝑥0,then 𝑔(𝑥0) = 𝑓(𝑥1) = 𝑔(𝑥1). By the mean value theorem (Theorem 4.1), there existssome �� between 𝑥0 and 𝑥1 such that

𝑔(𝑥1)− 𝑔(𝑥0) = 𝑝′(��)(𝑥1 − 𝑥0) = 𝑓 ′′(��)(𝑥1 − 𝑥0)− 2𝑎2(𝑥1 − 𝑡) = 0

which implies that

𝑎2 =1

2𝑓 ′′(��)

Setting 𝑥 = 𝑥1 − 𝑥0 in (4.50) gives the required result.

4.57 For any 𝑥1 ∈ 𝑆, define 𝑔 : 𝑆 → ℜ by

𝑔(𝑡) = 𝑓(𝑡) + 𝑓 ′[𝑡](𝑥1 − 𝑡) +1

2𝑓 ′′[𝑡](𝑥1 − 𝑡)2 +

1

3!𝑓 (3)[𝑡](𝑥1 − 𝑡)3 + . . .

+1

𝑛!𝑓 (𝑛)[𝑡](𝑥1 − 𝑡)𝑛 + 𝑎𝑛+1(𝑥1 − 𝑡)𝑛+1

𝑔 is differentiable on 𝑆 with

𝑔′(𝑡) = 𝑓 ′[𝑡]− 𝑓 ′[𝑡] + 𝑓 ′′[𝑡](𝑥1 − 𝑡)− 𝑓 ′′[𝑡](𝑥1 − 𝑡) +1

2𝑓 (3)[𝑡](𝑥1 − 𝑡)2 − 1

2𝑓 (3)[𝑡](𝑥1 − 𝑥0)2 + . . .

+1

(𝑛− 1)!𝑓 (𝑛)[𝑡](𝑥1 − 𝑡)𝑛−1 +

1

𝑛!𝑓 (𝑛+1)[𝑡](𝑥1 − 𝑡)𝑛 − (𝑛+ 1)𝑎𝑛+1(𝑥1 − 𝑡)𝑛

All but the last two terms cancel, so that

𝑔′(𝑡) =1

𝑛!𝑓 (𝑛+1)[𝑡](𝑥1 − 𝑡)𝑛 − (𝑛+ 1)𝑎𝑛+1(𝑥1 − 𝑡)𝑛 =

(1

𝑛!𝑓 (𝑛+1)[𝑡]− (𝑛+ 1)𝑎𝑛+1

)(𝑥1 − 𝑡)𝑛

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Note that 𝑔(𝑥1) = 𝑓(𝑥1) and

𝑔(𝑥0) = 𝑓(𝑥0) + 𝑓 ′[𝑥0](𝑥1 − 𝑥0) +1

2𝑓 ′′[𝑥0](𝑥1 − 𝑥0)2 +

1

3!𝑓 (3)[𝑥0](𝑥1 − 𝑥0)3 + . . .

+1

𝑛!𝑓 (𝑛+1)[𝑥0](𝑥1 − 𝑥0)𝑛 + 𝑎𝑛+1(𝑥1 − 𝑥0)𝑛+1 (4.51)

is a polynomial approximation for 𝑓 near 𝑥0. If we require that 𝑎𝑛+1 be such that𝑔(𝑥0) = 𝑓(𝑥1) = 𝑔(𝑥1), there exists (Theorem 4.1) some �� between 𝑥0 and 𝑥1 suchthat

𝑔(𝑥1)− 𝑔(𝑥0) = 𝑔′(��)(𝑥1 − 𝑥0) = 0

which for 𝑥1 ∕= 𝑥0 implies that

𝑔′(��) =1

𝑛!𝑓𝑛+1[��]− (𝑛+ 1)𝑎𝑛+1 = 0

or

𝑎𝑛+1 =1

(𝑛+ 1)!𝑓𝑛+1[��]

Setting 𝑥 = 𝑥1 − 𝑥0 in (4.51) gives the required result.

4.58 By Taylor’s theorem (Exercise 4.57), for every 𝑥 ∈ 𝑆 − 𝑥0, there exists �� between0 and 𝑥 such that

𝑓(𝑥0 + 𝑥) = 𝑓(𝑥0) + 𝑓 ′[𝑥0]𝑥+1

2𝑓 ′′[𝑥0]𝑥2 + 𝜖(𝑥)

where

𝜖(𝑥) =1

3!𝑓 (3)[��]𝑥3

and

𝜖(𝑥)

𝑥2=

1

3!𝑓 (3)[��](𝑥)

Since 𝑓 ∈ 𝐶3, 𝑓 (3)[��] is bounded on [0, 𝑥] and therefore

lim𝑥→0∣𝑒(𝑥)𝑥2∣ = lim

𝑥→01

3!∣𝑓 (3)[��](𝑥)∣ = 0

4.59 The function 𝑔 : ℜ → 𝑆 defined by

𝑔(𝑡) = 𝑡x0 + (1 − 𝑡)x𝑔 is 𝐶∞ with 𝐷𝑔[𝑡] = x and 𝐷𝑘𝑔(𝑡) = 0 for 𝑘 = 2, 3, . . . . By Exercise 4.51, thecomposite function ℎ = 𝑓 ∘ 𝑔 is 𝐶𝑛+1. By the Chain rule

ℎ′(𝑡) = 𝐷𝑓 [𝑔(𝑡)] ∘𝐷𝑔[𝑡] = 𝐷𝑓 [𝑔(𝑡)](x)

Similarly

ℎ′′(𝑡) = 𝐷(𝐷𝑓 [𝑔(𝑡)](x)

)= 𝐷2𝑓 [𝑔(𝑡)] ∘𝐷𝑔[𝑡](x− x0)= 𝐷2𝑓 [𝑔(𝑡)](x)(2)

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and for all 1 ≤ 𝑘 ≤ 𝑛+ 1

ℎ(𝑘)(𝑡) = 𝐷(𝐷(𝑘−1)𝑓 [𝑔(𝑡)](x)(𝑘−1)

)= 𝐷𝑘𝑓 [𝑔(𝑡)] ∘𝐷𝑔[𝑡](x− x0)(𝑘−1)= 𝐷𝑘𝑓 [𝑔(𝑡)](x)(𝑘)

4.60 From Exercise 4.54, the Hessian of 𝑓 is

𝐻(x) = 2

(𝑎 𝑏𝑐 𝑑

)

and the gradient of 𝑓 is

∇𝑓(x) = (2𝑎𝑥1, 2𝑐𝑥2) with ∇𝑓((0, 0))

= 0

so that the second order Taylor series at (0, 0) is

𝑓(x) = 𝑓(0, 0) +∇𝑓(0, 0)x+1

22x𝑇

(𝑎 𝑏𝑐 𝑑

)x

= 𝑎𝑥21 + 2𝑏𝑥1𝑥2 + 𝑐𝑥22

Not surprisingly, we conclude that the best quadratic approximation of a quadraticfunction is the function itself.

4.61 1. Since 𝐷𝑓 [x0] is continuous and one-to-one (Exercise 3.36), there exists aconstant 𝑚 such that

𝑚 ∥x1 − x2∥ ≤ ∥𝐷𝑓 [x0](x1 − x2)∥ (4.52)

Let 𝜖 = 𝑚/2. By Exercise 4.43, there exists a neighborhood 𝑆 such that

∥𝐷𝑓 [x0](x1 − x2)− (𝑓(x1)− 𝑓(x2))∥ = ∥𝑓(x1)− 𝑓(x2)−𝐷𝑓 [x0](x1 − x2)∥ ≤ 𝜖 ∥x1 − x2∥for every x1,x2 ∈ 𝑆. The Triangle Inequality (Exercise 1.200) implies

∥𝐷𝑓 [x0](x1 − x2)∥ − ∥(𝑓(x1)− 𝑓(x2))∥ ≤ 𝜖 ∥x1 − x2∥Substituting (4.52)

2𝜖 ∥x1 − x2∥ − ∥(𝑓(x1)− 𝑓(x2))∥ ≤ 𝜖 ∥x1 − x2∥That is

𝜖 ∥x1 − x2∥ ≤ ∥(𝑓(x1)− 𝑓(x2))∥ (4.53)

and therefore

𝑓(x1) = 𝑓(x2) =⇒ x1 = x2

2. Let 𝑇 = 𝑓(𝑆). Since the restriction of 𝑓 to 𝑆 is one-to-one and onto, and thereforethere exists an inverse 𝑓−1 : 𝑇 → 𝑆. For any y1,y2 ∈ 𝑇 , let x1 = 𝑓−1(y1) andx2 = 𝑓−1(y2). Substituting in (4.53)

𝜖∥∥𝑓−1(y1)− 𝑓−1(y2)∥∥ ≤ ∥y1 − y2∥

so that ∥∥𝑓−1(y1)− 𝑓−1(y2)∥∥ ≤ 1

𝜖∥y1 − y2∥

𝑓−1 is continuous.

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3. Since 𝑆 is open, 𝑇 = 𝑓−1(𝑆) is open. Therefore, 𝑇 = 𝑓(𝑆) is a neighborhood of𝑓(x0). Therefore, 𝑓 is locally onto.

4.62 Assume to the contrary that there exists x0 ∕= x1 ∈ 𝑆 with 𝑓(x0) = 𝑓(x1). Letx = x1 − x0. Define 𝑔 : [0, 1]→ 𝑆 by 𝑔(𝑡) = (1− 𝑡)x0 + 𝑡x1 = x0 + 𝑡x. Then

𝑔(0) = x0 𝑔(1) = x1 𝑔′(𝑡) = x

Define

ℎ(𝑡) = x𝑇(𝑓(𝑔(𝑡)

)− 𝑓(x0))

Then

ℎ(0) = 0 = ℎ(1)

By the mean value theorem (Mean value theorem), there exists 0 < 𝛼 < 1 such that𝑔(𝛼) ∈ 𝑆 and

ℎ′(𝛼) = x𝑇𝐷𝑓 [𝑔(𝛼)]x = x𝑇𝐽𝑓 (𝑔(𝛼))x = 0

which contradicts the definiteness of 𝐽𝑓 .

4.63 Substituting the linear functions in (4.35) and (4.35), the IS-LM model can beexpressed as

(1− 𝐶𝑦)𝑦 − 𝐼𝑟𝑟 = 𝐶0 + 𝐼0 +𝐺− 𝐶𝑦𝑇𝐿𝑦𝑦 + 𝐿𝑟𝑟 =𝑀/𝑃

which can be rewritten in matrix form as(1− 𝐶𝑦 𝐼𝑟𝐿𝑦 𝐿𝑟

)(𝑦𝑟

)=

(𝑍 − 𝐶𝑦𝑇𝑀/𝑃

)

where 𝑍 = 𝐶0 + 𝐼0 +𝐺. Provided the system is nonsingular, that is

𝐷 =

∣∣∣∣ 1− 𝐶𝑦 𝐼𝑟𝐿𝑦 𝐿𝑟

∣∣∣∣ ∕= 0

the system can be solved using Cramer’s rule (Exercise 3.103) to yield

𝑟 =(1− 𝐶𝑦)𝑀/𝑃 − 𝐿𝑦(𝑍 − 𝐶𝑦𝑇 )

𝐷

𝑦 =𝐿𝑟(𝑍 − 𝐶𝑦)𝑇 − 𝐼𝑟𝑀/𝑃

𝐷

4.64 The kernel

kernel 𝐷𝐹 [(x0, 𝜽0)] = { (x, 𝜽) : 𝐷𝐹 [(x0, 𝜽0)](x, 𝜽) = 0 }is the set of solutions to the equation

𝐷𝐹 [x0, 𝜽0]

(x𝜽

)=

(𝐷x𝑓(x0, 𝜽0)x +𝐷𝜽𝑓(x0, 𝜽0)𝜽

𝜽

)=

(00

)

Only 𝜽 = 0 satisfies this equation. Substituting 𝜽 = 0, the equation reduces to

𝐷x𝑓(x0, 𝜽0)x = 0

which has a unique solution x = 0 since 𝐷x𝑓 [x0, 𝜽0] is nonsingular. Therefore thekernel of 𝐷𝐹 [x0, 𝜽0] consists of the single point (0,0) which implies that 𝐷𝐹 [x0, 𝜽0] isnonsingular (Exercise 3.19).

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4.65 The IS curve is horizontal if its slope is zero, that is

𝐷𝑦𝑔 = −1−𝐷𝑦𝐶

−𝐷𝑟𝐼

This requires either

1. unit marginal propensity to consume (𝐷𝑦𝐶 = 1)

2. infinite interest elasticity of investment (𝐷𝑟𝐼 =∞)

4.66 The LM curve 𝑟 = ℎ(𝑦) is implicitly defined by the equation

𝑓(𝑟, 𝑦;𝐺, 𝑇,𝑀) = 𝐿(𝑦, 𝑟)−𝑀/𝑃 = 0

the slope of which is given by

𝐷𝑦ℎ = −𝐷𝑦𝑓

𝐷𝑟𝑓

= −𝐷𝑦𝐿

𝐷𝑟𝐿

Economic considerations dictate that the numerator (𝐷𝑦𝑓) is positive while the de-nominator (𝐷𝑟𝐿) is negative. Preceded by a negative sign, the slope of the LM curveis positive. The LM curve would be vertical (infinite slope) if the interest elasticity ofthe demand for money was zero (𝐷𝑟𝐿 = 0).

4.67 Suppose 𝑓 is convex. For any x,x0 ∈ 𝑆 let

ℎ(𝑡) = 𝑓(𝑡x+ (1− 𝑡)x0

)≤ 𝑡𝑓(x) + (1− 𝑡)𝑓(x0)

for 0 < 𝑡 < 1. Subtracting ℎ(0) = 𝑓(x0)

ℎ(𝑡)− ℎ(0) ≤ 𝑡𝑓(x)− 𝑡𝑓(x0)

and therefore

𝑓(x)− 𝑓(x0) ≥ ℎ(𝑡)− ℎ(0)

𝑡

Using Exercise 4.10

𝑓(x)− 𝑓(x0) ≥ lim𝑡→0ℎ(𝑡)− ℎ(0)

𝑡= ��x𝑓 [x0] = 𝐷𝑓 [x0](x− x0)

Conversely, let x0 = 𝛼x1+ (1−𝛼)x2 for any x1,x2 ∈ 𝑆. If 𝑓 satisfies (4.29) on 𝑆, then

𝑓(x1) ≥ 𝑓(x0) +𝐷𝑓 [x0](x1 − x0)𝑓(x2) ≥ 𝑓(x0) +𝐷𝑓 [x0](x2 − x0)

and therefore for any 0 ≤ 𝛼 ≤ 1

𝛼𝑓(x1) ≥ 𝛼𝑓(x0 + 𝛼𝐷𝑓 [x0](x1 − x0)(1− 𝛼)𝑓(x2) ≥ (1 − 𝛼)𝑓(x0 + (1 − 𝛼)𝐷𝑓 [x0](x2 − x0)

Adding and using the linearity of 𝐷𝑓 (Exercise 4.21)

𝛼𝑓(x1) + (1 − 𝛼)𝑓(x2) ≥ 𝑓(x0) +𝐷𝑓 [x0](𝛼x1 + (1− 𝛼)x2 − x0) (4.54)

= 𝑓(x0) = 𝑓(𝛼x1 + (1− 𝛼)x2)

That is, 𝑓 is convex. If (4.29) is strict, so is (4.54).

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4.68 Since ℎ is convex, it has a subgradient 𝑔 ∈ 𝑋∗ (Exercise 3.181) such that

ℎ(x) ≥ ℎ(x0) + 𝑔(x− x0) for every x ∈ 𝑋

(4.31) implies that 𝑔 is also a subgradient of 𝑓 on 𝑆

𝑓(x) ≥ 𝑓(𝑥0) + 𝑔(x− x0) for every x ∈ 𝑆

Since 𝑓 is differentiable, this implies that 𝑔 is unique (Remark 4.14) and equal to thederivative of 𝑓 . Hence ℎ is differentiable at x0 with 𝐷ℎ[x0] = 𝐷𝑓 [x0].

4.69 Assume 𝑓 is convex. For every x,x0 ∈ 𝑆, Exercise 4.67 implies

𝑓(x) ≥ 𝑓(x0) +∇𝑓(x0)𝑇(x− x0

)𝑓(x0) ≥ 𝑓(x) +∇𝑓(x)𝑇

(x0 − x

)Adding

𝑓(x) + 𝑓(x0) ≥ 𝑓(x) + 𝑓(x0) +∇𝑓(x)𝑇(x0 − x

)+∇𝑓(x0)

𝑇(x− x0

)or

∇𝑓(x)𝑇(x− x0

) ≥ ∇𝑓(x0)𝑇(x− x0

)and therefore

∇𝑓(x)−∇𝑓(x0)𝑇x− x0 ≥ 0

When 𝑓 is strictly convex, the inequalities are strict.

Conversely, assume (4.32). By the mean value theorem (Theorem 4.1), there existsx ∈ (x,x0) such that

𝑓(x)− 𝑓(x0) = ∇𝑓(x)𝑇x− x0By assumption

∇𝑓(x)−∇𝑓(x0)𝑇 x− x0 ≥ 0

But

x− x0 = 𝛼x0 + (1 − 𝛼)x− x0 = (1 − 𝛼)(x − x0)

and therefore

(1 − 𝛼)∇𝑓(x)−∇𝑓(x0)𝑇x− x0 ≥ 0

so that

∇𝑓(x)𝑇x− x0 ≥ ∇𝑓(x0)𝑇x− x0 ≥ 0

and therefore

𝑓(x)− 𝑓(x0) = ∇𝑓(x)𝑇x− x0 ≥ ∇𝑓(x0)𝑇x− x0

Therefore 𝑓 is convex by Exercise 4.67.

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4.70 For 𝑆 ⊆ ℜ, ∇𝑓(𝑥) = 𝑓 ′(𝑥) and (4.32) becomes

(𝑓 ′(𝑥2)− 𝑓 ′(𝑥1)(𝑥2 − 𝑥1) ≥ 0

for every 𝑥1, 𝑥2 ∈ 𝑆. This is equivalent to

𝑓 ′(𝑥2)(𝑥2 − 𝑥1) ≥ 𝑓 ′(𝑥1)(𝑥2 − 𝑥1)0or

𝑥2 > 𝑥1 =⇒ 𝑓 ′(𝑥2) ≥ 𝑓 ′(𝑥1)𝑓 is strictly convex if and only if the inequalities are strict.

4.71 𝑓 ′ is increasing if and only if 𝑓 ′′ = 𝐷𝑓 ′ ≥ 0 (Exercise 4.35). 𝑓 ′ is strictly increasingif 𝑓 ′′ = 𝐷𝑓 ′ > 0 (Exercise 4.36).

4.72 Adapting the previous example

𝑓 ′′(𝑥) = 𝑛(𝑛− 1)𝑥𝑛 − 2 =

⎧⎨⎩

= 0 if 𝑛 = 1

≥ 0 if 𝑛 = 2, 4, 6, 𝑑𝑜𝑡𝑠

indeterminate otherwise

Therefore, the power function is convex if 𝑛 is even, and neither convex if 𝑛 ≥ 3 is odd.It is both convex and concave when 𝑛 = 1.

4.73 Assume 𝑓 is quasiconcave, and 𝑓(x) ≥ 𝑓(x0). Differentiability at x0 implies forall 0 < 𝑡 < 1

𝑓(x0 + 𝑡(x − x0) = 𝑓(x0) +∇𝑓(x0)𝑡(x− x0) + 𝜂(𝑡) ∥𝑡(x− x0)∥where 𝜂(𝑡)→ 0 and 𝑡→ 0. Quasiconcavity implies

𝑓(x0 + 𝑡(x− x0) ≥ 𝑓(x0)

and therefore

∇𝑓(x0)𝑡(x− x0) + 𝜂(𝑡) ∥𝑡(x − x0)∥ ≥ 0

Dividing by 𝑡 and letting 𝑡→ 0, we get

∇𝑓(x0)(x− x0) ≥ 0

Conversely, assume 𝑓 is a differentiable functional satisfying (4.36). For any x1,x2 ∈ 𝑆with 𝑓(x1) ≥ 𝑓(x2 for every x,x0 ∈ 𝑆), define ℎ : [0, 1]→ ℜ by

ℎ(𝑡) = 𝑓((1− 𝑡)x1 + 𝑡x2

)= 𝑓

(x1 + 𝑡(x2 − x1)

)We need to show that ℎ(𝑡) ≥ ℎ(1) for every 𝑡 ∈ (0, 1). Suppose to the contrary thatℎ(𝑡1) < ℎ(1). Then (see below) there exists 𝑡0 with ℎ(𝑡0) < ℎ(1) and ℎ′(𝑡0) < 0. Bythe Chain Rule, this implies

ℎ′(𝑡0) = ∇𝑓(x0)(x2 − x1) < 0

critical where x0 = x1 + 𝑡(x2 − x1). Since x2 − x0 = (1− 𝑡)(x2 − x1) this implies that

ℎ′(𝑡0) =1

1− 𝑡∇𝑓(x0)(x2 − x0) (4.55)

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On the other hand, since 𝑓(x0) ≥ 𝑓(x2), (4.36) implies

∇𝑓(x0)(x2 − x0) ≥ 0

contradicting (4.55).

To show that there exists 𝑡0 with ℎ(𝑡0) < ℎ(1) and ℎ′(𝑡0) < 0: Since 𝑓 is continuous,there exists an open interval (𝑎, 𝑏) with 𝑎 < 𝑡1 < 𝑏 with ℎ(𝑎) = ℎ(𝑏) = ℎ(1) andℎ(𝑡) < ℎ(1) for every 𝑡 ∈ (𝑎, 𝑏). By the Mean Value Theorem, there exist 𝑡0 ∈ (𝑎, 𝑡1)such that

0 < ℎ(𝑡1)− ℎ(𝑎) = ℎ′(𝑡0)(𝑡1 − 𝑎)which implies that ℎ′(𝑡0) > 0.

4.74 Suppose to the contrary that

𝑓(x) > 𝑓(x0) and ∇𝑓(x0)(x− x0) ≤ 0

critical Let x1 = −∇𝑓(x0) ∕= 0. For every 𝑡 ∈ ℜ+∇𝑓(x0)(x+ 𝑡x1 − x0) = ∇𝑓(x0)𝑡x1 +∇𝑓(x0)(x− x0)

≤ 𝑡∇𝑓(x0)x1

= −𝑡 ∥∇𝑓(x0)∥2 < 0

Since 𝑓 is continuous, there exists 𝑡 > 0 such that

𝑓(x+ 𝑡x1) > 𝑓(x0) and ∇𝑓(x0)(x+ 𝑡x1 − x0) < 0

contradicting the quasiconcavity of 𝑓 (4.36).

4.75 Suppose

𝑓(x) < 𝑓(x0) =⇒ ∇𝑓(x0)(x− x0) < 0

This implies that

−𝑓(x) > −𝑓(x0) =⇒ ∇− 𝑓(x0)(x− x0) > 0

and −𝑓 is pseudoconcave.

4.76 1. If 𝑓 ∈ 𝐹 [𝑆] is concave (and differentiable)

𝑓(x) ≤ 𝑓(x0) +∇𝑓(x0)𝑇 (x− x0)

for every x,x0 ∈ 𝑆(equation 4.30). Therefore

𝑓(x) > 𝑓(x0) =⇒ ∇𝑓(x0)𝑇 (x− x0) > 0

𝑓 is pseudoconcave.

2. Assume to the contrary that 𝑓 is pseudoconcave but not quasiconcave. Then,there exists x = 𝛼x1 + (1− 𝛼)x2, x1,x2 ∈ 𝑆 such that

𝑓(x) < min{𝑓(x1), 𝑓(x2)} (4.56)

Assume without loss of generality that

𝑓(x) < 𝑓(x1) ≤ 𝑓(x2)

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Pseudoconcavity (4.38) implies

∇𝑓(x)(x2 − x) > 0 (4.57)

Since x1 = (x− (1− 𝛼)x2)/𝛼

x1 − x =1

𝛼

(x− (1− 𝛼)x2 − 𝛼x

)= −1− 𝛼

𝛼(x2 − x)

Substituting in (4.57) gives

∇𝑓(x)(x1 − x) < 0

which by pseudoconcavity implies 𝑓(x1) ≤ 𝑓(x) contradicting our assumption(4.56) .

3. Exercise 4.74.

4.77 The CES function is quasiconcave provided 𝜌 ≤ 1 (Exercise 3.58). Since𝐷𝑥𝑖𝑓(x) >0 for all x ∈ ℜ𝑛++, the CES function with 𝜌 ≤ 1 is pseudoconcave on ℜ𝑛++.

4.78 Assume that 𝑓 : 𝑆 → ℜ is homogeneous of degree 𝑘, so that for every x ∈ 𝑆𝑓(𝑡x) = 𝑡𝑛𝑓(x) for every 𝑡 > 0

Differentiating both sides of this identity with respect to 𝑥𝑖

𝐷𝑥𝑖𝑓(𝑡x)𝑡 = 𝑡𝑛𝐷𝑥𝑖𝑓(x)

and dividing by 𝑡 > 0

𝐷𝑥𝑖𝑓(𝑡x) = 𝑡𝑘−1𝐷𝑥𝑖𝑓(x)

4.79 If 𝑓 is homogeneous of degree 𝑘

��x𝑓(x) = lim𝑡→0𝑓(x+ 𝑡x)− 𝑓(x)

𝑡

= lim𝑡→0𝑓((1 + 𝑡)x) − 𝑓(x)

𝑡

= lim𝑡→0

(1 + 𝑡)𝑛𝑓(x)− 𝑓(x)

𝑡

= lim𝑡→0

(1 + 𝑡)𝑛 − 1

𝑡𝑓(x)

Applying L’Hopital’s Rule (Exercise 4.47)

lim𝑡→0

(1 + 𝑡)𝑘−1

𝑡𝑓(x) = lim

𝑡→0𝑘(1 + 𝑡)𝑘−1

1= 𝑘

and therefore

��x𝑓(x) = 𝑘𝑓(x) (4.58)

4.80 For fixed x, define

ℎ(𝑡) = 𝑓(𝑡x)

By the Chain Rule

ℎ′(𝑡) = 𝑡𝐷𝑓 [𝑡x](x) = 𝑡𝑘𝑓(𝑡x) = 𝑡𝑘ℎ(𝑡) (4.59)

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using(4.40). Differentiating the product ℎ(𝑡)𝑡−𝑘

𝐷𝑡

(ℎ(𝑡)𝑡−𝑘

)= −𝑘ℎ(𝑡)𝑡−𝑘−1 + 𝑡−𝑘ℎ′(𝑡) = 𝑡−𝑘

(ℎ′(𝑡)− 𝑘𝑡ℎ(𝑡)) = 0

from (4.59). Since this holds for every 𝑡, ℎ(𝑡)𝑡−𝑘 must be constant (Exercise 4.38), thatis there exists 𝑐 ∈ ℜ such that

ℎ(𝑡)𝑡−𝑘 = 𝑐 =⇒ ℎ(𝑡) = 𝑐𝑡𝑘

Evaluating at 𝑡 = 1, ℎ(1) = 𝑐 and therefore

ℎ(𝑡) = 𝑡𝑘ℎ(1)

Since ℎ(𝑡) = 𝑓(𝑡x) and ℎ(1) = 𝑓(x), this implies

𝑓(𝑡x) = 𝑡𝑘𝑓(x) for every x and 𝑡 > 0

𝑓 is homogeneous of degree 𝑘.

4.81 If 𝑓 is linearly homogeneous and quasiconcave, then 𝑓 is concave (Proposition3.12). Therefore, its Hessian is nonpositive definite (Proposition 4.1). and its diagonalelements 𝐷2𝑥𝑖𝑥𝑖

𝑓(x) are nonpositive (Exercise 3.95). By Wicksell’s law, 𝐷2𝑥𝑖𝑥𝑗𝑓(x) is

nonnegative.

4.82 Assume 𝑓 is homogeneous of degree 𝑘, that is

𝑓(𝑡x) = 𝑡𝑘𝑓(x) for every x ∈ 𝑆 and 𝑡 > 0

By Euler’s theorem

𝐷𝑡𝑓 [𝑡x](𝑡x) = 𝑘𝑓(𝑡x)

and therefore the elasticity of scale is

𝐸(x) =𝑡

𝑓(𝑡x)𝐷𝑡𝑓(𝑡x)

∣∣∣∣𝑡=1

=𝑡

𝑓(𝑡x)𝑘𝑓(𝑡x) = 𝑘


𝐸(x) =𝑡

𝑓(𝑡x)𝐷𝑡𝑓(𝑡x)

∣∣∣∣𝑡=1

= 𝑘

that is

𝐷𝑡𝑓(𝑡x) = 𝑘𝑓(𝑡x)

By Euler’s theorem, 𝑓 is homogeneous of degree 𝑘.

4.83 Assume 𝑓 ∈ 𝐹 (𝑆) is differentiable and homogeneous of degree 𝑘 ∕= 0. By Euler’stheorem

𝐷𝑓 [x](x) = 𝑘𝑓(x) ∕= 0

for every x ∈ 𝑆 such that 𝑓(x) ∕= 0.

4.84 𝑓 satisfies Euler’s theorem

𝑘𝑓(x) =

𝑛∑𝑖=1

𝐷𝑖𝑓(x)𝑥𝑖

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Differentiating with respect to 𝑥𝑗

𝑘𝐷𝑗𝑓(x) =𝑛∑𝑖=1

𝐷𝑖𝑗𝑓(x)𝑥𝑖 +𝐷𝑗𝑓(x)

or

(𝑘 − 1)𝐷𝑗𝑓(x) =

𝑛∑𝑖=1

𝐷𝑖𝑗𝑓(x)𝑥𝑖 𝑗 = 1, 2, . . . , 𝑛

Multiplying each equation by 𝑥𝑗 and summing

(𝑘 − 1)𝑛∑

𝑗=1

𝐷𝑗𝑓(x)𝑥𝑗 =𝑛∑

𝑗=1

𝑛∑𝑖=1

𝐷𝑖𝑗𝑓(x)𝑥𝑖𝑥𝑗 = x′𝐻x

By Euler’s theorem, the left hand side is

(𝑘 − 1)𝑘𝑓(x) = x′𝐻x

4.85 If 𝑓 is homothetic, there exists strictly increasing 𝑔 and linearly homogeneous ℎsuch that 𝑓 = 𝑔 ∘ ℎ (Exercise 3.175). Using the Chain Rule and Exercise 4.78

𝐷𝑥𝑖𝑓(𝑡x) = 𝑔′(𝑓(𝑡x))𝐷𝑥𝑖ℎ(𝑡x) = 𝑡𝑔′(𝑓(𝑡x))𝐷𝑥𝑖ℎ(x)

and therefore

𝐷𝑥𝑖𝑓(𝑡x)

𝐷𝑥𝑗𝑓(𝑡x)=

𝑡𝑔′(𝑓(𝑡x))𝐷𝑥𝑖ℎ(x)

𝐷𝑥𝑗 𝑡𝑔′(𝑓(𝑡x))𝐷𝑥𝑗ℎ(x)

=𝐷𝑥𝑖ℎ(x)

𝐷𝑥𝑗

𝐷𝑥𝑗ℎ(x)

=𝑔′(𝑓(x)𝐷𝑥𝑖ℎ(x)

𝐷𝑥𝑗𝑔′(𝑓(x))𝐷𝑥𝑗ℎ(x)

=𝐷𝑥𝑖𝑓(x)

𝐷𝑥𝑗

𝐷𝑥𝑗𝑓(x)

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Chapter 5: Optimization

5.1 As stated, this problem has no optimal solution. Revenue 𝑓(𝑥) increases withoutbound as the rate of exploitation 𝑥 gets smaller and smaller. Given any positive ex-ploitation rate 𝑥0, a smaller rate will increase total revenue. Nonexistence arises frominadequacy in modeling the island leaders’ problem. For example, the model ignoresany costs of extraction and sale. Realistically, we would expect per-unit costs to de-crease with volume (increasing returns to scale) at least over lower outputs. Extractionand transaction costs should make vanishingly small rates of output prohibitively ex-pensive and encourage faster utilization. Secondly, even if the government weightsfuture generations equally with the current generation, it would be rational to valuecurrent revenue more highly than future revenue and discount future returns. Dis-counting is appropriate for two reasons

∙ Current revenues can be invested to provide a future return. There is an oppor-tunity cost (the interest foregone) to delaying extraction and sale.

∙ Innovation may create substitutes which reduce the future demand for the fertil-izer. If the government is risk averse, it has an incentive to accelerate exploitation,trading-off of lower total return against reduced risk.

5.2 Suppose that x∗ is a local optimum which is not a global optimum. That is, thereexists a neighborhood 𝑆 of x∗ such that

𝑓(x∗, 𝜽) ≥ 𝑓(x, 𝜽) for every x ∈ 𝑆 ∩𝐺(𝜽)

and also another point x∗∗ ∈ 𝐺(𝜽) such that

𝑓(x∗∗, 𝜽) > 𝑓(x∗, 𝜽)

Since 𝐺(𝜽) is convex, there exists 𝛼 ∈ (0, 1) such that

𝛼x∗ + (1− 𝛼)x∗∗ ∈ 𝑆 ∩𝐺(𝜽)

By concavity of 𝑓

𝑓(𝛼x∗ + (1− 𝛼)x∗∗, 𝜽) ≥ 𝛼𝑓(x∗, 𝜽) + (1− 𝛼)𝑓(x∗∗, 𝜽) > 𝑓(x∗, 𝜽)

contradicting the assumption that x∗ is a local optimum.

5.3 Suppose that x∗ is a local optimum which is not a global optimum. That is, thereexists a neighborhood 𝑆 of x∗ such that

𝑓(x∗, 𝜽) ≥ 𝑓(x, 𝜽) for every x ∈ 𝑆 ∩𝐺(𝜽)

and also another point x∗∗ ∈ 𝐺(𝜽) such that

𝑓(x∗∗, 𝜽) > 𝑓(x∗, 𝜽)


𝛼x∗ + (1− 𝛼)x∗∗ ∈ 𝑆 ∩𝐺(𝜽)

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By strict quasiconcavity of 𝑓

𝑓(𝛼x∗ + (1− 𝛼)x∗∗, 𝜽) > min{ 𝑓(x∗, 𝜽), 𝑓(x∗∗, 𝜽) } > 𝑓(x∗, 𝜽)

contradicting the assumption that x∗ is a local optimum. Therefore, if x∗ is localoptimum, it must be a global optimum.

Now suppose that x∗ is a weak global optimum, that is

𝑓(x∗, 𝜽) ≥ 𝑓(x, 𝜽) for every x ∈ 𝑆

but there another point x∗∗ ∈ 𝑆 such that

𝑓(x∗∗, 𝜽) = 𝑓(x∗, 𝜽)


𝛼x∗ + (1− 𝛼)x∗∗ ∈ 𝑆 ∩𝐺(𝜽)

By strict quasiconcavity of 𝑓

𝑓(𝛼x∗ + (1− 𝛼)x∗∗, 𝜽) > min{ 𝑓(x∗, 𝜽), 𝑓(x∗∗, 𝜽) } = 𝑓(x∗, 𝜽)

contradicting the assumption that x∗ is a global optimum. We conclude that everyoptimum is a strict global optimum and hence unique.

5.4 Suppose that x∗ is a local optimum of (5.3) in 𝑋 , so that

𝑓(x∗) ≥ 𝑓(x) (5.80)

for every x in a neighborhood 𝑆 of x∗. If 𝑓 is differentiable,

𝑓(x) = 𝑓(x∗) +𝐷𝑓 [x∗](x− x∗) + 𝜂(x) ∥x− x∗∥

where 𝜂(x)→ 0 as x→ x∗. (5.80) implies that there exists a ball 𝐵𝑟(x∗) such that

𝐷𝑓 [x∗](x− x∗) + 𝜂(x) ∥x− x∗∥ ≤ 0

for every x ∈ 𝐵𝑟(x∗). Letting x→ x∗, we conclude that

𝐷𝑓 [x∗](x− x∗) ≤ 0

for every x ∈ 𝐵𝑟(x∗).

Suppose there exists x ∈ 𝐵𝑟(x∗) such that

𝐷𝑓 [x∗](x− x∗) = 𝑦 < 0

Let dx = x−x∗ so that x = x∗ +dx. Then x∗−dx ∈ 𝐵𝑟(x∗). Since 𝐷𝑓 [x∗] is linear,

𝐷𝑓 [x∗](−dx) = −𝐷𝑓 [x∗](dx) = −𝑦 > 0

contradicting (5.80). Therefore

𝐷𝑓 [x∗](x− x∗) = 0

for every x ∈ 𝐵𝑟(x∗).

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5.5 We apply the reasoning of Example 5.5 to each component. Formally, for each 𝑖,let 𝑓𝑖 be the projection of 𝑓 along the 𝑖𝑡ℎ axis

𝑓𝑖(𝑥𝑖) = 𝑓(𝑥∗1, 𝑥∗2, . . . , 𝑥

∗𝑖−1, 𝑥𝑖, 𝑥

∗𝑖+1, . . . , 𝑥

∗𝑛)

𝑥∗𝑖 maximizes 𝑓𝑖(𝑥𝑖) over ℜ+, for which it is necessary that

𝐷𝑥𝑖𝑓𝑖(𝑥∗𝑖 ) ≤ 0 𝑥∗𝑖 ≥ 0 𝑥∗𝑖𝐷𝑥𝑖𝑓𝑖(𝑥

∗𝑖 ) = 0

Substituting

𝐷𝑥𝑖𝑓𝑖(𝑥∗𝑖 ) = 𝐷𝑥𝑖𝑓 [x∗]

yields

𝐷𝑥𝑖𝑓 [x∗] ≤ 0 𝑥∗𝑖 ≥ 0 𝑥∗𝑖𝐷𝑥𝑖𝑓 [x∗] = 0

5.6 By Taylor’s Theorem (Example 4.33)

𝑓(x∗ + dx) = 𝑓(x∗) +∇𝑓(x∗)dx+1

2dx𝑇𝐻𝑓 (x∗)dx+ 𝜂(dx) ∥dx∥2

with 𝜂(dx)→ 0 as dx→ 0. Given

1. ∇𝑓(x∗) = 0 and

2. 𝐻𝑓 (x∗) is negative definite

and letting dx→ 0, we conclude that

𝑓(x∗ + dx) < 𝑓(x∗)

for small dx. x∗ is a strict local maximum.

5.7 If x∗ is a local minimum of 𝑓(x), it is necessary that

𝑓(x∗) ≤ 𝑓(x)

for every x in a neighborhood 𝑆 of x∗. Assuming that 𝑓 is 𝐶2, 𝑓(x) can be approxi-mated by

𝑓(x) ≈ 𝑓(x∗) +∇𝑓(x∗)dx+1

2dx𝑇𝐻𝑓 (x∗)dx

where dx = x−x∗. If x∗ is a local minimum, then there exists a ball 𝐵𝑟(x∗) such that

𝑓(x∗) ≤ 𝑓(x∗) +∇𝑓(x∗)dx+1

2dx𝑇𝐻𝑓 (x∗)dx

or

∇𝑓(x∗)dx+1

2dx𝑇𝐻𝑓 (x∗)dx ≥ 0

for every dx ∈ 𝐵𝑟(x∗). To satisfy this inequality for all small dx requires that the

first term be zero and the second term nonnegative. In other words, for a point x∗ tobe a local minimum of a function 𝑓 , it is necessary that the gradient be zero and theHessian be nonnegative definite at x∗. Furthermore, by Taylor’s Theorem

𝑓(x∗ + dx) = 𝑓(x∗) +∇𝑓(x∗)dx+1

2dx𝑇𝐻𝑓 (x∗)dx+ 𝜂(dx) ∥dx∥2

with 𝜂(dx)→ 0 as dx→ 0. Given

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12

1

2

3

(1,2,3)

12

0

𝑥1

𝑥2

𝑥1

Figure 5.1: The strictly concave function 𝑓(𝑥1, 𝑥2) = 𝑥1𝑥2+3𝑥2−𝑥21−𝑥22 has a uniqueglobal maximum.

1. ∇𝑓(x∗) = 0 and

2. 𝐻𝑓 (x∗) is positive definite

and letting dx→ 0, we conclude that

𝑓(x∗ + dx) > 𝑓(x∗)

for small dx. x∗ is a strict local minimum.

5.8 By the Weierstrass theorem (Theorem 2.2), 𝑓 has a maximum 𝑥∗ and a minimum𝑥∗ on [𝑎, 𝑏]. Either

∙ 𝑥∗ ∈ (𝑎, 𝑏) and 𝑓 ′(𝑥∗) = 0 (Theorem 5.1) or

∙ 𝑥∗ ∈ (𝑎, 𝑏) and 𝑓 ′(𝑥∗) = 0 (Exercise 5.7) or

∙ Both maxima and minima are boundary points, that is 𝑥∗, 𝑥∗ ∈ {𝑎, 𝑏} whichimplies that 𝑓 is constant on [𝑎, 𝑏] and therefore 𝑓 ′(𝑥) = 0 for every 𝑥 ∈ (𝑎, 𝑏)(Exercise 4.7).

5.9 The first-order conditions for a maximum are

𝐷𝑥1𝑓(𝑥1, 𝑥2) = 𝑥2 − 2𝑥1 = 0

𝐷𝑥2𝑓(𝑥1, 𝑥2) = 𝑥1 + 3− 2𝑥2 = 0

which have the unique solution 𝑥∗1 = 1, 𝑥∗2 = 2. (1, 2) is the only stationary point of𝑓 and hence the only possible candidate for a maximum. To verify that (1, 2) satisfiesthe second-order condition for a maximum, we compute the Hessian of 𝑓

𝐻(x) =

(−2 11 −2

)

which is negative definite everywhere. Therefore (1, 2) is a strict local maximum of 𝑓 .Further, since 𝑓 is strictly concave (Proposition 4.1), we conclude that (1, 2) is a strictglobal maximum of 𝑓 (Exercise 5.2), where it attains its maximum value 𝑓(1, 2) = 3(Figure 5.1).

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5.10 The first-order conditions for a maximum (or minimum) are

𝐷1𝑓(𝑥) = 2𝑥1 = 0

𝐷2𝑓(𝑥) = 2𝑥2 = 0

which have a unique solution 𝑥1 = 𝑥2 = 0. This is the only stationary point of 𝑓 . Sincethe Hessian of 𝑓

𝐻 =

(2 00 2

)

is positive definite, we deduce (0, 0) is a strict global minimum of 𝑓 (Proposition 4.1,Exercise 5.2).

5.11 The average firm’s profit function is

Π(𝑘, 𝑙) = 𝑦 − 1

2𝑘 − 𝑙− 1

6

and the firm’s profit maximization problem is

max𝑘,𝑙

Π(𝑘, 𝑙) = 𝑘1/6𝑙1/3 − 1

2𝑘 − 𝑙 − 1

6

A necessary condition for a profit maximum is that the profit function be stationary,that is

𝐷𝑘Π(𝑘, 𝑙) =1

6𝑘−5/6𝑙1/3 − 1

2= 0

𝐷𝑙Π(𝑘, 𝑙) =1

3𝑘1/6𝑙−2/3 − 1 = 0

which can be solved to yield

𝑘 = 𝑙 =1

9

The firm’s output is

𝑦 =1

9

1/6 1

9

1/3

=1

3

and its profit is

Π(1

3,

1

3) =

1

3− 1

2

1

9− 1

9− 1

6= 0

5.12 By the Chain Rule

𝐷x(ℎ ∘ 𝑓)[x∗] = 𝐷ℎ ∘𝐷x𝑓 [x∗] = 0

Since 𝐷ℎ > 0

𝐷x(ℎ ∘ 𝑓)[𝑥∗] = 0 ⇐⇒ 𝐷x𝑓 [x∗] = 0

ℎ ∘ 𝑓 has the same stationary points as 𝑓 .

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5.13 Since the log function is monotonic, finding the maximum likelihood estimators isequivalent to solving the maximization problem ( Exercise 5.12)

max𝜇,𝜎

log𝐿(𝜇, 𝜎) = −𝑇2

log 2𝜋 − 𝑇 log 𝜎 − 1

2𝜎2

𝑇∑𝑡=1

(𝑥𝑡 − 𝜇)2

For (��, ��2) to solve this problem, it is necessary that log𝐿 be stationary at (��, ��2),that is

𝐷𝜇 log𝐿(��, ��2) =1

��2

𝑇∑𝑡=1

(𝑥𝑡 − ��) = 0

𝐷𝜎 log𝐿(��, ��2) = −𝑇��

+1

��3

𝑇∑𝑡=1

(𝑥𝑡 − ��)2 = 0


�� = �� =1

𝑇

𝑇∑𝑡=1

𝑥𝑡

��2 =1

𝑇

𝑇∑𝑡=1

(𝑥𝑡 − ��)2

5.14 The gradient of the objective function is

∇𝑓(x) =

(−2(𝑥1 − 1)−2(𝑥2 − 1)

)

while that of the constraint is

∇𝑔(x) =

(2𝑥12𝑥2

)

A necessary condition for the optimal solution is that these be proportional that is

∇𝑓(𝑥) =

(−2(𝑥1 − 1)−2(𝑥2 − 1)

)= 𝜆

(2𝑥12𝑥2

)= ∇𝑔(x)


𝑥1 = 𝑥2 =1

1 + 𝜆

which includes an unknown constant of proportionality 𝜆. However, any solution mustalso satisfy the constraint

𝑔(𝑥1, 𝑥2) = 2

(1

1 + 𝜆

)2= 1

This can be solved for 𝜆

𝜆 =√

2− 1

and substituted into (5.80)

𝑥1 = 𝑥2 =1√2

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5.15 The consumer’s problem is

maxx≥0𝑢(x) = 𝑥1 + 𝑎 log 𝑥2

subject to 𝑔(x) = 𝑥1 + 𝑝2𝑥2 −𝑚 = 0

The first-order conditions for a (local) optimum are

𝐷𝑥1𝑢(x∗) = 1 ≤ 𝜆 = 𝐷𝑥1𝑔(𝑥

∗) 𝑥1 ≥ 0 𝑥1(1− 𝜆) = 0 (5.81)

𝐷𝑥2𝑢(x∗) =

𝑎

𝑥2≤ 𝜆𝑝2 = 𝐷𝑥2𝑔(𝑥

∗) 𝑥2 ≥ 0 𝑥2

(𝑎

𝑥2− 𝜆𝑝2

)= 0 (5.82)

We can distinguish two cases:

Case 1 𝑥1 = 0 in which case the budget constraint implies that 𝑥2 = 𝑚/𝑝2.

Case 2 𝑥1 > 0 In this case, (5.81) implies that 𝜆 = 1. Consequently, the first inequal-ity of (5.82) implies that 𝑥2 > 0 and therefore the last equation implies 𝑥2 = 𝑎/𝑝2with 𝑥1 = 𝑚− 𝑎.

We deduce that the consumer first spends portion 𝑎 of her income on good 2 and theremainder on good 1.

5.16 Suppose without loss of generality that the first 𝑘 components of y∗ are strictlypositive while the remaining components are zero. That is

𝑦∗𝑖 > 0 𝑖 = 1, 2, . . . , 𝑘

𝑦∗𝑖 = 0 𝑖 = 𝑘 + 1, 𝑘 + 2, . . . , 𝑛

(x∗,y∗) solves the problem

max 𝑓(x)

subject to g(x) = 0

𝑦𝑖 = 0 𝑖 = 𝑘 + 1, 𝑘 + 2, . . . , 𝑛

By Theorem 5.2, there exist multipliers 𝜆1, 𝜆2, . . . , 𝜆𝑚 and 𝜇𝑘+1, 𝜇𝑘+2, . . . , 𝜇𝑛 such that

𝐷x𝑓 [x∗,y∗] =𝑚∑𝑗=1

𝜆𝑗𝐷x𝑔𝑗[x∗,y∗]

𝐷y𝑓 [x∗,y∗] =

𝑚∑𝑗=1

𝜆𝑗𝐷y𝑔𝑗[x∗,y∗] +

𝑛∑𝑖=𝑘+1

𝜇𝑖𝑦𝑖

Furthermore, 𝜇𝑖 ≥ 0 for every 𝑖 so that

𝐷y𝑓 [x∗,y∗] ≤𝑚∑𝑗=1

𝜆𝑗𝐷y𝑔𝑗 [x∗,y∗]

with

𝐷𝑦𝑖𝑓 [x∗,y∗] =𝑚∑𝑗=1

𝜆𝑗𝐷𝑦𝑖𝑔𝑗 [x∗,y∗] if 𝑦𝑖 > 0

5.17 Assume that x∗ = (𝑥∗1, 𝑥∗2) solves

max𝑥1,𝑥2

𝑓(𝑥1, 𝑥2)

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subject to

𝑔(𝑥1, 𝑥2) = 0

By the implicit function theorem, there exists a function ℎ : ℜ → ℜ such that

𝑥1 = ℎ(𝑥2) (5.83)

and

𝑔(ℎ(𝑥2), 𝑥2) = 0

for 𝑥2 in a neighborhood of 𝑥∗2. Furthermore

𝐷ℎ[𝑥∗2] = −𝐷𝑥1𝑔[x∗]

𝐷𝑥2𝑔[x∗]

(5.84)

Using (5.41), we can convert the original problem into the unconstrained maximizationof a function of a single variable

max𝑥2

𝑓(ℎ(𝑥2), 𝑥2)

If 𝑥∗2 maximizes this function, it must satisfy the first-order condition (applying theChain Rule)

𝐷𝑥1𝑓 [𝑥∗] ∘𝐷ℎ[𝑥∗2] +𝐷𝑥2𝑓 [x∗] = 0

Substituting (5.42) yields

𝐷𝑥1𝑓 [𝑥∗](−𝐷𝑥1𝑔[x

∗]

𝐷𝑥2𝑔[x∗]

)+𝐷𝑥2𝑓 [x∗] = 0

or

𝐷𝑥1𝑓 [𝑥∗]𝐷𝑥2𝑓 [x∗]

=𝐷𝑥1𝑔[x

∗]

𝐷𝑥2𝑔[x∗]

5.18 The consumer’s problem is

maxx∈𝑋

𝑢(x)

subject to p𝑇x = 𝑚

Solving for 𝑥1 from the budget constraint yields

𝑥1 =𝑚−∑𝑛

𝑖=2 𝑝𝑖𝑥𝑖𝑝1

Substituting this in the utility function, the affordable utility levels are

��(𝑥2, 𝑥3, . . . , 𝑥𝑛) = 𝑢

(𝑚−∑𝑛


, 𝑥2, 𝑥3, . . . , 𝑥𝑛

)(5.85)

and the consumer’s problem is to choose (𝑥2, 𝑥3, . . . , 𝑥𝑛) to maximize (5.85). Thefirst-order conditions are that ��(𝑥2, 𝑥3, . . . , 𝑥𝑛) be stationary, that is for every good𝑗 = 2, 3, . . . , 𝑛

𝐷𝑥𝑗 ��(𝑥2, 𝑥3, . . . , 𝑥𝑛) = 𝐷𝑥1𝑢(x∗)𝐷𝑥𝑗

(𝑚−∑𝑛


)+𝐷𝑥𝑗𝑢(x

∗) = 0

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which reduces to

𝐷𝑥1𝑢(x∗)(−𝑝1

𝑝𝑗) +𝐷𝑥𝑗𝑢(x

∗) = 0

or

𝐷𝑥1𝑢(x∗)

𝐷𝑥𝑗𝑢(x∗)

=𝑝1𝑝𝑗

𝑗 = 2, 3, . . . , 𝑛

This is the familiar equality between the marginal rate of substitution and the priceratio (Example 5.15). Since our selection of 𝑥1 was arbitrary, this applies between anytwo goods.

5.19 Adapt Exercise 5.6.

5.20 Corollary 5.1.2 implies that x∗ is a global maximum of 𝐿(x,𝝀), that is

𝐿(x∗,𝝀) ≥ 𝐿(x,𝝀) for every x ∈ 𝑋which implies

𝑓(x∗)−∑𝜆𝑗𝑔𝑗(x

∗) ≥ 𝑓(x)−∑𝜆𝑗𝑔𝑗(x) for every x ∈ 𝑋

Since g(x∗) = 0 this implies

𝑓(x∗) ≥ 𝑓(x)−∑𝜆𝑗𝑔𝑗(x) for every x ∈ 𝑋

A fortiori

𝑓(x∗) ≥ 𝑓(x) for every x ∈ 𝐺 = {x ∈ 𝑋 : g(x) = 0 }5.21 Suppose that x∗ is a local maximum of 𝑓 on𝐺. That is, there exists a neighborhood𝑆 such that

𝑓(x∗) ≥ 𝑓(x) for every x ∈ 𝑆 ∩𝐺But for every x ∈ 𝐺, 𝑔𝑗(x) = 0 for every 𝑗 and

𝐿(x) = 𝑓(x) +∑𝜆𝑗𝑔𝑗(x) = 𝑓(x)

and therefore

𝐿(x∗) ≥ 𝐿(x) for every x ∈ 𝑆 ∩𝐺5.22 The area of the base is

Base = 𝑤2 = 𝐴/3

and the four sides

Sides = 4𝑤ℎ

= 4

√𝐴

3

√𝐴

12

=4𝐴

16

=2𝐴

3

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5.23 Let the dimensions of the vat be 𝑤 × 𝑙 × ℎ. We wish to

min𝑤,𝑙,ℎ

Surface area = 𝐴 = 𝑤 × 𝑙 + 2𝑤ℎ+ 2𝑙ℎ

subject to 𝑤 × 𝑙 × ℎ = 32

The Lagrangean is

𝐿(𝑤, 𝑙, ℎ, 𝜆) = 𝑤𝑙 + 2𝑤ℎ+ 2𝑙ℎ− 𝜆𝑤𝑙ℎ.The first-order conditions for a maximum are

𝐷𝑤𝐿 = 𝑙+ 2ℎ− 𝜆𝑙ℎ = 0 (5.86)

𝐷𝑙𝐿 = 𝑤 + 2ℎ− 𝜆𝑤ℎ = 0 (5.87)

𝐷ℎ𝐿 = 2𝑤 + 2𝑙− 𝜆𝑤𝑙 = 0 (5.88)

𝑤𝑙ℎ = 32

Subtracting (5.45) from (5.44)

𝑙 − 𝑤 = 𝜆(𝑙 − 𝑤)ℎ

This equation has two possible solutions. Either

𝜆 =1

ℎor 𝑙 = 𝑤

But if 𝜆 = 1/ℎ, (5.44) implies that 𝑙 = 0 and the volume is zero. Therefore, we concludethat 𝑤 = 𝑙. Substituting 𝑤 = 𝑙 in (5.45) and (5.46) gives

𝑤 + 2ℎ = 𝜆𝑤ℎ

4𝑤 = 𝜆𝑤2

from which we deduce that

𝜆 =4

𝑤


𝑤 + 2ℎ =4

𝑤𝑤ℎ = 4ℎ

which implies that

𝑤 = 2ℎ or ℎ =1

2𝑤

To achieve the required volume of 32 cubic metres requires that

𝑤 × 𝑙 × ℎ = 𝑤 × 𝑤 × 1

2𝑤 = 32

so that the dimensions of the vat are

𝑤 = 4 𝑙 = 4 ℎ = 2

The area of sheet metal required is

𝐴 = 𝑤𝑙 + 2𝑤ℎ+ 2𝑙ℎ = 48

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5.24 The Lagrangean for this problem is

𝐿(x, 𝜆) = 𝑥21 + 𝑥22 + 𝑥23 − 𝜆(2𝑥1 − 3𝑥2 + 5𝑥3 − 19)

A necessary condition for x∗ to solve the problem is that the Lagrangean be stationaryat x∗, that is

𝐷𝑥1𝐿 = 2𝑥∗1 − 2𝜆 = 0

𝐷𝑥2𝐿 = 2𝑥∗2 + 3𝜆 = 0

𝐷𝑥3𝐿 = 2𝑥∗3 − 5𝜆 = 0

which implies

𝑥∗1 = 𝜆 𝑥∗2 = −3

2𝜆 𝑥∗2 =

5

2𝜆 (5.89)

It is also necessary that the solution satisfy the constraint, that is

2𝑥∗1 − 3𝑥∗2 + 5𝑥∗3 = 19

Substituting (5.89) into the constraint we get

2𝜆+9

2𝜆+

25

2𝜆 = 19𝜆 = 19

which implies 𝜆 = 1. Substituting in (5.89), the solution is x∗ = (1,− 32 , 52 ). Since theconstraint is affine and the objective (−𝑓) is concave, stationarity of the Lagrangeanis also sufficient for global optimum (Corollary 5.2.4).

5.25 The Lagrangean is

𝐿(𝑥1, 𝑥2, 𝜆) = 𝑥𝛼1𝑥1−𝛼2 − 𝜆(𝑝1𝑥1 + 𝑝2𝑥2 −𝑚)

The Lagrangean is stationary where

𝐷𝑥1𝐿 = 𝛼𝑥𝛼−11 𝑥1−𝛼2 − 𝜆𝑝1 = 0

𝐷𝑥2𝐿 = 1− 𝛼𝑥𝛼1 𝑥1−𝛼−12 − 𝜆𝑝2 = 0

Therefore the first-order conditions for a maximum are

𝛼𝑥𝛼−11 𝑥1−𝛼2 = 𝜆𝑝1 (5.90)

1− 𝛼𝑥𝛼1𝑥1−𝛼−12 = 𝜆𝑝2 (5.91)

𝑝1𝑥1 + 𝑝2𝑥2 −𝑚 (5.92)

Dividing (5.48) by (5.49) gives

𝛼𝑥𝛼−11 𝑥1−𝛼2

1− 𝛼𝑥𝛼1 𝑥(1−𝛼)−12

= 𝑝1𝑝2

which simplifies to

𝛼𝑥2(1− 𝛼)𝑥1

=𝑝1𝑝2

or 𝑝2𝑥2 =(1− 𝛼)

𝛼𝑝1𝑥1

Substituting in the budget constraint (5.50)

𝑝1𝑥1 +(1 − 𝛼)

𝛼𝑝1𝑥1 = 𝑚

𝛼+ (1 − 𝛼)

𝛼𝑝1𝑥1 = 𝑚

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so that

𝑥∗1 =𝛼

𝛼+ (1− 𝛼)

𝑚

𝑝1

From the budget constraint (5.92)

𝑥∗2 =(1− 𝛼)

𝛼+ (1− 𝛼)

𝑚

𝑝2


𝐿(x, 𝜆) = 𝑥𝛼11 𝑥

𝛼22 . . . 𝑥

𝛼𝑛𝑛 − 𝜆(𝑝1𝑥1 + 𝑝2𝑥2 + ...+ 𝑝𝑛𝑥𝑛)

The first-order conditions for a maximum are

𝐷𝑥𝑖𝐿 = 𝛼𝑖𝑥𝛼11 𝑥

𝛼22 . . . 𝑥

𝛼𝑖−1𝑖 . . . 𝑥𝛼𝑛

𝑛 − 𝜆𝑝𝑖 =𝛼𝑖𝑢(𝑥)

𝑥𝑖− 𝜆𝑝𝑖 = 0

or

𝛼𝑖𝑢(𝑥)

𝜆= 𝑝𝑖𝑥𝑖 𝑖 = 1, 2, . . . , 𝑛 (5.93)

Summing over all goods and using the budget constraint

𝑛∑𝑖=1

𝛼𝑖𝑢(𝑥)

𝜆=𝑢(𝑥)

𝜆

𝑛∑𝑖=1

𝛼𝑖 =

𝑛∑𝑖=1

𝑝𝑖𝑥𝑖 = 𝑚

Letting∑𝑛

𝑖=1 𝛼𝑖 = 𝛼, this implies

𝑢(x)

𝜆=𝑚

𝛼


𝑝𝑖𝑥𝑖 =𝛼𝑖𝛼𝑚

or

𝑥∗𝑖 =𝛼𝑖𝛼

𝑚

𝑝𝑖𝑖 = 1, 2, . . . , 𝑛


𝐿(x, 𝜆) = 𝑤1𝑥1 + 𝑤2𝑥2 − 𝜆(𝑥𝜌1 + 𝑥𝜌2 − 𝑦𝜌).

The necessary conditions for stationarity are

𝐷𝑥1𝐿(x, 𝜆) = 𝑤1 − 𝜆𝜌𝑥𝜌−11 = 0

𝐷𝑥2𝐿(x, 𝜆) = 𝑤2 − 𝜆𝜌𝑥𝜌−12 = 0

or

𝑤1 = 𝜆𝜌𝑥𝜌−11

𝑤2 = 𝜆𝜌𝑥𝜌−12

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which reduce to

𝑤1𝑤2

=𝑥𝜌−11

𝑥𝜌−1 2

𝑥𝜌−12 =𝑤2𝑤1𝑥𝜌−11

𝑥𝜌2 =

(𝑤2𝑤1

) 𝜌𝜌−1

𝑥𝜌1

Substituting in the production constraint

𝑥𝜌1 +

(𝑤2𝑤1

) 𝜌𝜌−1

𝑥𝜌1 = 𝑦𝜌(1 +

(𝑤2𝑤1

) 𝜌𝜌−1

)𝑥𝜌1 = 𝑦𝜌

we can solve 𝑥1

𝑥𝜌1 =

(1 +

(𝑤2𝑤1

) 𝜌𝜌−1

)−1𝑦𝜌

𝑥1 =

(1 +

(𝑤2𝑤1

) 𝜌𝜌−1

)−1/𝑝𝑦

Similarly

𝑥2 =

(1 +

(𝑤1𝑤2

) 𝜌𝜌−1

)−1/𝑝𝑦

5.28 Example 5.27 is flawed. The optimum of the constrained maximization problem(ℎ = 𝑤/2) is in fact a saddle point of the Lagrangean. It maximizes the Lagrangean inthe feasible set, but not globally.

The net benefit approach to the Lagrange multiplier method is really only applicablewhen the Lagrangean (net benefit function) is concave, so that every stationary pointis a global maximum. This requirement is satisfied in many standard examples, suchas the consumer’s problem (Example 5.21) and cost minimization (Example 5.28). Itis also met in Example 5.29. The requirement of concavity is not recognized in thetext, and Section 5.3.6 should be amended accordingly.

5.29 The Lagrangean

𝐿(x, 𝜆) =

𝑛∑𝑖=1

𝑐𝑖(𝑥𝑖) + 𝜆

(𝐷 −

𝑛∑𝑖=1

𝑥𝑖

)(5.94)

can be rewritten as

𝐿(x, 𝜆) = −𝑛∑𝑖=1

(𝜆𝑥𝑖 − 𝑐𝑖(𝑥𝑖)

)+ 𝜆𝐷 (5.95)

The 𝑖th term in the sum is the net profit of plant 𝑖 if its output is valued at 𝜆. Therefore,if the company undertakes to buy electricity from its plants at the price 𝜆 and instructseach plant manager to produce so as to maximize the plant’s net profit, each manager

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will be induced to choose an output level which maximizes the profit of the companyas a whole. This is the case whether the price 𝜆 is the market price at which thecompany can buy electricity from external suppliers or the shadow price determinedby the need to satisfy the total demand 𝐷. In this way, the shadow price 𝜆 can be usedto decentralize the production decision.


𝐿(𝑥1, 𝑥2, 𝜆1, 𝜆2) = 𝑥1𝑥2 − 𝜆1(𝑥21 + 2𝑥22 − 3)− 𝜆2(2𝑥21 + 𝑥22 − 3)

The first-order conditions for stationarity

𝐷𝑥1𝐿 = 𝑥2 − 2𝜆1𝑥1 − 4𝜆2𝑥1 = 0

𝐷𝑥2𝐿 = 𝑥1 − 4𝜆1𝑥2 − 2𝜆2𝑥2 = 0

can be written as

𝑥2 = 2(𝜆1 + 2𝜆2)𝑥1 (5.96)

𝑥1 = 2(2𝜆1 + 𝜆2)𝑥2 (5.97)

which must be satisfied along with the complementary slackness conditions

𝑥21 + 2𝑥22 − 3 ≤ 0 𝜆1 ≥ 0 𝜆1(𝑥21 + 2𝑥22 − 3) = 0

2𝑥21 + 𝑥22 − 3 ≤ 0 𝜆2 ≥ 0 𝜆2(2𝑥21 + 𝑥22 − 3) = 0

First suppose that both constraints are slack so that 𝜆1 = 𝜆2 = 0. Then the first-orderconditions (5.96) and (5.97) imply that 𝑥1 = 𝑥2 = 0. (0, 0) satisfies the Kuhn-Tuckerconditions. Next suppose that the first constraint is binding while the second constraintis slack (𝜆2 = 0). The first-order conditions (5.96) and (5.97) have two solutions,𝑥1 =

√3/2, 𝑥2 =

√3/2, 𝜆 = 1/(2

√2) and 𝑥1 = −√3/2, 𝑥2 = −√3/2, 𝜆 = 1/(2

√2),

but these violate the second constraint. Similarly, there is no solution in which the firstconstraint is slack and the second constraint binding. Finally, assume that the bothconstraints are binding. This implies that 𝑥1 = 𝑥2 = 1 or 𝑥1 = 𝑥2 = −1, which pointssatisfy the first-order conditions (5.96) and (5.97) with 𝜆1 = 𝜆2 = 1/6.

We conclude that three points satisfy the Kuhn-Tucker conditions, namely (0, 0), (1, 1)and (−1,−1). Noting the objective function, we observe that (0, 0) in fact minimizesthe objective. We conclude that there are two local maxima, (1, 1) and (−1,−1), bothof which achieve the same level of the objective function.

5.31 Dividing the first-order conditions, we obtain

𝐷𝑘𝑅(𝑘, 𝑙)

𝐷𝑙𝑅(𝑘, 𝑙)=𝑟

𝑤− 𝜆(𝑠− 𝑟)

(1− 𝜆)𝑤Using the revenue function

𝑅(𝑘, 𝑙) = 𝑝(𝑓(𝑘, 𝑙))𝑓(𝑘, 𝑙)

the marginal revenue products of capital and labor are

𝐷𝑘𝑅(𝑘, 𝑙) = 𝐷𝑦𝑝(𝑦)𝐷𝑘𝑓(𝑘, 𝑙)

𝐷𝑙𝑅(𝑘, 𝑙) = 𝐷𝑦𝑝(𝑦)𝐷𝑙𝑓(𝑘, 𝑙)

so that their ratio is equal to the ratio of the marginal products

𝐷𝑘𝑅(𝑘, 𝑙)

𝐷𝑙𝑅(𝑘, 𝑙)=𝐷𝑘𝑓(𝑘, 𝑙)

𝐷𝑙𝑓(𝑘, 𝑙

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The necessary condition for optimality can be expressed as

𝐷𝑘𝑓(𝑘, 𝑙)

𝐷𝑙𝑓(𝑘, 𝑙)=𝑟

𝑤− 𝜆

1− 𝜆𝑠− 𝑟𝑤

whereas the necessary condition for cost minimization is (Example 5.16)

𝐷𝑘𝑓(𝑘, 𝑙)

𝐷𝑙𝑓(𝑘, 𝑙)=𝑟

𝑤

The regulated firm does not use the cost-minimizing combination of inputs.

5.32 The general constrained optimization problem

maxx𝑓(x)

subject to g(x) ≤ 0can be transformed into an equivalent equality constrained problem

maxx,s𝑓(x)

subject to g(x) + s = 0 and s ≥ 0through the addition of nonnegative slack variables s. Letting g(x, s) = g(x) + s, thefirst-order conditions a local optimum are (Exercise 5.16)

𝐷x𝑓(x∗) =∑𝜆𝑗𝐷x𝑔𝑗(x

∗, s∗) =∑𝜆𝑗𝐷x𝑔𝑗(x

∗)

0 = 𝐷s𝑓(x∗) ≤∑𝜆𝑗𝐷s𝑔𝑗(x, s) = 𝝀 (5.98)

s ≥ 0 𝝀𝑇 s = 0 (5.99)

Condition (5.98) implies that 𝜆𝑗 ≥ 0 for every 𝑗. Furthermore, rewriting the constraintas

s = −g(x)the complementary slackness condition (5.99) becomes

g(x) ≤ 0 𝝀𝑇g(x) = 0

This establishes the necessary conditions of Theorem 5.3.

5.33 The equality constrained maximization problem

maxx𝑓(x)

subject to g(x) = 0

is equivalent to the problem

maxx𝑓(x)

subject to g(x) ≤ 0−g(x) ≤ −0

By the Kuhn-Tucker theorem (Theorem 5.3), there exists nonnegative multipliers𝜆+1 , 𝜆

+2 , . . . , 𝜆

+𝑚 and 𝜆−1 , 𝜆

−2 , . . . , 𝜆

−𝑚 such that

𝐷𝑓(x∗) =∑𝜆+𝑗 𝐷𝑔𝑗[x

∗]−∑𝜆−𝑗 𝐷𝑔𝑗[x

∗] = 0 (5.100)

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with

𝜆+𝑗 𝑔𝑗(x) = 0 and 𝜆−𝑗 𝑔𝑗(x) = 0 𝑗 = 1, 2, . . . ,𝑚

Defining 𝜆𝑗 = 𝜆+𝑗 − 𝜆−𝑗 , (5.100) can be written as

𝐷𝑓(x∗) =∑𝜆𝑗𝐷𝑔𝑗[x

∗]

which is the first-order condition for an equality constrained problem. Furthermore, ifx∗ satisfies the inequality constraints

𝑔(x∗) ≤ 0 and 𝑔(x∗) ≥ 0

it satisfies the equality

𝑔(x∗) = 0

5.34 Suppose that x∗ solves the problem

maxxc𝑇x subject to 𝐴x ≤ 0

with Lagrangean

𝐿 = c𝑇x− 𝝀𝑇𝐴x

Then there exists 𝝀 ≥ 0 such that

𝐷x𝐿 = c𝑇 − 𝝀𝑇𝐴 = 0

that is, 𝐴𝑇𝝀 = c. Conversely, if there is no solution, there exists x such that 𝐴x ≤ 0and

c𝑇x > c𝑇0 = 0

5.35 There are two binding constraints at (4, 0), namely

𝑔(𝑥1, 𝑥2) = 𝑥1 + 𝑥2 ≤ 4

ℎ(𝑥1, 𝑥2) = −𝑥2 ≤ 0

with gradients

∇𝑔(4, 0) = (1, 1)

∇ℎ(4, 0) = (0, 1)

which are linearly independent. Therefore the binding constraints are regular at (0, 4).


𝐿(x, 𝜆) = 𝑢(x)− 𝜆(p𝑇x−𝑚)

and the first-order (Kuhn-Tucker) conditions are (Corollary 5.3.2)

𝐷𝑥𝑖𝐿[x∗, 𝜆] = 𝐷𝑥𝑖𝑢[x∗]− 𝜆𝑝𝑖 ≤ 0 x∗𝑖 ≥ 0 𝑥∗𝑖 (𝐷𝑥𝑖𝑢[x

∗]− 𝜆𝑝𝑖) = 0 (5.101)

p𝑇x∗ ≤ 𝑚 𝜆 ≥ 0 𝜆(p𝑇x∗ −𝑚) = 0 (5.102)

for every good 𝑖 = 1, 2, . . . ,𝑚. Two cases must be distinguished.

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Case 1 𝜆 > 0 This implies that p𝑇x = 𝑚, the consumer spends all her income.Condition (5.101) implies

𝐷𝑥𝑖𝑢[x∗] ≤ 𝜆𝑝𝑖 for every 𝑖 with 𝐷𝑥𝑖𝑢[x

∗] = 𝜆𝑝𝑖 for every 𝑖 for which 𝑥𝑖 > 0

This case was analyzed in Example 5.17.

Case 2 𝜆 = 0 This allows the possibility that the consumer does not spend all herincome. Substituting 𝜆 = 0 in (5.101) we have 𝐷𝑥𝑖𝑢[x

∗] = 0 for every 𝑖. At theoptimal consumption bundle x∗, the marginal utility of every good is zero. Theconsumer is satiated, that is no additional consumption can increase satisfaction.This case was analyzed in Example 5.31.

In summary, at the optimal consumption bundle x∗, either

∙ the consumer is satiated (𝐷𝑥𝑖𝑢[x∗] = 0 for every 𝑖) or

∙ the consumer consumes only those goods whose marginal utility exceeds thethreshold 𝐷𝑥𝑖𝑢[x

∗] ≥ 𝜆𝑝𝑖 and adjusts consumption so that the marginalutility is proportional to price for all consumed goods.

5.37 Assume x ∈ 𝐷(x∗). Then there exists �� ∈ ℜ such that x∗ + 𝛼x ∈ 𝑆 for every0 ≤ 𝛼 ≤ ��. Define 𝑔 ∈ 𝐹 ([0, ��]) by 𝑔(𝛼) = 𝑓(x∗ + 𝛼x). If x∗ is a local maximum, 𝑔has a local maximum at 0, and therefore 𝑔′(0) ≤ 0 (Theorem 5.1). By the chain rule(Exercise 4.22), this implies

𝑔′(0) = 𝐷𝑓 [x∗](x) ≤ 0

and therefore x /∈ 𝐻+(x∗).

5.38 If x is a tangent vector, so is 𝛽x for any nonnegative 𝛽 (replace 1/𝛼𝑘 by 𝛽/𝛼𝑘 inthe preceding definition. Also, trivially, x = 0 is a tangent vector (with x𝑘 = x∗ and𝛼𝑘 = 1 for all 𝑘). The set 𝑇 of all vectors tangent to 𝑆 at x∗ is therefore a nonemptycone, which is called the cone of tangents to 𝑆 at x∗.

To show that 𝑇 is closed, let x𝑛 be a sequence in 𝑇 converging to some x ∈ ℜ𝑛. Weneed to show that x ∈ 𝑇 . Since x𝑛 ∈ 𝑇 , there exist feasible points x𝑚𝑛 ∈ 𝑆 and 𝛼𝑚𝑛

such that

(x𝑚𝑛 − x∗)/𝛼𝑚𝑛 → x𝑛 as 𝑚→∞For any 𝑁 choose 𝑛 such that

∥x𝑛 − x∥ ≤ 1

2𝑁

and then choose 𝑚 such that

∥x𝑚𝑛 − x∗∥ ≤ 𝑁 and ∥(x𝑚𝑛 − x∗)/𝛼𝑚𝑛 − x𝑛∥ ≤ 1

2𝑁

Relabeling x𝑚𝑛 as x𝑁 and 𝛼𝑚𝑛 as 𝛼𝑁 we have we have constructed a sequence x𝑁 inS such that ∥∥x𝑁 − x∗∥∥ ≤ 𝑁and ∥∥(x𝑁 − x∗)/𝛼𝑁 − x∥∥ ≤ ∥∥(x𝑁 − x∗)/𝛼𝑁 − x𝑛∥∥ + ∥x𝑛 − x∥ ≤ 1

1𝑁

Letting 𝑁 →∞, x𝑁 converges to x∗ and (x𝑁 − x∗)/𝛼𝑁 converges to x, which provesthat x ∈ 𝑇 as required.

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5.39 Assume x ∈ 𝐷(x∗). That is, there exists �� such that x∗ + 𝛼x ∈ 𝑆 for every𝛼 ∈ [0, ��]. For 𝑘 = 1, 2, . . . , let 𝛼𝑘 = ��/𝑘. Then x𝑘 = x∗ + 𝛼𝑘x ∈ 𝑆, x𝑘 → x∗ and(x𝑘 − x∗)/𝛼𝑘 = (x∗ + 𝛼𝑘x− x∗)/𝛼𝑘 = x. Therefore, x ∈ 𝑇 (x∗).

5.40 Let dx ∈ 𝑇 (x∗). Then there exists a feasible sequence {x𝑘} converging to x∗ and asequence {𝛼𝑘} of nonnegative scalars such that the sequence {(x𝑘−x∗)/𝛼𝑘} convergesto dx. For any 𝑗 ∈ 𝐵(x∗), 𝑔𝑗(x

∗) = 0 and

𝑔𝑗(x𝑘) = 𝐷𝑔𝑗 [x

∗](x𝑘 − x∗) + 𝜂𝑗∥∥x𝑘 − x∗∥∥

where 𝜂𝑗 → 0 as k→∞. This implies

1

𝛼𝑘𝑔𝑗(x

𝑘) =1

𝛼𝑘𝐷𝑔𝑗 [x

∗](x𝑘 − x∗) + 𝜂𝑗∥∥(x𝑘 − x∗)/𝛼𝑘

∥∥Since x𝑘 is feasible

1

𝛼𝑘𝑔𝑗(x

𝑘) ≤ 0

and therefore

𝐷𝑔𝑗 [x∗]((x𝑘 − x∗)/𝛼𝑘) + 𝜂𝑖

∥∥(x𝑘 − x∗)/𝛼𝑘∥∥ ≤ 0

Letting 𝑘 →∞ we conclude that

𝐷𝑔𝑗[x∗](dx) ≤ 0

That is, dx ∈ 𝐿.

5.41 𝐿0 ⊆ 𝐿1 by definition. Assume dx ∈ 𝐿1. That is

𝐷𝑔𝑗 [x∗](dx) < 0 for every 𝑗 ∈ 𝐵𝑁 (x∗) (5.103)

𝐷𝑔𝑗 [x∗](dx) ≤ 0 for every 𝑗 ∈ 𝐵𝐶(x∗) (5.104)

where 𝐵𝐶(x∗) = 𝐵(x∗) − 𝐵𝑁 (x∗) is the set of concave binding constraints at x∗. Byconcavity (Exercise 4.67), (5.104) implies that

𝑔𝑗(x∗ + 𝛼dx) ≤ 𝑔𝑗(x∗) = 0 for every 𝛼 ≥ 0 and 𝑗 ∈ 𝐵𝐶(x∗)

From (5.103) there exists some 𝛼𝑁 such that

𝑔𝑗(x∗ + 𝛼dx) < 0 for every 𝛼 ∈ [0, 𝛼𝑁 ] and 𝑗 ∈ 𝐵𝑁 (x∗)

Furthermore, since 𝑔𝑗(x∗) < 0 for all 𝑗 ∈ 𝑆(x∗), there exists some 𝛼𝑆 > 0 such that

𝑔𝑗(x∗ + 𝛼dx) < 0 for every 𝛼 ∈ [0, 𝛼𝑆] and 𝑗 ∈ 𝑆(x∗)

Setting �� = min{𝛼𝑁 , 𝛼𝑆} we have

𝑔𝑗(x∗ + 𝛼dx) ≤ 0 for every 𝛼 ∈ [0, ��] and 𝑗 = 1, 2, . . . ,𝑚

or

x∗ + 𝛼dx ∈ 𝐺 = {x : 𝑔𝑗(x) ≤ 0, 𝑗 = 1, 2, . . . ,𝑚 } for every 𝛼 ∈ [0, ��]

Therefore dx ∈ 𝐷. We have previously shown (Exercises 5.39 and 5.40) that 𝐷 ⊂ 𝑇 ⊂𝐿.

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5.42 Assume that g satisfies the Quasiconvex CQ condition at x∗. That is, for every𝑗 ∈ 𝐵(x∗), 𝑔𝑗 is quasiconvex, ∇𝑔𝑗(x∗) ∕= 0 and there exists x such that 𝑔𝑗(x) < 0.Consider the perturbation dx = x − x∗. Quasiconvexity and regularity implies thatfor every binding constraint 𝑗 ∈ 𝐵(x∗) (Exercises 4.74 and 4.75)

𝑔𝑗(x) < 𝑔𝑗(x∗) =⇒ ∇𝑔𝑗(x∗)𝑇 (x− x∗) = ∇𝑔𝑗(x∗)𝑇dx < 0

That is

𝐷𝑔𝑗[x∗](dx) < 0

Therefore, dx ∈ 𝐿0(x∗) ∕= ∅ and g satisfies the Cottle constraint qualification condition.

5.43 If the binding constraints 𝐵(x∗) are regular at x∗, their gradients are linearlyindependent. That is, there exists no 𝜆𝑗 ∕= 0, 𝑗 ∈ 𝐵(x∗) such that∑

𝑗∈𝐵(x∗)

𝜆𝑗∇𝑔𝑗[x∗] = 0

By Gordan’s theorem (Exercise 3.239), there exists dx ∈ ℜ𝑛 such that

∇𝑔𝑗[x∗]𝑇dx < 0 for every 𝑗 ∈ 𝐵(x∗)

Therefore dx ∈ 𝐿0(x∗) ∕= ∅.5.44 If 𝑔𝑗 concave, 𝐵𝑁 (x∗) = ∅, and AHUCQ is trivially satisfied (with dx = 0 ∈ 𝐿1).For every 𝑗, let

𝑆𝑗 = {dx : 𝐷𝑔𝑗 [x∗](dx) < 0 }

Then

𝐿1(x∗) =

⎛⎝ ∩

𝑖∈𝐵𝑁 (x∗)

𝑆𝑖

⎞⎠∩⎛

⎝ ∩𝑖∈𝐵𝐶(x∗)

𝑆𝑖

⎞⎠

where 𝐵𝐶(x∗) and 𝐵𝑁 (x∗) are respectively the concave and nonconcave constraintsbinding at x∗. If 𝑔𝑗 satisfies the AHUCQ condition, 𝐿1(x∗) ∕= ∅ and Exercise 1.219implies that

𝐿1 =

⎛⎝ ∩

𝑖∈𝐵𝑁 (x∗)

𝑆𝑖

⎞⎠∩⎛

⎝ ∩𝑖∈𝐵𝐶(x∗)

𝑆𝑖

⎞⎠

Now

𝑆𝑖 = { dx : 𝐷𝑔𝑗[x∗](dx) ≤ 0 }

and therefore

𝐿1 =∩

𝑗∈𝐵(x∗)

𝑆𝑖 = 𝐿

Since (Exercise 5.41)

𝐿1 ⊆ 𝑇 ⊆ 𝐿and 𝑇 is closed (Exercise 5.38), we have

𝐿 = 𝐿1 ⊆ 𝑇 ⊆ 𝐿which implies that 𝑇 = 𝐿.

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5.45 For each 𝑗 = 1, 2, . . . ,𝑚, either

𝑔𝑗(x∗) < 0 which implies that 𝜆𝑗 = 0 and therefore 𝜆𝑗𝐷𝑔𝑗 [x

∗](x− x∗) = 0

or

𝑔𝑗(x∗) = 0 Since 𝑔𝑗 is quasiconvex and 𝑔𝑗(x) ≤ 0 = 𝑔(x∗), Exercise 4.73 implies that

𝐷𝑔𝑗[x∗](x− x∗) ≤ 0. Since 𝜆𝑗 ≥ 0, this implies that 𝜆𝑗𝐷𝑔𝑗 [x

∗](x − x∗) ≤ 0.

We have shown that for every 𝑗, 𝜆𝑗𝐷𝑔𝑗 [x∗](x − x∗) ≤ 0. The first-order condition

implies that

𝐷𝑓 [x∗](x − x∗) =∑𝑗

𝜆𝑗𝐷𝑔𝑗 [x∗](x− x∗) ≤ 0

If

∇𝑓(x∗) ≤∑𝜆𝑗∇𝑔𝑗(x∗) x∗ ≥ 0

(∇𝑓(x∗)− 𝜆𝑗∇𝑔𝑗(x∗)

)𝑇x∗ = 0

The first-order conditions imply that for every x ∈ 𝐺, x ≥ 0 and

(∇𝑓(x∗)− 𝜆𝑗∇𝑔𝑗(x∗)

)𝑇x ≤ 0

and therefore (∇𝑓(x∗)− 𝜆𝑗∇𝑔𝑗(x∗)

)𝑇(x− x∗) ≤ 0

or

∇𝑓(x∗)𝑇 (x − x∗) ≤∑𝜆𝑗∇𝑔𝑗(x∗)𝑇 (x− x∗) ≤ 0

5.46 Assuming 𝑥𝑑 = 𝑥𝑑 = 0, the constraints become

2𝑥𝑐 ≤ 30

2𝑥𝑐 ≤ 25

𝑥𝑐 ≤ 20

The first and third conditions are redundant, which implies that 𝜆𝑓 = 𝜆𝑚 = 0. Com-plementary slackness requires that, if 𝑥𝑐 > 0,

𝐷𝑥𝑐𝐿 = 1− 2𝜆𝑓 − 2𝜆𝑙 − 𝜆𝑚 = 0

or 𝜆𝑙 = 12 . Evaluating the Lagrangean at (0, 1/2, 0) yields

𝐿

(x,

(0,

1

2, 0

))= 3𝑥𝑏 + 𝑥𝑐 + 3𝑥𝑑

− 1

2(𝑥𝑏 + 2𝑥𝑐 + 3𝑥𝑑 − 25)

=25

2+

5

2𝑥𝑏 +

3

2𝑥𝑑

This basic feasible solution is clearly not optimal, since profit would be increased byincreasing either 𝑥𝑏 or 𝑥𝑑.

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Following the hint, we allow 𝑥𝑑 > 0, retaining the assumption that 𝑥𝑏 = 0. We mustbe alert to the possibility that 𝑥𝑐 = 0. With 𝑥𝑏 = 0, the constraints become

2𝑥𝑐 + 𝑥𝑑 ≤ 30

2𝑥𝑐 + 3𝑥𝑑 ≤ 25

𝑥𝑐 + 𝑥𝑑 ≤ 20

The first constraint is redundant, which implies that 𝜆𝑓 = 0. If 𝑥𝑑 > 0, complementaryslackness requires that

𝐷𝑥𝑑𝐿 = 3− 3𝜆𝑙 − 𝜆𝑚 = 0

or

𝜆𝑚 = 3(1− 𝜆𝑙) (5.105)

The requirement that 𝜆𝑚 ≥ 0 implies that 𝜆𝑙 ≤ 1. Substituting (5.105) in the secondfirst-order condition

𝐷𝑥𝑐𝐿 = 1− 2𝜆𝑙 − 𝜆𝑚 = 1− 2𝜆𝑙 − 3(1− 𝜆𝑙) = −2 + 𝜆𝑙

implies that

𝐷𝑥𝑐𝐿 = −2 + 𝜆𝑙 < 0 for every 𝜆𝑙 ≤ 1

Complementary slackness then requires implies that 𝑥𝑐 = 0.

The constraints now become

𝑥𝑑 ≤ 30

3𝑥𝑑 ≤ 25

𝑥𝑑 ≤ 20

The first and third are redundant, so that 𝜆𝑓 and 𝜆𝑚 = 0. Equation (5.105) impliesthat 𝜆𝑙 = 1.

Evaluating the Lagrangean at this point (𝜆 = 0, 1, 0), we have

𝐿(𝑥, (0, 1, 0)) = 3𝑥𝑏 + 𝑥𝑐 + 3𝑥𝑑

− (𝑥𝑏 + 2𝑥𝑐 + 3𝑥𝑑 − 25)

= 25 + 2𝑥𝑏 − 𝑥𝑐

Clearly this is not an optimal solution, An increase in 𝑥𝑏 is indicated. This leads usto the hypothesis 𝑥𝑏 > 0, 𝑥𝑑 > 0, 𝑥𝑐 = 0 which was evaluated in the text, and in factlead to the optimal solution.

5.47 If we ignore the hint and consider solutions with 𝑥𝑏 > 0, 𝑥𝑐 ≥ 0, 𝑥𝑑 = 0, theconstraints become

2𝑥𝑏 + 2𝑥𝑐 ≤ 30

𝑥𝑏 + 2𝑥𝑐 ≤ 25

2𝑥𝑏 + 𝑥𝑐 ≤ 20

These three constraints are linearly dependent, so that any one of them is redundantand can be eliminated. For example, 3/2 times the first constraint is equal to the sum of

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the second and third constraints. The feasible solution 𝑥𝑏 = 0, 𝑥𝑐 = 5, 𝑥𝑑 = 10, wherethe constraints are linearly dependent, is known as a degenerate solution. Degeneracyis a significant feature of linear programming, allowing the theoretical possibility of abreakdown in the simplex algorithm. Fortunately, such breakdown seems very rare inpractice. Degeneracy at the optimal solution indicates multiple optima.

One way to proceed in this example is to arbitrarily designate one constraint as redun-dant, assuming the corresponding multiplier is zero. Arbitrarily choosing 𝜆𝑚 = 0 andproceeding as before, complementary slackness (𝑥𝑑 > 0) requires that

𝐷𝑥𝑑𝐿 = 3− 2𝜆𝑓 − 𝜆𝑙 = 0

or

𝜆𝑙 = 3− 2𝜆𝑓 (5.106)

Nonnegativity of 𝜆𝑙 implies that 𝜆𝑓 ≤ 32 .

Substituting (5.106) in the second first-order condition yields

𝐷𝑥𝑐𝐿 = 1− 2𝜆𝑓 − 2𝜆𝑙

= 1− 2𝜆𝑓 − 2(3− 2𝜆𝑓 )

= −5 + 2𝜆𝑓 < 0 for every 𝜆𝑓 ≤ 3

2

Complementary slackness therefore implies that 𝑥𝑐 = 0, which takes us back to thestarting point of the presentation in the text, where 𝑥𝑏 > 0, 𝑥𝑐 = 𝑥𝑑 = 0.

5.48 Assume that (c1, 𝑧1) and (c2, 𝑧2) belong to 𝐵. That is

c1 ≤ 0 𝑧1 ≥ 𝑧∗c2 ≤ 0 𝑧2 ≥ 𝑧∗

For any 𝛼 ∈ (0, 1),

c = 𝛼c1 + (1 − 𝛼)c2 ≤ 0 𝑧 = 𝛼𝑧1 + (1 − 𝛼)𝑧2 ≤ 𝑧∗

and therefore (c, 𝑧) ∈ 𝐵. This shows that 𝐵 is convex. Let 1 = (1, 1, . . . , 1) ∈ ℜ𝑚.Then (c− 1, 𝑧 + 1) ∈ int 𝐵 ∕= ∅. There 𝐵 has a nonempty interior.

5.49 Let (c, 𝑧) ∈ int 𝐵. This implies that c < 0 and 𝑧 > 𝑧∗. Since 𝑣 is monotone

𝑣(c) ≤ 𝑣(0) = z∗ < 𝑧

which implies that (c, 𝑧) /∈ 𝐴.

5.50 The linear functional 𝐿 can be decomposed into separate components, so thatthere exists (Exercise 3.47) 𝜑 ∈ 𝑌 ∗ and 𝛼 ∈ ℜ such that

𝐿(c, 𝑧) = 𝛼𝑧 − 𝜑(c)

Assuming 𝑌 ⊆ ℜ𝑚, there exists (Proposition 3.4) 𝝀 ∈ ℜ𝑚 such that 𝜑(c) = 𝝀𝑇 c andtherefore

𝐿(c, 𝑧) = 𝛼𝑧 − 𝝀𝑇 c

The point (0, 𝑧∗ + 1) belongs to 𝐵. Therefore, by (5.75),

𝐿(0, 𝑧∗) ≤ 𝐿(0, 𝑧∗ + 1)

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which implies that

𝛼𝑧∗ − 𝝀𝑇0 ≤ 𝛼(𝑧∗ + 1)− 𝝀𝑇0

or 𝛼 ≥ 0. Similarly, let { e1, e2, . . . , e𝑚 } denote the standard basis for ℜ𝑚 (Example1.79). For any 𝑗 = 1, 2, . . . ,𝑚, the point (0− e𝑗 , 𝑧∗) (which corresponds to decreasingresource 𝑗 by one unit) belongs to 𝐵 and therefore (from (5.75))

𝑧∗ − 𝝀𝑇 (0− e𝑗) = 𝑧∗ + 𝜆𝑗 ≥ 𝑧∗ − 𝝀𝑇0 = 𝑧∗

which implies that 𝜆𝑗 ≥ 0.

5.51 Let

c = 𝑔(x) < 0 and 𝑧 = 𝑓(x)

Suppose 𝛼 = 0. Then, since 𝐿 is nonzero, at least one component of 𝝀 must be nonzero.That is, 𝝀 ≩ 0 and therefore

𝝀𝑇 𝑐 < 0 (5.107)

But (c, 𝑧) ∈ 𝐴 and (5.74) implies

𝛼𝑧 − 𝝀𝑇 c ≤ 𝛼𝑧∗ − 𝝀𝑇0

and therefore 𝛼 = 0 implies

𝝀𝑇 𝑐 ≥ 0

contradicting (5.107). Therefore, we conclude that 𝛼 > 0.

5.52 The utility’s optimization problem is

max𝑦,𝑌≥0

𝑆(𝑦, 𝑌 ) =

𝑛∑𝑖=1

∫ 𝑦𝑖

0

(𝑝𝑖(𝜏) − 𝑐𝑖)𝑑𝜏 − 𝑐0𝑌

subject to 𝑔𝑖(y, 𝑌 ) = 𝑦𝑖 − 𝑌 ≤ 0 𝑖 = 1, 2, . . . , 𝑛

The demand independence assumption ensures that the objective function 𝑆 is concave,since its Hessian

𝐻𝑆 =

⎛⎜⎜⎝𝐷𝑝1 0 . . . 0 0

0 𝐷𝑝2 . . . 0 00 . . . 𝐷𝑝𝑛 00 . . . 0 0

⎞⎟⎟⎠

is nonpositive definite (Exercise 3.96). The constraints are linear and hence convex.Moreover, there exists a point (0, 1) such that for every 𝑖 = 1, 2, . . . , 𝑛

𝑔𝑖(0, 1) = 0− 1 < 0

Therefore the problem satisfies the conditions of Theorem 5.6. The optimal solution(y∗, 𝑌 ∗) satisfies the Kuhn-Tucker conditions, that is there exist multipliers 𝜆1, 𝜆2, . . . , 𝜆𝑚such that for every period 𝑖 = 1, 2, . . . , 𝑛

𝐷𝑦𝑖𝐿 = 𝑝𝑖(𝑦𝑖)− 𝑐𝑖 − 𝜆𝑖 ≤ 0 𝑦𝑖 ≥ 0 𝑦𝑖(𝑝𝑖(𝑦𝑖)− 𝑐𝑖 − 𝜆𝑖) = 0 (5.108)

𝑦𝑖 ≤ 𝑌 𝜆𝑖 ≥ 0 𝜆(𝑌 − 𝑦𝑖) = 0

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and that capacity be chosen such that

𝐷𝑌 𝐿 = 𝑐0 −𝑛∑𝑖=1

𝜆𝑖 ≤ 0 𝑌 ≥ 0 𝑌

(𝑐0 −

𝑛∑𝑖=1

𝜆𝑖

)= 0 (5.109)

where 𝐿 is the Lagrangean

𝐿(𝑦, 𝑌, 𝜆) =𝑛∑𝑖=1

∫ 𝑦𝑖

0

(𝑝𝑖(𝜏) − 𝑐𝑖)𝑑𝜏 − 𝑐0𝑌 −𝑛∑𝑖=1

𝜆𝑖(𝑦𝑖 − 𝑌 )

In off-peak periods (𝑦𝑖 < 𝑌 ), complementary slackness requires that 𝜆𝑖 = 0 and there-fore from (5.108)

𝑝𝑖(𝑦𝑖) = 𝑐𝑖

assuming 𝑦𝑖 > 0. In peak periods (𝑦𝑖 = 𝑌 )

𝑝𝑖(𝑦𝑖) = 𝑐𝑖 + 𝜆𝑖

We conclude that it is optimal to price at marginal cost in off-peak periods and chargea premium during peak periods. Furthermore, (5.109) implies that the total premiumis equal to the marginal capacity cost

𝑛∑𝑖=1

𝜆𝑖 = 𝑐0

Furthermore, note that

𝑛∑𝑖=1

𝜆𝑖𝑦𝑖 =∑Peak

𝜆𝑖𝑦𝑖 +∑

Off-peak

𝜆𝑖𝑦𝑖

=∑𝑦𝑖=𝑌

𝜆𝑖𝑦𝑖 +∑𝜆𝑖=0

𝜆𝑖𝑦𝑖

=∑𝑦𝑖=𝑌

𝜆𝑖𝑌

=

𝑛∑𝑖=1

𝜆𝑖𝑌 = 𝑐0𝑌

Therefore, the utility’s total revenue is

𝑅(𝑦, 𝑌 ) =

𝑛∑𝑖=1

𝑝𝑖(𝑦𝑖)𝑦𝑖

=𝑛∑𝑖=1

(𝑐𝑖 + 𝜆𝑖)𝑦𝑖

=

𝑛∑𝑖=1

𝑐𝑖𝑦𝑖 +

𝑛∑𝑖=1

𝜆𝑖𝑦𝑖

=𝑛∑𝑖=1

𝑐𝑖𝑦𝑖 + 𝑐0𝑌 = 𝑐(𝑦, 𝑌 )

Under the optimal pricing policy, revenue equals cost and the utility breaks even.

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Chapter 6: Comparative Statics

6.1 The Jacobian is

𝐽 =

(𝐻𝐿 𝐽𝑇g𝐽g 0

)

where 𝐻𝐿 is the Hessian of the Lagrangean. We note that

∙ 𝐻𝐿(x0) is negative definite in the subspace 𝑇 = {x : 𝐽gx = 0 } (since x0 satisfiesthe conditions for a strict local maximum)

∙ 𝐽g has rank 𝑚 (since the constraints are regular).

Consider the system of equations(𝐻𝐿 𝐽𝑇g𝐽g 0

)(xy

)=

(00

)(6.28)

where x ∈ ℜ𝑛 and y ∈ ℜ𝑚. It can be decomposed into

𝐻𝐿x+ 𝐽𝑇g y = 0 (6.29)

𝐽gx = 0 (6.30)

Suppose x solves (6.30). Multiplying (6.29) by x𝑇 gives

x𝑇𝐻𝐿x+ x𝑇 𝐽𝑇g y = x𝑇𝐻𝐿x+ (𝐽gx)𝑇y = 0

But (6.30) implies that the second term is 0 and therefore x𝑇𝐻𝐿x = 0. Since 𝐻𝐿 ispositive definite on 𝑇 = {x : 𝐽gx = 0 }, we must have x = 0. Then (6.29) reduces to

𝐽𝑇g y = 0

Since 𝐽g has rank 𝑚, this has only the trivial solution y = 0 (Section 3.6.1). We haveshown that the system (6.38) has only the trivial solution (0,0). This implies that thematrix 𝐽 is nonsingular.


𝐿 = 𝑓(x)− 𝝀𝑇(g(x)− c)

By Corollary 6.1.1

∇𝑣(c) = 𝐷c𝐿 = 𝝀

6.3 Optimality implies

𝑓(x1, 𝜽1) ≥ 𝑓(x, 𝜽1) and 𝑓(x2, 𝜽2) ≥ 𝑓(x, 𝜽2) for every x ∈ 𝑋In particular

𝑓(x1, 𝜽1) ≥ 𝑓(x2, 𝜽1) and 𝑓(x2, 𝜽2) ≥ 𝑓(x1, 𝜽2)

Adding these inequalities

𝑓(x1, 𝜽1) + 𝑓(x2, 𝜽2) ≥ 𝑓(x2, 𝜽1) + 𝑓(x1, 𝜽2)

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Rearranging and using the bilinearity of 𝑓 gives

𝑓(x1 − x2, 𝜽1) ≥ 𝑓(x1 − x2, 𝜽2)

and

𝑓(x1 − x2, 𝜽1 − 𝜽2) ≥ 0

6.4 Let 𝑝1 denote the profit maximizing price with the cost function 𝑐1(𝑦) and let 𝑦1 bethe corresponding output. Similarly let 𝑝2 and 𝑦2 be the profit maximizing price andoutput when the costs are given by 𝑐2(𝑦).

With cost function 𝑐1, the firms profit is

Π = 𝑝𝑦 − 𝑐1(𝑦)

Since this is maximised at 𝑝1 and 𝑦1 (although the monopolist could have sold 𝑦2 atprice 𝑝2)

𝑝1𝑦1 − 𝑐1(𝑦1) ≥ 𝑝2𝑦2 − 𝑐1(𝑦2)

Rearranging

𝑝1𝑦1 − 𝑝2𝑦2 ≥ 𝑐1(𝑦1)− 𝑐1(𝑦2) (6.31)

The increase in revenue in moving from 𝑦2 to 𝑦1 is greater than the increase in cost.

Similarly

𝑝2𝑦2 − 𝑐2(𝑦2) ≥ 𝑝1𝑦1 − 𝑐2(𝑦1)

which can be rearranged to yield

𝑐2(𝑦1)− 𝑐2(𝑦2) ≥ 𝑝1𝑦1 − 𝑝2𝑦2Combining the previous inequality with (6.31) yields

𝑐2(𝑦1)− 𝑐2(𝑦2) ≥ 𝑐1(𝑦1)− 𝑐1(𝑦2) (6.32)

6.5 By Theorem 6.2

𝐷wΠ[w, 𝑝] = −x∗ and 𝐷𝑝Π[w, 𝑝] = 𝑦∗

and therefore

𝐷𝑝𝑦(𝑝,w) = 𝐷2𝑝𝑝Π(𝑝,w) ≥ 0

𝐷𝑤𝑖𝑥𝑖(𝑝,w) = −𝐷2𝑤𝑖𝑤𝑖Π(𝑝,w) ≤ 0

𝐷𝑤𝑗𝑥𝑖(𝑝,w) = −𝐷2𝑤𝑖𝑤𝑗Π(𝑝,w) = 𝐷𝑤𝑖𝑥𝑗(𝑝,w)

𝐷𝑝𝑥𝑖(𝑝,w) = −𝐷2𝑤𝑖𝑝Π(𝑝,w) = −𝐷𝑤𝑖𝑦(𝑝,w)

since Π is convex and therefore 𝐻Π(w, 𝑝) is symmetric (Theorem 4.2) and nonnegativedefinite (Proposition 4.1).

6.6 By Shephard’s lemma (6.17)

𝑥𝑖(𝑤, 𝑦) = 𝐷𝑤𝑖𝑐(𝑤, 𝑦)

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Using Young’s theorem (Theorem 4.2),

𝐷𝑦𝑥𝑖[w, 𝑦] = 𝐷2𝑤𝑖𝑦𝑐[w, 𝑦]

= 𝐷2𝑦𝑤𝑖𝑐[w, 𝑦]

= 𝐷𝑤𝑖𝐷𝑦𝑐[w, 𝑦]

Therefore

𝐷𝑦𝑥𝑖[w, 𝑦] ≥ 0 ⇐⇒ 𝐷𝑤𝑖𝐷𝑦𝑐[w, 𝑦] ≥ 0

6.7 The demand functions must satisfy the budget contraint identically, that is

𝑛∑𝑖=1

𝑝𝑖𝑥𝑖(p,𝑚) = 𝑚 for every p and 𝑚

Differentiating with respect to m

𝑛∑𝑖=1

𝑝𝑖𝐷𝑚𝑥𝑖[p,𝑚] = 1

This is the Engel aggregation condition, which simply states that any additional incomebe spent on some goods. Multiplying each term by 𝑥𝑖𝑚/(𝑥𝑖𝑚)

𝑛∑𝑖=1

𝑝𝑖𝑥𝑖𝑚

𝑚

𝑥𝑖(p,𝑚)𝐷𝑚𝑥𝑖[p,𝑚] = 1

the Engel aggregation condition can be written in elasticity form

𝑛∑𝑖=1

𝛼𝑖𝜂𝑖 = 1

where 𝛼𝑖 = 𝑝𝑖𝑥𝑖/𝑚 is the budget share of good 𝑖. On average, goods must have unitincome elasticities.

Differentiating the budget constraint with respect to 𝑝𝑗

𝑛∑𝑖=1

𝑝𝑖𝐷𝑝𝑗𝑥𝑖[p,𝑚] + 𝑥𝑗(𝑝,𝑚) = 0

This is the Cournot aggregation condition, which implies that an increase in the priceof 𝑝𝑗 is equivalent to a decrease in real income of 𝑥𝑗𝑑𝑝𝑗 . Multiplying each term in thesum by 𝑥𝑖/𝑥𝑖 gives

𝑛∑𝑖=1

𝑝𝑖𝑥𝑖𝑥𝑖𝐷𝑝𝑗𝑥𝑖[p,𝑚] = −𝑥𝑗

Multiplying through by 𝑝𝑗/𝑚

𝑛∑𝑖=1

𝑝𝑖𝑥𝑖𝑚

𝑝𝑗𝑥𝑖𝐷𝑝𝑗𝑥𝑖[p,𝑚] = −𝑝𝑗𝑥𝑗

𝑚

𝑛∑𝑖=1

𝛼𝑖𝜖𝑖𝑗 = −𝛼𝑗

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6.8 Supermodularity of Π(x, 𝑝,−w) follows from Exercises 2.50 and 2.51. To showstrictly increasing differences, consider two price vectors w2 ≥ w1

Π(x, 𝑝,−w1)−Π(x, 𝑝,−w2) =𝑛∑𝑖=1

(−𝑤1𝑖 )𝑥𝑖 −𝑛∑𝑖=1

(−𝑤2𝑖 )𝑥𝑖

=

𝑛∑𝑖=1

(𝑤2𝑖 − 𝑤1𝑖 )𝑥𝑖

Since w2 ≥ w1, w2 −w1 ≥ 0 and∑𝑛

𝑖=1(𝑤2𝑖 − 𝑤1𝑖 )𝑥𝑖 is strictly increasing in x.

6.9 For any 𝑝2 ≥ 𝑝1, 𝑦2 = 𝑓(𝑝2) ≤ 𝑓(𝑝1) = 𝑦1 and 𝑐(𝑦1, 𝜃) − 𝑐(𝑦2, 𝜃) is increasing in 𝜃and therefore −(𝑐(𝑓(𝑝2), 𝜃)− 𝑐(𝑓(𝑝1), 𝜃)) is increasing in 𝜃.

6.10 The firm’s optimization problem is

max𝑦∈ℜ+

𝜃𝑝𝑦 − 𝑐(𝑦)

The objective function

𝑓(𝑦, 𝑝, 𝜃) = 𝜃𝑝𝑦 − 𝑐(𝑦)

is

∙ supermodular in 𝑦 (Exercise 2.49)

∙ displays strictly increasing differences in (𝑦, 𝜃) since

𝑓(𝑦2, 𝑝, 𝜃)− 𝑓(𝑦1, 𝑝, 𝜃) = 𝜃𝑝(𝑦2 − 𝑦1)− (𝑐(𝑦2)− 𝑐(𝑦1))

is strictly increasing in 𝜃 for 𝑦2 > 𝑦1.

Therefore (Corollary 2.1.2), the firm’s output correspondence is strongly increasing andevery selection is increasing (Exercise 2.45). Therefore, the firm’s output increases asthe yield increases. It is analogous to an increase in the exogenous price.

6.11 With two factors, the Hessian is

𝐻𝑓 =

(𝑓11 𝑓12𝑓21 𝑓22

)

Therefore, its inverse is (Exercise 3.104)

𝐻−1𝑓 =

1

Δ

(𝑓22 −𝑓12−𝑓21 𝑓11

)

where Δ = 𝑓11𝑓22− 𝑓12𝑓21 ≥ 0 by the second-order condition. Therefore, the Jacobianof the demand functions is(

𝐷𝑤1𝑥1 𝐷𝑤2𝑥1𝐷𝑤1𝑥2 𝐷𝑤2𝑥2

)=

1

𝑝𝐻−1

𝑓 =1

𝑝Δ

(𝑓22 −𝑓12−𝑓21 𝑓11

)

Therefore

𝐷𝑤1𝑥2 = −𝑓21𝑝Δ

{< 0 if 𝑓21 > 0

≥ 0 otherwise

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