20. homogeneous and homothetic functionsboydj/mathecon/math20.pdf · 2020. 11. 16. ·...

20. Homogeneous and Homothetic

Functions

11/10/20

Homogeneous and homothetic functions are closely related, but are used in differentways in economics. We start with a look at homogeneity when the numerical valuesthemselves matter.

This happens with production functions. You should be familiar with the idea ofreturns to scale. Let’s focus on constant returns to scale. A production function exhibitsconstant returns to scale if, whenever we change all inputs in a given proportion, outputchanges by the same proportion.

In equations, let f : Rm+ → R be a production function. It exhibits constant returns

to scale iff(tx) = tf(x)

for every x ∈ Rm+ and t > 0.

2 MATH METHODS

20.1 Homogeneous Functions

Homogeneous Function. On Rm+ , a real-valued function is homogeneous of degree γ if

f(tx) = tγf(x)

for every x ∈ Rm+ and t > 0.

The degree of homogeneity can be negative, and need not be an integer. Constantreturns to scale functions are homogeneous of degree one. If γ > 1, homogeneousfunctions of degree γ have increasing returns to scale, and if 0 < γ < 1, homogeneousfunctions of degree γ have decreasing returns to scale.

◮ Example 20.1.1: Cobb-Douglas Production. Consider the Cobb-Douglas productionfunction defined on R

2+ by f(x, y) = Axαyβ with A,α, β > 0.

Thenf(tx, ty) = A(tx)α(ty)β = tα+βAxαyβ = tα+βf(x, y)

showing that f is homogeneous of degree α+β. If α+β < 1, f has decreasing returnsto scale, if α+β = 1, f has constant returns to scale, and if α+β > 1, f has increasingreturns to scale. ◭

20. HOMOGENEOUS AND HOMOTHETIC FUNCTIONS 3

20.2 Examples of Homogeneous Functions

Homogeneous functions arise in both consumer’s and producer’s optimization prob-lems. The cost, expenditure, and profit functions are homogeneous of degree one inprices. Indirect utility is homogeneous of degree zero in prices and income. Further,homogeneous production and utility functions are often used in empirical work.

The Cobb-Douglas functions u(x) = A∏m

i=1 xγi

i with γi > 0 are homogeneous ofdegree

∑i γi on R

m+ . The constant elasticity of substitution function f(x) = [δx−ρ

1 +(1−δ)x−ρ

2 ]ν/ρ for ν > 0 and ρ > −1, ρ 6= 0 is homogeneous of degree ν. The Leontieffunction u(x) = min xi is homogeneous of degree 1.

Any monomial,

a

m∏

i=1

xγi

i

is homogeneous of degree γ =∑m

i=1 γi on Rm+ . A sum of monomials of degree

γ is homogeneous of degree γ, the sum of monomials of differing degrees is nothomogeneous.

Additional examples of homogeneous functions include:

p·x, x3/2 + 3x1/2y,√x + y + z

x8 + 5x4y4 + 10x3y5 + 7xy7 + 13y8, Q(x) = xTAx.

The following are not homogeneous:

sin xyz, x2 + y3, 3xyz + 3x2z− 3xy

xy + yz + xz

x + z2, exp(x2 + y2).

Most quasi-linear utility functions, such as u(x) = x1 + x1/22 are not homogeneous of

any degree. The linear term means that they can only be homogeneous of degreeone, meaning that the function can only be homogeneous if the non-linear term is alsohomogeneous of degree one. Non-linear cases that are homogeneous of degree onerequire at least three goods. One example is

x1 + (x2x3)1/2

which is homogeneous of degree one.

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20.3 Cones and Homogeneity

Homogeneous functions require us to know f(tx) when we know f(x). That means thattx ∈ dom f whenever x ∈ dom f. Sets with that property are called cones, and conesare the natural domain of homogeneous functions.

Cone. Let V be a vector space. A set C ⊂ V is a cone if for every x ∈ C and t > 0,tx ∈ C. Equivalently, C is a cone if tC ⊂ C for all t > 0.

Cones include the positive orthant Rm+ , the strictly positive orthant Rm

++, any vectorsubspace of Rm, any ray in R

m and any set of non-negative linear combinations of acollection of vectors {x1, . . . , xi}. The set {(x, y) : x, y ≥ 0, y ≤ x} is as example of thelast as it can also be written {x = t1(1, 0) + t2(1, 1) : t1, t2 ≥ 0}.

We can redefine homogeneity to apply to functions defined on any cone.

Homogeneous Function. Let C be a cone in a vector space V. A function f : C → R ishomogeneous of degree γ if

f(tx) = tγf(x)

for every x ∈ Rm and t > 0.

Restricting the domain of a homogeneous function so that it is not all of Rm allowsus to expand the notation of homogeneous functions to negative degrees by avoidingdivision by zero.

E.g., 1/‖x‖2 is homogeneous of degree −1 on the cone Rm++. Restricting the domain

also allows us to consider f(x)/g(x) where f is homogeneous of degree γ1 and g ishomogeneous of degree γ2, both on R

m++. The quotient is homogeneous of degree

γ1 − γ2 on Rm++.


20.4 Derivatives of Homogeneous Functions

When a function is homogeneous, homogeneity is inherited by the derivative, but withthe degree reduced by one.

Theorem 20.4.1. Let C ⊂ Rm be an open cone and f : C → R is C1 and homogeneous

of degree γ. Then the derivative Df is homogeneous of degree (γ− 1).

Proof. By homogeneity, f(tx) = tγf(x). Taking the x derivative of that equation, weobtain tDf(tx) = tγDf(x). Dividing by t, we find that Df(tx) = t(γ−1) Df(x).

We assumed C was open so that Df could be defined at all points of C.The converse fails. If Df is homogeneous of degree β, we cannot conclude that f is

homogeneous of degree (β+1). For example, let m = 2 and consider f(x) = 1+x1x2,which is not homogeneous of any degree.

A quick calculation shows that (Dxf)(x) = (x2, x1) which is homogeneous of degreeone, even though f is not homogeneous. Fortunately, addition of a constant is the mainthing that goes wrong with the converse when Df is homogeneous of degree β forβ 6= −1.

The case β = −1 can suffer from two other types of complications. The first involveslogarithmic functions. Suppose f(x) = b lnφ(x) where φ is homogeneous of degreeone with φ > 0. Then

Df =b

φ(x)Dφ(x),

which is homogeneous of degree minus one.The β = −1 case has a second type of complication when there is more than one

variable. In Rm with m > 1, functions can be homogeneous of degree zero without

being constant. One such function is g(x) = x1/(x1 + x2). Its derivative is

Dg =1

(x1 + x2)2(x2,−x1) ,

which is clearly homogeneous of degree minus one.

6 MATH METHODS

20.5 Homogeneity and Indifference Curves

The fact that derivatives of homogeneous functions are also homogeneous has conse-quences for the shape of indifference curves and isoquants. It implies that the marginalrate of substitution and marginal rate of technical substitution are constant along raysthrough the origin. That is, they are homogeneous of degree zero.

To see this letu be homogeneous of degreeγ and consider MRSij = (∂u/∂xi)/

(∂u/∂xj).We calculate

MRSij(tx) =∂u∂xi

(tx)∂u∂xj

(tx)=

tγ−1 ∂u∂xi

(x)

tγ−1 ∂u∂xj

(x)=

∂u∂xi

(x)∂u∂xj

(x)= MRSij(x).

Of course, the calculation for the marginal rate of technical substitution is essentiallythe same.

Consequences of this include that facts that income expansion paths and scale ex-pansion paths are rays through the origin whenever the original production or utilityfunction is homogeneous.


20.6 Euler’s Theorem

The second important property of homogeneous functions is given by Euler’s Theorem.

Euler’s Theorem. Let f : Rm++ → R be C1. Then f is homogeneous of degree γ if and

only if[

Dxf(x)]

x = γf(x), that is

m∑

i=1

xi∂f

∂xi(x) = γf(x).

Proof. Define ϕ(t) = f(tx). By the chain rule, dϕ/dt =[

Df(tx)]

x.If f is homogeneous of degree γ, we can also write ϕ(t) = tγf(x) so that dϕ/dt =

γtγ−1f(x). Setting t = 1 we obtain (Df)·x = γf(x).Conversely, if

[

Dxf(x)]

x = γf(x),

dϕ

dt=

[

Dxf(tx)]

x =1

t

[

Dxf(tx)]

(tx) =γf(tx)

t=

γϕ(t)

t.

It follows thatd(lnϕ)

dt=

1

ϕ

dϕ

dt=

γ

t.

Integrating from t = 1 to t, we obtain lnϕ(t) − lnϕ(1) = γ ln t. Taking the expo-nential yields ϕ(t) = ϕ(1)tγ. This can be rewritten f(tx) = tγf(x), showing that f ishomogeneous of degree γ.

Gradients and Euler’s Theorem. Euler’s Theorem is sometimes written using the gradient,∇f = DfT , which is a column vector. In that case, the condition can be writtenx·∇xf = γf(x). Either way, it means that

∑

i

xi∂f

∂xi= γf(x).

8 MATH METHODS

20.7 Wicksteed’s Theorem

Suppose f is a constant returns to scale production function. If the firm is a price-takerin both input and output markets, profits are pf(x) −w·x where p is the output priceand w the vector of factor prices.

To maximize profit we set its derivative equal to zero.

pDf(x∗) = wT .

This tells us that for each good i, the value of marginal profit must be equal to the factorprice:

p∂f

∂xi= wi.

What about profits? Euler’s Theorem provides the answer. By Euler’s Theorem,

f(x) =∑

i

xi∂f

∂x i.

We multiply by p to obtain

pf(x) =∑

i

xip∂f

∂x i.

Then substitute the first order conditions to find

pf(x) =∑

i

xiwi = w·x,

showing that revenue equals cost and that profit is zero.If the production function was homogeneous of degree γ, 0 < γ < 1, the same

argument would show profit is positive, while if γ > 1, profit would be negative.


20.8 Homogeneity and Monotonicity

We define three types of monotonicity for real-valued functions of m variables.

Monotonicity. A real-valued function f defined on a subset of Rm is monotonic if

f(x) ≥ f(y) whenever x ≥ y. It is strictly monotonic is f(x) > f(y) whenever x ≫ y.Finally, it is strongly monotonic if f(x) > f(y) whenever x > y.

For functions on R, strong and strict monotonicity are the same.We saw earlier that the homogeneous of degree zero function g(x) = x1/(x1 + x2)

is not monotonic. This is normal for such functions. Indeed, Euler’s Theorem can beused to show that functions that are homogeneous of degree zero cannot be monotonicwhen there are two or more variables.

Theorem 20.8.1. Any function f ∈ C1(Rm++) for m > 1 that is homogeneous of degree

zero is not monotonic.

Proof. We prove this by applying Euler’s Theorem,

0 =∑

i

xi∂f

∂xi.

If every ∂f/∂xi > 0, the right-hand side is positive. This is impossible as it must bezero, so there has to be j with ∂f/∂xj < 0,. The function f cannot be monotonic.

However, it is possible to combine a logarithmic form with a homogeneous of degreezero form to get an increasing function with a derivative that is homogeneous of degreeminus one. The function

f(x1, x2) =x1

x1 + x2+ ln(x1 + x2)

is such a case.

10 MATH METHODS

20.9 Cardinal Functions vs. Ordinal Functions

The terms cardinal and ordinal are often used when discussing utility functions. How-ever, they are not properties of the functions themselves, but of our interpretationof them. When we say a production function is cardinal, we mean that the outputnumbers have direct meaning. Two tons of steel and three tons of steel are differentthings.

In contrast, with utility functions what is important is not the particular level of utility,but rather comparisons of utility levels—which is more and which is less.

To that end, we say that utility functions u, v : X → R are equivalent utility functionsif u(x) ≥ u(y) if and only if v(x) ≥ v(y) for all x,y ∈ X.

Theorem 20.9.1. Equivalence between utility functions from X → R is an equivalencerelation.

Proof. The relation is reflexive. A function u : X → R is equivalent to itself becauseu(x) ≥ u(y) if and only if u(x) ≥ u(y). Yes, that was pretty trivial.

The relation is symmetric. If u is equivalent to v, then v is trivially equivalent to u.Here u(x) ≥ u(y) if and only if v(x) ≥ v(y), so v(x) ≥ v(y) if and only if u(x) ≥ u(y).

Finally, the relation is transitive. This is barely harder to show. If u is equivalent tov and v is equivalent to a function w then u(x) ≥ u(y) if and only if v(x) ≥ v(y) ifand only if w(x) ≥ w(y), so u(x) ≥ u(y) if and only if w(x) ≥ w(y), showing that u isequivalent to w. This proves transitivity.

A property of a function f : X → R is ordinal if it shared by all equivalent utilityfunctions.


20.10 Two Types of Equivalence

You’ll notice that this treatment is slightly different from that in Simon and Blume(section 20.3) with an inequality rather than an equals sign. The reason is that if weused equality, as they do, the utility functions u(x, y) = x + y and v(x, y) = −x − ywould be equivalent functions. However, as utility functions, they are completelydifferent. For u, more is better, for v, more is worse. Our definition maintains the senseof utility.

If two utility functions are equivalent in our sense, they are equivalent in the Simonand Blume sense. As the above example shows, the converse is false.

Theorem 20.10.1. If u and v are equivalent utility functions on X, then u(x) = u(y) ifand only if v(x) = v(y) for all x,y ∈ X.

Proof. Now u(x) = u(y) if and only if both u(x) ≥ u(y) and u(y) ≥ u(x). Byequivalence, that holds if and only if both v(x) ≥ v(y) and v(y) ≥ v(x). The last pair ofstatements hold if and only if v(x) = v(y).

12 MATH METHODS

20.11 Monotonic Transformations

Monotonic transformations will allow us to characterize all equivalent utility functions.

Monotonic Transformation. Let I be an interval in the real line. A function φ : I → R

is a monotonic transformation of I if φ is strictly increasing on I. Moreover, if f is areal-valued function of m variables and φ is a monotonic transformation on its image,we say

φ ◦ f : x 7→ φ(

f(x))

is a monotonic transformation of f.

If φ is differentiable with φ′ > 0, then φ is a monotonic transformation.


20.12 Monotonic Transformations and Utility Equivalence

One important result is that two utility functions are equivalent if and only if there aremonotonic transformations of each into the other.

Theorem 20.12.1. Let u, v : X → R. Then u and v are equivalent utility functions on X

if and only if there is a monotonic transformation φ with u = φ ◦ v.

Proof. If: Suppose u = φ ◦ v. Because φ is strictly increasing, v(x) ≥ v(y) if and onlyif u(x) = φ

(

v(x))

≥ φ(

v(y))

= u(x), showing that the two functions are equivalent.Only If: Suppose u and v are equivalent. Let u ∈ ranu. There is x ∈ X with

u = u(x). Define φ(u) = v(x).The function v is well-defined.1 If we had picked any other y with u(y) = u, we

would have v(x) = v(y) by Theorem 20.10.1. This shows that the definition of φ(u)depends only on u, not on our choice of x with u(x) = u.

Now v(x) = φ(

u(x))

for all x ∈ X. We need only show that φ is increasing.Suppose u0 < u1. Take xi with u(xi) = ui, i = 0, 1. Then u(x1) > u(x0). By thedefinition of equivalence, v(x1) ≥ v(x0). By Theorem 20.10.1, v(x1) = v(x0) if and onlyif u(x0) = u(x1). Since u(x0) 6= u(x1), v(x0) 6= v(x1), showing that v(x1) > v(x0). Thenφ(u1) = v(x1) > v(x0) = φ(u0). This shows that φ is increasing and completes theproof.

Thus on R+, the functions x, x2, x7, ex, and ln x are all monotonic transformationsof f(x) = x, and so all equivalent. Similarly, xy, xy + (xy)4, log xy, and (xy)1/2 are allequivalent on R

2+.

When utility is differential, one property preserved by monotonic differentiable trans-formations is the marginal rate of substitution. Suppose v(x) = φ

(

u(x))

. Then by theChain Rule,

MRSvij =

∂v∂xi

∂v∂xj

=φ′ ∂u

∂xi

φ′ ∂u∂xj

=∂u∂xi

∂u∂xj

= MRSuij .

1 “Well-defined” is a mathematics term meaning that the definition is unambiguous.

14 MATH METHODS

20.13 Homothetic Functions

Homothetic functions are the ordinal equivalent of homogeneous functions.

Homothetic Function. Let C be a cone. A function f : C → R is homothetic if for everyx,y ∈ C and t > 0, f(x) ≥ f(y) if and only if f(tx) ≥ f(ty).

One consequence of the definition of homotheticity is that f is equivalent to g definedby g(x) = f(tx).

Any homogeneous utility function is also homothetic. And since homotheticity isan ordinal property, any increasing transformation of a homogeneous utility function ishomothetic too.

However, not all homothetic preferences have a homogeneous utility representation.Lexicographic preferences are homothetic, but cannot be represented by any utilityfunction—homogeneous or otherwise. If we require that preferences are also mono-tonic and continuous, we can represent homothetic preferences by a homogeneousutility function.

For a utility function, homotheticity means that preferences are invariant under scalarmultiplication in the sense that the set of indifference curves is unchanged when allconsumption bundles are multiplied by the same positive number. More precisely,preferences are invariant under homothetic transformations centered on the origin.

Homothetic preferences include commonly used functional forms such as Cobb-Douglas utility and constant elasticity of substitution utility.

One special type of homothetic utility is homogeneous utility, where multiplyingthe consumption bundle by a scalar multiplies utility by some power of that scalar.The Homothetic Representation Theorem will show that monotonic continuous andhomothetic preferences can be represented by a homogeneous utility function.

Homotheticity in economics is based on comparing positive scalar multiples of vec-tors. By restricting our attention to consumption sets that are cones, we ensure thatscalar multiplication is always possible. Such scaling preserves the shapes of objects,including indifference surfaces. It only changes their scale.


20.14 Homothetic and Homogeneous Functions

One remarkable fact is that every continuous homogeneous utility functions is a mono-tonic transformation of a homogeneous utility function.

The key part of the theorem is separated in a lemma, as it is of general use.

Lemma 20.14.1. Suppose x∗ ∈ Rm++ and u is continuous and increasing on R

m+ . If

x ∈ Rm+ , there is a unique α ≥ 0 with u(αx∗) = u(x).

Proof. Let x ∈ Rm+ . Choose α such that αx∗ ≫ x ≥ 0 and define A+ = {α ∈ [0, α] :

u(αx∗) ≥ u(x)} and A− = {α ∈ [0, α] : u(x) ≥ u(αx∗)}. By continuity, both A+ andA− are closed sets. Further, they obey A+ ∪ A− ⊃ [0, α]. Now 0 ∈ A− and α ∈ A+,so both are non-empty. Since intervals are connected, A+ ∩ A− is non-empty. Byconstruction, u(x) = u(αx∗) whenever α ∈ (A+ ∩A−). Transitivity and the fact that uis increasing imply there is only one such α.

16 MATH METHODS

20.15 Homothetic Representation Theorem

We are now ready to prove the Homothetic Representation Theorem.

Homothetic Representation Theorem. Supposeu : Rm+ → R is increasing, homothetic,

and continuous on Rm+ . Then u(x) is a monotonic transformation of a homogeneous of

degree one function v.

Proof of Theorem. Apply Lemma 20.14.1 to x∗ = e. Then for every x ∈ Rm+ , there

is a α ≥ 0 with u(αe) = u(x). In that case, define v(x) = α, so that u(

v(x)e)

= u(x).

b x

bαe

b αe

Figure 20.15.1: The position of the intersection of the indifference curve with the diagonaldetermines the unique α with u(x) = u(αe).

Now v(x) ≥ v(y) if and only if

u(x) = u(

v(x)e)

≥ u(

v(y)e)

= v(y).

It follows that v is utility equivalent to u.We know u(x) = u

(

v(x)e)

. By homotheticity of u, u(tx) = u(

tv(x)e)

. But u(tx) =u(

v(tx)e)

, sou(

tv(x)e)

= u(

v(tx)e)

. (20.15.1)

so tv(x) = v(tx) because Lemma 20.14.1 shows that equation (20.15.1) has a uniquesolution. It follows that v is homogeneous of degree one.

Since u and v are equivalent, by Theorem 20.12.1, there is a monotonic transforma-tion φ with u(x) = (φ ◦ v)(x).

Although homothetic functions are related to homogeneous functions, there aredifferences. Nothing like Euler’s Theorem need hold for functions that are merelyhomothetic. To see this, consider f(x) =

∑ℓ αℓ ln xℓ. Then

[

Dxf(x)]

x =∑

ℓ αℓ, whichcannot be written in the desired form.


20.16 Homotheticity and Marginal Rates of Substitution

When preferences are described by a smooth utility function, we can also describehomotheticity by saying that the marginal rate of substitution is the same anywhere ona given ray through the origin. The marginal rates of substitution are constant alongany ray through the origin. This means that the shape of the indifference curves arepreserved under scalar multiplication. All slopes remain unchanged.

Theorem 20.16.1. Let C ⊂ Rm be an open cone and f : C → R with Df(x) ≫ 0 on C.

Suppose f : C → R is homothetic and differentiable. Then MRSkℓ(x) = MRSkℓ(tx) forall t > 0 and x ∈ C.

The requirement that the marginal rate of substitution exists rules out cases wherewe are dividing by zero.

The proof is easy if we can write f = F ◦φ, F monotonic, φ homogeneous of degreeone, and both F and φ differentiable. Use the Chain Rule to compute MRSkℓ(x) =(∂φ/∂xk)

/

(∂φ/∂xℓ). By Theorem 20.4.1, both derivatives are homogeneous of degreezero, and so is the marginal rate of substitution. Below, we prove this in a slightly moregeneral fashion.

Proof. Consider the upper contour sets U(x) = {y : f(y) ≥ f(x)}. By homotheticity,U(tx) = tU(x). Homogeneity of the inner product then shows that if p supports U(x),it also supports U(tx).

Since f is differentiable, p and Df(x) are proportional by Theorem 21.22.1. ThusDf(x) and Df(tx) are also proportional. The factors of proportionality cancel whenconstructing the marginal rates of substitution, so MRSkℓ(x) = MRSkℓ(tx).

There is a converse, which we will state, but not prove.

Theorem 20.16.2. Suppose f : Rm++ → R is C2, Df ≫ 0, and MRSkℓ is homogeneous

of degree zero in x for every k and ℓ. Then f is homothetic.

Proof. A proof may be found in Lau.2 Lau uses the slightly weaker assumption thatthere is some j with ∂f/∂xj 6= 0, in which case the MRS condition must be restated towork around the fact that MRSkℓ may not be defined for all pairs k and ℓ. Lau doesnot do this, but the replacement for the MRS condition is that for all k and ℓ, there arehomogeneous of degree zero functions gkℓ such that ∂f/∂xk = gkℓ × (∂f/∂xℓ). WhenDf ≫ 0, this is equivalent to the marginal rates of substitution being homogeneous ofdegree zero in x.

2 Lemma 1 in Lau, Lawrence J. (1969) Duality and the structure of utility functions J. Econ. Theory, 1,374–396.

18 MATH METHODS

21. Concave and Quasiconcave Functions

11/12/20

NB: Problems 14, 18 and 22 from Chapter 19 and problems 1, 11, and 18 fromChapter 20 are due on Thursday, November 19.

Convex, concave, and similar functions arise naturally in economics. They includethe indirect utility function, cost function, expenditure function, and profit function.Moreover, concavity is usually assumed of utility as it ensures a diminishing (or at leastnon-increasing) marginal rate of substitution. It often applies to production.

21. CONCAVE AND QUASICONCAVE FUNCTIONS 19

21.1 Convex and Concave Functions

Recall that in any vector space, the line segment between x and y is given by ℓ(x,y) ={(1 − t)x + ty : 0 ≤ t ≤ 1}, and that a set S is convex if it contains ℓ(x,y) wheneverx,y ∈ S.

Convex and Concave Functions. Let S be a convex set.• A function f : S → R is convex if for all x,y ∈ S and 0 ≤ t ≤ 1, f

(

tx + (1 − t)y)

≤tf(x) + (1 − t)f(y).

• A function f : S → R is concave if for all x,y ∈ S and 0 ≤ t ≤ 1, f(

tx+ (1− t)y)

≥tf(x) + (1 − t)f(y).

When the inequalities are strict for 0 < t < 1, we say the function is strictly convexor strictly concave.

One consequence is that for convex functions, every chord of the graph of thefunction lies on or above the graph. For concave functions, every chord lies on orbelow the graph.

Convex

b

b

Concave

b

b

Figure 21.1.1: The left panel shows a convex function, where every chord connecting anytwo points of the graph lies above the graph.

The right panel illustrates a concave function, and every chord connecting any two pointsof the graph lies below the graph.

◮ Example 21.1.2: Convex Functions. Any linear function f(x) = p·x is both concaveand convex, as is the generic affine function f(x) = a + p·x. The function f(x) =∑m

ℓ=1 x2ℓ is convex while f(x) =

∑mℓ=1 x

1/2ℓ is concave. The function f(x) = ex is

convex, while f(x) = ln x is concave. ◭

20 MATH METHODS

21.2 Basic Properties of Concave and Convex Functions

Several easily established properties of convex and concave functions are collectedtogether without proof in Theorem 21.2.1.

Theorem 21.2.1.

1. A function f is (strictly) convex if and only if −f is (strictly) concave.2. A positive scalar multiple of a concave (convex) function is concave (convex).3. The sum of two concave (convex) functions is concave (convex).4. If φ is concave (convex) and weakly increasing on R and f is a concave (convex)

function, then φ ◦ f is concave (convex).5. The pointwise limit of a sequence of concave (convex) functions is concave (con-

vex).6. The infimum (supremum) of a sequence of concave (convex) functions is concave

(convex).

Proof. You should be able to prove these yourself.


21.3 Support Property

Supporting Hyperplane. We say a vector p supports a set S at x0 if either p·x ≥ p·x0

for every x ∈ S or p·x ≤ p·x0 for every x in S.

Of course, the set H = {x : p·x = p·x0} is a hyperplane. So supporting the setmeans that the set is on one side of the hyperplane. Moreover, they necessarily touchat x0.

That means we can restate the definition in terms of half-spaces. A vector p supportsS at x0 if and only if S is contained in the one of the two half-spaces H+(p,p·x0) andH−(p,p·x0).

Let f : S → R be a function where S ⊂ Rm. The upper contour set is defined by

U(y) = {x ∈ S : f(x) ≥ f(y)}. The lower contour set is defined by L(y) = {x ∈ S :f(x) ≤ f(y)}.

◮ Example 21.3.1: A Convex Upper Contour Set. Figure 21.14.2 illustrates an uppercontour set for u : R2

+ → R given by u(x, y) = xy.

p

U(x0)

bx0

p

p·x=p·x

0

Figure 21.3.2: The shaded area is the upper contour set U(x0) for u(x, y) = xy. The half-space H+(p, x0) is the hatched area when x0 = (1, 1) and p = (1, 1). The vector p supportsU(x0) at x0.

◭

22 MATH METHODS

21.4 Support Property Theorem

The Support Property Theorem shows that a differentiable function f is concave if andonly if Df(x0) supports the upper contour set at x0 for every x0 ∈ dom f. It also showsthat f is convex if and only if Df(x0) supports the lower contour set at x0 for everyx0 ∈ dom f.

Support Property Theorem. Suppose f : U → R is C1 where U is an open convex setU ⊂ R

m. The function f is concave if and only if

f(y) ≤ f(x) + Df(x)(y − x) (21.4.1)

for all x,y ∈ U. The function f is convex if and only if

f(y) ≥ f(x) + Df(x)(y − x) (21.4.2)

for all x,y ∈ U.

Proof (Only If). First suppose f is concave and take ε with 0 < ε < 1.

f(

x + ε(y − x))

= f(

(1 − ε)x + εy)

≥ εf(y) + (1 − ε)f(x).

We can rearrange to obtain

f(

x + ε(y − x))

− f(x) ≥ ε[

f(y) − f(x)]

.

Dividing by ε > 0 and letting ε → 0 yields

Df(x)(y− x) ≥ f(y) − f(x).

When f is convex, the only change to this part of the proof is that all three inequalitiesmust be reversed.

Proof Continues...


21.5 Support Property, continued

Proof (If). Now suppose the inequality

f(y) ≤ f(x) + Df(x)(y − x) (21.4.1)

is satisfied for all x,y ∈ U. Replace x by x′ = x + (1 − α)(y − x) = αx + (1 − α)ywhere 0 ≤ α ≤ 1. Then

f(y) ≤ f(x′) + αDf(x′)(y − x) ≤ f(x′) − αDf(x′)(x− y). (21.5.3)

Rewrite the support equation (21.4.1) by replacing x with x′ and y with x. This yields

f(x) ≤ f(x′) + (1 − α)Df(x′)(x− y). (21.5.4)

Then multiply equation (21.5.3) by (1 − α), equation (21.5.4) by α, and add themtogether. The Df(x′) terms cancel, leaving

αf(x) + (1 − α)f(y) ≤ f(x′) = f(

αx + (1 − α)y)

establishing concavity. The proof for the convex case is the same, but with everyinequality reversed.

One consequence of the Support Property Theorem is that if f is concave, the Df(x0)supports the upper contour set U(x0) at x0. To see this, suppose f(y) ≥ f(x0). Thenequation (21.4.1) implies

0 ≤ f(y) − f(x0) ≤ Df(x0)(y − x0),

soDf(x0)y ≥ Df(x0)x0. This impliesU(x0) ⊂ H+(

Df(x0), x0

)

as in Figure Fig:SupportedSet.

24 MATH METHODS

21.6 The Support Property in R

By the Support Property Theorem, a differentiable function is concave if and only ifequation (21.4.1) holds. This important inequality can be taken as the definition ofconcavity for differentiable functions.

When f is a function on a subset of the real line, we can use the right-hand side ofequation (21.4.1) to define a line, y = f(x0) + f′(x0)(x − x0). This line is tangent tothe graph of f at the point

(

x0, f(x0))

, and the graph of the function is in the lowerhalf-space that the tangent line defines. The tangent line supports the graph of f in thesense that the graph lies within one of the half-spaces defined by the tangent.

The tangent line supports both the graph and subgraph of f at(

x0, f(x0))

. Thesubgraph is sub f{(x, y) : f(x) ≤ y} at

(

x0, f(x0))

.

x

y

p

b

−f∗(p)

sub f

x0

f

Figure 21.6.1: The tangent line at x0 has the equation y = f(x0) + f′(x0)(x− x0). Because fis concave, the tangent line supports the subgraph of f. The graph is never above the tangentline and touches it at

(

x0, f(x0))

. Let p = f′(x0). The vector p = (p,−1) is perpendicularto the tangent line. Its vertical intercept is the negative of the concave conjugate functionf∗(p) = px0 − f(x0).


21.7 Support Property in Rm

In Figure 21.6.1 p = f′(x0) is the slope of the tangent line. We now rewrite the equationof the tangent in a way that that shows it is a hyperplane in R

2. Thus

(p,−1)(

xy

)

= px0 − f(x0). (21.7.5)

The vector p = (p,−1) is perpendicular to the tangent line.The right-hand side of equation (21.7.5) is not zero unless tangent goes through the

origin. It tells us how much the tangent line is offset from the origin. That value is calledthe concave conjugate function and is denoted f∗(p) = px0 − f(x0) when p = f′(x0). Infact, f∗(p) is the negative of the vertical intercept of the tangent line. The equation ofthe tangent line then becomes

(p,−1)(

xy

)

= f∗(p)

and the supergradient inequality is

(p,−1)(

xf(x)

)

≥ f∗(p). (21.7.6)

More generally, if f is a function of m variables, we can consider its graph in Rm+1,

and the picture is much the same.

(Df(x0),−1)(

xf(x)

)

≤ (Df(x0),−1)(

x0

f(x0)

)

where p = Df(x0).

26 MATH METHODS

21.8 Supporting Non-Differentiable Functions

We can often support concave functions in the same way even if they aren’t differen-tiable. However, there may be more than one vector p that supports them in such acase.

x

y

b

−f∗(p)

x0

f

Figure 21.8.1: As you can see, there are many lines that support the function f at(

x0, f(x0))

.The function has slope +1 to the left of x0 and −1 to the right. Any line with a slope between−1 and +1 that is tangent to the graph of f at

(

x0, f(x0))

will satisfy the support property.

Supergradient and Subgradient. If f is a concave function on a convex set U ⊂ Rm, and

p ∈ Rm satisfies

f(y) ≤ f(x) + p·(y− x), (21.8.7)

for all x,y ∈ U, we call p a supergradient of f at x. It is a type of generalized derivative.There’s also one for convex functions, called a subgradient. A vector p is a subgradientof f at x if

f(y) ≥ f(x) + p·(y− x), (21.8.8)

for all x,y ∈ U.

The Support Function Theorem tells us that when f is differentiable and concave(convex) Df(x0) is the only supergradient (subgradient) of f at x0.


21.9 Supergradients and Optimization

When f is concave (convex) the first order conditions are not only necessary for anoptimum, they are sufficient too.

Theorem 21.9.1. Let U ⊂ Rm be convex.

(a) Suppose 0 is a supergradient of f at x∗. Then x∗ is a global maximizer of f overU. In particular, this applies if f is differentiable and Df(x∗) = 0.

(b) Suppose 0 is a subgradient of f at x∗. Then x∗ is a global minimizer of f over U.In particular, this applies if f is differentiable and Df(x∗) = 0.

Proof. Setting p = 0 in the supergradient inequality (21.8.7), yields f(y) ≤ f(x∗) forall y ∈ U.

This even works on some concave functions that aren’t differentiable, as in Figure21.8.1 where a horizontal line (p = 0) supports the function at x0. That means x0 is amaximum, as is also clear from the graph.

This result is extremely useful in economics, as we are faced with a convex feasibleset U and a concave utility function we wish to maximize or convex function suchas w·z needing to be minimized. In either case, the first order necessary conditionsbecome sufficient for optimization.

We can go a bit further than this for differentiable functions.

Theorem 21.9.2. Let U ⊂ Rm be convex.

(a) Suppose f is a concave C1 function on U and x∗ obeys Df(x∗)(y− x∗) ≤ 0 for ally ∈ U. Then x∗ is a global maximizer of f on U.

(b) Suppose f is a convex C1 function on U and x∗ obeys Df(x∗)(y − x∗) ≥ 0 for ally ∈ U. Then x∗ is a global minimizer of f on U.

Proof. For the concave case, if Df(x∗)(y − x∗) ≤ 0, then f(y) ≤ f(x∗) + Df(x∗)(y −x∗) ≤ f(x∗). The convex case is similar.

◮ Example 21.9.3: Decreasing Functions. Suppose f is a decreasing C1 function of onevariable on an interval [a, b]. Then f′(a)(x−a) ≤ 0 for all x ∈ [a, b] because x−a ≥ 0and f′(a) ≤ 0. It follows that a is a maximizer of f on [a, b] ◭

It works in Rm too. Here’s an example for R2.

◮ Example 21.9.4: Global Minimum via Support. Let f(x, y) = −x2 − y2 on the set[0, 4] × [0, 3] ⊂ R

2. Then Df = (−2x,−2y). We will show that x∗ = (4, 3) is aminimizer. Now Df(x∗) = (−8,−6) and Df(x∗)(x−x∗) = (−8,−6)·(x−4, y−3). Hereboth x ≤ 4 and y ≤ 3, so Df(x∗)(x− x∗) ≥ 0, showing that x∗ is a global minimum. ◭

28 MATH METHODS

21.10 Hessian Convexity Tests: Necessity

The support property can be used to show that the Hessian D2f(x) is negative semidef-inite when f is concave and positive semidefinite when f is convex.

Theorem 21.10.1. Suppose f : U → R is C2 on an open convex set U ⊂ Rm. If f is

concave, then the Hessian D2f(x) is negative semidefinite for all x ∈ U. If f is convex,then the Hessian D2f(x) is positive semidefinite for all x ∈ U.

Proof. We will prove the concave case, the convex case is similar, with inequalitiesreversed. By the Support Property Theorem (equation 21.4.1)

f(y) ≤ f(x) + Df(x)(y− x) = f(x) −Df(x)(x− y)

reversing x and y in equation (21.4.1) yields

f(x) ≤ f(y) + Df(y)(x− y).

Adding the equations together and simplifying, we obtain

0 ≤(

Df(y) −Df(x))

(x− y).

Now set y = x + h. We then have

(

Df(x + h) −Df(x))

h ≤ 0.

Divide by ‖h‖ and let ‖h‖ → 0, obtaining hTD2f(x)h ≤ 0. In other words, D2f(x) isnegative semidefinite for all x ∈ U.


21.11 Hessian Convexity Tests: Necessity and Sufficiency

When f is C2, we can use the Hessian matrix and Taylor’s formula to determine whetherf is concave or convex.

Theorem 21.11.1. Suppose f : U → R is C2 on an open convex set U ⊂ Rm.

1. The function f is concave if and only if D2f(x) is negative semidefinite for all x ∈ U.2. If D2f(x) is negative definite for all x ∈ U, f is strictly concave.3. The function f is convex if and only if D2f(x) is positive semidefinite for all x ∈ U.4. If D2f(x) is positive definite for all x ∈ U, f is strictly convex.

Proof. Taylor’s formula tells us that

f(y) = f(x) + Df(x)(y− x) + (y− x)TD2f(z)(y − x). (21.11.9)

for some z on the line segment between x and y.(1) If D2f is negative semidefinite on U, this implies the support property, so f is

concave by the Support Property Theorem. Conversely, if f is concave, Proposition21.10.1 shows that D2f(x) is negative semidefinite on U.

(2) If f is negative definite on U, equation (21.11.9) yields f(y) < f(x) +Df(x)(y− x)for all y 6= x. Repeating the calculations in the Support Property Theorem for y 6= xand 0 < α < 1, shows that f is strictly concave.

Parts (3) and (4) are similar.

30 MATH METHODS

21.12 Convexity and Concavity: Determinant Tests

Recall the following definitions

Definite and Semidefinite Matrices. Recall that an L× L symmetric matrix A is

(a) Positive semidefinite if xTAx ≤ 0 for all x ∈ RL,

(b) Positive definite if xTAx < 0 for all x 6= 0,(c) Negative semidefinite if xTAx ≤ 0 for all x ∈ R

L,(d) Negative definite if xTAx > 0 for all x 6= 0, and(e) Indefinite if there are x and y with xTAx > 0 and yTAy < 0.

The support property can be used to show that D2f(x) is negative semidefinite whenf is concave and positive semidefinite when f is convex.

If A is a matrix, we will use Ak to denote a generic kth-order principal minor of A.One example of a non-leading principal minor for m = 6 and k = 3 is

∣

∣

∣

∣

∣

a11 a13 a16

a31 a33 a36

a61 a63 a66

∣

∣

∣

∣

∣

where the 2nd, 4th, and 5th rows and columns have been deleted. Keep in mind thatthere are usually many kth-order principal minors.

We can rewrite our results relating convexity and definiteness by applying the usualdeterminant tests to the Hessian. That yields the following theorem.

Theorem 21.12.1. Suppose f : U → R is C2 on an open convex set U ⊂ Rm. Let

H(x) = D2f(x) denote the Hessian of f.

1. The function f is convex if and only if every kth-order principal minor obeys

Hk(x) ≥ 0 for all x ∈ U.2. The function f is concave if and only if every kth-order principal minor obeys

(−1)kHk(x) ≥ 0 for all x ∈ U.3. Suppose the leading principal minors obey Hk(x) > 0 for k = 1, . . . ,m and all

x ∈ U. Then f is strictly convex.4. Suppose the leading principal minors obey (−1)kHk(x) > 0 for k = 1, . . . ,m and

all x ∈ U. Then f is strictly concave.


21.13 Using the Determinant Test

◮ Example 21.13.1: Cobb-Douglas Utility. Consider the Cobb-Douglas utility functionon R

2++ defined by u(x) = xα1 x

1−α2 with 0 < α < 1. The Hessian is

H(x) =

[

α(α− 1)xα−21 x1−α

2 α(1 − α)xα−11 x−α

2

α(1 − α)xα−11 x−α

2 (1 − α)(−α)xα1 x−α−12

]

.

Clearly, detH = 0. The two first-order minors are α(α − 1)xα−21 x1−α

2 < 0 and (1 −α)(−α)xα1 x

−α−12 < 0, which means that u is concave on R

2++. Concavity on R

2+ then

follows from continuity.Note that u is not strictly concave because αu(0) + (1 − α)u(e) = 1 − α = u

(

(1 −α)e

)

. ◭

◮Example 21.13.2: CES Utility. Another example is the constant elasticity of substitutionutility function u(x) = [x−ρ

1 + x−ρ2 ]−1/ρ where ρ > −1, ρ 6= 0. The Hessian is

H(x) = (1 + ρ)[x−ρ

1 + x−ρ2 ]−( 1+2ρ

ρ)

x2+ρ1 x2+ρ

2

( −x22 x1x2

x1x2 −x21

)

.

The two first-order minors are both negative, while the second-order minor is 0. Thusu is concave on R++. ◭

32 MATH METHODS

21.14 Example: Upper and Lower Contour Sets

Upper contour sets are convex for concave functions while lower contour sets areconvex for convex functions. It is often the case that the lower contour set of a concavefunction is not convex. This happens in the figure below.

◮ Example 21.14.1: Upper and Lower Contour Sets. The left side Figure 21.14.2 illus-trates an upper contour set for u : R2

+ → R given by u(x, y) = x1/3y2/3. The right sideshows the lower contour set for the same function. It is not convex.

U(y)

bx

Convex

L(y)

bx

Not Convex

Figure 21.14.2: The shaded area in the left panel is the upper contour set U(y) for u(x, y) =x1/3y2/3 and y = (1, 1).

The shaded area in right panel is the lower contour set for the same function. The dashedline segment would be in L(y) if L(y) were convex. The line segment goes outside L(y), sothe lower contour set is not convex.

◭


21.15 Convexity of Upper and Lower Contour Sets

In Example 21.14.2, the upper contour set was convex. This was no accident. Rather,it is a consequence of using a concave function.

Theorem 21.15.1. Let U ⊂ Rm be a convex set.

1. If f : U → R is a concave function, then the upper contour set U(y) is convex.2. If f : U → R is a convex function, then the lower contour set L(y) is convex.

Proof. Let x, x′ ∈ U(y). Then f(x), f(x′) ≥ f(y). By concavity of f

f(

(1 − t)x + tx′)

≥ (1 − t)f(x) + tf(x′) ≥ (1 − t)f(y) + tf(y) = f(y),

showing that the convex combination (1 − t)x + tx′ ∈ U(y) whenever x, x′ ∈ U(y).The proof of (2) is similar.

Contour Sets are Ordinal. Notice that U(x) and L(x) remain unchanged when a mono-tonic transformation is applied to f. This is because f(x) ≥ f(y) if and only if(φ ◦ f)(x) ≥ (φ ◦ f)(y) when φ is a monotonic transformation.

34 MATH METHODS

21.16 Quasiconvex and Quasiconcave Functions

Quasiconvexity and quasiconcavity are the ordinal versions of convexity and concavity.They are defined in terms of the upper and lower contour sets, respectively. We justsaw that upper and lower contour sets are unaffected by monotonic transformations,and hence ordinal.

Quasiconvexity and Quasiconcavity. Let f : S → R.

1. The function f is quasiconcave on S if and only if the upper contour set U(y) ={x ∈ S : f(x) ≥ f(y)} is a convex set for every y ∈ S.

2. The function f is quasiconvex on S if and only if the lower contour set L(x) ={x ∈ S : f(x) ≤ f(y)} is a convex set for every y ∈ S.

From the definition, it is easy to see that f is quasiconvex if and only if −f is quasi-concave.

There are other ways to characterize quasiconvexity and quasiconcavity. One is thefollowing.

Theorem 21.16.1. Let f : S → R where S is a convex set.

1. The function f is quasiconcave if and only if for all t obeying 0 ≤ t ≤ 1, f(tx +(1 − t)y) ≥ min{f(x), f(y)}.

2. The function f is quasiconvex if and only if for all t obeying 0 ≤ t ≤ 1, f(tx+ (1−t)y) ≤ max{f(x), f(y)}.

Of course, if f : S → R is quasiconcave or quasiconvex on S, S must be a convex setbecause if x′, x′′ ∈ S, convex combinations of x′ and x′′ are in either {x ∈ S : f(x) ≥min[f(x′), f(x′′)]} (f quasiconcave) or in {x ∈ S : f(x) ≤ max[f(x′), f(x′′)]} (f quasiconvex).

In Theorem 21.15.1 we saw that convex functions are quasiconvex and concavefunctions are quasiconcave. The relationship is only one-way. Indeed, when f : R → R,every monotone function is simultaneously quasiconvex and quasiconcave! In contrast,the only functions that are both convex and concave on R are the affine functions.


21.17 Quasiconvexity on the Real Line

For f defined on a convex subset of R, we can completely characterize the quasiconcaveand quasiconvex functions.

Theorem 21.17.1. Let I be an interval in R. Suppose f : I → R.

1. The function f is quasiconcave if and only f is either monotone or there exists anumber x∗ such that f is weakly increasing when x ≤ x∗ and weakly decreasingwhen x ≥ x∗.

2. The function f is quasiconvex if and only if f is either monotone or there exists anumber x∗ such that f is weakly decreasing when x ≤ x∗ and weakly increasingwhen x ≥ x∗.

Proof. It is enough to prove the quasiconcave case as −f is quasiconcave if f isquasiconvex.

We first prove sufficiency. As noted above, monotone functions are quasiconcave.We need only consider the case where x∗ exists. Let y ∈ I be given and let x1 and x2

be arbitrary points in the upper contour set U(y) with x1 < x2. We must show that theinterval [x1, x2] ⊂ U(y).

If x∗ is not in [x1, x2], then f is either always weakly decreasing or always weaklyincreasing on [x1, x2]. Either way, for any z ∈ [x1, x2], f(z) ≥ min{f(x1), f(x2)} ≥ f(y), soz ∈ U(y). This shows that [x1, x2] ⊂ U(y).

If x∗ ∈ [x1, x2], f is weakly increasing on [x1, x∗] and weakly decreasing on [x∗, x2].

Again, for any z ∈ [x1, x2], f(z) ≥ min{f(x1), f(x2)} ≥ f(y), so z ∈ U(y) and hence[x1, x2] ⊂ U(y). In other words, f is quasiconcave.

We now turn to necessity. Suppose that f is quasiconcave on I, but not monotone.We prove this part by contradiction. If x∗ does not exist, we can find x1, x2, and x3

in I such that x1 < x2 < x3 and f(x2) < min{f(x1), f(x3)}. If f(x1) ≤ f(x3), set y = x1,otherwise take y = x3. Then the upper contour set U(y) includes x1 and x3, but notx2. It is not convex, contradicting the fact that f is quasiconcave.

To sum up, quasiconcave functions defined on a real interval are either monotonic orsingle-peaked, while quasiconcave functions are either monotonic or single-troughed.Either the peak or trough may be an interval.

36 MATH METHODS

21.18 Strict Quasiconvexity and Quasiconcavity

One consequence of Proposition 21.16.1 is that a function is quasiconcave if for everyx,y with f(x) ≥ f(y) and every t, 0 ≤ t ≤ 1, we have f

(

(1 − t)x+ ty)

≥ f(y). We canuse this criterion for quasiconcavity to define strict quasiconcavity.

Strict Quasiconcavity and Strict Quasiconvexity. Let f : S → R where S is a convex set.

1. The function f is strictly quasiconcave on S if and only if for every x,y ∈ S withf(x) ≥ f(y) and every t with 0 < t < 1, we have f

(

(1 − t)x + ty)

> f(y).2. The function f : S → R is strictly quasiconvex if and only if for every x,y ∈ S

with f(x) ≥ f(y) and every t with 0 < t < 1, we have f(

(1 − t)x + ty)

< f(y).

It is easy to see that any increasing transformation of a convex (or quasiconvex) func-tion is quasiconvex and any increasing transformation of a concave (or quasiconcave)function is quasiconcave.

Not all quasiconcave functions are transformations of concave functions. Quasicon-cave functions that are a monotonic transformation of a concave function are calledconcavifiable. Similarly, a quasiconvex function that is a monotonic transformation of aconvex function is called convexifiable. There are quasiconcave functions that are notconcavifiable.1

◮ Example 21.18.1: Quasiconvex does Not Mean Convex. Consider f(x, y) = (x2 +y2)1/3. The function g(x, y) = x2 + y2 is convex (the Hessian is 2I2). Raising it to the1/3-power is an increasing transformation. Thus f is quasiconvex.

However, f is not convex. To see this, we compute f(0, 0) = 0 and f(1, 1) = 2. Thenf(t, t) = 2t2 < 2t = (1 − t)f(0, 0) + tf(1, 1) for 0 < t < 1. If it were convex, theleft-hand side would be larger than the right. Instead, it is smaller. ◭

A strictly quasiconcave (quasiconvex) function can have at most one maximizer(minimizer).

Theorem 21.18.2. Let f : S → R where S is a convex set. Suppose f is strictly qua-siconcave there are x0, x1 ∈ S so that for all y ∈ S, f(y) ≤ f(xi), i = 0, 1. Thenx0 = x1.

Proof. Since both xi maximize f over S, f(x1) = f(x0). If x0 6= x1, define x2 =12x0 + 1

2x1. By strict quasiconcavity, f(x2) > f(x0), showing that x0 is not a maximum.

This contradiction shows that x0 = x1.

A similar result holds for minima of a strictly quasiconvex function.

1 This has been known since de Finetti (1949). The problem of finding whether a given quasiconcavefunction is an increasing transformation of a concave function was posed (in terms of level sets) by Fenchel(1953). Although some results are known, it still does not have a completely satisfactory solution.


21.19 Support Property Theorem II

There are also support properties that characterize quasiconvex and quasiconcavefunctions.

Support Property Theorem II. Suppose f is C1 on U, an open convex subset of Rm.The function f is quasiconcave if and only if for all x,y ∈ U

f(y) ≥ f(x) implies[

Df(x)]

(y − x) ≥ 0. (21.19.10)

The function f is quasiconvex if and only if for all x,y ∈ U

f(y) ≥ f(x) implies[

Df(x)]

(y − x) ≤ 0.

then f is quasiconcave.

Proof (Only if). Here f is quasiconcave. Suppose f(y) ≥ f(x). Consider (1−ε)x+εy =x + ε(y − x) for 0 < ε < 1. By quasiconcavity,

f(

x + ε(y − x))

≥ f(x) sof(

x + ε(y − x))

− f(x)

ε≥ 0.

Let ε → 0 to obtain [Df(x)](y− x) ≥ 0, which establishes the result. (Only If)

Proof Continues...

38 MATH METHODS

21.20 Support Property Theorem II, Part II

Proof (If). Now suppose that for all x,y ∈ U; f(y) ≥ f(x) implies [Df(x)](y− x) ≥ 0.We must show that f is quasiconcave.

We do this by contradiction. If f is not quasiconcave, there are x0, x1 and 0 < t0 < 1with f(x1) ≥ f(x0) and f

(

(1 − t0)x0 + t0x1

)

< f(x0). Define

xt = (1 − t)x0 + tx1.

We can then write f(xt0 ) < f(x0).Take an interval I = [t1, t2] ⊂ [0, 1] with t0 ∈ I and f(xt) ≤ f(x0) for all t ∈ I, and

f(xt1) = f(xt2) = f(x0).

Now if t ∈ I with 0 < t < 1, f(x1) ≥ f(x0) ≥ f(xt). But then by equation (21.19.10),

Df(xt)(x0 − xt) ≥ 0 and Df(xt)(x1 − xt) ≥ 0. (21.20.11)

Of course, x0 − xt = t(x0 − x1) and x1 − xt = (1 − t)(x1 − x0). We substitute intoequation (21.20.11) to obtain

−tDf(xt)(x1 − x0) ≥ 0 and (1 − t)Df(xt)(x1 − x0) ≥ 0.

Since −t < 0 and (1 − t) > 0, we must have

Df(xt)(x1 − x0) = 0 (21.20.12)

for all t ∈ I, 0 < t < 1.Now 0 < f(x0) − f(xt0 ) = f(xt1

) − f(xt0 ), so by the Mean Value Theorem, there ist3 ∈ (t1, t0) with

0 < f(x0) − f(xt0) = Df(xt3 )(xt1− xt0) = (t0 − t1)Df(xt3 )(x1 − x0).

But this contradicts equation (21.20.12). That contradiction shows that f is quasicon-cave.


21.21 Maximization and Quasiconcavity

We can use Support Property II for f to show that any point solves a maximizationproblem.

Theorem 21.21.1. Suppose f is quasiconcave and C1 on a convex open set U and thatp = Df(x). Then x maximizes f(x) under the constraints p·x ≤ p·x and x ∈ U.

Proof. We prove this by contradiction. Suppose to the contrary there is x ∈ U withp·x ≤ p·x and f(x) > f(x). Then for ε > 0 small enough x − εp ∈ U (because U isopen) and f(x − εp) > f(x) (by continuity). By Support Property II, p·(x− εp) ≥ p·x.But then, p·x ≥ p·x + ε‖p‖2 > p·x. This contradicts our original supposition and soproves the theorem.

Since the constraint x ∈ U cannot bind when U is open, x maximizes f over all x withp·x ≤ p·x. By Theorem , hT [D2f(x)]h ≤ 0 for all h obeying p·h = 0 or equivalentlythe bordered Hessian

B =

[

0 pT

p D2f

]

has bordered principal minors that obey (−1)n−1An ≥ 0 for n = 3, . . . ,m + 1. Thisprovides a second-derivative test for quasiconcavity.

40 MATH METHODS

21.22 Uniqueness of Supports

When f is differentiable, we can show that the derivative is the only vector (up to scalarmultiplication) that can support the upper (or lower) contour set. More precisely, wesay p supports a set S at x0 if p·x ≥ p·x0 whenever x ∈ S. We apply this idea to theupper contour set {x : f(x) ≥ f(x0)} at x0. It is important to realize that this result doesnot require quasiconcavity. However, if the function is not quasiconcave, the uppercontour set may not have supports.

Theorem 21.22.1. Suppose f is differentiable at x0 with Df(x0) 6= 0 and that p supports{x : f(x) ≥ f(x0)} at x0. Then p = αDf(x0) for some α 6= 0.

Proof. Let z = Df − (p·Df/‖p‖2)p. Then p·z = 0 and Df·z = ‖Df‖2 −|p·Df|2/‖p‖2. By the Cauchy-Schwartz inequality, either p is proportional to Df (andwe are done) or Df·z > 0.

In the case where Df·z > 0, the first-order Taylor approximation f(x0) + αDf·z >f(x0) shows

f(x0 + αz) > f(x0)

for small α > 0. Continuity of f then yields

f(x0 + αz− εp) > f(x0)

for ε > 0 small. Apply Support Property II to obtain

p·x0 ≤ p·(x0 + αz− εp)

= p·x0 − ε‖p‖2

< p·x0.

This contradiction shows Df·z > 0 is impossible. Therefore p is proportional toDf(x0).


21.23 Bordered Hessian Test for Quasiconcavity

Since λp = Df, we can instead use the bordered Hessian with Df. The borderedprincipal minors are multiplied by λ2, so the signs are unchanged. This yields thefollowing theorem.

Theorem 21.23.1. Suppose f : U → R is a C2 function on some open convex setU ⊂ R

m with m > 1. Consider the bordered Hessian

H =

[

0 Df

DfT D2f

]

=

0 ∂f∂x1

. . . ∂f∂xm

∂f∂x1

∂2f∂x2

1. . . ∂2f

∂x1 ∂xm

......

. . ....

∂f∂xm

∂2f∂x1 ∂xm

. . . ∂2f∂x2

m.

1. If the bordered leading principal minors Hk obey (−1)n−1Hn > 0 on U forn = 3, . . . ,m + 1, then f is quasiconcave on U.

2. If all non-trivial bordered leading principal minors are negative on U, then f isquasiconvex on U.

3. If f is quasiconcave on U, then every kth order bordered principal minor Hk

obeys (−1)n−1Hn ≥ 0 on U for n = 3, . . . ,m + 1.4. If f is quasiconvex on U, then all non-trivial bordered principal minors are non-

positive on U.

Proof. For the quasiconcave case, see Arrow and Enthoven (1961). Applying theresult to −f yields the quasiconvex case.

November 13, 2020

Copyright c©2020 by John H. Boyd III: Department of Economics, Florida InternationalUniversity, Miami, FL 33199

20. homogeneous and homothetic functionsboydj/mathecon/math20.pdf · 2020. 11. 16. ·...

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