real analysis contents - moothathu.files.wordpress.com · real analysis t.k.subrahmonian moothathu...

REAL ANALYSIS

T.K.SUBRAHMONIAN MOOTHATHU

Contents

1. The space of real numbers 2

2. A few remarks about sets 5

3. Metric spaces: preliminaries 10

4. Continuous functions 15

5. Completeness 21

6. Compactness 23

7. Connectedness 27

8. Convergence of a real sequence 31

9. Convergence of a real series 38

10. Sequences of functions 43

11. Series of real functions 50

12. Differentiation: preliminaries 53

13. The Mean value theorem 57

14. LHospital’s rule and Taylor’s theorem 61

15. Integration: preliminaries and properties 66

16. Fundamental theorem of calculus and further properties of integration 72

17. Weierstrass approximation theorem 78

18. Convex functions 82

Very briefly speaking, we will discuss seven mutually overlapping C’s in our course: Convergence,

Continuity, Completeness, Compactness, Connectedness, Calculus, and Convexity.

For further reading: (i) T.M. Apostol, Mathematical Analysis, 2002 (chapters 1-9).

(ii) R.G. Bartle and D.R. Sherbert, Introduction to Real Analysis, 1992.

(iii) R.R. Goldberg, Methods of Real Analysis, 1976 (chapters 1-10).

(iv) W.J. Kaczor and M.T. Nowak, Problems in Mathematical Analysis, I - III, 2000, 2001, 2003.

(v) W. Rudin, Principles of Mathematical Analysis, 1976 (chapters 1-7, and half of chapter 8).1

2 T.K.SUBRAHMONIAN MOOTHATHU

1. The space of real numbers

We will suppose that the student has a preliminary knowledge of Real Analysis from the graduate

class (for instance, it will be assumed that the student knows what it means to say a sequence of

real numbers is bounded, or Cauchy, or convergent). The following topic is left to the student

for self-study: construction of real numbers from rational numbers (for instance, as the space of

equivalence classes of asymptotic Cauchy sequences in Q). We have the following fundamental fact

(see Theorem 1.19 of Rudin):

R is a totally ordered field having the least upper bound property, and Q is a subfield of R.Various terms appearing in the above statement are briefly explained below.

Definition: Let X be a nonempty set. An order relation ≤ on X is said to be a partial order if it

satisfies the following:

(i) (reflexivity) x ≤ x for every x ∈ X.

(ii) (antisymmetry) If x, y ∈ X are such that x ≤ y and y ≤ x, then x = y.

(iii) (transitivity) If x, y, z ∈ X are such that x ≤ y and y ≤ z, then x ≤ z.

If ≤ is a partial order on X, then (X,≤) is called a partially ordered set. A partial order ≤ on

X is said to be a total order (or a linear order) if for every x, y ∈ X we have either x ≤ y or y ≤ x.

If ≤ is a total order on X, then (X,≤) is called a totally ordered set.

Definition: Let (X,≤) be a totally ordered set. Let A ⊂ X be a nonempty set. We say A is bounded

above in X if there is b ∈ X with a ≤ b for every a ∈ A; and in this case we say b is an upper bound

of A in X. Moreover, an upper bound b of A in X is said to be a least upper bound (or supremum)

of A in X if b ≤ c for any upper bound c of A in X (and we write b = sup(A)). Similarly, we may

define the following notions for A: being bounded below in X, a lower bound of A in X (if A is

bounded below), and a greatest lower bound (or infimum) of A in X (written inf(A), if it exists).

We say (X,≤) has the least upper bound property if every nonempty set A ⊂ X which is bounded

above in X has a least upper bound in X.

Definition: (X,+, ·) is said to be a field if X is a nonempty set, and ‘+’ and ‘·’ are binary operations

(called addition and multiplication respectively) on X such that (i) (X,+) is an abelian group

with additive identity denoted as 0, (ii) (X \ {0}, ·) is an abelian group with multiplicative identity

denoted as 1, and necessarily 1 = 0, and (iii) multiplication distributes over addition, i.e., x(y+z) =

xy + xz for every x, y, z ∈ X (where we have suppressed the ‘·’ symbol). If (X,+, ·) is a field and

Y ⊂ X is such that (Y,+, ·) is a field with the same additive and multiplicative identities as that

of X, then Y is called a subfield of X.

Definition: (X,+, ·,≤) is called a totally ordered filed if (X,+, ·) is a field, (X,≤) is a totally ordered

set, and if the following two axioms are true: (i) if x, y, z ∈ X are with y < z (which means y ≤ z

and y = z), then x+ y < x+ z, and (ii) if x, y ∈ X are such that 0 < x and 0 < y, then 0 < xy.

REAL ANALYSIS 3

Definition: For A,B ⊂ R and x ∈ R, let A+B = {a+b : a ∈ A and b ∈ B}, A+x = {a+x : a ∈ A}and xA = {xa : a ∈ A}.

[101] (i) [Archimedean property of R] If x, y ∈ R and x > 0, then there is n ∈ N with nx > y.

(ii) For every real number ε > 0, there is n ∈ N with 1/n < ε.

(iii) If x > 0, then {mx : m ∈ Z} intersects every interval J ⊂ R having length > x.

(iv) Let a < b be reals. Then (a, b) ∩Q and (a, b) \Q are nonempty.

Proof. (i) If there is no such n ∈ N, then the set A = {nx : n ∈ N} must be bounded above by y,

and hence z := sup(A) must exists in R by the least upper bound property of R. As x > 0, we

have z − x < z, and hence z − x cannot be an upper bound for A. Therefore there is n ∈ N with

nx > z − x. But then (n+ 1)x > z, a contradiction to the choice of z.

(ii) Apply (i) with x = ε and y = 1.

(iii) Obvious, but make sure that it is obvious.

(iv) By (i), there is n ∈ N with n(b − a) > 1, or equivalently 1/n < b − a. Now note by (iii) that

both X := {k/n : k ∈ Z} and X +√2 must intersect (a, b). �

Exercise-1: (i) sup(−A) = − inf(A) and inf(−A) = − sup(A) for A ⊂ R.(ii) sup{xn+yn : n ∈ N} ≤ supn∈N xn+supn∈N yn and inf{xn+yn : n ∈ N} ≥ infn∈N xn+infn∈N yn

for sequences (xn) and (yn) in R.(iii) The set A := {x ∈ Q : x2 < 2} is nonempty and bounded above in Q, but A does not have a

least upper bound in Q. Thus Q does not have the least upper bound property.

[Hint : (iii) If y := sup(A) exists in Q, then either y2 < 2 or y2 > 2. Then by [101], there is n ∈ Nwith (y + 1

n)2 ≤ y2 + 2y+1

n < 2 or (y − 1n)

2 ≥ y2 − 2yn > 2 respectively, a contradiction.]

Definition: Consider a positive real number x, i.e., x ∈ (0,∞).

(i) For each n ∈ N, there is a unique real number y > 0 such that yn = x (Theorem 1.21 of Rudin),

and we denote this y as x1/n (and y is called the unique positive nth root of x).

(ii) For k, n ∈ N, we define xk/n := (x1/n)k = (xk)1/n.

(iii) For a ∈ (0,∞), the element xa is defined as sup{xk/n : k, n ∈ N and k/n < a}. This definitionis meaningful because the set A := {xk/n : k, n ∈ N and k/n < a} is bounded above (by 1 if x ≤ 1,

and by xr for any rational r > a if x > 1), and hence has a supremum by the least upper bound

property of R.(iv) x0 = 1 by convention and x−a := 1/xa for a ∈ (0,∞).

Remark: (i) 00 is undefined. (ii) If x < 0, then even to define x1/2, we need to go to the field C.

While dealing with inequalities, the following simple observations will be used often:


Exercise-2: (i) For a ∈ R, we have a = 0 ⇔ |a| ≤ ε for every ε > 0 ⇔ |a| ≤ 1/n for every n ∈ N ⇔|a| ≤ 1/n for infinitely many n ∈ N.(ii) For a, b ∈ R, we have a ≤ b ⇔ a ≤ b + ε for every ε > 0 ⇔ a ≤ b + 1/n for every n ∈ N ⇔a ≤ b+ 1/n for infinitely many n ∈ N.(iii) For a, b, c ∈ R, we have |a− b| ≤ c ⇔ a ≤ b+ c and b ≤ a+ c.

Remark: While negating a sentence, (i) ‘ ∀ ’ becomes ‘ ∃ ’, and vice versa, (ii)‘for infinitely many

n ∈ N’ becomes ‘for all but finitely many n ∈ N’, and vice versa, (iii) ‘ ≤ ’ becomes ‘ > ’, and vice

versa, and so on. One should also consider the overall meaning while negating a sentence.

Exercise-3: [Verify] (i) Let A ⊂ R and (xn) be a sequence in R. Then, the negation of “There is

a ∈ A such that a ≤ xn for all but finitely many n ∈ N” is “For every a ∈ A, we have that a > xn

for infinitely many n ∈ N”.(ii) Let A be a collection of subsets of R, Then, the negation of “For every n ∈ N and ε > 0, there

is A ∈ A such that n < |a| < n+ ε for infinitely many a ∈ A” is “There exist n ∈ N and ε > 0 with

the property that for every A ∈ A, the set {a ∈ A : n < |a| < n+ ε} is finite”.

Some commonly used identities and inequalities are the following:

[102] (i) If x, y ∈ R and n ∈ N, then (x + y)n =∑n

j=0nCjx

n−jyj , where nCj =n!

j!(n− j)!. In

particular, if y > 0 and n ∈ N, then (1+y)n = 1+ny+n(n− 1)y2

2+· · ·+yn so that (1+y)n ≥ 1+ny

for n ∈ N, and (1 + y)n ≥ n(n− 1)y2

2for n ≥ 2.

(ii) If x, y ∈ R and n ≥ 2, then xn − yn = (x− y)∑n−1

j=0 xn−1−jyj .

(iii) If n ≥ 2 and x1, . . . , xn ∈ R, then (x1 + · · ·+ xn)2 =

∑nj=1 x

2j + 2

∑1≤j<k≤n xjxk.

(iv) [Bernoulli inequality] (1 + x)n ≥ 1 + nx for every real number x > −1 and n ∈ N.(v) [Lagrange’s identity] If n ∈ N and x1, . . . , xn, y1, . . . , yn ∈ R, then(∑n

j=1 xjyj)2 +

∑1≤j<k≤n(xjyk − xkyj)

2 = (∑n

j=1 x2j )(

∑nk=1 y

2k).

(vi) [Cauchy-Schwarz inequality] Let x = (x1, . . . , xn) and y = (y1, . . . , yn) be members of Rn.

Then∑n

j=1 |xjyj | ≤ ∥x∥∥y∥, where ∥x∥ is the Euclidean norm of x defined as ∥x∥ = (∑n

j=1 x2j )

1/2.

(vii) If x1, . . . , xn ∈ R, then (∑n

j=1 xj)2 ≤ n

∑nj=1 x

2j .

(viii) [Minkowski’s inequality] Let x = (x1, . . . , xn) and y = (y1, . . . , yn) be members of Rn. Then

∥x+ y∥ ≤ ∥x∥+ ∥y∥, where ∥x∥ is the Euclidean norm of x defined as ∥x∥ = (∑n

j=1 x2j )

1/2.

(ix) If x ∈ R\Q and n ∈ N, then there exist p ∈ Z and q ∈ {1, . . . , n} such that |x− p

q| < 1

nq≤ 1

q2.

Proof. (i), (ii), and (iii) can be proved using induction on n.

(iv) Use induction on n. The assertion is true for n = 1. Assuming it to be true for n, note that

(1+ x)n+1 = (1+ x)(1+ x)n ≥ (1+ x)(1+nx) = 1+ (n+1)x+nx2 ≥ 1+ (n+1)x, where the first

inequality is justified by the induction assumption and the fact that 1 + x > 0.

REAL ANALYSIS 5

(v) S1 := (∑n

j=1 xjyj)2 =

∑nj=1 x

2jy

2j + 2

∑1≤j<k≤n xjyjxkyk by (iii), and

S2 :=∑

1≤j<k≤n(xjyk − xkyj)2 =

∑1≤j<k≤n(x

2jy

2k + x2ky

2j − 2xjykxkyj). Therefore,

S1 + S2 =∑n

j=1 x2jy

2j +

∑1≤j<k≤n(x

2jy

2k + x2ky

2j ) =

∑nj=1

∑nk=1 x

2jy

2k = (

∑nj=1 x

2j )(

∑nk=1 y

2k).

(vi) This follows from (v).

(vii) Apply Cauchy-Schwarz to (x1, . . . , xn) and (1, . . . , 1).

(viii) ∥x+y∥2 =∑n

j=1(xj+yj)2 = ∥x∥2+∥y∥2+2

∑nj=1 xjyj . Moreover,

∑nj=1 xjyj ≤

∑nj=1 |xjyj | ≤

∥x∥∥y∥ by Cauchy-Schwarz. Hence ∥x+ y∥2 ≤ ∥x∥2 + ∥y∥2 + 2∥x∥∥y∥ = (∥x∥+ ∥y∥)2.

(ix) For each j ∈ {0, 1, . . . , n}, let mj ∈ Z be such that jx−mj ∈ [0, 1). Applying the Pigeonhole

principle to these n+ 1 numbers in [0, 1) and the n intervals [i− 1

n,i

n) for 1 ≤ i ≤ n, we may find

0 ≤ j < k ≤ n such that |(kx −mk) − (jx −mj)| <1

n. Hence |x − mk −mj

k − j| < 1

n(k − j). Take

p = mk −mj and q = k − j. �

2. A few remarks about sets

Exercise-4: Let f : X → Y be a map between two sets X and Y . Then,

(i) f(∪

j∈J Aj) =∪

j∈J f(Aj) for subsets Aj ⊂ X and any indexing set J .

(ii) f(A∩B) ⊂ f(A)∩f(B) for A,B ⊂ X, but the inclusion can be strict. However, if f is injective,

then f(∩

j∈J Aj) =∩

j∈J f(Aj) for subsets Aj ⊂ X and any indexing set J .

(iii) Let A ⊂ X. In general, there is no inclusion relation between f(X \A) and Y \ f(A); however,the two sets are equal when f is bijective.

(iv) Even if A,B ⊂ X are disjoint sets, f(A) ∩ f(B) can be non-empty. However, if f is injective

and Aj ⊂ X are pairwise disjoint for j ∈ J , then f(Aj)’s are also pairwise disjoint.

On the other hand, the operation of taking pre-image commutes with the set theoretic operations

of union, intersection and complementation:

Exercise-5: Let f : X → Y be a map between two sets X and Y . For Z ⊂ Y , let f−1(Z) = {x ∈X : f(x) ∈ Z}. Also, let f−1(y) = f−1({y}) = {x ∈ X : f(x) = y} for y ∈ Y . Then,

(i) f−1(∪

j∈J Zj) =∪

j∈J f−1(Zj) for subsets Zj ⊂ Y and any indexing set J .

(ii) f−1(∩

j∈J Zj) =∩

j∈J f−1(Zj) for subsets Zj ⊂ Y and any indexing set J .

(iii) f−1(Y \ Z) = X \ f−1(Z) for Z ⊂ Y .

(iv) If Zj ⊂ Y are pairwise disjoint for j ∈ J , then f−1(Zj)’s are pairwise disjoint for j ∈ J . In

particular, f−1(y) ∩ f−1(z) = ∅ for any two distinct points y, z ∈ Y .

Exercise-6: Let f : X → Y be a map between two sets X and Y .

(i) For A ⊂ X, we have A ⊂ f−1(f(A)), and equality holds when f is injective.

(ii) For Z ⊂ Y , we have f(f−1(Z)) ⊂ Z, and equality holds when f is surjective.


Exercise-7: Let f : X → Y be a finite-to-one map between two sets X and Y , i.e., f−1(y) is a finite

set (possibly empty) for each y ∈ Y . Let An ⊂ X be such that An+1 ⊂ An for every n ∈ N. Thenf(∩∞

n=1An) =∩∞

n=1 f(An). [Hint : To prove ‘⊃’, consider y ∈∩∞

n=1 f(An). As f is finite-to-one,

there is x ∈ X such that x ∈ An for infinitely many n ∈ N and f(x) = y. As An’s are decreasing,

we get that x ∈∩∞

n=1An.]

Definition: Let X be a set, (xn) be a sequence in X, and A ⊂ X. We say (i) (xn) is eventually

in A if xn ∈ A for all but finitely many n ∈ N (i.e., if there is n0 ∈ N such that xn ∈ A for every

n ≥ n0), and (ii) (xn) is frequently in A if xn ∈ A for infinitely many n ∈ N.

Exercise-8: Let X be a set, (xn) be a sequence in X, and A ⊂ X. Then,

(i) (xn) fails to be eventually in A ⇔ (xn) is frequently in X \A.(ii) (xn) fails to be frequently in A ⇔ (xn) is eventually in X \A.

Exercise-9: Let X be a set and An ⊂ X for n ∈ N. Then,(i) {x ∈ X : x ∈ An for infinitely many n ∈ N} =

∩∞k=1

∪∞n=k An.

(ii) {x ∈ X : x ∈ An for for all but finitely many n ∈ N} =∪∞

k=1

∩∞n=k An.

(iii)∪∞

k=1

∩∞n=k An ⊂

∩∞k=1

∪∞n=k An, but the inclusion can be strict.

(iv) If (An) is either increasing or decreasing, then the inclusion becomes an equality in (iii), and

both sides are equal to either∪∞

n=1An or∩∞

n=1An respectively.

Exercise-10: Let f : X → Y and g : Y → Z be maps between sets.

(i) If g ◦ f : X → Z is injective, then f is injective.

(ii) If g ◦ f : X → Z is surjective, then g is surjective.

(iii) Even if X = Y = Z and g ◦ f is bijective, f need not be surjective and g need not be injective.

[Hint : (iii) Let f, g : N → N be f(n) = n+ 1, g(1) = 1, and g(n) = n− 1 for n ≥ 2.]

Exercise-11: Let X,Y be nonempty sets. Then, there exists a surjective map f : X → Y ⇔ there

exists an injective map g : Y → X. [Hint : (i) Given a surjective f , define g(y) to be any point in

f−1(y). Given an injective g, define f on g(Y ) as f(g(y)) = y, and define f on X \g(Y ) arbitrarily.]

[103] [Cantor’s theorem] Let X be a nonempty set. Then there does not exist any surjective map

f : X → P(X), where P(X) denotes the power set of X, i.e., the collection of all subsets of X

including X and the empty set ϕ.

Proof. Suppose, for a contradiction, that there exists a surjective map f : X → P(X). Let A =

{x ∈ X : x /∈ f(x)}. Since f is surjective, there is x ∈ X with f(x) = A. Now if x ∈ A, then the

definition of A implies x /∈ f(x) = A, a contradiction. On the other hand, if x /∈ A = f(x), then

the definition of A implies x ∈ A, again a contradiction. �

[104] [Schroeder-Bernstein theorem] Let X,Y be nonempty sets. Then the following are equivalent:

(i) There are injective maps f : X → Y and g : Y → X.

REAL ANALYSIS 7

(ii) There are surjective maps f : X → Y and g : Y → X.

(iii) There is a bijection F : X → Y .

Proof. In view of Exercise-11, it suffices to show (i) ⇒ (iii). So assume f : X → Y and g : Y → X

are injective maps. We may assume that neither f nor g is surjective. Let X1 = X \ g(Y ) and

Y1 = Y \ f(X). Next, let X2 = g(Y1) and Y2 = f(X1). In general for n ∈ N, let Yn+1 = f(Xn) and

Xn+1 = g(Yn). Also, let X ′ = X \∪∞

n=1Xn and Y ′ = Y \∪∞

n=1 Yn. We observe that:

(1) Xn’s are pairwise disjoint for n ∈ N; and similarly, Yn’s are pairwise disjoint for n ∈ N.(2) For n ∈ N, f maps Xn bijectively onto Yn+1, and g maps Yn bijectively onto Xn+1.

(3) X ′ is the set of all x ∈ X for which there is an infinite sequence (an)∞n=0 in X ∪ Y such that

x = a0 = g(a1), a1 = f(a2), a2 = g(a3), a3 = f(a4), . . .. Similarly, Y ′ is the set of all y ∈ Y

for which there is an infinite sequence (bn)∞n=0 in X ∪ Y such that y = b0 = f(b1), b1 = g(b2),

b2 = f(b3), b3 = g(b4), . . .. Consequently, f maps X ′ bijectively onto Y ′.

Define F : X → Y as F (x) = f(x) if either x ∈ X ′ or x ∈ X2n−1 for some n ∈ N, and

F (x) = g−1(x) if x ∈ X2n for some n ∈ N. Note that F (X ′) = Y ′, F (X2n−1) = Y2n and

F (X2n) = Y2n−1 for n ∈ N. By the observations above, it follows that F : X → Y is a bijection. �

[105] [Zorn’s lemma] Let (F ,≤) be a nonempty partially ordered set. If every chain in F (i.e.,

every totally ordered subset of F) has an upper bound in F , then (F ,≤) has a maximal element.

Proof. Zorn’s lemma is equivalent to Axiom of choice, and we accept it as an axiom of our Set

Theory. See for instance the book T.J. Jech, The Axiom of Choice for more details. �

[106] Let X,Y be nonempty sets and suppose there does not exist any injective map from Y to X.

Then there exists an injective map f : X → Y .

Proof. Let F = {(A, f) : A ⊂ X and f : A→ Y is injective}. Fix x0 ∈ X, y0 ∈ Y , define f :

{x0} → Y as f(x0) = y0, and note that ({x0}, f) ∈ F . Thus F is nonempty. For (A1, f1), (A2, f2) ∈F , define (A1, f1) ≤ (A2, f2) if A1 ⊂ A2 and f2|A1 = f1. Then≤ is a partial order on F . If {(Aj , fj) :

j ∈ J} is a chain in F , then by putting B =∪

j∈J Aj and defining h : B → Y as h(x) = fj(x) for

x ∈ Aj , we may see that (B, h) ∈ F is an upper bound for the chain {(Aj , fj) : j ∈ J}. Hence

by Zorn’s lemma, F has a maximal element, say (A, f). Note that f : A → Y is injective by the

definition of f , and f cannot be surjective as otherwise f−1 : Y → X will be injective (which is

against the hypothesis). If A = X, then by choosing x1 ∈ X \A and y1 ∈ Y \ f(A), we may define

A1 = A ∪ {x1} and an injective map f1 : A1 → Y extending f by putting f1(x1) = y1. Then

(A1, f1) ∈ F and this will contradict the maximality of (A, f). Therefore, we must have A = X.

Thus f : X → Y is injective. �

Definition: [Cardinality of sets] The notion of cardinality of a set X, denoted as card(X), is defined

by the following conditions:


(i) IfX is a finite set, then card(X) = |X| := number of elements in X. In particular, card(ϕ) = 0,

where ϕ denotes the empty set. Also, we write card(ϕ) < card(Y ) for every nonempty set Y .

(ii) If X,Y are nonempty sets, then we write card(X) ≤ card(Y ) if there is an injective map

f : X → Y (or equivalently, if there is a surjective map g : Y → X). Note by [106] that one of the

following must hold for any two sets X,Y : card(X) ≤ card(Y ) or card(Y ) ≤ card(X).

(iii) For nonempty sets X,Y , we write card(X) = card(Y ) if there is a bijection f : X → Y .

Note by [104] that card(X) = card(Y ) iff card(X) ≤ card(Y ) and card(Y ) ≤ card(X). We write

card(X) < card(Y ) if card(X) ≤ card(Y ) and card(X) = card(Y ) .

(iv) A set X is said to be countable1 if card(X) ≤ card(N), i.e., if either X is empty or there is an

injective map f : X → N (or there is a surjective map g : N → X). If a set X is not countable,

then X is said to be an uncountable set (and in this case, necessarily card(N) < card(X) by [106]).

Remark: IfX,Y are sets, then by [104] and [106], exactly one of the following must hold: card(X) <

card(Y ), card(X) = card(Y ), or card(Y ) < card(X).

Exercise-12: For a nonempty set X, the following are equivalent: (i) X is countable.

(ii) There exist a countable set Y and an injective map f : X → Y .

(iii) There exist a countable set Y and a surjective map g : Y → X.

[Hint : (iii) ⇒ (i): There is a surjection f : N → Y . Then g ◦ f : N → X is surjective.]

[107] (i) f : N2 → N given by f(k,m) = 2k−1(2m − 1) is a bijection. Hence g : N3 → N given

by g(k,m, n) = f(f(k,m), n) is a bijection. In general, for each m ∈ N, we can find a bijection

h : Nm → N. In particular, Nm is countable for every m ∈ N.(ii) Finite product of countable sets is countable.

(iii) Countable union of countable sets is countable.

(iv) Subsets of a countable set are countable. Hence supersets of an uncountable set are uncountable.

Proof. (i) Check the claims.

(ii) Let X1, . . . , Xm be countable sets. If all of them are nonempty, then there are surjections

fj : N → Xj , and then f : Nm →∏m

j=1Xj given by f = (f1, . . . , fm) is surjective. This shows that∏mj=1Xj is countable by Exercise-12.

(iii) Let Xj be countable sets for j ∈ J , where J is a countable indexing set. Assume they are

nonempty. Then there are surjections fj : N → Xj and g : N → J . Now∪

j∈J Xj is countable by

Exercise-12 because h : N2 →∪

j∈J Xj defined as h(k,m) = fg(k)(m) is surjective.

(iv) If X ⊂ Y and Y is countable, then the inclusion map i : X → Y is injective, and hence X is

countable by Exercise-12. �

1As per our definition, finite sets are also countable. However, some authors reserve the term ‘countable’ for

infinite countable sets, and they refer to finite sets as ‘at most countable sets’.

REAL ANALYSIS 9

[108] (i) Z and Q are countable. More generally, Zm and Qm are countable.

(ii) P(N), {0, 1}N, and NN are uncountable.

(iii) If Xn are sets with |Xn| ≥ 2 for n ∈ N, then∏∞

n=1Xn is uncountable.

(iv) (0, 1), (a, b) (for reals a < b), R, and R \Q are uncountable.

(v) Let Xn be countable sets and fix an ∈ Xn for n ∈ N, then A := {(xn) ∈∏∞

n=1Xn : xn =

an for all but finitely many n ∈ N} is countable.

(vi) Let PF (N) = {all finite subsets of N} and P∞(N) = P(N) \PF (N) = {all infinite subsets of N}.Then PF (N) is countable, and hence P∞(N) is uncountable.

Proof. (i) Z is countable because f : N2 → Z given by f(k,m) = k−m is surjective. Now Z×N is

countable by [107](ii), and therefore Q is countable by the surjectivity of g : Z × N → Q given by

g(m,n) = m/n. By [106], Zm and Qm are also countable.

(ii) P(N) is uncountable because card(N) < card(P(N)) by [103]. The map f : {0, 1}n → P(N) givenby f((xn)) = {n ∈ N : xn = 1} is bijective, and hence {0, 1}N is uncountable. Being a superset of

{0, 1}N, the set NN is also uncountable.

(iii) Since |Xn| ≥ 2, there is a natural injective map f : {0, 1}N →∏∞

n=1Xn.

(iv) Note that f : {0, 1}N → (0, 1) given by f((xn)) = 0.y1y2y3 · · · , where decimal expansion on

the right side is defined by yn = 3 if xn = 0 and yn = 5 if xn = 1, is injective, and hence (0, 1) is

uncountable. As x→ (b−a)x+a is a bijection from (0, 1) to (a, b), the interval (a, b) is uncountable.

Being a superset of (0, 1), the set R is uncountable. Finally, as R is uncountable and Q is countable,

the set R \Q must be uncountable.

(v) To see A is countable, write A as a countable union of countable sets as A =∪∞

k=1Ak, where

Ak := {(xn) ∈∏∞

n=1Xn : xn = an for every n ≥ k + 1}. Here, each Ak is countable because Ak is

bijective with the countable set∏k

n=1Xn.

(vi) PF (N) =∪∞

n=1 P({1, . . . , n}), a countable union of finite sets, and hence countable by [107](iii).

Or, note that PF (N) is bijective with {(xn) ∈ {0, 1}N : (xn) is eventually 0}, and apply (v). �

Exercise-13: (i) The collection F = {A ⊂ N : A ∩ (A+ k) is finite for every k ∈ N} is uncountable.

(ii) There exists an uncountable collection Γ ⊂ P∞(N) such that A∩B is finite for any two distinct

members A,B ∈ Γ.

[Hint : (i) F contains an infinite set: consider {n2 : n ∈ N} or {2n : n ∈ N}. Also, if A ∈ F , then

P(A) ⊂ F . (ii) Write Q = {r1, r2, . . .}. For each x ∈ R, choose an increasing sequence (nk) of

natural numbers such that (rnk) → x and put Ax = {nk : k ∈ N}. Let Γ = {Ax : x ∈ R}.]

Exercise-14: A real number a is said to be algebraic if there is a non-constant polynomial p with

integer coefficients (equivalently, with rational coefficients) such that p(a) = 0. Note that every

rational number is algebraic, and moreover many irrational numbers (example:√2,

√3) are also


algebraic. If a real number is not algebraic, then it is called a transcendental number. It is known

that e, π are transcendental but it is still an open problem whether e+π and eπ are transcendental!

(i) The collection of all real polynomials with integer coefficients is countable because there is a

natural injective map from this collection to the countable set∪∞

m=1 Zm.

(ii) {x ∈ R : x is algebraic} is countable, and hence {x ∈ R : x is transcendental} is uncountable.

3. Metric spaces: preliminaries

Certain topological notions such as compactness and connectedness are needed in Real Analysis.

We believe that it is better to introduce these notions in the general setting of a metric space rather

than in the restrictive setting of the real line. Accordingly, we will go through a little bit of metric

space theory before doing more of Real Analysis.

Definition: Let X be a nonempty set. A function d : X ×X → [0,∞) is called a metric on X if for

every x, y, z ∈ X we have the following:

d(x, y) = 0 iff x = y;

d(x, y) = d(y, x); and

d(x, z) ≤ d(x, y) + d(y, z) (called the triangle inequality).

If d is a metric on X, then (X, d) is called a metric space. Standard examples are: any nonempty

subset of Rm with the Euclidean metric dE given by dE(x, y) = ∥x− y∥ := (∑m

j=1 |xj − yj |2)1/2 for

x = (x1, . . . , xm) and y = (y1, . . . , ym) in Rm, and any nonempty set X with the discrete metric

given by d(x, x) = 0 and d(x, y) = 1 for x = y.

Definition: Let (X, d) be a metric space. For x ∈ X and r > 0, the subset B(x, r) := {y ∈ X :

d(x, y) < r} is called the open ball of radius r centered at x. We say U ⊂ X is open in X if for every

x ∈ U there is r > 0 with B(x, r) ⊂ U . Note that if y ∈ B(x, r), then for any δ ∈ (0, r − d(x, y)),

we have B(y, δ) ⊂ B(x, r), and hence B(x, r) is open in X. Consequently, U ⊂ X is open in X iff

U is a union of open balls in X. We say F ⊂ X is closed in X if X \ F is an open set in X. Note

that the subsets ∅, X are both open and closed in X.

Example: In (R2, dE), open balls are open discs, geometrically. On the other hand, open balls in

R2 w.r.to the metric d∞((u1, u2), (v1, v2)) := max{|u1 − v1|, |u2 − v2|} are open squares with sides

parallel to the axes, and open balls in R2 w.r.to the metric d1((u1, u2), (v1, v2)) := |u1−v1|+|u2−v2|are open squares with sides parallel to the lines y = x and y = −x.

Example: (i) If a < b are reals, then (a, b) is open but not closed in (R, dE), and [a, b] is closed but

not open in (R, dE). (ii) N and Z are closed in (R, dE). (iii) {(x, y) ∈ R2 : xy = 1} is closed in

(R2, dE). (iv) In a discrete metric space, every subset is both open and closed. (v) [√2,√5] ∩Q is

both open and closed in Q with respect to dE . (vi) {1} ∪ {n+1

k: n, k ∈ N} is closed in (R, dE).

Exercise-15: Let (X, d) be a metric space. Then,

REAL ANALYSIS 11

(i) Finite intersection of open sets and arbitrary union of open sets are open in X.

(ii) Finite union of closed sets and arbitrary intersections of closed sets of X are closed in X.

(iii) In general, an arbitrary intersection of open subsets of X need not be open in X, and an

arbitrary union of closed subsets of X need not be closed in X.

(iv) {x} is closed in X for every x ∈ X. Hence finite subsets are closed in X by (ii).

(v) Let Z ⊂ Y ⊂ X. Then Z is open in (Y, d) iff there is an open set U ⊂ X such that Z = Y ∩U .

Consequently, Z is closed in (Y, d) iff there is a closed set A ⊂ X such that Z = Y ∩A.[Hint : (i) If U1, . . . , Un ⊂ X are open and if x ∈

∩nj=1 Uj , then there are rj > 0 with B(x, rj) ⊂ Uj

for 1 ≤ j ≤ n. If we take r = min{r1, . . . , rn}, then B(x, r) ⊂∩n

j=1 Uj . (ii) This follows from (i) by

considering the complement of sets. (iii) In (R, dE), note that [a, b] =∩∞

n=1(a− 1/n, b+ 1/n) and

(a, b) =∪∞

n=k[a+ 1/n, b− 1/n], where k is any natural number with a+ 2/k < b. (v) If a ∈ Y and

r > 0, note that {y ∈ Y : d(a, y) < r} = Y ∩ {x ∈ X : d(a, x) < r}. Now if A is open in Y , then A

is a union of open balls in Y . Take U to be the union of corresponding open balls in X.]

Definition: Let (X, d) be a metric space. For a nonempty set A ⊂ X, define its diameter as

diam(A) = sup{d(a, b) : a, b ∈ A}. Also define diam(A) = 0 if A = ∅. We say A ⊂ X is a bounded

subset of X if diam(A) <∞. For example, finite subsets of X are always bounded. If X itself is a

bounded subset of X, then we say d is a bounded metric on X.

Example: (i) In (R, dE), we have diam((a, b)) = diam([a, b]) = b − a, and diam((a,∞)) =

diam((−∞, b)) = diam(R) = ∞. In particular, R is not a bounded subset of (R, dE). We may also

note that a nonempty subset A ⊂ (Rm, dE) is bounded iff sup{∥x∥ : x ∈ A} <∞.

(ii) On any nonempty set X, the discrete metric is a bounded metric (it is bounded by 1), and

hence every subset of a discrete metric space is bounded. In particular, R is a bounded set w.r.to

the discrete metric on R.(iii) {(x, y) ∈ R2 : x+ y ≤ 1} is an unbounded subset of (R2, dE).

(iv) If (X, d) is a metric space and B(x, r) is an open ball in X, then diam(B(x, r)) ≤ 2r < ∞and hence B(x, r) is bounded. The inequality diam(B(x, r)) ≤ 2r can be strict: if the metric

under consideration is the discrete metric, then for every r ∈ (0, 1) we see B(x, r) = {x} and hence

diam(B(x, r)) = diam({x}) = 0 < 2r.

Exercise-16: Let (X, d) be a metric space. Then,

(i) A subset A ⊂ X is bounded iff there are x ∈ X and r > 0 such that A ⊂ B(x, r).

(ii) Finite union of bounded subsets of X is again a bounded subset of X.

[Hint : (i) If M := diam(A) < ∞ and if x ∈ A, then A ⊂ B(x,M + 1). Conversely, if A ⊂ B(x, r),

then d(a, b) ≤ d(a, x) + d(x, b) for every a, b ∈ A, and hence diam(A) ≤ 2r < ∞. (ii) Enough

to prove for two bounded sets, for then we may argue inductively. Let A1, A2 ⊂ X be bounded,

Mj = diam(Aj), and xj ∈ Aj for j = 1, 2. If r =M1+M2+d(x1, x2)+1, then A1∪A2 ⊂ B(x1, r).]


A technique for transferring metric: IfX is a nonempty set, (Y, d′) is a metric space, and f : X → Y

is an injective function, then the metric on Y can be pulled back using f to define a metric d on

X, where d is defined as d(a, b) := d′(f(a), f(b)). For example, d(a, b) := |2a − 2b| defines a metric

on the set R, and d(a, b) := |1/a− 1/b| defines a (bounded) metric on the set (1,∞).

Definition: Let (X, d) be a metric space. We say a sequence (xn) in X converges to a point x ∈ X

if for every r > 0, there is n0 ∈ N such that xn ∈ B(x, r) (i.e., d(x, xn) < r) for every n ≥ n0. If this

happens, we write (xn) → x. If (xn) is a sequence in X and if (nk) is a strictly increasing sequence

of natural numbers, then (xnk)∞k=1 is said to be a subsequence of (xn). Note that if (xn) → x, then

every subsequence of (xn) also converges to x.

Remark: Note that a sequence ((xn, yn)) in R2 converges to (x, y) ∈ R2 w.r.to the Euclidean metric

dE iff (xn) → x and (yn) → y. More generally, a sequence converges in Rm iff each coordinate

sequence converges in R.

Definition: Let (X, d) be a metric space. A sequence (xn) in X is said to be: (i) a Cauchy sequence

if for every ε > 0, there is n0 ∈ N such that d(xm, xn) < ε for every m,n ≥ n0, (ii) a bounded

sequence if {xn : n ∈ N} is a bounded set, and (iii) eventually constant if there are x ∈ X and

n0 ∈ N such that xn = x for every n ≥ n0.

Example: (i) If (xn) is a sequence of rational numbers converging to an irrational number w.r.to

dE , then (xn) is a Cauchy sequence in (Q, dE), but (xn) is not a convergent sequence in (Q, dE).(ii) In any metric space X, if x ∈ X and xn ∈ B(x, 1/n) for every n ∈ N, then (xn) → x. More

generally, if (rn) is a sequence of positive reals decreasing to 0, and if xn ∈ B(x, rn) for every n ∈ N,then (xn) → x. (iii) If we consider R with the discrete metric, then every sequence is bounded,

but the sequence (1/n) is neither convergent nor Cauchy. In fact, if (xn) is a sequence in a discrete

metric space, then (xn) is Cauchy ⇔ (xn) is convergent ⇔ (xn) is eventually constant.

Exercise-17: Let (X, d) be a metric space, and (xn) be a sequence in X.

(i) (xn) can converge to at most one point of X.

(ii) If (xn) is convergent, then (xn) is Cauchy.

(iii) If (xn) is Cauchy, then (xn) is bounded.

(iv) If (xn) is Cauchy and a subsequence (xnk) converges to some x ∈ X, then (xn) → x.

(v) Let an be positive reals with∑∞

n=1 an < ∞ (for example, an = 1/2n). If d(xn+1, xn) ≤ an for

every n ∈ N, then (xn) is Cauchy.

(vi) If there is δ > 0 such that d(xm, xn) ≥ δ for every m = n, then no subsequence of (xn) is

Cauchy, and hence no subsequence of (xn) is convergent.

[Hint : Imitate the proofs that you already know from elementary Real Analysis. (i) If x, y ∈ X

are distinct, and if r := d(x, y)/3, then B(x, r) and B(y, r) are disjoint, and hence it is not possible

for (xn) to be in both the balls eventually. (iii) Let n0 ∈ N be such that d(xm, xn) < 1 for every

REAL ANALYSIS 13

m,n ≥ n0. By taking m = n0 we see {xn : n > n0} ⊂ B(xn0 , 1). And {x1, . . . , xn0} is a finite set.

(v) d(xn+k, xn) ≤∑k−1

j=0 d(xn+j+1, xn+j) ≤∑k−1

j=0 an+j ≤∑∞

i=n ai, which is the tail of a convergent

series of positive reals, and hence can be made arbitrarily small by taking n large enough.]

Definition: Let (X, d) be a metric space, and A ⊂ X.

(i) The closure of A in X, denoted as A, is defined as the smallest closed subset of X containing A.

It is well-defined since A is equal to the intersection of all closed subsets of X containing A. Note

that A = A; and A is closed in X ⇔ A = A. How to determine A is mentioned in [109] below.

(ii) The interior of A in X, denoted as int(A), is defined as the largest open subset of X contained

in A. It is well-defined since int(A) is equal to the union of all open subsets of X contained in A.

Note that int(int(A)) = int(A); and A is open in X ⇔ int(A) = A.

(iii) A point x ∈ X is said to be a boundary point of A in X if B(x, ε) intersects both A and X \Afor every ε > 0. Let ∂A denote the collection of all boundary points of A in X.

(iv) A point x ∈ X is said to be a limit point of A in X if B(x, ε) ∩ (A \ {x}) = ∅ for every ε > 0.

In text books, you may find other names for a limit point: cluster point, accumulation point, etc.

(v) A point x ∈ A is called an isolated point of A in X if there is ε > 0 such that B(x, ε)∩A = {x}.

[109] Let (X, d) be a metric space, and A ⊂ X. Then,

(i) int(A) = {x ∈ A : ∃ ε > 0 with B(x, ε) ⊂ A} = {x ∈ X : x is an interior point of A}.(ii) A = X \ int(X \A).(iii) A = {x ∈ X : B(x, ε) ∩A = ∅ ∀ ε > 0} = {x ∈ X : ∃ a sequence (xn) in A converging to x}.(iv) ∂A = A ∩X \A. Moreover, we have the disjoint union A = int(A) ⊔ ∂A.(v) x ∈ X is a limit point of A ⇔ there is a sequence (xn) in A \ {x} converging to x. Moreover,

we have the disjoint union A = {isolated points of A} ⊔ {limit points of A in X}. Consequently,

we also have A = A ∪ {limit points of A in X} (which may not be a disjoint union).

(vi) If x ∈ X is a limit point of A, then B(x, ε) ∩A is an infinite set for every ε > 0.

(vii) A is open in X ⇔ no sequence in X \A converges to a point of A.

Proof. (i) Clear. (ii) For Y ⊂ X, note that Y is a closed subset of X containing A iff X \ Y is

an open subset of X contained in X \A.

(iii) By (i) and (ii), A = X \ int(X \A) = X \ {x ∈ X : ∃ ε > 0 with B(x, ε) ⊂ X \A} = {x ∈ X :

B(x, ε) ∩A = ∅ for every ε > 0}. Varying ε over 1/n’s, we may deduce the second equality.

(iv) This follows from (iii), (i), and the definition of ∂A.

(v) This follows from (iii), and the definition of an isolated point and a limit point.

(vi) If B(x, ε) ∩ A \ {x} = {a1, . . . , ak}, choose 0 < δ < min{d(x, aj) : 1 ≤ j ≤ k}. Then,

B(x, δ) ∩A \ {x} = ∅, a contradiction.


(vii) Let Z = X \A, and note that A fails to be open in X ⇔ Z fails to be closed in X ⇔ A∩Z = ∅.Now apply the sequential characterization of closure in (iii) to Z. �

Example: (i) In (R, dE), we have that (a, b) = [a, b] and int([a, b]) = (a, b).

(ii) Consider (R, dE), and let A = {0}∪{1/n : n ∈ N}∪ [3, 4)∪{x ∈ Q : 7 < x ≤ 8}. Then, int(A) =(3, 4), ∂A = {0} ∪ {1/n : n ∈ N} ∪ {3} ∪ {4} ∪ [7, 8], and A = {0} ∪ {1/n : n ∈ N} ∪ [3, 4] ∪ [7, 8].

Also, {isolated points of A} = {1/n : n ∈ N} and {limit points of A in R} = {0} ∪ [3, 4] ∪ [7, 8].

(iii) If X is a discrete metric space and A ⊂ X, then A = A = int(A) (since every subset is both

closed and open in X), and A has no limit points in X (since the only convergent sequences in X

are the eventually constant sequences).

(iv) Define a metric d on R as d(x, y) = 0 if x = y, and d(x, y) = |x| + |y| if x = y. Note that

B(0, ε) = (−ε, ε) for ε > 0 and B(x, ε) = {x} for x ∈ R \ {0} and 0 < ε < |x|. Hence (0, 1) = [0, 1)

and int([0, 1]) = (0, 1] in (R, d).

Definition: A subset Y of a metric space (X, d) is said to be dense in X if Y = X, equivalently if

Y ∩ U = ∅ for every nonempty open set U ⊂ X. For example, Q and R \Q are dense in R, but Nand Z are not dense in R. The only dense subset of a discrete metric space (X, d) is X. If d is the

metric on R in Example (iv) above, then R and R \ {0} are the only dense subsets of R.

Example: Let y ∈ R \ Q and A = Z + Zy = {m + ny : m,n ∈ Z}. We claim that A is dense

in R. Observe that if a ∈ A ∩ (0, ε), then the subset {ja : j ∈ Z} of A intersects every open

interval of R having length ≥ ε. Hence it suffices to show that A ∩ (0, ε) = ∅ for every ε > 0. For

each n ∈ N choose mn ∈ Z with mn + ny ∈ [0, 1). By Pigeonhole principle, there are k = n with

0 ≤ (mn + ny) − (mk + ky) < ε. Taking p = mn −mk and q = (n − k), we have p + qy ∈ A and

0 ≤ p+ qy < ε. Moreover, p+ qy = 0 since y is irrational and q = 0.

Exercise-18: Let (X, d) be a metric space and A ⊂ X be nonempty.

(i) For x ∈ X, define the distance from x to A as dist(x,A) = inf{d(x, a) : a ∈ A}. Then,

A = {x ∈ X : dist(x,A) = 0} and int(A) = {x ∈ X : dist(x,X \A) > 0}.(ii) If there is δ > 0 such that d(a, b) ≥ δ for any two distinct points a, b ∈ A, then A has no limit

points in X, and hence A is closed in X.

(iii) A is dense in X ⇔ A ∩ B(x, r) = ∅ for every x ∈ X and r > 0 ⇔ for every x ∈ X, there is a

sequence (an) in A converging to x.

[Hint : (ii) If a ∈ A, then A ∩ B(a, δ/2) = {a}. If x ∈ X \ A, then B(x, δ/2) can contain at most

one point of A, and therefore B(x, r) for a sufficiently small r > 0 is disjoint from A.]

[110] (i) [Nested interval theorem] Let [a1, b1] ⊃ [a2, b2] ⊃ · · · be a decreasing sequence of closed

intervals in R. Then∩∞

k=1[ak, bk] = ∅. If in addition (bk−ak) → 0, then∩∞

k=1[ak, bk] is a singleton.

(ii) Let Ck’s be m-dimensional closed cubes in Rm with C1 ⊃ C2 ⊃ · · · (and edges parallel to the

coordinate axes). Then∩∞

k=1Ck = ∅. If in addition, diam(Ck) → 0, then∩∞

k=1Ck is a singleton.

REAL ANALYSIS 15

(iii) Every bounded sequence in Rm has a convergent subsequence (w.r.to dE).

Proof. (i) The set A = {ak : k ∈ N} is bounded above by b1 and hence x := sup(A) exists in R by

the least upper bound property. By choice, ak ≤ x for every k ∈ N. Since each bk is an upper bound

for A, we also have x ≤ bk for every k ∈ N. Thus x ∈∩∞

k=1[ak, bk]. If in addition (bk − ak) → 0,

then clearly∩∞

k=1[ak, bk] cannot contain any other element.

(ii) This can be deduced from (i) by arguing coordinatewise since an m-dimensional closed cube is

a product of m closed intervals.

(iii) First suppose m = 1, and consider a bounded sequence (xn) in R. Choose a closed interval

J1 = [a1, b1] with {xn : n ∈ N} ⊂ J1. Let c = (a1 + b1)/2. Then either [a1, c] or [c, b1] must contain

xn’s for infinitely many n ∈ N. Call this subinterval J2 and note that diam(J2) = diam(J1)/2. Next

divide J2 into two closed intervals of equal length and let J3 be one of the two intervals containing

xn’s for infinitely many n ∈ N. Repeat the process. There is a unique element x ∈∩∞

k=1 Jk by (i).

Let n1 ∈ N be the smallest such that xn1 ∈ J1. In general, let nk+1 ∈ N be the smallest such that

nk+1 > nk and xnk+1∈ Jk+1. Since diam(Jk) → 0, it follows that (xnk

) → x.

For the general case, consider a bounded sequence (xn) in Rm, choose an m-dimensional closed

cube C1 with {xn : n ∈ N} ⊂ C1, and let n1 = 1. Assume we have chosen xn1 , . . . , xnk. To choose

xnk+1, write Ck as a union of 2m closed cubes having side-length half of that of Ck, and note that

one of these 2m closed cubes, say Ck+1, must contain xn for infinitely many n ∈ N. Let nk+1 > nk

be the smallest with xnk+1∈ Ck+1. By (ii),

∩∞k=1Ck is a singleton, say {x}. Since xnk

∈ Ck and

diam(Ck) → 0, it follows that (xnk) → x.

Another way to argue the general case is to write xn = (xn,1, . . . , xn,m) and to apply the one-

dimensional case to each coordinate, choosing subsequences m-times. �

4. Continuous functions

Definition: Let (X, d), (Y, d′) be metric spaces, and f : X → Y be a function. We say:

(i) f is continuous at x ∈ X if for every ε > 0, there is δ > 0 such that f(B(x, δ)) ⊂ B(f(x), ε);

and f is a continuous function if it is continuous at each point of X.

(ii) f is uniformly continuous if for every ε > 0 there is δ > 0 such that f(B(x, δ)) ⊂ B(f(x), ε) for

every x ∈ X.

(iii) f is Lipschitz continuous if there is λ > 0 such that d′(f(a), f(b)) ≤ λd(a, b) for every a, b ∈ X.

(iv) f is a homeomorphism if f : X → Y is a bijection and both f and f−1 are continuous (and we

say X and Y are homeomorphic if there is at least one homeomorphism f : X → Y ).

Notation: Let C(X,Y ) = {f : X → Y : f is continuous}.

[111] Let (X, d), (Y, d′) be metric spaces, and f : X → Y be a function. Then the following are

equivalent: (i) f ∈ C(X,Y ), i.e., f is continuous.


(ii) (f(xn)) → f(x) in Y whenever (xn) → x in X.

(iii) f(A) ⊂ f(A) for every subset A ⊂ X.

(iv) f−1(Z) is closed in X for every closed set Z ⊂ Y .

(v) f−1(V ) is open in X for every open set V ⊂ Y .

Proof. (i) ⇒ (ii): Let (xn) → x in X and ε > 0. Choose δ > 0 such that f(B(x, δ)) ⊂ B(f(x), ε).

As (xn) → x, there is n0 ∈ N such that xn ∈ B(x, δ) for every n ≥ n0. Then f(xn) ∈ f(B(x, δ)) ⊂B(f(x), ε) for every n ≥ n0.

(ii) ⇒ (i) (optional): Let ε > 0 be given. If f(B(x, 1/n)) is not included in B(f(x), ε) for any

n ∈ N, then choose xn ∈ B(x, 1/n) such that f(xn) /∈ B(f(x), ε) for each n ∈ N. Now (xn) → x,

but d′(f(x), f(xn)) ≥ ε for every n ∈ N, a contradiction.

(ii) ⇒ (iii) can be deduced using [109](iii).

(iii) ⇒ (iv): Let A = f−1(Z). Since f(f−1(Z)) ⊂ Z and Z is closed, we get by applying (iii) that

f(A) ⊂ f(A) ⊂ Z = Z, and hence A ⊂ f−1(Z) = A, which implies f−1(Z) is closed.

(iv) ⇒ (v): This is clear by taking complements since f−1(Y \ V ) = X \ f−1(V ).

(v) ⇒ (i): Let V = B(f(x), ε). Since f−1(V ) is open in X by (v), and since x ∈ f−1(V ), there

must exist δ > 0 such that B(x, δ) ⊂ f−1(V ). Then f(B(x, δ)) ⊂ B(f(x), ε). �

Example and Remark: (i) A corollary of [111] is that the composition of two continuous functions

is again a continuous function. Moreover, if X is a metric space and f, g : X → R are continuous,

it may be deduced (using [111](ii) and Exercise-27) that af + bg (for a, b ∈ R), the product fg, andf/g (when g is non-vanishing) are continuous. Any polynomial f : R → R is a sum of products of

the constant map and finitely many copies of the identity map, and hence is continuous.

(ii) If x, y ∈ Rn, then |xj−yj | ≤ ∥x−y∥, and hence the projection πj : Rn → R to the jth coordinate

is (Lipschitz) continuous. Any linear map f : Rn → Rm may be written as f = (f1, . . . , fm),

where fj : Rn → R is linear. Note that each fj is a linear combination of projections, and hence

continuous. Thus every linear map f : Rn → Rm is continuous. Consequently, every linear bijection

f : Rn → Rn is a homeomorphism: f and f−1 are continuous because they are linear. A direct

proof that every linear map f : Rn → Rm is Lipschitz continuous is as follows: if e1, . . . , en ∈ Rn are

the standard basis vectors and λ :=∑n

j=1 ∥f(ej)∥, then ∥f(x)− f(y)∥ = ∥∑n

j=1(xj − yj)f(ej)∥ ≤∑nj=1 |xj − yj |∥f(ej)∥ ≤

∑nj=1 ∥x− y∥∥f(ej)∥ = λ∥x− y∥.

(iii) We may identify anm×n real matrix with an element of Rm×n = Rmn. With this identification,

we may see that the determinant function det : Rn×n → R is a finite linear combination of product

of projections and hence continuous. Consequently, {A ∈ Rn×n : det(A) = 0} is open in Rn×n.

Moreover, {A ∈ Rm×n : rank(A) ≥ r} is open in Rm×n for each r because if rank(A) ≥ r, then

REAL ANALYSIS 17

(after a reordering of the coordinates) A has an r × r submatrix with non-zero determinant, and

this determinant remains non-zero if the entries of A are perturbed only a little.

(iv) If A ∈ Rn×n, then the characteristic polynomial p of A is defined as p(x) = det(xI − A), and

deg(p) = n ≥ 1. Since p has at most n roots in R, for any ε > 0 we may choose x ∈ (−ε, ε) withp(x) = 0 and then det(A− xI) = (−1)np(x) = 0. This shows that {A ∈ Rn×n : det(A) = 0} is also

dense in Rn×n.

[112] Let f : (X, d) → (Y, d′) be a function between two metric spaces.

(i) Suppose f is uniformly continuous. Then f takes Cauchy sequences to Cauchy sequences.

Moreover, if (an), (bn) are sequences in X with d(an, bn) → 0, then d′(f(an), f(bn)) → 0.

(ii) f fails to be uniformly continuous ⇔ there exist ε > 0 and sequences (an), (bn) in X such that

limn→∞ d(an, bn) = 0 and d′(f(an), f(bn)) ≥ ε for every n ∈ N.(iii) If f is Lipschitz continuous, then f takes bounded sequences to bounded sequences.

(iv) Suppose that f is a continuous bijection. Then f is a homeomorphism ⇔ f(U) is open in Y

for every open set U ⊂ X ⇔ f(A) is closed in Y for every closed set A ⊂ X.

(v) Assume f is a homeomorphism, and let A ⊂ X. Then f |A : A→ f(A) is also a homeomorphism.

(vi) Let f ∈ C(X,Y ), and A ⊂ X be with A = X. If g ∈ C(X,Y ) and g|A = f |A, then g = f .

Proof. (i)-(v) are left to the student. (vi) Let x ∈ X = A. Then there is a sequence (an) in A

with (an) → x. Now g(x) = limn→∞ g(an) = limn→∞ f(an) = f(x). �

Example: (i) By considering f : (0,∞) → (0,∞) given by f(x) = 1/x and the sequence (1/n), we

see that a continuous function need not take Cauchy sequences to Cauchy sequences (in fact, here

the image sequence is not even bounded).

(ii) If d is the discrete metric on a set X, then every function f : (X, d) → (R, dE) is uniformly

continuous (since we may take δ = 1/2). In particular, if d is the discrete metric on R, then the

identity map I : (R, d) → (R, dE) is uniformly continuous. By considering the sequence (n), we see

that a uniformly continuous function need not take bounded sequences to bounded sequences.

(iii) f : (R, dE) → (R, dE) given by f(x) = x2 takes Cauchy sequences to Cauchy sequences (hint :

Cauchy sequences are bounded and |x2 − y2| = |x+ y||x− y|), and bounded sequences to bounded

sequences, but f is not uniformly continuous because |f(n + 1/n) − f(n)| ≥ 2 for every n ∈ N.If g(x) = x, then g(x)g(x) = f(x), and this shows that the product of two uniformly continuous

(even Lipschitz continuous) functions need not be uniformly continuous.

(iv) f : [0, 1] → [0, 1] defined as f(0) = 0 and f(x) =√x for x > 0 is uniformly continuous (∵

if |x − y| < δ := ε2, then |√x − √

y|2 ≤ |√x +

√y||

√x − √

y| = |x − y| < ε2) but not Lipschitz

continuous (∵ |f(1/n2)− f(0)||1/n2 − 0|

= n for every n ∈ N).

Exercise-19: Let (X, d) be a metric space.


(i) Let z ∈ X, and A ⊂ X be nonempty. Define f, g : X → R as f(x) = d(x, z) and g(x) =

dist(x,A). Then |f(x) − f(y)| ≤ d(x, y) and |g(x) − g(y)| ≤ d(x, y) for every x, y ∈ X. Thus f, g

are Lipschitz continuous and hence uniformly continuous.

(ii) Let Y, Z ⊂ X be disjoint closed sets. Then there is a continuous map f : X → [0, 1] such that

f |Y ≡ 0, f |Z ≡ 1 and f(x) ∈ (0, 1) for every x ∈ X \ (Y ∪ Z).(iii) Let Y, Z ⊂ X be such that Y ∩ Z = ∅ = Y ∩ Z. Then there are disjoint open sets U, V ⊂ X

such that Y ⊂ U and Z ⊂ V .

(iv) If (xn) → x and (yn) → y in X, then (d(xn, yn)) → d(x, y). Hence d : (X ×X, d) → (R, dE) iscontinuous, where d is a metric on X ×X defined as d((x, y), (x′, y′)) = d(x, x′) + d(y, y′).

[Hint : (i) d(x, z) ≤ d(x, y) + d(y, z) and hence f(x)− f(y) = d(x, z)− d(y, z) ≤ d(x, y). Similarly,

f(y) − f(x) ≤ d(y, x) = d(x, y). Given ε > 0, there is a ∈ A with d(y, a) < g(y) + ε, and

hence g(x) ≤ d(x, a) ≤ d(x, y) + d(y, a) < d(x, y) + g(y) + ε, yielding g(x) − g(y) < d(x, y) + ε;

similarly g(y) − g(x) < d(x, y) + ε. Thus |g(x) − g(y)| < d(x, y) + ε. (ii) If Y = ∅ = Z, take

f ≡ 1/2. If Y = ∅ = Z, take f(x) =dist(x, Y )

1 + dist(x, Y ). If Y = ∅ = Z, take f(x) =

1

1 + dist(x,Z).

If Y and Z are nonempty, take f(x) =dist(x, Y )

dist(x, Y ) + dist(x,Z). (iii) Assume Y, Z = ∅. Let

f : X → R be f(x) = dist(x, Y ) − dist(x, Z). Take U = f−1((−∞, 0)) and V = f−1((0,∞)). (iv)

|d(x, y)− d(xn, yn)| ≤ |d(x, y)− d(xn, y)|+ |d(xn, y)− d(xn, yn)| ≤ d(x, xn) + d(y, yn).]

Remark: Let (X, d) be a metric space and ε > 0. By Exercise-19(i), {x ∈ X : d(x, z) < δ} and

{x ∈ X : dist(x,A) < ε} are open in X for each z ∈ X and each nonempty set A ⊂ X.

Exercise-20: Let f : R → R be a continuous function.

(i) If there is a ∈ R \Q such that f(x) = f(x+ 1) = f(x+ a) for every x ∈ R, then f is constant.

(ii) If f is uniformly continuous and f(Z) is a bounded set, then f is bounded.

(iii) Even if f is bounded, f need not be uniformly continuous.

[Hint : (i) f(x) = f(x+m+na) for every x ∈ R and m,n ∈ Z. Also, {m+na : m,n ∈ Z} is dense in

R. (ii) Assume f(Z) ⊂ [−M,M ]. Choose k ∈ N such that |f(a)− f(b)| ≤ 1 whenever |a− b| ≤ 1/k.

For any x ∈ R, there exist x0, x1, . . . , xk ∈ R such that x0 ∈ Z, xk = x, and |xj − xj−1| ≤ 1/k for

1 ≤ j ≤ k. Hence |f(x)| ≤M+k for every x ∈ R. (iii) Consider a continuous function f : R → [0, 1]

such that f(n) = 0 and f(n+ 1/n) = 1 for every natural number n ≥ 2.]

Definition: Let X be a metric space, a ∈ X be such that a ∈ X \ {a}, and f be a function from

either X or X \ {a} to R (it is not necessary that f should be defined at a). We say limx→a f(x)

exists if there is y ∈ R such that limn→∞ f(xn) = y for every sequence (xn) in X \ {a} converging

to a. If this holds, we write limx→a f(x) = y. Note that limx→a f(x) = y ⇔ for every ε > 0, there

is δ > 0 such that |y − f(x)| < δ for every x ∈ B(a, δ) \ {a}.

REAL ANALYSIS 19

Example: Let f : R2 \ {(0, 0)} → R be f(x, y) =x2 + y

|x|+ |y|. Then limn→∞ f(1/n, 0) = 0 = 1 =

limn→∞ f(0, 1/n) and hence lim(x,y)→(0,0) f(x, y) does not exist.

Definition: Let −∞ ≤ u < v ≤ ∞, a ∈ (u, v) and f be a function from either (u, v) or (u, v) \ {a}to R. (i) We say f has left limit equal to y ∈ R at a and we write f(a−) = y if limn→∞ f(xn) = y

for any sequence (xn) in (u, a) converging to a (this is equivalent to the condition that for every

ε > 0 there is δ > 0 such that f((a− δ, a)) ⊂ (y − ε, y + ε)).

(ii) Similarly we say f has right limit equal to y ∈ R at a and we write f(a+) = y if limn→∞ f(xn) =

y for any sequence (xn) in (a, v) converging to a.

(iii) If f(a−) and f(a+) exist as real numbers but f is discontinuous at a (i.e., if either f(a−) =f(a+), or f(a−) = f(a+) but f is not defined at a, or f(a−) = f(a+) = f(a)), then we say f has

a simple discontinuity at a.

(iv) For y ∈ R, note that f(a−) = f(a+) = y ⇔ limx→a f(a) = y.

Example: (i) Let f : R → R be f(0) = 0 and f(x) = 1/x for x = 0. Then 0 is a point of

discontinuity but it is not a simple discontinuity because f(0−) and f(0+) do not exist as real

numbers. (ii) Let f : R → R be f(x) = 0 for x ∈ R \ Q, f(0) = 1, and f(p/q) = 1/q whenever

p ∈ Z \ {0}, q ∈ N, and p, q are coprime. Then f is continuous at every point of R \Q, but f has a

simple discontinuity at every x ∈ Q because f(x−) = f(x+) = 0 < f(x).

Seminar topic: Exercises 16,17, and 19 on p.100 of Rudin.

[113] (i) Let X ⊂ R be a bounded or unbounded interval and f : X ⊂ R be a monotone function.

Then A := {x ∈ X : f is discontinuous at x} is countable (possibly empty).

(ii) If f : R → R is a continuous bijection, then f is a homeomorphism.

(iii) If f : R → R is continuous and satisfies f(x + y) = f(x) + f(y) for every x, y ∈ R, then for

c := f(1) we have that f(x) = cx for every x ∈ R (i.e., f must be an R-linear map).

(iv) If 0 < θ < π/2, then 0 < sin θ < θ.

(v) The functions sinx and cosx are continuous on R with sin 0 = 0 and cos 0 = 1.

(vi) limθ→0sin θ

θ= 1.

Proof. (i) Without loss of generality assume f is increasing. Then f(x−) and f(x+) exist and

f(x−) ≤ f(x) ≤ f(x+) for every x ∈ X. Hence, x ∈ A ⇔ f(x−) < f(x+). Define g : A → Q as

g(x) = any rational number in (f(x−), f(x+)). Since f(x+) ≤ f(y−) for x ≤ y in X, the map g is

strictly increasing and hence injective. Therefore A is countable.

(ii) Being a continuous bijection from R to R, the function f must be strictly monotone, i.e., either

strictly increasing or strictly decreasing. Hence f takes open intervals to open intervals. Now if

U ⊂ R is a nonempty open set, then U is a union of open intervals, and consequently f(U) is open

in R. Thus f is a homeomorphism by [112](iv).


(iii) Since f(0) = f(0 + 0) = f(0) + f(0), we must have f(0) = 0. For x ∈ R, as 0 = f(0) =

f(x − x) = f(x) + f(−x), we must also have f(−x) = −f(x). For x ∈ R and n ∈ N, note that

f(nx) = f(x+ · · ·+ x) = f(x) + · · ·+ f(x) = nf(x). Now consider p, q ∈ N. Then pf(1) = f(p) =

f(q · pq ) = qf(pq ) and hence f(pq ) =

pqf(1). By continuity, f(x) = cx for every real number x ≥ 0

(where c = f(1)), and then f(−x) = −f(x) = −cx = c(−x) for x ≥ 0 as well.

(iv) Consider the points A = (0, 0), B = (cos θ, 0), C = (cos θ, sin θ), and D = (1, 0) in R2. Note

that ABC is a right-angled triangle. Since the length |AC| = 1, we get sin θ = |BC|/|AC| = |BC|Now |BC| < the length of the circular arc DC, which means exactly that sin θ < θ.

(v) Since sin(−x) = − sinx, we get by (iv) that | sinx| < |x| for 0 < |x| < π/2, which implies

sinx is continuous at 0 with sin 0 = 0. Then the identity cos(2x) = 1 − 2 sin2 x implies cosx is

continuous at 0 with cos 0 = 1. Now using the identities sin(x + y) = sinx cos y + sin y cosx and

cos(x+ y) = cosx cos y− sinx sin y, we may see that sinx and cosx are continuous at every x ∈ R.

(vi) Suppose 0 < θ < π/2. Let P = (0, 0), Q = (1, 0), S = (cos θ, sin θ) and T = (1, tan θ) in R2.

Note that T is the point where the line passing through P and S intersects the line x = 1. Hence

the area of the circular sector PQS is < the area of the triangle PQT . This means θ/2 < (tan θ)/2.

Combining with (iv), we see cos θ <sin θ

θ< 1. Since cos θ → 1 as θ → 0+ by (v), we get by

Sandwich property that limθ→0+sin θ

θ= 1. Since sin(−θ) = − sin θ, the result follows. �

Example: Let {xj : j ∈ J} be a Hamel basis of the vector space R over Q (then necessarily J

is uncountable) and suppose that xj0 = 1 for some j0 ∈ J . Define f : R → R as follows: for

x =∑

j∈F αjxj ∈ R (where F ⊂ J is finite), f(x) = αj0 if j0 ∈ F , and f(x) = 0 if j0 /∈ F . In other

words, f : R → R is the unique Q-linear map with f(xj0) = 1 and f(xj) = 0 for every j = j0. Then

f(x+ y) = f(x) + f(y) for every x, y ∈ R and f(x) = x for every x ∈ Q. The function f cannot be

continuous by [113](iii).

Remark: Let X be a metric space and f : X → R be a function. We say f is lower semi continuous

if {x ∈ X : f(x) > y} = f−1((y,∞)) is open in X for every y ∈ R, and upper semi continuous

if {x ∈ X : f(x) < y} = f−1((−∞, y)) is open in X for every y ∈ R. For example, the indicator

function 1A : X → R of a subset A ⊂ X (defined as 1A(x) = 1 if x ∈ A and 1A(x) = 0 if x ∈ X \A)is lower semi continuous when A is open in X, and upper semi continuous when A is closed in X.

For a function f : X → R, the following can be verified:

(i) f is continuous ⇔ f is both lower semi continuous and upper semi continuous.

(ii) f is lower semi continuous ⇔ for every x ∈ X and ε > 0, there is δ > 0 with f(B(x, δ)) ⊂(f(x)− ε,∞) ⇔ lim infn→∞ f(xn) ≥ f(x) whenever (xn) → x in X.

(iii) f is upper semi continuous ⇔ for every x ∈ X and ε > 0, there is δ > 0 with f(B(x, δ)) ⊂(−∞, f(x) + ε) ⇔ lim supn→∞ f(xn) ≤ f(x) whenever (xn) → x in X.

REAL ANALYSIS 21

5. Completeness

Definition: A metric space (X, d) is complete if every Cauchy sequence in X is convergent in X.

Example: (i) (Q, dE) is not complete since a sequence of rational numbers converging to an irrational

number is a Cauchy sequence in Q but does not converge in Q. (ii) Every discrete metric space is

complete since every Cauchy sequence there is eventually constant. More generally, let (X, d) be

a metric space with for which there is a δ > 0 such that d(x, y) ≥ δ for any two distinct points

x, y ∈ X. Then every Cauchy sequences in X is eventually constant, and hence (X, d) is complete.

[114] (i) (Rm, dE) is complete.

(ii) Let (X, d) be a complete metric space and A ⊂ X. Then the metric space (A, d) is complete

⇔ A is closed in X.

(iii) Let (X, d) be a metric space, A ⊂ X be dense in X, and (Y, d′) be a complete metric space. If

f : (A, d) → (Y, d′) is uniformly continuous, then there is a unique uniformly continuous function

f : (X, d) → (Y, d′) with f |A = f .

(iv) If (X, d) and (Y, d′) are complete metric spaces, then X × Y is a complete metric space w.r.to

the metric d defined as d((x1, y1), (x2, y2)) = d(x1, x2) + d′(y1, y2); and also w.r.to the metric d∞

defined as d∞((x1, y1), (x2, y2)) = max{d(x1, x2), d′(y1, y2)}.

Proof. (i) Let (xn) be a Cauchy sequence in Rm. By Exercise-17(iii), every Cauchy sequence

is bounded. Hence by [110](ii), there is a subsequence (xnk) converging to some x ∈ Rm. By

Exercise-17(iv), (xn) → x in Rm.

(ii) ⇒: If x ∈ A, there is a sequence (xn) in A converging to x by [109]. Then (xn) is a Cauchy

sequence in A, and hence by the completeness of A, should converge to some y ∈ A. By the

uniqueness of the limit (Exercise-17(i)), we get x = y ∈ A, and thus A is closed. ⇐: If (xn) is

Cauchy in (A, d), then (xn) → x for some x ∈ X by the completeness of X. Then x ∈ A = A by

[109] and the closedness of A.

(iii) Consider x ∈ X. Choose a sequence (an) in A converging to x. By [112](i), (f(an)) is Cauchy in

the complete space (Y, d′). Define f(x) = limn→∞ f(an). If (bn) is another sequence in A converging

to x, then limn→∞ f(bn) = limn→∞ f(an) by the second assertion of [112](i). Hence the value f(x)

is independent of the particular choice of a sequence in A converging to x. Clearly f |A = f . Given

ε > 0, choose δ > 0 such that d′(f(a), f(b)) < ε/3 for every a, b ∈ A with d(a, b) < δ. Now if

u, v ∈ X are with d(u, v) < δ, choose r > 0 with d(u, v) + 2r < δ. There are a ∈ A ∩ B(u, r) and

b ∈ A ∩ B(v, r) such that d′(f(u), f(a)) < ε/3 and d′(f(v), f(b)) < ε/3. Also d′(f(a), f(b)) < ε/3

because d(a, b) < d(x, y) + 2r < δ. Therefore, d′(f(u), f(v)) < ε by triangle inequality. Thus f is

uniformly continuous. Uniqueness of the extension follows by [112](vi).


(iv) We give an outline and leave the details to the student. If ((xn, yn)) is a Cauchy sequence in

X × Y , then (xn) is Cauchy in X and (yn) is Cauchy in Y . Since X,Y are complete, there are

x ∈ X and y ∈ Y such that (xn) → x and (yn) → y. Then ((xn, yn)) → (x, y) in X × Y . �

To contrast with [115] below, note that∩∞

n=1[n,∞) = ∅ in R.

[115] Let (X, d) be a complete metric space. Let (An) be a sequence of nonempty closed subsets

of X with A1 ⊃ A2 ⊃ · · · and limn→∞ diam(An) → 0. Then∩∞

n=1An is a singleton.

Proof. Let an ∈ An. If m,n ≥ k, then am ∈ Am ⊂ Ak and an ∈ An ⊂ Ak and hence d(am, an) ≤diam(Ak). From this observation it follows that (an) is Cauchy since diam(Ak) → 0. Since X is

complete, (an) converges to some a ∈ X. As am ∈ Am ⊂ An for every m ≥ n and An is closed,

we get a ∈ An for every n. Thus a ∈∩∞

n=1An. If there is b = a in this intersection, then for

δ := d(a, b) > 0, we will get that diam(An) ≥ δ for every n ∈ N, a contradiction. �

Definition: Let (X, d) be a metric space. We say a subset A ⊂ X is totally bounded if for every

ε > 0, there are finitely many points x1, . . . , xk ∈ X such that A ⊂∪k

j=1B(xj , ε).

Remark: Every totally bounded subset of a metric space is bounded; but the converse is not true

- consider R as a subset of itself w.r.to the discrete metric. However, we have the following fact:

Exercise-21: (i) Every bounded subset of (Rm, dE) is totally bounded (Heine-Borel theorem and

Arzela-Ascoli theorem depend on this fact).

(ii) Let (X, d) be a metric space, and ∅ = A ⊂ X. Then A is totally bounded⇔ A is totally bounded

⇔ For every ε > 0, there are finitely many points a1, . . . , ak ∈ A such that A ⊂∪k

j=1B(aj , ε).

[Hint : (i) Suppose m = 2 and ε > 0. Any bounded subset of R2 can be put inside a square. Any

square in R2 can be covered with finitely many squares of diameter (length of diagonal) < 2ε,

and any square of diameter < 2ε can be covered by an open disc of radius ε. (ii) Assume A is

totally bounded. Given ε > 0, there are x1, . . . , xk ∈ X such that A ⊂∪k

j=1B(xj , ε/2). Assume

A∩B(xj , ε/2) = ∅ for every j. If we pick aj ∈ A∩B(xj , ε/2), then B(xj , ε/2) ⊂ B(aj , ε) and hence

A ⊂∪k

j=1B(aj , ε). Conversely, if A ⊂∪k

j=1B(aj , ε/2), then A ⊂∪k

j=1B(aj , ε/2) ⊂∪k

j=1B(aj , ε).]

The notion of total boundedness has the following characterization in terms of Cauchy sequences.

[116] (i) Let (X, d) be a metric space and A ⊂ X. Then A is totally bounded ⇔ every sequence in

A has a Cauchy subsequence.

(ii) If X is a complete and totally bounded metric space, then every sequence in X has a convergent

subsequence.

Proof. (i) ⇒: Let (xn) be a sequence in A. Cover A with finitely many open balls of radius 1/2.

One of them, say B1, must satisfy xn ∈ B1 for infinitely many n ∈ N. Let n1 ∈ N be the smallest

with xn1 ∈ B1. At the (k+1)th step, cover A with finitely many open balls of radius 2−(k+1). One

of them, say Bk+1, must satisfy xn ∈ Bk+1 ∩ Bk for infinitely many n ∈ N. Let nk+1 > nk be the

REAL ANALYSIS 23

smallest integer with xnk+1∈ Bk+1 ∩ Bk. Then d(xnk+1

, xnk) ≤ diam(Bk) ≤ 2−(k−1), and hence

(xnk) is Cauchy by Exercise-17(v).

⇐: If A is not totally bounded, then there is ε > 0 such that A cannot be covered with finitely

many open balls of radius ε. Inductively choose x1 ∈ A and xn+1 ∈ A \∪n

j=1B(xj , ε). Then

d(xm, xn) ≥ ε for every m = n, and hence the sequence (xn) in A has no Cauchy subsequence.

(ii) If (xn) is a sequence in X, then it has a Cauchy subsequence (xnk) by (i), and then this Cauchy

subsequence must converge in X by the completeness of X. �

Seminar topic: If X is a complete metric space and f : X → X is a contraction, then f has a

unique fixed point (see Theorem 4.48 in Apostol).

6. Compactness

In set theory, the simplest kind of set is a finite set. In the theory of metric spaces, the analogue

of a finite set is a compact set. It can be expressed in many equivalent ways.

Definition: Let (X, d) be a metric space and A ⊂ X. We say:

(i) A is sequentially compact if every sequence in A has a subsequence converging to some x ∈ A.

(ii) A is limit point compact (or has the Bolzano-Weierstrass property) if every infinite subset of A

has a limit point in A.

(iii) A is compact if every open cover U of A in X has a finite subcover, i.e., if U = {Uj : j ∈ J} is

a collection of open subsets of X with A ⊂∪

j∈J Uj , then there is a finite subcollection {Ujk : 1 ≤k ≤ p} of U such that A ⊂

∪pk=1 Ujk .

(iv) A has the Lebesgue number property if for any open cover U = {Uj : j ∈ J} of A in X, there

is δ > 0 (called a Lebesgue number) such that for any subset Y ⊂ A with diam(Y ) < δ, there is

j ∈ J with Y ⊂ Uj . For example, every discrete metric space has the Lebesgue number property

since any δ ∈ (0, 1) works.

Example: Let X be a metric space. Then every finite subset of X is compact. Moreover, if (xn) → x

in X, then A := {x} ∪ {xn : n ∈ N} is a compact subset of X.

Non-example: Let d be the discrete metric on R. (i) (R, dE) and (R, d) are not compact since the

open cover {(−n, n) : n ∈ N} has no finite subcover. (ii) (R, dE) and (R, d) are not sequentially

compact since the sequence 1, 2, 3, . . . has no convergent subsequence. (iii) (R, dE) and (R, d) are

not limit point compact since the infinite subset N has no limit points in the full space. (iv) (R, dE)does not have the Lebesgue number property: if we consider the open cover U = {(−∞,

3

2)} ∪

{(n, n+1+1

n) : n ∈ N} of R, then the sequence (Yn) of sets defined as Yn = [n+1, n+1+

1

n] has

the property that diam(Yn) → 0, but no Yn is contained in any member of U .


[117] [Compactness theorem for metric spaces] Let (X, d) be a metric space and A ⊂ X. Then the

following are equivalent: (i) A is compact.

(ii) Every open cover of A using open balls has a finite subcover.

(iii) A is limit point compact (has the Bolzano-Weierstrass property).

(iv) A is sequentially compact.

(v) A is totally bounded and (A, d) is complete.

(vi) A is totally bounded and has the Lebesgue number property.

Proof. (i) ⇒ (ii) is trivial. (ii) ⇒ (iii): Assume Y ⊂ A has no limit points in A. We will show Y

is finite. For each a ∈ A, there is an open ball Ba in X centered at a such that Y ∩Ba ⊂ {a}. Now

{Ba : a ∈ A} is a collection of open balls covering A. Let {Baj : 1 ≤ j ≤ k} be a finite subcover.

Then Y ⊂ {aj : 1 ≤ j ≤ k}.

(iii) ⇒ (iv): Consider a sequence (xn) in A. We may assume (xn) has no constant subsequences.

Then in particular, Y := {xn : n ∈ N} must be infinite, and hence Y has a limit point x ∈ A by

(iii). Let n0 ∈ N be such that xn = x for every n ≥ n0. By [109](vi), inductively we may choose

n0 < n1 < n2 < · · · such that xnk∈ B(x, rk), where rk = min{1/k, d(x, xnk−1

)}. Then (xnk) → x.

(iv) ⇒ (v): By (iv), every sequence in A has a Cauchy subsequence, and hence A is totally bounded

by [116](i). Next, if (xn) is a Cauchy sequence in A, then a subsequence converges to some x ∈ A

by (iv), and hence (xn) → x by Exercise-17(iv). This shows that (A, d) is complete.

(v) ⇒ (vi): Let U = {Uj : j ∈ J} be an open cover for A. If U has no Lebesgue number w.r.to A,

then there is a sequence (Yn) of subsets of A such that diam(Yn) → 0 and Yn is not a subset of Uj

for every n ∈ N and every j ∈ J . Let yn ∈ Yn. By [116](i), (yn) has a Cauchy subsequence. So

after a relabeling, we may assume (yn) itself is Cauchy. Since (A, d) is complete, (yn) converges to

some a ∈ A. Let j0 ∈ J be such that a ∈ Uj0 , and r > 0 be such that B(a, 2r) ⊂ Uj0 . Choose n ∈ Nlarge enough so that yn ∈ B(a, r) and diam(Yn) < r. Then Yn ⊂ B(a, 2r) ⊂ Uj0 , a contradiction.

(vi) ⇒ (i): Let U = {Uj : j ∈ J} be an open cover for A. Let δ > 0 be a Lebesgue number for U .Since A is totally bounded, A can be covered with finitely many balls B1, . . . , Bp of diameter < δ

(take radius = δ/3). Since δ is a Lebesgue number, there are jk ∈ J with Bk ⊂ Ujk for 1 ≤ k ≤ p.

Then A ⊂∪p

k=1Bk ⊂∪p

k=1 Ujk . Thus A is compact. �

Any of the equivalent conditions in [117] may be used to tackle compact metric spaces. In the

rest of this section, we present some of the frequently used properties of compact metric spaces.

[118] (i) Let (X, d) be a metric space, and K ⊂ X be compact. Then K is closed and bounded in

X, and every closed subset A of K is compact.

(ii) [Heine-Borel theorem] K ⊂ (Rm, dE) is compact ⇔ K is closed and bounded.

(iii) If A ⊂ Rm is a non-compact set, then there exists an unbounded continuous function f : A→ R.

REAL ANALYSIS 25

(iv) If A ⊂ Rm is a nonempty closed set and x ∈ Rm, then there is a ∈ A with dE(x, a) = dist(x,A).

Proof. (i) If x ∈ K, then there is a sequence (xn) in K converging to x. As K is sequentially

compact, a subsequence of (xn) must converge to a point y ∈ K, and then by the uniqueness of the

limit, we get x = y ∈ K. Thus K is closed. Next, K is totally bounded and hence bounded. If A

is closed in K, and (xn) is a sequence in A, then a subsequence converges to some x ∈ K since K

is sequentially compact. Then x ∈ A = A, and this shows A is (sequentially) compact.

(ii) ⇐: K is totally bounded by Exercise-21(i). As K is closed in the complete space Rm, the space

(K, dE) is complete by [114](ii). Hence K is compact by [117].

(iii) If A is unbounded, take f : A → R to be f(x) = ∥x∥ = dE(0, x). If A is bounded but not

closed, let z ∈ Rm \A be a limit point of A, and define f : A→ R as f(x) = 1/dE(x, z).

(iv) Let r = dist(x,A), andK = A∩B(x, r + 1). ThenK is closed and bounded, and hence compact

by (ii). Also, dist(x,K) = dist(x,A) = r. Choose a sequence (an) in K with (dE(x, an)) → r. By

the compactness of K, a subsequence (ank) converges to some a ∈ K ⊂ A. Then dE(x, a) = r. �

Remark: (i) In the Heine-Borel theorem, the metric under consideration is dE . If we use the metric

d = min{1, dE}, then R is closed and bounded in (R, d), but is not compact since the open cover

{(−n, n) : n ∈ N} of R has no finite subcover. (ii) Sequential compactness is a purely intrinsic

characterization of compactness in metric spaces (it does not refer to the super set). Therefore, if

(X, d) is a metric space and K ⊂ Y ⊂ X, then K is compact in (Y, d) ⇔ K is compact in (X, d).

[119] Let (X, d) be a compact metric space, (Y, d′) be a metric space, and f ∈ C(X,Y ). Then,

(i) f(X) is compact, and f is uniformly continuous.

(ii) If f : X → Y is a continuous bijection, then f is a homeomorphism.

(iii) If (Y, d′) = (R, dE), then f is bounded, and attains its maximum and minimum (i.e., there are

a, b ∈ X such that f(a) = min f(X) and f(b) = max f(X).

Proof. (i) Use sequential compactness to see f(X) is (sequentially) compact. Given ε > 0, cover

f(X) with finitely many balls B1, . . . , Bk of radius ε/3. Then U = {f−1(Bj) : 1 ≤ j ≤ k} is an

open cover for X. If δ > 0 is a Lebesgue number for U , then for every a, b ∈ X with d(a, b) < δ,

there is j ∈ {1, . . . , k} with {a, b} ⊂ f−1(Bj), and hence d′(f(a), f(b)) ≤ diam(Bj) < ε. Thus f

is uniformly continuous. Another proof for uniform continuity is: given ε > 0, choose δx > 0 with

f(B(x, δx)) ⊂ B(f(x), ε), use the compactness of X to extract a finite subcover {B(xj , δxj/2) : 1 ≤j ≤ k} of the cover {B(x, δx/2) : x ∈ X}, and check that δ := min{δxj/2 : 1 ≤ j ≤ k} works.

(ii) Let A ⊂ X be closed. By [112](iv), it suffices to show f(A) is closed in Y . By [118](i), A is

compact, and hence f(A) is compact by (i). Again by [118](i), f(A) is closed in Y .

(iii) f(X) is compact by (i), and hence f(X) is bounded. Let y = inf f(X) and z = sup f(X). Let

(an), (bn) be sequences in X with (f(an)) → y and (f(bn)) → z. By the sequential compactness of


X, there exist subsequences (ank) and (bmk

) converging to two points (say) a, b ∈ X respectively.

Then f(a) = y and f(b) = z by continuity. �

Exercise-22: Let f : R → R be a continuous function.

(i) If ∃ u, v ∈ R with limx→−∞ f(x) = u and limx→∞ f(x) = v, then f is uniformly continuous.

(ii) If f is periodic, i.e., if there is a > 0 such that f(x + a) = f(x) for every x ∈ R, then f is

uniformly continuous, bounded, and attains its maximum and minimum.

[Hint : (i) Let ε > 0. Choose r > 0 such that |u − f(x)| < ε/2 for x < −r and |v − f(x)| < ε/2

for x > r. Also, by the uniform continuity of f on the compact interval [−(r + 1), r + 1], there

is δ ∈ (0, 1) such that |f(a) − f(b)| < ε/2 for every a, b ∈ [−(r + 1), r + 1] with |a − b| < δ. (ii)

Given ε > 0, choose δ ∈ (0, a) such that |f(s) − f(t)| < ε for every s, t ∈ [0, 2a] with |s − t| < δ.

Now if x, y ∈ R are with x ≤ y < x + δ, then choosing m ∈ Z with x + ma ∈ [0, a], note that

y+ma ∈ [0, 2a] and |(x+ma)− (y+ma)| < δ so that |f(x)−f(y)| = |f(x+ma)−f(y+ma)| < ε.]

Remark: Let (X, d), (Y, d′) be metric spaces and f : X → Y be a function. If f is continuous, verify

that its graph G(f) := {(x, f(x)) : x ∈ X} is closed inX×Y w.r.to the metric d((x1, y1), (x2, y2)) :=

d(x1, x2) + d′(y1, y2). In the other direction, even if X is compact and G(f) is closed in X × Y ,

f need not be continuous: consider f : [0, 1] → R defined as f(0) = 0 and f(x) = 1/x for x > 0.

But if Y is compact and G(f) is closed in X × Y , then f is continuous. Proof : Let (xn) → x in

X. If (f(xn)) does not converge to f(x), then there exist δ > 0 and a subsequence (xnk) such that

d′(f(x), f(xnk)) ≥ δ for every k ∈ N. As Y is compact, by passing onto a subsequence we may

assume (f(xnk)) converges to some y ∈ Y . Then d′(f(x), y) ≥ δ, and hence (x, y) /∈ G(f). But

(x, y) ∈ G(f) since (xnk, f(xnk

)) → (x, y). Thus G(f) cannot be closed, a contradiction.

Remark: Let X be a compact metric space and f : X → R be a function. We may verify that: (i)

if f is lower semi continuous, then f is bounded below and attains its minimum, and (ii) if f is

upper semi continuous, then f is bounded above and attains its maximum.

Definition: Let (X, d) be a metric space, and Y, Z ⊂ X be nonempty. The distance between Y

and Z is defined as dist(Y, Z) = inf{d(y, z) : y ∈ Y and z ∈ Z}. Note Y := {(x, y) ∈ R2 : x >

0 and xy = 1} and Z := {(x, y) ∈ R2 : y = 0} are disjoint closed subsets of R2, but dist(Y,Z) = 0.

Exercise-23: Let (X, d) be a metric space, and Y, Z ⊂ X be nonempty.

(i) If Y is compact, there is y ∈ Y such that dist(y, Z) = dist(Y,Z).

(ii) If Y and Z are compact, then there are y ∈ Y and z ∈ Z such that d(y, z) = dist(Y, Z).

(iii) Suppose Y is compact and Z is closed. Then Y ∩ Z = ∅ ⇔ dist(Y, Z) = 0.

[Hint : (i) Let (yn) be a sequence in Y such that dist(Y, Z) = limn→∞ dist(yn, Z). Since Y is se-

quentially compact, a subsequence (ynk) converges to some y ∈ Y . By the continuity of the distance

function, dist(Y, Z) = limk→∞ dist(ynk, Z) = dist(y, Z). (ii) Let (yn), (zn) be sequences in Y, Z re-

spectively such that dist(Y, Z) = limn→∞ d(yn, zn). Since Y is sequentially compact, a subsequence

REAL ANALYSIS 27

(ynk) converges to some y ∈ Y . As Z is also sequentially compact, replacing (nk) with a further

subsequence, assume (znk) converges to some z ∈ Z. Then dist(Y,Z) = limk→∞ dist(ynk

, znk) =

dist(y, z). (iii) ⇐: Assume dist(Y, Z) = 0. By (i), there is y ∈ Y with 0 = dist(Y, Z) = dist(y, Z).

This implies y ∈ Z by Exercise-18(i) since Z is closed.]

[120] Let (X, d) be a metric space. (i) Let K ⊂ X be compact, U ⊂ X be open, and K ⊂ U . Then

there is δ > 0 with B(K, δ) ⊂ U , where B(K, δ) :=∪

x∈K B(x, δ), the δ-neighborhood of K.

(ii) Let U ⊂ X be open, and (Kn) be a decreasing sequence of compact sets in X such that∩∞n=1Kn ⊂ U . Then there is n0 ∈ N such that Kn ⊂ U for every n ≥ n0.

(iii) If (Kn) is a decreasing sequence of nonempty compact sets in X, then∩∞

n=1Kn = ∅.

Proof. (i) For each x ∈ K, there is δx > 0 with B(x, 2δx) ⊂ U . The open cover {B(x, δx) : x ∈ K}of K has a finite subcover {B(xj , δxj ) : 1 ≤ j ≤ k}. Let δ = min{δxj : 1 ≤ j ≤ k}. If z ∈ B(K, δ),

then there is y ∈ K with d(y, z) < δ, and then there is j ∈ {1, . . . , k} with d(xj , y) < δxj . Hence

d(xj , z) < δxj + δ ≤ 2δxj , implying z ∈ B(xj , 2δxj ) ⊂ U . This shows B(K, δ) ⊂ U .

(ii) Note that K1 \ U , being a closed subset of the compact set K1, is compact. Let Un = X \Kn,

which is open in X and Un ⊂ Un+1. Since∩∞

n=1Kn ⊂ U , we see X \ U ⊂∪∞

n=1 Un. In particular

{Un : n ∈ N} is an open cover for the compact set K1 \U . Since Un’s are also increasing, we deduce

that there is p ∈ N such that K1 \ U ⊂ Up. Then Kp \ U ⊂ K1 \ U ⊂ Up = X \Kp, which implies

Kp \ U = ∅, and hence Kp ⊂ U . Therefore Kn ⊂ Kp ⊂ U for every n ≥ p. Another proof : Else

there exist a strictly increasing sequence (nj) of natural numbers and points xj ∈ Knj \U ⊂ K1 \Ufor every j ∈ N. Since K1 \ U is compact, by passing onto a subsequence we may assume (xj)

converges to some x ∈ K1 \ U . Now, for each j ∈ N, we have that xnm ∈ Knm ⊂ Knj for every

m ≥ j, and hence x ∈ Knj for every j ∈ N. But∩∞

j=1Knj =∩∞

n=1Kn since Kn’s are decreasing.

Hence we deduce that x ∈∩∞

n=1Kn ⊂ U , a contradiction.

(iii) If∩∞

n=1Kn = ∅, then by applying (ii) with U = ∅, we get Kn = ∅ for all sufficiently large

n ∈ N, a contradiction. A direct proof : If∩∞

n=1Kn = ∅, then letting Un = X \Kn, we see that (Un)

is an increasing sequence of open sets with X =∪∞

n=1 Un. As K1 is compact, there must exist some

p ∈ N with K1 ⊂ Up = X \Kp, which is not possible because Kp is nonempty and Kp ⊂ K1. �

7. Connectedness

Definition: Let (X, d) be a metric space. (i) We say X is connected if the only clopen subsets (i.e.,

subsets which are both closed and open) of X are ϕ and X. Otherwise X is said to be disconnected.

A subset A ⊂ X is said to be connected if (A, d) is connected. Note that X is connected iff every

continuous function f : X → {0, 1} is constant (∵ f−1(0) and f−1(1) are clopen subsets of X).

(ii) A pair (U, V ) of open subsets of X is said to be a disconnection (or separation) for a subset A

if A∩U = ∅, A∩ V = ∅, A ⊂ U ∪ V , and A∩U ∩ V = ∅. Note that A ⊂ X fails to be a connected


set iff A has a disconnection (U, V ) in X. Evidently, if (U, V ) is a disconnection for A ⊂ X, then

the sets A ∩ U and A ∩ V are clopen in A. Also note that for a disconnection (U, V ) of A, it can

happen that U ∩ V = ∅ since we require only that A ∩ U ∩ V = ∅.

[121] For a nonempty subset A ⊂ (R, dE), the following are equivalent: (i) A is connected.

(ii) If x ≤ y are points in A, then [x, y] ⊂ A.

Thus R and all (bounded or unbounded) intervals are connected. On the other hand, {0, 1}, N,Z, Q, R \Q are not connected.

Proof. (i) ⇒ (ii): If x < z < y are such that x, y ∈ A but z /∈ A, then taking U = (−∞, z) and

V = (z,∞), we may see that (U, V ) is a disconnection for A in R,

(ii) ⇒ (i): Consider a continuous function f : A→ {0, 1}, and we have to show f is constant. Let

x < y be two points in A, and we need to show f(x) = f(y). By (ii), we have [x, y] ⊂ A. Since f

restricted to the compact set [x, y] is a uniformly continuous function to the set {0, 1}, there is δ > 0

with the following property: whenever u, v ∈ [x, y] are with |u − v| < δ, then |f(u) − f(v)| < 1/2

and consequently f(u) = f(v). Choose finitely many points x = x0 < x1 < · · · < xn = y such that

|xj−1 − xj | < δ for every j ∈ {1, . . . , n}. Then f(x) = f(x0) = f(x1) = · · · = f(xn) = f(y). �

Definition: Let X be a metric space. A continuous map ψ : [a, b] → X is called a path in X from

ψ(a) to ψ(b). After pre-composing ψ with an affine map (i.e., a map of the form x 7→ c1x + c2),

if necessary we may assume that [a, b] = [0, 1]. We say X is path connected if for every x, y ∈ X,

there is a path ψ : [0, 1] → X such that ψ(0) = x and ψ(1) = y.

Example: (i) A finite metric space is (path) connected iff it is a singleton. (ii) Rm is path connected

for everym ∈ N, and Rm\F is path connected for everym ≥ 2 and every finite subset F ⊂ Rm. (iii)

The union and intersection of two (path) connected sets need not be connected; the intersection

of the upper and lower halves of the unit circle is a discrete set with two elements. (iv) The

intersection of a decreasing sequence (An) of (path) connected subsets need not be connected;

consider An := [−1, 1] × [−1/n, 1/n] \ {(0, 0)} in R2. (v) Let Y = {(x, sin 1/x) : x ∈ (0, 1]} and

Z = {0} × [−1, 1]. The subset Y = Y ∪ Z of R2 is connected by [122](iv) below, but is not path

connected since there is no path in Y from a point of Y to a point Z. As Y is path connected, we

also note that the closure of a path connected subset need not be path connected.

[122] (i) Continuous image of a (path) connected space is (path) connected.

(ii) Every path connected metric space is connected.

(iii) If X,Y are (path) connected, then X × Y is (path) connected.

(iv) If X is a metric space and Y ⊂ X is connected, then any Z ⊂ X with Y ⊂ Z ⊂ Y is connected;

and in particular, Y is connected.

REAL ANALYSIS 29

(v) If (Kn) is a decreasing sequence of nonempty compact connected sets in a metric space X, then∩∞n=1Kn is nonempty, compact, and connected.

Proof. (i) Let X,Y be metric spaces and f ∈ C(X,Y ) be surjective. If ψ : [0, 1] → X is a path in

X, then f ◦ ψ is a path in Y . Also, if (U, V ) is a disconnection for Y , then (f−1(U), f−1(V )) is a

disconnection for X. It follows that Y is (path) connected whenever X is.

(ii) Let X be path connected. If X is not connected, there is a nonempty clopen set Y ( X. Let

y ∈ Y , z ∈ X \ Y and ψ : [0, 1] → X be a (continuous) path with ψ(0) = y and ψ(1) = z. Then

ψ−1(Y ) is a nonempty, proper clopen subset of [0, 1]; a contradiction because [0, 1] is connected.

(iii) If ψ1 : [0, 1] → X and ψ2 : [0, 1] → Y are paths, then ψ := (ψ1, ψ2) : [0, 1] → X × Y is a path

in X × Y , and it follows that X × Y is path connected if whenever X and Y are. Next suppose X

are Y are connected, and consider a continuous function f : X ×Y → {0, 1}. We need to show f is

constant. Note that f is constant on X×{y} for each y ∈ Y because X is connected. Similarly, f is

constant on {x}×Y for each x ∈ Y because Y is connected. Hence f(x1, y1) = f(x2, y1) = f(x2, y2)

for any (x1, y1), (x2, y2) ∈ X × Y .

(iv) Any continuous f : Z → {0, 1} must be constant because it is constant on the dense subset Y .

(v) K :=∩∞

n=1Kn is nonempty and compact by [120](iii). If K is not connected, there is a pair

(U, V ) of nonempty open sets in X forming a disconnection for K. Since K ⊂ U ∪V , there is k ∈ Nsuch that Kn ⊂ U ∪V for every n ≥ k by [120](ii). Fix x ∈ K and suppose x ∈ U . Since x ∈ Kk∩Uand Kk ⊂ U ∪ V , it follows by the connectedness of Kk that Kk ⊂ U . Then K ⊂ Kk ⊂ U , which

contradicts the earlier assumption that (U, V ) is a disconnection for K. �

Exercise-24: (i) Let X be a metric space with at least two elements in it. If X is connected, then

there is a continuous surjection f : X → [0, 1] and hence X must be uncountable.

(ii) Let X be a metric space. If Yj ⊂ X is (path) connected for every j ∈ J and∩

j∈J Yj = ∅, thenY :=

∪j∈J Yj is (path) connected. In particular, if ∅ = Y1 ⊂ Y2 ⊂ · · · are (path) connected subsets

of X, then∪∞

n=1 Yn is (path) connected.

[Hint : (i) Let a, b ∈ X be distinct and define f : X → [0, 1] as f(x) =d(a, x)

d(a, x) + d(b, x). Then f is

continuous and hence f(X) is connected. Also f(a) = 0 and f(b) = 1. Therefore f(X) = [0, 1]. (ii)

Suppose Yj ’s are connected. Let z ∈∩

j∈J Yj and consider a continuous function f : Y → {0, 1}.Then f restricted to each Yj is the constant f(z) because Yj is connected.]

[123] (i) [Intermediate value theorem] Let f : [a, b] → R be continuous and y ∈ R be such that

either f(a) ≤ y ≤ f(b) of f(b) ≤ y ≤ f(a). Then there is x ∈ [a, b] with f(x) = y.

(ii) If f : [a, b] → R is a continuous map such that either [a, b] ⊂ f([a, b]) or [a, b] ⊃ f([a, b]), then

there is x ∈ [a, b] with f(x) = x.

(iii) If f : (a, b] → (a, b] is a continuous surjection, then then there is x ∈ (a, b] with f(x) = x.


(iv) Let S1 = {z ∈ C : |z| = 1} (the unit circle), and f : S1 → R be continuous. Then there is

z ∈ S1 with f(z) = f(−z), and in particular f is not injective.

(v) If f : S1 → S1 is continuous and injective, then f is a homeomorphism.

(vi) If f : R → S1 is continuous and injective, then f cannot be surjective.

Proof. (i) f([a, b]) is connected by [121] and [122](i), and hence y ∈ f([a, b]) by hypothesis.

(ii) Apply Intermediate value theorem to g : [a, b] → R defined as g(x) = f(x)− x.

(iii) If f(b) = b, then there is c ∈ (a, b) with f(c) = b. Now g : [c, b] → R defined as g(x) = f(x)−xis continuous, g(c) ≥ 0 and g(b) ≤ 0. By (i), there is x ∈ [c, b] with 0 = g(x) = f(x)− x.

(iv) Let g : [−1, 1] → S1 be g(x) = x + i√1− x2 (the upward projection), and h : [−1, 1] → R be

h(x) = f(g(x)) − f(−g(x)). Then h is continuous, and h(1) = f(1) − f(−1) = −h(−1). Hence

by Intermediate value theorem, there is x ∈ [−1, 1] with h(x) = 0. Letting z = g(x), we see

0 = h(x) = f(z)− f(−z) and hence f(z) = f(−z).

(v) Let a, b ∈ S1 be distinct, and A1, A2 be the two closed arcs of S1 having end points a and b.

Then A1 ∩A2 = {a, b} and A1 ∪A2 = S1. Since f is injective, f(a) = f(b). By injectivity and the

intermediate value property, f(A1) and f(A2) must be the two closed arcs of S1 having end points

f(a) and f(b). Hence f(S1) = f(A1) ∪ f(A2) = S1. Thus f is surjective. Since S1 is compact, it

follows by [119](ii) that f is a homeomorphism.

(vi) Let U = f((0,∞)) and V = f((−∞, 0)). By continuity and injectivity, U and V must be

disjoint open arcs of S1. If f is surjective, then we must have S1 \ {f(0)} = U ∪ V , which implies

(U, V ) is a disconnection for S1 \ {f(0)}. This is not possible because S1 \ {f(0)} is connected.

Another proof : If f is also surjective, then {f((x, x+1)) : x ∈ R} is an open cover for the compact

set S1, and hence there are x1, . . . , xk ∈ R with S1 =∪k

j=1 f((xj , xj + 1)) = f(∪k

j=1(xj , xj + 1)),

which contradicts the injectivity of f because∪k

j=1(xj , xj + 1) = R. �

Example: Let f : [0, 1] → [0, 1] be f(x) = 1 − x/2 for 0 ≤ x ≤ 1/2 and f(x) = (1 − x)/2 for

1/2 < x ≤ 1. Then f is decreasing and injective, but f has no fixed points. Let g : [0, 1] → [0, 1]

be g(x) = (6x+ 1)/4 for 0 ≤ x ≤ 1/2 and g(x) = (6x− 3)/4 for 1/2 < x ≤ 1. Then g is surjective,

but has no fixed points. Contrast these examples with Exercise-25(i) below.

Exercise-25: (i) If f : [a, b] → [a, b] is increasing, then f has a fixed point.

(ii) If f : [a, b] → R is a monotone function having the intermediate value property, then f is

continuous. On the other hand, the function f : R → R defined as f(0) = 0 and f(x) = sin(1/x)

for x = 0 has the intermediate value property, but f is not continuous at 0.

(iii) Let m ≥ 2 and A ⊂ Rm be countable. Then Rm \A is path connected.

(iv) Let m ∈ N, and U ⊂ Rm be a nonempty open set. If U is connected, then U is path connected.

(v) There does not exist any continuous map f : R → R with f(Q) ⊂ R \Q and f(R \Q) ⊂ Q.

REAL ANALYSIS 31

[Hint : (i) Suppose f has no fixed points. Let A = {x ∈ [a, b] : x < f(x)}. Then a ∈ A. So

y := sup(A) exists in [a, b]. If f(y) < y, then there is x ∈ A∩ (f(y), y), and hence f(y) < x < f(x),

a contradiction since x < y. If y < f(y), then y < b, and for x ∈ (y, f(y)) ∩ (y, b], we have

f(x) < x < f(y), a contradiction. (ii) If f is increasing, then f(x−) ≤ f(x+) ≤ f(y−) ≤ f(y+)

for a ≤ x < y ≤ b. (iii) Consider x = y in Rm \ A. As there are uncountably many lines in Rm

passing through x, we may choose one such line L1 disjoint with A. Similarly, choose a line L2 in

Rm passing through y, disjoint with A, and not parallel to L1. Then L1 and L2 intersect, and this

gives a path in Rm \A from x to y. (iv) In fact, we may show U is polygonally connected, i.e., any

two points in U can be joined by a path consisting of finitely many line segments. For this, fix

x ∈ U and show that the set Y := {y ∈ U : there is a polygonal path in U from x to y} is both

open and closed in U . (v) Otherwise, f(R) is both countable and connected, and hence f must be

a constant. This contradicts the assumption that f(Q) ⊂ R \Q and f(R \Q) ⊂ Q.]

Exercise-26: (i) [0, 1] is not homeomorphic to either [0, 1]2 or S1.(ii) [0, 1) is not homeomorphic to any of the following: (0, 1), [0, 1], S1, and S1 \{w} for any w ∈ S1.(iii) R is not homeomorphic to any of the following: S1, [0,∞), and R2.

(iv) S1 is not homeomorphic to either [0, 1]2 or the sphere S2 = {(x, y, z) ∈ R3 : x2 + y2 + z2 = 1}.[Hint : (i) [0, 1] \ {1/2} is not connected, but [0, 1]2 \ {y} is connected for every y ∈ [0, 1]2. Use

[112](v). (In some cases, use the fact that continuous image of a compact set is compact).]

8. Convergence of a real sequence

We assume that the student has some familiarity with real sequences and series. We will revise

these topics with some extra details. Recall that we have already defined the convergence of a

sequence in a metric space.

Definition: Let (xn) be a real sequence.

(i) We say limn→∞ xn = ∞ if for every M > 0, there is n0 ∈ N such that xn ≥M for every n ≥ n0.

Similarly, we say limn→∞ xn = −∞ if for every M > 0, there is n0 ∈ N such that xn ≤ −M for

every n ≥ n0.

(ii) lim infn→∞ xn := supn∈N inf{xk : k ≥ n} = limn→∞ inf{xk : k ≥ n}.(iii) lim supn→∞ xn := infn∈N sup{xk : k ≥ n} = limn→∞ sup{xk : k ≥ n}.

Remark: By definition, −∞ ≤ lim infn→∞ xn ≤ lim supn→∞ xn ≤ ∞ for any real sequence (xn).

Note that (i) (xn) → x ⇔ lim infn→∞ xn = x = lim supn→∞ xn, and (ii) (xn) is a convergent

sequence ⇔ −∞ < lim supn→∞ xn ≤ lim infn→∞ xn <∞.

Example: (i) If (xn) is the sequence 0, 1, 0, 1, 0, 1, . . ., then lim infn→∞ xn = 0 = 1 = lim supn→∞ xn,

and hence (xn) is not convergent. (ii) If (xn) is an enumeration of all rational numbers, then

lim infn→∞ xn = −∞ and lim supn→∞ xn = ∞.


Exercise-27: (i) For a real sequence (xn), we have (xn) → 0 ⇔ (|xn|) → 0.

(ii) Let (xn) → x and (yn) → y in R. Then (axn + byn) → ax+ by for every a, b ∈ R.(iii) If (xn) → x and (yn) → y in R, then (xnyn) → xy.

(iv) If (xn) → x in R \ {0}, then (1/xn) → 1/x.

(v) Let xn > 0 and (xn) → 0. Then limn→∞ x1/kn = 0 for each k ∈ N.

(vi) Let (xn) be a real sequence, a = lim infn→∞ xn and b = lim supn→∞ xn (where a, b ∈ [−∞,∞]).

Then, there are subsequences (xmk) and (xnk

) with a = limk→∞ xmkand b = limk→∞ xnk

. Also, if

a, b ∈ R, then for every ε > 0, there is n0 ∈ N such that a− ε ≤ xn ≤ b+ ε for every n ≥ n0.

(vii) Let a < b be real numbers. Then there exists a real sequence (xn) such that lim infn→∞ xn = a,

lim supn→∞ xn = b and limn→∞(xn+1 − xn) = 0. Moreover if (xn) is any such sequence, then for

each c ∈ (a, b), there must exist a subsequence (xnk) converging to c.

[Hint : (ii) |(ax + by) − (axn + byn)| ≤ |a||x − xn| + |b||y − yn|. (iii) As a convergent sequence

is bounded, there is M > 0 with x, y, xn, yn ∈ [−M,M ] for every n ∈ N. Now |xy − xnyn| =|xy − xny + xny − xnyn| ≤ M(|x− xn|+ |y − yn|). (iv) There is n0 ∈ N such that |xn| ≥ |x|/2 for

every n ≥ n0, and hence |1/x − 1/xn| = |xn − x|/|xxn| ≤ 2|xn − x|/|x|2 for every n ≥ n0. (v) If

xn < εk, then x1/kn < ε. (vi) If xn < a−ε for infinitely many n ∈ N, then lim infn→∞ xn ≤ a−ε < a, a

contradiction; and similarly, if xn > b+ε for infinitely many n ∈ N, then lim supn→∞ xn ≥ b+ε > b,

a contradiction. (vii) Let ε > 0 be such that (c − ε, c + ε) ⊂ (a, b). Choose n2 > n1 > n0 such

that |xn+1 − xn| < ε for every n ≥ n0, and xn1 < c < xn2 . Then there is n1 ≤ n3 < n2 with

xn3 ≤ c ≤ xn3+1, and so |c− xn3 | ≤ |xn3+1 − xn3 | < ε.]

Exercise-28: (i) Let (xn) be an increasing real sequence which is bounded above. Then (xn)

converges to x := sup{xn : n ∈ N}. Similarly, if (xn) is a decreasing real sequence which is bounded

below, then (xn) converges to x := inf{xn : n ∈ N}.(ii) Let (xn) → x and (yn) → y in R. If xn ≤ yn for every n ∈ N, then x ≤ y.

(iii) [Sandwich theorem] Let xn ≤ yn ≤ zn for every n ∈ N and suppose there is x ∈ R such that

(xn) → x and (zn) → x. Then (yn) → x. More generally, if limn→∞ xn = x = limn→∞ zn and if yn

is between xn and zn (i.e., either yn ∈ [xn, zn] or yn ∈ [zn, xn]) for each n ∈ N, then limn→∞ yn = x.

(iv) If (xn) → 1 and xn > 0 for every n ∈ N, then (xyn) → 1 for every y ∈ R.[Hint : (iv) Clear for y = 0. For y ∈ Z \ {0}, this follows by Exercise-27(iii) and Exercise-27(iv).

For a general y ∈ R \ {0}, choose k ∈ Z with k ≤ y ≤ k + 1. Note that either xkn ≤ xyn ≤ xk+1n

(when xn ≥ 1) or xk+1n ≤ xyn ≤ xkn (when 0 < xn < 1) for each n ∈ N. Apply (iii).]

[124] (i) If x > 0, then limn→∞ 1/nx = 0.

(ii) If x > 0, then limn→∞ x1/n = 1.

(iii) limn→∞ n1/n = 1.

(iv) If |x| < 1, then limn→∞ xn = 0.

(v) If |x| > 1 and f is a real polynomial, then limn→∞f(n)

xn= 0.

REAL ANALYSIS 33

(vi) If x > 0, then limn→∞ xn/n! = 0.

(vii) If x > 0 and (yn) → 0, then limn→∞ xyn = 1.

(viii) Let (xn) → x and (yn) → y. Suppose x > 0 and xn > 0 for every n ∈ N. Then (xynn ) → xy.

(ix) Let x1 ≥ x2 ≥ · · · ≥ 0, and sn =∑n

j=1 xj for n ∈ N. If sup{sn : n ∈ N} <∞, then (nxn) → 0.

(x) If 0 ≤ x ≤ 1 and sn =∑n

j=1

1

jx, then (sn) → ∞. In particular, limn→∞(1+1/2+· · ·+1/n) = ∞.

Proof. Recall from [102](i) that (1 + ε)n ≥ 1 + nε and (1 + ε)n ≥ n(n− 1)ε2

2for ε > 0 and n ≥ 2.

(i) Given ε > 0, choose n0 > (1/ε)1/x. Then 1/nx < ε for every n ≥ n0.

(ii) First suppose x ≥ 1 and note that x1/n ≥ 1 for every n ∈ N. Given ε > 0, choose n0 ∈ N with

1 + n0ε > x. Then x < 1 + nε ≤ (1 + ε)n and hence x1/n < 1 + ε for every n ≥ n0. If x < 1, apply

the above to 1/x.

(iii) Clearly n1/n ≥ 1 for every n ∈ N. Given ε > 0, choose n0 ≥ 2 with(n0 − 1)ε2

2> 1. Then

n <n(n− 1)ε2

2≤ (1 + ε)n, or n1/n < 1 + ε for every n ≥ n0.

(iv) If x = 0, there is δ > 0 with 1/|x| > 1 + δ. Given ε > 0, choose n0 ∈ N with 1 + n0δ > 1/ε.

Then 1/|x|n > (1 + δ)n ≥ 1 + nδ > 1/ε or |x|n < ε for every n ≥ n0.

(v) It suffices to show that limn→∞ nk/xn = 0 for each k = 0, 1, 2, . . .. For k = 0, this follows

from (iv) proved above. Now fix k ∈ N, and choose δ > 0 such that (1 + δ)k = |x|. Observe that

(1 + δ)n ≥ n(n− 1)δ2

2for n ≥ 2. Letting an = n/(1 + δ)n, we see that |an| = an ≤ 2

(n− 1)δ2→ 0

as n→ ∞. Hence limn→∞ |nk/xn| = limn→∞ akn = 0.

(vi) If k > x, thenxk+n

(k + n)!=xk

k!× xn

(k + 1) · · · (k + n)≤ xk

k!× xn

kn→ 0 as n→ ∞ by (iv).

(vii) Let ε > 0. Since limk→∞ x−1/k = limk→∞(1/x)1/k = 1 = limk→∞ x1/k by (ii), there is k ∈ Nsuch that |1− x−1/k| < ε and |1− x1/k| < ε. Since (yn) → 0, there is n0 ∈ N such that |yn| ≤ 1/k

for every n ≥ n0. Then for every n ≥ n0, either x−1/k ≤ xyn ≤ x1/k or x1/k ≤ xyn ≤ x−1/k, and

hence |1− xyn | < ε.

(viii) Note that |xy − xynn | ≤ |xy − xyn| + |xyn||1 − xznn |, where zn := yn − y. By Exercise-28(iv),

|xy − xyn| = |xy||1− (xn/x)y| → 0, and this also implies (xyn) is bounded. Hence it remains to show

that (xznn ) → 1. By hypothesis, there are 0 < a < b such that xn ∈ [a, b] for every n ∈ N. Hence

either azn ≤ xznn ≤ bzn or bzn ≤ xznn ≤ azn for each n ∈ N. Moreover, limn→∞ azn = 1 = limn→∞ bzn

by (vii). Hence limn→∞ xznn = 1 by the Sandwich theorem.

(ix) Let M = sup{sn : n ∈ N}. Since (sn) is increasing and converging to M , we have that

M−sn → 0 as n→ ∞. Now note that 2nx2n ≤ 2(xn+1+· · ·+x2n) ≤ 2(M−sn), and (2n+1)x2n+1 ≤2(n+ 1)x2n+1 ≤ 2(xn+1 + · · ·+ x2n+1) ≤ 2(M − sn).


(x) Letting xn =1

nx, we note that x1 ≥ x2 ≥ · · · ≥ 0 and (nxn) does not converge to 0. Hence we

conclude by (ix) that limn→∞ sn = sup{sn : n ∈ N} = ∞. �

Remark: Fix t ∈ (0, 1), let xn = tn, and yn = 1/n. Then (xn) → 0, (yn) → 0, and (xynn ) → t.

Contrast this with [124](viii).

Exercise-29: (i) If p(x) =∑k

j=0 ajxj and q(x) =

∑mj=0 bjx

j are polynomials with degrees 0 ≤ k < m,

then limn→∞p(n)

q(n)= 0.

(ii) If p(x) =∑k

j=0 ajxj , q(x) =

∑kj=0 bjx

j and k = deg(p) = deg(q) ∈ N, then limn→∞p(n)

q(n)=akbk

.

(iii) If p(x) =∑k

j=0 ajxj and k = deg(p) ∈ N, then limn→∞

p(n+ c)

p(n)= 1 for every c ∈ R.

[Hint : (i)p(n)

q(n)=

a0/nm + a1/n

m−1 + · · ·+ ak/nm−k

b0/nm + b1/nm−1 + · · ·+ bm. (ii)

p(n)

q(n)=

a0/nk + a1/n

k−1 + · · ·+ akb0/nk + b1/nk−1 + · · ·+ bk

.

(iii) If we put q(x) = p(x+ c), then q(x) =∑k

j=0 bjxj with bk = ak so that (ii) applies.]

Exercise-30: (i) If 0 ≤ cj < 1 for j = 0, 1, . . . , k − 1, then∏k−1

j=0(1− cj) ≥ 1−∑k−1

j=0 cj .

(ii)k! nCk

nk≥ 1− k(k − 1)

2nfor n ∈ N and 0 ≤ k ≤ n.

[Hint : (i) Use induction on k. (ii) Note thatk! nCk

nk=

n!

nk(n− k)!=

∏k−1j=0(1− j/n) and apply (i).]

[125] [The constant e] Let an =∑n

k=0

1

k!, bn = (1 +

1

n)n, and cn = (1 +

1

n)n+1 for n ∈ N. Then,

(i) (an) and (bn) are strictly increasing, and (cn) is strictly decreasing.

(ii) bn < cn = bn(1 +1

n) and an − 3

2n≤ bn ≤ an ≤ 3 for every n ∈ N.

(iii) (an), (bn), and (cn) are convergent with the same limit, and this common limit is denoted by

the symbol e, i.e., e :=∑∞

n=0

1

n!= limn→∞(1 +

1

n)n = limn→∞(1 +

1

n)n+1.

(iv) a2 = 212 < e ≤ 3, and (1 +

1

n)n < e < (1 +

1

n)n+1 for every n ∈ N.

Proof. (i) Clearly (an) is strictly increasing. For the case of (bn) and (cn), we will use the Bernoulli

inequality [102](i) that (1 + y)n ≥ 1 + ny for y > −1. We have

bn+1

bn=

[n(n+ 2)

(n+ 1)2

]n(1 +

1

n+ 1) = (1− 1

(n+ 1)2)n(1 +

1

n+ 1) ≥ (1− n

(n+ 1)2)(1 +

1

n+ 1)

=(n2 + n+ 1)(n+ 2)

(n+ 1)3> 1, and hence (bn) is strictly increasing. Similarly, since

(1 +1

n+ 1)cncn+1

=

[(n+ 1)2

n(n+ 2)

]n+1

= (1 +1

n2 + 2n)n+1 > (1 +

1

(n+ 1)2)n+1 ≥ 1 +

1

n+ 1,

we get that (cn) is strictly decreasing.

(ii) bn =∑n

k=0

nCk

nk=

∑nk=0

∏k−1j=0(1− j/n)

k!≤

∑nk=0

1

k!= an ≤ 1 + 1 +

∑n−1j=1

1

2j≤ 3. Moreover,

by Exercise-30(ii), bn ≥∑n

k=0

1− k(k−1)2n

k!= an − an−2

2n≥ an − 3

2n.

REAL ANALYSIS 35

Statements (iii) and (iv) follow from (i) and (ii). �

[126] [Exponential and Logarithm functions] (i) Let bn, cn be as in [125]. If n ∈ N and n ≤ y ≤ n+1,

then bn+1(1 +1

n+ 1)−1 = (1 +

1

n+ 1)n ≤ (1 +

1

y)y ≤ cn, and hence limR∋y→∞(1 +

1

y)y = e by

Sandwich property and [125]. Consequently, limy→∞(1 +1

y)xy = ex for every x ∈ R by [124](viii).

In particular, limn→∞(1 +x

n)n = limn→∞(1 +

1

n/x)n = ex for every x ∈ R.

(ii) The exponential function exp : R → R defined as exp(x) = ex is continuous by [124](viii), it

is strictly increasing and injective with limx→∞ ex = ∞ and limx→∞ e−x = limx→∞ 1/ex = 0, and

(thus) it maps R onto (0,∞).

(iii) Since ea = limn→∞(1 +1

n)na, we get eaeb = ea+b for every a, b ∈ R by Exercise-27(iii)

(alternately, check ea+b = eaeb, for a, b ∈ Q and then use continuity). Also e0 = 1. Thus the

exponential function is a group isomorphism from (R,+) to ((0,∞), ·).(iv) The logarithm function log : (0,∞) → R is defined as the inverse of the exponential function,

i.e., log y = x if ex = y for y ∈ (0,∞). Since the exponential function is a strictly increasing

continuous bijection from R onto (0,∞), it follows that the logarithm function is a strictly increasing

continuous bijection from (0,∞) onto R with limy→0 log y = −∞ and limy→∞ log y = ∞.

(v) Being the inverse of the exponential function, the logarithm function is a group isomorphism

from ((0,∞), ·) to (R,+). We note that log 1 = 0, log(uv) = log u+log v and log(u/v) = log u−log v

for every u, v ∈ (0,∞). In particular, log(1/v) = log 1 − log v = − log v and log(vn) = n log v for

every v ∈ (0,∞) and n ∈ N.(vi) n log(1 +

1

n) < 1 < (n+ 1) log(1 +

1

n) for every n ∈ N by the inequality in [125](iv).

Proof. The statements are self-explanatory. �

Exercise-31: (i) e is irrational.

(ii) If f : R → R is a continuous function such that f(1) = e and f(x + y) = f(x)f(y) for every

x, y ∈ R, then f(x) = ex for every x ∈ R.(iii) ex ≥ 1 + x for every x ∈ R.

(iv) limy→∞yk

ey= 0 for every k ∈ N, and hence limy→∞

p(y)

ey= 0 for every polynomial p.

(v) limx→∞log x

x1/k= 0 for every k ∈ N. In particular, limx→∞

log x

x= 0 = limx→∞

log x√x.

[Hint : (i) Let an =∑n

k=0

1

k!. If possible, let there be p, q ∈ N with e = p/q. Then q!

∑∞n=q+1

1

n!=

q!(e − aq) ∈ N. But q!∑∞

n=q+1

1

n!=

∑∞n=1

1

(q + 1) · · · (q + n)<

∑∞n=1

1

(q + 1)n=

1

q≤ 1 by

[131](i), a contradiction. (ii) Show that f(x) = ex for x ∈ Q and then use continuity. Another

argument is as follows. If x = 0, then f(x) = f(x/2 + x/2) = (f(x/2))2 ≥ 0 and 0 = e = f(1) =

f(x + (1 − x)) = f(x)f(1 − x), and hence f(x) > 0. Define g : R → R as g(x) = log f(x) and use

[113](iii) to conclude g(x) = x for every x ∈ R. (iii) ex > 0 ≥ 1 + x for x ≤ −1. If x > −1, then


ex = limn→∞(1+x

n)n ≥ 1+x by [126](i) and Bernoulli inequality. (iv) limn→∞

nk

en= 0 by [124](v).

If n ≤ y ≤ n+ 1, thennk

en+1≤ yk

ey≤ (n+ 1)k

en. (v) Put y = log x in (iv) and use Exercise-27(v).]

Remark: (i) e is a transcendental real number with value e = 2.71828 · · · . (ii) For x > 0 and y ∈ R,we have x = elog x and hence xy = ey log x. Take this as the definition of xy from now onwards.

[127] [Euler’s constant γ] (i) Let un = 1+1

2+ · · ·+ 1

n− 1− log n and vn = 1+

1

2+ · · ·+ 1

n− log n =

un +1

nfor n ≥ 2. Then un < un+1 < vn+1 < vn. Moreover u2 = 1 − log 2 = log(e/2) > 0. Hence

there is a constant γ ∈ (0,∞) such that limn→∞ un = γ = limn→∞ vn (in fact γ = 0.5772 · · · ).

(ii) limn→∞

∑nk=1 1/k

log n= 1.

(iii) If (xn) → x in R, then limn→∞∑n

k=1

xkk log n

= x.

Proof. (i) We have un+1 − un =1

n− log(n+ 1) + log n =

1

n− log(

n+ 1

n) =

1

n− log(1 +

1

n) > 0 by

[126](vi). Similarly, vn+1 − vn =1

n+ 1− log(1 +

1

n) < 0 by [126](vi).

(ii) By (i),

∑nk=1 1/k

log n− 1 =

∑nk=1 1/k − log n

log n→ γ

∞= 0.

(iii) Putting cn,k =1

k log nfor n ∈ N and 1 ≤ k ≤ n, verify that the assumptions of Toeplitz

theorem mentioned below are satisfied; use (ii). �

Seminar topic: Toeplitz theorem, Stolz theorem, and applications (see Section 2.3 of W.J. Kaczor

and M.T. Nowak, Problems in Mathematical Analysis, I ). Toeplitz theorem says the following. Let

(xn) → x in R. For n ∈ N and 1 ≤ k ≤ n, let cn,k ∈ R be such that limn→∞ cn,k = 0 for every

k ∈ N, limn→∞∑n

k=1 cn,k = 1, and sup{∑n

k=1 |cn,k| : n ∈ N} <∞. Then limn→∞∑n

k=1 cn,kxk = x.

[128] (i) Every real sequence (xn) has a monotone subsequence.

(ii) If (xn) → x in R, then(x1 + · · ·+ xn

n

)→ x.

(iii) If (xn+1 − xn) → x in R, then (xnn) → x.

(iv) If xn ∈ R and xk+n ≤ xk + xn ∀ k, n ∈ N, then limn→∞xnn

= inf{xnn

: n ∈ N} ∈ [−∞,∞).

(v) If yn ∈ (0,∞) and yk+n ≤ ykyn for every k, n ∈ N, then limn→∞ y1/nn = inf{y1/nn : n ∈ N}.

Proof. (i) Let A = {k ∈ N : xn ≥ xk for every n ≥ k}. If A is infinite, then write A = {n1 < n2 <

· · · }, and we see (xnk) is increasing. If A is finite, then there is n1 ∈ N such that n1 > k for every

k ∈ A. Since n1 /∈ A, there is n2 > n1 such that xn2 < xn1 . Since n2 /∈ A, there is n3 > n2 such

that xn3 < xn2 , and so on. Then (xnj ) is a decreasing subsequence.

REAL ANALYSIS 37

(ii) Given ε > 0, choose k ∈ N such that |x− xn| < ε/2 for every n ≥ k. Then choose n0 ≥ k such

that

∑kj=1 |x− xj |

n< ε/2 for every n ≥ n0. Then for every n ≥ n0, we have that

|x−∑n

j=1 xj

n| = |

nx−∑n

j=1 xj

n| ≤

∑kj=1 |x− xj |

n+

∑nj=k+1 |x− xj |

n<ε

2+

(n− k)ε

2n< ε.

(iii) Let yn = xn − xn−1, where x0 := 0. Then by (ii),xnn

= (∑n

j=1 yj)/n→ x.

(iv) It suffices to show that lim supn→∞xnn

≤ inf{xmm

: m ∈ N}. Fix m ∈ N, and let M =

max{|x1|, . . . , |xm|}. Given n > m, write n = km + r with k ∈ N and 1 ≤ r ≤ m. Then

xn = xkm+r ≤ kxm + xr ≤ kxm +M . Sincen−m

m≤ k ≤ n

m, we deduce that

xnn

≤ xmm

+M

nif

xm ≥ 0, andxnn

≤ (n−m)xmnm

+M

nif xm < 0. In either case, it follows that lim supn→∞

xnn

≤ xmm

.

As this holds for every m ∈ N, we get lim supn→∞xnn

≤ inf{xmm

: m ∈ N}.

(v) Proof similar to that of (iv). Or, apply (iv) to xn := log yn. �

Remark: (i) We may also use [128](i) to prove that every bounded real sequence has a convergent

subsequence because a bounded monotone real sequence must be convergent. (ii) The converse of

[128](ii) is false: consider the sequence 1,−1, 1,−1, . . ..

[129] (i) If xn > 0 for all but finitely many n ∈ N, then we have

lim infn→∞xn+1

xn≤ lim infn→∞ x

1/nn ≤ lim supn→∞ x

1/nn ≤ lim supn→∞

xn+1

xn; and in particular

limn→∞ x1/nn = limn→∞

xn+1

xnwhenever the limit on the right hand side exists.

(ii) limn→∞(1/n!)1/n = 0.

(iii) limn→∞ n/(n!)1/n = e and limn→∞ nn/n! = ∞.

Proof. (i) Let r = lim supn→∞ x1/nn and s = lim supn→∞

xn+1

xn. If s < ∞, then for any t ∈ (s,∞),

there is k ∈ N such that xn+1 ≤ txn for every n ≥ k. So xn ≤ tn−kxk for every n ≥ k, and hence

r = lim supn→∞ x1/nn ≤ lim supn→∞(tn−kxk)

1/n = t. This shows r ≤ s.

(ii) Let xn =1

n!, and use (i) after noting that xn+1/xn = 1/(n+ 1) → 0.

(iii) Let xn = nn/n!. Then xn+1/xn = (n+1)n/nn = (1+1/n)n → e and hence limn→∞ x1/nn = e by

(i). Since e > 2, it follows that xn > 2n for all sufficiently large n ∈ N, and therefore (xn) → ∞. �

Exercise-32: (i) For each k ∈ N, we have limn→∞(√n+ k −

√n) = 0.

(ii) limn→∞(√n2 + n− n) = 1/2.

(iii) If x1 = 1 and xn+1 =√2 + xn for n ∈ N, then limn→∞ xn = 2.

[Hint : (i) Multiply the numerator and denominator with√n+ k+

√n. (ii) Multiply the numerator

and denominator with√n2 + n+n. (iii) Verify that 1 ≤ xn ≤ 2 for every n ∈ N. Also x2n+1−x2n =

2 + xn(1 − xn) ≥ 0 since 1 ≤ xn ≤ 2. Thus (xn) is an increasing sequence in [1, 2]. Letting

x = limn→∞ xn ∈ [1, 2], note that x =√2 + x and hence x2 − x− 2 = 0, or (x− 2)(x+ 1) = 0.]


9. Convergence of a real series

We may consider a real series either in the form∑∞

n=1 xn or in the form∑∞

n=0 xn; and moreover

the former series can be converted into the latter series when necessary by taking x0 = 0.

Definition: Let xn ∈ R for n ∈ N.(i) We say that the real series

∑∞n=1 xn converges to x ∈ R and write

∑∞n=1 xn = x if limn→∞ sn = x,

where sn = x1+ · · ·+xn for n ∈ N (we call (sn) the sequence of partial sums). If limn→∞ sn = ±∞,

then we write∑∞

n=1 xn = ±∞, and we say the series∑∞

n=1 xn diverges to ±∞. If (sn) does not

converge to any real number, then we say the series∑∞

n=1 xn is not convergent (or the series is

divergent). For example, if xn = (−1)n, then the series∑∞

n=1 xn is divergent (even though it may

appear to converge to zero).

(ii) We say∑∞

n=1 xn is absolutely convergent if∑∞

n=1 |xn| <∞, i.e., if sup{∑n

j=1 |xj | : n ∈ N} <∞.

(iii) We say∑∞

n=1 xn is unconditionally convergent if there is x ∈ R such that∑∞

n=1 xσ(n) = x

for every permutation (bijection) σ : N → N. Note that if n ∈ N and x1, . . . , xn ∈ R, then∑nj=1 xσ(j) =

∑nj=1 xj for every permutation σ : {1, . . . , n} → {1, . . . , n}; however, the addition of

infinitely many real numbers is is not commutative in general (we will see examples soon). A real

series∑∞

n=1 xn is unconditionally convergent precisely when the ‘infinite addition’ is commutative

and the sum is finite.

Remark: If∑∞

n=1 xn = x and∑∞

n=1 yn = y, then∑∞

n=1(axn + byn) = ax+ by.

[130] Let∑∞

n=1 xn be a real series.

(i) If∑∞

n=1 xn is convergent (i.e. if it converges to some real number), then limn→∞ xn = 0, but

the converse is false.

(ii) If xn ≥ 0 for every n ∈ N, then∑∞

n=1 xn is convergent ⇔ the sequence (sn) of partial sums is

bounded, where sn = x1 + · · ·+ xn.

(iii) If∑∞

n=1 xn is absolutely convergent, then∑∞

n=1 xn converges to some x ∈ R.(iv) If

∑∞n=1 xn is absolutely convergent, then

∑∞n=1 xn is unconditionally convergent.

Proof. (i) Let sn = x1 + · · ·+ xn and suppose (sn) → x, Then xn+1 = sn+1 − sn → 0. To see that

the converse is false, consider the harmonic series∑∞

n=1

1

n, which diverges to ∞ by [124](ix).

(ii) If xn’s are non-negative, then (sn) is an increasing sequence.

(iii) Let sn =∑n

j=0 xj and tn =∑n

j=0 |xj |. Since∑∞

n=1 xn is absolutely convergent, the increasing

sequence (tn) is bounded above and hence converges. In particular, (tn) is a Cauchy sequence. Now

|sn+k − sn| = |∑k

j=1 xn+j | ≤∑k

j=1 |xn+j | = tn+k − tn = |tn+k − tn|, and therefore (sn) is also a

Cauchy sequence. Then (sn) must converge to some x ∈ R by the completeness of R.

REAL ANALYSIS 39

(iv) Let σ : N → N be a permutation, sn =∑n

j=1 xj , and s′n =

∑nj=1 xσ(j). By (iii), (sn) → x for

some x ∈ R. So it suffices to show |sn−s′n| → 0. Let ε > 0. As∑∞

n=1 xn is absolutely convergent, we

may choose k ∈ N with∑∞

n=k+1 |xn| < ε. Next choose m ∈ N such that {1, . . . , k} ⊂ σ({1, . . . ,m}).Then for every n ≥ m, we have |sn − s′n| ≤

∑∞n=k+1 |xn| < ε, and thus |sn − s′n| → 0. �

[131] (i) [Geometric series] If |x| < 1, then∑∞

n=0 xn =

1

1− xand

∑∞n=1 x

n =x

1− x. Applying this

to ±x2, we also have: if |x| < 1, then∑∞

n=0 x2n =

1

1− x2and

∑∞n=0(−1)nx2n =

1

1 + x2.

(ii) If x > 1, then∑∞

n=1

1

nxis convergent; in fact,

∑∞n=1

1

nx≤ 2x

2x − 2<∞.

(iii) If k ∈ N, then∑∞

n=1

1

n(n+ k)=

∑kj=1

1

jk. In particular,

∑∞n=1

1

n(n+ 1)= 1.

(iv) Let (yn) → y in R, let k ∈ N, and xn = yn − yn+k. Then∑∞

n=1 xn = (y1 + · · ·+ yk)− ky.

Proof. (i) 1 + x+ x2 + · · ·+ xn−1 =1− xn

1− x→ 1

1− x, and

∑∞n=1 x

n = x∑∞

n=0 xn =

x

1− x.

(ii) Let sn =∑n

j=1

1

jx. Since (sn) is a strictly increasing sequence of positive reals, it suffices to

show (s2n+1) is bounded above by2x

2x − 2. Note that s2n+1 = 1+

∑nj=1

1

(2j)x+∑n

j=1

1

(2j + 1)x<

1 + 2∑n

j=1

1

(2j)x= 1 + 21−xsn < 1 + 21−xs2n+1 and hence s2n+1 <

1

1− 21−x=

2x

2x − 2.

(iii) Let xn =1

n(n+ k)=

1

k(1

n− 1

n+ k). Then for n > k,

∑nj=1 xj =

1

k(∑n

j=1

1

j−

∑nj=1

1

j + k) =

1

k(∑k

j=1

1

j−

∑kj=1

1

n+ j).

(iv) For n > k, note that∑n

j=1 xj = (y1 + · · ·+ yk)− (yn+1 + · · ·+ yn+k). �

Example: Let∑∞

n=1 xn be the series 1− 1+1/2− 1/2+1/3− 1/3+1/4− 1/4+ · · · . The sequence

of partial sums is 1, 0, 1/2, 0, 1/3, 0, 1/4, 0, . . . and hence∑∞

n=1 xn = 0. We claim that for each

r ∈ (0,∞), there is a permutation σ : N → N such that∑∞

n=1 xσ(n) = r. We will use the fact

that∑∞

n=k 1/n = ∞ for each k ∈ N. Let n0 = 0 and let n1 ∈ N be the smallest such that∑n1j=n0+1 1/j =

∑n1j=1 1/j > r. Then

∑n1j=n0+1 1/j ∈ (r, r + 1]. Let n2 > n1 be the smallest such

that (∑n1

j=n0+1 1/j)− 1 +∑n2

j=n1+1 1/j > r. Then (∑n1

j=n0+1 1/j)− 1 +∑n2

j=n1+1 1/j ∈ (r, r + 1/2]

because n2 ≥ 2. Let n3 > n2 be the smallest such that (∑n1

j=n0+1 1/j)− 1+ (∑n2

j=n1+1 1/j)− 1/2+

(∑n3

j=n2+1 1/j) > r. Then (∑n1

j=n0+1 1/j)− 1+ (∑n2

j=n1+1 1/j)− 1/2+ (∑n3

j=n2+1 1/j) ∈ (r, r+1/3]

because n3 ≥ 3. Proceed like this and verify that (∑n1

j=n0+1 1/j) − 1 + (∑n2

j=n1+1 1/j) − 1/2 +

(∑n3

j=n2+1 1/j)− 1/3 + · · · = r.

[132] (i) [Riemann rearrangement theorem] Let∑∞

n=1 xn be a real series which is convergent but

not absolutely convergent, and let −∞ ≤ a ≤ b ≤ ∞. Then there is a permutation σ : N → N such

that for tn :=∑n

j=1 xσ(j), we have that lim infn→∞ tn = a and lim supn→∞ tn = b.

(ii) A real series∑∞

n=1 xn is absolutely convergent ⇔∑∞

n=1 xn is unconditionally convergent.


Proof. (i) We leave the proof to the student as a reading assignment; see Theorem 3.54 of Rudin.

The proof uses some arguments similar to those in the Example above.

(ii) The implication ‘⇒’ is already proved in [130](iv), and ‘⇐’ follows from (i) above. �

Remark: (i) A real series∑∞

n=0 xn which is convergent but not absolutely/unconditionally conver-

gent is said to be conditionally convergent. (ii) The result [132](ii) is also true for series in Rm.

However, if X is an infinite dimensional complete normed space, then any absolutely convergent

series in X is unconditionally convergent, but the converse does not hold.

Exercise-33: (i) [Comparison test] If a real series∑∞

n=1 yn is absolutely convergent and if |xn| ≤ |yn|for all sufficiently large n ∈ N, then

∑∞n=1 xn is absolutely convergent.

(ii) [Limit comparison test] If∑∞

n=1 yn is absolutely convergent and if lim supn→∞|xn||yn|

<∞, then∑∞n=1 xn is absolutely convergent.

[Hint : (ii) There are M > 0 and n0 ∈ N with |xn| ≤M |yn| for every n ≥ n0. So (i) applies.]

Exercise-34: (i)∑∞

n=2

1

logn= ∞. (ii)

∑∞n=1

log n

n2<∞.

(iii)∑∞

n=1 sin(1/n2) and

∑∞n=1

sin(1/n)

1/nare absolutely convergent, but

∑∞n=1 sin(1/n) = ∞.

[Hint : (i) By Exercise-31(v), there is k ∈ N such that log n ≤ n and hence1

log n≥ 1

nfor every

n ≥ k. Apply Comparison test. (ii) By Exercise-31(v), there is k ∈ N such that log n ≤√n for

every n ≥ k. Now use [131](ii) and the Comparison test. (iii) By [113](vi), there is k ∈ N such that

sin(1/n2) ≤ 2/n2 and 1/(2n) ≤ sin(1/n) ≤ 2/n for every n ≥ k. Apply Comparison test.]

Definition: An expression∑∞

n=0 cnxn with cn ∈ R is called a (real) power series (centered at 0). The

radius of convergence R of∑∞

n=0 cnxn is defined asR = sup{r ≥ 0 : (cnr

n) is a bounded sequence} ∈

[0,∞]. For example, the radius of convergence of∑∞

n=0

xn

Rnis R if R ∈ (0,∞).

[133] Let R ∈ [0,∞] be the radius of convergence of a real power series∑∞

n=0 cnxn. Then,

(i) 1/R = lim supn→∞ |cn|1/n (where 1/0 = ∞ and 1/∞ = 0 by convention).

(ii) If cn = 0 for all but finitely many n, then lim infn→∞|cn+1||cn|

≤ 1/R ≤ lim supn→∞|cn+1||cn|

, and

consequently 1/R = limn→∞|cn+1||cn|

whenever the limit exists.

(iii) If x ∈ R and |x| < R, then∑∞

n=0 cnxn is absolutely convergent and hence convergent.

(iv) If x ∈ R and |x| > R, then∑∞

n=0 cnxn does not converge.

(v) If x ∈ R and |x| = R, then∑∞

n=0 cnxn may or may not converge.

(vi) The radius of convergence R′ of the power series∑∞

n=1 ncnxn−1 satisfies R′ = R.

Proof. (i) Let C = lim supn→∞ |cn|1/n. If C < r < ∞, then there is n0 ∈ N such that |cn| ≤ rn for

every n ≥ n0. Hence (cn(1/r)n) is bounded, which implies R ≥ 1/r, or 1/R ≤ r. Thus 1/R ≤ C.

REAL ANALYSIS 41

If 0 < s < C, then choose t ∈ (s, C). Since t < C, we have that |cn| > tn for infinitely many n ∈ N.Hence (cn(1/s)

n) is unbounded, which implies R ≤ 1/s, or 1/R ≥ s. Thus 1/R ≥ C.

(ii) This follows from (i) and [129](i).

(iii) Let |x| < r < R. Then M := sup{|cn|rn : n ≥ 0} < ∞ by the definition of R. Letting

yn =M(|x|/r)n, we see that |cnxn| ≤ yn for every n ≥ 0 and∑∞

n=0 yn <∞ as |x|/r < 1. Therefore∑∞n=0 cnx

n is absolutely convergent by the Comparison test, Exercise-33.

(iv) If |x| > R, then (cnxn) is unbounded, and in particular (cnx

n) does not converge to 0. Hence

by [130](i), the series∑∞

n=0 cnxn cannot be convergent.

(v) The radius of convergence is 1 for both∑∞

n=1 xn/n and

∑∞n=1 x

n/n2 (check). The second power

series converges at x = 1 but the first one does not converge at x = 1.

(vi) If 0 < r < R′, then (ncnrn−1) is a bounded sequence, which implies (cnr

n) is bounded, and

hence r ≤ R. Thus R′ ≤ R. If 0 < s < R, choose t ∈ (s,R). Then there is M > 0 with |cn|tn ≤M

for every n ≥ 0. Now n|cn|sn−1 ≤ Mn(s/t)n

s→ 0 as n→ ∞ by Exercise-29(ii) and hence (ncns

n−1)

is bounded, which implies R′ ≥ s. Thus R ≤ R′. We may prove this result also using part (i). �

Exercise-35: If R denotes the radius of convergence, then

(i) R = 1 for∑∞

n=0 nkxn for each k ∈ Z.

(ii) R = 1 for∑∞

n=0 p(n)xn, where p is any non-constant real polynomial.

(iii) R = 1/c for∑∞

n=0 cnxn for each c > 0.

(iv) R = ∞ for∑∞

n=0 xn/n! (where 0! = 1 by convention).

(v) R = 0 for∑∞

n=0 n!xn.

(vi) R = 1/e for∑∞

n=1 cnxn if cn :=

1n + 2n + · · ·+ nn

1! + 2! + · · ·+ n!.

[Hint : For (i)-(v), use [133](ii) after making suitable observations. (i) If k ∈ N, then (n+1)k/nk =

(1 + 1n)

k → 1 as n → ∞ and (n + 1)−k/n−k = nk/(n + 1)k = (1 − 1n+1)

k → 1 as n → ∞. Also

(n + 1)0/n0 = 1. (ii) |p(n + 1)|/|p(n)| → 1 as n → ∞ because p(n + 1) = p(n) + q(n) for some

polynomial q with deg(q) < deg(p). (vi) Note thatnn

n× n!≤ cn ≤ n× nn

n!. Also

n

(n!)1/n→ e by

[129](iii) and (n1/n) → 1. Hence (c1/nn ) → e by Sandwich property. Now use [133](i).]

[134] Consider a real series∑∞

n=0 cn (where note that we may write a real series∑∞

n=1 cn also in

the form∑∞

n=0 cn by taking c0 = 0).

(i) [Root test] Let λ = lim supn→∞ |cn|1/n. If λ < 1, then∑∞

n=0 cn is absolutely convergent and

hence convergent. If λ > 1, then∑∞

n=0 cn does not converge. If λ = 1, then we cannot say anything.


(ii) [Ratio test] Assume cn = 0 for all sufficiently large n. If lim supn→∞|cn+1||cn|

< 1, then∑∞

n=0 cn

is absolutely convergent and hence convergent. If lim infn→∞|cn+1||cn|

> 1, then∑∞

n=0 cn does not

converge. If lim infn→∞|cn+1||cn|

≤ 1 ≤ lim supn→∞|cn+1||cn|

, then we cannot say anything.

Proof. Let R be the radius of convergence of the real power series∑∞

n=0 cnxn.

(i) R = 1/λ by [133](i). Evaluate∑∞

n=0 cnxn at x = 1 and apply [133].

(ii) Note that lim infn→∞|cn+1||cn|

≤ 1/R ≤ lim supn→∞|cn+1||cn|

by [133](ii). Now evaluate∑∞

n=0 cnxn

at x = 1 and apply [133]. �

Example: (i) Let a > 0, λ > 0, and consider the series∑∞

n=0 λnna. Letting cn = λnna, we note

that |cn+1|/|cn| = λ(1 + 1/n)a → λ as n → ∞. Hence by the Ratio test,∑∞

n=0 λnna is absolutely

convergent (and consequently (λnna) → 0) when λ < 1, and∑∞

n=0 λnna does not converge when

λ > 1. If λ = 1, then∑∞

n=0 λnna =

∑∞n=0 n

a = ∞.

(ii) Let 0 < a < b < 1. Define cn = an for even n, and cn = bn for odd n. Then the Ratio test is

not applicable for the series∑∞

n=1 cn because lim infn→∞cn+1

cn= 0 < 1 < ∞ = lim supn→∞

cn+1

cn.

However, lim supn→∞ c1/nn = max{a, b} = b < 1, and hence

∑∞n=1 cn converges by the Root test.

Remark: By[129](i) and [134], we observe that the Root test is applicable wherever the Ratio test

is applicable. Also, as above, the Root test may be applicable even when the Ratio test is not

applicable. But when both tests are applicable, very often the Ratio test is easier to apply.

[135] [Dirichlet’s test] Let x1 ≥ x2 ≥ x3 ≥ · · · ≥ 0, limn→∞ xn = 0, and (yn) be a real sequence

such that there is M > 0 with |∑n

j=1 yj | ≤M for every n ∈ N. Then∑∞

n=1 xnyn converges.

Proof. Let sn =∑n

j=1 xjyj and tn =∑n

j=1 yj . It suffices to show (sn) is Cauchy. Given ε > 0,

choose n0 ∈ N such that xn0 <ε

2M. Then for n0 < k < n, we have |sn − sk| = |

∑nj=k+1 xjyj | =

|∑n

j=k+1 xj(tj − tj−1)| = |∑n

j=k+1 xjtj −∑n−1

j=k xj+1tj | = |∑n−1

j=k+1(xj −xj+1)tj +xntn−xk+1tk| ≤∑n−1j=k+1(xj − xj+1)M + xnM + xk+1M = 2xk+1M ≤ 2xn0M < ε. �

Exercise-36: (i) [Leibniz test/Leibnitz test] If x1 ≥ x2 ≥ x3 ≥ · · · ≥ 0 and limn→∞ xn = 0, then the

alternating series∑∞

n=1(−1)nxn and∑∞

n=1(−1)n+1xn are convergent.

(ii) If x > 0, then 1− 1/2x + 1/3x − 1/4x + · · · converges.

[Hint : For (i) apply [135]. For (ii), apply (i).]

Remark: Let xn = (−1)n/n and yn = (−1)n+1/n. Then∑∞

n=1 xn and∑∞

n=1 yn are convergent

by Exercise-36(i), but∑∞

n=1max{xn, yn} and∑∞

n=1min{xn, yn} are not convergent. Contrast this

with the behavior of convergent real sequences while taking maximum or minimum of nth terms.

Exercise-37: If x1 ≥ x2 ≥ x3 ≥ · · · ≥ 0 and limn→∞ xn = 0, then

(i)∑∞

n=1 xn sin(nθ) is convergent for every θ ∈ R.

REAL ANALYSIS 43

(ii)∑∞

n=1 xn cos(nθ) is convergent for every θ ∈ R\2πZ (but diverges when θ ∈ 2πZ and xn = 1/n).

[Hint : Recall the trigonometric identities 2 sinx sin y = cos(x− y)− cos(x+ y) and 2 cosx sin y =

sin(x+y)− sin(x−y). (i) Let yn = sin(nθ). If θ ∈ 2πZ, then yn = 0. If θ /∈ 2πZ, then sin(θ/2) = 0.

From 2|∑n

j=1 sin(jθ) sin(θ/2)| = |∑n

j=1(cos(jθ−θ/2)−cos(jθ+θ/2))| = | cos(θ/2)−cos(n+θ/2)| ≤2, we get |

∑nj=1 sin(jθ)| ≤ 1/ sin(θ/2) for every n ∈ N. So [135] applies. (ii) Let yn = cos(nθ). From

2|∑n

j=1 cos(jθ) sin(θ/2)| = |∑n

j=1(sin(jθ + θ/2)− sin(jθ − θ/2))| = | sin(n+ θ/2)− sin(θ/2)| ≤ 2,

we get |∑n

j=1 cos(jθ)| ≤ 1/ sin(θ/2) for every n ∈ N. Again apply [135].]

Exercise-38: (i) [Product of two series] Let∑∞

n=0 xn = x,∑∞

n=0 yn = y and zn =∑n

j=0 xjyn−j for

n ≥ 0. Also assume∑∞

n=0 xn or∑∞

n=0 yn is absolutely convergent. Then∑∞

n=0 zn = xy.

(ii) If |x| < 1, then∑∞

n=0(n+ 1)xn =1

(1− x)2.

[Hint : (i) See Theorem 8.46 of Apostol or Theorem 3.50 of Rudin. See also Example 3.49 of Rudin.

(ii) We have∑∞

n=0 xn =

1

1− xfor |x| < 1. Take the product of this series with itself and apply (i).]

Exercise-39: Let yn > 0 for n ∈ N.(i) If

∑∞n=1 yn < M <∞, then there are xn > 0 such that (xn) → ∞ and

∑∞n=1 xnyn < M .

(ii) If∑∞

n=1 yn = ∞, then there are xn > 0 such that (xn) → 0 and∑∞

n=1 xnyn = ∞.

[Hint : Let s =∑∞

n=1 yn and s < t < M . There is p1 ∈ N such that∑

n>p1yn < (t − s)/2, and

hence∑p1

n=1 yn+2∑

n>p1yn < t. Applying the same reasoning to the series

∑p1n=1 yn+2

∑n>p1

yn,

find p2 > p1 such that∑p1

n=1 yn + 2∑p2

n=p1+1 yn + 22∑

n>p2yn < t. Proceed like this. Put xn = 1

for 1 ≤ n ≤ p1 and xn = 2k for pk < n ≤ pk+1. (ii) Let p1 ∈ N be such that∑p1

n=1 yn ≥ 1. Since∑n>p1

yn = ∞, there is p2 > p1 such that (1/2)∑p2

n=p1+1 yn ≥ 1. Similarly, as∑

n>p2yn = ∞,

there is p3 > p2 such that (1/22)∑p3

n=p2+1 yn ≥ 1. Proceed like this. Put xn = 1 for 1 ≤ n ≤ p1

and xn = 1/2k for pk < n ≤ pk+1.]

Seminar topic: (i) Infinite product of real/complex numbers (see Section 8.26 of Apostol). (ii) If

pn is the nth prime, then∑∞

n=1

1

pn= ∞.

10. Sequences of functions

Definition: Let (X, d), (Y, d′) be metric spaces and fn : X → Y be functions for n ∈ N. We say:

(i) (fn) converges pointwise to a function f : X → Y if (fn(x)) → f(x) in Y for each x ∈ X.

(ii) (fn) converges uniformly to a function f : X → Y if for every ε > 0 there is n0 ∈ N such that

d′(f(x), fn(x)) < ε for every n ≥ n0 and every x ∈ X.

(iii) (fn) is uniformly Cauchy if for every ε > 0 there is n0 ∈ N such that d′(fm(x), fn(x)) < ε for

every m,n ≥ n0 and every x ∈ X.

Exercise-40: Let (X, d), (Y, d′) be metric spaces and fn : X → Y be functions for n ∈ N.(i) (fn) does not converge uniformly to a function f : X → Y ⇔ ∃ ε > 0, a strictly increasing

sequence (nk) in N, and a sequence (xk) in X such that d′(f(xk), fnk(xk)) ≥ ε for every k ∈ N.


(ii) (fn) fails to be uniformly Cauchy ⇔ there exist ε > 0, two strictly increasing sequences (mk)

and (nk) in N, and a sequence (xk) in X such that d′(fmk(xk), fnk

(xk)) ≥ ε for every k ∈ N.(iii) If (fn) → f uniformly for some function f : X → Y , then (fn) → f pointwise and (fn) is

uniformly Cauchy.

Example: (i) Let fn : R → R be fn(x) = x/n for n ∈ N, and f ≡ 0. Clearly (fn) → f uniformly

on [−M,M ] for any real number M > 0. By Exercise-40(i), the convergence is not uniform on Rbecause fn(n) = 1.

(ii) Let fn : [0, 1] → R be fn(x) = xn for n ∈ N, and let f : [0, 1] → R be f(1) = 1 and f(x) = 0

for 0 ≤ x < 1. Then (fn) → f pointwise. This shows that the pointwise limit of a sequence of

continuous functions need not be continuous even if the domain is compact. Letting xn = (1/2)1/n,

we see |f(xn)− fn(xn)| = 1/2 for every n ∈ N, and hence by Exercise-40(i), (fn) does not converge

to f uniformly on [0, 1], or even on [0, 1). Next we observe that (fn) → f uniformly on [0, c] for

each c ∈ (0, 1) because xn ≤ cn → 0 for x ∈ [0, c].

(iii) Let fn : [0, 1] → R be fn(x) =xn

1 + xnfor n ∈ N. Let f : [0, 1] → R be f(x) = 0 for x < 1 and

f(1) = 1/2. Then (fn) → f pointwise. If xn = (1/2)1/n, then |fn(xn)− f(xn)| = fn(xn) = 1/2 for

every n ∈ N and hence (fn) does not converge uniformly to f . On the other hand, if c ∈ (0, 1),

then |fn(x)| ≤ xn ≤ cn for every x ∈ [0, c], and hence (fn) → f uniformly on [0, c].

(iv) Let fn : [0, 1] → R be fn(x) = nx(1 − x)n for n ∈ N. Then fn(0) = fn(1) = 0. Moreover,

(fn(x)) → 0 for each x ∈ (0, 1) by [124](v). Thus (fn) → 0 pointwise. Recalling that ex =

limn→∞(1 +x

n)n, we note that fn(1/n) = (1 − 1

n)n → e−1, and hence there is n0 ∈ N such that

|fn(1/n)| ≥1

2efor every n ≥ n0. This shows that (fn) does not converge uniformly to 0.

(v) Let fn : R → R be fn(x) =nx

2n. If k ∈ N, then |fn(x)| ≤

nk

2nfor x ∈ [−k, k], and hence (fn) → 0

uniformly on [−k, k] by [124](v). So (fn) → 0 pointwise on R. But the convergence is not uniform

on R because fn(n) ≥ 1 for every n ≥ 2.

(vi) Let fn : R → R be fn(x) =1

(1 + x2)1/n. Then (fn) → 1 pointwise. For a fixed n ∈ N,

limx→∞ fn(x) = 0, and hence there is xn ∈ R with fn(x) < 1/2. Thus |1− fn(xn)| > 1/2 for every

n ∈ N, and therefore (fn) does not converge uniformly.

(vii) Let f, fn : (0,∞) → R be f(x) = log x and fn(x) = log(x +1

n). Then (fn) → f pointwise.

Since |f(x+1

n)− f(x)| = log(1 +

1

nx), we see |fn(1/n2)− f(1/n2)| = log(1 + n) → ∞ as n→ ∞,

and hence the convergence is not uniform.

(viii) Let fn : [0, 1] → R be fn(x) = limk→∞(cos(n!πx))2k, and let A = [0, 1] ∩Q. Then (fn) → 1A

pointwise (see 7.4 of Rudin), but the convergence is not uniform because fn(1

2 · n!) = 0.

REAL ANALYSIS 45

Remark: For Examples (ii) and (iii), the failure of uniform convergence can also be deduced using

the fact that the limit function is not continuous; see [136](i) below.

Remark: Let X be a metric space, f, fn, g, gn : X → R be functions, and suppose (fn) → f and

(gn) → g pointwise. Then the following are easy to verify: (i) (afn + bgn) → af + bg pointwise for

every a, b ∈ R and (fngn) → fg pointwise, (ii) (fn ◦ h1) → f ◦ h1 and (h2 ◦ fn) → h2 ◦ f pointwise

for any function h1 : X → X and any continuous function h2 : R → R. However, some of the

corresponding statements are no longer true for uniform convergence; see Exercise-41.

Definition: Let (X, d), (Y, d′) be metric spaces and fn : X → Y be functions for n ∈ N. We say:

(i) (fn) converges locally uniformly to a function f : X → Y if for each x ∈ X, there is r > 0 such

that (fn) → f uniformly on B(x, r).

(ii) (fn) converges uniformly on compact subsets of X to a function f : X → Y if for each compact

set K ⊂ X, (fn) → f uniformly on K.

For example, if fn : R → R is fn(x) = x/n, then (fn) → 0 locally uniformly on R and (fn) → 0

uniformly on compact subsets of R even though the convergence is not uniform on the whole of R.

The two notions of convergence defined above are generally considered for continuous functions.

[136] Let (X, d), (Y, d′) be metric spaces and fn ∈ C(X,Y ) for n ∈ N.(i) If (fn) converges uniformly or locally uniformly to a function f : X → Y , then f ∈ C(X,Y ).

(ii) If (fn) converges locally uniformly to a function f : X → Y , then (fn) → f uniformly on

compact subsets of X. The converse holds if X is locally compact, i.e., if X satisfies the property

that for each x ∈ X, there is r > 0 with B(x, r) compact (example: X = Rm).

(iii) If X is locally compact and (fn) converges to a function f : X → Y uniformly on compact

subsets of X, then f ∈ C(X,Y ).

Proof. (i) Fix a ∈ X and ε > 0. We need to fined δ > 0 with f(B(a, δ)) ⊂ B(f(a), ε). Since

(fn) → f locally uniformly, there is r > 0 such that (fn) → f uniformly on B(a, r). Choose n ∈ Nwith d′(f(b), fn(b)) < ε/3 for every b ∈ B(a, r), and then use the continuity of fn to choose δ ∈ (0, r)

such that fn(B(a, δ)) ⊂ B(fn(a), ε/3). Then for every b ∈ B(a, δ), we see by triangle inequality

that d′(f(a), f(b)) ≤ d′(f(a), fn(a)) + d′(fn(a), fn(b)) + d′(fn(b), f(b)) < ε/3 + ε/3 + ε/3 = ε.

(ii) Assume (fn) → f locally uniformly. Let K ⊂ X be compact. For each x ∈ K, there is rx > 0

such that (fn) → f uniformly on B(x, rx). Then {B(x, rx) : x ∈ K} is an open cover for the

compact set K, and hence there are finitely many points x1, . . . , xk ∈ K with K ⊂∪k

j=1B(xj , rxj ).

Since (fn) → f uniformly on each B(xj , rxj ), it follows that (fn) → f uniformly on∪k

j=1B(xj , rxj )

and hence on K. To prove the converse, assume X is locally compact and (fn) → f uniformly

on compact subsets of X. Fix x ∈ X and choose r > 0 with B(x, r) compact. Since (fn) → f

uniformly on B(x, r) by assumption, (fn) → f uniformly on the subset B(x, r) as well.


(iii) This follows from (i) and (ii). �

[137] Let (X, d), (Y, d′) be metric spaces and fn ∈ C(X,Y ) for n ∈ N.(i) If (fn) converges uniformly to a function f : X → Y , then limn→∞ fn(xn) = f(x) whenever

(xn) → x in X. The converse is not true.

(ii) If (fn) is uniformly Cauchy on a dense subset A of X, then (fn) is uniformly Cauchy on X.

Proof. (i) Note that f ∈ C(X,Y ) by [136](i). Let ε > 0 be given. Since (f(xn)) → f(x) and since

(fn) → f uniformly, we may choose k ∈ N such that d′(f(x), f(xn)) < ε/2 for every n ≥ k and

d′(f(a), fn(a)) < ε/2 for every n ≥ k and every a ∈ X. Then d′(f(x), fn(xn)) ≤ d′(f(x), f(xn)) +

d′(f(xn), fn(xn)) < ε/2 + ε/2 = ε for every n ≥ k.

To see that the converse is not true, consider fn : R → R defined as fn(x) = x/n for n ∈ N. If

(xn) → x in R, then (xn) is bounded, and hence (fn(xn)) → 0 = f(x). But we know that (fn) does

not converge to 0 uniformly.

(ii) Since A = X, we see sup{d′(fm(x), fn(x)) : x ∈ X} = sup{d′(fm(x), fn(x)) : x ∈ A} for every

m,n ∈ N. This implies (fn) is uniformly Cauchy on X. �

Definition: Let X be a metric space and F be a family of functions from X to R. We say that

(i) F is pointwise bounded if ∀x ∈ X, there is Mx > 0 such that |f(x)| ≤Mx ∀ f ∈ F .

(ii) F is uniformly bounded if there is M > 0 such that |f(x)| ≤M ∀ f ∈ F and ∀x ∈ X.

Example: (i) If fn : R → R is fn(x) = x/n, then the family F := {fn : n ∈ N} is pointwise bounded

but not uniformly bounded. However F restricted to [−k, k] is uniformly bounded for any k ∈ N.(ii) Let fn : [0, 1] → [0, n] be any continuous function with f(0) = 0, f(1/(n+1)) = n and f(x) = 0

for 1/n ≤ x ≤ 1. Then the family F := {fn : n ∈ N} is pointwise bounded but not uniformly

bounded even though the domain space [0, 1] is compact.

Exercise-41: Let X be a metric space, fn, f, gn, g ∈ C(X,R), and suppose (fn) → f and (gn) → g

uniformly. Then, (i) (afn + bgn) → af + bg uniformly for every a, b ∈ R.(ii) If each fn is bounded, then f is bounded and (fn) is uniformly bounded.

(iii) If fn’s and gn’s are bounded, then (fngn) → fg uniformly.

(iv) (fn ◦ h1) → f ◦ h1 uniformly for any function h1 : X → X, and (h2 ◦ fn) → h2 ◦ f uniformly

for any uniformly continuous function h2 : R → R.(v) The following example shows that some of the assumptions above are necessary. Let f, fn, h2 :

R → R be f(x) = x, fn(x) = x + 1/n, and h2(x) = x2. Then (fn) → f uniformly, and h2 is

continuous. However, as |n2 − (n + 1/n)2| ≥ 2 for every n ∈ N, the convergences (f2n) → f2 and

(h2 ◦ fn) → h2 ◦ f are not uniform.

[Hint : (ii) If k ∈ N is such that |f(x) − fn(x)| ≤ 1 for every n ≥ k and every x ∈ X, and

if M := sup{|fk(x)| : x ∈ X}, then |f(x)| ≤ M + 1 and |fn(x)| ≤ M + 2 for every x ∈ X

REAL ANALYSIS 47

and n ≥ k. (iii) |f(x)g(x) − fn(x)gn(x)| = |f(x)g(x) − f(x)gn(x) + f(x)gn(x) − fn(x)gn(x)| ≤|f(x)||g(x)− gn(x)|+ |f(x)− fn(x)||gn(x)|.]

Definition: Let X be a compact metric space, and define the supremum metric d∞ on C(X,R) :={f : X → R : f is continuous} as d∞(f, g) = sup{|f(x) − g(x)| : x ∈ X} for f, g ∈ C(X,R) (note

that ‘d∞(f, g) < ε’ means that the graph of g lies within the ε-tube around the graph of f).

[138] Let X be a compact metric space and d∞ be the supremum metric on C(X,R). Then,(i) For f, f1, f2, f3, . . . ∈ C(X,R), we have limn→∞ d∞(f, fn) = 0 ⇔ (fn) → f uniformly.

(ii) A sequence (fn) in C(X,R) is Cauchy with respect to d∞ ⇔ (fn) is uniformly Cauchy.

(iii) A family F ⊂ C(X,R) is uniformly bounded ⇔ F is a bounded subset of (C(X,R), d∞).

(iv) The metric space (C(X,R), d∞) is complete.

(v) Let A ⊂ X be a dense subset and f : A→ R be a function. Assume there is a sequence (fn) in

C(X,R) converging to f uniformly on A. Then there is f ∈ C(X,R) with f |A = f .

Proof. (i), (ii), and (iii) are easy to check.

(iv) Let (fn) in C(X,R) be Cauchy. Since |fm(x)− fn(x)| ≤ d∞(fm, fn), it follows that (fn(x)) is

Cauchy in R for each x ∈ X. Since R is complete, we may define f : X → R as f(x) = limn→∞ fn(x).

Let ε > 0. Since (fn) is uniformly Cauchy by (ii), there is n0 ∈ N such that |fm(x)− fn(x)| < ε for

every m,n ≥ n0 and every x ∈ X. Fixing n ∈ N and letting m→ ∞, we get that |f(x)−fn(x)| ≤ ε

for every n ≥ n0 and every x ∈ X by the continuity of the function y 7→ |y − fn(x)| from X to R(with n ∈ N and x ∈ X fixed). This shows that (fn) → f uniformly. Therefore, f ∈ C(X,R) by

[136](i), and moreover (fn) → f with respect to d∞ by (i).

(v) (fn) is uniformly Cauchy on A, and hence uniformly Cauchy onX by [137](ii). The completeness

of C(X,R) proved in (iv) implies that there is f ∈ C(X,R) such that (fn) → f uniformly on X,

and hence on A. By the uniqueness of the limit function, we get f |A = f . �

Remark: (i) Similarly, (C(X,C), d∞) is complete when X is a compact metric space. (ii) Let

f : (0, 1) → R be continuous and suppose there is a sequence (fn) of polynomials converging

uniformly to f on (0, 1). Then by [138](v), there is f ∈ C([0, 1],R) with f |(0,1) = f .

Example: Let (εn) be a strictly decreasing sequence in (0, 1) converging to 0. For n ∈ N, let

fn : [0, 1] → [0, 1] be the continuous function defined by the following conditions: fn(x) = 0 for

x ∈ [0, εn+1], fn(x) = 1 for x ∈ [εn, 1], and the graph of fn is linear in [εn+1, εn]. Then for every

k < n, fk(εn) = 0 and fn(εn) = 1 so that d∞(fk, fn) ≥ 1. Therefore (fn) does not have any

convergent subsequence with respect to d∞. This shows that C([0, 1], [0, 1]) is not (sequentially)

compact in (C([0, 1],R), d∞) even though C([0, 1], [0, 1]) is closed and bounded in (C([0, 1],R), d∞).

To describe compact subsets of (C([0, 1],R), d∞), we need the notion of an equicontinuous family.


Definition: Let (X, d), (Y, d′) be metric spaces. A family of functions F from X to Y is said to

be equicontinuous if for every ε > 0 there is δ > 0 such that f(BX(x, δ)) ⊂ BY (f(x), ε) for every

f ∈ F and every x ∈ X. For example, if fn : [0, 1] → R is fn(x) = x/n, then F := {fn : n ∈ N} is

an equicontinuous family.

Remark: Let F be a family of functions from a metric spaces X to R. Then (i) F fails to be an

equicontinuous family ⇔ there exist ε > 0, sequences (an) and (bn) in X, and a sequence (fn) in Fsuch that limn→∞ d(an, bn) = 0 and |fn(an)−fn(bn)| ≥ ε for every n ∈ N. (ii) If {F = {fn : n ∈ N},then F fails to be an equicontinuous family ⇔ there exist ε > 0, a strictly increasing sequence (nk)

in N, and sequences (ak) and (bk) in X such that limk→∞ d(ak, bk) = 0 and |fnk(ak)− fnk

(bk)| ≥ ε

for every k ∈ N. (iii) If F is equicontinuous, then each f ∈ F is uniformly continuous; but the

converse is false; for example, consider F := {fn : n ∈ N}, where fn : [0, 1] → R is fn(x) = xn, and

note that |fn(1)− fn(bn)| = 1/2 for every n ∈ N for bn := (1/2)1/n.

Remark: When a family F of functions is equicontinuous, all functions in F behave like a single

function - this intuitive idea can be seen in [139](ii) and [140] below.

[139] Let (X, d) be a compact metric space, and (fn) be a sequence in C(X,R) converging pointwiseto some f ∈ C(X,R).(i) [Dini’s theorem] If either f1 ≤ f2 ≤ f3 ≤ · · · or f1 ≥ f2 ≥ f3 ≥ · · · , then (fn) → f uniformly.

(ii) If {fn : n ∈ N} is equicontinuous, then (fn) → f uniformly.

Proof. (i) Assume f1 ≤ f2 ≤ f3 ≤ · · · and put gn = f − fn. Then g1 ≥ g2 ≥ g3 ≥ · · · ≥ 0 and

(gn) → 0 pointwise. If the convergence is not uniform, there exist ε > 0, a strictly increasing

sequence (nk) in N, and a sequence (xk) in X such that gnk(xk) ≥ ε for every k ∈ N. Since X is

compact, by passing onto a subsequence we may suppose that (xk) converges to some x ∈ X. For

any fixed k ∈ N and every m ≥ k we have gnk≥ gnm ≥ 0 and hence gnk

(xm) ≥ gnm(xm) ≥ ε. This

implies gnk(x) ≥ ε by the continuity of gnk

since (xm) → x. Thus gnk(x) ≥ ε for every k ∈ N. This

contradicts the fact that (gn) → 0 pointwise.

(ii) Let f0 = f and note that {fn : n ≥ 0} is also equicontinuous. Given ε > 0, choose δ > 0

such that fn(B(x, δ)) ⊂ B(fn(x), ε/3) for every n ≥ 0. Since X is compact, there are finitely many

points x1, . . . , xk ∈ X such that X =∪k

j=1B(xj , δ). Using pointwise convergence of (fn) at the

points xj , we may choose n0 ∈ N such that |f(xj)− fn(xj)| < ε/3 for every n ≥ n0 and 1 ≤ j ≤ k.

Now consider x ∈ X and choose j ∈ {1, . . . , k} with x ∈ B(xj , δ). Then for every n ≥ n0, we see

that |f(x)− fn(x)| ≤ |f(x)− f(xj)|+ |f(xj)− fn(xj)|+ |fn(xj)− fn(x)| < ε/3+ ε/3+ ε/3 = ε. �

Remark: [139](ii) remains true if R is replaced with any metric space Y .

[140] Let (X, d) be a compact metric space, and F ⊂ C(X,R) be equicontinuous and pointwise

bounded. Then, (i) F is uniformly bounded, i.e., F is a bounded subset of (C(X,R), d∞).

REAL ANALYSIS 49

(ii) F is also a totally bounded subset of (C(X,R), d∞).

Proof. (i) By equicontinuity, find δ > 0 such that |f(x) − f(y)| < 1 for every x, y ∈ X with

d(x, y) < δ and every f ∈ F . Choose x1, . . . , xk ∈ X with X =∪k

j=1B(xj , δ). Let M > 0 be such

that |f(xj)| ≤M for every f ∈ F and 1 ≤ j ≤ k. Then |f(x)| ≤M +1 for every f ∈ F and x ∈ X.

(ii) Let ε > 0. Since F is uniformly bounded by (i), and since bounded subsets of R are totally

bounded, there are t1, . . . , tq ∈ R such that∪

f∈F f(X) ⊂∪q

j=1(tj−ε/5, tj+ε/5). By the equiconti-

nuity of F , there is δ > 0 such that |f(x)−f(y)| < ε/5 for every x, y ∈ X with d(x, y) < δ and every

f ∈ F . Choose x1, . . . , xp ∈ X with X =∪p

i=1B(xi, δ) by the compactness of X. Since {x1, . . . , xp}and {t1, . . . , tq} are finite sets, we may find a finite subcollection G ⊂ F such that for each f ∈ Fthere is g ∈ G with the following property: f(xi) ∈ (tj−ε/5, tj+ε/5) iff g(xi) ∈ (tj−ε/5, tj+ε/5) foreach i ∈ {1, . . . , p}. If g is chosen for f in this manner, then for any x ∈ X, by choosing i ∈ {1, . . . , p}with x ∈ B(xi, δ), we see that |f(x) − g(x)| ≤ |f(x) − f(xi)| + |f(xi) − g(xi)| + |g(xi) − g(x)| <ε/5 + 2ε/5 + ε/5 = 4ε/5 and hence d∞(f, g) ≤ 4ε/5 < ε. Thus F ⊂

∪g∈G B(g, ε). �

[141] [Arzela-Ascoli theorem] Let (X, d) be a compact metric space and F ⊂ (C(X,R), d∞), where

d∞ is the supremum metric. Then F is compact ⇔ F is equicontinuous and pointwise bounded.

Proof. ⇒: Since compact subsets are bounded, F is bounded with respect to d∞, and this implies Fis pointwise bounded. To show F is equicontinuous, consider ε > 0. The compactness of F implies Fis totally bounded, and therefore we may choose f1, . . . , fk ∈ F with F ⊂

∪kj=1B(fj , ε/3). Choose

δ > 0 such that |fj(x)− fj(y)| < ε/3 for every x, y ∈ X with d(x, y) < δ and every j ∈ {1, . . . , k}.Now for every x, y ∈ X with d(x, y) < δ and every f ∈ F , if we choose fj with d∞(f, fj) < ε/3,

then we see that |f(x)− f(y)| < |f(x)− fj(x)|+ |fj(x)− fj(y)|+ |fj(y)− f(y)| < ε/3 + ε/3 + ε/3

(in fact, this argument shows that any totally bounded family F ⊂ C(X,R) is equicontinuous).

⇐: 2 Being a closed subset of the complete space C(X,R), the space F is complete. In view of

[117] and Exercise-21, it remains to show F is totally bounded. And this is proved in [140](ii). �

Remark: Let (X, d) be a compact metric space and (fn) be an equicontinuous and pointwise

bounded sequence in (C(X,R), d∞). Then by [141], a subsequence of (fn) converges to some

f ∈ C(X,R) uniformly (i.e., with respect to d∞).

Example: (i) A crucial part in the proof of [141] is the fact that bounded subsets of R are totally

bounded. We may also replace R with C or Rm in [141], but not with an arbitrary complete

metric space. Consider N with the discrete metric, and let fn : [0, 1] → N be fn ≡ n. Then

F := {fn : n ∈ N} ⊂ C([0, 1],N) is equicontinuous and pointwise bounded, but F is not compact

because d∞(fk, fn) ≥ 1 for every k = n.

2For a slightly different proof of Arzela-Ascoli theorem, see my notes Functional Analysis.


(ii) Let F = {fn : n ∈ N} ⊂ C([0, 1],R), where fn(x) = xn. Since F is not equicontinuous, F is

not compact by [141].

(iii) F = {fa : a ∈ R} ⊂ C([0, 1],R), where fa(x) = sin(x + a). Then F is clearly pointwise

bounded. Also note that |fa(x) − fa(y)| = |(sinx cos a + cosx sin a) − (sin y cos a + cos y sin a)| ≤| sinx− sin y|+ | cosx− cos y|, which implies F is equicontinuous. Hence F is compact by [141].

Recall (from algebra): Let R be a commutative ring with unity. We say J ⊂ R is an ideal of R

if J is a subgroup of (R,+) and RJ ⊂ J . Note that any ideal of R containing the multiplicative

identity 1 must be equal to R. Among ideals J = R, those which are maximal with respect to

inclusion are called maximal ideals.

Exercise-42: Let X be a compact metric space. (i) With respect to pointwise operations, C(X,R)is a real vector space and a commutative ring with unity, where the multiplicative identity is the

constant function 1. Since C(X,R) contains constant functions, every ideal of C(X,R) is a vector

subspace of C(X,R).(ii) For ∅ = A ⊂ X, let JA = {f ∈ C(X,R) : f |A ≡ 0}. Then JA is a proper ideal of C(X,R) (hencea proper vector subspace), and JA is closed in (C(X,R), d∞). Moreover, JA1 ⊂ JA2 if A1 ⊃ A2.

(iii) Let Ja := J{a} = {f ∈ C(X,R) : f(a) = 0} for a ∈ X. Then Ja is a maximal ideal of C(X,R).(iv) If J is a maximal ideal of C(X,R), then J = Ja for some a ∈ X.

[Hint : (iii) Let J be an ideal of C(X,R) properly containing Ja. To show J = C(X,R), it suffices to

show 1 ∈ J . Let f ∈ J \Ja. Then f(a) = 0. We may suppose f(a) = 1 after a scalar multiplication.

Then (1− f)(a) = 0 so that 1− f ∈ Ja ⊂ J and hence 1 = f +(1− f) ∈ J . (iv) Assume J = Ja for

every a ∈ X. Being maximal, J cannot be properly contained in any Ja. Hence there are fa ∈ J

with fa(a) = 0 for each a ∈ X. Choose open neighborhoods Ua of a ∈ X with fa not vanishing in

Ua. As X is compact, there are a1, . . . , ak ∈ X with X ⊂∪k

j=1 Uaj . Then f :=∑k

j=1 f2aj ∈ J as J

is closed under products and sums, and moreover 1/f ∈ C(X,R) because f is non-vanishing on X.

Hence 1 = (1/f)f ∈ J , which implies J = C(X,R), a contradiction.]

11. Series of real functions

Definition: Let X be a metric space (or just a set), fn : X → R be functions for n ∈ N and let

sn : X → R be sn = f1 + · · ·+ fn.

(i) We say the series∑∞

n=1 fn converges at x ∈ X if the real series∑∞

n=1 fn(x) is convergent,

equivalently if the sequence (sn(x)) converges.

(ii) We say∑∞

n=1 fn is absolutely convergent at x ∈ X if∑∞

n=1 |fn(x)| <∞.

(iii) We say∑∞

n=1 fn converges uniformly to a function f : X → R if (sn) → f uniformly on X.

(iv) We say∑∞

n=1 fn is uniformly Cauchy if the sequence (sn) is uniformly Cauchy, i.e., if for every

ε > 0 there is k ∈ N such that sup{|sn(x)− sm(x)| : x ∈ X} < ε for every n,m ≥ k.

Remark: We may also consider∑∞

n=0 fn (with n starting from 0), and can have similar definitions.

REAL ANALYSIS 51

Example: (i) Let fn : R → R be fn(x) = x2/n2. Then∑∞

n=1 fn converges absolutely at each x ∈ Rand hence

∑∞n=1 fn converges at each x ∈ R. But the convergence is not uniform on R: defining

f : R → R as f(x) =∑∞

n=1 fn(x) and sn = f1 + · · · + fn, we see that |f(n + 1) − sn(n + 1)| ≥|fn+1(n+ 1)| = 1 for every n ∈ N.

(ii) Let fn : R → R be fn ≡ (−1)n

n. Then

∑∞n=1 fn converges uniformly on R by Leibniz test

(Exercise-36). However,∑∞

n=1 fn is not absolutely convergent at any x ∈ R.(iii) Let fn : (0,∞) → R be fn(x) = xlogn. Note that xlogn = (elog x)logn = (elogn)log x = nlog x.

Hence∑∞

n=1 fn converges at x ∈ (0,∞) ⇔ log x < −1 ⇔ x < 1/e.

(iv) Let fn : [0, 1] → R be fn(x) =x

(1 + x)n. Then fn ≥ 0 and

∑∞n=0 fn converges (absolutely and)

pointwise to 0 for x = 0 and to 1 + x for x > 0. Since the limit function is not continuous, the

convergence is not uniform.

Exercise-43: (i) Let X be a metric space and fn ∈ C(X,R) for n ∈ N.(i) If there is a dense subset A of X such that

∑∞n=1 fn converges uniformly on A, then

∑∞n=1 fn

converges uniformly on X to some f ∈ C(X,R).(ii) Assume X is compact, fn ≥ 0 for every n ∈ N, and there is f ∈ C(X,R) such that

∑∞n=1 fn

converges pointwise to f on X. Then∑∞

n=1 fn converges uniformly to f on X.

[Hint : Let sn = f1 + · · · + sn. (i) By [137](ii), (sn) is uniformly Cauchy on X and converges

uniformly to a function f : X → R. By [136](i), f is continuous. (ii) Let gn = f − sn =∑∞

k=n+1 fk.

Then gn’s are continuous, g1 ≥ g2 ≥ · · · and (gn) → 0 pointwise. Apply Dini’s theorem [139](i).]

[142] [Weierstrass M-test] Let X be a metric space (or a set), and fn : X → R be functions for

n ∈ N. Suppose there exist Y ⊂ X, k ∈ N, and Mn ≥ 0 for n ∈ N such that∑∞

n=1Mn < ∞ and

|fn(y)| ≤Mn for every n ≥ k and every y ∈ Y . Then the series∑∞

n=1 fn converges absolutely and

uniformly on Y .

Proof. Clearly,∑∞

n=1 |fn(y)| ≤∑k

n=1 |fn(y)| +∑∞

n=k+1Mn < ∞ for each y ∈ Y , and absolute

convergence implies convergence of the series on Y . Let f : Y → R be f(y) =∑∞

n=1 fn(y). To

establish uniform convergence on Y , put sn = f1+ · · ·+fn and consider ε > 0. Choose n0 ≥ k with∑∞n=n0+1Mn < ε. Then for every n ≥ n0 and y ∈ Y , we have |f(y)− sn(y)| ≤

∑∞m=n+1 |fm(y)| ≤∑∞

m=n+1Mm < ε. Thus (sn) → f uniformly on Y . �

Example: (i) Let fn : [0,∞) → R be fn(x) =xn

enx. If x ∈ [0, 1/2], then

x

ex≤ 1

2. If x ∈ [1/2, 3],

thenx

ex≤ x

1 + x=

1

1 + 1/x≤ 1

1 + 1/3=

3

4. If x ∈ [3,∞), then

x

ex≤ x

x2/2=

2

x≤ 2

3. Letting

λ = max{1/2, 3/4, 2/3}, we see that∑∞

n=1 λn <∞ and fn(x) ≤ λn for every x ∈ [0,∞) and every

n ∈ N. Hence by [142],∑∞

n=1 fn converges uniformly on [0,∞).

(ii) Let fn : [0,∞) → R be fn(x) =nx

en. If k ∈ N and n ≥ (k + 3)!, then en ≥ nk+3

(k + 3)!≥ nk+2, and

hence for every x ∈ [0, k] we have that fn(x) ≤nk

nk+2=

1

n2. Therefore by [142],

∑∞n=1 fn converges


uniformly on [0, k] for each k ∈ N. However, fn(n) ≥ 1 for every n ≥ 3 and hence∑∞

n=1 fn does

not converge uniformly on [0,∞).

[143] (i) Let∑∞

n=0 cnxn be a real power series with radius of convergence R ∈ (0,∞]. Then, for

each r ∈ (0, R), the series∑∞

n=0 cnxn converges absolutely and uniformly on [−r, r]. Consequently,

f : (−R,R) → R defined as f(x) =∑∞

n=0 cnxn is continuous.

(ii) ex =∑∞

n=0

xn

n!for every x ∈ R, where 0! := 1.

Proof. (i) Fix r ∈ (0, R), and letMn = |cn|rn. Then∑∞

n=0Mn <∞ by [133](iii). Since |cnxn| ≤Mn

for every x ∈ [−r, r] and n ≥ 0,∑∞

n=0 cnxn converges uniformly on [−r, r] by Weierstrass M-test.

Let sn(x) =∑n

k=0 ckxk, which is a polynomial and hence continuous. Since (sn) → f uniformly on

[−r, r], f is continuous on [−r, r] by [136](i). Hence f is continuous on (−R,R).

(ii) The radius of convergence of∑∞

n=0

xn

n!is ∞ by Exercise-35(iv). Hence we may define f : R → R

as f(x) =∑∞

n=0

xn

n!. Then f is continuous by (i), and f(1) =

∑∞n=0

1

n!= e by [125](iii). Keeping

in mind (i) and Exercise-38(i), we note for x, y ∈ R that f(x)f(y) = (∑∞

k=0

xk

k!)(∑∞

m=0

ym

m!) =∑∞

n=0

∑nk=0

xk

k!

yn−k

(n− k)!=

∑∞n=0

1

n!(∑n

k=0nCkx

kyn−k) =∑∞

n=0

(x+ y)n

n!= f(x + y). Hence

f(x) = ex by Exercise-31(ii). �

Remark: It may be noted that by considering e as a real number, we showed above that ex =∑∞n=0

xn

n!without using the notions of differentiation and integration. In some text books, the

discussion of exponential function starts by defining exp(x) =∑∞

n=0

xn

n!, but then one has to do

some work to show exp(x) = ex for the special real number e ∈ R (often, this work is skipped).

Exercise-44: (i) Let X be a metric space and fn : X → R be functions such that f1 ≥ f2 ≥ · · · ≥ 0

and tn := sup{fn(x) : x ∈ X} → 0 as n→ ∞. Then∑∞

n=1(−1)nfn converges uniformly on X.

(ii) Let fn : [0, 1] → R be fn(x) = xn(1 − x). Then∑∞

n=1(−1)nfn converges absolutely and

uniformly on [0, 1], but∑∞

n=1 |fn| does not converge uniformly on [0, 1].

[Hint : (i) By Leibniz test,∑∞

n=1(−1)nfn converges pointwise to some f : X → R. By assumption,

t1 ≥ t2 ≥ · · · ≥ 0 and limn→∞ tn = 0. Let rn(x) =∑∞

k=n+1(−1)kfk(x). Note that r2n+1 = f2n+2 +∑∞j=1(−f2n+2j+1+f2n+2j+2) so that 0 ≤ r2n+1 ≤ f2n+2, and hence sup{|r2n+1(x)| : x ∈ X} ≤ t2n+2.

Similarly, −f2n+1 ≤ r2n ≤ 0 and hence sup{|r2n(x)| : x ∈ X} ≤ t2n+1. So (rn) → 0 uniformly on X.

(ii) Verify absolute convergence for x = 1 and x < 1 separately. Check that f1 ≥ f2 ≥ · · · ≥ 0. Also,

sup{fn(x) : x ∈ [0, 1]} = f(n

n+ 1) =

nn

(n+ 1)n+1=

1

(n+ 1)(1 + 1/n)n→ 0. Hence

∑∞n=1(−1)nfn

converges uniformly on [0, 1] by (i). But the pointwise limit of∑∞

n=1 |fn| is the discontinuous

function f : [0, 1] → R given by f(1) = 0 and f(x) = (1− x)∑∞

n=1 xn = x for x < 1.]

REAL ANALYSIS 53

Remark: We will discuss more about sequences and series of functions after mentioning a few facts

about differentiation and integration. See also the problems in Chapter 3 of W.J. Kaczor and M.T.

Nowak, Problems in Mathematical Analysis II, 2001.

12. Differentiation: preliminaries

Definition: Let U ⊂ R be open and f : U → R be a function.

(i) We say f is differentiable at a ∈ U if limx→af(x)− f(a)

x− aexists as a real number; i.e., if there

is some y ∈ R with the property that for every ε > 0 there is δ > 0 such that B(a, δ) ⊂ U and

|f(x) − f(a) − y(x − a)| < ε|x − a| for every x ∈ B(a, δ) \ {a}. Stated in a slightly different

manner, f is differentiable at a ∈ U if limh→0f(a+ h)− f(a)

hexists as a real number; i.e. if there

is some y ∈ R with the property that for every ε > 0 there is δ > 0 such that B(a, δ) ⊂ U and

|f(a+ h)− f(a)− yh| < ε|h| for every h ∈ R with 0 < |h| < δ.

(ii) If f is differentiable at a ∈ U , then the derivative f ′(a) of f at a is defined as f ′(a) =

limx→af(x)− f(a)

x− a= limh→0

f(a+ h)− f(a)

h. Note that f is differentiable at a with f ′(a) = y ⇔

limn→∞f(xn)− f(a)

xn − a= y for every sequence (xn) in U \ {a} converging to a.

(iii) We say f is differentiable in U if f is differentiable at each point of U .

Remark: (i) Geometrically,f(x)− f(a)

x− ais the slope of the line joining (a, f(a)) and (x, f(x)).

Hence f ′(a) is the slope of the tangent line to the graph of f at (a, f(a)).

(ii) We may also think off(x)− f(a)

x− aas a quantity measuring the rate of change of f near a, and

hence f ′(a) is the rate of change of f at a. If we think of f(x) as the position of an object at time

x, then f ′(a) is the velocity of that object at time a (and the second derivative f ′′(a) := (f ′)′(a) is

the acceleration of the object at time a).

(iii) For a function f : [a, b] → R, the differentiability of f at the end points a and b are defined by

considering the one-sided limits limx→a+f(x)− f(a)

x− aand limx→b−

f(x)− f(b)

x− b.

Example: Let b ∈ R and f : R → R be f(x) = |x − b|. Then f is continuous. But f is not

differentiable at b because limn→∞f(b+ 1/n)− f(b)

(b+ 1/n)− b= 1 = −1 = limn→∞

f(b− 1/n)− f(b)

(b− 1/n)− a. It

may be checked that f is differentiable at every a ∈ R \ {b}. More generally, if b1, . . . , bk ∈ R are

finitely many points, then the continuous function f : R → R defined as f(x) =∑k

j=1 |x − bj | isdifferentiable at a ∈ R iff a ∈ R \ {b1, . . . , bk}.

[144] Let U ⊂ R be open, f : U → R be a function, and a ∈ U . Then,

(i) [Caratheodory lemma] f is differentiable at a ⇔ there is a function fa : U → R such that

f(x) − f(a) = (x − a)fa(x) for every x ∈ U and fa is continuous at a (fa may not be continuous

at other points of U). If such a function fa exists, then it is necessarily unique and is called the

Caratheodory function of f at a.


(ii) If f is differentiable at a, then f is continuous at a.

(iii) If f is differentiable at a and f is non-vanishing in a neighborhood of a, then 1/f is differentiable

at a and (1/f)′(a) =−f ′(a)(f(a))2

.

Proof. (i) ⇒: Define fa : U → R as fa(a) = f ′(a) and fa(x) =f(x)− f(a)

x− afor x = a. ⇐:

Note thatf(x)− f(a)

x− a= fa(x) for x ∈ U \ {a}. Hence by the continuity of fa at a, we obtain

limx→af(x)− f(a)

x− a= fa(a) (and thus f ′(a) = fa(a)). This proof also shows why fa is unique.

(ii) Let fa be the Caratheodory function for f at a. If (xn) → a in U , then (fa(xn)) → fa(a) by

the continuity of fa at a, and hence f(xn)− f(a) = (xn − a)fa(xn) → 0 as n→ ∞.

(iii) We may suppose f is non-vanishing in U . If fa is the Caratheodory function for f at a, then

1/f(x)−1/f(a) =−(x− a)fa(x)

f(x)f(a)and hence

1/f(x)− 1/f(a)

x− a=

−fa(x)f(x)f(a)

→ −f ′(a)(f(a))2

as x→ a. �

Remark: (i) Let f : R → R be differentiable at a ∈ R. Then limh→0f(a+ h)− f(a− h)

2h=

limh→0f(a+ h)− f(a)

2h+ limh→0

f(a− h)− f(a)

−2h=f ′(a)

2+f ′(a)

2= f ′(a). On the other hand, if

f : R → R is any function with f(x) = f(−x) for every x ∈ R, then limh→0f(a+ h)− f(a− h)

2h= 0

at a = 0, but f may not even be continuous at 0; for example consider the indicator function 1Q.

(ii) There exists f ∈ C(R,R) which is nowhere differentiable; see Theorem 7.18 of Rudin.

Exercise-45: Let f : R → R be differentiable at 0. (i) If f(0) = 0, then limx→0f(x) + f(−x)

x= 0.

(ii) If yn < 0 < xn and limn→∞ yn = 0 = limn→∞ xn, then limn→∞f(xn)− f(yn)

xn − yn= f ′(0).

[Hint : (i)f(x) + f(−x)

x=f(x)− f(0)

x− f(−x)− f(0)

−xsince f(0) = 0. (ii) Let tn =

xnxn − yn

so

that 1 − tn =−yn

xn − yn. Then we have that |f ′(0) − f(xn)− f(yn)

xn − yn| ≤ |tn||f ′(0) −

f(xn)− f(0)

xn| +

|1− tn||f ′(0)−f(yn)− f(0)

yn| → 0 as n→ ∞ because |tn| ≤ 1 and |1− tn| ≤ 1.]

Exercise-46: Let f : R → R be differentiable and suppose for every ε > 0, there is δ > 0 such that

|f(a+ h)− f(a)

h− f ′(a)| < ε for every a ∈ R and every h ∈ (−δ, δ) \ {0}. Then f ′ is continuous.

[Hint : Let gn : R → R be gn(x) =f(x+ 1/n)− f(x)

1/n. Then gn’s are continuous, and (gn) → f ′

uniformly by the hypothesis. Now apply [136](i).]

[145] Let U ⊂ R be open, and f, g : U → R be differentiable at a ∈ U . Then,

(i) [Linearity] cf, f + g are differentiable at a with (cf)′(a) = cf ′(a) and (f + g)′(a) = f ′(a)+ g′(a).

(ii) [Product rule] fg is differentiable at a with (fg)′(a) = f ′(a)g(a) + f(a)g′(a).

(iii) If g is non-vanishing in U , then f/g is differentiable at a with (f/g)′(a) =f ′(a)g(a)− f(a)g′(a)

(g(a))2.

REAL ANALYSIS 55

Proof. (ii) Note that f(x)g(x) − f(a)g(a) = (f(x)g(x) − f(a)g(x)) + (f(a)g(x) − f(a)g(a)) =

(x− a)fa(x)g(x) + (x− a)f(a)ga(x) and use the continuity of fa, g, and ga at a.

(iii) Use [144](iii), and apply (ii) to f and 1/g. �

Notation and Definition: Let U ⊂ R be open and f : U → R be differentiable. We may denote f ′

also as f (1), and we may write f ′(x) also asdf(x)

dx. The (k + 1)th derivative of f is defined as the

derivative of the kth derivative of f for k ∈ N, if such a derivative exists. The second and third

derivatives of f are denoted respectively as f ′′ and f ′′′, or respectively as f (2) and f (3). In general,

if f is k-times differentiable, the kth derivative of f is denoted as f (k), and we may write f (k)(x)

also asdkf(x)

dxk. We say f is a Ck-map if f is k-times differentiable and the kth derivative f (k) is

continuous. Thus for example, a C0-map is just a continuous map, and a C1-map is a differentiable

map whose derivative is continuous. We say f is a C∞-map if f is infinitely often differentiable.

Example: For n = 0, 1, 2, . . ., let fn : R → R be fn(x) = xn. Then f0 ≡ 1, f1(x) = x, and we

have f ′0 ≡ 0 and f ′1 ≡ 1 by the definition of differentiability. Since f2 = f1f1, we get f ′2(x) =

f ′1(x)f1(x) + f1(x)f′1(x) = 2x by the product rule. Inductively, assuming f ′n(x) = nxn−1, we see

that f ′n+1(x) = (fnf1)′(x) = f ′n(x)f1(x) + fnf

′1(x) = nxn + xn = (n + 1)xn. By linearity of

differentiation, we deduce that if f : R → R is a non-constant polynomial f(x) =∑n

k=0 akxk, then

f is differentiable on R with f ′(x) =∑n

k=1 kakxk−1. In particular, f ′ is again a polynomial with

deg(f ′) = deg(f)− 1. If f is a constant polynomial, then f = cf0 for some c ∈ R and hence f ′ ≡ 0

by linearity. From these observations, we also conclude that if f : R → R is a polynomial of degree

n, then the nth derivative f (n) is constant, and hence f (m) ≡ 0 for every m ≥ n+ 1.

[146] (i) [Chain rule] Let U, V ⊂ R be open, f : U → V and g : V → R be functions. If f is

differentiable at a point a ∈ U and g is differentiable at f(a), then g ◦ f : U → R is differentiable

at a and (g ◦ f)′(a) = g′(f(a))f ′(a).

(ii) If f : R → R is a constant, then f ′ ≡ 0. If f : R → R is f(x) = xn for some n ∈ N, thenf ′(x) = nxn−1. If f : R → R is the polynomial f(x) =

∑nk=0 akx

k, then f ′(x) =∑n

k=1 kakxk−1.

(iii) If f : R \ {0} → R is f(x) = x−n for some n ∈ N, then f ′(x) = −nx−(n+1).

(iv)d sinx

dx= cosx,

d cosx

dx= − sinx, and

dex

dx= ex. Moreover, if f : R → R is differentiable, then

def(x)

dx= ef(x)f ′(x). In particular,

decx

dx= cex.

(v) If f : (a, b) → (u, v) and g : (u, v) → (a, b) are differentiable with (g ◦ f)(x) = x for every

x ∈ (a, b), then g′(f(x)) =1

f ′(x)and in particular f ′(x) = 0 for every x ∈ (a, b).

Proof. (i) Let fa and gf(a) be the Caratheodory functions of f at a and g at f(a). Then,

g(f(x))− g(f(a)) = (f(x)− f(a))gf(a)(f(x)) = (x− a)fa(x)gf(a)(f(x)), and hence

limx→ag(f(x))− g(f(a))

x− a= limx→a fa(x)gf(a)(f(x)) = fa(a)gf(a)(f(a)) = f ′(a)g′(f(a)).


(ii) This was shown in the Example above.

(iii) Let g(x) = xn so that f(x) = 1/g(x). Now apply (ii) and [144](iii).

(iv) Recall the identity sinx − sin y = 2 cos(x+ y

2) sin(

x− y

2) and the fact limx→0

sinx

x= 1 from

[113](vi). Hence limh→0sin(x+ h)− sinx

h= limh→0

cos(x+ h/2) sin(h/2)

h/2= cosx, where we used

also the continuity of cosx at 0. Thusd sinx

dx= cosx. Since cosx = sin(π/2 − x), we deduce by

chain rule thatd cosx

dx= − cos(π/2−x) = − sinx. Next, we get by [143](ii) that limh→0

eh − e0

h=

limh→0(1/h)(1 +h

1!+h2

2!+h3

3!+ · · · − 1)

= limh→0(1

1!+h

2!+h2

3!+ · · · ) = 1. Hence limh→0

ea+h − ea

h= ea(limh→0

eh − e0

h) = ea. The final

assertion follows by the Chain rule.

(v) By the Chain rule, g′(f(x))f ′(x) = 1 for every x ∈ (a, b). �

Remark: One should not writeg(f(x))− g(f(a))

x− a=g(f(x))− g(f(a))

f(x)− f(a)× f(x)− f(a)

x− ain order to

prove the Chain rule [146](i) because it can happen that f(x) = f(a) for x arbitrarily close to a.

Example: Let f, g : R → R be f(0) = 0 = g(0), f(x) = x sin(1/x) and g(x) = x2 sin(1/x). Clearly,

f and g are continuous on R and differentiable on R \ {0} with f ′(x) = sin(1/x)− (1/x) cos(1/x),

g′(x) = 2x sin(1/x) − cos(1/x). Nowf(x)− f(0)

x− 0= sin(1/x), which does not tend to any limit as

x → 0, and therefore f is not differentiable at 0. Since|g(x)− g(0)|

|x− 0|= |x sin(1/x)| ≤ |x|, we get

g′(0) = 0. But g′ is not continuous at 0 because cos(1/x) does not tend to any limit as x→ 0.

Definition: Let (X, d) be a metric space, f : X → R be a function, and x0 ∈ X. We say f has a

local minimum (respectively, local maximum) at x0 if there is an open neighborhood U of x0 in X

such that f(x) ≥ f(x0) (respectively, f(x) ≤ f(x0)) for every x ∈ U .

[147] (i) If f : (a, b) → R is a function having either a local minimum or a local maximum at a

point c ∈ (a, b) and if f is differentiable at c, then f ′(c) = 0.

(ii) [Darboux theorem - intermediate value property of the derivative] Let f : [a, b] → R be differ-

entiable. If either f ′(a) < y < f ′(b) or f ′(a) > y > f ′(b), then there is c ∈ (a, b) with f ′(c) = y.

Proof. (i) Assume that f has a local minimum at c (if f has a local maximum at c, the proof is

similar). Let ε > 0 be such that (c− ε, c+ ε) ⊂ (a, b) and f(x) ≥ f(c) for every x ∈ (c− ε, c+ ε).

Then for 0 < h < ε, we have f(c± h) ≥ f(c) and thereforef(c− h)− f(c)

−h≤ 0 ≤ f(c+ h)− f(c)

h.

Now letting h→ 0, we obtain that f ′(c) = 0.

(ii) Assume f ′(a) < y < f ′(b), and define g : [a, b] → R as g(x) = f(x)−yx. Then g is differentiable

and g′(a) < 0 < g′(b). Since g is continuous and [a, b] is compact, g attains its minimum at some

REAL ANALYSIS 57

c ∈ [a, b]. Since limh→0+g(a+ h)− g(a)

h= g′(a) < 0 and limh→0+

g(b− h)− g(b)

−h= g′(b) > 0,

there is ε > 0 such that g(a+h) < g(a) and g(b−h) < g(b) for 0 < h < ε. Consequently, c /∈ {a, b}.Therefore c ∈ (a, b) and so by (i), 0 = g′(c) = f ′(c)− y. �

Remark: (i) Let f : [1, 2] → R be f(x) = x2. Then f attains its minimum at 1 and maximum at

2, but f ′(1) = 0 and f ′(2) = 0. (ii) Let g : [0, 1] → R be g = 1[0,1/2], the indicator function of

[0, 1/2]. Then g does not have the intermediate value property. Hence g is not the derivative of any

differentiable function f : [0, 1] → R. (iii) It is a fact that for any function f : R → R, there are

g, h : R → R having the intermediate value property such that f = g + h. Choosing f such that f

is not a derivative, at least one of g, h is not a derivative. Thus the intermediate value property is

not sufficient for a function to be a derivative.

13. The Mean value theorem

[148] (i) [Rolle’s theorem] Let f ∈ C([a, b],R). If f is differentiable in (a, b) and f(a) = f(b), then

there is c ∈ (a, b) with f ′(c) = 0.

(ii) [Generalized mean value theorem] Let f, g ∈ C([a, b],R), and assume f and g are differentiable

in (a, b). Then there is c ∈ (a, b) with (f(b)− f(a))g′(c) = (g(b)− g(a))f ′(c).

(iii) [Mean value theorem] Let f ∈ C([a, b],R), and assume f is differentiable in (a, b). Then there

is c ∈ (a, b) with f(b)− f(a) = (b− a)f ′(c).

Proof. (i) Since f is continuous and [a, b] is compact, f attains its minimum and maximum in [a, b].

Let u = min f([a, b]) and v = max f([a, b]). If u = v, then f is constant, and hence f ′ ≡ 0 in (a, b).

If u < v, then at least one of them must differ from f(a) = f(b). Suppose u < f(a) = f(b). Then

f(c) = u for some c ∈ (a, b). By [147](i), f ′(c) = 0.

(ii) Let h : [a, b] → R be h(x) = (f(b) − f(a))g(x) − (g(b) − g(a))f(x). Then h is continuous on

[a, b] and differentiable in (a, b). Moreover h(a) = f(b)g(a) − g(b)f(a) = h(b). Hence by Rolle’s

theorem, there is c ∈ (a, b) with 0 = h′(c) = (f(b)− f(a))g′(c)− (g(b)− g(a))f ′(c).

(iii) Apply (ii) with g(x) = x. Another, a more geometric, proof is as follows. Let g : [a, b] → Rbe the function whose graph is the line segment joining (a, f(a)) and (b, f(b)). Thus g(x) =

f(a) +f(b)− f(a)

b− a(x − a). Let h : [a, b] → R be h = f − g. Then h is continuous on [a, b],

differentiable in (a, b), and h(a) = 0 = h(b). Hence by Rolle’s theorem, there is c ∈ (a, b) with

h′(c) = 0, i.e., with f ′(c) = g′(c). But g′(c) =f(b)− f(a)

b− a, the slope of the line segment joining

(a, f(a)) and (b, f(b)). �

Remark: Note that Rolle’s theorem is a special case of Mean value theorem.

[149] (i) Let f : [a, b] → R be differentiable with f ′ ≡ 0. Then f is constant.

(ii) Let f, g : [a, b] → R be differentiable with f ′ = g′. Then f − g is constant.


(iii) Let f : [a, b] → R be differentiable. Then f is increasing (decreasing) ⇔ f ′ ≥ 0 (f ′ ≤ 0).

(iv) If f : [a, b] → R is a C1-map, i.e., if f is differentiable with f ′ continuous, then f is Lipschitz

continuous on [a, b]. In particular, every polynomial f : [a, b] → R is Lipschitz continuous on [a, b].

(v) Let f : [a, b] → R be a C1-map. Let u = min{f ′(c) : c ∈ [a, b]} and v = max{f ′(c) : c ∈ [a, b]}.Then u(y − x) ≤ f(y)− f(x) ≤ v(y − x) for every x < y in [a, b].

Proof. (i) Let x < y be points in [a, b]. By Mean value theorem, there is c ∈ (x, y) with f(x)−f(y) =(x− y)f ′(c). Since f ′(c) = 0, we get f(x) = f(y).

(ii) Apply (i) to f − g.

(iii) Suppose f is increasing and consider c ∈ [a, b]. Then for every x ∈ [a, b] \ {c}, we havef(x)− f(c)

x− c≥ 0. Letting x → c, we get f ′(c) ≥ 0. Conversely assume f ′ ≥ 0. For x < y in [a, b],

there is c ∈ (x, y) by Mean value theorem with f(y)− f(x) = f ′(c)(y− x). Since f ′ ≥ 0 and y > x,

it follows that f(y) ≥ f(x), and thus f is increasing.

(iv) Since f ′ is continuous and [a, b] is compact, there is λ > 0 such that |f ′(c)| ≤ λ for every

c ∈ [a, b]. Now if x < y are points in [a, b], there is c ∈ (x, y) with f(y) − f(x) = (y − x)f ′(c) by

Mean value theorem, and hence |f(x)− f(y)| ≤ λ|x− y|.

(v) Follows by Mean value theorem. �

Example: (i) We claim that tanx > x for 0 < x < π/2. Let 0 < a < b < π/2 and f : [a, b] → R

be f(x) = tanx. Note that f ′(x) =cos2 x+ sin2 x

cos2 x=

1

cos2 x. Hence for any x ∈ (a, b), there is

y ∈ (a, x) by Mean value theorem with tanx = tanx− tan 0 =x

cos2 y. This proves the claim.

(ii) Let f : R → R be differentiable with f ′ = f and f(0) = 1. Then f(x) = ex for every x ∈ R. Tosee this, consider g(x) := f(x)e−x. By product rule and [146](iv), g′(x) = f ′(x)e−x − f(x)e−x = 0

as f ′ = f . Hence g is constant by [149](i). Therefore, g(x) = g(0) = 1 for every x ∈ R.

Exercise-47: (i) Let f : R → R be a differentiable map with limx→∞ f ′(x) = 0 and f(n) = 0 for

every n ∈ N. Then limx→∞ f(x) = 0.

(ii) Let f : R → R be a C2-map with f(1/n) = 1/n2 for every n ∈ N. Then f ′′(0) = 2.

(iii) If f : [0, 1] → R is a differentiable map with f(0) = 1 and f(1) = 1/e, then there is c ∈ (0, 1)

with f(c) = f ′(c).

(iv) If f : [1, e] → R is differentiable, there is c ∈ (1, e) with f(e)− f(1) = cf ′(c).

(v) If f : [1,√3] → R is differentiable, there is c ∈ (1,

√3) with f(

√3)− f(1) =

f ′(c)

c.

[Hint : (i) If n < x ≤ n+ 1, there is cn ∈ (n, x) with f(x) = f(x)− f(n) = f ′(cn)(x− n) by Mean

value theorem. (ii) Let g(x) = f(x) − x2. By Rolle’s theorem, there are cn ∈ (1

n+ 1,1

n) with

g′(cn) = 0, and then there are yn ∈ (cn+1, cn) with g′′(yn) = 0. By continuity, g′′(0) = 0. (iii)

Letting g(x) = e−xf(x), we see g(0) = 1 = g(1). Apply Rolle’s theorem to g. For (iv) and (v),

apply Generalized mean value theorem to f and log x, and to f and x2 respectively.]

REAL ANALYSIS 59

Exercise-48: (i) Let f : R → R be differentiable and suppose there are M > 0 and a > 0 such that

|f ′(x)| ≤Mx whenever x ≥ a. Then limx→∞f(x)

x2+δ= 0 for each δ > 0.

(ii) Let f : [a, b] → R be differentiable with f(a) = 0. If there is M > 0 such that |f ′(x)| ≤M |f(x)|for every x ∈ [a, b], then f ≡ 0.

(iii) Let f : [−1, 1] → [−1, 1] be a differentiable surjective map with |f ′| ≤ 1. Then either f = I or

f = −I, where I denotes the identity map.

[Hint : (i) Consider x > a. By Mean value theorem, there is c ∈ (a, x) with f(x)−f(a) = f ′(c)(c−a)and hence |f(x)| ≤ |f(a)| + Mc(c − a) ≤ |f(a)| + Mx2. (ii) Let 0 < δ <

1

2Mbe such that

|f(x) − f(y)| < 1 whenever |x − y| < δ. Consider x0 ∈ (a, a + δ). By Mean value theorem and

hypothesis, there is x1 ∈ (a, x0) with |f(x0)| ≤ |f ′(x1)||x0−a| < M |f(x1)|δ < |f(x1)|/2. In general,

there is xn ∈ (a, xn−1) with |f(xn−1)| ≤ |f(xn)|/2. Hence |f(x0)| ≤ |f(xn)|/2n <1

2nas f(a) = 0

and |f(xn)− f(a)| < 1. Thus f ≡ 0 in [a, a+ δ]. Now repeat the argument with a+ δ in the place

of a, and so on. (iii) Observe that |f(x)− f(y)| ≤ |x− y| for every x < y in [−1, 1] by Mean value

theorem. Hence f−1({−1, 1}) = {−1, 1} by surjectivity. If f(−1) = −1 and f(1) = 1, then we get

f(s) = s for every s ∈ (−1, 1) by applying the initial observation to [−1, s] and [s, 1]. If f(−1) = 1

and f(1) = −1, then similarly we get f(s) = −s for every s ∈ (−1, 1).]

[150] Let f : (a, b) → R be differentiable and suppose f ′(c) = 0 for every c ∈ (a, b). Then,

(i) Either f ′(c) > 0 for every c ∈ (a, b), or f ′(c) < 0 for every c ∈ (a, b). Consequently, f is either

strictly increasing or strictly decreasing (respectively) on (a, b).

(ii) There exist −∞ ≤ u < v ≤ ∞ such that f((a, b)) = (u, v).

(iii) The inverse g : (u, v) → (a, b) of f is differentiable and g′(f(c)) = 1/f ′(c) for every c ∈ (a, b).

Proof. (i) By Darboux theorem, either f ′ > 0 or f ′ < 0. Suppose f ′ > 0 on (a, b). If x < y are

points in (a, b), then by Mean value theorem, there is c ∈ (x, y) with f(y)−f(x) = (y−x)f ′(c) > 0,

and this shows f is strictly increasing. If f ′ < 0, we may see similarly that f is strictly decreasing.

(ii) This is because f is continuous and strictly monotone.

(iii) Note that g is continuous because f - being a strictly monotone continuous function - takes

open intervals to open intervals. Fix c ∈ (a, b), let z = f(c), and fc be the Caratheodory function

of f at c. Since f is injective by (i) and since f(x)− f(c) = (x− c)fc(x), it follows that fc(x) = 0

for every x = c. For y = f(x) ∈ (u, v) \ {z}, we have g(y)− g(z) = x− c =f(x)− f(c)

fc(x)=

y − z

fc(g(y)).

Moreover, by the continuity of g and the continuity of fc at c, the function h : (u, v) → R defined

as h(y) =1

fc(g(y))is continuous at z with h(z) =

1

fc(g(z))=

1

fc(c)=

1

f ′(c). Thus by [144](i), h is

the Caratheodory function of g at z, and g is differentiable at z with g′(z) = h(z) =1

f ′(c). �

Remark: Let f : [−1, 1] → R be f(x) = x3. Then f is differentiable and strictly increasing, but f ′

vanishes at 0. Contrast this with [149](iii) and [150](i).


[151] (i) If g : (0,∞) → R is g(x) = log x, then g is differentiable and g′(x) = 1/x for x ∈ (0,∞).

(ii) Let y > 0 and f : [0,∞) → R be f(x) = xy. Then f is differentiable in (0,∞) with f ′(x) = yxy−1

for x ∈ (0,∞). If y ≥ 1, then f is also differentiable at 0, where f ′(0) = 0 for y > 1 and f ′(0) = 1

for y = 1. If 0 < y < 1, then f is not differentiable at 0. In particular, the function x 7→ x1/k is

not differentiable at 0 for any integer k ≥ 2.

(iii) Let X ⊂ R be nonempty connected, f : X → R be a C2-map, and c ∈ int(X) be with f ′(c) = 0.

If f ′′(c) > 0, then c is a local minimum of f . If f ′′(c) < 0, then c is a local maximum of f .

Proof. (i) Let f : R → (0,∞) be f(y) = ey. Then the hypothesis of [150] holds, and the inverse of

f is g. Writing x = ey, we get by [150](iii) that g′(x) = g′(f(y)) = 1/f ′(y) = 1/ey = 1/x.

(ii) f(x) = xy = ey log x for x > 0, and hence f ′(x) = ey log x(y/x) = xy(y/x) = yxy−1 by (i)

and the Chain rule [146](i). If y > 1, thenf(x)− f(0)

x= xy−1 → 0 as x → 0. If y = 1, then

f(x)− f(0)

x= 1 for x > 0. If 0 < y < 1, then

f(x)− f(0)

x=

1

xy−1→ ∞ as x→ 0.

(iii) Suppose f ′′(c) > 0. Choose a, b ∈ X such that a < c < b and f ′′ > 0 in (a, b). Applying [151](i)

to f ′, we see that f ′ is strictly increasing in (a, b). Since f ′(c) = 0, it follows that f ′ < 0 on (a, c)

and f ′ > 0 on (c, b). Now applying [150](i) to f , we get that f is strictly decreasing on (a, c) and

f is strictly increasing on (c, b). Hence c must be a local minimum of f . �

Example: Let f, g : R → R be f(x) = x3 and g(x) = x1/3. Then f and g are inverses of each other,

both and f and g are strictly increasing, f is differentiable on R, g is continuous on R, and g is

differentiable on R \ {0}. But g is not differentiable at 0 by [151](ii). Contrast this with [150](iii).

The problem is due to the fact that f ′(0) = 0.

Exercise-49: Let a > 0 and f : [a,∞) → (0,∞) be differentiable with limx→∞xf ′(x)

f(x)= 0. Then for

every b > 1, we have that limx→∞f(bx)

f(x)= 1. [Hint : Let g(x) = log f(x), and note that it suffices

to show limx→∞(g(bx)− g(x)) = 0. By Mean value theorem, for each x ∈ [a,∞) there is c ∈ (x, bx)

with g(bx)−g(x) = (b−1)xg′(c). And |xg′(c)| = |xf ′(c)||f(c)|

≤ |cf ′(c)||f(c)|

→ 0 as x→ ∞ because c > x.]

[152] [Tangent to a path] Let f, g : [a, b] → R be C1-maps and let ψ : [a, b] → R2 be the path

defined as ψ = (f, g). Let c ∈ (a, b), and suppose (f ′(c), g′(c)) = (0, 0). Then the tangent to the

path ψ at ψ(c) is the line (y − g(c))f ′(c) = (x− f(c))g′(c), whose slope is g′(c)/f ′(c).

Proof. Since (f ′(c), g′(c)) = (0, 0) and f ′, g′ are continuous, there is an open neighborhood U ⊂(a, b) of c such that either f ′ or g′ is non-vanishing in U . Consider t ∈ U \ {c}. Then (f(t), g(t)) =(f(c), g(c)) by the choice of U and Rolle’s theorem. The equation of the line passing through

(f(c), g(c)) and (f(t), g(t)) is (y − g(c))(f(t) − f(c)) = (x − f(c))(g(t) − g(c)). This equation

may be written using the Caratheodory functions of f and g at c as (y − g(c))(t − c)fc(t) =

REAL ANALYSIS 61

(x− f(c))(t− c)gc(t), or equivalently as (y− g(c))fc(t) = (x− f(c))gc(t). Letting t→ c, we get the

equation of the tangent line to ψ at ψ(c) as (y − g(c))f ′(c) = (x− f(c))g′(c). �

Example: Let ψ : [0, 2π] → R2 be ψ(t) = (f(t), g(t)) := (√2 cos t,

√2 sin t). Note that ψ(π/4) =

(1, 1), and (f ′(π/4), g′(π/4)) = (−√2 sin(π/4),

√2 cos(π/4)) = (−1, 1) = (0, 0). The equation of

the tangent line at (1, 1) to the circle specified by ψ is (y − g(π/4))f ′(π/4) = (x− f(π/4))g′(π/4),

or (y − 1)(−1) = (x− 1) · 1, which simplifies to x+ y = 2, whose slope is −1.

14. LHospital’s rule and Taylor’s theorem

To evaluate certain limits, LHospital’s rule is quite useful3. In [153] below, we present two

situations where LHospital’s rule holds. More such situations are indicated by Examples afterwards.

[153] (i) [LHospital’s rule for0

0indeterminate form] Let a < b be reals and f, g : (a, b) → R be

differentiable functions such that f ′ is non-vanishing in (a, b), and limx→a f(x) = 0 = limx→a g(x).

If the limit l := limx→ag′(x)

f ′(x)exists in [−∞,∞], then limx→a

g(x)

f(x)= l.

(ii) [LHospital’s rule for∗∞

indeterminate form: valid even when the numerator does not tend to

any limit] Let ∞ ≤ a < b ≤ ∞ and f, g : (a, b) → R be differentiable with f ′ is non-vanishing in

(a, b) and limx→a f(x) = ∞. If the limit l := limx→ag′(x)

f ′(x)exists in [−∞,∞], then limx→a

g(x)

f(x)= l.

Proof. (i) Extend f and g continuously to [a, b) by putting f(a) = 0 = g(a). Note that f is non-

vanishing in (a, b) by Rolle’s theorem because f(a) = 0 and f ′ is non-vanishing in (a, b). Let (xn)

be any sequence in (a, b) converging to a. By the Generalized mean value theorem, there are cn ∈

(a, xn) such that (g(xn) − g(a))f ′(cn) = (f(xn) − f(a))g′(cn), which simplifies tog(xn)

f(xn)=g′(cn)

f ′(cn)since f(a) = 0 = g(a) and since f ′ is non-vanishing on (a, b). Now note that (xn) → a and (cn) → a

as n→ ∞. The geometric meaning is the following: by [152],g′(x)

f ′(x)is the slope of the tangent line

at (f(x), g(x)) to the path ψ := (f, g), and limx→ag(x)

f(x)= limx→a

g(x)− g(a)

f(x)− f(a)is the slope of the

tangent at (f(a), g(a)) = (0, 0).

(ii) We need to prove two things: (1) If l < t < ∞, then there is u ∈ (a, b) such thatg(x)

f(x)< t

for every x ∈ (a, u), and (2) If −∞ < t < l, then there is u ∈ (a, b) such thatg(x)

f(x)> t for

every x ∈ (a, u). We prove only (1), and the proof of (2) is similar. Let l < s < t. Then

there is w ∈ (a, b) such thatg′(x)

f ′(x)< s for every x ∈ (a,w). Since limx→a f(x) = ∞, there

is v ∈ (a,w) with f(x) > max{0, f(w)} for every x ∈ (a, v). Now consider x ∈ (a, v). By

the Generalized mean value theorem, there is c ∈ (x,w) withg(x)− g(w)

f(x)− f(w)=

g′(c)

f ′(c)< s. Since

f(x) − f(w) > 0, we get g(x) − g(w) < s(f(x) − f(w)). Dividing by f(x) > 0 and rearranging,

3LHospital is a french name - written also as LHopital - and is pronounced roughly as ‘lopital’.


we haveg(x)

f(x)< s +

g(w)− sf(w)

f(x). Since limx→a f(x) = ∞ and s < t, we conclude that there is

u ∈ (a, v) such thatg(x)

f(x)< t for every x ∈ (a, u). �

Remark: [153] is stated for the case where the limit is taken at the left end point a of (a, b). Clearly,

similar results hold for the case where the limit is taken at the right end point b of (a, b). Combining

both cases, LHospital’s rule can be stated for the case where the limit is taken at a point c ∈ (a, b).

Moreover, LHospital’s rule can also be formulated for indeterminate forms such as ∞−∞, 0 · ∞,

1∞, 00, ∞0, etc. (often reduced to the previous cases by the application of log and exponential

functions). We omit the details, but some illustrative examples are presented below.

Notation: The notation ‘=L’ means ‘the equality is obtained by an application of LHospital’s rule’.

Example: (i) [∞−∞ form] limx→0+(1

x− 1

(ex − 1)) = limx→0+

ex − 1− x

xex − x

=L limx→0+ex − 1

ex + xex − 1=L limx→0+

ex

2ex + xex= 1/2. (Since ex = 1 +

x

1!+x2

2!+x3

3!+ · · ·

by [143], we may also evaluate the above limit as limx→0+(1

x− 1

(ex − 1)) = limx→0+

ex − 1− x

x(ex − 1)=

limx→0+x2/2 + x3/6 + · · ·x2 + x3/2 + · · ·

= limx→0+1/2 + x/6 + · · ·1 + x/2 + · · ·

= 1/2.)

(ii) [0 · (−∞) form] limx→0+ x log x = limx→0+log x

1/x=L limx→0+

1/x

−1/x2= limx→0+(−x) = 0.

(iii) [00 form] limx→0+ xx = limx→0+ e

x log x = 1 by (ii) and the continuity of ex.

(iv) [1∞ form] limx→0+(cosx)1/x2

= e−1/2 because (cosx)1/x2= e

log(cosx)

x2 , and

limx→0+log(cosx)

x2=L limx→0+

− sinx

2x cosx=L limx→0+

− cosx

2 cosx− 2x sinx= −1/2.

(iv) [∞0 form] limx→∞(x2 + 4)1/ log(x3+5) = e2/3 because (x2 + 4)1/ log(x

3+5) = elog(x2+4)/ log(x3+5),

and limx→∞log(x2 + 4)

log(x3 + 5)=L limx→∞

2x/(x2 + 4)

3x2/(x3 + 5)= limx→∞

2x3 + 10

3x3 + 12x= 2/3.

Exercise-50: Let us write g ≪ f if limx→∞g(x)

f(x)= 0. Then we have:

log(log x) ≪√log x≪ log x≪

√x≪ x

log x≪ x≪ x log x≪ x2 ≪ ex.

[Hint : log x ≪√x because limx→∞

log x√x

=L limx→∞1/x

1/(2√x)

= limx→∞2√x

= 0; x2 ≪ ex

because limx→∞x2

ex=L limx→∞

2x

ex=L limx→∞

2

ex= 0; or use Exercise-31.]

Remark: There are situations where LHospital’s rule should not be applied. (i) When limx→a f(x) =

0 = limx→a g(x) does not hold. For example, it is false to conclude limx→0x2

x2 + 1= limx→0

2x

2x= 1;

the correct conclusion is limx→0x2

x2 + 1= 0. (ii) When the condition f ′(c) = 0 for every c ∈ (a, b)

REAL ANALYSIS 63

is not satisfied or when the limit limx→ag′(x)

f ′(x)does not exist. For example, the correct way to

evaluate limx→∞x

x+ sinxis limx→∞

x

x+ sinx= limx→∞

1

1 + (sinx/x)= 1. But if we blindly

apply LHospital’s rule, limx→∞x

x+ sinx= limx→∞

1

1 + cosx, and the latter limit does not exist.

Remark: A variant of [153](i) has an easier proof as indicated in the hint of Exercise-51(i) below;

this also gives an insight why [153](i) is true.

Exercise-51: (i) Let f, g : [a, b] → R be such that f(a) = 0 = g(a), f(x) = 0 for every x ∈ (a, a+ ε)

for some ε > 0, and f, g be differentiable at a with f ′(a) = 0. Then limx→a+g(x)

f(x)=g′(a)

f ′(a).

(ii) Let f, g : R → R be f(x) = sinx and g(x) = x2 for x ∈ Q and g(x) = 0 for x ∈ R \ Q. Then

limx→0+g(x)

f(x)= 0.

(iii) If f : (0,∞) → R is differentiable and l := limx→∞(f(x) + f ′(x)) exists in [−∞,∞], then

limx→∞ f(x) = l. Consequently, limx→∞ f ′(x) = 0 if l ∈ R.

[Hint : (i)g(x)

f(x)=

(g(x)− g(a))/(x− a)

(f(x)− f(a))/(x− a). (ii) Apply (i) after noting that f(0) = 0 = g(0),

f(x) = 0 for x ∈ (0, π/2), f ′(0) = 1, and g′(0) = 0. (iii) limx→∞ f(x) = limx→∞f(x)ex

ex=L

limx→∞(f(x) + f ′(x))ex

ex= limx→∞(f(x) + f ′(x)) = l.]

We will see that a sufficiently nice function f : [a, b] → R can be approximated by polynomials.

For a differentiable f , Taylor expansion will give local approximations (as presented below); and

for a continuous f , Weierstrass approximation theorem will give global approximations (this will

be proved a little later). Note that if we write a polynomial p in the form p(x) =∑n

k=0 ck(x− a)k

(with a ∈ R fixed), then p(a) = c0 and p(k)(a) = k!ck for 1 ≤ k ≤ n, where p(k) denotes the kth

derivative of p. This observation motivates the following definition.

Definition: Let X ⊂ R be an interval, a ∈ X, n ∈ N, and f : X → R be a function which is

n-times differentiable at a. Then the nth Taylor polynomial of f at a is defined as P[f,a,n](x) =∑nk=0

f (k)(a)

k!(x − a)k = f(a) +

f ′(a)

1!(x − a) +

f ′′(a)

2!(x − a)2 + · · · + f (n)(a)

n!(x − a)n. Note that

f (k)(a) = P(k)[f,a,n](a) for 0 ≤ k ≤ n, where f (0) = f and f (k) is the kth derivative of f for 1 ≤ k ≤ n.

We will show below that if f is (n + 1)-times differentiable in (a, b) and f (n) is continuous on

[a, b], then P[f,a,n](b) is a good approximation for f(b) for points b > a near a (and similarly for

points b < a near a), and the error in the approximation can be expressed in terms of f (n+1).

[154] [Taylor’s theorem] Let a < b, f : [a, b] → R be a function, n ∈ N, and assume f (n) is

continuous on [a, b] and f (n) is differentiable on (a, b). Then there is c ∈ (a, b) such that

f(b) = f(a) +∑n

k=1

f (k)(a)

k!(b− a)k +

f (n+1)(c)

(n+ 1)!(b− a)n+1 = P[f,a,n](b) +

f (n+1)(c)

(n+ 1)!(b− a)n+1.


Proof. Let P = P[f,a,n], and M ∈ R be such that f(b) = P (b) +M(b − a)n+1. We have to show

that M =f (n+1)(c)

(n+ 1)!for some c ∈ (a, b). Define g : [a, b] → R as g(x) = f(x)−P (x)−M(x−a)n+1.

Then g(n) is differentiable in (a, b) and g(n) is continuous on [a, b]. Since f (k)(a) = P (k)(a), we may

see inductively that g(k)(a) = 0 for 0 ≤ k ≤ n. Also, g(b) = 0 by the choice of M . Since g(a) =

0 = g(b), there is c1 ∈ (a, b) with g′(c1) = 0 by Rolle’s theorem. Since g′(a) = 0 = g′(c1), there is

c2 ∈ (a, c1) with g′′(c2) by Rolle’s theorem. Proceeding like this, we may find c := cn+1 ∈ (a, cn)

with g(n+1)(c) = 0. But g(n+1)(c) = f (n+1)(c)− 0− (n+1)!M by the definition of g and by the fact

that deg(P ) = n. Hence f (n+1)(c)/(n+ 1)! =M , as required. �

Example: (i) [Approximate value of√2] Let f(x) = x1/2 for x > 0. Then f ′(x) =

x−1/2

2and

f ′′(x) =−x−3/2

4. Applying Taylor’s theorem with a = 1, b = 2 and n = 1, we may find c ∈ (1, 2)

with√2 = f(b) = f(1)+ f ′(1)(2− 1)+

f ′′(c)

2(2− 1)2 =

3

2− c−3/2

8. Since 1 ≤

√2 ≤ 3

2(as 2 ≤ 9/4)

and c ∈ (1, 2), we have −1 ≤ −c−3/2 ≤ −2−3/2 ≤ −1

3, and hence

3

2− 1

8≤

√2 ≤ 3

2− 1

24.

(ii) [Approximate value of e] Let f(x) = ex. Applying Taylor’s theorem with a = 0, b = 1 and

n = 3, we may find c ∈ (0, 1) with e = f(b) =∑3

k=0

f (k)(0)

1!(1 − 0)k +

f (4)(c)

4!(1 − 0)4 =

8

3+ec

24.

Since 1 = e0 ≤ ec ≤ e1 ≤ 3, we deduce that8

3+

1

24≤ e ≤ 8

3+

3

24, which gives 2.7 ≤ e ≤ 2.8.

[155] [Taylor series] (i) Let f : [a, b] → R be a C∞-map, i.e., an infinitely often differentiable map.

Assume there is M > 0 such that |f (n)(x)| ≤ M for every x ∈ [a, b] and every n ∈ N. Then

f(x) =∑∞

n=0

f (n)(a)

n!(x− a)n for every x ∈ [a, b], where the series converges to f uniformly in [a, b]

(and similar result holds for [b, a] when b < a).

(ii) sinx = x− x3

3!+x5

5!− x7

7!+ · · · and cosx = 1− x2

2!+x4

4!− x6

6!+ · · · for every x ∈ R, where the

convergence of the two series are uniform on [−r, r] for each r > 0.

(iii) Let a ∈ R, r > 0, and f : (a− r, a+ r) → R be a C∞-map, Assume there is M > 0 such that

|f (n)(x)| ≤ Mn!

rnfor every x ∈ (a− r, a+ r) and every n ∈ N. Then f(x) =

∑∞n=0

f (n)(a)

n!(x− a)n

for every x ∈ (a−r, a+r), and the series converges to f uniformly in [a−t, a+t] for every t ∈ (0, r).

Proof. (i) Note that the nth partial sum of the series∑∞

n=0

f (n)(a)

n!(x− a)n is precisely P[f,a,n](x).

Clearly, P[f,a,n](a) = f(a) for every n ∈ N. Now if x ∈ (a, b], then by [153], there is c ∈ (a, x) such

that f(x) = P[f,a,n](x) +f (n+1)(c)

(n+ 1)!(x− a)n+1, and hence |f(x)−P[f,a,n](x)| ≤

M(b− a)n+1

(n+ 1)!→ 0 as

n→ ∞ independent of x by [124](vi).

(ii) Fix r > 0. Check that the functions sinx and cosx satisfy the assumptions of (i) with [a, b] =

[0, r] (and similarly with [b, a] = [−r, 0]), and M = 1.

REAL ANALYSIS 65

(iii) Let t ∈ (0, r). We will show that the series converges uniformly to f on [a, a+ t]; and similar

argument works for [a−t, a]. As before, note that nth partial sum of the series∑∞

n=0

f (n)(a)

n!(x−a)n

is precisely P[f,a,n](x), and P[f,a,n](a) = f(a) for every n ∈ N. Now, if x ∈ (a, a+ t], then by [153],

there is c ∈ (a, x) such that f(x) = P[f,a,n](x)+f (n+1)(c)

(n+ 1)!(x−a)n+1, and hence |f(x)−P[f,a,n](x)| ≤

Mtn+1

rn+1→ 0 as n→ ∞ independent of x because 0 < t < r. �

Remark: The series expansions of sinx and cosx in [155](ii) are called the Taylor series of sinx

and cosx. The Taylor series of ex presented in [143](ii) can also be derived using [155](iii).

Exercise-52: (i) Let X ⊂ R be a non-degenerate interval, n ∈ N, and f : X → R be a function with

f (n+1) ≡ 0. Then f is a polynomial of degree ≤ n.

(ii) Let f : R → R be twice differentiable with M := sup{|f ′′(x)| : x ∈ R} <∞ and limx→∞ f(x) =

0. Then limx→∞ f ′(x) = 0.

(iii) Let f : [0, 1] → R be infinitely often differentiable. Assume f(1/n) = 0 for every n ∈ N and

M := sup{|f (n)(x)| : x ∈ [0, 1] and n ∈ N} <∞. Then f ≡ 0 on [0, 1].

[Hint : Fix a ∈ X, apply Taylor’s theorem and note that the error term is zero by assumption.

Hence f(x) = P[f,a,n](x) for x ∈ X. Another proof is by induction on n, with [149](i) providing the

starting step. (ii) For a < b, there is c ∈ (a, b) with f(b) = f(a) + f ′(a)(b − a) + f ′′(c)(b − a)2/2

by [154] and hence |f ′(a)| ≤ |f(b)− f(a)|b− a

+M(b− a)

2. Given ε > 0, there is a0 ∈ R such that

|f(b) − f(a)| < ε2 for b > a ≥ a0 since limx→∞ f(x) = 0. For every a ≥ a0, taking b = a + ε, we

get |f ′(a)| ≤ ε2

ε+Mε

2= (1 +

M

2)ε. (iii) By continuity, f(0) = 0. Between two zeroes of f , there

should be a zero of f ′. Hence f ′(0) = 0 by the continuity of f ′. In general, f (k)(0) = 0 for every

k ∈ N ∪ {0}. Now by [154], for each b ∈ (0, 1], there is c ∈ (0, b) with f(b) = 0 +f (n+1)(c)

(n+ 1)!(b − 0)

and hence |f(b)| ≤ Mb

(n+ 1)!for every n ∈ N, which implies f(b) = 0.]

Exercise-53: (i) log(1+x) = x− x2

2+x3

3− x4

4+ · · · for |x| < 1/2, and the series converges uniformly

on [−t, t] for every t ∈ (0, 1/2) (we remark that this series expansion holds for |x| < 1 by [167](v),

but with our present tools we can manage only the case |x| < 1/2).

(ii) cosx ≥ 1− x2

2for every x ∈ R.

[Hint : (i) Apply [155](iii) with a = 0 and r = 1/2 to f(x) := log(1+x). For n ∈ N we have f (n)(x) =

(−1)n+1(n− 1)!

(1 + x)n. So |f (n)(x)| ≤ (n− 1)!

(1/2)n≤ n!

(1/2)nfor |x| < 1/2. Also

f (n)(0)

n!=

(−1)n+1

n. (ii) If

|x| ≥ π ≥ 3, then it is clear. If x ∈ [−π, π] \ {0}, then applying Taylor’s theorem with a = 0 and

n = 2, find c between 0 and x with cosx = 1− x2

2+x3 sin c

6. Also x3 and sin c have the same sign.]


15. Integration: preliminaries and properties

In this section, we will discuss the basic theoretical aspects of Riemann integration.

Definition: (i) A partition P of a closed interval [a, b] is a finite sequence x0, x1, . . . , xk of points in

[a, b] with a = x0 ≤ x1 ≤ · · · ≤ xk−1 ≤ xk = b. Usually we will just say “P = {x0 ≤ x1 ≤ · · · ≤ xk}is a partition of [a, b]”, where it is implicitly understood that x0 = a and xk = b.

(ii) If P,Q are partitions of [a, b] and if P ⊂ Q as sets, then we will say Q is a refinement of P .

(iii) If P = {x0 ≤ x1 ≤ · · · ≤ xk} is a partition of [a, b], then the norm (or mesh) ∥P∥ of P is

defined as ∥P∥ = max{xj − xj−1 : 1 ≤ j ≤ k}.(iv) If f : [a, b] → R is a bounded function and P = {x0 ≤ x1 ≤ · · · ≤ xk} is a partition of [a, b],

then the lower Riemann sum L(f, P ) and the upper Riemann sum U(f, P ) of f with respect to P

are L(f, P ) :=∑k

j=1 inf{f(x) : xj−1 ≤ x ≤ xj} × (xj − xj−1), and

U(f, P ) :=∑k

j=1 sup{f(x) : xj−1 ≤ x ≤ xj} × (xj − xj−1).

Clearly, L(f, P ) ≤ U(f, P ). If f ≥ 0, then L(f, P ) and U(f, P ) provide respectively a lower and

an upper approximation to the area of the region between the graph of f and the x-axis.

[156] Let f : [a, b] → R be a bounded function, and M > 0 be such that |f | ≤M .

(i) Let P,Q be partitions of [a, b], and assume Q is a refinement of P obtained by adding n more

points to P , where n ∈ N. Then,L(f, P ) ≤ L(f,Q) ≤ L(f, P ) + 2Mn∥P∥, and U(f, P )− 2Mn∥P∥ ≤ U(f,Q) ≤ U(f, P ).

(ii) If P1, P2 are any two partitions of [a, b], then L(f, P1) ≤ U(f, P2).

Proof. (i) It is enough to prove the result for the case n = 1, for then the general case will follow

by a repeated application of this case. So consider a partition P = {x0 ≤ x1 ≤ · · · ≤ xk} of

[a, b], and Q be a refinement of P obtained by adding one extra point c ∈ (xj−1, xj) for some

j ∈ {1, . . . , k}. For this j, let u = inf{f(x) : x ∈ [xj−1, xj ]}, v1 = inf{f(x) : x ∈ [xj−1, c]} and

v2 = inf{f(x) : x ∈ [c, xj ]}. Observe that 0 ≤ vi − u ≤ 2M for i = 1, 2. Since L(f,Q)− L(f, P ) =

v1(c − xj−1) + v1(xj − c) − u(xj − xj−1) = (v1 − u)(c − xj−1) + (v2 − u)(xj − c), it follows that

0 ≤ L(f,Q)−L(f, P ) ≤ 2M(xj −xj−1) ≤ 2M∥P∥. A similar proof gives the estimates for U(f,Q).

(ii) Let Q = P1 ∪P2, which is a common refinement of P1 and P2. Then by (i), we have L(f, P1) ≤L(f,Q) ≤ U(f,Q) ≤ U(f, P2). �

Definition: Let f : [a, b] → R be a bounded function. The lower Riemann integral L(f) and the

upper Riemann integral U(f) of f are defined as

L(f) = sup{L(f, P ) : P is a partition of [a, b]}, andU(f) = inf{U(f, P ) : P is a partition of [a, b]}.Note that L(f) ≤ U(f) by [156](ii). If L(f) = U(f), then we say f is Riemann integrable. If f

is Riemann integrable and if y ∈ R is such that z = L(f) = U(f), then we write∫ ba f(x)dx = z,

REAL ANALYSIS 67

and we say the integral of f over [a, b] is equal to z. In the expression∫ ba f(x)dx, the function f

is sometimes called the integrand. Sometimes we will just write∫ ba fdx instead of

∫ ba f(x)dx for

notational simplicity.

Notation: Let R[a, b] denote the collection of all Riemann integrable functions f : [a, b] → R. Keep

in mind that R[a, b] is a subcollection of the collection of all bounded functions from [a, b] to R.

[157] [Necessary and sufficient condition for Riemann integrability] Let f : [a, b] → R be a

bounded function. Then f ∈ R[a, b], i.e., f is Riemann integrable ⇔ for every ε > 0, there is

a partition P of [a, b] with 0 ≤ U(f, P ) − L(f, P ) ≤ ε. Here, note that U(f, P ) − L(f, P ) =∑kj=1 diam(f([xj−1, xj ]))(xj − xj−1) if P = {x0 ≤ x1 ≤ · · · ≤ xk}.

Proof. ⇒: Let z = L(f) = U(f). By the definition of L(f) and U(f), we may choose two partitions

P1 and P2 of [a, b] with z − ε/2 < L(f, P1) ≤ z ≤ U(f, P2) < z + ε/2. Let P = P1 ∪ P2, which is

a common refinement of P1 and P2. Then z − ε/2 < L(f, P ) ≤ z ≤ U(f, P ) < z + ε/2 by [156](i)

and by the definition of L(f) and U(f). Hence 0 ≤ U(f, P )− L(f, P ) ≤ ε.

⇐: 0 ≤ U(f)− L(f) ≤ ε for every ε > 0 by the given condition, and hence L(f) = U(f). �

Example: Let A = Q ∩ [0, 1] and f : [0, 1] → R be f = 1A. Then U(f, P ) − L(f, P ) = 1 for any

partition P of [0, 1]. Hence by [157], f is not Riemann integrable.

[158] (i) Let f ∈ R[a, b] and g : [a, b] → R be a bounded function. If there is a constant M > 0

such that |g(x)− g(y)| ≤M |f(x)− f(y)| for every x, y ∈ [a, b], then g ∈ R[a, b].

(ii) If f : [a, b] → R is a bounded monotone function, then f ∈ R[a, b].

(iii) If f : [a, b] → R is continuous, then f ∈ R[a, b].

(iv) If f : [a, b] → R is a function such that f ≡ c on (a, b), then f ∈ R[a, b] and∫ ba fdx = c(b− a).

In particular,∫ ba 0dx = 0.

Proof. (i) Apply [157] after noting the following: if P = {x0 ≤ x1 ≤ · · · ≤ xk} is any partition of

[a, b], then U(g, P )− L(g, P ) =∑k

j=1 diam(g([xj−1, xj ]))(xj − xj−1) ≤M(U(f, P )− L(f, P )).

(ii) Let s < t be real numbers with s ≤ f ≤ t. Given ε > 0, choose a partition P = {x0 ≤x1 ≤ · · · ≤ xk} of [a, b] with ∥P∥ <

ε

t− s. Since f is a monotone function, it may be noted

that∑k

j=1 diam(f([xj−1, xj ])) = diam(f [a, b]) ≤ t − s, and hence we have U(f, P ) − L(f, P ) ≤∑kj=1 diam(f([xj−1, xj ]))× ∥P∥ ≤ (t− s)× ε

t− s= ε. Therefore, f ∈ R[a, b] by [157].

(iii) Let ε > 0 be given. Since f is uniformly continuous by the compactness of [a, b], we may

choose δ > 0 such that |f(x) − f(y)| < ε

b− afor every x, y ∈ [a, b] with |x − y| < δ (we may

assume a < b to avoid trivialities). Let P = {x0 ≤ x1 ≤ · · · ≤ xk} be a partition of [a, b] with

∥P∥ < δ. Then diam(f([xj−1, xj ])) ≤ ε

b− afor 1 ≤ j ≤ k, and hence U(f, P ) − L(f, P ) =

68 T.K.SUBRAHMONIAN MOOTHATHU∑kj=1 diam(f([xj−1, xj ]))(xj − xj−1) ≤

ε

b− a

∑kj=1(xj − xj−1) =

ε

b− a× (b − a) = ε. Therefore,

f ∈ R[a, b] by [157].

(iv) If f ≡ c on [a, b], then L(f, P ) = U(f, P ) = c(b − a) for any partition P of [a, b], implying

the required result. In the general case, let s < c < t be with s ≤ f ≤ t on [a, b], and consider

ε > 0. Let a < u < v < b be with max{u− a, b− v} < ε

2(t− s), and choose a partition P0 of [u, v].

Then for the partition P = {a} ∪ P0 ∪ {b} of [a, b], we have L(f, P ) ≤ c(b − a) ≤ U(f, P ) with

U(f, P ) − L(f, P ) ≤ (t − s)[(u − a) + (b − v)] ≤ ε, which implies f ∈ R[a, b] by [157]. As we also

have L(f, P ) ≤∫ ba fdx ≤ U(f, P ) and ε > 0 is arbitrary, we conclude

∫ ba fdx = c(b− a). �

Definition: Let f : [a, b] → R be a bounded function and P = {x0 ≤ x1 ≤ · · · ≤ xk} be a partition

of [a, b]. Suppose a finite set T = {t1, . . . , tk} ⊂ [a, b] with k elements in it is a tag for P in the sense

that tj ∈ [xj−1, xj ] for 1 ≤ j ≤ k. Then S(f, P, T ) :=∑k

j=1 f(tj)(xj − xj−1) is called a Riemann

sum of f with respect to P . Note that

(i) L(f, P ) ≤ S(f, P, T ) ≤ U(f, P ).

(ii) L(f, P ) = inf{S(f, P, T ) : T is a tag for P}, and U(f, P ) = sup{S(f, P, T ) : T is a tag for P}.

[159] Let f ∈ R[a, b]. Then for every ε > 0, there is a δ > 0 such that the following are true for

every partition P of [a, b] with ∥P∥ < δ:

(i) 0 ≤ U(f, P )− L(f, P ) ≤ ε and L(f, P ) ≤∫ ba fdx ≤ U(f, P ).

(ii) |∫ ba fdx− S(f, P, T )| ≤ ε for any Riemann sum S(f, P, T ) of f with respect to P .

Consequently, for any f ∈ R[a, b] and any sequence (Pn) of partitions of [a, b] with (∥Pn∥) → 0,

the following are true:∫ ba fdx = limn→∞ L(f, Pn) = limn→∞ U(f, Pn) = limn→∞ S(f, Pn, Tn) for

any choice of tags Tn of Pn.

Proof. (i) Let ε > 0. Choose a partition P0 = {x0 ≤ x1 ≤ · · · ≤ xk} of [a, b] using [157] with

0 ≤ U(f, P0) − L(f, P0) ≤ ε/2. Let M > 0 be with |f | ≤ M , and put δ =ε

8kM. Now consider a

partition P of [a, b] with ∥P∥ < δ, and we claim that U(f, P )−L(f, P ) ≤ ε. Let Q = P0∪P , which is

a common refinement of P0 and P . Then U(f,Q)−L(f,Q) ≤ U(f, P0)−L(f, P0) ≤ ε/2. Moreover,

since Q is obtained by adding at most k − 1 points to P (∵ P0 contains at most k − 1 points in

(a, b)), L(f,Q)− 2M(k− 1)∥P∥ ≤ L(f, P ) ≤ U(f, P ) ≤ U(f,Q)+ 2M(k− 1)∥P∥ by [156](i). Also,

∥P∥ < δ =ε

8kM. Hence we deduce that U(f, P )−L(f, P ) ≤ U(f,Q)−L(f,Q)+4M(k− 1)∥P∥ ≤

ε/2 + ε/2 = ε, proving the claim. Finally note that we already know L(f, P ) ≤∫ ba fdx ≤ U(f, P ).

(ii) This is a consequence of (i) because L(f, P ) ≤ S(f, P, T ) ≤ U(f, P ). �

[160] Let f, g ∈ R[a, b]. Then,

(i) [Linearity] f + g ∈ R[a, b] and∫ ba (f + g)dx =

∫ ba fdx+

∫ ba gdx.

Moreover, cf ∈ R[a, b] and∫ ba (cf)dx = c

∫ ba fdx for every c ∈ R.

(ii) fg ∈ R[a, b], where fg(x) := f(x)g(x).

REAL ANALYSIS 69

(iii) If g is non-vanishing in [a, b] and 1/g is bounded, then 1/g, f/g ∈ R[a, b].

(iv) If f ≥ 0, then∫ ba fdx ≥ 0. If f ≤ g, then

∫ ba fdx ≤

∫ ba gdx.

(v) If s, t ∈ R are with s ≤ f ≤ t, then s(b− a) ≤∫ ba fdx ≤ t(b− a).

(vi) |f | ∈ R[a, b] and |∫ ba fdx| ≤

∫ ba |f |dx.

(vii) max{f, g},min{f, g} ∈ R[a, b]. So, f+ := max{f, 0} and f− := −min{f, 0} belong to R[a, b].

Proof. (i) Let ε > 0 be given. By using [157] and by considering a common refinement of two

partitions, we may find a partition P of [a, b] such that 0 ≤ U(f, P ) − L(f, P ) ≤ ε/2 and 0 ≤U(g, P )− L(g, P ) ≤ ε/2. Observe that

L(f, P ) + L(g, P ) ≤ L(f + g, P ) ≤ U(f + g, P ) ≤ U(f, P ) + U(g, P ).

This implies 0 ≤ U(f + g, P )−L(f + g, P ) ≤ ε, and hence f + g ∈ R[a, b] by [157]. Moreover, it

also implies∫ ba (f + g)dx =

∫ ba fdx+

∫ ba gdx.

To prove the assertions about cf , observe the following: if P is a partition of [a, b] and c ≥ 0,

then L(cf, P ) = cL(f, P ), U(cf, P ) = cU(f, P ), L(−f, P ) = −U(f, P ), and U(−f, P ) = −L(f, P ).

(ii) Let M > 0 be such that |f | ≤ M . Then |f2(x) − f2(y)| = |f(x) + f(y)||f(x) − f(y)| ≤2M |f(x)− f(y)| for every x, y ∈ [a, b]. Hence f2 ∈ R[a, b] by [159](i). Combining this observation

with (i), and using the identity fg = [(f + g)2 − (f − g)2]/4, we deduce that fg ∈ R[a, b].

(iii) Since 1/g is bounded, there is δ > 0 such that |g(x)| ≥ δ for every x ∈ [a, b]. Hence we have

| 1

g(x)− 1

g(y)| = |g(y)− g(x)|

|g(x)||g(y)|≤ |g(x)− g(y)|

δ2for every x, y ∈ [a, b]. Consequently, 1/g ∈ R[a, b] by

[159](i). Then f/g = f × (1/g) ∈ R[a, b] by (ii).

(iv) If f ≥ 0, then U(f, P ) ≥ L(f, P ) ≥ 0 for any partition P of [a, b] and hence∫ ba fdx ≥ 0. If

f ≤ g, then g − f ≥ 0, and hence by linearity∫ ba gdx−

∫ ba fdx =

∫ ba (g − f)dx ≥ 0.

(v) This follows from (iv) and [159](iv).

(vi) Since ||f(x)| − |f(y)|| ≤ |f(x) − f(y)|, we get |f | ∈ R[a, b] by [159](i). Since f ≤ |f | and−f ≤ |f |, we get ±(

∫ ba fdx) ≤

∫ ba |f |dx by (iv) and hence |

∫ ba fdx| ≤

∫ ba |f |dx.

(vii) Note max{f, g} =f + g

2+

|f − g|2

and min{f, g} =f + g

2− |f − g|

2. Now use (i) and (v). �

[161] (i) Let f ∈ R[a, b] and [u, v] ⊂ [a, b]. Then f |[u,v] ∈ R[u, v].

(ii) If f ∈ R[a, b], then∫ ba fdx =

∫ ca fdx+

∫ bc fdx for every c ∈ [a, b].

(iii) Let f : [a, b] → R be a function. If there is c ∈ [a, b] such that f |[a,c] ∈ R[a, c] and f |[c,b] ∈ R[c, b],

then f ∈ R[a, b].

(iv) Let f ∈ R[a, b] and (cn) be a sequence in (a, b) with (cn) → b. Then∫ ba fdx = limn→∞

∫ cna fdx.

Proof. (i) Let ε > 0. By [157], there is a partition P of [a, b] with 0 ≤ U(f, P ) − L(f, P ) ≤ ε.

By refining P , we may suppose u, v ∈ P . Letting g = f |[u,v] and Q = P ∩ [u, v], we note that

U(g,Q)− L(G,Q) ≤ U(f, P )− L(f, P ) ≤ ε. Hence g = f[u,v] ∈ R[u, v] by [157].


(ii) The result is trivial for c ∈ {a, b}. So assume a < c < b. Let (Pn) be a sequence of partitions of

[a, b] such that c ∈ Pn for every n ∈ N and (∥Pn∥) → 0. Let Tn be a tag for Pn for each n ∈ N. Then∫ ba fdx = limn→∞ S(f, Pn, Tn) by [159]. Now let P ′

n = Pn∩[a, c], P ′′n = Pn∩[c, b], T ′

n = Tn∩[a, c], andT ′′n = Tn∩ [c, b]. Also, let f1 = f[a,c], f2 = f[c,b]. Clearly S(f, Pn, Tn) = S(f1, P

′n, T

′n)+S(f2, P

′′n , T

′′n ).

Also,∫ ca fdx = limn→∞ S(f1, P

′n, T

′n) and

∫ bc fdx = limn→∞ S(f1, P

′′n , T

′′n ) by [159].

(iii) Let f1 = f |[a,c] and f2 = f |[c,b]. Consider ε > 0. By (i), there are partitions P1 of [a, c] and P2

of [c, b] with 0 ≤ U(fi, Pi) − L(fi, Pi) ≤ ε/2 for i = 1, 2. Then P := P1 ∪ P2 is a partition of [a, b]

and U(f, P )− L(f, P ) =∑2

i=1(U(fi, Pi)− L(fi, Pi)) ≤ ε. Hence f ∈ R[a, b] by [157].

(iv) If |f | ≤M , then |∫ ba fdx−

∫ cna fdx| = |

∫ bcnfdx| ≤

∫ bcn

|f |dx ≤M(b− cn) → 0. �

Definition: (i) If X ⊂ R is an interval, let µ(X) denote its length. In particular, if X is as a

singleton, then µ(X) = 0. (ii) A function f : [a, b] → R is called a step function if f =∑k

j=1 cj1Xj ,

where k ∈ N, cj ∈ R, and X1, . . . , Xk are pairwise disjoint subintervals of [a, b]. In other words, a

step function is a finite linear combination of indicator functions of pairwise disjoint intervals.

[162] (i) Let f : [a, b] → R be a step function given by f =∑k

j=1 cj1Xj , where cj ∈ R andX1, . . . , Xk

are pairwise disjoint subintervals of [a, b]. Then f ∈ R[a, b] and∫ ba fdx =

∑kj=1 cjµ(Xj).

(ii) Let f : [a, b] → R be a bounded function, and P be a partition of [a, b]. Then there are step

functions g, h : [a, b] → R such that g ≤ f ≤ h, L(f, P ) =∫ ba gdx, and U(f, P ) =

∫ ba hdx.

(iii) If f ∈ R[a, b], then∫ ba fdx = sup{

∫ ba gdx : g is a step function with g ≤ f} = inf{

∫ ba hdx :

h is a step function with f ≤ h}.(iv) If f ∈ R[a, b] and g : [a, b] → R is a function such that Y := {y ∈ [a, b] : f(y) = g(y)} is a finite

set, then g ∈ R[a, b] and∫ ba gdx =

∫ ba fdx.

Proof. (i) First suppose f = 1X , where X ⊂ [a, b] is an interval. Then there are three possibilities:

X = [a, b], or [a, b] \X is an interval, or [a, b] \X is a disjoint union of two intervals. In each case,

we get f ∈ R[a, b] with∫ ba fdx = µ(X) by [161](iii), [161](ii), and [158](iv). In the general situation

where f =∑k

j=1 cj1Xj , the linearity of integration given by [160](i) yields the required result.

(ii) Write P = {x0 ≤ x1 ≤ · · · ≤ xk}. If xj−1 = xj for some j, then the interval [xj−1, xj ] reduces to

a singleton and does not contribute anything to L(f, P ) or U(f, P ). So assume x0 < x1 < · · · < xk.

Let Xj = [xj−1, xj) for 1 ≤ j ≤ k − 1 and Xk = [xk−1, xk]. Also, let uj = inf(f([xj−1, xj ])),

vj = sup(f([xj−1, xj ])), g =∑k

j=1 uj1Xj , and h =∑k

j=1 vj1Xj . Check that this works.

(iii) This follows from (ii), [160](iv), and [159].

(iv) Let cy = g(y)−f(y) for y ∈ Y , and h =∑

y∈Y cy1{y}. Then h is a step function and g = f +h.

By (i), we have h ∈ R[a, b] and∫ ba hdx = 0. Hence by the linearity [160](i), g ∈ R[a, b] and∫ b

a gdx =∫ ba fdx+

∫ ba hdx =

∫ ba fdx. �

Exercise-54: (i) If f ∈ R[a, b] is with f ≥ 0 and [u, v] ⊂ [a, b], then∫ ba fdx ≥

∫ vu fdx.

REAL ANALYSIS 71

(ii) If f ∈ C([a, b],R) is such that f ≥ 0 and∫ ba fdx = 0, then f ≡ 0.

(iii) If f ∈ R[a, a+1] is an increasing function, then f(a) ≤∫ a+1a f(x)dx ≤ f(a+ 1). For example,

log n ≤∫ n+1n log xdx ≤ log(n+ 1) for every n ∈ N.

(iv) If f ∈ R[a, a+ 1] is a decreasing function, then f(a+ 1) ≤∫ a+1a f(x)dx ≤ f(a). For example,

1

n+ 1≤

∫ n+1n

1

xdx ≤ 1

nfor every n ∈ N.

[Hint : (i) Let g : [a, b] → R be g = 1[u,v]. Then fg ∈ R[a, b] and f ≥ fg so that∫ ba fdx ≥

∫ ba fgdx =∫ v

u fdx. (ii) Otherwise, by the continuity of f we may find u < v in [a, b] and δ > 0 such that

f(x) ≥ δ for every x ∈ [u, v]. Then∫ ba fdx ≥

∫ vu fdx ≥ δ(v − u) > 0, a contradiction. (iii) Note

that f(a) ≤ f(x) ≤ f(a+ 1) and f(a) =∫ a+1a f(a)dx, etc.]

Definition: A ⊂ R is said to be a null set4 if for every ε > 0, there is a sequence (Xn) of intervals

such that A ⊂∪∞

n=1Xn and∑∞

n=1 µ(Xn) ≤ ε, where µ(Xn) denotes the length of the interval Xn.

Exercise-55: If A ⊂ R is a countable set, then A is a null set.

(ii) If Ak ⊂ R are null sets for k ∈ N, then A :=∪∞

k=1Ak is also a null set.

[Hint : (i) Write A = {a1, a2, . . .}. Given ε > 0, let Xn be an interval containing an with µ(Xn) <

ε/2n. (ii) Given ε > 0, choose intervals Xk,n such that Ak ⊂∪∞

n=1Xk,n and∑∞

n=1 µ(Xk,n) ≤ ε/2k.

Then A ⊂∪∞

k=1

∪∞n=1Xk,n and

∑∞k=1

∑∞n=1 µ(Xk,n) ≤

∑∞k=1 ε/2

k = ε.]

[163] Let f : [a, b] → R be a bounded function, and A = {x ∈ [a, b] : f is not continuous at x}.(i) If A is a finite set, then f ∈ R[a, b].

(ii) [Characterization of Riemann integrable functions] f ∈ R[a, b] ⇔ A is a null set.

Proof. (i) After dividing [a, b] into finitely many closed intervals and considering each such closed

subinterval separately, we may suppose in view of [161](iii) that A is a singleton and that this

singleton is an end point of [a, b]. Without loss of generality, assume A = {a}. Let M > 0 be such

that |f | ≤ M , and choose c ∈ (a, b) with c − a <ε

4M. Now consider ε > 0. Since g := f |[c,b]

is continuous, there is a partition P of [c, b] with 0 ≤ U(g, P ) − L(g, P ) ≤ ε/2 by [157]. Then

Q := {a} ∪ P is a partition of [a, b] and U(f,Q) − L(f,Q) ≤ diam(f([a, c]))(c − a) + U(g, P ) −L(g, P ) ≤ 2M × ε

4M+ ε/2 = ε. Hence f ∈ R[a, b] by [157].

(ii) This proof is left to the student as a reading assignment; see Theorem 7.48 of Apostol, or see

my notes Measure Theory. �

Example: (i) Let f : [0, 1] → R be f(0) = 0 and f(x) = sin(1/x) for x > 0. Then 0 is the only

point of discontinuity of f and hence f ∈ R[0, 1] by [163].

(ii) Let f : [0, 1] → R be f(0) = 0 and f(x) = 1/n for 1/(n+ 1) < x ≤ 1/n and n ∈ N. Then f is

an increasing bounded function and hence f ∈ R[0, 1] by [158](ii) even though f is not continuous

at infinitely many points.

4What we have defined as a ‘null set’ is nothing but ‘a set whose Lebesgue measure is zero’.


Definition: If f ∈ R[a, b], we define∫ ab f(x)dx := −

∫ ba f(x)dx.

Exercise-56: Let f ∈ R[a, b] and u, v, w ∈ [a, b] be points placed in an arbitrary order. Then,

(i)∫ wu f(x)dx =

∫ vu f(x)dx+

∫ wv f(x)dx. (ii) |

∫ vu fdx| ≤ |

∫ vu |f |dx|.

16. Fundamental theorem of calculus and further properties of integration

The Fundamental theorem of calculus stated below says roughly that differentiation and integra-

tion are inverse operations of each other.

[164] [Fundamental theorem of calculus] Let f ∈ R[a, b].

(i) Define F : [a, b] → R as F (t) =∫ ta f(x)dx. Then F is Lipschitz continuous. Moreover, if f is

continuous at a point s ∈ [a, b], then F is differentiable at s and F ′(s) = f(s).

(ii) If there is a differentiable function F : [a, b] → R with F ′ = f , then∫ ba fdx = F (b)− F (a).

Proof. (i) Let M > 0 be with |f | ≤ M . We claim that F is Lipschitz continuous with M as a

Lipschitz constant. Consider points s < t in [a, b]. Then F (t) − F (s) =∫ ta f(x)dx −

∫ sa f(x)dx =∫ t

s f(x)dx by [161](ii) or Exercise-56. Hence |F (t) − F (s)| ≤∫ ts |f |dx ≤

∫ ts Mdx = M(t − s) =

M |t − s|. This proves the claim. Next suppose f is continuous at a point s ∈ [a, b]. Given

ε > 0, choose δ > 0 such that |f(t) − f(s)| < ε for every t ∈ [a, b] with |t − s| < δ. Note that

F (t)− F (s)

t− s=

∫ ts f(x)dx

t− sand f(s) =

∫ ts f(s)dx

t− sfor every t ∈ [a, b] \ {s}. Hence for any t ∈ [a, b]

with 0 < |t− s| < δ, we have |F (t)− F (s)

t− s− f(s)| =

|∫ ts (f(x)− f(s))dx|

|t− s|≤

|∫ ts |f(x)− f(s)|dx|

|t− s|≤

|∫ ts εdx|

|t− s|= ε. Therefore F is differentiable at s with F ′(s) = f(s).

(ii) Let ε > 0. By [159], there is δ > 0 such that |∫ ba f(x)dx− S(f, P, T )| ≤ ε for every partition P

of [a, b] with ∥P∥ < δ and any tag T of P . Select a partition P = {x0 < x1 < · · · < xk} of [a, b]

with ∥P∥ < δ. Then by Mean value theorem, there are tj ∈ (xj−1, xj) with F (xj)− F (xj−1)

= F ′(tj)(xj − xj−1) = f(tj)(xj − xj−1). Letting T = {t1, . . . , tk}, we note that S(f, P, T ) =∑kj=1 f(tj)(xj−xj−1) =

∑kj=1(F (xj)−F (xj−1)) = F (b)−F (a). Hence |

∫ ba f(x)dx−(F (b)−F (a))| =

|∫ ba f(x)dx− S(f, P, T )| ≤ ε. Since ε > 0 is arbitrary, we deduce

∫ ba f(x)dx = F (b)− F (a). �

Remark: In [162](i), the function F may fail to be differentiable at points where f is discontinuous.

Let f : [0, 1] → R be f = 1[0,1/2], and F : [0, 1] → R be F (t) =∫ t0 f(x)dx. Then F (t) = t for

x ∈ [0, 1/2] and F (t) = 1/2 for t ∈ [1/2, 1]; hence F is not differentiable at 1/2.

Definition: Let f : [a, b] → R be a function. We say a function F : [a, b] → R is a primitive

or antiderivative of f if F ′ = f . Note that if F1, F2 : [a, b] → R are antiderivatives of f , then

(F1 − F2)′ = 0, and hence F1 − F2 is a constant.

REAL ANALYSIS 73

Example: (i)ecx

cis an antiderivative of ecx when c = 0, (ii) sinx is an antiderivative of cosx, (iii)

− cosx is an antiderivative of sinx, (iv)xn+1

n+ 1is an antiderivative of xn for n ∈ Z \ {−1} (when

n < 0, we have to assume x = 0), (v) for x > 0, log x is an antiderivative of1

x, (vi) for x < 0,

log(−x) is an antiderivative of1

x, (vii) for x > 0, x log x− x is an antiderivative of log x.

Exercise-57: (i)∫ ba e

cxdx =ecb − eca

cfor c = 0.

(ii) For n ∈ Z \ {−1},∫ ba x

ndx =bn+1 − an+1

n+ 1.

(iii) If 0 < a < b, then∫ ba

1

xdx = log b− log a. In particular,

∫ b1

1

xdx = log b for b > 1.

(iv)∫ n1 log xdx = n log n− n+ 1 for every integer n ≥ 2.

(v) limn→∞∑n

j=1

1

n+ j= log 2.

[Hint : For (i)-(iv), apply [164](ii) by using an antiderivative. (v) Let f : [0, 1] → R be f(x) =1

1 + xand Pn = {0 < 1/n < 2/n < · · · < n/n}. Then

∑nj=1

1

n+ j=

∑nj=1

1/n

1 + j/n=∑n

j=1 f(j/n)(1/n) = S(f, Pn, Tn) →∫ 10 fdx by [159]. Since F ′ = f for F (x) := log(1 + x), we

get by [164](ii) that∫ 10 fdx = F (1)− F (0) = log 2.]

Remark: Let vn = 1+1

2+· · ·+ 1

n−log n for n ∈ N. We know from [127] that (vn) converges. Here is

another argument for the same. Since vn−vn+1 = log(n+1)− log n− 1

n+ 1=

∫ n+1n (

1

x− 1

n+ 1)dx,

we get 0 ≤ vn − vn+1 ≤∫ n+1n (

1

n− 1

n+ 1)dx =

∫ n+1n

1

n(n+ 1)dx =

1

n(n+ 1)≤ 1

n2, which implies

(vn) is a Cauchy sequence and hence convergent.

[165] (i) [Mean value property of the Riemann integral] If f ∈ C([a, b],R), then there is c ∈ (a, b)

with∫ ba fdx = f(c)(b− a).

(ii) If f, g ∈ C([a, b],R), then there is c ∈ (a, b) with g(c)∫ ba fdx = f(c)

∫ ba gdx.

(iii) If f, g ∈ C([a, b],R) and g ≥ 0, then there is c ∈ [a, b] with∫ ba fgdx = f(c)

∫ ba gdx.

Proof. (i) Define F : [a, b] → R as F (t) =∫ ta f(x)dx. Then F is differentiable with F ′ = f by

[164](i), and hence∫ ba fdx = F (b)−F (a) by [164](ii) or by the definition of F . Applying the Mean

value theorem to F , we may find c ∈ (a, b) with∫ ba fdx = F (b)−F (a) = F ′(c)(b−a) = f(c)(b−a).

(ii) Define F,G : [a, b] → R as F (t) =∫ ta f(x)dx and G(t) =

∫ ta g(x)dx. Then F,G are differentiable

with F ′ = f and G′ = g by [164](i). By the Generalized mean value theorem, there is c ∈ (a, b)

with G′(c)(F (b)− F (a)) = F ′(c)(G(b)−G(a)), which means g(c)∫ ba fdx = f(c)

∫ ba gdx.

(iii) If∫ ba gdx = 0, then g ≡ 0 by Exercise-54(ii), and hence

∫ ba fgdx =

∫ ba 0dx = 0. In this case, we

may take any c ∈ [a, b]. Next suppose∫ ba gdx > 0. Since f is continuous and [a, b] is compact, there

are a0, b0 ∈ [a, b] with f(a0) = s := inf(f([a, b])) and f(b0) = t := sup(f([a, b])). Since s ≤ f ≤ t


and g ≥ 0, we have sg ≤ fg ≤ tg, and hence s ≤∫ ba fgdx∫ ba gdx

≤ t. By the Intermediate value theorem,

there is c between a0 and b0 with f(c) =

∫ ba fgdx∫ ba gdx

. �

Exercise-58: (i) If f ∈ C([a, b],R) is such that∫ vu f(x)dx = 0 for every [u, v] ⊂ [a, b], then f ≡ 0.

(ii) If f ∈ C([0, 1],R), then limn→∞∫ 10 f(x

n)dx = f(0).

[Hint : (i) Enough to show f ≡ 0 on (a, b). Fix c ∈ (a, b) and choose a < un < c < vn < b

with vn − un → 0. By [165](i), there are cn ∈ (un, vn) with f(cn)(vn − un) =∫ vnunf(x)dx =

0, and hence f(c) = limn→∞ f(cn) = 0. (ii) Let M > 0 be with |f | ≤ M . Fix ε ∈ (0, 1).

Then |∫ 11−ε f(x

n)dx| ≤∫ 11−ε |f(x

n)|dx ≤ Mε. Also there are cn ∈ (0, 1 − εn) by [165](i) with∫ 1−ε0 f(xn)dx = f(cnn)(1− ε) → f(0)(1− ε) since 0 ≤ cnn ≤ (1− ε)n → 0.]

Exercise-59: (i) If f, g ∈ C([a, b],R), then there is c ∈ (a, b) with g(c)∫ ca f(x)dx = f(c)

∫ bc g(x)dx.

(ii) If f ∈ C([a, b],R), then there is c ∈ (a, b) with∫ ca f(x)dx = f(c)(b− c).

(iii) If f ∈ C([a, b],R), then there is c ∈ (a, b) with∫ bc f(x)dx = f(c)(c− a).

[Hint : (i) Let z =∫ ba g(x)dx. Define F : [a, b] → R as F (t) = (

∫ ta f(x)dx)(

∫ bt g(x)dx) =

(∫ ta f(x)dx)(z −

∫ ta g(x)dx). Then F (a) = 0 = F (b). Also, F is differentiable with F ′(t) =

f(t)∫ bt g(x)dx−g(t)

∫ ta f(x)dx by [164](i) and the product rule. Apply Rolle’s theorem. (ii) Let F :

[a, b] → R be F (t) = (t− b)∫ ta f(x)dx. Then F is differentiable with F ′(t) =

∫ ta f(x)dx+(t− b)f(t)

by [164](i), and F (a) = 0 = F (b). Apply Rolle’s theorem. (iii) Let z =∫ ba f(x)dx, and F : [a, b] → R

be F (t) = (t − a)∫ bt f(x)dx = (t − a)(z −

∫ ta f(x)dx). Then, F ′(t) =

∫ bt f(x)dx − (t − a)f(t) by

[164](i), and F (a) = 0 = F (b). Apply Rolle’s theorem.]

Notation: (i) We will write f |ba to mean f(b) − f(a). (ii) The expression ‘∫fdx = F + c’ (where

c ∈ R is a constant) means F ′ = f .

[166] (i) [Change of variable formula] Let f ∈ C([a, b],R), and g : [u, v] → [a, b] be a C1-map.

If g(u) = a and g(v) = b, then∫ vu f(g(x))g

′(x)dx =∫ ba f(y)dy. Similarly,

if g(u) = b and g(v) = a, then∫ vu f(g(x))g

′(x)dx = −∫ ba f(y)dy.

(ii) [Integration by parts] Let f, g : [a, b] → R be C1-maps. Then∫ ba fg

′dx = fg|ba −∫ ba f

′gdx.

(iii) [Integration by parts - abstract form] If f, g are C1 maps defined on some interval in R, then∫fg′ = fg +

∫f ′g + c, where c ∈ R is any constant.

Proof. (i) Note that both f and (f ◦ g)g′ are continuous and hence Riemann integrable. Define

F : [a, b] → R as F (t) =∫ ta f(y)dy. Then F

′ = f by [164](i). Since f(g(x))g′(x) = F ′(g(x))g′(x) =

(F ◦ g)′(x) by the Chain rule, we deduce by [164](ii) that∫ vu f(g(x))g

′(x)dx = F (g(v))− F (g(u)).

Now if g(u) = a and g(v) = b, then∫ vu f(g(x))g

′(x)dx = F (b) − F (a) =∫ ba f(y)dy. Similarly, if

g(u) = b and g(v) = a, then∫ vu f(g(x))g

′(x)dx = F (a)− F (b) = −∫ ba f(y)dy.

(ii) Since (fg)′ = f ′g + fg′, we get fg′ = (fg)′ − f ′g. Also∫ ba (fg)

′dx = fg|ba by [164](ii). �

REAL ANALYSIS 75

Remark: (i) The change of variable rule in [166](i) may be remembered as the following formal

rule: if y = g(x), then dy = g′(x)dx so that f(y)dy = f(g(x))g′(x)dx. (ii) We may use [166](i) to

transform an integral over [u, v] to an integral over [a, b] or vice versa; and in the latter case, in order

to choose the points u and v without confusion, the function g should be injective (see Example

(iii) below). Therefore, in [166](i), sometimes it is also assumed that g′ is non-vanishing (to ensure

that g is injective and monotone), but the conclusion of [166](i) holds even without this extra

assumption. (iii) We may use [166](iii) to find antiderivatives in some cases: taking f(x) = log x

and g(x) = x, we have∫log xdx = x log x −

∫1dx + c′ = x log x − x + c, where c′, c ∈ R; hence

F (x) := x log x− x satisfies F ′(x) = log x.

Example: (i)∫ π/20 sinx cosxdx =

∫ 10 ydy =

y2

2|10 = 1/2 by the substitution y = sinx.

(ii)∫ 20

2x

1 + x2dx =

∫ 51

1

ydy = log y|51 = log 5 by the substitution y = 1 + x2.

(iii)∫ 1/

√2

0

1√1− y2

dy =∫ π/40 dx = π/4 by putting y = sinx because dy = cosxdx =

√1− y2dx.

(iv) Let s =∫ π/20 (sinx)exdx. Integrating with parts, we have s = (sinx)ex|π/20 −

∫ π/20 (cosx)exdx.

Also,∫ π/20 (cosx)exdx = (cosx)ex|π/20 +

∫ π/20 (sinx)ex = (cosx)ex|π/20 + s. Hence we see that 2s =

(sinx− cosx)ex|π/20 = eπ/2 + 1, and therefore∫ π/20 (sinx)exdx = s = (eπ/2 + 1)/2.

Exercise-60: Let f ∈ C(R,R). Then, (i)∫ ba f(x+ u)dx =

∫ b+ua+u f(x)dx for every u ∈ R.

(ii) If ∃ s > 0 with f(x+ s) = f(x) for every x ∈ R, then∫ a+sa f(x)dx =

∫ s0 f(y)dy for every a ∈ R.

[Hint : (i) The substitution y = x+u yields∫ ba f(x+u)dx =

∫ b+ua+u f(y)dy =

∫ b+ua+u f(x)dx. (ii) Writing

a = ms+ t with m ∈ Z and 0 ≤ t < s, we have∫ a+sa f(x)dx =

∫ s+tt f(x+ms)dx =

∫ s+tt f(x)dx =∫ s

t f(x)dx+∫ s+ts f(x)dx =

∫ st f(x)dx+

∫ s0 f(x+ s)dx =

∫ st f(x)dx+

∫ t0 f(x)dx =

∫ s0 f(x)dx.]

Exercise-61: Let a < b, n ∈ N, and suppose f : [a, b] → R is a Cn+1-map. Then,

(i)∫ ba f

(k)(x)(b− x)k−1

(k − 1)!dx =

f (k)(a)

k!(b− a)k +

∫ ba f

(k+1)(x)(b− x)k

k!dx for 1 ≤ k ≤ n.

(ii) [Integral form of Taylor’s theorem] f(b) = f(a)+∑n

k=1

f (k)(a)

k!(b−a)k+

∫ ba f

(n+1)(x)(b− x)n

n!dx.

[Hint : (i) Fix 1 ≤ k ≤ n. Let u(x) = f (k)(x), v(x) =−(b− x)k

k!so that v′(x) =

(b− x)k−1

(k − 1)!.

Integration by parts says∫ ba uv

′dx = uv|ba −∫ ba u

′vdx. (ii) f(b) − f(a) =∫ ba f

′(x)dx by [164](ii).

Now apply (i) repeatedly.]

Remark: (i) Exercise-61(ii) does not involve any unknown point c ∈ (a, b), and this is useful in

some situations. However, the assumption on f here is slightly stronger than that in [154].

(ii) Let fn ∈ C([0, 1],R) be defined by the following conditions fn(x) = 0 for x ≥ 1

n, fn(0) = 0,

fn(1

2n) = 2n, and the graph of fn is linear on each of the intervals [0,

1

2n] and [

1

2n,1

n]. Then

(fn) → 0 pointwise, but∫ 10 fndx = 1 9 0 =

∫ 10 0dx. Contrast this with [167](i) below.


(iii) Write Q ∩ [0, 1] = {r1, r2, . . .}. Let f, fn : [0, 1] → R be f = 1Q∩[0,1] and fn = 1{r1,...,rn}.

Then fn ∈ R[0, 1] either by [162](i) or [163](i). Also (fn) → f pointwise. But f /∈ R[0, 1] because

U(f, P ) − L(f, P ) = 1 for every partition P = {x0 < x1 < · · · < xk} of [0, 1] (or use [163](ii)).

Contrast this with [167](ii) below.

[167] (i) Let f, fn ∈ C([a, b],R) and suppose (fn) → f uniformly. Then∫ ba fdx = limn→∞

∫ ba fndx.

(ii) Let (fn) be a sequence in R[a, b] converging uniformly to a function f : [a, b] → R. Then

f ∈ R[a, b] and∫ ba fdx = limn→∞

∫ ba fndx.

(iii) Let f, g, fn : [a, b] → R be functions for n ∈ N. Assume fn’s are C1-maps such that (fn) → f

and (f ′n) → g uniformly. Then f is differentiable and f ′ = g.

(iv) Suppose the radius of convergence R of a power series∑∞

n=0 cnxn is > 0. Let f : (−R,R) → R

be f(x) =∑∞

n=0 cnxn. Then f is differentiable and f ′(x) =

∑∞n=1 ncnx

n−1 for every x ∈ (−R,R).Moreover,

∫ x0 f(y)dy =

∑∞n=0

cnn+ 1

xn+1 for every x ∈ (−R,R).

(v) log(1 + x) = x− x2

2+x3

3− x4

4+ · · · =

∑∞n=1

(−1)n+1

nxn for |x| < 1.

Proof. (i) Let d∞ be the supremum metric on C([a, b],R). Then we see |∫ ba fdx −

∫ ba fndx| =

|∫ ba (f − fn)dx| ≤

∫ ba |f − fn|dx ≤

∫ ba d∞(f, fn)dx = d∞(f, fn)(b− a) → 0 as n→ ∞.

(ii) As in Exercise-41(ii), we may see f is bounded. Note that the supremum metric d∞ can

be defined also on {all bounded functions from [a, b] to R}. Consider ε > 0, and choose k ∈ Nwith d∞(f, fk) <

ε

3(b− a). Since fk ∈ R[a, b], there is a partition P of [a, b] with U(fk, P ) −

L(fk, P ) ≤ ε

3by [157]. Since U(f, P ) ≤ U(fk, P ) +

ε

3and L(f, P ) ≥ L(fk, P ) −

ε

3, we deduce

that U(f, P ) − L(f, P ) ≤ ε

3+ε

3+ε

3= ε, and hence f ∈ R[a, b] by [157]. Now as in (i), we see

|∫ ba fdx−

∫ ba fndx| ≤

∫ ba |f − fn|dx ≤

∫ ba d∞(f, fn)dx = d∞(f, fn)(b− a) → 0 as n→ ∞.

(iii) g is continuous by uniform convergence. Define G : [a, b] → R as G(t) =∫ ta g(x)dx. Applying

(i) to the sequence (f ′n) and using [164](ii), we obtain G(t) = limn→∞∫ ta f

′n(x)dx = limn→∞(fn(t)−

fn(a)) = f(t)− f(a). Thus f −G is a constant. Therefore, f ′ = G′ = g by [164](i).

(iv) By [133](vi), the radius of convergence of∑∞

n=1 ncnxn−1 is equal toR. Let g(x) =

∑∞n=1 ncnx

n−1

and fn(x) =∑n

k=0 ckxk for n ∈ N. Fix r ∈ (0, R). Then (fn) → f and (f ′n) → g uniformly on

[−r, r]. Also, fn’s are C1-maps (being polynomials). Hence f ′ = g on [−r, r] by (iii). To prove

the last assertion, define F : (−R,R) → R as F (x) =∑∞

n=0

cnn+ 1

xn+1, where the definition is

justified by [133](vi). By what is already proved, F ′ = f . Hence∫ x0 f(y)dy = F (x)− F (0) = F (x)

by [164](ii) since F (0) = 0.

(v) Let f, F : (−1, 1) → R be f(x) =1

1 + xand F (x) = log(1 + x). For |x| < 1, putting

t = −x, we see f(x) =1

1− t=

∑∞n=0 t

n =∑∞

n=0(−1)nxn, and hence the radius of convergence

REAL ANALYSIS 77

of this power series is ≥ 1. Since F ′ = f and F (0) = 0, we get by [164](ii) and part (iv) that

F (x) = F (x)− F (0) =∫ x0 f(y)dy =

∑∞n=0

(−1)n

n+ 1xn+1 =

∑∞n=1

(−1)n+1

nxn for |x| < 1. �

Example: Let fn : [0, 1] → R be fn(x) =sin(nx)

n. Then (fn) → 0 uniformly. But f ′n(x) = cosnx,

and hence (f ′n) does not converge even pointwise; consider the point π/4 ∈ [0, 1].

We can also define integration for unbounded functions and functions on unbounded intervals in

certain cases. For this, we need the notion of an improper integral defined below.

Definition: (i) Let −∞ ≤ a < b < ∞ and f : (a, b] → R be a function with f ∈ R[t, b] for

every t ∈ (a, b). If l := limt→a

∫ bt f(x)dx exists in [−∞,∞], then the value of the improper integral∫ b

a f(x)dx is defined to be l. In addition if l ∈ R, then we say the improper integral∫ ba fdx converges.

(ii) Let −∞ < a < b ≤ ∞ and f : [a, b) → R be a function with f ∈ R[a, t] for every t ∈ (a, b).

If l := limt→b

∫ ta f(x)dx exists in [−∞,∞], then the value of the improper integral

∫ ba f(x)dx is

defined to be l. In addition if l ∈ R, then we say the improper integral∫ ba fdx converges.

(iii) Let −∞ ≤ a < b ≤ ∞ and f : (a, b) → R be a function. If there exist c ∈ (a, b) and

l1, l2 ∈ [−∞,∞] such that∫ ca fdx = l1 in the sense of (i),

∫ bc fdx = l2 in the sense of (ii), and if

l := l1 + l2 is well-defined in [−∞,∞], then the value of the improper integral∫ ba f(x)dx is defined

to be l. In addition if l ∈ R, then we say the improper integral∫ ba fdx converges. It may be noted

the existence or value of the improper integral∫ ba fdx is independent of the particular choice of

a point c ∈ (a, b) because: if c1 < c2 are points in (a, b), then∫ c2a fdx =

∫ c1a fdx +

∫ c2c1fdx and∫ b

c2fdx =

∫ bc1fdx−

∫ c2c1fdx.

Exercise-62: (i)∫∞0 e−xdx = 1. (ii)

∫ 10 log xdx = −1. (iii)

∫ 10

1

xdx = ∞.

[Hint : (i)∫ t0 e

−xdx = 1 − e−t → 1 as t → ∞. (ii) If t ∈ (0, 1), then we have∫ 1t log xdx =

(x log x−x)|1t = −1− t log t+ t→ −1 as t→ 0+ because limt→0+ t log t = lims→∞− log s

s= 0. (iii)

If t ∈ (0, 1), then∫ 1t

1

xdx = log x|1t = log 1− log t→ ∞ as t→ 0+.]

Exercise-63: (i) If 0 < y < 1, then∫ 10

1

xydx =

1

1− y. In particular,

∫ 10

1√xdx = 2.

(ii) If 0 < y < 1, then∫∞1

1

xydx ≥

∫∞1

1

xdx = ∞.

(iii) If y > 1 and a > 0, then∫∞a

1

xydx =

1

(y − 1)ay−1< ∞. In particular,

∫∞1

1

xydx =

1

y − 1for

y > 1, and∫∞a

1

x2dx =

1

afor a > 0.

[Hint : (i)∫ 1t

1

xydx =

x1−y

1− y|1t =

1− t1−y

1− y→ 1

1− yas t → 0+. (ii)

∫ t1

1

xdx = log t − log 1 → ∞ as

t→ ∞. (iii)∫∞a

1

xydx =

x1−y

1− y|ta =

a1−y − t1−y

y − 1→ 1

(y − 1)ay−1as t→ ∞.]

Exercise-64: (i) Let f : R → (−π/2, π/2) be f(x) = tan−1 x. Then f ′(x) =1

1 + x2.


(ii)∫∞−∞

1

1 + x2dx = π.

[Hint : (i) If g(x) := tanx, then g(f(x)) = x, and so the Chain rule yields g′(f(x))f ′(x) = 1.

Moreover, g′(x) =cos2 x+ sin2 x

cos2 x= 1 + tan2 x, and hence g′(f(x)) = 1 + x2. (ii) By (i),∫ t

0

1

1 + x2dx = f(t)−f(0) = tan−1 t→ π/2 as t→ ∞, and therefore

∫∞0

1

1 + x2dx = π/2. Similarly,∫ 0

−∞1

1 + x2dx = limt→−∞(− tan−1 t) = π/2.]

Remark: (i) Let f : R → R be a function. If the improper integral∫∞−∞ fdx exists, then we may

verify that∫∞−∞ fdx = limt→∞

∫ t−t f(x)dx, but the converse is false; consider f(x) = x for example.

(ii) Let f : (0, 1) → R be an increasing function for which the improper integral∫ 10 fdx exists.

Since∫ k/n(k−1)/n fdx ≤ f(k/n)

n≤

∫ (k+1)/nk/n fdx, we have

∫ 1−1/n0 fdx ≤

∑n−1k=1 f(k/n)

n≤

∫ 11/n fdx.

Letting n → ∞, we obtain∫ 10 fdx = limn→∞

∑n−1k=1 f(k/n)

n. If f(x) = log x, then

∑n−1k=1 f(k/n)

n=

(1/n) log((n− 1)!

nn−1) = (1/n) log(

n!

nn) = log(

(n!)1/n

n). Also

∫ 10 fdx = −1 by Exercise-62(ii). Hence

we obtain, e−1 = limn→∞(n!)1/n

n, a result that we proved earlier in a different manner in [129](iii).

(iii) Let k ∈ N and f : [k,∞) → [0,∞) be a decreasing function for which the improper integral∫∞k fdx exists. Since f(n + 1) =

∫ n+1n f(n + 1)dx ≤

∫ n+1n f(x)dx ≤

∫ n+1n f(n)dx = f(n), we have∑∞

n=k+1 f(n) ≤∫∞k fdx ≤

∑∞n=k f(n). Hence

∫∞1 fdx < ∞ ⇔

∑∞n=k f(n) < ∞. Moreover, fixing

y > 1 and taking f(x) =1

xy, we obtain that

∑∞n=k+1

1

ny≤

∫∞k

1

xydx ≤

∑∞n=k

1

ny, and consequently∑∞

n=k+1

1

ny≤ 1

(y − 1)ky−1≤

∑∞n=k

1

nyby Exercise-63(iii).

Remark: [Riemann-Stieltjes integral] Let f ∈ R[a, b] and g : [a, b] → R be a C1-map. If we define

the Riemann-Stieltjes integral∫ ba fdg in some natural manner, then for any partition P = {x0 ≤

x1 ≤ · · · ≤ xk} of [a, b] with ∥P∥ sufficiently small and any choice of points tj ∈ [xj−1, xj ], we

should have∫ ba fdg ∼

∑kj=1 f(tj)(g(xj)− g(xj−1)). Now observe that fg′ ∈ R[a, b] by [158](iii) and

[160](ii), and that if we choose tj ∈ [xj−1, xj ] with g(xj) − g(xj−1) = g′(tj)(xj − xj−1) by Mean

value theorem, then∑k

j=1 f(tj)(g(xj) − g(xj−1) =∑k

j=1 f(tj)g′(tj)(xj − xj−1) ∼

∫ ba fg

′dx. So we

define the Riemann-Stieltjes integral∫ ba fdg as

∫ ba fdg :=

∫ ba f(x)g

′(x)dx whenever f ∈ R[a, b] and

g : [a, b] → R is a C1-map. For example, if f(x) = x and g(x) = x2, then∫ 10 fdg =

∫ 10 2x2dx = 2/3.

We remark that∫ ba fdg can also be defined with weaker assumptions on g, using partitions P of

[a, b]; see Chapter 6 of Rudin and Chapter 7 of Apostol.

Seminar topics/for further reading: (i) Improper integrals, (ii) Riemann-Stieltjes integral.

17. Weierstrass approximation theorem

Observe that the theory of power series (radius of convergence, etc.) we discussed earlier for power

series of the form∑∞

n=0 cnxn (centered at 0) also applies to power series of the form

∑∞n=0 cn(x−a)n

REAL ANALYSIS 79

(centered at a general point a ∈ R). Thus, if R > 0 is the radius of convergence, then the power

series∑∞

n=0 cn(x−a)n fails to converge when |x−a| > R, and∑∞

n=0 cn(x−a)n converges absolutely

and uniformly for x ∈ [a− r, a+ r] for each r ∈ (0, R).

Definition: Let U ⊂ R be open. A function f : U → R is said to be real analytic if for each

a ∈ U , there is r > 0 such that (a − r, a + r) ⊂ U and f is represented by a convergent power

series of the form∑∞

n=0 cn(x − a)n in (a − r, a + r). If this holds, then [167](iv) implies that f

is infinitely often differentiable and that the nth derivatives of f in (a − r, a + r) is obtained by

term-by-term differentiation of the power series n-times; consequently, cn =fn(a)

n!for every n ≥ 0.

In other words, the nth partial sum of the power series must be the nth Taylor polynomial of f

at a. Moreover, if∑∞

n=0 cn(x − a)n is the power series representing f in (a − r, a + r) ⊂ U , then

by putting pn(x) =∑n

k=0 ck(x − a)k, we see that the sequence (pn) of polynomials converge to f

uniformly in [a− t, a+ t] for each t ∈ (0, r).

Remark: In Complex Analysis, the following are true: f is differentiable ⇒ f is a C∞-map (i.e.,

infinitely often differentiable) ⇒ f can be represented locally by a power series. These implications

are false in Real Analysis; to see that the second implication is false, see Exercise-65 below.

Exercise-65: Let f : R → R be f(x) = 0 for x ≤ 0 and f(x) = e−1/x2for x > 0. Then,

(i) For each n ∈ N, f (n)(x) = 0 for x ≤ 0. Moreover, there is a sequence (pn) of polynomials such

that deg(pn) = 3n and f (n)(x) = pn(1/x)e−1/x2

for every x > 0 and n ∈ N.(ii) f is a C∞-map, but f does not have a power series representation in any neighborhood of 0,

and therefore f is not real analytic.

[Hint : (i) Check the result for n = 1; for example f ′(x) =2e−1/x2

x3for x > 0. Now assume the result

for n. Then f (n+1)(0) = limx→0+f (n)(x)− f (n)(0)

x− 0= limx→0+

pn(1/x)e−1/x2

x= limy→∞

ypn(y)

ey2=

0 by Exercise-31(iv). Moreover, for x > 0, f (n+1)(x) =[p′n(1/x)(−1/x2) + pn(1/x)(2/x

3)]e−1/x2

=

pn+1(1/x)e−1/x2

, where pn+1(x) := 2x3pn(x)− x2p′n(x) so that deg(pn+1) = deg(pn) + 3 = 3n+ 3.

(ii) f is a C∞-map by (i). If f has a power series representation∑∞

n=0 cnxn in a neighborhood U

of 0, then cn =f (n)(0)

n!= 0 for every n ≥ 0 by (i), implying f ≡ 0 on U , a contradiction.]

From Exercise-65, we learn that it may not be possible to approximate a C∞-map using its Taylor

polynomials. We will now discuss two methods to approximate f ∈ C([a, b],R) by polynomials.

Idea of the first method: If g : {−1, 0, 1} → R is a function, its average isg(−1) + g(0) + g(1)

3,

which can be thought of as a weighted average∑1

k=−1w(k)g(k) with equal weights w(−1) =

w(0) = w(1) = 1/3. If we use the weight sequence (1/100, 98/100, 1/100), which is ‘concentrated’

at the middle place, then∑1

k=−1w(k)g(k) ∼ g(0). This is also true for infinite sums. Let g : Z → Rbe non-zero only at finitely many places, and w : Z → R be a weight function, i.e, w(n) ≥ 0 for every

n ∈ Z and∑

n∈Zw(n) :=∑∞

n=1w(−n) +∑∞

n=0w(n) = 1. Assume w is ‘concentrated’ at 0, say


w(0) = 98/100, w(±1) = 1/100, and w(n) = 0 for all other n ∈ Z. Then∑

n∈Zw(n)g(n) ∼ g(0).

Moreover,∑

n∈Zw(k − n)g(n) ∼ g(k) for each k ∈ Z because the weight function n 7→ w(k − n) is

‘concentrated’ at k. These observation can be extended to integrals since an infinite sum is nothing

but a discrete version of the integral. Let g : R → R be a continuous function which is zero outside

a compact interval [a, b], and let w : R → R be a weight function, i.e., w ≥ 0 and∫∞−∞w(x)dx = 1

(or w(x) ≥ 0 for every x ∈ [a, b] and∫ ba w(x)dx = 1). Assume w is ‘concentrated’ at 0. Then∫∞

−∞w(x)g(x)dx ∼ g(0), and∫∞−∞w(y − x)g(x)dx ∼ g(y) for each y ∈ R.

Definition: Let f, g : R → R be functions such that the integral∫∞−∞ f(y − x)g(x)dx is defined for

each y ∈ R. Then5 the function f ∗ g : R → R defined as f ∗ g(y) =∫∞−∞ f(y − x)g(x)dx is called

the convolution of f and g.

Exercise-66: (i) Let w : R → R be a polynomial, and g ∈ C(R,R) be such that g vanishes outside

a compact interval [a, b]. Then w ∗ g : R → R defined as w ∗ g(y) =∫∞−∞w(y − x)g(x)dx =∫ b

a w(y − x)g(x)dx for y ∈ R is also a polynomial.

(ii) For n ∈ N, let wn : R → R be wn(x) = cn(1−x2)n, where cn > 0 is chosen with∫ 1−1w(x)dx = 1.

Then cn ≤√n. Consequently, for each δ ∈ (0, 1), (wn) → 0 uniformly on [−1,−δ] ∪ [δ, 1] (thus wn

gets more and more ‘concentrated’ at 0 as n→ ∞).

[Hint : (i) If k = deg(w), then we may find polynomials p0, p1, . . . , pk such that w(y− x) = p0(x) +

p1(x)y + · · · + pk(x)yk for every y ∈ R. Then

∫ ba w(y − x)g(x)dx =

∑kj=0(

∫ ba pj(x)g(x)dx)y

j =∑kj=0 cjy

j , where cj :=∫ ba pj(x)g(x)dx. (ii) (1 − x2)n ≥ 1 − nx2 by Bernoulli inequality. Hence

1

cn= 2

∫ 10 (1−x

2)ndx ≥ 2∫ 1/

√n

0 (1−x2)ndx ≥ 2∫ 1/

√n

0 (1−nx2)dx =4

3√n≥ 1√

n. Now if δ ≤ |x| ≤ 1,

then |wn(x)| ≤√n(1− δ2)n → 0 as n→ ∞.]

Idea of the second method: Let f ∈ C([0, 1],R). Think of a game, for example, shooting an arrow

at a target. Fix x ∈ [0, 1] and assume that the probability of a hit is x and the probability of

a miss is 1 − x. The probability of k hits from n trials is nCkxk(1 − x)n−k. Suppose f( kn) is

the reward point for k hits out of n trials. Then the expected reward from n trials is En(x) =∑nk=0 f(

kn)

nCkxk(1−x)n−k. Now if n is very large, then we may expect approximately nx hits and

n(1−x) misses from n trials. Therefore, the expected reward from n trials should be approximately

equal to f(nxn ) = f(x). That is, we expect the convergence (En) → f as n → ∞. Here, note that

En’s are polynomials.

Exercise-67: [Facts from Probability Theory - left to the student as a reading assignment] (i) For the

binomial distribution b(n, k, x) = nCkxk(1− x)n−k, the mean is nx and the variance is nx(1− x).

(ii) [Chebychev’s inequality] If a random variable X has mean µ and variance σ2, then for any

t > 0, the probability Pr(|X − µ| ≥ t) ≤ σ2

t2.

5The student is encouraged to learn more about the convolution of two functions by self-study.

REAL ANALYSIS 81

[168] [Weierstrass approximation theorem] Let f ∈ C([a, b],R). Then there is a sequence (pn) of

polynomials converging to f uniformly on [a, b]; equivalently, there is a sequence (pn) of polynomials

with limn→∞ d∞(f, pn) = 0, where d∞ is the supremum metric on C([a, b],R).

Proof. We may transfer the problem to [0, 1] by composing f with the homeomorphism x 7→a + (b − a)x from [0, 1] to [a, b] whose inverse homeomorphism is x 7→ (x − a)/(b − a), where

note that both the homeomorphism and its inverse are polynomials of degree 1. In other words,

we may suppose [a, b] = [0, 1] and f ∈ C([0, 1],R).

First proof : Let c0 = f(0), c1 = f(1) − f(0), and g : [0, 1] → R be g(x) = f(x) − c0 − c1x.

Then g ∈ C([0, 1],R) and g(0) = 0 = g(1). It suffices to produce a sequence (pn) of polynomials

converging to g uniformly on [0, 1], for then by taking qn(x) = pn(x) + c0 + c1x, we may see

that (qn) → f uniformly on [0, 1]. Since g(0) = 0 = g(1), we may extend g continuously to the

whole of R by putting g(x) = 0 for x ∈ R \ [0, 1]. For n ∈ N, let wn : R → R be wn(x) =

cn(1 − x2)n, where cn is chosen with∫ 1−1wndx = 1. By Exercise-66(i), pn : R → R defined as

pn(y) =∫∞−∞wn(y − x)g(x)dx =

∫ 10 wn(y − x)g(x)dx is a polynomial for each n ∈ N.

Consider ε > 0. Since g is uniformly continuous on R, there is δ ∈ (0, 1) such that |g(x)−g(y)| <ε/2 for every x, y ∈ R with |x − y| ≤ δ. Let M > 0 be such that |g| ≤ M . By Exercise-66(ii),

(wn) → 0 uniformly on [−1,−δ]∪[δ, 1], and therefore there is n0 ∈ N such that |wn(x)| <ε

8M(1− δ)for every x ∈ [−1,−δ] ∪ [δ, 1] and every n ≥ n0. Fix n ≥ n0 and y ∈ [0, 1], and make the

following two observations: (i) g(y) =∫ 1−1wn(x)g(y)dx since

∫ 1−1wn(x)dx = 1, and (ii) by putting

x = t + y, we have pn(y) =∫ 10 wn(y − x)g(x)dx =

∫ 1−y−y wn(−t)g(t + y)dt =

∫ 1−1wn(t)g(t + y)dt =∫ 1

−1wn(x)g(x + y)dx since g ≡ 0 on R \ [0, 1] and wn(−t) = wn(t). Hence for every n ≥ n0 and

y ∈ [0, 1], we see that |pn(y)− g(y)| ≤∫ 1−1wn(x)|g(x+ y)− g(y)|dx

≤ 2M∫ −δ−1 wn(x)dx+

ε

2

∫ δ−δ wn(x)dx+ 2M

∫ 1δ wn(x)dx

≤ 2Mε

8M(1− δ)(1− δ) +

ε

2

∫ 1−1wn(x)dx+ 2M

ε

8M(1− δ)(1− δ) =

ε

4+ε

2+ε

4= ε.

Second proof : Let ε > 0. Choose δ > 0 such that |f(x) − f(y)| < ε

2for every x, y ∈ [0, 1] with

|x − y| < δ. Let M > 0 be with |f | ≤ M , and choose n0 ∈ N with2M

n0δ2≤ ε

2. Let En be the

polynomial En(x) =∑n

k=0 f(kn)

nCkxk(1 − x)n−k for n ∈ N. Fixing n ≥ n0 and x ∈ [0, 1], let

Γ1 = {0 ≤ k ≤ n : |x − k/n| < δ} and Γ2 = {0, 1, . . . , n} \ Γ1. As∑n

k=0nCkx

k(1 − x)n−k =

(x+1−x)n = 1, we have |f(x)−En(x)| = |∑n

k=0(f(x)−f(kn))

nCkxk(1−x)n−k| ≤ S1+S2, where

Sj :=∑

k∈Γj|f(x)− f( kn)|

nCkxk(1− x)n−k for j = 1, 2. Now, S1 ≤

∑nk=0

ε

2nCkx

k(1− x)n−k =ε

2by the definition of Γ1 and the choice of δ. Moreover, we see by Exercise-67 that

S2 ≤ 2M ×Pr(|kn− x| ≥ δ) = 2M ×Pr(|k−nx| ≥ nδ) ≤ 2M

nx(1− x)

n2δ2≤ 2M

nδ2≤ 2M

n0δ2≤ ε

2. Thus

|f(x)− En(x)| ≤ε

2+ε

2= ε for every n ≥ n0 and every x ∈ [0, 1]. �

Remark: For a third proof of [168], using Fourier series, see Theorem 11.17 of Apostol.


Exercise-68: (i) If f : [a, b] → [u, v] is Riemann integrable and g ∈ C([u, v],R), then g ◦ f ∈ R[a, b].

(ii) If f ∈ R[a, b] and f ≥ 0, then f1/n ∈ R[a, b] for every n ∈ N.(iii) Let f ∈ C([0, 1],R) and assume

∫ 10 f(x)x

ndx = 0 for every n ∈ N ∪ {0}. Then f ≡ 0.

[Hint : (i) Since fk ∈ R[a, b] for every k ∈ N ∪ {0}, it follows that p ◦ f ∈ R[a, b] for every

polynomial p. By [168], choose a sequence (pn) of polynomials converging uniformly to g. Then

(pn ◦ f) → g ◦ f uniformly. So by [167](ii), g ◦ f ∈ R[a, b]. (ii) Apply (i) with g(x) := x1/n. (iii)

We see∫ 10 f(x)p(x)dx = 0 for every polynomial p. Then by [168] and [167](ii), we deduce that∫ 1

0 f2dx = 0. Since f2 ≥ 0, it follows that f2 ≡ 0, and hence f ≡ 0.]

18. Convex functions

Recall from [121] that if X ⊂ R is connected, then [x, y] ⊂ X for any two points x < y in X.

Definition: Let X ⊂ R be a nonempty connected set and f : X → R be a function.

(i) We say f is convex if f(tx+ (1− t)y) ≤ tf(x) + (1− t)y for every x, y ∈ X and every t ∈ [0, 1]

(equivalently, for every x = y in X and every t ∈ (0, 1)). When X has at least two points, we say

f is strictly convex if f(tx+ (1− t)y) < tf(x) + (1− t)y for every x = y in X and every t ∈ (0, 1).

Note that the convexity condition of f means geometrically that the graph of f between x and y

lies below the line segment joining (x, f(x)) and (y, f(y)).

(ii) We say f is concave if f(tx+ (1− t)y) ≥ tf(x) + (1− t)y for every x, y ∈ X and every t ∈ [0, 1]

(equivalently, for every x = y in X and every t ∈ (0, 1)). When X has at least two points, we say

f is strictly concave if f(tx+ (1− t)y) > tf(x) + (1− t)y for every x = y in X and every t ∈ (0, 1).

Note that f is (strictly) concave iff −f is (strictly) convex.

Example: (i) Fix any integer n ≥ 2. Let f, g : [0, 1] → R be f(x) = xn and g(x) = x1/n. Then f is

strictly convex and g is strictly concave. (ii) Let f : [0, 1] → R be f(0) = 1 and f(x) = x for x > 0.

Then f is convex, but is not continuous at 0. Contrast this with [169] below.

[169] Let X ⊂ R be a nonempty connected set and f : X → R be a convex function. Then,

(i) For points x < y < z in X, we havef(y)− f(x)

y − x≤ f(z)− f(x)

z − x≤ f(z)− f(y)

z − y(where the

inequalities are strict if f is strictly convex).

(ii) f is continuous at every c ∈ int(X). In particular, f is continuous if X is open in R.Analogous results hold for concave functions - with inequalities reversed in (i).

Proof. (i) Choose t ∈ (0, 1) such that y = tx + (1 − t)z. Then y − x = (1 − t)(z − x) and

z− y = t(z−x). Also by convexity, f(y)− f(x) ≤ tf(x)+ (1− t)f(z)− f(x) = (1− t)(f(z)− f(x))

and f(z)− f(y) ≥ f(z)− tf(x)− (1− t)f(z) = t(f(z)− f(x)).

(ii) Consider c ∈ int(X), and we wish to show f is continuous at c. Choose points a, b ∈ X with

a < c < b. Let C1 =f(c)− f(a)

c− aand C2 =

f(b)− f(c)

b− c. For x, y ∈ X with a < x < c < y < b,

REAL ANALYSIS 83

we get by repeated application of (i) that C1 ≤ f(c)− f(x)

c− x≤ C2, and C1 ≤ f(y)− f(c)

y − c≤ C2, or

equivalently, C1(c− x) ≤ f(c)− f(x) ≤ C2(c− x) and C1(y − c) ≤ f(y)− f(c) ≤ C2(y − c). This

implies that f is continuous at c. In fact, we get that f is ‘locally Lipschitz’ at the point c. �

[170] Let X ⊂ R be a nonempty connected set and f : X → R be a differentiable function. Then

f is (strictly) convex ⇔ f ′ is (strictly) increasing. Consequently, when f is twice differentiable, we

have the following implications: f is convex ⇔ f ′′ ≥ 0; and f is strictly convex whenever f ′′ > 0.

Proof. ⇒: Assume f is convex, and consider a < b in X. Choose x ∈ (a, b). By [169](i), we havef(x)− f(a)

x− a≤ f(b)− f(a)

b− a≤ f(b)− f(x)

b− x. Letting x → a and x → b in the first and second

inequalities respectively, we get f ′(a) ≤ f(b)− f(a)

b− a≤ f ′(b). Thus f ′(a) ≤ f ′(b).

Next suppose f is strictly convex, and consider a < b in X. Fix c ∈ (a, b). Choosing x ∈ (a, c)

and y ∈ (c, b), we getf(x)− f(a)

x− a<f(c)− f(a)

c− a<f(b)− f(c)

b− c<f(b)− f(y)

b− yby repeated use of

[169](i). Letting x→ a and y → b, we obtain f ′(a) ≤ f(c)− f(a)

c− a<f(b)− f(c)

b− c≤ f ′(b).

(ii) ⇒ (i): Consider x < y in X, t ∈ (0, 1), and let a = tx+(1−t)y. Then x < a < y. By Mean value

theorem, there are c1 ∈ (x, a) and c2 ∈ (a, y) such thatf(a)− f(x)

a− x= f ′(c1) and

f(y)− f(a)

y − a=

f ′(c2). Since c1 < c2 and f ′ is (strictly) increasing, we deduce thatf(a)− f(x)

a− x(<) ≤ f(y)− f(a)

y − a.

Note that a − x = (1 − t)(y − x) and y − a = t(y − x). Hencef(a)− f(x)

1− t(<) ≤ f(y)− f(a)

t, or

equivalently t(f(a)− f(x))(<) ≤ (1− t)(f(y)− f(a)), which implies f(a)(<) ≤ tf(x)+ (1− t)f(y).

For the last assertion, use [149](iii) and [150](i). �

Remark: (i) If f : [−1, 1] → R is f(x) = x4, then f is twice differentiable and strictly convex, but

f ′′ vanishes at 0. (ii) Recall that f is (strictly) concave iff −f is (strictly) convex. Therefore, we

get the following from [170]:

[171] Let X ⊂ R be a nonempty connected set and f : X → R be twice differentiable. Then f

is (strictly) concave ⇔ f ′ is (strictly) decreasing. Consequently, when f is twice differentiable, we

have the following implications: f is concave ⇔ f ′′ ≤ 0; and f is strictly concave whenever f ′′ < 0.

Example: (i) By [170], ex is strictly convex on R. (ii) If f(x) = log x, then f ′(x) =1

xis strictly

decreasing on (0,∞), and hence by [171], log x is strictly concave. (iii)d(x log x)

dx= 1 + log x and

d(x log x− x)

dx= log x are strictly increasing, and hence by [170], x log x and x log x−x are strictly

convex on (0,∞). (iv) Fix y ∈ (0,∞), and let f : (0,∞) → R be f(x) = xy. Then by [170] and

[171], f is strictly convex for y > 1, and strictly concave for 0 < y < 1.


Definition: A finite linear combination∑n

k=1 tkxk of points x1, . . . , xk in R (or in Rm) is said to be

a convex combination if tk ≥ 0 for 1 ≤ k ≤ n and∑n

k=1 tk = 1.

Exercise-69: (i) Let X ⊂ R be a nonempty connected set, x1, . . . , xn ∈ X, t1, . . . , tn ∈ [0, 1], and

assume∑n

k=1 tk = 1. If f : X → R is a convex function, then f(∑n

k=1 tkxk) ≤∑n

k=1 tkf(xk).

Similarly, if f : X → R is a concave function, then f(∑n

k=1 tkxk) ≥∑n

k=1 tkf(xk).

(ii) [Geometric mean≤Arithmetic mean] If x1, . . . , xn ∈ (0,∞), then (∏n

k=1 xk)1/n ≤ x1 + · · ·+ xn

n.

(iii) Let p, q > 0 be such that1

p+

1

q= 1. Then xy ≤ xp

p+yq

qfor every x, y ∈ (0,∞).

[Hint : (i) Let f be convex. Assume the result for n− 1, and suppose tn = 1. Put sk = tk/(1− tn)

for 1 ≤ k ≤ n− 1, and y =∑n−1

k=1 skxk. Then f(y) ≤∑n−1

k=1 skf(xk) by induction assumption. Also

f((1− tn)y + tnxn) ≤ (1− tn)f(y) + tnf(xn) by the convexity of f . Combine the two inequalities.

(ii) Since log x is a concave function, we get by (i) that log(x1 + · · ·+ xn

n) ≥ 1

n

∑nk=1 log xk =

1

nlog(

∏nk=1 xk) = log((

∏nk=1 xk)

1/n). Now apply the exponential function to both extremes. (iii)

Since log x is concave, we have log(xp

p+yq

q) ≥ 1

plog(xp) +

1

qlog(yq) = log x+ log y = log(xy).]

Exercise-70: (i) If f ∈ C([a, b],R) is strictly convex, then f attains its minimum at a unique point.

(ii) Let X ⊂ R be a nonempty connected set and f : X → R be a C2-map. If f is convex, then

f(y)− f(x) ≥ f ′(x)(y − x) for every x, y ∈ X.

(iii) Let f, g : R → R be convex, and suppose g is increasing. Then g ◦ f is convex. In particular,

if f : R → R is convex, then ef is convex.

(iv) If f : R → R is convex and bounded above, then f is constant (contrast this with the fact that

the convex function ex is bounded below on R and bounded above on (−∞, 0)).

[Hint : (i) If f attains its minimum at two points x < y, then by convexity, f is constant on [x, y],

a contradiction to strict convexity. (ii) Assume x < y. By Taylor’s theorem, there is c ∈ (x, y)

with f(y) = f(x) + f ′(x)(y − x) +f ′′(c)

2(y − x)2. Also, f ′′ ≥ 0 by [170]. (iii) Let x, y ∈ R

and z = tx + (1 − t)y, where t ∈ [0, 1]. Since f is convex and g is increasing, we get g(f(z)) ≤g(tf(x)+(1−t)f(y)). Also, the right hand side is≤ tg(f(x))+(1−t)g(f(y)) as g is convex. (iv) If f is

not constant, then there are x, y ∈ R with f(y)−f(x) > 0. Choose zn ∈ R with y =(n− 1)x+ zn

n.

By convexity, f(y) ≤ (n− 1)f(x) + f(zn)

n. Hence f(y) + (n − 1)(f(y) − f(x)) ≤ f(zn) for each

n ∈ N. Therefore, f cannot be bounded above.]

Seminar topics: (i) [Holder’s inequality] Let p, q > 1 be with1

p+

1

q= 1. Then

∑nk=1 |xkyk| ≤

(∑n

k=1 |xk|p)1/p(∑n

k=1 |yk|q)1/q for every x1, . . . , xn, y1, . . . , yn ∈ R. (ii) Convex sets in Rm.

*****

real analysis contents - moothathu.files.wordpress.com · real analysis t.k.subrahmonian moothathu...

Documents