Notes on Analysis

Introduction

to

Applied AnalysisDefinitions, Theorems and Problems

SUBRATA PAUL

FALL - 2014

1

Contents

1. Real Number System 12. Basic Topology 33. Problems on Basic Topology 53.1. Rudin’s Solution 54. Metric Space 75. Problems on Metric Space 116. Numerical Sequences and Series 466.1. Sequences 466.2. Series 477. Rudin’s Solution: 497.1. Problems from other sources 558. Compactness 578.1. Problems 589. Continuity 739.1. Completing a Metric Space with an Isometry 749.2. Continuous Function on Compact Sets 749.3. Proofs of the Theorems 769.4. Problems: 7710. Differentiation of Single Variable 9310.1. Some Problems 9411. Sequences and Spaces of Continuous Functions 9511.1. Space of Continuous Functions 9511.2. Compact Subset of Continuous Function Space 9511.3. Proof of the Theorems: 9711.4. Problems on Sequence of Functions 9812. Differential Calculus and Banach Space 12713. Problems on Differentiation 13014. List of Problems 14714.1. Problems in Norm and Metric 14714.2. Cauchy and Compactness 14814.3. Compactness 15014.4. Continuity 15114.5. Sequence of Functions 15314.6. Differentiation 156

1. Real Number System

Definition 1.1. Let S be a set. An order on S is a relation, denoted by <, with the

following two properties

(i) If x ∈ S and y ∈ S then one and only one of the statements

x < y, x = y, y < x

is true.

(ii) If x, y, z ∈ S, if x < y and y < z then x < z

An order set is a set S in which an order is defined.

Definition 1.2. Suppose S is an order set, and E ⊂ S. If there exists an β ∈ S such

that x ≤ β for every x ∈ E, we say that E is bounded above, and call β an upper bound

of E. Lower bounds are defined in the same way.

Definition 1.3. Suppose S is an ordered set, E ⊂ S, and E is bounded above. Suppose

that there exists an α ∈ S with the following properties:

(i) α is an upper bound of E.

(ii) If γ < α then γ is not an upper bound of E. Then α is called the least upper

bound of E or the supremum of E, and we write,

α = supE

The greatest lower bound, or infimum, of aset E which is bounded below is defined

in the same manner: The statement

α = inf E

means that α is a lower bound of E and that no β with β > α is a lower bound

of E.

Definition 1.4. An ordered set S is said to have the least-upper-bound property if the

following is true:

If E ⊂ S, E is not empty, and E is bounded above, then supE exists in S.

Theorem 1.1. Suppose S is an ordered set with the least-upper-bound property, B ⊂ S,

B is nonempty, and B is bounded below. Let L be the set of all lower bounds of B. Then,

α = supL

exists in S, and α = inf B. In particular inf B exists in S.

Theorem 1.2. There exists an ordered field R which has the least-upper-bound property.

Moreover, R contains Q as a subfield.1

Theorem 1.3. (a) Archimedean property: If x ∈ R, y ∈ R and x > 0, then there is

a positive integer n such that

nx > y

(b) If x ∈ R, y ∈ R, and x < y, then there exists a p ∈ Q such that x < p < y

Theorem 1.4. For every real x > 0 and every integer n > 0 there is one and only one

real y such that yn = x.

2

2. Basic Topology

Definition 2.1. If there exists a 1-1 mapping of A onto B, we say that A and B can be

put in 1-1 correspondence, or that A and B have the same cardinal number, or briefly,

that A and B are equivalent, and we write A ∼ B. This relation clearly has the following

properties

• It is reflexive: A ∼ A

• It is symmetric: If A ∼ B, then B ∼ A

• It is transitive: If A ∼ B and B ∼ C, then A ∼ C.

Any relation with these properties is called an equivalent relation.

Definition 2.2. For any positive integer n let Jn be the set whose elements are integers

1, 2, . . . . . . , n; let J be the set consisting of all positive integers. For any set A, we say:

(a) A is finite if A ∼ Jn for some n (the empty set is also considered to be finite).

(b) A is infinite if A is not finite.

(c) A is uncountable if A is neither finite nor countable.

(d) A is at most countable if A is finite or countable.

Countable sets are sometimes called enumerable, or denumerable.

Definition 2.3. By a sequence, we mean a function f defined on the set J of all positive

integers. If f(n) = xn, for n ∈ J , it is customary to denote the sequence f by the symbol

{xn} or sometimes by x1, x2, . . . . . . The values of f , that is , the elements xn are called

the terms of the sequence. If A is a set and if xn ∈ A for all n ∈ J , then {xn} is said to

be a sequence in A, or a sequence of elements of A.

Theorem 2.1. Every infinte subset of a countable set A is countable.

Theorem 2.2. Let En, n = 1, 2, 3, . . . . . . be a sequence of countable sets, and put

S =∞⋃n=1

En

Then S is countable.

Corollary 1. Suppose A is at most countable, and, for evey α ∈ A,Bα is at most

countable. Put

T =⋃α∈A

Bα

Then T is at most countable.

Theorem 2.3. Let A be a countable set, and let Bn be the set of all n−tuples (a1, . . . . . . , an)

where ak ∈ A(k = 1, . . . . . . , n), and the elements a1, . . . . . . , an need not be distinct. Then

Bn is countable.

Corollary 2. The set of all rational numbers is countable.3

Theorem 2.4. Let A be the set of all sequences whose elements are the digits 0 and 1.

This set A is uncountable.

Proof 1. Let E be a countable subset of A, and let E consist of the sequence s1, s2, . . . . . . .

We construct a sequence s as follows. If the nth digit sn is 1, we let the nth digit of s be

0, and vice versa. Then the sequence s differs from every member of E in at least one

place; hence s /∈ E. But clearly s ∈ A, so that E is a proper subset of A.

We have shown that every countable subset of A is a proper subset of A. It follows

that A is uncountable (for otherwise A would be a proper subset of A, which is absurd.)

Theorem 2.5. If p is a limit point of a set E, then every neighborhood of p contains

infinitely many points of E.

Corollary 3. A finite point set has no limit point.

Theorem 2.6. A set E is open if and only if its complement is closed.

4

3. Problems on Basic Topology

3.1. Rudin’s Solution

Problem 3.1. Prove that the set of all algebraic numbers is countable.

Solution.

Definition 3.1. A complex number z is said to be algebraic if there are integers a0, a1, . . . . . . , an,

not all zero, such that

a0zn + a1z

n−1 + . . . · · ·+ an−1z + an = 0.

Let AN be the set of all complex numbers satisfying one of the equation just listed

with

n+ |a0|+ |a1|+ . . . · · ·+ |an| = N

which means the absolute sum of the coefficient is less than N . The set AN is finite

because each equation has only a finite set of solutions and there are only finitely many

equations satisfying this condition since for each coefficient can take only a finite different

values (< N). Since all the set AN are finite, they are countable and therefore⋃∞N=1 AN is

either finite or countable. Since all the rational numbers are algebraic the set of algebraic

numbers are countable.

Problem 3.2. Is the set of all irrational real numbers countable?

Solution. No. If it were, the set of all real numbers, being the union of the rational and

irrational numbers, would be countable.

Problem 3.3. Construct a bounded set of real numbers with exactly three limit points.

Solution.

A = { 1

n} ∪ {1 +

1

n}+ {2 +

1

n}

The set A is bounded. The only limit points are 0, 1, 2. 0 is a limit point because any

neighborhood of 0 contains a real number and by Archimedean property there is N such

that 1/n 6= 0 is in the neighborhood. Similarly 1 and 2 are also limit points. Now we

claim that any other point can not be a limit point. Take any point a + 1m

in A where

a = 0, 1 or 2. Take r = 1m− 1

m+1and observe that the neighborhood Br(a + 1

m) does

not contain any point from A other than itself and hence a+ 1m

can not be a limit point.

Since the point was chosen arbitrarily, we can conclude that the only three limit points

of A are 0,1,2.

Problem 3.4. Let E ′ be the set of all limit points of a set E. Prove that E ′ is closed.

Prove that E and E have the same limit points. Do E and E ′ always have the same limit

points?

Solution. See Problem 165

Problem 3.5. Let A1, A2, . . . . . . be subsets of a metric space.

(a) If Bn =⋃ni=1 Ai, prove that Bn =

⋃ni=1Ai, for n = 1, 2, 3, . . . . . .

(b) If B =⋃∞i=1 Ai, prove that B ⊃

⋃∞i=1Ai

Show by example, that the inclusion can be proper.

Solution. (a) Fist we will show that if E,F are subsets of a metric space then

E ∪ F = E ∪ F . Assume x ∈ A ∪ B. If x ∈ A or x ∈ B then it is obvious that

x ∈ A∪B ⊂ A ∪B. Now take x ∈ A′\A. So Bε(x)∩A\{x} 6= ∅ for any neighbor-

hood Bε(x). Therefore Bε∩(A∪B \{x}) 6= ∅ and therefore x ∈ (A∪B)′ ⊂ A ∪B.

Same argument applies if x ∈ B′. Therefore A ∪B ⊂ A ∪B.

Now take x ∈ A ∪B. If x ∈ A ∪ B then obviously x ∈ A ∪ B. Assume that

x ∈ A ∪B \ (A ∪ B). Then there is a sequence {xn} such that xn → x. Then

either A or B or both contains infinitely many points of {xn}. Without loss of

generality if A contains infinite number of points of {xn} then there is a sub-

sequence {xn} ∩ A that converges to x and hence x ∈ A′. Which implies that

x ∈ A′ ∪B′. Hence x ∈ A ∪B and therefore A ∪B ⊂ A ∪B.

That’s prove that A ∪B = A ∪B. By induction we see the result.

(b) Let x ∈⋃∞i=1Ai. If x ∈ Ai for any i then x ∈ B and we are done. If not then

there is i such that x ∈ A′i. Then there is a sequence {xn} such that xn → x.

But {xn} ⊂ B and therefore x is a limit point of B and hence x ∈ B. Hence

B ⊂⋃∞i=1 Ai.

Let Ai = {1i} for i ∈ N and then B = {1

i: i ∈ N}. Observe that Ai = { i

i}

and 0 /∈⋃∞i=1 but 0 ∈ B because 0 is a limit point of B.

6

4. Metric Space

Definition 4.1. Let X be an arbitrary nonempty set. A metric on X is a function

d : X ×X → R

with the following properties

(a) d(x, y) ≥ 0 for all x, y ∈ X and d(x, y) = 0 if and only if x = y.

(b) d(x, y) = d(y, x) for all x, y ∈ X(c) d(x, y) ≤ d(x, z) + d(z, y) for all x, y, z ∈ X

A metric space (X, d) is a set X equipped with a metric d.

Definition 4.2. Let (X, dX) be a metric space and Y ⊂ X. The metric subspace (Y, dY )

or (X, dX) is defined by setting dY to be the restriction of dX to the points in Y .

Definition 4.3. If X and Y are sets then the Cartesian product X × Y is the set of

ordered pairs (x, y) with x ∈ X and y ∈ Y . If dX and dY are metrics on X and Y

respectively, then the product metric dX×Y on X × Y is defined by

dX×Y ((x1, y1), (x2, y2)) = dX(x1, x2) + dY (y1, y2)

for all x1, x2 ∈ X and y1, y2 ∈ Y .

Definition 4.4. A norm on a linear space X (over a scalar field F) is a function ‖.‖ :

X → R with the following properties:

(a) ‖x‖ ≥ 0 for all x ∈ X;

(b) ‖λx‖ = |λ| ‖x‖ for all x ∈ X and scalar λ ∈ F

(c) ‖x+ y‖ ≤ ‖x‖+ ‖y‖ for all x, y ∈ X(d) ‖x‖ = 0 implies x = 0

Theorem 4.1. A normed linear space (X, ‖.‖) is a metric space (X, d) with the norm-

induced metric

d(x, y) = ‖x− y‖

for x, y ∈ X. Moreover, the metric is translation invariant and homogeneous i.e. for any

x, y, z ∈ X, d(x+ z, y + z) = d(x, y) and for any scalar λ ∈ F, d(λx, λy) = |λ| d(x, y)

Definition 4.5. A sequence (xn) in metric space (X, d) converges to x ∈ X if for every

ε > 0 there is an N ∈ N such that d(xn, x) < ε for all n ≥ N . The sequence is Cauchy if

for every ε > 0 there is an N ∈ N such that d(xm, xn) < ε for all m,n ≥ N

Definition 4.6. A metric space (x, d) is complete if every Cauchy sequence in X con-

verges to a limit in X. A subset Y of X is complete if the metric subspace (Y, d) is

complete. A normed linear space that is complete with respect to the norm-induced

metric is called a Banach space.7

Definition 4.7. Let X be a normed linear space. If (xn) is a sequence in X, then the

series∞∑n=1

converges to x ∈ X if the sequence of partial sum (sn) given by sn =n∑k=1

xk

converges to x.

Definition 4.8. Let A be a nonempty subset of metric space (X, d). The diameter of A

is

diamA = sup{d(x, y) : x, y ∈ A}.

The set A is bounded if its diameter is finite. The distance (x,A) of a point x ∈ X from

the set A is d(x,A) = inf{d(x, y) : y ∈ A}

Definition 4.9. A topology of a nonempty set X is a collection T of subsets of X such

that

(a) ∅, X ∈ T ;

(1) if Gα ∈ T for α ∈ A, then ∪α∈AGα ∈ T ;

(2) if Gi ∈ T for i = 1, 2, . . . . . . , n, then ∩ni=1Gi ∈ TWe call the pair (X, T ) a topological space.

Definition 4.10. A subset A of (X, T ) is a closed set if and only if its complement,

AC = X \ A, is open.

Definition 4.11. Let (X, d) be a metric space. The open ball, Br(a), of radius r > 0

and center a ∈ X is the set

Br(a) = {x ∈ X : d(x, a) < r}.

The closed ball Br, is the set

Br(a) = {x ∈ X : d(x, a) ≤ r}.

Definition 4.12. A subset G of a metric space X is open if for every x ∈ G there exists

an r > 0 such that Br(x) ⊂ G. A subset F of X is closed if its complement F c = X \ Fis open.

Theorem 4.2. Let (X, d) be a metric space. The set of all open sets is a topology on X

called the metric topology.

Theorem 4.3. Let (X, d) be a metric space.

(a) The empty set ∅ and the whole set X are both open and closed.

(b) A finite intersection of open sets is open.

(c) An arbitrary union of open sets is open.

(d) A finite union of closed sets is closed.

(e) An arbitrary intersection of closed sets is closed.

Definition 4.13. Let (X, T ) be a topological space. For any x ∈ X, a set U ⊂ X is a

neighborhood of x is there exists an open set G ⊂ U with x ∈ G.8

Definition 4.14. Let (X, T ) be a topological space. A sequence (xn) ⊂ X converges

to a limit x ∈ X if for every neighborhood U of x there is an N ∈ N such that n ≥ N

implies xn ∈ U .

Definition 4.15. Let (X, T ) be a topological space. All points and sets below are

elements and subsets of X.

(a) A point x is a limit point of the set A is every neighborhood of p contains a point

y 6= x such that y ∈ A.

(b) If x ∈ A and x is not a limit point of A, then x is an isolated point of A.

(c) A point x is an interior point of A if there is a neighborhood U of x such that

U ⊂ A.

(d) A is perfect if A is closed and if every point of A is a limit point of A.

Theorem 4.4. A subset F of a topological space (X, T ) is closed if and only if every

convergent sequence of elements in F converges to a limit in F . In other words, if

xn → x ∈ X and xn ∈ F for all n, then x ∈ F .

Theorem 4.5. A subset of a complete metric space is complete if and only if it is closed.

Definition 4.16. Let (X, T ) be a topological space. The closure A of a set A ⊂ X is

the smallest closed set containing A.

Definition 4.17. Let (X, T ) be a topological space. A subset A ⊂ X is dense in X if

A = X.

Definition 4.18. A topology T on X is called Hausdorff if every pair of distinct points

x, y ∈ X has a pair of nonintersecting neighborhoods. In other words, there exists

neighborhoods Vx of x and Vy of y such that Vx ∩ Vy = ∅.

Definition 4.19. A subset B of a topology T is a base for T if for every G ∈ T there is

a collection of sets Bα ∈ B such that G =⋃αBα.

Definition 4.20. A collection N of neighborhoods of a point x ∈ X is called a neigh-

borhood base for x if for each neighborhood V of x there is a neighborhood W ∈ N such

that W ⊂ V .

Definition 4.21. A topological space (X, T ) is first countable if every x ∈ X has a

countable neighborhood base, and second countable if T has a countable base.

Definition 4.22. A topological space (X, T ) is separable if it contains a countable, dense

subset.

Definition 4.23. A Hamel basis (or algebraic basis ) B of a linear space X is maximal

linearly independent set of vectors such that every x ∈ X can be written uniquely as a

finite linear combinition of vectors from B.9

Definition 4.24. Let X be a separable Banach space. A sequence (xn) is a Schauder

basis of X is for every x ∈ X there exists a unique sequence of scalars (cn) such that

x =∞∑n=1

cnxn.

10

5. Problems on Metric Space

Problem 5.1. Show that d(x, y) = ‖x− y‖ is a metric on Rn where ‖·‖ is the standard

Euclidean norm on Rn.

Solution. d(x, y) ≥ 0 since ‖x‖ ≥ 0 by the definition of Euclidean norm.

If d(x, y) = 0 then ‖x− y‖ = 0 and hence by definition of norm, x− y = 0⇒ x = y.

d(x, y) = ‖x− y‖ = ‖ − 1(y − x)‖ = | − 1|‖y − x‖ = d(y, x)

and hence the commutativity of metric is proved. We now want to prove the triangle

inequality.

d(x, y) = ‖x− y‖ = ‖(x− z)− (y − z)‖ ≤ ‖x− z‖+ ‖y − z‖ = d(x, z) + d(y, z)

Hence d(x, y) is indeed a metric.

11

Problem 5.2. Determine which of the following define metrics on R.

(a) d1(x, y) = (x− y)2,

(b) d2(x, y) =√|x− y|,

(c) d3(x, y) = |x2 − y2|,(d) d4(x, y) = |x− 2y|,(e) d5(x, y) = |x− y| /[1 + |x− y|].

Solution. (a) Consider to points x = 5 and y = −5 and z = 0 and observe that

d1(x, y) = (5 + 5)2 = 25 and , d1(x, z) = d1(y, z) = 25, and d1(x, y) � d1(x, z) + d1(y, z)

That is triangle inequality does not hold. Hence d1 is not a metric.

(b) d2(x, y) ≥ 0 and d2(x, y) = 0⇒√|x− y| = 0⇒ |x− y| = 0⇒ x = y.

Commutativity is trivial because d2(x, y) =√|x− y| =

√|y − x| = d2(y, x)

The triangle inequality comes from the triangle inequality of the absolute value

function,

|x− y| ≤ |x− z|+ |y − z|

⇒ |x− y| ≤ |x− z|+ |y − z|+ 2√|x− z|

√|y − z|

⇒(√|x− y|

)≤(√|x− z|+

√|y − z|

)2

⇒d2(x, y)2 ≤ (d2(x, z) + d2(y, z))2

⇒d2(x, y) ≤ d2(x, z) + d2(y, z)

(c) Assume x = 5 and y = −5 then d3(x, y) = |52 − (−5)2| = 0 though x 6= y. Hence

d3 is not a metric.

(d)

12

Problem 5.3. For f, g : [0, 1]→ R, let d1(f, g) = supx∈[0,1] {x2 |f(x)− g(x)|}, d2(f, g) =

supx∈[0,1] |f(x)− g(x)|, and d3(f, g) =∫ 1

0|f(x)− g(x)| dx. Prove that d1, d2, and d3 de-

fine metrics on the linear space of C([0, 1]) of continuous functions f : [0, 1]→ R.

Solution. Claim: d1 is a metric on C([0, 1]).

Since x2 ≥ 0 and |f(x)− g(x)| ≥ 0 we have d1(f, g) ≥ 0.

If d1(f, g) = 0 then x2 |f(x)− g(x)| = 0 for all x ∈ [0, 1]. If x 6= 0 then we must have

f(x) = g(x). Since f and g are equal at all the points and both of the functions are

continuous f(0) = g(0). Therefore f = g for all x ∈ [0, 1].

Commutativity is obvious because x2 |f(x)− g(x)| = x2 |g(x)− f(x)|.Now we will prove the triangle inequality:

Let h ∈ C([0, 1]). By the triangle inequality of real numbers under usual metric we know,

for all x ∈ [0, 1]

|f(x)− g(x)| ≤ |f(x)− h(x)|+ |h(x)− g(x)| .

Lety =d1(f, h) + d1(g, h)

= supx∈[0,1]

{x2 |f(x)− h(x)|}+ supx∈[0,1]

{x2 |h(x)− g(x)|}

≥{x2 |f(x)− h(x)|}+ {x2 |h(x)− g(x)|}, ∀x ∈ [0, 1]

≥{x2 |f(x)− g(x)|}Therefore y is an upper bound of x2 |f(x)− g(x)| and hence y ≥ supx∈[0,1]{x2 |f(x)− g(x)|},

i.e.

d1(f, g) ≥ d1(f, h) + d1(h, g).

Claim: d2 is a metric on C([0, 1]).

It is very similar to the proof for d1.

Claim: d3 is a metric on C([0, 1]).

Since |f(x)− g(x)| ≥ 0 for all x ∈ [0, 1],

d3(f, g) =

∫ 1

0

|f(x)− g(x)| dx ≥ 0.

Let h(x) = |f(x)− g(x)| and so, h(x) ≥ 0 for all x ∈ [0, 1]. If d3(f, g) ≥ 0 then∫ 1

0h(x)dc = 0. If h(x) 6= 0 for at least one point by the continuity of h in [0, 1] and the

property of Riemann Integral∫ 1

0h(x)dc > 0 and hence h(x) = 0 for all x ∈ [0, 1] which

implies f = g.

Commutativity is obvious because of commutativity of real numbers with usual metric.

Let h ∈ C([0, 1]). By the triangle inequality of real numbers under usual metric we know,

for all x ∈ [0, 1]

|f(x)− g(x)| ≤ |f(x)− h(x)|+ |h(x)− g(x)| .13

Problem 5.5. Consider the convergence of sequences (xn) = (ξn,1, ξn,2, ξn,3, · · · ) to a

point y = (η1, η2, η3, · · · ) in `2. Show that if xn → y, then ξn,i → ηi for i = 1, 2, 3, · · · .But, construct an example that shows that ξn,i → ηi for i = 1, 2, 3, · · · does not imply

that xn → y. (This is similar to the idea of pointwise convergence not guaranteeing

metric-convergence for sequences of functions.) Consider the Hilbert Cube subspace of

`2 defined by,

H ={x = (ξ1, ξ2, ξ3, · · · ) : x ∈ `2 and |ξi| ≤ 1/i, i = 1, 2, 3, · · ·

}.

Show that a sequence (xn) in H converges to a point y = (η1, η2, η3, · · · ) if and only if

ξn,i converges to ηi for i = 1, 2, 3, · · · .

Solution. Solution to 1(a) Let ε > 0 be given. Since xn → y, there exists N ∈ N such

that d(xn, y) ≤ ε for all n ≥ N .

d(xn, y) ≤ ε

⇒

√√√√ ∞∑i=1

|ξn,i − ηi|2 ≤ ε

⇒∞∑i=1

|ξn,i − ηi|2 ≤ ε2

⇒|ξn,i − ηi|2 ≤ ε2, ∀i = 1, 2, 3, . . . . . .

⇒|ξn,i − ηi| ≤ ε ∀i = 1, 2, 3, . . . . . .

From above we conclude that, for any given ε > 0 there is N such that |ξn,i − ηi| ≤ ε

whenever n > N for all i = 1, 2, 3, . . . . . . and hence ξn,i → ηi for i = 1, 2, 3, . . . . . .

Solution to 1(b) Consider the sequence (xn) such that

ξn,i =

1 if n = i

0 if n 6= i

Fix an i, then ξn,i → 0 because for any ε, |ξn,i − 0| = 0 < ε for all n > i. But d(xn, 0) = 1

for all n and hence xn does not converge to 0.

Solution to 1(c) Since Hilbert space is also an `2 space by the proof in 1(a) we know

if xn → y then ξn,i → ηi for all i = 1, 2, 3, . . . . . . .

Lemma 1. If s =∑∞

i=1 xi converges then for given ε > 0 there exists an M such that∣∣∑∞i=M+1 xi

∣∣ < ε.

Proof: The sum convergence if the sequence of partial sum sn =∑n

i=1 xi converges to

s that is for given ε > 0 there is M such that |s− sn| < ε for all n ≥ M . We take

|s− sM | < ε which implies∑∞

i=M+1 si < ε

16

Problem 5.6. Let p ≥ 1 be a fixed real number. The real space `p is defined such that

x ∈ `p is a sequence of real numbers x = (ξj) = (ξ1, ξ2, ξ3, · · · ) such that∑∞

j=1 |ξj|p <∞.

For x, y ∈ `p, where x = (ξj) and y = (ηj), let d(x, y) be defined by

d(x, y) =

(∞∑j=1

|ξj − ηj|p)1/p

.

Show that `p with d defined above is a metric space. (Hint: If(∑∞

j=1 |ξj − ηj|p)1/p

is

finite, then d is a map to R and properties (a) and (b) are clearly satisfied. To show this is

finite, use Minkowski’s inequality, which is also useful in proving the triangle inequality.

You do not need to prove Minkowski’s inequality.)

Solution. Since y = (ηi) ∈ `p,∑∞

j=1 |ηj|p < ∞ and hence

∑∞j=1 |−ηj|

p < ∞. Also

x = (ξj) ∈ `p implies∑∞

j=1 |ξj|p <∞. By Minkowski inequality we get,

d(x, y) =

(∞∑j=1

|ξj − (−ηj)|p)1/p

=

(∞∑j=1

|ξj + ηj|p)1/p

≤

(∞∑j=1

|ξj|p)1/p

+

(∞∑j=1

|ηj|p)1/p

<∞

d(x, y) = 0 implies(∑∞

j=1 |ξj − ηj|p)1/p

= 0 ⇒ |ξj − ηj| = 0 for all j and hence ξj = ηj

for all j. Therefore x = y.

Commutativity of d follows from commutativity of real numbers under usual metric. Now

we will prove the triangle inequality. To do that let z = (ζj) ∈ `p

d(x, y) =

(∞∑j=1

|ξj − ηj|p)1/p

=

(∞∑j=1

|ξj − ζj + ζj − ηj|p)1/p

≤

(∞∑j=1

||ξj − ζj|+ |ζj − ηj||p)1/p

≤

(∞∑j=1

|ξj − ζj|p)1/p

+

(∞∑j=1

|ζj − ηj|p)1/p

=d(x, z) + d(z, y)

Another way to show this is to prove that

‖x‖ =

(∞∑j=1

|ξj|p)1/p

is a norm. Then by theorem 1 we can say that d is a metric in `p.

18

Theorem 1: A normed linear space (X, ‖.‖) is a metric space (X, d) with the norm-induced

metric d(x, y) = ‖x− y‖ for x, y ∈ X. Moreover, the metric is translation invariant and

homogeneous, i.e. for any x, y, z ∈ X, d(x+ z, y + z) = d(x, y) and for any scalar λ ∈ F,

d(λx, λy) = |λ| d(x, y).

19

Problem 5.8. Clearly `1 ⊂ `∞. Is `1 dense in `∞? Is `1 closed in `∞? Justify your

answers.

Solution. (a) Consider the constant sequence x = (1) ∈ `∞. If y = (ηj) ∈ `1 then∑∞j=1 |ηj| < ∞. By the elementary analysis we know that ηj → 0. Therefore for

any y ∈ `1, d(x, y) ≥ 0 for all y ∈ `1. Hence a ball centered at x ∈ `∞ with radius

0 < ε < 1 does not contain any element form `1. We conclude that `1 is not dense

in `∞

(b) Consider the sequence xn = ξn,j so that ξn,j = 1j

if j ≤ n otherwise ξn,j = 0. This

sequence converges point-wise to y = 1n. Indeed the sequence (xn) converges in

`∞ because for any given ε > 0 there is N ∈ N such that 1n< ε for all n > N and

so ‖xn − y‖∞ = 1n+1

< ε for all n ≥ N . But since∑∞

n=11n

does not converge, the

limit point y /∈ `1 and hence `1 is not closed.

21

Problem 5.9. Prove that if a subsequence of a Cauchy sequence in a metric space

converges, then the full sequence converges to the same limit.

Solution. Consider the Cauchy sequence (xn) in the metric space (X, d). By definition

there exists an N1 ∈ N such that d(xn, xm) < ε2

for all n,m > N1. Now suppose the

subsequence (xnk) of (xn) converges to x ∈ X and hence for the given ε there exists an

N2 ∈ N such that for all nk > N2, d(xnk , x) < ε2. Therefore,

d(xn, x) ≤ d(xn, xnk) + d(xn,k, x) < ε

for all n ≥ N = max(N1, N2). Hence the sequence (xn) converges to the same limit of

the subsequence.

22

Problem 5.10. Let (xn) and (yn) be Cauchy sequences in a metric space (X, d). Show

that (d(xn, yn)) converges whether or not (xn) and (yn) converge.

Solution. Let ε > 0 be given, Since (xn) is a Cauchy sequence, there exists an N1 ∈ Nsuch that d(xn, xm) < ε

2for all m,n > N1.

Since (yn) is a Cauchy sequence, there exists an N2 ∈ N such that d(yn, ym) < ε2

for all

n,m > N2.

d(xn, yn) ≤ d(xn, xm) + d(xm, ym) + d(yn, ym)

Which implies

d(xn, yn)− d(xm, ym) ≤ d(xn, xm) + d(yn, ym) <ε

2+ε

2= ε

Similarly we can show that

d(xm, ym)− d(xn, yn) < ε

and hence

|d(xn, yn)− d(xm, ym)| < ε

for all n,m > N = max(N1, N2) Which implies that d(xn, yn) is a Cauchy sequence in Rand hence is convergent.

23

Problem 5.11. Let (X1, d1) and (X2, d2) be two complete metric spaces. Is the product

metric space X1 ×X2 complete?

Solution. Take a Cauchy sequence ((xn, yn)) in X × Y . For given ε > 0 there exists an

M such that n,m ≥M implies

dX×Y ((xn, yn), (xm, ym)) ≤ ε.

Since dX(xn, xm) ≥ 0 and dy(yn, ym) ≥ 0 we get,

dX×Y ((xn, yn), (xm, ym)) ≤ ε

⇒dX(xn, xm) + dY (yn, ym) ≤ ε

⇒dX(xn, xm) ≤ ε and dY (yn, ym) ≤ ε

Therefore, the sequences (xn) and (yn) in the metric space X and Y respectively are

Cauchy sequences. Since X and Y are complete theses sequences converge. Say xn → x

and yn → y for some x ∈ X and y ∈ Y .

Let ε > 0 be given. xn → x implies there exists an N1 ∈ N so that dX(xn, x) ≤ ε2

whenever n ≥ N1. yn → y implies that there exists an N2 ∈ N so that dY (yn, y) ≤ ε2.

Take N = max(N1, N2).

dX×Y ((xn, yn), (x, y)) = dx(xn, x) + dy(yn, y) ≤ ε

2+ε

2= ε

for all n ≥ N . Therefore the sequence (xn, yn) converges to (x, y). Since (xn, yn) was

chosen to be an arbitrary Cauchy sequence we can conclude that X × Y is a complete

metric space.

24

Problem 5.12. Prove that a metric space X is complete if and only if the intersection

of every descending sequence of closed balls whose radii approach zero consists of a single

point.

Solution. Let (Bn) be a descending sequence of closed balls and xn be the center of Bn

for all n ∈ N. Since diam(Bn)→ 0, for given ε > 0 there exists N such that diam(Bn) < ε

for all n ≥ N . For m > n > N , Bm ⊂ Bn and so xm ∈ Bn and hence d(xn, xm) ≤ ε.

So, (xn) is a Cauchy sequence and hence converges to x ∈ X (say) since the space X is

complete.

Now we want to show that x ∈ Bn for all n. If not consider there is anm for which x /∈ Bm.

Since we are considering a nested sequence of closed balls, xi ∈ Bm, ∀i ≥ m. So the

subsequence (xi, i ≥ m) is contained in Bm. A subsequence of a convergent sequence

converges and since the subsequence is entirely contained in Bm and Bm is closed set, by

definition, the subsequence has to converge to a point in Bm which contradicts the fact

that subsequence of a convergence sequence converges to the same limit point. Hence

x ∈ Bn for all n and therefore x ∈⋂∞n=1Bn.

Suppose there is another point y 6= x in⋂∞n=1Bn. But by the definition of descending

sequence of closed balls limn→∞ diam(Bn)→ 0. If x 6= y, d(x, y) > 0 say β = d(x, y) and

hence diam (⋂∞n=1) ≥ β which contradicts the definition of descending sequence of closed

balls.

Now we consider that intersection of every descending sequence of closed balls whose

radii approach zero consists of a single point and want to prove that X is complete.

Let (xn) be a Cauchy sequence in X. First we construct a descending sequence of closed

balls.

Since (xn) is Cauchy there exists an N1 such that for all m,n1 ≥ N1, d(xn1 , xm) < 12.

Consider a closed ball B1 = B(xn1 , 1) = {y ∈ X : d(y, xn1) ≤ 1}. So, xm ∈ B1 for all

m ≥ n1.

Now take ε = 14. So ∃N∗ such that d(xn2 , xm) < 1

4for all m > n2 ≥ N∗. Take N2 =

max(N1, N∗) and that gives, d(xn1 , xn2) <12

and d(xn2 , xm) < 14

for all m ≥ N2. Construct

a closed ball B2 = B(xn2 ,12). We claim that B2 ⊂ B1. Take any point xk ∈ B2 and observe

that,

d(xn1 , xk) ≤ d(xn1 , xn2) + d(xn2 , xk) ≤1

2+

1

4< 1

which implies xk ∈ B1 and hence B2 ⊂ B1.

By induction we can contract such a sequence of closed balls. Suppose d(xnk , xm) < 12k

for all m > nk ≥ Nk and consider the closed ball Bk = B(xnk ,

12k−1

)and so xm ∈ Bk for

25

all m > Nk. Again since (xn) is Cauchy there is N∗ so that ∀m > nk+1 ≥ N∗ we have

d(xnk+1, xm) < 1

2k+1 . Take Nk+1 = max(Nk, N∗) to assert both d(xnk , xnk+1

)< 1

2kand

d(xnk+1, xm) < 1

nk+1 for all m > Nk+1. Now construct the closed ball Bk+1 = B(xnk+1, 1

2k).

We claim that Bk+1 ⊂ Bk and it is direct consequence of the triangle inequality.

Therefore we have a descending sequence of closed balls Bk whose radii 12k

approaches

zero and by the hypothesis the intersection of all the closed balls consists of a single point,

say (x ∈ X). We want to show now that x is the limit point of the subsequence (xnk).

Since x ∈ Bk for all k and xnk is the center of Bk we can write d(xnk , x) < 12k

. By

comparison d(xnk , x) → 0 as k → ∞ and hence xnk converges to x. From elementary

analysis we know if a subsequence of a Cauchy sequence converges then the sequence

itself converges to the same point (Proof as lemma is given bellow)

Since (xn) is chosen to be arbitrary Cauchy sequence we can assert that any Cauchy

sequence in X converges and hence X is complete.

Lemma. If a subsequence of a Cauchy sequence converges then the sequence converges.

Proof: Let (xn) is a Cauchy sequence in the metric space (X, d) and (xnk) is a subsequence

that converges to x ∈ X. Let ε > 0 be given. Since the sequence is Cauchy, there exists

an N1 such that d(xn, xm) < ε2

for all m,n ≥ N1. Also since the subsequence converges

to x, there is K such that d(xnk , x) ≤ ε2

for all k ≥ K. Let N = max(N1, K). By triangle

inequality,

d(xn, x) ≤ d(xn, xnk) + d(xnk , x) < ε

since nk ≥ k > N . Therefore (xn) converges to x.

26

Problem 5.13. For f, g : [0, 1]→ R, let d1(f, g) = supx∈[0,1] {x2 |f(x)− g(x)|}, d2(f, g) =

supx∈[0,1] |f(x)− g(x)|, and d3(f, g) =∫ 1

0|f(x)− g(x)| dx. Prove that (C([0, 1]), d1) is

not complete but (C([0, 1], d2) is complete (this requires recalling some useful results from

elementary analysis that we will revisit in more generality in lecture 4). Since these

metrics can be defined by norms, one choice of norm produces a Banach space, while the

other does not (this happens on infinite dimensional spaces where norms are not always

equivalent). If we use d3 is the resulting metric space complete?

Solution. (1) To show that (C([0, 1]), d1) is not complete we have to show that there

is a Cauchy sequence that does not converge in C([0, 1]).

Consider the sequence of (fn) continuous functions

fn = (1− x)n

Then,

d(fn, fm) = supx∈[0,1]

{x2 |(1− x)n − (1− x)m|} = supx∈[0,1]

{x2 |(1− x)n|} = supx∈[0,1]

{x2(1− x)n}

At x = 0, d(fn, fm) = 0 and for x ∈ (0, 1], (1 − x)n → 0, so (fn) is a Cauchy

sequence. The sequence of function converges to a function f given by f(0) = 0

and f(x) = 1 if x ∈ (01, ], which is not a continuous function and hence f /∈C([0, 1]). We conclude that (C([0, 1]), d1) is not complete.

(2) Consider any Cauchy sequence (fn) in C([0, 1]). Let ε > 0 be given. There exists

an N ∈ N such that d2(fn, fm) < ε for all n,m > N . For each x ∈ [0, 1] we have

that

|fn(x)− fm(x)| ≤ d2(fn, fm) < ε.

So, (fn(x)) is a Cauchy sequence in R. Since R is complete, for each x ∈ [0, 1]

(fn(x)) converges. We define point-wise limit function f(x) = limn→∞

fn(x) for each

x ∈ [0, 1]. If we can show that fn → f uniformly, then from elementary analysis, f

is continuous on [0, 1] and d(fn, f) has meaning. We use the result from example

1.25 in the text that says:

Example 1.25 If {xn,α ∈ R|n ∈ N, α ∈ A} is a set of real number indexed by the

natural numbers N and an arbitrary set A, then

supα∈A

[lim infn→∞

xn,α

]≤ lim inf

n→∞

[supα∈A

xn,α

]Use the result to obtain,

supx∈[0,1]

= supx∈[0,1]

|fn(x)− f(x)| = limm→∞

|fn(x)− fm(x)| ≤ lim infm→∞

supx∈[0,1]

|fn(x)− fm(x)| .

This implies that

supx∈[0,1]

|fn(x)− f(x)| ≤ lim infm→∞

d2(fn, fm).

27

Since (fn) is Cauchy in (C([0, 1]), d2), there exists N such that n,m > N implies

d2(fn, fm) < ε, which implies fn → f uniformly. Moreover, this proves that

fn → f in (C([0, 1])).

28

Problem 5.14. Prove that a discrete metric space X is separable if and only if X is

countable. (A discrete metric space is any set X with the discrete metric d(x, y) = 0 if

x = y and d(x, y) = 1 if x 6= y.)

Solution. By the definition of separable space X to be separable has to contain a count-

able, dense subset. In the discrete metric space no proper subset is dense in X because

if we take a proper subset A of X then there is a point a ∈ X \ A and since all the con-

vergent sequence in discrete metric space are the constant sequence there is no sequence

in A that converges to a /∈ A. So, the only dense subset of X is the space itself and by

definition, to be separable it has to be countable.

If X is countable then X is the countable dense subset of X and hence it is separable.

29

Problem 5.15. Prove that `∞ is not separable, but `p with 1 ≤ p < +∞ is separable.

Prove that `p with 1 ≤ p ≤ +∞ is complete.

Solution. (1) Let y = (ηj) be a sequence of 0’s and 1’s. Then y ∈ `∞. With y we

associate the real number y with the decimal expansion of 0’s and 1’s given by

y = 0.η1η2η3 . . . . . . . By the previous problem, there are an uncountable number

of such y ∈ `∞. Furthermore, the metric on `∞ restricted the subset of such y is

the discrete metric, so by one of the examples from class `∞ cannot be separable.

(2) Let A denote the set of all sequences of the form y = (η1, η2, . . . . . . , ηn, 0, 0, . . . . . . )

where ηi ∈ Q for each i. By the proposition that Cartesian product of countable

sets is countable we have that A is countable. We want to show that A is dense

in `p. Suppose x = (ξi) be as sequence in `p. Therefore∑∞

i=1 |ξi|p is finite. A

series converges if the sequence of the partial sum sn =∑n

j=1 |ξj|p converges to

the sum of the series. Therefore the sequence∑∞

i=n+1 |ξi|p = s−

∑ni=1 |ξi|P → 0.

Therefore for any given ε > 0 there is N ∈ N such that∑∞

i=n+1 |ξi|p < εp

2for all

n ≥ N .

Consider a sequence y ∈ A such that y = (η1, η2, . . . . . . , ηN , 0, 0, . . . . . . ) where

|ηi − ξi| < εp√2 p√N

for all i = 1 . . . . . . N . Such a sequence exists because Q is dense

in R. Observe that,

d(x, y) =

(∞∑i=1

|ξi − ηi|p)1/p

=

(N∑i=1

|ξi − ηi|p +∞∑

i=N+1

|ξi − ηi|p)1/p

<

(εp

2+εp

2

)1/p

= ε

Therefore for any given ε, any ball Bε(x) contains a point y of A and hence A is

dense in `p.

We conclude that `p is separable.

(3) We consider two cases. First we will prove completeness of `p for 1 ≤ p <∞ and

then we will prove for `∞.

Let (xn) be a Cauchy sequence in `p where xn = (ξn,i). Let ε > 0 be given.

There exists an N ∈ N such that d(xn, xm) < ε for all n,m ≥ N . That is(∞∑i=1

|ξn,i − ξm,i|p)1/p

< ε

Which implies,

(5.1)∞∑i=1

|ξn,i − ξm,i|p < εp

Since |ξn,i − ξm,i| ≥ 0 we must have,

|ξn,i − ξm,i|p < εp ⇒ |ξn,i − ξm,i| < ε

for all i whenever n,m ≥ N . Therefore for a fixed i, (ξn,i) is a Cauchy sequence in

R and hence convergence. Let xn,i → ηi for all i. Construct a sequence y = (ηi).30

Claim: y ∈ `p. The equation (5.1) implies that for any k

k∑i=1

|ξn,i − ξm,i|p < εp

Since the sum is finite and true for all m,n ≥ N we can take the limit m → ∞to get,

(5.2)k∑i=1

|ξn,i − ηi|p < εp

Since the above inequality is true for every k we can take k → ∞ and that

implies the sequence∑∞

i=1 |ξn,i − ηi|p < ∞ and hence xn − y ∈ `p . Writing

y = (y − xn) + xn and using Minkowski’s inequality we can write

∞∑i=1

|ηi|p <∞

and hence y ∈ `p.Now we want to prove that xn → y. To show that we use the equation (5.2).

d(xn, y) =

(∞∑i=1

|ξn,i − ηi|p)1/p

< ε

for all n ≥ N . Hence xn → y in `p showing that `p space is complete.

Now we will prove `∞ is complete. Let (xn) be any Cauchy sequence in `∞ where

xn = (ξn,j). Consider any ε > 0. There exists N such that n,m > N implies

d`∞(xm, xn) = supj∈N|ξm,j − ξn,j| < ε.

Thus, for each fixed j ∈ N, and n,m > N , |ξm,j − ξn,j| < ε, so for fixed j, (ξn,j)

is a Cauchy sequence of real numbers. By the completeness of R, these sequences

converge for each j. Let (ξj) denote the sequence x defined by the term-wise limit

of each of these Cauchy sequences of real numbers. We show that x ∈ `∞ and

xn → x. For each j, we have the following inequalities

|ξj| − |ξn,j| ≤ |ξj − ξn,j| ≤ |ξj − ξm,j|+ |ξj − ξn,j| .

There exists anN such that for n,m > N we have |xij − ξm,j| < ε2

and |ξj − ξn,j| <ε2, so |ξj| − |ξn,j| ≤ ε. Fix any n > N , and observe that xn ∈ `∞ implies there

exists real number cn such that |ξn,j| < cn for each j, so

|ξj| ≤ ε+ cn.

The bound does not depend on j, so x = (ξj) ∈ `∞. We also have that d(x, xn) < ε

for n > N from the previous inequality since ε does not depend on j, which shows

xn → x. Since (xn) was an arbitrary Cauchy sequence, `∞ is complete.

31

Problem 5.16. Let E ′ be the set of all limit points of a set E ⊂ X where (X, d) is a

metric space. Prove that E ′ is closed. Prove that E and E have the same limit points.

Do E and E ′ always have the same limit points?

Solution. (1) If the set of limit point of E ′ is empty then we have nothing to prove.

Let x be a limit point of E ′. For any given ε > 0 the open ball Bε(x) contains a

point y 6= x of E ′. So y is a limit point of E. d(x, y) < ε and so ε − d(x, y) > 0.

The ball centered at y and radius ε − d(x, y) contains a point z ∈ E and so

d(y, z) < ε− d(x, y). So,

d(x, z) ≤ d(x, y) + d(y, z) < d(x, y) + ε− d(x, y) = ε.

Which implies that for any ε > 0 the open ball Bε(x) contains a point of E and

hence x is a limit point of E. Therefore we conclude that x ∈ E ′ and hence E ′ is

closed.

(2) Suppose x is a limit point of E. Then there is a sequence (xn) ⊂ E such that

xn → x. Since E ⊂ E we have (xn) ⊂ E and together with the fact that xn → x

in E we conclude that x is a limit point of E.

Now assume x is a limit point of E. For any given ε > 0 the open ball Bε(x)

contains a point of y ∈ E. If y ∈ E then we have nothing to prove if not then

y ∈ E ′. Similarly above we can show that there is a point zıE such that d(x, z) < ε

and hence Bε(x) contains a point of E other than itself. Hence x is a limit point

of E. Therefore we conclude that E and E have the same limit points.

(3) NO! Consider X = R with usual metric and E = 1, 12, . . . . . . 1

n. . . then E ′ = {0}.

The limit only limit point of E is 0 and E ′ does not have any limit point.

32

Problem 5.17. Let A1, A2, A3, . . . be subsets of a metric space.

(a) If Bn = ∪ni=1Ai, prove that Bn = ∪ni=1Ai for n = 1, 2, 3, . . ..

(b) If B = ∪∞i=1Ai, prove that ∪∞i=1Ai ⊂ B. Show, by an example, that this inclusion can

be proper.

Solution.

33

Problem 5.18. Is every point of every open set E ⊂ R2 a limit point of E? Answer the

same question for closed sets in R2.

Solution: Every point of an open set is a limit point of E because for each x ∈ E and

there exists r > 0 such that Br(x) ⊂ E and each of these balls contains an infinite number

of points on E ∩ R2.

It is not true for closed set. Consider the singleton set E = {(0, 0)}. It does not have

any limit point that is E ′ = ∅ and clearly E 6= ∅

34

Problem 5.19. Let Eo denote the set of all interior points of a set E.

(a) Prove that Eo is always open.

(b) Prove that E is open if and only if Eo = E.

(c) If G ⊂ E and G is open, prove that G ⊂ Eo.

(d) Prove that the complement of Eo is the closure of the complement of E.

(e) Do E and E always have the same interiors?

Solution. (1) If x ∈ Eo then, by definition of interior point, there is an neighborhood

of x that is contained entirely in E. Assume that the ε ball Bε(x) ⊂ E. Take

any point y ∈ Bε(x) and x 6= y. The open ball centered at y and radius β = ε−d(x, y) > 0, written as Bβ(y) is entirely contained in Bε(x) and hence Bβ(y) ⊂ E.

Since y is an arbitrary point, we conclude that all the point in Bε(x) are interior

points of E, so Bε(x) ⊂ Eo. This implies that x is an interior point of Eo. Since

x was chosen to be arbitrary, we conclude that Eo is an open set.

(2) First assume that E is open. By definition, for every point there is an open neigh-

borhood that is entirely contained in E. Then all the points of E are interior

points that is if x ∈ E then x ∈ Eo or in other words E ⊂ Eo and by definition

of a set of interior point Eo ⊂ E and hence E = Eo.

Now assume that E = Eo. By proof (a) Eo is open and hence E.

(3) Take any x ∈ G. Since G is open there is a ball Bε(x) contained in G that

is Bε(x) ⊂ G ⊂ E. Which tell us that for every point of G there is a open

neighborhood that is entirely contained in E and hence all the points of G are

interior points of E and hence G ⊂ Eo.

(4) We have proved that Eo is open. So in the topological space induced by the

metric, we can write Eo as arbitrary union of open intervals Gα where α ∈ Awhere A is an index set. So, Eo =

⋃α∈AGα by De-Morgan’s law,

(Eo)c =⋂α∈A

Gcα.

Since Gα is open by definition Gcα is closed. Observe that Gα ⊂ E0 ⊂ E for all

α ∈ A and so, Ec ⊂ Gcα for all α ∈ A. Therefore Gc

α is a closed set that contains

Ec and⋂α∈A is the smallest closed set containing Ec and by definition is the

closure of Ec. Hence we conclude that the complement of Eo is the closure of the

complement of E.

(5) NO! Consider the set E = (−1, 0) ∪ (0, 1) in R with usual metric. The closure,

E = [−1, 1]. Observe that 0 is not an interior point of E because 0 /∈ E. But 0 is

an interior point of E.

35

Problem 5.20. Show that Rk is separable.

Solution. We want to show that there is a countable dense subset of Rk. Consider the

subset Qk.

Claim1: Qk is countable.

Since finite Cartesian product of countable subset is countable we immediately have Qk

is countable.

Claim 2: Qk is dense in Rk.Let x ∈ Rk be any point of the form x = (x1, x2, . . . . . . , xk). We want to show that every

ball centered at x contains a point of Qk. For any ε > 0 consider the ball Bε(x). Since Qis dense in ≈x≈ there is a rational number ξj in (xi − ε2

k, xi + ε2

k) and so |xi − ξi| < ε2

k

for all 1 ≤ i ≤ k. Let y = (ξ1, ξ2, . . . . . . , ξk) ∈ Qk. Then,

d(x, y) =

√√√√ k∑i=1

(xi − ξi)2 ≤√kε2

k= ε.

This implies there any ε ball around x contains a point of Qk and hence Qk is dense in

Rk

36

Problem 5.21. Prove that every separable metric space has a countable base.

Solution. Let A = {xi}∞i=1 denote a countable dense subset of X. For each n ∈ N and

r ∈ Q+ let Vn,r = Br(xn). Note that {Vn,r} is countable.

Consider any nonempty open set G ⊂ X. For each x ∈ G, there exists εx > 0 such

that Bεx ⊂ G. The open ball B εx2

(x) contains a point xm for some m, and there exists

r ∈ Q+ such that r < εx2

implying Br(xm) ⊂ Bεx ⊂ G. Thus, G can be written as the

union of balls of rational radii centered at the points of a countable dense subset. This

implies {Vn,r} is a countable base for the metric-topology.

Problem 13.5 A metric space (X, d) is separable if and only if it is second countable.

Solution: Since a metric space is also a topological space, the above shows that separa-

ble implies second countable. Now we will prove that if X is second countable then it is

separable.

Since X is second countable, by definition, the metric topology has a countable base

B =⋃Gα with Gα 6= ∅. Now form a subset E taking one element from each Gα. So E is

countable. Let x ∈ X be any point and for any ε > 0, Bε be a neighborhood of x. Since

Bε(x) is open, by the definition of countable base there is a collection aα ⊂ B such that

Bε(x) =⋃αGα. Therefore Bε(x) contains a point of E and hence E is a countable dense

subset of X and therefore X is separable.

37

Problem 5.22. Prove that any uncountable set with the discrete metric is not second

countable.

Solution. Let (X, d) be an uncountable metric space with the discrete metric. Every

point of X is also an open set, so every point is a neighborhood. So the metric topology

is uncountable and hence does not have a countable base implying that it is not second

countable.

38

Problem 5.23. Show that a Schauder basis (fn)∞n=0 of C([0, 1]) may be constructed from

”tent” functions.

39

Problem 5.24. Let E be the set of all x ∈ [0, 1] whose decimal expansion contains only

the digits 4 and 7. Is E countable? Is E dense in [0, 1]? Is E compact? Is E perfect?

Solution. (1) E is not countable.

To see this we use same argument as cantor diagonalize argument. Assume that

E is countable. Then we have an ordering of the elements of E : {xn}n∈N. Write

out the decimal expansion of xn as yn1 yn2 . . . . . . y

nj . . . where ynj represents the jth

digit of xn. Let z have the decimal expansion given by zj = 4 if yjj = 4 and zj = 7

if yii = 4. Then z ∈ E but z 6= xn for any n, contradiction.

(2) E is not dense in [0, 1]

0 ∈ [0, 1] however B0.1(0) does not contain any point of E because B0.1(0) =

(0, 0.1) and all the point inside the ball has decimal expansions beginning with a

0 and thus are not in E. Therefore 0 is not a limit point of E so E is not dense

in [0, 1].

(3) E is compact:

If x /∈ [0, 1] then r = min{d(x, 0), d(x, 1)} > 0 then Nr(x) ∩ E = ∅. Now suppose

x ∈ [0, 1] \E. Then the decimal expansion of x has some digit which is not a 4 or

a 7. Let d1d2 . . . . . . denote the decimal expansion of x. Let k be the lowest index

such that dk 6= 4 or 7.

Now let r = 110k+2 . Then Br(x) only contains elements which either have dk as

their kth digit or they have 9 as their k+1th digit. Both of these conditions ensure

that the element of Br(x) is not an element of E. Therefore x is not a limit point

of E. Therefore E contains all of its limit points and is closed.

Since E ⊂ [0, 1], it is bounded. By Heine-Borel theorem E is compact.

(4) E is not perfect:

The answer to this question kinda depends on how one interpret E. If it is allowed

to have points like 0.4 = 0.4000000 . . . which can be written with or without 0’s

then E is not perfect because 0.4 is not a limit point because B0.1(0.4) does not

contain any points of E except for 0.4 because every point in (0.39, 0.41) either

has a 3 as its first digit or has a 0 as its second digit and has non-zero digits

following that 0 if it is not exactly 0.4. Therefore E is not perfect.

On the other hand, if you interpret this question such that 0.4000 . . . is not in-

cluded in E and the only elements in E are infinite expansions where all of the

infinite decimal places are either 4 or 7 then every point in E is a limit point

because:

For every ε > 0 there exists an N such that 110N

< ε. Then for a given x ∈ E we

want to find a y ∈ E such that y ∈ Bε(x) and y 6= x. We construct such a y by

letting its first N digits be the same in x. Then let the N + 1st digit be different

from x(if x has a 4 then y has a 7 and vice versa) and let all the other digits be40

same. Then

|x− y| = 3

10N+1<

1

10N< ε

so y ∈ Bε(x) so since ε was arbitrary, x is a limit point of E. So this proves that

all the points of E which are infinite sequence of 4’s and 7’s are limit points of E

so if this is all of E then E is perfect.

41

Problem 5.25. Below, A and B are sets in metric space (X, d).

(a) If A and B are disjoint closed sets prove that they are separated.

(b) Fix p ∈ X, δ > 0, define A = {q ∈ X : d(p, q) < δ} and B = {q ∈ X : d(p, q) > δ}.Prove that A and B are separated.

(c) Prove that every connected space with at least two points is uncountable.

Solution. (1) Since A and B are closed, A = A and B = B. Also given that

A ∩B = ∅. So, A ∩B = A ∩B = ∅ and A ∩B = A ∩B = ∅. Hence A and B are

separated.

(2) Note that A and B are disjoint open sets. Let E = Bc = X \B. Since E is closed

and A ⊂ E, by the definition that closure is the smallest closed subset containing

the set we have A ⊂ E. Since E ∩ B = ∅ we have A ∩ B = ∅. Similarly we can

show that A ∩B = ∅. Therefore, A and B are separated.

(3) Consider x, y ∈ X with x 6= y, and d(x, y) > 0. For δ ∈ (0, α), since X is

connected there exists z ∈ X such that d(x, z) = δ or else X is separated by (a)

and (b) with point p = x or y. Thus there exists some subset of X that can be put

with one -one correspondent with [0, α] ⊂ R. Which implies X is uncountable.

42

Problem 5.26. Let A and B be separated subsets in Rn, suppose a ∈ A and b ∈ B and

define p(t) = (1− t)a + tb for t ∈ R. Put A0 = p−1(A) and B0 = p−1(B) (thus t ∈ A0 if

and only if p(t) ∈ A).

(1) Prove that A0 and B0 are separated subsets of R.

(2) Prove that there exists t0 ∈ (0, 1) such that p(t0) /∈ A ∪B.

(3) Prove that every convex subset of Rk is connected.

Solution. (1) We prove by contradiction. First observe that A0 6= ∅ because 0 ∈ A0

as a ∈ A and also 1 ∈ B0 for b ∈ B implies B0 6= ∅. Assume that A0 and B0 are

not separated then one of the followings is true

A0 ∩B0 6= ∅

A0 ∩B0 6= ∅

A0 ∩B0 6= ∅

First we take that A0 ∩ B0 6= ∅. Let t′ ∈ A0 ∩ B0 which implies t′ ∈ A0 and

t′ ∈ B0. By the definition of the given function t′ ∈ A0 ⇒ p(t′) ∈ A and

t′ ∈ B0 ⇒ p(t′) ∈ B. So A ∩ B 6= ∅ which contradicts the fact that A and B are

separated. So, we can draw a conclusion that A0 and B0 are disjoint.

Now assume that A0 ∩ B0 6= ∅. Let t′ ∈ A0 ∩ B0. Then t′ ∈ B0 and t′ is a limit

point of A0 and so there is a sequence (tn) in A0 that converges to t′. That is for

given ε > 0 there exists an N ∈ N such that |tn − t′| < ε2c

where c = max(‖a‖, ‖b‖)whenever n ≥ N . Now construct a sequence p(tn) in A. We want to show that

p(tn) converges to p(t′).

For given ε > 0, n ≥ N implies,

‖p(tn)− p(t′)‖ =‖(1− tn)a+ tnb− (1− t′)a− t′b‖

≤‖(t′ − tn)a‖+ ‖(tn − t′)b‖

=‖a‖ |tn − t′|+ ‖b‖ |tn − t′|

≤‖a‖ ε2c

+ ‖b‖ ε2c

≤ ε2

+ε

2

=ε

Therefore p(tn) → p(t′) and so p(t′) ∈ A. Together with p(t′) ∈ B we conclude

that p(t′) ∈ A ∩ B and hence A ∩ B 6= ∅. Which is a contradiction to the fact

that A and B are separated. A0 ∩B0 = ∅Similarly, we can show that A0 ∩B0 = ∅. Hence A0 and B0 are separated.

(2) We prove by contradiction. Assume p(t) ∈ A ∪ B for all t ∈ [0, 1], then [0, 1] ⊂A0∪B0. Let E = [0, 1]∩A0 and F = [0, 1]∩B0. None that E and F are nonempty

43

since 0 ∈ A0 and 1 ∈ B0. Also, [0, 1] = E ∪ F , but [0, 1] is connected whereas E

and F are separated by part (a), a contradiction.

(3) By the definition of convex set if E is convex then for all t ∈ [0, 1], p(t) ∈ E. If E

is not connected then there are non-empty subsets A,B ⊂ E such that E = A∪Band A and B are separated set. According part (b) there exists t0 ∈ (0, 1) such

that p(t0) /∈ A ∪B = E which is contradiction to the definition of convex set.

44

Problem 5.27. Give an example of matrix space that is separable but not second count-

able.

Solution. The space R` defined by the topology to be the collection of set of the form

[a, b). R` is first countable because for any element x ∈ R consider the basis [x, 1n) which

is countable. Also the set of rational number is dense in R` which is countable and hence

R` is separable. But we will show that R` is not second countable.

Let B be a basis for R`. Choose for each x, an element Bx ∈ B such that x ∈ Bx ⊂[x, x + 1). If x 6= y, then Bx 6= By, since x = inf Bx and y = inf By. Therefore, B must

be uncountable.

Problem 5.28. Any open set in R can be written as countable union of open intervals.

Solution. Suppose O be an open subset of real numbers. For each rational number r,

define the interval Ir to be the largest open interval containing r that is a subset of O (it is

possible of course that if r is not inO then Ir is empty, so in this case we simply ignore this

r). More particularly, if we define ar = inf{x : [x, r] ⊆ O} and br = sup{x : [r, x] ⊆ O},then Ir = (ar, br). Since the rationals are a countable set, the collection {Ir : r ∈ Q} is a

countable collection of intervals.

Of course they need not be disjoint. However it is true that if s is rational and x ∈ Irthen Is = Ir. This follows by noting that since Is is the largest open interval containing

s that is a subset of O, and since s ∈ Ir, an open interval that is a subset of O, Ir ⊂ Is.

Since then we have that r ∈ Is the same argument shows that Is ⊆ Ir. It is also true that

if s and r are rational and s /∈ Ir then Is ∩ Ir = ∅. To see this suppose that x ∈ Is ∩ Ir.Since Is and Ir are both open intervals, so is Ir ∪ Is. Since Ir is the largest open interval

containing r and a subset of O, Ir ∪ Is ⊆ Ir. But this implies that s ∈ Ir contradicting

our hypothesis. Therefore, we conclude that Is∩Ir = ∅. Together, this implies that given

two intervals Is and Ir that they are either disjoint or identical.

By removing duplicates in {Ir : r ∈ A}, we arrive at a sequence of disjoint intervals

{Ik}∞k=1 such that⋃r∈Q Ir =

⋃∞k=1 Ik. It remains only to show that O =

⋃∞k=1 Ik. Since

by construction, each Ik ⊆ O,⋃∞k=1 Ik ⊆ O. Now let x ∈ O. Since O is open there is a

natural number n such that (x − 1n, 1 + 1

n) ⊆ O. By the density or the rationals in the

reals, there is a rational number r ∈ (x− 1n, x+ 1

n), and so by construction the largest open

interval containing r that is a subset of O will have to include the interval (x− 1n, x+ 1

n).

Hence x ∈ Ir for this r, and so x ∈⋃∞k=1 Ik, and we conclude that O =

⋃∞k=1 Ik.

45

6. Numerical Sequences and Series

6.1. Sequences

(1) Suppose {an} and {bn} are sequences of real numbers such that an ≤ bn for all

n ≥ N and some V ∈ N. If an → a and bn → b then a ≤ b

(2) Let {an} be a sequence such that an → a. If a > 0, then an > 0 for all n > N

and some N ∈ N(3) Suppose an ≤ bn ≤ cn ∀n ≥ N and some n ∈ N. If an → l and cn → l then

bn → l

Theorem 6.1. Let {sn} and {tn} be two sequences of complex number such that sn → s

and tn → t for some complex number s and t. Let z be any complex number. Then,

(1) sn + tn → s+ t

(2) zsn → zs

(3) sntn → st

(4) 1sn→ 1

s

Theorem 6.2. Let {xn} be a monotonic sequence. {xn} is convergent if and only if it is

bounded.

Definition 6.1. Let {xn} be a sequence in R.

We say {xn} converges to ∞ and we write xn →∞ if and only if given any M > 0 there

is n ∈ N such that xn > M for al n > N .

We say {xn} converges to −∞ and we write xn → −∞ if and only if given any M > 0

there is n ∈ N such that xn < M for al n > N .

Definition 6.2. {xn} be a sequence of real numbers, bn = sup{xk|k ≥ n}if{xn} is

bounded above and an = inf{xk|k ≥ n} if {xn} is bounded bellow. Note that bn and an

are non increasing and non decreasing sequences and limn→∞

bn and limn→∞

an exist in R.

lim supn

xn =

limn→∞

bn if{xn}isboundedabove

∞ otherwise

Theorem 6.3. Let {xn} be a sequence of real numbers. Let E be the set of all y in Rsuch that y is a limit of a subsequence of {xn}. Then,

lim supxn = supEand lim inf xn = inf E

Here inf E and supE are in the extended metric R

Theorem 6.4. Let {xn} be a sequence of real numbers. xn → x for some x ∈ R if

lim supxn = x = lim inf xn

Theorem 6.5. If {xn} and {yn} are sequences in R such that xn ≤ yn for all n ≥ N

and some N ∈ N, then,46

(1) lim supxn ≤ lim inf yn

(2) lim inf xn ≤ lim inf yn

Theorem 6.6. (1) If p ≥ 0 then limn→∞

1np

= 0

(2) If p > 0 then limn→∞

n√p = 1

(3) limn→∞

n√n = 1

(4) If p > 0 and α is a real number, then limn→∞

nα

(1+p)n= 0

(5) If |x| < 1, then limn→∞

|x|n = 0

(6) If |x| > 0 then limn→∞

|x|n =∞ if x > 1 and does not exists if x ≤ −1

6.2. Series

Definition 6.3. Let {an} be a sequence of complex numbers. The formal sum a1 + a2 +

a3 + . . . · · · + an + . . . . . . which can be written as∑∞

k=1 ak is called an infinite series or

simply a series.

With the above infinite series we associate another sequence called the sequence of

partial sums of the series denoted by {sn} and defined by sn =∑n

k=1.

We say the series∑∞

k=1 ak converges to a real number s or has the sums iff the associated

sequence {sn} of partial sum converges to s. Otherwise that is if {sn} diverges we say

that the series∑∞

k=1 ak diverges.

Theorem 6.7. The series∑∞

n=1 an converges if and only if for every ε there is a positive

integer N such that if m > n > N then |∑m

j=n aj| < ε.

Theorem 6.8. If∑an converges then an → 0

Note that this theorem doesn’t day if an → 0 then the sum converges rather the contra-

positive is true that is if an doesn’t tends to zero then the sum doesn’t converge.

Theorem 6.9. A series of non negative terms converges if and only if its sequence of

partial sums is bounded.

Theorem 6.10 (Comperison Test). (1) Let |an| < cn, ∀n ≥ N0, where N0 is some

fixed integer, and if∑cn converges, then

∑an converges.

(2) Let an ≥ dn ≥ 0, ∀n ≥ N0, and, if∑dn diverges, then

∑an diverges.

Theorem 6.11. The series∑∞

k=0 xk is convergent if and only if |x| < 1. If |x| < 1 then∑∞

k=0 xk = 1

1−x

Theorem 6.12 (Cauchy Condensation Test). Suppose {an} is a non increasing sequence

of nonnegative terms then∑∞

n=1 an converges if and only if 2na2n converges.

Theorem 6.13.∑

1np

converges if p > 1 and diverges if p ≤ 147

Theorem 6.14. If P > 1,∞∑n=2

1

n (log n)p

converges; if p ≤ 1, the series diverges.

Theorem 6.15 (Root Test). Given∑an, put α = lim sup

n→∞n√an. Then

(a) if α < 1,∑an converges;

(b) if α > 1,∑an diverges;

(c) if α = 1, the test gives no information.

Theorem 6.16 (Ratio Test). The series∑an

(a) converges if lim supn→∞

∣∣∣an+1

an

∣∣∣ < 1,

(b) diverges if∣∣∣an+1

an

∣∣∣ ≥ 1 for n ≥ n0, where n0 is some fixed integer.

Definition 6.4. Given a sequence {cn} of complex numbers, the series

∞∑n=0

cnzn,

is called a power series. The numbers cn are called the coefficients of the series; z is a

complex number.

Theorem 6.17 (Radius of convergence). Given the power series∑cnz

n, put

α = lim supn→∞

n√|cn|, R =

1

α,

(If α = 0, R = +∞; if α = +∞, R = o.) Then∑cnz

n converges if |z| < R and

diverges if |z| > R

Theorem 6.18 (Summation by Parts). Given two sequences {an} and {bn}, put

An =n∑k=0

ak

if n ≥ 0; put A−1 = 0. Then if 0 ≤ p ≤ q, we have

q∑n=p

anbn =

q−1∑n=p

An(bn − bn+1) + Aqbq − Ap−1bp

Theorem 6.19. Suppose

(a) the partial sums An of∑an form a bounded sequence;

(b) b0 ≥ b1 ≥ b2 ≥ . . . . . . i.e. ({bn} is a decreasing sequence);

(c) limn→∞

bn = 0.

Then∑anbn converges.

Theorem 6.20 (Alternating series test). Suppose

(a) |c1| ≥ |c2| ≥ |c3| ≥ . . . i.e. {|cn|} is a decreasing sequence48

(b) c2m−1 ≥ 0, c2m ≥ 0, m = 1, 2, 3 . . . . . . , i.e. alternating series

(c) limn→∞

cn = 0.

Then∑cn converges.

7. Rudin’s Solution:

Problem 7.1. Prove that convergence of {sn} implies convergence of {|sn|}. Is the

converse true?

Solution. Let ε > 0 be given. Since {sn} converges, it is a Cauchy sequence and so there

exists natural number N such that n,m ≥ N implies |sn − sm| < ε. We have

||sn| − |sm|| ≤ |sn − sm| < ε

for all n,m ≥ N . Hence the sequence {|sn|} is also a Cauchy sequence and therefore

converges in the complete metric R.

Problem 7.2. Calculate limn→∞

(√n2 + n− n

)Solution. √

n2 + n− n =n√

n2 + n+ n=

1√1 + 1

n+ 1

It follows that the limit is 12

Problem 7.3. If s1 =√

2, and

sn+1 =√

2 +√sn, (n = 1, 2, 3, . . . . . . )

prove that {sn} converges, and that sn < 2 for n = 1, 2, 3, . . . . . .

Solution. S1 =√

2 < 2. Assume that sn < 2 then sn+1 <√

2 + 2 = 2. Therefore by

mathematical induction sn < 2 for all n = 1, 2, 3, . . . . . . .

It is obvious that s1 =√

2 <√

2 +√s1 = s2. Now assume that s1 < s2 < . . . · · · < sn < 2

then,

sn =√

2 +√sn−1 <

√2 +√sn = sn+1

So by mathematical induction {sn} is an increasing sequence.

{sn} being increasing and bounded sequence of nonnegative terms it converges.

Problem 7.4. Find the upper and lower limits of the sequence {sn} defined by

s1 = 0, s2m =s2m−1

2, s2m+1 =

1

2+ s2m.

Solution. Construction: We compute some of the terms of the sequence

s1 = 0, s2 = 0, s3 =1

4, s5 =

3

4, s6 =

3

8, s7 =

7

8, s8 =

7

16.

49

We see the sequence have two subsequences corresponding to the even and odd numbered

terms

0,1

2,3

4,7

8,15

16, . . . · · · ≡ s2m−1 =

2m + 1

2m= 1− 1

2m

0,1

4,3

8,

7

16,15

32≡ s2m =

2m − 1

2m+1=

1

2− 1

2m

Solution: We shall prove by induction that

(7.1) s2m =1

2− 1

2m, s2m+1 = 1− 1

2m

for m = 1, 2, . . . . . . . Immediate computation shows that s2 = 0 and s3 = 12. Now assume

that the both formulas hold for m ≤ r. Then

s2r+2 =1

2s2r+1 =

1

2

(1− 1

2r

)=

1

2− 1

2r+1

Therefore the first of (7.1) relations is true for all m and to see the second we use the

given equation

s2m+1 =1

2+ s(2m) =

1

2+

1

2− 1

2m= 1− 1

2m

So the subsequence of even terms converges to 12

and the subsequence of the odd terms

converge to 1.

Now we want to show that there are no other sub-sequential limits. For any x other than

1 and 12, choose r = 1

2min{|x− 1| , |−1/2|}. Note that sn ∈ Br(x) for at most finitely

many n ∈ N. Thus 12

and 1 are the only subsequential limits of {sn}, so therefore,

lim supn→∞

sn = 1, lim infn→∞

sn =1

2.

Problem 7.5. For any tow real sequences {an} and {bn} prove that

lim supn→∞

(an + bn) ≤ lim supn→∞

an + lim supn→∞

bn

provided the sum of the right is not the form ∞−∞

Solution. Since the um of the right hand side is not of the form∞−∞, if one term is −∞then both of the term have to be bounded above by M1(say). Let M be any real number.

Without loss of generality assume that lim supn→∞

an = −∞ which implies that there is a

subsequence {ank} such that ank → −∞ that is for every M ∈ R there is an integer N1

such that nk ≥ N1 implies ank ≤ M −M1. Therefore ank + bnk ≤ M −M1 + M1 = M .

Which implies that ank + bnk → −∞ and so lim supn→∞

(an + bn) = −∞.

Now suppose one of the term in the right side is∞ then other term has to be bounded bel-

low by M1(say). Without loss of generality assume that lim supn→∞

an =∞. So, there exists

a subsequence {ank} of {an} such that ank →∞. Let M be any given real number. There

exists n ∈ N such that ank ≥M−M1 for all nk ≥ N . Hence ank+bnk ≥M−M1+M1 = M .

So there is a subsequence of ank + bnk →∞ and hence lim supn→∞

(an + bn) =∞.

Now assume that both terms in the right are finite. Suppose lim supn→∞

an = α and

50

lim supn→∞

bn = β. Fix any ε > 0. There exist natural numbers N1 and N2 such that

n ≥ N1 implies an < α + ε2

and n ≥ N2 implies bn < β + ε2. Let N := max{N1, N2}.

Then n ≥ N implies

an + bn < α + β + ε

Since ε > 0 is arbitrary, therefore,

an + bn ≤ α + β = lim supn→∞

an + lim supn→∞

bn

for all but finitely many n. Thus every sub-sequential limit of {an+bn} is bounded above

by lim supn→∞

an + lim supn→∞

bn, and in particular,

lim supn→∞

(an + bn) = lim supn→∞

an + lim supn→∞

bn

Problem 7.6. Investigate the behavior of∑an if

(a) an =√n+ 1−

√n;

(b) an =√n+1−

√n

n;

(c) an = ( n√n− 1)

n;

(d) an = 11+zn

, for complex values of z.

Solution. (a)

an =(√n+ 1−

√n)(√n+ 1 +

√n)

(√n+ 1 +

√n)

=1

(√n+ 1 +

√n)≥ 1

2√n+ 1

We know that the series∑

1√n

diverges and so 12

∑1√n+1

. By the comparison test

the given series∑an diverges.

(b)

|an| =1

n(√n+ 1 +

√n)≤ 1

n2√n≤ 1

n3/2

By comparison test∑an converges.

(c) We use the root test. n√an = n

√n − 1 and thus lim sup

n→∞n√an = 0 < 1. So by the

root test the series converges.

(d) If |z| ≤ 1, then |an| ≥ 12, so that an does not tend to zero and hence

∑an

diverges. If |z| > 1 then |an| =∣∣ 1

1+zn

∣∣ ≤ 1zn

= (1z)n. Note that 1

|z| < 1 and so∑(1z

)nconverges and by the comparison test

∑an converges.

Problem 7.7. Prove that the convergence of∑an implies the convergence of

∑ √ann

, if

an ≥ 0.

Solution. Since(√

an − 1n

)2 ≥ 0, it follows that∣∣∣∣√ann∣∣∣∣ ≤ ∣∣∣∣12

(an +

1

n2

)∣∣∣∣ =1

2

(an +

1

n2

)Since

∑an and

∑1n2 both converges we conclude by the comparison test that the series∑ √

ann

converges.

51

Problem 7.8. If∑an converges, and if {bn} is a monotonic and bounded, prove that∑

anbn converges.

Solution. Since {bn} is a bounded monotonic sequence it converges to b(say). Define,

cn =

b− bn if {bn}ismonotonically increasing.

bn − b if {bn}ismonotonically decreasing.

Then {cn} is monotonically decreasing sequence and lim cn = 0. Therefore∑ancn con-

verges.

If {bn} is monotonically decreasing, we have, for all N ∈ NN∑n=1

anbn =N∑n=1

ancn + b

N∑n=1

an.

Similarly if {bn} is increasing we have, for all N ∈ N,

N∑n=1

anbn =N∑n=1

ancn − bN∑n=1

an.

In both cases, both the term in the right side converge and hence∑anbn converges.

Problem 7.9. Suppose that the coefficients of the power series∑anz

n are integers,

infinitely many of which are distinct form zero. Prove that the radius of convergence is

at most 1.

Solution. Since cn ≥ 1 which implies n√cn ≥ 1 for all but finitely many n. So, α =

lim supn→∞

≥ 1 and therefore the radius of convergence R = 1α≤ 1

Problem 7.10. Suppose an > o, sn = a+ 1 + · · ·+ an, and∑an diverges,

(a) Prove that∑

an1+an

diverges.

(b) Prove that aN+1

sN+1+ . . . · · ·+ aN+k

sN+k≥ 1− sN

sN+kand hence deduce that

∑ans2n

converges.

(c) Prove that ans2n≤ 1

sn−1− 1

snand deduce that

∑ans2n

converges.

(d) What can be said about∑ an1 + nan

, and∑ an

1 + n2an

Solution. (a) We assume to the contrary that∑

an1+an

converges. That means thatan

1+an→ 0 which implies 1

1an

+1→ 0 which is possible only if 1

an→ ∞ ⇔ an → 0.

So there exists an integer N such that an < 1 for all n ≥ N . Then,

an −an

1 + an=

a2n

1 + an<

an1 + an

which implies

0 < an <an

1 + an.

So, by comparison test∑an converges which is a contradiction and hence

∑an

1+an

does not converge.52

(b) Since an > 0 we must have sN+k ≥ sN+i for all 1 ≤ i ≤ k and therefore

aN+1

sN+1

+ . . . · · ·+ aN+k

sN+k

≥ aN+1 + . . . . . . aN+k

sN+k

=sN+k − sNsN+k

=1− sNsN+k

Since sn does not converge for a fixed N we can choose k large enough so that

sNsN+k

≥ 12

and so the partial sumN+k∑k=N+1

aksk≥ 1

2which is true for any choice of N .

Therefore the sequence of partial sum of {ansn} does not form a Cauchy sequence

and so does not converge.

(c) We observe that for n ≥ 2,

1

sn−1

− 1

sn=sn − sn−1

snsn−1

=an

snsn−1

≥ ans2n

Also,∞∑n=2

(1

sn−1

− 1

sn

)=

1

s1

− 1

s∞=

1

a1

.

By the comparison test since∣∣∣∣ans2n

∣∣∣∣ ≤ 1

sn−1

− 1

sn

and the series of terms in the right side converges∑

ans2n

also converges.

(d) The series∑

an1+nan

may be either divergent or convergent.

If the sequence {nan} is bounded above that is if there exists M ∈ R such that

nan ≤M for all n then ∣∣∣∣ an1 + nan

∣∣∣∣ ≥ 1

1 +Man

Since∑an diverges, by comparison test

∑an

1+nanalso diverges.

If {nan} is bounded bellow that is if there exists ε > 0 such that an ≥ ε for all n,

then,

nan + εnan ≥ ε+ εnan ⇒nan

1 + nan≥ ε

1 + εwhich implies, ∣∣∣∣ an

1 + nan

∣∣∣∣ ≥ ε

1 + ε

1

n.

Again, by comparison test the series does not converge.

However the series∑

an1+nan

converges. For example let an = 1n2 if n is not a

perfect square and an = 1√n

if n is a perfect square. For a nonsquare term,

an1 + nan

=1

n2 + n

53

also for a square integer n,

an1 + nan

=1√

n+ n2

in both of the cases∣∣∣ an

1+nan

∣∣∣ ≤ 1n2 and hence by comparison test the series con-

verges.

For the second series, ∣∣∣∣ an1 + n2an

∣∣∣∣ =1

1an

+ n2≤ 1

n2.

Hence by comparison test, the series∑

an1+n2an

converges.

Problem 7.11. Suppose an > 0 and∑an converges. Put

rn =∞∑m=n

am.

(a) Prove thatamrm

+ . . . · · ·+ anrn

> 1− rnrm

if m < n, and deduce that∑

anrn

diverges.

(b) Prove thatan√rn

< 2(√rn −

√rn+1)

and deduce that∑

an√rn

converges.

Solution. (a) Obviously rm > rk for all k > m since an > 0. Therefore akrk> ak

rm.

amrm

+ . . . · · ·+ anrn

>am + · · ·+ an

rm=rm − rnrm

= 1− rnrm.

Since the series∑an converges and suppose

∑an = s, then the sequence of partial

sum sn converges to s. Fix an m ∈ N, there exists N such that |s− sn−1| < rm2

for all n > N . But

|s− sn−1| <rm2⇒ rn

rm<

1

2.

Therefore, for any m we can find an n such thatn∑

k=m

anrn> 1

2. Hence

∑anrn

is not

convergent.

(b)an√rn

(√rn +

√rn+1) = an + an

√rn+1√rn

< 2an = 2(rn − rn+1)

Therefore,an√rn

< 2(√rn −

√rn+1).

54

Since∑an convergent the sequence of partial sum sn → s and so rn → 0 as

n → ∞. Therefore∞∑n=1

(√rn −

√rn+1) =

√r1. Hence the series

∑an√rn

converges

by the comparison test.

Problem 7.12. Suppose {pn} is a Cauchy sequence in a metric space X, and some

subsequence {pn} converges to a point p ∈ X. Prove that the full sequence {pn} converges

to p.

7.1. Problems from other sources

Problem 7.13. Prove that every convergent sequence has a monotone subsequence.

Solution. Consider the convergent sequence an that converges to a. So there is an

integer N1 such that |an − a| < 1 for all n ≥ N1. Select the b1 = max{a1, . . . aN1−1}.By induction there is an integer Nj such that |an − a| < 1

jfor all n ≥ Nj. Select

bj = max{aNj−1, . . . . . . aNj−1

}. By construction {bn} is a decreasing subsequence of {an}.

Theorem 7.1 (Leibniz’s Rule for Alternating Series). If {an} is a monotonic decreasing

sequence with limit 0, then the alternating series∞∑n=1

(−1)n+1an converges. If S denotes

its sum and sn denotes its nth partial sum, we also have the inequalities

0 < (−1)n(S − sn) < an+1

Proof 2. The subsequence of partial sum s2k is increasing because

s2k+2 − s2k = a2k+1 − a2k+2 > 0.

Similarly the subsequence of partial sums s2k−1 is decreasing, because,

s2k+1 − s2k−1 = −a(2k) + a2k+1 < 0.

Observe that,

s2k = a1 − (a2 − a3)− . . . . . . (a2k−2 − a2k−1)− a2k < a1

and so the sequence {s2k} is also bounded above by a1. Therefore s2k converges. Now

s2k−1 = s2k + a2k

and it follows that

limk→∞

s2k−1 = limk→∞

s2k + limk→∞

ak = limk→∞

s2k.

Since {s2n} and {s2n−1} converges tot he same limit, sn converges.

Since s2k is increasing and s2k−1 is decreasing, we have

s2k < s2k+2 ≤ S, and S ≤ s2k+1 < s2k−1, ∀k ≥ 1.

We have,

0 < S − s2k ≤ s2k+1 − s2k = a2k+1 ⇒ |S − s2k| ≤ a2k+1

55

and,

0 < s2k−1 − S ≤ s2k−1 − s2k = a2k ⇒ |S − s2k−1| ≤ a2k.

Putting these last two equations together gives the result

|S − sn| ≤ an+1, ∀n ∈ N.

Problem 7.14. Give an example of a sequence of positive numbers so that an →∞, but

the series∞∑n=1

(−1)n+1an diverges.

Solution. Consider the sequence

an =

1n

if n is odd

0 if n is even

Obviously, ∑an =

∑ 1

nwhich does not converge.

Another example could be,

an =

1n

if n is odd

1n2 if n is even

56

8. Compactness

In this section we define compactness of a subset of a metric space. For subsets of real

number compactness is simply being closed and bounded. We start with definitions and

theorems.

Sequentially Compact: A subset K of a metric space (X, d) is sequentially compact if

every sequence in K has a convergent subsequence whose limit belongs to K. If K = X,

then X is said to be sequentially compact and every sequence in X has convergent sub-

sequence.

Theorem 8.1 (Heine-Borel Theorem). A subset of Rn is sequentially compact if and only

if it is closed and bounded.

Theorem 8.2 (BolzanoWeierstrass theorem). Every bounded sequence in Rn has a con-

vergent subsequence.

Definition 8.1. Let K be a subset of a topological space (X, T ). Let {Gα}α∈Λ denote

an arbitrary set of open sets in T . If K ⊂⋃α∈ΛGα, then we call {Gα}α∈Λ is an open

cover of A.

Definition 8.2 (Compact Set). If every open cover of K has a finite subcover, then K

is compact.

Definition 8.3 (ε− net). Let K be a subset of a metric space (X, d). For a fixed ε > 0,

we call {xα}α∈A ⊂ X an ε-net of K if the family of open balls {Bε(xα)}α∈A is an open

cover of A. If {xα} is finite, then we call {xα} a finite ε-net of A.

Definition 8.4. The subset K of a metric space (X, d) is totally bounded if it has a finite

ε-net for every ε > 0.

Theorem 8.3. A subset of a metric space is sequentially compact if and only if it is

complete and totally bounded.

Theorem 8.4. A subset of a metric space is compact if and only if it is sequentially

compact.

Lemma 1. A sequentially compact metric space is separable.

Definition 8.5. A subset K of a topological space (X, T ) is precompact if its closure is

compact.

Theorem 8.5. A closed subset of a compact space is compact.

Corollary 4. Every subset of a compact space is precompact.57

8.1. Problems

Problem 8.1. Let f : A ⊂ R → R have the property that for every x ∈ A there is an

ε > 0 such that

f(t) > ε if t ∈ (x− ε, x+ ε) ∩ A.

Show that if the set A is compact, then there is some c > 0 such that f(x) > c for all

x ∈ A.

Solution. Observe that⋃x∈A (Bεx(x)) covers A. Since A is compact there is a finite

subcover, say,⋃ni=1Bεi(xi). (Here we write εxi as εi for notational simplicity). In each

ball Bi = Bεi(xi) that for each 1 ≤ i ≤ n, we have, f(t) > εi if t ∈ (xi − εi, xi + εi).

Take c = min{εi}ni=1. Let x be any point in A. x ∈ Bi for some 1 ≤ i ≤ n. Therefore,

f(x) > εi ≥ c. So, we conclude that f(x) > c for all x ∈ A.

58

Problem 8.2. Use compactness to show that if a function f : R→ R is locally increasing

at every point in R, then f must be increasing. Here, locally increasing at a point x means

there exists δ > 0 such that f(s) < f(x) < f(t) whenever x− δ < s < x < t < x+ δ.

Solution.

59

Problem 8.3. Let (X, d) be a metric space. Let {K1, K2, . . . , Kn} be a set of compact

subsets of X. Prove that ∪ni=1Ki is a compact subset of X. Let {Kα} be a set of compact

subsets of X. Prove that ∩αKα is a compact subset of X.

Solution. Let G =⋃α∈ΛGα be a cover for Y =

⋃ni=1 Ki. Since Ki ⊂ Y , for each

1 ≤ i ≤ n, the set⋃α∈ΛGα ∩Ki is a cover for Ki. Since each Ki is compact there is a

finite subcover Gi =⋃kij=1Ki ∩ Gj of Ki for each i. Take Gf =

⋃ni=1Gi. We claim that

Gf is the finite subcover of Y corresponding to the cover G.

Let x ∈ Y , then x ∈ Ki for some 1 ≤ i ≤ n consequently x ∈ Gi and hence x ∈ Gf .

Therefore Y ⊂ Gf showing that Gf is a cover.

Since each of Gi is a finite subcover, Gf being a finite union of them is also finite collec-

tion of sets that covers Y .

Hence Y =⋃ni=1Ki is compact.

60

Problem 8.4. Let (X, d) be a metric space with the discrete metric d. What subsets of

X are compact?

Solution. Only the finite subsets in a discrete metric are compact. A finite subset of

discrete metric is complete and also totally bounded because for any ε > 0 the ε− net is

finite and hence compact.

Now we want to prove that a compact set A in discrete metric is finite. If not, let (xn)

be a sequence such that xn 6= xk for any n 6= k, infiniteness of the set A make it possible

to form such sequence. The sequence (xn) has no convergent subsequence since only con-

vergence subsequence in discrete metric is the constant sequence which is a contradiction.

Therefore we conclude that only a finite subset in a discrete metric is compact.

61

Problem 8.5. Describe an open cover of the open unit square {(x, y) : 0 < x < 1, 0 < y < 1}that has no finite subcover.

Solution. The open cover

G =⋃n∈N

B1− 1n+1

((0, 0))

has no finite subcover. For an instant suppose it has a finite subcoverGf =⋃n∈AB1− 1

n+1((0, 0)).

Since A is finite there is a maximum say m. Any point in R2, 1 − 1m+2

distance away

from the origin does not contained in Gf and hence Gf can not be a cover.

62

Problem 8.6. Let (xn) be a sequence in a metric space (X, d) that converges to a limit

x. Show the set {x1, x2, x3, · · · } is compact.

Solution.

63

Problem 8.7. Show that the product of two compact metric spaces with the product

metric is a compact metric space.

64

Problem 8.8. Let a, b ∈ R and a < b. Show that any closed ball in C([a, b]) is not com-

pact with the metric d(f, g) = supx∈[a,b] |f(x)− g(x)| for f, g ∈ C([a, b]). Hint: Sequential

compactness and compactness are the same on a metric space (theorem from lecture 2)

and there are some really nice examples of sequences with no convergent subsequence

within the unit ball centered at zero from which we can easily extend these examples to

any closed ball centered at any point (i.e., continuous function) by simply translating and

scaling these examples as necessary. The so-called “tent” functions of Example 2.13 are

one such sequence. The sequence {sin(2nπx/(b− a))} is another (easy) one (you can also

use {sin(nπx)}). Also, the sequence {([x− a]/b)n} works. None of these sequences con-

verge with respect to this metric even though you can find the pointwise limits. Can you

prove this for each one? What this is saying is that a sequence of continuous functions

taken from a closed and bounded subset of continuous functions on a compact space is

not enough to guarantee that this subset is itself compact. Additionally, we can show that

C([a, b]) is complete (lecture 4), so the issue here is that the space is not totally bounded.

Once we show that C([a, b]) is complete, then any of the sequence examples shows the

space is not totally bounded (why? see a specific theorem in lecture 2).

65

Problem 8.9. Prove that every compact metric space K has a countable base, and that

K is therefore separable. Hint: For every positive integer n, there are finitely many

neighborhood of radius 1/n whose union covers K.

66

Problem 8.10. Theorem 2.36 in Rudin

Prove that if {Kα} is a collection of compact subsets of a metric space X such that the

intersection of every finite subcollection of {Kα} is nonempty, then ∩Kα is nonempty.

(Hint: Try a proof by contradiction by assuming there is some member of {Kα}, say

K1, such that no point in K1 belongs to every Kα, then create an open cover of K1

using complements of Kα. Show that this becomes false (in R, for example) if the word

“compact” is replaced by either “closed” or by “bounded.”

Solution. Fix a member K1 of {Kα} and put Gα = Kcα. Assume that no point of

K1 belongs to every Kα. Then, there are finitely many indices α1, α2, . . . , αn such that

K1 ⊂ Gα1 ∪ . . . · · · ∪ Gαn . So for any x ∈ K1 we have x ∈ ∪ni=1Gαi and therefore

x /∈ ∩ni=1Gcαi

= ∩ni=1Kαi . But this means that

K1 ∩Kα1 ∩ · · · ∩Kαn

is empty, in contradiction to our hypothesis.

Consider Kα = [α,∞). Let Kα1 , . . . . . . , Kαn be a finite sub-collection and let β =

max(α1, . . . . . . αn) then [β,∞) ⊂ Kαi for all i = 1, . . . , n and hence the intersection

of any subcollection is not empty. But ∩Kα = ∅. If not let β ∈ ∩Kα, and let r > β where

r ∈ R. Observe that β /∈ Kr which is a contradiction. Hence ∩Kα is empty.

Consider the bounded Kn = (0, 1n). Any finite intersection Kn1 ∩ · · · ∩Knm is not empty

because it contains Km+1 where m = max{n1, . . . . . . nm}. Let 0 < x < 1 is in the in-

tersection ∩Kn. There is m such that 1m< x and hence Km does not contain x. Hence

∩Kn = ∅.

Corollary 5. If {Kn} is a sequence of nonempty compact sets such that Kn ⊃ Kn+1(n =

1, 2, 3 . . . ), then ∩∞1 Kn is not empty.

67

Problem 8.11. Let E be the set of all x ∈ [0, 1] whose decimal expansion contains only

the digits 4 and 7. Is E compact?

Solution. E is not countable:

Assume E is countable. Then we have an ordering of the elements of E: {xn}n∈N. Write

out the decimal expansion xn as dn1dn2 . . . . . . . Then let z have decimal expansion given

by zi = 4 if dii = 7 and zi = 7 if dii = 4. Then z ∈ E but z 6= zn for any n, contradiction.

E is not dense in [0, 1]:

0 ∈ [0, 1] however B0.1(0) ∩ E = ∅ because B[0.1](0) = (0, 0.1) and all of these have

decimal expansions beginning with a 0 and thus are not in E. Therefore 0 is not a limit

point of E so E is not dense in [0, 1].

E is compact:

First we claim that E is closed. If x 6= [0, 1] then let r = min{d(x, 0), d(x, 1)} > 0 then

Br(x) ∩ E = ∅. Now suppose x ∈ [0, 1] \ E. Then the decimal expansion of x has some

digit which is not a 4 or a 7. Let d1, d2, . . . . . . denote the decimal expansion of x. Let

k be the lowest index such that dk 6= 4 or 7. Now let r = 110k+2 . Then Br(x) only

contains elements which either have dk as their kth digit or they have 9 as their n + 1st

digit. Both of these conditions ensure that the element of Br(x) is not an element of E.

Therefore x is not a limit point of E. Therefore E contains all of its limit points and is

closed.

Since E ⊂ [0, 1], it is bounded. By Heine-Borel theorem E is compact.

E is not perfect because B0.01(0.4) does not contain any point of E other than 0.4 and

hence 0.4 is not a limit point.

68

Problem 8.12. Let X be a metric space in which every infinite subset has a limit point.

Prove that X is separable and compact. Hint: Once we have that X is separable, then

from the lecture 1 examples, we have the existence of a countable base, so any open cover

of X has a countable subcover. Now prove by contradiction assuming that the countable

subcover does not have a finite subcover.

69

Problem 8.13. (a) If X is a compact metric space and if {pn} is a Cauchy sequence

in X, then {pn} converges to some point of X

(b) In Rk every Cauchy sequence converges.

Solution. Supporting Materials

(a) If E is the closure of a set E in a metric space X, then

diam E = diam E.

(b) If Kn is a sequence of compact sets in X such that Kn ⊃ Kn+1 (n = 1, 2, . . . . . . )

and if

limn→∞

diam Kn = 0,

then⋂∞

1 Kn consists of exactly one point.

If {pn} is a sequence in X and if EN consists of the points pN , pN+!, pN+2, . . . . . . , it is

clear from the definition of Cauchy sequence and diameter of a set that {pn} is a Cauchy

sequence if and only if

limn→∞

diam EN = 0

Proof

(a) Let {pn} be a Cauchy sequence in the compact space X. For N = 1, 2, 3, . . . . . . ,

let EN be the set consisting of pN , pN+1, . . . . . . . Then

(8.1) limn→∞

diam EN = 0.

Being a closed subset of the compact space X, each EN is compact. Also EN ⊃EN+1, so that EN ⊂ EN+1.

Therefore there is a unique p ∈ X which lies in every EN .

Let ε > 0 be given. By (8.1) there is an integer N0 such that diam En < ε if

N ≥ N0. Since p ∈ EN , it follows that d(p, q) < ε for every q ∈ EN , hence for

every q ∈ EN . In other words, d(p, pn) < ε if n ≥ N0. Hence pn → p.

(b) Let {xn} be a Cauchy sequence in Rk. Define EN as above, with xi in place of pi.

For some N , diam EN < 1. The range of {xn} is the union of EN and the finite

set {x1, . . . . . . xN−1}. Hence {xn} is bounded. Since every bounded set of Rk has

compact closure in Rk the result follows from (a).

70

Problem 8.14. Let Q be a metric space with metric d(p, q) = |p− q|. Let E be the set

of all p ∈ Q such that 2 < p2 < 3. Show E is closed and bounded in Q but not compact.

Is E open in Q?

Solution. Let x ∈ Q \ E. We need to show that x is not a limit point of E. Since

x ∈ Q \ E either x2 ≤ 2 or x2 ≥ 3. Since ±√

2 and ±√

3 are not rational x2 6= 2 and

x2 6= 3 in Q. Considering the space R take,

r = min{d(x,√

2), d(x,√

3), d(x,−√

2), d(x,−√

3)} > 0.

Then Br(x) ⊂ Q \ E and therefore Br(x) ∩ E = ∅ and hence x is not a limit point of E.

This proves that E is closed.

Fix a point p ∈ E then d(x, p) ≤ 2√

3 for all x ∈ E and hence E is bounded.

Construct a sequence {xn} in E so that xn →√

2. Construction of such sequence is

possible since Q is dense in R. Since any subsequence of a convergent sequence conver-

gence to the same point, all the convergent subsequence converges in√

2 but√

2 /∈ E

and hence E is not sequentially compact and hence not compact.

Another way to show that E is not compact could be as follows:√

3 is a limit point

of E because any neighborhood of it contains a rational less than√

3 since Q is dense in

R. However√

3 /∈ E since it is irrational. Therefore E as a subset of R is not closed.

Then by the Heine-Borel theorem E is not compact as a subset of R. Then E is not

compact relative to Q ⊂ R by Rudin theorem 2.33.

Finally we prove that E is open. For any q ∈ E, either q ∈ (√

2,√

3) or (−√

3,−√

2).

Without loss of generality we assume q ∈ (√

2,√

3). So,√

2 < q <√

3 therefore

r = min{∣∣√3− q

∣∣ , ∣∣√2− q∣∣} > 0. Then Br(q) ⊂ E and hence q is an interior point.

Since q was arbitrary, E is open.

71

Problem 8.15. Rudin Theorem 2.33

Suppose K ⊂ Y ⊂ X. Then K is compact reltive to X if and only if K is compact

relative to Y .

Solution. Suppose K is compact relative to X, and let {Vα} be a collection of sets, open

relative to Y , such that K ⊂⋃α Vα. There are sets Gα open relative to X, such that

Vα = Y ∩Gα, for all α; and since K is compact relative to X, we have

(8.2) K ⊂n⋃i=1

Gαi

for some choice of finitely many indices α1, . . . , αn. Since K ⊂ Y , from (8.2) we have

(8.3) K ⊂n⋃i=1

Vαi .

This proves that K is compact relative to Y .

Conversely, suppose K is compact relative to Y , let {Gα} be a collection of open subsets

of X which covers K, and put Vα = Y ∩ Gα. Then (8.3) will hold for some choice of

α1, . . . , αn; and since Vα ⊂ Gα, (8.3) implies (8.2)

72

9. Continuity

Continuous Function: Let (X, dX) and (Y, dY ) be a metric spaces. Any function

f : X → Y is continuous at x0 ∈ X if for every ε > 0 there is a δ > 0 such that

dX(x, x0) < δ implies dY (f(x), f(x0)) < ε. The function f is continuous on X if it is

continuous at every point in X. If f is not continuous at x, then we say f is discontinuous

at x.

Continuity in Topological Space: Let (X, T ) and (Y,S) be topological spaces. A

function f : X → Y is continuous at x ∈ X if for each neighborhood W of f(x) there

exists a neighborhood V of x such that f(V ) ⊂ W . We say that f is continuous on X if

it is continuous at every x ∈ XSequential continuity in Topology: Let (X, T ) and (Y,S) be topological spaces. A

function f : X → Y is sequentially continuous at x ∈ X if for every sequence (xn) that

converges to x ∈ X, the sequence (f(xn)) converges to f(x) ∈ Y .

Continuity and Sequential Continuity are same:

Theorem Let (X, dX) and (Y, dY ) be metric spaces. A function f : X → Y is continuous

at x ∈ X if and only if it is sequentially continuous at x

Another way to state this theorem is same as Theorem 4.2 of Baby Rudin

Definition: Let (X, dX) and (Y, dY ) be metric spaces; suppose E ⊂ X, f maps E into

Y , and p is a limit point of E. We write f(x)→ q as x→ p, or

limx→p

f(x) = q

if there is a point q ∈ Y with the following property: For each ε > 0 there exists a δ > 0

such that dY (f(x), q) < ε for all points x ∈ E for which 0 < dX(d, p) < δ.

Theorem: Let (X, dx) and (Y, dY ) be metric spaces; suppose E ⊂ X, f maps E into Y ,

and p is a limit point of E. Then limx→p

f(x) = q if and only if limn→∞

f(pn) = q for every

sequence {pn} in E such that limn→∞

pn = p where pn 6= p.

Proof: Suppose limn→∞

f(x) = q holds. Now choose any sequence {pn} so that pn 6= p

that is {pn} is not a constant sequence and pn → p as n → ∞. We want to show that

limn→∞

f(pn) = q.

Let ε > 0 be given. Since limx→p

f(x) = q, there exists a δ > 0 such that dY (f(x), q) < ε

if x ∈ E and 0 < dX(x, p) < δ. Also since pn → p, there exists an N ∈ N such that

dX(pn, p) < δ for all n > N . Thus for all n > N , we have dY (f(pn), q) < ε, which implies

limn→∞

f(pn) = q.

Conversely suppose limn→∞

f(fn) = q for every sequence {pn} in E such that limn→∞

pn = p

where pn 6= p but limx→p

f(x) 6= q. Then there exists some ε such that for every δ > 0 there

exists a point x ∈ E (depending on δ), for which dY (f(x), q) ≥ ε but 0 < dX(x, p) < δ.

Let δn = 1n. Take a point p1 ∈ E such that 0 < dX(p1, p) < δ1 but dY (f(p1), q) > ε1 and

similarly choose p2 ∈ E such that 0 < d(p2, p) < δ but dY (f(p1), q) > ε2. We construct a

sequence in this way. Note that we guarantee existence of such points pj because p is a

73

limit point of E. Now consider the sequence {pn} and note that pn → p by construction

but d(f(pn), q) > εn for all n which is a contradiction to the fact that limn→∞

f(pn) = q for

all {pn}.Theorem: Let (X, T ) and (Y,S) be two topological spaces and f : X → Y . Then f is

continuous on X if and only if f−1(G) ∈ T for every G ∈ S.

Corollary: A function f is continuous if and only if the inverse image of closed sets is

closed.

Uniform Continuity: Let (X, dX) and (Y, dY ) be metric spaces. A function f : X → Y

is uniformly continuous on X if for every ε > 0 there exists δ > 0 such that, dX(x, y) < δ

implies dY (f(x), f(y)) < ε for all x, y ∈ X.

9.1. Completing a Metric Space with an Isometry

Definition: Let (X, dX) and (Y, dY ) be metric spaces. A map ` : X → Y which satisfies

dY (`(x1), `(x2)) = dX(x1, x2), ∀x1, x2 ∈ X

is called an isometry or isometric embedding of X into Y . An isometry which is onto is

called a metric space isomorphism. We say that X and Y are isometric if there exists an

isomorphism ` : X → Y .

Definition: A metric space (X, d) is the completion of (X, d) if the following conditions

are satisfied:

(a) there exists an isometric embedding ` : X → X;

(b) the image space `(X) is dense in X;

(1) the space (X, d) is complete.

Theorem: Every metric space has a completion and the completion is unique up to

isomorphism

9.2. Continuous Function on Compact Sets

Theorem: Let f : K → Y be continuous on K, where K is a compact metric space and

Y is any metric space. Then f(K) is compact.

Theorem: Let f : K → Y be a continuous function on a compact set K. Then f is

uniformly continuous.

Theorem: Let K be a compact metric space and f : K → R a continuous, real-valued

function. Then f is bounded on K and attains its maximum and minimum. That is,

there are points x, y ∈ K such that

f(x) = infz∈ Kf(z), f(y) = sup

z∈Kf(z)

74

Theorem: A function f : X → R on a metric space (X, d) is Lipschitz continuous on X

if there is a constant C ≥ 0 such that

|f(x)− f(y)| ≤ Cd(x, y), ∀x, y ∈ X

75

9.3. Proofs of the Theorems

Theorem: Let f : K → Y be continuous on K, where K is a compact metric space and

Y is any metric space. Then f(K) is compact.

Proof: Consider any open cover {Gα}α∈Λ ∈ Y of f(K). Since inverse image of a open

set is open under continuous map f−1(Gα) is open in K for all α ∈ Λ. For any x ∈ K,

f(x) ∈ Gα for some α ∈ Λ and so x ∈ f−1(Gα). Therefore, {f−1(Gα)}α∈Λ is an open

cover of K. Since K is compact, there exists a finite subcover, which we denote by

{f−1(Gi)}ni=1. We want to show that {Gi}ni=1 is a finite subcover of f(K). Take y ∈ f(K)

be any arbitrary point and so there is x ∈ K such that f(x) = y and let x ∈ f−1(Gi)

for any 1 ≤ i ≤ n and therefore y = f(x) ∈ Gi. Since y was taken to be arbitrary we

conclude that f(K) ⊆⋃ni=1 Gi and hence Gα has a finite subcover. Therefore f(K) is

compact.

Theorem: Let f : K → Y be a continuous function on a compact set K. Then f is

uniformly continuous.

Proof: Suppose f is not uniformly continuous. Then there is an ε > 0 such that for all

δ > 0, there are x, y ∈ K with dK(x, y) < δ and dY (f(x), f(y)) ≥ ε. Taking δ = 1n

for

n ∈ N, we find that there are sequence (xn) and (yn) in K such that

dK(xn, yn) <1

n, dY (f(xn), f(yn)) ≥ ε

Since K is compact there are convergent subsequences of (xn) and (yn) which we denote

by (xni) and (ynj). Since dK(xn, yn) < 1n

the subsequences converges to the same limit

(taking n → ∞, d(x, y) → 0). But the sequence (f(xn)) and (f(yn)) either diverge or

converge to different limits. This contradicts the continuity of f .

76

9.4. Problems:

Problem 9.1. Let X and Y be metric spaces and A ⊂ X. Suppose f maps A into

Y and p is a limit point of A. We say that f has a limit at p and write f(x) → q as

x → p, or limx→p f(x) = q if there is a point q ∈ Y with the property that for every

ε > 0 there exists a δ > 0 such that for all points x ∈ E with 0 < dX(x, p) < δ implies

dY (f(x), q) < ε. Prove the following theorems.

(a) limx→p f(x) = q if and only if limn→∞ f(pn) = q for every sequence (pn) ⊂ E such

that pn 6= p for all n and pn → p.

(b) If f has a limit at p, then the limit is unique.

(c) The function f is continuous at p if and only if limx→p f(x) = f(p).

Solution. Solution:(a) First assume that limx→p

f(x) = q holds. Now choose any sequence

{pn} so that pn 6= p that is {pn} is not a constant sequence and pn → p as n → ∞. We

want to show that limn→∞

f(pn) = q.

Let ε > 0 be given. Since limx→p

f(x) = q, there exists a δ > 0 such that dY (f(x), q) < ε

if x ∈ E and 0 < dX(x, p) < δ. Also since pn → p, there exists an N ∈ N such that

dX(pn, p) < δ for all n > N . Thus for all n > N , we have dY (f(pn), q) < ε, which implies

limn→∞

f(pn) = q.

Conversely suppose limn→∞

f(fn) = q for every sequence {pn} in E such that limn→∞

pn = p

where pn 6= p but limx→p

f(x) 6= q. Then there exists some ε such that for every δ > 0 there

exists a point x ∈ E (depending on δ), for which dY (f(x), q) ≥ ε but 0 < dX(x, p) < δ.

Let δn = 1n. Take a point p1 ∈ E such that 0 < dX(p1, p) < δ1 but dY (f(p1), q) > ε1 and

similarly choose p2 ∈ E such that 0 < d(p2, p) < δ but dY (f(p1), q) > ε2. We construct a

sequence in this way. Note that we guarantee existence of such points pj because p is a

limit point of E. Now consider the sequence {pn} and note that pn → p by construction

but d(f(pn), q) > εn for all n which is a contradiction to the fact that limn→∞

f(pn) = q for

all {pn}.

Solution (b): Since f has a limit at p is same as saying every sequence {pn} in X that

converges to p the sequence {f(pn)} converges to some point q ∈ Y . From elementary

analysis we know that limit of a convergent sequence is unique and hence limit of {f(pn)}is unique and so limit of f at p is unique.

Solution (c): limx→p

f(x) = f(p) holds if and only if for any ε > 0 there exists a δ > 0 such

that dX(x, p) < δ implies dY (f(x), f(p)) < ε and this implies continuity of f at p ∈ E.

77

Problem 9.2. Let f : R2 → R be defined

f(x1, x2) =

x21x2x41+x22

, (x1, x2) 6= (0, 0),

0, (x1, x2) = (0, 0).

Show that limx→0 f(x,mx) = 0 for every m ∈ R but f is discontinuous at (0, 0). Interpret

this result.

Solution. If x 6= 0 then f(x,mx) = mx3

x4+m2x2= mx

x2+m2 . If m = 0 then f(x,mx)=0. If

m 6= 0 then x2 +m2 ≥ m2 ∀ x, but mx→ 0 as x→ 0, which shows

limx→0

f(x,mx) = 0, ∀m ∈ R

However f(x,mx2) tells a different story. Assume m 6= 0 and x 6= 0, then (x,mx2) →(0, 0) as x→ 0 but

f(x,mx2) =m

1 +m2

and therefore

limx→0

f(x,mx2) 6= 0.

Hence f is not continuous at (0, 0)

78

Problem 9.3. Let C1([a, b]) be the metric space of continuously differentiable functions

on [a, b] with metric

d(f, g) = supa≤x≤b

|f(x)− g(x)|+ supa≤x≤b

|f ′(x)− g′(x)| .

Let D : C1([a, b])→ C([a, b]) be defined by D(f) = f ′. Prove that D is continuous.

Solution. Let ε > 0 be given. Fix a function f ∈ C1([a, b]).

d0(D(f), D(g)) =d0(f ′, g′)

= supa≤x≤b

|f ′(x)− g′(x)|

≤ supa≤x≤b


|f ′(x)− g′(x)| = d(f, g)

Therefore taking δ = ε we can conclude that, for given ε > 0, there exists a δ > 0 such

that d(f, g) < δ implies d0(D(f), D(g)) < ε. Since f was arbitrary it infers that D is

continuous.

79

Problem 9.4. Let C([a, b]) be the metric space of continuous functions on [a, b] with

metric

d(f, g) =

∫ b

a

|f(t)− g(t)| dt.

Define T : C([a, b])→ R by

T (f) =

∫ b

a

f(t) dt.

(a) Prove that T is continuous using sequential definition of continuity.

(b) Prove that T is continuous using δ − ε definition.

Solution. Solution (a): Let {fn} be any sequence in C([a, b]) so that fn → f and

fn 6= f for all n. Since fn → f , for given ε > 0 there is N ∈ N such that d(fn, f) < ε for

all n > N . Observe that, for all n > N ,

|T (fn)− T (f)| =∣∣∣∣∫ b

a

fn(t)− f(t)

∣∣∣∣ ≤ ∫ b

a

|fn(t)− f(t)| = d(fn, f) < ε

Which implies that the sequence T (fn) converges to T (f) and by definition, T is sequen-

tially continuous and hence continuous by Theorem 1.

Solution (b): Let f be any continuous function in C([a, b]).Let ε > 0 be given. Since

|T (f), T (g)| ≤ d(f, g) taking δ = ε we can conclude that d(f, g) < δ implies |T (f), T (g)| <ε. So, T is continuous at f ∈ C([a, b]). Since f is arbitrary we conclude that T is contin-

uous in C([a, b])

80

Problem 9.5. Let f and g be continuous maps of a metric space X into a metric space

Y and let E be a dense subset of X.

(a) Prove that f(E) is dense in f(X).

(b) If g(x) = f(x) for all x in a dense subset E of X, prove that g(x) = f(x) for all

x ∈ X.

Solution. Solution (a): Take any y ∈ f(X). For any ε > 0 the ball Bε(y) is open in

Y and f−1(Bε(y)) is open in X since f is continuous in X. Since E is dense in X there

exists x ∈ f−1(Bε(y)) ∩ E. So f(x) ∈ Bε(y). So f(E) ∩ Bε(y) 6= ∅ and contains a point

of f(E), for any y ∈ f(X) and hence f(E) is dense in f(X).

Alternative: Lemma: For all A ⊂ X, f(A) ⊂ f(A)

Proof of Lemma: We only need to show that f(A′) ⊂ f(A). Consider x ∈ A′ and we want

to show that f(x) ∈ f(A). Consider any neighborhood W of f(x). Since f is continuous

there exists a neighborhood of V of x in X such that V ⊂ f−1(W ). Since x is a limit

point of A there is a point y ∈ A ∩ V such that y 6= x which implies f(y) ∈ W ∩ f(A).

Since W was arbitrary we can conclude that every neighborhood of f(x) contains a point

of f(A) and hence f(x) ∈ f(A). Its prove the lemma.

Since E is dense in X, we have E = X and hence f(X) = f(E) ⊂ f(E) showing that

f(E) is dense in f(X).

Solution(b): Consider the function h : X → R defined by h(x) = dY (f(x), g(x)). First

we show that h is continuous. Let x0 ∈ X be any point and ε > 0 be given,

|h(x)− h(x0)| = |dY (f(x), g(x))− dY (f(x0), g(x0))|

= |dY (f(x), g(x)) + dY (f(x0), g(x))− dY (f(x0), g(x))− dY (f(x0), g(x0))|

≤ |dY (f(x), g(x)) + dY (f(x0), g(x))|+ |dY (f(x0), g(x)) + dY (f(x0), g(x0))|

≤ |dY (f(x), f(x0))|+ |dY (g(x), g(x0))|

Since f : X → Y is continuous, for given ε there exists a δ1 > 0 such that dX(x, x0) < δ1

implies dY (f(x), f(x0)) < ε2. Again since g : X → Y is continuous there exists a δ2 > 0

such that dX(x, x0) < δ2 implies dY (g(x), g(x0)) < ε2. Therefore, taking δ = min(δ1, δ2)

we get, d(x, x0) < δ implies

|h(x)− h(x0)| < ε

2+ε

2= ε.

Therefore h is continuous.

Since f(x) = g(x) in E we have h(x) = 0 in E. Take any point y ∈ X \ E. For any

δ > 0, Bδ(y) contains a point z ∈ E since E is dense in X. Assume that h(y) 6= 0 that

is h(y) ≥ εy. So, ∀δ we have |h(y) − h(z)| > εy which contradicts the fact that h is

continuous. So h(y) = 0 and since y was arbitrary we can conclude that h(x) = 0 for all

x ∈ X and therefore f(x) = g(x) for all x ∈ X.81

Problem 9.6. Let A be a set and x a point in a metric space (X, d). Define the “distance

from x to A” to be

d(x,A) = infy∈A

d(x, y).

(a) Show that for fixed A, the function f : X → R (with the usual metric on R) given by

f(x) = d(x,A) is continuous. (b) Show {x : d(x,A) = 0} = A.

82

Problem 9.7. Suppose f is a mapping from metric space X to metric space Y . Prove

that a function is continuous if and only if f−1(A) is closed in X for every closed set

A ⊂ Y .

Solution. First assume that f is continuous and we want to show that if A is closed

then f−1(A) is closed. Since A is closed Ac is open and since f is continuous, f−1(Ac) is

open. But

f−1(Ac) = {x ∈ X|f(x) ∈ Ac} = {x ∈ X|f(x) /∈ A} = {x ∈ X|x /∈ f−1(A)} =(f−1(A)

)cshows that (f−1(A))

cis open and hence f−1(A) is closed.

Now we take that f−1(A) is closed whenever A is closed in Y and want to show that

f is continuous. Let B be any open set in Y . So Bc is closed. By the hypothesis f−1(Bc)

is closed. But we have shown that f−1(Bc) = (f−1(B))c

and hence (f−1(B))c

is closed

which implies f−1(B) is open. Since B was arbitrary we conclude that for every open set

B the preimage f−1(B) is open and hence f is continuous.

83

Problem 9.8. Assume f : R → R is continuous and define A = {x : f(x) = 0}. Show

A is closed.

Solution.

A = {x : f(x) = 0} = f−1 ({0})

But the set {0} being a singleton set is closed and by the above problem A = f−1 ({0})is closed.

84

Problem 9.9. Suppose f is a continuous bijection of a compact metric space X onto a

metric space Y . Prove that the inverse map f−1 defined on Y by f−1(f(x)) = x (where

x ∈ X) is a continuous mapping of Y onto X.

Solution. Let A ⊂ X be any closed subset. Since X is compact being a closed subset

A is compact. Since f is continuous it maps a compact set to a compact set and hence

f(A) is compact and therefore is closed. But(f−1)−1

(A) = {y ∈ Y : f−1(y) ∈ A} = {y ∈ Y : y ∈ f(A)} = f(A)

Since A is arbitrary we conclude that the inverse image of a closed set A under the

function f−1 is closed and hence f−1 is continuous.

85

Problem 9.10. Let (X, d) be a metric space and F,G ⊂ X with F closed, G open,

and F ⊂ G. Show that there is a continuous function f : X → R such that 0 ≤f(x) ≤ 1, f(x) = 1 for x ∈ F , and f(x) = 0 for x ∈ Gc. Hint: Consider the function

f(x) = d(x,Gc)/[d(x,Gc) + d(x, F )]. This result is called Urysohn’s lemma and is useful

in showing the existence of partitions of unity.

Solution. First we would see the construction. f(x) = 1 for x ∈ F and f(x) = 0 for

x ∈ Gc. So we can see this as a parallel line to x axis over F and the x axis itself over Gc.

To make it continuous we have to draw a line form (Gc, 0) (thinking Gc to be the corner

point and) to (F, 1). The equation of the line would look like y = x−GcF−Gc = d(x,Gc)

d(x,Gc)+d(x,F ).

Take

f(x) =d(x,Gc)

d(x,Gc) + d(x, F )

For all x ∈ F , d(x, F ) = 0 and we claim that d(x,Gc) > 0. Since x ∈ F ⊂ G and G

is open, there is ε such that Bε(x) ⊂ G and so d(x, y) ≥ ε for all y ∈ Gc and hence

d(x,Gx) ≥ ε > 0. Therefore we have

f(x) =d(x,Gc)

d(x,Gc) + d(x, F )=d(x,Gc)

d(x,Gc)= 1, ∀x ∈ F

Now take x ∈ Gc. For any y ∈ F since y ∈ G and G is open using same argument above

we conclude that d(x, y) ≥ ε and hence d(x, F ) ≥ ε > 0. Therefore,

f(x) =0

d(x, F )= 0

Now we want to show that f is continuous. (The idea is to use the fact that distance

metric is continuous and composition is also continuous but have to fill up with details

to show the denominator is not equal to zero)

86

Problem 9.11. Let f and g be continuous maps of a metric space X into a metric space

Y and let E be a dense subset of X. (a) Prove that f(E) is dense in f(X). (b) If

g(x) = f(x) for all x in a dense subset E of X, prove that g(x) = f(x) for all x ∈ X.

Explain why these results are important for our everyday experience with functions and

numbers.

87

Problem 9.12. (a) Show that the requirement in the definition of uniform continuity

can be rephrased as follows, in terms of diameters of sets: For every ε > 0 there exists a

δ > 0 such that diam(f(E)) < ε for all E ⊂ X with diam(E) < δ. (b) Let E be a dense

subset of a metric space X, and let f be a uniformly continuous function into a complete

metric space Y defined on E. Prove that f has a continuous extension from E to X in

two ways using part (a) and then using Cauchy sequences.

88

Problem 9.13. Let f : R → R be a continuous function that is periodic with period

T > 0. (a) Show that f has a finite maximum and minimum values on R. (b) Show that

f is uniformly continuous on R.

89

Problem 9.14. Let X be a metric space.

(a) Call two Cauchy sequences (pn), (qn) in X equivalent if

limn→∞

d(pn, qn) = 0.

Prove that this is an equivalence relation.

(b) Let X∗ be the set of all equivalence class so obtained. If P ∈ X and Q ∈ X, (pn) ∈ Pand (qn) ∈ Q, define

∆(P,Q) = limn→∞

d(pn, qn).

Note that by an exercise from lecture 1 that this limit exists. Show that the number

∆(P,Q) is unchanged if (pn) and (qn) are replaced by equivalent sequences, and hence

that ∆ is a distance function in X.

(c) Prove that the resulting metric space X is complete.

(d) For each p ∈ X, there is a Cauchy sequence all of whose terms are p; let Pp be the

element of X which contains this sequence. Prove that ∆(Pp, Pq) = d(p, q) for all

p, q ∈ X. In other words, the mapping φ defined by φ(p) = Pp is an isometry of X

into X.

(e) Prove that φ(X) is dense in X, and that φ(X) = X if X is complete. By part (d),

we may identify X and φ(X) and thus regard X as embedded in the complete space

X. We call X the completion of X.

90

Problem 9.15. Let (X, d) be a metric space and suppose f is uniformly continuous on

the compact subsets E1, . . . , En of X. (a) Prove that f is uniformly continuous on ∪ni=1Ei.

(b) Give an example of a function f that is continuous on R and a sequence of compact

intervals E1, E2, E3, . . . on each of which f is uniformly continuous, yet f is not uniformly

continuous on ∪∞i=1Ei.

91

Problem 9.16. Prove that if f is a continuous mapping of a metric space X into a

metric space Y and E ⊂ X is connected, then f(E) is connected.

92

10. Differentiation of Single Variable

Definition 10.1. Let f be defined on [a, b]. For any x ∈ [a, b] form the quotient

φ(t) =f(t)− f(x)

t− x(a < t < b, t 6= x)

and define

f ′(x) = limt→x

φ(t),

provided this limit exists.

We thus associate with this function f a function f ′ whose domain is the set of points x

at which the limit exists; f ′ is called the derivative of f .

If f ′ is defined as a point x, we say that f is differentiable at x. If f ′ is defined at every

point of a set E ⊂ [a, b], we say that f is differentiable on E.

Theorem 10.1. Let f is defined on [a, b]. If f is differentiable at a point x ∈ [a, b], then

f is continuous at x.

Theorem 10.2. Suppose f and g are defined on [a, b] and are differentiable at a point

x ∈ [a, b]. Then f + g, fg and f/g are differentiable at x, and

(a) (f + g)′(x) = f ′(x) + g′(x);

(b) (fg)′(x) = f ′(x)g(x) + f(x)g′(x);

(c)(fg

)′(x) = g(x)f ′(x)−g′(x)f(x)

g2(x)

In (c) we assume of course that g(x) 6= 0.

Theorem 10.3 (Chain Rule). Suppose f is continuous on [a, b], f ′(x) exists at some

point x ∈ [a, b], g is defined on an interval I which contains the range of f , and g is

differentiable at the point f(x). If

h(t) = g (f(t)) (a ≤ t ≤ b),

then h is differentiable at x, and

h′(x) = g′(f(x))f ′(x)

Definition 10.2. Let f be a real function defined on a metric space X. We say that f

has a local maximum at a point p ∈ X if there exists δ > 0 such that f(q) ≤ f(p) for all

q ∈ X with d(p, q) < δ

Theorem 10.4. Let f be defined on [a, b]; if f has a local maximum at a point x ∈ (a, b),

and if f ′(x) exists, then f ′(x) = 0.

Theorem 10.5 (Generalized Mean Value Theorem). If f and g are continuous real

functions on [a, b] which are differentiable in (a, b), then there is a point x ∈ (a, b) at

which

[f(b)− f(a)]g′(x) = [g(b)− g(a)]f ′(x).93

Theorem 10.6. If f is a real continuous function on [a, b] which is differentiable in

(a, b), then there is a point x ∈ (a, b) at which

f(b)− f(a) = (b− a)f ′(x)

Theorem 10.7. Suppose f is differentiable in (a, b).

(a) If f ′(x)] ≥ 0 for all x ∈ (a, b) , then f is monotonically increasing.

(b) If f ′(x) = 0 for all x ∈ (a, b), then f is constant.

(c) If f ′(x)] ≤ 0 for all x ∈ (a, b) , then f is monotonically decreasing.

Theorem 10.8 (Similar to IVT). Suppose f is a real differentiable function on [a, b] and

suppose f ′(a) < λ < f ′(b). Then there is a point x ∈ (a, b) such that f ′(x) = λ

Theorem 10.9 (L’Hospital’s Rule). Suppose f and g are real and differentiable in (a, b),

and g′(x) 6= 0 for all x ∈ (a, b), where −∞ ≤ a ≤ b ≤ +∞. Suppose

f ′(x)

g′(x)→ A, as x→ a

If f(x)→ 0 and g(x)→ 0 as x→ a or if g(x)→ +∞ as x→ a, then

f(x)

g(x)→ A as x→ a.

The consequence of the mean value theorem which is also valid for vector valued func-

tion is as follows:

Theorem 10.10. Suppose f is a continuous mapping of [a, b] into Rk and f is differen-

tiable in (a, b). Then there exists x ∈ (a, b) such that

|f(b)− f(a)| ≤ (b− a) |f ′(x)| .

10.1. Some Problems

Problem 10.1. content...

94

11. Sequences and Spaces of Continuous Functions

Definition 11.1. A sequence of bounded, real-valued functions (fn) on a metric space

(X, d) converges uniformly to a function f if

limn→∞

‖fn − f‖∞ = 0.

Another representation can be

limn→∞

supx∈X|fn(x)− f(x)| = 0.

Theorem 11.1. Let (fn) be a sequence of bounded, continuous, real-valued functions on

a metric space (X, d). If fn → f uniformly, then f is continuous.

11.1. Space of Continuous Functions

Definition 11.2. Let (X, d) be a metric space. We define the set of continuous, real-

valued functions f : X → R by C(X). We let Cb(X) denote the space of bounded contin-

uous functions on X.

Theorem 11.2. Let K be a compact subset of a metric space (X, d). The space C(K) is

complete.

Definition 11.3. The support, supp(f) of a function f : X → R on a metric space (X, d)

is the closure of the set on which f is nonzero,

supp(f) = {x ∈ X : f(x) 6= 0}.

We say that f has compact support if supp(f) is a compact subset of X, and denote

the space of continuous functions on X with compact support by Cc(X). We denote the

closure of Cc(X) by C0(X).

Theorem 11.3 (Weierstrass Approximation Theorem). Let a < b be real numbers. The

set of polynomials on [a, b] is dense in C([a, b]).

11.2. Compact Subset of Continuous Function Space

Definition 11.4 (Equicontinuity). Let F be a family of functions from a metric space

(X, dX) to a metric space (Y, dY ). The family F is equicontinuous if for every x ∈ X and

ε > 0 there is a δ > 0 such that d(x, y) < δ implies d(f(x), f(y)) < ε for all f ∈ F . If δ

can be chosen independently of x, then F is called uniformly equicontinuous.

Theorem 11.4. An equicontinuous family of functions from a compact metric space to

a metric space is uniformly equicontinuous.95

Theorem 11.5 (Arzela-Ascoli). Let K be a compact subset of metric space (X, d). A

subset of C(K) is compact if and only if it is closed, bounded, and equicontinuous.

Theorem 11.6. If X is a compact metric space and (fn) is a sequence of continuous

functions from X to metric space Y that converges uniformly to f , then {fn} is equicon-

tinuous on X.

96

11.3. Proof of the Theorems:

Theorem: Let (fn) be a sequence of bounded, continuous, real-valued functions on a

metric space (X, d). If fn → f uniformly, then f is continuous.

Proof: Let ε > 0 be given. Since fn → f uniformly, there exists N ∈ N such

that |fn(x)− f(x)| < ε3

for all n ≥ N and for all x ∈ X. Let x0 be an arbitrary

point in X. Since fN is continuous, there exists δ > 0 so that d(x, x0) < δ implies

|fN(x)− fN(x0)| < ε3. Therefore, d(x, x0) ≥ δ implies,

|f(x)− f(x0)| ≥ |f(x)− fN(x)|+ |fN(x)− fN(x0)|+ |fN(x0)− f(x)| < ε

3+ε

3+ε

3= ε

Therefore f is continuous in X.

Theorem: An equicontinuous family of functions from a compact metric space to a

metric space is uniformly equicontinuous.

Proof: Suppose that K is a compact metric space, and F is a family of functions

f : K → Y that is not uniformly equicontinuous. We will prove that F is not equicon-

tinuous.

Since F is not uniformly equicontinuous, there is an ε > 0, such that for every n ∈ Nthere are points xn, yn ∈ K and a function fn ∈ F with

(11.1) d(xn, yn) <1

n, and d(fn(yn), fn(xn)) ≥ 2ε.

Since K is compact, the sequence (xn) has a convergent subsequence, which are also

denoted by (xn). Suppose that xn → x as n→∞. Then d(xn, yn) < 1n

implies yn → x as

well. Hence, for all δ > 0, there are points xn, yn such that d(xn, x) < δ and d(yn, x) < δ.

But from (11.1), we must have either d(fn(xn), fn(x)) ≥ ε or d(fn(yn), fn(x)) ≥ ε, so Fis not equicontinuous at x.

97

11.4. Problems on Sequence of Functions

Problem 11.1. Prove the sequence ( xn+x2

) converges uniformly on R with the usual

metric.

Solution. Fix x ∈ R then limn→∞

xn+x2

= 0 and so fn = xn+x2

converges pointwise to 0.

To see uniform convergence we let

Mn = supx∈R

∣∣∣∣ x

n+ x2

∣∣∣∣Let g(x) = x

n+x2then g′(x) = n−x2

(n+x2)2which gives the critical value x =

√n which gives

the maximum value 12√n. So, ∣∣∣∣ x

n+ x2

∣∣∣∣ ≤ 1

2√n

For given ε we can find N ∈ N such that 12√N< ε by Archimedean property to conclude

that n ≥ N implies∣∣ xn+x2

∣∣ < ε for all x ∈ R. Hence the given sequence is uniformly

convergence.

98

Problem 11.2. (a) Define what it means for a sequence of functions to be uniformly

Cauchy and prove that a uniformly convergent sequence of functions is uniformly Cauchy.

(b) Prove that every uniformly convergent sequence of bounded functions is uniformly

bounded. Here, we say that a sequence of real-valued functions (fn) is uniformly bounded

if there exists M > 0 such that ‖fn‖∞ ≤M . (c) Show there exists a sequence of functions

that converges pointwise to the function that is identically 0, but each function in the

sequence is unbounded Hint: Consider the sequence {x/n} on [0,∞).

Solution. (a) A sequence of function (fn(x)) in a subset A of a metric space (X, d)

is called uniformly Cauchy if for any ε > 0 there exists N ∈ N such that n,m ≥ N

implies d(fn(x), fm(x)) < ε for all x ∈ A.

The another way of defining the uniform Cauchy criteria is using the natural

norm i.e. the infinity norm. A sequence of real valued function (fn) on a metric

space (X, d) is said to be uniformly Cauchy Let the sequence (fn) of functions

converges to f uniformly. Let ε > 0 be given. There exists an N ∈ N such that

d(fn(x), f(x)) < ε2

for all x ∈ A whenever n ≥ N . Now take n,m ≥ N , then

d(fn(x), fm(x)) ≤ d(fn(x), f(x)) + d(fm(x), f(x)) <ε

2+ε

2= ε

for all x ∈ A. Therefore (fn) is uniformly Cauchy.

(b) Since (fn) is a sequence of bounded functions for each n there is Mn such that

‖fn‖∞ ≤M . Since the sequence is uniformly convergence it is uniformly Cauchy

and so for ε = 1 there is N ∈ N such that d(fm(x), fN(x)) < 1 for all m ≥ N .

Take M = 1 + max{M1, . . . . . .MN}.

99

Problem 11.3.

(a) Prove that if the sequences of real-valued functions (fn) and (gn) converge uniformly

on A ⊂ X, then (fn+gn) converges uniformly on A. (b) If fn and gn are also bounded for

each n ∈ N, prove that (fngn) converges uniformly on A. (c) Show that the boundedness

is necessary in (b) by constructing an example where (fngn) does not converge uniformly

(although it must converge pointwise).

Solution. Let ε > 0 be given. Since (fn) converges uniformly to f(say), there exists an

N1 ∈ N such that |fn(x)− f(x)| < ε2

for all n ≥ N1 and for all x ∈ A.

Also, since (gn) converges uniformly to g(say), there exists anN2 ∈ N such that |gn(x)− g(x)| <ε2

for all n ≥ N2 and for all x ∈ A. Take N = max{N1, N2}, then n ≥ N implies

|(fn + gn) (x)− (f + g)(x)| ≤ |fn(x)− f(x)|+ |gn(x)− g(x)| < ε

2+ε

2= ε

for all x ∈ A. Hence (fn + gn) converges uniformly.

Since the function gn is bounded, there is Mn ∈ R such that |gn| ≤ Mn for all x ∈ A.

Then, |gN2(x)− g(x)| < ε implies |g(x)| ≤ |gN2(x)|+ ε ≤MN2 + ε = Q(say). Since (fn) is

a uniformly convergent sequence of bounded function, the sequence is uniformly bounded

(Rudin Problem 1) by P(say). Now, since fn → f uniformly, there exists N1 ∈ N such

that n ≥ N1 implies |fn(x)− f(x)| < ε2Q

for all x ∈ A. Similarly, for gn → g uniformly,

there exists N2 ∈ N such that |gn(x)− g(x)| < ε2P

for all x ∈ A.

|(fngn)(x)− (fg)(x)| = |fn(x)gn(x)− fn(x)g(x) + fn(x)g(x)− f(x)g(x)|

≤ |fn(x)| |gn(x)− g(x)|+ |g(x)| |fn(x)− f(x)|

<Pε

2P+Q

ε

2Q= ε,

for all x ∈ A. Therefore, fngn converges uniformly.

Let fn(x) = 1x

and gn(x) = 1n

for each n on the interval (0, 1). Then the sequences (fn)

converges to f(x) = 1/x uniformly because it is a constant sequence. Also (gn) converges

uniformly to g(x) = 0 uniformly because the functions are independent of x. We claim

the the product hn = (fngn)(x) = 1xn

converges pointwise but not uniformly.

(hn) converge point wise because for a fixed x and for given ε > 0 by Archimedean

Property we can pick N ≥ 1εx

so that for all n ≥ N we have

|hn(x)− 0| =∣∣∣∣ 1

xn

∣∣∣∣ < 1

Nx< ε

So the sequence converges to 0 pointwise. However for any given N we can choose

0 < x < 1N

so that ∣∣∣∣ 1

Nx− 0

∣∣∣∣ =1

Nx> 1,

so the convergence can not be uniform.

100

Problem 11.4. Let (X, dX) and (Y, dY ) be metric spaces, A ⊂ X, and (fn) a sequence

of continuous functions from A to Y that converges to a function f : A→ Y uniformly on

A. Given any x ∈ A, prove that limn→∞ fn(xn) = f(x) for every sequence (xn) in A with

xn → x. Construct an example showing the converse is not true, but if A is compact,

then the converse is true.

101

Problem 11.5. (General Arzela Ascoli Theorem) If X is a compact metric space, (fn) ⊂C(K), (fn) is pointwise bounded and equicontinuous on X, prove that

(a) (fn) is uniformly bounded on X, and

(b) f(n) contains a uniformly convergent subsequence.

Hint: For (a) use compactness to get the uniform bound. For (b), use the fact that

a compact set is separable to extract a countable dense subset. Then prove that on the

countable set the sequence has a convergent subsequence. Prove this subsequence converges

uniformly on X.

Solution. (a) Since (fn) is an equicontinuous family of function on a compact metric

space it is uniformly equicontinuous. That is for given ε > 0 there is a δ > 0 such

that d(x, y) < δ implies dY (f(x), f(y)) < ε for all x ∈ X. Since X is compact

there is a finite δ − net, say xi : 1 ≤ i ≤ m. Since (fn) is pointwise bounded

fn(xi) < Mi for 1 ≤ i ≤ m. Now take M = max(M1,M2, . . . . . . ,Mm) + ε. We

claim that M is a bound of fn(x) independent of x. Take any x ∈ X, there is xi

for some i so that d(x, xi) < δ which implies dY (fn(x), fn(xi)) < ε and therefore

|fn(x)| ≤ |fn(xi)|+ ε < M

(b) SinceX is compact it is sequentially compact and hence is separable. AlsoX being

separable has a countable dense subset, say, E. Let the elements of the countable

set E are {x1, x2, . . . . . . }. The numerical sequence {fn(x1)}∞n=1 is bounded, so by

Bolzano-Weierstrass it has a convergent subsequence, say,

102

Problem 11.6. Let (X, d) be a metric space and suppose K ⊂ X is compact, fn → f on

K (and all functions are real-valued and continuous), and fn(x) ≥ fn+1(x) for all x ∈ Kand n ∈ N. Prove that fn → f uniformly on K. Construct an example showing the

necessity of compactness.

Solution. Set gn = fn− f . Then gn is continuous, gn → 0 pointwise, and gn ≥ gn+1. We

have to prove that gn → 0 uniformly on K.

Let ε > 0 be given. Let Kn be the set of all x ∈ K with gn(x) ≥ ε. The set

Yn = {gn(x) ∈ gn(K) : g(x < ε)} is open. Take any point g(x0) ∈ Yn then there is

a neighborhood Bδ(g(x0)) with δ < ε − g(x0) > 0 is contained entirely in Yn and hence

the set {gn(x) ∈ gn(K) : g(x ≥ ε)} is closed. By continuity the inverse image Kn is also

closed. Since Kn is a closed subset of a compact set K, Kn is itself compact.

Since gn ≥ gn+1, we have Kn+1 ⊂ Kn. Fix x ∈ K. Since gn(x) → 0, we see that

x /∈ Kn if n is sufficiently large. Thus x ∈⋂Kn. Since x was arbitrary it implies that⋂

Kn = ∅ Hence KN is empty for some N . It follows that 0 ≤ gn(x) < ε for all x ∈ Kand for all n ≥ N . This proves the theorem.

Compactness is necessary. For example if

fn(x) =1

nx+ 1(0 < x < 1;n = 1, 2, 3, . . . . . . )

then fn(x)→ 0 monotonically in (0, 1), but the convergence is not uniform. To see that

take x = 1n

and so fn(x) = 12.

103

Problem 11.7. Is the set of functions{x2 + n

1+nx}∞n=1

taking R to R (with the usual

metric) equicontinuous on [0, 1]?

Solution. .

Process 1 The set [0, 1] is compact. The sequence of function (fn) =(x2 + n

n+1x)

is point-

wise convergent to f = x2 + x as n → 0. Also for a fixed x, fn is monotonically

decreasing. Hence according to above problem the sequence of function is uni-

formly convergent and hence equicontinuous.

Process 2 ∣∣∣∣x2 +n

n+ 1x− y2 − n

n+ 1y

∣∣∣∣ ≤ ∣∣x2 − y2∣∣+

n

n+ 1|x− y|

≤ |x− y| |x+ y|+ |x− y|

= (|x+ y|+ 1) |x− y|

≤ 3 |x− y|Let ε > 0 be given. Taking δ < ε

3we can conclude that, for any x ∈ [0, 1],

|x− y| < δ implies |fn(x)− fn(y)| < ε for all n. Hence (fn) is equicontinuous.

104

Problem 11.8. Compute and plot the Bernstein polynomials for 1/(1 + x) on [0, 1] of

degrees 1, 2, and 3.

105

Problem 11.9. Assume that (fn) is a sequence of functions with fn : [a, b] → R for all

n where we take the usual metric on [a, b] and R.

(a) Assume that {fn} is uniformly Lipschitz continuous on [a, b] with constant L. Prove

that {fn} is equicontinuous on [a, b].

(b) Assume that {fn} is uniformly Lipschitz continuous on [a, b] with constant L and

further that (fn(x)) is a uniformly bounded sequence of numbers for some a ≤ x ≤ b.

(1) Prove that (fn) has a subsequence that converges uniformly on [a, b]. (2) Prove

the second assumption is needed with an example.

(c) Assume that (fn) is a bounded subset of C([a, b]). Show that the sequence of functions

(Fn) from [a, b] into R defined by

Fn(x) =

∫ x

a

fn(x) ds

contains a subsequence that converges uniformly on [a, b]. (This is quite a remark-

able result when considering the sequence of functions (sin(nx)) as an example of a

bounded sequence in C([a, b]) that fails to have any convergent subsequence.)

Solution. Definition: A function f : X → R on a metric space (X, d) is Lipschitz

continuous on X if there is a constant C ≥ 0 such that

|f(x)− f(y)| ≤ Cd(x, y), ∀x, y ∈ X

(a) Since (fn) is uniformly Lipschitz continuous there is a constant L ≥ 0 such that

|fn(x)− fn(y)| ≤ Ld(x, y), ∀x, y ∈ X, ∀n

So, if L = 0 then the functions are the constant functions and there is nothing to

prove. So we consider more interesting case L > 0.

Let ε > 0 be given. Taking 0 < δ < εL

we can conclude that, for any x ∈ [a, b],

d(x, y) < δ implies |fn(x)− fn(y)| ≤ L εL

= ε for all n. Therefore (fn) is equicon-

tinuous.

(b) Since (fn) is bounded equicontinuous family of functions, (fn) is precompact and

hence has a convergent subsequence. The simple constant sequence (fn) = (n)

has uniform Lipschitz constant 0, but clearly fails to converge even pointwise.

(c) Let M be the uniform bound for the set {fn}.

|Fn(x)| =∣∣∣∣∫ x

a

fn(s)ds

∣∣∣∣ ≤ ∫ x

a

|fn(s)| ds ≤M(x− a) ≤M(b− a)

Therefore, (Fn) is uniformly bounded. Also for any x, y ∈ [a, b]

|Fn(x)− Fn(y)| =∣∣∣∣∫ x

a

fn(s)ds−∫ y

a

fn(s)ds

∣∣∣∣ ≤ ∫ x

y

|f(s)| ds ≤M

∫ x

y

ds ≤M |x− y| .

So, {Fn} is equicontinuous. Now by part (b), (Fn) has a subsequence that con-

verges uniformly.

106

Problem 11.10. Suppose K = {x1, . . . , xn} is a finite space with the discrete metric d.

Show C(K) is linearly isomorphic to the finite-dimensional space Rn with the ∞-norm

(which is just the maximum norm in finite dimensions). Hint: Any function f : K → Ris just a mapping of each xi to some point yi ∈ R, so the function can be identified with

a point y = (y1, . . . , yn) ∈ Rn where f(xi) = yi. This is a linear isometry that is also a

metric space isomorphism with the metric induced by the ∞-norm.

107

Problem 11.11. Suppose f is a real-valued continuous function with dom(f) = R. Let

fn(x) = f(nx) for n ∈ N and any x ∈ R. If {fn} is equicontinuous, prove that f must be

a constant function on [0,∞).

Solution. Suppose f is not constant then there are x, y ∈ [0,∞) with x 6= y and f(x) 6=f(y). Let ε = |f(x)− f(y)| > 0. Then |fn(x/n)− fn(y/n)| = ε for all n ∈ N. When n

is large enough x/n, y/n ∈ [0, 1] and |x/n− y/n| → 0. So for any δ > 0 there is N such

that |x/n− y/n| < δ for all n ≥ N but |fn(x/n)− fn(y/n)| = ε which is contradiction

to the fact that {fn} is equicontinuous.

108

Problem 11.12. Suppose that (fn) is an equicontinuous sequence of functions on a

compact set K and (fn) converges pointwise on K. Prove that (fn) converges uniformly

on K.

Solution. Let ε > 0 be given.

Since (fn) is equicontinuous, for a fixed x ∈ K, there exists a δ > 0 such that dK(x, y) < δ

implies dY (fn(x), fn(y)) < ε3

for all n ∈ N. Since K is compact, it is totally bounded

and there is a finite δ − net, that covers K. For any x ∈ K there exists xi such that

x ∈ Bδ(xi) where xi ∈ δ − net. Therefore, dK(x, xi) < δ and hence by equicontinuity

dY (fn(x), fn(xi)) <ε3.

Since fn → f is pointwise convergent for a fixed xi ∈ K, fn(xi) converges to f(xi)

and so (fn(xi)) is Cauchy sequence. There exists an Ni such that n,m ≥ Ni implies

dY (fn(xi), fm(xi)) < ε/3.

Now take N = maxNi then for all xi in the set of δ − net. So for any xi,

n,m > N ⇒ dY (fn(xi), fm(xi)) < ε/3

For any n,m > M we have,

dY (fn(x), fm(x)) ≤ dY (fn(x), fn(xi))+dY (fn(xi), fm(xi))+dY (fm(xi), fm(x)) ≤ ε

3+ε

3+ε

3= ε

for all x ∈ K, showing that (fn) is uniformly Cauchy and since Y is complete, (fn) is

uniformly convergent.

109

Problem 11.13. LetX and Y be two metric spaces and A ⊂ X. Let (fn) be a sequence of

functions from A to Y that converges to f pointwise on A. Prove the following theorems.

(a) Let Mn = supx∈A dY (fn(x), f(x)). Prove that fn → f uniformly if and only if Mn →0.

(b) If fn is continuous on A for each n, then f is continuous on A. This is a generalization

of the theorem in the notes.

(c) Suppose Y is complete, then fn converges uniformly on A if and only if for every ε > 0

there exists an integer N such that m ≥ N , n ≥ N , x ∈ E implies dY (fn(x), fm(x)) <

ε. Construct an example showing the necessity of Y being complete.

(d) Suppose Y is complete, and fn converges uniformly on A. Let x be a limit point of A

and suppose that limt→x fn(t) = yn for each n. Then (yn) converges and limt→x f(t) =

limn→∞ yn. In other words, prove limt→x limn→∞ fn(t) = limn→∞ limt→x fn(t). Con-

struct an example showing the necessity of Y being complete.

110

Problem 11.14. Prove that if f ∈ C([0, 1]) and∫ 1

0f(x)xn dx = 0 for all n ∈ N, then

f(x) = 0 on [0, 1]. Hint: If∫ 1

0f 2(x) dx = 0, then a result from elementary real analysis

states that f ≡ 0. Consider a sequence of polynomials that converge uniformly to f and

take the product with f to get a uniformly convergent sequence to f 2.

111

Problem 11.15. Let (X, d) be a metric space and pick any fixed but arbitrary a ∈ X.

Define the real-valued function fp for each p ∈ X by

fp(x) = d(x, p)− d(x, a),

where x ∈ X.

(a) Prove that ‖fp‖∞ ≤ d(a, p) and fp ∈ C(X).

(b) Prove that

‖fp − fq‖∞ = d(p, q)

for all p, q ∈ X.

(c) Let Φ(p) = f(p), so that Φ is an isometry of X onto Φ(X) ⊂ C(X). Let Y be the

closure of Φ(X) in C(X). Show that Y is complete which implies X is isometric to a

dense subset of a complete metric space Y . Note: This is a different proof of what we

previously showed in lecture 3 examples about defining the closure of a metric space.

112

Problem 11.16. Let fn(x) = (−1)n xn

n.

(a) Show that∞∑n=1

fn(x) converges uniformly on [0, 1].

(b) Show that∞∑n=1

|fn(x)| converges uniformly on [0, 1).

(1) Show that∞∑n=1

|fn(x)| does not converge uniformly on [0, 1).

Solution.

Theorem 11.7 (Leibniz’s Rule for Alternating Series). If {an} is a monotonic decreasing

sequence with limit 0, then the alternating series∞∑n=1

(−1)n+1an converges. If S denotes

its sum and sn denotes its nth partial sum, we also have the inequalities

0 < (−1)n(S − sn) < an+1

(a) For each x ∈ [0, 1] the series∞∑n=1

(−1)n xn

nconverges pointwise since it is an alternat-

ing series of decreasing positive terms approaching to 0. Define F (x) =∞∑n=1

fn(x)

for each x ∈ [0, 1]. To show that the series converges uniformly we need to show

that the uniform norm

‖n∑j=1

(−1)jxj

j− F (x)‖∞ → 0

as n→∞. By the Leibniz’s rule for alternating series we have,∣∣∣∣∣n∑j=1

(−1)jxj

j− F (x)

∣∣∣∣∣ < |fn(x)|

for each x ∈ [0, 1] and there fore,

‖n∑j=1

(−1)jxj

j− F (x)‖∞ < ‖fn(x)‖∞ =

1

n→∞, as n→∞

Therefore the series converges uniformly.

(b) Fix x ∈ [0, 1). We use ratio test.∣∣∣∣ |fn+1(x)||fn(x)|

∣∣∣∣ =nx

n+ 1→ x, as n→∞

, so for x ∈ [0, 1), the series converges pointwise.

(c) Fix an m ∈ N. Choose x ∈ [0, 1) close enough to 1 so that x2m > 12. So, xj > 1

2

for all m+ 1 ≤ j ≤ 2m. Then

∞∑n=m+1

xn

n>

2m∑n=m+1

xn

n>

1

2

2m∑n=m+1

1

n>

1

2

2m∑n=m+1

1

2m=

1

2

1

2mm =

1

4.

Therefore, however large m we choose there is an x ∈ [0, 1) for which∞∑

n=m+1

xn

n> 1

4

and hence the series does not converge uniformly.113

Problem 11.17 (Rudin 1). Prove that every uniformly convergent sequence of bounded

functions is uniformly bounded.

Solution. Let (fn) be a uniformly convergent sequence of bounded functions on a metric

space X. So, by Cauchy Criteria, there exists an N ∈ N such that for all m,n ≥ N we

have

|fn(x)− fm(x)| < 1

for all x ∈ X. Since each of the functions (fn) are bounded there is Mn ∈ R such that

|fn(x)| < Mn for all x ∈ X. Thus if n ≥ N , then

|fn(x)| = |fn(x)− fN(x) + fN(x)| ≤ |fn(x)− fN(x)|+ |fN(x)| < 1 +MN .

for all x ∈ X. Let M = max{1 +MN ,M1, . . . . . .MN−1}. Then |fn(x)| ≤M for all n ∈ Nand all x ∈ X.

114

Problem 11.18 (Rudin 2). If {fn} and {gn} converge uniformly on a set E, prove that

{fn + gn} converges uniformly on E. If in addition, {fn} and {gn} are sequences of

bounded functions, prove that {fngn} converges uniformly on E.

Solution. See This Problem

Problem 11.19. Construct sequences {fn} and {gn} which converges uniformly on some

set E, but such that {fngn} does not converge uniformly on E.

Solution. See This Problem

115

Problem 11.20. Consider

f(x) =∞∑n=1

1

1 + n2x

For what values of x does the series converge absolutely? On what intervals does it

converge uniformly? On what intervals does it fail to converge uniformly? Is f continuous

whatever the series converges? Is f bounded?

Solution. • Where it converges: This series converges for all x except 0 and

x = − 1n2 , n = 1, 2, . . . . . . . For x = 0, f(x) =

∞∑n=1

1 = ∞. For x = − 1n2 the nth

term is not defined. For all other values of x the series converges.

By Weierstrass M-test the series converges uniformly on the interval [δ,∞) if

δ > 0, since on that interval

1

1 + n2x≤ 1

n2δ.

Likewise, the series converges uniformly on (−∞,−δ] except at the points x =

− 1n2 , since for n ≥

√2δ

we have,

116

Problem 11.21. Rudin Chapter 7, Problem 5:

Let

fn(x) =

0 x < 1

n+1

sin2 πx

1n+1≤ x ≤ 1

n

0 1n< x.

Show that {fn} converges to a continuous function, but not uniformly. Use the series∑fn

to show that absolute convergence, even for all x, does not imply uniform convergence.

Solution. The limit of fn(x) is zero. If x ≤ 0 or x ≥ 1, then fn(x) = 0 for all n, and so

this assertion is obvious. If 0 < x < 1, then fn(x) = 0 for all n ≥ 1x, and so once again

the assertion is obvious.

The convergence is not uniform, since, no matter how large n is taken, there is a point

x, namely x = 12n+ 1

2

, for which fn(x) = 1.

The series converges to 0 for x ≤ 0 and x ≥ 1 and to sin2 πx

for 0 < x < 1. Since the

terms are nonnegative, the series obviously converges absolutely. Since the sum is not

continuous at 0, the series does not converge uniformly on any interval containing 0.

117


Prove that the series∞∑n=1

(−1)nx2 + n

n2

converges uniformly in every bounded interval, but does not converge absolutely for any

value of x.

Solution. The series is the sum of two series

x2

∞∑n=1

(−1)n

n2+

n∑n=1

(−1)n

n

The first of these converges both uniformly and absolutely on any bounded interval [a, b]

because by Weierstrass M test ∣∣∣∣x2 (−1)n

n2

∣∣∣∣ ≤ M2

n2

where M = max(|a| , |b|) and∞∑n=1

M2

n2 converges.

The second sum is independent of x and converges, hence it converges uniformly in

x. Being a sum of two uniformly convergence sequence the resulted sum∞∑n=1

(−1)n x2+nn2

also uniformly converges (Rudin Exercise 7.2).

Since∣∣∣(−1)n x

2+nn2

∣∣∣ ≥ 1n

for any x, and∑

1n

does not converge we conclude that the series

does not converge absolutely.

118


For n = 1, 2, 3, . . . . . . , x real, put

fn(x) =x

1 + nx2

Show that {fn} converges uniformly to a function f , and that the equation f ′(x) =

limn→∞

f ′n(x) is correct if x 6= 0, but false if x = 0.

Solution. Since (1−√nx)2 ≥ 0 implies 1 + nx2 ≥ 2

√nx (Schwarz inequality) we have

|fn(x)| ≤∣∣∣∣ x

1 + nx2

∣∣∣∣ ≤ |x|2√n |x|

=1

2√n

if x 6= 0. Also note that fn(0) = 0 for all n. Therefore fn(x) tends to uniformly to 0.

Now f ′n(x) = 1−nx2(1+nx2)2

, which tends to 0 if x 6= 0, though f ′n(0) = 1 for all n

119

Problem 11.24. Rudin Chapter 7, Problem 8: If

I(x) =

0 x ≤ 0

1 x > 0

if {xn} is a sequence of distinct points of (a, b), and if∑|cn| converges, prove that the

series

f(x) =∞∑n=1

cnI(x− xn) (a ≤ x ≤ b)

converges uniformly, and that f is continuous for every x 6= xn.

Solution. If∑n

|cn| converges then for any m ≥ n,∣∣∣∣∣m∑j=n

cjI(x− xj)

∣∣∣∣∣ ≤m∑j=n

|cj| , ∀x ∈ [a, b]

By Weierstrass M test, the series converges uniformly on [a, b].

Let ε > 0 be given. There exists N such that∣∣∣∣∣∑j≥N

cjI(x− xj)

∣∣∣∣∣ < ε

3, ∀y ∈ [a, b].

If x 6= xn for any n then each of the functions and hence the sum∑j<N

cjI(y−xj) is continu-

ous at x, so there exists δ > 0 such that |x− y| < δ ⇒

∣∣∣∣∣ ∑j<N cjI(x− xj)−∑j<N

cjI(x− xj)

∣∣∣∣∣ <ε3. Each of the functions cjI(x− xj) are continuous because the only discontinuity occurs

at xj.

Then we see that , if |x− y| < δ,

|f(x)− f(y)| ≤

∣∣∣∣∣∑j<N

cjI(x− xj)−∑j<N

cj(y − xj)

∣∣∣∣∣+∣∣∣∣∣∑j≥N

cjI(x− xj)

∣∣∣∣∣+∣∣∣∣∣∑j≥N

cjI(y − xj)

∣∣∣∣∣ < ε.

Thus, f is continuous at x.

120

Problem 11.25. Rudin Chapter 7, Problem 9

Let {fn} be a sequence of continuous functions which converges uniformly to a function

f on a set E. Prove that

limn→∞

fn(xn) = f(x)

for every sequence of point xn ∈ E such that xn → x and x ∈ E. Is the converse of this

true?

Solution. Let ε > 0 be given. Since fn converges uniformly on E, there exists an N1 ∈ Nsuch that for all n ≥ N1, |fn(x)− f(x)| < ε

2for all x ∈ E. Also since fn is continuous for

all n, there exits δ > 0 such that |fn(x)− fn(y)| < ε2

whenever |x− y| < δ. Now since

xn → x, there exists an N2 ∈ N such that |xn − x| < δ whenever n ≥ N2. Now taking

N = max(N1, N2) we get,

|fn(xn)− f(x)| ≤ |fn(xn)− fn(x)|+ |fn(x)− f(x)| < ε

for all n ≥ N and therefore limn→∞

fn(xn) = f(x).

121

Problem 11.26. Let fn(x) = (1+xn)1n for x ≥ 0. Does the sequence converge pointwise

on E = [0,∞)? If so, find the limit function. Check whether or not the convergence is

uniform on E.

Solution. If 0 ≤ x ≤ 1 then

fn+1(x)

fn(x)=

(1 + xn+1)1

1+n

(1 + xn)1/n≤(

1 + xn+1

1 + xn

) 1n

≤ 11/n = 1

Therefore {fn(x)} is monotonically decreasing for 0 ≤ x ≤ 1. Now if x > 1 then we write

fn(x) = x(1 +(

1n

)n)

1n . Since 0 < 1

x< 1 we conclude that

fn+1

fn=x(1 + 1

xn+1 )1

n+1

x(1 + 1xn

)1n

≤ 1

Therefore, the for x ∈ [0,∞) fn is monotonically decreasing and bounded bellow by x

because 1 + xn ≥ xn ⇒ (1 + xn)1/n ≥ x for all x ≥ 0. Hence the sequence converges

pointwise for each x ∈ [0,∞).

For x > 1,

limn→∞

(1 + xn)1/n = x limn→∞

(1 +1

xn)1/n = x.

For x < 1,

limn→∞

(1 + xn)1/n = 1.

Now to determine uniform convergence. First consider the case x > 1

Mn = supx∈[1,∞)

|(1 + xn)1/n − x|

d

dx(1 + xn)1/n − x = (1 + xn)1/n−1xn−1 − 1 =

(1 + xn)1/nxn

x(1 + xn)− 1

Observe that,

(1 + xn)n = xn2

+

(n

1

)(xn)n−1 + . . . · · ·+ 1

(1 + xn)n ≥ xn2

+ (xn)n−1

xn(1 + xn)n ≥ xn(xn2

+ (xn)n−1)

xn(1 + xn)n ≥ xn2

(xn + 1)

x(1 + xn) ≥ xn(1 + xn)1/n

xn(1 + xn)1/n

x(1 + xn)≤ 1

xn(1 + xn)1/n

x(1 + xn)− 1 < 0

Therefore the function is decreasing and hence its maximum occurs at x = 1. Therefore

Mn = 21/n − 1→ 0 and so fn → x uniformly.122

Now consider the case when x ≤ 1

Mn = supx∈[0,1]

|(1 + xn)1/n − 1|.

Since derivative of (1 +xn)1/n−1 is (1 +xn)1/n−1xn−1 which is positive for 0 ≤ x ≤ 1 and

hence the increasing. So maximum occurs at x = 1. So, Mn = 21/n − 1 → 0. Therefore

for x < 1 we get fn → 1 uniformly. We conclude that the given sequence of function

uniformly convergent to,

f(x) =

x if x ≥ 1

1 if 0 ≤ x ≤ 1

123

Problem 11.27. Let {fn} be a sequence of continuous real functions on R, and that

fn → f uniformly on R. Let gn(x) = fn(x+ 1n) for all n = 1, 2, . . . . . . , and x ∈ R. Show

that {gn} converges uniformly to f on any bounded interval [a, b].

Solution. Since fn is continuous on a compact set [a, b], we can assert that fn is uniformly

continuous on [a, b]. Hence for given ε > 0 there is δ > 0 such that

|x− y| < δ ⇒ |fn(x)− fn(y)| < ε/2

By Archimedean property for any δ > 0 there is N1 ∈ N such that 1n< δ for all n ≥ N1.

So,

|(x+1

n)− x| < 1

n< δ ∀n ≥ N1

Hence for given ε > 0 there is N1 ∈ N such that,

n ≥ N1 ⇒ |(x+1

n)− x| < δ ⇒ |fn(x+

1

n)− f(x)| < ε/2

Since fn → f uniformly there is N2 ∈ N such that

|fn(x)− f(x)| < ε/2 ∀n ≥ N2, ∀x ∈ [a, b]

Take N = max(N1, N2), then we get,

|gn(x)− f(x)| = |fn(x+1

n)− f(x)|

≤ |fn(x+1

n)− fn(x)|+ |fn(x)− f(x)|

≤ ε

2+ε

2= ε

∀n ≥ N and for all x ∈ [a, b].

Therefore {gn} is uniformly convergent to f on [a, b]

124

Problem 11.28. Consider the series

f(x) =∞∑n=1

1

1 + n2 |x|.

(a) For what values of x does the series converge?

(b) Characterize the subsets of R on which the series converges uniformly.

(c) Ia f continuous on its domain?

(d) Is f bounded on its domain?

Solution.

(a) For x = 0 the given series diverges.

For x 6= 0 we get∞∑n=1

1

1 + n2|x|=

1

|x|

∞∑n=1

1

n2 + 1|x|

Since 1n2+ 1

|x|< 1

n2 by comparison test we can say that the given series converges.

(b) For any intervals of the form [a, b] with either a > 0 or b < 0 we get,

| 1

1 + n2|x|| ≤ 1

1 + n2c≤ 1

n2cforc = max(|a|, |b|)

Since∑∞

n=11n2c

<∞ by Weierstrass M test the series convergence uniformly.

Now consider the interval (0, b]

Suppose the series converges uniformly in (0, b]. Then the sequence of partial sum sn =∑nn=1

11+n2|x| converges uniformly. By Cauchy criterion of uniform convergence there

exists N ∈ N such that

|sn − sm| <1

3⇒ |

n∑k=m+1

1

1 + n2|x|| < 1

3for ∀n > m ≥ N, ∀x ∈ (0, b]

But this is not true for x = 1N2 . Hence the series is not uniformly convergent on the

interval of the form (0, b].

(c) 11+n2|x| is a sequence of continuous functions. So (by the corollary after the theorem

of uniform convergence and continuity) f is continuous on the intervals where the series

uniformly converges to f .

(d) The function f is not bounded. For example limx→0

f(x) =∞.

125

Problem 11.29. For each positive integer n, let fn(x) = nn2+xn

. Show that the sequence

{fn} converges uniformly to f(x) ≡ 0 on E = [0,∞)

126

12. Differential Calculus and Banach Space

Definition 12.1. A linear map or linear operator T between real (or complex) linear

spaces X, Y is a function T : X → Y such that

T (λx+ µy) = λTx+ µTy, ∀ λ, µ ∈ R (or C) and x, y ∈ X.

A linear map T : X → X is called a linear transformation of X, or a linear operator on

X. If T : X → Y is one-to-one and onto, then we say that T is nonsingular or invertible,

and define the inverse map T−1 : Y → X by T−1y = x if and only if Tx = y, so that

TT−1 = IY , T−1T = IX , where IY and IX are identity maps.

Definition 12.2. Let X and Y be two normed linear spaces with norms ‖·‖X and ‖·‖Y ,

respectively. A linear map T : X → Y is bounded if there exists a constant M ≥ 0 such

that

‖Tx‖Y ≤M ‖x‖X , ∀ x ∈ X.

If no such constant exists, then we say that T is unbounded. If T : X → Y is a bounded

linear map, then we define the operator norm ‖T‖ of T by

‖T‖ = inf {M : ‖Tx‖Y ≤M ‖x‖X ∀ x ∈ X} .

We denote the set of all linear maps T : X → Y by L(X, Y ), and the set of all bounded

linear maps T : X → Y by B(X, Y ). When the domain and range spaces are the same,

we write L(X) and B(X).

Theorem 12.1. A linear map is bounded if and only if it is continuous.

Theorem 12.2 (Bounded linear transformation). Let X be a normed linear space and Y

a Banach space. If M is a dense linear subspace of X and T : M ⊂ X → Y is a bounded

linear map, then there exists a unique bounded linear map T : X → Y such that T x = Tx

for all x ∈M and∥∥T∥∥ = ‖T‖.

Definition 12.3. Two linear spaces X and Y are linearly isomorphic if there is a one-

to-one, onto linear map T : X → Y . If X and Y are normed linear spaces and T, T−1

are bounded linear maps, then X and Y are topologically isomorphic. If T also preserves

norms, meaning ‖Tx‖Y = ‖x‖X for all x ∈ X, then X and Y are isometrically isomorphic.

Theorem 12.3. If X is a normed linear space and Y is a Banach space, then B(X, Y )

is a Banach space with respect to the operator norm.

Definition 12.4. A scalar-valued linear map from a linear space X to R is called a

linear functional or linear form on X. The space of linear functionals on X is called the

algebraic dual space of X, and the space of continuous linear functionals on X is called

the topological dual space of X.

Theorem 12.4 (Hahn-Banach). If Y is a linear subspace of a normed linear space X

and ψ : Y → R is a bounded linear functional on Y with ‖ψ‖ = M , then there exists127

a bounded linear functional φ : X → R such that φ restricted to Y is equal to ψ and

‖φ‖ = M .

Definition 12.5. A function f : (a, b)→ X from an open interval (a, b) of real numbers

into a Banach space X is differentiable at t ∈ (a, b) with derivative f ′(t) ∈ X if the

following limit exists in X:

f ′(t) = limh→0

f(t+ h)− f(t)

h.

The function f is continuously differentiable in (a, b) if f ′ : (a, b)→ X is continuous.

Theorem 12.5. If f : (a, b) → X is differentiable in (a, b) and f ′ = 0, then f is a

constant function.

Theorem 12.6 (Mean value). If f is continuously differentiable in an open interval that

contains the closed, bounded interval [a, b], with values in a Banach space, then

‖f(b)− f(a)‖ ≤M(b− a) where M = supa≤t≤b

‖f ′(t)‖ .

Definition 12.6. A map f : U ⊂ X → Y whose domain U is an open subset of a

Banach space X and whose range is a Banach space Y is differentiable at x ∈ U if there

is a bounded linear map A : X → Y such that

limh→0

‖f(x+ h)− f(x)− Ah‖Y‖h‖X

= 0.

We call the map A the Frechet derivative and write A = f ′(x). If f is differentiable at

each point of U , then f ′ : U → B(X, Y ) is the map that assigns to each point x ∈ U

the bounded linear map f ′(x) : X → Y that approximates f near x. We say that f is

continuously differentiable at x if the map f ′ is continuous at x, where the domain U

is equipped with the norm on X and the range B(X, Y ) is equipped with the operator

norm.

Definition 12.7. Let X and Y be Banach spaces, and f : U ⊂ X → Y , where U is an

open subset of X. The directional derivative of f at x ∈ U in the direction h ∈ X is

given by

δf(x;h) = limt→0

f(x+ th)− f(x)

t.

If this limit exists for every h ∈ X, and f ′G(x) : X → Y defined by f ′G(x)h = δf(x;h) is a

linear map, then we say that f is Gateaux differentiable at x, and we call f ′G the Gateaux

derivative of f at x.

Theorem 12.7. Suppose that f : U ⊂ X → Y is a Gateaux differentiable function from

an open subset U of a Banach space X to a Banach space Y . If x, y ∈ U , and the line

segment {tx+ (1− t)y : 0 ≤ t ≤ 1} connecting x and y is contained in U , then

‖f(x)− f(y)‖Y ≤M ‖x− y‖X where M = sup0≤t≤1

‖f ′G(tx+ (1− t)y)‖ .

128

Theorem 12.8. Suppose that f : U ⊂ X → Y is a Gateaux differentiable function from

an open subset U of a Banach space X to a Banach space Y . If the Gateaux derivative

f ′G : U ⊂ X → B(X, Y ) is continuous at x ∈ U , then f is Frechet differentiable at x and

f ′(x) = f ′G(x).

Theorem 12.9 (Chain Rule). Suppose that X, Y, and Z are Banach spaces, and f : U ⊂X → Y and g : V ⊂ Y → Z, where U and V are open subsets of X and Y , respectively. If

f is differentiable at x ∈ U and g is differentiable at f(x) ∈ V , then g ◦f is differentiable

at x and

(g ◦ f)′(x) = g′(f(x))f ′(x).

Definition 12.8. If f : X×Y → Z is a differentiable map on the product of two Banach

spaces, then we have

f(x+ h, y + k) = f(x, y) + Ah+Bk + o(h, k)

for suitable linear maps A : X → Z and B : Y → Z. We call A and B the partial

derivatives of f with respect to x and y, respectively, and denote them by A = Dxf(x, y)

and B = Dyf(x, y).

Theorem 12.10 (Inverse function). Suppose that f : U ⊂ X → Y is a differentiable

map from an open subset U of a Banach space X to a Banach space Y . If f is continu-

ously differentiable in U and f ′(x) has a bounded inverse at x ∈ U , then there are open

neighborhoods V ⊂ U of x and W ⊂ Y of f(x) such that f : V → W is a one-to-one,

onto continuous map with continuous inverse f−1 : W → V . Moreover, the local inverse

is continuously differentiable at f(x) and

(f−1)′(f(x)) = [f ′(x)]−1.

Theorem 12.11 (Implicit function theorem). Suppose that X, Y , and Z are Banach

spaces, and F : U ⊂ X × Y → Z is a continuously differentiable map defined on an open

subset U of X × Y . If (x0, y0) ∈ U is a point such that F (x0, y0) = 0, and DyF (x0, y0) :

Y → Z is a one-to-one, onto, bounded linear map, then there is an open neighborhood

V ⊂ X of x0, an open neighborhood W ⊂ Y of y0, and a unique function f : V → W

such that

F (x, f(x)) = 0 ∀ x ∈ V.

The function f is continuously differentiable, and

f ′(x) = −[DyF (x, f(x))]−1DxF (x, f(x)).

129

13. Problems on Differentiation

Problem 13.1. Define the discontinuous real-valued function f : R2 → R by

f(x, y) =

xyx2+y2

, if(x, y) 6= (0, 0),

0, else.

prove that (the Gateaux derivatives) Dxf(x, y) and Dyf(x, y) exist everywhere in R2 even

though f is discontinuous at (0, 0).

Solution. For (x, y) 6= (0, 0), the existence of the derivative follows from the standard

rules of differentiation. By symmetry, we can easily determine both partial derivatives

with one computation.

Dxf(0, 0) = limt→0

f(t, 0)− f(0, 0)

t= lim

t→0

0

t= Dyf(0, 0).

Thuse Gateaux derivatives Dxf and Dyf exists everywhere even though f is clearly not

continuous at (0, 0) because the limit lim(x,y)→(0,0)

f(x, y) along x axis is 0 but the same limit

along x = y line is 12.

130

Problem 13.2. Suppose that f : A ⊂ Rn → R where A is an open set and the partial

derivatives D1f, . . . , Dnf are bounded in A. Prove that f is continuous in A.

Solution. Consider any ε > 0 and x = (x1, x2, . . . . . . , xn) ∈ A ⊂ Rn.

Since the derivatives are bounded, let

M = maxy∈A,1≤i≤n

|Dif(y)| .

Choose δ = ε(M+1)n

. Consider any y ∈ Bδ(x) ∩ A, and for 0 ≤ i ≤ n let

y(i) = (y1, . . . , yi, xi+1, . . . , xn)

So, y(0) = x and y(n) = y

|f(y)− f(x)| =∣∣f(y(n))− f(y(0))

∣∣≤∣∣f(y(n))− f(y(n−1))

∣∣+∣∣f(y(n−1))− f(y(n−1))

∣∣+ dotsdots+∣∣f(y(1))− f(y(0))

∣∣=

n∑i=1

∣∣f(y(i))− f(y(i−1))∣∣

By the Mean Value Theorem for 1-D functions, we have that there exists ci in the

interval defined by xi and yi such that∣∣f (y(i))− f(yi−1)

∣∣ = |f(y1, . . . , yi−1, xi, . . . , xn)− f(y1, . . . , yi, xi+1, . . . , xn)| = |Dif(ci)(xi − yi)| < Mδ

Since the bound is independent of i we get,

|f(y)− f(x)| < M

M + 1ε < ε.

Therefore f is continuous.

131

Problem 13.3. Suppose that f : A ⊂ Rn → R is differentiable, where A is an open set

and that f has a local maximum at a point x ∈ A. Prove f ′(x) = 0.

Solution. Define g(t) = x + ty, from R to Rn, where y ∈ Rn and t ∈ R. Note that

g is nothing more than a parameterized line in Rn that passes through the local maxi-

mum point x ∈ A of the function f . Clearly the function g is differentiable on R. Now

g(0) = x ∈ A. Since A is open and g is continuous there exists a δ > 0 such that g(t) ∈ Afor any t ∈ (−δ, δ).

We define h(t) = f(g(t)). Note that h : R → R has a local maximum at t = 0 since f

has a local maximum at g(0) = x. If t is sufficiently close to 0 so that g(t) ∈ A, then h

is differentiable function of t by the Chain Rule, and

h′(t) = f ′(g(t))g′(t) = f ′(x+ ty)y

Hence h has a local maximum at t = 0 and is a differentiable real-valued function of one

variable, we have that

0 = h′(0) = f ′(x)y

Since y ∈ Rn is arbitrary, we have that f ′(x) = 0.

132

Problem 13.4. Suppose that f : A ⊂ Rn → R is differentiable, where A is an open

connected set and f ′(x) = 0 for every x ∈ A. Prove that f is constant in A.

Solution. Let x ∈ A be any point. Define y(i) = (y1, . . . , yi, xi+1, . . . , xn) where y =

(y1, y2, . . . , yn) ∈ A ⊂ Rn. Define the function hi : R→ R as

hi(z) = f(y1, . . . , yi−1, z, xi+1, . . . , xn)

So, hi(yi) = f(y(i)) and hi(xi) = f(y(i−1)) Since f is differentiable in A it is obvious that

hi is differentiable in the interval defined by xi and yi.

133

Problem 13.5. Rudin Exercise 9.17

Let f : R2 → R2 be defined by

f(x, y) =

(f1(x, y)

f2(x, y)

)=

(ex cos y

ex sin y

).

(a) Show that the range of f is R2\ {(0, 0)}.(b) Show that the Jacobian of f is non-zero on R2. Thus, every point of R2 has a

neighborhood in which f is 1-1, but f is not 1-1 on R2.

(c) Let (x, y) = (0, π/3) and (u, v) = f(x, y). Let g = f−1 denote the continuous inverse

of f defined in a neighborhood of (u, v) such that g(u, v) = (x, y). Find an explicit

formula for g, compute f ′(x, y), g′(u, v) and show that g′(f(x))f ′(x) = IR2 .

(d) What are the images under f of lines parallel to the coordinate axes?

Solution. (a) If (u, v) 6= (0, 0), chose x = ln√u2 + v2. Notice that

∣∣∣ u√u2+v2

∣∣∣ ≤ 1 and

so it is possible to choose y so that

cos y =u√

u2 + v2

then

sin y =v√

u2 + v2.

It is obvious from the equations defining x and y that u = ex cos y and v = ex sin y.

Hence every point except (0, 0) is the range of f . The point (0, 0) is not in the

range since

u2 + v2 = e2x > 0

for any point (u, v) = f(x, y).

(b) The Jacobian of f(x, y) is[ex cos y −ex sin y

ex sin y ex cos y

]= e2x 6= 0.

However, since f(x, y) = f(x, y + 2π), it follows that f is not one-one in R2.

(c) By our definition, b = f(0, π/3) =(

12,√

32

). We can therefore take y = arctan

(vu

)near b, the arctangent being between −π/2 to π/2. Thus we have

g(u, v) =(

ln√u2 + v2, arctan

(vu

)).

We then have,

g′(u, v) =

[u

u2+v2v

u2+v2

−vu2+v2

uu2+v2

]When we take u = ex cos y and v = ex sin y, we find that

g′(f(x, y)) =

[e−x cos y e−x sin y

−e−x sin y e−x cos y

]134

Therefore

g′(f(x, y))f ′(x, y) =

[1 0

0 1

](d) The family of lines x = c parallel to y-axis maps to the family of concentric circles

centered at origin with radius ec. The lines y = c parallel to the x axis maps to

rays starting from origin with the angle c− 2nπ for n = 0, 1, . . . . . . with positive

horizontal axis.

135

Problem 13.6. If X = Y = Z = R in the implicit function theorem, interpret the

theorem and its proof graphically.

136

Problem 13.7. Define f : R3 → R by f(x, y, z) = x2y+ex+z. Show that f(0, 1,−1) = 0,

Dxf(0, 1,−1) 6= 0, and that there exists a differentiable function g in some neighborhood

of (1,−1) in R2 such that g(1,−1) = 0 and f(g(y, z), y, z) = 0. Find Dyg(1,−1) and

Dzg(1,−1).

137

Problem 13.8. Suppose f ≥ 0, f is continuous on [a, b], and∫ baf(x) dx = 0. Prove that

f(x) = 0 for all x ∈ [a, b].

138

Problem 13.9. Let p and q be positive real numbers such that 1/p + 1/q = 1. Prove

the following statements.

(a) If u ≥ 0 and v ≥ 0, then

uv ≤ up

p+vq

q,

and equality holds if and only if up = vq.

(b) If f, g ∈ R([a, b]), f, g ≥ 0, and∫ bafp = 1 =

∫ bagq, then

∫ bafg ≤ 1.

(c) If f, g,∈ R([a, b]), then∣∣∣∣∫ b

a

fg

∣∣∣∣ ≤ [∫ b

a

|f |p]1/p [∫ b

a

|g|q]1/q

.

139

(1) For f ∈ R([a, b]) define ‖f‖2 =[∫ b

a|f |2]1/2

. Suppose f, g, h ∈ R([a, b]). Prove the

triangle inequality

‖f − h‖2 ≤ ‖f − g‖2 + ‖g − h‖2 .

140

(2) Show that for any f ∈ R([a, b]) and ε > 0 that there exists g ∈ C([a, b]) such that

‖f − g‖2 < ε.

141


Suppose that f is a real valued function defined in an open set E ⊂ Rn, and that the

partial derivatives D1f, . . . . . . , Dnf are bounded in E. Prove that f is continuous in E.

Solution. Fix x ∈ E and ε > 0. Since E is open, there is an open ball S0 ⊂ E, with

center at x and radius r. Let M = maxx∈E{D1f(x), . . . . . . , Dnf(x)}. Choose

δ = min{r, ε

(n+ 1)M}.

The open ball S ⊂ E centered at x and radius δ. Suppose y − x =∑hjej with |h| < δ.

Put v0 = 0 and vk = h1e1 + . . . · · ·+ hkek, for 1 ≤ k ≤ n. Then

f(y)− f(x) =n∑j=1

[f(x+ vj)− f(x+ vj−1)] .

Since vk < δ for 1 ≤ k ≤ n and since S is convex, the segments with end points x+ vj−1

and x + vj lie in S. Since vj = vj−1 + ejhj, the mean value theorem shows that the jth

summand is equal to

jj(Djf)(x+ vj−1 + θjhjej)

for some θj ∈ (0, 1).

|f(y)− f(x)| ≤n∑j=1

M |hj| ≤n∑j=1

Mδ < ε.

Which shows that |y − x| < δ implies |f(y)− f(x)| < ε and hence f is continuous.

142


Suppose f is differentiable real function in an open set E ⊂ Rn, and that f has a local

maximum at a point x ∈ E. Prove that f ′(x) = 0.

Solution. We will first prove for single variable case.

Definition 13.1. Let f be a real function defined on a metric space X. We say that f

has a local maximum at a point p ∈ X if there exists δ > 0 such that f(q) ≤ f(p) for all

q ∈ X with d(p, q) < δ.

Choose δ in accordance with above definition of local maximum, so that

a < x− δ < x < x+ δ < b.

If x− δ < t < x, thenf(t)− f(x)

t− x≥ 0.

Letting t→ x, we see that f ′(x) ≥ 0.

If x < t < x+ δ, thenf(t)− f(x)

t− x≤ 0,

which show that f ′(x) ≤ 0. Hence f ′(x) = 0

Let u be any vector in Rn. Consider the real valued function φ(t) = f(x + tu) defined

near t = 0. The function φ(t) is differentiable by the Chain rule, and has a maximum

at t = 0. Therefore φ′(0) = 0 which implies, by chain rule, f ′(x)(u) = 0. Since u is

arbitrary, it follows by the definition of zero linear transformation that f ′(x) = 0.

143


If f is differentiable mapping of a connected open set E ∈ Rn into Rm, and if f ′(x) = 0

for every x ∈ E, prove that f is constant in E.

Solution. Pick some point x0 ∈ E and define,

C = {x ∈ E : f(x) = f(x0)}.

C is open.

Let x ∈ C. Then f(x) = f(x0). Since x ∈ E and E is an open set, there is a neighborhood

Bε(x) ⊂ E for some ε > 0. Let y ∈ Bε(x). So, [x, y] ⊂ E and hence by Mean Value

Theorem,

|f(x)− f(y)| ≤ ‖f ′(z)‖ |x− y|

for some z ∈ [x, y]. By the given hypothesis, |f(x)− f(y)| ≤ 0 ⇒ f(x) = f(y) = f(x0)

and hence y ∈ C. Therefore Bε(x) ⊂ C and hence C is open.

C is closed.

Let x ∈ E \ C. Therefore f(x) 6= f(x0). Since E is open Bε(x) ⊂ C for some ε > 0.

Let y ∈ Bε(x). So, [x, e] ⊂ E and using Mean Value Theorem we can show that f(x) =

f(y) 6= f(x0) and therefore y ∈ E \ C. Hence Bε(x) ∩ C = ∅ which implies that x can

not be a limit point of C. Since x is arbitrarily chosen we conclude that C is closed.

Since E is connected the only subset of E that is open and closed is the set E itself.

Hence C = E and therefore f(x) = f(x0) for all x ∈ E and therefore f is constant.


Suppose f is a differentiable mapping of R1 into R3 such that |f(t)| = 1 for every t. Prove

that f ′(t).f(t) = 0

Solution.

|f(t)| = 1⇒ f(t).f(t) = 1

Differentiating and using product rule

f ′(t).f(t) = 0

144


Define f(0, 0) = 0 and

f(x, y) =x3

x2 + y2, if (x, y) 6= (0, 0).

(a) Prove that D1f and D2f are bounded function in R2.

(b) Let u be any unit vector in R2. Show that the directional derivative (Duf)(0, 0)

exists, and that its absolute value is at most 1.

(c) Let γ be a differentiable mapping of R1 into R2, with γ(0) = (0, 0) and |γ′(0)| > 0.

Put g(t) = f(γ(t)) and prove that g is differentiable for every t ∈ R1. If γ ∈ C1,

prove that g ∈ C1.

(1) In spite of this, prove that f is not differentiable at (0, 0).

Solution. For (x, y) 6= (0, 0),

D1f =x4 − 3x2y2

(x2 + y2)2 ≤3x2(x2 + y2)

(x2 + y2)2 ≤3x2

x2 + y2≤ 3

|D2f | =∣∣∣∣ −2x3y

(x2 + y2)2

∣∣∣∣ ≤ x2

x2 + y2≤ 1.

Also,

D1f(0, 0) = limt→0

f(t, 0)− f(0, 0)

t= lim

t→0

t− 0

t= 1

and

D2f(0, 0) = limt→0

f(0, t)− f(0, 0)

t= 0.

Hence, as asserted, f(x, y) is continuous.

Let u = (a, b) be an unit vector. Then,

Duf(0, 0) = limt→0

f(ta, tb)− f(0, 0)

t= lim

t→0

t3a3

t(t2a2 + t2b2)=

a3

a2 + b2.

Since |u| = a2 + b2 = 1 this implies |a| ≤ 1. Hence we have

Duf(0, 0) = a3

and

|Duf(0, 0)| =∣∣a3∣∣ ≤ 1

Since both the partial derivatives are continuous in R2 possibly at origin, the function

f is differentiable every where except possibly at origin. and So by the chain rule g is

differentiable everywhere except possib

145


Show that the system of equations

3x+ y − z + u2 = 0

x− y + 2z + u = 0

2x+ 2y − 3z + 2u = 0

can be solved for x, y, u in terms of z; for x, z, u in terms of y; for y, z, u in terms of x;

but not for x, y, z in terms of u.

Solution. We note if (f, g, h)(x, y, z, u) = (0, 0, 0), then f − g − h = 0. Expand-

ing this gives u2 − 3u = 0, so we must have u = 0 or u = 3. Hence, there is no

nonempty open interval I ⊂ R with a γ : I → R3, γ(u) = (x(u), y(u), z(u)) satisfying

(f, g, h)(x(u), y(u), z(u), u) = (0, 0, 0) as we would have I ⊂ {0, 3}. That is, we cannot

solve for x, y, z in terms of u. For the others, by the implicit function theorem the claim

reduces to checking that various 3×3 submatrices of F ′ are invertible, where F : R4 → R3

by F = (f, g, h). We have

F ′(x, y, z, u) =

3 1 −1 2u

1 −1 2 1

2 2 −3 2

And the matrices to consider are, for x, y, u in terms of z,3 1 2u

1 −1 1

2 2 2

which has determinant 8u− 12. As points we invert around have values u = 0 or u = 3,

this is invertible on both cases and so we may solve for x, y, u in terms of z. For solving

for x, z, u in terms of y, the matrix to consider is3 −1 2u

1 2 1

2 −3 2

which has determinant 21− 14u. As before, this is invertible for u = 0 or u = 3. Lastly,

for solving for y, z, u in terms of x, the matrix to consider is 1 −1 2u

−1 2 1

2 −3 2

with determinant 3− 2u. As before, this is invertible for u = 0 or u = 3.

146

14. List of Problems

14.1. Problems in Norm and Metric

Problem 1Show that d(x, y) = ‖x− y‖ is a metric on Rn where ‖·‖ is the standard

Euclidean norm on Rn.

Problem 2Determine which of the following define metrics on R.

(a) d1(x, y) = (x− y)2,

(b) d2(x, y) =√|x− y|,

(c) d3(x, y) = |x2 − y2|,(d) d4(x, y) = |x− 2y|,(e) d5(x, y) = |x− y| /[1 + |x− y|].

Problem 3 For f, g : [0, 1] → R, let d1(f, g) = supx∈[0,1] {x2 |f(x)− g(x)|}, d2(f, g) =

supx∈[0,1] |f(x)− g(x)|, and d3(f, g) =∫ 1

0|f(x)− g(x)| dx. Prove that d1, d2, and d3 de-

fine metrics on the linear space of C([0, 1]) of continuous functions f : [0, 1]→ R.

Problem 4 Show that Rn with d(x, y) = ‖x− y‖∞ is a metric space, where ‖x‖∞ =

supxi |xi| for i = 1, 2, . . . , n.

Problem 5 Consider the convergence of sequences (xn) = (ξn,1, ξn,2, ξn,3, · · · ) to a point

y = (η1, η2, η3, · · · ) in `2. Show that if xn → y, then ξn,i → ηi for i = 1, 2, 3, · · · . But,

construct an example that shows that ξn,i → ηi for i = 1, 2, 3, · · · does not imply that

xn → y. (This is similar to the idea of pointwise convergence not guaranteeing metric-

convergence for sequences of functions.) Consider the Hilbert Cube subspace of `2 defined

by,

H ={x = (ξ1, ξ2, ξ3, · · · ) : x ∈ `2 and |ξi| ≤ 1/i, i = 1, 2, 3, · · ·

}.

Show that a sequence (xn) in H converges to a point y = (η1, η2, η3, · · · ) if and only if

ξn,i converges to ηi for i = 1, 2, 3, · · · .

Problem 6 Let p ≥ 1 be a fixed real number. The real space `p is defined such that

x ∈ `p is a sequence of real numbers x = (ξj) = (ξ1, ξ2, ξ3, · · · ) such that∑∞

j=1 |ξj|p <∞.

For x, y ∈ `p, where x = (ξj) and y = (ηj), let d(x, y) be defined by

d(x, y) =

(∞∑j=1

|ξj − ηj|p)1/p

.

Show that `p with d defined above is a metric space.

Problem 7 Let `∞ be the space defined by real-valued bounded sequences, i.e., x ∈ `∞

implies x = (ξi) = (ξ1, ξ2, · · · ) such that |zetai| ≤ cx for all j ∈ N where cx ≥ 0 is some

real number that may depend on x but not on j. For x, y ∈ `∞, where x = (ξi) and147

y = (ηi), show that

d(x, y) = supj∈N|ξj − ηj|

defines a metric on `∞.

Problem 8 Clearly `1 ⊂ `∞. Is `1 dense in `∞? Is `1 closed in `∞? Justify your an-

swers.

14.2. Cauchy and Compactness

Problem 9 Prove that if a subsequence of a Cauchy sequence in a metric space converges,

then the full sequence converges to the same limit.

Problem 10 Let (xn) and (yn) be Cauchy sequences in a metric space (X, d). Show

that (d(xn, yn)) converges whether or not (xn) and (yn) converge.

Problem 11 Let (X1, d1) and (X2, d2) be two complete metric spaces. Is the product

metric space X1 ×X2 complete?

Problem 12 Prove that a metric space X is complete if and only if the intersection

of every descending sequence of closed balls whose radii approach zero consists of a single

point.

Problem 13 For f, g : [0, 1] → R, let d1(f, g) = supx∈[0,1] {x2 |f(x)− g(x)|}, d2(f, g) =

supx∈[0,1] |f(x)− g(x)|, and d3(f, g) =∫ 1

0|f(x)− g(x)| dx. Prove that (C([0, 1]), d1) is

not complete but (C([0, 1], d2) is complete (this requires recalling some useful results from

elementary analysis that we will revisit in more generality in lecture 4). Since these met-

rics can be defined by norms, one choice of norm produces a Banach space, while the

other does not (this happens on infinite dimensional spaces where norms are not always

equivalent). If we use d3 is the resulting metric space complete?

Problem 14 Prove that a discrete metric space X is separable if and only if X is count-

able. (A discrete metric space is any set X with the discrete metric d(x, y) = 0 if x = y

and d(x, y) = 1 if x 6= y.)

Problem 15 Prove that `∞ is not separable, but `p with 1 ≤ p < +∞ is separable.

Prove that `p with 1 ≤ p ≤ +∞ is complete.

Problem 16 Let E ′ be the set of all limit points of a set E ⊂ X where (X, d) is a

metric space. Prove that E ′ is closed. Prove that E and E have the same limit points.

Do E and E ′ always have the same limit points?148

Problem 17 Let A1, A2, A3, . . . be subsets of a metric space.

(a) If Bn = ∪ni=1Ai, prove that Bn = ∪ni=1Ai for n = 1, 2, 3, . . ..

(b) If B = ∪∞i=1Ai, prove that ∪∞i=1Ai ⊂ B. Show, by an example, that this inclusion can

be proper.

Problem 18 Is every point of every open set E ⊂ R2 a limit point of E? Answer the same

question for closed sets in R2.

Problem 19 Let Eo denote the set of all interior points of a set E.

(a) Prove that Eo is always open.

(b) Prove that E is open if and only if Eo = E.

(c) If G ⊂ E and G is open, prove that G ⊂ Eo.

(d) Prove that the complement of Eo is the closure of the complement of E.

(e) Do E and E always have the same interiors?

Problem 20 Show that Rk is separable.

Problem 21 Prove that every separable metric space has a countable base.

Problem 22 Prove that any uncountable set with the discrete metric is not second count-

able.

Problem 23 Show that a Schauder basis (fn)∞n=0 of C([0, 1]) may be constructed from

”tent” functions.

Problem 24 Let E be the set of all x ∈ [0, 1] whose decimal expansion contains only

the digits 4 and 7. Is E countable? Is E dense in [0, 1]? Is E compact? Is E perfect?

Problem 25 Below, A and B are sets in metric space (X, d).

(a) If A and B are disjoint closed sets prove that they are separated.

(b) Fix p ∈ X, δ > 0, define A = {q ∈ X : d(p, q) < δ} and B = {q ∈ X : d(p, q) > δ}.Prove that A and B are separated.

(c) Prove that every connected space with at least two points is uncountable.

Problem 26 Let A and B be separated subsets in Rn, suppose a ∈ A and b ∈ B and

define p(t) = (1− t)a + tb for t ∈ R. Put A0 = p−1(A) and B0 = p−1(B) (thus t ∈ A0 if

and only if p(t) ∈ A).

(1) Prove that A0 and B0 are separated subsets of R.

(2) Prove that there exists t0 ∈ (0, 1) such that p(t0) /∈ A ∪B.

(3) Prove that every convex subset of Rk is connected.

Problem 27 Give an example of matrix space that is separable but not second countable.149

Problem 28 Any open set in R can be written as countable union of open intervals.

14.3. Compactness

Problem 1 Let f : A ⊂ R → R have the property that for every x ∈ A there is an ε > 0

such that

f(t) > ε if t ∈ (x− ε, x+ ε) ∩ A.

Show that if the set A is compact, then there is some c > 0 such that f(x) > c for all

x ∈ A.

Problem 2

Use compactness to show that if a function f : R → R is locally increasing at every

point in R, then f must be increasing. Here, locally increasing at a point x means there

exists δ > 0 such that f(s) < f(x) < f(t) whenever x− δ < s < x < t < x+ δ.

Problem 3 Let (X, d) be a metric space. Let {K1, K2, . . . , Kn} be a set of compact

subsets of X. Prove that ∪ni=1Ki is a compact subset of X. Let {Kα} be a set of compact

subsets of X. Prove that ∩αKα is a compact subset of X.

Problem 4 Let (X, d) be a metric space with the discrete metric d. What subsets of

X are compact?

Problem 5 Describe an open cover of the open unit square {(x, y) : 0 < x < 1, 0 < y < 1}that has no finite subcover.

Problem 6 Let (xn) be a sequence in a metric space (X, d) that converges to a limit

x. Show the set {x1, x2, x3, · · · } is compact.

Problem 7 Show that the product of two compact metric spaces with the product metric

is a compact metric space.

Problem 8 Let a, b ∈ R and a < b. Show that any closed ball in C([a, b]) is not compact

with the metric d(f, g) = supx∈[a,b] |f(x)− g(x)| for f, g ∈ C([a, b]).

Problem 9 Prove that every compact metric space K has a countable base, and that

K is therefore separable.

Problem 10 Prove that if {Kα} is a collection of compact subsets of a metric space

X such that the intersection of every finite subcollection of {Kα} is nonempty, then ∩Kα

is nonempty. (Hint: Try a proof by contradiction by assuming there is some member of

{Kα}, say K1, such that no point in K1 belongs to every Kα, then create an open cover150

of K1 using complements of Kα. Show that this becomes false (in R, for example) if the

word “compact” is replaced by either “closed” or by “bounded.”

Problem 11 Let E be the set of all x ∈ [0, 1] whose decimal expansion contains only

the digits 4 and 7. Is E compact?

Problem 12 Let X be a metric space in which every infinite subset has a limit point.

Prove that X is separable and compact.

Problem 13

14.4. Continuity

Problem 1 Let X and Y be metric spaces and A ⊂ X. Suppose f maps A into Y and p

is a limit point of A. We say that f has a limit at p and write f(x) → q as x → p, or

limx→p f(x) = q if there is a point q ∈ Y with the property that for every ε > 0 there

exists a δ > 0 such that for all points x ∈ E with 0 < dX(x, p) < δ implies dY (f(x), q) < ε.

Prove the following theorems.

(a) limx→p f(x) = q if and only if limn→∞ f(pn) = q for every sequence (pn) ⊂ E such

that pn 6= p for all n and pn → p.

(b) If f has a limit at p, then the limit is unique.

(c) The function f is continuous at p if and only if limx→p f(x) = f(p).

Problem 2 Let f : R2 → R be defined

f(x1, x2) =

x21x2x41+x22

, (x1, x2) 6= (0, 0),

0, (x1, x2) = (0, 0).

Show that limx→0 f(x,mx) = 0 for every m ∈ R but f is discontinuous at (0, 0). Interpret

this result.

Problem 3 Let C1([a, b]) be the metric space of continuously differentiable functions on

[a, b] with metric

d(f, g) = supa≤x≤b


|f ′(x)− g′(x)| .

Let D : C1([a, b])→ C([a, b]) be defined by D(f) = f ′. Prove that D is continuous.

Problem 4 Let C([a, b]) be the metric space of continuous functions on [a, b] with metric

d(f, g) =

∫ b

a

|f(t)− g(t)| dt.

Define T : C([a, b])→ R by

T (f) =

∫ b

a

f(t) dt.

151

(a) Prove that T is continuous using sequential definition of continuity.

(b) Prove that T is continuous using δ − ε definition.

Problem 5 Let f and g be continuous maps of a metric space X into a metric space Y

and let E be a dense subset of X.

(a) Prove that f(E) is dense in f(X).

(b) If g(x) = f(x) for all x in a dense subset E of X, prove that g(x) = f(x) for all

x ∈ X.

Problem 6 Let A be a set and x a point in a metric space (X, d). Define the “distance

from x to A” to be

d(x,A) = infy∈A

d(x, y).

(a) Show that for fixed A, the function f : X → R (with the usual metric on R) given by

f(x) = d(x,A) is continuous. (b) Show {x : d(x,A) = 0} = A.

Problem 7 Suppose f is a mapping from metric space X to metric space Y . Prove

that a function is continuous if and only if f−1(A) is closed in X for every closed set

A ⊂ Y .

Problem 8 Assume f : R → R is continuous and define A = {x : f(x) = 0}. Show

A is closed.

Problem 9 Suppose f is a continuous bijection of a compact metric space X onto a

metric space Y . Prove that the inverse map f−1 defined on Y by f−1(f(x)) = x (where

x ∈ X) is a continuous mapping of Y onto X.

Problem 10 Let (X, d) be a metric space and F,G ⊂ X with F closed, G open, and

F ⊂ G. Show that there is a continuous function f : X → R such that 0 ≤ f(x) ≤ 1,

f(x) = 1 for x ∈ F , and f(x) = 0 for x ∈ Gc.

Problem 11 Let f and g be continuous maps of a metric space X into a metric space

Y and let E be a dense subset of X. (a) Prove that f(E) is dense in f(X). (b) If

g(x) = f(x) for all x in a dense subset E of X, prove that g(x) = f(x) for all x ∈ X.

Explain why these results are important for our everyday experience with functions and

numbers.

Problem 12 (a) Show that the requirement in the definition of uniform continuity can be

rephrased as follows, in terms of diameters of sets: For every ε > 0 there exists a δ > 0

such that diam(f(E)) < ε for all E ⊂ X with diam(E) < δ. (b) Let E be a dense subset

of a metric space X, and let f be a uniformly continuous function into a complete metric

space Y defined on E. Prove that f has a continuous extension from E to X in two ways152

using part (a) and then using Cauchy sequences.

Problem 13 Let f : R→ R be a continuous function that is periodic with period T > 0.

(a) Show that f has a finite maximum and minimum values on R. (b) Show that f is

uniformly continuous on R.

Problem 14 Let X be a metric space.

(a) Call two Cauchy sequences (pn), (qn) in X equivalent if

limn→∞

d(pn, qn) = 0.

Prove that this is an equivalence relation.

(b) Let X∗ be the set of all equivalence class so obtained. If P ∈ X and Q ∈ X, (pn) ∈ Pand (qn) ∈ Q, define

∆(P,Q) = limn→∞

d(pn, qn).

Note that by an exercise from lecture 1 that this limit exists. Show that the number

∆(P,Q) is unchanged if (pn) and (qn) are replaced by equivalent sequences, and hence

that ∆ is a distance function in X.

(c) Prove that the resulting metric space X is complete.

(d) For each p ∈ X, there is a Cauchy sequence all of whose terms are p; let Pp be the

element of X which contains this sequence. Prove that ∆(Pp, Pq) = d(p, q) for all

p, q ∈ X. In other words, the mapping φ defined by φ(p) = Pp is an isometry of X

into X.

(e) Prove that φ(X) is dense in X, and that φ(X) = X if X is complete. By part (d),

we may identify X and φ(X) and thus regard X as embedded in the complete space

X. We call X the completion of X.

Problem 15 Let (X, d) be a metric space and suppose f is uniformly continuous on the

compact subsets E1, . . . , En of X. (a) Prove that f is uniformly continuous on ∪ni=1Ei.

(b) Give an example of a function f that is continuous on R and a sequence of compact

intervals E1, E2, E3, . . . on each of which f is uniformly continuous, yet f is not uniformly

continuous on ∪∞i=1Ei.

Problem 16 Prove that if f is a continuous mapping of a metric space X into a met-

ric space Y and E ⊂ X is connected, then f(E) is connected.

14.5. Sequence of Functions

Problem 1 Prove the sequence ( xn+x2

) converges uniformly on R with the usual metric.

Problem 2 (a) Define what it means for a sequence of functions to be uniformly Cauchy

and prove that a uniformly convergent sequence of functions is uniformly Cauchy. (b)

Prove that every uniformly convergent sequence of bounded functions is uniformly bounded.153

Here, we say that a sequence of real-valued functions (fn) is uniformly bounded if there

exists M > 0 such that ‖fn‖∞ ≤ M . (c) Show there exists a sequence of functions that

converges pointwise to the function that is identically 0, but each function in the sequence

is unbounded Hint: Consider the sequence {x/n} on [0,∞).

Problem 3 (a) Prove that if the sequences of real-valued functions (fn) and (gn) converge

uniformly on A ⊂ X, then (fn + gn) converges uniformly on A. (b) If fn and gn are also

bounded for each n ∈ N, prove that (fngn) converges uniformly on A. (c) Show that

the boundedness is necessary in (b) by constructing an example where (fngn) does not

converge uniformly (although it must converge pointwise).

Problem 4 Let (X, dX) and (Y, dY ) be metric spaces, A ⊂ X, and (fn) a sequence of

continuous functions from A to Y that converges to a function f : A → Y uniformly on

A. Given any x ∈ A, prove that limn→∞ fn(xn) = f(x) for every sequence (xn) in A with

xn → x. Construct an example showing the converse is not true, but if A is compact,

then the converse is true.

Problem 5 (General Arzela Ascoli Theorem) If X is a compact metric space, (fn) ⊂ C(K),

(fn) is pointwise bounded and equicontinuous on X, prove that

(a) (fn) is uniformly bounded on X, and

(b) f(n) contains a uniformly convergent subsequence.

Problem 6 Let (X, d) be a metric space and suppose K ⊂ X is compact, fn → f on K

(and all functions are real-valued and continuous), and fn(x) ≥ fn+1(x) for all x ∈ K and

n ∈ N. Prove that fn → f uniformly on K. Construct an example showing the necessity

of compactness.

Problem 7 Is the set of functions{x2 + n

1+nx}∞n=1

taking R to R (with the usual metric)

equicontinuous on [0, 1]?

Problem 8 Compute and plot the Bernstein polynomials for 1/(1 + x) on [0, 1] of de-

grees 1, 2, and 3.

Problem 9 Assume that (fn) is a sequence of functions with fn : [a, b] → R for all n

where we take the usual metric on [a, b] and R.

(a) Assume that {fn} is uniformly Lipschitz continuous on [a, b] with constant L. Prove

that {fn} is equicontinuous on [a, b].

(b) Assume that {fn} is uniformly Lipschitz continuous on [a, b] with constant L and

further that (fn(x)) is a uniformly bounded sequence of numbers for some a ≤ x ≤ b.

(1) Prove that (fn) has a subsequence that converges uniformly on [a, b]. (2) Prove

the second assumption is needed with an example.154

(c) Assume that (fn) is a bounded subset of C([a, b]). Show that the sequence of functions

(Fn) from [a, b] into R defined by

Fn(x) =

∫ x

a

fn(x) ds

contains a subsequence that converges uniformly on [a, b]. (This is quite a remark-

able result when considering the sequence of functions (sin(nx)) as an example of a

bounded sequence in C([a, b]) that fails to have any convergent subsequence.)

Problem 10 Suppose K = {x1, . . . , xn} is a finite space with the discrete metric d. Show

C(K) is linearly isomorphic to the finite-dimensional space Rn with the ∞-norm (which

is just the maximum norm in finite dimensions).

Problem 11 Suppose f is a real-valued continuous function with dom(f) = R. Let

fn(x) = f(nx) for n ∈ N and any x ∈ R. If {fn} is equicontinuous, prove that f must be

a constant function.

Problem 12 Suppose that (fn) is an equicontinuous sequence of functions on a com-

pact set K and (fn) converges pointwise on K. Prove that (fn) converges uniformly on

K.

Problem 13 Let X and Y be two metric spaces and A ⊂ X. Let (fn) be a sequence

of functions from A to Y that converges to f pointwise on A. Prove the following theo-

rems.

(a) Let Mn = supx∈A dY (fn(x), f(x)). Prove that fn → f uniformly if and only if Mn →0.

(b) If fn is continuous on A for each n, then f is continuous on A. This is a generalization

of the theorem in the notes.

(c) Suppose Y is complete, then fn converges uniformly on A if and only if for every ε > 0

there exists an integer N such that m ≥ N , n ≥ N , x ∈ E implies dY (fn(x), fm(x)) <

ε. Construct an example showing the necessity of Y being complete.

(d) Suppose Y is complete, and fn converges uniformly on A. Let x be a limit point of A

and suppose that limt→x fn(t) = yn for each n. Then (yn) converges and limt→x f(t) =

limn→∞ yn. In other words, prove limt→x limn→∞ fn(t) = limn→∞ limt→x fn(t). Con-

struct an example showing the necessity of Y being complete.

Problem 14 Prove that if f ∈ C([0, 1]) and∫ 1

0f(x)xn dx = 0 for all n ∈ N, then f(x) = 0

on [0, 1]. Hint: If∫ 1

0f 2(x) dx = 0, then a result from elementary real analysis states that

f ≡ 0. Consider a sequence of polynomials that converge uniformly to f and take the

product with f to get a uniformly convergent sequence to f 2.

Problem 15 Let (X, d) be a metric space and pick any fixed but arbitrary a ∈ X. Define155

the real-valued function fp for each p ∈ X by

fp(x) = d(x, p)− d(x, a),

where x ∈ X.

(a) Prove that ‖fp‖∞ ≤ d(a, p) and fp ∈ C(X).

(b) Prove that

‖fp − fq‖∞ = d(p, q)

for all p, q ∈ X.

(c) Let Φ(p) = f(p), so that Φ is an isometry of X onto Φ(X) ⊂ C(X). Let Y be the

closure of Φ(X) in C(X). Show that Y is complete which implies X is isometric to a

dense subset of a complete metric space Y . Note: This is a different proof of what we

previously showed in lecture 3 examples about defining the closure of a metric space.

14.6. Differentiation

Problem 1 Define the discontinuous real-valued function f : R2 → R by

f(x, y) =

xyx2+y2

, if(x, y) 6= (0, 0),

0, else.

prove that (the Gateaux derivatives) Dxf(x, y) and Dyf(x, y) exist everywhere in R2 even

though f is discontinuous at (0, 0).

Problem 2 Suppose that f : A ⊂ Rn → R where A is an open set and the partial

derivatives D1f, . . . , Dnf are bounded in A. Prove that f is continuous in A.

Problem 3 Suppose that f : A ⊂ Rn → R is differentiable, where A is an open set

and that f has a local maximum at a point x ∈ A. Prove f ′(x) = 0.

Problem 4 Suppose that f : A ⊂ Rn → R is differentiable, where A is an open connected

set and f ′(x) = 0 for every x ∈ A. Prove that f is constant in A.

Problem 5 Let f : R2 → R2 be defined by

f(x, y) =

(f1(x, y)

f2(x, y)

)=

(ex cos y

ex sin y

).

(a) Show that the range of f is R2\ {(0, 0)}.(b) Show that the Jacobian of f is non-zero on R2. Thus, every point of R2 has a

neighborhood in which f is 1-1, but f is not 1-1 on R2.156

(c) Let (x, y) = (0, π/3) and (u, v) = f(x, y). Let g = f−1 denote the continuous inverse

of f defined in a neighborhood of (u, v) such that g(u, v) = (x, y). Find an explicit

formula for g, compute f ′(x, y), g′(u, v) and show that g′(f(x))f ′(x) = IR2 .

Problem 6 If X = Y = Z = R in the implicit function theorem, interpret the theorem

and its proof graphically.

Problem 7 Define f : R3 → R by f(x, y, z) = x2y + ex + z. Show that f(0, 1,−1) = 0,

Dxf(0, 1,−1) 6= 0, and that there exists a differentiable function g in some neighborhood

of (1,−1) in R2 such that g(1,−1) = 0 and f(g(y, z), y, z) = 0. Find Dyg(1,−1) and

Dzg(1,−1).

Problem 8 Suppose f ≥ 0, f is continuous on [a, b], and∫ baf(x) dx = 0. Prove that

f(x) = 0 for all x ∈ [a, b].

Problem 9 Let p and q be positive real numbers such that 1/p + 1/q = 1. Prove the

following statements.

(a) If u ≥ 0 and v ≥ 0, then

uv ≤ up

p+vq

q,

and equality holds if and only if up = vq.

(b) If f, g ∈ R([a, b]), f, g ≥ 0, and∫ bafp = 1 =

∫ bagq, then

∫ bafg ≤ 1.

(c) If f, g,∈ R([a, b]), then∣∣∣∣∫ b

a

fg

∣∣∣∣ ≤ [∫ b

a

|f |p]1/p [∫ b

a

|g|q]1/q

.

157

Notes on Analysis

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