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REAL ANALYSIS

Problems

CLAUDIA TIMOFTE

REAL ANALYSIS

Problems

To my students

Preface

The present book is a collection of problems in differential calculus. The book

is based on some lectures I delivered for a number of years at the Faculty of Physics

of the University of Bucharest and covers the curriculum on differential calculus for

the students of the first year of this faculty.

The material follows the textbook [34]. Each chapter contains a brief review of

the corresponding theoretical results, worked out examples and proposed problems.

Since the ”learning-by-doing” method is a successful one, the student is encouraged

to solve as many exercises as possible. The basic prerequisites for studying dif-

ferential calculus using this book are undergraduate courses in linear algebra and

one-variable calculus.

The second volume will cover important topics, such as Fourier series and integral

calculus. The reader interested in acquiring a more profound knowledge of differen-

tial calculus is referred to some very good textbooks and monographs recommended

at the end of these notes.

It is my hope that this book will serve as an useful outlook for the students of

the first year of the Faculty of Physics of the University of Bucharest.

I would like to thank to my students for their continuous questions, comments

and suggestions, which helped me to improve the content of these notes.

Claudia Timofte

7

Contents

1 Real Numbers 11

1.1 Elements of Topology of the Real Line. Elementary Inequalities . . . 11

1.2 Number Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

2 Multidimensional Spaces 29

2.1 Metric Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

2.2 Elements of Topology . . . . . . . . . . . . . . . . . . . . . . . . . . 33

2.3 Sequences in Metric Spaces . . . . . . . . . . . . . . . . . . . . . . . 39

2.4 Inner Product Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

2.5 Normed Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

3 Series of Real and Complex Numbers 55

4 Limits and Continuity 69

4.1 Limits of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

4.2 Continuous Functions . . . . . . . . . . . . . . . . . . . . . . . . . . 76

5 Multivariable Differential Calculus 89

5.1 Derivable Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

5.2 Differential Operators . . . . . . . . . . . . . . . . . . . . . . . . . . 111

5.3 Taylor’s Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

5.4 Implicit Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125

5.5 Local Extrema . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131

9

10 CONTENTS

6 Sequences and Series of Functions 145

6.1 Sequences of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . 145

6.2 Series of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149

6.3 Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152

6.4 Taylor Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156

Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .161

Chapter 1

Real Numbers

1.1 Elements of Topology of the Real Line. ElementaryInequalities

We shall begin with a quick review of some results and ideas from one-variable

calculus. More precisely, we shall recall some results concerning the topolology of

the real line and some useful elementary inequalities. To fix the notation, let us

recall some important sets we often encounter in mathematics; to identify them,

they have acquired special names and notational conventions.

One of these is the empty or the void set ∅, which is the unique set containing

no elements. Special sets of numbers include:

a) the set of natural numbers, N = {0, 1, 2, . . . , n, . . .};

b) the set of natural numbers, without zero, N∗ = {1, 2, . . . , n, . . .};

c) the set of integer numbers, Z = {0,± 1,± 2, . . . ,±n, . . .};

d) the set of rational numbers, Q =

{m

n| m,n ∈ Z, n = 0

};

e) the set of the real numbers R, also called the number line, the number axis,

the real line or the the real axis, having in mind its geometrical representation

(for more details, see [34]).

In what follows, we shall consider (R,+, ·,≤) as being a completely ordered commu-

tative field.

11

12 REAL ANALYSIS

Definition 1.1 For any real number a, the absolute value or the modulus of a is

defined as follows:

|a| ={a, if a ≥ 0,

−a, if a < 0.

As can be seen from the above definition, the absolute value of a is always positive.

Definition 1.2 Let a, b ∈ R, with a < b.

a) The set [a, b] = {x ∈ R | a ≤ x ≤ b} is called the closed interval with end

points a and b.

b) The set (a, b) = {x ∈ R | a < x < b} is said to be the open interval with end

points a and b.

c) The set [a, b) = {x ∈ R | a ≤ x < b} is called the half-open (semi-open,

half-closed) interval with end points a and b.

All the above sets are bounded intervals. We can also consider the following

unbounded intervals: (a,∞) = {x ∈ R | x > a}; [a,∞) = {x ∈ R | x ≥ a};(−∞, b) = {x ∈ R | x < b}; (−∞, b] = {x ∈ R | x ≤ b}; (−∞,∞) = R.

Let us mention that if we take a, b ∈ R, with a = b, we obtain degenerate

intervals. For instance, (a, a) = ∅ (the empty set) and [a, a] = {a} (a singleton).

Definition 1.3 Let a ∈ R, r ∈ R, r > 0. The set B(a, r) = (a− r, a+ r) is called

the open interval centered at the point a and having the radius r.

Definition 1.4 Let a ∈ R, r ∈ R, r ≥ 0. The set B(a, r) = [a − r, a + r] is called

the closed interval centered at the point a and having the radius r.

Remark 1.5 If in the above definition we take r = 0, then B(a, 0) = {a}.

Definition 1.6 A set A ⊆ R is said to be an open set if for any x ∈ A there exists

r > 0 such that B(x, r) ⊆ A.

Remark 1.7 The empty set is considered to be an open set.

Proposition 1.8 Any centered open interval is an open set.

REAL NUMBERS 13

Definition 1.9 A set B ⊆ R is called a closed set if R \B is an open set.

Remark 1.10 R and ∅ are simultaneously open and closed sets.

Definition 1.11 (The Extended Real Number Line R) Let −∞ and ∞ be two dif-

ferent elements which are not real numbers. They are called points at infinity or

improper numbers. Let R = R ∪ {−∞} ∪ {∞}. We extend to R the usual order

relation defined on R by putting −∞ < x < ∞, for any x ∈ R, and −∞ < ∞. The

set R, endowed with the above order relationship, is called the extended real line.

Remark 1.12 Any nonempty subset of R possesses in R a supremum and an infi-

mum.

Definition 1.13 For some pairs of elements from R, we extend the usual operations

of addition and multiplication in R as follows:

1) a+∞ = ∞+ a = ∞, ∀a ∈ R, a = −∞;

2) a+ (−∞) = (−∞) + a = −∞, ∀a ∈ R, a = ∞;

3) a · ∞ = ∞ · a = ∞, ∀a > 0;

4) a · (−∞) = (−∞) · a = −∞, ∀a > 0;

5) b · ∞ = ∞ · b = −∞, ∀b < 0,

6) b · (−∞) = (−∞) · b = ∞, ∀b < 0;

7)a

∞=

a

−∞= 0, ∀a ∈ R.

Remark 1.14 We do not define:

∞+ (−∞), (−∞) +∞, 0 · (±∞), (±∞) · 0, ±∞±∞

.

The elements of R are called finite real numbers. Therefore, if a ∈ R, then

−∞ < a <∞.

The element −∞ is called minus infinity or negative infinity, while the element ∞is called plus infinity or positive infinity.

14 REAL ANALYSIS

The extended real number system is also called the closed real line and it is

denoted by R = [−∞,+∞]. Intervals in R are sets I ⊆ R of the form (a, b), [a, b], (a, b]

or [a, b), where a, b ∈ R and a < b. So, R = [−∞,+∞].

If a < b, a, b ∈ R, then

(a, b), (a,∞), (a,∞], (−∞, a), [−∞, a), R, R,

are open intervals in R.

If a, b ∈ R, with a ≤ b, then [a, b] is a closed interval in R.

Remark 1.15 The set (R,≤) is a totally ordered set.

Exercise 1.16 Show that the absolute value has the following properties:

1) |a| =√a2;

2) |a| = 0 if and only if a = 0;

3) |a+ b| ≤ |a|+ |b|;

4) | − a| = |a|;

5) ||a| − |b|| ≤ |a− b|, for any a, b ∈ R.

Exercise 1.17 Let A1, A2, . . . , Ak ∈ R be open sets. Prove thatk∩

i=1Ai is an open

set.

Exercise 1.18 Let {Ai}i∈I be an arbitrary family of open sets in R. Prove that∪i∈I

Ai is also an open set.

Exercise 1.19 Prove that any open interval (a, b) is an open set.

Exercise 1.20 Show that any closed interval [a, b] is a closed set.

Exercise 1.21 Let B1, B2, . . . , Bk ⊆ R be closed sets. Prove thatk∪

i=1Bi is also a

closed set.

Exercise 1.22 Let {Bi}i∈I be an arbitrary family of closed sets in R. Prove that∩i∈I

Bi is also a closed set.

REAL NUMBERS 15

Hint: use the fact that

R \k∪

i=1

Bi =k∩

i=1

(R \Bi)

and

R \∩i∈I

Bi =∪i∈I

(R \Bi).

Exercise 1.23 If a1, a2, . . . , ak ∈ R∗+ and a1 ·a2 · · · ak = 1, show that a1+a2+ · · ·+

ak ≥ k.

Hint. Use Use mathematical induction..

Exercise 1.24 Let

Ak =x1 + x2 + · · ·+ xk

k, Gk = k

√x1 · x2 · · ·xk,

and

Hk =k

1x1

+ 1x2

+ · · ·+ 1xk

be the arithmetic, geometric and the harmonic mean of the positive real numbers

x1, x2, . . . , xk. Prove that Ak ≥ Gk ≥ Hk.

Exercise 1.25 Prove that for any nonnegative real numbers x1, x2, . . . , xk, we have

x1 + x2 + · · ·+ xkk

≥ k√x1 · x2 · · ·xk

and the equality holds true if and only if x1 = x2 = · · · = xk. In other words, the

arithmetic mean is greater than or equal to the geometric mean and they are equal

if and only if all the numbers xi are equal.

Hint. Use mathematical induction.

Exercise 1.26 Let x1, x2, . . . , xk be positive numbers. Show that

max{x1, x2, . . . , xk} ≥ x1 + x2 + · · ·+ xkk

≥

k√x1x2 · · ·xk ≥ k

1x1

+ 1x2

+ · · ·+ 1xk

≥ min{x1, x2, . . . , xk}

The equality is obtained if and only if x1 = x2 = · · · = xk.

16 REAL ANALYSIS

The above inequalities can be generalized to power means and weighted power

means.

Exercise 1.27 (Bernoulli’s Inequality) Prove that for any a, b ∈ R, with a ≥ −1

and b ≥ 1, we have

(1 + a)b ≥ 1 + a b.

Hint. Use the fact that the function f : [−1,∞) → R, defined by f(x) = (1 + x)b −bx− 1, is monotone and attains a minimum equal to zero at the point x = 0.

Exercise 1.28 Prove that for any a ∈ R, with a ≥ −1 and any n ∈ N∗, we have

(1 + a)k ≥ 1 + k a.

Hint. Take b = k in Bernoulli’s inequality.

Exercise 1.29 Prove that for any a1, a2, . . . , ak ∈ R, with ai ≥ −1 for any i and

having all the same sign, we have

(1 + a1) (1 + a2) · · · (1 + ak) ≥ 1 + a1 + a2 + · · ·+ ak.

Hint. Use mathematical induction.

Exercise 1.30 Prove that for any a1, a2, . . . , ak ∈ R∗+, with k ≥ 2, we have

a1a2

+a2a3

+ · · ·+ ak−1

ak+aka1

≥ k.

Hint. Use Exercise 1.25.

Exercise 1.31 Prove that for any a1, a2, . . . , ak ∈ R∗+, with ai > 0 and such that

a1 · a2 · · · ak = 1, we have

(1 + a1) (1 + a2) · · · (1 + ak) ≥ 2k.

Solution. Using the inequality

1 + ai2

≥√ai,

we obtain immediately the result.

REAL NUMBERS 17

Exercise 1.32 Prove that for any x > 0 and any k ∈ N, we have

xk

1 + x+ x2 + · · ·+ x2k≤ 1

2k + 1.

Solution. Using the inequality Ak ≥ Gk, we get

xk =2k+1

√1 · x · x2k ≤ 1 + x+ x2 + · · ·+ x2k

2k + 1.

Exercise 1.33 Let ak be an arithmetic progression containing positive terms. Show

that√a1 ak ≤

√k ≤ k

√a1 · a2 · · · ak ≤ a1 + ak

2.

Solution. The first inequality can be immediately proven by induction. The second

one is an easy consequence of the geometric-arithmetic mean inequality.

Exercise 1.34 Show that

√k ≤ k

√k! ≤ k + 1

2, ∀k ∈ N∗.

Solution. Use Exercise 1.33.

Exercise 1.35 Let a, b, c > 0. Show that

bc

a+ac

b+ab

c≥ a+ b+ c

and1

a+

1

b+

1

c≥ 1√

bc+

1√ca

+1√ab.

Solution. Since a2 + b2 + c2 ≥ ab + bc + ca, it follows that b2 c2 + a2 c2 + a2 b2 ≥a b c (a+ b+ c), which leads to the first desired result. The second inequality can

be proven in a similar way.

18 REAL ANALYSIS

1.2 Number Sequences

We start this section with by a brief summary of the main concepts and results

about number sequence. This summary is not intended to be a substitute for the

textbook [34] that the student is advised to read carefully.

Definition 1.36 A map f : N → R is called a sequence of real numbers. Denoting

ak = f(k), a sequence may be symbolized by (ak)k∈N or, simply, by (ak)k. For

shortness, the notation (ak) ⊆ R is also used. The element ak is called the general

term of the sequence (ak)k.

Definition 1.37 Let A be a nonempty set and (ak)k∈N be a given sequence in A.

A subsequence of (ak)k∈N is a sequence of the form (akp)p∈N, where (kp) ⊆ N is a

strictly increasing sequence, for any p ∈ N.

Definition 1.38 A sequence (ak)k∈N of real numbers is called bounded if there exists

a real number M ≥ 0 such that |ak| ≤M, ∀k ∈ N.

Example 1.39 For instance, the sequence ak = (−1)k is bounded, since |ak| ≤ 1,

for all k ∈ N.

Remark 1.40 If there exist two real numbers m and M such that m ≤ ak ≤M , for

any k ∈ N, then the sequence (ak) is said to be bounded from above and from below.

M is called the upper bound and m the lower bound. A sequence (ak is bounded if

and only if (ak) has an upper and a lower bound.

Definition 1.41 A sequence (ak) is said to be unbounded if for any positive number

M , there exists kM ∈ N such that |akM | > M .

Example 1.42 If q > 1, the sequence ak = qk is unbounded.

Definition 1.43 A sequence (ak)k∈N of real numbers is called convergent if there

exists a ∈ R such that for any neighborhood V of a, there exists nV ∈ N such that

ak ∈ V , for any k ≥ kV .

The element a ∈ R is called the limit of the sequence (ak)k∈N and we shall symbolize

this as follows:

a = limk→∞

ak.

REAL NUMBERS 19

Proposition 1.44 A convergent sequence of real numbers has only one limit.

Proposition 1.45 Let (ak)k∈N be a sequence of real numbers. The sequence (ak)k∈N

is convergent to a ∈ R if and only if for any positive number ε, there exists kε ∈ Nsuch that

|ak − a| < ε, ∀k ≥ kε.

Proposition 1.46 Let (ak)k∈N and (bk)k∈N be two given sequences of real numbers

with |ak| ≤ bk, for any k ∈ N. If we assume that bk → 0, then ak → 0. As a

consequence, if (ak)k∈N, (bk)k∈N and (ck)k∈N are three given sequences of real numbers

with ak ≤ bk ≤ ck, for any k ∈ N, and if we assume that (ak)k∈N and (ck)k∈N converge

to a certain limit l ∈ R, then (bk)k∈N is convergent to l, as well.

Remark 1.47 If (ak)k∈N and (bk)k∈N are convergent sequences of real numbers,

then:

limk→∞

(ak + bk) = limk→∞

ak + limk→∞

bk,

limk→∞

(ak bk) = ( limk→∞

ak) ( limk→∞

bk)

and if limk→∞

bk = 0 and bk = 0 for any k, then

limk→∞

(ak/bk) = ( limk→∞

ak)/( limk→∞

bk).

Proposition 1.48 A convergent sequence is bounded.

Proposition 1.49 Any subsequence of a convergent sequence is convergent to the

same limit.

Definition 1.50 A sequence (ak)k∈N of real numbers is called increasing if ak ≤ak+1, for all k ∈ N and strictly increasing if ak < ak+1, for all k ∈ N. The sequence

(ak)k∈N is called decreasing if ak ≥ ak+1, for all k ∈ N and strictly decreasing if

ak > ak+1, for all k ∈ N.

Some authors use another terminology; more precisely, they call a sequence (ak)k∈N

of real numbers nonincreasing if ak ≥ ak+1, ∀k ∈ N. A sequence (ak)k∈N of real

numbers is called nondecreasing if ak ≤ ak+1, ∀k ∈ N.

20 REAL ANALYSIS

Proposition 1.51 a) Any sequence of real numbers which is nondecreasing and

bounded above is convergent to its least upper bound.

b) Any sequence of real numbers which is nonincreasing and bounded below is

convergent to its greatest lower bound.

Theorem 1.52 (Bolzano-Weierstrass) A bounded sequence always contains a con-

vergent subsequence.

Definition 1.53 A sequence (ak)k∈N of real numbers is said to be a fundamental

sequence or a Cauchy sequence if for any positive number ε, there exists kε ∈ N such

that

|ap − am| < ε, ∀p,m ≥ kε.

Proposition 1.54 Any convergent sequence is fundamental.

Proposition 1.55 Any Cauchy sequence is bounded.

Proposition 1.56 Any Cauchy sequence in R is convergent.

Theorem 1.57 Let (ak)k∈N be a sequence of real numbers. The sequence (ak)k∈N is

convergent if and only if it is a fundamental one.

Definition 1.58 a) A sequence (ak)k∈N ⊆ R is said to be convergent to ∞ if for

any M ∈ R, there exists kM ∈ N such that

ak ≥M, ∀k ≥ kM .

b) A sequence (ak)k∈N ⊆ R is said to be convergent to −∞ if for any M ∈ R,there exists kM ∈ N such that

ak ≤M, ∀k ≥ kM .

Remark 1.59 Let (ak)k∈N ⊆ R.

a) If (ak)k∈N is increasing, then it is convergent in R and

limk→∞

ak = supk∈N

ak (in R);

REAL NUMBERS 21

b) If (ak)k∈N is decreasing, then it is convergent in R and

limk→∞

ak = infk∈N

ak (in R).

Let (ak)k≥1 ⊆ R. Consider the sequence (bk)k≥1 defined by

bk = infj≥k

aj , k ≥ 1.

Obviously, this sequence is increasing, and, therefore, it converges to its supremum.

So, there exists

l = limk→∞

bk = supk≥1

bk. (1.1)

In a similar manner, the sequence (ck)k≥1, defined by

ck = supj≥k

aj , k ≥ 1,

is decreasing, and, therefore, it converges to its infimum. So, there exists

L = limk→∞

ck = infk≥1

ck. (1.2)

The element l ∈ R defined by (1.1) is called the limit inferior of the sequence

(ak)k and it is denoted by lim infk→∞

ak or by limk→∞

ak.

The element L ∈ R defined by (1.2) is called the limit superior of the sequence

(ak)k and it is denoted by lim supk→∞

ak or by limk→∞

ak.

Hence, the inferior and, respectively, superior limits of the sequence (ak)k≥1 are:

lim infk→∞

ak = supk≥1

infj≥k

aj

and

lim supk→∞

ak = infk≥1

supj≥k

aj .

Obviously,

−∞ ≤ lim infk→∞

ak ≤ lim supk→∞

ak ≤ +∞

and these quantities always exist.

22 REAL ANALYSIS

Also, it is easy to see that a sequence of real numbers (ak)k≥1 is convergent in Rif and only if

lim infk→∞

ak ∈ R,

lim supk→∞

ak ∈ R

and

lim infk→∞

ak = lim supk→∞

ak.

So, if a sequence (ak)k≥1 ⊆ R is convergent in R, then

limk→∞

ak = limk→∞

ak = limk→∞

ak.

Let us notice that if (ak)k≥1 is a bounded sequence of real numbers, then its limit

superior can be viewed as being the largest limit of any subsequence of (ak)k≥1. Of

course, a similar statement is true for the limit inferior. In fact, the limit superior of

an arbitrary sequence can be viewed as being the largest subsequential limit of all

its convergent subsequences, including those possible subsequences tending to +∞or −∞. Similarly, the limit inferior is the smallest subsequential limit. This remark

is better described in the following theorem:

Theorem 1.60 Let (ak)k be an arbitrary sequence. Let l and, respectively, L be its

limit inferior and, respectively, superior. Let (akp) be an arbitrary subsequence of

(ak) such that limp→∞

akp exists, it is +∞ or −∞. Then,

l ≤ limp→∞

akp ≤ L.

Moreover, there exists a subsequence (akp) of (ak) such that

L = limp→∞

akp

and there exists a subsequence (a ′kp) of (ak) such that

l = limp→∞

a ′kp.

REAL NUMBERS 23

Remark 1.61 If (ak)k ⊆ R+, then:

limk→∞

ak+1

ak≤ lim

k→∞k√ak ≤ lim

k→∞k√ak ≤ lim

k→∞

ak+1

ak.

Moreover, if there exists limk→∞

ak+1

ak, then

limk→∞

ak+1

ak= lim

k→∞k√ak.

Exercise 1.62 Prove that the sequence

ak =k2 + 1

k, k ≥ 1

is strictly increasing.

Solution. Sincek2 + 1

k= k +

1

k,

we obtain

ak+1 − ak = 1− 1

k(k + 1)> 0.

Therefore, ak+1 > ak, i.e. our sequence is strictly increasing.

Exercise 1.63 Show that the sequence

ak =k

2k, k ≥ 1

is strictly decreasing and find its limit.

Solution. Since

ak+1 =k + 1

2 kak < ak,

for k > 1, it follows that the sequence (ak)k is strictly decreasing. Since, obviously,

(ak)k is bounded from below by 0, it follows that (ak)k is convergent. Let a =

limk→∞

ak. Then, we get a =1

2a, which means that a = 0.

24 REAL ANALYSIS

Exercise 1.64 Let a1 =√2 and

ak+1 =√2 +

√ak, ∀k ≥ 1.

Show that (ak)k is convergent and find its limit.

Solution. By induction, we can prove that 0 < ak < 2 and, since

a2k+1 − a2k =√ak −

√ak−1,

that (ak)k is monotone. Hence, (ak)k is convergent. Denoting by l its limit, we get

l =√2 +

√l, which, using Cardan’s formula, leads to

l =1

3

(3

√1

2

(79 + 3

√249

)+

3

√1

2

(79− 3

√249

)− 1

).

Exercise 1.65 Let a ∈ (−1, 1) and

ak =k∑

j=1

j aj , ∀k ≥ 1.

Prove that (ak)k is convergent and has the limit equal toa

(1− a)2.

Exercise 1.66 Let a1 = 1,

ak = k (ak−1 + 1).

Show that

limk→∞

k∏j=1

(1 +

1

aj

)= e ( Euler’s number).

Exercise 1.67 Let (ak)k≥1 be a convergent sequence having the limit equal to a.

Prove the existence and find the value of the following limit:

l = limk→

1

k

k∑i=1

xi.

Prove a similar result for the geometric mean of the first k terms of a sequence of

strictly positive real numbers.

REAL NUMBERS 25

Exercise 1.68 Let (ak)k≥1 be given by the following recurrence formula:a1 = a2 = 1,

ak+1 =1

ak−1 +1

ak

, ∀k ≥ 2.

Show that this sequence is convergent and find its limit.

Exercise 1.69 Let (ak)k≥1 be defined by the following recurrence formula:a0 = 0,

ak+1 =1

2

(a+ a2k

), ∀k ≥ 0,

where a ∈ [0, 1]. Prove that this sequence is convergent and find its limit.

Exercise 1.70 Test the following sequences for convergence:

a) ak =ln k!

kp, p ≥ 2;

b) ak =k∑

i=1

(−1)i−1 1

i.

Exercise 1.71 Test the following sequences for convergence:

ak =k∑

i=1

1

i− ln k ( Euler’s sequence ).

Solution. We prove that the sequence (ak)k is monotone and bounded. To this end,

using Lagrange’s formula for the function ln : [k, k + 1] → R, for k > 0, it follows

that there exists ξ ∈ (k, k + 1) such that ln (k + 1)− ln k = 1/ξ. Then, since

1

k + 1<

1

ξ<

1

k,

we get1

k + 1< ln (k + 1)− ln k <

1

k.

26 REAL ANALYSIS

Now, since ak+1 − ak = 1k+1 − ln (k + 1) + ln k < 0, it follows that (ak)k is strictly

decreasing. On the other hand,

ak = 1+1

2+1

3· · ·+1

k−ln k > (ln 2−ln 1)+(ln 2−ln 1)+· · ·+(ln(k + 1)− ln k)−ln k =

ln(k + 1)− ln k > 0.

Thus, the sequence (ak)k is bounded from below, and, hence, convergent.

Exercise 1.72 Prove that the sequence (ak)k, defined by

ak =cos 1

1 · 2+

cos 2

2 · 3+ · · ·+ cos k

k · (k + 1), k ≥ 1,

is convergent.

Solution. It is not difficult to see that (ak)k is a Cauchy sequence in R. Hence, (ak)k

is convergent.

Exercise 1.73 Prove that the sequence (ak)k, defined by

ak =k!

(2k + 1)!!, k ≥ 1,

is convergent and find its limit.

Solution. It is not difficult to see that

ak+1 =k + 1

2k + 3ak < ak,

for k > 1. Thus, it follows that the sequence (ak)k is strictly decreasing. Since,

obviously, (ak)k is bounded from below by 0, it follows that (ak)k is convergent. Let

a = limk→∞

ak. Then, we get a =1

2a, which means that a = 0.

Exercise 1.74 Let (ak) be a sequence in R. Show that ak → 0 if and only if

|ak| → 0.

Exercise 1.75 If ak → a in R and p ∈ N, prove that akp → ap.

REAL NUMBERS 27

Solution. We prove the result by induction on p. The conclusion is obviously true

for p = 1. Suppose that akp−1 → ap−1. Then, ak

p = akp−1 · ak → ap−1 · a = ap.

Exercise 1.76 Let ak =5k3 + 4k + 2

4k3 + 7k2 + 5, for k ∈ N. Prove that ak → 5/4 as k → ∞.

Exercise 1.77 Find limk→∞

(√k2 + 4k − k).

Exercise 1.78 Let ak =sin(4 + πk

2)

k3, for each k ≥ 1. Show that ak → 0 as

k → ∞.

Hint. Use Proposition 1.46.

Exercise 1.79 Let a sequence (ak) in R be defined as follows: a1 = 0 and 5ak+1 =

3ak + 4, for any k ≥ 1. Show that (ak) is convergent.

Solution. The sequence ak is bounded and increasing, and, hence, convergent. Let

us denote by l its limit. Passing to the limit in the recurrence formula, we get l = 2.

Exercise 1.80 Let a1 = −10 and ak+1 = 1−√1− ak. Prove that (ak)k is conver-

gent and find its limit.

Hint. Show that ak < 0, for any k ≥ 1 and (ak)k is increasing.

Exercise 1.81 Compute

limk→∞

1 · 1! + 2 · 2! + · · · k · k!(k + 1)!

.

Hint. Let us notice that k · k! = (k + 1)!− k!, for any k ≥ 1. Then,

limk→∞

1 · 1! + 2 · 2! + · · · k · k!(k + 1)!

= limk→∞

(k + 1)!− 1

(k + 1)!= 1.

Exercise 1.82 Let (ak)k be a sequence which converges to a. Prove that

limk→∞

a1 + a2 + · · · akk

= a.

28 REAL ANALYSIS

Exercise 1.83 Let (ak)k be a positive sequence which converges to a. Prove that

limk→∞

k√a1 · a2 · · · ak = a.

Exercise 1.84 Show that the sequence (ak)k, for k ≥ 1, defined by

a”k = 1 +1

2. . .++

1

k,

is divergent.

Exercise 1.85 Test the following sequence for convergence

ak = k(−1)k .

Solution. Obviously,

infj≥k

aj = 0 and supj≥k

aj = ∞.

Therefore,

limk→∞

ak = 0 and limk→∞

ak = ∞.

Hence, the sequence (ak)k is divergent.

Exercise 1.86 Find the limit superior and the limit inferior for the following se-

quences:

1) ak = (−1)k; 2) ak = k (−1)k;

3) ak = k + k (−1)k; 4) ak = (−1)k(a+

1

k

), a ≥ 0.

Chapter 2

Multidimensional Spaces

We start this chapter with a brief review of some useful theoretical results about

metric spaces.

2.1 Metric Spaces

Definition 2.1 Let X = ∅. A map d : X ×X → R is called a metric or a distance

on X if:

1) d(x, y) = 0 ⇔ x = y, ∀x, y ∈ X;

2) d(x, y) = d(y, x), ∀x, y ∈ X;

3) d(x, z) ≤ d(x, y) + d(y, z), ∀x, y, z ∈ X (the triangle inequality). The pair

(X, d) is called a metric space. Often, d is omitted and one just writes X for a

metric space if it is clear from the context what metric is used.

Remark 2.2 Any distance takes only positive values, i.e. d(x, y) ≥ 0, for all x, y ∈X. Hence, d : X ×X → R+.

Indeed, it is not difficult to see that we have:

0 = d(x, x) ≤ d(x, y) + d(y, x) = 2d(x, y) ⇒ d(x, y) ≥ 0, ∀x, y ∈ X.

Exercise 2.3 Let X = ∅. Show that the map d : X ×X → R+, defined by

d(x, y) =

{1, x = y,

0, x = y,

29

30 REAL ANALYSIS

is a distance on X, called the discrete metric. In this case, the space (X, d) is called

a discrete metric space or a space of isolated points.

Exercise 2.4 Show that on R, the map d : R×R → R+ defined by d(x, y) = |x− y|is a metric.

Exercise 2.5 Prove that on Rn the following maps are distances:

d(x, y) =

(n∑

i=1

(xi − yi)2

)1/2

( the Euclidean distance);

d1(x, y) =n∑

i=1

|xi − yi| (the Manhattan distance);

d∞(x, y) = max1≤i≤n

|xi − yi| (the Chebyshev distance).

The metric d∞ is sometimes denoted by du and it is called the uniform metric on

Rn. On the real line R, the three preceding metrics, d, d1 and du, agree.

Exercise 2.6 Let a, b ∈ R, a < b and let

C([a, b]) = {f : [a, b] → R | f is continuous on [a, b]}.

Show that the map d : C([a, b])× C([a, b]) → R+, defined by

d(f, g) = maxx∈[a,b]

|f(x)− g(x)|,

is a distance on C([a, b]). Therefore, (C([a, b]), d) is a metric space.

Exercise 2.7 Prove that the map

d(f, g) =

b∫a

|f(x)− g(x)| dx

is a metric on C([a, b]).

MULTIDIMENSIONAL SPACES 31

Exercise 2.8 Let (X, d) be a metric space and let d1 : X ×X → R+ defined by

d1(x, y) = ln(1 + d(x, y)), ∀x, y ∈ X.

Prove that d1 defines a new metric on X.

Solution. We notice that due to the fact that d is a distance, i.e. d(x, y) ≥ 0,∀x, y ∈M , d1 is well defined. Let us verify that it satisfies all the three axioms in Definition

2.1.

(i) We have to prove that d1(x, y) = 0 ⇔ x = y. Obviously, d1(x, y) = 0 ⇔ln(1 + d(x, y)) = 0 ⇔ 1 + d(x, y) = 1 ⇔ d(x, y) = 0 ⇔ x = y.

(ii) We have to prove that d1(x, y) = d1(y, x). But d(x, y) = d(y, x), and hence,

ln(1 + d(x, y)) = ln(1 + d(y, x)), i.e. d1(x, y) = d1(y, x).

(iii) We shall prove that for any x, y, z ∈ X, we have d1(x, z) ≤ d1(x, y)+d1(y, z).

Indeed, using the monotonicity of the ln and the fact that d satisfies the triangle

inequality, we have

d1(x, z) = ln(1 + d(x, z)) ≤ ln(1 + d(x, y) + d(y, z)) ≤

ln(1 + d(x, y) + d(y, z) + d(x, y)d(y, z)) =

ln((1 + d(x, y))(1 + d(y, z)) = ln(1 + d(x, y)) + ln(1 + d(y, z)) =

d1(x, y) + d1(y, z).

Hence, d1 is a metric on X.

Exercise 2.9 Prove that on R, the map

d(x, y) =√|x− y|

is a distance.

Exercise 2.10 Show that on R, the map

d(x, y) =|x− y|

1 + |x− y|

is a distance.

32 REAL ANALYSIS

Exercise 2.11 Let (X, d) be a metric space. Prove that

d1(x, y) =d(x, y)

1 + d(x, y)

is also a distance on X.

Hint. Prove first thata

1 + a≤ b

1 + bif 0 ≤ a ≤ b.

Exercise 2.12 Define d(x, y) = | arctanx− arctan y|, for any x, y ∈ R. Show that

d is a metric on R and compute d(√3,−1).

Exercise 2.13 Let d(x, y) = |x2 − y2|, for any x, y ∈ R. Show that d is not a

distance on R.

Exercise 2.14 Let d be the Euclidean metric on R2. Compute d(x, y) for x = (1, 2)

and y = (3, 1) and draw the open ball Br(a) for a = (0, 0) and r = 2.

Exercise 2.15 Let d1 be the Manhattan (taxicab) metric on R2. Compute d1(x, y),

for x = (3, 2) and y = (4, 1), and sketch the open ball Br(a) for this metric when

a = (0, 0) and r = 2.

Exercise 2.16 Let du be the uniform metric on R2. Compute du(x, y) when x =

(3, 2) and y = (4, 1) and draw Br(a) for this metric for a = (0, 0) and r = 2.

Exercise 2.17 Let d be the usual metric on R2 and 0 be the origin. We define ρ

on R2 as follows:

ρ(x, y) =

{d(x, y), if x and y are collinear with 0,

d(x, 0) + d(0, y), otherwise.

Prove that ρ is a metric on R2, called the Greek airline metric.

Exercise 2.18 Let ρ be the Greek airline metric on R2 and let x = (1, 2) and

y = (3, 4). Compute ρ(x, y) and sketch Br(a) for this metric when a = (0, 0) and

r = 2.

Exercise 2.19 Let (X, d) be a metric space and x, y, z ∈ X. Prove that

|d(x, z)− d(y, z)| ≤ d(x, y).


2.2 Elements of Topology

Definition 2.20 Let (X, d) be a metric space. Also, let x ∈ X and r > 0. The set

B(x, r) = {y ∈ X | d(x, y) < r}

is called the open ball with center at the point x and having the radius r.

Definition 2.21 Let (X, d) be a metric space. Also, let x ∈ X and r ≥ 0. The set

B(x, r) = {y ∈ X | d(x, y) ≤ r}

is called the closed ball with center at the point x and having the radius r.

Remark 2.22 If in the above definition we take r = 0, then B(x; 0) = {x}.

Definition 2.23 Let (X, d) be a metric space, a ∈ X and r > 0. The set

S(a, r) = {x ∈ X | d(a, x) = r}

is called the sphere centered at the point a and having the radius r.

Remark 2.24 The empty set is considered to be an open set.

Proposition 2.25 Any open ball B(x, r) is an open set.

Definition 2.26 A set E ⊆ X is called a closed set if X \ E is an open set.

Proposition 2.27 Any closed ball B(x, r) is a closed set.

Definition 2.28 Let (X, d) be a metric space and A, V ⊆ X. The set V is called a

neighbourhood of A if there exists an open set D such that A ⊆ D ⊆ V . If A = {x},V is called a neighbourhood of the point x. We shall denote by V(A) the collection

of all the neighbourhoods of A and by V(x) the collection of all the neighbourhoods

of x.

Definition 2.29 A set D ⊆ X is said to be an open set if for any x ∈ D there

exists r > 0 such that B(x, r) ⊆ D.

34 REAL ANALYSIS

Definition 2.30 Let (X, d) be a metric space and A ⊆ X. A point x ∈ X is called

an interior point of A if A ∈ V(x).

We shall denote by Int A or by A the set containing all the interior points of A.

Let us notice that the interior of a set depends upon the distance we are working

with. Usually, in what follows, if not otherwise stated, our examples will be taken

in the Euclidean space Rn, endowed with the standard distance.

Definition 2.31 Let (X, d) be a metric space and A ⊆ X. A point x ∈ X is called

an adherent point of A if for any V ∈ V(x), V ∩A = ∅. The set

A = {x ∈ X | x is an adherent point of A}

is called the adherence or the closure of A.

Example 2.32 If we take X to be the Euclidean space R and if we consider A =

(0, 1) ⊂ R, then A = [0, 1].

Definition 2.33 Let (X, d) be a metric space and A ⊆ X. A point x ∈ X is

called a limit point or an accumulation point or a cluster point of A if for any

V ∈ V(x), (V \ {x}) ∩A = ∅. The set

A ′ = {x ∈ X | x is a limit point of A}

is called the derived set of A.

Example 2.34 If we take X to be the Euclidean space R and if A = (0, 1) ⊂ R,then A ′ = [0, 1].

Example 2.35 Let X = R2, with its usual metric, and let A = ((0, 1)× (0, 1)) ∪{(2, 7)} ⊆ R2. Then A ′ = [0, 1]× [0, 1] and A = ([0, 1]× [0, 1]) ∪ {(2, 7)}.

Definition 2.36 Let (X, d) be a metric space and A ⊆ X.

∂A = A ∩ CA

is called the boundary of the set A and

ExtA = Int (X \A)

is called the exterior of A.


Example 2.37 In the Euclidean space R, endowed with the standard distance, if

A = (0, 1) ⊂ R, then ∂A = {0, 1} and Ext A = (−∞, 0) ∪ (1,∞).

Definition 2.38 A point which is not a limit point for A is called an isolated point

of A. A set A which contains only isolated points is said to be discrete. We shall

denote by ′A the set of all the isolated points of A.

Example 2.39 N is a discrete set in R.

Definition 2.40 Let (X, d) be a metric space and A ⊆ X. The number diam A =

sup{d(x, y) | x, y ∈ A} is called the diameter of the set A.

Definition 2.41 Let (X, d) be a metric space and A ⊆ X. The set A is called

bounded if diam A <∞.

Definition 2.42 Let (X, d) be a metric space, x ∈ X and A ⊆ X. The number

d(x,A) = inf{d(x, y) | y ∈ A} is called the distance from x to the set A.

Definition 2.43 Let (X, d) be a metric space and A,B ⊆ X. The number d(A,B) =

inf{d(a, b) | a ∈ A, b ∈ B} is called the distance between the sets A and B.

Definition 2.44 A metric space (X, d) is called compact if for any family of open

sets (Dα)α∈A, Dα ⊆ X, such that X =∪

α∈ADα, there exists a finite set J ⊆ A such

that X =∪

α∈JDα.

Definition 2.45 A set K ⊆ X in a metric space (X, d) is called compact if for any

family of open sets (Dα)α∈A, Dα ⊆ X, such that K ⊆∪

α∈ADα, there exists a finite

set J ⊆ A such that K ⊆∪

α∈JDα.

Example 2.46 Let us consider the metric space (R, d), where d(x, y) = |x − y|.Then, if a, b ∈ R, with a < b, the closed interval [a, b] is a compact set. Also, notice

that R is not compact and every finite set in (R, d) is compact.

Theorem 2.47 (Heine-Borel) Let (X, ∥ · ∥) be a finite dimensional space and K ⊆X. Then, K is compact if and only if K is closed and bounded.

36 REAL ANALYSIS

Definition 2.48 A metric space (X, d) is said to be connected if X cannot be re-

presented as the union of two open, nonempty and disjoint sets.

Definition 2.49 A set A ⊆ X in a metric space (X, d) is said to be connected if

there exists no pair of open sets D1, D2 ⊆ X such that A ⊆ D1∪D2, A∩D1∩D2 = ∅and A ∩Di = ∅, for all i = 1, 2.

If a set A is not connected, it is said to be disconnected.

Remark 2.50 Let us consider the metric space (R, d), where d(x, y) = |x − y|. A

nonempty set A ⊆ R is connected if and only if A is an interval. As a consequence,

R is connected, the open interval (a, b), with a, b ∈ R, a < b, is also connected, but

Q is not a connected set.

Definition 2.51 Let (X, d) be a metric space and A ⊆ X. The number

diam A = sup{d(x, y) | x, y ∈ A}

is called the diameter of the set A. Notice that diam A ∈ [0,∞] and diam (A) =

diam A.

Definition 2.52 Let (X, d) be a metric space and A ⊆ X. The set A is called

bounded if diam A <∞.

In a metric space, any ball (open or closed) with a finite radius is a bounded

set. As a matter of fact, we could define a bounded set as being a set which is

contained in a ball having a finite radius. For instance, the balls B(0, 3) and B(0, 3)

are bounded sets in R3.

Definition 2.53 Let (X, d) be a metric space, x ∈ X and A ⊆ X. The number

d(x,A) = inf{d(x, y) | y ∈ A}

is called the distance from the point x to the set A.

Definition 2.54 Let (X, d) be a metric space and A,B ⊆ X. The number

d(A,B) = inf{d(a, b) | a ∈ A, b ∈ B}

is called the distance between the sets A and B.


Exercise 2.55 If A = N, show that Int A = ∅, A = ∂A = A, A ′ = ∅, ′A = A.

Also, for A = Q, prove that Int A = ∅, A = ∂A = A′= R, ′A = ∅.

Exercise 2.56 Show that for A = R, we have Int A = A, A = A = A′= R,

′A = ∅, ∂R = ∅.

Exercise 2.57 For A = [0, 2) ∪ {3} ∪ {7}, show that Int A = (0, 2), A = [0, 2] ∪{3} ∪ {7}, A ′ = [0, 2], ∂A = {0} ∪ {2} ∪ {3} ∪ {7}, ′A = {3} ∪ {7}.

Exercise 2.58 Show that in R2 endowed with the discrete metric, the open ball

B(x, r) contains only one point, its center x, if 0 < r < 1, and contains all the

points of R2, if r ≥ 1.

Exercise 2.59 Prove that in R2 endowed with the Chebyshev distance the open ball

B(x, r) consists of all the points in the interior of a square centered at the point x

and with sides parallel to the coordinate axes and of length equal to 2r.

Exercise 2.60 Prove that in R2 endowed with the Euclidean distance the open ball

B(x, r) consists of all the points of the disk centered at the point x and having the

radius equal to r.

Exercise 2.61 Let X = {−1}∪ [0, 3) be regarded as a subspace of R endowed with

the Euclidean metric. Find, in this subspace, the open balls B1(−1), B1(0), and

B2(0).

Solution. Following the definition of the Euclidean distance, we get B1(−1) = {−1},B1(0) = [0, 1), and B2(0) = {−1} ∪ [0, 2).

Exercise 2.62 Let (X, d) be a metric space and A ⊆ X, A = ∅. Then, A is open if

and only if A ∈ V(x), ∀x ∈ A.

Solution. Let us assume that A is open. Then, x ∈ A ⊆ A, ∀x ∈ A. Since

A is open, it follows immediately that A ∈ V(x). Conversely, let us assume that

A ∈ V(x), ∀x ∈ A. Then, for any x ∈ A, there exists Dx open such that x ∈ Dx ⊆ A.

So,

A =∪x∈A

{x} ⊆∪x∈A

Dx ⊆ A.

38 REAL ANALYSIS

Therefore,

A =∪x∈A

Dx,

and since Dx are open sets, it follows that A is open, too.

Exercise 2.63 Let (X, d) be a metric space and x ∈ X. The set V(x) possesses thefollowing properties:

(i) if V1 ∈ V(x) and V1 ⊆ V2, then V2 ∈ V(x);(ii) if V1, V2 ∈ V(x), then V1 ∪ V2 ∈ V(x), V1 ∩ V2 ∈ V(x).

In fact, the union of any collection of neighbourhoods is also a neighbourhood. Also,

the intersection of any finite number of neighbourhoods is a neighbourhood, too.

Exercise 2.64 Let X = R, endowed with the standard distance. If A = [0, 1] ⊂ R,then Int A = (0, 1).

Exercise 2.65 Let X = R2 equipped with its usual metric and A = Q2. Show that

Int A = ∅.

Hint. There exists no open ball in R2 that contains only points with both coordinates

rational.

Exercise 2.66 Identify the interior, the closure and the set of all the accumulation

points of each of the following sets: (i) the integers Z in R; (ii) the rational

numbers Q in R; (iii) the complex numbers with rational real and imaginary parts

in C; (iv) the set A = [0, 1) ∪ {3} in R.

Exercise 2.67 Let (X, d) be a metric space and A,B ⊆ X. Show that:

(i) Int A ⊆ A, A ⊆ A, A ′ ⊆ A;

(ii) A ∪B = A ∪B, (A ∪B) ′ = A ′ ∪B ′; A ∩B ⊆ A ∩B, (A ∩B) ′ ⊆ A ′ ∩B ′;

(iii) the interior of A is the largest open set contained in A; the set A is the

smallest closed set which contains A;

(iv) A is open ⇔ A = Int A;

(iv) A = A ∪A ′; A = A ∪ ∂A; ∂A = A \ IntA.


Exercise 2.68 Let (X, d) be a metric space and A be a nonempty subset of X.

Prove that for any x, y ∈ X, we have |d(x,A)− d(y,A)| ≤ d(x, y). Also, show that

if A,B are nonempty subsets of X, then d(A,B) = d(A,B).

Exercise 2.69 1) Let A be a connected set and B a set such that A ⊆ B ⊆ A.

Prove that B is connected.

2) If A is a connected set, then A is also connected.

3) Let (Ai)i∈I be a family of connected sets such that∩i∈I

Ai = ∅. Prove that

A =∪i∈I

Ai is a connected set.

Exercise 2.70 Let (X, d) be a metric space and A be a nonempty subset of X.

Prove that for any x, y ∈ X, we have

|d(x,A)− d(y,A)| ≤ d(x, y).

Exercise 2.71 Let (X, d) be a metric space and A,B be nonempty subsets of X.

Prove that

d(A,B) = d(A,B).

2.3 Sequences in Metric Spaces

Definition 2.72 Let (X, d) be a metric space and (xk)k ⊆ X. The sequence (xk)k

is called convergent if there exists x ∈ X such that, for any V ∈ V(x), there exists

kV ∈ N such that xk ∈ V , for any k ≥ kV . We write x = limk→∞

xk.

Remark 2.73 Any metric space has the following property: for any x, y ∈ X with

x = y, there exist U ∈ V(x) and V ∈ V(y) such that U ∩ V = ∅. A space with such

a property is called separated or Hausdorff.

Remark 2.74 A convergent sequence in a metric space has only one limit.

Remark 2.75 We emphasize that the limit of a convergent sequence must be an

element of the space X. For example, if we take X to be the open interval (0, 1) ⊂ Rendowed with the usual distance d(x, y) = |x− y|, then the sequence (xk)k≥2 defined

by xk = 1/k is not convergent, since 0 /∈ X.

40 REAL ANALYSIS

Definition 2.76 A sequence (xk)k in a metric space (X, d) is called a fundamental

or a Cauchy sequence if for any ε > 0 there exists kε ∈ N such that d(xk, xp) <

ε, ∀k, p ≥ kε.

Proposition 2.77 A convergent sequence in a metric space (X, d) is also a Cauchy

sequence.

Remark 2.78 The converse implication in Proposition 2.77 is, in general, not true.

As a counterexample, let (R, d), with d(x, y) = | arctan y−arctanx|. It is not difficult

to see that the sequence (xk), with xk = k, is fundamental, but not convergent.

Definition 2.79 A metric space (X, d) is called a complete space if in this space

any Cauchy sequence is convergent.

Remark 2.80 A subspace M of a complete metric space (X, d) is complete if and

only if the set M is closed in X.

Example 2.81 (i) The metric space (R, d), with d(x, y) = |x − y|, is complete.

Also, the Euclidean space Rn and the unitary space Cn (endowed with the usual

Euclidean distances) are complete.

(ii) The space lp, with 1 ≤ p <∞, and the space l∞ are complete.

Example 2.82 (i) In X = (0, 1), the sequence xk = 1/k, with k ≥ 2, is Cauchy,

but it is not convergent (in X).

(ii) The set of all rational numbers Q with the usual metric given by d(x, y) =

|x− y|, for x, y ∈ Q, is not complete.

Definition 2.83 A map f : (X, d) → (Y, ρ) is called an homeomorphism if f is a

bijection and both f and f−1 are continuous. If such a map exists, the metric spaces

(X, d) and (Y, ρ) are called homeomorphic.

Definition 2.84 Let (X, d) and (Y, ρ) be metric spaces. A map φ : X → Y is

called an isometry or a distance-preserving map if for any x, y ∈ X, we have

ρ (φ(x), φ(y)) = d(x, y).

Definition 2.85 Two metric spaces X and Y are said to be isometric if there exists

a bijective isometry from X to Y .


Definition 2.86 A completion of a metric space (X, d) is a pair consisting of a

complete metric space (X, d) and an isometry φ : X → X such that φ(X) is dense

in X.

Remark 2.87 Every metric space has a completion. Such a completion is unique

up to isometry.

Exercise 2.88 Let (xk)k be a convergent sequence in a metric space (X, d). If there

exists A ⊆ X and k0 ∈ N such that xk ∈ A, for any k ≥ k0, then limk→∞

xk ∈ A.

Exercise 2.89 Let (xk)k be a sequence in a metric space (X, d) and let x ∈ X.

Prove that x = limk→∞

xk if and only if for all ε > 0, there exists kε ∈ N such that

d(xk, x) < ε, for all k ≥ kε (ε-criterion).

Exercise 2.90 Let C([0, 1]) be the space of all the continuous real-valued functions

on [0, 1], endowed with the metric d(f, g) =

∫ 1

0|f(t)−g(t)|dt. Show that (C([0, 1]), d)

is not a complete metric space.

Exercise 2.91 Let M be a nonempty subset of a metric space (X, d). Show that

x ∈M if and only if there exists a sequence (xk)k in M convergent to x. Also, prove

that M is closed if and only if the limit x of any convergent sequence (xk)k ⊆ M

belongs to M .

Exercise 2.92 (i) Any Cauchy sequence in a metric space is bounded.

(ii) If a Cauchy sequence (xk)k possesses a subsequence (xkk) which is conver-

gent to x ∈ X, then (xk)k is convergent to x.

(iii) Let (zk)k ⊆ C and z ∈ C. Then, zk → z if and only if Re zk → Im z and

Im zk → Im z.

(iv) Let (xk), (yk) ⊆ Rn such that xk → x, yk → y, with x, y ∈ Rn. If λ, µ ∈ R,then λxk + µ yk → λx+ µ y.

Exercise 2.93 Prove that the sequence(1, 12 , . . . ,

1k , . . .

)is not in l1, but it is in

any lp, for p > 1.

Exercise 2.94 Show that any isometry is injective.

42 REAL ANALYSIS

Exercise 2.95 Let (X, d) be a metric space and M a dense subset of X with the

property that every Cauchy sequence in M is convergent in X. Show that X is

complete.

2.4 Inner Product Spaces

Definition 2.96 Let K be a commutative field and 1K be its multiplicative identity.

Also, let (V,+) be a group. If there exists a map · : K× V → V such that

1) α · (x+ y) = α · x+ α · y, ∀α ∈ K, ∀x, y ∈ V ,

2) (α+ β) · x = α · x+ β · x, ∀α, β ∈ K, ∀x ∈ V ,

3) α · (β · x) = (αβ) · x, ∀α, β ∈ K, ∀x ∈ V ,

4) 1K · x = x, ∀x ∈ V ,

then V is called a linear space or a vector space over the field K. The elements of

V are called vectors and the elements of K are called scalars.

Often, boldface letters are used for denoting vectors.

Remark 2.97 If V is a vector space, then the group (V,+) is commutative.

Definition 2.98 Let X be a K-linear space, with K = R or K = C. A map ⟨·, ·⟩ :X × X → K is called a scalar product or a dot or inner product if the following

three axioms are satisfied:

1) ⟨x, y⟩ = ⟨y, x⟩, ∀x, y ∈ X;

2) ⟨αx+ βy, z⟩ = α ⟨x, z⟩+ β ⟨y, z⟩, ∀α, β ∈ K, ∀x, y, z ∈ X;

3) ⟨x, x⟩ ≥ 0, ∀x ∈ X and ⟨x, x⟩ = 0K ⇒ x = 0X .

The pair (X, ⟨·, ·⟩) is called a linear space with scalar product or a pre-Hilbertian

space.

If K = R, we have a real scalar product and, hence, a real pre-Hilbertian space. In

this case, the first axiom becomes ⟨x, y⟩ = ⟨y, x⟩, for all x, y ∈ X. If K = C, wehave a complex scalar product and, hence, a complex pre-Hilbertian space.


Let us notice that we can replace axiom 3) in the above definition by the following

one: ⟨x, x⟩ > 0, ∀x ∈ X \ {0X}. Also, notice that the nonnegativity condition

∀x ∈ X, ⟨x, x⟩ ≥ 0 makes sense, because ⟨x, x⟩ ∈ R, for any x ∈ X.

An alternative definition, encountered especially in physics, would require to

consider the inner product to be linear in the second argument rather than in the

first one. In this case, the first argument becomes conjugate linear.

Definition 2.99 Let K = R or K = C and let X be a K-linear space. If X is

finite dimensional and pre-Hilbertian, then X is said to be an Euclidean space. An

Euclidean space X is said to be real if K = R and complex if K = C.

As typical examples, we shall consider the real Euclidean space Rn and the complex

Euclidean space Cn.

Proposition 2.100 (Cauchy-Bunyakovsky-Schwarz Inequality) Let K = R or C and

let X be a pre-Hilbertian space. Then, the following inequality holds true:

|⟨x, y⟩|2 ≤ ⟨x, x⟩ ⟨y, y⟩, ∀x, y ∈ X. (2.1)

Remark 2.101 In (2.1), we have an equality if and only if the vectors x and y are

linear dependent, i.e. there exists λ ∈ K such that x = λ y.

Remark 2.102 If we take X = Rn with the usual Euclidean scalar product, then,

from (2.1), for any x = (x1, . . . , xn), y = (y1, . . . , yn) ∈ Rn, we obtain the following

well-known inequality:

∣∣∣ n∑i=1

xiyi∣∣∣ ≤ (

n∑i=1

x2i

)1/2( n∑i=1

y2i

)1/2

,

called Schwarz inequality.

For X = C([a, b]) and the standard scalar product defined above, from (2.1) we

get the well-known Cauchy-Schwarz inequality:

∣∣∣ b∫a

f(x)g(x) dx∣∣∣ ≤

b∫a

f2(x) dx

1/2 b∫a

g2(x) dx

1/2

.

44 REAL ANALYSIS

The space of square-summable sequences l2 is an inner product space, with the

scalar product given by

⟨x, y⟩ =∞∑i=1

xiyi for x = (xi)i, y = (yi)i in l2.

Then, in this case, for any two sequences x = (xi)i and y = (yi)i in l2, we get

∣∣∣ ∞∑i=1

xiyi∣∣∣ ≤ ( ∞∑

i=1

|xi|2)1/2( ∞∑

i=1

|yi|2)1/2

.

Definition 2.103 Let X be a pre-Hilbertian space. Two vectors x, y ∈ X are called

orthogonal if ⟨x, y⟩ = 0. We shall denote this by x ⊥ y.

Definition 2.104 Let x and y be nonzero vectors in Rn. The angle between x and

y is the unique number θ ∈ [0, π], given by

cos θ =⟨x, y⟩√

⟨x, x⟩√⟨y, y⟩

.

This definition makes sense, since, by Cauchy-Bunyakovsky-Schwarz inequality,

−1 ≤ ⟨x, y⟩√⟨x, x⟩

√⟨y, y⟩

≤ 1.

Let us notice that for θ =π

2, we get ⟨x, y⟩ = 0, which justifies Definition 2.103.

Exercise 2.105 Prove that the set of all ordered n-tuples of real numbers

Rn = {(x1, x2, . . . , xn) | xi ∈ R, i = 1, . . . , n},

endowed with two algebraic operations, the vector addition + : Rn × Rn → Rn,

defined by

x+ y = (x1 + y1, . . . , xn + yn), ∀x = (x1, . . . , xn), y = (y1, . . . , yn) ∈ Rn

and the scalar multiplication · : R× Rn → Rn, defined by

α · x = (αx1, . . . , αxn), ∀α ∈ R, x ∈ Rn,


is a real linear space.The origin of Rn, denoted by 0, is defined as being 0 = (0, . . . , 0).

For brevity, we shall omit the symbol ”·” and we shall write simply αx instead of

α · x. Moreover, when there is no danger of confusion, we shall not denote vectors

by boldface letters.

The standard basis of Rn is given by the set of vectors {e1, . . . , en}, wheree1 = (1, 0, . . . , 0),e2 = (0, 1, . . . , 0),........................en = (0, 0, . . . , 1).

(2.2)

Any vector x = (x1, x2, . . . , xn) can be uniquely written as

x =n∑

i=1

xiei.

Exercise 2.106 Show that Rn is a real linear space and Cn is a complex linear

space.

Exercise 2.107 If ⟨·, ·⟩ : X ×X → K is a scalar product, then

1) ⟨0, x⟩ = ⟨x, 0⟩ = 0, ∀x ∈ X;

2) ⟨x, αy + βz⟩ = α ⟨x, y⟩+ β ⟨x, z⟩ , ∀α, β ∈ K, ∀x, y, z ∈ X.

Exercise 2.108 Let X = Rn, K = R. Prove that the map ⟨·, ·⟩ : Rn × Rn → Rdefined by

⟨x, y⟩ =n∑

i=1

xiyi

is a scalar product on Rn. Therefore, (Rn, ⟨·, ·⟩) is a (real) pre-Hilbertian space.

Exercise 2.109 Let X = Cn, K = C. Show that the map ⟨·, ·⟩ : Cn × Cn → Cdefined by

⟨x, y⟩ =n∑

i=1

xiyi

is a scalar product on Cn. Thus, (Cn, ⟨·, ·⟩) is a (complex) pre-Hilbertian space.

46 REAL ANALYSIS

Exercise 2.110 Let X = C([a, b]) = {f : [a, b] → R | f continuous on [a, b]}. Prove

that the map ⟨·, ·⟩ : X ×X → R defined by

⟨f, g⟩ =b∫

a

f(x) g(x) dx (2.3)

is a scalar product. So, (C, ⟨·, ·⟩) is a pre-Hilbertian space.

Exercise 2.111 Let x = (1,−2, 3) and y = (−2, 0, 4) be vectors in Rn. Compute

⟨x, y⟩, where the inner product ⟨·, ·⟩ is defined in Exercise 2.108

Exercise 2.112 Let X be a pre-Hilbertian space. Show that if x ⊥ y, then y ⊥ x,

λx ⊥ y and x ⊥ λ y, for any λ ∈ K.

Exercise 2.113 Let x, y be two given vectors in Rn. Prove that if ⟨z, x⟩ = ⟨z, y⟩,for any z ∈ Rn, then x = y.

2.5 Normed Spaces

Definition 2.114 Let K = R or K = C and let X be a K-linear space. A map

∥ · ∥ : X → R is called a norm if:

1) ∥x∥ = 0 ⇒ x = 0X ;

2) ∥αx∥ = |α| ∥x∥, ∀α ∈ K, ∀x ∈ X;

3) ∥x+y∥ ≤ ∥x∥+∥y∥, ∀x, y ∈ X (the triangle inequality). The pair (X, ∥·∥)is called a normed space.

Remark 2.115 Any norm takes only positive values, i.e. ∥x∥ ≥ 0, for all x ∈ X.

Thus, ∥ · ∥ : X → R+. Also, as a consequence of the second axiom, we have ∥x∥ =

0 ⇔ x = 0X .

Example 2.116 For p ∈ [1,∞), the space of p-summable sequences lp is defined

as being the collection of all the sequences x = (xi)∞i=1 for which

∞∑i=1

|xi|p <∞.


The space lp is a normed space with respect to the norm

∥x∥p =( ∞∑

i=1

|xi|p)1/p

.

Proposition 2.117 Any normed space is a metric one.

Remark 2.118 If a metric d on a vector space X satisfies the following properties:

1) d(x, y) = d(x+ z, y + z), for any x, y, z ∈ X (translation invariance);

2) d(αx, αy) = |α| d(x, y), ∀α ∈ K,∀x ∈ X (homogeneity),

then we can define a norm on X by ∥x∥ = d(x, 0).

Definition 2.119 Two norms ∥ · ∥ and ∥ · ∥ ′ are called equivalent if there exist

α, β > 0 such that

∥x∥ ≤ α ∥x∥ ′ ≤ β ∥x∥, ∀x ∈ X.

We write ∥ · ∥ ∼ ∥ · ∥ ′.

Remark 2.120 Let us notice that two norms on a linear space will generate the

same topology if and only if they are equivalent.

Remark 2.121 Let X be a normed space with dimKX = finite. One can prove

that in such a finite dimensional normed space, any two norms are equivalent.

Definition 2.122 Let X be a normed space. We shall say that X has Heine-Borel’s

property if every bounded sequence in X possesses a convergent subsequence, whose

limit belongs to X.

Remark 2.123 A normed space X has Heine-Borel’s property if and only if X is

finite dimensional.

Definition 2.124 A normed space (X, ∥·∥) which is complete is said to be a Banach

space.

Example 2.125 For any fixed real number p ≥ 1, the Banach space Lp([a, b]) is

the completion of the normed space of all the continuous real-valued functions on

[a, b], with the norm given by

∥f∥p =(∫ b

a|f(x)|p dx

)1/p

.

48 REAL ANALYSIS

Definition 2.126 A pre-Hilbertian space (X, ⟨·, ·⟩) which is complete is called a

Hilbert space.

Example 2.127 The Euclidean spaces Rn and Cn are Banach spaces. In fact, any

finite dimensional normed space X is a Banach space.

Definition 2.128 Let X be a normed space. A subspace Y of the normed X is a

linear subspace endowed with the norm induced from the one in X.

So, if (X, ∥·∥) is a normed space and Y ⊂ X is a linear subspace, then (Y, ∥·∥) is anormed space.

Example 2.129 (Weierstrass Approximation) The space of polynomials P (x) is a

dense subspace of C([0, 1]). Also, the set of all continuous functions C([0, 1]) is a

dense subspace of L1([0, 1]). So, we can approximate any integrable function by

continuous functions.

Exercise 2.130 Prove that the following maps are norms on Rn:

a) ∥x∥1 =n∑

i=1

|xi|;

b) ∥x∥2 =

√√√√ n∑i=1

x2i (the Euclidean norm);

c) ∥x∥∞ = max1≤i≤n

|xi|;

d) ∥x∥p =(

n∑i=1

|xi|p)1/p

, 1 ≤ p <∞.

Exercise 2.131 Show that the following maps are norms on Cn:

a) ∥x∥1 =n∑

i=1

|xi|;

b) ∥x∥2 =

√√√√ n∑i=1

|xi|2;

c) ∥x∥∞ = max1≤i≤n

|xi|.


Exercise 2.132 Prove that the following map is a norm on C([a, b]):

∥f∥ =

√∫ b

af2(x) dx.

Exercise 2.133 Show that any scalar product determines a norm.

Solution. Indeed, from a given inner product, we can naturally define a norm by:

∥x∥ =√⟨x, x⟩. (2.4)

So, any pre-Hilbertian space is a normed space. Let us remark that the map ∥ · ∥ :

X → R+ is well-defined, since ⟨x, x⟩ ≥ 0, for any x ∈ X. We shall prove that (2.4)

verifies all the axioms in Definition 2.115. Indeed, we have:

1) ∥x∥ = 0 ⇔√⟨x, x⟩ = 0 ⇔ ⟨x, x⟩ = 0 ⇔ x = 0X ;

2) ∥αx∥ =√⟨αx, αx⟩ =

√αα⟨x, x⟩ =| α |

√⟨x, x⟩ = |α| ∥x∥, ∀α ∈ K, ∀x ∈ X;

3) ∥x + y∥2 = ⟨x + y, x + y⟩ = ∥x∥2 + ∥y∥2 + ⟨x, y⟩ + ⟨x, y⟩ = ∥x∥2 + ∥y∥2 +2Re⟨x, y⟩ ≤ ∥x∥2 + ∥y∥2 + 2|⟨x, y⟩|. Hence, ∥x + y∥2 ≤ ∥x∥2 + ∥y∥2 + 2|⟨x, y⟩|.Using (3.4), we get ∥x+ y∥2 ≤ ∥x∥2 + ∥y∥2 + 2∥x∥ ∥y∥ = (∥x∥+ ∥y∥)2, which gives

∥x+ y∥ ≤ ∥x∥+ ∥y∥.

Exercise 2.134 For m, p ∈ N, we introduce the space of matrices Mm,p of all the

m × p matrices with complex entries. Then, prove that Mm,p is an inner product

space with the trace inner product defined as

⟨A,B⟩ =m∑i=1

p∑j=1

aij bij .

The norm generated by this inner product on Mm,p is called the Hilbert-Schmidt or

the Frobenius norm of matrices:

∥A∥ =( m∑

i=1

p∑j=1

|aij |2)1/2

.

Remark 2.135 The converse implication in Exercise 2.134 is false. Most real and

complex normed vector spaces do not have inner products. For instance, in R2, the

norm

∥x∥∞ = max{|x1|, |x2|}

50 REAL ANALYSIS

is not generated by a scalar product. A normed space (X, ∥ · ∥) is a pre-Hilbertian

one if and only if for any x, y ∈ X, a certain relationship, called the parallelogram

identity, holds true:

∥x+ y∥2 + ∥x− y∥2 = 2(∥x∥2 + ∥y∥2).

If the norm satisfies this identity, the associated inner product is given by the so-

called polarization identity. In the real case, this identity, which generates the scalar

product, is:

⟨x, y⟩ = 1

4

(∥x+ y∥2 − ∥x− y∥2

).

In the complex case, the polarization identity is given by

⟨x, y⟩ = 1

4

(∥x+ y∥2 − ∥x− y∥2 + i∥x+ iy∥2 − i∥x− iy∥2

).

Additionally, the inner product generated by a given norm is unique, as a consequence

of the polarization identity.

Exercise 2.136 Let X be a real pre-Hilbertian space and let x, y ∈ X. Prove that

∥x+ y∥ = ∥x∥+ ∥y∥ ⇔ ⟨x, y⟩ = ∥x∥ · ∥y∥

and

⟨x, y⟩ = ∥x∥ · ∥y∥ ⇔ ∃α ∈ R such that y = αx.

Exercise 2.137 Let X be a pre-Hilbertian space and let us consider k vectors in

X, x1, . . . , xk, which are two-by-two orthogonal, i.e. xi ⊥ xj = 0, for any i = j.

Prove that ∥∥∥∥∥k∑

i=1

xi

∥∥∥∥∥2

=k∑

i=1

∥xi∥2 (Pythagoras).

Exercise 2.138 Let x, y ∈ Rn. Prove that x ⊥ y if and only if ∥x+ y∥ = ∥x− y∥.

Exercise 2.139 Show that any normed space is a metric one.

Solution. Let (X, ∥ · ∥) be a normed space. If we define d : X × X → R+ by

d(x, y) = ∥x− y∥, it is easy to check that this map is indeed a distance on X. The

metric d is said to be induced by the norm ∥ · ∥. Hence, (X, d) is a metric space.


Exercise 2.140 Prove that in Rn we have

∥ · ∥1 ∼ ∥ · ∥2 ∼ ∥ · ∥∞.

Exercise 2.141 Prove that if X and Y are normed spaces, then the product space

X × Y is also a normed space. For instance, the map

∥(x, y)∥X×Y = ∥x∥X + ∥y∥Y

is a norm on X × Y . Notice that one can also consider other norms, such as

∥(x, y)∥X×Y = max{∥x∥X , ∥y∥Y }

or

∥(x, y)∥X×Y =√∥x∥2

X+ ∥y∥2

Y.

Exercise 2.142 Let X be a compact topological space and let us consider the space

C(X) = {f : X → K | f is continuous on X},

endowed with the uniform norm, i.e.

∥f∥u = supx∈X

|f(x)| <∞.

Prove that C(X) is a Banach space.

Exercise 2.143 Prove that the space of all bounded sequences (real or complex)

x = (xk)k≥1 with the norm

∥x∥ = supk

∥xk∥

is a Banach space.

Exercise 2.144 Let X be the space of all the continuous real functions on [0, 1]

and let d : X ×X → R+ defined by

d (f, g) =

∫ 1

0|f(x)− g(x)|dx.

Prove that d is a metric on X, but the space (X, d) is not complete.

52 REAL ANALYSIS

Exercise 2.145 Show that the space c of convergent sequences in C, endowed with

the supremum norm, is a Banach space.

Exercise 2.146 Prove that the space l∞ of bounded sequences (with the supremum

norm) is a Banach space.

Exercise 2.147 Show that the space of all the continuous real functions on [0, 1],

with the norm

∥f∥L1 =

∫ 1

0|f(x)|dx

given by the Riemann integral of the absolute value, is not a complete normed space.

Exercise 2.148 (Minkowski Inequality in lp) Let p ∈ [1,∞). Prove that for any

two sequences x, y ∈ lp, we have

( ∞∑i=1

|xi + yi|p)1/p

≤( ∞∑

i=1

|xi|p)1/p

+( ∞∑

i=1

|yi|p)1/p

.

Exercise 2.149 (i) Show that the set c of convergent sequences and the set c0 of

sequences converging to zero are closed subspaces of l∞.

(ii) Let p ∈ [1,∞). Prove that the set of p-summable sequences lp is a closed

subspace of l∞ and a dense subspace of c0.

Exercise 2.150 Let us consider a Banach space X and let Y be a linear subspace

of X. Prove that Y is Banach if and only if Y is closed in X.

Exercise 2.151 Show that the space l∞ of all bounded sequences of real numbers,

with the supremum norm, is a Banach space.

Exercise 2.152 Prove that the space of polynomials P (x) on the interval [0, 1] is

not a Banach space with respect to the supremum norm.

Exercise 2.153 Let x = (1, 1, 0) and y = (2, 0, 0). Compute ⟨x, y⟩, ∥x∥, ∥y∥ and

the angle between x and y.

Exercise 2.154 Let x, y ∈ Rn. Prove that x ⊥ y if and only if ∥x+ y∥ = ∥x− y∥.

Exercise 2.155 Show that any normed space is a metric one.


Solution. Let (X, ∥ · ∥) be a normed space. Then, if we define d : X ×X → R+ by

d(x, y) = ∥x− y∥,

it is easy to verify that this map is indeed a distance on X. The metric d is said to

be induced by the norm ∥ · ∥. Hence, (X, d) is a metric space.

Exercise 2.156 Let x = (x1, x2) ∈ R2. Prove that

∥x∥1 = |x1|+ |x2|,

∥x∥2 =√x21 + x22,

and

∥x∥∞ = max{| x1 |, | x2 |}

are norms in R2 and draw the corresponding unit balls and unit circles.

Exercise 2.157 Show that in R3, the set of all the vectors whose Euclidean norm

is a given constant forms the surface of a sphere, the set of vectors whose 1-norm is

a given constant forms the surface of a cross polytope and the set of vectors whose

∞-norm is a given constant forms the surface of a hypercube.

Exercise 2.158 Prove that in R2, the following norms are equivalent:

∥ · ∥1 ∼ ∥ · ∥2 ∼ ∥ · ∥∞.

Hint. It is not difficult to see that

∥x∥∞ ≤ ∥x∥2 ≤ ∥x∥1 ≤√2∥x∥2 ≤ 2∥x∥∞, for all x ∈ Rn.

Exercise 2.159 Prove that if X and Y are normed spaces, then the product space

X × Y is also a normed space.

Hint. The map

∥(x, y)∥X×Y = ∥x∥X + ∥y∥Y

is a norm on X × Y . Let us notice that we can also consider other norms, such as

∥(x, y)∥X×Y = max{∥x∥X , ∥y∥Y }

54 REAL ANALYSIS

or

∥(x, y)∥X×Y =√∥x∥2

X+ ∥y∥2

Y.

A similar result holds true for the product of n normed spaces.

Chapter 3

Series of Real and ComplexNumbers

In what follows, we shall consider that X = R or X = C.

Definition 3.1 Let (xk)k≥0 ⊆ X. If

Sk =k∑

i=0

xi, (3.1)

then the pair ((xk)k≥0, (Sk)k≥0) is called a series (the series associated to the sequence

(xk)n≥0).

Instead of the above pair notation, we shall use a simpler one:

∑k≥0

xk (3.2)

Sk is called the kth partial sum of the series (3.2), while xk is said to be its general

term.

Definition 3.2 If there exists S ∈ X such that

S = limk→∞

Sk,

then the series (3.2) is called convergent. In this case, S is said to be its sum.

55

56 REAL ANALYSIS

If the series∑k≥0

xk converges to S, we shall write:

S =∞∑k=0

xk.

Definition 3.3 A series which is not convergent is called divergent.

Definition 3.4 A series∑k≥0

xk is called absolutely convergent if the series∑k≥0

|xk|

is convergent.

Definition 3.5 A series which is convergent, but not absolutely convergent, is called

semi-convergent or conditionally convergent.

Proposition 3.6 If the series∑k≥0

xk is absolutely convergent, then the series∑k≥0

xk

is convergent. The converse implication is, in general, not true.

Example 3.7 The series∑k≥1

(−1)k

kis only semi-convergent.

Proposition 3.8 If the series∑k≥0

xk is convergent, then the sequence (xk)k is con-

vergent and

limk→∞

xk = 0.

Remark 3.9 The converse implication in Proposition 3.8 is false.

Example 3.10 The harmonic series∑k≥1

1

kis divergent, despite the fact that

xk =1

k→ 0.

Proposition 3.11 (Cauchy’s Criterion) The series∑k≥0

xk is convergent in X if

and only if for any ε > 0, there exists kε ∈ N such that

|xk+1 + · · ·+ xk+p| < ε, ∀k ≥ kε, ∀p ≥ 1. (3.3)

SERIES 57

Proposition 3.12 (Comparison Test) Let (xk)k ⊆ C and (yk)k ⊆ R+. We assume

that there exists k0 ∈ N such that

|xk| ≤ yk, ∀ k ≥ k0.

If the series∑k≥0

yk is convergent, then the series∑k≥0

xk is absolutely convergent.

Corollary 3.13 Let xk, yk ∈ R such that

0 ≤ xk ≤ yk, ∀ k ∈ N.

(i) If∑k≥0

yk is convergent, then∑k≥0

xk is convergent.

(ii) If∑k≥0

xk is divergent, then∑k≥0

yk is also divergent.

(iii) If xk, yk ∈ R such that xk, yk > 0 and there exists limk→∞

xkyk

= L ∈ (0,∞),

then the series∑k≥0

xk and∑k≥0

yk have the same nature.

The next test is known as Cauchy’s Test or the Root Test.

Theorem 3.14 Let (xk)k ⊆ C and

L = lim supk→∞

k

√|xk|.

Then, we have:

(i) if L < 1, the series∑k

xk is absolutely convergent;

(ii) if L > 1, the series∑k

xk is divergent.

Corollary 3.15 If there exists

L = limk→∞

k

√|xk|,

then:

58 REAL ANALYSIS

(i) if L < 1, the series∑k


(ii) if L > 1, the series∑k

xk is divergent;

(iii) if L = 1, we cannot decide upon the nature of the series∑k

xk.

The next test is known as D’Alembert’s Test or the Ratio Test.

Theorem 3.16 Let (xk)k≥0 ⊆ C.

(i) If lim supk→∞

∣∣∣xk+1

xk

∣∣∣ < 1, then the series∑k≥0

xk is absolutely convergent.

(ii) If lim infk→∞

∣∣∣xk+1

xk

∣∣∣ > 1, then the series∑k≥0

xk is divergent.

Corollary 3.17 If there exists

L = limk→∞

∣∣∣xk+1

xk

∣∣∣,then:

(i) if L < 1, the series∑k≥0


(ii) if L > 1, the series∑k≥0

xk is divergent;

(iii) if L = 1, we cannot decide upon the nature of the series∑k≥0

xk.

Proposition 3.18 (Abel-Dirichlet’s Test) Let (xk)k ⊆ C and (yk)k ⊆ R+. We

assume that there exists M > 0 such that

∣∣∣ k∑i=0

xi∣∣∣ ≤M, ∀ k ≥ 0.

If the sequence (yk)k is nonincreasing and

limk→∞

yk = 0,

then the series∑k0

xk yk is convergent.

SERIES 59

Proposition 3.19 (Leibniz) If xk ≥ xk+1 ≥ 0, ∀ k ∈ N and limk→∞

xk = 0, then the

series∑k≥0

(−1)kxk is convergent. A series of this form is called a Leibniz series or

an alternating series.

Proposition 3.20 (Condensation Test) Let (xk)k≥0 ⊆ R+ be a nonincreasing se-

quence. Then, the series∑k≥0

xk is convergent if and only if the series∑k≥0

2kx2k is

convergent.

Proposition 3.21 (Raabe-Duhamel’s Test) Let xk > 0, ∀ k ∈ N, and let

Rk = k

(xkxk+1

− 1

).

If there exists

L = limk→∞

Rk,

then:

(i) if L > 1, the series∑k≥0

xk is convergent;

(ii) if L < 1, the series∑k≥0

xk is divergent;

(iii) if L = 1, we cannot decide upon the nature of the series∑k≥0

xk.

Definition 3.22 Let∑k≥0

xk and∑k≥0

yk be series of real (or complex) numbers. Let

us consider the series∑k≥0

zk, with

zk =∑

i+j=k

xiyj .

This series is called the Cauchy product series of the series∑k≥0

xk and∑k≥0

yk.

60 REAL ANALYSIS

Theorem 3.23 (Mertens) Let∑k≥0

xk and∑k≥0

yk be convergent series of real (com-

plex) numbers. If at least one of our two series is absolutely convergent, then the

Cauchy product series∑k≥0

zk, with

zk =∑

i+j=k

xiyj

is convergent and∞∑k=0

zk = (∞∑k=0

xk)

( ∞∑k=0

yk

).

Theorem 3.24 (Cauchy) Let∑k≥0

xk and∑k≥0

yk be absolutely convergent series of

real (or complex) numbers. Then, the Cauchy product series∑k≥0

zk is absolutely

convergent.

Remark 3.25 In general, a product series of two convergent series is not conver-

gent. For instance, if we take

xk = yk =(−1)k√

k,

then the Cauchy product series∑k≥0

zk is divergent.

Exercise 3.26 Show that the series∑k≥1

1

k (k + 1)

is convergent and its sum is equal to 1.

Solution. Since the general term of our series can be written as

xk =1

k (k + 1)=

1

k− 1

k + 1,

by induction we can prove that the k-th partial sum Sk is

Sk = 1− 1

k + 1.

SERIES 61

Therefore, since

limk→∞

Sk = 1,

our series is convergent and its sum is equal to 1.

Exercise 3.27 Prove that the series∑k≥2

ln

(1− 1

k2

)

is convergent and its sum is equal to − ln 2.

Solution. Since the general term of our series can be written as

xk = ln

(1− 1

k2

)= ln(k − 1) + ln(k + 1)− 2 ln k,

by induction we can show that the k-th partial sum Sk is

Sk = lnk + 1

2 k.

Thus, since

limk→∞

Sk = − ln 2,

our series is convergent and its sum is equal to − ln 2.

Exercise 3.28 Find the sum of the series∑k≥1

arctan1

k2 + k + 1.

Solution. it is not difficult to see that

1

k2 + k + 1=

1

k− 1

k + 1

1 +1

k

1

k + 1

.

On the other hand, we know that

arctanx− arctan y = arctanx− y

1 + xy.

62 REAL ANALYSIS

Therefore, the general term of our series can be written as follows:

xk = arctan1

k− arctan

1

k + 1.

By induction,

Sk = arctan 1− arctan1

k + 1,

which converges toπ

4. Hence, our series is convergent and its sum is equal to

π

4.

Exercise 3.29 Compute the sum of the following series:

a)∑k≥1

1

4k2 − 1;

b)∑k≥1

1

k (k + p), p ∈ N∗;

c)∑k≥1

k −√k2 − 1√

k (k + 1).


1

kis divergent.

Solution. We have

|S2k − Sk| =1

k + 1+

1

k + 2+ · · ·+ 1

2k≥ k

1

2k=

1

2.

Therefore, condition (3.3) is not satisfied and, so, our series is divergent.

Exercise 3.31 If∑k≥0

xk and∑k≥0

yk are convergent series in X and if α, β ∈ K, show

that the series ∑k≥0

(αxk + βyk)

is convergent and∞∑k=0

(αxk + βyk) = α∞∑k=0

xk + β∞∑k=0

yk.

SERIES 63

Exercise 3.32 Let (zk)k ⊆ C,

zk = xk + i yk.

Show that ∑k≥0

zk is convergent ⇔∑k≥0

xk and∑k≥0

yk are convergent.

Exercise 3.33 Let (xk)k ⊆ Rn, xk = (x1k, x2k, . . . , x

nk). Show that∑

k≥0

xk is convergent ⇔∑k≥0

xik, i = 1, n, are convergent.

Exercise 3.34 (The geometric series) Let q ∈ R such that |q| < 1. Prove that the

series∑k≥0

qk is convergent and

∞∑k=0

qk =1

1− q.

If |q| ≥ 1, show that the series∑k≥0

qk is divergent.

Exercise 3.35 If z ∈ C and |z| < 1, show that the series∑k≥0

zk is convergent and

∞∑k=0

zk =1

1− z.

Solution. Let q ∈ R such that |q| < 1.

Sk = 1 + q + · · ·+ qk =1− qk+1

1− q.

Therefore,

limk→∞

Sk =1

1− q.

If |q| ≥ 1, then

|xk| = |qk| ≥ 1.

Hence, xk 9 0 and the series is divergent.

64 REAL ANALYSIS


k!

(2k)!is convergent.

Solution. The series is convergent since

k!

(2k)!=

1

(k + 1) · · · (2k)<

1

2k

and the geometric series∑k≥1

1

2kis convergent.


1√k(k + 1)

is divergent.

Solution. The given series is divergent because

1√k(k + 1)

>1

k + 1

and the series∑k≥1

1

k + 1is divergent.


1

k + 1and

∑k≥1

1

5k + 2have the same nature.

Solution. The conclusion follows immediately because

limk→∞

5k + 2

k + 1= 5 ∈ (0,∞).

Exercise 3.39 Test the following series for convergence:

∑k≥1

k3 + 1

k6 + k + 3.

Solution. By comparing with the series∑k≥1

1

k3, which is convergent, it turns out

that the given series is convergent.


(−1)k1

kpis convergent for p > 0 and diver-

gent for p ≤ 0.

SERIES 65

Exercise 3.41 Show that the series

∑k≥1

1

kα

is convergent for α > 1 and divergent for α ≤ 1. This series is known as the

generalized harmonic series or Riemann’s series.

Solution. Using the condensation test, the series∑k≥1

1

kαis convergent if and only if

the series∑k≥0

2k1

(2k)αis convergent. But

2k1

(2k)α=

(1

2α−1

)k

.

If α > 1, then1

2α−1< 1. Hence, our series is convergent.

If α = 1, then

(1

20

)k

= 1 9 0. Therefore, our series is divergent.

If α < 1, then kα < k and1

kα>

1

k. So, the given series is divergent.


∑k≥1

1

k (1 + a+ · · ·+ ak), a > 0.

Solution. We have to distinguish between three cases: a = 1, a > 1 and 0 < a < 1.

For a = 1, the general term of our series becomes1

k (k + 1)and, obviously, the series

is convergent.

For a > 1, we have

1

k (1 + a+ · · ·+ ak)<

1

ak, for any k ≥ 1.

Since the geometric series∑k≥1

1

akis convergent, having the ratio

1

a< 1, using the

comparison test we conclude that our series is convergent.

66 REAL ANALYSIS

In the case 0 < a < 1, we shall compare our series with the classical harmonic series,

which is divergent. Indeed, we have:

limk→∞

1

k (1 + a+ · · ·+ ak)1

k

= limk→∞

1

1− ak+1

1− a

= 1− a ∈ (0,∞).

Hence, in this case, our series is divergent.


∑k≥1

k!

kk.

Solution. Applying the ratio test, we get

limk→∞

∣∣∣xk+1

xk

∣∣∣ = limk→∞

(k + 1)! kk

k!(k + 1)k+1= lim

k→∞

kk

(k + 1)k=

1

e< 1.

Hence, our series is absolutely convergent.

Exercise 3.44 Determine the nature of the series∑k≥1

2k(1− 1

k)k

2.

Solution. Applying the root test, we get

limk→∞

k

√|xk| = lim

k→∞2(1− 1

k)k =

2

e< 1.

Therefore, our series is absolutely convergent.


ak

k!, defined for a > 0, is convergent.

Solution. Applying the ratio test, we get

limk→∞

∣∣∣xk+1

xk

∣∣∣ = limk→∞

ak+1

(k + 1)!

k!

ak= lim

k→∞

a

k + 1= 0 < 1.

Thus, our series is absolutely convergent.

SERIES 67


1

(ln k)kis convergent.


(ln k)k is divergent.

Exercise 3.48 Test the following series for convergence

∑k≥1

(2k − 1)!!

(2k + 2)!!.

Solution. Using Raabe-Duhamel’s test, we obtain

limk→∞

k

(xkxk+1

− 1

)= lim

k→∞k

(2k + 4

2k + 1− 1

)=

3

2> 1.

So, our series is convergent.

Exercise 3.49 Using the ratio and the root test, discuss the nature of the following

series:

a)∑k≥1

ak

k!, a > 0; b)

∑k≥1

ak

kb, a, b > 0; c)

∑k≥2

1

(ln k)k;

d)∑k≥1

k!

kk, e)

∑k≥1

(2 + sin k)k

(1− 2

k)k2

.

Exercise 3.50 Compute the sum of the following series:

a)∑k≥2

1

k(k2 − 1);

∑k≥1

1

k(k + 1)(k + 2); c)

∑k≥1

ln(k + 1)2

k (k + 2).


a)∑k≥0

1

3k + k; b)

∑k≥1

√k + 1−

√k

k; c)

∑k≥1

tan1

k.

Exercise 3.52 Decide upon the nature of the following series:

a)∑k≥1

sin1

k; b)

∑k≥2

1

k ln k; c)

∑k≥1

sin k

2k.

68 REAL ANALYSIS


a)∑k≥1

(−1)k

(3k + 1)2; b)

∑k≥1

1

k +√k + 1

;

c)∑k≥1

k!

a(a+ 1) · · · (a+ k), a > 0;

d)∑k≥2

sin k + 1

k(ln k)2; e)

∑k≥2

e−kβ , β ∈ R.

Exercise 3.54 Discuss, in terms of p and q, the nature of the following series:

∑k≥2

1

kp (ln k) q.

Exercise 3.55 Decide upon the nature of the following series:

a)∑k≥2

√k + 2−

√k − 2

n2; b)

∑k≥1

(ak + b

ck + d

) k

, a, b, c, d > 0;

c)∑k≥1

(√(k + 1)(k + a)− k) k, a > 0; d)

∑k≥1

1

k!

(k

a

) k

, a > 0.

Chapter 4

Limits and Continuity

4.1 Limits of Functions

Definition 4.1 Let (X, d) and (Y, ρ) be metric spaces, M ⊆ X, a ∈M′and

f :M → Y.

A number l ∈ Y is called the limit of the function f at the point a if for any V ∈ V(l),there exists U ∈ V(a) such that, for any x ∈ U ∩M \{a}, it follows that f(x) ∈ V .

We shall use the following notation l = limx→a

f(x).

Remark 4.2 If a function f possesses a limit l at a given point a, then this limit

is unique.

Theorem 4.3 Let (X, d) and (Y, ρ) be metric spaces, M ⊆ X, a ∈M′. Also, let

f :M → Y . The following statements are equivalent:

(i) l = limx→a

f(x);

(ii) ∀ ε > 0 ∃ δε > 0 such that ∀x ∈ M, x = a, with d(x, a) < δε, it follows that

ρ(f(x), l) < ε;

(iii) ∀ (xn)n ⊆M, xn = a, such that xn → a, it follows that f(xn) → l.

Theorem 4.4 (Cauchy’s Criterion) Let f : E ⊆ Rn → Rm and let a ∈ E′. Then,

there exists l = limx→a

f(x) if and only if for any ε > 0, there exists Uε ∈ V(a) such

that for any x′, x

′′ ∈ Uε ∩E \ {a}, it follows that d(f(x′), f(x

′′)) < ε.

69

70 REAL ANALYSIS

Definition 4.5 Let f : E ⊆ Rn → Rm and let u ∈ Rn, u = 0, a ∈ E ′. The function

f possesses the limit l ∈ Rm at the point a in the direction of the vector u if there

exists

l = limh→0

f(a+ hu) ∈ Rm.

We shall denote the directional limit of f along the vector u by

lu = limx→au

f(x).

Notice that, for the existence of the above limit, we need that a+hu ∈ E, for h→ 0.

Remark 4.6 If the global limit limx→a

f(x) exists, then all the directional limits lu

exist, for all u = 0.

The converse implication is, in general, false. As a concrete example, let us consider

the function f : R2 \ {(0, 0)} → R, defined by

f(x, y) =x2y

x4 + y2.

It is not difficult to see that lim(x,y)→(0,0)

f(x, y) doesn’t exist, but all the directional

limits limu for all u = 0 exist.

Remark 4.7 Usually, directions are taken to be normalized, although the above

definition works for arbitrary vectors u = 0. So, in such a case we shall work with

unit vectors (also called by some authors direction vectors), i.e. with vectors with

∥u∥ = 1.

Remark 4.8 We recall now the concept of iterated limits for a two-variable func-

tion. Let f : A ⊆ R2 → R and let (a, b) be an accumulation point of A. The

limits

l12 = limx→a

(limy→b

f(x, y)

)and

l21 = limy→b

(limx→a

f(x, y))

are called iterated or repeated limits.

LIMITS AND CONTINUITY 71

Remark 4.9 Let f : A ⊆ R2 → R, (a, b) ∈ A′. If there exists

lim(x,y)→(a,b)

f(x, y) = l

and there exists

limx→a

(limy→b

f(x, y)) = l12,

then l = l12. So, if the global limit and one of the iterated limits exist at a given

point, then the values of these limits must be equal.

Let us notice that if the repeated limits exists, but are not equal, then the global

limit doesn’t exist.

Example 4.10 Let

f(x, y) =x sin

1

x+ y

x+ y.

Then,

limx→0

(limy→0

f(x, y)

)= lim

x→0sin

1

x

does not exist and

limy→0

(limx→0

f(x, y)

)= lim

y→01 = 1.

Also, the global limit lim(x,y)→(0,0)

f(x, y) does not exist.

Example 4.11 If

f(x, y) =2x2

x2 + y2,

then

limx→0

(limy→0

f(x, y)

)= lim

x→0

2x2

x2= 2

and

limy→0

(limx→0

f(x, y)

)= lim

y→00 = 0.

Thus, the global limit lim(x,y)→(0,0) f(x, y) again does not exist, despite the fact that

both of the iterated limits exist.

72 REAL ANALYSIS

Exercise 4.12 Let f : Rn → Rm, f = (f1, . . . , fm) and l = (l1, . . . , lm) ∈ Rm. Show

that

limx→a

f(x) = l ⇔ limx→a

fj(x) = lj , j = 1,m.

Exercise 4.13 Let f, g : E ⊆ Rn → R and let a ∈ E′. If we assume that lim

x→af(x)

and limx→a

g(x) exist and that c is an arbitrary constant, then

a) limx→a

(f ± g)(x) = limx→a

f(x)± limx→a

g(x); b) limx→a

(c f)(x) = c limx→a

f(x);

c) limx→a

(fg)(x) = limx→a

f(x) limx→a

g(x); d) limx→a

f(x)

g(x)=

limx→a

f(x)

limx→a

g(x),

provided that g(x) = 0 on E and limx→a

g(x) = 0.

Exercise 4.14 Let f, g : E ⊆ Rn → R and let a ∈ E′. If we assume that

d(f(x), l) ≤ g(x), ∀x ∈ E \ {a} and if limx→a

g(x) = 0, show that

limx→a

f(x) = l.

Exercise 4.15 Let f : E ⊆ Rn → C, f = u+ iv and let a ∈ E′. Then,

limx→a

f(x) = l ⇔

limx→a

(Ref)(x) = Re l,

limx→a

(Imf)(x) = Im l.

Exercise 4.16 Let f1, f2 : E ⊆ Rn → R and let a ∈ E′. If we assume that

limx→a

f1(x) = l, limx→a

f2(x) = l

and if

f1(x) ≤ φ(x) ≤ f2(x), ∀x ∈ U ∩E \ {a}, U ∈ V(a),

then

limx→a

φ(x) = l.

Exercise 4.17 Let f(x, y) =x2 − y2

x2 + y2, ∀(x, y) ∈ R2 \ {(0, 0)}. Prove that the limit

of this function, for (x, y) → (0, 0), doesn’t exist.


Solution. If we consider the sequence (xk, yk) =

(1

k,α

k

), with α > 0, which is an

admissible one, i.e. (xk, yk) → (0, 0), for k → ∞, we get

limk→∞

f(xk, yk) =1− α2

1 + α2.

So, the limit of f(xk, yk) for k → ∞ depends on α and, hence, the limit of f(x, y)

for (x, y) → (0, 0) doesn’t exist.

Exercise 4.18 Let f(x, y) =x2y2

x2 + y2, ∀(x, y) ∈ R2 \ {(0, 0)}. Prove that

lim(x,y)→(0,0)

f(x, y) = 0.

Solution. It is easy to see that

x2 ≤ x2 + y2, y2 ≤ x2 + y2, ∀(x, y) ∈ R2.

Therefore, we obtain

|f(x, y)| ≤ x2 + y2.

Obviously, since

lim(x,y)→(0,0)

(x2 + y2) = 0,

we get

lim(x,y)→(0,0)

f(x, y) = 0.

Exercise 4.19 Decide if the following limit exists:

l = lim(x,y)→(0,0)

x2y

x2 + y2cos

1

x2 + y2.

Solution. Since

lim(x,y)→(0,0)

x2y

x2 + y2= 0

and the cosine function is bounded, the limit l exists and it is equals to 0.

Exercise 4.20 Compute the directional limit at the point a = (1, 2, 3), in the di-

rection u = (1, 1, 1), of the function f : R3 → R, defined by f(x, y, z) = x2 + y + yz.

74 REAL ANALYSIS


l = limh→0

f(a+ hu) = limh→0

[(1 + h)2 + (2 + h) + (2 + h)(3 + h)] = 9.

Exercise 4.21 Let f : R2 \ {(0, 0)} → R defined by

f(x, y) = xyx− y

x2 + y2.

Test this function for the existence of directional limits at the point (0, 0).


0 ≤∣∣∣xy x− y

x2 + y2

∣∣∣ ≤ | x2y |x2 + y2

+| xy2 |x2 + y2

≤| y | + | x | .

Thus, there exists

lim(x,y)→(0,0)

f(x, y) = 0

and, hence, there exist lu, for all u = 0.

Exercise 4.22 Prove that limx→0

sin1

xdoesn’t exist.

Exercise 4.23 Show that the limit lim(x,y)→(0,0)

xy

x2 + y2doesn’t exit.

Exercise 4.24 Prove that

lim(x,y)→(0,0)

xy√x2 + y2

= 0.

Exercise 4.25 Show that the limit lim(x,y)→(0,0)

x− y

x+ ydoesn’t exist, but the iterated

limits exist and

limx→0

(limy→0

f(x, y)) = limy→0

( limx→0

f(x, y)).

Exercise 4.26 Prove that the limit lim(x,y)→(0,0)

(x+y) sin1

xsin

1

yexists, but we don’t

have iterated limits.


Exercise 4.27 Compute the directional limits in the direction u = (1, 2), at the

point (0, 0), for the function

f(x, y) =xy

x2 + y2.

Exercise 4.28 Compute the directional limits, at the point (0, 0), in the direction

(1, 2) for the function

f(x, y) =x2y

x2 + y2.


lim(x,y)→(0,0)

(x2 + y2) sin1

x2 + y2= 0.


f(x, y) = x sin1

y+ y sin

1

x.

Prove that the global limit lim(x,y)→(0,0)

f(x, y) exists, but the iterated limits don’t.

Exercise 4.31 Let f : R3 \ {(0, 0, 0)} → R defined as follows:

f(x, y, z) =x2 − y2 + z2

x2 + y2 + z2.

Prove that the limit lim(x,y,z)→(0,0,0)

f(x, y, z) doesn’t exist.

Exercise 4.32 Let f : R2 \ {(0, 0)} → R defined as follows:

f(x, y) =x2y2

x2y2 + (x− y)4.

Prove that the global limit lim(x,y)→(0,0)

f(x, y) doesn’t exist, but the iterated limits

exist.

Exercise 4.33 Compute the directional limits at the point (π/4, 1), in the direction−→u = (

√2/2,

√2/2), for the function

f(x, y) = (x2 + x cos y, ln sinx

y).

76 REAL ANALYSIS

Exercise 4.34 Compute the directional limits at the point (1, 1), in the direction−→u = (

√2/2,

√2/2), for the function

f(x, y) = (x2 + y2, 2x y).


f(x, y) =1− cos(x3 + y3)

x2 + y2.

Test this function for the existence of directional limits at the point (0, 0).

4.2 Continuous Functions

Definition 4.36 Let (X, d) and (Y, ρ) be metric spaces, a ∈ X. Also, let f : X →Y . The function f is said to be continuous at the point a if for any V ∈ V(f(a)),there exists U ∈ V(a) such that f(U) ⊆ V .

Definition 4.37 Let (X, d) and (Y, ρ) be metric spaces and f : X → Y . The

function f is said to be continuous on X if it is continuous at each point a ∈ X.

Theorem 4.38 Let (X, d) and (Y, ρ) be metric spaces, a ∈ X. Also, let f : X → Y .

The following statements are equivalent:

(i) f is continuous at the point a;

(ii) for any ε > 0, there exists δε > 0 such that for any x ∈ X with d(x, a) < δε, it

follows that ρ(f(x), f(a)) < ε;

(iii) ∀(xn)n ⊆ X such that xn → a, it follows that f(xn) → f(a).

Remark 4.39 Let f : R → R. The following statements are equivalent:

(i) f is continuous;

(ii) the inverse image of any open subset of R is an open subset in R;

(iii) the inverse image of any closed subset of R is a closed subset of R.

Remark 4.40 a) Let f : Rn → Rm. Then, f is continuous if and only if the

inverse image of any open subset of Rm is an open subset in Rn.

b) If f : Rn → Rm and g : Rm → Rp are continuous maps, then the map h = g ◦ f :

Rn → Rp is continuous, too.


Definition 4.41 Let f : E ⊆ Rn → Rm and let u ∈ Rn, u = 0, a ∈ E. The function

f is said to be continuous at the point a in the direction of the vector u if there exists

limh→0

f(a+ hu) = f(a).

Notice that, for the existence of the above limit, we need that a+hu ∈ E, for h→ 0.

Remark 4.42 If f is continuous at the point a, then f is continuous at the point

a in any direction u = 0.

The converse implication is, in general, false.

Remark 4.43 Usually, as in the case of directional limits, we shall work with unit

vectors, i.e. we shall consider that ∥u∥ = 1.

Definition 4.44 Let (X, d) and (Y, ρ) be two metric spaces. A function f : (X, d) →(Y, ρ) is called uniformly continuous on X if for any ε > 0, there exists δε > 0 such

that for any x ′, x ′′ ∈ X with d(x ′, x ′′) < δε, it follows that ρ(f(x′), f(x ′′)) < ε.

Remark 4.45 If f is uniformly continuous on X, then f is continuous on X. The

converse implication is, in general, false.

Indeed, it is easy to see that the function f : R → R, f(x) = x2 is not uniformly

continuous on R, despite the fact that it is obviously continuous on R.

Remark 4.46 1) Let f : R → R. If f is bounded and monotone, then f is

uniformly continuous on R.

2) If f : R → R is uniformly continuous on R, then ∃ a, b ≥ 0 such that

| f(x) |≤ a | x | + b,

for any x ∈ R.

Definition 4.47 Let K = R,C and X and Y be two vector spaces over K. A map

T : X → Y is called linear if

T (αx+ βy) = αT (x) + βT (y), ∀x, y ∈ X, α, β ∈ K.

Let us notice that for a linear map,

T (0) = 0.

78 REAL ANALYSIS

Exercise 4.48 Let K = R or K = C and let X be a normed space over K. Also,

let f, g : X → K. If f and g are continuous and α, β ∈ K, show that αf + βg, f g,

f/g, for g = 0, are continuous, too.

Exercise 4.49 Let f : A ⊆ Rn → Rm and a ∈ A ′. Prove that f is continuous at

the point a if and only if the limit limx→a

f(x) exists and equals f(a). Notice that if a

is an isolated point for A, then f is continuous at a.

Exercise 4.50 Let f : A ⊆ Rn → Rm, f = (f1, . . . , fm) and let a ∈ A ′. Prove that

f is continuous at the point a if and only if f1, . . . , fm are continuous at the point

a.

Exercise 4.51 If f : A ⊆ Rn → C and a ∈ A′, then f is continuous at the point a

if and only if Re (f) and Im (f) are continuous at the point a.

Exercise 4.52 If (X, ∥ · ∥) is a normed space, show that the map ∥ · ∥ is continuous.

Exercise 4.53 Prove that the function f : (0,∞) → R, defined by

f(x) =2x

1 + x+ x

is uniformly continuous on R.

Solution. Taking x1, x2 ∈ (0,∞), we have

∣∣∣f(x1)− f(x2)∣∣∣ = ∣∣∣ 2x1

1 + x1+ x1 −

2x21 + x2

− x2∣∣∣

≤| x1 − x2 |∣∣∣1 + 2

(x1 + 1)(x2 + 1)

∣∣∣ < 3 | x1 − x2 | .

Therefore, for any ε > 0, there exists δε =ε

3> 0 such that for any x1, x2 ∈ (0,∞)

with | x1 − x2 |< δε, we obtain

| f(x1)− f(x2) |< 3 | x1 − x2 |< 3δε = ε.

Thus, f is uniformly continuous on (0,∞).


Exercise 4.54 Show that the function f : (0, 3) → R, defined by

f(x) =1

x2,

is not uniformly continuous on (0, 3).

Solution. To prove that f is not uniformly continuous on the interval (0, 3), it suffices

to prove that there exists ε0 > 0 and there exist two sequences (xn) and (yn), with

d(xn, yn) < δ, but with ρ(f(xn), f(yn)) ≥ ε0, for any n ≥ n0. Indeed, taking ε0 = 1

and

xn =1√n, yn =

1√n+ 1

,

it follows that | xn − yn |→ 0, while | f(xn)− f(yn) |= 1 = ε0 > 0. So, the function

f is not uniformly continuous (0, 3).

Exercise 4.55 (Heine-Cantor) Let (X, d) and (Y, ρ) be metric spaces. If (X, d) is

compact and f : (X, d) → (Y, ρ) is continuous on X, then f is uniformly continuous

on X.

Solution. Let us suppose that f is continuous, but not uniformly continuous on X.

Therefore, there exists ε0 > 0 such that for any δ > 0 there exist xδ, yδ ∈ X with

d(xδ, yδ) < δ and ρ(f(xδ), f(yδ)) ≥ ε0. In particular, for any n ∈ N∗, there exist

xn, yn ∈ X such that

d(xn, yn) <1

n(4.1)

and

ρ(f(xn), f(yn)) ≥ ε0. (4.2)

Since X is compact, there exists a subsequence (xnk) of the sequence (xn) and there

exists an element x ∈ X such that

limk→∞

xnk= x. (4.3)

In a similar manner, there exist a subsequence (ynk) of the sequence (yn) and an

element y ∈ X such that

limk→∞

ynk= y. (4.4)

80 REAL ANALYSIS

Since

d(ynk, x) ≤ d(ynk

, xnk) + d(xnk

, x),

it follows immediately that x = y.

Using the continuity of f , (4.3) and (4.4), it follows that there exists n0 ∈ N such

that

ρ(f(xnk), f(x)) <

ε04

and

ρ(f(ynk), f(x)) <

ε04

for any k ≥ n0. Thus,

ρ(f(xnk), f(ynk

)) <ε02, ∀k ≥ n0,

which contradicts (4.2). Hence, f is uniformly continuous on X.

Remark 4.56 Let us notice that a contraction mapping between two metric spaces

is uniformly continuous.

Exercise 4.57 Let I ⊆ R be an open interval and f : I → R be derivable on I. If

its derivative is bounded on I, then f is uniformly continuous on I.

Solution. Let M > 0 such that | f ′(x) |≤ M , for all x ∈ I. Using Lagrange’s

theorem, it follows that for any x, y ∈ I there exists θ between x and y such that

f(x)− f(y) = f ′(θ)(x− y).

So,

| f(x)− f(y) |≤M | x− y |, for all x, y ∈ I.

Let ε > 0 and δε =ε

M. Therefore, for any x, y ∈ I with | x− y |< δε, we obtain

| f(x)− f(y) |≤Mε

M=M,

which implies that f is uniformly continuous on I.


Exercise 4.58 Show that the function f : R → R defined by

f(x) = 7x+ cosx,


Solution. The function f is obviously continuous and it is also uniformly continuous

on R, since it is derivable and its derivative is bounded.

Exercise 4.59 Let f : K ⊆ R → R, with K compact. If f is continuous on K, then

f is bounded on K and attains its maximum and minimum values on K, i.e. there

exist a, b ∈ K such that f(a) = inf f(K) and f(b) = sup f(K).

Solution. Since K is compact and f : K → R is continuous, it follows that f(K) is

compact in R, i.e. it is bounded and closed. Let m = inf f(K) and M = sup f(K).

Therefore, m,M ∈ f(K) = f(K) and, hence, there exist a, b ∈ K such that m =

f(a) and M = f(b).

Exercise 4.60 Let f : (X, d) → (Y, ρ) be a continuous function. If K ⊆ X is

compact, then f(K) is compact. If A ⊆ X is connected, then f(A) is connected.

Solution. Let (Fα)α∈A be a family of open sets such that

f(K) ⊆∪α∈A

Fα.

It follows that

K ⊆ f−1(f(K)) ⊆ f−1(∪α∈A

Fα) =∪α∈A

f−1(Fα).

Since f is continuous and (Fα)α∈A is a family of open sets, it follows that f−1(Fα) ∈D. Using the fact that K is compact, there exists J finite, J ⊆ A such that

K ⊆∪α∈J

f−1(Fα)

and, so,

f(K) ⊆∪α∈J

Fα.

Thus, f(K) is compact.

82 REAL ANALYSIS

Exercise 4.61 Let X/K, K = R or K = C, be a compact space and let us consider

the space

C(X) = {f : X → K | f is continuous on X},

endowed with the uniform norm, i.e.

∥f∥u = supt∈X

| f(t) |<∞.

Prove that C(X) is a Banach space.

Exercise 4.62 Let f : (X, d) → (Y, ρ) be a continuous function. If A ⊆ X is

connected, then f(A) is connected. So, connectedness is preserved by continuous

maps. This result can be considered as being a generalization of the intermediate

value theorem.

Solution. We shall argue by contradiction. Let us suppose that f(A) is not a

connected set. Then, there exist D1, D2 ⊆ Y open sets such that

f(A) ⊆ D1 ∪D2,

f(A) ∩D1 ∩D2 = ∅,

f(A) ∩Di = ∅, ∀i = 1, 2.

Since f is continuous, it follows that f−1(Di) ∈ D, ∀i = 1, 2. Therefore,

A ⊆ f−1(D1) ∪ f−1(D2),

A ∩ f−1(D1) ∩ f−1(D2) = ∅,

A ∩ f−1(Di) = ∅, ∀i = 1, 2.

So, A is not connected, which contradicts our hypotheses. Thus, our assumption is

false and, hence, f(A) is connected.

Exercise 4.63 Let (X,D) be a topological space and f : (X,D) → (R,DR). If f is

continuous on X, prove that f(X) is an interval.

Solution. The result is an easy consequence of Exercise 4.62.


Exercise 4.64 Show that every product of a family of connected spaces is also

connected.

Exercise 4.65 Let X and Y be two normed spaces. If T : X → Y is linear and

continuous at a point a ∈ X, then T is continuous on X.

Solution. Let x0 ∈ X. Since T is continuous at the point a, it follows that for any

ε > 0 there exists δε > 0 such that for any x ∈ X with ∥x− a∥ < δε, we have

∥T (x)− T (a)∥ < ε.

Let t ∈ X, with ∥t− x0∥ < δε, and x = a+ (t− x0). Obviously, x− a = t− x0 and

∥x− a∥ = ∥t− x0∥ < δε. Since T is continuous at the point a and ∥x− a∥ < δε, we

get

∥T (x)− T (a)∥ < ε.

From the linearity of T , we obtain

T (x) = T (a+ (t− x0)) = T (a) + T (t)− T (x0).

Thus,

T (x)− T (a) = T (t)− T (x0),

which implies that

∥T (t)− T (x0)∥ < ε,

for any t with ∥t− x0∥ < δε. Hence, T is continuous at any point x0 ∈ X.

Exercise 4.66 Let X and Y be two normed spaces and T : X → Y be a linear

map. Show that T is continuous on X if and only if there exists M ≥ 0 such that

∥T (x)∥ ≤M∥x∥, ∀x ∈ X.

Solution. Let us prove the implication (i) ⇒ (ii). This is equivalent, in fact, to

proving that non (ii) ⇒ non (i). Thus, let us suppose that for any M ≥ 0, there

exists xM ∈ X such that

∥T (xM )∥ > M∥xM∥.

84 REAL ANALYSIS

In particular, taking M = n ∈ N∗, it follows that there exists xn ∈ X such that

∥T (xn)∥ > n∥xn∥.

So,

∥T(

1

n∥xn∥xn

)∥ > 1, ∀n ∈ N∗,

which means that

T (1

n∥xn∥xn) 9 0.

On the other hand,

limn→∞

1

n∥xn∥xn = 0.

Hence, since

T

(1

n∥xn∥xn

)9 T (0),

it follows that T is not continuous at the point 0.

(ii) ⇒ (i). Let a ∈ X. Since T is linear, we get

∥T (x)− T (a)∥ = ∥T (x− a)∥.

Therefore, there exists M > 0 with

∥T (x− a)∥ ≤M∥x− a∥,

which means that

∥T (x)− T (a)∥ ≤M∥x− a∥.

Hence, T is continuous at any point a ∈ X.

Exercise 4.67 Let Y be a normed space. If T : Rn → Y is linear, then T is

continuous on Rn.

Solution. Let {e1, . . . , en} be the canonical basis of Rn and x = (x1, . . . , xn) ∈ Rn.

Therefore,

x =n∑

i=1

xiei.


Using the linearity of T , we obtain

∥T (x)∥ = ∥T (n∑

i=1

xiei)∥ = ∥n∑

i=1

xiT (ei)∥ ≤

n∑i=1

∥xiT (ei)∥ =n∑

i=1

|xi|∥T (ei)∥ ≤ ∥x∥n∑

i=1

∥T (ei)∥.

Denoting by M =n∑

i=1

∥T (ei)∥, we get

∥T (x)∥ ≤M∥x∥, for any x ∈ Rn.

Thus, the map T is continuous on Rn.

Exercise 4.68 Let

f(x) =

√1− cos x2

1− cos x, x ∈ [−π/2, π/2] \ {0},

a, x = 0.

Find the values of a for which f is continuous.

Exercise 4.69 Let

f(x) =

6sin a(x− 1)

x− 1, x ∈ [0, 1),

−a+ 5x, x ∈ [1, 2].

Find the values of a for which f is continuous.

Exercise 4.70 Test the following function for continuity:

f(x, y) =

xy

x2 + y2, (x, y) = (0, 0),

0, (x, y) = (0, 0).

86 REAL ANALYSIS


f(x, y) =

xy√x2 + y2

, (x, y) = (0, 0),

0, (x, y) = (0, 0).


f(x, y) =

(x2 + y2) sin

1

x2 + y2, (x, y) = (0, 0),

0, (x, y) = (0, 0).


f(x, y) =

x2y

x2 + y2, (x, y) = (0, 0),

0, (x, y) = (0, 0).

Exercise 4.74 Let

f(x, y) =

| x |α| y |β

(x2 + y2)γ, (x, y) = (0, 0),

0, (x, y) = (0, 0).

Prove that f is continuous ⇔ α+ β > 2γ.


f(x, y) =

x2

y, y = 0,

0, y = 0.

Exercise 4.76 Let

f(x, y) =

1− cos(x3 + y3)

x2 + y2, (x, y) = (0, 0),

0, (x, y) = (0, 0).

Test this function for continuity and directional continuity at the point (0, 0).


Exercise 4.77 Let

f(x, y) =

x2y2

x2y2 + (x2 − y)2, (x, y) = (0, 0),

0, (x, y) = (0, 0).


Exercise 4.78 Let

f(x, y) =

|x| y2√x2 + y2

, (x, y) = (0, 0),

0, (x, y) = (0, 0).



f(x, y) =

|x| sin 1

x2 + y2, (x, y) = (0, 0),

0, (x, y) = (0, 0).

Exercise 4.80 Show that the function f : R → R, given by

f(x) =√1 + x2,


Exercise 4.81 Prove that the function f : (0, π/2) → R, defined by

f(x) = sinx,

is uniformly continuous on (0, π/2).

Exercise 4.82 Show that the function f : (0,∞) → R, defined by

f(x) = x+x

1 + x,

is uniformly continuous on (0,∞).

Chapter 5

Multivariable DifferentialCalculus

5.1 Derivable Functions

In what follows, we shall work only with real Banach spaces.

Definition 5.1 Let E and F be Banach spaces, ∅ = U ⊆ E, U open, a ∈ U . Also,

let f : U → F . The function f is said to be derivable or Frechet differentiable at

the point a if there exists f ′(a) ∈ L(E,F ) such that

limx→ ax = a

∥f(x)− f(a)− f ′(a) ◦ (x− a)∥∥x− a∥

= 0. (5.1)

f ′(a) is called the derivative of the function f at the point a.

Remark 5.2 If we put

ωf,a(x) =

f(x)− f(a)− f ′(a) ◦ (x− a)

∥x− a∥, ∀x ∈ U \ {a},

0F , x = a,

then the condition (5.1) becomes

limx→a

ωf,a(x) = 0F .

89

90 REAL ANALYSIS

So, f is derivable at the point a if there exist f ′(a) ∈ L(E,F ) and ωf,a : U → F

such that

f(x) = f(a) + f ′(a) ◦ (x− a) + ∥x− a∥ωf,a(x), ∀x ∈ U (5.2)

and

limx→a

ωf,a(x) = 0F . (5.3)

Theorem 5.3 (Uniqueness of the Derivative) If f : U ⊆ E → F is derivable at the

point a ∈ U , then there exists only one linear mapping f ′(a) ∈ L(E,F ) such that

(5.1) holds true.

Notice that due to the uniqueness of the derivative f ′(a) ∈ L(E,F ), we can define

a map

f ′ : U → L(E,F ).

Also, notice that the domain of the linear mapping f ′(a) is the entire space E,

despite the fact that the function f is defined only locally about the point a.

Proposition 5.4 If the function f : U ⊆ E → F is derivable at the point a ∈ U ,

then f is continuous at the point a.

Exercise 5.5 If f : U ⊆ E → F is constant on U , then f is derivable on U and

f ′(a) = 0, ∀a ∈ U .

Solution. Obviously, we have f(x) = f(a) + 0 ◦ (x − a), ∀x ∈ U . Therefore, f is

derivable at the point a and f ′(a) = 0.

Exercise 5.6 If f ∈ L(E,F ), then f is derivable at any point a ∈ E and f ′(a) =

f, ∀a ∈ E.

Solution. Since f is linear, then

f(x)− f(a)− f(x− a)

∥x− a∥=

0

∥x− a∥= 0, ∀x ∈ E \ {a}.

So, f is derivable at any point a ∈ E and f ′(a) = f, ∀a ∈ E.

MULTIVARIABLE DIFFERENTIAL CALCULUS 91

Exercise 5.7 (Linearity of the Derivative) Let E and F be Banach spaces, U ⊆ E,

U open, a ∈ U . Also, let f, g : U → F . If f and g are differentiable at the point a,

then the sum f + g is differentiable at the point a and

(f + g) ′(a) = f ′(a) + g ′(a).

Moreover, if α ∈ R, then αf is differentiable at the point a and

(αf) ′(a) = αf ′(a).

Theorem 5.8 (Chain Rule) Let E,F,G be Banach spaces, ∅ = U ⊆ E, ∅ = V ⊆ F ,

U, V open. If f : U → V is derivable at the point a ∈ U and g : V → G is derivable

at the point f(a) , then g ◦ f is derivable at the point a and

(g ◦ f) ′(a) = g ′(f(a)) ◦ f ′(a).

Remark 5.9 If f ′ is continuous on U , we shall say that f is continuously diffe-

rentiable on U . We shall use the following notation:

C1(U,F ) = {f : U → F | f is continuously differentiable on U}.

Definition 5.10 Let E,F be Banach spaces, ∅ = U ⊆ E, U open, f : U → F, a ∈U . The function f is said to be twice (two-times) differentiable at the point a if:

1) ∃V ∈ V(a), V open, such that f is derivable on V ;

2) f ′ : V → L(E,F ) is derivable at the point a.

The derivative of f ′ at the point a is denoted by f ′′(a) and it is called the second

derivative of f at the point a.

Remark 5.11 If the function f is two-times differentiable on U and the map

f ′′ : U → L(E,L(E,F ))

is continuous on U , then we shall say that f is twice continuously differentiable on

Uand we shall use the notation:

C2(U,F ) = {f : U → F | f is twice continuously differentiable on U}.

92 REAL ANALYSIS

Remark 5.12 f ′′(a) ∈ L(E,L(E,F )) ≃ L2(E,E;F ).

By usual induction, we can define

Cn(U,F ) = {f : U → F | f is n-times continuously differentiable on U}.

Putting

C0(U,F ) = {f : U → F | f is continuous on U},

we can define

C∞(U,F ) =∩n∈N

Cn(U,F ).

A function f ∈ C∞(U,F ) is said to be infinitely (indefinitely) derivable on U .

Definition 5.13 Let X be a vector space and let A ⊆ X. The set A is said to be

convex if for any x, y ∈ A the segment [x, y] ⊆ A, i.e. for any x, y ∈ A and for any

t ∈ [0, 1], we have

(1− t)x+ ty ∈ A.

Definition 5.14 Let A ⊆ X be a convex set. The function f : A → R is called

convex if for any x, y ∈ A and for any t ∈ [0, 1], it follows that

f((1− t)x+ ty) ≤ (1− t)f(x) + tf(y).

Proposition 5.15 Let E be a Banach space. If f : [a, b] → E is continuous on

[a, b], derivable on (a, b) and if there exists M ≥ 0 such that ∥f ′(t)∥ ≤ M , for any

t ∈ (a, b), then

∥f(b)− f(a)∥ ≤M(b− a).

Theorem 5.16 (The first theorem of finite increments) Let f : U ⊆ E → F ,

∅ = U open and convex. If f is derivable on U , then

∥f(v)− f(u)∥ ≤ ∥v − u∥ supx∈[u,v]

∥f ′(x)∥, ∀u, v ∈ U.

Corollary 5.17 Let f : U ⊆ E → F , where ∅ = U is open and convex. If f is

derivable on U and if there exists M ≥ 0 such that ∥f ′(x)∥ ≤M , for any x ∈ U , it

follows that

∥f(v)− f(u)∥ ≤M∥v − u∥,∀u, v ∈ U.


Theorem 5.18 (The second theorem of finite increments) If f : U ⊆ E → F ,

∅ = U open and convex, is derivable on U , then

∥f(b)−f(a)−f ′(c)◦(b−a)∥ ≤ ∥b−a∥ supx∈[a,b]

∥f ′(x)−f ′(c)∥, ∀a, b ∈ U, c ∈ [a, b].

We consider now the case in which

E = Rn, F = Rm.

Exercise 5.19 Let ∅ = U ⊆ Rn be an open set and a ∈ U . Show that if the functions

f, g : U → R are derivable at the point a, then f ± g, λf, λ ∈ R, f g, f/g, g = 0,

are derivable at the point a. Moreover, if f is injective and continuous on U and

f ′(a) = 0, prove that the inverse function f−1 is derivable at the point b = f(a) and

(f−1)′(f(a)) =

1

f ′(a).

Proposition 5.20 (Chain Rule) Let U, V ⊆ R be open nonempty sets and let f :

U → R and g : V → R such that f(U) ⊆ V . If f is derivable at the point a ∈ U and

g is derivable at the point b = f(a) ∈ V , then g ◦ f is derivable at the point a and

(g ◦ f) ′(a) = g ′(f(a))f ′(a).

Exercise 5.21 Let f : R → R,

f(x) =

x2 sin

1

x, x = 0,

0, x = 0.

Prove that f is derivable on R, but f is not a C1 function on R.


2 arctanx+ arcsin2x

1 + x2= π sgn x, ∀ | x |≥ 1.

94 REAL ANALYSIS

Let us consider now briefly another important case, i.e. the case in which E = Rn

and F = Rm. So, let ∅ = U ⊆ Rn be an open set, f : U → Rm and a ∈ U . If

f = (f1, . . . , fm),

then f is derivable at the point a if and only if each component fi, with i = 1,m, is

derivable at the point a ( the componentwise nature of differentiability).

Example 5.23 The function f : R2 → R3 defined by

f(x, y) = (x2 + y2, xy, x3 + y3)

is obviously differentiable on R2, since its three components are composed of ele-

mentary functions.

Definition 5.24 Let us define

pi : Rn → R, pi(x) = xi, ∀x = (x1, . . . , xn) ∈ Rn

pi are called the canonical projections.

Remark 5.25 pi are linear and continuous. Thus, they are are derivable.

Theorem 5.26 Let f : U ⊆ Rn → R, U open and nonempty. If f is derivable at the

point a ∈ U , then f is partial derivable with respect to each variable xi, 1 ≤ i ≤ n,

at the point a and

f ′(a) =n∑

i=1

∂f

∂xi(a) pi.

Proposition 5.27 Let ∅ = U ⊆ Rn, U open, f : U → Rm derivable at the point

a ∈ U . The matrix associated to the linear map f ′(a) : Rn → Rm in the canonical

bases of Rn and Rm, called the Jacobi matrix, is

Mf (a) =

∂f1∂x1

(a) · · · ∂f1∂xn

(a)

..............................

∂fm∂x1

(a) · · · ∂fm∂xn

(a)


If m = n, the determinant

Jf (a) = det Mf (a)

is called the Jacobian of the function f at the point a.

If f : U ⊆ Rn → Rn, U open and nonempty and if f = (f1, . . . , fn), we shall

denote the Jacobian of the vector valued function f at the point a by

Jf (a) =∂(f1, ..., fn)

∂(x1, . . . , xn)(a).

Exercise 5.28 Compute the Jacobi matrix at the point (1, 1) for the function f :

R2 → R2 defined by

f(x, y) = (x2 + y2, 2xy).


Mf (1, 1) =

2 2

2 2

and its Jacobian is Jf (1, 1) = 0.

Theorem 5.29 Let U be an open nonempty set in Rn and f : U → R. Then,

f ∈ C1(U) if and only if f is derivable with respect to each variable xi and∂f

∂xi:

U → L(R,R), 1 ≤ i ≤ n, are continuous.

Usually, we denote the canonical projections pi by

pi = dxi

and we call dxi the differential of xi. So, f′(a) ∈ L(Rn,R) and

f ′(a) =n∑

i=1

∂f

∂xi(a) dxi, (5.4)

where

∂f

∂xi(a) = lim

xi→ai

f(a1, . . . , ai−1, xi, ai+1, . . . , an)− f(a1, . . . , ai−1, ai, ai+1, . . . , an)

xi − ai.

96 REAL ANALYSIS

In fact, {p1, . . . ,pn} is the canonical basis for L(Rn,R) and, so, (5.4) is just the

unique representation of the linear map f ′(a) in this basis. The partial derivatives∂f

∂xi(a) are exactly the coefficients of f ′(a) in the canonical basis.

Since f ′(a) ∈ L(Rn,R) ≃ Rn, the derivative f ′(a) can be also regarded as being the

vector

f ′(a) ≃ (∂f

∂x1(a), . . . ,

∂f

∂xn(a)).

Remark 5.30 Let ∅ = U ⊆ Rn, U open, f : U → R derivable at the point a ∈ U .

Then, there exists a unique vector y ∈ Rn such that

f ′(a)(x) = ⟨x, y⟩.

This vector is denoted by gradaf and it is called the gradient of the function f at

the point a.

Exercise 5.31 Let f : R2 → R be defined by

f(x, y) =√x2 + y2.

Prove that f is not differentiable at the point (0, 0).

Solution. If f would be differentiable at the point (0, 0), then its partial derivatives

at the point (0, 0) would exist. In this case,

∂f

∂x(0, 0) = lim

x→0

f(x, 0)− f(0, 0)

x− 0= lim

x→0

|x|x

and∂f

∂y(0, 0) = lim

y→0

f(0, y)− f(0, 0)

y − 0= lim

y→0

|y|y.

But these limits don’t exist and, hence, f is not differentiable at the point (0, 0).

Exercise 5.32 Define f : R2 → R as follows:

f(x, y) =

xy

x2 + y2, (x, y) = (0, 0),

0, (x, y) = (0, 0).

Test it for differentiability at the point (0, 0).


Solution. If f would be differentiable at the point (0, 0), then f would be continuous

at the point (0, 0), i.e. lim(x,y)→(0,0)

f(x, y) = f(0, 0). But lim(x,y)→(0,0)

f(x, y) doesn’t

even exist and, hence, f is not differentiable at the point (0, 0). Still, the partial

derivatives∂f

∂x(0, 0) and

∂f

∂x(0, 0) exist and are equal to zero.

Exercise 5.33 Test the following function for differentiability:

f(x, y) =

x2y

x2 + y2, (x, y) = (0, 0),

0, (x, y) = (0, 0).

Solution. Obviously, f is differentiable on R2 \ {(0, 0)} and

f ′(a, b) = (∂f

∂x(a, b),

∂f

∂y(a, b)) = (

2ab3

(a2 + b2)2,a2(a2 − b2)

(a2 + b2)2),

for any (a, b) = (0, 0).

Let us see now what happens at the origin. It is easy to see that f is continuous

at the point (0, 0) and its first partial derivatives exist. Indeed,

∂f

∂x(0, 0) = lim

x→0

f(x, 0)− f(0, 0)

x− 0= lim

x→0

0

x= 0

and, in a similar manner,∂f

∂y(0, 0) = 0. Therefore, if the function f would be

differentiable at the point (0, 0), then the derivative of f at this point should be the

zero map and

lim(x, y) → (0, 0)

f(x, y)− f(0, 0)− f ′(0, 0) ◦ (x− 0, y − 0)

∥(x− 0, y − 0)∥= 0,

i.e.

lim(x, y) → (0, 0)

x2y

x2 + y2√x2 + y2

= 0.

But this limit doesn’t exist and, hence, f is not differentiable at the origin.

98 REAL ANALYSIS

Exercise 5.34 Define f : R2 → R as follows:

f(x, y) =

1 x = y = 0,

0 otherwise.

Show that f is not continuous at the point (0, 0), but both partial derivatives∂f

∂x(0, 0) and

∂f

∂x(0, 0) exist.

Solution It is easy to see that if we let (x, y) to approach the origin along the line

x = y, then

lim(x,y)→(0,0)

f(x, y) = 1 = f(0, 0)

and this implies the fact that f is not continuous at the origin. However, both

partial derivatives exist at the origin. Indeed,

∂f

∂x(0, 0) = lim

x→0

f(x, 0)− f(0, 0)

x− 0= lim

x→0

0

x= 0

and, in a similar manner,∂f

∂y(0, 0) = 0.

Exercise 5.35 Let f : R2 → R be defined by

f(x, y) = ln(1 + 2x2 + 3y2).

Compute f ′(1, 2).


f(x, y) =

xy√x2 + y2

, (x, y) = (0, 0),

0, (x, y) = (0, 0).

Exercise 5.37 Test the following function, defined on R2, for differentiability:

f(x, y) =

(x2 + y2) sin

1

x2 + y2, (x, y) = (0, 0),

0, (x, y) = (0, 0).



f(x, y) =

x2y3

x2 + y2, (x, y) = (0, 0),

0, (x, y) = (0, 0).

Partial derivatives of higher order.

Let f : U ⊆ Rn → R. Then,

f ′′(a) ∈ L2(Rn,Rn;R).

Theorem 5.39 (Schwarz) Let f : U ⊆ Rn → R, ∅ = U open. If f is two-times

derivable at the point a ∈ U , then

∂2f

∂xi∂xj(a) =

∂2f

∂xj∂xi(a), ∀i, j = 1, n,

f ′′(a) =n∑

i,j=1

∂2f

∂xi∂xj(a) pi ⊗ pj .

(5.5)

In (5.5),

∂2f

∂xi∂xj(a) = lim

xi→ai

∂f

∂xj(a1, . . . , xi, . . . , an)−

∂f

∂xj(a1, . . . , ai, . . . , an)

xi − ai

and pi ⊗ pj is the tensor product of the linear maps pi and pj , defined by

pi ⊗ pj : Rn × Rn → R, (pi ⊗ pj)(x, y) = pi(x)pj(y) = xiyj .

In fact, {pi ⊗ pj}1≤i,j≤n is the canonical basis of L2(Rn,Rn;R). So, (5.5) is the

unique representation of the bilinear map f ′′(a) in this basis. The mixed partial

derivatives∂2f

∂xi∂xj(a) are the coefficients of f ′′(a) in the canonical basis.

Just as the Jacobi matrix was defined for representing (in the canonical bases)

the first derivative f ′(a) of a function f : U ⊆ Rn → Rm which is derivable at the

point a ∈ U , in the case of the second derivative, for a function f taking values in

R, we have the following analogue representation of the second derivative:

100 REAL ANALYSIS

Theorem 5.40 Let ∅ = U ⊆ Rn be an open set and f : U → R be a twice differen-

tiable function on U . Then, the second derivative of f at the point a is determined,

in the canonical basis, by the Hessian matrix:

Hf (a) =

∂2f

∂x21(a) · · · ∂2f

∂x1∂xn(a)

.................................

∂2f

∂xn∂x1(a) · · · ∂

2f

∂x2n(a)

Exercise 5.41 Let f : R3 → R be given by

f(x, y, z) = xy2z3.

Then,

Hf (x, y, x) =

0 2yz3 3y2z2

2yz3 2xz3 6xyz2

3y2z2 6xyz2 6xy2z

.

Remark 5.42 If f ∈ C2(U), then

∂2f

∂xi∂xj(a) =

∂2f

∂xj∂xi(a), ∀i, j = 1, n.

Remark 5.43 The existence of the mixed partial derivatives of the second order for

a given function doesn’t imply their symmetry.

As a counterexample, it is not difficult to see that for the function f : R2 → R,defined by:

f(x, y) =

xy

x2 − y2

x2 + y2, (x, y) = (0, 0),

0, (x, y) = (0, 0),

one has∂2f

∂x∂y(0, 0) = ∂2f

∂y∂x(0, 0).


Indeed, let us compute∂2f

∂x∂y(0, 0). We get:

∂2f

∂x∂y(0, 0) = lim

x→0

∂f

∂y(x, 0)− ∂f

∂y(0, 0)

x− 0.

But∂f

∂y(x, 0) = lim

y→0

f(x, y)− f(x, 0)

y − 0= x

and∂f

∂y(0, 0) = lim

y→0

f(0, y)− f(0, 0)

y − 0= 0.

So,∂2f

∂x∂y(0, 0) = lim

x→0

x− 0

x− 0= 1.

On the other hand, computing∂2f

∂y∂x(0, 0), we obtain:

∂2f

∂y∂x(0, 0) = lim

y→0

∂f

∂x(0, y)− ∂f

∂x(0, 0)

y − 0.

But∂f

∂x(0, y) = lim

x→0

f(x, y)− f(0, y)

x− 0= −y

and∂f

∂x(0, 0) = lim

x→0

f(x, 0)− f(0, 0)

x− 0= 0.

So,∂2f

∂y∂x(0, 0) = lim

y→0

−y − 0

y − 0= −1.

Hence,∂2f

∂x∂y(0, 0) = ∂2f

∂y∂x(0, 0).

As already mentioned, for simplicity, we shall use the notation:

pi = dxi.

102 REAL ANALYSIS

Also, we shall usually omit to write explicitly the symbol ”⊗ ”. So, we have:

f ′′(a) =n∑

i,j=1

∂2f

∂xi∂xj(a) dxi dxj .

For the particular case n = 2, we get

f ′′(a, b) =∂2f

∂x2(a, b) dx2 +

∂2f

∂x∂y(a, b) dx dy +

∂2f

∂y∂x(a, b) dy dx+

∂2f

∂y2(a, b) dy2.

Remark 5.44 For m ≥ 2, by induction, we get

∂mf

∂x1 · · · ∂xm(a) =

∂mf

∂xσ(1) · · · ∂xσ(m)(a),

for any permutation σ : {1, . . . ,m} → {1, . . . ,m}, and

f (m)(a) =n∑

i1,...,im=1

∂mf

∂xi1 · · · ∂xim(a) pi1 ⊗ · · · ⊗ pim .

Exercise 5.45 Verify that the function f : R2 → R, defined by f(x, y) = ex cos y is

a solution of Laplace’s equation, i.e.

∂2f

∂x2+∂2f

∂y2= 0.

Solution. Indeed, since f ∈ C2(R2), we have

∂2f

∂x2= ex cos y

and∂2f

∂y2= −ex cos y.

Therefore, their sum is zero.

Exercise 5.46 Compute the second derivative at the point (1, 1) of the function

f : R2 → R, defined by:

f(x, y) = ex2+y2 .


Solution. Obviously, f ∈ C2(R2) and it is not difficult to compute its second order

partial derivatives at the point (1, 1). As a result, we get

f ′′(1, 1) = 6 e2 dx2 + 4 e2 dx dy + 4 e2 dy dx+ 6 e2 dy2.

Exercise 5.47 Compute the first and the second differential for the function f :

R2 → R, f(x, y) = x3 + y3 + ln (2 + x2 + y2) at the point (1, 1).

Exercise 5.48 Compute the second derivative at the point (2, 2) of the function

f : R2 → R2, defined by

f(x, y) = (x4 + y4, x2 + y2).

Exercise 5.49 For the function f : R2 → R, defined by:

f(x, y) =

xy sin

x2 − y2

x2 + y2, (x, y) = (0, 0),

0, (x, y) = (0, 0),

prove that∂2f

∂x∂y(0, 0) = ∂2f

∂y∂x(0, 0).

Exercise 5.50 Compute the first and the second differential at the point (1, 1) for

the function f : R2 → R, defined by

f(x, y) = sin (x2 + y2).

Definition 5.51 Let U be a nonempty open set in Rn and let f : U → Rm. Also,

let u ∈ Rn, u = 0. If there exists

δf(a, u) = limt→0

f(a+ tu)− f(a)

t∈ Rm,

then δf(a, u) is called the derivative of the function f , at the point a, in the direction

u.

104 REAL ANALYSIS

Sometimes, we shall use similar notation, i.e.

δf(a, u) =∂f

∂u(a)

or

δf(a, u) = f ′u(a).

Usually, a direction in Rn is specified by a unit vector, i.e. a vector u ∈ Rn with

∥u∥ = 1. So, often, we shall consider that the vectors u are taken to be normalized,

although the definition above works for arbitrary (even zero) vectors.

Theorem 5.52 Let U be a nonempty open set in Rn. Let f : U → Rm and u ∈Rn, u = 0. If f is derivable at the point a ∈ U , then f is derivable at the point a in

any direction u = 0 and

δf(a, u) = f ′(a)(u), ∀u = 0.

We focus on the particular case in which U is an open nonempty set in Rn and

f : U ⊆ Rn → R. If a ∈ U , the function f is called Gateaux differentiable at the

point a if it is derivable at a in any direction u ∈ Rn. The number δf(a, u) is

called the Gateaux differential of f at the point a in the direction u and the map

df(a) : Rn → R, defined by

df(a)(u) = δf(a, u)

is the Gateaux derivative of the function f at the point a.

Let us notice that we cannot establish a connection between the continuity of

a given function and its directional derivability. For instance, on one hand, the

function f : R2 → R, defined by

f(x, y) =√x2 + y2

is continuous at the point (0, 0), but it is not Gateaux differentiable at this point.

On the other hand, if we consider the function f : R2 → R, defined by

f(x, y) =

x2

y, y = 0,

0, y = 0,


it is not difficult to see that f is Gateaux differentiable at the point (0, 0), but it is

not continuous at this point.

It is worth noting that the Gateaux derivative is not, in general, linear, unlike

the Frechet derivative. Moreover, let us mention that a similar notion can be defined

for functions between more general spaces, such as locally convex topological vector

spaces or Banach spaces.

Exercise 5.53 Let U be a nonempty open set in Rn and let fi : U → R, i = 1, n,

be derivable at the point a ∈ U in the direction u. Show thatn∑

i=1

cifi, where ci ∈ R,

is derivable at the point a in the direction u and

∂

∂u

(n∑

i=1

cifi

)(a) =

n∑i=1

ci∂fi∂u

(a).

Exercise 5.54 Let U be a nonempty open set in Rn. If f, g : U → R are derivable

at the point a ∈ U in the direction u, then f g is derivable at the point a in the

direction u and∂

∂u(fg)(a) =

∂f

∂u(a)g(a) + f(a)

∂g

∂u(a).

Exercise 5.55 Let U be a nonempty open set in Rn. Show that if f : U → Rm, f =

(f1, . . . , fm), then f is derivable at the point a ∈ U in the direction u if and only if

f1, . . . , fm are derivable at the point a in the direction u.

Remark 5.56 Let f : U ⊆ Rn → R, U = ∅, U open, a = (a1, . . . , an) ∈ U . We

know that

∂f

∂xj(a) = lim

xj→aj

f(a1, . . . , xj , . . . , an)− f(a1, . . . , aj , . . . , an)

xj − aj.

If we take

t = xj − aj ,

then∂f

∂ej(a) = lim

t→0

f(a+ tej)− f(a)

t=

∂f

∂xj(a),

where

ej = (0, . . . , 1, . . . , 0).

106 REAL ANALYSIS

Hence, partial derivatives can be seen as particular directional derivatives along the

standard basis vector directions.

Remark 5.57 Let us suppose that f : Rn → R is derivable at a given point a. Then,

δf(a, u) = f ′(a)(u) = ∇f(a) · u, ∀u = 0.

So, the slopes in all directions are known from slopes in n directions (given by the

partial derivatives).

Exercise 5.58 If f : R2 → R is defined by

f(x, y) = 4− 2x2 − y2

and

u = − 1√2(1, 1),

then∂f

∂u(1, 1) = 3

√2.

Exercise 5.59 Let U be a nonempty open set in Rn and let f : U → Rm. Also, let

u ∈ Rn, u = 0. Show that if f is derivable at the point a in the direction u, then f

is derivable at a in the direction −u and

∂f

∂(−u)(a) = −∂f

∂u(a).

More generally, if f is derivable at the point a in the direction u, then f is derivable

at a in the direction αu, α ∈ K and

∂f

∂(αu)(a) = α

∂f

∂u(a).

Exercise 5.60 Let U be a nonempty open set in Rn and let f : U → Rm. Also, let

u ∈ Rn, u = 0. Prove that if f is derivable at the point a in the direction u, then f

is continuous at a in the direction u.


Exercise 5.61 Let us consider the function f : R2 → R, defined by:

f(x, y) =

x2y

x2 + y2, (x, y) = (0, 0),

0, (x, y) = (0, 0).

Prove that f is not differentiable at the point (0, 0), despite the fact that it possesses

directional derivatives at this point with respect to any direction u = 0.

Exercise 5.62 Compute the directional derivative of the function f : R2 → R2,

defined by

f(x, y) = (x2 + y, x+ y2)

at the point a = (2, 3), in the direction of the vector u = (1, 1).

Exercise 5.63 Let f : R2 → R, defined by:

f(x, y) =

xy2

x2 + y2, (x, y) = (0, 0),

0, (x, y) = (0, 0),

a) Prove that f is continuous at the point (0, 0).

b) Compute the partial derivatives of f at (0, 0).

c) Compute the directional derivatives of the function f at the point (0, 0) along

any direction u with ∥u∥ = 1.

d) Prove that f is not differentiable at the point (0, 0).

Proposition 5.64 Let ∅ = U ⊆ Rn, U open, f : U → Rm derivable at the point

a ∈ U . The matrix associated to the linear map f ′(a) : Rn → Rm in the canonical

bases of Rn and Rm is

Mf (a) =

∂f1∂x1

(a) · · · ∂f1∂xn

(a)

................................

∂fm∂x1

(a) · · · ∂fm∂xn

(a)

∈ M(m,n;R).

108 REAL ANALYSIS

If m = n, the determinant

Jf (a) = det Mf (a)

was called the Jacobian of the function f at the point a and it was also denoted by

Jf (a) =∂(f1, . . . , fn)

∂(x1, . . . , xn)(a).

Theorem 5.65 (Chain Rule) Let E,F,G be Banach spaces, U ⊆ E, V ⊆ F , U, V

nonempty open sets. If f : U → V is derivable at the point a ∈ U and g : V → G is

derivable at the point f(a) , then g ◦ f is derivable at the point a and

(g ◦ f) ′(a) = g ′(f(a)) ◦ f ′(a).

Proposition 5.66 Let f : U ⊆ Rn → Rm, g : V ⊆ Rm → Rk, U, V nonempty open

sets. If f is derivable at a ∈ U and g is derivable at b = f(a) ∈ V , then

Mg◦f (a) =Mg(b) ·Mf (a). (5.6)

Corollary 5.67 Let U, V be nonempty open subsets of Rn, f : U → Rn, f =

(f1, . . . , fn) derivable at the point a ∈ U , g : V → Rn, g = (g1, . . . , gn) derivable

at the point b = f(a) ∈ V . Then, the map h = g ◦ f : U → Rn, h = (h1, . . . , hn), is

derivable at the point a and

∂(h1, . . . , hn)

∂(x1, . . . , xn)(a) =

∂(g1, . . . , gn)

∂(f1, . . . , fn)(b)

∂(f1, . . . , fn)

∂(x1, . . . , xn)(a).

Exercise 5.68 Let g : R → R be an arbitrary C1 function and let

F (x, y) = g(x2 + y2).

Compute∂F

∂xand

∂F

∂y.

Solution. If we denote by u(x, y) = x2 + y2, then∂F

∂x=dg

du

∂u

∂x=dg

du2x = g ′ 2x,

∂F

∂y=dg

du

∂u

∂y=dg

du2y = g ′ 2y.


Exercise 5.69 Let

F (x, y) = xy + g(x+ y, x3 + y3),

where g : R2 → R is an arbitrary given function of class C1. Compute∂F

∂xand

∂F

∂y.

Solution. If we denote by

u(x, y) = x+ y

and by

v(x, y) = x3 + y3,

then∂F

∂x= y +

∂g

∂u

∂u

∂x+∂g

∂v

∂v

∂x,

i.e.∂F

∂x= y +

∂g

∂u+ 3x2

∂g

∂v.

In a similar manner, we get

∂F

∂y= x+

∂g

∂u+ 3y2

∂g

∂v.

Exercise 5.70 Let

F (x, y, z) = xyz + g(x+ y + z, x2 + y2 + z2, xyz),

where g : R3 → R is an arbitrary given function of class C1. Compute∂F

∂x,∂F

∂yand

∂F

∂z.

Solution. If we denote

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

v(x, y, z) = x2 + y2 + z2

and

w(x, y, z) = xyz,

then∂F

∂x= yz +

∂g

∂u

∂u

∂x+∂g

∂v

∂v

∂x+∂g

∂w

∂w

∂x,

110 REAL ANALYSIS

i.e.∂F

∂x= yz +

∂g

∂u+ 2x

∂g

∂v+ yz

∂g

∂w.


∂F

∂y= xz +

∂g

∂u+ 2y

∂g

∂v+ xz

∂g

∂w.

and∂F

∂z= xy +

∂g

∂u+ 2z

∂g

∂v+ xy

∂g

∂w.

Exercise 5.71 Let

F (x, y) = xy + g(x2 + y2, xy),

where g : R2 → R is an arbitrary given function of class C2. Compute∂2F

∂x2,∂2F

∂y2

and∂2F

∂x∂y.

Solution. If we denote by

u(x, y) = x2 + y2

and by

v(x, y) = xy,

then∂2F

∂x2= 4x2

∂2g

∂u2+ 4xy

∂2g

∂u∂v+ y2

∂2g

∂v2+ 2

∂g

∂u,

∂2F

∂x∂y= 1 + 4xy

∂2g

∂u2+ (2x2 + 2y2)

∂2g

∂u∂v+ xy

∂g

∂v2+∂g

∂u

and∂2F

∂y2= 4y2

∂2g

∂u2+ 4xy

∂2g

∂u∂v+ x2

∂2g

∂v2+ 2

∂g

∂u.

Exercise 5.72 Compute the first and the second differential for the function

F (x, y) = (x2 + y2)u(x+ y),

where u : R → R is an arbitrary function of class C2.


Exercise 5.73 Let

F (x, y) = xyu(x+ y, xy),

where u : R2 → R is a given function of class C2. Compute the first and the second

differential of F .

Exercise 5.74 Let F (x, y, z) = (x2 + y2 + z2)u(x + y + z, xyz), where u ∈ C2.

Compute the first and the second differential of F .

Exercise 5.75 Let F (x, y) = u(x+ y, xy, x2 + y2), where u ∈ C2. Compute

∆F =∂2F

∂x2+∂2F

∂y2.

Exercise 5.76 Prove that the function

z(x, y) = f(x+ φ(y)),

where f, φ ∈ C2(R), satisfies the equation:

∂z

∂x

∂2z

∂x∂y=∂z

∂y

∂2z

∂x2.

Exercise 5.77 Prove that the function

g(x, t) = φ(x− at) + ψ(x+ at),

where φ,ψ are given functions of class C2 and a > 0, satisfies the equation:

∂2g

∂t2= a2

∂2g

∂x2,

called the equation of the vibrating string.

5.2 Differential Operators

Definition 5.78 Let ∅ = U ⊆ Rn, U open. Also, let f : U ⊆ Rn → R be a scalar

field, f ∈ C1(U) and a ∈ U . The gradient of the function f at the point a is defined

as being

grad f : U → Rn, (grad f)(a) = gradaf.

112 REAL ANALYSIS

Thus, the gradient of f at the point a can be written as

(grad f)(a) =

(∂f

∂x1(a), . . . ,

∂f

∂xn(a)

).

If we introduce the nabla symbol ∇, we have

∇f(a) =(∂f

∂x1(a), · · · , ∂f

∂xn(a)

)or

∇f(a) =n∑

i=1

∂f

∂xi(a)ei.

The differential operator

∇ =

(∂

∂x1, · · · , ∂

∂xn

)is called the Hamiltonian operator. So, in the n-dimensional Euclidean space Rn,

this operator can be written as

∇ =n∑

i=1

∂

∂xiei

or, using the Einstein summation notation, as

∇ =∂

∂xiei

Remark 5.79 Let f : U ⊆ Rn → R be differentiable at the point a ∈ U and let

u ∈ Rn be a unit vector. Then, f is derivable at the point a in the direction u and

∂f

∂u(a) = ⟨∇f(a), u⟩ = ∥∇f(a)∥ cos θ,

where θ is the angle between the vectors ∇f(a) and u. Therefore, the rate of change

of the function f in the direction u depends on u and varies from −∥∇f(a)∥ (when

θ = π, i.e. u points in a direction opposite to ∇f(a)) to ∥∇f(a)∥ (when θ = 0, i.e.

u points in the same direction as ∇f(a)). The vector ∇f(a) points in the direction

of the greatest rate of increase of the function f at the point a and the greatest rate

of change is exactly the magnitude of this vector, i.e. ∥∇f(a)∥.


Exercise 5.80 The gradient ∇f is orthogonal to the level surface f(x) = C.

Solution. Since ∇f · dr = df and f(x) = C, it is not difficult to see that ∇f · dr = 0.

Exercise 5.81 Compute the gradient of the scalar field f : R3 → R, defined by

f(x, y, z) = 2x+ y2 − cos z.

Solution. The gradient of f is the vector ∇f = (2, 2y, sin z).

Exercise 5.82 Compute the gradient of the scalar field f : R3 → R, defined by

f(x, y, z) = x y z.

Solution. The gradient of f is the vector ∇f = (y z, x z, x y).

In what follows, we shall use the notation −→v to denote a vector v. Notice the

little abuse of notation, since up to now we didn’t use this arrow notation when

referring to vectors in Rn.

Exercise 5.83 Let r be the length of the position vector −→r = −→0 of a point (x, y, z)

in R3, i.e.

r = ∥−→r ∥ = 0.

Compute grad

(1

r

).

Solution. We have

grad

(1

r

)= −

−→rr3.

More general, it is easy to see that

gradf(r) =f ′(r)

r−→r ,

for any f ∈ C1(U), U ⊆ R∗+.

Definition 5.84 Let ∅ = U ⊆ Rn, U open and f : U ⊆ Rn → R be a scalar field

such that f ∈ C2(U). The Laplacian of the function f is defined as being:

∆f =∂2f

∂x21+ · · ·+ ∂2f

∂x2n.

114 REAL ANALYSIS

So, the Laplace operator is a second order differential operator defined as the sum

of all the unmixed second partial derivatives:

∆ =n∑

i=1

∂2

∂x2i.

Remark 5.85 A function f is said to be harmonic if ∆f = 0.

Exercise 5.86 Let f : R2 \ {(0, 0)} → R be defined by

f(x, y) = ln(x2 + y2).

Prove that f satisfies Laplace’s equation, i.e.

∆f = 0.


∂2f

∂x2=

2(y2 − x2)

(x2 + y2)2.

In a similar manner,∂2f

∂y2=

2(x2 − y2)

(x2 + y2)2.

Hence,

∆f =∂2f

∂x2+∂2f

∂y2= 0.

This proves that our function f is harmonic.

Definition 5.87 Let ∅ = U ⊆ Rn, U open and −→v : U ⊆ Rn → Rn be a vector field

such that −→v ∈ C1(U). The divergence of the vector −→v is defined as being

div−→v =∂v1∂x1

+ · · ·+ ∂vn∂xn

.

Remark 5.88 A vector field −→v is called solenoidal (or, sometimes, incompressible)

if div −→v = 0 (no sources or sinks).


Exercise 5.89 Let −→r = (x, y, z) ∈ R3, −→r = −→0 and

r = ∥−→r ∥,

i.e.

r =√x2 + y2 + z2.

Prove that the divergence of the field

−→E =

q−→rr3

is zero. Hence,−→E is a solenoidal field.

Solution. Let−→E = (Ex, Ey, Ez), where

Ex =qx

r3, Ey =

qy

r3, Ez =

qz

r3.

Since, by the chain rule,∂r3

∂x= 3xr,

it is easy to see that∂Ex

∂x= q

r3 − 3x2r

r6.

In a similar manner,∂Ey

∂y= q

r3 − 3y2r

r6

and∂Ez

∂z= q

r3 − 3z2r

r6.

Therefore,

div−→E = 0.


div−→r = 3.

Exercise 5.91 Compute the divergence of the vector field

−→v =(x y z, x+ y + z, x2 + y2 + z2

).

116 REAL ANALYSIS

Solution. The divergence of −→v is the scalar field div−→v = y z + 1 + 2 z.

Definition 5.92 Let ∅ = U ⊆ R3, U open and −→v : U → R3 be a vector field,−→v ∈ C1(U). The differential operator

rot −→v =

(∂v3∂x2

− ∂v2∂x3

,∂v1∂x3

− ∂v3∂x1

,∂v2∂x1

− ∂v1∂x2

)is called the curl or the rotation of the vector −→v .

We shall also denote by curl −→v the rotation of the vector field −→v .The curl of −→v can be interpreted as the symbolic vector product of the Hamil-

tonian operator ∇ by the vector field −→v :

curl −→v = ∇×−→v ,

i.e.

curl −→v =

∣∣∣∣∣∣∣∣∣∣∣∣∣

−→e1 −→e2 −→e3

∂

∂x1

∂

∂x2

∂

∂x3

v1 v2 v3

∣∣∣∣∣∣∣∣∣∣∣∣∣Remark 5.93 A vector field −→v is called irrotational if curl −→v = 0 (irrotational,

nonturbulent flow).


curl−→r = 0

and

curl (f(r)−→r ) = 0,

for any f ∈ C1(U), U ⊆ R∗+.

Exercise 5.95 Compute the curl of the vector field

−→v =(x y z, x+ y + z, x2 + y2 + z2

).


Exercise 5.96 Let−→V =

−→V0 + −→ω × −→r be the velocity field in a rigid solid. Show

that

curl−→V = 2−→ω .

and

div−→V = 0.

Remark 5.97 If −→v is an irrotational field in a simply connected neighborhood U of

a point x, then in this neighborhood, −→v is a potential field, i.e. −→v is given in terms

of the gradient of a scalar field Φ:

−→v = −∇Φ.

Exercise 5.98 Let U ⊆ R3 be a nonempty open set and f, g : U → R, f, g ∈ C1(U),

α, β ∈ R. Prove that:

a) grad (α f + β g) = α grad f + β grad g; b) grad (f g) = g grad f + f grad g;

c) grad (f/g) =ggrad f − fgrad g

g2, g(x) = 0, ∀x ∈ U .


div (α−→u + β−→v ) = α div −→u + β div −→v ,

for all α, β ∈ R, ∀−→u ,−→v ∈ C1.


div (φ−→v ) = φ div −→v + (grad φ) · −→v ,

for all φ, −→v ∈ C1.


div (−→u ×−→v ) = −→v · rot −→u −−→u · rot −→v ,

for all −→u ,−→v ∈ C1. As a consequence, if −→u and −→v are irrotatational vector fields,

then −→u ×−→v is solenoidal.

118 REAL ANALYSIS


div (grad φ) = △φ,

for all φ ∈ C2.


rot (grad φ) = 0,

for all φ ∈ C2.


rot (α−→u + β−→v ) = α rot −→u + β rot −→v , ∀α, β ∈ R,

for all −→u ,−→v ∈ C1.


rot (φ−→v ) = φ rot −→v −−→v × grad φ,

for all φ, −→v ∈ C1.


rot (rot −→u ) = grad (div −→u )−△−→u ,

for any −→u ∈ C2.


div (rot −→v ) = 0,

for any −→v ∈ C2.


5.3 Taylor’s Formula

Theorem 5.108 Let ∅ = I ⊆ R be an open interval and f : I → R. If f possesses

derivatives up to the order p + 1 on I, then for any a, x ∈ I, a = x, there exists ξp

between a and x such that:

f(x) =p∑

k=0

f (k)(a)

k!(x− a)k +

f (p+1)(ξp)

(p+ 1)!(x− a)p+1. (5.7)

(5.7) is called Taylor’s formula for the function f at the point a.

If we denote by

Tp(x, a) =p∑

k=0

f (k)(a)

k!(x− a)k

the Taylor’s polynomial of degree p and by

Rp(x) =f (p+1)(ξp)

(p+ 1)!(x− a)p+1

the p− th remainder of Taylor’s formula in Lagrange’s form, (5.7) becomes:

f(x) = Tp(x, a) +Rp(x). (5.8)

Theorem 5.109 Let ∅ = I ⊆ R be an open interval and f : I → R, f ∈ Cp+1(I).

Then, for any x0, x ∈ I, we have

f(x) = Tp(x, x0) +1

p!

∫ x

x0

(x− t)pf (p+1)(t) dt.

Rp(x) =1

p!

∫ x

x0

(x− t)pf (p+1)(t) dt

is called the p− th remainder in the integral form.

Remark 5.110 If in Theorem 5.107 we expand about the origin, i.e. we take a = 0,

we get the so-called Maclaurin’s formula.

120 REAL ANALYSIS

Proposition 5.111 Let ∅ = I ⊆ R be an open interval and f : I → R, f ∈C(p+1)(I). If [a, a+ h] ⊂ I, then

f(a+ h) = f(a) + f ′(a)h+f ′′(a)

2!h2 + · · ·+ f (p)(a)

p!hp+

1

p!

∫ 1

0(1− t)pf (p+1)(a+ th)hp+1 dt.

Let us write down Maclaurin’s series for some important elementary functions.

1) For the function f : R → R defined by f(x) = ex, we have

ex =∞∑k=0

xk

k!.

2) For f : R → [−1, 1] defined by f(x) = cos x, we have

cos x =∞∑k=0

(−1)kx2k

(2k)!.

3) For f : R → [−1, 1], f(x) = sin x.

sin x =∞∑k=0

(−1)kx2k+1

(2k + 1)!.

Exercise 5.112 Let f : R → R, defined by

f(x) = x3ex.

Prove that

f(x) =∞∑k=0

xk+3

(k)!.

Exercise 5.113 Let f : (−1, 1] → R be defined by

f(x) = ln (1 + x).

Prove that

f(x) =∞∑k=0

(−1)kxk+1

k + 1.


Exercise 5.114 Let f : (−3, 3) → R, defined by

f(x) =x

9 + x2.

Prove that

f(x) =∞∑k=0

(−1)kx2k+1

9k+1.

Exercise 5.115 For f : (−1, 1) → R, given by

f(x) =1

1− x,

show that

f(x) =∞∑k=0

xk.

Exercise 5.116 Expand in Maclaurin’s series the function f : (−1, 1) → R, definedby

f(x) =1

1 + x.

Exercise 5.117 Expand in Maclaurin’s series the function f : (−1, 1) → R, definedby

f(x) =1

1− x2.


(1 + x)m = 1 +mx+m(m− 1)

2!x2 +

m(m− 1)(m− 2)

3!x3 + · · · . (5.9)

This expansion, called the binomial expansion, is valid for −1 < x < 1.

Exercise 5.119 Using the binomial expansion, deduce the expression for the clas-

sical kinetic energy from the formula of total relativistic energy.

Solution. The total relativistic energy of a particle of mass m and velocity v is given

by

E = mc2(1− v2

c2

)−1/2

.

122 REAL ANALYSIS

From (5.9), with x = −v2/c2 and m = −1/2, we have

E = mc2 +1

2mv2 +

3

8mv2

v2

c2+ · · · .

So,

E = mc2 +1

2mv2

(1 +

3

4

v2

c2+ · · ·

).

The first term, mc2 is the rest mass energy. Then,

Ekinetic =1

2mv2

(1 +

3

4

v2

c2+ · · ·

).

If the particle velocity is much smaller than the velocity of light, i.e. if v ≪ c, the

expression in the parentheses reduces to unity and we see that we obtain the classical

result for the kinetic part of the total relativistic energy.

Theorem 5.120 Let E be a Banach space, U an open nonempty convex subset of

E and f : U → R. If f is m + 1-times differentiable on U , then, for any a, x ∈ U ,

there exists ξ between a and x such that:

f(x) =m∑p=0

1

p!f (p)(a) (x− a)p +

1

(m+ 1)!f (m+1)(ξ) (x− a)m+1. (5.10)

(5.10) is called Taylor’s formula for the function f at the point a.

Corollary 5.121 If E = Rn, we have:

f(x) = f(a) +n∑

i=1

∂f

∂xi(a)(xi − ai) +

1

2!

n∑i,j=1

∂2f

∂xi∂xj(a)(xi − ai)(xj − aj) + · · ·

+1

m!

n∑i1,...,im=1

∂mf

∂xi1 . . . ∂xim(a)(xi1 − ai1) . . . (xim − aim)

+1

(m+ 1)!

n∑i1,...,im+1=1

∂m+1f

∂xi1 . . . ∂xim+1

(ξ)(xi1 − ai1) . . . (xim+1 − aim+1)

or

f(x1, . . . , xn) = f(a1, . . . , an) +n∑

j=1

∂f

∂xj(a1, . . . , an)(xj − aj)


+1

2!

n∑j,k=1

∂2f

∂xj∂xk(a1, . . . , an)(xj − aj)(xk − ak)

+ · · ·+ 1

m!

∑j1+···+jn=m

∂mf

∂xj11 . . . ∂xjnn(a1, . . . , an)(x1 − a1)

j1 . . . (xn − an)jn

+1

(m+ 1)!

∑k1+···+kn=m+1

∂m+1f

∂xk11 . . . ∂xknn(ξ)(x1 − a1)

k1 . . . (xn − an)kn .

Remark 5.122 In the particular case n = 2, we get:

f(x, y) = f(x0, y0) +1

1!

(∂f

∂x(x0, y0)(x− x0) +

∂f

∂y(x0, y0)(y − y0)

)

+1

2!

(∂2f

∂x2(x0, y0)(x− x0)

2 + 2∂2f

∂x∂y(x0, y0)(x− x0)(y − y0) +

∂2f

∂y2(x0, y0)(y − y0)

2

)+· · ·

+1

m!

(∂mf

∂xm(x0, y0)(x− x0)

m + C1m

∂mf

∂xm−1∂y(x0, y0)(x− x0)

m−1(y − y0

)+ · · ·

+∂mf

∂ym(x0, y0)(y − y0)

m)

+1

(m+ 1)!

(∂m+1f

∂xm+1(ξ, η)(x− x0)

m+1 + C1m+1

∂m+1f

∂xm∂y(ξ, η)(x− x0)

m(y − y0) + · · ·

+∂m+1f

∂ym+1(ξ, η)(y − y0)

m+1

),

where ξ = x0 + t(x− x0),

η = y0 + t(y − y0),t ∈ (0, 1).

Remark 5.123 Let E be a Banach space, U an open nonempty convex subset of

E, f : U → R, f ∈ Cm+1(U). Then, for any a, x ∈ U , we have:

f(x) =m∑p=0

1

p!f (p)(a) (x− a)p + εf,a(x) ∥x− a∥m, (5.11)

with

limx→a

εf,a(x) = 0.

124 REAL ANALYSIS

Exercise 5.124 Write Taylor’s formula of the third order for the function f : U →R, f(x, y) = ln(1 + x+ y), about the point (0, 0). Here,

U = {(x, y) ∈ R2 | x+ y > −1}.

Solution. Obviously, f is four times differentiable on the convex set U and, comput-

ing the needed derivatives, we obtain

ln(1 + x+ y) = x+ y − 1

2(x+ y)2 +

1

3(x+ y)3 − 1

4(x+ y)4(1 + ξ + η)−4,

where (ξ, η) = (tx, ty), with t ∈ (0, 1).

Exercise 5.125 Write Taylor’s formula of the third order for the function f : R2 →R, f(x, y) = ex+y, about the point (1,−1).

Solution. The function f is four times differentiable on the convex set R2. Computing

the needed derivatives, we obtain

ex+y = 1 +1

1!(x+ y) +

1

2!(x+ y)2 +

1

3!(x+ y)3 +

eξ+η

4!(x+ y)4,

where (ξ, η) = (1 + tx,−1 + ty), with t ∈ (0, 1).

Exercise 5.126 Write Taylor’s formula of the order 2 about the point (x0, y0) =

(0, 0) for the function f : R2 → R, defined by f(x, y) = ex sin y.

Exercise 5.127 Write Taylor’s formula of the order 2 at the point (x0, y0) =

(0,π

2

)for the function f : R2 → R, defined by f(x, y) = e2x cos y.

Exercise 5.128 Write Taylor’s formula of the order 2 at the point (x0, y0) =

(0,π

2

)for the function f : R2 → R, defined by f(x, y) = cosx cos y.

Exercise 5.129 Write Taylor’s formula of the order 2 at the point (x0, y0) = (1, 1)

for the function f : (0,∞)× R → R, defined by

f(x, y) = xy.


5.4 Implicit Functions

Definition 5.130 Let E,F be Banach spaces, U a nonempty open subset of E, V

a nonempty open subset of F . A function f : U → V is called an homeomorphism

if:

1) f is bijective;

2) f : U → V is continuous on U ;

3) f−1 : V → U is continuous on V .

Example 5.131 The function f : R → R defined by f(x) = x3 is an homeomor-

phism on R.

Definition 5.132 Let E and F be Banach spaces, U a nonempty open subset of E,

V a nonempty open subset of F . A function f : U → V is called a C1-diffeomorphism

if:

1) f is bijective;

2) f : U → V , f ∈ C1(U);

3) f−1 : V → U , f ∈ C1(V ).

Example 5.133 The function f : (0, 1) → (0, 3) defined by f(x) = 3x is a C1-

diffeomorphism.

Theorem 5.134 (Local Inversion Theorem) Let E and F be Banach spaces, U

a nonempty open subset of E, f : U → F , f ∈ C1(U) and a ∈ U such that

f ′(a) ∈ Isom (E,F ). Then, there exist V ⊆ E, W ⊆ F open sets such that a ∈ V ⊆U, b = f(a) ∈W and f : V →W is a C1-diffeomorphism.

Corollary 5.135 Let U be a nonempty open subset of Rn, f : U → Rn, f ∈ C1(U)

and a ∈ U such that∂(f1, . . . , fn)

∂(x1, . . . , xn)(a) = 0.

Then, there exist V,W ⊆ Rn open sets such that a ∈ V ⊆ U, b = f(a) ∈ W and

f : V →W is a C1-diffeomorphism.

126 REAL ANALYSIS

Example 5.136 The simplest example of a diffeomorphism in Rn is offered by the

translation operator. Indeed, if we fix a ∈ Rn, then the map Ta : Rn → Rn, defined

by

Ta(x) = x+ a, ∀x ∈ Rn,

is a C1-diffeomorphism.

Example 5.137 Let U = (0,∞)× (0, 2π) ⊆ R2 and V = R2 \ {[0,∞)× {0}} ⊆ R2.

For the function f : U → V , defined in polar coordinates by

f(ρ, θ) = (ρ cos θ, ρ sin θ),

we have:∂(f1, f2)

∂(ρ, θ)|(ρ,θ) = ρ.

It is easy to see that f ∈ C1(U). Also, for any ρ ∈ (0,∞), f ′(ρ, θ) ∈ Isom (U, V ).

So, f is a C1-diffeomorphism on U .

Theorem 5.138 Let E,F be Banach spaces, ∅ = U ⊆ E × F , U open, (a, b) ∈

U, h : U → F continuous on U , h(a, b) = 0,∂h

∂y: U → L(F, F ) continuous,

∂h

∂y(a, b)

bijective. Then, there exist a number r > 0 and a function φ : B(a, r) → F such

that:

1) (x, φ(x)) ∈ U, ∀x ∈ B(a, r);

2) φ is continuous on B(a, r);

3) h(x, φ(x)) = 0, ∀x ∈ B(a, r);

4) φ(a) = b;

5) φ unique.

Moreover, if h is derivable at the point (a, b), then φ is derivable at the point a and

φ ′(a) = −(∂h

∂y(a, b)

)−1

◦ ∂h∂x

(a, b). (5.12)

Corollary 5.139 (The case E = F = R) Let ∅ = U ⊆ R2, U open, (a, b) ∈

U, h : U → R continuous on U , h(a, b) = 0,∂h

∂y: U → L(R,R) ≃ R continuous,

∂h

∂y(a, b) = 0. Then, there exist r > 0 and φ : (a− r, a+ r) → R such that:


1) (x, φ(x)) ∈ U, ∀x ∈ (a− r, a+ r);

2) φ is continuous on (a− r, a+ r);

3) h(x, φ(x)) = 0, ∀x ∈ (a− r, a+ r);

4) φ(a) = b;

5) φ unique.

Moreover, if h is derivable at the point (a, b), then φ is derivable at the point a and

φ ′(a) = −

∂h

∂x(a, b)

∂h

∂y(a, b)

. (5.13)

Corollary 5.140 (The case E = Rn, F = Rm) Let ∅ = W ⊆ Rn × Rm, W open,

(a, b) ∈ W, f : W → Rm, f = (f1, . . . , fm), fi = fi(x1, . . . , xn, y1, . . . , ym), i = 1,m

such that:

1) fi ∈ C1, i = 1,m;

2) fi(a, b) = 0, i = 1,m;

3)∂(f1, . . . , fm)

∂(y1, . . . , ym)(a, b) = 0.

Then, there exist ∅ = U × V ⊆ W , U × V open, with (a, b) ∈ U × V , and gi : U →R, i = 1,m, gi ∈ C1 such that the system:

f1(x1, . . . , xn, y1, . . . , ym) = 0,

........................................,

fm(x1, . . . , xn, y1, . . . , ym) = 0

is equivalent to the following one:

y1 = g1(x1, . . . , xn),

............................,

ym = gm(x1, . . . , xn),

(x1, . . . , xn) ∈ U.

128 REAL ANALYSIS

Moreover,

bj = gj(a1, . . . , an), j = 1,m

and

∂yj∂xi

(x) = −

∂(f1, . . . , fm)

∂(y1, . . . , yj−1, xi, yj+1, . . . , ym)(x, g(x))

∂(f1, . . . , fm)

∂(y1, . . . , ym)(x, g(x))

,

for any x = (x1, ..., xn) ∈ U, i = 1, n, j = 1,m, g = (g1, . . . , gm).

The above corollary gives conditions for the solvability of a system of functional

equations and offers formulas for calculating the partial derivatives of the implicitly

specified functions.

Example 5.141 As an application of the implicit function theorem, consider the

so-called triple product rule, also known as the cyclic chain rule, the cyclic relation,

or Euler’s chain rule, which relates the partial derivatives of three interdependent

variables. This rule finds its applications, for instance, in thermodynamics, where

frequently three variables can be related by a function of the form F (x, y, z) = 0, so

each variable is given as an implicit function of the other two variables. For example,

an equation of state for a fluid might relate in this manner the temperature, the

pressure and the volume.

Let F : D ⊂ R3 → R, D nonempty open set , F ∈ C1(D) and let us assume that

∂F

∂x(x, y, z) · ∂F

∂y(x, y, z) · ∂F

∂z(x, y, z) = 0.

Since F is of class C1 on D and∂F

∂x(x, y, z) = 0, it follows that locally, around

the point (x, y, z), we can express x in terms of y and z, i.e. there exists X such

that

x = X(y, z)

and

F (X(y, z), y, z) = 0.

Therefore,∂

∂y[F (X(y, z), y, z)] = 0,


i.e.∂F

∂x

∂X

∂y+∂F

∂y= 0,

which implies that

∂X

∂y= −

∂F

∂y∂F

∂x

.


∂Y

∂z= −

∂F

∂z∂F

∂y

,∂Z

∂x= −

∂F

∂x∂F

∂z

.

So, if

F (x, y, z) = 0,

then, locally,∂X

∂y· ∂Y∂z

· ∂Z∂x

= −1. (5.14)

Formula (5.14) is known as the triple product rule.

Exercise 5.142 Compute the derivative of the function y = f(x) implicitly defined

by the equation

x2 + 4y2 + 2xy = 12

in the neighbourhood of the point (2, 1).

Solution. The function h(x, y) = x2 + 4y2 + 2xy − 12 satisfies all the conditions

of Corollary 5.138. Therefore, there exists a derivable function φ, defined on a

neighbourhood of the point x = 2, such that φ is unique, φ(2) = 1 and

φ ′(2) = −

∂h

∂x(2, 1)

∂h

∂y(2, 1)

= −1

2.

130 REAL ANALYSIS

Exercise 5.143 Consider the following system:x+ y + z = 1,

x3 − y3 + z3 = −1.

Prove that, in a neighbourhood of the point (1, 1,−1), we can uniquely determine y

and z as functions of x. Also, computedy

dx(1),

dz

dx(1),

d2y

dx2(1),

d2z

dx2(1).

Solution. If we take f = (f1, f2) : R× R2 → R2, defined byf1(x, y, z) = x+ y + z − 1,

f2(x, y, z) = x3 − y3 + z3 + 1,

it is easy to see that f ∈ C∞(R3), f(1, 1,−1) = (0, 0) and

∂(f1, f2)

∂(y, z)(1, 1,−1) = 6 = 0.

Therefore, using Corollary 5.139, it follows that there exists an open set U ⊆ R and

there exist two functions g1, g2 : U → R, of class C1, such that, locally,y = g1(x),

z = g2(x)

and

g1(1) = 1, g2(1) = −1.

So, on U , we have x+ g1(x) + g2(x) = 1,

x3 − g31(x) + g32(x) = −1.

So, by differentiation, we get

dg1dx

(1) = 0,dg2dx

(1) = −1.

In a similar manner, we obtain

d2g1dx2

(1) = 0,d2g2dx2

(1) = 0


Exercise 5.144 Compute the derivative of the function y = f(x) implicitly defined

by the equation x2 − y − cos y = 0 in a neighbourhood of the point (1, 0).

Exercise 5.145 Compute the first and second order partial derivatives for the func-

tion z=f(x,y) defined implicitly by the equation x3 + y3 + z3 − 3xyz = 0 in a neigh-

bourhood of the point (−1, 1, 0).

Exercise 5.146 Compute∂2f

∂x∂y(1, 0) for the function z = f(x, y) defined by the

equation x2 + y2 − z2 + xz − yz − 1 = 0 in a neighbourhood of the point (1, 0, 1).

Exercise 5.147 Compute the first and second order partial derivatives for the func-

tion z=f(x,y) defined implicitly by the equation x3 + y3 + z3 − 3xyz = 0 in a neigh-

bourhood of the point (−1, 1, 0).

Exercise 5.148 Compute the differential d f(0, 0) for the function z=f(x,y) defined

implicitly by the equation x+ y + z − sin xyz = 0 in a neighbourhood of the point

(0, 0, 0).

Exercise 5.149 Consider the following system:x+ yz − z3 = 1,

x3 − xz + y3 = 0.

Prove that, in a neighbourhood of the point (1,−1, 0), we can uniquely determine x

and y as functions of z. Computedx

dz(0),

dy

dz(0).

5.5 Local Extrema

Definition 5.150 Let E be a Banach space, U ⊆ E an open nonempty subset,

a ∈ U and f : U → R.

a) The function f attains a local minimum at the point a if there exists V ∈ V(a)such that f(x) ≥ f(a), for any x ∈ V ∩ U .

b) The function f possesses a local maximum at the point a if there exists

V ∈ V(a) such that f(x) ≤ f(a), for any x ∈ V ∩ U .

132 REAL ANALYSIS

We shall use the terms global or absolute maximum and, respectively, global or

absolute minimum, to refer to the overall maximum and minimum values of the

function on the range under consideration. Also, the term local is synonymous with

relative. Moreover, extremum is a term that will be used to denote both a maximum

and a minimum: a local extremum will be a local or relative maximum or minimum,

and a global extremum will be a global or absolute maximum or minimum.

Definition 5.151 Let E be a Banach space, U ⊆ E an open nonempty subset,

a ∈ U and f : U → R. The function f is said to attain a strict local minimum at

the point a if there exists V ∈ V(a) such that f(x) > f(a), for any x ∈ V ∩U \ {a}.Of course, f attains a strict local maximum at the point a if f(x) < f(a), for any

x ∈ V ∩ U \ {a}.

For instance, the point (0, 0) is a strict local minimum for the function f(x, y) =

x2 + y2, but not for g(x, y) = (x+ y)2.

It is not difficult to see that, exactly like in the one-dimensional case, we have

the following result:

Theorem 5.152 (Fermat) Let E be a Banach space, U ⊆ E an open nonempty

subset, a ∈ U and f : U → R. If f is derivable at the point a and a is a local

extremum for f , then f ′(a) = 0.

Remark 5.153 The points a at which f ′(a) = 0 are called stationary points.

In one-variable calculus, we know that if a function f : I → R defined on an open

interval I ⊆ R has a local extremum at a point a ∈ I, then either f ′(a) = 0 or f ′(a)

does not exist. A critical point is a point (belonging to I) where the derivative of

f is equal to zero or a point where the function is not differentiable. Analogously,

in multivariable calculus, a point for which the derivative of a function f : D → R,defined on an open set D ⊆ Rn, vanishes or at which the function is not differentiable

is said to be a critical point. Let us mention that in the multivariable case critical

points are also sometimes defined as being points where the Jacobian matrix of a

given function is not of maximum rank.

A stationary point is always a critical one, but the converse is not true, i.e. a

critical point is not always a stationary one. However, for a smooth function of


several real variables, these notions coincide, and, so, the condition that a point is

a critical one for f is equivalent to the fact all the partial derivatives of f must be

zero at that point.

Also, let us notice that exactly like in the one-dimensional case, one can prove

that if we consider a function f : Rn → R which is continuous on an open set U and

if S is a nonempty closed and bounded subset of U , then f has a maximum value

and a minimum value on S. Notice that this result needs no reference to derivatives.

Moreover, let us notice that global extrema of a given function f on a closed

and bounded set S can occur only at boundaries, non-differentiable points and sta-

tionary points. But, sometimes, dealing with the boundary of a given set and with

non-differentiable points requires a significant amount of work. Therefore, in this

paragraph, we shall work only with smooth functions defined on open sets and, in

such a case, as already mentioned, local extrema can be found among the stationary

(critical) points of the given function. Still, we have to keep in mind that this is

only a necessary condition, since not all the stationary points are local extrema. So,

a local extremum must occur at a stationary point, but the converse may not be

true. A stationary point which is not a local extremum is called a saddle point.

Definition 5.154 Let E be a real Banach space. h : E → R is called a quadratic

form if there exists h : E × E → R, bilinear, symmetrical and continuous such that

h(x) = h(x, x), ∀x ∈ E.

Example 5.155 If E = Rn, then

h(x) =n∑

i,j=1

aijxixj ,

with aij = aji, is a quadratic form.

Definition 5.156 The quadratic form h : Rn → R is called non-degenerate if

det(aij) = 0.

Definition 5.157 Let h : Rn → R be a quadratic form. Then:

134 REAL ANALYSIS

a) h is called positive definite if h(x) > 0, for any x ∈ Rn, x = 0;

b) h is called positive semi-definite if h(x) ≥ 0, for any x ∈ Rn;

c) h is called negative definite if h(x) < 0, for any x ∈ Rn, x = 0;

d) h is called negative semi-definite if h(x) ≤ 0, for any x ∈ Rn;

e) h is called indefinite if it is neither positive nor negative semi-definite.

Remark 5.158 If h : Rn → R is a positive definite and non-degenerate quadratic

form, then there exists m > 0 such that h(x) ≥ m∥x∥2, ∀x ∈ Rn.

Definition 5.159 Let f : U ⊆ Rn → R, U an open nonempty set, f ∈ C2(U) and

a ∈ U . The quadratic form

Hf (a)(x) = f ′′(a)(x, x), ∀x ∈ Rn

is called the Hessian of f at the point a.

Remark 5.160 Let f : U ⊆ Rn → R, U open nonempty set, f ∈ C2(U) and a ∈ U .

a) The Hessian Hf (a) is non-degenerate if and only if

det

(∂2f

∂xi∂xj(a)

)= 0.

b) The Hessian Hf (a) is positive semi-definite if and only if

n∑i,j=1

∂2f

∂xi∂xj(a)xixj ≥ 0, ∀x ∈ Rn.

c) The Hessian Hf (a) is negative semi-definite if and only if

n∑i,j=1

∂2f

∂xi∂xj(a)xixj ≤ 0, ∀x ∈ Rn.

Usually, in the definition of the Hessian of a given function f at a point a we shall

consider that f ′(a) = 0, i.e. the point a is a stationary one. We shall say that a

stationary point a ∈ U is non-degenerate if the Hessian Hf (a) is non-degenerate.


Theorem 5.161 Let U ⊆ Rn be an open nonempty convex set, f : U → R, f ∈C2(U) and a ∈ U a non-degenerate critical point.

a) If the Hessian Hf (a) is positive definite, then a is a strict local minimum

for f .

b) If the Hessian Hf (a) is negative definite, then a is a strict local maximum

for f .

Theorem 5.162 Let U ⊆ R2 be an open nonempty convex set, f : U → R, f ∈C2(U), a ∈ U a non-degenerate critical point. Let

D2(a) =

∣∣∣∣∣∣∣∣∣∣∣

∂2f

∂x2(a)

∂2f

∂x∂y(a)

∂2f

∂y∂x(a)

∂2f

∂y2(a)

∣∣∣∣∣∣∣∣∣∣∣= 0.

If D2(a) < 0, then a is not a local extremum for f . If D2(a) > 0, then a is a strict

local minimum if D1(a) =∂2f

∂x2(a) > 0 and a strict local maximum if D1(a) < 0.

Let us notice that if the point a is degenerate, then the test is inconclusive.

Theorem 5.163 (Sylvester) Let U ⊆ Rn be an open nonempty convex set, f : U →

R, f ∈ C2(U), a ∈ U a non-degenerate critical point, i.e. det

(∂2f

∂xi∂xj(a)

)= 0.

For 1 ≤ k ≤ n, let us consider the kth leading principal minor

Dk(a) =

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

∂2f

∂x21(a) · · · ∂2f

∂x1∂xk(a)

· · · · · · · · · · · · · · · · · · · · · ·

∂2f

∂xk∂x1(a) · · · ∂

2f

∂x2k(a)

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣.

Then:

a) if ∆1(a) > 0, ∆2(a) > 0, . . . ,∆n(a) > 0, then Hf (a) is positive definite and

a is a strict local minimum;

136 REAL ANALYSIS

b) if ∆1(a) < 0, ∆2(a) > 0, . . . , (−1)n∆n(a) > 0, then Hf (a) is negative definite

and a is a strict local maximum.

Notice that if Hf (a) is positive or negative semi-definite, we cannot decide upon the

nature of the point a using this method. The test is inconclusive and this is an open

case. Also, notice that if the Hessian Hf (a) is indefinite, then the point a is a saddle

point for f . If the point a is degenerate, no conclusion can be drawn.

Exercise 5.164 Find the local extrema of the function f : R2 → R, defined by

f(x, y) = x3 + 3xy2 − 15x− 12y.

Solution. Obviously, f ∈ C∞(R2). The critical points for the function f are the

solutions of the equation f′(x, y) = 0, i.e. solutions for the following system:

∂f

∂x(x, y) = 0,

∂f

∂y(x, y) = 0.

So, computing the partial derivatives of f , we have to solve the systemx2 + y2 = 5

xy = 2(5.15)

which, obviously, has four solutions: (1, 2), (2, 1), (−1,−2), (−2,−1). So, we have

four critical points and for deciding which of them are local extrema for the function

f we shall use Theorem 5.162. So, we have to compute the partial derivatives of the

second order for f . We obtain:

∂2f

∂x2= 6x,

∂2f

∂y2= 6x,

∂2f

∂x∂y=

∂2f

∂y∂x= 6y.

Let

D2(a) =

∣∣∣∣∣∣∣∣∣∣∣

∂2f

∂x2(a)

∂2f

∂x∂y(a)

∂2f

∂y∂x(a)

∂2f

∂y2(a)

∣∣∣∣∣∣∣∣∣∣∣.


Now, we shall take look at each critical point and we shall decide, using Theorem

5.162, if it is a local extremum or not.

For the point (1, 2), we obtain D2(1, 2) < 0 and, so, this point is not a local

extremum. Also, the point (−1,−2) is not a local extremum, because D2(−1,−2) <

0.

For the point (2, 1), we getD2(2, 1) > 0 andD1(2, 1) > 0. Therefore, this point is

a local minimum and fmin = f(2, 1) = −28. In a similar manner, we obtain that the

point (−2,−1) is a local maximum, because D2(−2,−1) > 0 and D1(−2,−1) < 0.

Moreover, fmax = f(−2,−1) = 28.


f(x, y, z) = x2 + y2 + z2 + 4x+ 2y − 8z.

Solution. It is not difficult to see that the only critical point of the function f is

(−2,−1, 4). For this point, D3(−2,−1, 4) > 0, D2(−2,−1, 4) > 0, D1(−2,−1, 4) >

0. Hence, the point (−2,−1, 4) is a local minimum. Moreover, fmin = −21.

Exercise 5.166 Show that the function f : R2 → R, defined by

f(x, y) = x2 + 2xy − 4x− y2,

has no extrema.

Solution. The point (1, 1) is the only stationary point of f . The Hessian of f at this

point is

Hf (1, 1) =

2 2

2 − 2

.Therefore, (1, 1) is a saddle point. The function f has no extrema.

Exercise 5.167 Determine the nature of the stationary points of the function f :

R2 → R, defined as being

f(x, y) = (x2 + y2)2ex2+y2 .

138 REAL ANALYSIS

Exercise 5.168 Find the local extrema for f : R2 → R, defined by f(x, y) =

x2 − (y − 2)2.

Exercise 5.169 Find the local extrema for f : R3 → R, defined by f(x, y, z) =

x2 + y2 + z2 − xy + x− 2z.


x4 + y4 + z4 + xyz.


2x2 + 2y2 + z2 + 2(xy + yz + x+ y + 3z).

Exercise 5.172 Find the local extrema for f : (0, ∞)2 → R, defined by

f(x, y) = xy +4

x+

2

y− 3.

Let us look now at the problem of finding local conditional extrema of a given

function f . Very often, we need to consider the problem of minimizing or maximizing

a function under one or several constraints imposed on its variables. Equivalently,

we have to optimize some function whose domain is a given level set. For instance,

we might want to maximize the geometric mean of n positive numbers subject to

the constraint that their arithmetic mean is 1 or we would like to know what shape

should a rectangular parallelepiped with a given volume have in order to minimize

its surface area. Also, similar questions are quite relevant for analyzing physical

processes, in which, for instance, an object is constrained on some track or surface

or in which various coordinates are related (like position and angle when a wheel

rolls without slipping).

In constrained optimization problems, Lagrange’s multipliers method, named af-

ter Joseph Louis Lagrange, is a mathematical tool that allows us to find the extrema

of a function of several variables subject to one or more constraints. This mathe-

matical approach was introduced by Lagrange in the framework of statics, in order

to determine the general equations of equilibrium for problems with constraints.

Definition 5.173 Let S be a nonempty subset of a topological space X, f : X → Rand a ∈ S. The point a is called a local conditional extremum for the function f if

a is a local extremum for the function f |S.


Theorem 5.174 (The existence of Lagrange’s multipliers) Let U be an open subset

of Rn×Rm, h : U → Rm, f : U → R, S = h−1(0) and c = (a, b) ∈ S such that h and

f are of class C1,∂h

∂y(c) : Rm → Rm is bijective and the point c is a local conditional

extremum for f . Then, there exists Λ ∈ L(Rm,R) such that c is a stationary point

for f − Λ ◦ h, i.e.f

′(c)− Λ ◦ h′

(c) = 0. (5.16)

In other words, the extrema of f on S can be found among the stationary points of

the auxiliary function f − Λ ◦ h.The map Λ being in L(Rm,R), it follows that there exists λj ∈ R such that

Λ =m∑j=1

λjqj ,

where qj : Rm → R, j = 1,m, are the canonical projections.

If hj = qj ◦ h, we obtain

∂f

∂xi(c)−

m∑j=1

λj∂hj∂xi

(c) = 0, 1 ≤ i ≤ n+m. (5.17)

The coefficients λj , j = 1,m, are called Lagrange’s multipliers associated to the

function f , the surface S and the point c.

By this method, we replace the problem of finding the stationary points of a

constrained function in n variables with m constraints to the problem of finding the

stationary points of an unconstrained function of n+m variables. Introducing, for

each constraint, a new unknown scalar variable, the Lagrange multiplier, we define

a new function (the Lagrangian) in terms of the original function, the constraints,

and the Lagrange multipliers. Then, we have to find the stationary points of this

auxiliary function L and to decide which of them is an extremum. The local extrema

of the Lagrangian will be exactly the local conditional extrema of the function f ,

subject to the constraints hj .

Algorithm for finding local conditional extrema

1) Write the Lagrangian L = f − Λ ◦ h.

140 REAL ANALYSIS

2) Solve the system

∂f

∂xi(c)−

m∑j=1

λj∂hj∂xi

(c) = 0, 1 ≤ i ≤ n+m,hj = 0, 1 ≤ j ≤ m (5.18)

and find the stationary points of the Lagrangian L.

3) Analyze the definiteness of the Hessian of the Lagrangian L and determine

the local extrema of L, which are exactly the local conditional extrema of f under

the constraints hj .

Let us notice that Lagrange’s method only provides candidates for extrema. We

must check if indeed the function f possesses an extremum and then we have to

carefully evaluate f at every critical point in order to decide upon the nature of all

theses critical points and to determine correctly the extrema.

Also, it is worth mentioning that Lagrange multipliers can fail to identify all the

conditional extrema if ∇h = 0. So, additionally to Lagrange’s method, we have to

check carefully the points with ∇h(x, y) = 0.

Exercise 5.175 Find the local extrema of the function f : R3 → R, f(x, y, z) =

x− 2y + 2z, under the constraint x2 + y2 + z2 = 9.

Solution. Let

F (x, y, z, λ) = x− 2y + 2z − λ(x2 + y2 + z2 − 9).

The stationary points of the Lagrangian F can be found by solving the following

system:

1− 2λx = 0

−2− 2λy = 0

2− 2λz = 0

x2 + y2 + z2 = 9

(5.19)

We get λ2 =1

4, i.e. λ = ±1

2. Hence, the stationary points of F are

(1,−2, 2;

1

2

)and

(−1, 2,−2;−1

2

).


For λ =1

2, the associated quadratic form is negative definite, i.e. the point

(1,−2, 2) is a local maximum for F and a local conditional maximum for the function

f , the maximum value being fmax = 9.

For λ = −1

2, the associated quadratic form is positive definite, i.e. the point

(−1, 2,−2) is a local minimum for F and, of course, a local conditional minimum

for f , the minimum value being fmin = −9.

Example 5.176 In information theory, the information entropy, introduced by

Claude Elwood Shannon in 1948, is a measure of the uncertainty associated with a

random variable.

The information entropy of a discrete random variable X, that can take the

values {x1, . . . , xn}, is

H(X) = −n∑

i=1

p(xi) log2 p(xi),

where p(xi) = Pr(X = xi) = pi is the probability mass function of X.

Alternatively, we can think of an experiment in which n events (outcomes), with

probabilities p1, p2, . . . , pn can occur. Of course,

0 < pi ≤ 1

and

p1 + · · ·+ pn = 1. (5.20)

The entropy associated to this experiment is

H(p1, . . . , pn) = −n∑

j=1

pj log2 pj . (5.21)

Obviously,

H(p1, . . . , pn) ≥ 0, ∀p1, . . . , pn.

Let us find the extrema of H, under the constraint (5.20). Let

F (p1, . . . , pn, λ) = H(p1, . . . , pn)− λ(p1 + · · ·+ pn − 1).

142 REAL ANALYSIS

Solving the system ∂F

∂p1= 0, . . . ,

∂F

∂pn= 0

p1 + · · ·+ pn = 1,

we obtain

p1 = p2 = · · · = pn =1

n,

for

λ = − 1

ln 2(1− ln n).

It is not difficult to see that the quadratic form associated to H is negative definite

at the point

a =

(1

n, . . . ,

1

n

),

which means that the entropy attains a maximum when the probabilities p1, . . . , pn

are equal (uncertainty is highest when all possible events are equiprobable). In other

words, the maximum entropy probability distribution is the uniform one. Notice that

maximal entropy means least information content.

Let us notice that by this method it is not always possible to find all the conditional

extrema of a given function (see, for instance, [19] and [34]), i.e. the Lagrange

equations ∇f(x, y) = λ∇h(x, y), h(x, y) = 0 do not give us all the conditional

extrema.

Exercise 5.177 Find the local extrema of the function f : [0,∞)3 → R, defined by

f(x, y, z) = xyz, subject to the constraint x+ y + z = 1.


f(x, y) = xy, subject to the constraint x2 + y2 = 1.

Exercise 5.179 Find the local extrema of the function f : (0,∞)3 → R, defined by

f(x, y, z) = xy + yz + zx, subject to the constraint xyz = 1.

Exercise 5.180 Find the local extrema of the function f : (0,∞)3 → R, given by

f(x, y, z) = x2 + y2 + z2, under the constraint x+ y + z = 1.


Exercise 5.181 Find the rectangle which is inscribed in the ellipse

x2

a2+y2

b2= 1, a > b > 0

and has the largest perimeter.


f(x, y, z) = x2 + y2 + z2, subject to the constraint

x2

a2+y2

b2+z2

c2= 1,

for a > b > c > 0.

Chapter 6

Sequences and Series ofFunctions

6.1 Sequences of Functions

Definition 6.1 Let us consider a sequence of functions (fk)k≥1, defined on a set

∅ = A ⊆ R:

fk : A→ R, k ≥ 1.

The sequence (fk)k≥1 is said to be pointwise convergent at the point a ∈ A if the

number sequence (fk(a))k≥1 is convergent.

The sequence (fk)k≥1 is said to be pointwise convergent on the set A if (fk)k≥1 is

convergent at any point a ∈ A. In this case, we can define a function f : A→ R, byputting

f(x) = limk→∞

fk(x), x ∈ A.

The function f is called the limit of (fk)k≥1 on A. We shall use the notation fk → f

to denote the fact that f is the limit of the sequence (fk)k.

We shall denote by Ac the set of all the points where (fk)k is pointwise convergent.

Example 6.2 Let fk : (−1, 1) → R defined by fk(x) = xn, for k ≥ 1. Then,

limk→∞

fk(x) = 0, i.e. (fk)k is convergent and Ac = (−1, 1).

145

146 REAL ANALYSIS

Example 6.3 Let fk : R → R given by

fk(x) =sin kx

k2, k ≥ 1.

It follows immediately that

limk→∞

fk(x) = 0,

i.e. (fk)k is convergent and Ac = R.

Example 6.4 If fk : (0,∞) → R is given by

fk(x) = (−1)k ekx, k ≥ 1,

then (fk)k is not convergent and Ac = ∅.

Definition 6.5 Let ∅ = A ⊆ R and fk : A → R, k ≥ 1. The sequence (fk)k≥1 is

said to be uniformly convergent on Au ⊆ A to f if for any ε > 0 there exists kε ∈ Nsuch that for any x ∈ Au,

|fk(x)− f(x)| < ε,

for any k ≥ kε.

We shall write fku−→ f to denote the fact that the sequence (fk)k≥1 is uniformly

convergent to f .

Proposition 6.6 Let fk : A ⊆ R → R, k ≥ 1, be a convergent sequence and let

f(x) = limk→∞

fk(x). Let Au ⊆ A. If

ak = supx∈Au

|fk(x)− f(x)|

and

limk→∞

ak = 0,

then (fk)l is uniformly convergent on Au.

SEQUENCES AND SERIES OF FUNCTIONS 147

Example 6.7 Let fk : [0, 1] → R, fk(x) = xk, k ≥ 1. Obviously, fk → f , where

f(x) =

0 if x ∈ [0, 1),

1 if x = 1.

On the other hand, one can see immediately that

ak = supx∈[0,1]

|fk(x)− f(x)| = 1,

i.e. ak 9 0. Therefore, the sequence (fk)k is not uniformly convergent to f on [0, 1].

Proposition 6.8 Let ∅ = A ⊆ R and fk : A → R, k ≥ 1. If the sequence (fk)k is

uniformly convergent to f on A and all the maps fk are continuous at a ∈ A, then

f is continuous at a.

Remark 6.9 Let us notice that if ∅ = A ⊆ R and the sequence fk : A → R, k ≥ 1,

is uniformly convergent to f on A and fk are continuous at a ∈ A, then

limx→a

limk→∞

fk(x) = limk→∞

limx→a

fk(x).

Example 6.10 The pointwise convergence does not preserve, in general, the proper-

ties of the functions fk. For instance, the pointwise limit of a sequence of continuous

functions may not be continuous. As an example, let us consider the sequence of

continuous functions

fk(x) = xk, x ∈ [0, 1],

which is pointwise convergent to the function

f(x) =

{0, if x ∈ [0, 1),1, if x = 1.

Obviously, f is discontinuous at the point x = 1.

Proposition 6.11 (Cauchy’s Criterion) Let ∅ = A ⊆ R and fk : A → R, k ≥ 1.

The sequence (fk)k is uniformly convergent on A if and only if for any ε > 0 there

exists kε ∈ N such that for any k ≥ kε, for any p ∈ N and for any x ∈ A, we have

|fk+p(x)− fk(x)| < ε.

148 REAL ANALYSIS

Remark 6.12 Let fk : [a, b] → R be a monotone sequence of functions which is

pointwise convergent to f on [a, b]. If fk and f are continuous, then (fk)k is uni-

formly convergent to f on [a, b].

Exercise 6.13 Let fk : R → R, defined by

fk(x) =3k + cos kx

k, k ≥ 1.

Show that (fk)k is uniformly convergent on R.

Solution. Obviously, fk → f , where f(x) = 3, for any x ∈ R. On the other hand,

one can see immediately that

ak = supx∈R

|fk(x)− f(x)| → 0,

i.e. the sequence (fk)k is uniformly convergent to f on R.

Exercise 6.14 Let fk : [0,∞) → R defined by

fk(x) =x

1 + 2knx.

Show that (fk)k is uniformly convergent.

Solution. The sequence (fk)k is pointwise convergent to f(x) = 0, for x ≥ 0. Also,

(fk)k converges uniformly to f , since

ak = supx∈[0,∞)

|fk(x)− f(x)| → 0.

Exercise 6.15 Test the following sequences of functions for convergence:

a) fk : (−1, 1) → R, fk(x) = xk, k ≥ 1;

b) fk : R → R, fk(x) =sin kx

k3, k ≥ 1;

c) fk : [0,∞) → R, fk(x) =x

k + x, k ≥ 1.

Exercise 6.16 Test the following sequences of functions for convergence:

a) fk : [0, 1] → R, fk(x) =x

k + x, k ≥ 1;

b) fk : (0, 1) → R, fk(x) =kx

k + x, k ≥ 1;

c) fk : R → R, fk(x) = arctan (kx), k ≥ 1.


6.2 Series of Functions

Definition 6.17 Let ∅ = A ⊆ R, fk : A→ R, k ≥ 1, and

Sk(x) =k∑

i=1

fi(x). (6.1)

The pair ((fk)k≥1, (Sk)k≥1) is said to be a series of functions (the series associated

to the sequence (fk)k≥1).

We use the notation∑k≥1

fk. Sk is called the kth partial sum of the series∑k≥1

fk and

fk is said to be its general term.

Definition 6.18 The series∑k≥1

fk(x) is said to be convergent at x ∈ A if the se-

quence (Sk(x))k≥1 is convergent. The limit S(x) of (Sk(x))k≥1 is called the sum of

the convergent series∑k≥1

fk(x) at the point x.

We shall use the following symbol to denote the sum of a convergent series:

S =∞∑k=0

fk.

Definition 6.19 A series which is not convergent is called divergent.

Definition 6.20 A series∑k≥0

fk is called absolutely convergent at x if the number

series∑k≥0

|fk(x)| is convergent.

Definition 6.21 The series∑k≥0

fk is called uniformly convergent on the set Au ⊆ A

if (Sk)k≥1 is uniformly convergent on Au.

Proposition 6.22 (Cauchy’s Criterion) The series∑k≥0

fk is uniformly convergent

on Au ⊆ A if and only if for any ε > 0 there exists kε ∈ N such that

|fk+1(x) + · · ·+ fk+p(x)| < ε, ∀k ≥ kε, ∀p ≥ 1, ∀x ∈ Au.

150 REAL ANALYSIS

Theorem 6.23 (Weierstrass) Let fk : A ⊆ R → R, k ≥ 1. If there exists ak ≥ 0

such that

|fk(x)| ≤ ak,

for any k and for any x ∈ A ′ ⊆ A and if∑k≥1

ak is convergent, then∑k≥1

fk is

absolutely and uniformly convergent on A ′ ⊆ A.

Theorem 6.24 Let ∅ = A ⊆ R and fk : A → R, k ≥ 1. If∑k≥1

fk is uniformly

convergent on A and fk are continuous at a ∈ A, then the sum of the series∑k≥1

fk

is continuous at a ∈ A.

Theorem 6.25 Let I = ∅ be an interval in R and fk : I → R, k ≥ 1. If fk are

continuous and the sequence (fk)k≥1 is uniformly convergent to f on I, then, for

any interval [α, β] ⊆ I, we have:

limk→∞

∫ β

αfk(x) dx =

∫ β

αf(x) dx.

Theorem 6.26 Let I = ∅ be an interval in R and fn : I → R, n ≥ 1. If fn are

continuous and∑n≥1

fn is uniformly convergent to f on I, then, for any interval

[α, β] ⊆ I, we have: ∑n≥1

∫ β

αfn(x) dx =

∫ β

αf(x) dx.

Theorem 6.27 Let I = ∅ be a bounded interval in R and fk : I → R, k ≥ 1,

such that fk are derivable and fk′ are continuous. If there exists x0 ∈ I such that

(fk(x0))k≥1 is convergent and if (f′k )k≥1 is uniformly convergent on I to g : I → R,

then there exists f : I → R such that (fk)k≥1 is uniformly convergent to f on I, f

is derivable and

f ′ = g.

Theorem 6.28 Let I = ∅ be a bounded interval in R and fk : I → R, k ≥ 1,

such that fk are derivable and f′k are continuous. If there exists x0 ∈ I such that∑

k≥1

fk(x0) is convergent and if∑k≥1

f′k is uniformly convergent on I to g : I → R,


then there exists f : I → R such that∑k≥1

fk is uniformly convergent to f on I, f is

derivable and

f ′ = g.

Example 6.29 Let us consider

fk(x) =sin kx

kp, for p > 1 and x ∈ R.

The series∑k≥1

fk is absolutely and uniformly convergent on R, because

∣∣∣fk(x)∣∣∣ = ∣∣∣sin kxkp

∣∣∣ ≤ 1

kp

and the series∑k≥1

1

kpis convergent for p > 1.


cos kx

x2 + k2is uniformly convergent on R.

Solution. Since ∣∣∣ cos kxx2 + k2

∣∣∣ ≤ 1

x2 + k2≤ 1

k2

and the number series∑k≥1

1

k2is convergent, it follows immediately that our series is

uniformly convergent on R.

Exercise 6.31 Test the following series of functions for convergence:

a)∑k≥1

fk, where fk : R → R, fk(x) =sin kx

k3;

b)∑k≥1

fk, where fk : R → R, fk(x) = arctan2x

x2 + k4;

c)∑k≥1

fk, where fk : [0,∞) → R, fk(x) =(−1)k

x+ k.

152 REAL ANALYSIS

6.3 Power Series

Definition 6.32 Let x0 ∈ R, ak ∈ R, k ≥ 0. A series of functions for which

fk(x) = ak(x− x0)k, k ≥ 0,

is called a power series on R. We shall denote such a power series by

∑n≥0

ak(x− x0)k. (6.2)

The above series is convergent for x = x0.

Theorem 6.33 (Cauchy-Hadamard) Let R ≥ 0 be defined as follows:

R =1

limk→∞

k√|ak|

,

with the convention that R = 0 if limk→∞

k√| ak | = ∞ and R = ∞ if lim

k→∞k√|ak| = 0.

Then:

1) if R > 0, the power series (6.2) is absolutely convergent for any x such that

|x− x0| < R and divergent for any x with |x− x0| > R;

2) if R > 0, the power series (6.2) is uniformly convergent over any compact

set [x0 − r, x0 + r] with 0 < r < R;

3) if R = 0, the series (6.2) is convergent only for x = x0.

Remark 6.34 The non-negative quantity R is called the radius of convergence for

the series (6.2).

The set

Dc = {x ∈ R | |x− x0| < R}

is called the interval or the set of convergence for the series (6.2). At the end points

of the interval of convergence, i.e. at the points x − x0 = R and x − x0 = −R, thepower series (6.2) may be both convergent or divergent.


Remark 6.35 If there exists limk→∞

∣∣∣ akak+1

∣∣∣, thenR = lim

k→∞

∣∣∣ akak+1

∣∣∣.Moreover, if there exists lim

k→∞

∣∣∣ akak+1

∣∣∣, thenR = lim

k→∞

∣∣∣ akak+1

∣∣∣.Example 6.36 The radius of convergence of the series

∑k≥0

k!xk is equal to 0. There-

fore, this series is convergent only for x = 0, i.e. Dc = {0}.

Example 6.37 The radius of convergence for the series∑k≥0

xk

k!is equal to ∞. Thus,

the set of convergence is Dc = R.

Exercise 6.38 Find the radius and the domain of convergence for the series

∑k≥1

(−1)k−1 (x− 5)k

k 3k.

Solution. It is easy to see that the radius of convergence for this series is equal to

3. Therefore, using Cauchy-Hadamard theorem, we see that our series is convergent

for x ∈ (2, 8) and divergent for x ∈ (−∞, 2) ∪ (8,∞). Let us analyze what happens

at the end points x = 2 and x = 8. For x = 2, our series becomes∑k≥1

(−1)2k−1

k,

which is a divergent series. For x = 8, our series becomes∑k≥1

(−1)k−1

kand, using

Leibniz criterion, it is a convergent series. So, the interval of convergence of the

given power series is

Dc = (2, 8].

Remark 6.39 Every convergent power series defines on its interval of convergence

Dc a function

f(x) =∞∑k=0

ak(x− x0)k.

154 REAL ANALYSIS

When two functions f and g are decomposed into power series about the same point

x0, the power series of their sum or difference can be obtained by termwise addition

or subtraction. More precisely, if:

f(x) =∞∑k=0

ak(x− x0)k

and

g(x) =∞∑k=0

bk(x− x0)k,

then

(f ± g)(x) =∞∑k=0

(ak ± bk)(x− x0)k.

Also, we can define the product of the power series f(x) and g(x) as being:

(f g)(x) =∞∑i=0

∞∑j=0

aibj(x− x0)i+j =

∞∑k=0

(k∑

i=0

aibk−i

)(x− x0)

k.

The sequence

ck =k∑

i=0

aibk−i

is called the convolution of the sequences (ak)k and (bk)k.

Remark 6.40 Usually, we shall work with power series about the point x0 = 0, i.e.

with series of the form ∑k≥0

akxk,

usually called Maclaurin series.

Definition 6.41 A power series

f(x) =∞∑k=0

ak(x− x0)k

is said to be invertible if there exists a power series g(x) such that f(x) g(x) = 1,

for any x ∈ Dc.


Remark 6.42 The power series∞∑k=0

akxk is invertible if and only if a0 = 0.

Example 6.43 The power series 1− x is invertible and its inverse is the geometric

series∑k≥0

xk.

Remark 6.44 If a power series having the radius R is uniformly convergent over

[x0 − r, x0 + r], 0 < r < R, then its sum defines a continuous function at any point

x with |x− x0| < R.

Remark 6.45 Let R be the radius of convergence for the series (6.2). Then, the

series∑k≥1

k ak(x − x0)k−1 (obtained by differentiating (6.2) with respect to x term

by term) has the same radius of convergence.

Remark 6.46 Let R > 0 be the radius of convergence for the series (6.2) and let f ,

defined on (x0−R, x0+R), be the sum of this series. Then, f is indefinite derivable

(term by term) on this interval.

Example 6.47 For any value of x between −1 and +1, we may write

1

1 + x2= 1− x2 + x4 − x6 + · · ·+ (−1)k x2k + · · · .

Integrating this series term by term between the limits 0 and x, where |x| < 1, we

find

arctanx =x

1− x3

3+ · · ·+ (−1)k

x2k+1

2k + 1+ · · · .

Since the new series converges also for x = 1, it follows that

π

4= 1− 1

3+

1

5− · · ·+ (−1)k

1

2k + 1+ · · · .

Exercise 6.48 Compute the radius and the domain of convergence for the following

series:

a)∑k≥0

xk

k!; b)

∑k≥1

xk

kk; c)

∑k≥0

k!xk; d)∑k≥1

xk

k2;

e)∑k≥1

xk

k 2k; f)

∑k≥1

(−1)k−1 (x− 5)k

k 3k; g)

∑k≥1

(x− 1)2k

k 9k.

156 REAL ANALYSIS

Exercise 6.49 Compute the sum of the series:

S(x) =∑k≥1

xk

k, x ∈ [−1, 1).

Exercise 6.50 Compute the sum of the series

S(x) =∑k≥1

(−1)k+1 xk

k, x ∈ (−1, 1].

6.4 Taylor Series

Definition 6.51 Let I be an open nonempty interval in R, f : I → R, f ∈ C∞(I)

and x0 ∈ I. The power series ∑k≥0

1

k!f (k)(x0)(x− x0)

k

is called the Taylor series of the function f at the point x0.

The function f is called analytic if for any x0 ∈ I, there exists r > 0 such that

the radius of convergence of the Taylor series of f at the point x0 is R ≥ r and

f(x) =∞∑k=0

1

k!f (k)(x0)(x− x0)

k, ∀x ∈ I, |x− x0| < r. (6.3)

This definition characterizes the class of the functions which can be locally approx-

imated by convergent power series.

Let us notice that every power series with a positive radius of convergence is

analytic on the interior of its region of convergence. Also, notice that sums and

products of analytic functions are analytic.

For an analytic function, the coefficients ak can be computed as

ak =f (k) (x0)

k!.

So, every analytic function can be locally represented by its Taylor series. The global

form of an analytic function is completely determined by its local behavior, i.e. if

f and g are two analytic functions defined on the same open nonempty interval I,

and if there exists an element c ∈ I such that f (k)(c) = g(k)(c), for all k ≥ 0, then

f(x) = g(x) for all x ∈ I.


Remark 6.52 If a function is analytic, then it is indefinitely derivable, but, in

the real case, the converse implication is not generally true, i.e. not any function

f ∈ C∞(I) is analytic.

Example 6.53 The function f : R → R, defined by

f(x) =

e−

1

x2 , x = 0,0, x = 0,

is indefinitely derivable on R, but it is not analytical.

Theorem 6.54 Let f : (a, b) → R be an indefinitely derivable function on the open

interval (a, b) and let x0 ∈ (a, b). If there exist R > 0 and M > 0 such that for any

x ∈ (x0 −R, x0 +R) ⊂ (a, b) and for any k ≥ 0,

|f (k)(x)| ≤Mk,

then f is analytic at the point x0.

Theorem 6.55 Let I be an open nonempty interval in R, f : I → R, f ∈ C∞(I)

and x0 ∈ I. Then, the function f can be expanded in a Taylor series

f(x) =∞∑k=0

1

k!f (k)(x0)(x− x0)

k,

(i.e.f is the sum of its associated Taylor series) if and only if

limk→∞

Rk(x) = 0.

Elementary functions

Using power series, we can rigorously define the most common used elementary

functions of real arguments.

1) A polynomial functions P (x) can be defined as being a power series with a

finite number of nonzero coefficients:

P (x) = a0 + a1x+ · · ·+ akxk,

158 REAL ANALYSIS

with ai ∈ R. The radius of convergence of this power series is ∞ and the polynomial

function is indefinitely derivable on R.

2) Rational functions are functions of the form

f(x) =P (x)

Q(x),

where P and Q are polynomial functions in x and Q is not the zero polynomial. The

domain of f is the set of all points x for which the denominator Q(x) is not zero.

3) The exponential function exp : R → R is defined by

exp(x) =∑k≥0

xk

k!.

This function is also denoted by ex and its radius of convergence is ∞.

4) The sine function sin : R → [−1, 1] is defined by

sin x =∑k≥0

(−1)kx2k+1

(2k + 1)!.

Its radius of convergence is ∞.

5) The cosine function cos : R → [−1, 1] is defined by

cos x =∑k≥0

(−1)kx2k

(2k)!.


6) The hyperbolic sine function sinh : R → R, defined by

sinh x =∑k≥0

x2k+1

(2k + 1)!.



7) The hyperbolic cosine function cosh : R → R, defined by

cosh x =∑k≥0

x2k

(2k)!.


8) ln (1 + x) =∑k≥0

(−1)kxk+1

k + 1, |x| < 1.

The above series is also convergent for x = 1 and we get

ln 2 = 1− 1

2+

1

3− 1

4+ · · ·

9) (1 + x)α = 1 +∑k≥1

α(α− 1) · · · (α− k + 1)

k!xk, α ∈ R, x ∈ (−1, 1).

The above series is called the binomial series. If α ∈ N, we get the well-known

Newton’s formula.

Exercise 6.56 Prove that the above elementary functions gave the following prop-

erties:

1) ex+y = ex · ey, ∀x, y ∈ R;

2) (ex)′= ex, ∀x ∈ R;

3) eix = cos x+ i sin x, ∀x ∈ R (Euler’s formula);

4) sin x =eix − e−ix

2i, ∀x ∈ R;

5) cos x =eix + e−ix

2, ∀x ∈ R;

6) sinh x =ex − e−x

2, ∀x ∈ R;

6) cosh x =ex + e−x

2, ∀x ∈ R;

160 REAL ANALYSIS

7) sin2 x+ cos2 x = 1, ∀x ∈ R;

8) cosh2 x− sinh2 x = 1, ∀x ∈ R;

9) (sin x)′= cos x, (cos x)

′= − sin x, ∀x ∈ R;

10) (sinh x)′= cosh x, (cosh x)

′= sinh x, ∀x ∈ R;

11) sin (−x) = − sin x, cos (−x) = cos x, ∀x ∈ R;

12) sinh (−x) = − sinh x, cosh (−x) = cosh x, ∀x ∈ R;

13) sin (x+ y) = sin x cos y + cos x sin y, ∀x, y ∈ R;

14) cos (x+ y) = cos x cos y − sin x sin y, ∀x, y ∈ R;

15) sinh (x+ y) = sinh x cosh y + cosh x sinh y, ∀x, y ∈ R;

16) cosh (x+ y) = cosh x cosh y + sinh x sinh y, ∀x, y ∈ R.

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