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An Introduction to Mathematical Analysis

5. Zailman University of Montreal

World Scientific Singapore'NewJerseyLondorrHongKong

Published by

World Scientific Publishing Co. Pte. Ltd. P O Box 128, Farrer Road, Singapore 912805 USA office: Suite IB, 1060 Main Street, River Edge, NJ 07661 UK office: 57 Shelton Street, Covent Garden, London WC2H 9HE

Library of Congress Cataloging-in-Publication Data Zaidman, Samuel, 1933-

Advanced calculus : an introduction to mathematical analysis / S. Zaidman. p. cm.

Includes bibliographical references (p. 171) and indexes. ISBN 9810227043 1. Mathematical analysis. I. Title.

QA300.Z285 1997 515»dc21 97-20207

CIP

British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library.

Copyright © 1997 by World Scientific Publishing Co. Pte. Ltd. All rights reserved. This book, or parts thereof, may not be reproduced in any form or by any means, electronic or mechanical, including photocopying, recording or any information storage and retrieval system now known or to be invented, without written permission from the Publisher.

For photocopying of material in this volume, please pay a copying fee through the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, USA. In this case permission to photocopy is not required from the publisher.

This book is printed on acid-free paper.

Printed in Singapore by Uto-Print

PREFACE

The present book is, as its title indicates, a presentation of some fundamental ideas related to the elementary real analysis.

For a better understanding of the material given here, a course in basic differential and integral calculus can be a good preliminary preparation.

The emphasis in our present text lies on the so-called rigorous method where everything (well, almost everything) is given a clear definition, a detailed statement, a complete, logically coherent proof.

The book starts with an exposition of the theory of real numbers; it is this theory which is taken as a basis for the concepts developed afterwards. We present real numbers as equivalence classes of Cauchy sequences of rational numbers.

This is different from what can be found in most recent books, where one prefers the axiomatic way or (sometimes) the method of "Dedekmd cuts." We consider the method of Cauchy sequences a very clear manner of introducing general, real numbers; furthermore, it has the advantage of being applicable in other situations, for instance in the theory of "metric spaces." Afterwards we give the usual, modern presentation, of such topics as: sequences of real numbers, infinite numerical series, continuous functions, derivatives and integration theory.

There are also two chapters of a peculiar type: the first one concerns convex functions, a class of functions which appear useful in many applications of functions, a class of functions which appear useful in many applications of calculus; the second one, about metric spaces, reflects a recent tendency to start presenting "topological" ideas from the very beginning of the undergraduate prest;ii uxii£, lupuiu^ i^di mathematical life.

The attentive reader will note probably the absence of most, "well-known" elementary functions, like sin x, cos rr, ex, log x, from the book. They are

V

vi Preface

never used here; their rigorous presentation would take many more pages and the student learns about them in any case in other places.

At the end of each Chapter we added some quite simple exercises, which, if solved, would help for a higher understanding of the subject matter previously treated.

We terminate the book with an Appendix concerning general concepts of logic and set theory which are used in the text. There is also an Index of Notations, an Index of Subjects and a short list of bibliographical references. The student should find them useful.

CONTENTS

Preface v

1. Numbers 1

2. Sequences of Real Numbers 29

3. Infinite Numerical Series 49

4. Continuous Functions 61

5. Derivatives 81

6. Convex Functions 103

7. Metric Spaces 109

8. Integration 131

Index 151

Index of Notations 157

Appendix (Logic, Set Theory and Functions) 163

Bibliography 171

vn

Chapter I

NUMBERS

Numbers are the basic building bricks of analysis; everything else is based on them.

In this Chapter we first present an informal discussion of natural numbers and negative integers. The subsequent discussion of rational and then of real numbers is much more elaborate and gives a real flavour of what the analysis is all about.

1.1 The Natural Numbers

We indicate in the following lines, a familiar discussion about the so called natural numbers. These are indicated by symbols 1,2,3, . . . and originate in counting finite collections of objects; the natural numbers are also called positive integers.

In the realm of these (mathematical) objects one considers two (algebraic) operations: addition (+) and multiplication (•). To any given pair of natural numbers each of these operations associates another natural number, in such a way that the following properties are true:

m + n = n + ra, m • n = n -m (the commutative laws) (m + n ) + p = m + (n + p ) , (m -n) • p = m • (n • p) (the associative laws)

m • (n + p) = m • n + m • p (the distributive laws) m -1 = m (1 is an identity with respect to multiplication)

We note also the following trichotomy law:

l

2 Advanced Calculus

Given any natural numbers m and n, one and only one of the following possibilities occurs:

(i) m = n; (ii) 777, = n + x for some natural number x;

(iii) n = m -f 7/ for some natural number y.

If (ii) holds we write m > n or n < m and we say that m is larger or greater than n and that n is smaller or less than m. If either (i) or (ii) holds, we write m > n or n < m and say that m is larger or equal to n and that n is less than or equal to m. Let us denote with N the set (collection) of all natural numbers.

A fundamental result that we consider valid in N is the so called principle of mathematical induction. By this we understand the following property:

(a) Let M be a subset of N such that: 1 G M and if n G M then n + 1 is also in M. It results: M = N.

An equivalent formulation goes as follows: ((3) Let V(n) be a property associated with the natural number n (for all

n G N). Assume that V(l) is a true statement, and that, ifV{n) is true, then V{n 4-1) is also true. It follows that V(n) is true for all n G N .

Remark. We can prove the equivalence of (a) and ((3). First, let us assume (a); then define M — {n G N,V(n) is true}. The statement in (/?) follows immediately.

On the other hand, let us assume now (/?); given M C N, we define the property (proposition) V(n) by the relation: n G M (that is, V(n) is true when n G M). The assumptions in (/?) now reads as follows: 1 G M; if n G M then n + 1 G M. The conclusion in (/?): n G M for all n G N. This means N c M , hence M = N. This is the conclusion in (a).

Example 1. Let us assume that p is an odd number (p = 2k + 1 for some k G N). Then pn = p-p- • -p (n times) is also an odd number (Vn G N).

Proof. It is obvious that the proposition V(n): "pn is an odd number" is true when n = 1 (this is our assumption!). So, let us assume also that pn is an odd number: pn = 2s + 1, for some s G N. It follows that p n + 1 = (2s + 1) -p = (25 + l)(2fc + 1) = 4*;* + 2fc + 2s + 1 = 2(2A;s 4- k + s) + 1, which is again an odd number.

Numbers 3

Example 2. Let h G N be given; then the inequality: (1 -f h)n > 1 + nh, Vn 6 N, holds true.

(This inequality will appear later on, in somewhat more general form).

Proof. The inequality is in fact an equality for n = 1. If we assume that (l + /i)n > 1+n/i it follows that (l + / i ) n + 1 = (l + /i)n(l + /0 > (l+nh)(l + h) = 1 + (n + l)fc 4- n/i2 > 1 4- (n + l)fc.

Corollary. For all n G N, 3 n > n.

In fact 3 n = (1 + 2)n > 1 + 2 • n > n, Vn G N.

Example 3. The equality: 1 + £ L i 8A; = (2n + l ) 2 , Vn G N is satisfied.

Proof. The case n = 1 reduces to the obvious identity: 1 -f 8 = (2 + l ) 2

which can be checked by everybody. Next, we shall assume the equality for n G N and deduce it for n-f 1. Thus:

n + l n 1 + ] T 8A; = 1 + ] T Sk + 8(n + 1) = (2n + l ) 2 + 8(n -f 1)

fc=l i b = l

= 4n2 + Yin + 9 = (2(n + 1) + l ) 2 .

Our last example deals with an elementary exercise in number theory:

Example 4. For all n G N the number: 4 2 n + 1 4- 3n+2 is a multiple of 13.

Proof. For n = 1 we have 43 + 33 = 91 = 7 • 13. Next, let us assume that: 4 2 n + 1 + 3 n + 2 = 13 • k (for some ke N). It results that:

4 2n+3 + 3 n + 3 = 4 2 n + l ^ 1 3 + 3) + 3 n + 2 . 3 = 1 3 . 4 2 n + l + 3 . ^2n+l + 3 n + 2 )

= 13 • 4 2 n + 1 + 39 • k

= 13(42n+1 + 3k),

that is another multiple of 13. In some problems it appears useful to state explicitly the so-called:

Well-ordering principle in N. If M is a nonempty subset of N, it has a smallest element.

(That is, there exists m G M such that m < n V n G M).

4 Advanced Calculus

Example 5. There is no number in N, say a G N, with the property that 0 < a< 1.

Remark. The number 0 (not yet introduced) is defined by the property: 0 + n = nV n G N. We assume 0 < 1.

Proof. In the contrary case, the set: A = {a G N, 0 < a < 1} is not empty. By the well-ordering principle, it has a smallest element, say k. Thus, k G A and k < a V a G A. In particular, 0 < k < 1. It follows that: 0 < k2 < k < 1. Therefore k2 G A and k2 < k which is not possible.

Remark . The principle of mathematical induction can be regarded as a consequence of the well-ordering principle.

Proof. Let therefore M C N, such that: 1 G M and if n G M then n-h 1 G M. Let us assume also that M ^ N. Then the set A = {n G N, n £ M} is non empty, so it has a smallest element a. As 1 G M, =» a ^ 1. Also, as 0 < a < 1 is impossible, it results that a > 1. As a is the smallest in A, a — 1 is no more in A, hence a — 1 belongs to M. But in this case (a — 1) + 1 = a belongs also to M. Again we found a contradiction, which completes the proof.

1.2 The Integer Numbers

We next continue the informal discussion of the concept of numbers, by introducing (again!) the number 0 (zero) and also the, so-called, negative integers, —1, —2, —3, The set of natural numbers, together with the number 0 and the negative integers form the set-denoted with Z - of all integer numbers.

We next define the operation of addition in Z: first, if — n is a negative integer while m G N we put

(i) — n + m = 0 if m = n (we say that — n is the additive inverse to n) (ii) —n -f m = —p if n = m + p (where p G N)

(iii) — n + m = p if m = n + p (where p G N)

(Note that, by the "trichotomy law" in N, one and only one of cases (i), (ii) and (iii) will occur).

Next we define: m + (— n) = -n + m; (— n) -f (-m) = —(n + m)\ 0 + m = m + 0 = m (where m G N); 0 + ( -n ) = ( -n ) + 0 = - n ; 0 + 0 = 0.

Numbers 5

Multiplication in Z is defined in the following way:

m • (-ra) = (-ra) • m = - ( r a • ra); (-ra) • (-ra) = ra • n;

0 • (-ra) = (-ra) -0 = 0-ra = ra-0 = 0-0 = 0(where ra,ra G N).

/£ can 6e proved that the commutative and associative laws (which are valid in N) hold also in Z, as the distributive law, and that a • 1 = a Va G Z.

(for instance: we know that a • 1 = a if a G N; if a = 0, 0 -1 = 0 was defined previously; if a = —ra, (-ra) • 1 = — (ra • 1) = — ra).

We complete definitions above: If a = —ra, where ra G N, we put —a = ra. Also, we put - 0 = 0. It results: a + ( -a ) = ( -a ) + a = 0 Va G Z.

We are ready now to point out the following important fact:

Proposition 1. Given a, b G Z £/ie equation a + x = b has a (unique) solution x G Z.

Proof. We shall first establish that there is at most a solution. Assume in fact x and y are solutions (in Z). Then a + x = a + y = b. So, a + x = a + y and (a -f x) - (a + y) — 0, hence x — y — 0, x = y. To find a solution we just take the integer x = (—a) + 6. Then a -f- x = a 4- (—a) + 6 = 04-6 = 6.

Remark. It is often convenient to use the notation: (—a) + 6 = 64- (—a) = 6 — a, for a, 6 G Z.

Then, the unique solution to the equation a 4- x = 6 is given as: x = 6 — a.

Remark. If the above found solution x = 6 — a is a positive integer, we say that 6 is larger (or greater) than a and we write: 6 > a on a < 6 (a is smaller than 6).

If the solution x = 6 — a is a negative integer, then the equation 6 + 7/ = a has the (unique) solution y = a — b which is a positive integer, and therefore a > 6.

Therefore 6 > 0 means that 6 G N; a < 0 means that 0 — a = — a G N, hence a = —(—a) is a negative integer. Note that, Va G Z, a ^ 0, we have a2 > 0. (for, if a G N => a2 G N; if a = (-6), 6 G N, => a2 = (-6)(-6) = 62 G N).

The trichotomy law holds in Z: if a G Z, then a > 0, or a = 0, or a < 0.

6 Advanced Calculus

1.3 Rational Numbers

The construction of rational numbers is a basic example of the concept of equivalence classes on a set.

Let us remember that given an abstract set X, one has an equivalence relation on X, denoted with ~ , if and only if the following holds:

x ~ x V:r G X(reflexive property); x ~ y <& y ~ :r(symmetric property);

x ~ y and y ~ z => x ~ ^(transitive property).

Next, for all x G X, let Ex = {y G X% y ~ x}. It is the "equivalence class" containing x. We get the following:

(i) x G Ex (due t o x ~ x). (ii) Ex = Ey iff x ~ y. In fact, if Ex = Ey, and because x G Ex, we find

x G Eyi hence x ~ y. Conversely, let us assume that x ~ y. We shall prove that Ex C Ey and Ey C Ex. Let z G Ex ; => z ~ x ~ y. Hence z ~ y and 2 G JE^. Let 2 G Ey\ =$> z ~ y ~ x, hence z ~ x, z E Ex.

(iii) If ^ x ^ Ey , it follows that ExnEy = $. In fact, otherwise, let z e Ex

and z € Ey. Therefore z ~ x, z ~ y, hence a; ~ y, hence 2? — Ey. (iv) The set of all distinct equivalence classes is a partition of X, that is

X is the union of the disjoint equivalence classes Ex. In fact, each x G X belongs to Ex, hence to their union. The converse inclusion:

U Ex C X is obvious.

Before introducing rational numbers, let us indicate why we need (any) new numbers, (why Z is not sufficient for everything?)

Consider the multiplicative equation: a-x = 6, where a, b G N. Sometimes, it has no solution in Z! For instance, if a = 2 and 6 = 3, we see that 2.1 < 3, 2.2 > 3, • • • 2.n > 3 Vn > 2. Also, 2.0=0, 2.(-n) = -2n , a negative integer. Thus, 2.x=3 where x G Z, is never true. Motivated by this observation, let S be the set of pairs of integers (a, 6), where b ^ 0. Therefore S = {(a, 6), a,6G Z and 6 ^ 0 } .

Define on S the relation (a, 6) ~ (c, d) iff a • d = b • c. /£ is an equivalence relation on S: First, the reflexive property: (a, 6) ~ (a, b) means a - b = b - a; if (a, b) ~ (c, d), then a - d = b - c. It follows that (c,d) ~ (a, 6) (it means c.b = d.a\). Finally, if we assume (a, b) ~ (c,d) and (c, d) ~ (e , / ) we obtain: a - d = b - c and c • / = d • e; by multiplication we get ( a - d - c - / = 6 - c - d - e ,

Numbers 7

hence c • d(a ■ f - b ■ e) = 0. If c ^ 0, we have af = be and (a, b) ~ (e, / ) . If c = 0 we have (a, 6) ~ (0, d) hence ad = 0 and a = 0; also (0, d) ~ (e, / ) , hence d ■ e = 0 and e = 0; obviously (0,6) = (0, / ) .

(We used here the property that Z has no zero-divisors: a • (3 € Z and a • /? = 0 => a = 0 or P = 0).

Accordingly, we can consider, for any couple (a, b) 6 S, the equivalence class £a,<, which is also denoted f or a/b and is called a rational number. One denotes with Q the set of all rational numbers.

Note that (a, b) ~ (ma, mb) Vm € Z. Hence a/6 = ma/mb where 6 ^ o and m e Z. There is an imbedding ofTL into Q: e € Z — f- e Q.

Next we define addition and multiplication in Q as follows:

-E(a,6) + -E(c,d) = E(ad+bc,bd)>E(a,b) ' -^(c,<2) = -^(ac.bd)-

In order to have well-defined operations (that is, independent of the couple (a,/?), P ^ 0 which "generates" the equivalence class Eia<p) = f) we have to establish the following result.

Theo rem 1. Assume that (a,b) ~ (a',b') and {c,d) ~ (c',d',)(b,b' # 0,d,d' ^ 0). It follows that: (ad + bc,bd) ~ (a'd' + 6'c',6'd') and (ac,6d) ~ (a'c',b'd').

Proof. The assumption implies: ab' = a'b,cd' = c'd. Let us multiply the first equality by d • d' and the second by b ■ b'. We obtain:

ab'dd' = a'bdd',cd'bb' = c'dbb'.Ii we add we get b'd'(ad + be) = bd(a'd' + c'6').This is the first equivalence.

Next, by straightforward multiplication we have ab'cd' = a'be'd, meaning (ac,6d)~(a'c ' ,6 'd ') □

Remark . The class E(0il) is a zero in Q: £?(o,6) + E{0<1) = E{a>b) for (a, b) € 5. InfactE ( o ,6 )+E ( 0 l i) = E(«.i+6.o,n) = E(«,6)- Next: The class £ ( M )

is a "unif in Q: E(a>6) .£( l i l ) = E(o,6) V(a,6) e 5. Any non-zero element in Q has a multiplicative inverse: in fact, let (a,b) € 5, (a,b) *> (0,1); this means: a-1 ^ 6-0 = 0, a # 0. Hence a ^ 0,6 # 0 and accordingly (6, a) € 5, £ ( 6 ,a ) € Q. Furthermore £(6,o) • £(a,5) = E(a<b) ■ E(b,a) = E(ab,ab) = £ (1>1).

It follows readily that the equation a ■ x = p,a £ Q,'p e Q,a ^ 0, has a solution in Q : x = p-a'1. (This solves the problem which motivated introduction of Q(a~Y is the multiplicative inverse to a € Q).)

E(a,b) + E(c,d) = -S(a<i+6c,6<i) >-E(a,6) - E(c,d) = -E(ac,6<i)-E(a,b) + Elcd) = Etad+bcbd),E(ab) ■ E(cd) = Elacbd).

ab'dd' = a'bdd!,cd'bb' = c'd66'.If we add we get ao aa = a oaa ,ca oo = c auu .11 we auu we gei b'd'(ad + be) = bd(a'd' + c'o').This is the first equivalence. u u \uuu ~r Ui. — uu\u, it T t y 1. J. mo 1a t u c m o t cuu iva i c i i t c .

8 Advanced Calculus

Let us define now the positivity relation in Q. First note that (a, b) ~ (—a, —6), for 6 / 0 . Then we say:

E{a,h) > 0 iff a • b > 0.

Again, we must check the following: (a, b) ~ (a', 6') and ab > 0 => a'6' > 0. This follows readily from : ab' = a'b : (we get ab' • a'b = (a'b)2 > 0, hence (ab) • (a'6') > 0).

We also point out the following properties: (O-from "order")

O • 1 : E ( 0 | 6 ) > 0 and £(c>d) > 0 => JE(af6) • E{Cyd) > 0 and E{a,b)

+ £(c,<£)>0, O • 2 : VJS(a|6) G Q we have £ (a>6) > 0 or £ (a>6) = 0 or - £ (a>6) > 0

(trichotomy law).

Proof. We have ab > 0 and cd > 0; then a b e d > 0 which gives E{a,b)'E{c4) = £(ac,6d) > 0. Also, £ ( a j 6)+£(C ) d) = E(ad+hcM) and the product (ad 4- 6c) • bd = abd2 -f 62cd is > 0 too. Next, if (a, 6) € S, we have ab > 0 or a& < 0 or ab = 0. In the last case, as b ^ 0, => a = 0, hence #(a,b) = 0Q. If a6 < 0, we get (—a) • 6 > 0. Also JE7( <«.,&) — "-^(a,6) (their sum is null). Hence -£(a,6) > 0.

We must also point out that the above defined operations of addition and multiplication in Q obey the same commutative, associative and distributive laws as those valid in Z. (the proofs are based on those properties in Z).

Finally, let us note that the previously introduced imbedding: Z —> Q (where a G Z —» -E(0,i) G Q) is one-to-one:

if E(aui) = ^(o2li) => ai • 1 = fl2 • l ,a i = a2).

The algebraic operations in Z are preserved by this correspondence: E(a>1)-h £(fc,l) = £ ( a + 6 , l ) , # ( a , l ) ' £(&,1) = ^(a5,l)» ^ e a s i l y S e e n *

An order relation on Q can also be defined: We say that £(a,6) > ^(c,d) ^ their difference: E(ad-bCjbd) is > 0, that is, if and only if (ad — bc)bd > 0. As usual, if x, y G Q we write y < rr when x > y.

1.4 Ordered Fields

It is possible to encompass the above seen properties of the set Q in the general algebraic concepts of a field, and of an ordered field.

Numbers 9

To start with, we say that a set F is a field if the following holds: Va, b G F , there are defined elements a + b G F and ab G F , called the sum and product of a and 6, subject to the following axioms:

Associative laws: a + (b + c) = (a + b) + c and a • (6 • c) = (a -6) • c Commutative laws: a + b = b + a,a-b = b-a Distributive laws: a'(b-\-c)=a-b + a-c

There is an element 0 G F such that a + 0 = a V a G F ( a "zero element"). There is an element 1 G F, 1 ^ 0, such that 1 a = a Va € F (identity element). For each a G F there is additive inverse, denoted ( -a) such that a 4- (—a) = 0. For each a G F, a / 0, there is an element of F , multiplicative inverse, denoted a - 1 , such that a • a~1 = 1.

Example 1. The smallest field consists of two elements, 0 and 1, where we put: 1 + 1 = 0 ,0-0 = 0-1 = 0,1 • 1 = 1,1 + 0 = 1.

Example 2. The set Q of rational numbers is a field. (It follows from the discussion in § 3; the additive inverse to F(mjn) is

F( — ra,n) and the multiplicative inverse to F(m,n) , m ^ 0 is E^n,m).) In a field we can define the substraction: a — 6, where a, 6 G F as: a — b =

a + (—6), and also the division: | by: f = a • ft-1. Note £/ie absence of "zero divisors": if a • 6 = 0 and 6 ^ 0 => a = 0 (in fact

ab = 0 gives (a • 6)6_1 = 0 • b'1 = 0 and (a • 6)&-1 = a(b • 6"1) = a • 1 = a, hence a = 0). Note that in this proof we use the following result:

a • 0 = 0 Va G F; in fact a • 0 = a • (0 + 0) = a • 0 + a • 0; hence a • 0 — a • 0 = a • 0 = 0; in general : a + a = a ^ a = 0.

We point out also the following result of unicity:

Proposition 1. In a field, there is only one zero and only one 1.

Proof. Let in fact be 0' with the same properties as 0; we get 0' + 0 = 0;, 0' + 0 = 0 hence 0; = 0. Also, if 1' is another (multiplicative) unit, we have obviously the equality 1' = 1 • V = 1, hence 1 ; = 1.

In the same vein, we see that the additive and multiplicative inverses are unique.

10 Advanced Calculus

Proof. Let a € F having Wi and W2 as additive inverses: thus a + w\ = a + W2 = 0 and then w^ + (a + w2) = (wi + a) + ^2 =^ wi + 0 = 0 + to2; hence i/>i = w2.

Let also a G F , a / 0, with wi and w2 as multiplicative inverses. We get: a.Wi = 1 = a - W2; hence (a • wi) • W2 = 1 • W2 = W2,(a - ^2)^1 = 1 • wi = wi, and (a • Wi) • 1 2 = (a • ^2) • î» which gives wi = W2.

Remark. The zero element cannot have multiplicative inverse: (otherwise O"1 -0 = 0 = 1).

Finally we note: ( - l ) - a = —aVa G F. In fact (—l)-a+a = (—l)-a+l-a = ( - 1 + 1) -a = 0 -a = 0.

The order properties of Q are also abstracted in the axioms of an ordered field. Therefore, we say that the field F is ordered if it contains a subset P (the "positive" elements of F) characterized by the following axioms ("order axioms").

0.1: x,y G P => xy G P and x + y e P. 0.2: Wx G F we have a; G P or x = 0 or — x G P (trichotomy law).

We then define in such a field P , the relation: a 6 iff a - 6 G P . It follows that: a < 0 i f f 0 - a G P that is -a G P; a > 0 iff a e P. Note also that: i / a / 0 , £ften a2 > 0 (m particular 1 = l2 > 0).

Proof. If a > 0, then a G P hence a • a = a2 G P and a2 > 0. If a < 0 we have —a G P , hence (—a) • (—a) G P . Next, note first that (—1) • (—1) = - ( - 1 ) = 1; hence ( -a ) • ( -a ) = (-1) - a - ( - l ) a = (-1) • (-1) -a2 = a2; again we obtained that 0? G P .

We point out the following properties:

1.) If a ac < be, in fact: b — a G P and c e P gives

(6 — a) • c = be — ac G P . 3.) If a < 6 => a + c < b + c Vc G P; in fact (6 + c) - (a + c) = b - a G P .

Finally, we define: a < 6 ^ a < 6 or a = 6. In most of the subsequent topics, the concept of absolute value is of funda

mental importance. It can be defined (as we now do), in any ordered field (in particular in Q).

Therefore, if F is an ordered field then Va G P , we put: \a\ = a if a > 0 and \a\ — — a if a < 0. We note the following properties: \a\ > 0; |a| = 0 <& a = 0 (if a > 0, |a| = a > 0, if a < 0, \a\ = -a > 0; hence, if |o| = 0 by 0-2 => a = 0);

Numbers 11

also: | - a\ = \a\ (if a > 0, -a < 0, | - a\ = -(-a) — a = |a|; if a < 0, then - a > 0, | - a\ = -a = |a|; also: |o • 6| = \a\ • \b\ (if a • 6 > 0 => a > 0 and b > 0, or a < 0 and 6 < 0; for if, say, a > 0 and 6 < 0 => a > 0 and - 6 > 0 =* a • (-6) = a • (-1) • b = - a • b > 0, hence a • 6 < 0. Therefore: \a- b\ = a • b = \a\\b\ = (—a) • (—b)\ if a • b < 0 and, say, a > 0, 6 < 0, we get: |a • 6| = —(a -6) = a • (—6) = \a\ • |6|, again. Next, let us note the following property:

x,c € F and \x\ < c <£> — c < x < c.

(In fact: if x < c and x > 0, |rrr | = x < c; if x < 0 and —c<x=>c> —x = \x\; conversely, if \x\ < c, then, if x > 0, => x < c and if x < 0, => - x < c, —c < x).

Corollary. Va G F , - | a | < a < |a|.

Remark. The important inequality: \a -f b\ < \a\ + |6|, Va, 6 € F can be obtained from the inequalities: — \a\ < a < |a|, — \b\ < b < \b\ by addition: we get — (|a| + |6|) < a + 6 < |a| + |6|, hence, from above, we have \a + b\ < |a| + |6|.

A simple corollary.

I M-I&l | < | a - 6 | , V a , 6 G F .

For, if we write a = (a — b) -f- 6, we get \a\ — \b\ < \a — 6|; interchanging a with b yields: |fe| - \a\ < \b - a\ = \a - b\. Therefore: | \a\ - \b\ \ < \a - b\.

Finally, let us note the following: ifxeF,x>0 and x2 = a2, a E F, then x = \a\. In fact x2 = \a\2 gives (x — \a\)(x + \a\) = 0; if x > 0, => x + \a\ > 0, hence

(x - |a|)(x + \a\)(x + | a | ) _ 1 = x - \a\ = 0,x = \a\

1.5 The Real Numbers

The basic motivation for introducing other kind of numbers is again, the existence of simple equations in Q without solution in Q. One such equation is

x2 = 3

Let us prove that no x G Q will do as solution. We have x = E^p, a), q ^ 0, p, q E Z. If x > 0, we have p • a > 0; as 25(Pfg) = JE5(_PJ_9), we may assume p € N, 0 G N. If x < 0 we have p • o < 0 and we may assume p = - m , g = n,


where m, n G N. In both cases, the equation x2 = 3 becomes E^v2^ — ^(3,1)' hence p2 • 1 = 3 • q2 or (—m)2 = 3 • n2 , which has the same form: m2 = 3 • n2 , with m, n G N. Thus, it remains to show that, Vm, n G N, the relation m2 = 3n2 is impossible. We shall now make use of some, (very) elementary number theory. Let d be the greatest common divisor of m and n. Therefore m = d- mi , n = d- n\ and now the greatest common divisor of mi and n\ is 1. The equation m2 = 3n2 becomes d2 • m2 = 3 - d2 - n{ which simplifies to m2 = 3n2. Therefore, we can say that m2 is a multiple of 3. As for the number mi itself: it is, necessarily, (again, using some "elementary number theory"), of the form 3r or 3r + 1 or 3r - 1. However, (3r ± l ) 2 is not a multiple of 3; it follows, accordingly, that mi = 3r. Therefore, 9r2 = 3n\ and n\ = 3r2. Again, using the same argument we find that n\ = 3s.

We see that the greatest common divisor of mi and n\ is at least 3. We obtained a contradiction which establishes our result.

Therefore, our next goal will be an enlargement of the field Q of rational numbers to another ordered field (the so called "real numbers"). In this new field equations of the form x2 = 2, x2 = 3 and other will always have solutions (not however the equation x2 = —1!). These new numbers will be again obtained as equivalence classes: classes of "Cauchy sequences" of rational numbers, under a natural equivalence relation.

We shall need a certain collection of preliminary definitions and results, concerning the (new) concept of the so called "rational sequence". (Here we enter the analysis proper; the previous discussions pertain rather to elementary algebra).

A basic idea is that of a sequence of elements in a set: suppose to have a mapping (function), from N (the natural numbers) into S — a non-empty set. We write N —+ S; also x = (xn) = (xi,£2,#3, • • •); xn is the nth term of the sequence. If S = Q we call x a rational sequence. We remark that it is possible that xn = xm for some n^m (thus x need not be an injective map). Also, it is very important to distinguish the sequence (xn) from the set {xn} — which is the image of N under x in S.

Examples

I) Let a € Q. Then xn = a Vn G N defines a constant sequence x = (a, 0 , 0 , . . . ) .

II) Let a G Q and m G N, m > 1. Define xn = a for n > m, and write down any xi , #2 • • -^m-i (in Q). Then x — (xn) is said to be ultimately constant, i.e. x = (x i ,x 2 , . . . x m _ i , a , a , a , . . . ) .

Numbers 13

III) x = (0,1,0,1 • • •) is a N-valued sequence. We can give the n t h term by the formula

_ ! + (-!)" Xn~ 2

In this case the range of (xn), {xn} = rr(N) = {0,1}. IV) x = (n) = (1,2,3,. . .) is again N-valued sequence and x(N) = N. V) x — (0, —1, —2, —3,...) is a Z-valued sequence. VI) x = (£(i,n)) is a rational sequence: (£(1,1), £(1,2), • • • £(i,n), • • •)• VII) Xfl — Tb* — 1 • 2 • 3 . . . n, called "factorial n", gives the sequence (n!).

Prom now on we shall be concerned with rational sequence: x — (xn), where xn € QVn G N. Important classifications of these sequences appear in the following definitions.

1. Bounded sequence.

(i) x = (xn) is bounded above iff 3M G Q, such that x n < M Vn G N. (ii) x is bounded below iff 3m G Q, such that xn > m Vn G N.

(iii) £ is called bounded iff it is bounded above and below; i.e. there is M > 0, such that \xn\ < M Vn G N. (in fact, if last inequality holds, then —M < xn < M Vn G N; conversely if m < x n < M Vn G N we have: \xn\ < max(|m|, \M\) as easily seen: if xn > 0, |rrn| = xn < M < |M|; if xn < 0, \xn\ = -xn < -m < \m\).

A sequence which is not bounded will be called unbounded.

2. Null sequence. It is a sequence (xn),N —> Q with the following property: for all e G Q,e > 0,3n0(e) G N, such that: \xn\ < e for n > n0(e). We shall use sometimes the notation "co" for the collection of all the null sequences in Q.

3. Cauchy sequence, x = (xn) is a "Cauchy sequence" iff: for all e > 0, e G <3,3n0(e) G N, such that \xn — xm\ < e for n > n 0 ,m > n0.

(We shall use the notation "C" for all Cauchy sequences in Q).

Let us give short look at examples (I)-(VII). Obviously, the constant sequence in (I) is bounded and belongs to C: it is not a null sequence, unless a = 0. The ultimately constant sequence in (II) has exactly the same properties, as easily seen. The sequence in (III) is bounded, but is neither a null sequence nor a Cauchy one. (for instance, if it is a null sequence we


must have \xn\ < e for n > no(e) where e is any positive rational. Take e = \ and we have an obvious contradiction; if it is a Cauchy sequence, we would have: \xn — x m | < | for n > no, which is contradicted by the relation \xn - x n + i | = 1 Vn G N).

The sequence of natural numbers is unbounded, (for, take any M G Q; we can assume M > 0, M — j , r , s G N. Take n = r + 1. Then: r + 1 > j (this amounts to s(r +1) > r, which follows immediately from the inequalities: (5 - r + 5 > 5 - r > 1 • r = r) . Therefore, no M > 0, M G Q can be an "upper bound" for all natural numbers. (We can say: VM G <2, Af > 0,3n G N, n > M - the "Archimedean property" in Q). For a similar reason, the sequence (n) is not a null sequence; (we have \n\ = n > 1 for all n G N; take e < 1). Also (n) ^ C: for: \n — m\ > 1 Vn G N, Vm G N. The sequence in (V) is unbounded (|xn | = n — 1 Vn G N). It is not a null sequence, and does not belong to C.

The sequence (^) is a null sequence: for, |^ | = ^ < e if n > no, where no G N and no > \ (existence of such no is a consequence of the Archimedean property). It is also a bounded sequence, for \xn\ = ^ < 1 Vn G N. It is even a Cauchy sequence! This can be seen, for example, from the obvious estimate

n m 1 1 2

< —I < e i f n and ra are > no, where no > - . n m e

The sequence (n!) is unbounded: (obviously n! > n Vn G N). It cannot be a null sequence, for the same estimate. We can see that it is not a Cauchy sequence as a corollary of following

Theorem 1. If (xn) G C, then (xn) is bounded.

Proof. Take e = 1; then, 3no G N, such that \xn — x m | < 1 for all n, m > n0 . Accordingly we get: \xn\ < \xn — Xm\ + |# m | < 1 + |xm | for all n, m > n0 . Take m — no, we find \xn\ < 1 4- |xn o | Vn > no- For the remaining n( l < n < no) there is a largest number of the finite set {|a;n|, 1 < n < no}, say H = \xp\ for some p G {1,2, . . . no}. If we let M = max(#, 1 -f |xno |) we see that

\xn\ < M Vn G N.

We are now quite prepared to introduce in C an equivalence relation. We say that x = (xn) and y = (i/n) are equivalent (write x ~ y) iff the

sequence (xn—yn) is a null sequence. We have to prove that it is an equivalence relation:

Numbers 15

x ~ x amounts to: (0) (the constant sequence: xn = 0 Vn G N) is a null sequence.

x ~ y <& y ~ x. This amounts to (xn — yn) is a null sequence iff (yn — xn) is a null sequence, which is obvious from the definition.

x ~ y and y ~ z => x ~ z.

In fact, we assume: (xn — yn) and (yn — zn) are null-sequences. We see that Xn — z<n = (xn — yn) + (yn — zn). Thus, it is sufficient to prove that:

Proposition 1. If a = (a n ) , 0 = (/3n) are null sequences, then (an + /3n) is again a null sequence.

The proof is simple: we have | a n | < f for n > no and \/3n\ < | for n > n\ (any e > 0,e G Q). Therefore | a n + /?n| < | a n | + |/3n | < e for n > max (n 0 ,n i ) .

Thus, we have this way obtained a partition of C in distinct equivalence classes, which we shall call the real numbers (denote E the set of all real numbers). If x = (xn) is a rational Cauchy sequence and Ex is a corresponding equivalence class, Ex is a notation for a real number (but sometimes we shall write x instead of Ex, in order to simplify notations).

Note the imbedding of Q into E : Va G Q -» E(a, a, a , . . . ) G E. Next, we define addition and multiplication in E, in the following way:

Ex+Ey = Ex+y, Ex-Ey = Ex.y (where, if x = (xn), y = (yn), x+y = (xn+2/n), x - y = (xn • yn))> This has a definite meaning: however it is not an obvious fact. We must prove that:

Theorem 2. Ifx,y£C=>x + y£C and x • y G C. Furthermore, if xi ~ x2, y\ ~ 2/2, then xi+yi ~X2 + 2/2 and xi • 2/1 ~ x2 • 2/2-

(here xi, x2, 2/1, 2/2 belong to C).

Proof. By assumption we have \xn—xm\ < f, if n, m > n i , and |2/n—2/m| < §, if n, ra > n2. Therefore: |(xn + 2/n) - (xm + 2/m)| < |z n - xm\ + |2/n -ym\ < e for n, m > max (ni ,n 2 ) . Also, due to relation: (xn - xm)2/n + Xmiyn - ym), we obtain the estimate \xn • yn - x m • 2/m| < kn - ^m||2/n| + km||2/n ~ 2/m| < # ( k n - xm\ + \yn ~ 2/m|) (existence of such a number H follows from Theorem 1). Let us choose now no G N, such that, for n,m>n0 => \xn-xm\ < jft a n d \yn-ym\ < air • We derive \xnyn-xmym\ < e for n, m > no-

Next, let £1 - x2 G c0, 2/1 - 2/2 € c0. It follows that (xi + 2/1) - (x2 + 2/2) = (xi — x2) + (2/1 - 2/2) € c0 (because of Proposition 1). As for the product: we


must show that (xi,ni/i,n - x2,n2/2,n) G c0. We can write: £i,n2/i,n - x2,ny2,n = (î,n - X2,n)yi,n + ^2,n(2/i,n - 2/2,n)- Then we shall apply Proposition 1 and also next

Proposition 2. If x — (xn) is bounded and z = (zn) G Co => (xn^n) G Co-

(It suffices to see the estimate: \xn - zn\ = \xn\ \zn\ < M\zn\ < e for n > no where M > \xn\ Vn G N and \zn\ < -^ for n > no).

This completely proves Theorem 2.

Therefore, the addition and multiplication in E are unambiguously defined. It is immediate that above operations enjoy the usual commutative, asso

ciative and distributive laws, which were verified in N, Z and Q. In order to define a zero in E, we take 0 = £7(0,0,...). (Note that if x G Co,

then x ~ (0 ,0,0. . . ) ; thus 6 = Ex when x G c0). (It is obvious that VT/ G C, y+Qi1) ~ y, hence Ey + E0 = £ y ) . Also, the class £7(i,i...) is an identity in E — as easily seen. In addition, we see that: #(1,1...) / ^(0,0...) a s (1> !»!•••) ^ co-

Our next result is somewhat more involved

Proposition 3. If Ex G E and Ex # 0, there exists Ey G E snc/i £/m£ Ex • Ey = Ey • Ex — -E7(i,i,...) (existence of the multiplicative inverse for any non-zero real numbers).

Proof. If Ex ^ 0 it follows that: x G C and re ^ c0. Let us prove now the

Lemma. / / (xn) G C and (xn) £ c0, there exist p G N and r > 0, r G Q, 5uc/i that \xn\ >rWn> p.

Proof of Lemma. In fact, suppose not. Then for all p G N and for all r > 0 in Q, we would have \xn\ < r for some n > p. Now, as (xn) G C, we find that \xn — xm\ < r for all n, ra > n0(3n0 G N such that . . . ) . Take p = n0 and choose n > p such that \xn\ < r. Hence, for m > p we obtain |^m| < \xm ~ Xn\ + \xn\ < r + r = 2r; therefore (xm) G c0, a contradiction.

Let us terminate now the proof of Proposition 3. Let us define a sequence (yn) = y by: yn = 1 for 1 < n < p, yn = ■£- for

n > p. Then: y G C In fact, for n, m > p, we have |?/n — 7/m| = | j ^ - | = \xm—Xn\ ^ |£rn_—£n_[ \xn\ \xm\ — T^ '

(^Here 0 means any sequence x G CQ

Numbers 17

Therefore, taking n0 > p such that |xm - xn\ < er2 for n, m > n0 , we get |2/n ~ 3/m| < e for n,m> n0 , hence (yn) G C.

Finally, we see that: Ex • Ey = Ex.y = B(Xllx2...a!p>i>i,i,...)-Whence Ex • Ey = Ey • Ex = £(1,1,1,...).

We have proved that E is a field, (the existence of the additive inverse is obvious). The imbedding Q —► E ( / (g) = -E(9>g,g...)) generates an isomorphism, Q -» /(<?)• (For example, /(p) = /(g) =* E(p...) = £ ( g . . . ) => (p) ~ (g), hence p = q. Also, /(p) • /(g) = ^(p...)^.. .) = E(pq,...) = f(P'Q)-- etc.)

Our next objective: to introduce an order relation on E in such a manner that it becomes an ordered field. Therefore, we start with a definition of positivity: (One would be tempted to say: Ex is positive if x = (xn) and xn > 0 Vn G N. However, the simple example where rrn = ^ Vn G N and ■^(i) = # is the zero in E shows that this is not the correct definition.) Thus, we put the following

Definition. The real number Ex is positive if 3r G Q, r > 0 and p G N such that xn>rVn> p.

Remark . The definition of positivity above concerns the ultimate behavior of the sequence (xn)\ what happens for x i , ^ •.. xv is completely irrelevant.

It is also essential to note the

Proposition 4. If X\ ~ xn), and xi,n >rVn> p-. Then, if x^ = (#2,n) we have: (xiyTl — X2,n) £ CQ\ accordingly, |rri>n — #2,n| < § for n > g. Therefore #i,n > T and X2,n — #i,n > — § for n > p 4- g. On addition we obtain: #2,n > § for n > p + g, that is i£X2 is also positive.

Next, we shall check validity of the order axioms 0.1 and 0.2 which were given in § 4. Let us call V the set of all positive reals. Assume Ex and Ey G V. We have xn > r > 0 (r G Q), Vn > pi and yn > 5 > 0 (s G <?), Vn > p2 . Then £n + yn > r + 5 > 0 for n > pi + p2 and xn.2/n > r • 5 > 0, Vn > pi + p2 . Accordingly, we find that Ex + Ey € V and Ex - Ey e V too. The proof of the "trichotomy law" 0.2 is deeper. Let E x ^ 0; this means (xn) ^ c0 and (xn) G C. We use again the "Lemma" which has been used in the proof of Proposition 3. Therefore, 3r > 0 and p G N, such that \xn\ > r Vn > p; also, since (xn) G C, 3n0 G N, such that \xn - xm\ < § for, n, m > n0 . Thus, for


n = no + p we have \xn\ > r; by the trichotomy law in Q, we obtain xn > 0 or xn < 0. In the first case, using also the lower bound x m — xn > — | for ra > n0 , one gets x n > r, x m — x n > — | , hence x m > | for all ra > no- Hence a; > 0 (that is i£x > 0). In the second case (xn < 0) we have — xn > r and x m — x n < | for m > n0 , hence x m < — | , —xm > | for m > n0 which means - x > 0 (that is Ex < 6).

We shall now prove the important "Archimedean property of E" (property A).

Proposition 5. For any x G E, x > 0 (that is x G 'P), £/iere zs an ra G N suc/i £/ia£ m> x.

Proof. The property A has already been established in the ordered field Q (it appeared in the proof of the unboundedness of N). Let now x — (xn), where (xn) G C. Therefore (xn) is bounded; 3M > 0, M G Q, such that xn < \xn\ < M Vn G N. Now, by Property A in Q, 3m G N, m > M + 1; hence m > xn + 1 Vn G N. Therefore J5(m) > £(x) or m > x- (if x < 0 we have immediately 1 > x).

Corollary 1. Let a, b G E w/iere a > 0. TTien, £/&ere exists n G N, stic/i £/m£ na > b.

In fact, 3n G N, n > £ = ba_1 (we may assume b > 0; if 6 < 0, 1 • a > b).

Corollary 2. For any x G E t/iere ea;i5t5 a unique integer n such that n < x < n + 1.

Proof. Uniqueness. Let m, n € Z, and m < n. Then n — ra G Z and n — ra > 0, therefore n — ra>l, m - f - l < n . If x G [n, n -h 1) fl [m, ra + 1) we obtain ra<x<ra-i-l<n<x<n + l , x < x , a contradiction.

Existence. Let x > 0 and 5 = {fc G N, k • 1 > x}. Hence S ^ </> (Proposition 5) and it has a smallest element ra : ra G £; hence m > x (we apply here the well-ordering principle for N). If ra = 1, then 0 < x < 1 and the assertion is proved with n = 0. If m > 1, then ra — l G N , m — 1 £ S. Hence ra — 1 < x and x < ra; ra — 1 < x < ra.

Let x < 0. There exists j G N, j > - x , that is j + x > 0. Hence, 3A; G N, k <j + x < k + 1. This means that fc — j < x < k — j 4-1.

Numbers 19

Remark. The unique integer found in Corollary 2 is denoted by [x]; (greatest integer < x, or "integer part" of x).

Corollary 3. Let w > 0. Then, Vx G E, 3n G Z, a; = nw + a, 0 < a < w.

Proof. Let n - [^]. Hence n < ^ < n + 1 and nw < a; < (n + l)w. If a = x — nit;, we see that 0 < a < w.

The following result states the so called "density" of Q in E. Precisely we have

Proposition 6. It a, b G E and a < b, there exists q G Q such that a < q < b.

Proof. We have a = E x , b = Ey with x, y e C and ?/n - rcn > r > 0 for all n > p. Also, 3n0 G N, such that \xn — xm\ < \ and \yn - ym\ < | for m, rc > n0. Define s = p + n0 and consider the rational number q = xs + | . If we £a&e n > 5, we obtain

r r r

(for: \xn - xs\ < \ gives xn - xs < \ and xn + \<xa + \\ also yn > xn + r iov n > p gives xs -ys < -r and |ys - j / n | < J implies ys < yn + £; hence z s < 2/s - r- < 2/n + \ - r, xs + § < ?/n - \.)

Actually above inequalities for n > 5 mean that: Ex < Eq < Ey which proves the result.

At this stage in our discussion we shall remember having defined the absolute value |x| in any ordered field. Thus, this concept has a meaning in E too and all the results which were proved in general are valid in E too. In particular, the "triangle inequality" is valid.

\x + y\ < W + |y| Vx,y eR.

The definitions concerning Q-valued sequences extend without difficulty to E - valued sequences: x = (xn), N -* E. This applies in particular to bounded sequences (\xn\ < M Vn G N, where M is some real number) null sequences (Ve G E, e > 0, 3n0 G N, such that \xn\ < e for n>n0) and Cauchy sequences (Ve G E, e > 0, 3n0 G N, such that \xn - xm\ < e for n, m > n0) .

We shall add here one more fundamental concept, that of a convergent sequence. Thus, if (xn) is a sequence N —> E, we say it is convergent (with

20 Advanced Caaculus

limit 1<ER) iff, 3i € R, such thatVe > 0, e € R, 3n0 6 N, suc/i that \xn-l\ < e /or all n>n0 (we write: lim xn = t or xn ^ l(n -» oo).)

Remark. The null sequences in Q are particular cases of convergent XXÎxlclxxS.. ±IltJ I1U.ll

sequences, where £ = 0. A very important property of R ("completeness") is given as follows:

Theorem 3. Any Cauchy sequence in R has a limit (is convergent) in R.

Proof. Let x € C(R) that is a; = (x»), z n € R, is a Cauchy sequence. P r o o f . ±jei x t \*\EL.) wia i is x = \xn), xn t » , is a ^ a u c i Using Proposition 6, we find, Vn € N, a gn € <?, such that

r - - < ■ ( ! < r +— ( t h a t is- r —a\< -) n n n

Let us take n0 € N, such that, given any e € Q, e > 0, we have

|x„ - xm\ <- for n,m> n0.

We obtain that We obtain that

1 e 1 \Qn - qm\ < \qn - xn\ + \xn - xm\ + \xm - qm\ < - + - + — < e for m^n^n-^ |g» - 9m | < \qn ~ xn\ + \xn-xm\ + \xm-qm\ <- + - + —<€ for m,n > m

(using also property A in R - that is Proposition 5). Thus q = {qn) e C{Q). Let £?, e R. We shall show that xn -» 25, (n -f oo). We have | i n - 25,| < | a ;„-«„ | + | 9 n - ^ , | < ^ + | « n - S g | < f + | g n - £ 7 , | f o r n > n where now c > 0 is any number in K. Therefore, it is sufficient to prove that \qn — Eq\ < | for n > m , that is - f < qn - Eq < § for large n.

(using also property A in R - that is Proposition 5). Thus q = (qn) e C(Q). T,Pt R. P » W P shall show that <r_ ->. K_(T) _> rv-,1! W P havP U - R I <T

| a ;„-«„ | + | 9 n - ^ , | < ^ + | « n - S g | < f + | g n - £ 7 , | f o r n > n where now c > 0 is any number in K. Therefore, it is sufficient to prove that \qn — Eq\ < | for is any number in K. Therefore, it is sufficient to prove that \qn — Eq\ < | for n > m , that is - f < qn - Eq < § for large n.

As e € R and e > 0, if we put § = Ez where « - («„) € C(Q), we have

I U U l l l g U 1 L > U ^ l U y V / l U J .* X XXX JJ.\i. U U U t J 1 U X I W L / U U l U l U l i KJ I . X U U U 1/ V S i ' M / — v l V / J .

Let 25, € R. We shall show that x„ - Eq(n -► oo). We have |z n - 15,| < | x „ - g „ | + | g n - 2 5 , | n where now c > 0 • 1 i _ TTD T * l x* *x •_ i32 • x x _ x l x I .. m I ^ f x*

that *_„ > r > D ^for snmp r ^ 0 in Q\ a«? snnn as -m. > m.n f= N Alsn as

Wnj t »-VV;5 w« nave \qn - c/m| ^. 2 ior m , n ^ nu max is - ^ < qn - q^ < g> m, n > ni. It follows that:

9n - 9m + 2 m > r - - = - for n > ri!, m > m0 + rii.

that zm > r > 0 (for some r > 0 in Q), as soon as m > m0 G N. Also, as (qn) G C(Q), we have |gn - gm| < \ for m, n > m, that is - § < gn - gm < | , m, n > n\. It follows that:

r r .

This means obviously that

£ ( g n i g n i . . . ) -E g + E z > 0 f o r n > n i

hence g n — E g > — - , for all 7i > 7ii . qn — h,q > — - , tor all n >ri\ .

Numbers 21

Next, from qn — gm < | and —zm < —r, m > mo, we get:

tfn -qm~ zm < - - , form > m 0 - h n i , n >nx.

Hence qn — Eq — Ez < 0 for n > ni , qn — Eq < | , n > n\. We have therefore proved: \qn — Eq\ < | for n > n i , hence: |xn — j£g| < e

for n > n\ + n.

Remark. In previous proof we have established the general fact: if (qn) e C(Q), then qn —> ^ g (n —► oo).

Thus, we are quite far in our study of numbers (real numbers). However, the equation x2 = 3, (or x2 = 2 for that matter) is still unsolved, although a solution in R does exist! (A complete proof will be given in next section.)

1.6 Real Numbers, the "Upper Bound Property", Solution of Equations xn = a.

We shall start with some preliminaries. A monotone sequence of real numbers: (#n), is defined by the property Xi < x %2 > X3 ... (decreasing sequence). We have

Theorem 1. / / (xn) is increasing and bounded above, or decreasing and bounded below, it is a Cauchy sequence.

Proof. First note that the statement has a meaning in Q as well as in E. Also, the first part implies the second: for if x\ > x^ > . . . > m, then 2/i < 2/2 < • • • < -rn, where yi = -x{. Thus |2/m - yn\ < e for n, m > n(e), that is | — Xm + xn\ < e for n, m > n(e). It will be sufficient to prove that: / / (xn) is increasing and is not in C, then it is not upper bounded.

In fact: if (xn) is not a Cauchy sequence, there is an eo E M, eo > 0 such that, Vp G N, 3 a couple np , mp in N, both > p, such that \xUp — Xmp\ > eo- Let us take p = 1; there are two numbers mi and n± in N such that 1 < mi < n\ and \xni — xmi\ = xni — xmi > eo- Next, take p = n\\ choose m,2 and 722 in N, such that n\ < m €0; we continue in the same manner: we find natural numbers m^, njt, such that 1 < mi < n\ < m2 < n2 < m3 < n3 . . . and xnk — xrnk > e0\/k = 1,2,3, — We thus obtained: xni > e0 + xTni; xn2 > e0 + xrYl2 > e0 + xni > 2e0 + xmi; #n3 > 0 + xm3 > 0 + %n2 > 3eo + x m i , In general, we see that, for any k € N, we have

Xnfc ^_ ACCQ 1 Xjxi\ '


Hence, the sequence (xn) is not upper bounded.

Corollary. If (xn) is a monotone increasing sequence in E and xn < L Vn G N, then xn has a limit, as n —► oo. 77&e same is true for monotone decreasing sequences in E, such that xn > I Vn G N.

Next we proceed to the

Theorem 2. (About nested intervals) Let Jn = [an, bn] be a closed interval on E (that is: [an,6n] = {x G E, an < x < bn}). Let us assume that h ^ h Q h • • • 2 In 2 • • •> a^d #&a£ lirnn-ôo(^n - «n) = 0. Then, there exists x G i , such that

CO

n=l

Proof. Let us make first the remark that: a^ < bj Vfc G N, j G N. In fact, we have dj < bj. Also, if k < j , a^ < a < bj\ ii k > j , a^ < bk < bj. (here we used the obvious monotonicity properties: a\ < a2 < a^ ..., b\ > h > fa > • • • , following from the inclusions i i D I2 2 h • • •)• Therefore, (an) is increasing and an < b\\/n G N. Also (6n) is decreasing and bn > aiVn G N. Let a = lim an and /? = lim bn. We have a = /?: for |a — /?| < |a: — an | + |an — M H- |^n — P\ < e if n > no(e). This is impossible when e = | | a — /?| and a ^ (3. Let us show also that: an < a < bn Vn G N (for, if, say, some ap is > a, then ag > ap > a for q > p and then \aq — a\ — aq — a > ap — a = €o, for all q > p, which contradicts aq —> a). Thus, we obtained: a G fl^Li Ai-

If 3î ^ a, Pi e f|i° n» we obtain: an < a < 6n, an < /3\ < 6n, and if, say, a < /?i, we get bn — an > /?i — a Vn G N, which contradicts:

lim (6n — an) = 0.

We have now everything needed in order to establish the fundamental "Least upper bound Theorem". We have to give first the necessary definitions:

Definition 1. A nonempty set 5 of real numbers is: upper bounded if there is L G E such that 5 < L Vs G 5; lower bounded if there is M G E such that s > M Vs G S; bounded, if 3M > 0 such that \s\ < M Vs G S.

Note. A nonempty set S is bounded if and only if it is both upper bounded and lower bounded.

Numbers 23

In fact, if \s\ < M, then -M < s < M Vs G S. Also, if Mx < s < Lx Vs G 5, we have: if 5 > 0, |s| = s < L\ and if 5 < 0, |s| = —5 < -M\\ thus, |s| < max ( L i , - M i ) .

A set which is not bounded will be called unbounded. Any number L such that 5 < L Vs G S is an upper bound for S. Any

number M such that s > M Vs G 5 is a lower bound for 5.

Definition 2. A number L is a least upper bound of the upper bounded set S if

1°. It is an upper bound for S. 2°. For any e > 0, e G M, the number L — e is not an upper bound for S.

Similarly, a number M is a greatest lower bound of the (lower bounded) set S if

1°. It is a lower bound for S. 2°. For any, e G R, e > 0, the number M 4- e is not a lower bound for 5.

Notation. L = L U • B S = sup 5 (sup - for "supremum"). M = G-L- B S = inf S (inf - for "infimum").

Again we have a fundamental result about least upper bounds and greatest lower bounds in the field R of all real numbers.

Theorem 3. (Existence of a least upper bound) / / the nonempty set S is upper bounded, there exists L G i , L = sup S.

Proof. Let us define a set U of real numbers as follows: U = {x G M, x is an upper bound for S}. Therefore U is not empty (as S is upper bounded). Take 5 G 5, put x\ = 5 — 1 and take any y\ G U. We see that x\ £ U (because s > x\ and s G 5). Let us denote by i i the interval \x\,y\\. If 2/1 = sup 5, we terminate here. If not, we consider the intervals \x\, \{x\ +2/1)] and [\{xi +2/1), 2/1].

If \(xi+yi) G £/we put x2 = xuy2 = |(xi-f-?/i). If | ( x i + y i ) £ C/we put x2 — \{x\ + yi), 2/2 = 2/1- In both cases, the interval 72 = [#2,2/2] is so chosen that X2 £ U but 7/2 € ?7. Next, again, applying the same procedure to J2 we find an interval 73 = [£3,2/3] where £3 ^ C7, 2/3 € 17 and y3 — £3 = |(T/2 — #2)-

We proceed inductively; we get intervals In = [xn,yn] where xn $. U but yn G £/; also h D I2 D J3 . . . , and yn - xn = ^{y\ - £1), Vn G N, (thus, lim

•* n—► oo

(3/n ~ En) = 0 - as 2n > 1 + n Vn G N).


Let us apply Theorem 2. We find L G E, {L} = |XLi Ai, that is:

^n < £ < 2/n Vn G N.

Then L = sup S (we prove this below!). We know in fact that L = lim xn = n—►oo

lim yn. We have yn G U Vn G N, which means: n—KX>

5 < 3/n Vs G 5, Vn G N.

This implies 5 < L, Vs G 5 (for, if so > £, => Vn > so > L Vn G N, 2/n — L > SQ — L > 0 Vn G N and accordingly we obtain that yn -f* L). Therefore L eU. Next, L = sup S (otherwise, for some e > 0, L — e eU too; however, as xn —> L, we have xn > L — e for n > n0; then xne C/ for n > no which contradicts the fact that xn £ U Vn G N). This proves the theorem.

Remark. The supremum of an upper bounded set S is unique.

For, if Li, L2 are two least upper bounds of S we get: L\ < L2{L2 is upper bound for S and L\ is a least upper bound!) and similarly, L2 < ^ i - Hence Li = Z/2-

Finally, we are now ready for the main existence results of solutions for equations xn — a (this was our main motivation so far!).

To make things more transparent we first present the particular case of the equation x2 = 2.

Theorem 4. Let A = {x G Q,x > 0, x2 < 2}; then A is nonempty and upper bounded; if s = sup A, it follows that s2 — 2.

Proof. We see that x = 1 belongs to A; also, # = 2 is an upper bound for A (if x2 < 2 then x < 2 - for x > 2 => rr2 > 4).

We shall establish following: if s2 < 2 then 5 is not an upper bound for A while if s2 > 2,5 zs no£ the least upper bound. Thus, let first assume that s2 < 2; we shall find h > 0, such that (s + h)2 < 2 (that is s + /i G A, s + h> s, impossible). We notice that s2 4- 2s • h + h? < s2 + 2s • /i + /i if h < 1. To have s2 + (25 + l ) / i < 2 we take 0 < h< |=^-.

Next, let us assume that s2 > 2; now we find k > 0 such that (5 — A;)2 > 2, A; < 5 (once we have this, using 5 — k < s and 5 = sup A, 3 £ G A such that 0 < s - f c < £ < s and then (5 - A;)2 < £2 < 2, 2 < £2 < 2, impossible).

In order to find A: > 0 such that (5 — A;)2 > 2, we write: (5 — A:)2 = s2—2sA;+A;2 > s2—2sA;; it is sufficient to have s2—2sA; > 2, that is —2sfc > 2—s2,

Numbers 25

k < ^-27^. (To say this another way, let us assume: s > 0, s2 > 2. Take 0 < k < ^ j 2 - Then we get 2sk - s2 + 2 < 0, (5 - A;)2 - k2 - 2 > 0, (s - A;)2 > k2 + 2 > 2; also, as ^ ^ < 5, we get 0 < k < s too).

Remark. The real positive solution of the equation s2 = 2 is unique, (if s2 = s2 = 2 and si > 0, s2 > 0, we obtain (51 - s2)(si + «s2) = 0, hence si - s2 = 0.)

Consider now the general case: the equation xn = a, n G N, a > 0. We have

Theorem 5. Let be A = {x G M, x > 0, xn < a}. Then A is nonempty and upper bounded. If (3 = sup A, then Pn = a.

Proof. A is nonempty: let in fact be k G N, k > K Then kn > k > \, hence pr < a, \ G A.

A is upper bounded: ifO < xn < a => x < max ( l ,a) ; in fact, if a < 1 and £ n < a =>> a; < 1; if a > 1 and rcn < a => x < a (because x>a>l^xn> an >a,xn> a).

Let us put P = sup A. Again, as in Theorem 4, we show that:

(i) if /3n < a, 3 /?' > /?, with P n < a too, thus contradicting the fact that P is an upper bound for A.

(ii) if pn > a, 3 /?' < /?, /?'n > a; then we contradict the fact that ft is the least upper bound for A.

So; let Pn < a; then a - /3n > 0 and h = ( 1 +a^C^n > 0.

Choose * G R, 0 < z < 1, 0 < z < h. Put ff = P + z. We show that (/3')n < a. We have in fact p'n = (p + z)n = /3n + n/3 n" 1z+ ^ ^ " " V + .. . + zn < pn + nPn~lz + I i ^ l / 3 ^ - 2 ^ + . . . = / ? " + ^ n / T " 1 + 2l!f l l^n-2 + . . . + i) =

/?n + z{{\ + £ ) n - /?n} < /3n + M(l + P)n -Pn} = Pn + a-pn = a. Next, let Pn > a; choose zeR, 0<z<l, zPn- npn~lz - n{n~1]pn'2z... - z

= /3n- z{(l + j3)n - Pn} >/3n-(l3n-a) = a.


Thus we have: (3n > a and for some ft > 0, ft < /?, again (ft)n > a. As (3 = sup A, 3£ G A, such that ft < £ < ft Then we obtain: ft71 < £n < /3n , a < / 3 n < £ n < a , a contradiction.

This settles the existence of a (obviously unique) positive solution of the equation xn = a, for a G 1 , a > 0, where n G N; we write x = tfa — a*.

Our main goal in the extension of Q to R (to solve those equations a;71 = a which have no solution in Q) is now attained.

1.7 Intervals on the Real Line

We present here in a systematic way the concept of an interval of R; this appears to be most useful in study of calculus. (Some intervals have been already seen previously). First, there are the following 4 kinds of "finite" (also called "bounded") intervals:

(i) [a, 6] = {x G R,a <x < &}; (ii) [a, &) = {x G R, a < x < &}; (iii) (a, b) = {x G R, a < x < b}\ (iv) (a, b] = {x G M, a < x < 6}.

The set in (i) is called a closed interval; the ones in (ii) and (iv) are semi-closed (or semiopen); the set in (iii) is an open interval. The points a and b are the endpoints of the corresponding interval. Note that (a, a) = [a, a) = (a, a] = (j), because a < a is impossible. On the other hand [a, a] = {a} (the set whose only element is a).

Next, there are four "half-lines": {x G E; x < c}; { x G l ; x < c}; {x G R; x > c}; {x G R; x > c}. Finally, R itself is regarded as an interval.

Let us introduce now two new symbols: -foo (plus infinity), and - c o (minus infinity) defined by the property - c o < x < +oo, Vx G R. With this notation we may write the half-lines above as unbounded intervals, where +oo or -oo or both appear as endpoints:

[c, +oo) = {x G R, c < x < +00} = {x G R, x > c} (c, +00) = {x G R,c < x < +00} = {x G R, x > c} (open unbounded

interval) (-00, c] = {x G R, -00 < x < c] = {x G R, x < c] (—00, c) = {x € R, —00 < x < c] — {x G R, x < c] (again, open unbounded interval) (-00, +00) = {x G R, - c o < x < +00} = R.

{Note that when +00 or —00 (neither of which is a real number) is used

Numbers 27

as an "endpoint" of an interval in E, it is always absent from the interval, therefore it is always adjacent to a parenthesis (never a square bracket).}

Thus, we defined this way 9 types of intervals of E; in particular the empty set qualifies as a "degenerate" interval; same thing for singletons.

The intervals of E (all of them) are characterized by a single property (called convexity):

Theorem 1. Let A be a nonempty subset o /E . Then A is an interval if and only if; Vx, y G A, the segment joining them is contained in A: that is

x,y E A, x <y ^ [x,y] C A.

Proof, a) One implication is quite obvious: let for instance A = [a, b). Let a < x < y < b] let x < z < y\ => a < z < b, z e A; similar reasonings hold for the other 8 types!

b) Next, assume A is nonempty and convex. There are four cases, according as A is bounded above (or not) and bounded below (or not).

1. A bounded below but not bounded above: Let a — inf A. We will show that A = (a, +oo) or A — [a,+oo). First, notice that A C [a, oo). Let then r > a. There exists x G A, a < x < r. Also, as A is not bounded above there exists y £ A, y > r. Hence x < r < y where x, y G A. By convexity => r G A. Hence, (a, + oo) C A C [a, oo).

2. A bounded above but not below; a similar reasoning gives that, with b = sup A

(-00,6) C A C (-00,6].

3. A bounded above and below; let a = inf A, b = sup A. Then A C [a, b]. Let a < r < b. There exists x e A, a < x < r and also y G A, r < y < b. Hence x < r < y, where x, y G A. It follows that r G A. Consequently we obtain (a, b) C A. Thus, A is one of the four kinds of bounded intervals.

4. A is neither bounded above nor below. Take any r G E. There are x, y in A, such that x < r and y > r. As A is convex => r G A. Hence E C A C E, A = (-00, +00).

1.8 Exercises

1. Prove that 2(1 + 2 + 3 + • • • + n) = n(n + 1), Vn G N.


2. By considering the set S = {1} U {x G N, x = 2a or x = 2b + 1} prove that every natural number is either even or odd.

3. If n2 = n • n and 2 = 1 + l(n G N), show that (n + l ) 2 = n2 + 2 • n 4-1. 4. Prove that: a • 6 G Z and a & = 0=>a = 0 o r 6 = 0. 5. Let i?(a,&) a n d -E(c,d) be rational numbers and -E(a>&) "E(c,d) = -^(0,1)• Prove

that in this case JS(0j6) = E(0,i) or E(Cid) = £(0,1) • 6. If r, s, £ belong to Q, if r > s and s > t, then r > t. 7. Let r, s, t £ Q and r < 5; it follows that r 4-1 < s + 1 . 8. In any ordered field, if 0 < a < 6, then &_1 < a - 1 . 9. In any ordered field a 4-1 > a.

10. In any ordered field we have: {x, |a; — a\ < b} — {xâ — b < x < a + b}. 11. In any ordered field, if \x-2\ < 1 then |(x — 2)(x4-5)| < 8. (in any ordered

field, the elements 2, 3, . . . etc. are defined by: 2 = 1 + 1, 3 = 2 4 - 1 , 4 = 3 + 1 and so on; we see that 0 < 1 < 2 < . . . ) .

12. Establish the following equalities: max {a, b} = | ( a + b + \a — 6|), min {a, b} = \{a + b- |a - 6|),Va,6 G M.

13. True or false? (explain): if (xn) and {xnyn) are bounded sequences in Q, then (yn) is also a bounded sequence (in Q).

14. Let (an), (6n) be sequences in Q and \an\ < bn\fn G N. Show that if (6n) is a null-sequence then (an) is also a null sequence.

15. Prove that the sequence (an) where an = | ( 1 — ^ ) Vn G N is a Cauchy sequence.

16. Show that isomorphism / , Q —► R given by /(^) = (q,q,q,...) is order preserving: p > q in <2 => /(p) > /(g) in R.

17. Let x be a fixed real number. If x < e for all e G R, e > 0, prove that x < 0.

18. Find inf and sup for the sets (in Q): { ( - l ) n } n E N , {^}neN; {^j} neN (give complete proofs!).

19. Let F c ^ b e bounded sets of real numbers. Prove that: inf E < inf F x < ?/. Prove that: sup E < inf F.

21. If / and J are intervals in R and I C\ J ^ <f> then / U J is also an interval in R.

22. True or false (explain): If 7, J are nonempty intervals such that / U J is again an interval, then 7 n J ^ 0.

23. True or false (explain): let A C R and Vx G A, 3y e A, such that x < y and [x,7/] C A; then A is an interval.

Chapter II

SEQUENCES OF REAL NUMBERS

In this chapter we again explain properties of sequences in ffi. As we have just seen, our discussion of the real number system was essentially based on the concept of a "Cauchy sequence" of rational numbers. In order to make the explicit construction of the above concept, we had to define and study numerical sequences and to explain some of their properties.

In the present chapter we continue that discussion in a slightly more methodical way.

I I . 1 Convergent Sequences

Thus, a sequence (xn) of real numbers is said to be convergent — with limit £ G R — if the following holds

Ve > 0, e G E, 3n(e) G N, such that \xn - £\ < e for n > n(e).

We write: lim xn — £ or xn —► £ (n —> oo) (if £ = 0 we have a "null sequence"). The first question to be considered: is it possible for a convergent sequence to have several limits? The answer is "no", and the proof is simple. Thus let us assume that: lim xn = £i, and lim xn — £2- We have accordingly the inequalities: \xn — £\\ < e for n > ri\ and \xn - £<i\ < £ for n>ri2 (where e > 0 is in M). Therefore, if n > max (ni, n2) (or if n > n\ + 712 which is even bigger!) we obtain

\£i - £i\ < \£i - xn\ + \xn - £2\ < 2e (We G E, e > 0) .

This inequality is not always possible (unless £\ — £2 = 0). In fact, if

29


\£i - h\ > 0 and e is taken = \\£i - £2\ we obtain \£x - £2\ < \\£i - £2\ which is obviously false. □

We say also that: (xn)J° is divergent, if $£ G E such that l imxn = £. Next, let us also remember the definition of a Cauchy sequence: (xn) G C(E)

(set of all Cauchy sequences of real numbers) if Ve > 0, 3n(e) G N, such that

\%n — xm\ < £ for n, ra > n(e).

This definition shows that the property C is somehow intrinsic to the sequence, whose terms get closer together when we go "far enough".

As we have seen in the previous chapter, the basic (completeness) property of E asserts that any Cauchy sequence in E is convergent (to a real number). (This dissuades us from trying to construct some "hyper real" numbers by repeating the construction of real numbers as equivalence classes of Cauchy sequences of rational numbers; in doing so we would obtain nothing new at all).

Note also the much more elementary converse property: every convergent sequence is also a Cauchy sequence.

For, let (xn), where xn —> £ (n —+ 00). Therefore \xn — £\ < | for n > n, and accordingly we have (Ve > 0):

\xn - Xm\ < \xn - £\ + \£ ~ Xm\ < S for, 71, 771 > ft .

Remark. It is possible and often quite useful to extend the notion of limit of a sequence to include the value +00 or —00 (which are not real numbers).

Thus, we say that: lim xn — +00 if, given any A > 0, 3n G N, such that for all n > n we have xn > A.

We also say that: lim xn = —00 if, given any A < 0, 3n G N, such that if 7i > n we have xn < A. (These sequences are divergent — see definition above).

Extending the definition for rational sequences we say that: a sequence (xn) in E is bounded if 3 L > 0, L G E, such that |rcn| < L for n > n.

Therefore, sequences which "converge" to +00 or to —00 cannot be bounded. Furthermore, as earlier proved for Cauchy sequences in Q, we can see that any Cauchy sequence (hence any convergent sequence) must also be bounded.

Next, let us dedicate a few lines to the extremely important concept of a subsequence of a given sequence. (Note that up to now we somehow managed

Sequences of Real Numbers 31

to avoid completely taking subsequences). Therefore we now give:

Definition 1. Let (xn) be a sequence of real numbers and then (rik) be a monotone strictly increasing sequence of natural numbers: ni < n^ < n% <

Define yk = xnk (k = 1,2,3...); one calls (yk) a subsequence of (xn). (This is also expressed by saying that "(xnk) is a subsequence of (#n)")-

Example. If (xn) is a sequence, (#2,#4j£6 • • •) or (x5,X6,X7,x$ . . . ) are subsequences.

Remark. If (n^) is a strictly increasing sequence of natural numbers then lim rik = H-oo.

In fact, it will be sufficient to show that Uk > kWk G N. This is obvious for k = 1; assuming that n& > k, we have rik+i >7ifc + l > f c + l.

The following simple result is often found useful:

Proposition 1. Let xn —► £ (n —» oo) and xnk be a subsequence. Then xnk —► £(k —>• oo).

Proof. Let yk = #nfc and e > 0. By assumption there is n G N, such that \%n — £\ < £ for 7i > n.

Next, let us choose I G N , such that k > k => nk > ft, (this is possible because n^ —> +00). Therefore we get that \yk - £\ = |^nfc — ^| < £ for A; > fc, that is: yk —> ^.

We shall now discuss again some typical examples of convergent sequences. The present arguments are more involved than those used for the examples given at 1.5.

(1) The sequence ( ( | ) n ) is a null sequence, in R. In fact, as seen in 1.1, we know the lower estimate: 3 n > nVn G N. Therefore ^- < ^ < e as soon as n > n0 where n0 G N and no > 7, (existence of such n0 G NVe > 0, e G R is assured by "Property A" which appeared in 1.5, Proposition 5).

(2) We explain now an extension of the previous example:

lim an = 0 for all a G R, 0 < a < 1.


In fact, if 0 < a < 1, then £ > 1, so £ = 1 + ft, where ft > 0; accordingly we can write that a = ~ ^ , ft > 0.

We must prove that Ve > 0, we get n+^n < £ for n > no- In order to do this, we shall use "Bernoulli's inequality"

(1 + x)n > 1 + nx for x>-l (Vn G N)

(this inequality is not completely new for us: we proved it in the particular case when x — ft G N (in Section 1.1), it is in fact true in the present situation: for 7i = 1 it is an obvious equality. Next, assume it is true for n = m and try to establish it for n = m - f 1 .

From (1 + x ) m > 1 + mx, by multiplication with the nonnegative number 1+s we obtain: ( l + x ) m + 1 > (l+mx)(l+x) = l+( ra+ l )x+mx 2 > l+ ( ra+l )x .

Therefore, we have , 1 + ^ w < j±^ and j ^ ^ < e if (1+nft) > j , 7ift > 7 —1, 71 > J^(J — 1). Thus, we take 7i0 > ^(7 — 1) and it follows that for n > 7i0, an <e.

(3) In our present example we show that

lim $a = 1 for a G E, a > 0. n—*oo

First, let us remember Theorem 5 in Chapter 1.6 which ensures the existence of the unique x G K, x > 0, solution of the equation xn = a where a > 0 is given. Note also that: a > 1 => a« > 1 (for, if a i < 1 =* (a- ) n < l n , a < 1); also, if a < 1 =» an < 1; (for, if a« > 1 =* (a~ ) n = a > l n = 1).

Now, if we are in the case a > 1, then | \ /a — 1| ^Va" — 1 and it will be sufficient to prove that y/a — 1 < e for 71 > n0(e), that is: a < (1 + e)n for n > 7io(e).

Actually, again by Bernoulli's inequality we have (1 + e)n > 1 + ne. Also, we get 1 + ne > a if n > g^, and 710(e) > g^ will do.

In the case of a < 1, \tfa — 1| becomes 1 — {/a; we look now for an estimate I- tya<e, l-e<tya. Note that 1 - e < ^ Ve > 0. Thus, it is sufficient to establish: ^ < tfa, that is a > (^ j ) 7 1 . Again, (1 + e)n > 1 + ne; hence, if we get, (1 + ne) > ^ we are done. Take accordingly no > - ( 7 — 1).

(4) Finally, we shall establish that: lim yfn = 1. n—+00

Proof. We now have tfn > 1 Vn G N. Given e > 0 we look for the inequality tfn — 1 < e which will follow from (1 -f £)n > n.


By the binomial theorem we have: (14- e)n = 1 + ne + n 'n2 x'e2 + . . . (other

positive terms). Therefore (1 + e)n > ne + n ^ n ~^£ 2 . It will be sufficient accordingly to get: n£+n(n~1' e2 > n, that is £+Ik=^£2 >

1, T 1 > ^ , n > ^ ^ ^ + 1. Thus, we can take n0(e) > ^ ^ + 1.

II.2 Fundamental Properties of Convergence

In this section we shall explain some basic properties of convergent sequences as they appear in the following theorem.

Theorem 1. Let (an), (bn) be two sequences in E, such that lim an = A and lim bn = B; then lim (an -f bn) exists and equals A + B.

(in short, lim (an + bn) = lim an 4- lim bn).

Proof. By assumption, for any e > 0, there exist natural numbers no and n\ such that

£ £

| a n - A | < - i f n > n 0 , \bn - B\ < - i f n > n i .

Let 7i2=max (no,ni). If n > ri2 then

|(on + 6„)- ( i4 + B) | - | (o n - i4 ) + (6 n - J3) | < | a n - A | + | 6 „ - B | < | + | = e. Hence, (an -I- 6n) converges to A -h B.

As an important preparatory step we show that

Proposition 1. A convergent sequence is bounded.

In fact we already pointed out at this result in Sec. 1 of this chapter; the proof there followed the implication arrows.

Convergent sequence => Cauchy sequence => bounded sequence.

However there is also a direct proof, which makes no use of the concept of Cauchy sequence.

We have, if (an) is a sequence and an —> A, that \an — A\ < 1 for all n > no, where n0 is some natural number. Hence \an\ = \an — A + A\ < \an — A\ + \A\ < 1 + \A\ if n>n0.


Let K = |ai | 4- |a2| + . . . | a n o_i | + 1 + |A|. Then \an\ < K for all n > 1. This completes the proof.

Theorem 2. Lei (an) , (bn) be sequences in R, swc/i £/m£ an —> A and bn —► 5 . 7%en an • 6n —► AB.

(In short, lim (an • 6n) = (lim an) • (lim 6n).

Proof. By previous proposition, \bn\ < K for all n > 1 (for some K > 0). Let £ be any positive number; there exist positive integers no and n\ such that

\an~A\<2K i f n - n ° ' l ^ - ^ l < 2 ( U | + i) i f n ^ n i -

Let ni — max(no,ni). Then, if n > n2, we have:

\anbn - AB\ = \{anbn - Abn) + {Abn - AB)\ < \bn\\an -A\ + \A\\bn - B\

This completes the proof.

Remark. From this Theorem it follows that: for any real number A,

lim(Aan) = A lim an .

In dealing with our next property of convergent sequences, we need the preparatory

Proposition 2. Let (bn) be a real sequence, lim bn = B and B ^ O . Then, there exists n* G N such that bn ^ 0 for all n > n*; in fact, \bn\ > \\B\ for all n>n*.

Proof. Let us take e = | | J B | and n* such that \bn — B\ < \\B\ if n > n*. Then it results:

|6n| = |(6n -B) + B\> \B\ - \bn -B\> \B\ - ±|B| = \\B\ > 0 .

Hence, the following holds true:

Theorem 3. Let (an), (bn) be real sequences, an —> A, bn —> B and B ^ 0.


Then, if bn ^ 0, the sequence j*- converges to ^ .

Remark . From Proposition 2, it will follow that bn ^ 0 for n > n*.

Proof. If we use Theorem 2, we only need to prove that ^ > -^(n —► oo). We have | £ - £ | = i^ f f f < ^ . ^ forn > n*, therefore | £ - ± | n*. Now, if e > 0 is given, we have \bn — B\ < £ ^p- for n > n\ and, accordingly

\K ~ i l < ]W^~ = £ for n - m a x (n i 'n*)-Our next result is

Proposition 3. If (xn) is convergent to x, it results that (\xn\) is convergent to \x\.

In fact, we use the inequality: \\xn\ — \x\\ < \xn — x\ and then the result is obvious.

Of importance is the following:

Theorem 4. Let (an), (bn) be convergent sequences, and an < bn for n sufficiently large. Then liman < lim 6n.

Proof. Let us assume that an < bn for n > no, and that A = lim an > B = lim bn. Then bn — an > 0 for n > n0 , and A — B > 0. We also have B — e < bn ni , A — e < an < A + e for n > n2 . Therefore bn — an max (ni, n2) , also 0 < frn — an max(n0, n i , n2).

On the other hand, B - A < 0 and B - A + 2e < 0 for 2e = ^ - . We obtain 0 < 0, a contradiction.

Next, we shall prove the following:

Theorem 5. Let (an) be a convergent sequence, such that an > OVn G N and lim an = a. T/ien; VA; G N, we have lim a£ = a*=.

Proof. First, let us remark that a > 0 (it follows from Theorem 4). If a = Owe have 0 < an < £k for n > no, and this entails: 0 < ak < £ for n > no.

If fc = 2 and a > 0, we have the simple reasoning: ^/a^ — y/a — j£^\-\

therefore \y/a~n~ — y/a\ < 'aV-al and this is less than e for \an — a\ < Sy/a.


For the general case (k G N, k > 2) and a > 0, we start with notations: y ^ = £n, -â = £. (we see that £n > 0, £ > 0).

This gives: £ - £k = an - a = (£n - Otf*"1 + # " 2 • £ + • • • + £*_1), and accordingly we obtain

\fa-VZ\ = \£n-£\ = -7Z=I \an - a\ k-2c , tfc-1 K^ + - ^ + .-.e

Next, we look for a lower bound of the expression: ££_1 + ££~2£ + • • • Z,k~l • The "simplest" way: ^ n" 1 + ££"2£ + • • • + ^ _ 1 > f*"1 = a~^. Thus the estimate

| \ ^ - \ / a | < - T 3 r | a n - a | , V n G N. a fc

We have \an — a\ < e • a~fc~ for n > n0, hence \tya^ —y/a\ < £, for n > n0. We terminate this section with the study of the following:

Example. Let xn — -Jp\ where P(n) = a0nk + aink~l + . . . + dk, Q{n) = b0nl + b^'1 + . . . bt, and Q{n) / 0, Vn G N; a0 + 0, 60 7 0.

(i)Iffc = ^wehave: P(n) = nfc(a0 + ^ + .. . + {£), Q(n) - nfc(60 + ^ + . . . £fr), hence we obtain

P(n)=za0 + ^ + ...^t Q(n) b0 +% + ...&'

Next we use Theorem 3 and Theorem 1 (this one extended to finite sums!), as well as the obvious limits: ^ —> 0(n —* oo), Vp > 1. It results: l i m EM. = M

n _ o o Q(n) bo*

(ii) If * > *, we have: %& = n / c - ^ » + | + - g = n*-* . ^

Now obviously lim nk~£ = +oo, lim a n = f11

Next we have

Proposition. Let (An), (/xn) 6e sequences in R, and An —► oo, /xn —> \i. Then if \i > 0, An • fin —■> oo; if fi < 0, An • //n —► —oo.

In fact, in first case we have: VA > 0, 3n G N, such that An > A for n > n. Also /x — e < fj,n < (1 + e for n > ni , and with e = ^, jin > |/x, n > n2. Therefore An • \in > \\x • A for n > max (n,ri2). Finally, taking any Ai > 0, choose A such that \\i • A > Ai.


In the second case (/i < 0), we have: An > A, for n > ft and /xn < /x + e, for n > ni . Take now e = — |/x, and obtain [in < |/x, n > n2. Consequently: AnMn < !Â> n > max {n,n2). Finally, for any Ai < 0, |/xA < Ax if A > ^ A 1 .

Using this results we get, for k > £, lim ^ 4 = +00 if a0 • b0 > 0 and

n 1^5{=} = - o o i f o o 6 d < 0 .

(iii) The last possibility arises when k < £. In this case, again 5^4 = nk~£an, and n*~* --> 0 while a n —► §&. Therefore now we get lim ^ 4 = 0.

Remark 1. Assume (an) —► 0. Then | a n | < £ for n > n0 and r^-r > - > A if e < -£- and 71 > n0 (e). Therefore T^T —► +00.

However, we cannot infer that -£- —► +00 or that -^- —» —00. (See for example: an = (1, - 1 , §, - I , . . . 1, - i . . . ) ) .

Remark 2. If a n —► 0 and /Jn -» 00, the sequence (a n • /3n) can converge to a finite limit or to +00.

(oin = ;jr, /?n = n, gives an /3n -► 0; an = £, /3n = n2 gives an /3n = n -► 00; «n = f , Pn = n gives an(3n = B -+ B).

II.3 Monotone Sequences

Monotone sequences of real numbers have been introduced already, in connection with the "upper bound property".

We previously proved that an increasing upper bounded sequence is a Cauchy sequence, hence it has a limit.

We also used this result in order to prove that any nonempty set of real numbers which is upper bounded has a (unique) least upper bound.

It is important to point out the following:

Theorem 1. If a monotone increasing sequence xn is bounded above and £ = sup{xn} ({xn} is the set of values take by (xn)), then lim xn = £.

In short: if rri < X2 < . . . xn... < L, then lim xn exists and = sup {xux2,xn,...}.

Proof. Let £ — sup {xn}. Then xn < £ for all n and moreover, for any


e > 0, there exists xno > £ — e. It follows that xn > xno > £ — e if n > no, that is

\xn — £\ = £ — xn < e for n > no .

Similarly one can prove the

Theorem 2. If a monotone decreasing sequence {xn) is bounded below, then it converges to inf {xi,£2>... x n , . . . } .

Remark. A classical example of a convergent monotone sequence is the sequence (xn) where xn = (1 + ^ ) n .

We shall prove that it is a strictly increasing and upper bounded sequence. The limit: lim (1 + ^ ) n is denoted as "e". It appears in many fundamental formulas of analysis.

(i) First, the monotonicity relation: xn < #n+i> Vn £ N, can be established in a rather elementary way, by simple use of a strict form of Bernoulli's inequality. (see below) This inequality in fact implies that, for n > 2

' n 2 - l \ n A n * 1 , 1 n-1 = 1 - — > l - n - - j = l - - = . \ nz J nz n n

Then consider the quotient ^ = ( g f ^ r = ^ S ^ - ^ = {2J^f~ ' i 2 ^ 1 > i^) * ^=1 = !• T h u s xn > *n-i for n = 2 , 3 , . . . .

(Here we used the "strict" Bernoulli's inequality: If x > — 1 and x / 0 , then (1 + x)n > 1 + nx for all n > 2: again, it is true for n = 2; assume it true for n = m; thus (1 + x ) m > 1 -f rare. Multiply by (1 4- x) which is > 0. We obtain: (1 + x ) m + 1 > (1 + mx)(l + x) = 1 + (m + l)a + mx2 > 1 4- (m + l):r if r r / 0 ) .

(ii) Second, we establish the upper boundedness condition: xn < 3 for all n e N. We use the Binomial theorem and obtain that

/ - . i \ - . i nyn — i ) / l >

3) n(n - l)(n - H l ) / l V n\ ( 1 A;! Vn / " ' n! Vft

K'-=)+-+sHX1


+ ... + J .(1 .I)(1 .2 n\\ nj\ n

, , 1 1 1 < 1 + 1 + 2 ! + - * ! + - + ^!

We also have the obvious lower estimate: n! = 1 • 2 . . . n > 2 n _ 1 , n G N, hence x n < l + H - | + . . . yhr + • • • + 2 ^ r = 1 + T ^ r < 1 + x = 3, Vn G N. (Here

we applied the elementary formula: 1 + x + x2 + .. .xn = 1~1X_X which can

be obtained as follows: put s = 1 4- x 4- . . . xn. Thus xs = x 4- x2 4- . . . xn+1, hence (1 — x)s = 1 — xn+1).

Let us now note a slight extension of Theorem 1.

Theorem 3. A monotone increasing sequence is either convergent (to a real number) or is convergent to +oo.

Proof. Let (xn) be a monotone increasing sequence. The case when it is upper bounded is covered by Theorem 1. Suppose then that it is not bounded from above. Therefore, for any M > 0, there is n0 G N such that ano > M. It follows that an > M if n > n0. Hence, in this case, lim an = 4-oo.

Example 1. Consider the sequence: xn = (1 4- ^ ; ) n . We have xn = 2/n where yn = (1 4- ^ ) 2 n - Actually, (yn) is a subsequence of the sequence ((1 4- ^ ; ) m ) , hence lim yn = e. From Theorem II.2.5, we derive lim y£ = lim xn = yfe.

Example 2. Consider the sequence: xn — (1 4- ; ^ ) 3 T T ' • We write rrn = [ ( l + ^ 2 ) n + 2 ' ( l + ^ 2 ) - 2 ] 3 .

Then (zn) = ((1 + ; r b ) n + 2 ) i s a subsequence of ((1 4- ^ ) m ) , and accord-

(use also Theorem H.2.2.).

ingly lim zn — e. Obviously lim (1 4- r ro ) 2 = 1. We obtain: lim xn = e3

Example 3. We examine the sequence (xn) where xn — (1 4- J ) n . If we know beforehand the existence of lim xn = L, we could take the

n—►oo subsequence (ym) — (x2m) and then we would have:

L = lim [ 1 4- - ) = lim [ 1 4- — ) = lim n—*oo \ u l m—►oo y 2771J ra—>oc m

= e2


Keeping in account our present knowledge, we note^1) the simple equality:

-!)*-H)"(^)*-K)"(^r(^r-From this relation we easily find that L = lim (1 + J ) n = e • e • 1 = e2 as

guessed previously.

Example 4. Let us define a sequence (an) in the following way:

a i = \ /2, a2 = V 2 + \/2, a3 = Y 2 + V 2 + v ^ , . . . , a n + i = ^ 2 + an .

We shall see that this is a bounded increasing sequence of real numbers. We have an > OVn G N; next, a\ < 2 and if we assume that an < 2 it

follows that an+1 = y/2 + an < y/2 + 2 = 2. Thus, an < 2 Vn G N. Then, we also have: a\ < a2 < . . . an < a n + i In fact, a2 > A/2 = ai

is obvious. If we assume that ftn+i — &n ^ 0 we derive: &n_f-2 — n+i — (2 -f a n + i ) - (2 + an) = an+1 - an > 0. Hence a n + 2 - a n + i > 0 too.

Therefore lim an = L > 0 will exist. From relation a n + i = y/2 + an , we get L = \/2 + L, hence L2 - L - 2 = 0. As L > 0, it follows that L = 2.

Example 5. Let ai > 0 and then define a sequence (an) by the inductive relations: a n + i = y ^ , n = 1,2, We shall prove that lim an exists and equals 1.

In fact, let us assume first that a\ > 1; it follows that an > l V n G N, and also that a2 = yfai < ai , and a n + i < anVn G N (otherwise one would have a n + i > an for some n G N, y ^ > an, an > a2, a n ( l - an) > 0, an < 1).

We have therefore a\ > a2 > . . . > an . . . > 1 Vn G N. Thus lim an = L > 1 will exist. It results also that L = v T , L2 = L, L = 1. Let now ai < 1. Denote bn = ^T' w e S e t ' 6 i > 1 a n d 6^+i = ^ T = vfc = y ^ T = ^ ' n = 1,2,. . . .

From above it follows: bn —> 1 and accordingly, an = ■£ ► 1 too.

Example 6. Find lim (2n + 1)^HFT (if it exists!). It is a subsequence of n—*oo

the sequence (n)« (rik = 2A; + 1), hence, it —► 1.

Example 7. Find lim $n + p — V^)> (P a n d & a r e natural numbers);

(Dlf we are clever enough; otherwise we do some research: library, friends, etc!


we put an =ky/n + p, bn =k^/n, thus a£ — &£ = P- O n t n e other hand a n - bn = (an - M ^ n - 1 + a*~2fc + a£_362 4- . . . &£_1), therefore we obtain an -bn = p[(n + p)~r~ + (n + p)~^n*" + (n + p)~*~n^ + .. . n - ^ " ] - 1 which obviously —► 0 as n —► oo.

Example 8. Let us assume: an < bn < cn, Vn G N and an —► L, cn —> L, it follows: 6n —► L too (this can be called the "sandwich rule").

In fact, Ve > 0, we find n(e) such that L — e < an , n > n, cn < L + e, n>n. Therefore we have: L - £ < bn < L + £ for n > n, which means 6n —► L.

Example 9. Let (an), (bn) be two sequences, where an -^ a and 6n —> +oo. It follows that (an + 6n) —► +oo.

In fact, take any M > 0, we have 6n > M for n > n, take also e = 1, we get an > a — 1 for n > l\ thus, for n > max (n,I) we obtain an + bn > M -ha — 1. Finally, if Mi > 0 is given, take M such that M + a - 1 > Mx.

Example 10. Let (an)i° be a sequence, such that \an — a n+i | < Acn, Vn G N with some c e (0,1). Show that it is a convergent sequence.

We shall see that it is a Cauchy sequence, let us evaluate \an — am | , assume m > n and put p = m — n; we have

| ^n — Gn+pl < |^n ~ &n+l \ + | a n + l ~ «-n+2|

+ . . . 4- | a n + p _! - an+p\ < A(cn + c n + 1 + . . . + c7 1^"1)

1 - c 1 - c

as cn —► 0, we get Ve > 0, j4^cn < e if n > n(e) hence \an - a n + p | < e for n > n(e), p G N.

Example 11. Let us take ax < a2 and define an by the recursion a n + 2 = \{an + a n + i ) . Prove that it is a convergent sequence.

We shall establish by induction the estimate | a n + i — an | < ^ r ^ 1 , Vn G N. Obvious for n = 1, we next have an+2 = | ( a n +i + a n) , hence a n + 2 — &n+i = | ( an - On+i), therefore | a n + 2 - a n +i | = | | a n - a n +i | < 2ir(a2 - ai) .

Finally, we apply above result (Example 10).


II.4 Theorem of Bolzano-Weierstrass (the B-W Theorem)

A famous result in analysis is the following simple statement:

Theorem 1. Every bounded sequence of real numbers has a convergent subsequence.

A fairly direct and somewhat surprising proof can be obtained using the following.

Main Lemma. Every sequence in E has a monotone subsequence.

The proof is based on the concept of a "peak point" of the given sequence. Precisely, the element a,k of the sequence is called a "peak point" if ak > an, Vn>k.

We shall now distinguish the two possibilities. (1) There is an index n G N, such that, Vn > n, no an is a peak point

(hence, there are only a finite number of peaks, and none after n) ("peaks terminate ultimately").

(2) For any k G N, 3n^ G N, n^ > A:, such that ank is a peak point ("peaks appear frequently").

Let us consider the first situation. Take n\ > n, such that an i is not a peak; therefore, 3n2 > n i , such that an2 > ani. Again, aU2 is not a peak, thus, 3ns > n2, such tht an3 > an2; and as we continue the same way, we obtain the subsequence ani, an2, an3,... such that ani < an2 < an3 . . . < . . . (an increasing sequence).

Next, let us examine the second possibility. Take n\ G N, ani is a peak point. Then, 3n2 > n i , an2 is also a peak

point; 3^3 > ri2, anz is also a peak point and so on. We obtain a subsequence of peak points, ani, an2, an3 ... and because of the definition of peak points we get ani > an2 > an3 > . . . , a decreasing subsequence.

Now, the proof of B-W theorem is immediate. If (an) is a bounded sequence of real numbers, it will possess a monotone

subsequence which is still bounded, and, as seen previously (§ 6, Chapter I), any such (sub)sequence is convergent.

Remark. If we have beforehand that a < an < /?, Vn G N, the limit of a convergent subsequence belongs to the same interval [a,/?].


II.5 Limsup and Liminf of a Bounded Sequence

As easily seen, a bounded sequence is not necessarily convergent. However, as indicated in B-W theorem, it has convergent subsequences.

Actually, there can be various subsequences with different limits. This situation is best handled by means of the concept of limit superior and limit inferior of any bounded sequence.

Let therefore (an)nGN be an arbitrary bounded sequence of real numbers. For each n, let An be the set { a n , a n + i , a n + 2 , . . . } .

Then, obviously An is again a bounded set in R; let us define numbers bn

and cn by relations

bn = sup An = s u p { a n , a n + i , . . . } , cn = inf An = inf {a n , a n +i , •..} •

Because of the inclusions: A\ D A 6n+i and cn < cn + i Vn G N.

Therefore we have two monotonical sequences: (&n)n£N which is decreasing and (cn)nGN which is increasing.

On the other hand, due to boundednes of the sequence (an), there is M > 0 such that -M < an < M, Vn G N.

This obviously implies relations: bn < M and cn > - M Vn G N, that is

- M <cn<bn<M, Vn G N .

We got this way two monotone bounded sequences, they both have limits. We denote

liminf an = lim cn, limsup an = lim bn ,

If follows accordingly

—M < liminf an < limsup bn < M

Examples . (i) For the sequence 1, - 1 , 1, - 1 , . . . = ( ( - l ) n + 1 ) , the set An is {-1,1},

Vn G N, so bn = 1 and cn = - 1 for all n, therefore limsup an = 1, liminf an = - 1 .

(ii) For the sequence 1, - 1 , 1 , 1 , 1 , We have An = {1} for n > 3, so bn = cn = 1 for n > 3, therefore limsup an = liminf an = 1.


At this point we can characterize convergent sequences using the above introduced concepts. We have

Theorem 1. Let (an)neN be a sequence such that lim an = L. Then (an) is bounded and limsup an = liminf an = L.

Proof. We only have to prove the equality. Taking any e > 0 we know that L — e < an < L + e for n > n{e) (where ft G N). It results: L — e < bn < L + e for n > n(e), hence L — e < lim bn < L + e that is: limsup an = L.

In a similar manner we obtain that liminf an = L. Conversely we also get

Theorem 2. Assume that (an)neN is a bounded sequence, such that limsup an — liminf an = L. It follows that lim an exists and = L.

Proof. Remember that bn = sup {an, a n + i . . . } and note that L = lim bn = inf bn.

Therefore, given any e > 0, 3 n\ G N, such that bni < L H- e. Then we see that

a>k < bni < L + e VA; > ni .

In a similar manner we find n2 G N, such that cH2 > L — e; this implies that

dk > cn2 > L — e for all fc > ri2 .

Thus, for A; > max (711,712) we get L — e < ak < L + e. We next establish the special situation of limsup and liminf of a bounded

sequence with respect to all possible limits of its subsequences. We have precisely the following

Theorem 3. Let (an)n6N be a bounded sequence in E. Define S = {x G E, 3(anfc), subsequence of (an)nGN, such that ank —► x}. Then limsup an and liminf an both belong to S and they are the largest (respectively the smallest) element of S.

Proof. (a) Let us prove for instance that limsup an G S. Let be (3 = limsup an-

Therefore /3 = lim bn where bn = sup ( a n , a n +i , . . . ) . Then, given any e > 0, we find bn < (3 + e for n > n, hence an < 6n < /3 + e for n. > ft.


Next, we note the following: VA; G N, 3nk > k, and ank > {3 — e. (In fact, otherwise, 3k0 G N such that Vn > fc0, we get an < f3 — e\ this gives bko+i < ft — £ < Pi contrary to relation f3 = inf bn).

Choose now e = 1; we have an < f3 + 1 for n > ft, and 3fli > 1, with an i > (3 — 1. For 7ii=max (n,ni) we get therefore

)S — 1 < o n i < /? + 1, and ni > 1 .

Next, with £ = | , choose n2 > n\ such that

Continuing in this way we construct a subsequence (anfc) such that

d-\<ank<p+\

Therefore aUk —> /?, hence (3 e S.

In a similar vay we see that a = liminf an also belongs to 5. Thus, the (a)-part is verified.

(b) The second part of the theorem says that a < x < (3Vx e S.

Let us prove, say, the inequality x < f3. Again we use the inequality (given any e > 0)

an < (3 + e for n > ft .

Now, if anfc —► x, we have antc < /? + e, hence a; < (3 + e (see Theorem 2.4 in this chapter). Hence x < (3 (for £ > / ? = > £ > / ? + £: for £ < r r — /?, a contradiction).

A similar proof shows that a < x, Vx G 5.

Our last result here gives a (slightly different) characterization of limsup and liminf for a bounded sequence.

Theorem 4. Let (an)n€N be a bounded sequence. Then (3 G E is the limit superior of (an) iff

(1) Ve > 0, 3n0(£) G N such that an < f3 + £ forn> n0(e). (2) \f£ > 0, VA; G N, 3nfc > fc, nfc G N and ank > /3 - £. Also a G E zs the limit inferior of (an) iff (3) Ve > 0, 3 n0 G N, snc/i £/ia£ an > a - e for U>UQ.


(4) Ve > 0, \/k G N, 3nk > k, nk G N and anjc < a + e.

Proof. We have seen in the proof of Theorem 3 that if P = limsup an , then (l)-(2) hold true.

Conversely, let us assume, say, that P G K satisfies conditions (l)-(2). Again, as in the proof of Theorem 3, we find a subsequence (ank) such that P = \imank. Thus, p e S.

Next, we shall see that /3 is the largest element in S. Otherwise, 3 /5i G 5, Pi > p. A certain subsequence (amfc) of (an) is convergent to P\. Now, if 0 < € < Pi — /?, we have P + e < Pi and amfc > P + e for k > k0. This property contradicts assumption (1) above. Therefore, we have established that from assumptions (l)-(2), it follows that P is the largest element in S. Using Theorem 3 we find that P— limsup an.

The proof for liminf an equivalent with (3)-(4) is similar and is left to the reader.

II .6 Exercises

1. Show that the sequence (an) where an = yjri2 + 1 — n is a null-sequence. 2. Let a£R, \a\ < 1. Show that lim n\a\n = 0. 3. If (an) is a sequence of real numbers, it is unbounded if and only if there

exists a subsequence (anfc) such that \ank\ > k, Wk G N. 4. Let (an) be a divergent sequence, then, V o E l , 3 e > 0 and a subsequence

(anfc) such that \ank — a\ > £, Vfc G N. 5. Prove that a sequence (an) in R is bounded if and only if every subsequence

of (an) has a convergent subsequence. 6. If (an) is a monotone sequence in R that has a bounded subsequence, then

(an) is convergent. 7. Let —+ 6 and an < 6n, "frequently" (that is, Vp G N, 3n > p,

with an < bn). Then a < b. 8. Show that the sequence (an) where an = (1 — | ) (1 — | ) . . . (1 — ^ - ) is

convergent. 9. Let (An) be a sequence of nonempty subsets of M, such that:

(i) Ai D A2 D A3 D . . . and (ii) Vx,y G A n , | x - 7 / | < ^. Let (an) be a real sequence, where an G An , Vn G N. Show that (an) is convergent.

10. Consider the sequence: | , | , | , | , - - . | , ^ j , — Show that its limit superior is 1 and its limit inferior is zero.


11. Find limsup an and liminf o„ for the following sequences: an = ( -1 ) " + i , an = I + L $ I , an = [n + ( - l ) n (2n + 1)]±, an = (_ l )n ( 2 + I).

12. True or false (explain): If (an) is any real sequence, then the sequence (1 + \an\) l has a convergent subsequence.

13. Construct a sequence that has convergent subsequences converging (respectively) to 0 and 1.

Chapter III

INFINITE NUMERICAL SERIES

Infinite series is one of the fundamental topics in analysis. The present chapter touches on only the underlying basics of series of numbers, a natural generalization of the concept of finite sums of real numbers.

III. l First Definitions, Convergence, Divergence

Let us consider an infinite sequence (an)°° of real numbers. We can form "partial sums" sn in the following way:

si = ai, 52 = ai + a<i, • • • 5n = ai 4- a2 H an .

If lim sn exists (= S) we say that we have an (infinite) convergent series n—*oo

oo CO

y] a>k , and that S = V^ a^ . fc=i fc=i

CO

We note the simple (necessary) condition for convergence of the series ^ ak-k=i

CO

Proposition 1. If ]T a^ is convergent => lim a^ = 0. fc=i k^°°

In fact, we note that a\ = $i, ak = Sk — Sk-i for k > 2. If lim s^ = 5, it k —*-oo

follows lim 5fc_i = 5 too, hence lim a^ = lim (s*; — Sk-i) = 0. A:—►CO fc—► CO fc—+CO

CO

Example 1. The series ^2 xk •> x a r ea^ number, is called the geometric fc=i

series.

49


We shall see that it converges for x such that \x\ < 1 and does not converge (is "divergent") for all remaining real numbers.

This is done via an explicit computation of the partial sums sn (which is not always possible!).

Thus, if sn = x1 + x2 H xn, then xsn - sn = x2 + x3 H xn + x n + 1 -(x + x2 -\ xn) = xn+1 —x = x(xn — 1); sn(x-l) = x(xn — l); if x ^ 1 we get sn = £i£_zi). On the other hand, obviously, for x = 1 we get sn = n. We know that for \x\ < 1, lim xn = 0; therefore lim sn = T ^ - , Ircl < 1. For x = 1,

n—►oo n—»-oo A x

5n —> +oo. In general, if |x| > 1, |xfc| = \x\k > 1 and apply Proposition 1. CO

Again, using Proposition 1, we see the divergence of the series ^ A;, CO

Next, we remark: CO

The condition: lim a^ = 0 is not sufficient for the convergence of JZ a&-

Maybe the most famous example of this situation is the so called harmonic series:

°° 1

We can establish this interesting (and not obvious) fact in at least two ways. ( l ) W e h a v e S 2 n - S n = ^ T + ^ + - - - ^ > n - ^ = | , V n € N .

Therefore, the sequence of partial sums (sn)neN is n ° t a Cauchy sequence. (2) We can get an interesting "lower bound" for the subsequence (s2n)neN-

Precisely, we have:

^ = E ^ = (1 + J) + u + 7) + ( ^ ^ ^ ^ + 1 1 1 1 k V " 2 7 ' I 3 ' 4 7 ' \5 + 6 + 7

k=l \ • \ x \ _ _ M I 2 4 271-1 _ n

+ ' 2 " - 1 + 1 + 2n~1 + 2 + " , 2 r l J > 2 + 4 + 8 " h " , 2n ~ 2

Therefore the subsequence ($2™)i° is unbounded and this precludes convergence of the original sequence (sn).

Next, we can point out two simple properties of convergent series.

CO CO CO

Theorem 1. / / 53 a^ = S and 53 && = T, the series 53 [ak ± &fc) are

Infinite Numerical Series 51

oo oo convergent too: £ (ak ±bk) = S±T. Also, for any A G R, £ (Aa*) = AS.

k=l k=l

In fact, we simply note that

(ai ± 61) + (a2 ± 62) + • • • (on ± 6n) = (ai + a2 + • • • an) ± (61 -I- 62 4- • • • bn) and Aai + Aa2 H Xan = A(ai H an) .

00 For series with nonnegative terms (J2 ak where ak > 0) we shall indicate

k=l some very simple, sufficient conditions for convergence and divergence.

00 Theorem 2. Let us assume: 0 < ak < bk, V/c G N. T/ien i/ ]T) && 25

k=l 00 00

convergent, the series ^ a^ is also convergent. If ^ a& is divergent, the k=l k=l

00 series ^ 6 is also divergent

k=i

Proof. If sn = a>\ + --an, tn = h + '-bn, we see that, because of conditions ak> 0, bk > 0, V/c G N, (sn) and (tn) are monotonically increasing sequences. For such sequences upper boundedness is equivalent to convergence. Hence, if (tn) —> T, we have sn < T Vn G N, hence lim sn = S < T exists.

n—►oo

Also, if (sn) is unbounded, and because tn > 5nVn G N, we find that (£n) is also unbounded. □

Theorem 3. Let us assume: ak > 0 Vk G N, bk > 0Vk G N and lim f*- = k-+oo k

00 00

L > 0. It follows that the series ]T) ak and ^ 6A: are both convergent or both k=i k=i

divergent.

Proof. In fact, we have f < f*- < ^ for k > k0 (with some k0 G N). Hence a& < ^bk for k > ko and a^ > |-&fc for k > ko.

OO

Then use Theorem 2 above and the (obvious) fact that the series ^ ck fc=i

00 and ^ Cfc are either both convergent or both divergent. □

k=ko

oo Example 2. Consider the series ^2 77p due t o t n e inequality 4= >

A : = l

^VA; G N, we find that the series is divergent. Note also the lower bound for


the partial sums

S-Tl+T2+-Tn>Tn=Vi

which shows directly tha t the sequence (sn)i° is unbounded, hence divergent.

oo E x a m p l e 3 . Consider the series £ £ where p e R and p > 2.

I. i

First consider the case p = 2, and note the estimate

(fcW<fc(fcVT)'Vfc = 1 ' 2 ' " ' -OO

Next, we shall establish convergence of the series £ -g^ryr, we can in fact fc=i

explicitly compute its partial sums sn: T T T ■. 1 1 1 , - ~ We have - ^ ^ = \ - j ^ , k = 1,2,---. Therefore ^ + ^3 + • • • +r> -

OO

r - 5 + | - I + " ^ - ^ T = 1 - ^ T - I t follows: £ WH) is convergent, OO

its sum S = lim sn = 1. Using this result we find convergence of f] X, hence n—»-oo 2 / L r LAJ o

OO

of E ^ - Finally, if p > 2 we get -^ < £ , V A; G N, hence the convergence of I

U l £_^ -T2 . Xl l lCUlJ' , IX / / ^ ^ WC £ 1

OO

£ ^r for all p > 2. fc=i

I II .2 T h e R a t i o T e s t and t h e n t h R o o t T e s t

We shall explain now two classical "tests" for convergence or divergence of We shall explain now two classical "tests" tor convergence or divergence series with nonnegative terms, due to d'Alembert and Cauchy.

T h e o r e m 1. (The Ratio test) . Suppose an>0 for all n. If lim sup ^ ± i < 71—>00 n

OO

1, then the series 5 > n is convergent. J / l imin f ^ - > 1 then the series is i n

divergent.

Proof . We shall apply Theorem 5.4 in Chapter II. If lim sup ^ * ± = q n0. VVV ULfciVV "l 7 V£ ^ ^ *i ^ -1- t * J L A , J - v * ^ AXJ.J.VJ. fl/(J V_ i. N U U ^ l l U U U / U \ l£ JLW1 ft/ ^ _ I t/(J .

Therefore we get ano+i < <?'ano; a n o + 2 < q'ano+1 < q2anQ\-- and anQ+v < ±iitJiciuit; wt; get txnn_}_x \ 1/ ( q'Pano Vp € N.


As 53(tf')r is convergent, we obtain 53 an0+P convergent, hence 53 a n is 1 P=I 1

convergent. In the second situation: liminf ^s± — r > 1, apply (3) in Theorem 5.4 (II)

and for r > r' > 1 we find n\ such that ^ ± 1 > r' for n > n\. As previously we find that, \/p G N

ani+p > (r')vani

CO CO

and because of divergence of the series 53( r ' )p w e obtain that 53 a n is also 1 1

divergent.

CO

Corollary. If an > 0 for all n and if lim ^ ± exists, then the series 53 an l

converges if the limit is < 1 and diverges if the limit is > 1.

Note that for both the series J ] ^ ( a divergent one) and 53 ~^ ( a convergent series), the limit in the Corollary exists and = 1. Thus either convergence or divergence is possible in this case.

CO CO 2

Example 1. The series 53 ;b is convergent; the series 53 frr is also l n" l

convergent.

In fact, a n + i / a n = (n+i)r n ' = ^+1 ~* ® (ôr t n e ^ r s t series). In the

second case, we have ^ ^ = &±£- -K = h ( — f -> i

Our next criterion is called Cauchy's root test:

Theorem 2. (The root test). Suppose an > 0 for all n. If lim sup nv / a ^ <

CO CO

1, then the series 53 an converges. If lim sup nyfa~^ > 1 £/ien t/ie series 53 a n l l

diverges.

Proof. If lim sup 7\fan~ = q < 1 we take q', q < q' < 1 and we find no € N such that y ^ < q' for n > no- Hence an < {q')n for n > no, and therefore CO

53 &n is convergent. l

If lim sup ny/a^ = r > 1 we take r', 1 < r1 < r. Apply now Theorem 5.4 (II) (2). Take ni > 1, such that n\fa~^[ > r'. Then, take n2 > ni , such that n\fa^ > r', continuing this way we find a subsequence (a,nk)kLi of (an) such


that aUk > r'nk. Hence, for (an) the necessary condition for convergence of oo ^ a n , namely "an —> 0, when n —► oo" fails. D 1

oo

Example 2. The series J2 (2^+i)n i s c o n v e r g e n t -n=l

In fact, if an = ( 2 ^ 1 )n» t n e n V^n = 2^+1 "* 2 an(* t h e r o o t t e s t aPP l i e s-

III.3 Cauchy's General Criterion of Convergence: Absolute and Conditional Convergence

A general, necessary and sufficient condition, for the convergence of the series 00 J2 an c a n be stated as follows. 1

Theorem 1. An infinite series J2an is convergent if, and only if, for any e > 0 there is a positive integer no such that

|an+i + a n + 2 H am | < e for any m > n > n0 .

Proof. In fact, as we already know, $ân is convergent iff (sn)neN is a convergent sequence (sn = ai + ^2 + • • • on).

Furthermore, (sn) is a convergent sequence iff it is a Cauchy sequence which means that Ve > 0, 3n0 G N, such that

\sm - sn\ < e for m> n> n0.

However: s m — sn = an+i + an+2 + • • • am if TO > ra. This ends the proof. □

Let us next define the concept of absolute convergence.

Definition 1. A series ^ a n is said to be absolutely convergent if the series ^2\an\ is convergent.

If ^2 an is convergent but not absolutely convergent, then it is said to be conditionally convergent.

Theorem 2. If a series is absolutely convergent it is convergent.

Proof. We shall apply Theorem 1. Let us assume Yl \an\ is convergent. Then, Ve > 0, 3n0 G N, such that

| | a n + i | + | a n + 2 | + • • • |am | | = |an+i| + lan+2| + • • • \am\ < e


when m> n >UQ.

On the other hand we have

| a n + i + an + 2 H am | < | a n + i | + | a n + 2 | H |am | < eforra > n > n0.

This implies convergence of £^ an D

Example 1. The series Y v n2

; zs absolutely convergent. 1

The series J2 n *s n o t absolutely convergent. It is however, a convergent 1

series, as follows from the next statement:

Theorem 3. Let (an)ne^ be a sequence such that a\ > a^ > • • • > an • • • and lim an = 0. JVeart, /e£ (bn)n€N 6e a sequence such that the partial sums: sn = b\ -h • • • bn form a bounded sequence. It follows that the series Y, anbn is convergent.

Remark. If an = ^-,6n = (—l)n, we obtain that Y n 1S convergent.

i

Proof. It consists of a skillful application of Theorem 1.

Consider the sum (where m > n) m m m m /] apbp = 2 J ap(sp - Sp-i) = 2 J apSp— 2 J CLpSp-i.

p = n + l p—n-\-l p = n + l p = n + l

In the second sum we make the substitution p = q + 1; we obtain

m m m—1 m—1

22 CLpbp — 2 J 0,qSq-2_^ aq-\-lSq = / - / (aq ~ aq+l)Sq + arnSm ~ Un+lSn-p = n + l q=n+l q=n q=n+l

Taking absolute values we then derive the estimate

m I m — 1

^ apbp\ < ^^ \aP - V f l l M + I a m 5 m | + | a n + l « n | -n+1 ' n+1

Using the monotonicity assumption we find : \ap — a p + i | = ap — a p + i . Also, 0 = lim an = inf {a n}, hence an > 0, VneN. Finally, 3M > 0, such

that |an | < M V n e N .


We obtain accordingly the inequality

771 I 7 7 1 — 1

/] apbp\ < M( 22 (aP ~ aP+i) + am + a n + i ) = n + l ' n+1

M ( a n + i - a n + 2 + a n + 2 - a n + 3 H h am_i - am 4- am 4- a n + i ) = 2 M a n + i .

Hence, as an —► 0, we find, given any e > 0, a number rio(e) G N, such that m I ^2 apbp\ < e for m> n> n0(e). D

n+l I

Corollary, (the alternating series test) Let (an) be a monotone decreasing CO

sequence of positive numbers and let lim an = 0. TAen £/ie series ^2(—l)n_1an l

is convergent.

The proof is obvious.

III.4 Summability CO

Consider a divergent series: Ylan- Therefore, the sequence of finite sums: (sn) I

where sn = a\ 4- a2 4- • • • an is not convergent. It can however happen that the sequence of arithmetic means: (tn) where

1 , tn = —(«1 + S2 H S n )

n

is a convergent sequence. Hence: tn -±T. CO

We can say in this case that Yl an — T in (C, l)-sense.

Example. Take an = ( - l ) n + 1 , n G N. The series £ ( - l ) n + 1 is divergent. 7 1 = 1

(We see that sn = 0 for n = 2p, sn = 1 for n = 2p 4-1). We compute the numbers tn. If n = 2p we get t2p = ^ ( « i 4- s2 + • • • s2p) = ^ = §. If n = 2p+ 1 we get t2 p + 1 = ^ ( p H - 1 ) = § £ • As £2p = ^ for all p G N, while £2p+i —► ^ as p —► oo, we see that lim £n = ^.

oo

Hence, we can say that (C, 1) XX~~l)n+1 = | . The series is Cesaro convergent

(precisely, Cesaro of order 1 convergent).


We shall prove the following

oo oo Theorem 1. / / Yân is convergent and s = J2an then it is also (C, 1)

l l convergent and

oo

( C , l ) ^ a n = 5. l

Proof. Obviously, it will suffice to establish the following: If (a;n)n6N is a sequence and an —► a and if f}n = ^ (a i +a2-\ a n ) , then f3n -> a too. Let us write in fact the difference /3n — a as the arithmetic mean fin — a =

${<*! + ■ • ■ an) - a = («,-«)+(»,-«)+■•■(«,,-«) _

Let us take any e > 0, and then n0(e:/2) such that n > no =>• |qn — <*| < f • Take any n > no; we have

_ _ (ai - a) 4- [pL2 ~ <*) + • •• (o np - a) , (^n0+i - a) + • • • + (a n - <*) n n

We obtain

\Pn ~Oi\< - ( a i - a + a2 - a + • • • |an o - a ) + • n n z

Take nx G N such that n>ni=> ^( |«i - a| H \ano - a\) < §. It follows, if n > max (no,ni), that

\Pn - a\ 2 fc_1, k = 1,2 • • •. It follows that oo

n < 2* "> ^ e ^ an(* consequently, the series S n i s convergent. For oo

the partial sums we have ^ -\ y < 1 + | + H ^ t r < X) 2^=r =

oo Y3X = 2, hence X) n ^ 3- ^ n t n e o t n e r hand, it is obvious that

2 o

oo

0


oo

(2) Let us compute the sum of the series: J2 k(k+v) ( w n e r e V € N). It

is convergent, due to the estimate: u^+v) < P"' Note t n e identity:

1TOJ = ? ( i " kk)' T h e P a r t i a l s u m s S *(FFP7 a r e e x P r e s s e d a s n n

In the second sum we put k+p = /, and we get J ] k(k+v) = v p+n I

Z ^ / p L x ^ 2 ^ n p + 1 p + 2 Z=p+1 J Let us take n > p. Then we can write:

£*-Lfc=i

p + n J

E 1 1 -^ k(k + p) p

1 1 1 1 1 + 2 + " " p + p~+T + P + 2

1 1 1 _ n p + 1 p+2

1 n

1 1

1 1 1 - - -2 p n + 1 n + 2

1 n + p

n + 1 n + p

= 5 n (nth-partial sum).

Accordingly: £ j ^ ) = ^ 5 n = ±(1 + \ + £).

(3) A similar computation will allow us to compute the sum of the se-oo

ries ^2 k(k+i)(k+2) (ê c o n v e r g e n c e follows from the obvious estimate l ^ l \

fc(Jk+l)(fc+2) ^ F > Note the identity

1 1 - ( - -

1 ■ ) =

1 1 fc(fc + l)(* + 2) fc + 2vfc * + l ' fc(fc + 2) (fc + l)(fc + 2)

= I[I _ _l_] - l l < 2Lfc A; + 2J

Accordingly we find 1

Lfc + 1 A; + 2 J *

^ S *(* + !)(*+ 2) .)(fc + 2) 2 ^ * fc + 2 n + 2 n + 1 n + 2

E x 1 1 1 r-—v 1 1 ^—\ 1 ^—s 1 ^—\ 1

u }k + l~k + 2 ) = 2 ^ k ~ 2 ^ k ~ ^ k + 2^k k=l k=l k=3 k=2 k=3 k=l

n + 1 1 1 1

n + 2 ' 2 n + 2


Hence sn->|(l + i ) - | = i a s n - > o o . We get accordingly that OO

fcE Jb(*+l)(*+2) = 4 ' fc=l

OO CO OO

(4) Let us consider the series: f) ak ■ bk where f\ al < oo and £ fc=0 fc=0 fc=0 fc=U fc=0 fc=0

6| < oo. Then, the series is absolutely convergent. We write in fact the estimate: \ak ■ bk\ = \ak\ ■ \bk\ < i(|afc|2 + \bk\2) = \{a\ + b2

k). The remaining of the proof is obvious. OO

H a ^ + o ^ j . ine remaining 01 tne prooi is obvious. oo

(5) Let us establish convergence of the series: £ # . fc=i fc=i

We shall use the ratio: (D'Alembert) test for convergence. The quotient 2=±i equals: $±$r • £ = („ + 1 ) ( ^ ) « • = ( T ^ F ' W e k » ™ that: lim T^pW exists and = \. As ( z T = m " + IVM is increasing we have a; > 2 V n f N hence „ J !-m T = sunrr > > 2 Thus * < * and the series above is convergent" '

(6) We shall prove that if ak > 0, VA: G N, the series f > f c and £ > * o o

(1 + a , ) - 1 are convergent or divergent simultaneously. Let us assume convergence oi 2âfc- -^rom relation: a/c(l + a^j < a^

oo we derive convergence of V a n ( l + a n ) _ 1 .

o Let us assume, conversely, that ^ ^ ( 1 + ^ ) * is convergent. This

entails- lim ak(l +ak)~^ = 0 °

Therefore, 3k~ e N, such that, for k > k~, ofc(l + a f c) - 1 < ±. Then we find 2ak < 1 + ak and a^ < 1 for k > k . Therefore, we see that again for k > k~ we have

< o flfc

(in fact, if afc = 0, we have nothing to prove. If ak > 0, we see that 1 < Y ^ «*• 1 + o* < 2 o ofc < 1, hence afc < ^ ) .

OO oo

From these reasonings we derive that £ > * is convergent.

The two series are simultaneously convergent, hence simultaneously divergent.


III .6 Exercises

uu 3

1. Show that the series ^ j ^ - is convergent. n=l

2. True or false? (explain): Vc G R, c > 0, the series ^ ^j- is divergent. n=0

0 0 2 3. Check for the convergence of the series: ^ (^+r)n •

n=l oo

4. Show that: if X] an is a convergent series of positive numbers and if l

oo (bn) is a bounded sequence, then the series Ylanbn is convergent.

I oo

5. Determine the nature of the series: ^ (y/n + 1 — y/n). n=l

Chapter IV

CONTINUOUS FUNCTIONS

In this chapter we consider functions between a set of real numbers and another set of real numbers. Informally, continuity in a point x0 means that the function f(x) is as near as we wish to f(xo) as soon as # is sufficiently close to XQ. Many functions show this property in any point (for instance all polynomials) but on the other hand it is very easy to find functions which are discontinuous at least at some points: think of the function f(x) which equals 0 at all points x except x — 0 where we have /(0) = 1.

IV. 1 Definitions, Algebra of Continuity

We shall give now some formal definitions. Let 5 C 1 be an arbitrary set of real numbers, and / be a function with

domain D{f) = S and with range in E. If a G S we say that / is continuous at x = a if the following holds:

if (#n)neN is a sequence in S and xn —> a, then f(xn) —> / (a ) .

Examples .

(1) Let S — E, f{x) — xVx G E. It is obviously continuous for all a G E. (2) Let S C E and f(x) = c\/x G S (a constant function). Then / is

continuous Vo G 5. (3) Let f{x) = 1 for x > 0, = 0 for x < 0. Here D(f) = E; / is continuous

for a > 0 and for a < 0. It is not continuous however ("discontinuous") for a = 0: If xn —» 0 and xn > 0, we get f(xn) = 1 —» 1 ^ /(0) .

61


(4) f{x) = \x\. This function, with domain R is continuous at any point x = a, for we have

n\ ~ M < |xn - a|; if xn -► a => \xn\ -> \a\.

It is often important to possess an alternative definition of the continuity in a point in (£, 8) terms. Precisely, we shall establish the following:

Theorem 1. Let f : S C R —» R. TTien / zs continuous at a £ S if and only if the following holds:

Vs > 0, 3 8(e) > 0, such that: x G S and |x - a| < 8 => | /(x) - / ( a ) | < e. (1.1)

Proof.

(a) Let us assume Property 1.1. Take (xn) a sequence in 5, xn —► a. We must prove that f{xn) —► /(a) , which means: Ve > 0, 3n(e), such that: n > n => \f(xn) — f(a)\ < e. Using (1.1) we find 8, such that: |rc — a| < * => \f(x) - /(o)l < *• As x n —► a, we find n(<5) such that n > n =$■ \xn — a\ < 8, hence \f{xn) — / (a ) | < e. (as 8 depends on e, n(£) also depends on e).

(b) Let us assume continuity for x = a and prove (1.1). If it does not hold, 3e0 > 0 such that, V£ > 0, 3xs G S, \x$ — a\ < 8 and \f(xs) — f(a)\ > e0. Take 8n — ^ and then xsn = x'n £ S, such that \x'n — a\ < ^ and \f(x'n) — / ( a ) | > £o- Therefore x'n —> a but /(x^) -/> / (a ) . This contradiction establishes the result.

Now we treat the "algebra of continuity". This simply means that various combinations of continuous functions are continuous.

Theorem 2. Suppose a G S C R and f \ S —> R, g : S -^ R; let c be any real number. If f, g are continuous at a, then so are the functions f + g, f • g and c • / .

Proof. The functions in question are defined on S by the formulas

(f + g)(x) = f(x) + g(x) (f ■ g)(x) = f(x) ■ g(x)

(cf)(x) = cf{x).

Continuous Functions 63

(that is, they are "pointwise" sum, product and scalar multiple). Let us take now any sequence in S, (xn), such that xn —> a. Then f(xn) —»

f(a) and g(xn) —► g(a) by continuity assumptions on / and g. Using properties of convergent sequences previously encountered, we obtain that:

( / + 9)(Xn) = f(*n) + 9(Xn) "> /(<*) + g{d) = ( / + 0)(tt).

This shows that / + g is continuous at a; similar proofs can by given for / • g and cf (actually cf is the particular case of / • g when g{x) = c Vrr G S).

Corollary. Every polynomial function p : R —> R is continuous at every point ofR.

Proof. We know that p(x) = CLQX71 + a\xn~x + • • • + an; the function 'u(x) — x\lx G E is continuous. Hence -a • it, u - u - u- - • and so on are all continuous Vx G R, and (u-u — -u)(x) = xp\ Vx G M. Therefore, simple

p t imes application of Theorem 2 and induction on n will give the result.

There is also a result concerning continuity of the quotient function f/g at any point a where both / and g are continuous and g(a) ^ 0.

Theorem 3. Suppose a G S C E and / , #, 5 —> E are continuous at a. Assume also that g(a) ^ 0. Then f/g is also continuous at a.

Proof. Remember that f/g, Si —> E is defined by (f/g)(x) = f{x)/g(x), Vx G Si, where ^(x) ^ 0 iff x G Si.

What we prove now is really continuity at a G Si, with respect to Si. Thus, if (xn) is a sequence in Si and xn —► a then (f/g)(xn) — f(xn)/g(xn) —► /(a)/(/(a) - (f/g)(a) (by Theorem 3 in II.2).

Corollary. Let r(x) = p{x)/q(x)1 where p and q are polynomials, and D(r) = {x G E, g(x) / 0}. (Such functions r(-) are ca/Zea7 "rational"). It follows that r(x) is continuous Va € D(r).

A final result in the same vein concerns composition of continuous functions. If / , S —> E and #, T —> E are two functions, we can define the composition gof ("one function followed by another one") with the sole condition: x G S => / (*) G T.

In this case we put: (gof)(x) = g(f(x)) with domain D(gQf) = S.

Theorem 4. / / / is continuous at a and g is continuous at f(a) then gof


is continuous at a.

Proof. If (xn) is a sequence in S and then f(xn) —> f(a) so that »(/(*»)) - <?(/(«)) = (9of)(a).

Example. If g(x) is continuous, then \g(x)\ is continuous.

IV.2 Limits and Continuity

One can profitably define continuity (and even more general concepts as right and left continuity) using the concept of limit of a function at a point.

To simplify somewhat the discussion we consider a function denned in the open interval (a, 6) with the possible exception of a point x$ G (a, 6).

We say that /(•) has a limit as x —» XQ if the following holds. For some number L, we have: We > 0, 38 > 0, such that: x G (a,b) and 0 < \x — xo\ < 8^\f(x)-L\<e.

We write: lim f(x) = L or f{x) -» L as a; —> XQ. We call L the limit of X—>XQ

f(x) as x —► zo-Next, we establish

Theorem 1. Let / (#) , (a, b) —» R and XQ G (a,6). Then / is continuous at XQ if and only if: lim f(x) = f(xo).

X—>XQ

Proof, (i) Let / be continuous at x0. Then, Ve > 0, 3 6(e) > 0, such that \x — XQ\ < 8 and x G (a,b) => \f(x) — f(xo)\ < e. This implies: lim f(x) =

X—>Xo

(ii) Let us assume:

lim f(x) = f(xQ) and x0 G (a, b). x—*xo

It results: We > 0, 38 > 0 such that 0 < \x - x0\ < 8 and x G (a, b) => \f(x) -f(xo)\ < e. On the other hand, if x = xo we also have \f(x) — f(xo)\ = 0 < e. Thus / is continuous at XQ-

Example. The function x+ = max{:r,0} is continuous on R.

This is easily seen as we have rr+ = ^(x + \x\). Therefore, the powers re™ are all continuous, and so is the function S(x) = A0 + ]Cfc=i Akx\.

Example. The function Sgnx = 1, for x > 0, = - 1 for x < 0, = 0 for


x = 0, can be represented as: A for x ^ 0, and is continuous Wx G E/{0}.

We now extend the notion of limit at a point to that of right-limit or left-limit of a function at a point:

We again consider function /(•) denned on the open interval (a, b) with the possible exception of a point x0 G (a, 6), or denned in [a, b) with possible exception of XQ — a.

Let us assume the following: For some number A E l w e have: \fe > 0, 3 6(e) > 0, such that

\f(x) - A\ < e if x0 < x < x0 + 6 and x G (a, 6).

We say in this case that /(•) converges to A as x tends to x0 from the right and write: lim f(x) — A — f(xo+). (right-limit of / at XQ).

A similar situation arises when we consider the left limit of a function in a point. Again, the point XQ can belong to the open interval (a, b) and / is defined in (a, b) with possible exception of x = XQ\ or else, XQ can be=b (the right end point of (a, 6). In both cases, assume: 31? G K such that Ve > 0, 3 8(e) with property that

\f(x) — B\ < e if XQ — 6 < x < XQ and x G (a, 6).

Then we say that f(x) converges to B from the left and write

lim f(x) = B = f(x0-). X—►£o—"

We call B the left limit of f(x) at x0-

Definition. Let f be defined on [a, b) and XQ G [a,b). If f(xo+) exists and = /(ô) we 5a?/ £/ia£ / is right continuous at Xo.

Also let / be defined on (a, b] and #o € (a>&]- If f(xo—) exists and = /(ô) we say that / is left continuous at XQ.

Proposition 1. Let f be defined on (a, b) and XQ G (a, b). Then f is continuous for x = XQ if and only if, f(xo+) and f(xo~) both exist and are equal to / (x 0 ) .

Proof.

(1) Assume / continuous at x0 G (a, b). Then lim f(x) — f(x0). This X—►Xo

obviously implies that f(x0+) and f(xQ—) both exist and = /(#o)-


(2) Conversely, let x0 € (a, b). We have, from definitions of right and left limits at x0, that Ve > 0, 3 6i(e) such that

\f(x) - f{x0)\ < e if x0 < x < x0 +6i,x e (a, 6), and then

Ve > 0, 3 62(£) such that

\f(x) - f(x0)\ < e if x0 - 62 < x < so, x € (a, 6).

Accordingly, if 5 = min(î,52) we find

1/0*0 _ / (ô) | < e for x0 - ^ < a; < x0 + ^,x 9 x0 .

However, \f(x) — /(#o)| < £ also for x = xo- Hence / is continuous at #o- D

One can express continuity at right or left end points of an interval in terms of the above defined one sided continuity.

Proposition 2. Let f be defined on [a, 6). Then f is continuous at a if and only if it is right-continuous at a.

Proof. Let us use Theorem 1 (IV. 1), where S = [a, b). Continuity of / at a means that

Ve > 0, 3 6(e) > 0 such that x G [a, b) and \x - a\ < 8 =>► \f(x) - f(a)\ < e.

This obviously becomes: a < x > \f(x) — f(a)\ < £ or a < x \f(x) - f(a)\ < £.

This property is equivalent to: /(a-h) exists and = / (a ) . □

Example 1. The function Sgnx = 1, for x > 0, = - 1 for x < 0, = 0 for x = 0, has right limit at 0 = 1 and left-limit at 0 = - 1 . It is not right continuous or left continuous for x = 0.

Example 2. The function x —► y/x is defined and continuous for x > 0 (this follows for instance from the "sequential" definition of continuity and Theorem 5 in Chapter II.2).

Example 3. The function x —► M s continuous for x > 0. We prove that it has no right limit at x = 0.


Otherwise, Ve > 0, 3 8(e) > 0 such that |^ — A\ < £ for 0 < x < <5, where A is the right-limit. It results: ^ < e + \A\ for 0 < x < S which fails to hold if x is small.

Note . One can extend the concept of the limit in a point of a function to encompass limits at +00 or at —00. This is often quite important. For instance, let us give the following:

Definition. Assume that the function f(x) is defined in an interval (a,+00). We say that f(x) converges (or tends) to the real number A as x converges (or tends) to +00, and write

lim f(x) = A or f(x) —► A as x —> +00 x—> + oo

if the following holds: Ve > 0, 3X(e) G (a, 00) such that, if x > X, then \f(x) - A\<e.

Similarly, one defines the concept of uf(x) converges to A as x converges to —00." We now assume / defined on (—00, a) and write: lim f(x) = A if

x—* — CO

\/e> 0, 3X(e) G (-00, a) such that \f(x)-A\ <e if x < X. Other related concepts pertain to the case when the limit itself is +00 or

—00. For instance, let / be an interval and a € I. Let / be defined on / — {a}. Suppose that V ft > 0, 3 8 > 0, such that 0 < \x-a\ < S and x £ I => f(x) > O. Then we say that f(x) converges to +00 as x —> a. We write: l im/(x) = +00.

x—*a Similarly, one defines the concepts: limf(x) = —00, limf(x) = +00, etc.

x—>a x—>a x>a

One also defines analogous concepts with a replaced by +00 or —00. For instance

lim f(x) = —00 means that x—*+oo

Vft > 0, 3T > 0, such that: x > T => f(x) < - f t . IV.3 Continui ty and Boundedness

Let /(•) be a real-valued function defined on a set S. The range of / is the set {f(x),x€S} = Rf.

If Rf is upper bounded we say that / is bounded above. This means that, for some constant N G R the estimate

f{x)<NVxeS (3.1)

holds true.


The supremum (least upper bound) of Rf is called the supremum of / on S. We denote it by

sup f{x). xeS

If Rf is bounded below, then we say that / is bounded below. The infimum (greatest lower bound) of Rf is called the infimum of / on S.

It is denoted by

inf f(x). x e s

If Rf is a bounded set, then we say that / is a bounded function on 5. This happens if and only if, for some positive constant M we have

| / (x) | < M f o r a l l x G S . (3.2)

If / is not bounded, then we say that it is unbounded.

Example. The function x —► ^ is unbounded on (0,1]; same holds for x —> x2 on E.

The function x —> -^TTI is bounded on E. A fundamental result is stated as follows.

Theorem 1. Let f be a continuous function on a closed, bounded interval [a, b]. Then f is bounded.

Proof. Let us assume that / is unbounded. This means that for any n G N, there exists some xn G [a, b] for which

| / ( * n ) | > n , r i G N . (3.3)

If we use Bolzano-Weierstrass theorem we see that there is a subsequence (xnk) that is convergent to a point xo G [a,6].

From continuity of / we derive: f{xntc) —> f(xo). Thus the sequence (/(xnfc))i° is convergent, hence bounded. On the other hand, because of (3.3) we find that

\f(xnk)\>nky fcGN, (3.4)


As ni < n2 < - • -rife -» +00, the sequence (f(xnk)) cannot be bounded. D

Definition. Let /(•) be a function defined on the set S and bounded from above. Suppose there exists a point £ G S such that

f(x)<f(OforxeS (3.5)

that is

/ ( 0 = sup f(x). x e s

Then we call £ a maximum point of f and we call /(£) the maximum of f on S.

Alternatively, we say that f has a maximum, and that it assumes (or attains) its maximum on S at the point £.

Similarly, we define a minimum point rj of f on S by

f{x)>f(r))VxeS. (3.6)

/ / (3.6) holds, then we say f(r]) is the minimum of f on S. We also say that f has a minimum, and that it assumes its minimum on S at r).

Another fundamental result is stated as

Theorem 2. Let f(x) be a continuous function on a closed, bounded interval [a, 6]. Then f has maximum and minimum on [a, 6]. (it attains its extreme values on [a, 6]).

Proof. By Theorem 1, / is bounded in [a, 6], hence it is upper bounded and lower bounded.

Let M — sup f(x)\ this means that f(x) < M Vx G [a,6], and that for a<x<6

any e > 0, there is a point x£ G [a, 6] such that f(x£) > M — e. Take a sequence en — £; the corresponding points: yn = xi form a

bounded sequence and we have

M - - < / W < M , V n G N , (3.7)

If we (again) apply Bolzano-Weierstrass theorem, we find a subsequence {ynk) which converges to some £ G [a,6]. We also have: f{ynk) -» f(Q and because


of (3.7)

M- — <f(ynk)<M, k = 1,2,3,. •• . (3.8)

As rii < n2 < • • • < rik - - • —► +oo, we find from (3.8) that

lim/(yn f c) = M.

Thus, we conclude that /(£) = M = sup/(x). In a similar way we can find [a,b]

rj G [a, b] such that f(rj) = m = inf / (#) . [a,b]

Examples .

(1) The function g(x) = j ^ with domain (—1,1) is unbounded. In fact,

if xn = - 1 + £ we get: p(xn) = ^ = n(2 - £) = 2n - 1.

(2) The function /i(x) = ^ ^ w ^ n domain (0,+oo) is bounded. In fact, for 0 < x < 1 we have h(x) < x2 + 1 < 2; for 1 < x < -hoo we have x2 < x3, hence h(x) < 1. Also h(x) > OWx > 0.

(3) Let / , E —» R be continuous and periodic with period w > 0 (this means: 3a; > 0 such that f(x + a;) = /(rr) Vx G E).

Then / is bounded on E and has maximum and minimum. In fact, if we consider / restricted to [0,u;] (domain oi f = [0,u;]) then / is

bounded there and has maximum and minimum (Theorems 1 and 2 above). Now, if x is any real number, using Corollary 3-Proposition 5 (1.5), we find

n G Z such that x = nuj-ha, where 0 < a < LJ. Then f(x) = f(noj + a) = f(a). Hence, if M > | /(£) | , V£ G [0,a;], we find that |/(a;)| < M. Let sup/(a;) =

/ (x i ) and inf/(a;) = /(#2)- Note also that the sets {/(a;),0 < a; < a;} and [0,w]

{/(x),a; G E} are identical; hence /(a:i) = sup/(a:) and f(x2) = mif(x). xGEt xGIR

The extreme values of / are actually attained in the interval [0,CJ].

IV.4 Intermediate Values

The following result appears "intuitively" plausible:

"If a continuous function on a closed interval has opposite signs at the endpoints, then it must be zero somewhere in between."


Theorem 1. / / / , [a, b] -> R is a continuous function such that f(a)• /(b) < 0 then there exists a point c G (a, b) such that f(c) = 0.

Proof. We can suppose / (a) > 0 and f(b) < 0 (otherwise consider - / ) .

Consider the set A = {x G [a, 6], f(x) > 0}. Then A is nonempty (a G A) and upper bounded. Let c = sup A. We say

that /(c) = 0. We need the following (interesting in itself):

Lemma. Let g(x) be continuous inx0 G (a,/3) and g(x0) > 0 (< 0). Then, 3 8 > 0, such that g(x) > 0 (< 0) for x0 — 8 < x < x0 + 8.

In fact, in the contrary case, Vn G N, 3xn G (x0 — ^, x0 + ^) where #(#n) < 0 (> 0). Consequently, xn —► x0 and by continuity g(xn) —> g(xo).

If ^(xn) < 0 => g(x0) < 0 contradicting #(:ro) > 0 and if g(xn) > 0 => g(x0) > 0, contradicting #(#0) < 0- Both contradictions prove the Lemma.

We now end the proof of Theorem 1. If c = sup A = 6, we find, Vn G N, numbers an G A, 6— < an < b. Hence

^n —* b, f(an) > 0; it implies f(b) = lim f(an) > 0, a contradiction. If c = sup A belongs to (a, b) we apply the Lemma. If /(c) > 0 we find a > c, a G A, thus contradicting definition of c. If /(c) < 0 we find f(x) < 0 for c — 8 < x < c. But we know that there

must be some a G A, a > c — 8, hence / (a) > 0. Again we have a contradiction. It follows necessarily: /(c) = 0. A first extension of this result can be stated as

Theorem 2. Let f be a continuous function on a closed, bounded interval [a, 6]. Assume that f(a) ^ f(b) and let C be any number lying between f(a) and f(b). Then, there exists at least one point c in the open interval (a, b), such that /(c) = C.

In other words, a continuous function taking values A and B at the end-points of an interval must also take any intermediate value at some intermediate point of the interval.

Proof. Let A = / (a ) , B = f(b) and, say, A < B. Let A < C < B. Take g(x) = f(x)-C. Then g(a) = A-C < 0, g(b) = B-C > 0 and g is continuous on [a,6]. Hence, 3c e (a,6), g(c) = 0. □


Corollary. In the assumuptions of the Theorem 2, if m — inf f{x) and a<x<.b

M = sup f(x), it follows that: {/(#), a < x < b} = [ra, M]. (It results from a<x<6

Theorem 2 (IV.4) and Theorem 2 (IV.3)).

Another extension, quite a simple consequence of previous results, can be stated as

Theorem 3. (General intermediate value theorem). / / / is an interval in E and f, I —► E is a continuous function, the range f(I) = {f(x),x G / } is also an interval in E.

Proof. It is sufficient to establish convexity of / ( / ) : let therefore r, s G / ( / ) , r < s. By assumption r = / (a ) , s = /(6), where a, b G / . As r ^ s => a we apply Theorem 2, we find, Vk G (r, s), that 3 c in (a, 6) or (6, a), such that /(c) = k. Thus k G / ( I ) .

Examples .

(1) Let P(x) = ao-f aiaH ha2P+ia;2p+1 be a polynomial of odd degree. We shall see, as a consequence of Theorem 1, that it always possesses a real root x (thus P(x) = 0). Take P(n), where n G N. We have obviously P(n) = n2p+1(a2p+i + ^ + ^ r 1 + • • • + d & r ) . We see that lim P(n) = +oo if a2p+i > 0 and

71 n n—f+cxD lim P(n) = —oo if a2P+i < 0.

n—>oo In a similar way we see that lim P(—n) = —oo if a2P+i > 0 and lim P

n—K» n—*oo

( -n ) = +oo if a2p+i < 0. (Here use the fact that ( - n ) 2 p + 1 = - n 2 p + 1 !). This shows that there are real numbers X\ and #2, such that P{x\) < 0, P{x2) > 0; hence, between them we can find x — a root of P{x).

(2) We shall see that: if f(x) is continuous on the closed bounded interval [a, 6], if / (a ) = /(&) and I G (ra,M) where ra = inf / (#) , M = sup/(x),

M l [a,6] then, 3^,77 G [a, 6], £ ^ r? and /(£) = /fa) = /.

Proof. We have 3 possibilities:

(i) f(a) = f(b) = /, hence nothing to prove.

(1)And, as J is an interval, the interval (a, b) or (b, a) is contained in I.


(ii) / (a) < /. Then / (a) /; therefore m E, such that / is continuous and lim / (x) = 0, lim / (#) = X—► + OO X—+■ — OO

0. Then / is bounded on E. If f(xi) < 0 for some x\ G E, / takes its minimum value. If f{x2) > 0 for some #2 G E, / takes its maximum value. In fact; first we have, \f(x)\ < 1 for x > A > 0, and |/(:r)| < 1 for x < -B, B > 0. Hence | / (x) | < 1 in [J4,+00) and in (—00, -B]. Furthermore, / is bounded on [—B,A]. Let m = inf/(x), and assume also that f(xi) < 0. Hence m < f(xi) < 0.

IR

On the other hand, f(x) > f(xi) for |rr| > X (as f(x) —► 0 at ±00). Take £ G [-X,X] such that f(x) = inf{/(x), |x| < X}. (Now, | ^ i | < X, for if lîI > X we would have f(xi) > f{x\) which is not possible). Therefore, f(x) < f(xi) < f(x) for |x| > X. Hence f{x) > f(x) Va; G R, / (x) = m = inf fix). A similar proof holds for the attainment of the maximum value of / .

IV.5 Monotonic Functions and Inverse Functions

Let S C E and / a function with domain S and range in E. It is said to be: increasing if x, y G S and x < y => f(x) < / (y); strictly increasing if: x, y G 5, x <y => f(x) < f(y); decreasing, if x, y e S, x < y, => f(x) > f(y); strictly decreasing if: x, y € S, x < y => f(x) > /(?/).

If / is either increasing or decreasing it is said to be monotone; if it is strictly increasing or strictly decreasing it is strictly monotone.

Examples . Let n G N. The function / given by f(x) = xn, x G [0,oo) —► [0,oo) is strictly increasing; so is the function g, E —► E, defined by g(x) = x2n~1 (prove it!). On the other hand, the function E —► E defined by x —> x2 is neither increasing nor decreasing (why ?). The function (0, +00) —► (0, +00) defined by x —* ^ is strictly decreasing (all these facts follow essentially from the properties of E as an ordered field). Every constant function is increasing (and decreasing) but not strictly. If a function is both


increasing and decreasing, then it is a constant function. Next, let us remember the definition of injectivity: / ; S —» E is injective if

xi, X2 € S and X\ ^ X2 => f(x) = / (a^) . It is obvious that strictly monotone functions are injective.

An interesting result establishes strict monotonicity of continuous injective functions denned on an interval of E. We state it in the form of

Theorem 1. Let I C E be an interval and / ; / —► E be continuous and injective. Then f is strictly monotone.

Proof. Remember that intervals / in E are characterized by "convexity": x, y E I => [x,y] C I.

Let / be injective, and take xi, a?2 G / , x\ < X2- It follows that f(x\) < f(x2) or f(xi) > /(X2). Assume (for instance) that f(x\) < /(a;2). We shall prove in this case that / is strictly increasing on / : if £3, X4 € / and x3 < X* => f(x3) < / (x 4 ) .

Let g, [0,1] -► E be defined by the relation g(t) = f{(l-t)xi +tx3)-/((l-t)x2+tx^). (note that (1—t)x\+tx3 G [£1,23] C / , (l—t)x2+tx4 G [#2,£4] C / , for 0 < t < 1.) Then g is continuous; also #(0) = / ( z i ) — f(x2) < 0.

If #(1) = f(x3) — /(X4) = 0 we get £3 = £4, impossible. If #(1) > 0 there exists s E (0,1), such that g(s) = 0.

This would give: (1 — s)xi + sx3 = (1 — s)x2 + SX4 (by injectivity of / ) . Hence (1 — s)(xi — X2) = s(x4 — x3) which is obviously impossible. Hence, necessarily, ^(1) < 0, which means: f(x3) — / (x 4 ) < 0. D

Next we shall establish an important property of monotonic functions: they have right-limits and left-limits.

Theorem 2. Let I C E be an interval and f; I —► E be monotone. Then, ifxo is an interior point of I, both the right and the left limits: f(xo+), f(xo~) exist; if XQ is the right end point of I, f(xo—) exists; if XQ is the left end point of I, f(xo+) exists.

Proof. First note that "xo is an interior point of J" means that, 3 6 > 0, such that (#0 — <5, XQ + 6) C I.

We shall first prove, that in such a point f(xo-) exists. We also assume that / is increasing (leaving to the reader the case of / decreasing).


Let us consider the set in R : A = {/(rr),rr0-<5 < x < rr0}. As /(rr) < /(rr0), we see that A is upper bounded. Let a = sup A. Then f(x) < a, for x G A and Ve > 0, 3 x G (rr0 — <5, rro) such that /(rr) > a — e. Hence /(rr) > a — e for rro > rr > rr and we obtain — e < /(rr) — a < 0, that is

|/(rr) — a\ < e for x < x < x0.

This means that f(xo—) exists and = a. Next, let us consider the set B = {/(rr),rro < x < xo + 8}. Note that, for

x G B we have /(rro) < /(^)? hence 5 is lower bounded. Let b = inf B. Then for x e B, f(x) > b and Ve > 0, 3rr G (rr0,rr0 + ^) s u c n t n a t / (#) < & + £•

Hence, 6 < f(x) < b + e, and |/(rr) — &| < £ for rr0 < x < x as /(rr) < /(rr). Therefore /(rr0+) exists too and = b. □

Note. Remember that / is continuous in rr0 iff f(x0—) = /(rr0+) = /(rr0).

The proof when rr0 is the right or the left end point of / is similar.

Remark. If / is only defined on / — {rro}, w e obtain again existence of right or left-hand limits provided / is monotonical and bounded (it follows now that the above considered sets A and B are automatically bounded).

Strictly monotonical functions are injective; therefore, their inverse functions always exist. If / is strictly monotonical, S C M —> f(S) C M, then we can define g on f(S) as follows: if y G / ( £ ) , 3 one and only one x G 5, such that /(rr) = y. Then g(y) = rr. The function g thus defined is called the inverse function of / . It is also denoted / _ 1 . Since g(y) = rr if y = /(rr), it follows that

g(f(x)) = rr; f(g(y)) = y; go f = identity; / 0 g = identity.

Now, we also see that g is strictly monotone. (For instance, assume / is strictly increasing. We shall see that g is also strictly increasing: otherwise, 32/i > 2/2 and g(yx) < g(y2). If y\ = / (# i ) , y* = f(x2), we get g{yx) = xx < g(y2) = rr2, hence => / (# i ) = yi < f{x2) = 2/2, a contradiction).

Theorem 3. Let /(rr) be a strictly monotone continuous function in a closed bounded interval [a, b]. Then its inverse g, defined on /([a, b]) is also continuous.


(Note that the image of [a, b] under the strictly increasing continuous function / is the closed interval [/(«),/(&)] and, if / is strictly decreasing, it is the closed interval [f(b),f(a)].)

Proof. For any e > 0 consider the continuous function h(x) = f(x + e) — f(x) defined on the interval [a, b — e\. Hence, by a previously proved result, it attains its minimum value 6 at a point x G [a, b — e]. Hence h(x) = f(x + e) - f(x) — 6 > 0 (as / is strictly increasing).

Now, take any two points rci, x2 in [a,b]. lix2 > Xi+e then f(x2) — f(x\) > f(xi + e) — f(xi) > 6 and if x\ > x2 + e then f(xi) — f(x2) > 6. Thus, if |xi - x2\ > e => |/(2;i) - /(a;2)| > S. Setting / (x i ) = yu f(x2) = y2, we conclude that, if \g(yi) — g{y2)\ > £ then |T/I — y2\ > 6. This means that if I2/1 — 2/21 < # then |<7(?/i) — #(2/2)I < £• Thus g is continuous. □

Remark . A completely different proof can be given as follows: If g is not continuous in /([a, 6]), 3y in this interval and a sequence yn —► y such that <7(yn) -» g(y)- This last property implies the following: 3e0 > 0 and a subsequence (2/nfc) convergent to y such that \g(ynk) — g{V)\ >£o VfcGN. Let xnk = g{ynk). By the Bolzano-Weierstrass theorem, a subsequence (xnfc ) will converge to x G [a,6]. Hence, as / is continuous, we get: f(xnki) —> / (x ) . On the other hand, (f(xnte )) = (7/nfc ), as subsequence of (ynk), will converge to y. Thus, y = f(x) hence x = g(y).

However, we previously had the lower bounds: \g(ynk )~g{y)\ — \%nk —x\ > SQ, VZ G N. This contradicts xnk —> x. D

IV.6 Uniform Continui ty

Let / , S —y E be a continuous function: Then, Ve > 0 and Vx G 5, 3 5 > 0 (depending on e and on x) such that y G 5 and |2/ — a;| < £ => | /(#) — f(y)\ < e.

Consider the set {£(e, x)} where e is fixed and x varies in 5. As S(e, x) > 0, we see that the set {6(e,x)}xes is bounded from below. Let 6(e) = mi6(e,x). If 6(e) > 0 we obtain the following property: Ve > 0, 3 6(e) > 0 such that, x G S and y G 5 with |y - x| < 5 => | / (z) - / (y) | < e.

This latter property is called uniform continuity of / on S. It appears that it refers to couples of points x and y: if x and y are sufficiently close, the values f(x) and f(y) are as close as we wish. Obviously, uniform continuity implies continuity at any point of 5; it suffices to think of x as fixed and y close to it, so that f(y) be as close to f(x) as we need.


However, there are examples of continuous functions which are not uniformly continuous: for instance, let 5 = E, f(x) = x2. Consider the sequences xn = n, yn = n + £. We obtain |xn - yn\ = £, | / (x n ) - f(yn)\ = \n2 - (n + ^ ) 2 | = 2 + 4j- > 2. It is now obvious that if e = 1, there is no 8 > 0 such that \x — y\ < 6 => \f{x) — f{y)\ < 1 (if such a 8 would exist, take ^ < 6; then \f(xn) — f(yn)\ < 1 which is not true!).

A similar example of continuous but not uniformly continuous function is given by g{x) = £, 0 < x < 1. (Try it!)

There is nevertheless a special situation where (global) continuity and uniform continuity are equivalent concepts. Let us state this as

Theorem 1. Let f{x) be a continuous function on the closed bounded interval [a, b]. Then f is uniformly continuous on [a, 6].

Proof. Let us assume / continuous on [a, b] but not uniformly continuous. This expresses itself as follows:

There exists SQ > 0 such that, \/6 > 0, 3 a couple x$, y$ in [a, 6], where \xs -ys\<6 but \f(xs) - f(ys)\ > e0.

Let us take 6n = £; if £n = z±, 7yn = yi, we get |£n - rjn\ < ± but l / U n ) - / ( ^ n ) | > e o .

Now, all £n belong to [a, 6]. We apply Bolzano-Weierstrass theorem: a subsequence (£np) is convergent towards £0 £ [a, b]. Besides: \rjnp — £0| < l^np " £ n j + |£np " & I < ^ + |£np " & | < \ + |£np ~ U T h u s J | im Ifc, = &

too. Let us use at this point the continuity of / in £o- It follows: lim /(£ n p) =

p—KX>

/(fo) = lim / ( r / n j . Hence lim |/(£».) - /(*?np)| = 0 which contradicts the p—>oo ' p-*oo

lower estimate: | /(£n p) - /(iyp)| > e0, Vp G N. D

Remark . An important class of uniformly continuous functions are the so called Lipschitz continuous functions on a set S. This means that, 3 M > 0, such that x,y G S =» | / (z) - / (y) | < M|x - y|.

If we take any e > 0, we can use 6 - jfc to get \f(x) - / (T/ ) | < e. Therefore / is uniformly continuous on S.

Example 1. Take f{x) = ^ = , 0 < x < 1. We see that: f{x) -

f(v) = i . _ . 1 = ^ - ^ I 1 = Ir^, , , / ^ - Therefore we


get \f(x) - f(y)\ < J*z£l, as 0 < a; < 1, 0 < y < 1. Hence / is Lipschitz continuous on [0,1].

Example 2. Let f(x), E -► E be defined by: f(x) = j ^ . We shall establish the upper estimate:

\f(x)-f(y)\<\x-yl Vrr, y G K .

We have in fact: ^ - ^ = ( 1 ^ a + y » ) = (y - s)( (1+a*)y(1+y») +

(l+xÂi+y^))- N e x t w e s e e t h a t (i+«41i+y») ~ I+SP" - *' (i+x41i+ya) -(^^Mpy < I . We obtain \f(x) - f(y)\ < \y - x\{\ + I) = \x - y\.

Example 3. Let / , S —► E, be uniformly continuous, and (xn) be a Cauchy sequence in S. Then (f(xn)) is also a Cauchy sequence.

In fact, we have: Ve > 0, 36(e) > 0, such that \x — y\ < 6, x, y G S => \f(x) — f(y)\ < e: also W > 0, 3n(e') such that \xn — xm\ < e' for n, r7i > n(e').

Take thus e > 0; we find n(6(e)) such that |xn — x m | < <5 for n, m > n. Then | / (# n ) — f(xm)\ < e for n, m > n(<5(£)) = n0(e). This proves that (f(xn)) is a Cauchy sequence.

IV.7 Exercises

1. Prove that the function / , E -► E defined by: f(x) = 1 if x G Q, /(re) = 0 if x ^ <9, is discontinuous at every point a G E.

2. Discuss the points of continuity and discontinuity of the "integer part" function: f(x) = [x] = greatest integer < x.

3. Let / , E —► E be defined by: f(x) = x if x is irrational, / (#) = 0 if rr is rational. Prove that / is continuous at x = 0 and discontinuous elsewhere.

4. Let 5 C E and / , g continuous functions at a G S. Define functions ft, &, 5 —► E by: ft(a:) = max{/(a;),^(a:)}, k(x) = mm{f(x),g(x)}. Prove that both functions are continuous at a.

5. True or false? (explain): (i) If / + g is continuous at a then both / and g are continuous at a; (ii) If | / | is continuous at a then / is continuous at a.

6. If / , [a, b] —* E is a continuous function, prove that there exists a continuous function / , E -> E, such that f(x) = f(x) Vx G [a, 6].

7. If / , E -» E and #, E —► E are both continuous functions and if, furthermore, / ( r ) = g(r), Vr € Q, then f = g.


8. Let / , R -+ E and Gf (graph of / ) be the set of couples {(x, / (x)) , x G R}. Suppose / is continuous and prove that: (xniyn) G G/, x n —► x, yn —► y => (x,2/) G G/.

9. Find a function / , R —► R, which is not continuous but has the above property.

10. Let / , (a, b) —► R and g, (a, 6) —► R be right-continuous functions. Show that f + g and / • #, (a, 6) —> R are also right-continuous.

11. Indicate an example of an (everywhere) discontinuous function, R —► R, and a non-constant, everywhere continuous function g, R —► R, such that their composition g o f is continuous everywhere.

12. True or false? (explain): If / C R is a bounded interval and / , / —> R is continuous everywhere, then / is bounded.

13. Show that a polynomial function p, R —> R is bounded if and only if it is a constant.

14. If / , g are two uniformly continuous functions, S —> R, then / -h # and | / | are again uniformly continuous. What about / • g?

15. Let S C R be a bounded set and / , S —> R be uniformly continuous. Then / is bounded.

Chapter V

DERIVATIVES

The concept needs no introduction; it is the main theme of differential calculus, one of the major discoveries in mathematics, and in science, in general.

V . l Definition of the Derivative

Consider a function / , defined on (XQ — r, xQ + r) for some r > 0. Consider also the quotient

f( \ -f ( \ g(x) = , which is well-defined for 0 < \x — x0\ < r,

x — Xo

(it is called the "difference-quotient" function associated with / ) . We say that / is differentiate at XQ ("has a derivative" at XQ) if (and only

if) lim g(x) exists. We denote this limit by ff(x0). X—►Xo

We note the following interesting

Theorem 1. Let / , (xo — r,#o + f) —► E; then f is differentiate at XQ if and only if there exists a function F, (XQ — r, xo + r) —► R, such that F is continuous for x — xo and the representation formula

f(x) - f(xo) = F(x)(x - x0) holds, Vx G (x0 - r, x0 ■+- r). (1.1)

In this case f'(x0) = F(x0).

81


Proof. (i) First, assume / to be differentiate at Xo- Let us define F(x) =

fW-f(x°) = g(x) for 0 < |x - x0 | < r, and F(x0) = f'(x0). Then lim F(x) = x xo X—>Xo

/ ' ( X Q ) — F(xo) so that F is continuous in x0. Next, from above, we see that f{x)-f(x0) = (x—x0)F(x) for 0 < |x—x0| <

r; moreover, if x = xo, the equality is also true.

(ii) Conversely, if (1.1) holds, we get for x ^ x0; / ( xx I ^ X o ) = F(x) -* F(x0)

as x —► x0 hence f'(xo) exists and f'(xo) = F(xo).

Remark. A function F with above properties is uniquely determined. In fact, for x ^ x0, we have Fix) = *W-HX°) and F(x0) must be lim A*)-/(*o)

7 ' U ' V ' X —Xo v u / x—»-xo x — x°

which is unique (We leave to the reader to show that the limit of a function in a point is uniquely determined).

Corollary. / / / has a derivative at xo, it is continuous at xo-

Proof. Obvious, as / (x) = f(xo) + (x — xo)F(x), hence lim / (x) = f(xo). X—>-Xo

Examples. (1) The function / : E — {0} —► E defined by f(x) = ^ is differentiate at

x0 G E/{0}, with f'(x0) = - \ .

In fact, given xo / 0, / is defined in ]x0 — xo, #o 4-x$ = 2xo[ — ]0, 2xo[ (for x0 > 0) or ]2x0, 0[ (for x0 < 0). We also have

f(x) - f(x0) = = (x0-x) = -(x - x0)-X Xo X • Xo X • XQ

Thus, (1.1) holds where F(x) = - ^ - , 0 < x < 2x0 (when, say x0 > 0). As x —► — - is continuous in x = xo we get / '(#o) exists and = —\.

(2) The function / : E -» E defined by: / (x) = xn is differentiate at any point x0 G E and /'(xo) = nx^~l.

We have in fact

/ (x) - /(xo) = x n - x£ = (re - xo )^ 7 1 - 1 + rrn-2x0 + xn~3x20 + • • • x^"1)

Now, the polynomial function

P(x) = x n _ 1 + xn~2x0 + xn _ 3xg + • • • x ^ - 1

Derivatives 83

is continuous for x = x0, and P(x0) = nx$~l — ff(x0). D

V.2 Algebra of Derivatives

Making good use of Theorem 1 (V.l) we can reduce proofs of the basic "laws of differentiation" to some simple algebraic manipulations.

Let us state the following

Theorem 1. Let f, g be two functions defined on ]XQ —r, XQ +r[ and with real values. If f and g are differentiate at xo, then so are f + g, af(a G M), and f • g; furthermore we have

(f + 9y(xo) = f'(x0)+g'(xo) (af)'(xo) = af(x0) (fg)'M = f(x0)g'(x0) + f(x0)g(xo)i

if moreover, f(x0) ^ 0, then j is defined on]x0—ri, x0+ri[ (for some ri > 0),

and (j) (xo) exists = —pK°\-

Proof. Using the representation formula (1.1), we find continuous functions in x0, F, G (both defined for 0 < \x — x0\ < r ) , such that

f(x) = f(x0) + (x-x0)F(x) g(x) = g(x0) + (x- x0)G(x).

It follows that (f+g)(x) = (f+g)(x0)+(x-x0)(F+G)(x), F+G continuous for x = x0. Thus ( / + g)'(x0) exists = (F + G)(x0) = f'(x0) + g'(x0).

The proof for af is similar.

Next, if f(x) = f(xo) + (x — x0)F(x), g{x) = g(x0) + (x — xo)G(x), it follows that

f(x)-g(x) = f(xo)g(xo)-h{x-xo)[g(x0)F{x) + f(xo)G(x) + (x-x0)F(x)G(x)].

The function g(x0)F(x) + f(x0)G(x) + (x - x0)F(x)G(x) = H{x) is obviously continuous for x = XQ and H(x0) = g(xo)F(xo) + f(x0)G(xo) = g(xo)f'(xo) + f(xo)g'(xo). This gives the rule of differentiation for the product of two differentiate functions.

Next assume f(x0) ^ 0. We know that / is continuous in x0; if we use "Lemma" of Theorem 1 (IV.4) we find an interval (x0 — ri,x0 + r*i) where f{x) # 0.


In this interval we have the obvious relations:

1 X l :[/(*) " /(*o)] = - 7 7 ^ 7 7 T T ( * " X°WX)' f(x) f(x0) f(x)f(x0)UK ' 7V uyJ f(x)f(x0y

Here we see that the function K(x) = - / ( X ) ^ X Q ) • F(x) is continuous in x0.

Therefore we obtain: ( j j (x0) exists and = -JSJ^J = - /a (so) • a

Example. Consider the function g(x) = rr~n = ^ , M/{0} -+ R. Using last theorem we see that, Vxo / 0,

nxn (T-n)f - ° - -vr-™-1

V.3 Composition of DifFerentiable Functions and the "Chain Rule"

Let us consider two functions / , g where / : (x0 — r, XQ + r) —» M while £ : (/(ô) - P,f(xo) + p) -> R and also | /(x) - / (x 0 ) | < p if |x - x0 | < r; accordingly the composition g o / , (:r0 — r, XQ + r) —► M is well-defined. We have now

Theorem 1. Assume that f'(xo) and g'(f(xo)) both exist. Then g o f is differentiate at Xo and the equality

{9 o f)'(xo) = g'(f(xo))f'(xo) holds true.

Proof. Let us write: h{x) = (go f)(x) = g(f(x)), x0 — r < x < x0 + r. We have:

f(x) - f(x0) = (x- xo)F(x), for \x - x0\ < r

and then

9(y) - 9(Vo) = (y~ Vo)G(y), for \y - y0\ < p, where y0 = / (x 0 ) ,

with F continuous in XQ and G continuous in y0.

We obtain therefore the representation formula for g o / :

g(f(x)) - g(f(x0)) = (f(x) - f(x0))G(f(x)) = (x - x0)F(x)G(f(x)),

Derivatives 85

where the function x —► F(x)G(f(x)) is obviously continuous at x0. This shows that the derivative

(<K/))'(*o) exists and - F(x0)G(f(x0)) = f(x0)g'{f(xQ)). □

V.4 Some More Remarks

Let us consider functions /(•) defined on the interval [x0,x0 + r[, and let us assume that the right-limit

lim f{x) ~ fixo) exists. x->x0 X — XQ X>XQ

(thus, we speak here about the right-limit at x0 of the "difference-quotient" function g(x) = x_x > whose definition domain is the interval ]xo, XQ +r[) .

This limit is the so called right-derivative of / at x0 and is denoted by

Similarly, if /(•) is a function ]XQ — r, XQ] —► E and the left-limit

lim f{x) ~ f M exists, x-+x0 X — XQ X<Xo

we say that / has left-derivative at x0 (denoted by /!_(x0)).

Example 1. f(x) = \x\ has right-derivative for x = 0, /+(0) = 1, and left-derivative f_(0) = - 1 .

Remark 1. A function / , ]x0 - r, x0 + r[ —► E is differentiable at a:0 iff f+{x0) and fL{x0) both exist and are equal. Therefore, f(x) — \x\ has no derivative at x = 0.

Remark 2. Let / , ]a, b [—► E be differentiable at any point of ]a, b[ and increasing there ("a* < x2 => /(a*) < /(rr2)"). Then /'(a;) > 0 Vrr € ]a, 6[. In fact,

//(X) = iim f(x + h)-f(x). here / (a. + h)_ f{x) > 0> h.—>o Ai

if ft > 0, hence ft (a;) > 0; also f'(x) = f'+{x) > 0. □

Definition. / / (aJo>/(zo)) w a point m G/ (graph of f), we define the


tangent line at (xo,/(xo)) as being given by the equation

y- f(x0) = f{x0)(x-xo)

(assuming existence of the derivative at x0).

Example 2. I f / ' (x 0 ) exists, then / ' (x 0) = lim f ^ + h^~J^~h\ I n h—>o 2ft 7i>0

fact, we can write / (so + fe) - / (x 0 - ft) = 1

2ft 2 next we use:

/ ( x 0 + ft) - / ( x 0 ) / ( x 0 ) - / ( x 0 - ft) ft ft

,. f{xo + ft) - f{x0) , , . w/ \ !imn i; = UM = / (*o) /i-fO ft h>0

and ,. /(so - ft) - / ( x 0 ) , , l i m n »; = f-{xo) = J (ô)-/i->o —a h>0

V.5 Rolle's Theorem and the Mean Value Theorem We present now two (classical)-fundamental-results in calculus.

Theorem 1. Let /(•) be continuous, [a,b] —► E, / ' (x) exists Vx G ]a, 6[ and / (a ) = f(b). Then, 3c G ]a, b[, such that / ' (c) = 0.

Proof. Let M = sup/(-) = / (x i ) and m = inf /(•) = /(X2), where xi , [a,6] iaM

x2 G [a, 6]. If m = M then / is constant and / '(c) = 0 for all c G ]a, 6[. Suppose m < M that is / (x x ) > / (x 2 ) . Since / (a) = /(6), not both of xi

and X2 can be endpoints of [a, 6], so at least one of them must be an internal point. For example if X2 G ]a, b[, we see that / (x) > /(X2) for all x "near" to x2. It follows that / ( a 2 l ^ a 2 ) > 0 if x2 < x < x2 + ft (for some "small" ft), and accordingly /+(x2) > 0; also we have ^Zx < 0 for X2 — ft < x < x2 and then f'_(x2) < 0. Therefore: / ' (x 2 ) = f'^xl) = fZ(x2) = 0.

A fundamental Corollary is the "mean value theorem", here stated as

Theorem 2. Let /(•) be continuous, [a, b] —> E and / ' (x) exists, Vx E]a, b[ . Then, there exists a point c G ]a, b[ such that

f{b)-f(a) = (b-a)f'(c).

Derivatives 87

Proof. Consider the auxiliary function #(•) defined in [a, b] by the expression:

g(x) = f{x) - / (a) - mbZf

aia)\x - a), [a,b] - K.

It satisfies obviously the conditions of Theorem 1 with g(a) = g(b) = 0. Therefore, 3c G ]a, b[, such that g'{c) = / '(c) - m~Ja

{a) = 0. D

Using above Theorem we obtain readily

Corollary 1. Let /(•) be continuous, [a, b] —► R, differentiate, ]a, 6[—> R, and / ' (#) = 0 Vx G ]a, 6[. T/ien / 25 a constant function.

In fact, for x £ ]o, 6] we get f(x) — f(a) = (x - a)/ '(£) = 0 (where a<£<x).

Corollary 2. Let /(•) be continuous, [a, 6] —* R, differentiate in }a, b[ and f'(x) > 0 Vrr € ]a, 6[ . 77ien /(a;) zs increasing on [a, 6]; furthermore, if f'(x) > 0 \fx e ]a, b[, f{x) is strictly increasing in [a, 6].

In fact, for a < x\ < x2 < b we have f(x2) — / (^ i ) = (^2 — #i)/ ' (c) where Xi < c < x2.

Corollary 3. Let /(•) be continuous in [a, b], differentiable in ]a, b[ and the derivative / '(•) be bounded on ]a, b[ . Then f is uniformly continuous on [a,b).

In fact, again we have, for a < x\ < x2 < b; f(x2) — f{x\) = (x2 —xi)f'(c), where x\ < c < x2. It follows that 1/(0:2) - / ( ^ i ) | = \xi - £ i | | / ' (c) | < M(x2 — xi), and | / (^ i ) — f(x2)\ < M\x\ — x2\ too. Therefore, the function /(•) is Lipchitz continouus, hence also uniformly continuous.

Example 1. Let / be continuous on [a, 6], differentiable in ]a, b[ , and A = lim f'{x) exists. Then / i ( a ) exists and = A.

X—Kl x>a

In fact, for a < x < 6, we have: f{x) - f(a) = (x - a)/ ' (£), a < £ < x. Hence f^~J{a) -A = / '(£) - A, and then

/ ( * ) - / ( * ) A{

= \f'(0-A\. x — a

We know (from the assumption), that Ve > 0, 36(e) > 0, such that \f'(x) -A\ < e for a < x < a + 6. Therefore |/ '(£) - A\ < e for a < x < a + 6


(as a < £ < x), and we obtain

x — a <e for a<x<a + 6(e)

which means precisely that f+(a) exists and = A.

Example 2. Let / be twice differentiable in the interval [a, b]. Assume that 3 a sequence (xn)^°, xne[a, 6], xn ^ xm, Vn ^ ra, such that xn ^ x e [a,b]W and f(xn) = 0 VneN. Then f(x) = f'(x) = f"{x) = 0.

First assertion follows from continuity, Next, we have

m = iim m - f M = 0. n->oo X — Xn

Furthermore, using Rolle's theorem, we find, in any interval ]x n , x n + i [ or ]rcn+i,xn[, a point fn such that / ' ( f n ) = 0. (and fn ^ x)

On the other hand, we obviously have: |£n — x\ < |£n — xn\ 4- \xn — x\ < |xn+i — xn\ + \xn — x\ —> 0 as n —» oo; thus lim £n = x. Consequently we find that

Example 3. The following inequality holds true

VTTx < 1 + - Vrr > 0.

In fact, let us consider the function p(x) — 1 + § — v l + x; its derivative is given by: (p(x) = \ - 2,\+x which is > 0 for x > 0 — as y/1 + x > 1 for x > 0. Therefore (Corollary 2) we get: <p(x) > (p(0) = 0, Mx > 0.

(note that here we are using the formula for the derivative of y/1 + x : (vT+~x)' = 2 i + x ? Vx > 0; one way to obtain this result — using only the definition of the derivative — is the following: l e t 9(x) = ^ - ( v T T ^ - ^ / ^ T ^ ) = V l + a H !V l + X Q , the "difference-quotient" function associated with f(x) = y/l + x\ then lim g(x) = . * ).

x—»xo 2vl+JCo

(x) Taking — if necessary — a subsequence, we may assume that xn < x Vn £ N or that x n > x Vn € N.

Derivatives 89

V.6 Differentiability of Inverse Functions

Let us consider a function / , (a, b) -» E, which is differentiable and strictly monotone. Let xo G (a,b) be a point where / '(ô) ¥" 0- Let g be the inverse function of / , precisely of the restriction of / to the closed interval [xo —r, XQ +r] which is contained in (a, b).

If m and M are the infimum and the supremum of / in [x0 — r, x0 + r], we know from Chapter IV, that the inverse function g is continuous, [m, M] —> [xo — r, xo + r]. We shall now prove

Theorem 1. If y0 = /(xo), £Ae derivative g'(yo) exists and the relation g'(yo) = jr^j holds true.

Proof. Let us take any sequence (yn) where yn G [m, M], yn ^ i/o> Vn G N and yn —► 2/0• Next, consider the difference-quotient g - iffi ° associated with g.

We have yn = / ( x n ) , x n = #(2/n), where x n G [x0 - r ,x 0 + r]; also yn ^ y0 => xn ^ xo, and we see also that lim x n = #(yo) = xo (using continuity of

n—»>oo

the inverse function g). We now write the simple relations:

g(yn)-g(yo)_ x n - x 0 _ 1

This implies:

yn ~ yo f(Xn) ~ / ( X 0 ) / ( * n ) - / ( « o )

,. ^(3/n) -0(3/0) A 1 lim ^ -^ exists and = n-*°° yn-yo / ' ( zo) '

It follows also that g'(yo) exists and = 777^7 • Q (In fact, things are a little more complicated: we need a fundamental result, which is now given for the first time:

Proposition. Let h be defined in the open interval (a, b) with the possible exception of a point xo G (a, 6). Then lim h(x) = A exists if and only if: for

x—*xo

any sequence (xn), where xn G (a, b), xn ^ xo Vn G N, and x n —► xo, we have lim h(xn) = A.

n—►00

(i) Let us assume first that lim h(x) — A. Then, We > 0, we have £(e) > 0 X—^XQ

such that \h(x) - A\ < e for 0 < |x - x0 | < 6(e).


Now, we also have 0 < \xn — x0\ < S for n > n(6(e)) — from the assumption that lim xn — XQ. Therefore we obtain: \h(xn) - A\ < e for n > n(8{e))\ we get lim h(xn) = A.

n—► oo

(ii) Let us assume, conversely, that lim h(xn) — A for any sequence (xn) where xn G (a,6), xn ^ x0, Vn G N and lim XJI — XQ. Assume also that: lim h(x) = A does not hold. This means: There exists £0 > 0 such that,

x—>xo V<5 > 0, 3xs G (a, 6), 0 < \xs — xo\ < 6 and \h(xs) — A\ > eo- Take 6n = ^ and denote rc± with £n; thus £n G (a, 6), 0 < |£n - x0 | < ^ and |/i(£n) - ^4| > £o, V n G N . "

We got a sequence (£n) where £n —» xo, £n 7 #o> while /i(£n) -^ -4, which is a contradiction.

The Proposition is therefore completely proven.

Example. Let us consider the function / given by: f(x) = x», x > 0. It is the inverse of the function: x = g(y) = yn, y > 0.

The derivative of g, g'(yo) will exist for all y0 > 0 and also: g'(yo) = ny^-1 ? 0 for y0 > 0.

Therefore, the above theorem gives that the derivative of / also exists, Vrr0 > 0 : f(x0) = ^ ^ y where y0 = f(x0) = xfi . It follows that: f'(x0) = ^=r = (^)4"1 .Vxo>0.

V.7 Local Extrema

The concepts of local (and also global) extrema of a function are often very useful and in the case of differentiable functions we can use derivatives to find these (points) of extremum.

Consider functions / , 5 c I R - > E . We consider interior points of 5, a point c G S such that (c — r, c -f r) C S for some r > 0, then the point c is called a local maximum of / iff f(x) < /(c) for all x G (c — 6,c + 6) for some S > 0; c is a local minimum of / iff f(x) > /(c) for all x G (c — <?>, c + <5), again, for some 6>0.

Then, a local extremum is a local maximum or a local minimum. A global extremum of / , S —> E is defined as a point c e S, such that /(c) < / (#) , Vx G 5 (global minimum) or /(c) > f(x), Wx G S (global maximum). (Note that a global extremum need not be interior to 5).

Example 1. Define / , [-1,1] -> E by f{x) = x. We see that / ( l ) =

Derivatives 91

1 > fix), V:r G [—1,1]. Hence x = 1 is a global maximum of / . Likewise, rr = —1 is a global minimum for / . Moreover, x = 1 and a; = — 1 are not "interior" points of S = D(f) (no interval ( l - £ , l + £) is contained in [—1,+1]; same for (—1 — S, —1 + 6)). Therefore, they are not local extrema.

Example 2. Define / , [-1,1] —► E by f(x) = \x\. We readily see now that x = 0 is a local minimum (and also a global minimum!). There are global maxima at x — — 1 and x = 1 but they are not local maxima.

Example 3. Define / , E -> E by f(x) = x2(x- §). Note that /(0) = 0 and f{x) < 0 for |x| < 1. Thus 0 is a local maximum. Consider next the behavior of / near x = 1. Write x = 1 + ft, so that / (x) - / ( l ) = / ( l + ft) - / ( l ) = (1 + ft)2(ft - | ) + \ = h + 2ft2 + ft3 - | - h - \h2 + \ = ft2(ft + | ) . Therefore f(x) — / ( l ) > 0 if ft > —|, that is a: — 1 > - | , £ > —|, and we found that a; = 1 is a local minimum.

Moreover, there are no global extrema (the function is unbounded on E, both above and below).

In above examples we looked for extrema while lacking a systematic method. In the third example we may note that the derivative f'(x) = Sx(x — 1) is zero for x = 0 and x = 1. This suggests that the derivative (if it exists) is zero in a local extremum. Geometric intuition suggests the same property. This is true in fact, as established in

Theorem 1. If c is a local extremum and f'(c) exists, then / ' (c) = 0.

Proof. We shall derive this result as a consequence of the following.

Lemma 1. Let us assume that the function f, S —» E, has a local maximum at x = c. Then

(i) If f is right differentiate at c it follows that f'+{c) < 0, (ii) If f is left differentiate at c, then /1(c) > 0.

Proof of Lemma. (i) As x = c is a local maximum, there is 6 > 0, such that (c — <5, c + 6) C S and f(x) < /(c) for \x-c\ < 6. Furthermore (this is a general property — very easy to verify — of right-limits of function): V sequence (xn) where c < xn < c + 6 and lima;n = c we have f'Ac) = lim ^(x^)~{(c); on the other hand we also

^ n—>oo Xn c

have that f(xn) - f(c) < 0 while xn - c> 0. Thus we obtain /+(c) < 0. (ii) It is very much the same!


Using above Lemma and the existence of / ' (c) we find (where c is a local maximum): / ' (c) = f+(c) < 0 and / ' (c) = f'_\c) > 0. Hence, / ' (c) = 0.

In the case where the local extremum c is a local minimum, we may note that x = c is a local maximum for the function —/; if / '(c) exists, then (-f)'(c) = - / ' ( c ) = 0, hence / '(c) = 0 too. □

Remark. The converse of this theorem is in general false. For instance, let us take the function / given by f(x) = x3, Vrr € E. Then / ' (0) = 0 but, as easily seen, x — 0 is not a local extremum for / .

The following is also an important concept:

Definition. Let f,S-+Rbea function and x = c be an interior point of S; we say that c is a critical point of f if f has a derivative at c and / ' (c) = 0.

From previous arguments we derive: all local extrema of a differentiable function are critical points; there are critical points which are not local extrema.

Example 4. Let us define a function / , E —► E by the formula: f(x) = x3(1 — x). Then f'{x) = Sx2 — 4x3, Vrr G E and accordingly we see that f(x) = 0 iff x = 0 or x = f. Thus, / has two critical points and we shall check if they are (or not) local extrema. As /(0) = 0, f(x) < 0 for x < 0 while f(x) > 0 for 0 < x < 1 we see that x = 0 is not a local extremum. Next, we have to examine the difference: / ( f ) — /(#) and try to see if it keeps a constant sign for x "near" f. We prefer however to use a general result, involving "second derivatives" of a function which provides a sufficient condition for a point to be a local extremum.

If / , ]a, b[ —> E has a derivative / ' for any point x0 G ]a, b[ we can consider the derivative function / ' , again ]a, b[ —> E. (For instance, for the above defined function: f(x) = x3(l-x) we found the derivative function: f'(x) = Sx2 -Ax3, \/x G E). Then, the second derivative in a point x0 G ]a, b[ ,f"(xo) will be defined as:

r{xo) = l i m / ' ( » ) - / ' ( «o ) . x^x0 X — Xo

We now state

Theorem 2. Let / , ]a, b[ —> E has a derivative / ' Vx0 G ]a, b[ ; let c G ]a, b[ such that: / ' (c) = 0, /"(c) exists and is < 0; then x = c is a local maximum. Furthermore, if / ' (c) = 0 while /"(c) exists and is > 0, then x = c is a local minimum.

Derivatives 93

Before giving the proof, let us apply this result in Example 4 above. We have / ' ( f ) = 0; also f"(x) = 6x - 12x2, Vx G E, hence / " ( f ) = f - f < 0. Therefore we can conclude, without further reasoning, that x = | is a local maximum.

Proof of Theorem 2. Let us assume that c G ]a, 6[ , / ' (c) = 0, /"(c) = Um / f( c +^" / #< c) < 0. This will imply: 35 > 0 such that f(c + h) - f'(c) > 0 for -8 < h < 0 and / ' (c + /i) - / ' (c) < 0 for 0 < h < 8. (otherwise, we can find a sequence (hn), hn < 0, Vn G N, lim hn = 0, and / ' (c + /in) - / ' (c) < 0; then £ [ / ' ( c + /*n) - /'(c)] > 0 and /"(c) > 0, a contradiction; similar reasoning for the second situation).

Therefore we can conclude, if / ' (c) = 0 and /"(c) < 0, that / ' ( c + h) > 0 for - 5 < h < 0 and / ' (c + h) < 0 for 0 < h < 6, or also: / ' (x) > 0 for c- 6 < x < c, f(x) < 0 for c < x < c + 8.

The mean value theorem gives: f(x) — f(c) — [x — c)/ '(^), a; < £ < c, hence / (x) — /(c) < 0 for c — 8 < x < c; in similar way we find that f{x) — /(c) < 0 for c < x < c + 8. Thus, x = c is a local maximum for / .

Finally, the situation where / '(c) = 0 while /"(c) > 0 can be reduced to previous one, considering the function —/.

Remark. The previous sufficient conditions for local extrema are not necessary: for instance, if f(x) = x4, Vx G E, we see that / '(0) = /"(0) = 0 but x = 0 is a local minimum.

V.8 Some More Examples

Example 1. (the "intermediate value theorem" for derivatives). Let us remember the "intermediate value property" of all continuous functions defined on an interval / : their image set: / ( / ) = {f(x),x G / } is again an interval of M. (see Theorem 3 (IV.4)).

It is an interesting fact that a similar property belongs to any function / which has a "primitive" F: that is f(x) — F'(x), Vx G I (an interval on E). In this case, if x\, x2 G / , x\ < x<i and, say, / (# i ) < £ < /(#2) there exists 7] G ]xi, x%[ such that /(ry) = £.

In fact: Let us define the function g, I —» E, by g(x) = F(x) — £ • x. We see that g is continuous on [x 1,3:2], hence it attains a global minimum at a point 77 G [xi,X2]. Now, 77 G ]xi, x 2 [ . (Otherwise, if say, rj = xx, we get: g{x)

xZgx[Xl) > 0 for x > xi, hence g'+{x{) > 0 and F'(xi) - £ > 0, / (x x ) - £ > 0


— a contradiction; if 77 = x2 we arrive at contradiction in similar way). Finally, now 77 — being interior to [xi, x2] will also be a local minimum: hence g'(rj) = 0, that is f(rj) - £ = 0.

(The case xi < x2 and f(xi) > f(x2) is handled in similar way).

Corollary. The function f(x) = sgn x = 1 for x > 0; = — 1 /or x < 0; = 0, /or 2; = 0, /ias no primitive in any interval (a, b) where a < 0, b > 0.

Example 2. (Cauchy's mean value theorem). We present here another extension of Rolle's theorem, involving two difFerentiable functions.

We consider therefore functions /(•), g(-) defined and continuous on the interval [a, 6], and difFerentiable in the open interval ]a, b[. From Rolle's theorem we derive:

f(b) - / (a) = (b- o ) / ; ( 0 , where a < £ < 6, #(&) — #(a) = (6 — a)gf(rj), where a < rj < b.

Hence (b - a)[f(b) — f(a)]g'(r)) = [g(b) — g(a)](b — a)/ '(£) and consequently we obtain

[f(b)-f(a)}g'(v) = l9(b)-9(a)}f'(0-

A simple argument permits us to find a number c e]a, b[ , such that

9'(c)[f(b) - f(a)] = f'(c)[g(b) - g(a)}. (8.1)

Precisely, we introduce the auxiliary function

<p{x) = g(x)[f(b) - f(a)} - f(x)[g(b) - g(a)}

Again </?(•) continuous in [a, b] and difFerentiable in ]a, b[. Furthermore, we see that </?(a) = g(a)f(b) — g(b)f(a) = <p(b). An application of Rolle's theorem gives the result.

Remark. If g(x) = x we obtain Rolle's result as a particular case.

Remark. It is obvious that (8.1) holds iff g'(c)[f(a) - f(b)] = f'(c)[g{a) -g(b)]. Thus, the condition a < b is not essential.

Derivatives 95

V.9 L'Hospital's Rule

L'Hospital theorems deal with calculus of various limits of the form:

(9.1) lim M , ,im Mt ,im M x-*a g(x) x- oo g(x) x-»—oo g(a;)

in the "indeterminate" case when: lim f(x) = lim g(x) = 0 or lim / (x) = i : ~ - , ^ x . lim g(x) = +00, etc. x-va

Under appropriate assumptions we can establish that, for instance

^m.^m^m r^m (,2) (the limit on right-hand side will exist and accordingly, the limit on left-hand side will exist too).

We shall present here two results of this form:

Theorem 1. Let us assume that both functions f(-) and a(-) are defined X l l ^ V / l C l l l J. • JUZ,U UiO UtOOUillbZ, VIVUiU UUUIO J U/IOUUOWIOO J \ I U1IUU1 y \ I Uil Z, UiZ,JblUZ,Ut

for 0 < \x - a\ < r, are differentiate in 0 < \x - a\ < r, and g'(x) ^ 0 in this jor u <» \x — a\ <^ r, are aijjerennaoie in u <. \x — a\ <* r, ana g \x) ^ u m mis interval, as well as g{x). Assume furthermore that lim f(x) = limg(x) = 0,

x—+a x-+a and that lim $ 4 exists = L; then, lim 4 4 = L too.

x-*a.9 \xl x-+a.9(x> The proof is quite simple. Both functions / and g are continuous for 0 < The proof is quite simple. Both functions / and g are continuous for 0 <

\x - a\ < r; let us define: f{a) = g{a) = 0. It follows that f(x) and g(x) are now continuous in a - r < x < a + r, and differentiable in ]a, a + r[ as well as in ]o - r, o[. in ]o - r, o[.

Take Xl such that a < Xl < a + r. Apply Cauchy's mean-value theorem in the interval [a, a*]. We get a point c,a<c<xx, such that [/(zi)-/(a)]5'(c) = [g(Xl) - g(a)]f'{c); using g'(x) ^ 0 in (a,Xl) we get the equality

f<;Xll~ffl = S or also 4^4 = 4 T 4 ' for some c e ("î) (9-3) g{xi)-g(a) q'lc) g(xi) g'(c) I(,Xl\ f{

fal = 4 T 4 or ***> f~M = 4 ? 4 . f°r s°m* c G (a,Xl) (9.3) g(x1)-g{a) g'(c) g(xi) g'(c)

Take now any sequence (x n ) f where a < xn < a + r, and xn -► a. Then, from a < cn < xn < a + r, we see that cn -► a too. Therefore, lim 4 j f 4 = L,

lake now any sequence (xn)^ wnere a < xn < a + r, ana inen, from a < cn < xn < a + r, we see that cn-> a too. Therefore, lim 4 = L,

and consequently, lim 4 H exists = L. We obtain: lim 4 4 exists = L (see

Remark after Examples 1 and 2). Next, we repeat the same reasonings for


x < a, and obtain that lim 4 f j exists = L. This proves the result. x < a

Example 1. Let us compute: lim x +\Tl_2X » w e a r e u nder t n e assump-

X—*1

tionof the above theorem; if f(x) = x4+x3+x2+x—4, g(x) = x3+x—2, we see that / ( l ) = #(1) = 0, / , g are differentiate for all x, and g'{x) = 3x2 + 1 ^ 0 for all x. Also, lim 4 M = lim 4a3+,3T%a+2a+1 =

' x —i $'(*) x _ i 3x + 1

5 2 *

Example 2. Let f(x) — x — 6x6 + 5a;7, #(a;) = (1 — x)2\ again / ( l ) = g(l) = 0, / , # are everywhere different iable, and g ' ^ ) — 2(# — 1), so that o'(a;) ^ 0 for a; ^ 1. However, it is not clear if lim 4 ) 4 = lim i-36*5+35s6

* v ' ^ ^ ' x->l 0 W x-+l 2(x-l) exists, as again lim(l — 36a;5 4- 35a;6) = 0.

x—>1

We then apply a second time L'HospitaTs rule, and we therefore obtain

,. f'(x) ,. f"{x) ,. -180a;4 + 210a;5 i r lim ^-f^ = lim J—T-!- = hm = 15. *—i g'{x) x->i g"(x) x—I 2

Remark . In the proof of Theorem 1 the following result has been used. "Let <p(x) be defined for XQ < x < x and lim tp(xn) = A exists for any

n—>-oo

sequence (a;n)£° where xn > XQ and xn —► a;o- Then, lim <p(a;) = <p(#o-f) X—>Io x>xo

exists and = A" (if this is false, 3e0 > 0 such that, Vn € N, 3xn G (x0, a;o + ^) where |</?(a;n) — A| > £0- Then we have a sequence (a;n), where a;n > xo, Xn —> #o a n d <^(a;n) is not convergent to A).

Example 3. We have

lim 7= = lim , = lim . = 0. x^0 y/x x^0 -^= x-+0 J l + x x>0 x>0 2 v x x>0

(we apply here the obvious one-sided L'Hospital theorem, where everything is given only for 0 < x < r).

Example 4. We have Q _l_ % l

lim —= — = lim —=— = lim 2y/^x = 2\/3.

Derivatives 97

Next, we present another form of L'Hospital's rule.

Theorem 2. Let us consider two functions /(•) and g(-), defined for 0 < \x — a\ < r, and differentiable, where g'(x) ^ 0, such that lim f(x) =

x—*a lim q(x) = -hoo. Assume also that lim £jjH exists=L. Then lim ^ l exists x->a V ' x->a 9 {*) x-+a 9(x) — L too.

Remark . From the assumptions that lim f(x) = lim g(x) = -hoo it fol-x —*a x—+a

lows that f(x) > 0 and g(x) > 0 for 0 < \x — a\ < r\ (with some r\ <r).

Proof of Theorem 2. As in Theorem 1 we consider separately the right f(x\ f(x\

limit lim —^—- and the left limit, lim —;—r. x->a g(x) x-*a g\x) x>a x<a

We know that: Ve > 0, 36(e) > 0, such that, if a < t < a + 6 then i£$-Li< e .

Let us consider now x and y, such that a < x < 7 / < a + r i . Note that g(x) — g(y) = (x — y)g'{0 ^ 0. We now apply Cauchy's mean-value theorem (see V.8) and find, for some z G (X,T/), the equality

f'(z) = f(x)-f(y) = / (x) l -gS ^(s) g(x)-g(y) g(x)i-Sbll

g{x)

(provided that x is sufficiently close to a, so that 1 — 4]4 ^ 0). This relation can also be written under the form:

f(x) _f(z)l-3M g(x) g'(z) i _ m

(provided that x is sufficiently close to a, so that 1 - jA ^ 0). Take now a fixed y<a + 8(a<y<a + 6<a + ri). It results

f(z)_L < £.

Next, let us write the following immediate identity:

six) T (m Ly-9M XL(l-9M 9(x)-L-{9'(z)-L)1-m+L[1-i%

- 1


We derive the estimate:

I/O*) 9(x)

II-<£-

11 Ml + |L|

1 _ g{y) 1 g(x)

1

Now, if a; > a is sufficiently close to a we obviously have:

i i - g(v)i

l»-» 4S-<2aad '"fS

»-«9 This will imply:

0(3) - L < 3e for a < x < a + 5i(e).

- 1 < 2\L\

V.10 Taylor's Formula

In this section we extend the mean-value theorem (see Sec. 5 of this chapter) to functions which possess higher-order derivatives. Precisely, let us assume that /(n)(-) is continuous on [a, b] and differentiable on ]a, b[. Thus, / ' , / " , . . . f^ exists and are continuous for a < x < b while / ( n + 1 ) exists for a < x < b (without necessarily being continuous). We then are able to formulate

Theorem 1. (Taylor's formula with Lagrange's remainder) Let f and its first n derivatives be continuous when a < x < b (where a < b) and let the (n + l)-st derivative /^n+1^(x) exist when a < x < b. Then, there is a number £, a < £ < b, such that

f(b) =/(a) + (b - a)f'(a) + ^ # / " ( a ) + . . . ^ ^ / ( n ) ( a ) 2! n<

(n + 1)! J *' The same formula holds in case b < a, all the inequalities then being reversed.

Remark. When n = 0 we recover the previously found mean value theorem.

Proof. The following proof makes use of some auxiliary functions together with application of Cauchy's mean-value theorem. Thus, let us put

F(x) = f(b)-f(x)-f(x)(b-x) TV.

-(b - x)n and

G(x) =

Derivatives 99

(b - x ) n + 1

( n + 1 ) ! Note that F(b) = G(b) = 0. Also, it is an "easy" exercise to see that

n\

We also have G\x) = -i±=^)1. Applying Cauchy's mean value theorem we obtain:

F(a) F{a) - F(b) F ' ( 0 , , . , 7^~\ ~ Tu~\—TUJA ~ 7 777\"> w n e r e £ 1S between a and b. G(a) G(a) - G(b) G'(£)

Hence F(a) = ^rn\G(a) and therefore

F(a)- if<n+l)(:)(b-on(b-ar+i _ (n+1)(c)(b-ar+l

n!

On the other hand we have:

F(a) = f(b) - f(a) - f'(a)(b -a) ^^-(b - a)n.

This gives the result. (This proof is valid in both cases, a < b or b < a, because it is based on

Cauchy's mean-value theorem which is left invariant by an interchange of a and 6).

Remark . The previously proved result can be given alternative formulations by simple notational changes. One such formulation occurs when b = a+h where ft may be either positive or negative. The number £ between a and a + ft may then be written in the form £ = a + 0ft, where 0 is some number between 0 and 1. Thus we now have

f(a + h) = f(a) + hf'(a) + ... ^ / ( n ) ( a ) + j ^ y / (n+1)(<* + Oh), 0 < 0 < 1.

Another form of Taylor's result is obtained taking b = x. We thus can write

/ (*) = f(a) + (x - a)f'(a) + ■■■ < £ Z ^ > > ( « ) + { " ~ ^ f(n+1)(0,

where £ lies between x and a.


The expression Rn+i = (n+i)! / ^ n + 1 H 0 *s called Lagrange's form of the remainder.

Various important applications of above result are known; we shall only explain a simple test for local extrema (see previous Sec. 7).

Theorem 2. Letn > 2 and let f^ be continuous on[a — r,a+r]. Suppose that fw{a) = 0 for 1 < h < n - 1, but f{n\a) ^ 0. Then

(i) if n is odd (n = 2p4-1), a is not a local extremum, (ii) if n is even (n = 2p), a is a local maximum if f^n\a) < 0 and a local

minimum if f^n\a) > 0.

Proof. We write Taylor's formula for x G [a — r, a + r];

/(*> = £ ^ / ( f c ) ( a ) + ^ > > ( c ) = / (a) + ^ > > ( c ) fc=0

where c is between a and :r. Consider now (i). We see that: f(x) — f(a) = ^ ^ / ( n ) ( c ) - Assume / ( n ) ( a ) > 0. It follows, from continuity of f{n)(x), that /(n)(c) > 0 for x close enough to a. Then we find: /(a;) > / (a) for x near a and # > a, f{x) < / (a ) , for rr near a and re < a. Thus x = a is not a local extremum. Similar reasoning holds if f^n\a) < 0.

Next consider case (ii) - (n is even). Again, if f^n\a) > 0 we have f^n\c) > 0 for x close to a; then f(x) — f(a) > 0 for all x close to a (right and left of a). Therefore x = a is a local minimum. Similarly, f{n){a) < 0 gives local maximum. □

V . l l . Exercises

1. Show that a right-differentiable function is right-continuous. 2. Consider the function / , E —> E which is defined by the formula: f{x) = x

if x G Q, / (#) = 0 if x ^ Q. Prove that, for # = 0, / has neither a right-derivative nor a left-derivative.

3. Let c € (a, b) and / , (a, 6) —► E be differentiate at c. Prove (by induction): for every n G N, the function g = fn is differentiable at c and g'(c) = nr-i(c)f'(c).

n

4. Consider ao, a i , . . . a n G E, c G E and p(x) = £] afc(# — c)k. Prove that

a* = £rP(fc)(c).

Derivatives 101

5. Look for local and (or) global extrema for the function / , E —► E, defined by: f(x) = x3 - 3x, x e E.

6. True or false? (Explain): if / , [1,3] —» E is a continuous function which is differentiable on (1,3), with derivative f'(x) = [f(x)]2 + 4, Vx G (1,3), then /(3) - / ( l ) = 5.

Chapter VI

CONVEX FUNCTIONS

This is a special class of functions, defined on convex subsets of the real line ("intervals"), having a simple geometric property which is stated below. Their importance in various fields of analysis is steadily growing.

VI. 1 Definitions and First Properties

Let I be an interval in E and / a function, I —► E. We say that / is convex on I iff the following holds:

f(Xx + (l-X)y)<Xf(x) + (l-X)f(y), Vx,y € / , VA G [0,1] (1.1)

Remark. The inequality above defining convexity has a simple geometrical meaning: If we consider the chord joining the points (x,f(x)) with (y,f(y)) in the plane we get for the equation of this chord:

Y = f(x) + f{y)-f{x\x-x). y -x

If X = Xx + (1 - X)y, 0 < A < 1, the corresponding Y is equal to Xf(x) + (1 — X)f(y). Therefore we can say that, for a convex function on / , the chord connecting (x, f(x)) and (y, f(y)) lies above or is on the graph of / .

Example. The function a; -> a:2, on R, is a convex function. In fact, we must establish the inequality

{Xx + (l-X)y)2 <Xx2 + {l-X)y\ Vx,y G / , V A e [ 0 , l ] .

103


This amounts to: (A2 - X)x2 + [(1 - A)2 - (1 - A)] y2 + 2A(1 - A) xy < 0 or also A(A - l)x2 + (1 - X)(-X)y2 + 2A(1 - A)xy < 0, that is A(A - l)(x2 + y2) + 2A(1 - A)xy < 0, A(A - l)(x2 + ?/2 - 2xy) < 0, A(A - l)(x - y)2 < 0, which is obvious. Another simple example of a convex function or E : f(x) — |x|, as it is readily seen.

We shall now establish the

Theorem 1. Let f be a convex function on the interval I. Then, if #i, x 2 , . . •, xn (n > 2) are points in I and Ai, A2,. . . , An are numbers in [0,1] such that Ai + A2 + . . . An = 1, the inequality

/(Ai xi + A2 x2 + . . . An xn) < Ai / (x i ) + A2 / (x 2 ) 4 - . . . An f(xn) (1.2)

/ioJds true.

Proof. The case n = 2 is obvious. We now assume the inequality to be valid when there are n points x; and n numbers A;, and then establish it for n + 1 terms. Thus, let x i , x 2 , . . . ,x n +i , points in I and /AI,/A2, .. -Mn+i,

n+l numbers in [0,1] such that ]£ /x» = 1. We write (assuming /zn+i ^ 1-which is

1 a trivial case)

/ Y ] fJLi Xi ) = f 1(1- /Xn+1) V l- Xi + /Xn+i Xn+1 I

<(i-^l)f(J2rJ^—x)

4- /*n+i /(xn+i) (by convexity of /!) n

Next note that J2 1_f^n = 1 so that from the induction hypothesis we get t= i

Accordingly one derives the inequality

(n+l \ n

/ , Mi xi] < (1 - Mn+1 Mn+1 / ( Z n + l ) n+l

i=l

Convex Functions 105

This ends the proof (If we look at /?) (7.1) — V(n) means inequality (2) for any (n + 1) numbers Xi and AJ).

Example. If / is a convex function one has

/&i + -i»))<;f;/W (1-3)

VI.2 Differentiable Convex Functions

For differentiable functions / , I —► R, where I is an interval, convexity is equivalent to / ' being increasing. We establish this useful result in the following two statements.

Theorem 1. Let I be an interval, f, I —> R be convex and the derivative f'(x) exist, Vx G I. Then, for X\,x2 G I, X\ < x2, we have f'{x\) < f'fa).

Proof. Let x\ < x < x2. We see the obvious equality x = *2J* x\ + *2-x x<1 anc* t n e n ? fr°m convexity, we obtain that f(x) < **_* f(xi) +

E ^ / ( x 2 ) . Therefore f(x) - f(Xl) < ( ^ - l ) / (* i ) + ^ / f o ) = ^ / ( * i ) + | ^ / ( z 2 ) . Thus:

^ 2 X —

a?2 _x_l-^ 2

/ (*) " / ( * 0 < /(»») ~ / (» i ) ( 2 1 } X — X\ X2 — X\

In a similar way we obtain also the inequality

/ ( * 2 ) - / ( * l ) < f(*2) - / (*) ( 2 2 ) 0 2 — X\ ~ X2 — X

Then we obtain (using the obvious property: u<p(x) < L, Vrr ^ rr0 =» lim (p(x)<L")r

X —»-Xo

r ( x i ) </ (^HZ(El )< / ' ( x 2 ) (2.3) #2 — X\

This proves Theorem 1.

Remark. The last two inequalities entail:

f(x2) - f(xi) > (x2 - a?i)/'(a;i), x2 > xx


and also

fix2) - /(ml) < (x2 - a*) f'(x2), f(Xl) - f(x2) > (Xl - x2) f'(x2).

If follows obviously that

f(x) > f(x0) + {x- x0)f'(x0), Vx, x0 e I (2.4)

This inequality has an important geometrical meaning:

Iff,I->R is differentiable and convex its graph lies above any of its tangent lines.

Theorem 2. Let / , J -+ R has an increasing derivative. Then f is convex.

Proof. Take xux2 G /, X, < x2 and A G (0,1). Then, 3x G (xux2) such that A = ^ - (take x = \xx + (1 - A)x2)- We must check the inequality: / (*) ^ ^Skffa) + f^-JM which can be written also as

X^ f f(x) + X l f(x) < X2 X f(xi) + Xl f{x2) X2 — Xi X2 — Xi X2— Xl x2 — Xi

or else: ^ f (f(x) - f(xi)) < ^^(f(x2) - f(x)). (2.5)

Application of the mean-value theorem transforms this last inequality into:

ix2-x)(x-xi) slff.^ ix-xi)(x2-x)xl.f . / (£1) < / ( « ) (2.0)

x2 — Xi x2 — Xl

(where xi < £1 < x < £2 < z2); this is however obvious.

Remark. As seen in a previous chapter, a differentiable function on an interval is increasing if and only if its derivative is nonnegative. We obtain accordingly the following:

Corollary. Let f, I ->■ R, has first and second derivative. Then f is convexifff"ix)>0,VxGI.

Example. Let f(x) = x?, (p > 1), on [0,+oo). Then fix) = p x""1 , f"{x) = pip - 1)XP~2 > 0. Hence / is convex on [0, +00).

Convex Functions 107

It follows: (taking x0 = 1 in (2.4) above) the inequality

xp > 1 4- p(x - 1), Vx > 0. (2.7)

We also obtain (from: / ( X l +n " x " ) < £ ( / ( z i ) 4 • • • 4 / (x n ) ) ) , the inequality

(xrt^y 1, x i , x 2 , . . . , £ n E [0, oo).

VI .3 Exercises

1. Let / , g be convex functions on the interval I. Then, VA E [0,1], the function A/ + (1 — X)g is again convex on I.

2. Among polynomials of odd degree n > 3 there are no convex functions on IR. True or false? (Explain).

3. Let f(x) = xip(x), g(x) — <p{\) where <p"(x) exists for every x > 0. Show that / is convex on (0, oo) if and only if g is convex on (0, oo).

Chapter VII

METRIC SPACES

In this chapter we define the general (and most useful) concept of a "metric space" and then treat in detail the real line with the natural metric introduced by means of the absolute value, as well as other important examples.

This permits discussions of various "topological" concepts: open and closed sets and their properties; interior, boundary and closure of sets, etc.

VII. 1 Metric Spaces, Open and Closed Sets

The general concept of a metric space pertains to set theory: precisely we have

Definition 1. A metric space is a pair (X,d) where X is a set and d is a function from the cartesian product X2 into R^ called distance function or metric and satisfying following conditions: d(x,y) > 0, Vx, y G X; d(x,y) = 0 if and only if x = y; d(x, y) = d(y, x) (symmetry property); d(x, z) < d{x, y) -f d(y,z), Vx, y, z € X. (this is called the "triangle inequality" because in a certain important instance it expresses precisely a classical property of triangles in elementary plane geometry).

Before proceeding to the discussion of the metric space (R;d) where d(x,y) = \x — j / | , Vx, y G R (our main goal in this chapter), we shall indicate however two special, somewhat fancy, examples of metric spaces, of a different nature (a more "exotic" nature).

Wx2 (Cartesian product of X with itself) means the set of all ordered couples (x,y) where x G X and y G X. A function from X2 into R, as usual, assigns to any such couple a uniquely determined, real number.

109


Example 1. Let X be any nonempty set. The so-called "discrete metric" on X2 assigns to any couple (x,y) with x / y, the number d(x,y) = 1, and to any couple (x, x) the number d(x, x) = 0. Most properties which we asked for are trivially verified. As for the "triangle inequality" it results true after a simple discussion. Take x, y, z in X. If they are all different, we get 1 < 2 — which is obvious. If, say, x — y and z is different, we get: d(x,y) = 0 < d(x,z) + d(z,y) = 2; d(x,z) = 1 < d(x,y) + d(y,z) = 0 + 1 = 1, d(y,z) = 1 < d(y,x) + d(x,z) = 0 + 1 = 1. If a; = y = z, we get 0 = 0 -+ 0, which is also obvious.

Example 2. Let X = Q — the set of rational numbers. For a, b G Q we put d(a^b) = \a — b$l + \a — 6 |) - 1 . Again, all properties are obvious, except for the triangle inequality. We need the following:

Lemma. If x, y, z are nonnegative numbers and x < y + z, then j ^ <

1+2/ ^ l + z'

Proof. This inequality amounts to relation x(l + y)(l + z) < y{\ + x){l + z) + z(\ + x)(l + y) which becomes x + xy + xz + rr /z < ?/ + ^2/ + 2/ + ^2/^ + z + zx + zy + xyz and then x<y + z + yz + yz + xyz which is obvious.

Therefore, if we put x — \a — 6|, y = \a — c|, z — |6 — c\ we get (after writing a — b = a — c + c — b and |a — 6| < \a — c\ + \c — b$:

\a — b\ \a~c\ \c~b\ l + |a-6| " l + |o-c| + 1 H- |c — 6|

This is in fact the required inequality for d(a, 6).

After this short excursion in some "strange" metric spaces, let us turn back to the set E of all real numbers, or even, the set Q of all rational numbers. Both are made matric spaces, if we put: d(x,y) = \x — y\. All properties of d are obvious; the triangle inequality is now: \x — z\ < \x — y\ + \y — z\ which follows from the equality: x — z = (x — y) + (y — z) combined with the main inequality for absolute values: |£ + 77| < |^| + |ry| which has been seen in a previous chapter (in any ordered field).

Now, we shall define, in any metric space (X,d), the open ball of center #o and radius r:

B(x0,r) = {xE X,d(x,x0) < r}

Metric Spaces 111

and the closed ball of center xo and radius r

C(xo,r) = {x G X,d(x,x0) < r}

If X is E with the above (natural) metric: d(x,y) = \x — y\, we have

B(x0,r) = {x G E, \x - x0\ < r} and

C(x0,r) = {x e E , | x - a ; o | < r}

Thus we obviously have

B(x0lr) = {x G E, -r < x - x0 < r} = {x G R,xo - r < x < x0 +r}

which is called an open symmetric interval with center at xo and radius r and also

C(x0,r) = {x e M, — r < x — x0 < r} = {x G M, x0 - r < x < x0 + r}

which is a closed symmetric interval with center xo and radius rS1^

(The reason for the names "open ball" and "closed ball" given in the general case, is best seen in the special case of 3-dimensional space with the natural distance induced by the theorem of Pythagoras) (see next Sec. VII.2).

We are now ready to introduce the fundamental topological concept of an open set in a metric space.

Definition 2. In a metric space (X,d) a set G C X is open if, for any x G G, there is e G E, e > 0 such that B(x,e) C G.

In the metric space E we can see the following examples:

(i) All the "open intervals77 are open sets. (Remember that we call open interval in E the following sets: (a, b) = {x G E,a < x < 6}; (-oo,+oo) = E; (a,+oo) = {x G R,x > a}; ( -00 ,6 ) = {x G E , £ < b}

(Dlf instead, X = Q with the same metric, we have, Vrro G Q

B(x0yr) = {x G Q, x0 - r < x < x0 + r } ; C{XQ,r) = {x G Q,x 0 - r < x < x0 + r }

Here XQ and x are rational, but r can be any positive real number. Thus, if r is not rational, C(x0lr) = B(x0,r)\


Let us prove now our assertion: First, the obvious case of (-00, +00) when any e > 0 is acceptable. Next, consider (a, 6) and take x$ £ (a, b). Let 0 < e < min(:ro — a, & — xo). Then: £ < XQ — a gives XQ — £ > a while £ a, hence (x0 — £,x0 + £) C (a, +00).

(ii) This does not mean that all sets are open! Take for instance H = (a, b) U {xo} where xo £ (a, 6). This is not open because the element x0 fails to satisfy the condition: If, say: Xo > b, any symmetric interval (xo — £,xo + £) will contain points between b and #0, which are not i n # .

(iii) Now, we can ask ourselves if other open sets in R (different from the open intervals just discussed) can be defined in a simple manner. In fact, this is so: for instance, let us define a set G by the relation

G = (1 ,2 )U(3 ,4)U(5 ,6)U. . . = U ~ = 1 ( 2 n - l , 2 n )

(a "countable" union of open intervals).

In order to establish "openness" of G, let us take xo £ G. It follows that XQ G (2n — l,2n) for some n € N. Using previous reasonings, we can find a symmetric interval (xo — £,#o + £) completely contained in (2n — l,2n), hence "a fortiori" contained in G.

We are now ready to state (and then to prove) the following general result:

Theorem 1. Let (X,d) be a metric space; then

(a) The total set X and the empty set (j) are open. (b) / / Gi, G2, . . . Gn are open sets in X, then so is C\^=1Gk-(c) / / {Ga}a£A is & collection of open sets in X (A being any "indexing

set") then G = U a G ^G a is also an open set in X.

Proof.

(a) That the whole space X is open is obvious. As for the empty set 0, we note that the "openness" condition refers to elements x of G. As 0 has no elements, there is no condition to check!

Metric Spaces 113

(b) Let us take any x0 G nj^Gjfc. This means that x0 G G& for all k = 1,2,.. .n. But every set G& is open; it follows that positive numbers £fc, & = l , . . . n exist, with property: B(xo,6k) C G*;. Take now e = min(£i,£2,- • .£*:)• Then £ > 0 and, for any k = 1,2,.. .n, B(x0,e) C £(#<)>£*:) C Gk. Thus 2?(x0,£) C n£=1Gfc, and n^=1Gfc is an open set. (it could happen also that n^Gk = <j>\ then we are in (a)).

(c) Let us take xo G G. Therefore XQ G Gao for some a0 € -A- Hence, as Ga o is open, there is some e > 0, such that B(#o,e) C Gao C U a G ^G a = G. Thus, G is open. □

Closely related to the concept of an open set is that of a closed set: A set F in a metric space (X, d) is closed iff its complement CF = {x G X, x £ F} is open.

Example 3. Take (again) the real numbers E as metric space. A closed interval in E, [a, b] is defined by

[a, b] = {x G E, a < x < b}

Then, it is a closed set in the above given sense. In fact, C {[a, b]} = {x G E, x < a or x > b} = (—oo, a) U (6, oo) which is open as previously seen.

Remark. Other closed sets in E are the following intervals: [a, +oo) = {x G E, x > a}, (-00,6] = {x G E, x < 6}; E itself (which is both closed and open).

Remark. One should not make the "common sense" confusion: "a non-open set is closed". For instance, the interval (a, b] = {x G E, a < x < b} is neither open nor closed.

In fact, b G (a,6], but, Ve > 0, (6 — e,b + e) is not contained in (a,b] (b + | ^ (a, 6]). Hence (a, 6] is not open. To prove that it is not closed, we note that C {(a,b]} = (-oo, a] U (6, oo) which is not open ("openness" fails at x — a!).

There is a result similar to Theorem 1 and pertaining to closed sets. It can be stated as

Theorem 2. In a metric space (X,d):

(a) X and (j) are closed sets.


(b) The union of a a finite number of closed sets is a closed set. (c) The intersection of any collection of closed sets is a closed set.

Proof. As C X = <j> and C <\> — X, we can apply (a) of Theorem 1. Next, let Fu ... F n be closed sets; then C (Fx U F2 . . . Fn)=C FiD C F2 ... n C F n

and apply (b) of Theorem 1.

Finally, let {F a} be an arbitrary collection of closed sets. Then: C(f lF a) = U ( C F a ) ; each CFa is open, and apply (c) of Theorem 1.

We close this section with the following result, which establishes a first connection between convergence of sequences and open sets.

Theorem 3. Let (xn)J°=1 be a sequence in the metric space (X, d). (xn G X Vn G N.) Then xn —► x (this means that d(xn,x) —> 0 as n —► ooj z/ and onfa/ i/, for every open set G containing x, there exists no G N, such that n > no => x n G G.

Proof. First note that if e > 0, then B{x, e) is an open set containing xS1^ Hence, 3no G N, such that n > no => x n G J3(x,e), that is d(xn,rc) < £ for all n > no, which means that

Conversely, let and let us take any open set G containing x. Then there exists e > 0 such that B(x,s) C G. Furthermore, since x n —► x there exists n0 G N such that xn G B{x, e) for n > n0, hence xn G G for n > n0. D

Corollary 1. If F is a closed set in (X,d) and xn —► x t^/iere x n G F Vn G N t/ienx G F .

In fact, otherwise, x G C F which is open. Hence, 3e > 0, such that B(x1e) C C F ; no xn enters in B(x,e), contradicting Theorem 3.

Example 4. Let F be a closed set in R which is bounded from above. Then sup F G F.

In fact, let a = sup F\ then, Vn G N, 3xn G F , such that, a - £ < x n < a. Therefore, xn -> a. From Corollary 1 => a G F .

(1)This is seen as follows: take y £ B(x,e) and then rj = e — d(x,y) > 0. We obtain, if z e B(y, 77), that d(z, x) < d(z, y) + d(y, x) < 77 + d(x, y) = e, hence J3(y, 77) C B(x, e).

Metric Spaces 115

VII.2 The Space Rn with the Natural Metric

Starting with the set E of all real numbers, we can form the "Cartesian product": Rn = R x R x . . . R, (n-times): this means that we consider (ordered) systems (#i, #2, • • • %n), where all the XiS are in E, and we define in this set an addition:

(XX1X2,...xn) + (yi,y2,.. .yn) = (^1 + 2/i,x2 +2/2, •• -xn + yn)

as well as a product with real numbers:

A • (#i,#2, • • .xn) — (Axi, AX2,... \xn), VA G E, Vx; G E, Vi = 1,2.. .n .

We have this way defined E n as a vector-space over E. Next, we define on E n a "scalar-product". Let a = (xuX2,.. .xn), (3 =

n (2/I>2/2J-•-2/n)- Then we put : (a , /?) = X) #;.?/i. (** *s a r ea^ number, Va,

i = l /? G E n ) .

We shall verify the following fundamental properties of the scalar-product:

(1) (a, a) > 0 with equality iff a = 9 (the zero element in E n , denoted with 0, is given as 6 = (0 ,0 . . . 0)).

(2) (a,/J) = ( / ? , a ) , V a , / J € R n . (3) (Aa + /i/3,7) = A(a,7) + M(/?,7), Va, / 3 , 7 G E n and A,/* G E.

n n In fact (a, a) = ^ a? >0 and J ] a? = 0 if and only if all a; are zero.

i=l i= l

(a, /3) = £ a, ft = ]T fta* = (/?, a) r = l i = l

(Aa + M/3, 7) = Yl(Xai + Vî = X^2 aili + ^ Z ^ = A(a' 7) + M& 7) i=l i= l i= l

Let us define also a "norm" on E n : (denoted ||a||). It is given by the

relation ||a|| = ( £ a? ) (meaning the positive square root).

For n = 1 we get ||a||Ri = \fa* = \a\ (the "absolute value of a").

We note the following properties of this norm:

(i) ||a|| > 0 and ||a|| = 0 & a = 0.


(ii) ||Aa|| = |A| ||a||, Va 6 Rn, VA e E. (iii) llz + vll^lMI + l l i /H.Vs.yeR".

n In fact: (i) follows from the relation ||a||2 = Y, a2 = (a, a ) . For (ii) we

t = i

note that ||Aa|| = (A2a2 + . . . X2a2n)i = (A2)*(a? + . . . a 2 )* = |A| ||a||. As for

inequality (iii) ("the triangle inequality"), we first prove the Cauchy-Schwarz inequality for scalar products:

( iv ) | (a , /3 ) |< | | a | | | | i 8 | | l Va > /3eR w . We have in fact, Vt E t , the relation (ta +/3, ta + (3) > 0 which means that

t2\\a\\2 + 2*(a,/3) + ||/?||2 > 0, Vt G E. This is not possible unless |(a,/3)|2 < ||a||2||/?||2 which is equivalent to (iv).

Next, in order to prove (iii) we write (with x, y G Mn) ||X + T/||2 = (x + V,x + y)< \\x\\2 + \\y\\2 + 2$x,y)\ < \\x\\2 + \\y\\2 + 2\\x\\ \\y\\ = (\\x\\ + \\y\$* and the result follows.

Finally, we can also define a "metric" on M.n. Precisely, we put, Va, (3 G Rn.

d(a,/3) = | | a - j 8 | | = £(<**-A)2

All properties of a distance follow from previously proved results. In particular, Va, /?, 7 G Mn, we have a — 7 = (a — /?) + (/? — 7), hence ||a - 7ll < ||a " 011 + ||/3 " 7ll and d{a, 7) < <*(<*, /?) + d(/J, 7).

Note also the useful inequality: Va = (ai . . . a n ) , |ai | < ||a||, 2 = 1, 2 , . . . n which is immediate.

VII.3 Further Examples and Properties

First we remark the following:

Proposition 1. If (X,d) is a metric space and Y C X, then (Y,d) is also a metric space.

The proof is obvious.

Next we give the following

Example 1. On the vector space M2, let us put: d((xi,x2), (2/1,2/2)) = |#i — Vi\ + \x2 — 2/21- We obtain again a metric space (its metric is not the natural metric!).

Metric Spaces 117

In fact, if d = 0 => xi = 2/1, x2 — y2• The symmetry is also obvious. Take now 3 points P = (xux2), Q = (2/1,2/2), R = (zi,z2). We obtain d(P,Q) = \xi-yi\ + \x2-y2\ < | ^ i - ^ i | + Ni-2/i | + k 2 - ^ 2 | + k 2 - 2 / 2 | = r f ( - P , ^ ) + ^ , Q ) . Thus, we also have the triangle inequality.

Remark 1. Consider (see Example 1 in VII.l), the "discrete metric" on any nonempty set X. Then, the open ball: B(x,e) = {y G X,d(y,x) < e} is the "singleton" set {x} if e < 1 and the whole space X if e > 1.

Our next example is a metric space whose elements are functions; this is very important: metric spaces (or, more generally, topological spaces) were introduced mainly for this reason: to treat functions as points in a (geometrical) space, and accordingly, to use geometrical ideas (conveniently extended) in order to handle various properties of functions arising in calculus, partial differential or integral equations, and elsewhere.

Example 2. Let S be any set, and then B(S) consists of all bounded real-valued functions defined on S:

s G S -> f(s) G R and sup|/(s) | < +00. ses

Note first that B(S) is a vector space over R: If / , g G B(S), f + #, defined by

(/ +0)(s) = / ( * )+0(a) ; Vse Sand A/, defined by : (A/)(s) = A/(s), Vs G 5, VA G R

are again bounded functions over 5. We can define a "norm" on B(S):

II/OIIB(S) = SUP | / (*) | s€S

(the zero element in B(S) is the function /(•) such that f(s) = 0, Vs 6 S). It is obvious that | | /(-)| | > 0 and | |/(-)| | = 0 iff f(s) = 0, Vs e 5 .

Next, VA e K, | |A/| |B ( S ) = sup , 6 5 |A/(s)| = sup s € 5 |A||/(a)| = |A|sup|/(s)| sE.S

(we use here: A C R and upper bounded, /x G R+ , => sup(/xA) = /x sup A for: Va G A, a < sup A, \i a < \i sup A, hence sup(// a) < \i sup A. Let

aeA us assume sup(/x A) < \i sup A. We obtain sup(^ • [i A) < ^ sup(/x A), sup A < - sup(/x A) < sup A, a contradiction.)


Finally ||/(-) 4- g(-)\\B(S) = SUPSG<? \f(s) + #(5)h a s w e obviously have: \f(s) + g(s)\ < \f(s)\ + g(s)\ < sup s € 5 \f(s)\ + sup s € 5 \g(s)\, Vs G S\ it follows that sup | /(s) + <7(s)| < sup s G 5 | / (s) |4-sup s E 5 \g(s)\, which means the "triangle inequality" for the norm in B(S).

II/ + 5HB(S)<II/IIB(S) + NIB(S).

Now, from the norm on a vector space we can always derive a metric on the same space: Put,

Vf,geB{S), d{f,g) = \\f-g\\.

(see the case of E n previously treated in the same way).

Note that a sequence (/n)i° in B(S) is convergent to / G B(S) in the above defined metric iff

sup|/n(s) - / ( s ) | -► 0, as n -► oo ses

It is the uniform convergence of /n(-) towards /(•).

Example 3. We give a new example of an open set G. Precisely, define G = {(^1,^2) G E2 , a < #i < b}. Take any point (£i,£2) € G\ we look for e > 0 such that, if (2/1,2/2) £ B ( ( r r i ,^) ,^) , then (2/1,2/2) £ G. We assume therefore that (2/1 — #i)2 4- (2/2 — #2)2 < £2, which implies (by previously seen inequality):

\Vi - Xi\ < y/(xi -yi)2 + (x2 ~2/2)2 < e, i = 1,2.

This gives, for z = 1,

—e < 2/1 — rex < £, hence 2/1 < #1 4- e, 2/1 > xi — £-

We know that X\ < b\ therefore we get 2/1 a if xi — e > a, xi — a > e. Accordingly, one obtains: a < 2/1 < b if e < min(xi - a,b — x\). This means that 2?((a;i,£2),e) C G for these values of £.

Remark. The set in E2 , H, defined by

H = {(xux2),x1 <0}U{0 ,0}

is not open because B((0,0),£) is not completely contained in H no matter how small we choose e.

Metric Spaces 119

Remark. We repeat the warning previously made. A nonopen set is not necessarily closed.

Take for instance, the set K = {{xi,x2),xi > 0}u{0,0}. It is not open, for similar reasons as in the set H above. It is not closed either (we can see this using Corollary 1 to Theorem 3 in VII.l. Take a sequence (xi,n,x2,n) where xi,n — \ , #2,n = 1, Vn G N. Then we see that (#i,n,#2,n) —> (0,1) as n —► oo a n d ( 0 , l ) ^ k ) .

VII.4 Some More Topological Concepts

Let (X,d) be a metric space, and let A C X(A G V(X), the collection of all the subsets of X).

Definition 1. The interior of A, (notation A°) is the set {UG, G open, GcA}.

Then, obviously, A0 C A.

Definition 2. The closure of A, (notation A) is the set {nF, F closed, FDA}.

Again, obviously, A C A.

Proposition 1. For any set A £V(X), we have: A° is open, A is closed.

Proof. Apply above definitions and Theorems 1, 2 in VII.l. □

Definition 3. The boundary of A is denoted by dA and is defined by the formula

dA = An(CA)

Thus, dA is a closed set in X.

We shall next prove

Proposition 2. For any set A G V(X); A is open iff A = A° and A is closed iff A = A.

Proof. One half of the result appears in Proposition 1 above. Next, let us assume that A is open. Then obviously, {UG, G open, G C A} is A itself; hence A0 = A.


Also, assume that A is closed. Then, {OF, F closed, F D A} is A itself, hence A — A. □

An important characterization of A° appears as

Proposition 3. A point xo G X belongs to A0 if and only if there exists e > 0, such that B(xo,e) C A.

Note . If x G A° we say that x is interior to A or an interior point of A.

Proof. Let us use formula

A°=U{G,Gopen , G c A } .

Then, if x0 G A°, it follows that x0 G G for some open set G C A. From the definition of open sets, it follows that, for some e > 0, the ball B(x0,e) is contained in G; thus, B(xo,e) C A°.

Conversely, let xo e X with the property that B(xo, e) C A for some e > 0. As B(xo,e) is also an open set, we have: XQ G B(XQ,E) and B(xo,e) C A°, hence x0 G A°. □

We next examine other relationships between the interior, closure and boundary of any set. We state therefore

Proposition 4. The following equalities hold true: A° — C {CA ); A — C [{CA)°]\ dA = A-A0 = {xeA8Liidx$ A°}.

Proof. We have first: ~CA = {flF; F closed, F D C A). Then C (C~A ) = C (OF), where F is closed and F D C A Hence: C (C A ) = U ( C F ) , where F is closed and F D C A which means obviously also: UG, where G is open and G C A. Thus we obtained:

C(C~A) = A°. Next, we have: ( C J 4 ) ° = {UG,G open, G C C A}. Therefore: C[(CA)°]

— C (UG), where G is open and G C C A. Hence C [(CA)°] = n ( C G ) , G open, G C C A, that is = {nF, F closed,

F D A} which is A. Finally: A - A0 = An C (A°) = An C A = dA. D

The following result is also of interest.

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Proposition 5. For any two subsets ofX, A and B we have AU B = AuB and(AnB)0 = A0 D B°.

Proof. Let us demonstrate first that (A n B)° = A° n B°. We shall use Proposition 3 to establish the double inclusion: Let x £ (An B)°, hence, 3 ball B(xye) which is contained in An B, hence in A and in B\ therefore x e A° nB°.

Conversely, assume x G A° n B°; there are balls B(x,£\) C A and B(x,e2) C B .

Accordingly, if e = min(ei,£2)» the ball B(x,e) C A n B, hence xe(AnB)0.

The other property (concerning closure of the union) is derived from Proposition 4. We have:

C(Au B) = C(A)nC(B) = (CA)°n( C B)° = [(C A)n (CB))° Therefore one gets:

AuB=C[(CA)n(C B)}° = C [(C[A u B})°] = AuB. □

An important counterpart to Proposition 3 is the following characterization of the closure:

Proposition 6. Let A C X. Then, a point x G X belongs to A if and only if, Ve > 0 , B(x,e)nA^(i).

Proof. Let x G A=C [(C A)°]; thus x <£ (C A)0, hence, V e > 0, B(x,e) n A / ''</>. Conversely, if V e > 0, B(x,e) n A ^ </>, it follows that x £ (C A)0, hence x G A.

The points of the boundary dA of any set have a similar property:

Proposition 7. Let A C X. Then x G dA iff, Ve > 0, B(x,e) n A ± <\> and B(x,e)n C A ^ </>.

Proof. Let x G dA = A n (C A); then apply Proposition 6. Conversely, if B(x, s)nA / (j) and B(x, e) n C A / <f>, Ve > 0, =» a; G A and a; G (7A, hence rr G A fl C i = d A

Finally, let us note


Proposition 8. A set A is closed, iff dA C A.

Proof. Assume dA C A; we shall prove that A C A. Take x G A; then B(x,s)CiA^ (/),¥£> 0. If x G A we are done; if not, then x £ C A; therefore B(x,e) n A / 0, B(x,e) D C A ^ 0 too; hence x G dA, a contradiction.

Conversely, let A = A. Then, consider x G <9A; it follows x G A = A hence <9AC A.

VII.5 Sequences and Completeness in Metric Spaces

In this section we first complete discussions in VII. 1 concerning connections between convergent sequences and topological objects in a metric space. Remember that a sequence (£n)i° in a metric space X is a mapping N —► X. The range of this mapping is a set {xn}^° which is contained in X.

As usual, we say that: limxn = x or xn —> x if, Ve > 0, 3n G N, such that d(xn ,x) < £, Vn > n; this property means that the sequence in R+, (£n)i°> where £n = d(xn,x) is a null-sequence (lim £n = 0). In particular, if X = E n , a sequence (^p)i° in E n is convergent to x G Mn iff \\x — xp\\ —> 0 (as p —> oo).

Let us first complete Corollary 1 to Theorem 3 in VII. 1. We prove the following

Proposition 1. A set F C X is closed if and only if for every sequence (xn)J° where {xn} C F and xn —► x, we have x G F too.

Proof.

(i) Let us assume that F is not closed. Then, 3xQ G F and x0 ^ F. Use now Proposition 6 in VII.4, and find, Vn G N, elements xn G i?(£0, ^) H F . Obviously then x n —► x0, # n G F , Vn G N and #0 ^ F , a contradiction. Therefore, under the assumption of the Proposition, F must be closed.

(ii) Conversely, if F is closed, and (xn)^° is a sequence with {xn} C F , #n —► #o, we find that xn G B(x0,e), Vn > n(e); hence £?(:ro,£)nF ^ 0, Ve > 0, so xo G F = F . (a slightly different proof of this part was given in Corollary 1 to Theorem 3, VII. 1).

We now come across the very useful concept of an accumulation point of a set in a metric space. We shall give the following

Metric Spaces 123

Definition 1. Let A C X. An element x G X is an accumulation point of A if there exists an injective sequence (xn) in A, such that lim xn = x.

Note. As usual, injectivity means that if n, m G N and n ^ m then XJI J1 xrn.

Example. Take X = R, A = (0,1) U {2}. Every point x G (0,1) is an accumulation point of A (take n0 such that no > jz^ [this implies x + ^ < 1 for n > n0] and then put xn = x + ^q^;, n G N; we have a:n G (0,1), Vn, and (xn)i° is an injective sequence, such that xn —> x.)

However, x = 2 is not an accumulation point of A. (Otherwise, let (xn)™ be an injective sequence in A, xn —> 2. It follows that for n > n, xn ^ 2, hence 0 < £ n < 1. On the other hand, we must have xn > 1 (say) for n > n\. This is impossible if n > max(ni,n).

There are also two more accumulation points for A; x = 0 and x = 1. For instance, take x n = ^ y , n > 1. Then (xn) is injective, xn —> 0, and x n G A, V n G N .

A similar reasoning holds for x = 1. We are by now ready for the proof of

Proposition 2. A set A in X is closed iff it contains all its accumulation points; furthermore the closure A of any set A C X, is the union between A and its accumulation points.

Proof, (a) First, let us assume that A is closed. If x is an accumulation point of A, 3 an (injective) sequence (xn) in yi, Xfi ^ x . Use of Proposition 1 then gives: x G A. The remaining of the proof is based on the following

Lemma. A point x G X is an accumulation point of A iff for any ball B(x,e), the intersection B{x,e) fl A contains at least one point y G A, y ^ x.

Proof of Lemma,

(i) Let x be an accumulation point of A. Let then {xn) be an injective sequence in A, xn —> x. Thus, xn ^ x for all n, with possible exception of one such number, say n0. Now, any ball i?(x,£) contains all xn for n > n(e). Hence, any xn with n > max(n0,n) belongs to B(x,e) fl A and is different from x.


(ii) Take the ball B(x,l) and then 2/1 ^ x, 2/1 G B(x , l ) , and 1/1 G A. Next, take the ball B(x,r2) where r2 = min( | , d(yi,x)), and then y2 G B(x,r2) 0 A, y2 ^ x. We have d(y2,x) < r2 < d(yux), hence 2/2 7 2/i- Now> take r3 = min( | , d(y2,x)), and 2/3 € # (x , r 3 ) , 2/3 ^ £, 2/3 e A.

Then d(y3,*) < r3 < d(y2,x), hence 2/3 7 2/2- Also, d(2/3,z) < d(2/2,£) < d(2/i,x), hence 2/3 / 2/i- Thus, all the elements (yn) which are found in this way, are mutually distinct. (The sequence of distances (d(yp,x))i° is strictly decreasing).

Furthermore, we had, by our choice, d(yp,x) < K Therefore, we found an injective sequence (2/p)?0 of elements in A — {#}, such that yp —► x. Therefore x is an accumulation point of A.

The Lemma is proved.

(b) We now continue the Proof of Proposition 2. Let us assume that A contains all its accumulation points. Let also x G A. Then, Ve > 0, B(x,e) n A ^ <j>. If x G A we stop here. If not, 3y G B{x,e) n A, y / x. Hence x is an accumulation point of A, so it belongs to A. We have A C A, hence A is closed.

(c) Finally, let us establish that

A = A U {accumulation points of A}.

Let x G A; if a: G A, we stop here. If not, any ball B(x,e) contains a point of A, different from x. Hence x is an accumulation point of A.

Conversely, if x G A, it is in A (obvious). If x is an accumulation point of A, we apply Lemma, hence x G A too. □

We now turn to the second topic of this section: the completeness property in metric spaces. It is not a new topic in this book: we previously defined and discussed the completeness property of real numbers: completeness means that any Cauchy sequence has a limit; we have established that the (ordered field) E of all real numbers posseses this quality. However, this is not an ubiquitous property; for instance, we can establish with a few simple reasonings that The (ordered field) Q of all rational numbers is not complete.

For, assume the contrary: ("any Cauchy sequence in Q has a limit in Q".) We can repeat now the reasonings in the Sec. 6 of Chapter I and obtain

Metric Spaces 125

(1) Any increasing and upper bounded sequence in Q is a Cauchy sequence. (2) If (j> ^ S C Q is an upper bounded set of rationals, there exists L G Q,

L = sup S. (3) Use now Theorem 4 in 1.6: the set A = {x G Q, x > 0, x2 < 2} is

nonempty and upper bounded. Its supremum 5 (which is unique) is rational (by 2), and s2 = 2.

(4) As we did in Sec. 5.1 (for the equation x2 = 3, which has no solution in Q) we can show that no s G Q exists such that s2 = 2. (In the contrary case we arrive at the equation m2 = 2n2, with m, n G N, which simplifies to 7rii2 = 2ni2 , where the greatest common divisor of mi and n\ is 1.

It simply follows that mi must be even. Then mi = 2ra2 and ni2=2ra22 ; again we find that n\ must be even too. We obtain a contradiction which proves the result.) □

The discussion which follows now ("completeness in general metric spaces") is based on the concept of a Cauchy sequence in a metric space.

If (X, d) is a metric space, a sequence (xn)^° where {xn} C X, is a Cauchy sequence if, roughly said, it is a sequence whose terms get closer together as n gets large. In formal parlance:

V£ > 0,3n(e) G N, such that, if n, m G N and n,ra > n(e) then d{xn^xrri) < e.

We now state the following

Definition 2. A metric space (X, d) is complete if every Cauchy sequence in X has a limit in X.

A simple but important result, is given as

Proposition 3. The space E n is complete.

Proof. Let in fact (xp)i° be a Cauchy sequence in E n . Each xp is therefore of the form xp — (xPii,xp^, • • • #P,n), where xpj G E, Vj = 1,2,... n. Furthermore, due to relation: d(x,y) = \\x — y\\, we find that, (xp)™ being a Cauchy sequence, Ve > 0, 3n(e), such that \\xp — xq\\ < e for p, q > n(e)\ let us use now the inequality (which has been mentioned earlier, at VII.2):

\xPyj -xqj\ < \\xp-xq\l Vj = 1,2, . . . n .


It follows readily that, M j = 1,2, . . . n , the sequence (in E), {xpj)^zl is a Cauchy sequence.

Hence, by the completeness property in E it follows that there are real numbers Xj, j = 1,2,... n, such that

\xpJ ~ XJ\ ~¥ ® (P ~* °°)-

Then it follows that \\xp — x\\ —» 0 (p —► oo) where x = (xi,x2 .. . # n ) , (this n

because of the simple estimate): ||a|| < ^2 la*l> â = {oL\...an) G E n ; one could prove this for instance, in the following way: introduce the vectors ei, e2 , . . . en where e* = (0 ,0 , . . . 1,...0) (1 appears at rank i). Then obviously, a = a\e\ + a2e2 4- . . . a n e n , Va = (ai . . . a n ) . From the triangle inequality extended (by induction) to n vectors in Mn, we have \\xi + . . . xn\\ < \\xi\\ + . . . | | x n | | ; therefore ||a|| < | |aiei | | + . . . | | a ne n | | = | a i | ||ei|| + . . . | a n | ||cn|| = | a i | + . . . | a n | . ) .

Thus xp —> x and E n is complete. D

We shall now explain another example of a complete metric space: it is denoted by C [0,1]; its elements are the real-valued continuous functions defined in [0,1].

To begin with these functions form a vector-space over R (the sum f + g is given by

U + 9)(x) = f(x) + g{x),Vxe[o,i\, and maps continuous functions again in continuous functions).

Furthermore, the product between a G E and / G C [0,1] is defined by formula (a • f)(x) = a • f(x), Vx G [0,1] and again af G C [0,1].

Next, we define a norm on C [0,1]: (see VII-2) | |/ | | = sup | / (x) | . 0<x<l

From well-known properties of continuous functions defined on compact intervals, we find, first of all, that | | / | | < -foo, V / G C [0,1]. Obviously ll/H > 0 and if | | / | | = 0 => f(x) = 0, \fx G [0,1], hence / = 6. (the zero element in C [0,1].)

Also, VA G E, |A/(x)| = |A||/(x)|, hence sup[0fl] \\f{x)\ = |A|sup[0|1] | / ( ^ ) | = |A| 11/11-

Finally, Vx G [0,1] we have

\Hx) + g(x)\<\f(x)\ + \g(x)\<\\f\\ + \\g\\.

Metric Spaces 127

Accordingly, | | / | | + \\g\\ is an upper bound for the set {\f{x) + g(x)\, x E [0,1]} whence | | / + g\\ < \\f\\ + \\g\\.

It follows, as seen already for the space E n that the function d(f,g) = | | / — #|| is a metric on C [0,1].

Furthermore, C[0,1] is a complete space under this metric: For, let (/n) be a Cauchy sequence in C [0,1]; then, Ve > 0, 3n(e), such that

||/n - /mil < e, if n ,m > n(e).

It follows that, V# E [0,1], the sequence of real numbers (/nO))i° is also a Cauchy sequence; there exists therefore, Vx E [0,1], a number f(x) E R, such that fn(x) -► / (x) , (n -> oo).

Let us write now \fn(x) — / m (x ) | < e, if n, m > n(e). It follows that l imôo \fn(x) - fm(x)\ — \f(x) - fm(x)\ < e if m > n(e), and this for all x€ [0,1].

Therefore we obtain that | | / - / m | | = s u p o ^ ^ \f(x) - fm(x)\ < e, if m > n(e); hence, / m —> / in the metric of C [0,1].

It remains to see why / belongs to C [0,1]. Take any xo E [0,1]. We have, VxE [0,1]:

f(x) - / O o ) = f(x) - fn(x) + fn(x) - fn(Xo) + fn(x0) - / ( x 0 ) .

Take ft E N, such that \fn(x) - f(x)\ < § for all x E [0,1]. Take also 6 > 0, such that |x - x0 | < 6, x E [0,1] => | /nO) — /nOo)| < f • It results,

I /O) - /(*o)| < | /(x) - h(x)\ + \h(x) - U(x0)\ + |/*Oo) - / (*o) | e e e . . . c

< 3 + - + - = e for |x - xo\ < 6.

Thus, / G C [0,1] too.

VII.6 Picard—Banach Fixed Point Theorem and Some Related Topics

In complete metric spaces there is a simple but fundamental result, with many far reaching applications. It is essentially an abstract version of a classical method of solving "differential equations" by means of so called successive approximations (a "differential equation" is an equation involving unknown functions and their derivatives, for instance y = x(y')3, y' = 2a;3?/5, etc. One looks for solutions y = y{x)).


We are now ready for statement and proof of Picard-Banach fixed point theorem.

Theorem 1. Let (X,d) be a complete metric space and then T be a mapping of X into itself Let us assume that, for some a G [0,1), the property

d(Tx,Ty) < ad(x,y), Vx,y e X (6.1)

holds true. Then T has a unique fixed point XQ £ X, that is a point XQ such that Txo = XQ .

Note. Mappings with Property (6.1) are called contraction mappings.

Proof. The uniqueness of fixed points is very obvious: If x±, x2 are two fixed points of the mapping T we obtain xi = Tx\, x2 = Tx2, hence d(#i, x2) < ad{xi,x2), 0 < (1 — a)d{xi,X2) < 0, => d(x\,x2) — 0, x\ = x2.

The proof for existence of the fixed point goes on as follows. Let xo £ X. UTxo = XQ we are done. If not, let x\ — Tx$. If Tx\ = x\ we stop here. If not, we continue, and form x2 = Tx\.

If after any n steps we have not found a fixed point, we obtained a sequence in X: (#n)i° where X\ — Tx0, x2 = Tx\,... xn — Txn-\, . . . and so on. Let us assume that xn —> x £ X. Then the sequence (Txn-i) is convergent to Tx (this obviously follows from: d(Txn-i,Tx) < ad(xri-i,x) —» 0 (n —► oo)). The equality xn = Tx n_i then implies x = Tx, hence x is (the) fixed point. It remains to show that the sequence (xn) is indeed a convergent sequence. We shall prove that it is a Cauchy sequence, and then use completeness of X. We first note that

d(xj+i,Xj) < ad(xj,Xj-\) < a2d(xj-i,Xj-2) < ...ajd(xi,x0), Vj > l , j E N.

Thus, if m > n > 1. we obtain

< d(xrn,xrn-i) + d(rcm_i,xm_2) + . . . + d(a;n+i,xn) < am_1d(a;i,a;o) + am~2d(zi,:Eo) + . . . + and{xi,x{)) < an(l + a + a2 + . . . a ^ " 1 ) ^ , ^ ) < (a71/1 ~ a)d(xux0).

As an —> 0 (n —> oo), it follows that (^m)J° is a Cauchy sequence in X. D

Metric Spaces 129

Remark 1. As seen from the previous proof, the unique fixed point has been found as the limit of every sequence (xn)i° consisting of iterates of a point x0, that is, defined by picking a point x0 and then setting xn = Txn-U

n = 1,2,....

Remark 2. There is a slight extension of above result.

Theorem 2. Let (X,d) be a complete metric space and let T; X —> X be such that Tn = TQTQ ... T (n-times) is a contraction mapping for some n > 1. Then T has a unique fixed point.

Proof. We know from previous Theorem that Tn has a unique fixed point x, say. Since Tn(Tx) = T(Tnx) = Tx (as Tnx = x), it follows that Tx is also a fixed point of T n , hence Tx must be equal to x, so a; is a fixed point for T.

Next, any fixed point x of T gives: Tnx = TîTx) = Tn~lx = Tn~2x... = x, hence it is also a fixed point of T n ; we obtain: x = x. □

We shall end this section with some comments and examples.

(1) Any continuous function / , [a,b] —» [a, 6], —oo < a 0, <p(b) = f(b) - b < 0. Hence, 3c 6 [a, 6], (p(c) = 0, that is c is a fixed point. This fixed point is not necessarily unique: take for instance f(x) = x2, 0 < x < 1. We get /(0) = 0, / ( l ) = 1; we have 2 fixed points.

(2) Any continuous and bounded function g(-), [0,oo) —► [0, oo) has some fixed point. Take in fact again the auxiliary function <p(x) = g{x) — x, on [0, oo). We have <p(0) = 0(0) - 0 > 0. Next, if g(x) < M for x > 0, we find (f(M + 1) = g(M + 1) - M - 1 < - 1 . Therefore, again by the intermediate value theorem we find c G [0, oo), ip(c) = 0.

(3) There are examples of continuous functions, [0, oo) —► [0, oo) without a fixed point: Take h(x) = x2 +1, x > 0. If, for some x > 0 we would have x2 H-1 = x, it would follow that (x - \)2 + \ = 0 which of course is not possible.


(4) There are continuous mappings, [0,1] —» [0,1] which have a unique fixed point but are not contraction mappings. Take in fact: f(x) = x2 — x + 1, 0 < x < 1. We have: / (#) = (1 - x) + x2 > 0, and / (#) — 1 - (# - x2) < 1. The equation / (#) = x becomes: (x — l ) 2 = 0 which has the unique solution x = 1.

However, / is not a contraction mapping on [0,1]. In fact, suppose the contrary: then, 3a G [0,1), such that \f{x) — f(y)\ < a\x — y\, V#, 2/ G [0,1]. Take y = x + h, h > 0 and small enough, x < 1. Then /(x+fc)-/(s) < a; as h —► 0 we obtain l/^x)! < a < 1, Vx G [0,1). On the other hand: f'(x) = 2rr - 1; | / ' (0) | = 1, and we have a contradiction.

VII .7 Exercises

1. Let (#1,3/1), (#2,2/2) be points in E2 (vector space). Define: d((#1,2/1), (#2,2/2)) = max(|#i — #2|,|2/i — 2/21)• Show that d is a distance function on M2.

2. Let (X, d) be a metric space and #, 2/ G X, # ^ y. Show that there are disjoint open sets Gi, G2, such that # G G\ and 2/ G G2.

3. Let p(#,y) be a distance function on the set X. Prove that /3(#,2/) = p(#,2/)(l + p(#,2/))_1 is also a distance function on X.

4. Show that any open set in the metric space X is the union of a family of open balls in X.

5. Show that any "Singleton" {#}, (where # is any point in the metric space X) is a closed subset of X.

6. Prove the following: If (#n)i° is a Cauchy sequence in the metric space X and (#niJ)i° is a subsequence convergent to # G X, then #n —> #.

7. If (#n)i° is convergent to # and ( # n p ) ^ i is a subsequence, it is also convergent to X.

8. Let (X, d) be a metric space, and # n —* #. The rf(#n, 2) —> d(#, 2), Vz G X.

Chapter VIII

INTEGRATION

In this Chapter we shall shortly discuss two concepts of integrals: the Newton integral arises in the problem of inversion of differentiation; the Riemann integral permits a precise definition of the geometrical concept of "area" under a curve. There are fundamental results relating the two concepts of integration.

VIII . l The Newton Integral

Let us assume that a function f(x) is given, defined on a real interval / and with real values. We look for a function F , / —> E, such that F'(x) — f(x), V i G / (see Example 1, Chapter V.8). Such a function F (if it exists) is called a primitive of / or also an indefinite integral of / on / .

For instance, if f(x) = x2 on J, then F(x) = \ is a primitive of / . Also in general, if F is a primitive of / on / and K is a constant function, then F + K is again a primitive of / — as easily seen. After these (very) simple remarks, we give the formal definition of "Newton integrability".

Let / be a function, I -» E; we say that it is Newton integrable on / if and only if it has a primitive function F . We write

F(x) = J f(x)dx. (1.1)

We can also define the definite Newton integral of / on the closed bounded interval [a, b],

If F is any primitive of / on [a, 6], then

(AT) f f(x)dx = F(b) - F(a). (1.2)

131


Remark. In the notation (N) j a f{x)dx for the definite Newton integral we can write instead of the letter x any other letter (£, s, etc). We call x a "mute" variable.

Important remark. It may appear from Definition (1.2) that the definite TV-integral is depending on the chosen primitive F. This however is false, as explained in the following.

Theorem 1. / / / is (N)-integrable on [a, b], its (N)-integral is independent of the chosen primitive.

Proof. As seen in Chapter V, a function with null derivative in an interval is necessarily constant.

Now, if F and G are two primitive functions for / , we have (F — G)' = f — f = 0, hence F = G + C , C a constant function. It follows that F(b) — F(a) = G(b) - G{a\ whence (N) Ja

& f{t)dt = F(b) - F(a) = G(b) - G(a). □

It is also worth to note the

Proposition 1. If f is (N)-integrable on [a,6] and f(x) > 0 on (ayb), then (N)fb

af(x)dx>0.

Proof. Let F' = f on [a, 6]. Then F(b) - F(a) = (b - a ) F ; ( 0 , a < £ < 6, that is F(b) - F(a) = (b - a)f(£) > 0, whence the result.

Remark. We saw in Chapter V (Example 1, V.8), that if a function / on the interval / has a primitive F on / , then / has necessarily the intermediate value property; in particular, the function [x] (integer part of x), for, say 0 < x < 3, is not (iV)-integrable.

We shall see later that all continuous functions on an interval are (N)-integrable.

VIII.2 The Darboux Integral

We now present the so called "Riemann integral" using the method of Darboux (some authors speak about "Darboux integral" altogether; the concepts are however fully equivalent, as we shall see later). Consider bounded functions / , [a, b] —► M, where - c o < a < b < +oo. Let m = inf / = inf {/(#),

Integration 133

x G [a, 6]}, M = sup / = sup {f(x),x G [a,b]}. It follows that m < f(x) < M, Vz G [a,6].

A subdivision <r of [a, 6] is a finite list of points, starting at a, increasing strictly and ending at b: a = {a = x0 < x\ < x2 • • • < z n - i < xn = b} (it is also called a partition of [a, b]). The xv(v = 0 ,1 ,2 , . . . n) are the points of the subdivision. The "trivial subdivision" a = {a = xo < x\ = b} is allowed, and if a < c < 6, then a = {a,c,b} is a "finer" subdivision of [a, b]. The effect of a (where n > 1) is to break up the interval [a, b] into n subintervals [ 1 < r < n. The length of the vth subinterval is accordingly: xv — xv-i, and the largest of these lengths is called the norm of the partition a:

\\a\\ = max (xv - x v _ i ) . v=l,2'--n

As an example, if we put xr — a + ^ ^ £ , then {x0,xi... xn} is a partition of [a, 6] into n subintervals, each of length ^ p , the norm of this partition is ^B-too.

The next step in the theory: we consider numbers Mv and rav, defined by Mv = sup {f{x),xv-i <x< xv}, mv = inf {f(x),xv-i < x < xv}, 1 < v < n. (supremum and infimum of / restricted to the subinterval [xv-i,xu]).

It is quite obvious from above definitions that

m <mv < Mv < M, l < i ; < n .

One defines now, given a partition a, the upper and lower sum of / for cr, as numbers (Darboux sums):

n Sa(f) = J2Mv(xv — xv-i) (the upper sum) and

l n

S(j(f) = J2mv(xv — Xv-i) (the lower sum), l

(using "elementary geometrical" words, in the case f(x) > 0, Va: G [a, 6], the upper sum represents the total area of the union of rectangles [xv-i, xv] x [0, Mv], while the lower sum represents the total area of the union of rectangles

[0,rav]; these are crude "rectangular" approximations to the "area" under the graph of / , {(#,2/), G M2, a < x <b, 0 <y < f(x)}.

A first simple result is now expressed as

Proposition 1. We have, for any partition a of [a, 6], the inequalities m(b - a) < sa(f) < SAf) < M{b - a).


Proof. We have in fact: sa(f) = Yli mv(%v —xv-i > mJ2i(xv —xv-i) = m(b-a) and also 5CT(/) = Y%M„(xv-xv-i) < MJ2i(xv-xv-i) = M(b-a).

D

Corollary. The sets {sa(f); cr is any partition of [a, b]} and {Sa(f); a is any partition of [a, b]} are bounded (indeed^ they are subsets of the interval [m(b-a), M(b-a)].

Accordingly, the following definition makes sense: The lower integral of / over [a, 6], J f = sup {s<r(/), a is a partition of [a,&]}. The upper integral of

/ over [a,b], J af = inf {5 a ( / ) , a is a partition of [a,b]}. It is quite obvious that, for any partition a of [a, 6]:

sAf)< J^/and76a/<5CT(/);

also, one sees that:

m(b-a)<Jbf<M(b-a);

(b-a)<JJ<M(b-a). m

The lower and upper integrals are equal in many cases (for instance, when / is continuous on [a, 6], as we later shall see). However, if f{x) = 1 for all irrational x in [a, b] and f(x) = 0 for all rational x in [a, 6], then, for any partition cr, Mv = 1 while mv = 0, v = 1,2, ...ra. Hence sa(f) = 0, Sa(f) = b — a, for any partition cr; accordingly: f / = 6 — a, J / = 0.

Another simple example: If / = const, on [a, 6], we obtain, for any partition cr, that vfiv — c — Mv, v = 1,2, . . . n . Hence sCT(/) = c(b — a) = S(T(f), and

71/= £ / = <#-«)• We can end with following

Definition. >1 bounded function f on [a, 6] is (Darboux) integrable iff

The (Darboux) integral of / over [a, 6] is the common value of the two (upper and lower) integrals previously defined. In writing: (D) f f (later on, if no confusion arises, we omit the "(D)").

Integration 135

VIII.3 The Darboux and the Newton Integrals

Further study of above concepts will permit us to find important classes of functions which are (D)-integrable and also to find a fundamental relation between the (D) integral and the (iV)-integral, which is most useful when actually computing a (Z})-integral.

We start with a concept of order among partitions of the interval [a, 6]. Precisely, if G\, a2 are partitions of [a, b] we say that 0*1 refines cr2 (0*1 is a refinement of 0*2) (it is written G\ y cr2 or 0*2 -< 0"i), when, for 0*1 = {rr0,x\ • • • xn}, &2 = {2/o,2/i,...2/m} it happens that {y0,2/i,... 2/m} C {x0,xu...xn} (assets).

Thus 0*1 y a2 means that all points in 0*2 are points in o\, but G\ can have other points too.

We see that a y 0*; if 0*1 y a2 and cr2 y 0*3, then 0*1 y 0*3. If 0*1 y o2 and a2 y 0*1, then ai ==0*2. If 0*1 >- 0*2, then ||o"i|| < ||o*2||- Most important in the subsequent discussion is the following.

Proposition 1. If a, a' are partitions of [a, b] and a y a', then sa(f) > sa,(f) andS0(f)<SAf)-

Proof. We can assume that 0* has r > 1 points not in 0*'; then we can arrive at a starting with a' and inserting one of these points at a time. It thus suffices to show that the above inequalities for the upper and lower sums hold true when a has 1 more point than a'. Thus, we can write: af = {XQ, #1, . . . xn} and 0* = {£0,2*1,. . .Xi_i,c,£i, . . .£n}. The new terms which appear when computing sa and Sa are consequently:

mi,»_i • (c - Xi-i) + miti(xi - c), and Mi};_i(c - Xt_i) + Miti(xi - c)

where ra^-i = inf {/(x), £;_i < x < c},

mi ti = inf {/(x), c < x < Xi)

Afi,»_i = SUp {/(£),£;_! < X < C>,

Mi}i = sup {f(x),c < x < Xi}.

It is obvious that both rai,i_i, and ra^ are > ra; = inf {f(x),Xi-i <x< Xi} and both Mi} i_i, M M are < Af* = sup {/(-r),x»_i < x < £;}.


Accordingly we obtain that mi,i_i(c — Xi -O+mi^Xi — c) > rrii(c — £;_i-f Xi - c ) = ra;(:r; — Xi_i), while Miii-i(c — xi-i) +Mi^Xi — c) < Mi(xi-Xi-i) and this obviously implies the result.

The above remarks afford the establishment of

Proposition 2. / / a±, cr2 are (arbitrary) partitions of [a, 6], £/ien sC T l(/)<SC T 2(/) .

Proof. Consider the partition a3 = a\Ua2 (^3 is "a common refinement" of both o~\ and cr?)- We see that <J3 >- G\ and 03 >~ <72- Prom above Proposition 1 we infer that sai(f) < sa3(f). However, from Proposition 1 in VIII.2 we also have sff3(f) < Sa3(f), and now again, Sa3(f) < ^^(Z)- Altogether we see that sai(f) < sa3(f) < Sa3(f) < Sa2(f), a n d o u r result is established. □

We have at this point everything which is needed for

Proposition 3. For any bounded function f on [a,b], we have the inequal-

Proof. If we fix the partition cr2, we have

So-(/) < Sa2(f),V partition a of [a, b].

Consequently, / / = sup {sa(f)} < Sa2(f). Now, if a2 varies among all

partitions, we see that inf {Sa2(f)} = / / > / / . □

We are now ready for the following result (a necessary and sufficient condition for the equality of the lower and upper integral of a bounded function on [a,b]).

Theorem 1. Let f be a bounded function on the closed bounded interval [a, 6]. Then f is (Darboux) integrable on [a, b] iff for any e > 0, there is a partition a of [a, 6], such that Sa(f) — sa(f) < e.

Proof, (a) Let us assume: f f = Jaf = fa f. Take any e > 0; by definition of sup and inf there exist partitions CTI, cr2 such that sai (/) > / /~~f and S f f 2 ( / ) < / » / + § •

Integration 137

Take a = o~\ U a2 and obtain sa(f) > s(7l(f) > fa f - § and Sa(f) < Sc(f) < Ib

a f + I hence S„(/) - , , , ( / ) < /ab / + f - / / / + f = e.

(b) Assume Ve > 0, 3 a partition a of [a,6] such that: S„(f) < sCT(/) + £. Now, / J < S„(/) and £ / > a a ( / ) . We get: / J < £ / + e, Ve > 0.

Use also Propostition 3 above and get f f — J / > 0. Hence we obtain, if

this difference is strictly positive, that: 0 < J f — f < £, for all e > 0 which

of course is impossible (take e = | ( / 0 / - / / ) ' )• E

We next apply previous theorem in order to obtain (D)-integrability on closed bounded intervals of continuous functions, as well as of monotone (even discontinuous) functions.

Theorem 2. / / / is a continuous function, [a,b] —» E, £/ien it is (D)-integrable on [a,b\.

Proof. We use uniform continuity of / on [a, b] (see Chapter IV.6). Given e > 0, take any partition a of [a, 6] where Xi - Xi-\ < 6(^£^), Vi = 1,2,... n. (the positive number 5 is so chosen as to have: x, y G [a, b] and |rr — ?/| < 6 => 1/0*0-/(y)l<^)-

If we now consider the difference Sa(f) — sa(f) = ]r)™=1(Afi — rni)(xi — £i_i) we see that it is < ^ r ^ S ^ i O ^ * ~~ xi-i) — £- (Vi = 1,2,.. .n, 3£;, r/i G [a;i_i,a;t], such that Mi = /(&) and m^ = f(Vi)-) Now, we can apply Theorem 1 above. □

Theorem 3. Any monotone function f on a closed bounded interval [a, b] is (D)-integrable over [a,b].

Proof. Suppose that / is monotone decreasing. It follows that, for any partition a of [a, 6], Mi — f(xi-i), mi = f{xi). Accordingly we get, assuming that max (xi — Xi-\) < 6:

t = l , 2 , . . . n

SM) - sAf)

= E W -m»)(a;i-Xi_i) i=l

= t [f(*i-l) ~ f(Xi)](Xi - X^) < 6 £ [/(Xi-!) - f(Xi)] i=l i=l

= 6[f(a)-f(b))<e HSKJ^J^


where e is a given positive number, (here f{a) ^ f(b): if / (a ) = /(&), then / is a constant function, hence it is integrable). Again, application of Theorem 1 proves that / is integrable on [a, b]. □

We proceed now by explaining some of the fundamental properties of the (Z})-integral, which will later concur in establishing the connection of the (Z))-integral with the previously introduced (iV)-integral (Theorem 8 below).

Theorem 4. Assume that f is (D)-integrable on [a, 6], and a < c < b. Then f is (D)-integrable on [a,c] and on [c, 6], and the relation

rb pc rb

/ / = / / + / / holds true. J a J a J c

Proof. Let e > 0; there exists a partition a = [xo,xi- -xn], ô = a, xn = 6, such that

n S*(f) - sff(f) = 5^(M* - rrii)(xi - x*_i) < e .

l

Write a' = a U {c}. As a' y a, we have: 5CT/(/) < 5 ^ / ) and sa>(f) > sa(f), whence 5 ^ ( / ) - s^(f) < Sa(f) ~ **(/) < e too.

Next, we note that a' = o~\ Uff2 where G\ = [rro, x i , . . . xp = c] is a partition of [a, c], while cr2 = [x p ,x p +i , . . . xn] is a partition of [c, 6]. Now, we see that:

v YSMi - ™i)(xi - *»-i) < <M/) - M / ) < £ • 2 = 1

and also n

£ (Mi - m O t e - afi_i) < 5 f f.(/) - *„,(/) < e . i=p+l

This shows that / is integrable on [a,c] and on [c, 6]. Also, we have fcf = faf = inf {S<r(/)} for all partitions a of [o,c]. Hence, there exists <JI — partition of [a, c], such that: Scri (/) < J c / + e.

In the same way we see that there exists a2 — partition of [c, 6], such that: ^ 2 (/) < fc f + e- But a\ U <J2 is a partition of [a, 6] and so we have

/ f<SaiUa2(f) = Sai(f) + Sa2(f)< f / + / / + 2e.

Integration 139

whence, since e > 0 is arbitrary, we obtain the inequality

rb pc rb

Ja Ja Jc

By using lower sums we obtain the reverse inequality, hence the proof is complete. □

We next establish that the (D)-integrable functions over [a, b] form a "linear space" over R (see Chapter VII.2,3), and also that the integral is a "linear functional" on this space.

Theorem 5. The set of (D)-integrable functions over [a, b) is a real vector space. Moreover, if f, g are (D)-integrable and A G M , then

pb pb pb pb pb

/ (f + 9)= / / + / 9, and / (\f) = \ / . J a J a. J a J a. J a.

Proof. If e > 0, there exist partitions cr, a' of [a, 6], such that

/

b pb

f + e, Sa,(g)< g + e _■ J a

If a" = a U <r', we get

SMf) <s„(f)< I / + £, SMg) < sAg) < [ 9 + J a J a

b £ .

Now, we obviously have, for x £ [xi_i,Xt], that sup(/ + g) 0 ) . J a J a J a

Therefore —Jh pb pb / (f + g)< f+ 5 + 2e, V e > 0 ,

J a J a. J a.


hence p~ pb pb

/ (f + g)< f+ 9-J a J a, J a

Next, using lower sums instead, we obtain that

pb pb pb

/ / + / < ? < / (f + g) J a J a, J_ n

b ~b

On the other hand, / ( / + g) < f a(f + g); it follows that

pb pb pb I*" pb pb

f+ g< / ( / + <?)< / ( / + »)< f+ 9-J a J a J_ a J a J a J a

(the integral is an additive "functional"). The proof that A/ is integrable if / is integrable, and the equality Ja Xf =

X fa / , are obtained in a similar way. Take first A > 0. (the case A = 0 is trivial). As seen in Chapter VII.3,

for any upper bounded set in E, A, we have, if A > 0, sup (AA) = A sup A; similarly, inf (AA) = A inf A. Hence, given a partition a of [a, 6], we get: E?=i sup (Xf(x))(Xl - Xi-!) = A5 a( /) ; hence: Sa(Xf) = XSa(f)-

[xi-i,Xi]

Next, / ' ( A / ) = inf {Sa(Xf} = inf {ASa(/)} = A inf {£ , ( / )} = A/ a6 / ,

similarly, J^(A/) = sup {sa(Xf)} = sup {Xsa(f)} = X sup {$*(/)} = X f* f. When A < 0, we note the relations: sup {AA} = A inf A, inf{AA} = A sup A.

(for instance: Va E A we have a > inf A; hence Xa < A inf A, and then: sup{Aa} = sup{AA} < A inf A. Next, Ve > 0, 3a£ G A, a£ < inf A + e; hence, a£A for A < 0, Xa£ > X inf A + Xe and therefore sup{AA} > Xa£ > A inf A + Ae. As e > 0 is arbitrary we obtain sup{AA} > A inf A, whence the equality). Therefore, given any partition a of [a,6], we have (in case A < 0!) Sa(Xf) = Xsa(f) and then it follows that f*(Xf) = inf{5a(A/)} = inf{A5(T(/)} = Asup{5(J(/)} = Xf f. Therefore, the integral is a homogeneous "functional". Theorem 5 is thus completely established.

Corollary. Assume that f and g are (D)-integrable on [a, b] and that f{%) > 9{%) or [a,b]. Then

t/ a J a

Integration 141

Note . This is a very useful result; it enables us to integrate on both sides of an inequality.

Proof. We have f(x) - g(x) > 0 on [a, 6]. From Theorem 5 we infer that / — g is integrable on [a, b] and that

pb pb pb

/ ( / - » ) = / / - g. Ja Ja Ja

On the other hand, from Sec. 2 of this chapter we see that: / integrable on [a, b] and m < f{x) < M, Vx G [a, 6] =* m(6 - a) < f* f < M(b - a). Therefore, if / - $ > 0 on [a, ft], =* Ja

6(/ - #) > 0 too. □

Theorem 6. / / / is (D)-integrable on [a,b], the same is valid for | / | ; furthermore, the inequality

/ / < / I/I I J a I «/ a

holds true.

Proof. We first state the following

Lemma. The inequality sup | / (x) | - inf \f{x)\ f(y). Then: \f(x)\ < f(x) - f(y) + | / (y) | sup |/(a:)|-sup / (*) + f(y) > sup | / ( x ) | - sup f(x) + inf f(y). It follows that: inf \f(y)\ > sup | / (a ; ) | - sup f(x)+ inf /(?/), which can also be written as: sup \f(x)\- inf \f(x)\ < sup f(x)- inf /(re).

(b) f(x) < f(y). Then: | / (x) | < f(y) - f(x) + \f(y)\ < f(y)-mi f(x) + | /(y) | ; hence we obtain for y fixed in [ a^ - i , ^ ] , that: sup \f{x)\ < f(y)-inf f(x) + |/(f/)|, whence | / (y) | > sup | / (x) | - /(</)+ inf f(x) > sup


\f\- sup /+inf / , and accordingly: inf | / | > sup |/|-sup / + inf / , that is, sup |/|— inf | / | 0 is given, there is a partition a such that Sa(f) — sa(f) < e. This proves integrability of | / | .

Next, note that, Vrr € [a,6], —\f{x)\ < f(x) < \f(x)\; hence previous corollary gives that - Ja | / | < Ja / < fa | / | which means in fact that \ Ja f\ <

Il\f\- a We next investigate the (JD)-integral as a function of the upper limit of

integration: Let / be (D)-integrable on [a,b]. Then (Theorem 4) it will be (D)-integrable on [a, x] for a < x < b. We define a function F(x) on [a, b] by: F(x) = / * / , a < x < b. We put F(a) = 0, that is f^ f = 0. Also, it is convenient to define for a < b: f^ f = ~ fa f-

Remark. Let / be integrable on [a, 6], a < 6, and a < c < b. Then / : / = / : / + / ; / = / ; / - / ; / a n d , accordingly/ : / + / ; / = / ; / .

We are now able to establish the continuity of F on [a, 6]. In fact, if c and x are in [a, b] we have, from above, the equality:

F(x) - F{c) = J* f.

If x > c we derive: \F(x) - F(c)\ < /cx | / | < sup \f\(x - c). If x < c we

have: F(x) - F(c) = - £ f, hence \F(x)- F(c)\ < £ | / | < sup | / | (c - x). In either case we will have the estimate \F(x) — F(c)\ < sup |/||rc — c\. This shows that F is (Lipschitz) continuous on [a, b].

The following is the "fundamental theorem of the integral calulus".

Theorem 7. Let f be (D)-integrable on [a,6] and continuous at c £ [a,b]. Then F'(c) exists and = /(c).

Proof. Let ft, such that 0 < |ft| < 6 and c + ft G [a,6]. It follows that F(c + h)- F(c) = C h / , hence {[F(c + h) - F(c)] = ± fc

c+h f and i [F(c + / i ) -F(c)]- / (c) = I / c

c + , l [ / - / (c ) ] . Therefore we obtain | I [F(c+/ i ) -F(c) ] -/ ( c ) ! ^ / ^ ! / - / ^ ! ^ sup |/(;r)-/(c)|. |A|= sup |/(aO-/(c)|.

x£[c,c+h] xG[c,c+/i] (if ft < 0, [c, c + ft] is actually [c 4- ft, c]).

Integration 143

As / is continuous for x = c, we have | /(x) - / (c) | < e if \x - c\ < 6(e) and x G [a,6]; thus if \h\ < 6(e), we get \x - c\ < \h\ < 6(e) and \\F(c + h)-F(c)} - f(c)\ < e. Thus, limF(*):f<c> = / (c). D

x—>c

By now we are ready for the final result of this section namely

Theorem 8. Let f be continuous on [a,b]. Then both the (N) and the (D)-integrals exist and are equal

Proof. As previously seen, the function F(x) = Jx f is a primitive of / on [a, 6]. Hence (N) f* f = F(b) - F(a). However, F(a) = 0, hence (N) fc f = F(b) = (D)f*f. This proves the result. □

VIII.4 Further Properties of the Integral

Consider a continuous function / on [a,6]. Let m — inf / , M = s u p / . We iaM [a,6]

know that: m(b - a) < fa f < M(b - a), hence m < ^ Ja / < M. On the other hand from properties of continuous functions on closed bounded intervals, we see that for some £ G [a, b]: /(£) = ^ f* / , that is: f* f = (b - a)f(£).

Actually, the point £ can be found to be interior to [a, b]: a < £ < b. This is obvious if / is constant. Otherwise, let us proceed as follows: Let F(x) = Ja

x / ; then we get F(b) = fa / , F(a) = 0, and we have, by the "mean value theorem" for differentiable functions, that, 3£ G (a, 6), such that F(b) - F(a) = (b- a)F'(t) = (b- a)f{Q. D

Next, let / be a continuous function on [a,b] and / = h' on [a,b]. Then fafz=fahf = h(b) - h(a) (this is Theorem 8 in previous section).

Example 1. Evaluate Ja t3: we "guess" that the function h(t) = \tA is such that h'(t) = t3. Hence we obtain: Ja t3 = | 6 4 — \aA.

Example 2. (the substitution method): Let / be continuous on [a,b],</> be continuous with a continuous derivative on [a,/?], (f>(t) G [a,b] for t G [a,/3], a = </>(a), b = 4>(P). Let g be defined on [a,/?] by: ^(t) = /(<£(*)) -4>'(t). Then:

In fact, let us define: F(x) = f* / , a < x {t)) ■ <j>'(t) = f(<f>(t)) ■ (j>'{t) = g(t). Therefore / f g = H(0) - H(a) = F{<KP)) ~ F(<K<*)) = W>) ~ Ha) = £ /•


Example 3. We shall prove that: /Qx xm(l-x)n = J* xn(l-x)m, where m, n € N. Apply in fact the substitution method, where 4>{t) = 1 - 1 , 0 < t < 1. If /Or) = z m ( l - x)n then fU(t)) = (1 - 0"1*", and 5(<) = - ( 1 - *)min; , ' ] / n s „ , , „ . . i m. r f l /• fO _ f l _ f l / ! furthermore, 0(1) = 0, 4>{0) = 1. Therefore J 1 f = f°g = -£g = / ^ ( l -t )m tn . (This is a special case of the following: if / is continuous on [a, 6], we have fb f = fb g where 3(1) = f(a + b-x), a < x <b. have / a f = f°g where 3(1) = / ( a + 6 - x), a < x < b.

In fact, \et<P(t) = a + b-t, a<t<b, a = b, 0 = a. Also f (<f>{t))<j>'(t) = -f(a + b-t); hence f* f = /fca -3 = fcg).

Example 4. Let / be continuous on [a, b] and /„ / > 0 for any subinterval [a,/?]c[a,6]. T h e n / > 0 o n [0,6].

In fact, otherwise, 3£ € [a, b], / ( 0 < 0. Because of continuity, we can find 6 > 0, such that / (*) < §/(£) for |z - £| < 0, a: G [a, 6]. If, say, £ € (a, 6), we get /£!"/ / < | / ( 0 • 28 = 5 / ( 0 < 0, a contradiction. Similar reasonings hold «r . , C — „ ~r C — h for £ = a or £ = 6.

Example 5. Let / be continuous on [a, 6] and / f / = 0, V subinterval [a,/3] of [a,b]. Then / = 0 on [a,b].

Otherwise, there exists, say, x0 € [a,b\, /(*«,) > 0. Again, 35 > 0, such that f{x) > i / ( x 0 ) , for \x - x0\ < 8, x € [a,6]. If, say x0 e (a,b) we get (taking 6 small enough!) £ " + / / > i /0ro) • 25 = «/(*„) > 0, a contradiction.

Example 6. Let / be continuous on [a, b] and <f> be continuously differ-itiable on [a,0\, with <j>{t) € [a, 6], V* 6 [a,/3]. Then tf(t) = /* ( t ) / is fferentiable on [a, /3] and JT'(t) = f{<f>{t)W(*)•

entiable on [a,0\, with <j>{t) € [a, 6], V* 6 [a,/3]. Then tf(t) = /* ( t ) / is differentiable on [a, /3] and JT'(t) = f{<f>{t)W(*)•

In fact, if F(x) = £ / , a < x < b, we have K{t) = F(<j>(t)), a < t < f3. Hence, by the "chain rule" we get: K'(t) = F'{<j>{t))4>'(t) = f(<f>{t)W{t)-

Example 7. If / is continuous on [-a,a], then / " a / = /Qa ( / + fv) where Example 7. If / is continuous on [-a, a]

r * » - • v , . . . n<xampie / . 11 j is continuous on [-a, aj, 1 ■ p . Y n m r . 1 0 7 Tf f io m n t i m m n c « T , J = Jo

a(f + fV) where

, / = J0a fv (as follows

with the substitution cj>{t) = -*)•)

We next explain another method which simplifies the actual computation of some integrals; it is called the "method of integration by parts". of some integrals; it is called the "method of integration by parts".

Example 8. Let / , 3 be functions with a continuous derivative on [a, 6]. Then /a

6 / • 3' = f(b)g(b) - f(a)g(a) - f* / ' • 3.

To prove this formula, we remember that (Chapter V.2, Theorem 1).

Integration 145

( / • g)' = f'-g + f-g', a continuous function. Accordingly / ( / • g)' = (f9)(b)-(f9)(a) = j'f'9 + Jb

af-9'.

Example 9. Let / be a function with a continuous second derivative on [a,6]. Then f*xf"{x) = bf'(b) - af(a) + / (a ) - /(&). In fact, / > ( / ' ) ' = bf'{b) - af'(a) - £ / ' = bf(b) - af'(a) - f(b) + f(a).

Our final example in this section is the "inequality of Cauchy-Schwar^.

Example 10. Let / , g be continuous function on [a, b]. Then the following holds:

f.'-WA(f.'Y-We can establish this (very) important result, by considering the function

<p(A), where A G M, defined by

V(A) = f\f + \g? = f P + 2\ ["f-g + X2 f g2. J a Ja J a J a

This is a polynomial of second degree in A, and </?(A) > 0 , V A E l . It follows, by "elementary algebra" that

/ > / > - ( / " ) ' -

VIII.5 The Riemann Integral

Here we shall explain Riemann's definition for integrals of bounded functions on [a, b] and then show equivalence with previously explained (-D)-integral.

Given a bounded function / on [a, &], and a partition a = {x0,xi,.. .xn} of this interval, a Riemann sum for / is an expression Yl7=i f(&)(xi ~~ xi-i)> where fi are arbitrarily chosen numbers in [xi-i,Xi\.

We say that / is (R)-integrable on [a, b] if there exists a real number I with the following property: Ve > 0, 38(e) > 0, such that

5Z/(&)(zi-St-i)-J i=l

< e

for all partitions a with ||<r|| = max (x{ — X{-i) < 6 and for any choice of Si ^ [Xi—1i Xi\'


Let us consider a sequence of partitions o~n which is obtained by division of [a, b] in n equal subintervals (of length ^ ^ each). If we choose & = a + fc^p, we see that & 6 [xi_i)Tl, xiin] — (& = Xijn) — and the corresponding Riemann sum becomes *=* t i /('<* + * ^ ) = 5 n .

We see readily that, provided f is (R)-integrable, we get I = lim Sn • (Ve > n—>-oo

0 we have \Sn - I\ < e if ||<7n|| = ^ < 6(e), hence for n > n0(e)). This proves also unicity of 7 (if 7i, I2 satisfy the condition for (i^)-integra-

bility then i i = lim Sn = I2, hence Ji =12)-By definition then, the above found number I is the (i^)-integral of / over

[a, 6], I = (* ) / .* / • Our aim now is to establish the major.

Theorem 1. A bounded function f on [a, b] is (R)-integrable if and only if it is (D)-integrable and then the equality:

pb nb

(D) f = (R) f. J a J a

holds true.

Proof, (a) Let us assume that / is (i?)-integrable. Therefore, Ve > 0, 36(e) > 0, such that, V partition <J, \\a\\ < 6, and for any choice of & G [ffi-i, xi\ we have

b

" + £. (R) f / - £ < £ / ( ^ ( ^ - ^ - i ) < ( # ) / / Ja i=1 Ja

Let now: Mi = sup{/(a:),a;i_i < x < Xi}. It follows that we can find numbers £! G f X%— \ , Xi ], such that / ( ^ ) > Mi — ^ - . Accordingly we obtain that

n n

i = i i = i

Therefore, for the (Darboux) upper sum Y17=i Mi(xi—Xi-i) = 5^(7) we obtain the estimate

M/)<X)/Ki)(^-^- i ) + e < W / / + 2e.

In a similar way, for the (D) lower sum Y^=i mi(%i ~ Xi-i) = 5<x(/) we get the (lower) estimate

sa(f) >(R) [ / - 2e, (i?) / / < a„(/) + 2e. J a J a

Integration 147

It follows that

/ f<Sa(f)<(R) I f + 2e<sa(f) + te< f / + 4e. J a J a J_

As e > 0 is arbitrary, we see the inequality

hence / is (jD)-integrable. Also, from above inequalities, we get:

rb Pb rb

(D) f f<(R) f f + 2e<(D) [ f + 4e, Ve>0. Ja Ja Ja

which implies their equality (that is, the (D) and the (i?)-integrals are equal).

(b) Now we assume that / is (D)-integrable.

We shall prove first the following:^1)

Lemma. Ve > 0, 36 > 0 such that, for any partition a with \\a\\ < 6 we have: Sa(/)-(!>) /Q

6 / < e.

In fact, from: (D) f f — inf{Sa(f)} it follows that V e > 0 , 3 a partition a' = {yo,Vi, ■ ■■Vm} of [a,b], such that

(D)j"f<S<7,(f)<(D)Jj+£-. (5.1)

Next, consider any partition a = {#o,£i5 • • • %n} with ||cr|| < 6 (where 6 > 0 will be conveniently chosen below); put a" = cr'Ucr; we see that a" y cr', hence

sMf)<sAf). (5.2)

We shall now compare Sa»(f) with Sa(f) = T,f=1Mi(xi — Xi-±). Let us denote a" = {zo, zi,... zr}. As a C cr", there are points Zj-i and Zk in a" such that Zj-i = Xi-i and zk = X{.

If k = j we get [xi-i,Xi] = [ZJ-I,ZJ] and then Mi(xi — #;_i) = M"(ZJ — z^) (where S„»(/) = £ J = 1 Mj'(z, - ^ _ 0 ) .

( i ) Compare with "Darboux's theorem" in Reference [1], p. 170.


If k > j we are in the situation: Xi-i = ^ - i < Zj < zj+1 < < zk-i < Zk = x%-

The points Zj, zj+1 • • • zk_x belong to the partition a" but not to the partition a. On the other hand, we note that, if Af = sup|/(x)| , then M{ < M,

Vi = 1,2,... n, and Mi(xz - x^) < 8M - where 8 > ||a||, Vt = 1,2,,. . n. Next, note the following: We assumed that a' = {yo,yi • • •

points; this has as a consequence that the situation k > j can occur at most m-times, (since a" has at most m additional points over partition a). It follows that

n

vz = i , z , . . .n, ana M^X» - :r;_ij < om - wnere o > ||a||, v i = i , z , , . . n . Next, note the following: We assumed that a' = {yo,yi- 'Vm} has m +

points; this has as a consequence that the situation k > j can occur at mo m-times, (since a" has at most m additional points over partition a). It follow that

n n

S„(/) = J ] Mi(xi - Xi-X) = Sx + 5 2

where Si corresponds to intervals [n-i.x*] = fo-i,*,-], while S2 corresponds to intervals [xi-UXi] = [zj-i,zk], k > j . (and there are at most m such intervals fx i - i .n l ) . intervals [Xi-i,Xi\).

Then r Sl<^2 Mi(Zi ~ Zi~^ and 52 - m6M '

the estimate Thus, we have the estimate

S (f) < S »(f) + mSM Sa(f) < Sa"(f) + m6M (5.3) i l ix* _ / r i \ 1 / r o \ TH7» „ U i „ ' Let us use now above ineaualities (5 1) and (5.2) We obtain

(D)[bf<S„„(f)<SAf)<(B)[bf + £-. (D) f<s,„(f)<SAf)<(D) / + | .

using also (5.3), we get Sa(f) < S„»(f) + m6M < (D) tf f + § + ce (D) Ja f < Sa(f) < (D) Ja f +1 + m6M < (D) fa f + e, provided

Therefore, using also (5.3), we get 5CT(/) < S>(/) + m<5M < (D) / a / + § + m<5M. hence (.D)/a / < S„(f) < (D) Ja f+^+m6M < (D) Ja f+e, provided S < _ £ _ (which is our choice of 6\). - Note that the partition a'-hence the number m - depends on e only. This proves Lemma.

In the similar manner we also prove that Ve > 0, 36{e) > 0 such that for any partition a with \\a\\ < 8 we get any partition a with ||<r|| ) /a6 / - S(T(f) < e, we

i /(6)(a;i - xt-i) < Sa(f\ and, accordingly (D) f* f - e

End of proof of (b). If we now consider any Riemann sum for the partition a, with \\a\\ < 6,

in such a way that both: Sa(f) - (D) fif<e and (D) £ / - sv{f) < e, we obtain sa(f) < YZ=i /(&)(«< - ««-i) < 5 f f(/), and, accordingly (£>) j " f - e in such a way that both: Sa{f) - (D) J„ / < e and (D) J' f - sa(f) < e, we obtain 3v(f) < E?=i f(Zi)(xi - x^) < 5CT(/), and, accordingly (D) f* f - e

Therefore, using also (5.3), we get SCT(/) ( / ) + mOM < (^j Ja / + | + mi5M. hence (D) Ja f < 5CT(/) < (D) fa f + ^+m6M < (D) Ja f+e, provided S < _ £ _ (which is our choice of 6\). - Note that the partition a'-hence the number m - depends on e only. This proves Lemma.

In the similar manner we also prove that Ve > 0, 36{e) > 0 such that for any partition a with \\a\\ < 8 we get

oints; this has as a consequence that the situation k > j can occur at most t-times, (since a" has at most m additional points over partition a). It follows lat

n C / £\ _ \ ^ 71/f /^ . ^, \ C i C

Integration 149

< sM) < E?=i /(fc)(*i " * i - i ) < SAf) < (D) £ f + e, hence | £ ? = 1 /(&) (xi — Xi-i) — (D) Ja f\ < e. This completely proves the theorem.

VIII.6 Exercises

1. Establish in a direct way (without using the "intermediate value theorem" for derivatives) that the function x —► sgn x, = 1 if x > 0; = — 1 if x < 0; = 0 if x = 0, is not (iV)-integrable on [—2,2].

2. Compute the (JV)-integral: (N) f\\x\. 3. If / has a derivative on [a, 6], then g = / • / ' is (TV)-integrable on [a, 6]. 4. Write down the upper and lower (D) sums for the function: x —» 75x — x3

on the interval [2,6] using the partition in 4 equal subintervals. 5. Compute the Riemann sums for the function: x —► x2, 0 < x < 6, taking

partitions [0, £, f,... &=p±,b] and points &,n = £ , i = 1,2,... n. Then, as n —> oo, find the value of the (i^)-integral: J0 x2.

6. Compute Ja / when / = const., using Riemann sums only. 7. Let be <f>(t) = f*{3 + x3)-2. Find 0'(O), <£'(1), 0'(a). 8. Let / be (I})-integrable on [a, 6] and (JD) Ja f > 1. Show that there exists

£€[a ,6] such that /(£) > bh-9. Let / , g be continuous functions on [a, 6] and Ja f > fa g, for any subin-

terval [a,/3] of [a, 6]. Show that / (x) > #(x), Vx G [a,6]. 10. Let / be continuous, [a, 6] —> M and / (x) > 0, Vx G [a, 6]. If / a / = 0, then

/ (x) = 0,VxG[a,6] .

INDEX

A

absolute value absolutely convergent accumulation point alternating series Archimedean property area

B

Bernoulli's inequality strict

bijective Bolzano-Weierstrass theorem bound (upper, lower, least upper,

greatest lower) boundary (of set) bounded (upper, lower)

sequence function

C

Cartesian product Cauchy sequence (in Q)

sequence (in E) in metric spaces

Cauchy's general convergence criterion for series n t h root test of convergence mean value theorem

Cauchy-Schwarz inequality for scalar products in E n

for integrals (C, 1) summability (of series) Cesaro convergence (of series) chain rule closed interval in E

set ball symmetric interval

closure (of set) complement (of set) completeness of E

of E n

in metric spaces composite function conditionally convergence conjunction constant function

151


continuity uniform global Lipschitz

continuous function contraction mapping contradiction contraposition, contrapositive convergent (series) convexity of sets in E convex function countable union (of open intervals) critical point

D

Darboux continuity integral integrability

decreasing sequence (in E) definite (Newton integral) density (of rationals) derivative, one sided, right, left difference (of sets) difference-quotient function differentiate function

convex function differentiability of inverse functions differential equation differential quotient disjunction discrete metric distance function

E

e empty set equivalence relation

class

equivalence (logical) even (number)

F

field ordered of real numbers

fixed point theorem,

frequently function

bijective bounded (above, below) codomain of constant continuous discontinuous domain of extreme values of final set of identity increasing, decreasing

strictly indefinite integral of initial set of injective (one-one) intermediate values of inverse limit of

left, right at +00, —oo

maximum, minimum point of monotonic

strictly periodic polynomial primitive of (i^)-integrable

Index 153

range of rational restriction of right, left continuous supremum, infimum of surjective (onto) unbounded value of

intersection interval

closed, semiclose open, semiopen unbounded

inverse function

G

global continuity maximum, minimum

greatest lower bound common divisor

L

Lagrange's remainder least upper bound L'Hospital rules limit of sequence

inferior, superior limit of function linear functional Lipschitz continuity local extremum

maximum, minimum lower bounded set lower sum

H

half-line harmonic (series)

I

image implication (logical) increasing sequence indexing set indefinite (D)-integrai induction infimum infinity injective integer part integers integral

Newton Riemann Darboux lower, upper

interior point of set

M

ma, , mapping of metric space

mean value theorem metric

o n E n

onC([0,l]) space

monotone sequence (in E)

N

natural metric numbers

negation nested intervals theorem norm on E n


of a partition Rolle's theorem normed linear space (B(S)). root (of polynomial) null sequence

in R S

O sandwich rule scalar product (in E n )

open ball set sequence set rational interval bounded in Q symmetric interval bounded above, below

operator ordered system convergent

convergent to +00, —00 P in E

in E n

partition injective refinement of increasing, decreasing

peak point of sequence monotone Picard-Banach (fixed point theorem) series (infinite) positive integers absolutely convergent

rationals alternating elements (in ordered field) real numbers

Cauchy's general criterion conditionally convergent

power set convergent proposition Pythagorean theorem

divergent

geometric

Q n t h root test

Q partial sum of quantifier ratio test

existential summability of universal set

empty R ordered E n space power rational numbers singleton real numbers subdivision restriction (of function) subsequence Riemann sum subset

Index 155

strict successive approximations supremum syllogism symmetry

uniform continuity uniform convergence unbounded union upper sum upper bounded

set sequence T

tangent line Taylor's formula theorem topological space total set transformation triangle inequality trichotomy law

V

vector space

W

well-ordering principle

U

ultimately


INDEX OF NOTATIONS

Symbol Page

N (natural numbers) 2 pn (nth power) 2 0 (zero) 4

z (integer numbers) 4 > , < (greater, smaller) 5 r**> (equivalence) 6 Ex (equivalence class of x) 6 (a, b) (pairs of integers) 6 Q (rational numbers) 7 E(a,b),%,o,/b (rational numbers) 7 a-1 (multiplicative inverse) 7 > 0 (in Q) 8 — (imbedding of Z into Q) 7,8 0,1 (in fields) 9 —a,a~x (in fields) 9 P (positive elements in a field) 10 \a\ (absolute value of a) 10, 11 (Xn) (sequence) 12 {Xn} (set) 12 n\ (factorial) 13 CO (null sequences) 13

c (Cauchy sequences) 13

157


Symbol Page

R (real numbers) 15

e (zero in E) 16 V (positive real numbers) 17 [n,n + 1) (real numbers x such that

n < x < n -f 1) 18 < , > (less or equal, greater or equal) 18 [x] (integer part of x) 19 limxn (limit of the sequence (xn)) 20 xn —► l(n —> oo) (xn converges to / as n —► 00) 20 C (E), C (Q) (Cauchy sequences in R,Q) 19,20 [a>n, bn] (real numbers x, such that

a>n < x < bn) 22 max(Li,—Mi) (greatest number from Li, —Mi) 23 L. U. £ . , G. L. B. (least upper bound, greatest

lower bound) 23 sup, inf (supremum, infimum) 23, 24 lim xn

n—+00 (limit of the sequence (xn)) 24

2/n "* (the sequence (yn) is not convergent to) 24

V« (n th root of a) 26 [a, 6], [a, 6), (a, 6), (a, 6] (finite intervals on E) 26 [c, 4-oo), (c, 00), (-00, c] (infinite intervals on E) 26 ( - 0 0 , c), ( - 0 0 , + 0 0 ) —00, +00 (minus infinity, plus infinity) 26 max{a, 6}, min{a, b} (largest of a and 6, smallest

of a and b) 28 V^rifc J (subsequence) 31 lim inf, lim sup (limit inferior, limit superior) 43

E afc>E (sum of infinite series) 49 ifc=i 1 («n) nGN (sequence)

(summation symbol) 50 54


Symbol Page

m m E ,E p=n+l n-f-1

oo

(finite summation symbols) 55, 56

( c , i ) , ( C D E ( -1

- l ) n + 1 , (Cesaro sums of series) 56

( C , l ) f > n 1

57 oo n

E,E 0 k=l

(infinite sum, finite sum) 58, 59

m (function) 61 D(f) (domain of function / ) 61 (e,6) (couple of positive numbers) 62 n(e),n(8) (natural number, depending

on e or 6) 62 D(r) (domain of function r) 63 / , S - > R (function from S into E) 63 # o / or g0f (composition of functions) 63 lim / (x) (limit of function / at x0) 64

/(rr) —► L as x — ► XQ

^ - l - , ^ ; _ j _ (positive part of x, its powers) 64 lim J(x)J{xo+) (limit of function / at xo, from

the right) 65 lim f(x)J{x0-

x—►xo— - ) (limit of / at rr0, from the left) 65

lim f(x), lim /(a;) X—► + OO x—► —oo

(limit of / at +oo (resp. at —oo) 67

* / (range of function / ) 67

/ ( / ) (image of I under / ) 72 inf/(z)

IR (infimum of set {f(x),x £ M}) 73

(*n f c j ) (subsubsequence of (xn)) 76 ^(e,x) (^-function of e and x) 76

G/ (graph of / ) 79

f(xo) (derivative of / in x0) 81

(x-n)'X0 (derivative of (p(x) = x~n in x0) 84

f'+Mf'Jxo) (right and left derivatives at xQ) 85


Symbol Page

/"(#) (second derivative at x) 88 (xn)i° (sequence) 88 lim ip(x) (right-limit of (p at xo '• <p(xo+)) 96

X—Ho x>x0

lim 4 4 (left-limit of ^ at a : (l)(a-)) 99 x-»-ap(x) y g

x<a / (n) (n th derivative of / ) 98 (X, d) (metric space X with distance d) 109 A"2 (Cartesian product of X with itself) 109 d(x,y) (distance between x and y) 109 (#, 7/) (ordered couple) 109 B(xo,r) (open ball of radius r, centered in xo) 110 C(x 0 , r ) (closed ball of radius r, centered in x0) 111 n£=1Gfc (finite intersection of sets Gk) 112

U G a (union of the family of sets Ga) 112 aEA 4> (the empty set) . 112 C F (the complement of the set F) 113 HFa (intersection of the family of sets Fa) 114 (xn)n=i (sequence) 114 Rn (the Cartesian product E x E x . . .

E — n times) 115 115 115 115 117 117 122

119 119 119 119

(<*,0) (scalar product)

e (zero element in E n) llll (norm, on Rn) B(S) (bounded function, S —► E) II / ( • ) \\B(S) (sup | / (s ) | for 5 e S) R+ (nonnegative real numbers) V(X) (power set of X, collection of all

subsets of X) A° (interior of set A) A (closure of set .4) d (boundary of sets)


A-A° (difference between sets A and A°) 120 e-% (basic vectors in En) 126 C[0,1] (continuous functions, [0,1] —► E) 126 nnn (n th power of the mapping T) 129 I f(x)dx (primitive of / ) 131 (N)Jb

af{x)dx (definite Newton integral of / on [a ,*]) 131, 132 I k II (norm of the partition a of [a, b]) 133

SAf),*M) (upper and lower Darboux sums of / ) 133

£/./!/ (lower and upper Darboux integrals of / ) 134

(*>)£f,fbaf ((Darboux)-integral of / on [a, b]) 134 0i y 02 > 02 -< 0i (refinement relations among partitions

of [a, b}) 135 r (fv(x) = / ( - x ) , reflection of / ) 144

(R)!baf (Riemann integral of / on [a, b]) 146

= > ■ (logical implication) 163 <£> (logical equivalence) 163 A,V (conjunction, disjunction) 163 P ' (negation of proposition p) 163 iff (if and only if, logical equivalence) 163 D, Q.E.D. (end of a proof) 165 V (for all) 165 3 (there exist) 165 $ (negation of 3) 165 3\, a1 (there exists one and only one) 165 € (belongs) 166

* (does not belong) 166 C,C,D (inclusion of sets) 166

C (complement-of sets) 167

n,u intersection, union-of sets) 167 A-B (difference of sets) 167 AxB (Cartesian product of sets) 167 f:X^Y (function, from X to Y) 168 x -» f(x) (x goes in f(x)) 168 f(X) (image of X under / ) 168


Symbol Page

f\A (restriction of / to the subset A of X) 168 / _ 1 (inverse function to / ) 168

A P P E N D I X

(Logic, Set Theory and Functions)

A . l Logic

One of the essential ideas of logic is that of a proposition: it is a statement which is either true or false, but not both.

Examples .

(1) It is a nice day. (2) Tomorrow will be raining. (3) The integer 53241 is divisible by 35 (is it false?).

In the following we denote various propositions with letters p, q, r, etc. Given p, q we may form several other propositions:

negation: not p denoted by p\ which is true when p is false and false when p is true. conjunction: p and g, denoted by p A q. disjunction: p or q, denoted by p V q. implication: p implies q, denoted by p => q. (Up is true then q is true). equivalence: p if and only if g, denoted: p <$ q (thus we now have both implications; p => q and q => p) (Sometimes we write iff (for if and only if).

Examples . Let p denote Proposition (1) and q denote Proposition (2). Then

(a) p' is: It is not a nice day; q' is: Tomorrow will not be raining. (b) p A q is: It is a nice day and tomorrow will be raining.

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(c) p => q (If it is a nice day then tomorrow will be raining) (however, the implication is false!). (d) Let p be: "n is odd"; q is: "n + 1 is even." We see that in fact, p => q. (e) Let p be ux > 0 and x < 0"; q is ux = 0". We see that p O q.

Methods of proof. In Analysis, a theorem appears usually as an implication H => C (H is the hypothesis of the theorem, C its conclusion). A proof consists precisely in establishing the above implication. We can use several methods, as for instance

1. The Direct proof This means that, starting with Proposition H we make various deductions — a chain of several intermediate implications that are easier to verify — and finish with C (thus, H => p\ => p p3 => C).

The mechanism for inferring H => C from such a chain is the "law of syllogism": if p => q and q => r then p => r.

Example. H is "n is even"; C is "n2 is even". The (direct) proof: n is even <& n = 2p (for some p € N) => n2 = 4p2 = 2(2p2) => n2 is even.

2. The proof by construction

Theorem. There are convergent series which are not absolutely convergent

Proof. Take the series 7 )°° , i = i £ . 3. The proof by induction (We gave several examples in 1-1).

4. The proof by contraposition We want to prove that p => q. We actually prove instead that q' => pf (this is the "contrapositive" of "p => g").

Example.

Theorem. / / n2 is even then n is even.

We assume n is odd (n = 2p+ 1) and find (directly) that n2 = 4p2 + 4p+ 1 is also odd.

5. T/ie proo/ 6?/ contradiction Again, we want to prove that p => q (if p is true then </ is true). We assume to the contrary that q is false. Hence, we now have p and q'. Using authorized

Appendix 165

logical steps we produce a proposition r such that both r and r' are true, which is impossible ("the law of the excluded middle").

Example. The equation x2 = 3 has no solution in Q.

The proof given in 1-5 is by contradiction (we obtain there, assuming that x2 — 3 has a solution in <2, a couple mi , n\ having greatest common divisor = 1 and > 3 at the same time, which, of course, is not possible).

Remark. We use notation □ (box) to point out the end of a proof, (unless we do this, otherwise, for instance by writing Q.E.D.).

Universal and existential quantifiers. These names refer to the symbol V which is read as "for all" and is called

the universal quantifier, and to the symbol 3 which is read "there exists" and is called the existential quantifier.

Example, x2 > 0, \f x e R; 3y e R, yA = 16.

Thus, in the first example we express a general result about real numbers; in the second example we may note that both y = 2 and y = — 2, are solutions to the equation y* = 16. Therefore, the symbol 3 means, more precisely, that "there exists at least one". One uses sometimes the symbol 3! or 3 1 to mean that "there exists one and only one". For instance: 3\x such that 3a; = 9. We use also $ (there exists no . . . ) , as the negation of 3.

Remark. Sometimes (quite often in fact!) it is necessary to negate a proposition involving V and 3. As a general rule, in doing so, we interchange V with 3 and also we negate any proposition following the quantifiers.

For example, the negation of: "3 a3 &V c such that p(a, b, c)" is V aV 63 c such that p'(a,6,c).

Example, ("revisited") (see the proof of Theorem 1 in 1.6). A Cauchy sequence of real numbers (xn) is defined by the following property: Ve > 0, 3n0 € N, such that V couples n, m, where n > n0 , m > n0 it follows that \%n Xm\ \ c.

Accordingly, the sequence (xn) is not a Cauchy sequence if the following holds: 3e0 > 0, Vn0 G N, 3 a couple (n,m) where n > n0 , m > 7i0, such that


A.2 Sets

In the "naive" language, a set is any collection of definite, distinguishable objects of our thought to be conceived as a whole. These objects are called the "elements" of the set or the members of the set. Often, the sets are denoted by capital letters and the elements by lower case letters. If A is a set then x G A means that x is an element of A (we also say that x belongs to A). We also use the symbol ^ to mean "does not belong". For instance 0 ^ N.

Example. If X is the set whose elements are the alphabetical letters a, b and c, we write X = {a,b,c}. (We employ often the curly brackets to enclose the elements of a set). Thus, here we can say that, b G X, / ^ X, 3 £ X.

Example. A set may have elements of very different nature: for instance X = {a,/3,5, my apple}.

Example. We cannot say that {2, 1, 5, 1, 4} is a set because we cannot distinguish between 1 and 1.

Example. The elements in a set, unless otherwise specified, need not be "ordered". Thus {1, 3, 5, 2} and {3, 5, 1, 2} represent the same set.

In general we cannot write down explicitely all the elements of a set (if there are too many of them!). For instance, this is true for the set N of all natural numbers. In this case we write N = {rr, x is a natural number} or also N = {1,2,3, . . .} the dots implying that we somehow know how to write down any missing element.

We shall use notations like {x G N, x > 3} to be read as "the set of all natural numbers which are greater than 3". Another similar example: {x G M, x2 = 1} is the same as the set {1, -1}.

We next define the basic notions of set theory: subsets, unions, intersections and complements, as well as some related ideas.

If A, B are sets, we say that A is contained in B (or A is a subset of B) if x G A => x G B. The corresponding notation: A C B. Another (equivalent) notation is: B D A ("B contains A", or "B is a superset of A").

Equality of sets is then defined by: A = B iff A C B and B C A, that is x G A <£► x G B (the elements of A are precisely those of B). We say that A is a proper (or strict) subset of B if A C B and A±B. If A is a (not proper) subset of B we write A C B or B D A. The empty set (denoted with (f>)

Appendix 167

appears useful in many occasions. It is the set which has no elements ("the statement x G </> is always false"). We see that </> = {x G A, x / x} for any set A. Thus: 0 C A for any set A.

Let now A, B be subsets of a set X. The union of A and B is the set A U B = {x G X, x G A or rr G # } . The intersection of A and B is the set A fl i3 = {a; G X, # G A and x £ B}; the intersection of a sequence of sets, (An) is n^=1An = {xeX,xe An , Vn G N}.

The sets A and B are called disjoint iff A n 5 = <j)\ the complement of A in X is the set CA = {x G X, x ^ A}, also written as X — A. The difference A - B is defined as {x G X, x G A and x £ B}. Thus A- B = An CB.

Let us point out some simple properties: (the proof is left for the student). AUB = BUA; ( A u £ ) ( J C = A u ( £ U C ) ; A nB = Bn A; ( A n s ) n c = A n ( £ n C ) ; A n (£ U C) = (A n B) u (A n C); A U (B n C) = (A U JB) n (A U C); A U CA = X; A n CA = (p; C{CA) = A; A C 5 ^ C B C CA; C(Au£) = (CA)n(C£); C(A D B) =CAUCB; AcB&AnB = A&A\JB = B&A-B = <f).

Often in mathematics we have to distinguish between sets and ordered sets. For instance an ordered pair, (in notation: (x,y)) will be equal to the ordered pair (x',yf) iff x = x' and y = y' (However, as previously defined, the sets {x,y} and {y,x} are equal; thus (1, 2) ^ (2, 1) but {1, 2} = {2, 1}).

Hence, for sets A, JB, the cartesian product of A and B (in that order), written A x B, is the set

AxB = {(x,y),xe A,ye B}.

For instance, if A = {1,2} and B = {3,4,5}, then A x B = {(1,3), (1,4), (1,5), (2,3),(2,4),(2,5)}.

If X is a set, then the power set of X, written V(X) is defined by V(X) = {A, A C X } . Thus: A G V(X) & A C X. If 5 C P(X) , then fl5 = {rr G X, x G A, VA G 5} and US = {x G I , a; G i for some A G 5} .


A.3 Functions

If X and Y are sets, a function from X to Y, / , assigns to each element x € X a (single) element y G Y, called the value of / at x and denoted / (x) .

A compact notation: f : X -*Y\ also: x -► /(x) , (x G X). (when / is given by an explicit expression we dispense with the letter / ; for instance, we can speak of the function R —> R defined by: x —» x3).

The set X is the domain of / (also called the initial set of / ) ; the set Y is the codomain of / (also called the final set of / ) . Other terms which, as a rule, are taken as synonymous with function are: map, mapping, operator, transformation.

We define equality of two functions / , g, from X into Y, by the equality: / (x) = g(x), Vx G X (here X is the domain of both / and g).

If / :X -> Y then we define f{X) = {/(a), x G X}. Thus f(X) C Y; it is called the image of X under / , or the range of / . The inclusion: f(X) C Y can be strict. If however, we have that f(X) = Y we speak about a surjective (onto) function. If / : X —> Y and A C X, then the function g : A —► Y defined by #(a) = / (a ) , Va G A is the restriction of f to A (denoted / ^ ) .

Example . Let X = {1,2,3}, Y = {2,3}. Then / described by: 1 -> 2, 2 —► 2, 3 —► 3 is function (surjective) from X into Y.

But g described by 1 —► 3, 3 —► 2 is not a function from X into Y (it is not defined for x = 2!). Also ft, defined by: 1 —► 2, 1 -> 3, 2 -> 3, 3 —► 2 is not a function as to x = 1 it assigns two elements 2 and 3.

A function / : X —► Y is called injective (also: one-one) if / (x i ) = / (x 2 ) => xi = x2 (equivalents, if xi # a;2 => f(xi) ^ /(ar2)).

Injective and surjective functions are called bijective. It is easy to see that the function Z —> Z given by n —* 2n is injective

but not surjective. On the other hand, the function / , R —> Y = {x G R, x > 0}, defined by / (x) = |x| is not injective but is surjective. Furthermore, the function R —* R given by: x —> 3x + 5 is bijective.

Next we shall state (and even prove) that bijective functions possess a (unique) inverse function.

Theorem. Let / : X —► Y 6e a bijective mapping. Then, there is a unique function g,Y -> X, sucft Jfta* g(f(x)) = x,\/x e X and f(g(y)) = y,Vy € Y. PTe ca// g fte inverse of f and write g = f~l.

Appendix 169

Proof. Since / is surjective, Vy E Y 3\x E X, such that f(x) = y. (The uniqueness follows from injectivity). Hence, we can define a function g\ Y -> X, such that g(y) = x. It follows that g{f{x)) = x, f(g(y)) — y.

To show uniqueness of g, let us assume that 3ft, Y —► X, such that h(f(x)) = x, Vx E X. Then, take any 7/ E F . There exists £ E X such that y = f(x) and so h(y) = h(f{x)) = x = g{f{x)) = #(?/), (Vj/ E Y), whence h — g. This ends the proof. □

Example. The inverse to the bijective function x —> 3x4-5 is y —> \{y — 5). We now end by giving the idea of a "function of a function" (as appears in

elementary calculus: for instance f(x) = yjx, g(t) = t3 + 1 , f{g(t)) = Vt3 -f-1) or, also called, of a "composite function".

Let X, Y, W be sets, and then consider mappings / : X —> Y and #, Y -* W. Then ft, X —> W, defined by ft(x) = g(f(x)) is called the composition of g and / and is usually written ft = gof. Thus: (gof)(x) = g(f(x)), Vrr E X.


BIBLIOGRAPHY

1. S. K. Berberian, A first course in real analysis, Springer-Verlag, 1994. 2. J. B. Conway, Functions of one complex variable (Second Edition),

Springer-Verlag, 1978. 3. G. Fox, P. Morales, J. M. Terrier, Lecture notes (in French), Universite

de Montreal, 1991. 4. A. Friedman, Advanced calculus, Holt, Rinehart and Winston, Inc., 1971. 5. A. Giroux, Lecture notes (in French), Universite de Montreal, 1992. 6. I. J. Maddox, Introductory mathematical analysis, Adam Hilger Ltd, 1977.

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