REAL ANALYSIS
REAL ANALYSIS FRANK MORGAN
&tERICAN MATHE.AfATICAL SOCIETY Pr()vi(iCIICC,
Rlludc Islalld
2000 Mathematics S'Ubject Classification. Primary 26-XX. :Front cover: The cover illustrates how continuous functions can converge nonuniform1y to a discontinuous function. It is based on Figure 17.1, page 76. The balls in the background illustrate an open cover, as in the definition"of the important concept of compactness (Chapter 9, page 41). Back cover: The author, in the Math Library outside his office at Williams College. Photo by Cesar Silva.
Cover design by Erin Murphy of the American Mathematical Society, on a suggestion by Ed Burger.
For additional information and updates on this book, visit www.ams.org/bookpages/real
Library of Congress Cataloging-in-Publication Data NIorgan, Frank. Real analysis / Frank NIorgan. p. em. Includes index. ISBN 0-8218-3670-6 (alk. paper) 1. NIathematical analysis. I. Title. QA300.NI714 515-dc22
2005
2005041221
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[email protected].
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2005 by the author. All rights reserved. Printed in the United States of America.
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The paper used in this book is acid-free and falls within the guidelines established to ensure permanence and durability. Visit the AMS home page at http://vww . ams . erg/ 10 9 8 7 6 5 4 3 2 1
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Contents
Preface
vii
Part I. Real Numbers and Limits Chapter 1.
Numbers and Logic
3
Chapter 2.
Infinity
9
Chapter 3.
Sequences
13
Chapter 4.
Functions and Limits
21
Part II. Topology Chapter 5.
Open and Closed Sets
27
Chapter 6.
Continuous Functions
33
Chapter 7.
Composition of Functions
35
Chapter 8.
Subsequences
37
Chapter 9.
Compactness
41
Chapter 10.
Existence of Maximum
45
Chapter 11.
Uniform Continuity
47
Chapter 12.
Connected Sets and the Intermediate Value Theorem
49
Chapter 13.
The Cantor Set and Fractals
53
-
Contents
vi
Part III. Calculus Chapter 14.
The Derivative and the Mean Value Theorem
61
Chapter 15.
The Riemann Integral
65
Chapter 16.
The Fundamental Theorem of Calculus
71
Chapter 17.
Sequences of Functions
75
Chapter 18.
The Lebesgue Theory.
81
Chapter 19.
Infinite Series
Chapter 20.
Absolute Convergence
89
Chapter 21a
Power Series
93
Chapter 22.
Fourier Series
99
Chapter 23a
Strings and Springs
105
Chapter 24.
Convergence of Fourier Series
109
Chapter 25.
The Exponential Function
111
Chapter 26.
Volumes of n-Balls and the Gamma Function
115
:E an
85
Part IV. Metric Spaces Chapter 27.
Metric Spaces
121
Chapter 28a
Analysis on Metric Spaces
125
Chapter 29.
CompaGtness in Metric Spaces
129
Chapter 30a
Ascoli's Theorem
133
Partial Solutions to Exercises
137
Greek Letters
147
Index
149
Preface
Our lives and the universe barely work, but that's OK; it's amazing and great that they work at all. I think it has something to do with math, and especially real analysis, the theory behind calculus, which barely works. Did you know that there are functions that are not the integral of their derivatives, and that a function can be increasing and have a negative derivative? But if you're a little careful you can get calculus to work. You'll see. The theory is hard, subtle. After Newton and Leibniz invented the calculus in the late 1600s, it took puzzled mathematicians two hundr'ed years, until the latter 1800s, to get the theory straight. The powerful modern approach using open and closed sets came only in the 1900s. Like many others, I found real analysis the hardest of the math major requirements; it took me half the semester to catch on. So don't worry: just keep at it, be patient, and have fun. This text is designed for students. It presents the theoretical intellectual breakthroughs which made calculus rigorous, but always with the student in mind. If a shortcut or some more advanced comments without proof provide better illumination, we take the shortcut and make the comments. The result is a complete course on real analysis that fits comfortably in one semester.
-
vii
Vlll
Preface
This text developed with my one-semester undergraduate analysis course at Williams College. I would like to thank my colleagues and students, especially Ed Burger, Tom Garrity, Kris Tapp, and Nasser Al-Sabah '05, and my editors Ed Dunne and Tom Costa. Frank Morgan Department of Mathematics and Statistics Williams College Williamstown, Massachusetts www.williams.edu/Mathematics/fmorgan
[email protected]
Part I
Real Numbers and Limits
Chapter 1
Numbers and Logic
1.1. Numbers. Calculus and real analysis begin with numbers: The natural numbers
N = {1, 2, 3, ... }. The integers
z={ ... , -3, -2, -1,0,1,2,3, ... } (Z stands for the German word Zahl for number). The rationals Q = {pjq in lowest terms: p E Z, q E N} = {repeating or terminating decimals}.
(Q stands for quotients). The reals
JR. = {all decimals} with the understanding that .999· · · = 1, etc. Reals which' are not rational are called irrational. Thus the set of irrationals is the complement of the set of rationals, and we write
{irrationals} = QC = JR. - Q.
1.2. Intervals in
IR~
For a < b, define intervals
[a,b] = {x E JR.: a:S x::; b}, a.----.b (a,b)
[a, 00) and so on.
= {x =
{x
< x < b}, ao--ob a::; x}, a~
E JR.: a E~:
-
Q
1. Numbers and Logic
4
1.3. IRn. We will consider the plane ~2 of pairs (x, y) of real numbers and more generally n-dimensional space ~n of n-tuples (Xl, X2, ... ,xn ) of real numbers. In these spaces, by rationals we'll mean n-tllples of rationals. There are fancier number systems, such as the complex numbers (which will make a solo appearance in, Chapter 25) and the much fancier "quaternions" (which maybe you'll see in some future course).
1.4. Distance. The (Euclidean) distance between real numbers X and y is given by Iy - xl. More generally, the distance between two vectors (Xi), (Yi) in ~n is given by
Iy - xl == ((YI
-
XI)2
+ (Y2 - X2)2 + ... + (Yn - x n )2)1/2.
Distance satisfies the triangle inequality, which says that the length of one side of a triangle with vertices x, Y, z is less than or equal to the sum of the lengths of the other two sides:
Ix - zl
$
Ix - yl + Iy - zl·
The Ancient Greeks discovered with dismay that not every real is rational, with the following example. 1.5. Proposition.
V2 is irrational.
Proof. Suppose that V2 equals pi q, in lowest terms, so that p and q have no common factors. Since 2 == p2 I q2 , 2q2 = p2.
Thus p must be even: p
== 2p'.
Hence 2q2
== 4p'2,
q2 == 2p /2,
which means that q must be even, a contradiction of lowest terms, since p and q both have 2 as a factor. 0 There is an easier example: 1.6. Proposition. loglO 5 is irrational. Proof. Suppose that loglO 5 equals plq, which means that lQP/q 1()P ==
== 5 or
5q .
Bllt lOP ends in a 0, while 5q ends in a 5, contradiction.
o
1~8~
Implication
5
Probably the easiest way to give an example of an irrational is just to write down a nonrepeating, nonterminating decimal like .01 001 0001 00001 000001 ....
although that's no number you've ever heard of.
1.7. Logic and ambiguous English. English is a treacherous language, and we have to be very of "or" in English:
careful~
For example, there are two meanings
(1) A crazy "exclusive" or, meaning "one or the other, but not both." I had lunch in the French Quarter, and they offered me a choice of black-eyed peas or greens, bllt they wouldn't give me both. (2) The sensible, mathematical "inclusive" or, meaning "one or the other or both." You can work on your real analysis homework day or night.
1.8. Implication. There are many ways to say that one statement A implies another statement B. The following all mean exactly the same thing: • If A, then B. • A implies B, written A
• A only if B. • B if A, written B
'¢=
=}
B.
A.
Here are some examples of such equivalent statements: • If a number n is divisible by four, then it is even. • n divisible by four implies n even. • n is divisible by four only if it is even (never if it is odd). • n is even if divisible by four.
Such an implication is true if B is true or if A is false (in which case we say that the implication is "vacuously true"). For example, "If 5 is even, then 15 is prime" is vacuously true. This is pure logic and has nothing to do with causation. (Indeed, from the point of view of causation, if 5 were even, then 15, as a multiple ·of 5, would be even and hence not prime.) Such an implication is false only if A is true and B is false. Such an implication is logically equivalent to its contrapositive: not B implies not A. This is the basis for proof by contradiction. We assume the negation of the B we are trying to prove, and arrive at a contradiction of the hypothesis A.
1. Numbers and Logic
6
Such an implication is logically distinct from its converse: B implies A. • If x is a Williams student, then x is a human being. True.
• If x is not a human being, then x is not a Williams student. Contrapositive, true. • If x is a human being, then x is a Williams student. Converse, false in this case.
If it happens that" the implication and its converse are both true, i.e., if A => B and B => A, then we say that A and B are logically equivalent or that (A {:} B). A if and only if B Example. Let S C Z. Consider the statements
A: All elements of S are even. (For all XES, x is even.) B: Some element of S is even. (There exists XES, such that x is even.) Does B => A? Certainly not in general. Does A => B? Not in general because if S is the empty set 0, then A is true (vacuously), while B is false!
1.9. Abbreviations. Understand but avoid abbreviations such as /\ and
V or
rv
not
3 exists \I for all.
1.10. Sets. A set A is completely determined by the elements it contains: A={X:XEA}. If all elements of A are elements of B, we say that A is a subset of B (A c B) or that B contains A (B =:) A). You have to be careful, because the word "contains" is used both for subsets and for elements, as determined by context. Proper containment (A C B and A =1= B) is denoted by A ~ B. Some texts denote ordinary containment by A ~ B to remind you that equality is allowed. We say that sets A, B intersect if AnB =1= 0. The symbol n~l An denotes the set of elements common to all of the sets AI, A 2 , A 3 , . .. :
n 00
x E
An
if and only if x E An for all n.
n=l
We define the complement of a set A C == {x: x ¢ A},
and the difference of two sets
A- B
== {x
E A:
x ¢:. B}.
7
Exercises 1
1.11. Functions. A function f: A ---7 B takes an input x from a domain A and produces an output f(x) in some range B. For most of this book we'll assume that A c ~n and B c~. The set of all outputs
f(A) = {f(x): x E A} c B is called the image of f. If the image is the whole range, then f is called onto or surjective. If f maps distinct points to distinct values, then f is called injective or one-to-one (1-1). If f is both injective and surjective, then f is called bijective or a 1-1 correspondence and f has an inverse function f-l: B ---7 A. If X is a subset of A, then f(X) is the corresponding set of values:
f(X) = {f(x): x EX}. Whether or not f- 1 exists as a function on points, for a set Y always take the inverse image
c B, you can
f- 1 y = {x: f(x) E Y}. For example, if f(x) = sinx, then f-l{O} = {a, ±1r, ±21r, ±31r, . .. }.
Exercises 1 1. Is the statement If x E CQ, then x 2 E N true or false for the following values of x? Why? a. x = 1/2. b. x=2. c. x
=..J2.
d. x = {12. 2. For which real values of x is the converse of the statement of Exercise 1 true? 3. Which of the following statements are true of the real numbers. a. For all x there exists a y such that y > x 2 . b. There exists a y such that for all x, y > x 2 . c. There exists a y such that for all x, y < x 2 . d. Va, b, c, 3x such that ax 2
+ bx + c = O.
1. Numbers.and Logic
8
4. Which of the following statements are true for all real numbers x. a. If x E (1,2], then x 2 E (1,4]. b. Ifx E (-1,2], then x 2 E (-1,4]. c. Ifx E (-1,2], then x 2 E (1,4]. 5. a. What is the length of a diagonal of a unit square (with vertices (0,0), (0,1), (1,0), (1, I))? b. What is the length of a long diagonal of a unit cube? c. What is the length of the longest diagonal of a unit hypercube in ~4? d. What is the length of the longest diagonal of a unit hypercube in
6. Proye that
~n?
-v'2 is irrational.
7. Prove that lOgIO 15 is irrational. 8. Find infinitely many nonempty sets of natural numbers
N ::J 8 1 ::J 82 ::J 8 3 ::J ... such that n~=l 8 n
=
0.
9. Identify each of the following functions f from
~
to
~
as injective (1-1),
surjective (onto), neither, or both (bijective).
== -x. f(x) = x 2 . f(x) == sin x.
a. f(x) b. c.
d. f(x)
== eX.
e. f(x)
== x 3 + x 2 .
10. Give a counterexample to one of the following four formulas for images and inverse images of.sets (the other three are true):
== !(XI ) U f(X2), f- 1(Y1 U Y2) == !-l(yI ) U f- 1(Y2), f(X 1 n X2) == !(X1) n f(X2), f- 1(Y1 n Y2) == f- 1(y1) n f- 1 (Y2).
f(X 1 U X2)
Chapter 2
Infinity
Although some infinite sets contain other infinite sets, there is a good intuition that in some sense all infinite sets are the same size. For example, although the even natural numbers are a proper subset of all the natural numbers, the two sets can be paired up by matching 2n with n: 2, 4, 1, 2,
6, 8, 3, 4,
10, 5,
...
Although many infinite sets are the same size as the natural numbers, it turns out that some really are much bigger in a mathematically precise sense. 2.1. Definition. A set is called countable if it is finite or it can be put in 1-1 correspondence with the natural numbers, Le., if its elements can be listed.
Examples of countable sets include the set 3N of positive multiples of 3 and the set Z of all integers (including the negative ones): N 3N Z 1 3 a 2 6 1 3 9-1 4 12 2 5 15 -2 It is clear that a subset of a countable set is countable because you can use the order of the list of the set to list the subset. It turns out that even the apparently larger set of all rationals is still countable. 2.2. Proposition. The set Q of rationals is countable.
-
o
2. InB.n.i ty
10
Proof. It suffices to show that the positive rationals are countable; then the lists of positive and negative rationals and zero can be interspersed on a single list as for Z above. Arrange all ratios in an infinite table like this:
1/1 2/1 3/1 4/1 5/1
1/2 '2/2 3/2 4/2 5/2
1/3 2/3 3/3 4/3 5/3
1/4 2/4 3/4 4/4 5/4
1/5 2/5 3/5 4/5 5/5
Now starting at the upper lefthand corner, move through the northeasterly diagonals of this table, starting with 1/1, then 2/1 and 1/2, then 3/1, 2/2, and 1/3, and so on.
/ /1/3/1/4/1/5/ 1/2 / / /2/3/2/4/2/5 2/1 2/2 1/1
//// 3/1
3/2
3/3
3/4
3/5
4/1
4/2
4/3
4/4
4/5
/5/1
5/2
5/3
5/4
5/5
///
//
When you hit a repetition (like 2/2 == 1/1), skip it. This process puts the rationals on a list:
1/1, 2/1, 1/2, 3/1, 1/3, 4/1, 3/2, 2/3, 1/4, 5/1, 1/5,
o
Therefore, the rationals are countable. 2.3. Proposition. The Cartesian product of two countable sets 0 1 x 02 === {(Cl' C2): Cl E 01 and C2 E
02}
is countable. A countable union of countable sets is countable.
(A "countable union" means "the union of countably many.") Proof. The proofs (Exercises 4 and 5) are very similar to the proof of Proposition 2.2. 0
2.5. Uncountable inEnities
11
By now you may be thinking that all sets are countable. Georg Cantor showed in the late 1800s that the reals ~ are not countable. This sounds very hard to prove. How could you show that there is no 1-1 correspondence between ~ and N? The proof is amazing. 2.4. Proposition. The set
~
of reals is uncountable.
Proof. We will assume that Proposition 2.4 is false and derive a contradiction. In particular we will assume that the reals can be listed, and then exhibit a real number missing from the list, the desired contradiction. Since the argument applies to any list, that will complete the proof. Suppose that the reals were countable. Then the positive reals would be countable and could be listed, for example: 1. 2. 3. 4. 5. 6.
657.853260 2.313333 3.141592 .000207 49.494949 .873257
. . . . . .
1b obtain a contradiction, we just have to show that some real Q (Greek letter alpha, see Table of Greek letters on page 147) has to be missing from this list. We'll construct such an Q by making its first decimal place different from the first decimal place of the first number on the list, by making its second decimal place different from the second decimal place of the second number on the list, and in general by making its nth decimal place different from the nth decimal place of the nth number on the list. To be specific, we'll make the nth decimal place 1, unless the nth decimal place of the nth number on the list is 1, in which case we'll make it 2. For our example, Q
==
.122111 ...
As promised, we have found a number Q missing from the list, because it differs from the nth number ~n the list in the nth decimal place. Since this argument applies to any list"the reals cannot be listed. 0
2.5. Uncountable ,infinities. More generally, one can say that two infinite sets have the same size if they can be put in 1-1 correspondence. It turns out that there is an infinite hierarchy of infinitely many different size infinities. What's bigger than JR.? Not ~2 or ~n; they turn out to be the same infinity as~. One bigger infinity is the set of all functions from ~ to~. Another is the set of all subsets of~. Indeed, the set of all subsets of
2. Infinity
12
any set X is always bigger than X, so this is one way to keep going. The proof, which you could find on the internet, is short but tricky.
Exercises 2 1. Prove whether or not the set S is countable.
== {irrationals}. b. S == {terminating decimals}.
a. S
c. S == [0, .001).
d. S
== CQ x CQ.
e. S == JR x Z. 2. Prove that the intersection of two countable sets is countable. Hint: There is a very short answer. 3. Prove whether or not the intersection and union of two uncountable sets must be uncountable. 4. Prove that the Cartesian product of two countable sets
A x B == {(a, b): a E A and b E B)} is countable. Hint: Imitate the proof of Proposition 2.2. 5. Prove that a countable union of countable sets is countable. Hint: Imitate the proof of Proposition 2.2.
Chapter 3
Sequences
Ivlany would say that the hardest theoretical concept in analysis is limit. What does it mean for a sequence of numbers to converge to some limit? There just is no easy answer.. Oh, we'll try to find one, but there will always be some sequences that a simple answer cannot handle. III the end, we'll be forced to do something a little complicated, and to make it worse, we'll follow tradition and use the Greek letter c (epsilon).
3.1. Discussion. The sequence (1)
1, 1/2, 1/3, 1/4, 1/5,
converges to O. The sequence of digits of 7[
(2)
3, 1, 4, 1, 5, 9, 2,
does not converge to anything; it just bounces around. The sequence
(3)
1, 2, 3, 4, 5, ....
diverges to infinity. Those sequences are easy. But sometimes it is hard to decide. What about the sequence (4)
1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, ... ?
Common agreement is that it does not converge, but to decide we really need a good definition of "converge." How about this:
First attempt at a definition. A sequence
converges to, say,
°
if the terms get closer and closer to 0.
-
13
3. Sequences
14
According to this definition, (4) does not converge. Good, but what about the following sequences:
(5)
8, 1, 4, 1/2, 2, 1/4, 1, 1/8, 1/2, 1/16, 1/4, 1/32, 1 1 1 1 1 1 , 1 , IS' 1 , 1 , ... 32 2 4 16
(6)
The terms of (5) are not getting "closer and closer to 0," but the sequence does converge to O. The terms of (6) are getting "closer and closer to 0/' but the sequence does not converge to O. (We will see that this sequence converges to 1.) We need a more precise definition. The terms need to get and stay arbitrarily close to zero or whatever limit value L, eventually. "Arbitrarily close" means as close as anyone could prescribe, Le., given any positive error, the terms eventually have to stay within that tolerance of error. Since the letter e is already taken by 2.71828... , mathematicians usually use the Greek letter epsilon c for the given error. And what do we mean by "eventually"? We mean that given the tolerance of error c, we can come up with a big number N, such that all the terms after aN are within c of the limit value L. That is, given E > 0, we can come up with an N, such that whenever n > N, every subsequent an is within E of L. Now that's a good definition, and here it is written out concisely: 3.2. Definition of convergence. A sequence an converges to a limit L
if given
E
> 0, there is some N, such that whenever n > N,
Ian - LI < c. Otherwise we say that the sequence diverges. Notice that order in the definition is very important. First comes the sequence an and the proposed limit L. Second comes the tolerance of error E, which is allowed to depend on the sequence. Third comes the N, which is allowed to depend on c. A sequence which diverges might diverge to infinity like (3) or diverge by oscillation like (2). Some other things in the definition do not matter, such as whether the inequalities are strict or not. For example, if you knew only that you could get Ian - Llless than or equal to any given c, you could take a new c' = c/2 and get
Ian - LI ::; E' < c,
15
3.5. Proposition
strictly less than £. Similarly, it suffices to get Ian - Llless than 3c, because you can take c' == £/3 and get - L I < 3£' == c.
Ian
So the final c could be replaced by any constant times c or anything that's small when c is small. 3.3. Example of convergence. Prove that an == 1/n2 converges to O.
First, let's think it through. Given E > 0, we have to see how big n has to be to guarantee that an == 1/n2 is within c of 0:
1/ n2 -
1
01 < c.
This will hold if 1/n < c, that is, if n > 1/ VE. SO we can just take N to be l/VE, and we'll have the following proof: 2
Given
E
> 0, let N == l/VE. Then whenever n > N,
Ian -
LI
==
11/n2 - 01 = 1/n2 < 1/N2 == £.
Notice how we had to work backwards to come up with the proof. 3.4. Bounded. A sequence an is bounded if there is a number IvI such that for all n, Ian I ::; IvI.
For example, the sequence an == sin n is bounded by 1 (and by any
IvI 2: 1). The sequence an == (-1)n/n 2 is bounded by 1. The s~quence an == n 2 is not bounded.
3.5. Proposition. Suppose that the sequence an converges. Then
(1) the limit is unique; (2) the sequence is bounded. Before starting the proof of (1), let's think about why a sequence cannot have two limits, 0 and 1/4 for example. It's easy for the terms an to get within 1 of both, or to get within 1/2 of both, but no better than within c == 1/8 of both (see Figure 3.1). Similarly, if a sequence had any two different limits L 1 < L2, you should get a contradiction when c == (1/2)(L 2 - L 1 ). I think I'm ready to write the proof.
1/8
1/8
-I------l:---o
~
1/4
Figure 3.1. A number cannot be closer than distance 1/8 to both 0 and 1/4.
16
3. Sequences
Proof of (1). Suppose that a sequence an converges to two different limits L 1 < L 2. Let E == (1/2)(L2 - L 1 ). By the definition of convergence, there is some N 1 such that whenever n > N 1 , Ian - L 1 < c. Similarly there is some N 2 such that whenever n > N 2 , Ian - L 2 < E. Choose n greater than Nl and N2. Then 1
1
L2 - L 1
:::;
Ian - L 1 1 + IL 2 - anI
< c + c == 2c == L 2 - L 1 , D
a contradiction.
The proof of (2) is easier. After a while the sequence is close to its limit L, and once an is within say 1 of L, lanl < ILl + 1. There are only finitely many other terms to worry about, and of course any finite set is bounded (by its largest element). I'm ready to write the proof: Proof of (2). Let an be a sequence converging to L. Choose N such that whenever n > N, Ian - LI < 1, so that lanl < ILl + 1. Let
IvI == max{ILI + 1, lanl, with n :::; N}. Then if n < N, an 5. M. an :::; IvI.
If n > N, an <
ILl +
1 :::; IvI.
So always D
3.6. Proposition. Suppose real sequences an, bn converge to a and b:
Then
(1) can
~ ca,
(2) an + bn ~ a + b, (3) anbn ~ ab, (4) anlbn ~ alb, assuming every bn and b is a nonzero real number. PREPARATION FOR PROOF. We'll prove (1) and (4), and leave (2) and (3) as Exercises 13 and 14. As usual, it pays to build the proof backwards. At the end of the proof of (1), we'll need to estimate
lean - cal =
which will hold if Ian - al <
E/lcl
lei Ian -
al
< c,
(unless c == 0). I see how to do the proof.
Proof. We may assume that c =1= 0, since that case is trivial (it just says that 0,0,0, ... -T 0). Since c is a fixed constant, given E > 0, since an -T a, we can choose N such that whenever n > N, Ian - al < c/lcl. Then lean - cal
so that
can ~ ca.
== Icllari -
al
< E,
3.6. Proposition
17
The proof of (4) is harder, so we start with a discussion. At the end of the proof, we'll need to estimate lan/bn - a/bl in terms of things we know are small: Ian - al and Ibn - bl. The trick is an old one that you first see in the proof of the quotient rule in calculus: go from an/b n to alb in two steps, changing one part at a time, from an/b n to a/b n to alb, to end up with some estimate like:
lan/bn - a/bl
lan/bn - a/bnl + la/bn- a/bl == Ian - al/lbnl + Ib - bnlla/bbnl· Ib - bnl are small, and la/bl is just a ~
We know that Ian - al and constant, but what about those l/lbn l? We need to know that bn is not too close to O. Fortunately since bn ----+ b, eventually Ibn I > Ibl /2 (as soon as bn gets within Ibl/2 of b). Then l/lbnl ::; 2/lbl. So the estimate can continue
::; Ian - al(2/lbl) + Ib - bn112a/b2 < £/2 + c/2 == c, if we just make sure that Ian - al(2/lbl) and Ib - bn112a/b21 are less than c/2 by making Ian - al < Elbl/4 and Ib - bnl < £lb2/4al (which we interpret as 1
no condition on bn if a == 0). Can we guarantee both of those conditions at the same time? We can find an N I to make the first one hold for n > N I , and we can find an N2 to make the second one hold for n > N 2 . To make both work, just take N to be the maximum of N 1 and N 2 . In general, you can always handle finitely many conditions. Here's the whole proof from start to finish. Since an ----+ a and we can choose N such that whenever n > N, the following hold:
bn
----+
b,
Ian - al < clbl/4, Ib - bnl < clb2/4al, and Ib - bnl < Ibl/2, which implies that l/lbn l $ 2/lblThen
lan/bn- a/bl ::; lan/bn - a/bnl + la/b n - a/bl == Ian - al/lbnl + Ib - bnlla/bbnl ~ Ian - al(2/lbl) + Ib - bn112a/ b2 < £/2 +c/2 == c, and consequently an/b n ----+ "a/b.
1
o
It would have been OK and simpler to start out by just requiring that
Ian - al < c, Ib - bn I < c, and Ib - bnl < Ibl/2, which implies that
l/lbnl $ 2/l bl.
18
3. Sequences
Then I.
anl bn - albl ::; janlbn - albnl + lalbn - albl
== Ian - al/lbnl + Ib - bnllalbbnl < e(2/Ibl) + cl2alb21 == OE, where 0 is the constant (2/Ibl)+ 12alb21_ Although we haven't made it come out quite as neatly at the end, we've still shown that we can make the error arbitrarily small by choosing n large enough, which is sufficient.
3.7. Rates of growth. Limits of many sequences can be determined just by knowing that for n large,
1/n2
«
1/n« 1 «lnn ~ Vii« n« n 2 ~ n 3 ~ 2n « en « IOn «n!
where f(n) « g(n) means that f becomes a negligible percentage of g, f (n) I9 (n) ----+ 0, so that in a limit as n ----+ 00 whenever you see f + g, or even clf + C2g, you can ignore the negligible f. For example, 4
lim n +ln(n+l) n-tOO V5n B + 16
=
~=~
lim n--+oo
y'5.
J5n B
3.8. Three famous limits.
V'2 == 2 l / n
(1)
(2)
rjii == n1/n == (e 1n n)l/n ==
(The exponent. (in n) In
(3)
----+
----+
0 because In n
(1 + 1/n)n
2°
== 1.
e(ln n)/n ----+
« ----+
eO
==
1.
n.) e.
(This is sometimes used as the definition of the number e, after you check that the limit exists. Exercise 25.6 derives it from another definition.)
3.9. Accumulation points. A point p is an accumulation point of a set S if it is the limit of a sequence of points of S - {p}. It is equivalent to require that every ball (p - r, p + r) about p intersect S - {p} (Exercise 19). For example, 0 is an accumulation point of {lin: n E N} because 0 == lim lin or because every. ball (interval) about .0 intersects {lin}. III this case the accumulation point is not in the original set. Every point of the unit interval [0, 1] is an accumulation point. In this case all of the accumulation points are in the set.
3.10. lRn • Almost everything in this chapter works for vectors in JRn as well as for points in JR. The exception is Proposition 3.6(4) because there is no way to divide vectors. Proposition 3.6(3) holds both for the dot product and for the cross product.
Exercises 3
19
Exercises 3 Does the sequence converge or diverge? IT it converges, what is the limit? 1. 1, 0, 1/2, 0, 1/4, 0, 1/8, 0, ... 2. 3, 3.1, 3.14, 3.141, 3.1415, 3.14159, ...
3. an
=
_ 4 . an -
1 + (-l)n/n. l+(-l)n
n
.
5. an = (-1)n(1-1/n).
8 . an =
10. an .
en
n 5+n- 5·
=
sinn. n
11. Prove that an
== l/n
12. Prove that an
= 1000/n3
converges to O. converges to O.
13. Prove 3.6(2). 14. Prove 3.6(3). 15. Prove that if an ::; bn
::;
en and lim an =
lim Cn
== L
then lim bn = L.
16. Prove or give a counterexample. Let an be a sequence such that an+l - an ~ O. Does an have to converge?
3. Sequences
20
17. A sequence an is called Cauchy if, given c > 0, there is an N such that whenever m, n > N, lam - ani < c~ Prove that if a sequence in JR is convergent, then it is Cauchy~ (Exercise 8~6 will prove the converse in JR, so that Cauchy gives a nice criterion for convergence without mentioning what the limit is.) 18~
Prove that a Cauchy sequence is
bounded~
19. Describe the set of accumulation points of the following sets: a~
the rationals;
b. the irrationals; c. [a, b); d~
the integers.
20. Prove that p is an accumulation point of S if and only if every ball B about p intersects S - {p}~ 21. Prove or give a counterexample: There are only countably many sequences with llmit O. 22. Prove or give a counterexample: a real increasing sequence al
< a2 < a3 < ...
converges if and only if the differences an+l - an converge to
o.
Chapter
4
Functions and Limits
Recall that for us a function f takes an input x from some domain E C JRn and produces an output f(x) E JR. For example, for f(x) == 1/x, the natural domain is E = lR - {O}. We want to extend our notion of limit from sequences to functions. Instead of the limit limn - HXl an of a sequence, such as limn - HXl lin = 0, we want to consider a limit limx-tp f(x) of a function, such as limx --+3 x 2 == 9, to give a trivial example. Instead of going way out in our sequence (n > N), we will be taking x close to some fixed target p, taking Ix - pi smalL To say how small, we'll use the Greek letter before epsilon E, namely delta b, and require that Ix - pi < b.
4.1. Definitions. We say limx-tp f(x) = a ("the limit as x approaches p of f(x) equals a") if, given E > 0, there exists a b > 0 such that
o < Ix - pi
< b =? If(x) - al < c.
Of course we have to assume that x lies in the domain E of f, and hence we need to assume that p is an accumulation point of E. Notice that by requiring that 0 < Ix - pi, we do not allow x to equal p. The limit depends only on values f(x) at nearby points, not on f(P). If it happens that lim x --+ p f(x) = f(p), or p is not an accumulation point of E, then f is called continuous at p. Examples of continuous functions include powers of x, sin x, cos x, eX, and combinations of continuous functions such as eX cos x
x2 + 1 (as long as the denominator is never 0).
4.2. Example. Consider the function f(x) = sin~, defined on E = JR-{O}, graphed in Figure 4.1. Then limx-to f (x) does not exist (Exercise 4).
-
21
22
4. Functions and Limits
0.6
0.8
-1
Figure 4.1. For f(x)
= sin~,
lim:z:- o f(x) does not exist.
4.3. Example. Consider the function f(x) == x sin~, defined on E == IR. {O}, graphed in Figure 4.2. Then limx-to f(x) == 0 (Exercise 3). If we define f(O) == 0, then f will be continuous. 4.4. Example. Consider the function f(x) on IR. which is 1 on rationals and 0 on irrationals. Then limf(x) never exists, and f is continuous nowhere. This function is called the characteristic function of the rationals and written
using the Greek letter chi. 4.5. Proposition. Iflim x -+ p f(x) exists, then it is unique. Proof. The proof is just like the proof that the limit of a sequence is unique. Suppose that liIDx--+p f(x) has two distinct values a, b. Choose 8 > 0 such that if 0 < Ix - pi < 8, then
If(x) -
al < Ib - al/2
and
If(x) -
bl < Ib - al/2.
Since for limits we always assume that p is an accumulation point of the domain, some xo satisfies 0 < Ixo - pi < 8. Then
Ib - al
~ If(xo) -
al + If(xo) - bl < Ib - al/2 + Ib - al/2 == Ib - ai,
4.6. Proposition
23
-1
-0.5
-0.5
Figure 4.2. For f(x)
= x sin ~,
lirnx---+o f(x)
= o.
o
the desired contradiction. 4.6. Proposition. Suppose that lirn x-+ p f(x) Then
(1) lim x-+ p cf(x) = ea, (2) lirnx-+p(f + g)(x) == a + b, (3) limx-+p(fg)(x) = ab, and (4) limx-+p(f/g)(x) == alb if b -I
==
a and lirn x--+ p g(x) = b.
o.
Proof. The proof is similar to the proof of 3.6; see Exercises 5 and 6.
D
24
4. Functions and Limits
Exercises 4 1. Give an example of a function f: IR. the integers. 2. Give an example of a function f: IR.
---+
---+
IR. which is continuous except at
IR.
whi~h
is continuous only at O.
3. Prove that for the function f(x) of Example 4.3, limx~o f(x)
== O.
4. Prove that for the function lex) of Example 4.2, limx-+o f(x) does not equal o. 5. Prove 4.6(2). 6. Prove 4.6(3).
7. Prove that the sum of two continuous functions is continuous. 8. Prove that the product of two continuous functions is continuous. 9. Give an example of a function which is continuous only at the integers. Graph it. 10. Give an example of two functions, both discontinuous at 0, whose sum is continuous at O. Give an example of two functions, both discontinuous at 0, whose product is continuous at O. 11. Consider the function I: IR. ----7 IR. which is 0 at irrationals and l/q at a rational p/q (in lowest terms with q positive). Where is I continuous?
Part II
To po log y
Chapter 5
Open and Closed Sets
Although sequences played the crucial role in the rigorous understanding of calculus developed in the 1800s, over the next fifty years mathematicians sharpened the theory using the concepts of open and closed sets. If it took mathematicians fifty years to get used to these more abstract ideas, you shouldn't worry if it takes you a couple of weeks. It's actually quite fun if you take your time and think about lots of examples.
5.1. Definition. A point p in ~n is a boundary point of a set S in ~n if every ball about p meets both S and its complement Se. The set of boundary points of S is called the boundary of S and written as. (That funny symbol, sometimes called del, is not a Greek letter. It is used for partial differentiation too.) It follows immediately that a set S and its complement Se have the same boundary. For example, consider a ball about a point a of radius r > 0:
B(a,r) ==
{Ix - al
~
r}
(see Figures 5.1 and 5.2). Its boundary is the sphere:
aB-(a,r) == {Ix -
al =
r}.
In this case the boundary of S is part of S. In JR.2 , we sometimes call the ball a disc and we usually call its boundary a circle rather than a sphere.
In JR.1 , the ball B(a, r) is just the interval [a - r, a + r] and its boundary is just the two endpoints.
-
27
5. Open and Closed Sets
28
30 B(a,r)
Figure 5.1. The boundary of the ball B(a, r) is the sphere. The closed ball includes its boundary.
10 B(a,r)
20 B(a,r)
•
Figure 5.2 . The boundary of a 2D ball or "disc" is a circle. boundary of a ID ball or interval is two points.
As another example, consider all of moved:
s == ~n
-
~n
•
a• r The
with one point, the origin, re-
{a}.
Its boundary is the single point {O}. In this case, the boundary of S is not part of S. The boundary of the rationals Q is all of JR.. The boundary of 1R is the empty set.
5.2. Definition. A set S in ~n is open if it contains none of its boundary points. A set S in JR.n is closed if it contains all of its boundary points. It follows immediately that a set S is open if and only if its complement Se is closed. Many a set contains just part of the boundary and hence is neither open nor closed. "Not closed" does not mean "open." These terms are not opposites!
29
5.3 Proposition
CLOSED
~-~ .
1-. , - - .
I
OPEN
i i i
, --
i
..
i
.!
I~r:-~~ __j
NEITHER Figure 5.3. Some sets are closed or open but most are neither.
For example, the ball B(a, r) is closed. IT you remove its boundary, the resulting set is open and is called an open ball, for which unfortunately we have no special symbol. ~n - {O} is open. The rationals Q are neither open nor closed. The reals JR are both open and closed. See Figure 5.3 for a few more examples. Proposition 5.3 gives nice direct characterizations of open and closed. A set is open if it includes a ball about every point; of course the ball has to get smaller as you get close to the boundary. A set is closed if it includes all accumulation points. See Figure 5.4. 5.3. Proposition. A set S in IRn is open if and only if about every point of S there is a ball completely contained in S. A set S is closed if and only if it contains all of its accumulation points. Proof. Suppose that about some point p of S there is no ball completely contained in S. Then every ball about p contains a point of Se as well as
5. Open and Closed Sets
30
Figure 5.4. An open set includes a ball about every point. A closed set includes all of its accumulation points.
the point pinS. Consequently, p is a boundary point of S, and S is not open. Conversely, suppose that S is not open. Then some point p of S is a boundary point, and there is no ball about p contained in S. Suppose that an accumulation point p of S is not contained in S. Then every ball about p contains a point of SC (namely, p) and a point of S (because p is an accumulation point of S). Therefore, p is a boundary point not contained in S, so that S is not closed. Conversely, suppose that S is not closed. Then SC contains some boundary point p of S, which is an accumulation point of S, so S does not contain all of its accumulation points. 0 5.4. Proposition. Any union of open sets is open. A finite intersection of open sets is open. Any intersection of closed sets is closed. A finite union of closed sets is closed.
Proof. To prove the first statement, suppose that x belongs to the union of open sets Uo.. Then x belongs to some U{3' By Proposition 5.3, some ball about x is contained in U{31 and hence in the union of all the Uo.. Therefore, by Proposition 5.3 again, the union is open. To prove the second statement, let U be the intersection of finitely many open sets Ui . Let p belong to U. Since p belongs to Ui , there is a ball B(p, ri) contained in Ui. Let r === min{ri}. Then B(p, r) is contained in every Ui and hence in U. Therefore U is open. Let C be an intersection of closed sets Co.. Then CC is the union of the open sets and hence open. Therefore C is closed. Similarly let C be a union of finitely many closed sets Ci . Then CC is the intersection of the open sets and hence open. Therefore C is closed. 0
C£, cf,
5.5. Definitions (see Figure 5.5). The interior of a set S, denoted int S or S, is S - 8S. The closure of S, denoted cIS or S, is S U 8S. An isolated point of S is the only point of S in some ball about it.
31
Exercises 5
.p
.p
Figure 5.5. A set, its interior, its closure, and an isolated point p.
5.6. Proposition. The interior of S is the largest open set contained in S, and the closure of S is the smallest closed set containing S.
o
Proof. Exercises 11 and 12.
5.7. Topology. In IR.n or in more general spaces, the collection of open sets is called the topology, which determines, as we'll soon see, continuity, compactness, connectedness, and other "topological" properties.
Exercises 5 1. Say whether the following subsets of IR. are open, closed, neither, or both. Give reasons.
a. [0,1); b. Z;
c. {x E~: sin x > O},
d. U~=2[1/n, 1). 2. Show by example that the intersection of infinitely many open sets need not be open. 3. Show by example that the union of infinitely many closed sets need not be closed. 4. Is all of 1R the only open set containing Q? Prove your answer correct. 5. If S == [0,1), what are
as, S,
and
8?
32
5. Open and Closed Sets
6. If S == Z, what are 7. If S
as, S,
= B(O, 1) in JR2,
and
what are
S?
as, S,
and
S?
8. Give an example of op"en sets U1 => U2 => ... such that the intersection Ui is closed and nonempty.
n
9. Give an example of nonempty closed sets 0 1 => O2 => ... such that the intersection Ci is empty.
n
10. Prove that every point of S is either an interior point or a boundary point. 11. Prove that the interior of S is the largest open set contained in S. (First prove that it is open.) 12. Prove that the closure of S is the smallest closed set containing S. 13. Prove that a boundary point of S is either an isolated point or an accumulation point. 14. Prove or give a counterexample: two disjoint sets cannot each be contained in the other's boundary. 15. A subset So of a set S is dense in S if every ball about every point of S contains a point of So. Are the rationals dense in the reals? Are rationals with powers of 2 in the denominator dense in the reals? Are the points with rational coordinates dense in JRn?
Chapter 6
Continuous Functions
Continuous functions are the bread and butter of calculus. We'll now give three equivalent definitions of continuous functions~ The first, our original defipition, uses the idea of limit. The second uses the concept of a convergent sequence. The third, modern definition uses the concept of open set. It is the shortest and the most abstract, and it takes a little while to get used to. 6.1. Proposition. Let f be a junction from JRn to JR. The following are
equivalent definitions of what it means for f to be continuous everywhere: (1) For every point p, given
E
Ix - pi < 8 =}
> 0, there exists 6 > 0, such that If(x) - f(P)1 <
E.
(2) For every point p, for every sequence X n - ? p, f(x n ) (3) For every open set U in JR, f-1U is open.
-)
f(p).
Proof. (1) {::} (2)~ Easily (1) =} (2) because if all points x near p have values f(x) near f(p), then certainly the X n for n large will have values f(x n ) near f(p). Conversely, suppose that (1) fails. This means that for some E > 0, no small 8 bound on Ix-pi, for example 8 == lin, guarantees that If(x) - f(P)1 :s; E. Thus there must be some sequence x n , with IX n -pi::; lin and with If(x n ) - j(p)1 2:: E, so that (2) fails.
(1) =} (3). Let p E f-1U. Then f(p) E U and hence some small ball B(f(p),E) C U. By (1), choose 8 > such that .
°
Ix - pI < 8 =}
If(x) - f(p)1 <
c,
and hence
Ix - pi ::; 8/2 =}
If(x) - f(p)1 <
E.
-
33
6. Continuous Functions
34
In other words,
B(p,8/2)
C
j-l B(j(p), c)
C
f- 1 U.
So we have found a ball about p contained in j- 1 U, which means that j- 1 U is open. (3) :::} (1). Since j-l of the open ball about j(p) of radius contains p, it contains some ball B(p, 8). Therefore,
Ix -
E
is open and
o
pi < 8 :::} Ij(x) - j(p)1 < E.
6.2. Remark. Proposition 6.1 holds for functions with any domain D C JR.n, with the understanding that "j-l(U) open" means that every point of j-l(U) has a ball about it whose intersection with D is contained in j-l(U). You cannot expect to get whole balls in IR,n if D itself is not open. You just have to pretend that D is the whole universe and ignore other points of JRn. Sometimes such a condition is called "relatively open." For example, the function f(x) == x is a continuous function from [0,2] C JR. into JR. The inverse image j-l(l, 3) == (1,2] is not an open subset of JR., but it is (relatively) open as a subset of [0,2] even at 2 because points of [0, 2] near 2 are contained in (1, 2]. By these agreements, a function is continuous at an isolated point of the domain.
Exercises 6 1. Use definition (2) to prove that the sum of two continuous functions is continuous. 2. Use definition (2) to prove that the product of two continuous functions is continuous. 3. Use each definition to prove that every function f: Z
-7
JR is continuous.
4. Prove that a function j: JRn -+ JR is continuous at a point p if and only if for every open set U containing !(p), f- 1 (U) contains a ball B(p, r) about p.
Chapter 7
Composition of Functions
Compo~itionis an important way to build more complicated functions out of simpler ones. Thus you can understand complicated functions by analyzing simpler ones.
7.1. Definition. The composition fog of two functions is the function obtained by following one with the other:
(f
0
g)(x) = f(g(x)).
For this to make sense, the image of 9 must be contained in the domain of f. Examples. If f(x)
(f
0
= sin x
and g(x)
g)(x) = sin(x 2),
(g 0 f)(x) = (sinx)2, If f(x) =
= x 2, then
usually written simply sinx 2 ; usually written sin2 x.
-IX and g(x,y) = x 2 +y2, then (f
0
g)(x)
=
Jx 2 + y2.
7.2. Theorem. The composition fog of two continuous junctions is continuous. Proof. We will give three proofs, using the three equivalent definitions of continuous. See Figure 7.1.
-
35
7. Composition of Functions
36
fog
Figure 7.1. The composition fog of two functions.
c-8 definition. Given a point p in the domain of g, let q == g(p) be the resulting point in the domain of I, so that (f a g)(p) == f(q)· Given c > 0, since f is continuous, we can choose Cl > a such that
Iy - ql < Cl
(1)
=}
I/(y) - l(q)1 < c.
Since 9 is continuous, we can choose 8 > 0 such that
Ix - pI < 8 =} Ig(x) -
g(P)1 <
Cl·
Now applying (1) with y == g(x) and q == g(p) implies that
If(g(x)) - f(g(p))1 < c, M
desired.
Sequence definition. Suppose X n -) x. Since 9 is continuous, g(x n ) -+ g(x). Then since f is continuous, f(g(x n )) -) I(g(x)), as desired. Open set definition4 Just note that (I a g)-l(U) == g-l(f-l(U)) (working backwards, we have to start with f). Suppose U is open. Since 1 is continuous, f-1(U) is open. Then since 9 is continuous, g-1(/-1(U)) is open, as 0 desired 4
Exercises 7 1. What is
1 a g(x)
if
a. f(x) == -IX and g(x) == x 2 ; b. f(x) == sin x and g(x) == sin- 1 x. 2. Show by counterexample that this converse of Theorem 7.2 is false: If the composition of two functions is continuous, then both are continuous.
Chapter 8
Su bsequences
We like sequences to converge, but most don't. Fortunately, most sequences do have subsequences which converge. 8.1. Discussion. Limits are a kind of ideal, and in general we want sequences that converge. It's a way to find maxima in calculus and in life. Sequences don't always converge, but there's still hope. The sequence (1)
1
1111111 1 , -1 , 1 , -1 , 1 , -1 , 116' -116' ...
2
2 4
4 8
8
does not converge, but the odd terms converge to 1, and the even terms converge to -1. These are called convergent subsequences, and these are good things. Given a sequence, a subsequence is some of the terms in the same order. One subsequence of
(2) would be
(3) obtained by using just some of the subscripts. The right way to say this, although it may seem a bit confusing at first, is that given a sequence an, we can consider a subsequence amn , with ml < m2 < m3 < ... , where above ml = 2, m2 = 3, m3 = 5, m4 = 7, m5 = 11, .... 8.2. Definition of subsequence. Given a sequence an, a subsequence amn consists of some of the terms in the same order.
Now given a sequence, what is the chance that it has a convergent subsequence? Of course if it is convergent, every subsequence, including the
-
37
38
8. Subsequences
original sequence, converges to the same limit. On the other hand, a divergent sequence like
(1)
1, 2, 3, 4, ...
has no convergent subsequence because every subsequence diverges to 00. The following is a delightful and important theorem due to Bolzano and Weierstrass, around 1840. 8.3. Theorem. Every bounded sequence in JR. has a convergent subsequence. Proof. We'll first do the CMe of nonnegative sequences and then treat the general CMe. So we consider a nonnegative sequence
Each an starts off with a nonnegative integer before the decimal point, followed by infinitely many digits (possibly 0) after the decimal point. Since the sequence an is bounded, some integer part D before the decimal place occurs infinitely many times. Throwaway the rest of the an. Among the infinitely many remaining an that start with D, some first decimal place d 1 occurs infinitely many times. Throwaway the rest of the an. Among the infinitely many remaining an that start with D.d1 , some second decimal place d 2 occurs infinitely many times. Throwaway the rest of the an. Among the infinitely many an that start with D.d 1d2, some third decimal place d 3 occurs infinitely many times. Keep going to construct L = D .d1 d2d3 .... We claim that there is a s~bsequence converging to L. Let amI be one of the infinitely many an which starts with D.d 1 . Let am2 be a later an which starts with D.d1d2. Let am3 be a later an which starts with D.d1d2d3. The a mn converge to L because once n > N they agree with L for at least N decimal places. Having treated the case of nonnegative sequences, we now consider a sequence with some negative terms. Since the sequence is bounded, we can translate it to the right to make it nonnegative, use the above argument to get a convergent subsequence, and then translate it back. 0 8.4. Corollary. An increasing sequence in JR. which is bounded above converges. (Similarly, a decreasing sequence bounded below converges. The sequence need not be strictly increasing or strictly decreasing.) Proof. Exercise 3.
o
Exercises 8
39
Exercises 8 1. Give an example of a sequence of real numbers with two different subsequences converging to two different limits.
2. Give an example of a sequence of real numbers with subsequences converging to each integer. 3. Prove Corollary 8.4. 4. Give an example of a sequence of real numbers with subsequences converging to every real number. 5. Prove that a set S c JR is closed and bounded if and only if every sequence of points in S has a subsequence converging to a point of S. 6. Prove that every Cauchy sequence of reals converges. (Use Exercises 3.17 and 3.18.) 7. The lim sup of a sequence is defined as the largest limit of any subsequence, or ±oo. (It always exists.) What is the lim sup of the following sequences? a. an
= (-I)n
b. 1, -1,2, -2,3, -3, ... c. an == -n2
d. an
= sin(n7r/6)
1 11
21
1 11
21
1 11
21
e. 2' 2' 2' 4' 4' 4' 8' 8' 8'··· f. {an} = Q n (0,1) 8. Similarly the lim in! of a sequence is defined M the smallest limit of any subsequence, or ±oo. Of course the lim inf is less than or equal to the lim sup. Give two examples .of sequences for which the lim inf equals the lim sup.
Chapter 9
Compactness
Probably the most important new idea you'll encounter in real analysis is the concept of compactness. It's the compactness of [a, b] that makes a continuous function reach its maximum and that makes the Riemann integral exist. For subsets of JRn, there are three equivalent definitions of compactness. The first, 9.2(1), promises convergent subsequences. The second, 9.2(2) brings together two apparently unrelated adjectives, closed and bounded. The third, 9.2(3), is the elegant, modern definition in terms of open sets; it is very powerful, but it takes~ while to get used to.
9.1. Definitions. Let S be a set in JRn. S is bounded if it is contained in some ball B(O, R) about 0 (or equivalently in a ball about any point). A collection of open sets {Uu } is an open cover of S if S is contained in UUuo A finite subcover is finitely many of the Uu which still cover S. Following Heine and Borel, S is compact if every open cover has a finite subcover. 9.2. Theorem. Compactness. The following are all equivalent conditions on a set S in JRn . (1) Every sequence in S has a subsequence converging to a point of S. (2) S is closed and bounded. (3) S is compact: every open cover has a finite subcover.
Criterion (1) is the Bolzano-Weierstrass condition for compactness, which you met for JR in Theorem 8.3. The more modern Heine-Borel criterion (3) will take some time to get used to. A nonclosed set such as (0, 1] is not compact because the open cover {(lin, Do)} has no finite subcover. An
-
41
9~
42
Compactness
unbounded set such as 1R. is not compact because the open cover {( -n, n)} has no finite subcover. This is the main idea of the first part of the proof.
Proof. We will prove that (3)
=}
(2) => (1) => (3).
(3) => (2). Suppose that S is not closed. Let a be an accumulation point not in 8~ Then the open cover {{Ix - al > lin}} has no finite subcover. Suppose that 8 is not bounded. Then the open cover {{Ixi < n}} has no finite subcover. (2) => (1). Take any sequence of points in 8 c JRn. First look at just the first of the n components of each point. Since 8 is bounded, the sequence of first components is bounded. By Theorem 8.3, for some subsequence, the first components converge. Similarly, for some further subsequence, the second components also converge. Eventually, for some subsequence, all of the components converge. Since 8 is closed, the limit is in 8.
(1) ==> (3). Given an open cover {Un}, first we find a countable subcover. Indeed, every point x of 8 lies in a ball of rational radius about a rational' point, contained in some Un. Each of these countably many balls lies in some Un~ Let {Vi} be that countable subcover. Suppose that {Vi} has no finite subcover. Choose Xl in 8 but not in VI. Choose X2 in 8 but not in Vi U V2. Continue, choosing X n in 8 but not in U{lti: 1 :::; i :::; n}, which is always possible because there is no finite subcover. Note that for each i, only finitely many X n (for which n < i) lie in Vi. By (1), the sequence X n has a subsequence converging to some X in S, contained in some Vi. Hence infinitely X n are contained in Vi, a contradiction. D
9.3. Proposition. A nonempty compact set 8 of real numbers has a largest element (called the maximum) and a smallest element (called the minimum).
Proof. vVe may assume that 8 has some positive numbers, by translating it to the right if necessary. Since 8 is bounded, there is a largest integer part D before the decimal place. Among the elements of S that start with D, there is a largest first decimal place dl. Among the elements of S that start with D.d 11 there is a largest second decimal place d2. Keep going to construct a = D.dld2d3.... By construction, a is in the closure of 8. Since S is closed, a lies in 8 and provides the desired maximum.
A minimum is provided by - max( -8).
D
Exercises 9
43
Exercises 9 1. Prove that the intersection of two compact sets is compact, using criterion (2). 2. Prove that- the intersection of two compact sets is compact, using criterion (1). 3. Prove that the intersection of two compact sets is compact, using critenon (3). 4. Prove that the intersection of infinitely many compact sets is compact. 5. Prove that the union of two compact sets is compact, using criterion (2). 6. Prove that the union of two compact sets is compact, using criterion (1). 7. Prove that the union of two compact sets is compact, using criterion (3). 8. Is the union of infinitely many compact sets always compact? Give a proof or counterexample. 9. Identify the class of sets in 1R. characterized by the following conditions and give two examples of such sets.
a. Every open cover has a finite subcover. b. Every open cover has a countable subcover. c. Some open cover has a finite subcover. d. Every closed cover has a finite subcover. 10. Prove directly that (1) implies (2). 11. (Kristin Bohnhorst). Prove directly that (3) implies (1). Hint: Assume that some (infinite) sequence X n has no convergent subsequence. Then every point p of S has an open ball Up around it which contains X n for only finitely many n. 12. Is the converse of Proposition 9.3 true? Give a proof or counterexample.
9. Compactness
44
13. Prove that if a nonempty closed set S of real numbers is bounded above, then it has a largest element. 14. Define the supremum sup S of a nonempty bounded set of real numbers as max S. Prove that sup S 2: s for all s in S and that sup S is the smallest number with that property. For this reason, sup S is often called the least upper bound. Similarly, the infimum inf S == min S is called the greatest lower bound.
S
.------------e
infS
sup S
(If S is not bounded above, sup S is defined to be +00. If S is empty, sup S is defined to be -00.)
Chapter 10
Existence of Maximum
One of the main theorems of calculus is that a continuous function on a bounded, closed interval [a, b] attains a maximum (and a minimum). It is the reason that the calculus method of finding maxima and minima works. It appears as Corollary 10.2 below. As we will now see, this important result depends on the fact that the interval is compact. 10.1. Theorem. A continuous image of a compact set is compact.
We will give two different proofs, one using the Bolzano-Weierstrass sequential characterization of compactness, and one using the Heine-Borel open cover characterization of compactness. These are very general argum.ent~, which hold in JR., in JR.n, and even in metric spaces (Chapter 29). We will not give a proof using the "closed and bounded" characterization of compactness because I do not know one. The two natural steps are false, as shown by Exercises 2 and 3. Somehow when the two notions are combined as in the sequential or open cover characterizations, the proof comes fast. Proof using sequential characterization of compactness. Let f be a continuous function, let K be a compact set, and let an be a sequence in f(K). Choose bn in K such that f(b n ) == an· Since K is compact, a subsequence bmn , converges to. some limit b in K. Since f is continuous, the corresponding subsequence a mn == f(b mn ) converges to f(b) in f(K). D Proof using the open cover characterization of compactness. Let {Uo} be an open cover of f(K). Then {f-IUo } is an open cover of K. Since K is compact, it has a finite subcover {!-IUOn } . Then {UOn } is the desired finite subcover of f(K). D
-
45
10. Existence of Maximum
46
10.2. Corollary (Existence of extrema). A continuous function on a nonempt: compact set K, such as a closed bounded interval [a, b], attains a maximum and a minimum. Proof. By Theorem 10.1, I(K) is compact. By Proposition 9.3, I(K) has a maximum and a minimum. 0 Remark. Corollary 10.2 depends crucially on the compactness of an interval [a, b] of real numbers. Note that it fails for the rationals. For example, on the interval of rationals from 0 to 2, the continuous function (x 2 - 2)2 has no minimum; the continuous function 1j(x 2 - 2) is not even bounded.
Exercises 10 1. Give an example of a function (not continuous) on [0,1] which has no maximum and no minimum. 2. Give an example of a continuous function on a closed (but unbounded) set A c lR that has no maximum. 3. Give an example of a continuous function on a bounded (but not closed) set A C JR that has no maximum. 4. Give a counterexample to the following statement: If I is continuous and S is compact, then I-IS is compact. 5. Prove that if K is compact and I and 9 are continuous, then (I 0 g)(K) is compact. What do you have to assume about the domains and images of I and g1 6. Prove or give a counterexample: A nonempty set K in 1R. is compact if and only if every continuous function on K has a maximum. 7. (Candice Corvetti). Prove or give a counterexample: A function I : 1R. ----+ 1R. is continuous if and only if the iII;lage of every compact set is compact.
Chapter 11
Uniform Continuity
For integration, it turns out that continuity is not quite strong enough. One needs a kind of "uniform continuity," independent of x. 11.1. Definition. A function f is uniformly continuous if, given there is a 0 > 0 such that
Iy - xl < 0 ==>
If(y) - f(x)1 <
E
> 0,
E.
The only difference from mere continuity is that the 0 appears before the x, and consequently cannot depend on x. The same 0 must work for all x. 11.2. Theorem. A continuous function on a compact set K is uniformly continuous.
This theorem fails on general sets. For example, f(x) == x 2 is continuous on 1R. but not uniformly continuous. To show that it is not uniformly continuous, we have to come up with an E for which no 0 works. In this case, any E will do; let's take E == 1. Now no matter how small 0 is, there are large x's within oof each other for which the values of x 2 differ by more than 1. Exercise 4 asks for the details. Geometrically, the problem is that the slope goes to 00. Likewise, f(x) == l/x is continuous on (0,1) but not uniformly continuous (Exercise 5). Proof of Theorem 11.2. Given c > 0, since K, there is a Ox > such that
°
(1)
f is continuous, for each x in
Iy - xl < Ox ==> If(y) - f(x)1 < E/2. open ball Ux == {y: Iy - xl < ox/ 2}. The
Consider the collection {Ux} of these open balls is an open cover of K. Since K is compact, it has a finite
-
47
lla Uniform Continuity
48
subcover {Uxi}a Let 0 == min{OXi/ 2 } > Da Suppose Iy - xl < Oa Since x lies in K, x lies in some UXj and Ix - xjl < oXj/2 Since Iy - xl < 0 ~ oXj/2, Iy,-xjl < OXja Therefore, by (1), a
If(x) - f(xj)1 < c/2 and
If(y) - f(xj)1 < c/2.
We conclude that
If(y) - f(x)1
~
If(x) - f(xj)1 + If(y) - f(xj)1 < c/2 + c/2 == Ca
D
Exercises 11 la Give another example of a continuous function on a closed set which is not uniformly continuousa 2a Give another example of a continuous function on a bounded set which is not uniformly continuousa 3a Prove that a composition of uniformly continuous functions is uniformly continuous.
== x 2 is not uniformly continuous on lia Hint: Take C == 1, any 0 > 0, x == 1/0, y == 1/0 + 0/2, and show that even though Iy - xl < 0, If(y) - f(x)1 2: c. 4. Prove that the function f(x)
5a Prove that the function f(x)
== l/x is not uniformly continuous on (0,1).
6a Prove Theorem 11.2 using the sequential (Bolzano-Weierstrass) criterion for compactness. A proof by contradiction might be easiest. 7a Prove that a uniformly continuous image of a Cauchy sequence is Cauchy. Show by counterexample that uniformly is necessary. 8. (Zan Armstrong). Prove or give a counterexample to the following converse of Theorem 11.2. A set S c 1R. is compact if every continuous function on S is uniformly continuous.
Chapter 12
Connected Sets and the Intermediate Value Theorem
The third of the big Cs requisite for calculus, after "continuous" and "compact," is "connected." The famous Intermediate Value Theorem follows easily. 12.1. Definition. A set 8 is connected if it cannot be separated by two disjoint open sets UI and U2 into two nonempty pieces 8 n U1 and 8 n U2 (such that 8 = (8 nUl) U (8 n U2)).
For example, the subset S of the reals defined by 8 = {O} U [1,3] is disconnected, as may be seen by taking UI = (-1/2,1/2), U2 = (1/2,4). The rationals Q are disconnected, as may be seen for example by taking UI = (-00, V2) and U2 = (V2, (0). A singleton is obviously connected. Intervals of reals are connected, although that requires proof: 12.2. Proposition. An interval I of real numbers is connected. Proof. Suppose that I can be separated by two disjoint open sets into two nonempty pieces In UI and In U2. Let ai E In Ui. We may suppose that al < a2. Let bl be the largest element of [aI, a2] - U2 (which is compact and nonempty because it has al in it). Let b2 be the smallest element of [b l , a2] - UI (which is compact and nonempty because it has a2 in it). Then bl < b2. Choose bl < b3 < b2. By choice of b2, the point b3 E UI and hence b3 t/:- U2, which contradicts the choice of bl 0
-
49
12. Connected Sets and the Intermediate Value Theorem
50
The converse also is true:
12.3. Proposition. Every connected subset 8 of lR is an interval (or a single point or empty). Proof. If 8 is not an interval (or a single point or empty), then there are points a, b in 8 and a point c between a and b not in 8. Then UI = (-00, c) and U2 = (c, (0) show that 8 is not connected. 0 12.4. Theorem. The continuous image of a connected set is connected. Proof. If f(8) is disconnected by UI , U2, then 8 is disconnected by I-lUI, (which are open because f is continuous). 0
I- I U2
12.5. The Intermediate Value Theorem. Let I be a continuous realvalued function on a closed and bounded interval [a, b]. Then f attains all values between f(a) and f(b). Proof. By 12.2 and 12.4, f([a, b]) is connected. By Proposition 12.3, it contains all values between f(a) and f(b). 0 12.6. Definition. First note that a set is. disconnected if and only if it can be separated by two disjoint open sets UI and U2 into two nonempty pieces 8 n UI and 8 n U2. A set 8 is totally disconnected if it has at least two points and for all distinct points PI, P2 in 8, the set 8 can be separated by two disjoint open sets U I and U2 into two pieces 8nUl and 8nU2 containing PI and P2, respectively. For example, the integers and the rationals are totally disconnected (Exercise 10). Note that every totally disconnected set is disconnected.
12.7. Proposition. A set of reals is totally disconnected if and only if it contains at least two points but no interval. Proof. Exercise 11.
o
Exercises 12
51
Exercises 12 1. Prove from the definition that the integers are disconnected.
2. Give a counterexample to the following statement: if tinuous and S is connected, then I-IS is connected.
I : lR -+
JR. is con-
3. a. Show that Q x Q C JR.2 is disconnected. b. Is Q x Q
C
JR.2 totally disconnected?
4. Is [0,1) x [0,1) C JR2 connected? 5. Is the unit circle {x 2
+ y2
==:
I} in lR2 connected?
6. Prove that your height (in inches) once equaled your weight (in pounds). 7. Prove that if f is a continuous function from JR. to Q, then
I
is constant.
8. What can you say about the continuous image of an interval [a, b]? 9. Let I be a continuous function from JR. to JR. such that j(X)2 are the possibilities for f? Prove your answer correct.
= x 2.
What
10. Prove that the integers Z and the rationals Q are totally disconnected. 11. Prove Proposition 12.7. 12. Do you think that there arl? any uncountable, closed, totally disconnected subsets of JR.? (Think about this question before discovering the answer in the next chapter.)
Chapter 13
The Cantor Set and Fractals
13.1. The Cantor set. You are now ready to see the Cantor set C, the protqtypical fractal and one of the most interesting and amazing sets in analysis. Here's how you construct it. Start with the closed unit interval as in Figure 13.1. Remove the open middle third (1/3,2/3), leaving two closed intervals of length 1/3. Remove the open middle third of each, leaving four closed intervals of length 1/9. Continue. At the nth step you have a set Sn consisting of 2n closed intervals of length 1/3 n . Let C = Sn.
n
As a -subset of the unit interval, C is bounded. As an intersection of closed sets, C is closed. Hence C is co"mpact.
C contains countably infinitely many boundary points of intervals, such as 0, 1, 1/3, 2/3, 1/9, 2/9, and so on. However, C contains lots more points:
(
)
.-----{
H
H
)----.
--{r-(
)--<~
)-
)--<)---e
--
}-
p< Kr(
}
Figure 13.1. The Cantor set C comes from the unit interval by successively removing middle thirds forever.
-
53
54
13. The Cantor Set and Fractals
Figure 13.2. Sierpinski's Carpet has dimension log3 8 http://en.wikipedia.org/wiki/Sierpinski_carpet.)
~
1.9. (From
13.2. Proposition. The Cantor set is uncountable. Proof. Represent elements of the Cantor set as base 3 decimals such as .021201222. . . . At the first step in the construction of the Cantor set we removed decimals with a 1 in the first decimal place. At the second step we removed remaining decimals with a 1 in the second decimal place. Hence C consists of base 3 decimals consisting just of Os and 2s, such as .020020222 ....
The rest of the proof is like the proof of Proposition 2.4 that the set JR of reals is uncountable. Suppose that C is countable, listed. Construct an element of C not on the list by making the first digit different from the first one on the list, the second digit different from the second one on the list, and so on. Therefore C is not countable. D
Although uncountable, the Cantor set has length or measure O. To see this, just note that SI has measure 2/3, that S2 has measure (2/3)2, and Sn has measure (2/3)n, which approaches o. Since C is contained in each Sn, its measure must be o. (Measure is a general word that applies in all dimensions, meaning length or area or volume. A rigorous treatment requires substantial mathematical development, as in a course on "measure theory.") We also note that the Cantor set is totally disconnected because between any two elements there are deleted intervals.
13.4. Theorem
55
Figure 13.3. The Ivlenger sponge has dimension log3 20 ~ 2.7. © Paul Bourke. http://astronomy.swin.edu.au/-pbourke/index.html
13.3. Dimension and fractals (informal discussion). The plane is 2-dimensional, and a region in the plane is 2-dimensional. A curve in the plane, however, is just I-dimensional. Likewise, a line or the interval [0,1] is I-dimensionaL A single point is O-dimensionaL (Some people say that the empty set has dimension -1, whatever that means.) What about the Cantor set? It seems higher dimensional than a point, but lower dimensional than the interval [0,1]. In f~ct, it has dimension log3 2 ~ .63, according to an advanced mathematical concept of "Hausdorff dimension." It is the prototypical fractal, or fractional dimensional set. Like many fractals, it has self similarity: it is made up of two smaller copies of itself, each 1/3 as large as the whole. There is a nice formula for the dimension of such a self-similar set: 13.4. Theorem. A self-similar set made up of p smaller copies of itself, each 1/ q as large as the whole, has dimension logq p.
56
13. The Cantor Set and Fractals
Figure 13.4. A random fractal landscape by R. F. Voss from The Fractal Geometry of Nature by Benoit Mandelbrot. Used by permission.
That the Cantor set has dimension log3 2 ~ .63 is one example. The unit interval is made up of n smaller copies, each lin as large, for dimension logn n = 1. A unit square region is made up of n 2 smaller copies, each lin as large, for dimension logn n 2 == 2. The proof of Theorem 13.4 is beyond the scope of this text.
13.5. More fractals. Figure 13.2 shows a planar fractal called Sierpinski's carpet. It is made up of 8 smaller copies of itself, each 1/3 as large. Hence by Theorem 13.4 its dimension is log3 8 ~ 1.9. Figure 13.3 shows a spatial fractal called the Menger sponge. It is made up of 20 smaller copies of itself, each 1/3 the size. Hence its dimension is log3 20 ~ 2.7. Benoit Mandelbrot, the Father of Fractals, discovered that fractals provide better models of physical reality than the smooth surfaces of calculus. Figure 13.4 from his famous book on The Fractal Geometry of Nature shows a random fractal mountain which looks much more realistic than the upsidedown paraboloids of calculus books.
Exercises 13
57
Figure 13.5. A fractal constructed by repeatedly removing everything but four corners each 1/4 as large. (From Morgan's Geometric Measure Theory book.)
Exercises 13 1. Show that every point of the Cantor set is an accumulation point. (Such sets are called perfect. Every nonempty perfect real set is uncountable.) 2. Construct a version of the Cantor set by removing middle fifths instead of middle thirds. Is it still compact, uncountable, totally disconnected, and of measure O? 3. Construct a version of the Cantor set by starting with [0,1], removing a middle interval of length 1/4, then removing two middle intervals of total length 1/8, then removing four middle intervals of total length 1/16. Is it still compact, uncountable, and totally disconnected? What is its measure? 4. What is the dimension of the fractal of Figure 13.5? 5. Give an example of a totally disconnected set S C [0,1] whose closure is the whole interval. 6. Give examples, if possible, of continuous maps of the Cantor set onto lR, (0,1), and [0 1], or say why not possible. 1
Part III
Calculus
Part III: Calculus. It is satisfying to see how naturally the theory of calculus follows from the basic concepts of real analysis, especially when you remember that after Newton and Leibniz invented the calculus in the late 1600s, it took mathematicians two hundred years to get the theory straight. This part focuses on ]Rl rather than ]Rn, although Chapters 17 and 18 hold for real-valued functions on ]Rn as well.
Chapter
14
The Derivative and the Mean Value Theorem
Differentiation and integration are the two distinguishing processes of calculus. The first major theorem about the derivative, the Mean Value Theorem, follows easily from the compactness of the interval [a, b], via Proposition 14.2 and Rolle's Theorem 14.3. We begin by recalling the definition of the derivative of a function on an interval in JR. 14.1. Definition. Let / be a real-valued function on an open interval (a, b). Define the derivative /' = d// dx by
f'(x) = lim f(t) - f(x) = lim !:If. t-+x t- x 6x-+O ~x If this limit exists, we say that I is differentiable at x. One important interpretation of the derivative is the slope of the graph. More generally, the derivative gives a rate of change. We assume the easy and familiar properties of the derivative and now state and prove the important theorems. We mention that if / is differentiable at x then / is continuous at x. 14.2. Proposition. If I is differentiable at a local interior minimum (or maximum) point x, then I' (x) = o. Proof. For t near a local minimum point x, the numerator in the definition of the derivative is nonnegative. For t > x, the denominator is positive, and hence I'(x) is a limit of nonnegative numbers. For t < x, the denominator is negative, and hence f'(x) is a limit of nonpositive numbers. The only D possibility is I'(x) = o. The proof is similar for a local maximum.
-
61
14. Tbe Derivative and tbe Mean Value Tbeorem
62
a
b
c
Figure 14.1. The Mean Value Theorem says that there is a point c where the instantaneous slope equals the average slope from a to b.
Remark. Proposition 14.2 is the basis for finding maxima and minima in calculus. You just have to check where the derivative is 0 (or does not exist) for interior extrema, and also the endpoints (or extreme cases such as x ---+ ±oo). 14.3. Rolle's Theorem. Suppose that f is continuous on [a, b] and differentiable on (a, b), and that f(a) = I(b). Then fOT some c E (a, b), I'(c) = o.
Proof. As a continuous function on a compact set, f has a maximum and a minimum. If the maximum occurs on the interior, by Proposition 14.2 f' vanishes (is zero) there. Likewise if the minimum occurs on the interior, f' vanishes there. Otherwise, the maximum and minimum both occur at the endpoints, f is constant, and I' vanishes everywhere. D 14.4. The Mean Value Theorem. Suppose that I is continuous on [a, b] and differentiable on (a, b). Then for some c E ( a, b),
f'(c) = f(bi
=~(a) .
Proof. See Figure 14.1. By horizontal scaling and translation, we may assume that [a, b] = [0,1]. Unless f(O) = f(l) (in which case the result follows immediately from Rolle's Theorem), by vertical scaling and translation we may assume that f(O) = 0 and f(l) = 1. Let g(x) = f(x) - x. Then g(O) = g(l) = 0, so by Rolle's Theorem, for some c E (a, b), 0 = g'(C) = f'(C) - 1. Therefore, f'(c) = 1 = f(b) - f(a) . D b-a
14.. 6. Tbe Cantor function
63
1 .-
0-
_0
1 Figure 14.2. The Cantor function f has derivative 0 except on the Cantor set of measure 0, but it is not constant.
The following theorem is the only reason you need the Mean Value Theorem to do calculus. 14.5. Corollary of the Mean Value Theorem. On an interval where is always 0, I is constant.
I'
Proof. Suppose a < b are any two points in the interval. By the Mean Value Theorem, f(b) == I(a). D
14.6. The Cantor function. Corollary 14.5 may sound obvious, but it is almost false. There is a nonconstant continuous function on [0,1] with derivative 0 except at a set of points of measure O. The set is the Cantor set and the function is called the Cantor function. It is defined as follows (see Figure 14.. 2) . Define f to be 0 at 0 and 1 at 1. On the middle third, defin~ I to be 1/2. On the middle thirds of the remaining two intervals, define I to be 1/4 and 3/4. On the middle thirds of the remaining four intervals, define I to be 1/8, 3/8, 5/8, and 7/8. Continue. I extends to a continuous function on [0,1]. On the (open) complement of the Cantor set, I is constant on intervals, and hence has derivative O.
64
14. Tbe Derivative and the Mean Value Tbeorem
Exercises 14 1. Check the Mean Value Theorem for the function f"(x) = x 3 on [0,1]. 2. Suppose that f: IR --7 IR s~tisfies f(O) = 0 and If'(x)1 :::; M. Prove that If(x)1 :::; Mlxl· Apply this to the function f(x) = sinx. 3. Discuss the logical chain of reasoning from compactness of [a, b] to Corollary 14.5. 4. Show that the Cantor function is a continuous map of the Cantor set onto [0, 1], solving part of Exercise 13.6. 5. Define a map f from C into [0,1] as follows. Given a point in the Cantor set, represent it by a base three decimal without Is in it, such as .0222022002 ... , change the 2s to Is to get something like .0111011001, and then interpret it as a base two decimal in [0,1]. Is f continuous? surjective? Is f related to the Cantor function?
Chapter 15
The Riemann Integral
This chapter defines the standard, Riemann integral of a function on a bounded interval [a, b] in lR and shows that the process works for every continuous function f on [a, b], using the fact that a continuous function on a compact set is uniformly continuous.
J: f(x) dx of a the area under the graph. The
15.1..The Riemann integral. The Riemann integral
function f over an interval [a, b] represent~ area may be approximated as in Figure 15.1 by chopping the interval up into narrow subintervals of perhaps variable thickness .6.x, approximating each subarea by a skinny rectangle of height f (x), thickness .6. x, and area .6.A == f(x)~x, and adding them up: A ~ Lf(x)Llx.
This approximating sum is called the Riemann sum. To get the exact area, we take the limit as the maximum thickness goes to 0, and call this the Riemann integral:
Jar f(x) dx b
=
lim L
f(x)Llx.
6X---7Q
The limit must be independen~ of the choice of subintervals and of the choice of where we evaluate f(x) in :each subinterval. If the limit exists, we say that f is integrable on [a, b] . If f (x) is a constant c, then
l
f is integrable and
b
f(x) dx
=
c(b - a).
-
65
15.
66
~be
lliemann Integral
y 8
4
1
2
3 x
Figure 15.1. The area under the graph of f may be approximated by the sum of areas of skinny rectangles of height f (x) and thickness 6.x.
Indeed, every Riemann sum
L j(x)f:1x = c L f:1x = c(b -
a).
This corresponds to the fact that the area of a rectangle of height c and width b - a is c(b - a).
15.2. Nonintegrable functions. One function which is not integrable on [0,1] is the characteristic function XQ of the rationals, which equals 1 on the rationals and a off the rationals. Indeed, no matter how small .6.x·, there .are Riemann sums equal to one, obtained by always choosing to evaluate XQ(x) at a rational, and there are Riemann sums equal to zero, obtained by always choosing to evaluate XQ(x) at an irrational. Another function which is not integrable on [0,1] is the function f(x) == l/ft because no matter how small .6.x, there are Riemann sums arbitrarily large, obtained by choosing to evaluate f(x) at a point very close to a on the first subinterval. In general, an integrable junction must be bounded (Exercise 3). Fortunately, every continuous function is integrable: 15.3. Theorem. Every continuous fUnction is integrable on [a, b]. Proof. By Exercise 8.6, it suffices to show that any sequence of Riemann sums with .6.x ----7 0 is Cauchy. By Theorem 11.2, the continuous function j on the compact set [a, b] is uniformly continuous: given c > 0, there exists 8 > a such that
l.6.xl < 8 :::} l.6.fl < E. Now.consider two Riemann sums both with l.6.xl < 8/2. Their subintervals intersect ,in smaller subintervals of thickness l.6.xl < 8/2. On each such (1)
subinterval, the values f(x) from the two Riemann sums come from points at distance at most 6 /2 from a point in the intersection and hence at distance
15.5. Proposition
67
at most 8 from each other. By (1), these values differ by at most E. Summing over the smaller subintervals, we see that the absolute value of the difference of the Riemann sums equals
IL~f ~xl
: :; L I~fl ~x:::; L
E ~x = E L
~x =
E(b - a).
Therefore the sequence is Cauchy as desired and we conclude that the function is integrable. D Remark. The whole truth is that a function on [a, b] is integrable if and only if it is bounded and its set of discontinuities has measure O.
15.4. Negative functions. Although for the motivating interpretation as area it is easier to think of I as positive, everything applies as well to functions allowed negative values. Of course if the function is negative, the integral will be negative. Some students like to remember this as, "Area below the x-axis counts negative." Normally we expect that a < b, but it is convenient to define the Riemann integral for a > b by
l Finally, instead of
b
f(x) dx =
I: f(x) dx,
a
f(x) dx.
we sometimes just write
15.5. Proposition. Suppose that intervals and that C is a constant.
J: C I == C J: I· b. J: I + == J: I + J:
-i
I
I: f·
and 9 are integrable on the relevant
a.
9
c.
g.
J: I == f: f + fb f· c
d.
II: fl :::; I: If I·
e.
II I 5: 9
and a < b, then
J: I :::; J:
g.
Proof. Most of these facts f~llow immediately from the corresponding facts about Riemann sums. For it, the Riemann sums for Clare just C times the Riemann sums for I. For b, the Riemann sums for I + 9 are sums of Riemann sums for f and for g. For c, Riemann sums for the second and third terms yield Riemann sums for the first. For d, the inequality holds for Riemann sums because the absolute value of a sum is less than or equal to the sums of the absolute values. For e, the inequality holds for Riemann sums with the same subintervals and the same choice of places to evaluate I(x) and g(x). D
68
15.. Tbe Riemann Integral
f
15.6. Corollary. Suppose that
(b - a) min f(x) ::; a-.5.x-.5.b
is continuous on [a, b]. Then
l
b
a
f(x) dx
~ (b -
a) max f(x). a~x-.5.b
Proof. By Proposition 15.5e, taking 9 to be the constant c,
l
b
f(x) dx
~
l
ma:xa<x
f(x) ==
b
cdx = c(b - a),
proving the second inequality. The proof of the first is similar.
o
Exercises -15
69
Exercises 15 1. In approximating J0 x dx, suppose you divide [0,4] into four unit subintervals. What are the possible values of Riemann sums? 4
2. Compute directly from the definition that
l
b
cdx = c(b - a).
3. Compute directly from the definition that
1 1
2
x dx = 1/3
by the following steps. Since by Theorem 15.3 the integral exists independent of choice of subintervals or points to evaluate f(x), we may choose to divide [0,1] into n subintervals of length D"x =: lin and evaluate at the right-hand endpoints. a. Show that the Riemann sums equal
Lf(x)~x =
t(k/n)2(1/n) = k~l
b. Use the formula L:~=l k 2 = 1
n(n+1/i)(n+l)
~ t k 2. n k~l
and take the limit as n
--7
co
2
to conclude that fo x dx == 1/3. 4. Prove that a nonnegative Riemann integrable function is bounded. 5. Prove that every Riemann integrable function is bounded. 6. Explain why the definition of the Riemann integral does not apply to unbounded intervals such as [0, co). (Sometimes such an "improper" integral is defined as limR-·.... oo foR f(x) dx.) 7. Is the function f: [0, 1]
~ ~
f(x) == integrable?
defined by
{o
for 0 :; x ::; 1/2, 1 for 1/2 < x ::; 1
Chapter 16
The Fundamental Theorem of Calculus
The fundamental Theorem of Calculus is most popular for its second part, which says that you can integrate just by antidifferentiating, instead of doing painful limits of Riemann sums. Both parts essentially say that integration and differentiation are opposites. It is quite remarkable that there is any relationship between integration and differentiation, between area and slope, between limits of Riemann sums and linllts of ratios of change. For the first part, to be able to differentiate after integrating, we treat the upper limit of integration b as a variable, and think of how the Riemann integral changes as b changes.
16.1. The Fundamental Theorem of Calculus. Let function on [a, b].
I.
tb J: f(x) dx == f(b).
II. If f(x) == F'(x), then
J: f(x) dx
=
F(x)l~
:=
f be a continuous
F(b) - F(a).
Proof of I. By the definition of the derivative,
d db
l a
b
f(x) dx
.
=
lim
lb+6b a
6b--toO b db fb +
f(x) dx - l b f(x) dx .6..b a
f(x) dx
~b
-
71
16. The Fundamental Theorem of Calculus
72
If ~b > 0, by Corollary 15.6, .
fbb+D..b
mm f(x) < -Ix-bl:::;l.6.bl -
f(x) dx
~b
- <
max
- Ix-bl:::;ILlbl
f(x).
If ~b < 0, the sign of both the numerator and the denominator change, the fraction remains unchanged, and the inequalities still hold. As ~b ---+ 0, the left-hand side and the right-hand approach j(b); hence so does the fraction. Therefore b fb+Llb f( ) d f(x) dx ~ lim Jb x x = f(b)
!!-1 db
as
db~O
a
~b
o
desired.
Proof of II. Note that by I,
:b (F(b)
-l
b
f(x) dX) = F'(b) - f(b) = f(b) - f(b) =
o.
By Corollary 14.5 to the Mean Value Theorem,
F(b)
-l
b
f(x) dx = C.
Plugging in b == a yields
F(a) == C. Therefore
l
b
f(x) dx = F(b) - F(a).
o
J:
16.2. Remark. Why don't we just define f(x) dx as F(b)-F(a)? First of all, we may not know of any antiderivative F. Second of all, how could we expect that to have anything to do with area or other applications? The only sound approach is to define area somehow (the Riemann integral) and then figure out an easy way to compute it.
16.3. Remark. There are amazing generalizations of the Fundamental Theorem to functions on domains in ~n and more general domains, which go by names such as Green's Theorem, Gauss's Theorem or the Divergence Theorem, and Stokes's Theorem.
Exercises 16
73
Exercises 16 1. Use the fundamental Theorem to compute that easier than direct computation from the definition? 2. Compute
tb f: x 2dx two ways, first using 16.1(11), then using 16.1(1).
3. Let F(x) == foX e- t2 dt. Compute F'(x) and F'(O). 4. What is
f01 x 2dx == 1/3. Is this
ia f: f(x) dx?
Chapter 17
Sequences of Functions
"What does it mean for a sequence of functions to converge to some limit function? There is an easy definition, just looking at one point at a time.
17.1. Definition (pointwise convergence). A sequence of functions fn converges to f pointwise on some domain E if for every x E E, the sequence of numbers fn(x) converges to f(x); i.e., for every x in E, given c > 0, there is an N, such that n
> N => Ifn(x) - f(x)1 < c.
For example, if fn(x) == x/n, then fn ~ i, where f(x) == 0, on R. As a second example, if fn(x) == x n , as in Figure 17.1, then on [0,1] in ~ f, where
f (x)
== {
o 1
for 0 ::; x < 1 for x == 1.
It is somewhat disturbing that this limit of continuous functions is not itself continuous. The problem is that although fn ---+ f pointwise, the closer the point x gets to 1, the longer it takes for fn(x) to converge to f(x). Indeed, how big you have to take n to make fn(x) close to f(x) goes to infinity as x approaches 1. We need a more uniform kind of convergence.
17.2. Definition (uniform convergence). A sequence of functions fn ~ f uniformly on some domain E if given c > 0, there is an N, such that n > N => Ifn(x) - f(x)1 < c for all x in E.
-
75
17. Sequences of Functions
76
0.8
0.6
0.4
0.2
o
0.2
0.6
0.4
0.8
x Figure 17.1. The continuous functions fn(x) to a discontinuous function.
== x n converge pointwise
The only difference in the two definitions is that in the second, N appears before x, and hence the same N must work for all x, whereas in the first definition, N is allowed to depend on x. It is amazing that such a seemingly small change can be so important. Uniform convergence is just what you need to get continuous limits: 17.3. Theorem. A uniform limit of continuous junctions is continuous. Proof. The idea of the proof is that by uniform convergence, we can handle all points near any particular point p by looking at one f n that is uniformly near f.
Given
E
> 0, choose N such that for all x in the domain D, n > N => Ifn(x) - f(x)1 < c.
Since fN+l is continuous, given a point p in D, we can choose 8 > 0 such that
17.5. Theorem
Then if Ix -
77
pi < 8,
If(x) - f(p)1
~
+ IfN+l(X) -
If(x) - fN+l(X)1
fN+l(P) I + IfN+l(p) - f(p)1
< G + E + E == 3E. Therefore
f
o
is continuous.
(It is important that by uniform convergence N does not depend on x because 8 depends on Nand 8 must not depend on x.)
17.4. Question. Suppose a sequence of continuous functions f n converges pointwise to a function f on [0, 1]. Is the limit of the integrals equal to the integral of the limit:
1In 1 1
lim
1
=
lim I ?
In other words, can we switch the limit and the integral? Example. Consider our problematic fn(x) == x n , which converges pointwise but not uniformly to 0 on (0,1). Then lim
{I (I [ xn+1 ] 1 1 Jo in = lim Jo x n dx = lim ~ +-i 0 = lim n + 1 =
0,
while
I~ooks
OK.
°
Example. On [0,1], let fn(x) be 0, except let fn(x) be n for < x < lin, as in Figure 17.2, so that the integral is always 1, while fn(x) converges pointwise to o. Then
1In 1
lim while
1 1
lim In =
= lim 1 = 1,
1 1
limO =
o.
So it does not always work. Fortunately, it does always work if the convergence is uniform:
17.5. Theorem. For a uniform limit of continuous functions on a bounded interval [a, b],
17. Sequences of Functions
78
3~
2
f3
f2
1-
fl
--x
1
Figure 17.2. FUnctions fn(x) with integral! and limit O.
Proof. Given c > 0, choose N such that if n > N,
lin - 11 < Ce
Then
Example. 1
1
lim n-+oo
because x 2
Jor (x
2
+ ex2 In) dx =
Jro x
2
dx
=
1/3
+ ex2 In ---+ x 2 uniformly. Indeed,
l(x 2 + ex2 In)
-
x2 1== lex2 Inl ~
eln,
which gets small at a rate independent of x.
17.6. Remark. Even for uniform convergence, the derivative of the limit need not equal the limit of the derivatives (Exercise 9), but something is still true, as we'll see for example in Theorem 21.7.
Exercises 17
79
Exercises 17 1. Give an example of a sequence fn of functions on Z which converge pointwise but not uniformly.
2. Give an example of a pointwise convergent sequence fn of functions on [0,1] for which
3. Consider the statement: If fn ----+ I uniformly, then f; ----+ f2 uniformly. First give a counterexample. Second add a simple hypothesis and prove the revised statement. 4. Prove or give a counterexample: If nonvanishing (never 0) functions fn ----+ f uniformly, then 1/fn --7 1/ f uniformly. 5. Consider continuous functions from [0, 1] to converging pointwise to counterexample. 6. Find limn -700 f~ 7. Find lim n -7oo
a:::; fl :::; f2 f. Must f
~
:::; f3 :::; ... be continuous?
Give a proof or a
4
e:
dx. Justify.
f12x 2 -(sinnx)/n dx. Justify.
8. A function is called Lipschitz with Lipschitz constant G if
If(x) - f(y)1
~
Glx -
yl
for all x, y in the domain. Let fn be a pointwise convergent sequence of Lipschitz functions on [0,1] with Lipschitz constant G. Prove that the sequence converges uniformly. 9. Show by counterexample that even if differentiable functions In converge uniformly to a differentiable function f on ~, it does not follow that the derivatives f~ converge uniformly to f'. In particular, describe such a sequence of functions fn converging to 0 such that f~(O) = 1 for every n.
Chapter 18
The Lebesgue Theory
The Riemann integral, however natural, has certain technical flaws and complications alleviated by the more general Lebesgue theory. Although the Lebesgue theory is properly the subject of a graduate course, we present some of the convenient results here without proof so that you can start taking advantage of them right away.
The Lebesgue integral is a generalization of the Riemann integral. For any Riemann integrable function, it gives the same answer. However, there are more integrable functions. The Lebesgue integral can ignore a countable set (or any set of length or measure 0). For example, in the 1 Lebesgue theory, f0 XQ == 0, as it should, since the rationals are a negligible, 1 countable set. Similarly, for the Cantor set C, f0 Xc == O. We also allow unbounded functions and unbounded intervals, so that for example
1 1
x- 1/ 2
= 2x 1 / 2 16 = 2,
and
1
00
x -2 -- -x -11 00 1
--
1.
1
For switching the limit and the integral, there is a stronger theorem with hypotheses that are easier to check: 18.1. Lebesgue's Dominated Convergence Theorem. Let fn be functions on a domain in ~ converging pointwise to a limit f. If there is a function g with finite integral such that each Ifni ~ g, then
J
limfn
= lim
J
fn.
-
81
18. The Lebesgue Theory
82
18.2. Corollary (Uniformly bounded functions on a bounded interval). Let fn be junctions on [a, b] converging pointwise to a limit f. Suppose that for some M > 0, every Ifni::; M. Then
J
limfn = lim
J
fn·
Proof. Apply Lebesgue's Dominated Convergence Theorem, with g(x) == M. Since the domain is bounded, M has finite integral. 0 For switching the order of integration in a double integral, the following theorem is very usefuL
18.3. Fubini's Theorem. For a double integral, it is OK to switch the order of integration if either order yields a finite answer when the integrand is replaced by its absolute value.
18.4. Caveat on measura~ility. Technically, Theorems 18.1-18.3 have an additional hypothesis, that the integrands be "measurable." Measurable functions include continuous functions, piecewise continuous functions, perhaps altered on co-qntable sets or sets of measure 0, and much more. They include all functions that you have ever heard of ~nd all functions that can arise in the real world. Indeed, there is a theory of mathematics (without the Axiom of Choice) in which all functions are measurable. So this is not something to worry about. We give one more very useful theorem for switching integration with respect to one variable :Ii with differentiation with respect to another variable t.
I:g?
18.5. Leibniz's Rule. Suppose that f(x, t) dx exists, that a(t), b(t), and f(x, t) are all continuously differentiable with respect to t, and that there is a function g(x) ~ I~{ (x, t) I with I:cW g(x, t) dx bounded. Then
dd
t
l
b
(t)
f(x, t) dx =
aCt)
We allow a ==
I
b
(t)
a8tf (x, t) dx + f(b(t), t) b'(t) -
x=a(t) -00
or b ==
+00.
f(a(t), t) a'(t).
Exercises 18
83
Exercises 18 1. Compute lim n -
HXJ
I12 x 2 -(sinnx)/n dx. 1+( l)n e -nX x2
2. Compute lim n --+ oo
II
3. Compute limn --+ oo
I01 nIx dx.
4.
00
-
Justify.
2
dx. Justify.
Justify.
ComputeI:~o I:~~ xy cos xy2 dx dy. Justify.
Answer: 3/400.
5. Compute
fl Jy=O
IIx~y ~22 e- x 2/ y dx dy.
Justify.
Answer: 1 - lie. d 6. Compute Cit
f2x==1
sinx2 t
-3;-
dx. J ustl.fy.
A-nswer: ~(sin4t - sint) for t 7. Compute
f:-
0, ~. for t ==
o.
it I;~l ~e-x2t dx at t == 1. Justify.
Answer: _e~01 ~ -.17478. 8. Compute
ddy
I
CXJ x~
2+ yx 2dx +y 2 at y
== O.
Justify.
Answer: -1/4. 9. Give a counterexample to the following converse to Lebesgue's Dominated Convergence Theorem: If In --7 f and In --7 I (all finite), then there exists 9 ~ Ifni with J 9 < 00.
I
I
Chapter 19
L: an
Infinite Series
Infinite sums or series turn out to be very useful and important because in the limit you can actually attain an apparently unattainable value or function by adding on tiny corrections forever. In theory, they are no harder than sequences, because to see if a series converges you just look at the sequence of subtotals or partial sum.s~ For example, we say that the infinite series 1
1
1
1
"2 + 4: + 8 + 16 + ... converges to 1 because the sequence of partial sums
1 3 7 15 16' ...
2' 4' 8' converges to 1.
19.1. Definition. The series L~==l an converges to L if given is an No such that
E
> 0 there
N
N > No
=}
Lan - L
n==l Otherwise the series diverges (perhaps to "by oscillation").
< c.
+00, perhaps to -00, or otherwise
Of course if a series converges, the terms must approach 0 (Exercise 12)~ The converse fails. Just because the terms approach 0, the series need not converge. For example, see the harmonic series below.
19.2. Proposition. Suppose that L~=l an converges to A and L~=l bn converges to B. Then L~=l can converges to cA and L~=l an +bn converges to A+B.
-
85
86
19. Infinite Series
L
an
Proof. This proposition follows immediately from similar facts about se0 quences, as you'll show in Exercise 12. We now summarize some of the famous convergence tests from
calculus~
19.3. Geometric series. A geometric series (obtained by repeatedly multiplying the initial term ao by a constant ratio r) (1)
converges to l~r if Irl < 1 and diverges if Irl ~ 1 (and ao -j=. o)~ For example, the series at the beginning of Chapter 19, a geometric series with initial term ao = 1/2 and ratio r = 1/2, converges to 1~~2 = 1.
Proof. If Irl ~ 1 (and ao -j=. 0), the terms do not approach 0, so the series diverges. Suppose Irl < 1. Let Sn denote the subtotal Sn == ao rSn ==
+ aorn ; + aorn + aorn +1 ;
+ aor + aor 2 + aor3 + aor + aor 2 + aor 3 +
- aor n +1 ;
(1 - r)Sn == ao Sn ==
As n
-7
00,
Sn
-7
ao(l - r n +1 ) . 1-r
o
l~r.
19.4. p-test. The p-series
L
~ converges if p
> 1 and diverges if p ::; 1.
For example, taking p == 1, the" harmonic series 1
1
1
1
1
L;;:= i+2+3+4+··· diverges, despite the fact that the terms approach O. As a second example, taking p == 2,
L
1
1
n 2 = 12
1
1
1
+ 22 + 32 + 42 + ...
converges, although we do not yet know what it converges to. We omit the proof of the p-test, which uses comparison with integrals. 19.5. Comparison test. If lanl converges, and its limit a ::; (3.
For example, L verges by the p-test.
1;n
2
::; bn
and
L bn
converges to {3, then
Lan
converges by comparison with L~, which con"
19.6. Alternating series
87
Proof. It suffices to show that the sequence of subtotals, which is obviously bounded by {3, is Cauchy. For No :::; M :s; N, N
M
L
an -
n==l
L
N
an
M
L
=
n==l
L
an::;
n==M+l
N
L
Ian I::;
n==M+l
N
bn
=
n=M+l
which is small because the sequence of subtotals of
2: bn
L
M
bn
-
n==l
is Cauchy.
L
bn
n=l
0
Remark. The first million terms do not affect whether or not a series converges, although of course they do affect what it converges to. For example,
2:~=2 (IO~~~ n ) n n large, n=
1010
I lOOn agIO
is
<
converges by comparison with -21 .
ClOg) 10
2:~=2 (!) n
=
k because for
The limit, however, is huge; just the single term when
10
= 1010000000000.
19.6. Alternating series. If the terms of a series and lanl decrease with limit 0,
lall
~
la21
~
la31·· .
-7
2: an
alternate in sign
0,
then the series converges. For example, the alternating harmonic series (-1) n+ 1
1·
1
1
1
L - ;;:-=1:-2+3-4+··· converges (to ln2).
Proof. We may assume that the first term is positive. The subtotals after an odd number of terms are decreasing and positive, and hence converge to a limit £1. The subtotals after an even number of terms are increasing and bounded above by the first term, and hence converge to a limit £2. Since their differences approach 0, £1 = £2 is the limit of the sequence of subtotals. 0
,
19. Infinite Series
88
L
an
Exercises 19 For exercises 1-11, prove whether or not the series converges. If you can, give the limit. 1.
lo + 1~0 + 10100 + ....
3. 2 + 4 + 8 + 16 + .... 4. L~=l n 1/n. 5.
L oo 1 n=l nvn·
6.
L oo 1 n=O n 3 +2'
7.
L oo 5inn n=l nr·
8.
L~=2 (lnl~n. Why did we start the series with ",,00
n
== 2 instead of n == I?
(-l)n
9. L...,n=2 n1/n .
10 11
",,00 sinn . L...,n=3 (n 2 +n-l)(ln n)'
",,00
. L...,n=l
6n 2 +89n+73 n 4 -213n .
12. Prove that if a series converges, the terms approach 13. Prove Proposition 19.2.
o.
Chapter 20
Absolute Convergence
20.1. Definitions. A series
L lanl
solute values diverges).
L an
converges.
converges absolutely if the series of abOtherwise it converges conditionally (or
For example, the alternating harmonic series converges conditionally. It follows from the Comparison Test 19.4 that absolute convergence implies convergence. A rearrangement of a series has exactly the same terms, but not necessarily in the same order. You might expect that order should not matter, and it does not if the series converges absolutely (Prop. 20.2), but it does matter if the series converges merely conditionally (Prop. 20.3). Here is an example: 111
1
1
111
111
1
1
1
1
1
1
1
1
1
2" - 2" + 2 - 2 + 3 - 3+"3 - 3+3 - 3+ 4 -4 + 4 - 4 + 4 - 4+ 4 - 4+··' with partial sums 111
2'
0,
2'
0,
3'
1
0,
3'
1
0,
3'
1
0,
4'
1
0,
1
4'
0,
1
1
4'
1
0,
4'
0,
converges to 0, but the rearrangement 1
1
1
1
1
1
1
1
1
1
1
1
1
1
2+2 - 2 - 2+3 +3 +3 - 3 -"3 - 3 + 4+4 + 4+ 4 + 4 - 4 - 4 - 4+·" with partial sums 1110~~121011313110 1 2' , 3' 3' '3' 3' , 4' 2' 4' '4' 2' 4' ,
2'
diverges by oscillation. 20.2. Proposition. If a series converges absolutely, then all rearrangements converge to the same limit.
-
89
20a Absolute Convergence
90
Proof. Let L an denote the original series, which converges absolutely to some limit L, and let L bn denote the rearrangementa Given E > 0, choose N 1 such that if M, N > N 1 , then N
N
(1)
L lanl < c/2
Lan - L
and·
n=M
< c/2.
n=l
Choose N 2 such that the first N 2 terms of L bn include the first N 1 terms of L an· Let N > N 2 . Choose N 3 such that the first N a terms of Lan include the first N terms of L: bn . Then N
2.: bn- L n=l
:s;
N
~
~
n=l
n=l
n=l
L bn- L an + L an - L
.
Since N 3 > N 1 , by (1) the last expression is at most E /2. Consider the middle expressione All of the terms of the first sum are included in the second sum, and the first N 1 terms of the second sum are included in the first sum. Hence by (1) the middle exPression is at most c/2a We conclude that N
Lbn - L
::;
E,
n==l
which means that the rearrangement converges to the same limit La
D
20.3. Proposition. Suppose that a series L an converges conditionally. Then its terms may be rearranged to converge to any given limit, or to diverge to ±oo, or to diverge by oscillationa Proof sketch. First we claim that the amount of positive stuff, the sum of the positive terms, must diverge to +00, and that the amount of negative stuff must diverge to -DOa IT both converged, the series would converge absolutely. If just one converged, the series would divergea Hence both must diverge.
Second, note that because the series converges, the terms approach
o.
Our rearrangements will keep the positive terms in the same order and the negative terms in the same order. To get big (positive) limits, we'll front load the positive termsa To get small (negative) limits, we'll front load the negative terms. To get a rearrangement with prescribed limit L, take positive terms until you first pass L heading right a Then take negative terms until you first pass L heading left. Then take positive terms until you pass L heading right a Then take negative terms until you pass L heading left. Since there is an infinite amount of positive stuff and of negative stuff, you can continue
20.5. Root test
91
forever. Since the terms go to 0, the amount by which you overshoot goes to O. Hence the rearrangement converges to L. To get a rearrangement which diverges to +00, take positive terms until you pass 1, then a negative term, then positive terms until you pass 2, then a negative term, then positive terms until you pass 3, and so on. To get a rearrangement which diverges to -00, take negative terms until you pass -1, then a positive term, then negative terms until you pass -2, then a positive term, then negative terms until you pass -3, and so on. To get a rearrangement which diverges by oscillation, take positive terms until you pass 1, then negative terms until you pass -1, then positive terms until you pass 2, then negative terms until you pass -2, and so on. D 20.4. Ratio test. Given a series
L an,
let p = lim
Ia~:l I (Greek letter
rho). If p < 1, then the series converges absolutely. If p > 1, then the series diverges. If p = 1 or the limit does not exist, the test fails; the series could converge absolutely, converge conditionally, or diverge.
For example, consider the series
To get to the nth term from the previous term, you multiply by lin, so the limiting ratio p = O. Consequently the series converges absolutely (to e). Proof of Ratio test. Exercise 10.
D
Although I don't usually teach the following test in calculus, it turns out to be of theoretical importance in real analysis. 20.5. Root test. Given a series L an, let p == lim sup \llanl. If p < 1, then the series converges absolutely. If p > 1, then the series diverges. If p == 1, the test fails; the series could converge absolutely, converge conditionally, or diverge. Proof. If p < 1, to give us some room to work, let p < (}" < 1 (a is the Greek letter sigma). Then for all large n, \II an I ::; a, i.e., lanl ::; an, so that the series converges absolutely by comparison with the geometric series. If p > 1, for all large n, lanl > 1, and the series diverges because the terms do not approach o. D
20. Absolute Convergence
92
Exercises 20 1. What are the possible values of rearrangements of the following series:
a.
,",,00
1
~oo
(-l)n.
~oo
(-l)n. ..jii ,
LJn=l 2n;
b. LJn=l~' c.
LJn=l
d. L:~=l
)n.
Prove whether the following series 2-9 converge absolutely, converge conditionally, or diverge. Give the limit if you can. 2. L:~=o 5~·
~oo
sn
4.
LJn=O n!·
5
,",,00
· LJn==O
~oo
(-5)n n! .
(-l)n
7.
LJn==O n I/5 •
8
~oo
(-l)n
· LJn==O l+l/n·
9·
~oo
LJn=O
nsinn en .
10. Prove the Ratio test 20.4. Hint: Use the proof of the root test as a modeL
Chapter 21
Power Series
Power series are an important kind of series of functions rather than just numbers. We begin with consideration of general series of functions.
21.1. Definitions. A series L fn of functions converges (pointwise or uniformly) if the sequence of partial sums converges (pointwise or uniformly). The series L in converges absolutely if L IfnI converges absolutely. There is a nice test due to Weierstrass .f~r uniform convergence, essentially by comparison with a series of constants. 21.2. Weierstrass M-test. Suppose Ifni ~ lVIn where each M n is a positive number and L M n converges. Then 2.: fn converges uniformly. Proof. For each x, L f n (x) converges by comparison with To prove the convergence uniform, note that N
L n=l
N
fn(x) - f(x)
=
L
00
fn(x) -
n~l
L
L M n to some f (x)
00
fn(x) =
n=l
L
00
fn(x) <
n=N+l
L
M n,
n=N+l
which is small for N large, independent of x.
D
For example, L si~rx converges uniformly on IR by Weierstrass lvI-test comparison with L~' By Theorem 17.3, the limit is continuous. 21.3. Definition. A power series is a series of x~ such as 001n
L -x n! n=O
L anxn of multiples of powers
1
1
2
6
== 1 + x + -x 2 + -x3 + ...
'
-
93
94
21. Power Series
which we will identify as eX in Chapter 25. A function defined by a convergent power series is called real analytic. Of course a power series is more likely to converge if x is small. You might expect a power series to converge absolutely for x small, converge conditionally for x medium, and diverge for x large. The truth is even a bit simpler.
21.4. Radius of convergence. A power series convergence 0 ::; R ::; 00, such that
(1) for (2) for
Ixl < R, Ixl > R,
L: anxn
has a radius of
the series converges absolutely; the series diverges.
For x = ±R, the series might converge absolutely, converge conditionally, or diverge. On every interval [-Ro, Ro] c (-R, R), the series converges uniformly and the series may be intearated term by term. Note that this result admits the possibilities that the series always converges (the case R = 00) or never converges unless x = 0 (the case R = 0).
Proof. To prove (1) and (2), it suffices to show that if the series converges at xo, then it converges absolutely for Ixl < Ixol. Since it converges at xo, the terms anx o converge to 0; in particular, lanxol ~ C. Therefore
and L: anxn converges absolutely by comparison with the geometric series L:Clx/xol n . Next we prove uniform convergence on [-Ro, Ro]. For some leeway, choose R 1 between Ro and R. Since the series converges for x = Rl; the terms anRJ. converge to 0; in particular, lanR11 ::; C. Hence for alllxl ::; Ro,
lanxnl 5:. C(Ro/ Rl)n. To apply the Weierstrass M-test, let M n = C(Ro/ R1)n; then L: M n is a convergent. geometric series. By Weierstrass, L: anx n converges uniformly on [-Ro, Ro].
In particular, for [a, b] c (-R, R), by Theorem 17.5, the integral of the series equals the limit of the integrals of the partial sums, which of course equals the limit of the partial sums of the integrals of the terms, which equals the series of the integrals of the terms.
21~lO
Taylor's formula
95
o 21.5. Hadamard formula. The radius of convergence R of a power series L an xn is given by the formula R==
1 lim sup lanl l / n
under the agreement that 1/0 == 00 and 1/00
'
== O.
Proof. Using the Root test (20.5), we have p == lim sup lanxnl l / n == IxIJiHence by the Root test, the series converges absolutely if Ixl ~ < 1, Le., if Ixl < R, and diverges if Ixl-h > 1, i.e~, if Ixl > R. Therefore, R is the radius of convergence~ D 21.6. Corollary. Given a power series L anxn , the "derived series" L nanxn- l has the same radius of convergence~ Proof. Multiplication by x (which does not depend on n) of course yields a series with the same radius of convergence: L nanxn ~ Since lim n l/n == 1, by Hadamard's theorem the radius of convergence is the same as for Lanxn . D 21.7. Differentiation of power series. Inside the radius oj convergence, a power series may be differentiated term by term:
(L:
n anX )
I
=
L:
n 1 nanx - .
Proof. On an interval [- Ra, Ra] c (- R, R), L anx n converges uniformly to some j (x) , and the derived series L nanxn - 1 converges.uniformly to some g(x). Since g can be integrated term by term, f~Ro g == f(x) - f( -Ro)~ By the Fundamental Theorem of Calculus, g(x) == f'(x), as desired. 0 21.8. Corollary. A real-analytic function is infinitely differentiable (COO) . 21.9. Corollary. Inside the radius of convergence, you can antidifferentiate a power series term by term~ _ /(11) j(n) 21.10. Taylor's formula. If f (x) == ~ LJanxn , then an - nr, where denotes the .nth derivative.
96
21. Power Series
Proof.
f(x) == ao f'(X) ==
== I'II(X) == f"(X)
+ alx + a2x2 + a3 x3 + al
+ 2a2x + 3a3x2 + 2a2 + 3 2a3x + a
, , a
••
,
31a3+·a.,
f(O)
==
ao;
I' (0) == Ial; I" (0) == 2a2; f'll (0)
= 3! a3;
and so ona
D
Caveat. Given a Cr:xJ function f, Taylor's formula yields an associated Taylor Seriesa Even if the domain of f is all of IR, the Taylor Series could have radius of convergence 0, and even if the series converges, it need not converge to the original function f a In a more advanced course, you will see examples of this strange behaviora
21.11. Power series about c =J o. Power series L anx n are best when x is near 0 (x ~ 0). For x ~ C, we could use L an(x - c)n a There are only minor changes to the theory. The interval of convergence now goes from c - R to C + R, and Taylor's formula becomes (1)
For example, I(x) == I/x has no power series about 0 because it is not even defined there. However, in terms of u = x-I,
(2)
!:. = _1__ = 1 - u + u 2 x
u 3 + u4 (for lui < 1) l+u == 1 - (x - 1) + (x - 1)2 - (x - 1)3 + (x - 1)4 - . . . (for Ix - 11 < 1).
Exercises 21 Ia aa Prove that
3 5 9 x 7 + x__ aa . X -x- +x- __ 31 51 71 91 has radius of convergence R == DOa (Hint: Use the Ratio test 20a4a) ba TRUE/FALSE. It follows that the series converges absolutely and uniformly on every bounded interval [-Ra, RaJ. 2. Consider f(x) == L~==l ~. a. Use the Ratio test to show that the radius of convergence R
== 1.
b. Use the Hadamard formula to show that R == 1. c.
It follows that the series converges uniformly on every interval C (-1,1). Use the Weierstrass M-test to show that it actually converges uniformly on [-1,1].
[-Ra, Ra]
Exercises 21
97
3. Continuing Exercise 2, consider g(x)
x;.
== xf'(x).
a. Compute that g(x) == L~=l b. TRUE/FALSE. By Corollary 21.6, g(x) has the same radius of convergence R == 1. c. Use the Ratio test and Hadamard formula to show that R == 1. d. It follows that the series for g(x) converges absolutely for Ixl < 1. At the endpoints of the interval of convergence 1 and -1, does the series converge absolutely, converge conditionally, or diverge? 4. Find the Taylor Series and its radius of convergence for
== eX; b. f(x) == x 2 . a. f(x)
5. Leibniz's formula for
a. Justify that for
7r.
Ixl < 1, _1_
1 +x
b. Justify that for
2
==
1_
x2 + x4
_
x6 + x8
Ixl < 1, x3
x5 ·
x7
x9
tan- 1 x == x - - + - - - + - - .... 3 5 7 9 (Note that it does not suffice to show.that this is the Taylor series for tan- 1 x; see the Caveat after 21.10.) c. Assuming that part b also holds for x famous formula for 7r:
==
1 (as it does), deduce Leibniz's
4 4 444 1f==---+---+-_ .... 1
.3
5
7
9
6. Check Taylor's formula 21.11(1) for the power series 21.11(2). 7. Find a series for l/x in powers of u == x - 2 by noting that 1 1 1 1 ----x 2 +u 2 1 + u/2 and using the formula for a geometric series. Check Taylor's formula 21.11(1).
Chapter 22
Fourier Series
22.1. Sines and cosines. The functions sinnx and cosnx on [-7r,1T] are beautiful oscillations representing pure tones. A majestic orchestral chord, represented by a much more complicated function, is made up of many such pure tones. The remarkable underlying mathematical fact is that every smooth function on [-1T, 7r] can be decomposed as an infinite series in terms of sines and cosines, called its Fourier series. T~s is in strong contrast to the fact that only real-analytic functions are given by Taylor series in powers of x. Apparently, for decomposition, sines and cosines work much better than powers of x. There is good reason for studying sines and cosines so much, starting in high schooL 22.2. Theorem (Fourier series). Every continuous, piecewise differentiable junction f(x) on [-1T,1T] is given by a Fourier series 00
(1)
f(x)
= A o + L(Ancosnx + Bnsinnx), n=l
where
A o ==
(2)
An
-1171" 21T
== -1171" 1T
f(x) dx,
-71"
cos nx dx,
-71"
1171" sin nx dx,
B n == -
1T
-71"
except that at the endpoints the series converges to the average of f( -1T) and
f(1T)·
.... 99
22. Fourier Series
100
Outside [-1T, 1T], the Fourier series just keeps repeating every 21T, so it doesn't always equal f, unless f is also 21T periodic. The proof of Theorem 22.2 is just beyond the scope of this text, although Theorem 25.1 at least proves that the series converges if f is smooth; it takes more work to prove that it converges to f.
22.3. The coefficients. The above formulas 22.2(2) for the Fourier coefficients An and B n , due to Euler, deserve comment. First note that Ao is just the average value of f(x), about which f varies or oscillates. The other formulas are best unde.rstood by comparison with formulas for the coefficients of a vector
(1)
v
==
VIeI
+ V2e2 + ... + vNeN
in ]RN, expressed as a linear combination of standard basis vectors en. Such basis vectors are orthonormal, that is, em · en == 0, except. that em . em == 1. It follows immediately that the coefficients V n satisfy
(2) by just taking the dot product ·of each side of (1) with en. Analogously, on. the space of continuous functions on [-7f, 1T], one can define a dot product by
(3)
=:;1jIT _~ f(x) g(x) dx,
f .9
integrating all products of values instead of just summing products of components. We want to write everything in terms of the functions cos nx and sin nx, which turn out to be orthonormal functions for this dot product (see Exercise 1). The formulas for the Fourier coefficients 22.2(2) just say that each Fourier coefficient is obtained by dotting the function f(x) with the associated orthonormal function, just like the corresponding formula (2) for vectors.
22.4. Example. For example, consider the piecewise smooth function f(x)
==
{o
if -1T ::; x
X(1T-X)
~ 0,
ifO~x~7f
of Figure 22.1. Exercise 3 uses 22.2(2) to compute the Fourier series:
(1)
f(x)
1T2 12
== -
2 cosnx L....J n
"
n even
-2
+ "L....J
n odd
4 sinnx. 1Tn
-3
(See Figure 22.2.) Fourier series have many serious real-world applications, as in the upcoming Chapter 24. Meanwhile, here is a more amusing application. The
22.5. Proposition
101
3
2
-3
-2
2
-1
Figure 22.1. The piecewise smooth function f(x) x ::; 1T".
-2
3
== x(7r -
x) for 0 ::;
2
Figure 22.2. Sum of the first few terms of the Fourier series for the function of Figure 22.1.
following Proposition 22.5 has many proofs; in ours, Fourier series make an unexpected appearance. The proposition gives the exact sum of the previously mysterious series L:(1/n2 ). By the p-test 19.4, we knew that this series converged, and now we finally find out the surprising answer to what it converges to (how did that 1T get in there?). 22.5. Proposition. The series L(1/n2 ) converges to
1T
2
/6.
Proof. Plugging x == 0 into the Fourier series 22.4(1) yields:
i
2
0::= 12 -
(
2 22
+ 422 + 622 + ... ) .
1 12
1 + 212 + 32 + ... )
Multiplying by 2 yields 2
o= so that L:(1/n2 )
== 1T2 /6.
1T
(3 -
(
D
22. Fourier Series
102
22.6. Perspective. Since a Fourier series is determined by the coefficients, Fourier series provide a way to encode a function by two sequences An, B n of numbers with physical significance corresponding to "frequencies." For a violin string, the predominant frequencies are nice multiples of each other, called harmonics, and produce a rich, harmonious sound. In many applications, it is often the case that certain frequencies. dominate, and the function can be well approximated by finitely many terms. Such mathematics can provide a way to encode or compress complicated data, sounds, or images. JPEG is exactly such a method for efficient storage of images. Dolby stereo filters out hissing noise by avoiding the frequencies responsible for such noise. The next chapter (23) will show how Fourier series can be used to solve important differential equations.
22.7. Tlie sine series. If f(x) is an odd function, which means that f( -x) == - f(x), it follows immediately (Exercise 5) from the formulas 22.2(2) for the coefficients that every An vanishes (is 0), so that the Fourier series consists entirely of sines.
Every function f(x) on (0,1T) can be extended to an odd function on (-1T,1T), whose sine series holds for the original function f(x) on (O,7T): 00
(1)
f(x) == LBnsinnx, n=l
where
(2)
21
B n == -
1T
7f
f(x) sinnx dx.
0
The new factor of 2 comes because we are integrating only from 0 to 1T; we wouldn't need it if we integrated the extension from -1T to 1T.
22.8. General intervals. Fourier series (22.2) immediately generalize from [-1T,1T] to [-L, L], just by replacing x with 1TX/L: (1)
22.8. General intervals
103
where 1 A o = 2£
(2)
An = - 1 2£ 1
B n = 2£
1£ 1 1
-L
f(x) dx,
L
n1rX f(x) COSy dx,
-L
L
-L
.
n1rX
f(X)SIDydx,
22~
104
Fourier Series
Exercises 22 1~ Verify that the functions cosx and sinx are orthonormal on [-1T,1T] for the dot product 22~3(3)~
2~
What is the Fourier series for f(x) == sin x on [-1T,1T]?
3. Compute the Fourier series for Example
22~4
using
22.2(2)~
Hint: Use integration by parts and notice that cos n1T and 2 if n is even. 4~
+ 1 is a if n
is odd
Show that the Fourier series for f(x) == x (-7[ ::; x ::; 1T) is given by 1 ~ 1. 1 . ) x==-2 ( i SIn x - 2 SIn 2x + "3 sIn 3x - . ~ ~
~
Taking x == 1T /2, obtain Leibniz's amazing formula for 1T:
4
4
4
4
7[==---+---+ .... 1 3 5 7
5. Show that if f(x) is an odd function, which means that f( -x) == -f(x), then the coefficients An vanish (are 0), and you get a Fourier series of sines. Similarly if f(x) is an even function, which means that f( -x) == f(x), then the coefficients B n vanish, and you get a Fourier series of cosines. 6. Show that the Fourier series for the function
a f(x)
< X < 0, for 0 :::; x :s; 1T/2,
for -1T
== x
{ 1T -
x
for 1T /2 :::; x :::; 7[,
is given by
f(x) == -1T - -4(1 - cos 2x + -12 cos 6x 8 1T 22 6
+ -1T2
+ - 12 cos lax + .. ~ ) 10
(2" 1SIn . x - 31 SIn ~ 3x 2 1
. + -5x1 2 SIn 5x -
~ ~
·) .
Chapter 23
Strings and Springs
Chapter 23 provides classic applications of Fourier series to vibrations of a string and to oscillations of a spring.
23.1. The vibrating string. In 1755 Daniel Bernoulli found a general solution y(x, t) to the differential equation for the small vertical displacement y of a-vibrating string as a function of 0 :::; x :::; 1T and time t. We'll assume that the string has unit density and tension and is fixed at its endpoints. The simplest solution was y(x, t) == sin x cos t, the curve y
== sinx oscillating in time, as in Figure 23.1.
Similarly there were simple higher frequency solutions
y(x, t) == sin nx cos nt, oscillating at multiples of the fundamental frequency, as in Figure 23.2.
3
-I
0
-I
Figure 23.1. The simplest vibration of a string.
-I
-
105
106
23. Strings and Springs
-I
-I
-I
Figure 23.2. The second simplest vibration of a string.
Finite linear combinations of solutions are solutions. For the general solution, Bernoulli proposed infinite linear combinations: 00
(1)
y(x,t) ==
I: Bnsinnxcosnt. n=l
At time t == 0, the initial shape is given by 00
(2)
y(x)O) ==
I: B n sinnx. n=l
Many mathematicians asked, "What about other initial shapes?" The surprising answer was that every nice function is given as such a Fourier series 22.7(1). This Fourier series for the initial shape (2) determines the contribution of the various frequencies of oscillation in the solution (1). To think about such questions in any detail as we began in Chapter 22 requires a basic understanding of real analysis. Bernoulli had no idea, and it took mathematicians such as Weierstrass over a hundred years to develop what you already know. Questions about the generality and convergence of Fourier series were a major impetus in this development.
23.2. Forced spring oscillations. The displacement x(t) of a unit mass on a spring with spring constant w2 and force -w 2 x satisfies Newton's famous law F ::=: mao The acceleration a is the second derivative d2 x/dt 2 , which Newton wrote as X, using a dot for differentiation with respect to time t. Thus F ::=: rna becomes .. 2 0, x+wx==
(1)
with well-known general solution
(2)
x ==
Cl
cos wt + C2 sin wt.
If a driving force f(t) is applied, the differential equation becomes the much harder
(3)
23.2. Forced spring oscillations
107
We suppose that the driving force f(t) is periodic with period 21T and seek a nice periodic solution. For example, suppose that a pulse as in Example 22.4 is applied every 21T seconds. Assume that x is given by a Fourier series 00
x == Ao + I:(A n cos nt + B n sin nt). n==l
Hoping to be able to use our knowledge of real analysis to justify our actions later, we differentiate the series term by term, 00
x == -
(4)
I: n (A 2
n
cos nt + B n sin nt),
n=I
and plug the result into equation (3) along with the Fourier series for f from Example 22.4. Equating coefficients of like terms yields the following equations for the coefficients: A o ==
(5)
n even: (w 2 n odd:
2 1f2 /12w ,
n 2 )An == -2/n 2 , B n == 0,
-
An == 0,
Hence, 2
2 (6) x(t) == 2 2 + 2( 2 1 w n w - n 2) cosnt + 1T
I: -
n even
I: 1fn3(w42 -
n odd
n 2) sin nt,
a Fourier series for our answer. Resonance. Notice that the largest terms correspond to n closest to w, the natural frequency of the spring. Forced vibrations close to the natural frequency cause huge oscillations, in a phenomenon called resonance. In one notorious case in 1940, the Tacoma Narrows suspension bridge, excited by winds at its natural frequency, oscillated ever more wildly until it tore itself apart. (Check it out on the web.) According to legend, soldiers break formation before crossing a bridge lest their cadence produce hazardous resonance. Justification. The justification will use some of. our strongest tools, the Weierstrass M-test (20.2) and Theorem 16.5 on integrating series term by term (moving the integral inside the summation sign). Start by defining x as what you get if you differentiate (6) twice term by term:
(7)
. I:
x ==
n even
2
2
w -n
2
cos nt -
I: n odd
(2
4
n w -n
2)
. Sin nt.
Note that for n large, the absolute value of the nth term is bounded by M n == 2/n 2 . Hence by the Weierstrass M-test (21.2), this Fourier series converges uniformly. Therefore by Theorem 17.5, it can be integrated term by term,
108
23. Strings and Springs
yielding (6) up to a linear term C 1 + C2 t, thus justifying the original termby-term differentiation~ l\tloreover, any solution x whose Fourier series when so twice differentiated is uniformly convergent must by the above process be our solution (6). The equating of like terms will be justified by Exercise 24.2.
Exercises 23 1. Find the Fourier series 22.7(1) in terms of sines for the function
f(x) = {x 1r -
x
for 0 ::; x :::; 1r/2, for 7[/2.:::; x :::; 1r.
2. A plucked string has initial shape cf(x) from Exercise 1. Give its solution 23.1(1) as a function of x and t. 3. Check that 23.2(2) satisfies 23.2(1). 4~
Derive equations 23.2(5).
5. Show that a periodic solution to 23~2(3), for f(t) a unit pulse from t to t = 1r /2 repeated with period 21r, is given by 1 x(t) =·-4 2+ W
6. Check that y(x, t)
00
L
sin n2'i ' cos nt
+ (1 -
n1r (2 W
n=1
== sin nx cos nt 2
8 y 8t 2
-
cos n2'ii') sin nt
n 2)
.
is a solution of the wave equation
82y 8x 2 ~
=
a
Chapter
24
Convergence of Fourier Series
Theorem 24.1 provides the promised sample of Fourier theory, relatively easy for you, utterly beyond Bernoulli and 18th Century mathematics. 24.1. Th~orem. Let I be a C 2 function on [-7r,7r] with I(7f) == I( -7r) and f' (7r) == II (-7r). Then the Fourier series for f converges uniformly. Proof. Let C
=::
max.{f"(x)}. By integration by parts,
An == -1j1f f(x) cos nx dx 1T
-1f
1 1 j1f f(x) sin nxl~1f - - fl(X) sin nx dx 7f n -1f 1 1 1 1 j1f I 0 + -2" fl(X) cos nxl~1f - -2" I I(X) cos nx dx 7r n 7r n -rr 11
=:: - -
1T
==
n
1 1 ==0+0-7r 2n
j7f I
/I
(x ) cosnxdx.
-iT
Hence, 1 1 2C IAncosnxl ~ -2"(27r)C == 2" 7rn n 2 Similarly, IBnsinnxl ~ 2C/n • Therefore by the Weierstrass M-test (21.2), the Fourier series converges uniformly. 0
24.2. Remarks. It takes a bit more work than we are prepared to do to show that the limit is actually f"
-
109
110
24. Convergence of Fourier Series
..
n
...
v
{/"
Il"'\
~
I~
~
A~
JrJ
r
r
'vi
/
tT
14
I ~
..... ~
I
V I
-;4
Figure 24.1. The Fourier approximations to f(x) = x overshoot by about 9% at the endpoints, the so-called Gibbs phenomenon.
IT f(1r) # f( -1r), the Fourier series, which according to Theorem 22.2 converges to (f(1r) + f( -1f))/2 at the endpoints, thus has a limit discontinuous at the endpoints. Hence, it cannot converge uniformly because a uniform limit of continuous functi
Exercises 24 1. In the proof of 24.1, verify that IBn sin nxl
::; 2C /n 2 .
2. Prove that if a series of the form 22.2(1) converges uniformly to f(x), then it must be the Fourier series for f(x). Hint: cos nx times the series converges uniformly to may be integrated term by term. 3. Use Exercise 2 to solve Exercise 22.2 effortlessly.
f (x) cos nx, and hence
Chapter 25
The Exponential Function
It is no small feat to define a new function. There are several ways to define the exponential function exp(x) or eX and derive its properties. The most obvious definition is just as powers of the constant e. For that approach, first you need to define e, perhaps as lim(l + 1/n)n. Then e2 == ~ x e, en == e x ex· .. X e (n times), e- n == lien, a rational power eP / q is the qth root of eP , but real powers are more complicated, and the various properties of eX do not. follow easily. Rigorous calculus books define the natural logarithm. first as an integral, and then define the exponential function as the inverse. Here we follow more sophisticated texts by defining the exponential function by a power series. You may be surprised at how easily all the desired properties follow. 25.1. Definition. Define the exponential function 00 xn x2 x3 exp(x)==~· ==l+x+-+-+···. L...J n! 2! . 31
n=O
Define e == exp(l) ==
L -\ ~ 2.718281828. n.
25.2. Proposition. The series defining exp(x) converges absolutely on all of~, uniformly on every bounded interval. The function is Coo and may be integrated and differentiated term by term. It is its own derivative.
exp(O) == 1 and exp(x) is positive for x 2:: O.
-
111
25. Tbe Exponential Function
112
Proof. For the Ratio test (20.4), p == lim Ixl/n == 0; so the radius of convergence is 00 and by Theorem 21.4 the series converges uniformly on bounded intervals. By 21.7-21.9, exp(x) is Coo and may be integrated and differentiated term by term, to yield exp'(x) == 0 + 1 + x + x 2 /2! + ... == exp(x). It follows immediately from the definition that exp(O) positive for x 2:: O. 25.3. Proposition. exp(-b) exp(x) is positive for all x.
=
==
1 and exp(x) is 0
e~(b)" In particular, exp(-l)
=
lie, and
Proof. This is the part of the development where you might think you would have trouble. Dividing 1 by an infinite series is quite a long division problem. But it turns out that there is a slick proof. Notice that by the product rule d dx (exp(x) exp( -x))
= exp(x) exp( -x) -
exp(x) exp( -x)
= O.
Hence exp(x) exp( -x) is a constant function: exp(x) exp( -x) == C. Plugging in x == 0, we see that C == 1. Hence exp( -x) == 1/ exp(x) , as desITed. 0 25.4. Proposition. exp(a + b)
==
exp(a) exp(b).
Proof. Since by the quotient rule ~ exp(a
dx exp(a+x)
+ x) =
exp(a
+ x) exp(x) -
exp(x)
exp(a + x) exp(x) = a exp2(x) ,
== Cexp(x). Plugging in x == 0 yields C
== exp(a),
as desITed.
0
25.5. Corollary. exp(n) == en; exp(P/q) == f(eP. Proof. By Proposition 25.4, exp(n) == exp(l) exp(I) ... exp(l) == en. Likewise, (exp(p/q))q == exp(p) == eP , which implies that exp(p/q) == eP/ q. IT we now define eT as exp(r) for all real numbers, this definition will agree with any others by continuity. 0 25.6. Corollary. (1) exp(x) is strictly increasing.
> 0, such that for x 2:: 0, exp(x) > cn xn . (3) Given C > 0 and n, for x large, exp(x) > Cx n . (2) For every n, there is an Cn
Proof. Conclusion (1) follows because the derivative exp(x) is positive. Conclusion (2) follows from the defining series, with Cn == l/n!. In particular, given C > 0, for x ~ 0, exp(x) > (l/(n + 1)!)xn+ 1 , which is greater than Cx n once x > C(n + I)!. 0
Exercises 25
113
25.7. The natural logarithm. It follows from Corollary 25.6(1) that exp(x) is a bijective map from ~ onto the positive reals. The inverse function from the positive reals onto ~ is called the natural logarithm. function, written In x in calculus books and log x in more advanced mathematics. Its properties, such as (log x)' == l/x, follow from the properties of exp(x). See Exercise 4. 25.8. Complex exponentials. Although we have dealt only with real series, all of our results hold for series of complex numbers z == x + iy, including the definition and properties of exp(x). Here i 2 == -1, i 3 == -i, 1.·4 == 1, ....
Exercises 25 1. Development of the sine and cosine functions. Define
x2 2! x3 sinx == x - -
x4 x6 x8 4! 6! 8! ' x5 x7 x9 + - - - + - _ .... 3! 5! 7! 9! a. Check that the radii of convergence are 00, so that these functions are defined for all x. cos x
== 1 - - + - - - + - - ...
== cosx and that sin(-x) == -sin(x). Show that (sin x)' == cos x and that (cos x)' == -sinx. Prove that sin2 x + cos 2 X == 1.
b. Show that cos(-x) c. d.
Hint: First show the derivative 0, so that sin2 x x = 0 to find C. Note that this implies that Isin xl
+ cos2 X ==
C. Then use
::; 1 and I cos x I ~ 1.
2. By plugging into the series for eX, verify Euler's identity
eiO == cos (J
+ i sin B.
3. Use Euler's identity and the fact that ei(a+b) == eiaeib to deduce that sin( a+ b)
== sin a cos b + cos a sin b,
cos(a + b) == cos a cos b - sin a sin b.
25. Tbe Exponential Function
114
4. Since the exponential function has a nonvanishing (never 0) continuous derivative, it follows from the Inverse Function Theorem (which did not quite make it into this book) that its inverse, log x, is continuously differentiable. a. If y
== logx, then x == eYe Differentiate implicitly to deduce that (log x)' == l/x.
b. Prove that log ex
== log c + log x.
5. Obtain a power series for log x in powers. of x -1 by integrating 21.11(2). Check Taylor's formula (21.11(1)) for this series. 6. Prove that e definition of e.)
== limn --+ oo (l + l/n)n. (This is sometimes used as the
Hint: It suffices to show that log(l
+ l/n)n == nlog(l + l/n)
-7
By the definition of the derivative,
1og'(I) -Therefore, taking 6.x
1·1m log(lA+ 6.x) • ux
.6. x--+ 0
== lin, nlog(l
+ lin)
~ log'(l) == 1.
1.
Chapter 26
Volumes of n-Balls and the Gamma Function
There is a wonderful formula (26.5) for the volume of the ball B(O, r) in 1rn / 2
volB(O,R) = (n/2)!R
~n:
n .
For example, the "volume" or area of a ball in ~2 is (;;~~!r2 = 7fr 2 • The volume of a ball in }R4 is ~1r2r4. However, according to this formula, the "volume" or length of a ball in ~1, which should be 2r, is (~)~~!r. What is (1/2)1? For our formula to work, we would need (1/2)1
== v:rr/2.
Fortunately, there is a nice extension of (x-l)! from integers to real numbers greater than 1, called the Gamma Function r(x). 26.1. Definition. For x 2:: 1, define the Gamma Function r(x) by
1
00
r(x) =
e-tt x - 1 dt.
This integral converges because et grows faster than any power of t. 26.2. Proposition. r(x + 1)
== x r(x).
Proof. Using integration by parts, dv == e-tdt , du == xtx - 1 , v == -e- t ,
J u dv
== uv -
J v du,
with u == t X ,
-
115
26. Volumes of n-Balls and tbe Gamma Function
116
26.3. Proposition. For every nonnegative integer n, r(n + 1)
== n!.
We can use this relationship to extend the definition of the factorial function to every nonnegative real number by xl == r(x + 1). Proof by induction. First we note that for n
1
00
r(0 + 1) = r(1)
=
== 0,
e- t dt = -e-tlo = -0 - (-1) = 1 = O!.
Second, assuming the result for smaller values of n, we see by Proposition 26.2 that r(n + 1) == nr(n) == n(n - I)! == nL 0 26.4. Proposition. r(1/2)
== ..Jff.
Proof. By making the substitution t == s2, dt
r(1/2) =
roo e- t C
10
1/ 2
=
dt
== 2s ds, we see that
roo e- s2 s-l 2s ds = JOO e- s2 ds.
io
-00
Now comes the famous trick-multiplying r{1/2) by itself and switching to polar coordinates with dA == r dr dB:
r(1/2) r(1/2)
=
1: 1: s2
e-
== JOO
ds
11
du
2
e _s2jOO e _u du ds
-00
=
u2 e-
-00
e-
s2
e-
1= io=o
= (2tr
u2
duds = 2
e_r r dr de
11 =
~
27f [- e_r
duds 2 ]
2
r=O
Therefore r(1/2)
u e-(s2+ 2)
==
7f.
0
== ..Jff.
In particular, (1/2)! == r(3/2) == (1/2) r(1/2) == ..Jff/2, as we hoped. Note too that our formula gives for the volume of a ball in ~3: 'Jr3/2
volB(O,r)
3
'Jr3/2
= (3/2)!r = (3/2)..Jff/2r3 =
4 37fr3
the correct, familiar formula.
0
26.5. Proposition. The volume of a ball in
~n
1fn/2
volB(O,R) = (n/2)!R
is given by
n .
Proof. We haven't defined volume; we will assume here that it can be computed as in calculus, by integrating over slices. We've already checked this formula for n == 1 and n == 2. We'll prove it by induction, assuming the result for n - 2. (By jumping two dimensions at a time, we can use a much
26~ 7~
Stirling's Approximation
117
simpler polar coordinates proof~) We view the nD ball {x~ + x~ + x~ + ... + x;' ::; R 2 } as a 2D ball {r 2 == + x~ ::; R 2 } of (n - 2)D balls of radius s satisfying s2 == x~ + ... + x;' == R 2 - r 2 and volume (by induction)
xi
7f(n-2)/2
7r(n/2)-1
n-2
n-2
((n/2) _ I)! s
-((n---2)-/2-)! s Hence, volB(O,R) =
==
if 7r
7r(n/2)-1
((nj2) _1)!sn-2 dA
1 lR 27r
(n/2)-1
((n/2) - I)!
= _ 7f(n/2)-1
8=0 27f
((n/2) - I)!
(R2 _ r 2){n-2)/2 r dr dB
r=O
[_~ (R2 _ r 2)n/2] R 2
7r(n/2)-1 1 Rn ----27r-((n/2) - 1)1 2 n/2
n/2
0
7r n / 2
== _ _ Rn~
0
(n/2)!
26.6. Corollary. The volume of the (n - I)D sphere in IRn is given by 7rn / 2 n volS(O, R) = n(nj2)!R -
1
.
For example, the circumference of a circle in ~2 is given by 21R
== 27fR.
Proof. The volume of the sphere is just the derivative of the volume of the ball (the greater the surface area, the faster the volume of the ball increases) . D
26.7. Stirling's Approximation. The Gamma function may be used to derive an excellent asymptotic approximation to n! n!
r-.J
J27rn(n/e)n.
(Technically this asymptotic symbol rv means that as n 1.) It follows that en « n!« nn. (For the definition of rates of growth, see
347~)
-7
00,
V2rn(~/e)n -7 n.
26. Volumes of n-Balls and tbe Gamma Function
118
Exercises 26 1. a. Use Propositions 26.2 and 26.4 to find a numerical value for (3/2)! == r(5/2). b. Check the formula for the volume of the ball B(O, r) in ~3. 2. What is the volume of the ball B(O, r) in ~5? in ~6?
3. What is the area or volume of the sphere (surface of the ball) 8(0, r) in }R3? in ~4?
in
~5?
4. Use Stirling's Approximation and your calculator to estimate that 10gIO 1000! 2567, so that roughly 1000! ~ 102567 . Try to estimate 1000! by any other means. 5. Use Stirling's Approximation to prove that en «n!
Part IV
Metric Spaces
Chapter 27
Metric Spaces
To do analysis, you don't need JRn , just a space with a way to measure distance, called a "metric." In JRn , the distance p(x, y) between two points x and y is just p(x, y) == Ix - yl (we're using the Greek letter rho for distance). This distance has three nice properties: (1) It is always positive, except that the distance from a point to itself is zero. (2) The distance from x to y is the same as the distance from y to x. (3) The triangle inequality holds: the distance between two points is less than or equal to the total distance via a third point. Any function that satisfies these three properties is called a metric. 27.1. Definition. A metric space is a set X with a distance or metric p(x, y) defined between every two points of X satisfying the following three properties: (1) Positivity: p(x, y) (2) Symmetry: p(y, x)
~
0, with equality if and only if y == x.
== p(x, y).
(3) Triangle inequality: p(x, z) ::; p(x, y)
+ p(y, z).
27.2. The taxicab metric on IR 2 • As our first new example, we define a new metric between every two points (Xl, X2) and (YI, Y2) in )R2 as the sum of the horizontal and vertical distances:
-
121
27. Metric Spaces
122
Figure 27.1. rrhe unit ball for the taxicab metric is a diamond.
This is called the .taxicab metric because it is the distance that a taxicab, confined to horizontal and vertical streets, has to travel to get from one point to the other. Notice that the unit ball in this metric, the set of all points within distance 1 of the origin, is a diamond, as pictured in Figure 27.1. The point (~, ~) for example is at distance 1 from the origin. It is easy to check that the taxicab metric does indeed satisfy properties (1)-(3) above.
27.3. The space 0[0, 1] of continuous functions on [0, 1] with the sup metric. Our second new example is the huge space 0[0,1] of continuous functions on [0,1], with the sup metric defined as the maximum or supremum of the absolute value of their difference:
p(!, g) == sup{I!(x) - g(x)l: x E [O,ll}· Notice that the elements or "points" of this space are functions. Under the sup metric, two functions are within c of each other if their values at all points are within c of each other. Again it is easy to check that this metric satisfies properties (1)-(3). (For continuous functions, the maximum is attained, so we do not really need the more general notion of supremum from Exercise 9.13.)
27.. 4. The space W of words. Let W be the space of all words or gibberish of any finite number of letters, such as "cucumber" or "rqxiibt". There are lots of different metrics you could define on this space. We will define the distance between two elements of W as the fewest number of changes it takes to get from one to the other, where in each change you can:
123
Exercises 27
(1) delete any letter, such as "apple" to "aple,"
(2) insert any letter, such as "sand" to "stand," (3) replace any letter with another letter, such as "rocket" to "qocket," or
(4) switch any two adjacent letters, such as "slat" to "salt." It's fun to think about examples. Some of the words in B("pat", 1) are "at," "spat," and "apt." How many others are there? Of course, B("pat", 1) contains lots of gibberish too, such as "jat" and "pxt.. " The distance between the words "math" and "money"
p( "math" , "money")
== 4,
because it takes four changes to get from "math" to "money" (math moth --t monh ----+ mone ----+ money).
-----+
Exercises 27 1. Is each of the following a metric on the space of continuous functions
from [0,1] to lR? If not, explain why not. a. p(!,g) b. p(!, g)
== 1!(1/2) - g(1/2)1. == sup{I!'(x) - g'(x)I}. 1
c. p(!, g) == fo I!(x) - g(x)1 dx.
d. p(!,g)
== fo1If(x) - g(x)1 2 dx.
e. p(!, g)
==
(
fo1 If(x) -
g(x)1 2 dx
) 1/2
.
2. How many words can you find in B("pat" ,1)? How many elements (counting words and gibberish) are there in B("pat", I)?
Chapter 28
Analysis on Metric Spaces
Analysis on metric spaces is very similar to analysis in JRn. Just replace the distance Ix - yj with the more general p(x, y). 28.1. Definition. A sequence an in a metric space X converges to .a limit L (an ----+ L) if, given c > 0, there is some N such that whenever n > N, p(a n , L) < c~
In 0[0,1], In ~ I under the sup metric therefore means that In ~ I n uniformly~ So fn(x) = x/n converges to I(x).== 0, but In(x) == x does not converge (uniformly) to f(x) == 0. Most of the results on sequences of Chapter 3 remain true in metric The exception is Proposition 3~6 because in a general metric space there is no addition, multiplication, or division~ spaces~
28.2. Definition. A sequence an in a metric space X is bounded if all the terms lie in some ball. A sequence an in a metric space X is Cauchy if, given c > 0, there is some N such that whenever m, n > N, p(a m , an) < c. Every convergent sequence is Cauchy (Exercise 3), but whether every Cauchy sequence converges depends on the metric space. It is true in the reals. It fails in the space of rationals, since 1, 1.4, 1~41, 1.414, ... , which converges to J2 in JR, does not converge in (Q, even though it is still Cauchy. The problem is that J2 is missing-the space is somehow not "complete." 28.3. Definition. A metric space X is complete if every Cauchy sequence converges (to a point of X) ~
-
125
28. Analysis on Metric Spaces
126
~ is complete, but Q is not complete (Exercises 16 and 17). C[0,1] is complete (Exercise 18), essentially because a uniform limit of continuous functions is continuous and therefore is not missing from the space. Our word space W is complete because every point is isolated (no other points within distance 1/2) and hence all Cauchy sequences are eventually constant-the only way for words to be very close together is for them to be identicaL
If a metric space X is not complete, one can form the "completion" by adding limits of all Cauchy sequences. Instead of defining the reals via decimals, some texts define the reals as the completion of the rationals. You have to be careful because different Cauchy sequences correspond to the same real number. Topology on a metric space X is very similar to topology in JRn.
28.4. Definitions. A point p in a metric space X is a boundary point of a set S if every ball B(p,r)
== {x
E
X: p(x,p) ::; r}
about p meets both Sand Se. The set of boundary points "of S is called the boundary of S and written as. A set S is open if it contains none of its boundary points. A set S in ~n is closed if it contains all of its boundary points. For example, a ball B(p, r) is closed; if you remove the boundary, the resulting set is open and is called an open ball. "The interior of a set S, denoted int S or S, is S - as. The closure of S, denoted cl S or S, is S uas. An isolated point of S is the only point of S in some ball about it. Every ball about an accumulation point of S contains infinitely many points of S. All of the results of Chapters 5-7 continue to hold in metric spaces.
Exercises 28
127
Exercises 28 1. In the space 0[0,1]' does the sequence In converge or diverge (under the sup metric)? If it converges, what is the limit? a. In(x)
== x 2 /n;
b. fn(x)
== x 2n .
2. Suppose that a sequence an converges in a metric space. Prove that the . limit is unique and that the sequence is bounded~ 3. Prove that a convergent sequence in a metric space is Cauchy. 4. Prove that a set S in a metric space is open if and only if the complement Se is closed. 5. Prove that a set S in a metric space is open if and only if about every point of S there is a ball completely contained in S. 6. Prove that a set S in a metric space is closed if and only if it contains all of its accumulation points~ 7. Say whether the set is open, closed, neither, or both in the space 0[0,1].
{f(1/2) < 6}; b. {f(1/2) ::; 6}; c. B(O, 1). a~
8. Prove the following statements in a metric space. Any union of open sets is open. A finite intersection of open sets is open~ Any intersection of closed sets is closed. A finite union of closed sets is closed. 9. Prove that in the word space W every singleton is open, and hence every set is open and closed.
10~ If S
== {O ::; f(1/2) < 1} C 0[0,1]'
what are
as, S,
and 8?
11. Prove that the interior of S is the largest open set contained in S. 12. Prove that the closure of S is the smallest closed set containing S~
128
28. Analysis on Metric Spaces
13. Prove that a is an accumulation point of S if and only if a is the limit of a sequence of other points in S. 14. Prove that a boundary point of S is either an isolated point or an accumulation point. 15. Prove that every point of S is either an isolated point or an accumulation point. (Why does this not imply Exercise 14?) 16. Show that Q is not complete. 17. Prove that JR is complete. 18. Prove that C[O, 1] is complete. 19. Do Proposition 6.1 and proof generalize to real-valued functions on a metric space?
Chapter 29
Compactness in Metric Spaces
This is the moment when metric space topology takes a nasty turn. In a general metric space, the three characterizations of compactness of Theorem 9.2 are no longer equivalent: a closed and bounded set need not satisfy the .other two. First we will give two natural examples of this phenomenon. Then Theorem 29.3 explains that the other equivalences continue to hold. 29.1. Example. Let]Roo denote the space of all real vectors
with infinitely many components, within finite distance of the origin: JRoo == {(Xl, X2, X3, • .. ): xi
+ x~ + x~ + .·. < oo}
with metric
p(x, Y) =
(:L)Xi - Yi)2f/2 .
This space is a lot like ]Rn. The unit ball B(O,I) is closed and bounded, but unlike ]Rn, the unit ball B(O,I) admits sequences with ~o convergence subsequences. For example, let el =
(1,0,0,0,
),
e2 ==
(0, 1, 0, 0,
),
== (0, 0, 1, 0,
)
eg
and so on. This sequence has no convergent subsequence because it has no Cauchy subsequence because all terms are distance J2 apart. The problem is that ]Roo is infinite dimensional.
-
129
130
29. Compactness in Metric Spaces
29.2. Example. Similarly, consider the unit ball B(O, 1) in C[O, 1]. It is closed and bounded, but admits sequences (of functions) with no (uniformly) convergent subsequences. (Recall that convergence under the sup metric of C[O, 1] means uniform convergence.) For example, let fn(x) == x n , as in Figure 17.1. Then any subsequence converges pointwise to
f(x)
==
{o °::; 1
for x < 1, for x == 1,
which is therefore the only possible limit; but no subsequence can converge uniformly because a uniform limit would have to be contiriuous, and f is not continuous. Fortunately, although in metric spaces closed and bounded does not yield convergent subsequences, the stronger two characterizations of compactness remain equivalent:
29.3. Theorem. For a set K in a metric space X, the first two conditions are equivalent and imply the third.
(1) Every sequence in K has a subsequence converging to a point of K.
(2) K is compact: every open cover has a finite subcover. (3) K is closed and bounded.
In Exercise 1 you will prove that (1) => (3). We will prove first that (2) ::::} (1) and second that (1) ::::} (2), which is a bit harder (and false in still more general topological spaces than metric spaces). Our earlier proof of (1) ::::} (2) in JRn (Theorem 9.2) no longer works because an open cover of a general set need not have a countable subcover because there need not be a dense countable set like the rationals.
Proof that (2) ::::} (I). Assume that some sequence an in K has no subsequence converging to a point of K. Then every point x of K has a little open ball Ux around it containing only finitely many an (counting multiplicity, Le., containing an for only finitely many values of n, even if lots of the an happen to be equal). By hypothesis (2), finitely many such Ux cover all of K. Hence, there are only finitely many an, a contradiction, since an is an infinite sequence. D Proof that (1) ::::} (2). Let {Un} be an open cover of K. First we claim that for some E > 0, for every x E K, B(x, E) is contained in some Un. Otherwise, let X n be a sequence of points in K such that B(x n ,l/n) is contained in no Uo . By hypothesis (1), some subsequence converges to a point a in K. Some B(a, r) is contained in some Un. Hence, once p(x n , a) < r/2, B(xn ,r/2) is contained in Un, a contradiction once l/n < r/2.
131
Exercises 29
Now as long as possible choose disjoint balls of radius E/2 centered at points Yn in K. This cannot go on forever to produce an infinite sequence because such a sequence Yn could not have a convergent subsequence because it is not Cauchy. Hence, after choosing SOme YN, for every Y in K, B(y, E/2) intersects some B(Yn,E./2). Hence, the finite collection {B(Yn, E)} of balls twice as large covers K. Choose UOn containing B(Yn, E). Then {UOn } is the desired finite subcover of K. D 1
Remark. All of the results of Chapters 10-12 continue to hold in metric spaces. See for example Exercise 29.3.
Exercises 29 1. Prove that 29.3(1) ::::} 29.3(3). 2. Here is a trivial "artificial" example of a closed and bounded set which is not compact. Let X be the integers with metric p(m, n) == 1, except that p(n, n) == o. Check that p is a metric. Show that X is closed and bounded, but not compact. 3. Prove that in a metric space, a continuous real-valued function on a nonempty compact set attains a maximum. (Cf. Corollary 10.2.) . 4. Do Theorem 11.2 and its proof generalize to a metric space?
Chapter 30
Ascoli's Theorem
To solve problems, one needs compactness in order to extract from a sequence of approximate solutions an exact solution in the limit. For sophisticated problems, the desired solution is not just a number, but rather a function, perhaps describing an economically ideal schedule of production or the most efficient shape of an airplane wing~ Unfortunately, as we have seen, even the closed unit ball in the space
0[0,1] of real-valued continuous functions on the unit interval under the sup metric is not compact~ Fortunately, Ascoli's Theorem provides useful compact spaces of functions. The main hypothesis is a uni~orm kind of continuity called equicontinuity. See if you can spot the difference from uniform continuity: 30.1. Definition. A set of real-valued functions f on a domain D C JR is equicontinuous if, given E > 0, there is a > such that
a
Iy - xl < a =} If(y) -
°
f(x)1 <
The difference from uniform continuity is that x but also of f: the same a works for all f~
E.
a is independent not only of
Note how equicontinuity fails for our previous Example 29~2 (fn(x) = As n approaches infinity, x n increases rapidly near 1, and for given > 0, a approaches 0; a is not independent of which fn we consider.
xn)~ E
30.2. Ascoli's Theorem. Let F be a closed, bounded, equicontinuous subset of 0 [0, 1]. Then F is compact
-
133
30. Ascoli's Theorem
134
Proof. Let in be a sequence of functions in F. Note that for any fixed point p, in(P) is a bounded sequence of real numbers and hence has a convergent subsequence. Similarly, if q is another point, we can take a subsubsequence converging at q as well as at p. Indeed, given any finite number of points, it is easy to find .a subsequence converging at those points. Begin by taking a subsequence 81 that converges at 0, 1/2, and I, Le., at all multiples of 1/2, to some limit values h(O), h(I/2), h(I). By throwing away the beginning of the subsequence, we may assume that all values are within 1/2 of the limit values at those three points. Next take a subsequence" 82 of 1.')1 that converges at all multiples x of 1/4 to some limit values h(x), such that all values are within 1/4 of the limit values at those five points.
In general, take a subsequenc~ SN of SN-l that converges at all multiples x of 1/2 N to some limit values h(x), such that all values are within 1/2 N of the limit values at those 2N + 1 points. Now we are ready to construct the desired subsequence of in. As the first term, 91, choose the first term of 81. For 92, choose a later term of 82. In general, for 9N, choose a still later term of 8 N . We claim that the sequence gn is uniformly Cauchy. Given E > 0, by equicontinuity choose a > 0 such that (1) Choose N such that
1/2 N <
(2) Suppose m, n
a
and
2/2 N < E/3.
> N and let x be any point. Choose k/2 N such that
(3) Then
19m(x) - 9n(x)1
~ 19m(X) - gm(k/2 N )1
< E/3 + 2/2 N
+ 19m(k/2N ) - 9n(k/2 N ) I + 19n(k/2N ) + E/3 < E/3 + E/3 + E/3 == E.
9n(x)1
(The estimates on the first and third terms follow from (1) and (3). The estimate on the second term follows from the construction of the 9n and
(2).) We have shown that the sequence 9n is uniformly Cauchy. In particular, the 9n converge pointwise to some limit h. Since the sequence is uniformly Cauchy, the convergence is uniform, and the limit h is continuous. Thus the sequence 9n, a subsequence of the original sequence in! converges in the
Exercises 30
135
space C[O,lJ (under the sup metric). Since F is closed, the limit h lies in F. Therefore F is compact. 0
30.3. The Ascoli-Arzela Theorem. Arzela generalized Ascoli's Theore;rn from a bounded interval to any compact metric space. Much work in analysis has centered on finding nice spaces of functions with good compactness properties.
Exercises 30 1. Give an example of an equicontinuous sequence of functions in C[O, 1]
which has no convergent subsequence.
2. Give an example of a bounded sequence of functions in C[O, 1] which has no convergent subsequence. 3. (Saroj Bhattarai and Brian Simanek). Prove the converse of Ascoli's Theorem.
Hint: For the hard part, equicontinuity, use proof by contradiction: assume that for some E > 0 there are points in [O,lJ and functions in f such that IX n - Ynl < lin but If(x n ) - f(Yn)1 ~ E. 4. Prove that the set of "Lipschitz" functions in C[O, 1J satisfying
If(x) - f(y)1
~
Glx -
yl
for some fixed constant C is equicontinuous and closed. 5. Existence of shortest paths. Let p, q be two points on a smooth compact connected surface S in JR3 (such as a curvy sphere or torus). A path from p to q in S of length L parameterized by arclength is in particular a function f: [O,lJ --> S c JR3 for which
If(x) - f(y)1 ~ Llx -
yl·
a. Prove that given a sequence of such paths with lengths converging to the infimum length L o, there is a convergent subsequence (in the sup metric). b. Assuming (as is true) that length is "lowersemicontinuous," i.e., that length lim fn ::; lim inf length fn, conclude that there is a shortest path from p to q.
Partial Solutions to Exercises
Chapter 1. 1.1. F, T, T, T. 1.2. Hint: It's easiest to answer, "For all x except... " 1.3. a, c. 1.9a. bijective.
Chapter 2. 2.1a. S is uncountable; otherwise JR., as the union of two countable sets, S and (Q, would be countable.
Chapter 3. 3.1. Converges to O.
-
137
138
Partial Solutions to Exercises
3.11. Given E > 0, choose N
>
liE. Then if n
> N,
Ian - 01 == Illn - 01 == lin < liN < E. 3.19a. JR. 3.21. False.
Chapter 4. 4.1. There are lots of examples, such as the greatest integer less ,than or equal to x or the characteristic function X71 of the integers. (Note that the exercise asks for a function defined on all of JR.)
f and g are continuous. By definition, given a point p, lim f(x) == f(p) and lim g(x) == g(p).
4.7. Suppose that
x~p
x~p
Hence, by Proposition 4.6(2), lim(f + g)(x)
x~p
Therefore
== f(P) + g(p)
=
(f
+ g)(P).
f + 9 is continuous.
Chapter 5. 5.1. Neither, closed, open, open. 5.5. {O, I}, (0,1), [0,1]. 5.10. Since "or" is always inclusive, it suffices to show that if p in S is not a boundary point, then p is an interior point. But if pinS is not a boundary point, then pES - as == int S. 5.11. First note that the interior of S is an open subset of S. Indeed, if p E int S, then by definition p r;. as, and hence some small ball B I about p is contained in S. A smaller ball B2 about·p is contained in S - as because if a boundary point of S were contained in B2, a point of Se would lie in B I . Hence, by definition int S == S - as is open. Now let T be any open subset of S. T cannot contain any points of as because a ball about such
Partial Solutions to Exercises
139
a point contains points not in S and hence not in T. Hence, T is contained in S - as == int S. Therefore, int S is the largest open set contained in S. 5.13. Suppose that pEaS is not an isolated point of S. Then a ball of radius lin about p contains another point an of S, and p is the limit of the sequence an. Therefore p is an accumuIati,?n point of S.
Chapter 6. 6.1. Suppose that 1 and 9 are continuous. Let converging to a point p. By 6.1(2),
I(x n )
-7
I(p)
and
g(xn )
Xn
-7
be a sequence of points
g(p).
Hence, by Proposition 3.6(2),
(/ + g)(xn ) = Therefore, by 6.1(2),
I(x n ) + g(x n )
1+ 9
-7
1(P) + g(P)
=
(I + g)(p).
is continuous.
6.3. The idea is that nearby integers have nearby values of f because there aren't any nearby integers!
Consider an integer p. To check definition (1), note that given any c, you can take 8 = 1, and then Ix - pI < 8 implies that x = p, so that I/(x) - f(P)1 == a < c. To check definition (2), note that any sequence of integers approaching p is eventually just p, p, p, . . .. Hence, the corresponding sequence of values is eventually just 1(P), 1(P), I(p), ... , which of course converges to f(p)· To check definition (3), just note that 1-1U contains {p}, which is the intersection of a small ball about p with the domain Z.
Chapter 7. 7.1.
1 0 g(x)
is x for part b, but not for part a.
Chapter 8. 8.3. Let an be an increasing sequence in R which is bounded above. Since it is increasing, it is bounded below by at. By Theorem 8.3, it has a convergent subsequence. Since the original sequence is increasing, its terms lie between
Partial Solutions to Exercises
140
terms of the subsequence and, therefore, it also converges to the same limit as the subsequence. 8.6. Let an be a Cauchy sequence. By Exercise 3.18, an is bounded. By Theorem 8.3, an has a subsequence converging to a limit L. We claim that an converges to L. ~Given £ > 0, choose N so that for some m > N, lam - LI < £/2 and for all m, n > N, Ian - ami < £/2. Then for all n > N,
Ian - LI ::; Ian - amI + lam - LI < £/2 + £/2 == £. 8.7. 1, +00, -00, 1, 2, 1.
Chapter 9. 9.1. Let S and T be compact subsets of rr;g,n. Since Sand T are both closed, S nTis closed. Since Sand T are both bounded, S nTis bounded. Therefore S nTis compact. 9.14. Immediately, from the definition, sup S ~ s for all s in S. If a < sup S, then there is a point of S larger than a, and hence a point of S larger than a, so that a fails to satisfy a ~ s for all sinS.
Chapter 10. 10.5. Since 9 is continuous, g(K) is compact by Theorem 10.1. Likewise since f is continuous, f(g(K)) == (f 0 g)(K) is compact. Of course, you need to assume that g(K) is contained in" the domain of f.
Chapter 11. 11.3. Suppose that f and 9 are uniformly continuous. Since continuous, given £ > 0, we can choose 01 > 0 such that
f
is uniformly
Since 9 is uniformly continuous, we can now choose 0 > 0 such that whenever Iy - xl < 0, then Ig(y) - g(x)1 < 01, which in turn implies that If(g(y)) f(g(x))1 < £, by (*) with Yl == g(y) and Xl == g(x). Thus
Iy - xl < 0 =}
I(f 0 g)(y) - (f 0 g)(x)1 < £,
Partial Solutions to Exercises .
141
and fog is uniformly continuous.
Chapter 12. 12.1. The disjoint open sets UI = (-00,1/2) and U2 = (1/2, (0) separate the integers into two nonempty pieces. 12.7. Suppose that f takes on two different values, Yl and Y2. Choose an irrational number Y3 between YI and Y2. Then the open sets UI = (-00, Y3) and U2 = (Y3, (0) separate f(R) into two nonempty pieces. This contradicts Theorem 12.4, which says that the continuous image of a connected set is connected. 12.8. By Proposition 12.2, Theorem 12.4, and Proposition 12.3, it must be an interval. By Theorems 10.1 and 9.2, it must be a closed interval. 12.11. First suppose that S is totally disconnected. By definition, S has at least two points. If S contained an interval, a separation of two points of the interval would prove the interval disconnected, a contradiction. Conversely, suppose that S has two points but no interval. Let PI, P2 be distinct points of S. Since S does not contain an interval, there is a point P3 between PI and P2 not in S. Then the open sets UI = (-00,P3) and U2 = (P3, (0) provide the required separation to show that S is totally disconnected.
Chapter 13. 13.1. Let a be a point of the Cantor set. Let an be one of the endpoints of the interval of Sn containing a. (If one of the endpoints is a, choose the other one.) Then an is in C and a is the limit of the sequence an because the length of the intervals goes to o.
Chapter 14. 14.3. Since [a, b] is compact (Theorem 9.2), every continuous image is compact (Theorem 10.1), and hence every continuous function has a maximum (Corollary 10.2). This fact is used in the proof of Rolle's Theorem (14.3), to find a place where the derivative vanishes. Rolle's Theorem leads immediately to the Mean Value Theorem (14.4) and finally to Corollary 14.5,
Partial Solutions to Exercises
142
which says that on an interval where I' is always 0, f is constant. This .final result will be a key ingredient in the proof of the FUndamental Theorem of Calculus (16.1).
Chapter 15. 15.1. [6, 10]. 15.4. Let I be a nonnegative function with Riemann integral equal to A. Chop the interval up into identical subintervals with 6.x small enough to guarantee that every Riemann sum is less than A + 1. Since every contribution is nonnegative, each subinterval contributes at most A + 1 to the Riemann sum. Therefore, on every subinterval I is bounded above by (A + l)/6.x. 15.6. An unbounded interval cannot be chopped up into finitely many small intervals. 15.7. Yes. There will always be just one subinterval on which the chosen I(x) could be 0 or 1, and as the subinterval width shrinks, the effect becomes negligible. Similarly, finitely many discontinuities are OK.
Chapter 16. 16.4. - I(a).
Chapter 17. 17.3. A simple hypothesis is that 17.8. Given
£
I
be bounded.
> 0, choose 8 such that
Ix - yl < 8 ==> If(x) -
l(y)1 < £/3.
Second, choose M such that liM < 8 and such that
e/M <
n > N => I/n(k/M) - I(k/M)I < £/3
£/3. Third, choose
(k = 0,1,2, ... , M).
N
Partial Solutions to Exercises
143
Now suppose that n > N. Given x, choose k such that and, hence, by the Lipschitz condition
Ix -
k/MI < l/M,
lfn(x) - fn(k/M) I < e/M < £/3. Now
Ifn(x) - f(x)1 ::; Ifn(x) - fn(k/M) I + Ifn(k/M) - f(k/lvI) I + If(k/M) - f(x)1
< e/M + £/3 + c/3 < £/3 + £/3 + c/3
==
c.
Chapter 18. 18.4. Compute by switching the order of integration. Justify by Fubini's Theorem because the integral of the absolute value is at most the integral of (7[/3)(10)(1) = 107[/3, which integral is (10)(7[/3)(107[/3) < (XL 18.8. In justifying the use of Leibniz's Rule, note that for y near 0,
a
1
IBy x2 + y2
I= I(x 2 +2yy2)2 I::; x14 = g(x),
whose integral even from 1 to 00 is finite.
Chapter 19. 19.1. Converges to 1/9.
Chapter 20. 20.1. 1, -1/3, [-00, +00]' +00.
Chapter 21. 21.1b. TRUE.
144
Partial Solutions to Exercises
Chapter 22. 22.2. sin x
+ a + a + ... , as you would guess.
(If you graph the functions sin x and cos nx or sin nx, you might see how things cancel to make the integral of the product vanish; or you could just believe the statement in 22.3 that "cos nx and sin n~ ... turn out to be orthonormal functions," see also Section 22.7; or later you could just use Exercise 24.2. 22.3. By integration by parts,
1 1 7r
o
1 xcos(nx) = 2[cos(n7r) - 1], 7r
o
7r
1 o
1 7r
o
n
7r xsin(nx) == - - cos(n7r), n
27r x 2 cos(nx) = 2 cos(n7r), n
7r2
x 2 sin(nx) = - - cos(n1r) n
2
+ a [cos(n7r) - 1]. n
Now use formulae 22.2(2).
Chapter 23. 23 • 1 • i7r (~Sin x - ..!... sin 3x + l52 sin 5x - ... ) 1~ 32 23.2. c~
(-& sin x cos t - b sin 3x cos 3t + b sin 5x cos 5t -
... )
23.5. First compute the Fourier series for f(t):
I. -+ 4
Chapter 24. No solutions.
L
oo
n=l
sin
n27r
cos nt + (1 - cos n7r
n;) sin nt
.
Partial Solutions to Exercises
145
Chapter 25. 25.1a. Easiest to use ratio test. 25 5
,",00
(_1)n+1 (x-1)n -
• • L...tn=1
n
-
(x _ 1) _
{x-1)2 2
+
(x-1)3 _ . _. 3
Chapter 26. 26.1a. r(5/2) = (3/2) r(3/2) = (3/2)(1/2) r(1/2) = (3/4)J7[.
Chapter 27. 27.1. Two are metrics. 27.2. Alexandra Constantine found 67. Here are a few. First "pat." Then by rule (1) "at." By rule (2) "spat," "phat," "pate," and "path." By rule (3) bat, cat, fat, hat, mat, rat, sat, tat, vat, pet, pit, pot, put, pad, pal, pan, par, paw, pax, payBy rule (4) "apt." Including gibberish, there are
1(pat)
+ 3 + (4 x 26 - 3) + 3 x 25 + 2 = 182.
(The -3 is from double counting ppat, paat, and patt.)
Chapter 28. 28.1. One converges and one diverges. 28.2. Suppose that there are two different limits L 1 and £2, and let p( £1, L2). Choose N such that n
Then
> N ==>
p(a n , L 1 )
< £/2 and
p(an , L2)
< £/2.
£
=
Partial Solutions to Exercises
146
a contradiction. To show that the sequence is bounded, choose N such that n
> N => p(an,L) < 1.
Let M = max{l, p(an, L): n ::; N}. Then all of the terms lie inside B(L, M). 28.7. open, closed, closed. 28.16. 1, 1.4, 1.414, ... (converging to
.J2 in lR) does not converge in Q.
28.17. A Cauchy sequence is contained in a ball, which is compact. Hence, some subsequence converges· to a limit L. Since the original sequence is Cauchy, it must converge to L. 28.18. Let fn be a Cauchy sequence of functions in 0[0,1]. For each point x in [0,1], fn(x) is a Cauchy sequence of real numbers, which converges to a limit f(x) because R is complete. In other words, fn converges to f
pointwise. To prove uniform convergence, given £ > 0, choose N such that m, n > N implies that Ifm(x) - fn(x)1 < £/2 for all x. Now given any x, choose m > N (depending on x) such that Ifm(x) -.·f(x)1 < £/2. It follows that for n > N, Ifn(x) - f(x)1 < £, uniform convergence. Finally, as a uniform limit of continuous functions, f is continuous. 28.19. yes.
Chapter 29. 29.3. The proofs of Theorem 10.1 and Corollary 10.2 hold in metric spaces.
Chapter 30. No solutions.
Greek Letters
{3
A B
1
r
Q
8 tl £ E ( TJ ()
Z H
e
I K K A A J.L M /.,
alpha beta gamma delta epsilon zeta eta theta iota ~ppa
lambda mn
N nu xi ~ 0 omicron 0 7f II pi p P rho a E sigma T T tan v T upsilon
phi X X chi 'ljJ W psi w n omega 1/
t.......I
-
147
Index
1-1 correspondence. ;
conlposition of functions, 35
abhfc\'iation~.
connected. -19 CtonstafJtitJ{~, Alexandra. 14[,
conditional convergencp. 89 6
con\~erJt("nCt). 89
abMJI ute
accllllltllntio.. point. 18. 12ti Al...Sabah, Nasger~ viii alternal ing Arnlstroll~.
serit.~.
87
Zan... 8 133. 135
A8Coli'~ Thoorem~
bull 8(a ,ooL 27 l
Bernou)Ji~ Danit'l~
105. ]09 Bhat.tarai. Saroj, 135 bije(;th..:.. ; Bohnhorst-, Kristin. 43
Holzano ·\\·ci~rstrti."'iSl 38. -IL 45. 48. 130 b()ul1da.r~·.
27 ~ 12•• a. :JM. 4 L 66~ 125 Burger. Ed! iv. viii
h()und~cl, 1
(.'[0, 1]. 122. 125. 127, 128. )30, 1:J3. u:i calculus! 59 (:anlor function! 6:1 Cantor set t 53. 81 CRntOt, C;""()fg, 11 Cflrtt.~il\.n prodU(·l. 10 Cauchy sequence~ 20, 39.~8, 66.125.134 characteristic function x~ 22, 81 . circle,27 closoo ~t8~ 28. 29. 12fi cl~ed sc..."ls. unions and inh'"rst"Ctions. 30
closure t 30. 126 compacl~
41.
45~
129 8t.i coJuplernenl. 3! 6. 27
(·ontains. ti cont inuous. 2 J l·ontinuou~ runctiuJ~t 33 cont raposit )\"f", !) converKe, l·t~ 80S. 93. 125 convergf\ absolutely. X!), 9:) ("onvt'rK(! {'onditioually! ~9 cun\'ergc! p()intvl.. i!k~1 75 cotlvtJrge unifortnl,Y. 75, 125 c(tnvers('. ti ('oc\'et ti. ('andice. 46 Costa. Torn. "iii countable~ 9 couulabl~ sulJcuver.. 42 countable- union. 10
dl~USC. 32 derivalive,6.
differcuct~. (i
difft-rt.'utiable. 61 differentiation of poV1icr M'riCfi, 95 dimension. 55 disc. 2; disconnected, totally, :,0 distance. 4. 121
dh-crge. 14. K5, g1) dOlnain.7 ()f.tluinalt!d (~ot,~rg:ence Theorenl. tsl ()unn~, FA. viii
(~(}In parisou t t:"St"
complet~..
125
t.arnail addrt'tis. ,"iii EnglL'ih, r.
-
149
In de;
150
~quicontinllous!
133
Euler's identity, 113 exponential funct.ion, 111 Fourier c.oefficients t 100 Fourier series, 9'9, 102, 109 frEt.Ctals, 53 Fubini'H Theorem, 82 functions, 7 FUndamental Theorem of Calculus, 71
Gamma Function r(x), 115 Garrity, Tom. viii geometric series, 86 Gibbs phenomenon, 110 greatest lower bound, 44 Greek let.ters, 147
Hadamard formula, 95 harmonic series, 86, 87 Heine---Borel, 41 ~ 45. 130 image,7 implication, 5 infimum,44 infinite sets, 9 injective, 7 integers,3 integrable, 6S. 67 interior, 30, 126 Interlned.iate 'lalue Theorem, 50 intersect, 6 intersections, 30, 43 interval, 27. 49 interval,,! 3 Inverse FUnction Tbeorem, 114 inverse image, 7, 8 irrationals, 3, 4. 8 isolated. point, 30, 126
JPEG.. 1Q2 least upper bound, 44 Lebesgue integral. 81 Lebesgue',. Dominated Convergence Theorem. 81 Leibniz, vii, 60 Leibniz's formula for 11", 97. 104 Leibniz's Rule, 82 length t 8 limit, 1. 13. 14, 18, 125 limit of function, 21 limit, unique, 15 linl inf, 39 limsup.39 Lipschitz. constant, 79 Lipschitz function, 79, 135
135 logarithm, 113 logic, 5 lowersemicontinuous. 135 Mandelbrot., Benoit, 56 maximum, 42, 44-.46, 61 Mean Value Theorem, 62, 63, 72 measurabiUtYt 82 measure, 54 Menger sponae, 56 metric space, 119, 121, 125, 129 metric, EucUdean, 4 metric, .up, 122, 125 metric, taxicab, 121 minimum, 42, 46, 61 Murphy, Erin, iv nat.ural numbers, 3 Newton, vii, 60, 106 nonintegrable functions, 66 o~t4>one, 7 onto, 7 open cover, iv, 41, 130 open sets, vii, 28, 29, 126 open sets, unions and intersections, 30 u or",5
p--test, 86
power series, 93 radius of convergence, 94 range, 7 rates of growtb, 18 ratio test, 91 rationals, 3, 29 real analytic, 94, 95 real numbers, I, 3 rMrTangement, 89 relatively open, 34 resonance, 107 Riemann integral, 65 Riemann sum, 65 root test. 95 self-similar, 55 sequence, 13, 125 sequence of functions. 75, 125 series, 86 series E(1/n 2 ), 101 sets. 6 shortest paths, 135 Sierpinaki's ~..arpett 56 Silva, Cesar, iv Simanek, Brian, 135 sine and cosine functions, 113 sphere, 27
Index
151
springs. 106 Stirlin~(s Approxin13tion to n~, 11 i Stok~'s ·rht~renl. ;2 strings. 105 subsequence, 37
subMet.
lj
supremunl. 4-.1 surjective, 7 SV,'ih:h linlit and integraL 7;, 81 s\\'itch order of integrat.ion. S2
Tacollla Narrows suspension bridge, 107 Tapp. Kris, viii
Taylor's fortnula. 95,96 topology, 21i.. 31. 126 trianttlt.' inequality, ·t. 121
uncountable t 1 L 54
uniform continuity, 4.7 unions, 30, 43 'IOIUnlCS of tl-balls. 115
Voss, R. F .• 56 W8\-e
equation. 108
wcbpagc. i\f. viii W~ierstrBSS .\1 ·tt.-sl, 93 word8 W, 122, 126. 127
