REAL ANALYSIS
REAL ANALYSIS FRANK MORGAN
AmmiucArr MATHEMATICAL SOCIETY
Providence, Rhode Island
2000 Mathematics Subject Classzficatzon. Primary 26-XX.
Front cover: The cover illustrates how continuous functions can converge nonuniformly 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 Morgan, Frank. Real analysis / Frank Morgan. p. cm. Includes index. ISBN 0-8218-3670-6 (alk. paper) 1. Mathematical analysis. I. Title.
QA300.M714 2005 515-dc22
2005041221
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Contents
Preface
vii
Part I. Real Numbers and Limits Chapter
1.
Numbers and Logic
Chapter 2. Infinity Chapter
3.
Sequences
Chapter 4. Functions and Limits
3 9 13
21
Part H. 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.
41
Compactness
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
v
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 E a7
85
Chapter 20.
Absolute Convergence
89
Chapter 21.
Power Series
93
Chapter 22.
Fourier Series
99
Chapter 23.
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
Part IV. Metric Spaces Chapter 27.
Metric Spaces
121
Chapter 28.
Analysis on Metric Spaces
125
Chapter 29.
Compactness in Metric Spaces
129
Chapter 30.
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 hundred 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
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.... I. The integers
Z = {... , -3, -2, -1, 0, 1, 2, 3, ... } (Z stands for the German word Zahl for number). The rationals Q _ {p/q in lowest terms: p E Z, q E N} _ {repeating or terminating decimals}. (Q stands for quotients). The reals I[8 = {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} _ Q6 = IR - Q.
1.2. Intervals in R. For a < b, define intervals
[a,b]={xEIR:a20=1.
n n = nl/n = (elnn)1/n = e(Inn)/n
-
e° = 1.
(The exponent (Inn)/n --40 because Inn N), we will be taking x close to some fixed target p, taking Ix - pl small. To say how small, we'll use the Greek letter before epsilon e, namely delta b, and require that I x - PI < S.
4.1. Definitions. We say limx_,p 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
0 < Ix-pI 0:
B(a,r) _ {Ix - al < r} (see Figures 5.1 and 5.2). Its boundary is the sphere:
8B(a, r) =fix - al = r}. In this case the boundary of S is part of S. In R2, we sometimes call the ball a disc and we usually call its boundary a circle rather than a sphere. In II81, 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
3D B(a,r)
Figure 5.1. The boundary of the ball B(a, r) is the sphere. The closed ball includes its boundary.
2D B(a,r)
1 D B(a,r)
a r Figure 5.2. The boundary of a 2D ball or "disc" is a circle. The boundary of a 1D ball or interval is two points.
As another example, consider all of R' with one point, the origin, removed:
S=R--{0}. Its boundary is the single point {0}. In this case, the boundary of S is not part of S. The boundary of the rationals Q is all of R. The boundary of R is the empty set.
5.2. Definition. A set S in R' is open if it contains none of its boundary points. A set S in R7z 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 So 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!
5.3 Proposition
29
CLOSED
OPEN
NEITHER
Figure 5.3. Some sets are closed or open but most are neither.
For example, the ball B (a, r) is closed. If you remove its boundary, the resulting set is open and is called an open ball, for which unfortunately we have no special symbol. R' - {0} is open. The rationals Q are neither open nor closed. The reals R 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 R'ti 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 So 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 p in S. 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 So (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 So contains some boundary point p of S, which is an accumulation point of S, so S does not contain all of its accumulation points.
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 Ua. Then x belongs to some Up. By Proposition 5.3, some ball about x is contained in Up, and hence in the union of all the Ua. 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 U2. 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 Ca. Then CC is the union of the
open sets CC, 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 CP, and hence open. Therefore C is closed.
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 cl S or S, is S U ,9S. An isolated point of S is the only point of S in some ball about it.
Exercises 5
31
.
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. Proof. Exercises 11 and 12.
5.7. Topology. In 1l? 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 T1 are open, closed, neither, or both. Give reasons.
a. [0,1); b. 7L;
_c. {xEIIB: sinx>0}, 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 IR the only open set containing Q? Prove your answer correct.
5. If S = [0,1), what are 8S, S, and S?
32
5. Open and Closed Sets
6. If S = 7G, what are 8S, S, and S?
7. If S = B(0,1) in lR2, what are 8S, S, and S? 8. Give an example of open sets U1 D U2 D n UZ is closed and nonempty.
such that the intersection
9. Give an example of nonempty closed sets C1 D C2 D intersection n Cz is empty.
such that the
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 RI?
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 definition, 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 function from Rn to R. The following are equivalent definitions of what it means for f to be continuous everywhere:
(1) For every point p, given E > 0, there exists S > 0, such that
Ix - pI < S=> If(x)-f(p)I 0, no small 5 bound on Ix-pI, for example S = 1/n, guarantees that If (X) - f (p) I < E. Thus there must be some sequence x, with Ixn -PI < 1/n and with If (xn) - f (p) I > 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 S > 0 such that
Ix - pI < S= If(x)-f(p)I IR satisfies f (0) = 0 and I f'(x) I < M. Prove that I f (x) I < MIx1. 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.
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 5.
.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 11 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.
15.1. The Riemann integral. The Riemann integral fQ f (x) dx of a function f over an interval [a, b] represents the area under the graph. The area may be approximated as in Figure 15.1 by chopping the interval up into narrow subintervals of perhaps variable thickness Ax, approximating each subarea by a skinny rectangle of height f (x), thickness Ax, and area AA = f (x) .x, and adding them up:
A :: E f (x) Ax. 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: fb
f (x) dx = lim E f (x) Ox. The limit must be independent 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 f is integrable and fb
f (x) dx = c(b - a). 65
15. The Riemann Integral
66
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 Ox.
Indeed, every Riemann sum
f(x)Ax=c>i x=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 0 off the rationals. Indeed, no matter how small Ax, 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) _
1// because no matter how small Ax, there are Riemann sums arbitrarily large, obtained by choosing to evaluate f (x) at a point very close to 0 on the first subinterval. In general, an integrable function 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 Ax - 0 is Cauchy. By Theorem 11.2, the continuous function f on the compact set [a, b] is uniformly continuous: given e > 0, there exists 5 > 0 such that (1)
JAXI 1, for all large n, IanI > 1, and the series diverges because the terms do not approach 0.
20. Absolute Convergence
92
Exercises 20 1. What are the possible values of rearrangements of the following series: a.
F°°
1 .
n=1 2n 1
b. 1:0 1 00
C. Ln=1
(an)n .
(-1)n . rn-
I
d. En= 1 ,' 1
Prove whether the following series 2-9 converge absolutely, converge conditionally, or diverge. Give the limit if you can.
i
2. E0n=0 5n 3.
00 n=0
4. E00 n=0
(-1)n 5n 5n I,
5. E00 o (-5 )n
i
6. E00 n=0 nn
00 (-1)n . 7. En=0 771/5
(-1)n 00 8. En=0 1+1/n'
9. V'00
n=0
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 E fn of functions converges (pointwise or uniformly) if the sequence of partial sums converges (pointwise or uniformly). The series E fn converges absolutely if > IfnI converges absolutely. There is a nice test due to Weierstrass for uniform convergence, essentially by comparison with a series of constants.
21.2. Weierstrass M-test. Suppose Ifn I < Mn where each Mn is a positive number and E Mn converges. Then E fn converges uniformly. Proof. For each x, E fn(x) converges by comparison with E Mn to some f (x) To prove the convergence uniform, note that 00 00 N N fn(x)
- > fn(x)
fn(x) - f(x)
?1m
n=1
n=1
n=N+1
E Mn,
n=N+1
which is small for N large, independent of x. For example, comparison with
Sly converges uniformly on R by Weierstrass M-test . By Theorem 17.3, the limit is continuous.
21.3. Definition. A power series is a series E anxn of multiples of powers of x, such as 00 xn=1+x+-x2+6x3+...
n! 93
21. Power Series
94
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 E anxn has a radius of convergence 0 < R < oo, such that (1) for IxI < R, the series converges absolutely;
(2) for IxI > R, 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 integrated term by term. Note that this result admits the possibilities that the series always converges (the case R = oo) 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 IxI < Ixol. Since it converges at xo, the terms anx0 converge to 0; in particular, I anx0 j < C. Therefore lanxnl < CIx/xoln,
and E anxn converges absolutely by comparison with the geometric series >CIx/xo1n. Next we prove uniform convergence on [-Ro, Ro]. For some leeway, choose R1 between Ro and R. Since the series converges for x = Rl; the terms anRl converge to 0; in particular, I anRl I < C. Hence for all IxI < Ro,
Ianxn) < C(Ro/Rl)".
To apply the Weierstrass M-test, let Mn = C(Ro/Rl)n; then E Mn is a convergent, geometric series. By Weierstrass, E anxn 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.
21.10 Taylor's formula
95
For example,
[x++++j
1/2'
x2
J
x3
x4
11/2 o
1
1 1 1 2+222+233+244+...
21.5. Hadamard formula. The radius of convergence R of a power series E anxn is given by the formula 1
lim sup I an I
1/n'
under the agreement that 1/0 = oo and 1/oo = 0.
Proof. Using the Root test (20.5), we have p = limsup IanxnI1/n = lxl R Hence by the Root test, the series converges absolutely if x1 < 1, i.e., if x < R, and diverges if x R > 1, i.e., if lx > R. Therefore, RRis the radius of convergence.
21.6. Corollary. Given a power series E anxn, the "derived series" E nanxn-1 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: > nanxn. Since lim n1/n = 1, by Hadamard's theorem the radius of convergence is the same as for anxn.
21.7. Differentiation of power series. Inside the radius of convergence, a power series may be differentiated term by term:
(ax7) = > nanxn,-1
Proof. On an interval [-Ro, Ro] C (-R, R), E anxn converges uniformly to some f (x), and the derived series > nanxn-1 converges .uniformly to some g(x). Since g can be integrated term by term, f xRo g = f (x) - f (-Ro). By the Fundamental Theorem of Calculus, g(x) = f'(x), as desired.
21.8. Corollary. A real-analytic function is infinitely differentiable (C°°). 21.9. Corollary. Inside the radius of convergence, you can antidifferentiate a power series term by term.
21.10. Taylor's formula. If f (x) _ > anxn, then an denotes the.nth derivative.
where f (n)
21. Power Series
96
Proof. f(x) = ao + aix + a2x2 + a3x3 + , f(0) = ao; f'(0) = 1a1; f'(x) = a1 + 2a2x + 3a3x2+--- , f"(x) = 2a2 + 3.2a3x + , f"(0) = 2a2; f-'(0) 3!a3+--- , f'11(x) = = 3! a3; and so on.
Caveat. Given a C°° function f , Taylor's formula yields an associated Taylor Series. 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. In a more advanced course, you will see examples of this strange behavior.
21.11. Power series about c
0. Power series j anx' are best when c, we could use E an (x - c)". There are only x is near 0 (x ti 0). For x minor changes to the theory. The interval of convergence now goes from c - R to c + R, and Taylor's formula becomes an =
(1)
f (n) (c) n1
For example, f (x) = 1/x has no power series about 0 because it is not even defined there. However, in terms of u = x - 1, (2)
1 = X
1 1+u
= 1 - u + u2 - u3 + u4 -...
(for Jul < 1)
=1-(x-1)+(x-1)2-(x-1)3+(x-1)4-...
(for Ix-11 °°_1 . a. Use the Ratio test to show that the radius of convergence R = 1. b. Use the Hadamard formula to show that R = 1.
It follows that the series converges uniformly on every interval [-Ro, Ro] C (-1, 1). Use the Weierstrass M-test to show that it actually
c.
converges uniformly on [-1, 1].
Exercises 21
97
3. Continuing Exercise 2, consider g(x) = x f'(x).
E'
a. Compute that g(x) = 1 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 jxj < 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
a. f(x) = eS; b. f(x) = x2.
5. Leibniz's formula for 7r.
a. Justify that for lx i < 1, 1 1
1+x2
-x2+x4-x6+x8-...
b. Justify that for I xI < 1, x5 x3 tan lx=x-3 + - 7 -} 5
x7
x9
-...
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 = 1 (as it does), deduce Leibniz's famous formula for 7r: 4
4
4
4
4 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 1/x in powers of u = x - 2 by noting that
1' x
1
2+u
_
1
1
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, 7r] 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 [-7r, 7r] can be decomposed as an infinite series in terms of sins and cosines, called its Fourier series. This 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 function f (x) on [-ir, -7r] is given by a Fourier series 00 (1)
f(x) = Ao + E (An cos nx + Bn sinnx),
where 1
Ao = 1(2)
An = Bn
1
ff
J
f (x) dx, cos nx dx,
fsinnxdx,
except that at the endpoints the series converges to the average of f (-ir) and f(70 99
22. Fourier Series
100
Outside [-ir, -7r], the Fourier series just keeps repeating every 27r, so it doesn't always equal f, unless f is also 21r 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 Bn, due to Euler, deserve comment. First note that A0 is just the average value of f (x), about which f varies or oscillates. The other formulas are best understood by comparison with formulas for the coefficients of a vector
v = vlel + v2e2 + ... +. vNeN
(1)
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 vn satisfy (2)
by just taking the dot product of each side of (1) with en. Analogously, on. the space of continuous functions on [-7r, 7r], one can define a dot product by 1
fg=
(3)
7r
f
7r
f (x) g(x) dx, 7r
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 0
if-ir 0 and n, for x large, exp(x) > Cx".
Proof. Conclusion (1) follows because the derivative exp(x) is positive. Conclusion (2) follows from the defining series, with e,, = 1/n!. In particular, given C > 0, for x > 0, exp(x) > (1/(n + 1)!)xn+1, which is greater than Cxt once x > C(n + 1)!.
Exercises 25
113
25.7. The natural logarithm. It follows from Corollary 25.6(1) that exp(x) is as bijective map from R onto the positive reals. The inverse function from the positive reals onto R 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)' = 11x, 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 i2 = -1, i3 = -i,
i4 = 1.....
Exercises 25 1. Development of the sine and cosine functions. Define x2
cosx=
2
+
x4
x6
4! x5
6!
sinx=x - x33 + 5 -
x7 --
+
x8 8!
-,
+x99 - .
a. Check that the radii of convergence are oo, so that these functions are defined for all x.
b. Show that cos(-x) = cosx and that sin(-x) sin(x). c. Show that (sin x)' = cosx and that (cosx)' sin x. 2 d. Prove that sin x + cost x = 1. Hint: First show the derivative 0, so that sin 2 x + cost x = C. Then use
x=0tofind C. Note that this implies that I sin x < 1 and I cos x < 1. 2. By plugging into the series for ex, verify Euler's identity ez0
= cos 0 + i sin 0.
3. Use Euler's identity and the fact that ei(a+b) = eiaeib to deduce that sin(d + b) = sin a cos b + cos a sin b,
cos(a + b) = cos a cos b - sin a sin b.
25. The 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 = log x, then x = ey. Differentiate implicitly to deduce that (log x)' = 1/x. b. Prove that log cx = log c + log x. 5. Obtain a power series for logx in powers of x - 1 by integrating 21.11(2). Check Taylor's formula (21.11(1)) for this series. 6.
Prove that e = limn. + (1 + 1/n)'. (This is sometimes used as the
definition of e.)
Hint: It suffices to show that
log(1 + 1/n)n = n log(1 + 1/n) -p 1. By the definition of the derivative, log(1 + Ox) log' lim g( 1) = Ix->O Ox Therefore, taking Ox = 1/n, nlog(1 + 1/n) -* log'(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(0, r) in I[8": IT-/2
vol B (0, R) =
(n/2)
Rn.
For example, the "volume" or area of a ball in R2 is (222), r2 = 'irr2. The volume of a ball in R4 is 2ir2r4. However, according to this formula, the "volume" or length of a ball in R1, which should be 2r, is (%2), r. What is (1/2)! ? For our formula to work, we would need (1/2)! = -vfir/2.
Fortunately, there is a nice extension of (x-1)! from integers to real numbers greater than 1, called the Gamma Function r(x).
26.1. Definition. For x > 1, define the Gamma Function r(x) by
r (x) =
f
et1 dt.
This integral converges because et grows faster than any power of t.
26.2. Proposition. r(D; + 1) = xr(x).
Proof. Using integration by parts, f u dv = uv - f v du, with u = t', dv = e-tdt, du f= xtx-1, v = -e-t, 00
00
r(x + 1) =
J0
e-ttX dt = -e-ttx1o 00 +
J0
xe-ttx-1 dt = 0 + xr(x).
a
115
26. Volumes of n-Balls and the Gamma Function
116
26.3. Proposition. For every nonnegative integer n, F(n + 1) = n!.
-
We can use this relationship to extend the definition of the factorial function to every nonnegative real number by x! = F(x + 1).
Proof by induction. First we note that for n = 0,
I'0+1 =I'1 =
e-tdt=-e-t°°- 0- (-1) =1=0!.
Second, assuming the result for smaller values of n, we see by Proposition 26.2 that F(n + 1) = n r (n) = n(n - 1)! = n!.
26.4. Proposition. F(1/2) _ V'7r. Proof. By making the substitution t = s2, dt = 2s ds, we see that
F(1/2) =
f
e-tt-1/2 dt =
es2s ds = J 00 eds. J z
z
0
Now comes the famous trick-multiplying r(1/2) by itself and switching to polar coordinates with dA = r dr dB:
r(1/2) I'(1/2) = 00
jO27r
=
:;;:
:1::
=
oo
=oJr=o
1
z
e-T r dr dO = 21r --e-T 2
°°
z
= 7r. Jn
Therefore 1'(1/2) = V?-r.
In particular, (1/2)! = 1'(3/2) = (1/2)1'(1/2) = //2, as we hoped. Note too that our formula gives for the volume of a ball in R3: X3/2
x.3/2
4
vo1B(0,r) =
(3/2)!r3 the correct, familiar formula.
26.5. Proposition. The volume of a ball in I18' is given by rn/2
volB(0,R) = (n/2)[Rn 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 {x2 + x2 + x2 +
+
x2 < R2} as a 2D ball {r2 = xi + x2 < R2} of (n - 2)D balls of radius s satisfying s2 x3 + + xn = R2 - r2 and volume (by induction) ?r(n,-2)/2
n-2 = ((n/2) - 1)!' n-2 7r(n/2)-1
((n - 2)/2)!
s
Hence,
(n/2)-1 vol B (0, R) =
Ifsn-2 dA
((n/2) - 1)t 7r(n/2)-1
((n/2) - 1)! ((n/2) - 1)!
j 27r
R
(R2
- r2)(n.-2)/2r dr dO
=0 r=0 1 (R2 - r2)f/2
-2
R
n/2
1 Rn
, (n/2)-1
_ ((/2) n
2r
- 1) 2/2 n = (n/2) Rn
0
7rn/2
27r
.
26.6. Corollary. The volume of the (n - 1)D sphere in 1R is given by vol S(0, R) = n
7r
n/2
(n/2)
Rn
For example, the circumference of a circle in R2 is given by 2R = 27rR.
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).
26.7. Stirling's Approximation. The Gamma function may be used to derive an excellent asymptotic approximation to n! n!
27rn(n/e)n.
(Technically this asymptotic symbol - means that as n -> oo, 1.) It follows that e n monh -+ mone -' money).
Exercises 27 Is each of the following a metric on the space of continuous functions from [0, 1] to R? If not, explain why not. 1.
a. p(f, g) = If (1/2) - g(1/2) I b. p(f, g) = sup{If'(x) - g'(x)I}.
c. p(f, 9) = fo I f (x) - 9(x) I dx.
d. P(f 9) = f o I f (x) - 9(x) I2 dx. e. p(f, g) _
How many words can you find in B ("pat" ,1)? How many elements (counting words and gibberish) are there in B ("pat", 1)? 2.
Chapter 28
Analysis on Metric Spaces
Analysis on metric spaces is very similar to analysis in R. 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 e > 0, there is some N such that whenever n > N, p(an,L) < E.
In C[0,1], fn -f f under the sup metric therefore means that fn -p f uniformly. So fn(x) = x/n converges to f (x).= 0, but fn(x) = xn does not converge (uniformly) to f (x) = 0. Most of the results on sequences of Chapter 3 remain true in metric spaces. The exception is Proposition 3.6 because in a general metric space there is no addition, multiplication, or division. 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 e > 0, there is some N such that whenever m, n > N, p(a,n, an) < E. 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
in IR, does not converge in Q, even though it is still Cauchy.
The problem is that
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. 11
Topology on a metric space X is very similar to topology in R.
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 S and So. The set of boundary points of S is called the boundary of S and written 8S. A set S is open if it contains none of its boundary points. A set S in Rn 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 - S. The closure of S, denoted cl S or S, is S U 8S. 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 C[O, 1], does the sequence f, converge or diverge (under the sup metric)? If it converges, what is the limit? a. f,,(x) = x2 /n; b. fn(x) = x2n
2. Suppose that a sequence a,, 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 So 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 C[O, 1].
a. If (1/2) < 6}; b. {f(1/2) < 6};
c. B(0,1).
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 = {0 < f (1/2) < 1} C C[0,1], what are 8S, S, and S? 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 R 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 ]R°° denote the space of all real vectors (X1, x2, X3.... ),
with infinitely many components, within finite distance of the origin:
R' _{(mil, x2, X3, ...) : X1
X2
X 3+-'' < oo}
with metric
1/2
P(x, y)
y'')2)
This space is a lot like R'. The unit ball B(0,1) is closed and bounded, but unlike R", the unit ball B(0,1) admits sequences with no convergence subsequences. For example, let
el = (1,0,0,0,...), e2 = (0,1,0,0,...), e3 = (0, 0,1, 0, ... )
and so on. This sequence has no convergent subsequence because it has no apart. The problem Cauchy subsequence because all terms are distance is that R°° is infinite dimensional. 129
29. Compactness in Metric Spaces
130
29.2. Example. Similarly, consider the unit ball B(0, 1) in C[0,1]. It is closed and bounded, but admits sequences (of functions) with no (uniformly) convergent subsequences. (Recall that convergence under the sup metric of
C[0,1] means uniform convergence.) For example, let fn (x) = x7L, as in Figure 17.1. Then any subsequence converges pointwise to
f(X)-
] 0 for0 0, there is a 5 > 0 such that
y - xI < b= If(y)-f(x)I
