Charles Chapman Pugh
Real Mathetnatical Analysis
With 133 Illustrations
�Springer
Charles Chapman Pugh Mathematics Department University of California at Berkeley
Berkeley, CA 94720-3840 USA
Editorial Board s ruder
Mathematics Department San Francisco State
F. W. Gehring
K.A. Ribet
Mathematics Department
Mathematics Department
East Hall
University of California,
University of Michigan
University
Berkeley
San Francisco, CA 94132
Ann Arbor, MI 48109
Berkeley. CA 94720-3840
USA
USA
USA
Mathematics Subject Classification (2000): 26-01
Library of Congress Cataloging-in-Publication Data Pugh, C.C. (Charles Chapman), 194Q-Real mathematical analysis/Charles Chapman Pugh. p.
em.- (Undergraduate texts in mathematics)
Includes bibliographical references and index. ISBN 0-387-95297-7 (alk. paper)
L Mathematical analysis.
QA300.P994 515
2001
L Title.
IL Series. 2001032814
dc21
Printed on acid-free paper © 2002 Springer-Verlag New York. Inc. All rights reserved. This work may not be translated or copied in whole or in part without the written permission of the publisher (Springer-Verlag New York, Inc., 175 Fifth Av enue, New York,
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scholarly analysis. Use in connection with any form of information storage and retrieval,
electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed is forbidden. The use of general descriptive names, trade names, trademarks, etc., in this publication, even if the former are not especially identified, is not to be taken as a sign that such names, as understood by the Trade Marks and Merchandise Marks Act, may accordingly be used freely by anyone. Production managed by Michael Koy; manufacturing supervised by Erica Bresler. Formatted from the author's MI'JEX2e files by 'JEXniques, Inc., Cambridge, MA. Printed and bound by Maple-Vail Book Manufacturing Group, York, PA. Printed in the United States of America. 9
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ISBN 0-3R7-95297-7 Springer-Verlag
2 SPIN 10838675
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A member of BertelsmannSpringer Science+ Business Media GmbH
To the students who have encouraged me -especially A. W., D.H., and M.B.
Preface
Was plane geometry your favorite math course in high school? Did you like proving theorems? Are you sick of memorizing integrals? If so, real analysis could be your cup of tea. In contrast to calculus and elementary algebra, it involves neither formula manipulation nor applications to other fields of science. None. It is pure mathematics, and [ hope it appeals to you, the budding pure mathematician. Berkeley, California, USA
CHARLES CHAPMAN PUGH
Contents
1
Real Numbers
1 2 3 4 5* 6*
Preliminaries Cuts . . . . . Euclidean Space . Cardinality . . . Comparing Cardinalities The Skeleton of Calculus Exercises . . . . . .
.
2
A Taste of Topology
1 2 3 4 5 6* 7*
Metric Space Concepts Compactness Connectedness Coverings . . . Cantor Sets Cantor Set Lore Completion Exercises . . .
.
.
1 1 LO 21 28 34 36 40
51 51 76 82 88
95 99 108 115
Contents
X
3
1 2 3
4
139
Functions of a Real Variable
139
Differentiation . . . . Riemann Integration Series . . Exercises
1 54 1 79 1 86
Function Spaces I
2 3 4 5 6* 7* 8*
Uniform Convergence and C0[a, b] Power Series . . . . . Compactness and Equicontinuity in C0 . Uniform Approximation in C0 Contractions and ODE's . . . . . . . . Analytic Functions . . . . . . . . . . . Nowhere Differentiable Continuous Functions . Spaces of Unbounded Functions Exercises . . . . . .
.
5 Multivariable Calculus l Linear Algebra . . 2 Derivatives . . . . Higher derivatives . 3 4 Smoothness Classes . Implicit and Inverse Functions 5 6* The Rank Theorem 7* Lagrange Multipliers Multiple Integrals . . 8 Differential Forms . . 9 LO The General Stokes' Formula . 1 1 * The Brouwer Fixed Point Theorem . Appendix A: Perorations of Dieudonne . Appendix B : The History of Cavalieri 's Principle Appendix C: A Short Excursion into the Complex Field . . . . Appendix D: Polar Form Appendix E: Determinants Exercises . . . . . . . . . .
201
20 1 21 1 213 217 228 235 240 248 25 1 267
267 27 1 279 284 286 290 296 300 313 325 334 337 338 339 340 342 345
xi
Contents 6
363
Lebesgue Theory
1
2
3 4
5
6 7 8 9
Outer measure Measurability Regularity . Lebesgue integrals Lebesgue integrals as limits . Italian Measure Theory . . . Vitali coverings and density points Lebesgue's Fundamental Theorem of Calculus . Lebesgue's Last Theorem . . . . . . Appendix A: Translations and Nonmeasurable sets . Appendix B: The Banach-Tarski Paradox . . . Appendix C: Riemann integrals as undergraphs Appendix D: Littlewood's Three Principles Appendix E: Roundness . Appendix F: Money . Suggested Reading Bibliography Exercises . .
.
.
.
.
.
.
.
.
Index
.
.
.
.
363 367 37 1 376 383 387 391 396 40 1 407 409 409 41 1 412 413 414 415 417 431
1
Real Numbers
l
Preliminaries
Before discussing the system of real numbers it is best to make a few general remarks about mathematical outlook. Language
By and large, mathematics is expressed in the language of set theory. Your first order of business is to get familiar with its vocabulary and grammar. A set is a collection of elements. The elements are members of the set and are said to belong to the set. For example. N denotes the set of natural numbers, 1, 2, 3 , The members of N are whole numbers greater than or equal to I. Is 10 a member of N? Yes, 1 0 belongs to N. Is rr a member of N? No. Is 0 a member of N? No. We write .
.
.
.
x EA
and
yrf.B
to indicate that the element x is a member of the set A and y is not a member of B. Thus, 68 1 9 E N and .J2 rf. N. We try to write capital letters for sets and small letters for elements of sets. Other standard sets have standard names. The set of integers is denoted by Z, which stands for the German word zahlen. (An integer is a positive
2
Chapter 1
Real Numbers
whole number, zero, or a negative whole number.) Is v'2 E Z? No. v'2 f!J. Z. How about - 1 5 ? Yes, -15 E Z. The set of rational numbers is called Q, which stands for ''quotient." (A rational number is a fraction of integers, the denominator being nonzero.) Is v'2 a member of Q? No, Ji does not belong to Q. Is JT a member of Q? No_ Is 1 .4 1 4 a member of Q? Yes. You should practice reading the notation "{x E A :"as "the set of x that belong to A such that." The empty set is the collection of no elements and is denoted by 0. Is 0 a member of the empty set? No, 0 r:J_ 0. A singleton set has exactly one member. It is denoted as {x } where x is the member. Similarly if exactly two elements x and y belong to a set, the set is denoted as {x. y } . I f A and B are sets and each member o f A also belongs to B then A i s a subset of B and A is contained in B. We write t
A c B. Is N a subset of Z? Yes. Is it a subset of Q? Yes. If A is a subset of B and B is a subset of C , does it follow that A is a subset of C ? Yes. Is the empty set a subset of N? Yes, 0 c N. Is I a subset of N? No, but the singleton set { 1 } is a subset of N. Two sets are equal if each member of one belongs to the other. Each is a subset of the other. This is how you prove two sets are equal: show that each element of the first belongs to the second, and each element of the second belongs to the first. The union of the sets A and B is the set A U B, each of whose elements belongs to either A, or to B, or to both A and to B. The intersection of A and B is the set A n B each of whose elements belongs to both A and to B If A n B is the empty set then A and B are disjoint. The symmetric difference of A and B is the set AD..B each of whose elements belongs to A but not to B, or belongs to B but not to A . The difference of A to B is the set A \ B whose elements belong to A but not to B. See Figure 1 . A class is a collection of sets. The sets are members of the class For example we could consider the class£ of sets of even natural numbers. Is the set {2, 1 5 } a member of£? No. How about the singleton set {6}? Yes. How about the empty set? Yes, each element of the empty set is even. When is one class a subclass of another? When each member of the former belongs also to the latter. For example the class T of sets of positive integers divisible by 10 i s a subclass of£, the class of sets of even natural _
t W hen �orne mathematicians write A c 8 they mean that A is not adopt this convention_ We accept A cA.
a
subset of B, but A f= B- We do
Section 1
Preliminaries
3
Figure 1 Venn diagrams of union, intersection, and differences.
numbers, and we write T c £ . Each set that belongs to the class T also belongs to the class £. Consider another example. Let S be the class of singleton subsets of N andV be the class of subsets of N each of which has exactly two elements. Thus { 1 0} E Sand {2, 6} E V. IsS a subclass ofV? No. The members of S are singleton sets and they are not members ofV. Rather they are subsets of members ofV. Note the distinction, and think about it. Here is an analogy. Each citizen is a member of his or her country - I am an element of the USA and Tony Blair is an element of the UK. Each country is a member of the United Nations. Are citizens members of the UN? No, countries are members of the UN. In the same vein is the concept of an equivalence relation on a set S. It is a relation s "' s' that holds between some members s , s' E S and it satisfies three properties : For all s , s', s" E S (a) S"'S. (b) s s' implies that s' s. (c) s "' s' ,....., s" implies that s "'"' s". The equivalence relation breaks S into disjoint subsets called equiva lence classest defined by mutua] equivalence: the equivalence class con taining s consists of all elements s' E S equivalent to s and is denoted [s ] . The element s is a representative of its equivalence class. See Figure 2. Think again of citizens and countries. Say two citizens are equivalent if they are citizens of the same country. The world of equivalence relations is egalitarian: I represent my equivalence class USA just a s much as does the President. rv
rv
Truth
When is a mathematical statement accepted as true? Generally, mathemati cians would answer "Only when it has a proof inside a familiar mathematical t The phrase "equivalence class" is standard and widespread, although it would be more consistent with the idea that a class is a collection of sets to refer instead to an "equivalence set."
4
Chapter
Real Numbers
s
[t J
I
•
[s)
=
1
[s']
Figure 2 Equivalence classes and representatives.
framework."A picture may be vital in getting you to believe a statement. An analogy with something you know to be true may help you understand it. An authoritative teacher may force you to parrot it. A formal proof, how ever, is the ultimate and only reason to accept a mathematical statement as true. A recent debate in Berkeley focused the issue for me. According to a math teacher from one of our local private high schools, his students found proofs in mathematics were of little value, especially compared to "convincing arguments."Besides, the mathematical statements were often seen as obviously true and in no need of formal proof anyway. I offer you a paraphrase of Bob Osserman ' s response. But a convincing argument is not a proof. A mathemati cian generally wants both, and certainly would be less likely to accept a convincing argument by itself than a formal proof by itself. Least of all would a mathematician accept the pro posal that we should generally replace proofs with convincing arguments. There has been a tendency in recent years to take the notion of proof down from its pedestal. Critics point out that standards of rigor change from century to century. New gray areas appear all the time. Is a proof by computer an acceptable proof? Is a proof that is spread over many journals and thousands of pages, that is too long for any one person to master, a proof? And of course, venerable Euclid is full of flaws, some filled in by Hilbert, others possibly still lurking.
Section
I
Preliminaries
5
Clearly it is worth examining closely and critically the most ba sic notion of mathematics, that of proof. On the other hand, it is important to bear in mind that all distinctions and niceties about what precisely constitutes a proof are mere quibbles compared to the enormous gap between any generally accepted version of a proof and the notion of a convincing argument. Compare Euclid, with all his flaws to the most eminent of the ancient exponents of the convincing argument - Aristotle. Much of Aristotle's reasoning was brilliant, and he certainly convinced most thoughtful people for over a thousand years. In some cases his analyses were exactly right, but in others, such as heavy objects falling faster than light ones, they turned out to be totally wrong. In contrast, there is not to my knowledge a single theorem stated in Euclid's Elements that in the course of two thousand years turned out to be false. That is quite an astonishing record, and an extraordinary validation of proof over convincing argument. Here are some guidelines for writing a rigorous mathematical proof. See also Exercise 0. 1 . Narne each object that appears in your proof. (For instance, you might begin your proof with a phrase, "consider a set X, and elements x , y that belong to X ," etc.) 2. Draw a diagram that captures how these objects relate, and extract logical statements from it. Quantifiers precede the objects quantified: see below. 3 . Proceed step-by-step. each step depending on the hypotheses, previ ously proved theorems, or previous steps in your proof. 4. Check for "rigor" : all cases have been considered. all details have been tied down, and circular reasoning has been avoided. 5 . Before you sign off on the proof, check for counter-examples and any implicit assumptions you made that could invalidate your reasoning. Logic
Among the most frequently used logical symbols in math are the quantifiers V and 3. Read them always as "for each" and "there exists." Avoid reading V as "for all," which in English has a more inclusive connotation. Another common symbol is => . Read it as "implies."
Chapter
Real Numbers
6
I
The rules of correct mathematical grammar are simple: quantifiers appear at the beginning of a sentence, they modify only what follows them in the sentence, and assertions occur at the end of the sentence. Here is an example. (1)
For each integer n there is a prime number p which is greater than n. In symbols the sentence reads
Vn E Z 3p E P
such that
p
>
n,
where P denotes the set of prime numbers. (A prime number is a whole number greater than 1 whose only divisors in N are itself and 1 .) In English, the same idea can be re-expressed as (2)
Every integer is less than some prime number
or
A prime number can always be found
(3)
which is greater than any given integer.
These sentences are correct in English grammar, but disastrously WRONG when transcribed directly into mathematical grammar. They translate into disgusting mathematica1 gibberish: Vn E Z n < p 3p E P (WRONG 2) 3p E P p>n \:In E Z. (WRONG 3) Moral Quantifiers first and assertions last. In stating a theorem, try to apply the same principle. Write the hypothesis first and the conclusion second. See Exercise 0.
The order in which quantifiers appear is also important. Contrast the next two sentences in which we switch the position of two quantified phrases. (4)
(Vn E N)
(Vm E N)
( 3p E P)
such that
( nm < p ) .
(5)
(Vn E N)
(3p E P)
such that
(Vm E N)
(nm
0 and each x E [a , b] there is a 8 > 0 such that
t E [a , b] and I t - x i See Figure 1 8.
c, contrary to the fact that c is an upper bound for X. Thus, c = b, b E X, and the values of f form a bounded subset of R 0
A continuous function f defined on an interval [a , b ) takes on absolute minimum and absolute maximum values: fo r some x0 , x 1 E [a , b] and for all x E [a , b], f (xo ) ::::: f (x ) ::::: f (xi ) . 23 Theorem
Proof Let M = 1. u. b . f (t) as t varies in [a , b] . By Theorem 22, M exists. Consider the set X = {x E [a , b] : 1. u. b. Vx < M} where, as above, Vx is the set of values of f ( t) as t varies on La , x].
Case l . f (a ) = M . Then f takes on a maximum at a and the theorem is proved. Case 2. f ( a) < M. Then X I 0 and we can consider the least upper bound of X , say c. If f ( c ) < M, we choose f' > 0 with E < M f ( c) . By continuity, there exists a 1J > 0 such that i f - c i < 1J implies i f ( t ) - f ( c ) i < E . Thus, 1. u. b. Vc < M . If c < b this implies that there exist points t to the right of c at which 1. u. b. V1 < M, contrary to the fact that c is an upper bound of such points. Therefore, c = h which implies that M < M, a contradiction. Having arrived at a contradiction from the supposition that f (c) < M, we duly conclude that f (c) = M , so f assumes 0 a maximum at c. The situation with minima is similar.
A continuous function defined on an in terval [a , b] takes on (or "achieves," "assumes, " or "attains ") all inter mediate values: if f (a ) = a, f (b ) = /3. and Y is given. a ::::; Y ::::; {3, then there is some c E [a , b] such that f(c) = Y . The same conclusion holds if 24 Intermediate Value Theorem
f3 ::::: y ::::: a .
The theorem is pictorially obvious. A continuous function has a graph that is a curve without break points. Such a graph can not jump from one height to another. [t must pass through all intermediate heights. ::::;
Y } and c = 1 . u. b . X. Now c exists because X is non-empty (it contains a ) and it is bounded above (by b). We claim that f ( c ) = Y , as shown in Figure 20. To prove it, we just eliminate the other two possibilities, f ( c ) < Y and f ( c ) > Y , by showing that each leads to a contradiction. Suppose that f ( c ) < Y and take E = Y - f (c) . Continuity gives 8 > 0 such that I t - c l < 1J implies i f ( t ) - f (c) i < E . That is, Proof Set X =
{x E [a , b) : 1. u. b . Vx
t E (c - li, c + li)
:::::}
f (t)
O * , what are C and D ? (b) If X < o * , what are they? (c) Prove that x uniquely determines y . 1 1 . Prove that there exists n o smallest positive real number. Does there exist a smallest positive rational number? Given a real number x , does there exist a smallest real number y > x ? 1 2. Let b = I . u. b. S, where S is a bounded nonempry subset of JR. (a) Given E > 0 show that there exists an s E S with b-E �
S
� b.
(b) Can s E S always be found so that b - E < s < b? (c) If x = A l B is a cut in Q, show that x = I. u. b . A. 13. Prove that -J2 E lR by showing that x x = 2 where x = A l B is the cut in Q with A = {r = Q : r � 0 or r 2 < 2}. [Hint: Use Exercise 1 2. See also Exercise 1 5, below.] 14. Given y E JR. n E N, and E > 0, show that for some l) > 0, if I u - y I < l) then I u n - y n I < E. [Hint: Prove the inequality when n = 1 , n = 2, and then do induction on n using the identity ·
15. Given x > y
n
=
x.
0 and n E N, prove that there is a unique
y >
0 such that
That is, y = ::/X exists and is unique. [Hint: Consider
y = I . u. b. {s E lR : s n � x } . Then use Exercise 1 4 to show that y n can be neither < x nor > x .] 1 6. Prove that real numbers correspond bijectively to decimal expansions not terminating in an infinite strings of 9's, as follows. The decimal expansion of x E lR is N .x 1 x2 where N is the largest integer N ) , x2 is the largest integer � x. x1 is the largest integer � l O (x � I OO(x - (N + x 1 j 1 0) ) , and so on. •
•
•
-
Exercises
43
(a) Show that each Xk is a digit between 0 and 9. (b) Show that for each k there is an f_ � k such thatx t f= 9. (c) Conversely, show that for each such expansion N .x 1 x2 • • not terminating in an infinite string of 9's, the set N N � N � { ' + 10 ' + lO
+
x2
l OO '
...}
is bounded and its least upper bound is a real number x with decimal expansion N.x1 x2 • • • • (d) Repeat the exercise with a general base in place of 1 0. 1 7. Formulate the definition of the greatest lower bound (g. I.b.) of a set of real numbers. State a g.I.b. property of lR and show it is equivalent to the l.u.b. property of JR. 1 8. Let f : A ._ B be a function. That is, f is some rule or device which, when presented with any element a E A, produces an element b = f (a) of B. The graph of f is the set S of all pairs (a , b) E A x B such that b = j (a) . (a) If you are given a subset S c A x B , how can you tell if it is the graph of some function? (That is, what are the set theoretic properties of a graph?) (b) Let g : B ._ C be a second function and consider the composed function g o f : A ._ C. Assume that A = B = C = [0, 1 ] , draw A x B x C as the unit cube i n 3-space, and try to relate the graphs of f, g , and g o f in the cube. 1 9. A fixed point of a function f : A -- A is a point a E A such that f (a) = a . The diagonal of A x A is the set of all pairs (a , a) in A
X
A.
(a) Show that f : A ._ A has a fixed point if and only if the graph of f intersects the diagonal. (b) Prove that every continuous function f : [0, l ] -- [0, l ] has at least one fixed point. (c) Ts the same true for continuous functions f : (0, 1 ) ._ (0, l) ?t (d) Ts the same true for di scontinuous functions? 20. Given a cube in JRm , what is the largest ball it contains? Given a ball in JRm , what is the largest cube it contains? What are the largest ball and cube contained in a given box in JRm ? t A question posed in thi� manner means that, as well as answering the question wtth a "yes" or a ''no." you should give a proof if your answer is "yes" or a specific counter-e>�ample if your answer is "no." Also. to do this exercise you should read Theorems 22. 23. 24.
44
Chapter 1
Real Numbers
2 1 . A rational number p 1 q is dyadic if q is a power of 2, q = 2k for some nonnegative integer k. For example 0, 3j8, 3/ 1 , - 3/256, are dyadic rationals, but 1 / 3 , 5/ 1 2 are not. A dyadic interval is [a , b] where a = pj2k and b = (p + 1 ) / 2k . For example, [.75, 1] is a dyadic interval but [ 1 , 1r ] , [0, 2], and [ . 2 5 , . 75] are not. A dyadic cube is the
product of dyadic intervals having equal length. The set of dyadic rationals may be denoted as Q2 and the dyadic lattice as Q2 . (a) Prove that any two dyadic squares (i.e., planar dyadic cubes) of the same size are either identical or intersect only along their edges. (b) Show that the same intersection property is true for dyadic cubes in ]Rm , when "edge" is interpreted as " (m - 1 )- dimensional face." 22. (a) Given E > 0, show that the unit disc contains finitely many dyadic squares whose total area exceeds 1r - E , and which in tersect each other only along their boundaries. *(b) Show that the assertion is false for E < JT /2 if we demand that the dyadic squares be disjoint. (c) Formulate (a) in dimension m = 3 and m � 4. *(d) Do the analysis with squares and discs interchanged. That is, given E > 0 prove that finitely many disj oint discs can be drawn inside the unit square so that the total area of the discs exceeds 1 - E. *23 . Let b (R) and s (R) b e the number of integer unit cubes i n JRm that intersect the ball and sphere of radius R, centered at the origin. (a) Let m = 2 and calculate the limits
s (R) R--.oo b(R) .
hm --
and
1.
R--.oo
s (R) 2 b(R)
lill -- .
(b) Take m � 3 . What exponent k makes the limit
s (R) k . llm
R---+ 00
-
b(R)
interesting? (c) Let c(R ) be the number of integer unit cubes that are contained in the ball of radius R , centered at the origin. Calculate 1.
R---+ oo
lm
c(R)
-
b(R)
Exercises
45
(d) Shift the ball to a new, arbitrary center (not on the integer lattice) and re-calculate the limits. *24. Let f (k, m) be the number of k-dimensional faces of the m-cube. See Table 1 .
k=O k = I
k=2 k=3 k=4
...
II
m = I 2
I
m =2
I
4 4 I
0
0
0
0
0
...
...
I
m =3 8
I2 6 I
0
...
I
m=4
m =
f ( O , 4)
f (O , 5)
...
/ ( 1 . 4)
f ( l , 5) /(2. 5 )
...
5)
...
f ( 3 . m)
j(3, m + I)
f (4, m )
f (4, m
/ ( 2, 4) / ( 3 , 4)
!(4 . 4)
...
/ (3 ,
5
/ (4, 5 )
...
m
m+ l
j (O, m + I )
...
f (O. m) f ( l , m)
f( I , m + I)
...
j ( 2 , m)
f ( 2, m + I )
...
...
...
+
1)
Table 1 f (k, m) is the number of k-dimensional faces of the m -cube.
(a) Verify the numbers in the first three columns. (b) Calculate the columns m = 4, m = 5, and give the formula for passing from the mth column to the (m + l ) s1 • (c) What would an m = 0 column mean? (d) Prove that the alternating sum of the entries in any column is 1 . That is, 2 - 1 = 1 , 4 - 4 + I = 1 , 8 - 1 2 + 6 - 1 = 1 , and in general L ( - I ) k f (k . m ) = 1 . This alternating sum is called the Euler characteristic. 25. Prove that the interval [a , b] in IR is the same as the segment [a , b ] in IR I . That is,
E IR : a � x � b} ={y E IR : 3 s , t E [0, 1 ] , {x
s+
t = 1 , and y = sa + tb} .
[Hint: How do you prove that two sets are equal?] 26. A convex combination of W I , . . . , wk E IRm is a vector sum
such that St + · · · + sk = I and 0 � s 1 , , sk .:::: 1 . (a) Prove that if a set E i s convex then E contains the convex combination of any finite number of points in E. (b) Why i s the converse obvious? •
•
•
46
27.
Chapter
Real Numbers
1
(a) Prove that the ellipsoid E = ( (x .
xz y . z) E JR3 : 2 a
+
yz zz +2 bz x
::S
1}
is convex. [Hint: E is the unit ball for a different dot product. What is it? Does the Cauchy-Schwarz inequality not apply to all dot products?] (b) Prove that all boxes in JRm are convex. 28. A function f : (a , b) ---+ lR is a convex function if for all x , y E (a , b) and all s , t E [0. 1 ] with s + t = 1 ,
f (sx + ty)
::S
sf (x)
+ tf (y) .
(a) Prove that f is convex if and only if the set S of points above its graph is convex in JR2 . The set S is { (x , y) : f (x) ::S y}. *(b) Prove that a convex function i s continuous. (c) Suppose that f is convex and a < x < u < b. The slope a of the line through (x . f (x ) ) and (u , f (u)) depends o n x and u , a = a (x , u) . Prove that a increases when x increases. and a increases when u increases. (d) Suppose that f is second-order differentiable. That is. f is differentiable and its derivative f ' is also differentiable. As is standard, write ( / ' ) ' = f". Prove that f is convex if and only if f " (x) ::::=: 0 for all x E (a . b) . (e) Formulate a definition of convexity for a function f : M ---+ lR where M C JR.m is a convex set. [Hint: Start with m = 2.j *29. Suppose that E is a convex region in the plane bounded by a curve c. (a) Show that C has a tangent line except at a countable number of points. [For example, the circle has a tangent line at all its points. The triangle has a tangent line except at three points. and so on.] (b) Similarly, show that a convex function has a derivative except at a countable set of points. *30. Suppose that a function f : [a , b ] ---+ lR is monotone increasing: i.e . . X J ::S Xz =} f (x J ) ::S f (xz ) . (a) Prove that f is continuous except at a countable set of points. [Him: Show that at each x E (a . b) , f has right limit f (x +) and a left limit f (x -) . which are limits of f (x +h ) as h tends to 0 through positive, and negative values respectively. The jump
Exercises
47
of f at x is f (x + ) f (x - ) . Show that f is continuous at x if and only if it has zero jump at x . At how many points can f have jump :::: 1 ? At how many points can the jump be between 1 /2 and l ? Between l /3 and 1 /2?1 (b) [s the same assertion true for a monotone function defined on all of ffi.? 3 1 . Find an exact formula for a bijection f : N x N --+ N. Is one -
12 + "
f (i, j ) = j+( 1 + 2+ · +(i + j -2)) =
"
2
" (2 . 3) . 2 J - -J+ ? 2
J +z
32. Prove that the union of denumerably many sets
Bb each of which is countable, is countable. How could it happen that the union is finite? *33. Without using the Schroeder-Bernstein Theorem, (a) Prove that [a , b] "' (a , b J "' (a , b) . (b) More generally, prove that if C is countable then
ffi. \ C
ffi. U C.
(c) [nfer that the set of irrational numbers has the same cardinality as ffi., ffi. \ 0 be given. Think of E as fixed
Exercises
49
and consider the sets A (8 ) = { u E [a , b] : if x , t E [a , u ] and lx - t l < 8 then l f (x ) - j (t ) l
A = U A (8 ) .
0
Using the least upper bound principle, prove that b E A . Infer that f is uniformly continuous. The fact that continuity on la , b ] implies uniform continuity is one of the important, fundamental principles of continuous functions.] *41 . Define injections f : N � N and g : N � N by f(n) = 2n and g (n) = 2n . From f and g, the Schroeder-Bernstein Theorem produces a bijection N � N. What is it? *42. Let (an ) be a sequence of real numbers. It is bounded if the set A = {a 1 , az . . . } is bounded. The limit supremum. or lim sup. of (an ) as n � oo is
.
(
lim sup an = nlim sup ak -+ oo k ?:. n n � oo (a) (b) (c) (d)
)
If the sequence (an ) is bounded, why does the lim sup exist? If sup{ an } = oo, what is lim sup an ? If limn ---+ oo an = - oo , how should we define lim sup an ? When is it true that lim sup(an + bn ) :S: lim sup an + lim sup bn n ---+ oo
lim sup can n�oo
=
n � oo
n � oo
c lim sup a n ? n -+ oo
(e) Define the limit infimum, or lim inf. of a sequence of numbers, and find a formula relating it to the limit supremum. **43 . The unit ball with respect to a norm II II on JR. 2 is { v E 1R. 2 : ll v ll :S: l } . (a) Find necessary and sufficient geometric conditions on a subset of JR.2 that it be the unit ball for some norm. (b) Give necessary and sufficient geometric conditions that a subset be the unit ball for a norm arising from an inner product. (c) Generalize to JR.m . [You may find it useful to read about closed sets in the next chapter, and to consult Exercise 38 there.]
50
Real Numbers
Chapter
1
44. Assume that V is an inner product space whose inner product induces a norm as l x l = .J[X:XJ. (a) Show that I I obeys the parallelogram law
for all x , y E V . * (b) Show that any norm obeying the parallelogram law arises from a unique inner product. [Hints: Define the prospective inner product as x y x y (x , y) =
I
;
12 - I
;
12
Checking that ( , ) satisfies the inner product properties of symmetry and positive definiteness is easy. Also it is immediate that lx 1 2 = ( x , x ) , so ( , ) induces the given norm. Checking bilinearity is another story. (i) Let x , y , z E V be arbitrary. Show that the parallelogram law implies
(x + y , z) + (x - y, z )
=
2(x , y ) ,
and infer that (2x , z ) = 2 (x , z) . For arbitrary u , v E V set x = i 0 b e given. According t o Theorem 7, N€ (fp) is open in N . By (iii), pre (N€ ( fp ) ) is open in M. and p belongs to it, so there is some M� (p) C fPre(N€ (/p)). Thus.
d(x , p) < and
8 :::::}
f i s continuous. See Figure 30.
d(fx , fp) < E , D
I hope that you find the closed and open set characterizations of continuity elegant. Note that no explicit mention is made of the metric. The open set condition is purely topological. It would be perfectly valid to take as a definition of continuity that the pre-image of each open set is open. In fact this is exactly what's done in general topology.
66
A
Taste of Topology M
li!!ll
Chapter 2
N
I j ( MoJ>)
Figure 30 The E , 8
-
condition for a continuous function
f
: M
�
N.
1 1 Corollary A homeomorphism f : M � N bijects the collection of open sets in M to the collection of open sets in N. It bijects the topologies. Proof Let V be an open set in N . By Theorem 1 0, since f is continuous, the pre-image of V is open in M. Since f is a bijection, this pre-image U = { p E M : f (p) E V } is exactly the image of V by the inverse 1 bijection, U = f - (V). The same thing can be said about f - 1 since f - 1 is also a homeomorphism. That is, V = f (U) . Thus, sending U to f(U) bijects the topology o f M t o the topology o f N. D Because of this corollary, a homeomorphism is also c alled a topological equivalence. In general, continuous maps do not need to send open sets to open sets. For example, the squaring map x �---+ x 2 from lR to lR sends the open interval ( - 1 , 1 ) to the non-open interval [0, 1 ) . See also Exercise 34. Closure, Interior, and Boundary
Let S be a subset of the metric space M . Its closure, interior, and boundary are defined as follows: closure S = n K where K ranges through the collection of all closed sets that contain S. Equivalently,
S = {x E M : if K is closed and S
c
K then x E K } .
interior int ( S) = U U where U ranges through the collection of all open sets contained in S. Equivalently, int S
boundary
=
[x E M : for some open
as = s n sc .
U
C
S, x
E U}.
Section
I
Metric Space Concepts
67
Since the intersection of closeds is closed, S and a s are closed. By con struction, S is the smallest closed set that contains S because it is contained in every closed set that contains S. Similarly the interior of S is the largest open set contained in S, and the boundary is the closure minus the interior. Alternate notations for closure, interior, and boundary are cl(S)
s
s 0
int(S)
S = bd (S)
as
=
fr (S) .
The last stands for "frontier.'' Several useful facts about closure, interior, and boundary are presented in the exercises. See Exercise 92 in particular. 12 Proposition S
=
lim S.
Proof By Theorem ?, lim S is a closed set. It contains S, because for each s E S, the constant sequence (s , s, . . . ) converges to s . By minimality of the closure, S c lim S . On the other hand, lim S consists o f limits o f S, and because S is closed it must contain these limits, so lim S c S, 0
and the two sets are equal . Inheritance
Suppose that S is a subset of the metric space M and al so of the metric space N. If S is open or closed in M, is the same true when S is judged in N? Not necessarily. The interval (a , b) is closed in itself, as is any metric space, but it is not closed in JR. It fails to contain its limits a , b in JR. Even more striking is the situation concerning the set S
=
{x
EQ
:
- -J'i < X
0
3N such that k ,
n
:::: N
::::}
d(pk o Pn )
< E.
A
74
Taste of Topology
Chapter 2
The terms of a Cauchy sequence "bunch together" as n � oo Each convergent sequence (pn ) is Cauchy. For if (Pn ) converges to p as n � oo then, given E > 0, there is an N such that for all n :::=: N , E
d (pn , p) < 2 · By the triangle inequality, if k,
n
:::=: N
then
so convergence ==} Cauchy. Theorem 1 .5 states that the converse is true in the metric space ffi. . Every Cauchy sequence in ffi. converges to a limit in R In the general metric space, however, this need not be true. For example, consider the metric space Q of rational numbers, equipped with the standard distance d (x , y) = l x - y I , and consider the sequence ( rn )
= ( 1 .4 , 1 .4 1 ,
1 .4 14, 1 .4 1 42, . . . ) .
0, choose N > - log1 0 E . If k, n ;:: N then Irk - rn l � < E . Nevertheless, (rn ) refuses to converge in Q. After all, as a sequence in ffi. it converges to J2. and if it also converges to some r E Q, then by uniqueness of limits in R r = J2, something we know is false. In brief, convergence =} Cauchy but not conversely A metric space M is complete if each Cauchy sequence in M converges to a limit in M. Theorem 1 .5 implies that ffi. is complete. It is Cauchy. Given
w- N
26 Theorem ffi.m
E
>
is complete.
Proof Let ( Pn ) be a Cauchy sequence in ffi.m . Express Pn in components as
Pn
= (P in , · · ·
•
Pm n ) .
Because ( pn ) is Cauchy, each component sequence (Pjn )n EN is Cauchy. Completeness of ffi. implies that the component sequences converge, and 0 therefore the vector sequence converges. 27 Theorem
metric space.
Every closed subset of a complete metric space is a complete
Proof Let A be a closed subset of the complete metric space M and let (pn ) be a Cauchy sequence in A . It is of course also a Cauchy sequence in M and therefore it converges to a limit p in M . Since A is closed, p E A . 0
Section
I
28 Corollary space.
Metric Space Concepts
75
Every closed subset ofEuclidean space is a complete metric
Proof Obvious from the previous theorem and completeness of ]Rm
0
Boundedness
A subset S of a metric space M is bounded if for some p E M and some r > 0,
A set which is not bounded is unbounded. For example, the elliptical region 4x 2 + y 2 < 4 is a bounded subset of JR2 , while the hyperbola xy = 1 is unbounded. Distinguish the word "bounded" from the word "finite." The first refers to physical size, the second to the number of elements. The concepts are totally different. Also, boundedness has little connection to the existence of a boundary - a clopen subset of a metric space has empty boundary, but some clopen sets are bounded. It is easy to check that the terms of a Cauchy sequence ( Pn) in M form a bounded subset of M . Simply take E = 1 , and apply the Cauchy definition: there is an N such that for all n, m :=::: N, d (p1 , Pm ) < 1 . Choose r > 1
+ max{ d ( p J . P2 ) , . . . . d(p 1 . PN H -
By the triangle inequality, the Cauchy sequence lies entirely i n Mr ( P J ) . Since a convergent sequence is Cauchy, its terms also form a bounded set. Boundedness is not a topological property. For example, consider ( 1 , 1 ) and JR. They are homeomorphic. although ( - 1 . 1 ) i s bounded and lR is unbounded. The same example shows that completeness is not a topological property. A function from M to another metric space N is a bounded function if its range is bounded. That is. there exist q E N and r > 0 such that -
Note that a function can be bounded even though its graph is not. For example, x � sin x is a bounded function lR --+ lR although its graph, { (x , y) E JR2 : y = sin x } , is an unbounded subset of JR2 •
Chapter 2
A Taste of Topology
76
2
Compactness
Compactness is the single most important concept in real analysis. It is what reduces the infinite to the finite. A subset A of a metric space M is (sequentially) compact if every se quence (an ) in A has a subsequence (a n k ) that converges to a limit in A. The empty set and finite sets are trivial examples o f compact sets For a sequence (an ) contained in a finite set repeats a term infinitely often, and the corresponding constant subsequence converges. Compactness is a good feature of a set. We will develop criteria to decide whether a set is compact. The first is the most often used, but beware ! its converse is generally false. 29 Theorem
Every compact set is closed and bounded.
A is a compact subset of the metric space M and that p is a limit of A. There is a sequence (an) in A converging to p. By compactness, some subsequence (ank ) converges to some q E A, but every Proof Suppose that
subsequence of a convergent sequence converges to the same limit as does the mother sequence, so q = p and p E A . Thus, A is closed. To see that A is bounded, choose and fix any point p E M. Either A is bounded or else for each n E N there is a point an E A such that d ( p an) � n . Compactness implies that some subsequence (ank ) con verges. Convergent sequences are bounded, which contradicts the fact that d( p ank ) --* oo as k � oo . Therefore (an) can not exist and A is bounded. .
.
30 Theorem Proof Let
The closed interval [a , b]
c
0
lR is compact.
(xn ) be a sequence in [a , b] . Let C =
{x E [a . b] : Xn
0, Xn lies in (c E, c + E) only finitely often, which implies that c + E E C, contrary to c being an upper bound for C. Hence some subsequence of (xn ) does converge to c, and 0 [a , b] is compact. -
To pass from lR to !Rm , think about compactness in terms of Cartesian products. 31 Theorem
The Cartesian product of two compact sets is compact.
Compactness
Section 2
77
M and B C N are com pact. There exists a subsequence an k that converges to some point a E A as k --+ oo . The subsequence (bnk ) has a sub-subsequence (bn kct> ) that con verges to some b E B as e --+ oo. The sub-subsequence (an 1 c , ) continues to converge to the point a . Thus Proof Let (an . bn ) E
as e --+
oo.
A x B be given where A
C
This implies that A x B is compact.
32 Corollary
0
The Cartesian product of m compact sets is compact.
A 1 x A 2 x · · · x A m = A 1 x ( A 2 x · · · x A m ) and perform 0 induction on m . (Theorem 3 1 handles the bottom case m = 2.)
Proof Write
33 Corollary
A box [a� . bt J
x
· · ·
x [am , bm J is compact.
Proof Obvious from Theorem 30 and the previous corollary.
0
An equivalent formulation of these results is the 34 Bolzano-Weierstrass Theorem
convergent subsequence.
Any bounded sequence in ffi.m has a
Proof A bounded sequence is contained in a box, which is compact, and
therefore the sequence has a subsequence that converges to a limit in the box. 0 Here is a simple fact about compacts. 35 Theorem
A closed subset of a compact set is compact.
Proof If A is a closed subset of the compact set C and if (a n ) is a sequence
of points in A, then clearly (an ) is also a sequence of points in C, so by compactness of C, there is a subsequence (a n k ) converging to a limit p E C . 0 Since A i s closed, p lies i n A, which proves that A i s compact. Now we come to the first partial converse to Theorem 29 . 36 Reine-Borel Theorem
pact.
Every closed and bounded subset of!Rm is com
ffi.m be closed and bounded. Boundedness implies that A is contained in some box, which is compact. Since A is closed, Theorem 35 0 implies that A is compact.
Proof Let
A
c
Chapter 2
A Taste of Topology
78
The Heine-Borel Theorem states that closed and bounded subsets of Euclidean space are compact, but it is vital to remember tbat a closed and bounded subset of a general metric space may fail to be compact. For example, tbe set S of rational numbers in [0, l j is a closed, bounded, noncompact subset of Q. The set S is noncompact because it contains sequences which have no subsequences that converge in S . Ten Examples of Compact Sets
1. 2. 3. 4. 5. 6.
7.
8. 9. 1 0.
Any finite subset of a metric space: for instance. the empty set. Any closed subset of a compact set. The union of finitely many compact sets. The Cartesian product of finitely many compact sets. The intersection of arbitrarily many compact sets. The unit ball in JR3 . The boundary of a compact set. for instance the unit 2-sphere in JR3 . The set {x E lR : 3n E N and x = 1 /n } U {0} . The Hawaiian earring. See page 53. The Cantor set. See Section 5 .
Nests of Compacts
If A 1 ::J A2 ::J · · · ::J An sets. Its intersection is
::J
An+!
n An = { p See Figure 36.
:
::J
. . . then
for each n ,
(An) is a nested sequence of p
E An }
.
Figure 36 A nested sequence of sets.
Compactness
Section 2
79
For example, we could take A n to be the disc { z E JR2 : lz l � 1 /n } . The intersection of all the sets A n is then the singleton {0}. On the other hand, if A n is the ball {z E JR3 : lz l � 1 + 1 / n } then n A n is the closed unit ball B3 .
The intersection of a nested sequence of compact non-empty sets is compact and non-empty. 37 Theorem Proof Let
( A n) be such a sequence. By Theorem 29, A n is closed. The intersection of closed sets is always closed. Thus, nA n is a closed subset of the compact set A 1 , and is therefore compact. It remains to show that the intersection is non-empty. A n is non-empty, so for each n E N we can choose an E A n . The sequence (an) lies in A 1 since the sets are nested. Compactness of A 1 implies that (an) has a subsequence (ank ) converging to some point p E A 1 . The limit p also lies in the set A 2 since except possibly for the first term, the subsequence (ank ) lies in A 2 and A 2 is a closed set. The same is true for A 3 and for all the sets in the nested sequence. Thus, p E nA n and nAn is shown to be non-empty. D The diameter of a non-empty set S d (x . y) between points of S. 38 Corollary
c
M is the supremum of the distances
If in addition to being nested, non-empty, and compact, the
sets A n have diameter that tends to 0 as n point.
--*
oo,
then A
= nAn is a single
e
N, A is a subset of A k . which implies that A has diameter zero. Since any distinct points lie at positive distance from each other, A consists of at most one point, while by Theorem 37 it consists of at least one point. See also Exercise 27. D Proof For each
k
A nested sequence of non-empty noncompact sets can have empty in tersection. For example, the open discs with center ( l j n , 0) on the x-axis and radius 1 In are nested. n e N , but contain no common point. (Their closures do intersect at a common point, the origin. ) See Figure 37. Continuity and Compactness
Next we discuss how compact sets behave under continuous transforma tions.
If f : M --* N is continuous and A is a compact subset of then fA is a compact subset of N . That is, the continuous image of a compact is compact.
39 Theorem M
A Taste of Topology
80
Chapter 2
Figure 37 This nested sequence has empty intersection.
(bn) is a sequence in fA. For each n E N choose a point an E A such that f (an ) = bn . By compactness of A there exists a subsequence (ank ) that converges to some point p E A . By continuity of f Proof Suppose that
it
follows that
bn k = j (ank ) --+ fp E fA .
Thus, any sequence (bn ) in fA has a subsequence converging to a limit in 0 fA , and fA is compact. From Theorem 39 follows the natural generalization of the minimax theorem in Chapter I which concerned continuous real-valued functions defined on an interval [a . b] . See Theorem 23 in Chapter l .
A continuous real-valued function defined on a compact set is bounded; it assumes maximum and minimum values.
40 Corollary
Proof Let
:
M
--+ lR be continuous and let A be a compact subset of M. Theorem 39 implies that fA is a compact subset of !R, so by Theorem 29 it f
is closed and bounded. Thus, the greatest lower bound. v, and least upper bound, V, of f A exist and belong to fA ; there exist points p , P E A such 0 that for all a E A , v = fp ::: fa ::: f P = V . Homeomorphisms and Compactness
A homeomorphism is a hi-continuous bijection. Originally. compactness was called hi-compactness. Thi s is reflected in the next theorem.
Compactness
Section 2
4 1 Theorem If M
is compact and M is homeomorphic to compact. Compactness is a topological property.
Proof If f : M
f M is compact.
--+
N
81
N
then N is
is a homeomorphism then Theorem 39 implies that 0
It follows that [0, 1 ] and IR are not homeomorphic. One is compact and the other is not.
42 Theorem If M is compact then a continuous bijection f : M --+ N is a homeomorphism - its inverse bijection f-1 : N --+ M is continuous. N.
Since f is a bijection Pn = f- 1 (qn) and p = f- 1 (q) are well defined points in M. To check continuity of f-1 we must show that Pn --+ p. I f ( Pn ) refuses to converge to p then there is a subsequence ( Pn k ) and a 8 > 0 such that for all k. d ( pn k • p) :::: 8 . Compactness of M gives a sub-subsequence ( Pn k ) that converges to a point p * E M as l --+ oo . Necessarily. d(p, p * ) :::: 8 , which implies that p =!= p * . Since f is continuous Proof Suppose that
qn --+ q in
f ( Pnk (l) ) --+ f( p * )
as l --+ oo. The limit of a convergent sequence is unchanged by passing to a subsequence, and so f( Pn w) = qn km --+ q as l --+ 00 . Thus, f ( p* ) = q = f( p ). contrary to f being a bijection. It follows that Pn --+ p and therefore that f-1 is continuous. 0 If M is not compact then Theorem 42 can fail. For example, the bijec tion f : [0, 2rr ) --+ IR2 defined by j (x ) = (cos x , sin x) is a continuous bijection onto the unit circle in the plane, but it is not a homeomorphism. This useful example was discussed on page 58. Not only does this f fail to be a homeomorphism, but there is no homeomorphism at all from [0, 2rr ) to S 1 . The circle is compact while [0, 2rr ) is not. Therefore they are not homeomorphic. Absolute Closedness
Not only is a compact space M closed in itself. as is every metric space. but it is also a closed subset of each metric space in which it is embedded. More precisely, we say that h : M --+ N embeds M into N if h is a homeo morphism from M onto hM. (The metric on h M is the one it inherits from N.) Topologically M and h M are equivalent. A property of M that holds
A Taste of Topology
82
Chapter 2
also for every embedded copy of M is an absolute or intrinsic property of M. 4 3 Theorem A
compact is absolutely closed and absolutely bounded.
Proof Obvious from Theorems 29 and 39.
D
For example, no matter how the circle is embedded in JR\ it is a closed and bounded set. See also Exercises 28 and 87. Uniform Continuity and Compactness
In Chapter 1 we defined the concept of uniform continuity for functions of a real variable. The definition in metric spaces is analogous. A function f : M .....,_ N is uniformly continuous if for each E > 0 there exists a 8 > 0 such that
p, q 44 Theorem
continuous.
=>
dN (jp, fq) < E . Every continuous function defined on a compact is uniformly E M
and dM (p, q) < 8
f : M .....,_ lR is continuous, M is compact, but f is not uniformly continuous. Then there is some E > 0 such that no matter how small 8 is, there exist points p, q E M with d (p, q) < 8 but l fp - fq l � E . Take 8 = 1 jn and let Pn · qn be sequences of points in M such that d(pn , qn) < 1 / n while l f ( Pn ) - f(qn ) l � E. Compactness of M implies that there is a subsequence Pn k which converges to some p E M as k .....,_ oo. Since d(pn , qn ) < 1 jn .....,_ 0 as n .....,_ oo, (qn k ) converges to the same limit as does ( Pn k ) as k .....,_ oo ; namely, qnk .....,_ p. Continuity implies that f( Pn k ) --+ fp and f(qnk ) --+ fp. lf k is large then Proof Suppose not, and
I J ( Pnk) - f(qnk ) j
�
i f( Pn k ) - Jp j + l fp - f(qnk ) j < E ,
contrary to the supposition that l f(Pn ) - f(qn) l
� E for all n .
D
Theorem 44 gives a second proof that continuity implies uniform conti nuity on an interval [a , b] . For [a , b] is compact.
3
Connectedness
As another application of these ideas, we consider the general notion of connectedness. Let A be a subset of M, a metric space. If A is neither the empty set nor is it M, then A is a proper subset of M. Recall that if A is both closed and open in M it is said to be clopen. The complement of a clopen set is clopen The complement of a proper subset is proper.
Connectedness
Section 3
83
If M has a proper clopen subset A , M is disconnected. For there is a separation of M into proper, disjoint clopen subsets,
M = A u Ac . (The notation u indicates disjoint union.) M is connected if it is not discon nected - it contains no proper clopen subset. Connectedness of M does not mean that M is connected to something, but rather that M is one unbroken thing. See Figure 38. N
N
Figure 38 M and N illustrate the difference between being connected and
being disconnected.
is connected, f : M -+ N is continuous, and f is onto then N is connected. The continuous image of a connected is connected.
45 Theorem If M
Proof Simple ! If A is a clopen proper subset of N then, according to the open and closed set conditions for continuity, fPre (A) is a clopen subset of M. Since f is onto and A =/= 0, fP'e (A) =/= 0. Similarly, pre (N ) =/= 0. Therefore fP'e (A) is a proper clopen subset of M. contrary to M being connected. It follows that A can not exist and N is connected. D 46 Corollary If M
is connected and M is homeomorphic to N then connected. Connectedness is a topological property.
Proof
N is the continuous image of M .
N
is D
Every contin uous real-valued function defined on a connected domain has the interme diate value property. 47 Corollary (Generalized Intermediate Value Theorem)
Proof Assume that f : M -+ lR is continuous and M is connected. If f assumes values a < f3 in lR and if it fails to assume some value Y with a < Y < {3, then
M
=
{x E M
:
f (x)
Y }
i s a separation of M , contrary to connectedness. 48 Theorem lR
is connected.
D
84
Chapter 2
A Taste of Topology
Proof If U c ffi. is open, closed, and non-empty, we claim that U = ffi.. By Theorem 9, U is the countable disjoint union of open intervals whose
endpoints do not belong to U . If (a , b) is one of these intervals and b < oo , then closedness of U implies that b E U , a contradiction, so b = oo . Similarly, a = - oo , and U = ffi.. Since ffi. contains no proper clopen set, it is connected. 0 49 Corollary (Intermediate Value Theorem for ffi.) Every function f : ffi. � IR has the intermediate value property.
continuous
Proof Immediate from the Generalized Intermediate Value Theorem and connectedness of JR. 0
Thus we have
a
second proof of the Intermediate Value Theorem 1 .24.
50 Corollary The following metric spaces are connected: the intervals
(a , b), [a , b], the circle, and all capital letters of the Roman alphabet.
Proof The interval (a , b) is homeomorphic to ffi., while [a, b] is the con tinuous image of ffi. under the map whose graph is shown in Figure 39.
The circle is the continuous image of ffi. under the map t �---+ (cos t , sin t). Connectedness o f the letters A , . . . , Z is equally clear. 0 f(x) = b
f(x) = a
a
Figure 39 The function
b
f surjects ffi. continuously to [a . b].
Connectedness properties give a good way to distinguish non-homeo morphic sets. Example The union of two disjoint closed intervals is not homeomorphic
to a single interval. One set is disconnected and the other is connected
Connectedness
Section 3
85
[a , b ] is not homeomorphic to the circle S 1 • For removal of a point x E {a , b) disconnects [a , b] while the circle remains connected upon removal of any point. More precisely, suppose that h : [a , b] --+ S 1 is a homeomorphism. Choose a point x E (a , b) , and consider X = [a , b] \ {x } . The restriction of h to X is a homeomorphism from Example The closed interval
X onto Y , where Y is the circle with one point, hx , removed. But X is disconnected, while Y is connected. Hence h can not exist and the segment is not homeomorphic to the circle.
Example The circle is not homeomorphic to the figure eight. Removing
any two points of the circle disconnects it. but this is not true of the figure eight. Or, removing the crossing point disconnects the figure eight, but removing no point disconnects the circle. Example The circle is not homeomorphic to the disc. For removing two points disconnects the circle but does not disconnect the disc.
As you can see, it is useful to be able to recognize disconnected subsets S of a metric space M. By definition, S is a disconnected subset of M if it is disconnected when considered in its own right as a metric space with the metric it inherits from M ; it has a separation S = A u B such that A and B are proper closed subsets of S. According to the inheritance principle, closedness in S can be analyzed in M. The facts that no point of A is a limit of B and no point of b is a limit of A imply that the closure of B in M misses A and the closure of A in M misses B . (Note: "in M.") In symbols, An B
= 0 = A n B.
The separated sets A , B must b e disjoint but they need not b e closed i n M nor need their closures in M be disjoint. Example The punctured interval X =
=
La . b] \ { c} is disconnected if a < b. For X [a , c) u (c, b] is a separation of X . Observe that the closures in lR of the separated sets have non-empty intersection. c
k lie outside Un k which contradicts their convergence to p. Thus, at some finite stage the process of choosing points an and sets Un terminates, and U reduces to a finite subcovering { U1 , , Un ) of A , which implies that A i s covering compact. 0 •
•
•
•
Upshot In light of Theorem 54, the term "compact" may now be applied
equally to any set obeying ( a) or (b).
Chapter 2
A Taste of Topology
92
Figure 45 The point
an k
is so near p that the neighborhood M'A ( ank ) engulfs p.
Total Roundedness
The Heine-Borel Theorem states that a subset of �m is compact if and only if it is closed and bounded. In more general metric spaces, such as Q, the assertion is false. But what if the metric space is complete? It is still false. For example the discrete metric on N makes N closed and bounded in N but it is noncompact: after all, what subsequence of { l , 2, 3 , . . ) converges? But mathematicians do not quit easily. The Heine-Borel Theorem ought to generalize beyond �m somehow. Here is the concept we need: a set A C M is totally bounded if for each f > 0 there exists a finite covering of A by €-neighborhoods. No mention is made of a covering reducing to a subcovering. How close total boundedness is to the worthless fact that every metric space has a finite open covering ! .
56 Theorem (Generalized Reine-Borel Theorem) A subset ofa complete
metric space is compact if and only if it is closed and totally bounded.
Proof Let A be a compact subset of M. Therefore it is closed. To see that it is totally bounded. let f > 0 be given and consider the €-neighborhood covering of A , { M.,x : x E A } .
Compactness of A implies that this covering reduces to a finite subcovering and therefore A is totally bounded. Conversely, assume that A is a closed and totally bounded subset of the complete metric space M. We claim that A is sequentially compact: any
Coverings
Section 4
93
sequence (an ) in A has a subsequence that converges in A. Set Ek = 1 / k, k = 1 , 2, . . . . Since A is totally bounded we can cover it by finitely many E I -neighborhoods ME J ( q i ) , . . . , ME ! (qm ) . B y the pigeon-hole principle, terms of the sequence an lie in at least one of these neighborhoods infinitely often, say it is ME1 (P I ) - Choose an ] E AI = A n ME ! (P I ) -
Any subset of a totally bounded set is totally bounded, so we can cover A I by finitely many E2 -neighborhoods. For one of them, say ME2 (p2 ) , an lies in A 2 = AI n ME2 ( p2 ) infinitely often. Choose an2 E A 2 with n 2 > n i . Proceeding inductively, cover A k - I by finitely many Ek -neighborhoods, one of which, say MEk ( Pk ) , contains terms of the sequence (an ) infinitely often. Then choose ank E A k = A k - I n MEk (Pk ) with nk > n k I · Then (an k ) is a subsequence of (an ) - It is Cauchy because for k, .e 2:: K , -
2 . diam A K ::: 2EK = . K Completeness of M implies that (an k ) converges to some p E M and since A is closed, p E A. Hence A is compact. D 57 Corollary A
totally bounded.
metric space is compact if and only if it is complete and
Proof Every compact metric space M is complete. This is because, given a Cauchy sequence (Pn ) in M, compactness implies that some subsequence converges in M, and if a Cauchy sequence has a convergent subsequence
then the mother sequence converges too. As observed above, compactness immediately gives total boundedness. Conversely, assume that M is complete and totally bounded. Any metric D space is closed in itself. By Theorem 56, M is compact. Perfect Metric Spaces
A metric space M is perfect if M' = M : each p E M is a cluster point of M . Recall that M clusters at p if each Mr p is an infinite set. For example [a , b] is perfect and Q is perfect. N is not perfect; none of its points are cluster points. 58 Theorem
able.
Every non-empty, perfect, complete metric space is uncount
94
A Taste of Topology
Chapter 2
Proof Suppose not: assume M is non-empty, perfect. complete, and count
able. Since finite. Say
M consists of cluster points it must be denumerable and not
M = {XI . X2 , . . . } is a list of all the elements of M. Define Mr P = { q E M : d(p, q )
_:::
r}.
It i s the "closed neighborhood" of radius r at p . If x fj. Y we say that Y excludes x. Proceeding as in the proof of Theorem 56 we inductively choose a nested sequence of these closed neighborhoods to exclude more and more points in the sequence (xn). They nest down to a point in M that is different from all the Xn , a contradiction. Specifically take Yl = x2 , r1 = min ( l , d (x1 , x2) /2 ), and set �
YI = MY) ( yJ ) .
Then Y1 excludes x1 . Since M clusters at y 1 , infinitely many Xn lie in Mr1 (y t), so we can choose Y2 E Mrt (y J ) such that Y2 f. x2. The choice of
r2 = min { l /2 , d ( y2, x2) , ( r 1 - d (yt . Y2 ))}
causes Y2 ure 46.
�
= Mr2 (Y2) to be a subset of Y1 and to exclude x 1 , x 2 . See Fig-
.. ,
•
Figure 46 The exclusion of successively more points of the sequence (xn ) .
Nothing stops u s from continuing indefinitely, and we get a nested se quence of closed neighborhoods Yn
�
= Mrn ( Yn )
•
Cantor Sets
Section 5
95
such that Yn excludes x � o . . . , Xn , and has radius rn :::: 1 / n . Thus the center points Yn form a Cauchy sequence. Completeness of M implies that lim Yn = y E M
n---HXJ
exists. Since the sets Yn are closed and nested, y E Yn for each n . Does Yn equal x1 ? No, for Y1 excludes x 1 • Does it equal Xz? No, for Y2 excludes x2 . In fact, for each n, y ¥= Xn . The point y is nowhere in the supposedly complete list of elements of M, a contradiction. 0 59 Corollary lR
and [a , b] are uncountable.
Proof lR is complete and perfect, while
plete, and perfect. Neither is empty.
[a , b] is compact, therefore com 0
60 Corollary A
non-empty peifect complete metric space is everywhere uncountable in the sense that each r-neighborhood is uncountable.
Proof The r/2-neighborhood Mr ;z (p) is perfect: it clusters at each of its
points. The closure of a perfect set is perfect. Thus, Mr;z(p) is perfect. Being a closed subset of a complete metric space, it is complete. According to Theorem 58, Mr;z (p) is uncountable. Since Mr;z (p) C Mr p, Mr P is uncountable. 0
5
Cantor Sets
Cantor sets are fascinating examples of compact sets that are maximally disconnected. (To emphasize the disconnectedness, one sometimes refers to a Cantor set as "Cantor dust.") Here is how to construct the standard Cantor set. Start with the unit interval [0, 1 ] and remove its open middle third, ( 1 /3 . 2/3 ) . Then remove the open middle third from the remaining two intervals, and so on. This gives a nested sequence C 0 :::) C 1 :::) C 2 :::) . . . where C 0 = [0, 1 ] , C 1 is the union of the two intervals [0. 1 /3] and [2/3, 1 ] , C 2 is the union of four intervals [0, 1 /9], [2/9, 1 /3], [2/3. 7 /9], and [8/9, 1], C 3 is the union of eight intervals, and so on. See Figure 47. In general e n is the union of 2n intervals, each of length 1 j3 n . Each e n is compact. The standard middle thirds Cantor set is the nested intersection
We refer to C as "the" Cantor set. Clearly it contains the endpoints of each of the intervals comprising e n . Actually, it contains uncountably many more points than these endpoints ! There are other Cantor sets defined by
96
Chapter 2
A Taste of Topology c"
c'
----
c'
-----
c•
- -
c5
---
--
- An endpoint
-- --
c
Figure 47 The construction of the standard middle thirds Cantor set C .
removing, say, middle fourths, pairs o f middle tenths, etc. All Cantor sets tum out to be homeomorphic to the standard Cantor set. See Section 6. A metric space M is totally disconnected if each point p E M has arbitrarily small clopen neighborhoods. That is, given E > 0 and p E M, there exists a clopen set U such that
pEu
c
MEP·
For example, any discrete space i s totally disconnected. S o i s Q.
61 Theorem The Cantor set is a compact, non-empty, peifect, and totally disconnected metric space.
Proof The metric on C is the one it inherits from .IR, the usual distance ::J
£, the set of endpoints of the e n -intervals, C is non empty and infinite. It is compact because it is the intersection of compacts. To show C is perfect and totally disconnected, take any x E C and any E > 0. Fix n so large that 1 f3 n < E . The point x lies in one of the 2n intervals I of length 1 /3 n comprising e n . The set of C -endpoints in I , E n I , i s infinite and contained i n the interval (x - E , x + E) . Thus C clusters at x and C is perfect. See Figure 48.
lx - y l . Because C
I X-E
X
x + E
Figure 48 The endpoints of C cluster at x .
The interval I i s closed in lR and therefore in e n . The complement J = n e \ I consists of finitely many closed intervals and is therefore closed too.
Thus
Cantor Sets
Section 5
cc n n
97
u
cc n J) exhibits C as the disjoint union of two clopen sets, C n I being a subset of 0 (x - E , x + E) which contains x . Hence C is totally disconnected. c
62 Corollary
The Cantor set is uncountable.
Proof Being compact, C is complete, and by Theorem 58, every complete, 0 perfect, non-empty metric space is uncountable.
A more direct way to see that the Cantor set is uncountable involves a geometric coding scheme. Take the code: 0 = left and 2 = right. Then
Co = left interval = [0, 1 /3] C2 = right interval = [2/3, 1 ] , and C 1 = Co U C2 . Similarly, the left and right subintervals of Co are coded Coo and Coz, while the left and right subintervals of Cz are Czo and Czz.
This gives
The intervals that comprise C 3 are specified by strings of length 3. For instance, Czzo is the left subinterval of Czz. In general an interval of e n is coded by an address string of n symbols. each a 0 or a 2. Read it like a zip code. The first symbol gives the interval's gross location (left or right), the second symbol refines the location. the third refines it more. and so on Imagine now an infinite address string w = w 1 wzw3 . . . of O's and 2's. Corresponding to w. we form a nested sequence of intervals
the intersection of which is
a
point p = p (w) E C. Specifically,
where win = w 1 wn truncates w to an address of length n . See Theo rem 37. As we have observed, any infinite address string defines a point in the Cantor set. Conversely, any point p E C has an address w = w (p ) : its first n symbols a = win are specified by the interval Ca of en in which p lies. A second point q has a different address, since there is some n for which p and q lie in distinct intervals Ca and Cp of e n . In sum, the Cantor set is in one-to-one correspondence with the collection of addresses. Each address w defines a point p (w) E C and each point p E C has a unique address w (p) . •
•
•
A Taste of Topology
98
Chapter 2
If each 2 is replaced by l , an address w becomes a base-two expansion of some x E [0, l ] . There are uncountably many x E [0, l ] , uncountably many base-two expansions, and uncountably many addresses w, all of which re establishes the uncountability of C. In the same vein, w can be interpreted directly as the base-three expansion of p . See Exercise 1 6 in Chapter 1 . In Section 6 we will make more use of this geometric coding. If S c M and S = M then S is dense in M. For example, Q is dense in ffiL The set S is somewhere dense if there exists an open non-empty set U c M such that S n U :::) U . If S is not somewhere dense then it is nowhere dense.
63 Theorem
The Cantor set contains no interval.
Proof Suppose not: C contains some interval (a . b ) . Choose n such that l f3 n < b - a . The Cantor set is contained in a set e n consisting of finitely many closed intervals. all shorter than b - a . Hence en can not contain D (a , b) , and neither can C . 64 Corollary
The Cantor set is nowhere dense in �-
Proof If C is dense in an open set U then U c C . Since C is closed, U C C . Any non-empty open set U C � contains an interval (a , b ) , but C never contains an interval Hence U = 0 and C is nowhere dense. D The existence of an uncountable nowhere dense set is astonishing. Even more is true: the Cantor set is a zero set: it has "outer measure zero." By this we mean that, given any E > 0, there is a countable covering of C by open intervals (a k o bk) . and the total length of the covering is
(Outer measure is one of the central concepts of Lebesgue Theory. See Chapter 7.) After all, C is a subset of which consists of closed intervals, each oflength 1 f3 n . If n is large enough then 2n f3 n < E . Enlarging each of these closed intervals to an open interval keeps the sum of the lengths < E , and it follows that C is a zero set. If we discard subintervals of [0, 1] in a different way, we can make a fat Cantor set one that has positive outer measure. Instead of discarding the middle thirds of intervals at the nth stage in the construction, we discard only the middle 1 In ! portion. The discards are grossly smaller than the remaining intervals. See Figure 49. The total amount discarded from [0, 1 ]
en ,
-
2n
Cantor Set Lore
Section 6*
99
is < 1 , and the total amount remaining, the outer measure of the fat Cantor set, is positive. See Exercise 3 .32.
fat Cantor set, the gap intervals occupy a progressively smaller proportion of the Cantor set intervals.
Figure 49 In forming
6*
a
Cantor Set Lore
In this section, we explore some arcane features of Cantor sets. Although the continuous image of a connected set is connected, the con tinuous image of a disconnected set may well be connected. Just crush the disconnected set to a single point. Nevertheless, I hope you find the fol lowing result striking, for it means that the Cantor set C is the universal compact metric space, of which all others are merely shadows.
Given a compact non-empty metric space M, there is a continuous surjection of C onto M.
65 Cantor Surjection Theorem
See Figure 50. Exercise 107 suggests a direct construction of a continuous surjection C -+ [0, 1], which is already an interesting fact. The proof of Theorem 65 involves a careful use of the address notation from Section 5 and the following simple lemma about dividing a metric space M into small pieces. A piece of M is any compact, non-empty subset of M. c
- -
Figure 50
a surjects C onto M.
Chapter 2
A Taste of Topology
1 00
66 Piece Lemma
A compact metric space M is the union of dyadically
many small pieces. Specifically, given E > 0, there exist 2k pieces of M, each with diameter ::5 E, whose union is M.
Proof (Recall that "dyadic" refers to powers of 2. ) Cover M with neigh
borhoods of radius E /2. By compactness, finitely many of them suffice, say Um cover M. Then M is the union of the m pieces U1 o .
.
.
•
each of diameter ::5 E . Choose n with m ::5 2 n and set U; = Um for m ::5 i ::5 2n . (That is, repeat the last piece in the list 2n m times.) Then M is the union of the pieces U1 , , U2n , and each has diameter ::5 E . 0 -
•
•
•
Let us denote the length of an address string a =
a1 . . . an by
la l = n . There are exactly 2n such a. We refer also to a as a word using the letters 0, 2. A dyadic filtration of M is a collection 1ft = {Ma } of pieces of M such that (a) a varies freely in the collection of all finite words formed with the letters 0, 2. (b) For each n E N, M = U lal =n Ma . (c) If a is expressed as a compound word a = f3 � then Ma c Mp . (d) max{diam Ma : l a l = n } � 0 as n � oo . We call 1ft a filtration because its n th level 11tn = {Ma : I a I = n } consists of many small sets. Think of the sets Ma as the holes in a sieve that filters a liquid. (The analogy is imperfect since the Ma may overlap.) As n � oo , we filter M more and more finely. 67 Dyadic Filtration Lemma
a dyadic filtration.
Every non-empty compact metric space has
Proof By Lemma 66 there is an integer n 1 and 2n 1 pieces. each with diam
eter ::5 1 , whose union is M. Since there are 2 n 1 dyadic words of length n 1 , we use them to label these pieces as Ma . Then
If Y is a word of length k < n 1, define My to be the union of all the pieces Ma such that Y = a l k. (Recall that a l k is the truncation of a to its first
Cantor Set Lore
Section 6*
k letters ) This defines the filtration 1/en for 1
101 :S
n
:S
n1
and makes the
nesting condition (c) automatic. In the same way, each Ma can be expressed as a union of zn 2 sub-pieces Ma =
U
l fJ I=n 2
( Ma ) fJ ,
each of diameter :S 1 /2. Define Ma fJ = (Ma ) fJ when 1 ,8 1 = n z . If Y is a word of length k < n z define Mar to be the union of all the Ma 13 with Y = ,B i k. See Figure 5 1 .
M
Figure 51 The piece Ma divided into sub-pieces Ma 13
.
This defines pieces corresponding to all words � of length :S and M=
U
l a l=n l
For each word � of length word
(
U
l fJ I=n 2
n1
+
( Ma )fJ
nz
)
=
U
IBI=n 1 +n 2
n1
+
nz,
Ma .
is expressed uniquely as a compound
�=
a ,B
and 1 ,8 1 = n z This extends the previous filtration to 1/en , 1 :S n :S n 1 + n z , and again (c) is automatic. Nothing stops us from passing to pieces of ever smaller diameter, which produces the desired D dyadic filtration 1/e = U11tn . where Ia I =
n1
Consider the intervals Ca that define the standard Cantor set C . Their intersections with C, Ca n C , give a dyadic filtration of C . An infinite address w = w1 w2 . specifies the precise location of a point p E C as •
•
p (w) =
n Cwln ·
n EN
1 02
Chapter 2
A Taste of Topology
As we saw in Section 5 , points of C are in one-to-one correspondence with these infinite words w. For each p E C there is an w with p = p ( w) , and for each w, the point p( w ) belongs to C . Proof of Theorem 65 We are given a compact, non-empty metric space
M, and we must find a continuous surjection a : C -+ M, C being the standard Cantor set. Let {Ma l be a dyadic filtration of M and define
where w is an infinite dyadic address. Then define a :
a (p)
C
-+
M by
= q (w)
where p = p (w) . Every point of M has at least one address. (It has several addresses if the pieces Ma overlap. ) Hence a is a surjection. and it remains to check continuity. Let E > 0 be given and choose n so that max{diam Ma : I a I = n J
< E.
Then choose � s o small that the intervals Ca i n e n lie farther apart than 8 ; i.e. 8 < l j3 n . If p, p' E C and I P - p' l < 8 then p, p ' belong to a common interval Ca with I a I = n, so their infinite addresses w (p ), w (p') both begin with the same string of l ength n, namely a. Therefore a (p) , a (p ' ) both belong to the same piece Ma, and
since
Ma has diameter
CXJ ,
Then
As n � term is
oo,
0.
the first and third terms tend to
Jd (pn , qn) - d (p� , q� ) J
By Lemma
77, the middle
d (pn , p�) + d (qn , q�) , Hence L = L ' and D is well :S
defined which also tends to 0 as n � oo . on M. The d distance on M is symmetric and satisfies the triangle inequal ity. Taking limits, these properties carry over to D on M, while positive definiteness follows directly from the co-Cauchy definition. (b) Think of each p E M as a constant sequence, p = (p, p, p, p , . . . ) . Clearly it is Cauchy and clearly the D-distance between two constant se quences p and q is the same as the d -distance between the points p and q . In this way M i s naturally a metric subspace of M . (c) Let (Pk he N be a Cauchy sequence in M. We must find Q E M to which Pk converges as k ---+ oo. (Note that (Pk ) is a sequence of equivalence classes, not a sequence of points in M , and convergence refers to D not d .) Because D is well defined w e can use a trick t o shorten the proof. We claim that there exists an element ( Pk ,n) E Pk such that for all m , n E N ,
1
(1)
d ( pk,m • Pk,n ) < k .
All the terms of this representative are closely bunched. not merely those in the tail. Choose and fix some sequence (p: n )neN in Pk . Since it is Cauchy, there is some N = N (k) such that for all n ::::: N. m.
l
* • P*n) < d ( Pk,m k" k. The sequence Pk. n = p£n+ N is co-Cauchy with (Pt, n ) and it satisfies ( 1 ). For each k, choose a representative ( Pk , n ) E Pk that satisfies ( 1 ), and define qn = Pn n · We claim that (qn) is a Cauchy sequence in M and that Pk converges to its equivalence class Q as k � oo. That is, M is complete. .
�
Section 7*
Completion
Let E > 0 be given. There exists
If k , .e =:::
N
=:::
111
3/E such that if k, .e
=::: N
N then then by ( 1 ) ,
d (qk. qe ) = d ( pk,k . P uJ .:S d ( pk,k. P k,n) + d (Pk,n · P e,n ) + d ( pe,n . P u ) .:S
1 1 k + d ( pk, n . Pe, n ) + f
.:S
2E
J
+ d ( Pk. n • Pt.n ) .
The inequality is valid for all n and the left-hand side, d (qk . qe ) . does not depend on n . The limit of d ( pk,n · p 1 . n ) as n -+ oo is D ( Pk - Pe ) , which we know to be < E j 3 . Thus, if k , .e =::: N then d (qk . qe ) < E and (qn ) is Cauchy. Similarly we see that Pk -+ Q as k -+ oo. For, given E > 0, we choose N =::: 2/E such that if k , n =::: N then d (qk . qn ) < E /2, from which it follows that d ( Pk.n • qn ) .:S d ( pk,n • Pk.k) + d ( pk,k . qn ) = d ( Pk.n • Pk.k ) + d ( qk . qn ) 1 E < - + - < E. 2 - k The limit of the left-hand side of this inequality, as n -+ Thus
oo,
is D ( Pk , Q ) .
D
and M is complete.
Uniqueness of the completion is not surprising, and is left as Exercise I 0 I . A different proof of the Completion Theorem is sketched in Exercise 4.37. A Second Construction of JR. from
Q
In the particular case that the metric space
M is Q, the Completion Theorem
leads to a construction of lR from Q via Cauchy sequences. Note, however, that applying the theorem as it stands involves circular reasoning, for its proof uses completeness of !R to define the metric D. Instead, we use only the Cauchy sequence strategy.
Chapter 2
A Taste of Topology
1 12
Convergence and Cauchyness for sequences of rational numbers are con cepts that make perfect sense without a priori knowledge of R Just take all E 's and 0, we have Pn :::=:: E for all n then P is positive. If P is positive, P is negative. Then we define P -< Q if Q !:._ is positive. Exercise 1 03 asks you to check that this defines an order on Q, consistent with the standard order < on Q in the sense that for all p , q E Q, p < q - 0 and N E N such that for all
Pm + E < qn ·
m, n
:::::
N,
It remai�s to check the least upper bound propeny. Let P be a non-empty subset of Q that is bounded above. We must find a least upper bound for P. We assert that the least upper bound for P is the equivalence class Q of the following Cauchy sequence ( qo , q1 , qz , . . . ) . (a) q0 is the smallest integer such that q0 is an upper bound for P. (b) q 1 is the smallest fraction with denominator 2 such that q 1 is an upper bound for P.
Section
(c)
7*
L l3
Completion
q2 is the smallest fraction with denominator 4 such that q2 is an upper bound for P.
(d) (e)
qn
is the smallest fraction with denominator 2n such that qn is an upper bound for P. Since P -:/= 0 we can choose P * = [(p:)] E P . For some N * and all
m , n � N*,
I P! - P : l
0 and an N such that for all n � N, qN
+E
0 and N such that for all m. n � N, Fix a k
� N with 1 j2 k
< E . Then for all
rm
X t-X Higher derivatives are defined inductively and written J = (J ) ' . If f (r ) (x ) exists then f is r1h order differentiable at x . If f (r ) (x) exists for each x E (a , b) then f is r1h order differentiable. If j C r l (x) exists for all r and all x then f is infinitely differentiable, or smooth. The zero-th derivative of f is f itself, j COJ (x) = f (x ) .
If f is r 1h order differentiable and r continuousfunction ofx E (a, b ).
9 Theorem
:=:
1 then j Cr- 1 ) (x) is a
Proof Differentiability implies continuity and j Cr- l l (x ) is differentiable.
D
10 Corollary
A smooth function is continuous. Each derivative ofa smooth
function is smooth and hence continuous.
Proof Obvious from the definition of smoothness and Theorem 9.
D
Smoothness Classes
If f is differentiable and its derivative function f ' (x) is a continuous func tion of x, then f is continuously differentiable, and we say that .f is of class C 1 . If .f is rth order differentiable and f (r ) (x) is a continuous function of x , then f is continuously r1h order differentiable, and we say that f is
1 48
Functions of a Real Variable
Chapter 3
of class cr . If f is smooth, then by the preceding coronary, it is of class c for all finite r and we say that f is of class c oo . To round out the notation, we say that a continuous function is of class C0• Thinking of c r as the set of functions of class c r , we have the regularity hierarchy, C O ::J C I ::J . ::J n c r = C oo . .
.
rEN
cr cr+ t
Each inclusion ::J is proper; there exist continuous functions that are not of class C 1 , C 1 functions that are not of class C2• and so on. For example, is of class C 0 but not of class C 1
f (x) = lx I
•
f (x) = X lx I is of class c I but not of class C 2 . f (x) = lx l 3 is of class C 2 but not of class C 3 Analytic Functions
A function that can be expressed locally as a convergent power series is analytic. More precisely, the function f : (a , b) ---+ IR is analytic if for each x E (a . b). there exists a power series and a /3 > 0 such that if I h i
0 and define
g (t)
= f (x + t) - P (t) - Rh r(+hl) t r +l
=
t r+ l
R(t) - R (h) r h
+l
lSI
Differentiation
Section 1
for O :=:: t :=:: h . Note that since P (t ) i s a polynomial o f degree r , p (t ) = 0 for all t, and
Also, g (O) = g ' (O) = · · = g < r l (O) = 0, and g (h ) = R (h ) - R (h) = 0. Since g 0 at 0 and h, the Mean Value Theorem gives a t 1 E (0, h) such that g ' (t 1 ) = 0. Since g ' (0) and g ' (tt ) = 0, the Mean Value Theorem gives a t2 E (0, t 1 ) such that g " {t2 ) = 0. Continuing, we get a sequence t 1 > t2 > · · · > t r + l > 0 such that g ( tk ) = 0. The ( r + 1 ) 81 equation, g (tr + d = 0, implies that
=
·
Thus, () = x + tr + 1 makes the equation in (c) true. If h is symmetric. 12 Corollary For each r E isfies lim e ( h ) / h r = 0.
N. th e
0. Thus f is strictly monotone. Differentiability implies continuity, so f is a homeomorphism 1 (a , b) � (c, d) . To check differentiability of f- at y E (c, d) , define Then y = f(x ) and �y = f(x + �x) - x = tl.f. Thus
�X �y
1
�yj �x
�fl �x
Since f is a homeomorphism, �x --* 0 if and only if �y --* 0. so the limit D of �f- 1 j �y exists and equals l jf ' (x) . If a homeomorphism f and its inverse are both of class c r , r :::: 1 , then f is a cr diffeomorphism. 14 Corollary If f : (a . b) --* (c, d ) is a homeomorphism of class c r . 1 :::: r :::: 00, and f ' =!=- 0 then f is a cr diffeomorphism.
Differentiation
Section 1
1 53
Proof If r = l , the formula ( f - 1 ) ' (y) = l jj ' (x) = 1 /f ' < f - 1 (y)) implies that (f - 1 ) ' (y) is continuous, so f is a C 1 diffeomorphism. Induction on D r :=:: 2 completes the proof. The corollary remains true for analytic functions: the inverse of an ana lytic function with non-vanishing derivative is analytic. The generalization of the inverse function theorem to higher dimensions is a principal goal of Chapter 5 . A longer but more geometric proof o f the one dimensional inverse func tion theorem can be done in two steps. (i) A function is differentiable if and only if its graph is differentiable. (ii) The graph of f - 1 is the reflection of the graph of f across the diag onal, and is thus differentiable. See Figure 63 .
f! b )
�-----.-------.-------.---
y = f(x) !
-
-
-
-
-
-
-
-
·
·
·
-
-
-
·
·
·
-
-
-
•
• •
·
..
�
/( a) � - - - -
� .. : .
•
- �- - - . -� l1
/(a )
-
. . - . . . . . . -� . . . X
- �- - - - � . b
f(x)
.
.
...
�- - - :
{(h)
Figure 63 A picture proof of the inverse function theorem in �
1 54
Functions of a Real Variable
2
Chapter 3
Riemann Integration
Let f : [a, b] � lR be given. Intuitively, the integral of f is the area under its graph; i.e., for f � 0,
l b f (x) ds
= area 1U
where 1U is the undergraph of f. 1U
= { (x , y)
: a ::=:: x ::S b and 0 ::=:: y ::=:: f (x) } .
The precise definition involves approximation. A partition pair consists of two finite sets of points P , T C [a, bl; P = {xo , Xn } and T {tt , tn }, interlaced as •
•
.
.
.
.
.
.
We assume the points x0 , sponding to f, P. T is
R (f, P, T) =
.
•
.
, Xn are distinct. The Riemann sum corre
n
L f ( ti ) !lxi = /Ctt ) Llxt + f (tz)!l xz + · · · + f (tn) !lxn
i =l where Llx i = xi - X i - I · The Riemann sum R is the area of rectangles which approximate the area under the graph of f. See Figure 64. Think of the points ti as "sample points." We sample the value of the function f at lj .
((1;)1
• • • • • •
• • • •
a
Figure 64 The area of the strip is f (ti ) !lxi ·
b
Riemann [ntegration
Section 2
1 55
The mesh of the partition P is the length of the largest subinterval [xi _ 1 , Xi]. A partition with large mesh is coarse; one with small mesh is fine. In general, the finer the better. A real number I is the Riemann integral of f over [a, b J if it satisfies the following approximation condition: VE > 0 38
mesh P
0 such that if P , T is any partition pair then 8 ::::} IR - II < E
l b f (x dx
where R = R (f, P , T) . If such an I exists it is unique, we denote it as
)
a
=I =
lim
mesh P--+0
R (f, P , T),
and we say that f is Riemann integrable with Riemann integral I. See Exercise 26 for a formalization of this limit definition. 15 Theorem
If f is Riemann integrable then it is bounded.
Proof Suppose not. Let I = J: f (x) dx . There is some 8 > 0 such that I R - I I < 1 for all partition pairs P , T with mesh P < 8 . Fix such a partition pair P = { xo . . . . , Xn }, T = {tt , . . . , tn } . If f is unbounded on [a, b] then there is also a subinterval [xio - t . XioJ on which it is unbounded. Choose a new set T' = {t� , . . . , t� } with t[ = ti for all i f:. i0 and choose t[0 such that l f(t[0 ) - f(ti0) I �Xi0 > 2 .
This is possible since the supremum of { 1 / .x; .
a
upper sum
b
\: : : u 7: : : 1------f . .
.
.
. .
.
. .
.
b
Figure 65 The lower sum, the Riemann sum, and the upper sum.
Functions of a Real Variable
1 58
Chapter 3
To prove Theorem 1 9 it is convenient to refine a partition P by adding more partition points ; the partition P' refines P if P' :=> P . Suppose first that P' = P U { w } where w E (xio- 1 • Xi0 ) . The lower sums for P and P' are the same except that mi0 l:l.xi0 in P) splits into two terms in P') . The sum of the two terms is at least as large as mi0 l:l.xio · For the infimum of over the intervals [Xio-1 • w] and [w , xi0] is at least as P ) . See Figure 66. large as mio · Similarly, P') ::=::: Repetition continues the pattern and we formalize it as the
L(f,
L (f,
f
U (f,
U (f,
Refinement Principle Refining a partition causes the lower sum to in
crease and the upper sum to decrease.
upper summand
refined upper summand
refined lower swnmand
lower summand
Figure 66 Refinement increases L and decreases U .
,
The common refinement P * of two partitions P, P' of [a b] is
P * = P u P' . According to the refinement principle
L (f,
P) ::::
L(f,
P*)
:::: U (f, P * ) :::: U (f, P') .
We conclude: each lower sum is less than or equal to each upper sum, the lower integral is less than or equal to the upper; and thus
(2)
A bounded function f : [a , b] --* JR. is Darboux integrable if and only if 'v'E > 0 3 P such that
f
U (f. P ) - L (f.
P) < E.
Proof of Theorem 19 Assume that i s Darboux integrable: the lower and upper integrals are equal. say their common value is 1 . Given E > 0 we must find 8 > 0 such that I R - / I < E whenever R = R (f, P , T) is a
Riemann Integration
Section 2
1 59
Riemann sum with mesh P < 8 . By Darboux integrability and ( 2 ) there is a partition P1 of La , b] such that
U1 - L I
E < 2
where U1 = U (f, P1 ) , L 1 = L(f, P1 ) . Fix 8 = Ej8Mn 1 where n 1 is the number of partition points in P1 . Let P be any partition with mesh P < 8 . (Since 8 « E , think of P1 a s coarse and P a s much finer.) Let P * be the common refinement P U P1 . By the refinement principle.
< L1 where L *
=
L (f, P * ) and U *
< U* L* -
=
< -
U1
U (f, P * ) . Thus,
u* - L*
< �2
{x; } and P * = {xJ l for 0 ::::: i ::::: n and 0 ::::: j ::::: n * . The sums U = L M; �x; and U * = I: Mj �xj are identical except for terms with
Write P =
for some i , j . There are at most n 1 - 2 such terms and each is of magnitude at most M8 . Thus,
U - U* Similarly, L *
-L
u-L
=
E < (n i - 2)2M8 < - . 4
< E /4, and so
(U - u * ) + (U * - L * ) + (L * - L )
< E.
Since I and R both belong to the interval [L , U], w e see that I R - II < E . Therefore f is Riemann integrable. Conversely, assume that f is Riemann integrable with Riemann integral I . By Theorem 1 5 , f is bounded. Let E > 0 be given. There exists a 8 > 0 such that for all partition pairs P , T with mesh P < 8, I R - I I < E I 4 where R = R (f. P , T ) . Fix any such P and consider L = L (f, P ) , U = U (f, P ) . There are choices of intermediate sets T = {t; }, T ' = such that each f (t; ) is so close to m; and each J 0 there is a countable covering of Z by open intervals (a i , bi ) such that 00
L bi - a i i =l
_:::: E .
The sum o f the series is the total length o f the covering. Think o f zero sets as negligible; if a property holds for all points except those in a zero set then one says that the property holds almost everywhere, abbreviated "a.e."
A function f : [a , b] --+ JR. is Riemann integrable if and only if it is bounded and its set ofdiscontinuity points is a zero set.
21 Riemann-Lebesgue Theorem
The set
D of di scontinuity points is exactly what its name implies, D = {x
E
La , b] : f is discontinuous at the point x } .
A function whose set of discontinuity points i s a zero set i s continuous almost everywhere. The Riemann-Lebesgue theorem states that a function is Riemann integrable if and only if it is bounded and continuous almost everywhere. Examples of zero sets are and only countably many of them; given any ::0: E . See Exercise 2.30.
E >
0, there are only finitely many at which the jump is
Functions of a Real Variable
1 64
Chapter 3
(a) Any subset of a zero set. (b) Any finite set. (c) Any countable union of zero sets . (d) Any countable set. (e) The middle-thirds Cantor set. (a) is clear. For if Zo c Z where Z is a zero set, and if E > 0 is given, then there is some open covering of Z by intervals whose total length is _:::: E ; but the same collection of intervals covers Z0, which shows that Z0 is also a zero set. (b) Let Z = { z � o . . . , Z n } be a finite set and let E > 0 be given. The intervals (z; E j2n , z; + E/2n ) , for i = I , . . . , n , cover Z and have total length E . Therefore Z is a zero set. In particular. any single point is a zero set. (c) This is a typical "E"/2n -argument." Let Z1 • Z2 , . . be a sequence of zero sets and z = u zj . We claim that z is a zero set. Let E > 0 be given. The set Z1 can be covered by countably many intervals (ai l , bi l ) with total length L (hi l - ai l ) _:::: E/2. The set Z2 can be covered by countably many intervals (ai2 . bi2) with total length L (hi2 - a;2) _:::: E / 4 . In general. the set Zi can be covered by countably many intervals (a;j . b;j ) with total length -
.
00
'"" /; such that b; - a; = E j2n . (Since l j 3 n < E j 2n , and I; has length l /3 n , this i s possible.) The total length of these 2 n intervals ( a; . b; ) is E . Thus is a zero set. In the proof of the Riemann-Lebesgue Theorem, it is useful to focus on the "size" of a discontinuity. A simple expression for this size is the
C
,
C
•
•
•
•
Section 2
Riemann Integration
1 65
oscillation of f at x , oscx (f) = lim sup f {t) - lim inf f (t ) . r-+ x r-+ x Equivalently, oscx (f) = lim diam f ([x - r, x + r ] ) r-+ 0
(Of course, r > 0.) [t is clear that f is continuous at x if and only if oscx ( f ) = 0. It is also clear that if I is any interval containing x in its interior then M1 - m 1 ::=: oscx (f)
where M1 and m 1 are the supremum and infimum of f (t) as t varies in See Figure 72.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
:x
..
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
I.
.
Figure 72 The oscillation of f at x is lim sup f (t ) - lim inf j (t ) . r-+ x r->- x Proof of the Riemann-Lebesgue Theorem The set D of discontinuity points of f : [a .
b]
-4
[- M, M] naturally filters itself as the countable union D =
U Dk
kEN
where Dk = {x E [a ,
b]
:
l oscx (f) ::=: k } .
According to (a), (c) above, D is a zero set i f and only i f each Dk i s a zero set. Assume that f is Riemann integrable and let E > 0, k E N be given. By Theorem 19 there is a partition P such that U -L =
L(M; - m ; ) ll.x;
0. Since 1/f ' is continuous and positive on [c, d] there is a constant K > 0 such that for all 0 E [c, d], 1/f'(O ) =:: K . The Mean Value Theorem implies that for all u , v E [c, d], there exists a 0 between u and v such that 1/f (u) - 1/f (v) = 1/f ' (O ) (u - v ) . Thus,
For any
s, t
e
1 1/f (u ) - 1/f (v) l � K l u - v i . [a , b], set u = 1/f - 1 (s) , v = 1/f - 1 (t). Then I s - t i � K J l/f- 1 (s) - 1/f - 1 (t) J ,
which is a Lipschitz condition on 1/f - 1 with Lipschitz constant K = K - 1 • By Corollary 30, f o 1/f is Riemann integrable. D Versions of the preceding theorem and corollary remain true without the hypotheses that 1/f bijects. The proofs are harder because l/t can fold infinitely often. See Exercise 39. In calculus you learn that the derivative of the integral is the integrand. This we now prove.
32 Fundamental Theorem of Calculus If f : [a . b]
lx f {t) dt
integrable then its indefinite integral F (x)
=
---+
lR
is Riemann
is a continuousfunction ofx. The derivative of F (x) exists and equals f (x) at all points x at which f is continuous. Proof #1 Obvious from Figure 76. Proof #2 Since f is Riemann integrable, it is bounded; say for all x . By Corollary 28 I F (y) - F (x) l =
l l y f (t) dt l .s:
M ly -
l f (x) l
_::::
M
xl .
Therefore, F is continuous : given E > 0, choose o < E f M, and observe that I Y - x i < o implies that I F (y) - F (x) l < Mo < E . In exactly the same way, if f is continuous at x then
Chapter 3
Functions of a Real Variable
172
F(xJ
X
a
h
x+h
Figure 76 Why does this picture give a proof of the Fundamental Theorem of Calculus?
F (x + h) - F (x) h
as
h
---+ 0. For if
=
!._h
l X
x
+
h
f (t) d t
---+
f (x)
m (x , h) = inf { f (s ) : Is - x l s l h l } M (x, h) = sup{ f (s) : Is - x i s lh l }
then
m (x , h)
= s
1
h ]
h
lx+h m (x , h) dt 1 lx+h f (t) dt lx+h M(x . h) t M (x , h) . s
x
h
x
d =
x
When f is continuous at x, m (x , h) and M(x , h) converge to h ---+ 0, and so must the integral sandwiched between them,
1
h (If h
1x+h f(t) dt. X
< 0 then t fx +h f (t) dt is interpreted as - t J:+h f(t) dt .) x
f (x)
as
D
The derivative ofan indefinite Riemann integral exists almost everywhere and equals the integrand almost everywhere.
33 Corollary
Riemann Integration
Section 2
173
Proof Assume that f : [a , b] ---+ JR. is Riemann integrable and F (x) is its
indefinite integral. By the Riemann-Lebesgue Theorem, f is continuous almost everywhere, and by the Fundamenta1 Theorem of Calculus, F ' (x) exists and equals f (x) wherever f is continuous. D A second version of the Fundamenta1 Theorem of Calculus concerns an tiderivatives. lf one function is the derivative of another, the second function is an antiderivative of the first.
Note When G is an antiderivative of of g : [a , b] ---+ JR., we have
G ' (x) = g (x) for all x E
[a , b], not merely for almost all x E [a , b] .
34 Corollary
Every continuous function has an antiderivative.
Proof Assume that f : [a , b] ---+ JR. is continuous. By the Fundamental Theorem of Calculus, the indefinite integral F (x) has a derivative every D where, and F ' (x) = f (x) everywhere. Some discontinuous functions have an antiderivative and others don't. Surprisingly, the wildly oscillating function
f (x)
=
{ 0. if x ::::: 00 --{ oi 00 rr
sm
x
if X >
has an antiderivative, but the j ump function
g (x) does not. See Exercise 42.
if X :::::
if x >
35 Antiderivative Theorem An antiderivative of a Riemann integrable function, if it exists, differs from the indefinite integral by a constant.
Proof We assume that f : [a , b] ---+ JR. is Riemann integrable, that G is an antiderivative of f , and we assert that for all x E [a , b],
G (x) =
{x f (t ) dt
+
C,
Functions of a Real Variable
1 74
where C is a constant. (In fact, C
a = Xo < and choose
tk E [xk - I . xk]
Chapter 3
= G (a) .) Partition [a , x]
X1
< · · · < Xn
as
= X,
such that
Such a tk exists by the Mean Value Theorem applied to the differentiable function G. Telescoping gives
G (x) - G (a)
=
n
n
k=l
k =l
L G (xk ) - G (xk - l ) = L f (tk ) D. xk ,
which i s a Riemann sum for f on the interval [a , x ] . Since f is Riemann integrable, the Riemann sum converges to F (x) as the mesh of the partition tends to zero. This gives G (x) - G (a) = F (x) as claimed. D
36 Corollary
are valid.
lb
Standard integral formulas, such as 2 b3 - a 3 x dx = , 3
a
Proof Every integral formula is actually a derivative formula, and the Anti derivative Theorem converts derivative formulas to integral formulas.
lx -1 dt .
D
In particular, the logarithm function is defined as the integral, log x
=
1
t
Since the integrand 1 j t is well defined and continuous when t > 0, log x is well defined and differentiable for x > 0. Its derivative is l j x . By the way, as is standard in post calculus vocabulary, log x refers to the natural logarithm, not to the base 1 0 logarithm. See also Exercise 1 6. An antiderivative of f has G ' (x) = f(x) everywhere, and differs from the indefinite integral F (x) by a constant. But what if we assume instead that H ' (x) = f (x) almost everywhere? Should this not also imply H (x) differs from F (x) by a constant? Surprisingly, the answer is "no."
There exists a continuous function H : [0, 1 ] ---+ lR whose derivative exists and equals zero almost everywhere, but which is not con stant. 37 Theorem
Riemann Integration
Section 2
175
Proof. The counter-example is the Devil's staircase function, also called the Cantor function. It is defined as H (x ) -
1 /2
if X E [ l j3 , 2 j 3]
1 /4
if X E [ l j9 , 2j9]
3j4
if X E [7 /9 , 8j 9 ]
See Figures 77, 78.
· · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · ·
3 4
2 I
· · · · · · · · · · · · · ·
I
4
· · · ·
�---�
�-
9 I
0
�
9 2
3
3
9
2
I
7
9 8
Figure 77 The devil' s staircase. On each discarded interval in the Cantor set construction, H is constant. Thus H is differentiable with derivative zero at all points of [0, 1 ] \ C, and since C is a zero set, this implies H ' (x) = 0 for almost every x . To show that H is continuous we use base 2 and base 3 arithmetic. [f x E [0, 1 ] has base 3 expansion x = (.x1x2 h then the base 2 expansion of y = H (x) is (-YI Y2 . . . h where •
Yi =
��
Xi 2
if 3k
such that Xk
•
•
=1
= 1 and !Jk < i such that xk = 1 if xi = 0 or xi = 2 and !Jk i such that Xk = 1 .
if Xi
l f 3 k apart, and this fact is preserved when we take sums, J strictly in creases. The proof that J'(x) = 0 almost everywhere requires more and deeper theory. Next, we justify two common methods of integration.
If f E R and g : [ c, d ] --+ [a , b] is a continuously differentiable bijection with g ' > 0 ( g is a C 1 diffeomorphism) then f (y) dy = f (g(x))g ' (x) dx.
38 Integration by substitution
lb
[d
Proof The first integral exi sts by assumption. By Corollary 3 1 , the com posite f o g E R, and since g ' is continuous, the second integral exists by Corollary 25 . To show that the two integrals are equal we resort again to Riemann sums. Let P partition the interval [ c, dl as C =
and choose
tk E
Xo < X I
0 , convergence o f L bk implies there i s a large N such
that for all m , n 2: N , L�=m bk n
< E . Thus n
n
and convergence of L ak follows from the CCC.
0
Example The series L sin(k) j2k converges since it is dominated by the geometric series L 1 j2 k . A series L ak converges absolutely if L lak I converges. The comparison test shows that absolute convergence implies convergence. A series that converges but not absolutely converges conditionally: L ak converges and L lak I diverges. See below. Series and integrals are both infinite sums. You can imagine a series as an improper integral in which the integration variable is an integer,
More precisely, given a series L ak o define f
f(x) See Figure 80. Then
=
ak tf k - 1 < x
00
:L ak = k =O
:
[0, oo) _:::
1 00 f (x) dx .
---7-
lR by setting
k.
O
The series converges if and only if the improper integral does. The natural extension of this picture is the
41 Integral Test Suppose that and and L ak is a given series.
J000 f(x) dx is a given improper integral
(a) If lak l .::S f (x) for all sufficiently large k and all x E (k - 1 , k] then convergence of the improper integral implies convergence of the series. (b) lf l f (x) l .::S ak for all sufficiently large k and all x E [k , k + 1 ) then divergence of the improper integral implies divergence of the series.
Functions of a Real Variable
1 82
0
2
4
5
k- l
0
2
4
5
k
Chapter 3
k+ l
k+ l
k+2
Figure 8 0 The pictorial proof o f the integral test.
Proof See Figure 80. (a) For some large N0 and all N 2: N0 we have
t
��+ 1
iak l
::=::
1 N f(x) dx 1 00 f (x) dx , ::=::
�
0
which is a finite real number. An increasing, bounded sequence converges to a limit, so the tail of the series L la k l converges, and the whole series L lak I converges. Absolute convergence implies convergence. 0 The proof of (b) is left as Exercise 73.
Example The p-series, L 1 j kP converges when p > 1 and diverges when p ::::: l . Case 1 . p > 1 . By the fundamental theorem of calculus and differentiation rules b 1 -P - I l fb 1 = )1 xP 1 - p -----* p - I as b ---+ oo . The improper integral converges and dominates the p-series, which implies convergence of the series by the integral test. Case 2. p ::=:: 1 . The p-series dominates the improper integral h
1 dx { lt x P
=
{
log b
b 1 -P -
if p = 1 1
if p < 1 . 1-p As b -----* oo, these quantities blow up, and the integral test implies divergence of the series. When p = 1 we have the harmonic series, which we have just shown to diverge.
Series
Section 3
1 83
The exponential growth rate of the series L ak is ot
= lim sup � . k-+ oo
42 Root Test Let a be the exponential growth rare of a series L ak . If
< 1 the series converges. root test is inconclusive.
ot
Proof If ot
if ot
> I the series diverges. and if ot =
1
the
< 1 , fix a constant {3 , ot < f3 < 1 .
Then for all large k, l ak i i / k ::::: {3 ; i.e.,
which gives convergence of L ak by comparison to the geometric series k L: P . If a >
1 , choose {3 , 1 < f3
00
>
k
1 the series diverges, and otherwise
Functions of a Real Variable
1 84
Proof lf p i.e. ,
R. Moreover R is given by the fonnula
R-
1
lim sup V"'cJ .
----===
k--+ 00
Proof Apply the root test to the series L ck x k . Then lim sup k--+ oo
If l x I
0 be given. There exists N such that m, n ::::: N imply
Also, for each x E [a , b] there exists an m =
i fm (X ) - f (x ) i If n :::::
m (x )
::::: N such that
E < 2·
N and x E [a , b] then i fn (X) - f (x ) i
S
l fn (X) - fm (x) (x ) l
E E < - + - = E.
2
+
l fm (x) (X ) - f (x ) l
2
Hence fn :::4 f. The function f is bounded. For !N is bounded and for all x , i fN (x ) - f (x ) i < E . Thus f E Cb . By Theorem 2, uniform con vergence implies d-convergence, d ( fn , f ) -4 0, and the Cauchy sequence Un ) converges to a limit in the metric space Cb . D The preceding proof is subtle. The uniform inequality dUn , f ) < E is derived by non-uniform means: for each x we make a separate estimate using an m (x ) depending non-uniformly on x . It is a case of the ends justifying the means. Let C 0 = C 0 ( [a , b], Jl{) denote the set of continuous functions [a , b J -4 Jl{_ Each f E C 0 belongs to cb since a continuous function defined on a compact domain is bounded. That is, C 0 c Cb . 4 Corollary C 0 is a closed subset of Ch · It is a complete metric space. Proof Theorem l implies that a limit in Cb of a sequence of functions in
lies in C 0 . That is, C 0 is closed in Cb . A closed subset of a complete D space is complete. C0
Uniform Convergence and C 0 [a , b]
Section 1
207
Just as it is reasonable to discuss the convergence of a sequence of func tions we can also discuss the convergence of a series of functions, L fk · Merely consider the nth partial sum Fn (x) =
n
L /k (x) . k=O
It is a function. If the sequence of functions ( Fn ) converges to a limit function F then the series converges, and we write F (x ) =
00
L fk (x ) . k=O
If the sequence of partial sums converges uniformly, then so does the series. If the series of absolute values L I /k (x ) I converges, then the series L A converges absolutely. 5 Weierstrass M-test If L Mk is a convergent series of constants and if fk E C satisfies I fk I ::::: Mk for all k, then L fk converges uniformly and b absolutely. Proof If
n
telescope as
>
m
then the partial sums of the series of absolute values
n
n
k=m+ l
k=m+ l
Since L Mk converges, the last sum is < E when m , n are large. Thus ( Fn ) is Cauchy in C . and by Theorem 3 it converges uniformly. 0
b
Next we ask how integrals and derivatives behave with respect to uniform convergence. Integrals behave better than derivatives. 6 Theorem The uniform limit of Riemann integrable functions is Riemann integrable, and the limit of the integrals is the integral of the limit,
lim
{ b fn (x ) dx
n ---?>- 00 Ja
=
b
limf (X) dx . n ---?>- 00 n J{a unif
Function Spaces
208
Chapter 4
In other words, R is a closed subset of Ch and the integral functional f 1---7 J: f(x) dx is a continuous map from R to JR. This extends the regularity hierarchy to
Theorem 6 gives the simplest condition under which the operations of taking limits and integrals commute. Proof Let fn E R be given and assume that fn ::::::t f as n --+ oo. By the Riemann-Lebesgue Theorem, fn is bounded and there is a zero set Zn such that fn is continuous at each x E fa , b ] \ Zn . Theorem 1 implies that f is continuous at each x E [a , b] \ U Zn , while Theorem 3 implies that f is bounded. Since U Zn is a zero set, the Riemann-Lebesgue Theorem implies that f E R. Finally
1 1b 1b
j( } x dx
:S
- 1b
l 1 1b
fn (x ) dx =
f (x) - fn (x ) dx
l f (x) - fn (x ) l dx :S d U, fn ) ( b - a ) --+ 0
l
as n --+ oo. Hence the integral of the limit is the limit of the integrals. 7 Corollary unifonnly,
1 1x
If fn
E R and fn
- 1x
Proof As above, f ( t ) dt
:::::4
0
f then the indefinite integrals converge
l
fn ( t ) dt :S d Un . f) (x - a ) :S d Un , f) ( b - a ) --+ 0
when n --+ oo.
0
8 Term by Term Integration Theorem A unifonnly convergent series of integrable functions L fk can be integrated tenn by tenn in the sense that
b1 �OO
fk (x) dx =
� 1b fk (x) 00
dx .
Uniform Convergence and C 0 La , b J
Section 1
209
F, converges uniformly to L fk . Each F, belongs to R since it is the finite sum of members of R. According to
Proof The sequence of partial sums
Theorem 6,
�1 Ti
b
fk (x) dx =
1
b
1 � fk (x) dx . b OO
F, (x) dx
�
This shows that the series L I: fk (x) dx converges to I: L fk (x) dx . D
The uniform limit of a sequence of differentiable functions is differentiable provided that the sequence of derivatives also converges uni formly. 9 Theorem
Proof We suppose that f, : [a , b] � lR is differentiable for each n and that f, ::::4 f as n � oo. Also we assume that f� ::::4 g for some function g. Then we show that f is differentiable and in fact f ' = g . We first prove the theorem with a major loss of generality: we assume that each f� is continuous. Then f� . g E R and we can apply the fundamental theorem of calculus and Corollary 7 to write
f (a) +
1x g(t) dt .
Since f, ::::4 f we see that f(x) = f (a) + I: g(t) dt and, again by the fundamental theorem of calculus, f' = g. In the general case the proof i s harder. Fix some x E [a , b] and define _
0
if x ::::: 0
on page 1 49. Near x = 0, e (x ) can not be expressed as a convergent power series. Power series provide the clean and unambiguous way to define functions, especially trigonometric functions. The usual definitions of sine, cosine, etc. involve angles and circular arc length, and these concepts seem less fundamental than the functions being defined. To avoid circular reasoning, as it were, we declare that by definition 00
k
exp x = " � � koft.O
k!
.
sm x
=
oo (- l ) k x 2k+ t L (2k + I ) ! k=O
oo ( - l txzk cos x = L --- k=O
(2k) !
We then must prove that these functions have the properties we know and love from calculus. All three series are easily seen to have radius of conver gence R = oo. Theorem 1 2 justifies term by term differentiation, yielding the usual formulas, exp' (x) = exp x
sin ' (x) = cos x
cos ' (x) = - sin x .
The logarithm has already been defined as the indefinite integral J( I I t dt. We claim that if lx I < 1 then log( l + x ) is given as the power series log ( I + x) =
oo
L
( - l ) k+l x k
k
Compactness and Equicontinuity in
Section 3
C0
213
To check this, we merely note that its derivative is the sum of a geometric series, (log( l
1
00
00
+ x))' x +1 -l = 1 - (-x) = '"'C-x)k = '"'c-llxk. k=O k=O =
-
LJ
LJ
The last is a power series with radius of convergence 1 . Since term by term integration of a power series inside its radius of convergence is legal. we integrate both sides of the equation and get the series expression for log ( l x ) as claimed.
+
3
Compactness and Equicontinuity in
C0
The Heine-Bore} theorem states that a closed and bounded set in IR.m is 0 compact. On the other hand, closed and bounded sets in C are rarely compact. Consider for example the closed unit ball B =
{f E C 0 ([ 0 , 1 ] , JR.)
:
ll f ll �
1 }.
fn (x) xn .
To see that B is not compact we look again at the sequence = It lies in B. Does it have a subsequence that converges (with respect to the converges to f in C 0 then metric d of C 0 ) to a limit in C 0 ? No. For if = lim Thus = 0 if x < 1 and f ( l ) = 1 , but this k-+ oo 0 function f does not belong to C • The cause of the problem is the fact that C 0 is infinite-dimensional. In fact it can be shown that if V is a vector space with a norm then its closed unit ball is compact if and only if the space is finite-dimensional. The proof is not especially hard. Nevertheless, we want to have theorems that guarantee certain closed and bounded subsets of C 0 are compact. For we want to extract a convergent subsequence of functions from a given sequence of functions. The simple condition that lets us go ahead is equicontinuity. A sequence of functions in C 0 is equicontinuous if
j(x)
C fn)
fnk(x).
VE > Is -
0 38
fnk
f (x)
>
0
such that
l fn (s ) - fn(t)l < E . t i and n E N The functions fn are equally continuous. The depends on E but it does not depend on n . Roughly speaking, the graphs of all the fn are similar. For < 8
=}
8
total clarity, the concept might better be labeled uniform equicontinuity, in contrast to pointwise equicontinuity, which requires
VE
> 0 and Vx E
l x - ti
[a , b]
< 8 and n E N
38
>
=}
0 such that
l fn (X ) - fn (t) i < E.
Function Spaces
214
Chapter 4
The definitions work equally well for sets of functions, not only sequences of functions. The set £ c C 0 is equicontinuous if VE > 0 315 > 0 such that < t5 and
�
l f(s ) - f(t) i < E . The crucial point i s that t5 does not depend on the particular f E £. It is valid for all f E £ simultaneously. To picture equicontinuity of a family £, Is - t l
f
E
£
imagine the graphs. Their shapes are uniformly controlled. Note that any finite number of continuous functions [a , b] --+ JR. forms an equicontinuous family so Figures 88 and 89 are only suggestive.
Figure 88 Equicontinuity.
Figure 89 Non-equicontinuity.
The basic theorem about equicontinuity is the 14 Arzela-Ascoli Theorem Any bounded equicontinuous sequence offunc tions in C 0 ( [a , b] , JR.) has a uniformly convergent subsequence.
Compactness and Equicontinuity in C 0
Section 3
215
Think of this as d compactness result. H (fn) i s the sequence of equicon tinuous functions, the theorem amounts to asserting that the closure of the set { fn : n E N} is compact. Any compact metric space serves just as well as [a , b], and the target space lR can also be more general. See Section 8.
[a , b] has a countable dense subset D = { d1 , d2 , } . For instance we could take D = Ql n [a , b] . Boundedness of Un) means that for some constant M, all x E [a , b], and all n E N, l fn (x) i :::; M. Thus Un (di )) is a bounded sequence of real numbers. Balzano-Weierstrass implies that some subsequence of it converges to a limit in JR, say Proof
•
.
•
The subsequence fi . k evaluated at the point dz is also a bounded sequence in JR, and there exists a sub-subsequence fz. k such that fz. k (d2 ) converges to a limit in JR, say fz . k (dz) --+ yz as k --+ oo . The sub-subsequence evaluated at d 1 still converges to y1 • Continuing in this way gives a nested family of subsequences !m , k such that
Um . k ) is a subsequence of Um - I , k ) j :::S m � fm . k (dj) --+ Yi as k
Choose k(m) 2:::
m large enough that if j
--+ oo . :::; m and k(m) :::; k then
The superdiagonal subsequence gm (x) = !m , k (m J (x) converges to a limit at each point x E D. We claim that 8m (x) also converges at the other points x E [a , bJ and that the convergence is uniform. It suffices to show that (gm ) is a Cauchy sequence in C 0 . Let E > 0 be given. Equicontinuity gives a 8 > 0 such that for all s, t
E [a , bJ ,
Choose J large enough that every x E [a , b ] lies in the 8-neighborhood of some di with j :::; J. Since D is dense and [a , b] is compact, this is possible. See Exercise 1 9. Since {d 1 , , d1 } is a finite set and gm (dj) converges for each di , there is an N such that for all i , m :::: N and all j :::; J , •
•
•
Function Spaces
216 If l ,
m
:::: N and x E [a , bJ , choose di with j dj - x i < 8 and j ::S J . Then
gm (x) - g (x) l e
Hence
Chapter 4
::S
j gm (x ) - gm (dj ) j + j gm (dj ) - gf (dj ) j + j ge (dj ) - 8t (X
E E E < - + - + - = E. 3 3 3
(8m ) is Cauchy in C 0 , it converges in C 0 , and the proof is complete.
0
Part of the preceding development can be isolated as the
Pointwise convergena of an equicontinuous sequence of functions on a dense subset of the domain propagates to uniform convergence on the whole domain. 15 Arzela-Ascoli Propagation Theorem
0
Proof This is the E /3 part of the proof.
The example cited over and over again in the equicontinuity world is the following:
16 Corollary Assume that fn : La , b] --+ lR is a sequence of differentiable functions whose derivatives are uniformly bounded. Iffor some Xo, fn (xo ) is bounded as n --+ oo, then the sequence (fn) has a subsequence that converges uniformly on [a , b].
Proof Let M be a bound for the derivatives I f� (x) I , valid for all n E N and x E [a . b] . Equicontinuity of Un ) follows from the Mean Value Theorem:
Is - t i
< 8
==>
l fn (s) - fn (t ) i = l f� (B ) j l s - t i
::S M8
for some e between s and t. Thus, given E > 0, the choice 8 = Ej (M + 1 ) shows that Un ) is equicontinuous. Let C be a bound for l fn (xo) l . valid for all n E N. Then
l fn (x) l
l fn (x) - fn (xo ) l + l fn (xo) l ::S M l b - a l + C ::S
::S
M lx - .\ol + C
shows that the sequence Un ) is bounded in C 0 . The Arzela-Ascoli theorem 0 then supplies the uniformly convergent subsequence. Two other consequences of the same type are fundamental theorems in the fields of ordinary differential equations and complex variables.
Uniform Approximation in
Section 4
C0
217
(a) A sequence o f solutions to a continuous ordinary differential equation in IRm has a subsequence that converges to a limit, and that limit is also a solution of the ODE. (b) A sequence of complex analytic functions that converges pointwise, converges uniformly (on compact subsets of the domain of definition) and the limit is complex analytic. Finally we give a topological interpretation of the Arzela-Ascoli theorem.
A subset £ c C 0 is com pact if and only if it is closed, bounded, and equicontinuous. 17 Heine-Borel Theorem in a function space
Proof Assume that £ is compact. By Theorem 2.56, it is closed and totally
bounded: given E > 0 there is a finite covering of £ by neighborhoods in C 0 that have radius E/3, say �1 3 ( fk ) , with k = 1 . . n . Each fk is uniformly continuous so there is a 8 > 0 such that .
If f E
Is - t l
< 8
=>
.
.
l fk (s ) - fk ( t) l
E < 3·
£ then for some k, f E �;J ( fk ) , and Is - t l < 8 implies
l f(s) - f (t) l
:S
l f (s) - fd s) i + i fk (s) - fk (t) l + l fk (t) - f (t) i
E E E < - + -+ - = E
3 3 3 Thus £ is equicontinuous. Conversely, assume that £ is closed, bounded, and equicontinuous If < fn ) i s a sequence i n £ then by the Arzela-Ascoli theorem, some subsequence < fn k ) converges uniformly to a limit. The limit lies in £ since £ is closed D Thus £ is compact.
4
Uniform Approximation in C0
Given a continuous but nondifferentiable function f, we often want to make it smoother by a small perturbation. We want to approximate f in C 0 by a smooth function g . The ultimately smooth function is a polynomial, and the first thing we prove is a polynomial approximation result 18 Weierstrass Approximation Theorem in C 0 ([a , bl , JR) .
The set ofpolynomials is dense
Density means that for each f E C 0 and each E > 0 there is a polynomial function p (x) such that for all x E La, b],
l f (x) - p(x) i < E . There are several proofs of this theorem, and although they appear quite
Function Spaces
218
Chapter 4
different from each other, they share a common thread: the approximating function is built from f by sampling the values of f and recombining them in some nice way. It is no loss of generality to assume that the interval [a, bj is [0 . 1 ] . We do so.
Proof #1 For each n E
N, consider the sum
Pn (x) =
t (:)ckx k (l - xt-k , k=O
- ).
where ck = f ( k j n) and G) is the binomial coefficient n ! / k ! (n k ! Clearly Pn is a polynomial. It is called a Bernstein polynomial. We claim that the nth Bernstein polynomial converges uniformly to f as n ._ oo. The proof relies on two formulas about how the functions
rk (x ) = (: ) x k (l - xt -k
shown in Figure 90 behave. They are n
(2)
n
(3)
:�:)k - nx ) 2 rk (x) = nx ( I
k=O
In terms of the functions
(6)k xk
- x).
rk we write
.....
Figure 90 The seven basic Bernstein polynomials of degree six.
(1
- x)
6 -k
,k=0
6.
Uniform Approximation in C 0
Section 4
n Pn (X) = .L::Ck rk (X ) k=O
219
n f (x ) = L f(x)rk (x) . k=O
Then we divide the sum Pn - f = L: Cck - f)rk into the terms where k/n is near x , and other terms where kIn is far from x . More precisely, given E > 0 we use uniform continuity of f on [0, I ] to find 8 > 0 such that i t - s l < 8 implies i f (t) - f(s) i < E/2. Then we set K1 = {k e {O, . . , n } : .
This gives
l � - x � < 8}
K2 = {0, . . . n } \ K 1
and
.
n
The factors i ck - f (x) i in the first sum are < E/2 since ck = f(kj n) and kjn differs from x by < 8 . Since the sum of all the terms rk is 1 and the terms are non-negative. the first sum is < E /2. To estimate the second sum, use (3) to write
nx ( I - x)
n 2 = L(k - nx) 2 rk (x) :::: L (k - nx) rk (x) k=O
:::: since k
e
L (n8) 2 rk (x) , k E K2
keK2
K2 implies that (k - nx ) 2 :::: (n8) 2 . This implies that
L rk (x ) -
0 then
(10)
f3n (t)
=t
0 as n --+
oo and 8 ::::;
ItI
::::;
1.
This is clear from Figure 9 1 . Proceeding more rigorously, we have
,
n n ----> oo (1 - 1 / n ) we see that for some constant c and all n ,
Since l fe = lim
See also Exercise 29. Hence if 8 ::::; I t I ::::;
1
then
due to the fact that Jn tends to oo more slowly than (1 - 8 2 ) - n as n --+ oo . This proves ( I 0) . From ( 1 0) we deduce that Pn =t f, as follows: Let E > O be given. Uniform continuity of f gives 8 > O such that l t l < 8 implies l f (x + t) - f(x) l < E /2. Since f3n has integral 1 on L- 1 , 1 ] we have
I Pn (X ) - f(x) l ::S
=
I L: (f(x + t) - f(x))f3n (t) dt '
L: l f(x f(x) l f3n (t) dt 1 l (x + t) - f(x) l f3n (t) dt + 1 =
+ t) -
l r i 0 we must find G E A such that for all x E M, F (x) - E < G (x) < F (x ) + E .
(1 1) First we observe that
(12)
f EA
=>
lfl E
A
where A denotes the closure of A in C 0 M. Let E > 0 be given. According to the Weierstrass approximation theorem, there exists a polynomial p (y) such that E ( 1 3) sup{ l p(y) - l y l l : I Y I � ll f ll l < 2 After all, I y I is a continuous function defined on the interval [- I f I , I f II ] . The constant term of p (y) i s at most E/2 since l p (O) - 101 1 < E/2. Let q (y) = p (y) - p (O) . Then q (y) is a polynomial with zero constant term and ( 1 3) becomes ( 14)
l q (y) - l y / 1 < E .
Lemma 2 1 states that A i s an algebra, so g E A . t Besides, i f x E M and y = f (x) then
l g (x) - l f (x) / 1 = l q (y) - I Y I I < E . Hence I f I E A = A as claimed in ( 1 2). Next we observe that if f, g belong to A, then max (f, g) and min(f, g) also belong to A. For max(f, g) = min(f, g) =
f+g If - gl + 2 2 f g If gl
;
�
-
_
t Since a function algebra need not contain constant functions. it was important that q has no constant term One should not expect that g = ao + u1 f + + an f n belongs to A. · · ·
Uniform Approximation in C 0
Section 4
225
Repetition shows that the maximum and minimum of any finite number of functions in A also belongs to A. Now we return to ( 1 1 ) . Let F E C 0 M and E > 0 be given. We are trying to find G E A whose graph lies in the E -tube around the graph of F. Fix any distinct points p, q E M. According to Lemma 20 we can find a function in A with given values at p, q , say Hpq E A satisfies Hpq (p)
=
F (p)
and
Hpq (q ) = F (q ) .
Fix p and let q vary. Each q E M has a neighborhood Uq such that x E Uq
( 1 5)
::::::>
F (x) - E < Hpq (x ) .
For Hpq (x ) - F (x ) + E is a continuous function of x which is positive at x = q . The function Hpq locally supersolves ( 1 1 ) . See Ftgure 92.
Hpq(p) = F(p)
q
Figure 92 In a neighborhood of q ,
Hpq supersolves ( l l ) in the sense of
(1 5). Compactness of M implies that finitely many of these neighborhoods Uq cover M, say Uq 1 , , Uq" . Define •
•
•
G p (X )
Then Gp E A and
=
max (Hpq 1 (x ) , . . . , Hpq" (x ) ) .
Function Spaces
226
G p (p)
( 1 6)
for all x
=
F (p)
and
F(x) - E
Chapter 4
G p (x)
Gp (x) < F (x) + E .
See Figure 94. By compactness, finitely many of these neighborhoods cover
p
Figure 94 Gp (p ) = F (p) and Gp supersolves { 1 1 ) everywhere.
Uniform Approximation in C 0
Section 4
227
Figure 95 The graph of G lies in the € -tube around the graph of
F
M,
say VPI ' - - - . VPm ' Set G (x) = min(Gp 1 (x) , . . . , G Pm (x ) ) . We know that G E A and ( 1 6), ( 1 7) imply ( 1 L ) . See Figure 95.
22 Corollary Any 2rr -periodic continuous function of x formly approximated by a trigonometric polynomial
T (x) =
E
0
lR can be uni
n
n
L ak cos kx + L bk sin kx
.
k=O k=O Proof Think of [0, 2rr) parameterizing the circle S 1 by X !---+ (cos X ' sin x ) . The circle is compact, and 2rr -periodic continuous functions o n lR become continuous functions on S 1 . The trigonometric polynomials on S 1 form an algebra T c C 0 S 1 that vanishes nowhere and separates points. The 0 Stone-Weierstrass theorem implies that T is dense in C 0 S 1 . Here is a typical application of the Stone-Weierstrass Theorem: Consider a continuous vector field F : � -+ JR2 where � is the closed unit disc in the plane, and suppose that we want to approximate F by a vector field that vanishes (equals zero) at most finitely often. A simple way to do so is to approximate F by a polynomial vector field G . Real polynomials in two variables are finite sums
P (x . y) =
n
L ciix ; yi
i , j =O
where the Cij are constants. They form a function algebra A in C 0 ( � , JR) that
Chapter 4
Function Spaces
228
separates points and vanishes nowhere. By the Stone-Weierstrass Theorem, A is dense in C 0 , so we can approximate the components of F = (F1 , F2 ) by polynomials F1 = P Fz = Q .
The vector field ( P , Q) then approximates F . Changing the coefficients of P by a small amount ensures that P and Q have no common polynomial factor and F vanishes at most finitely often.
5
Contractions and ODE 's
Fixed point theorems are of great use in the applications of analysis, includ ing the basic theory of vector calculus such as the general implicit function theorem. If f : M --+ M and for some p E M, f( p ) = p, then p is a fixed point of f . When must f have a fixed point? This question has many answers, and the two most famous are given in the next two theorems. Let M be a metric space. A contraction of M is a mapping f : M --+ M such that for some constant k < 1 and all x, y E M,
f(y)) � kd(x, y). 2 3 Banach Contraction Principle Suppose that f M d(f (x),
: --+ M is a con traction and the metric space M is complete. Then f has a unique fixed point p and for any x E M, f" (x) = f o f o · · · o f(x) � p as n --+ 00.
Brouwer Fixed Point Theorem Suppose that uous where Bm is the closed unit ball in Rm. p E Bm.
f
:
Bm --+ Bm
Then
f
is contin has a fixed point
The proof of the first result is easy, the second not. See Figure 96 to picture a contraction. Proof #1 of the Banach Contraction Principle Beautiful, simple, and dy
namical ! See Figure 96. Choose any xo claim that for al1 n E N
E M and define Xn = f" (xo). We
( 1 8) This is easy:
d(xn , Xn+l ) � kd(f(Xn-J ), f(X71 ) ) � k2 d(f(Xn-z ), f(Xn-1 )) n � � k d(xo , xi ) . ·
·
·
From this and a geometric series type of estimate, it follows that the se quence (xn ) is Cauchy. For let E > 0 be given. Choose N large enough
Section 5
Contractions and ODE's
\
� X
•
•
\
fi
•
'1
/
Figure 96
Xo •
•
I
y
• Jy
•
Xn
••
r
•
229
/ fM
M
� -----
•
p
"
�
f contracts M toward the fixed point p
that
kN < E -1 - k d(xo, xt) . Note that ( 1 9) needs the hypothesis k < 1 . If N ::::: ( 1 9)
::::
m
n then
d(Xm , Xn) ::::: d(Xm , Xm+l) + d(Xm+ l • Xm+2 ) + · · · + d(Xn-t . Xn ) ::::: k"'d(xo, X J ) + km +ld(xo, xt ) + + kn -ld(xo, Xt ) ::::: km (l + k + · · · + kn - m -l)d(xo, X t) k :::: kN ( L kl ) d(xo, xt ) = 1 _N k d(xo, Xt) < E . l=O Thus (xn) is Cauchy. Since M is complete, Xn converges to some p E M ·
·
·
oo
as n --* oo . Then
d(p, f(p )) ::::: d( p, Xn) + d(Xn , f(xn)) + d(f(xn ), f ( p )) ::::: d(p , Xn ) + kn d(xo, xt) + kd(xn , p ) 0 as n --* --*
00 .
Since d (p, f(p ) ) i s independent of n, d( p , f( p)) = 0 and p = f(p ) . This proves the existence of the fixed point. Uniqueness is immediate. After all, how can two points simultaneously stay fixed and move closer together? D
230
Function Spaces
Chapter 4
Proof #2 of the Banach Contraction Principle - sketch Choose any point xo E M and choose ro so large that f(M,.0 (xo)) C Mr0 (xo ) - Let B0 =
= t n (Bn - d - The diameter of Bn is at most kn diam(B0 ) , and this tends to 0 as n --+ oo . The sets Bn nest downward as n --+ oo and f sends Bn inside Bn +I · Since M is complete, this implies that n Bn is a single point, say p, and f ( p ) = p . 0
Mr (xo ) and Bn 0
Proof of Brouwer's Theorem in dimension one The closed unit 1 -ball is the interval L - 1 , 1 ] in JR. If f : [ - 1 , l ] --+ [ - 1 , 1 ] is continuous then
so is g(x ) = x - f(x). At the endpoints ± 1 , we have g( - 1 ) ::::= 0 ::::= g( l ) . By the intermediate value theorem, there is a point p E [- 1 , 1 ] such that 0 g(p) = 0. That is, f(p) = p.
The proof in higher dimensions is harder. One proof is a consequence of the general Stokes ' Theorem. and is given in Chapter 5 . Another depends on algebraic topology, a third on differential topology. Ordinary Differential Equations
The qualitative theory of ordinary differential equations (ODE's) begins with the basic existence/uniqueness theorem in ODE's, Picard's Theorem. Throughout, U is an open subset of m-dimensional Euclidean space :!Rm . In geometric terms. an ODE is a vector field F defined on U . We seek a trajectory Y of F through a given p E U , i.e., Y : (a , b) --+ U is differentiable and solves the ODE with initial condition p . (20)
Y ' (t) = F(Y (t)) and Y (O) =
p.
See Figure 97. In this notation w e think o f the vector field F defining at each x E U a vector F (x) whose foot lies at x and to which Y must be tangent. The vector Y ' (t) is (Y; (t) , . . . . Y� (t)) where Y 1 , Y m are the components of Y . The trajectory Y (t) describes how a particle travels with prescribed velocity F. At each time t, Y (t) is the position of the particle; its velocity there is exactly the vector F at that point. Intuitively, trajectories should exist because particles do move. The contraction principle gives a way to find trajectories of vector fields, or - what is the same thing to solve ODE ' s. We will assume that F satisfies a Lipschitz condition - there is a constant L such that for all points x, y E U I F(x) - F (y) l ::::: L l x - y j . .
Here, I
I refers to the Euclidean length of a vector.
F,
x,
.
.
•
y are all vectors
Contractions and ODE's
Section 5
23 1
Figure 97 Y is always tangent to the vector field
F.
in :!Rm . It follows that F is continuous. The Lipschitz condition is stronger than continuity, but still fairly mild. Any differentiable vector field with a bounded derivative is Lipschitz.
Given p E U there exists an F -trajectory Y (t) in This means that Y : (a . b) � U solves (20). Locally, Y zs
24 Picard's Theorem
U through unique.
p.
To prove Picard's Theorem it is convenient to re-express (20) as an inte gral equation and to do this we make a brief digression about vector-valued integrals. Let's recall four key facts about integrals of real valued functions of a real variable, y = f(x ) , a ::::: x ::::: b. (a) J: f (x) dx is approximated by Riemann sums R = L f (tk) D.xk. (b) Continuous functions are integrable. (c) If f'(x) exists and is continuous then J: f'(x) dx = f(b) - f(a) .
jJ:
l
(d) f (x) dx ::::: M(b - a) where M = sup J f (x) J . The Riemann sum R in (a) has a = x0 ::::: · · · ::::: Xk- 1 ::::: Xn = b and all the D.xk = Xk - Xk - 1 are small. Given a vector-valued function of a real variable
a
:::::
x
tk ::::: xk
:::::
· · ·
:::::
j(x ) = ( /1 (x) . . . . . fm (x)) .
:::::
1b f (x) dx (1b j, (x) dx , . . . , 1b fm (x) dx)) .
b, we define its integral componentwise as the vector of integrals =
Function Spaces
232
Chapter 4
J:
Corresponding to (a) - (d) we have (a') f(x) dx is approximated by R = (R 1 , . . . , Rm ), with Rj a Riemann sum for [j . (b') Continuous vector-valued functions are integrable. (c') If f'tx) exists and is continuous, then J: f'(x) dx = f(b) - f(a).
l
I J:
(d' ) f(x) dx ::=: M (b - a) where M = sup l f(x) l . (a'), (b' ), (c') are clear enough. To check (d') we write
R = =
L Rjej = L L [j (tk) D.. xkej j k L L [j (tk )ej D.. Xk = L f(tk) D.. xk k k j
where e1 , . . . , em is the standard vector basis for :!Rm . Thus,
IRI
:S
L l f(tk ) l D.. xk k
::=:
L M D.. xk = M (b - a ) . k
By (a' ), R approximates the integral, which implies (d'). (Note that a weaker inequality with M replaced by ,JfiiM follows immediately from (d). This we aker inequality would suffice for most of what we do - but it is inele gant.) Now consider the following integral version of (20),
Y ( t) = p +
(2 1 )
1 ' F(Y (s) ) ds.
A solution of (2 1 ) is by definition any continuous curve Y : (a. b) � U for which (2 1 ) holds identically in t E (a , b). By (b') any solution of (2 1 ) i s automatically differentiable and its derivative i s F ( Y (t)) . That is. any solution of (2 1 ) solves (20) . The converse is also clear, so solving (20) is equivalent to solving (2 1 ). Proof of Picard's Theorem Since
F is continuous. there exists a compact (p) and a constant M such that I F (x) I ::=: M for all
neighborhood N = N r > 0 such that
x E N . Choose r (22)
r M ::=:
r
and r L < 1 .
Consider the set (3 of all continuous functions respect to the metric
Y
[-r, r]
d( Y , a) = sup{I Y (t) - a (t) l : t E [- r. r ] }
�
N.
With
Contractions and ODE's
Section 5
233
the set e is a complete metric space. Given y E e. define
(Y ) (t) =
p
+ for F(Y (s)) ds.
Solving (2 1 ) is the same as finding Y such that (Y) = Y. That is, we seek a fixed point of . We just need to show that is a contraction of e . Does send e into itself? Given Y E e we see that (Y ) (t ) is a continuous (in fact differentiable) vector-valued function of t and that by (22),
l (Y) (t) - P i =
1 1 1 F (Y (s)) ds l .:s rM
Therefore, does send e into itself. contracts e because
d( (Y ) , (u )) = s�p � �
�
r .
1 1 t F (Y (s)) - F (u (s)) ds l
T
sup I F (Y (s)) - F(u (s) ) l
r
sup L I Y (s) - u (s) l � rLd(Y , u ) s
and T L < 1 by (22). Therefore has a fixed point Y . and (Y) = Y implies that Y (t) solves (2 1), which implies that Y is differentiable and solves (20). Any other solution u (t) of (20) defined on the interval [ - r, r] also solves (2 1 ) and is a fixed point of , ( u ) = u . Since a contraction mapping has a unique fixed point, Y = u , which is what local uniqueness means. 0 The F -trajectories define a flow in the following way: To avoid the pos sibility that trajectories cross the boundary of U (they "escape from U") or become unbounded in finite time (they "escape to infinity") we assume that U is all of JRm . Then trajectories can be defined for all time t E JR. Let Y (t , p) denote the trajectory through p. Imagine all points p E JRm moving in unison along their trajectories as t increases. They are leaves on a river, motes in a breeze. The point PI = Y (ti . p ) at which p arrives after time t i moves according to Y (t, pi). Before p arrives at P I . however, p 1 has already gone elsewhere. This is expressed by the flow equation
Y (t, PI ) = Y (t + t 1 , p).
See Figure 98. The flow equation is true because as functions of t both sides of the equation are F -trajectories through PI , and the F -trajectory through a point
Function Spaces
234
_ _
/
/
y
I
I
/
�
/
/
/
.,
.-
Chapter 4
_
PJ
_...
/
/
/ / ,. / P2
/
/
p
Figure 98 The time needed to flow from from p to
p2
times needed to flow from p to P I and from
is the sum of the to P2 ·
PI
is locally unique. It is revealing to rewrite the flow equation with different notation. Setting qJ1 (p) = Y (t, p) gives
(/Jt +s (p) = (/Jr ( (/Js (p)) for all t, s E ffiL
(/)1 is called the t-advance map. It specifies where each point moves after time t . See Figure 99. The flow equation states that t �---+ (/Jr is a group /
/
/
/
/
Jf
/
/
"
/ /
"
/ /
Jf
/
Jf
/
"
Jf
Figure 99 The t-advance map shows how a set A flows to a set qJ1 (A ) .
homomorphism from ffi. into the group o f motions o f ffi.m . In fact each (/)1 i s a homeomorphism o f ffi.m onto itself and its inverse i s qJ_1 • For (/J- r o (/)1 = ({Jo and q;0 is the time-zero map where nothing moves at all, ({Jo = identity map.
Analytic Functions
Section 6*
6*
235
Analytic Functions
Recall from Chapter 3 that a function f : (a , b) � JR. is analytic if it can be expressed locally as a power series. For each x E (a , b) there exists a convergent power series L ck h k such that for all x + h near x,
f (x + h )
00
=
:I:Ck hk . k =O
As we have shown previously, every analytic function is smooth but not every smooth function is analytic. In this section we give a necessary and sufficient condition that a smooth function be analytic. It involves the speed with which the r th derivative grows as r � oo. Let f : (a , b) � JR. be smooth. The Taylor series for f at x E (a , b) is
Let I = [x - a, x + a ] be a subinterval of (a , b) , a > 0, and denote by Mr the maximum of l !(r ) (t) l for t E I . The derivative growth rate of f on I is
.
a = hm sup
r ---> 00
Clearly,
r i t (x ) l f r !
::S:
�Mr !r
-.
:./Mr fr!, s o the radius o f convergence
R=
------,====
lim sup
r I J< r > (x) l r!
of the Taylor series at x satisfies 1 - ::S:
a
R.
In particular, if a is finite the radius of convergence of the Taylor series is positive. 25 Theorem If a a < 1 , th e interval I.
un
then the Taylor series converges uniformly to f
Function Spaces
236
Proof Choose 8 > 0 such that (a + 8)a
0 be given. Choose a sequence of points Pn E M and radii rn > 0 such that rn < 1 / n and
M2q ( p i ) c M"' (po ) M2r2 ( P2 ) C Mr1 (p d n G t M2rn ( Pn + d
C
Mrn (Pn ) n G t · · · n Gn .
See Figure 1 0 1 . Then
ME (po )
:J
Mr J (PI )
:J
Mrz (P2 )
:J
·
·
·
·
The diameters of these sets tend to 0 as n ---+ oo . Thus (pn) is a Cauchy sequence and it converges to some p E M, by completeness. The point p belongs to each set M rn ( Pn ) and therefore it belongs to each G n . Thus p E G n ME (p0 ) and G is dense in M. To check that M i s not thin, w e take complements. Suppose that M = UKn and Kn is closed. If each Kn has empty interior, then each G n = K� is open-dense, and
a contradiction to density of G.
0
245
Nowhere Differentiable Continuous Functions
Section 7*
Figure 101 The closed neighborhoods Mrn (pn ) nest down to
33 Corollary
and thin.
a
point.
No subset ofa complete non-empty metric space is both thick
Proof If S is both a thick and thin subset of M then M \
S is also both
thick and thin. The intersection of two thick subsets of M is thick, so
0 = S n (M \ S) is a thick subset of M. By Baire's theorem, this empty set D is dense in M, so M is empty.
To prove Theorem 32 we use two lemmas. 34 Lemma
The set P L ofpiecewise linear functions is dense in C 0 .
Proof If � :
---* IR is continuous and its graph consists of finitely many line segments in :IR2 then � is piecewise linear. Let f E C0 and E > 0 be given. Since [a , b] is compact, f is uniformly continuous, and there is 8 > 0 such that I t - s l < 8 implies l f (t) - f (s) l < E . Choose n > (b - a)/8 and partition [a , b] into n equal subintervals /i = [xi - l , xd, each of length < 8. Let � : [a , b] ---* :IR be the piecewise linear function whose graph consists of the segments joining the points (xi 1 , f (xi 1 )) and
[a, b]
_
_
(xi , f (xi )) on the graph of f. See Figure 102 . The value of � (t) for t E h lies between f (x i - d and f (x i ) . Both these numbers differ from f (t) by less than E . Hence, for all t E [a , b],
Function Spaces
246
Chapter 4
X;
Xi - 1
Figure 102 The piecewise linear function ¢ approximates the continuous
function f.
l f (t) - f/J (t) l < E. In other words, d (f, ¢) < E and P L is dense in C 0 .
D
35 Lemma If ¢ E P L and E > 0 are given then there exists a sawtooth function a such that II a II ::; E, a has period ::; E, and I
min{ l slope a l } > max{ l slope ¢ 1 } + - . Proof Let () = max{ l slope ¢ 1 } and choose
E
c large. The compressed saw tooth a (x ) = Ea0 (cx) has l l a ll = E , period rc = l jc, and slope s = ±Ec. D When c is large. rc ::; E. and l s i > () + l jE .
Proof of Theorem 32 For n E
Rn = { f E C 0 : Vx E [a , b L n = { f E C 0 : Vx E [a +
N define
�]
3h
> 0 such that
� · b ] 3 h < 0 such that
I D..: I I D..: I
>
n}
>
n}
1 G n = { f E C 0 : f restricted to any interval of length - is non-monotone} ,
D..
n
where f = f (x + h) - f (x) . We claim that each of these sets is open dense in C0 .
Section 7*
Nowhere Differentiable Continuous Functions
247
For denseness it is enough to prove that the closure of each contains P L , since by Lemma 34 the closure o f P L i s C 0 Let cp E P L and E > 0 be given_ According to Lemma 35, there is a sawtooth function a such that l l a l l � E , a has period < 1 / n , and _
min{ l slope a I }
>
max{ l slope c/J I } + n .
Consider the piecewise linear function f = cp + a . Its slopes are dom inated by those of a , and so they alternate in sign with period < 1 j2n . At any x E [a , b + 1 / n ] there is a rightward slope either > n or < - n _ Thus f E Rn . Similarly f E L n . Any interval I of length 1 / n contains in its interior a maximum or minimum of a , and so it contains a subinterval on which f strictly increases, and another on which f strictly decreases. Thus, f E Gn . Since d(j. cp) = E is arbitrarily small this shows that Rn , L n . and G n are dense in C 0 . Next suppose that f E Rn is given. For each x E [a . b - 1 /n ] there is an h = h (x) > 0 such that
Since f is continuous, there is a neighborhood Tx of constant v = v > 0 such that this same h yields
I
f (t + h) h
-
x
in [a . b] and a
I >n+v
f(t)
for all t E Tx . Since [a , b - 1 / n J is compact, finitely many of these neigh borhoods Tx cover it, say Tx1 , • • • , Txm . Continuity of f implies that for all t E T Xi ' (25)
I
f (t + h i ) - f (t) h I-
I
::: n + vi ,
where hi = h (xi ) and vi = v (xd . These m inequalities for points t in the m sets Tx, remain nearly valid if f is replaced by a function g with d (j g) small enough; (25) becomes (26) which means that g E Rn and Rn is open in C 0 • Similarly L n
is
open in C 0 •
Function Spaces
248
Chapter 4
Checking that G n is open is easier. If (jk ) is a sequence of functions in G� and fk :::::::t f then we must show that f E G� . Each fk is monotone on some interval h of length 1 In. There is a subsequence of these intervals that converges to a limit interval I . Its length is l 1 n and by uniform convergence, f is monotone on I . Hence G� is closed and G n is open . Each set Rn . L n . G n is open dense in C 0 . Finally, if f belongs to the thick set
then for each x E
[a, b] there is a sequence h n
I
f (x + hn) - f (x) hn
I
f= 0 such that
>
n.
The numerator of this fraction is at most 2 II f 11. so hn -4 0 as n -4 oo. Thus f is not differentiable at x . Also, f is non-monotone on every interval of length lin. Since each interval J contains an interval of length 1/n when D n is large enough, f is non-monotone on J. Further generic properties of continuous functions have been studied, and you might read about them in the books A Primer of Real Functions by Ralph Boas, Diffe rentiation of Real Functions by Andrew Bruckner, or A Second Course in Real Functions by van Rooij and Schikhof.
8*
Spaces of Unbounded Functions
How important is it that the functions we deal with are bounded, or have domain [a, b] and target JR? To some extent we can replace [a, b] with a metric space X and lR with a complete metric space Y . Let :F denote the set of all functions f : X -4 Y . Recall from Exercise 2.84 that the metric dy on Y gives rise to a bounded metric
p (y, y ) '
=
1
dy ( y , y ' ) . + dy ( y . y')
where y, y ' E Y . Note that < 1 . Convergence and Cauchyness with p respect to and dy are equivalent. Thus completeness of Y with respect to dy impliespcompleteness with respect to p . In the same way we give :F the metric d d(f, g ) = sup y (j(x) , g (x)) XEX
1 + dy (f(x), g (x ) )
Spaces of Unbounded Functions
Section 8*
249
A function f E :F is bounded with respect to dy if and only if for any constant function c, supx dy (f(x) , c) < oo; i.e., d(f, c) < I . Unbounded functions have d (f, c) = 1 .
In the space :F equipped with the metric d, Uniform convergence of (fn) is equivalent to d-convergence. Completeness of Y implies completeness of F. The set :Fb of bounded functions is closed in F. The set C 0 ( X , Y) of continuous functions is closed in F.
36 Theorem
(a) (b) (c) (d)
f = unif lim means that dy (fn (x) , f (x)) � 0, which means n -HXl that d(fn , f) -+ 0.
Proof (a)
(b) If Un) is Cauchy in :F and Y is complete then, just as in Section 1 , f(x) nlim -+ oo fn (x ) exists for each x E X. Cauchyness with respect to the metric d implies uniform convergence and thus d ( fn , f) -+ 0. (c) If fn E :Fb and d(fn , f) -+ 0 then supx dy (fn (x), f (x )) -+ 0. Since fn is bounded, so is f. D (d) The proof that C 0 is closed in :F is the same as in Section 1 . =
The Arzela-Ascoli theorem is trickier. A family £ C :F is uniformly equicontinuous if for each E > 0 there is a 8 > 0 such that f E £ and dx (x , t) < 8 imply dy ( J (x ) , f(t)) < E . If the 8 depends on x but not on f E £ then £ is pointwise equicontinuous. 37 Theorem
Pointwise equicontinuity implies uniform equicontinuity if X
is compact. Proof Suppose not. Then there exists E > 0 such that for each 8 = 1 / n we have points Xn , tn E X and functions fn E £ with dx (Xn , tn) < 1 /n
and
Xn
dy (fn (Xn ) , fn C tn)) 2: E . By compactness of X we may assume that -+ xo . Then tn -+ xo , which leads to a contradiction of pointwise D
equicontinuity at x0 .
38 Theorem Ifthe sequence offunctions fn : X -+ Y is uniformly equicon tinuous, X is compact, andfor each x E X, (fn ( ) ) lies in a compact subset of Y, then (fn) has a uniformly convergent subsequence. x
D. Then the proof of the Arzela Ascoli Theorem in Section 3 becomes a proof of Theorem 38.
Proof Being compact, X has a countable dense subset
D
250
Function Spaces
Chapter 4
The space X is u -compact if it is a countable union of compact sets, X = m UX i . For example Z, 0. (c) Does the existence of these non-unique solutions to the ODE contradict Picard's Theorem? Explain. * (d) Find all solutions with initial condition y (O) = 0. 33. Consider the ODE x ' = x 2 on R Find the solution of the ODE with initial condition xo . Are the solutions to this ODE defined for all time or do they escape to infinity in finite time?
Function Spaces
256
Chapter 4
Suppose that the ODE x' = f (x ) on lR is bounded, I f (x ) 1 ::::: M for all x . (a) Prove that no solution of the ODE escapes to infinity in finite time. (b) Prove the same thing if f satisfies a Lipschitz condition, or, more generally, if there are constants C, K such that (x ) C l x l + K for all x . (c) Repeat (a) and (b) with JRm i n place of JR. (d) Prove that if : JRm ---+ JR.m is uniformly continuous then the condition stated in (b) is true. Deduce that solutions of uni formly continuous ODE's defined on JR.m do not escape to in finity in finite time. * * 3 5 . (a) Prove Borel's Lemma : given any sequence whatsoever o f real numbers (ar ) there is a smooth function : lR ---+ lR such that J (O) = ar . [Hint: Try = L fJk (x)akx k f k ! where fJk is a well chosen bump function. ) (b) Infer that there are many Taylor series with radius of conver gence = 0. (c) Construct a smooth function whose Taylor series at every x has radius of convergence = 0. [Hint: Try L fJk (x) e (x + qk) where {qJ , qz , . . . } = Q.] *36. Suppose that T C (a . b) clusters at some point of (a , b) and that J, g : (a , b) ---+ lR are analytic. Assume that for all t E T, ( t ) =
34.
If
I :::::
f
f
f
R
R
f
g (t) .
(a) Prove that = g everywhere in (a , b). (b) What if and g are only C00? (c) What i f T i s an infinite set but its only cluster points are a and
f
f
b?
* * (d) Find a necessary and sufficiem condition for a subset Z c (a , b) to be the zero locus of an analytic function defined on (a , b), Z = {x E (a , b) : f (x ) = 0}. [Him: Think Taylor. The result in (a) is known as the Identity Theorem. It states that if an equality between analytic functions is known to hold for points of T then it is an identity, an equality that holds everywhere.] 37. Let M be any metric space with metric d. Fix a point p E M and for each q E M define the function = d(q , x) - d(p, x ) . (a) Prove that i s a bounded, continuous function of x E M , and that the map q �---+ J sends M isometrically onto a subset Mo of C 0 (M, JR.) .
f
fq
fq (x)
q
Exercises
257
(b) Since C 0 (M. JR.) is complete, infer that an isometric copy of M is dense in a complete metric space, the closure of M0 , and hence that we have a second proof of the Completion Theorem 2. 76. 38. As explained in Section 8, a metric space M is cr -compact if it is the countable union of compact subsets, M = U M; . (a) Why i s it equivalent to require that M is the monotone union of compact subsets,
i.e., M1 C M2 C . . . ? (b) Prove that a a -compact metric space is separable. (c) Prove that Z, f(x) for all x E R Prove that f (x ) > 0 for all x > 0. 8 . Suppose that f : [a , b] -4 JR. and the limits of f (x) from the left and the right exist at all points of [a . b]. Prove that f is Riemann integrable. 9. Let h : [0. 1 ) -4 JR. be a uniformly continuous function where [0, 1 ) i s the half open interval. Prove that there is a unique continuous map g : [0. 1 ] -4 JR. such that g (x ) = h (x) for all x E [0. 1 ) . 1 0. Assume that f : JR. -4 JR. is uniformly continuous. Prove that there are constants A , B such that I f (x) I :::::: A + B lx I for all x E JR.. 1 1 . Suppose that f (x) is defined on [ - 1 , 1 ] and that its third derivative exists and is continuous . (That is, f is of class . ) Prove that the series
C3
L (n ( f ( l jn ) - f ( - 1 /n )) - 2J ' (O) ) 00
n =O
1 2.
1 3.
converges. Let A C IR.m be compact, x E A. Let (xn ) be a sequence in A such that every convergent subsequence of (xn ) converges to x . (a) Prove that the sequence (xn ) converges . (b) Give an example to show if A is not compact, the result in (a) is not necessarily true. Let f : [0, 1 ] -4 JR. be continuously differentiable, with f ( O ) = 0. Prove that
1 /11 2 :::::: 1 \f'(x ) )2 dx
1 /1 1 = sup { l f (t ) l : 0 :::::: t _::: I } .
where 14. Let fn :
fn (0)
=
JR. JR. be 0 and I f� (x) I -4
differentiable functions, n = _::: 2 for al l n , x . Suppose that
nlim -+oo fn (x) = g (x)
for all x . Prove that g is continuous.
1 , 2,
. . . with
Function Spaces
260
Chapter 4
15.
Let X be a non-empty connected set of real numbers. If every element of X is rational, prove that X has only one element. 1 6 . Let k � 0 be an integer and define a sequence of maps fn lR lR as
:
�
xk
fn (X) = � x +n
n
= 1 , 2, . . . . For which values of k does the sequence converge
uniformly on JR? On every bounded subset of lR? 1 7 . Let f [0, 1] lR be Riemann integrable over [b , 1] for every b such that 0 < b (a) If f i s bounded, prove that f is Riemann integrable over [0, 1 ] . (b) What if f i s not bounded? 1 8 . (a) Let S and T be connected subsets of the plane JR2 having a point in common. Prove that S U T is connected. (b) Let { Sa } be a family of connected subsets of JR2 all containing the origin. Prove that u sa is connected. 19. Let f : lR --+ lR be continuous. Suppose that lR contains a countably infinite set S such that
:
� :::: 1 .
iq f
(x) dx = 0
if p and q are not in S. Prove that f is identically zero. 20. Let f lR lR satisfy f (x) :::: f (y) for x y. Prove that the set where f is not continuous is finite or countably infinite. 2 1 . Let (gn ) be a sequence of Riemann integrable functions from [0, 1 ] into lR such that l gn (x) l :::: 1 for all n , x . Define
:
�
::::
1x gn (t) dt.
Gn (X ) =
Prove that a subsequence of (G n ) converges uniformly. 22. Prove that every compact metric space has a countable dense subset. 23. Show that for any continuous function f [0, 1] lR and for any E > 0 there is a function of the form
:
g (x ) = for some
n
n
L Ck x k k =O
E for all x in [0, 1 ] . lR --+ lR having all three of the
E N, and l g (x) - f(x ) l
2. (b) = 1, (c) has derivatives of all orders. JR JR whose 25 . (a) Give an example of a differentiable function derivative is not continuous. (b) Let f be as in (a). If < 2 < prove that (x = 2 for some x E [0, 1 ] . 26. Let U C JRm be an open set. Suppose that the map h U --+ JRm is a homeomorphism from U onto JRm which is uniformly continuous. Prove that U = JRm . 27. Let be a sequence of continuous maps [0, 1 ] JR such that
f'(l) f
f : --+
f'(O)
f'(l)
f ) '
:
fn (x)
1 1 ' fn (x) dx I
(x)
:S:
K
J01 fn (x ) dx
for all n where K is a constant. Does converge to 0 as n � oo? Prove or give a counter-example. 40. Let E be a closed, bounded, and non-empty subset of IRm and let f : E ---* E be a function satisfying I f (y) < - y I for all x , y E E, x t=- y . Prove that there is one and only one point E E such that f(xo) = 4 1 . Let : 2.rr I ---* lR b e a continuous function such that
(x) - f
Xo.
f LO,
for all integers
I
n
�
1 2:n: f(x) sin(nx) dx
1.
=0
Prove that f is identically zero.
lx
x0
Exercises
263
u
42. Let E be the set of all real valued functions : [0, I ] -+ IR satisfying = 0 and y ) l � l x - y l for all x , y E [0, 1 ] . Prove that the function
u(O)
lu(x) - u (
11 ((u(x) 2 - u (x)) dx
0. Prove that is one-to one, onto, and it has a continuous inverse. 46. Show that [0, 1 ] cannot be written as a countably infinite union of disjoint closed sub-intervals. 47. Prove that a continuous function f : IR -+ IR which sends open sets to open sets must be monotonic. 48. Let f : [0, oo) -+ IR be uniformly continuous and assume that ..... oo
lim
b
r b f (x ) dx
Jo
x ..... oo
exists (as a finite limit). Prove that lim
f (x )
= 0.
49. Prove or supply a counter-example: If f and g are continuously differentiable functions defined on the interval 0 < x < which satisfy the conditions
1
x---> 0
lim
f(x )
x-+ 0
= 0 = lim g (x )
and
J ' (x ) . an d 1 f g and g ' never vant· sh , then 1 Im · -- = x ..... o g ' (x) of L' Hospital's rule. )
f(x)
x---> 0 g (x )
lim c.
(Th"IS
IS ·
=c a converse
264
Function Spaces
Chapter 4
50. Prove or provide a counter-example: If the function f from JR to lR has both a left and a right limit at each point of JR, then the set of discontinuities is at most countable. 5 1 . Prove or supply a counter-example: is a non-decreasing real val ued function on [0 , 1 1 then there is a sequence n = 1 , 2, . . . of continuous functions on [0, 1 ] such that for each in [0, 1 ] , lim =
IT f
n -+oo fn (X)
j(x). f
fn ,
x
52. Show that if is a homeomorphism o f [0, 1 ] onto itself then there is a sequence of polynomials (x ) , n = 1 , 2, . . . , such that � uniformly on [0. 1 ] and each is a homeomorphism of [0. 1 ] onto 1 itself. [Hint: First assume that is C .] 53. Let be a C 2 function on the real line. Assume that is bounded with bounded second derivative. Let A = supx and B = supx I !" (x } I · Prove that
Pn Pn f
f
sup X
54. Let
I J (x) l '
Pn
f
f I f (x) I
_:::: 2 -Ji:B .
( k)
f be continuous on lR and let n l 1 fn (X) = ;; L J X + -n · k=O
fn (x)
Prove that converges uniformly to a limit on every finite interval [a , b] . 5 5 . Let be a real valued continuous function on the compact interval [a , b ] . Given E > 0, show that there is a polynomial such that
f
p
p(a) = f(a), p ' (a) = 0, and l p(x) - f(x) l < E.
,
for all x E [ b] . 5 6 . A function : [0, 1 ] � lR i s said t o b e upper semicontinuous if, given x E [0, L J and E > 0, there exists a 8 > 0 such that l y 0 such that if u E V and l u i < � then
I Tu l
For any nonzero and
E V , set u
v
I Tvl
--
lvl
o ( Df ) x and so differentiating this product produces .
.
.
.
284
Chapter 5
Multivariable Calculus
(The meaning of f) ; needs clarification.) Differentiating again produces four terms, two of which combine. The general formula is
(D
(Dr g
0
f) x =
r
L �)Dkg ) f(x ) k= i
ll
0
(D11 n x
where the sum on Jl is taken as f.l runs through all partitions of { I , . . . , r } into k disjoint subsets. See Exercise 4 1 . The higher-order Leibniz rule i s left for you a s Exercise 42.
4
Smoothness Classes
A map f : U --+ �m is of class cr if it is r order differentiable at each p E U and its derivatives depend continuously on p. (Since differentia bility implies continuity, all the derivatives of order < r are automatically h continuous; only the rt derivative is in question.) If f is of class for all r, it is smooth or of class c oo . According to the differentiation rules, these smoothness classes are closed under the operations of linear combination, product, and composition . We discuss next how they are closed under limits. Let ( fk ) be a sequence of functions fk : U -4 �m . The sequence is (a) Uniformly cr convergent if for some C' function f : U -4 �m ,
th
cr
cr
as k -4 oo.
(b) Uniformly cr Cauchy if for each E > 0 there is an N such that for all k, f. � N and all x E U,
lfk(x) - fe(x) i < E l i (Dfk)x - (Dfdx l < E 19 Theorem
Uniform
cr convergence and Cauchyness are equivalent.
Proof Convergence always implies the Cauchy condition. As for the con verse, first assume that r = 1 . We know that fk converges uniformly to a
Section 4
Smoothness Classes
285
continuous function f, and the derivative sequence converges uniformly to a continuous limit Dfk =:::t G .
We claim that Df = G. Fix p E U and consider points q in a small convex 1 neighborhood of p. The C Mean Value Theorem and uniform convergence imply that as k � oo,
fk (q) - fk (P)
=
tt
f (q ) - f (p)
=
11
(Dfk)p+r (q- p J dt (q - p)
1 1 G (p
tt +
t (q - p)) d t (q - p) .
This integral of G is a continuous function of q that reduces to G (p ) when p = q . By the converse part of the C 1 Mean Value Theorem, f is differen tiable and Df = G. Therefore f is C 1 and fk converges C 1 uniformly to f as k ---+ oo, completing the proof when r = 1 . Now suppose that r 2: 2. The maps Dfk : U ---+ £ form a uniformly C' - 1 Cauchy sequence. The limit, by induction, is C' - 1 uniform; i.e., as
k ---+
00,
Ds (Dfk )
=:::t
D·' G
for all s ::S r - I . Hence fk converges C' uniformly to f as k completing the induction. The cr norm of a C' function f : U � IRm is
ll f ll ,
= max f sup xEU
I I , makes C' (U, �m)
normed vector space.
oo, D
l f (x ) l , . . . , sup II (D ' nx II }. xEU
The set of functions with l l f ll , < oo is denoted C (U. �m ) . 20 Corollary II
�
a
Banach space -
a
complete
Proof The norm properties are easy to check; completeness follows from Theorem 19. D
is a convergenT series of constants and if II fk II , ::S Mk for all k, then the series offunctions L fk converges in C' ( U , JRm ) to a function f. Term by term d(ffe rentiation of order ::S r is valid, D' f = L k D' fk · 21 cr M -test If "'E. Mk
Proof Obvious from the preceding corollary.
D
5
Chapter 5
Multivariable Calculus
286
Implicit and Inverse Functions
Let f : U ---+ JR.m be given, where U is an open subset of JRn x JRm. Fix attention on a point (xo , Yo) E U and write f (xo , Yo) = zo. Our goal is to solve the equation
f(x , y) = zo
(7)
near (xo, Yo ) . More precisely, we hope to show that the set of points (x , y) nearby (xo, Yo) at which f (x , y) = zo, the so-called zo-locus of f , is the graph of a function y = g(x). If so, g is the implicit function defined by (7) . See Figure 1 06. f= Zo
Yo
f = Zo
Figure 106 Near
(xo , Yo) the zo-locus of f is the graph of a function y = g (x ) .
Under various hypotheses we will show that g exists, is unique, and is differentiable. The main assumption. which we make throughout this section, is that
the m x m matrix
B=
[ afi (xoayj, Yo) ]
is invertible.
Equivalently the linear transformation that B represents is an isomorphism
]Rm ---+ ]Rm .
If the function f above is C, 1 .:::: r .:::: oo, then near (xo, Yo). the zo-locus of f is the graph of a unique ju."lction y = g (x). Besides, g is C. 22 Implicit Function Theorem
Proof Without loss of generality, we suppose that (x0 , y0) is the origin (0, 0) in JRn x JRm , and zo = 0 in JRm . The Taylor expression for f is
f (x , y) = Ax + By + R where A is the m
x
n matrix
Section
5
287
Implicit and Inverse Functions
[
(x A = aj; o. Yo)
]
a xj and R is sublinear. Then, solving j(x , y) = 0 for y = gx is equivalent to solving
y = - B -1 (Ax + R (x , y)) . In the unlikely event that R does not depend on y, (8) is an explicit formula (8)
for gx and the implicit function is an explicit function. In general, the idea is that the remainder R depends so weakly on y that we can switch it to the left hand side of (8), absorbing it in the y term. Solving (8) for y as a function of x is the same as finding a fixed point of
Kx : y 1--+ - B -1 ( A x + R (x , y)),
so we hope to show that Kx contracts. The remainder R is a C 1 function, and (DR) (O.O) = 0. Therefore if r is small and lx l , IY I ::::: r then
I B - 1 11
1 a R �:· y) I ::::: � ·
By the Mean Value Theorem this implies that
I Kx Cy i ) - Kx ( Yz) i ::::: I B - 1 II I R(x, yt ) - R (x , Yz) i ::::: I I B - 1 1 1
for
1 �; I IYt - Yz i
:::::
� IY1 - Yz i
lx l , IYI I , iYz i ::::: r . Due to continuity at the origin. if lx l ::::: T « r then I Kx CO) I :::::
I
2.
Thus, for all x E X , Kx contracts Y into itself where X is the T -neighborhood of 0 in JR.n and Y is the closure of the r -neighborhood of 0 in JR.m . See Fig ure 1 07 .
Figure 107
Kx contracts Y into itself.
Multivariable Calculus
288
Chapter 5
By the Contraction Mapping Principle, Kx has a unique fixed point g (x) in Y . This implies that near the origin, the zero locus of f is the graph of a function y = g(x) . It remains to check that g i s c r . First we show that g obeys a Lipschitz condition at 0. We have l gx l = I Kx (gx ) - K, ( O) + Kx CO) I � Lip( Kx ) l gx - 01 + I Kx CO) I 1 1 1 � (x , 0)) x l + x B1 g � (A + R I 1 2 2 lgx l + 2L lx l
where L condition
I B -1 11 11 A I I and l x I is small. Thus
g
satisfies the Lipschitz
l gx l �
4L l x l . In particular, g is continuous at x = 0. Note the trick here. The term l g x I appears on both sides of the inequality but since its coefficient on the r.h.s. is smaller than that on the l.h.s., they combine to give a nontrivial inequality. By the chain rule, the derivative of g at the origin, if it does exist, must satisfy A + B(Dg)0 = 0, so we aim to show that ( Dg)o = -B- 1 A . Since gx is a fixed point of Kx , we have g x = - B 1 A (x + R ) and the Taylor estimate for g at the origin is -
l g (x) - g (O) - (- B - 1 Ax ) l = I B- 1 R (x , g x) l � II B - ' II I R (x , gx ) l � II B - ' 11 e (x , gx ) ( l x l + l g x l ) � II B - ' II e (x , gx) ( l + 4L) I x l where e(x , y) --* 0 as (x, y) --* (0, 0). Since g x --* 0 as x --* 0, the error factor e(x . gx ) does tend to 0 as x --* 0, the remainder is sublinear with respect to x, and g is differentiable at 0 with (Dg)0 = - B-1 A .
All facts proved at the origin hold equally at points (x . y ) on the zero locus near the origin. For the origin is nothing special. Thus, g is differentiable at x and (Dg)x = - B; ' o Ax where
aj (x , gx ) ax
aj(x, gx) . ay Since gx is continuous (being differentiable) and f is C 1 , Ax and Bx are continuous functions of x . According to Cramer's Rule for finding the in verse of a matrix, the entries of B; 1 are explicit, algebraic functions of the Ax =
---
Bx =
entries of Bx . and therefore they depend continuously on x . Therefore g is C 1 • To complete the proof that g is c r , we apply induction. For 2 � r < oo, assume the theorem is true for r - I . When f is c r this implies that g is
Implicit and Inverse Functions
Section 5
289
cr- l . Because they are composites of cr- l functions, Ax and Bx are cr- l . Because the entries of B; 1 depend algebraically on the entries of Bx . B; 1 is also cr - l Therefore ( D g) X is cr- l and g is cr If f is c oo ' we have just D s hown that g is c r for all finite r, and thu s g is c oo . 0
0
Exercises 35 and 36 discuss the properties of matrix inversion avoiding Cramer's Rule and finite dimensionality. Next we are going to deduce the Inverse Function Theorem from the Implicit Function Theorem. A fair question is: since they tum out to be equivalent theorems, why not do it the other way around? Well, in my own experience, the Implicit Function Theorem is more basic and flexible. l have at times needed forms of the Implicit Function Theorem with weaker differentiability hypotheses respecting x than y and they do not follow from the Inverse Function Theorem. For example, if we merely assume that B = af (xo , Yo)fay is invertible, that af (x , y)jax is a continuous function of (x , y) , and that f is continuous (or Lipschitz) then the local implicit function of f is continuous (or Lipschitz). It is not necess ary to assume that f is of class C 1 • Just as a homeomorphism is a continuous bij ection whose inverse is continuous, so a cr diffeomorphism is a cr bij ection whose inverse is C. (We assume 1 � r � oo.) The inverse being C is not automatic. The example to remember is f (x) = x 3 . It is a C00 bijection JR. -4 JR. and is a homeomorphism but not a diffeomorphism because its inverse fails to be differentiable at the origin. Since differentiability implies continuity, every diffeomorphism is a homeomorphism. Diffeomorphisms are to cr things as isomorphisms are to algebraic things. The sphere and ellipsoid are diffeomorphic under a diffeomorphism JR.3 -4 JR. 3 , but the sphere and the surface of the cube are only homeomor phic, not diffeomorphic.
If m = n and f : U � JR.m is C , I � (Df) p is an isomorphism, then f is a C diffeomorphism from a neighborhood of p to a neighborhood of f(p). 23 Inverse Function Theorem r � oo, and if at some p E U,
f(x ) - y. Clearly F is C, F (p, fp ) = 0, and the derivative of F with respect to x at (p, fp) is (Df) p . Since (Df)p is Proof Set F (x ,
y)
=
an isomorphism, we can apply the implicit function theorem (with x and y interchanged ! ) to find neighborhoods Uo of p and V of fp and a C implici t function h : V � Uo uniquely defined by the equation
F (hy , y)
=
0.
Chapter 5
Multivariable Calculus
290
Then f (hy) = y, so f o h = idv and h is a right inverse of f . Except for a little fussy set theory, this completes the proof: f bijects U1 = {x E Uo : f x E V } onto V and its inverse is h , which we know to be a cr map. To be precise, we must check three things, (a) U1 is a neighborhood of p. (b) h is a right inverse of f u 1 • That is, f u o h = idv . 1 (c) h is a left inverse of f u1 • That is, h o f u1 = id u1 • See Figure 108.
l
l
l
l
h
Figure 108 f is a local diffeomorphism.
(a) Since f is continuous, Ut is open. Since p E U0 and fp E V , p belongs to U 1 · (b) Take any y E V . Since hy E Uo and f (hy) = y, we see that hy E U1 . Thus, f u1 o h is well defined and f u o h (y) = f o h (y) = y . 1 (c) Take any x E U1 . By definition of U1 , fx E V and there is a unique point h (fx) in Uo such that F (h (fx ) , fx) = 0. Observe that x itself is just such a point. It lies in Uo because it lies in U1 , and it satisfies F (x , fx) = 0 0 since F (x , fx) = fx - f x . By uniqueness of h, h (f (x)) = x .
l
l
Upshot If (Df)p is an isomorphism, then f is a local diffeomorphism
at p.
6*
The Rank Theorem
The rank of a linear transformation T : IRn � IRm is the dimension of its range. In terms of matrices, the rank is the size of the largest minor with nonzero determinant. If T is onto then its rank is m . If it is one-to-one, its rank is n . A standard formula in linear algebra states that rank T + nullity T = n
where nullity is the dimension of the kernel of T. A differentiable function f : U � IRm has constant rank k if for all p E U the rank of (Df)p is k.
The Rank Theorem
Section 6*
29 1
An important property of rank is that if T has rank k and I S - T I is small, then S has rank 2: k. The rank of T can increase under a small perturbation of T but it cannot decrease. Thus, if f is C1 and (Df) p has rank k then automatically ( Df) x has rank 2: k for all x near p. See Exercise 43 . The Rank Theorem describes maps of constant rank. It says that locally they are just like linear projections. To formalize this we say that maps f : A -4 B and g : C -4 D are equivalent (for want of a better word) if there are bijections a : A ----* C and f3 : B -4 D such that g = f3 o f o a - 1 . An elegant way to express this equation is as commutativity of the diagram f
A
B
g
c
D.
Commutativity means that for each a E A , {3(/(a )) = g (a (a )) . Following
the maps around the rectangle clockwise from A to D gives the same result as following them around it counterclockwise. The a. f3 are "changes of variable." If f, g are C and a, f3 are C diffeomorphisms, l :::::: r :::::: oo , then f and g are said to be cr equivalent, and we write f � g . As cr r maps, f and g are indistinguishable. 24 Lemma cr
on rank.
equivalence is an equivalence relation and it has no effect
Proof Since diffeomorphisms form a group, �r is an equivalence relation. Also. if g = f3 o f o a- 1 , then the chain rule implies that
Dg = D/3
o
o
Df
Da - 1 •
Since D/3 and Da - 1 are isomorphisms, Df and Dg have equ al rank.
P : !Rn -4 IRm
The linear projection P (x r
•
0
.
.
.
, Xn ) = (x r
•
.
.
[hxk OJ
, Xk o 0, . . . , 0)
.
has rank k. It projects !Rn onto the k-dimensional subspace IRk x 0. The matrix of P is 0
0
.
25 Rank Theorem Locally, a cr constant rank k map is cr equivalent to a linear projection onto a k-dimensional subspace.
Multivariable Calculus
292
Chapter 5
As an example, think of the radial projection n : JR3 \ {0} -+ S2 • where n (v) vj l v i . It has constant rank 2, and is locally indistinguishable from linear projection of JR3 to the (x . y )-plane. =
U -+ IR.m have constant rank k and let p E U be given. We will show that on a neighborhood of p , f � r P . Step 1 . Define translations of IR. n and IR.m by ---
Proof Let f :
r
'
: ]Rm -+ ]Rm
Z I-+ Z - fp I
I
The translations are diffeomorphisms of JR.n and !Rm and they show that f is cr equivalent to r ' 0 f 0 r . a cr map that sends 0 to 0 and has constant rank k. Thus, it is no loss of generality to assume in the first place that p is the origin in IR.n and fp is the origin in IR.m . We do so. Step 2. Let T : IR.n -+ IR.n be an isomorphism that sends 0 x !Rn - k onto the kernel of (D /)0 . Since the kernel has dimension n - k , there is such a T . Let T ' : IR.m -+ IR. m be an isomorphism that sends the image o f ( DJ)o onto JR.k x 0. Since ( DJ )o has rank k there is such a T'. Then f � r T ' o f o T . This map sends the origin i n IR.n t o the origin i n IR.m , its derivative at the origin has kernel 0 x JR.n -k, and its image JRk x 0. Thus, it is no loss of generality to assume in the first place that f has these properties. We do so. Step 3. Write --f (x , y )
=
Ux (x , y ) ,
fy (x ,
y) )
E IR
k
x
IR.m - k _
We are going to find a g � r f such that
g (x , 0) = (x , 0) . The matrix of (D /)0 is where A is k x k and invertible. Thus, by the Inverse Function Theorem. the map a : �--+ fx (x , 0)
x
is a diffeomorphism a : X -+ X ' where X, X' are small neighborhoods of the origin in JR.k . For x' E X'. set h (x' )
=
1 Jy (a - (x ' ) . 0) .
Secti on 6*
The Rank Theorem
This makes h a cr map
293
X' ---+ JR.m-k . and h ( a (x ) )
= fy (x . 0) .
The image of X x 0 under f is the graph of h . For f (X
X
0) = { f (x , 0) :
x
E X } = { { fx (x , 0) , fy (x , 0)) :
x
E X}
= { {fx (a - 1 (x ' ) . 0) , jy (a - 1 (x ') . 0) ) : x' E X'} = { (x' , h (x')) : x ' E X' } .
See Figure 1 09. y
X
X
Figure 109 The image of X
For (x ' ,
y ')
Y'
f
E X'
O"X
X'
x 0 is the graph of h.
x JR.m -k , define 1/f (x', y ' ) = ( a - 1 (x' ) , y ' - h (x ' ) ) .
Since l/f is the composite of cr diffeomorphisms,
(x' . y ' ) �---+ (x' , y ' - h (x')) �----* ( a - 1 (x ) , y ' - h (x') ) . '
it too i s a C diffeomorphism. (Alternately, you could compute the deriva tive of l/f at the origin and apply the Inverse Function Theorem.) We observe that g = l/f o f � f satisfies
r
g (x , 0) = l/f o ( fx (x , 0) , jy (x , 0 ) ) = ( a - 1 o fx ( x , 0) . Jy (x , 0) - h (fx (x , 0)) = (x , 0) . Thus, it is no loss of generality to assume in the first place that f (x , 0) = (x , 0) . We do so. Step 4. Finally, we find a local diffeomorphism q; in the neighborhood of O in JR.n so that f o q; is the projection map P (x , y ) = (x , 0) . The equation fx ( � , y ) - x = 0 defines � = � (x , y ) implicitly in a neighborhood of the origin; it is a cr map from JR.n into JR.k and has � (0, 0) = 0. For, at the origin, the derivative
294
Multivariable Calculus
Chapter 5
of fx (l; , y) - x with respect to � is the invertible matrix that ({J (x . y) = (Hx , y) , y )
hx k · We clrum
is a local diffeomorphism of JRn and G = f o (/) is P . The derivative of � (x , y ) with respect to x at the origin can be calculated from the chain rule (this was done in general for implicit functions) and it satisfies a F a� aF d F(�(x . y ) . x . y ) a� 0 = -+ hxk - hxk• a� ax dx ox ax -
[hxk
That is, at the origin a � ;ax is the identity matrix. Thus,
(D({J)o =
0
* l(n- k ) x (n- k )
J
which is invertible no matter what * is. Clearly
0
Multiple Integrals
Section 8
Yj }j - 1
301
R,J •
(sij· ti)
Figure 1 14 A grid and
a
sample point.
The lower and upper sums of a bounded function grid G are
f with respect to the
U (f, G) = L Mij I Rij i where mij and Mi1 are the infimum and supremum of f (s, t) as (s , t) varies over R1 . The lower integral is the supremum of the lower sums and the upper integral is the infimum of the upper sums. The proofs of the following facts are conceptually identical to the one dimensional versions explained in Chapter 3 : (a) If f i s Riemann integrable then it i s bounded. (b) The set of Riemann integrable functions R ----+ lR is a vector space R = R( R ) and integration is a linear map R � JR. (c) The constant function f = k i s integrable and its integral is k I R I . (d) If f g E R and f ::=:: g then ,
(e) Every lower sum is less than or equal to every upper sum, and con sequently the lower integral is no greater than the upper integral,
Chapter 5
Multivariable Calculus
302
(f) For a bounded function, Riemann integrability is equivalent to the equality of the lower and upper integrals, and integrability implies equality of the lower, upper, and Riemann integrals. The Riemann-Lebesgue Theorem is another result that generalizes nat urally to multiple integrals. It states that a bounded function is Riemann integrable if and only if its discontinuities form a zero set. First of all , Z c IR2 is a zero set if for each E > 0 there is a countable covering of Z by open rectangles Sk whose total area is < E ,
L I Sk l k
< E.
By the E f2k construction, a countable union of zero sets is a zero set. As in dimension one, we express the discontinuity set of our function f : R � IR as the union where DK is the set points z OSCz
K >0 E R at which the oscillation is �
f = lim diam ( f ( Rr (z))) � r ---+ 0
K.
That is,
K
where Rr (Z) is the r-neighborhood of z in R. The set DK is compact. Assume that f : R � IR is Riemann integrable. It is bounded and its upper and lower integrals are equal . Fix K > 0. Given E > 0, there exists � > 0 such that if G is a grid with mesh < � then U ( f,
G) - L ( f, G ) < E .
Fix such a grid G. Each R;j i n the grid that contains i n its interior a point of DK has M;j - m ij � K, where m;j and M;j are the infimum and supremum of f on Rij . The other points of DK lie in the zero set of gridlines X; x [c, d] and [a , b] x yj. Since U L < E , the total area of these rectangles with oscillation � K does not exceed E f K . Since K is fixed and E is arbitrary, D�< is a zero set. Taking K = 1 /2, 1 / 3 , . . shows that the discontinuity set D = U DK is a zero set. Conversely, assume that f is bounded and D is a zero set. Fix any K > 0. Each z E R \ DK has a neighborhood W = Wz such that -
.
sup{ f ( w ) : w E W }
-
inf{ f ( w) : w E W }
0 be given. Take any grid G that partitions R into squares Rii of radius r . (The smallness of r will be specified below.) Let Z ij be the center point of Rij and call
The Taylor approximation to cp on Rij is
0 consider
lr, p : u I-* p + rL(u)
where i s the linear inclusion map that sends JRk to the XA -plane. 1 sends to a cube in the XA -plane at p. As � 0, the cube shrinks to p. If 1 ascends, the Jacobians of t are
Ik
L
r
if 1 = A Thus, if 1 =F A then
fi d
x1
w (t) = Continuity of
if 1 =I A . (t )
= 0 and
/AdxA (l) = r k
fA implies that
1 /A (t (u)) du. [k
(20) which is how the value of w on small k-cells at
fA (p ) .
p determines the coefficient D
38 Corollary
If k > n then Qk (JRn ) = 0.
Proof There are no ascending k-tuples of integers in { 1 , ,
. . . , n}.
D
Moral A form may have many names, but it has a unique ascending name. Therefore if definitions or properties of a form are to be discussed in terms of a form's name, the use of ascending names avoids ambiguity. Wedge Products
L I a1dx1
L J b1dx1. L I, J aibJdxu
Let a be a k-form and f3 be an .e-form. Write them in their ascending presentations, a = and f3 = Their wedge product is the (k + .e) -form a 1\ f3 = where = (i t , ik ) . J = (h . . . , h_ ) , I J = (i , , . . . , ib h . . , h.) . and the sum is taken over all ascending J . The use of ascending presen tations avoids name ambiguity, although Theorem 39 makes the ambiguity moot. A particular case of the definition is
I
•
.
.
.
.
I,
dxt 1\ dxz dxo. 2)· =
.
.
Differential Forms
Section 9
319
The wedge product 1\ Qk x Qf ---+ Qk +f satisfies four natural conditions: (a) distributivity: (a + fJ) 1\ Y = a 1\ Y + fJ 1\ Y and Y 1\ (a + {J) = y 1\ a + y 1\ fJ. (b) insensitivity to presentations: a 1\ fJ = Lu aib J dX IJ for general presentations a = L a1dx1 and fJ = L b1dx 1. (c) associativity: a 1\ (fJ 1\ Y) = (a 1\ fJ) 1\ Y. (d) signed commutativity: fJ 1\ a = ( - l ) k£ a 1\ fJ, when a is a k-form and fJ is an l-form. In particular, dx 1\ dy = -dy 1\ dx.
39 Theorem
40 Lemma
:
The wedge product of basic forms satisfies
Proof #1 See Exercise 54. Proof #2 If I and J ascend then the lemma merely repeats the definition of the wedge product. Otherwise, let JT and p be permutations that make JT I and p l non-descending. Call the permutation of I J that is n on the first k terms and p on the last l. The sign of is sgn (n ) sgn(p), and
a
a
dxi i\dXJ = sgn (n ) sgn (p) dx I 1\dXpJ :n:
= sgn ( ) dxa ( IJ )
a
= dxiJ . D
Proof of Theorem 39 (a) To check distributivity, suppose that a = L a 1dx 1 and fJ = L b1dx1 , are k-forms, while Y = L c1dx1 is an £-form, and all sums are ascending presentations. Then
is the ascending presentation of a and
+ fJ
(this is the only trick in the proof),
T, J
I, J
l, J
which is a 1\ Y + fJ 1\ Y , and verifies distributivity on the left. Distributivity on the right is checked in a similar way. (b) Let L a1 dx1 and L b1dx1 be general, non-ascending presentations of a and {J . By distributivity and Lemma 40,
( LI aidxi ) 1\ ( L b1dx1 ) = L a1 b1 dx1 1\ dx1 = L a1b1 dxiJ �J J
�J
320
Multivariable Calculus
Chapter 5
L, cK dX K , L, b1dxJ . c�::::a[ dX[) 1\ ( L::> J CKdXJK ) L a l b JCK dX I JK . J
(c) By (b), to check associativity we need not use ascending presentations . Thus, if a = L, 1d 1 f3 = and Y = then
a x
a 1\ ({3
1\ Y )
=
I
=
,K
I.J.K
which equals (a 1\ {3 ) 1\ Y . (d) Associativity implies that it makes sense to write products d 1 1\ 1\ and 1\ Thus, 1\
xi
·
•
dxik
·
dxh
·
·
•
dxit .
dx1
and
dx1
as
It takes kf pair-transpositions to push each
dx1 1\ dx1
=
dxi past each dx i, which implies (- l lfdx1 1\ dx 1 .
Distributivity completes the proof of signed commutativity for the general a and {3 . D The Exterior Derivative
Differentiating a form is subtle. The idea, as with all derivatives, is to imagine how the form changes under small variations of the point at which it is evaluated. A 0-form is a smooth function f Its exterior derivative is by definition the functional on paths qJ : [0, 1 ] -+ �" ,
(x) .
41 Proposition
df : cp �--+ / (cp ( l ) ) - f (cp (O)) . df is a 1 -form and when n 2, it is expressed as =
df
In particular, d (x)
=
dx.
=
-aaxf dx + -aJay dy .
Proof When no abuse of notation occurs we use calculus shorthand and write fx = af/ax, /y = af/ay. Applied to ({J, the form w = fx d + /y d y produces the number w
(cp)
=
to
J
(
dx (t)
dy (t )
)
x
fy (cp (t)) ------;[( + /y (cp (t)) ------;[( dt.
By the Chain Rule, the integrand is the derivative of f o qJ(t), so the Fun damental Theorem of Calculus implies that w (({J) = f (cp ( 1 ) ) - / (({J (O) ) . Therefore, f = w as claimed. D
d
Section
9
Differential Forms
321
Definition Fix k � 1 . Let :L /Jdx1 be the ascending presentation of a k-form w. The exterior derivative of w is the (k + 1 ) -form
dw =
L dh
1\
dxJ .
I
The sum is taken over all ascending k-tuples I . Use of the ascending presentation makes the definition unambiguous, although Theorem 42 makes this moot. Since d/J is a 1 -form and dx1 is k-form, dw is indeed a (k + 1 ) -form. For example we get
d(fdx + gdy) = (gx - /y)dx 1\ dy . :
Qk
Q k+ l
satisfiesfour natural conditions. (a) It is linear: d(a + cf3) = da + cdf3. (b) It is insensitive to presentation: ifL fidxi is a general presentation ofw then dw = 'L_ dh 1\ dxJ. (c) It obeys a product rule: if a is a k-form and f3 is an £-form then d(a 1\ {3) = da 1\ f3 + (- l ) k a 1\ df3 .
42 Theorem
Exterior differentiation d
--+
(d) d 2 = 0. That is, d (dw) = Ofor all w E Q k . Proof (a) Linearity is easy and is left for the reader as Exercise 55. (b) Let n make n I ascending. Linearity o f d and associativity o f 1\ give
d( /J dx1 ) = sgn( n ) d ( /J dx:n: 1 ) =
sgn (n ) d ( /J )
1\
dx:n: 1 = d ( /J ) 1\ dx 1 •
Linearity of d promotes the result from simple forms to general ones. (c) The ordinary Leibniz product rule for differentiating functions of two variables gives
atg afg d(fg) = - dx + - dy ax ay = fx g dx + /yg dy + fgx dx + /gy dy which is gdf + fdg, and verifies (c) for 0-forms in JR2 . The higher dimensional case is simi lar. Next we consider simple forms a = fdx1 and f3 = gdx 1 . Then
d(a 1\ {3) = d (fg dx u ) = (gdf + fdg) 1\ dx u = (df /\ dxJ) /\ (gdx1 ) + (- l ) k (fdx1 ) A (dg /\ dx 1 ) = da 1\ f3 + ( - 1 / a 1\ df3.
Multivariable Calculus
322
Chapter 5
Distributivtty completes the proof for general a and {3 . The proof of (d) i s fun. We check i t first for the special 0-form x . B y Proposition 4 1 , the exterior derivative x i s dx and i n turn the exte rior derivative of dx is zero. For dx = l dx, d l = 0, and by definition, d (ldx) = d (l ) 1\ dx = 0. For the same reason, d(dx1 ) = 0. Next we consider a smooth function f : IR.2 � JR. and prove that d 2 f = 0. Since d 2 x = d 2 y = 0 we have
d 2 f = d < fx dx + fy dy) = d < fx ) 1\ dx + d (fy ) 1\ dy = < fxx dx + fxy dy ) 1\ dx + (/y x dx + fyydy ) 1\ dx =0
.
The fact that d 2 = 0 for functions easily gives the same result for forms. The higher-dimensional case is similar. D Pushforward and Pullback
According to Theorem 36, forms behave naturally under composition on the right. What about composition on the left? Let T : JR.n � JR.m be a smooth transformation. It induces a natural transformation T* : Ck (JR.n ) � Ck (JR.m ) , the pushforward of T , defined by
k Dual to the pushforward is the pullback T * : C k (JR.m ) � C (JR.n ) defined by T * : Y t--+ Y o T.
Thus, the pullback of Y E C k (JR.m ) is the functional on Ck (JR.n )
T * Y : cp
�---+
Y (cp o T ) .
The pushforward T* goes the same direction as T, from JR.n to IR.m , while the pullback T * goes the opposite way. The pushforward/pullback duality is summarized by the formula
Pullbacks obey the following four natural conditions. (a) The pullback transformation is linear and (S o T) * = T * o S * . (b) The pullback of a form is aform; in particular. T * (dyi) = d T1 and T * (f dy1 ) = T *f d T1, where d T1 = d1i1 1\ · · · 1\ d ijk .
43 Theorem
Differential Forms
Section 9
323
(c) The pullback preserves wedge products, T * (a 1\ {3) = T � 1\ T *fJ. (d) The pullback commutes with the exterior derivative, dT * = T * d. Proof (a) This is left as Exercise 56. (b) We rely on a nontrivial result in linear algebra, the Cauchy-Binet Formula, which concerns the determinant of a product matrix AB = C , where A i s k x n and B i s n x k . See Appendix E. In terms of Jacobians, the Cauchy-Binet Formula asserts that if the maps k k q; : JR � JR" and 1/f : JR" � JR are smooth, then the composite ¢ = k k 1/f o cp : JR -+ JR satisfies
a¢ _ au
L OX]o l/J J
acp 1
au
where the Jacobian o 1/ff o x 1 i s evaluated at x = cp ( u ) , and J ranges through all ascending k-tuples in { 1 , . . . , n } . Then the pullback of a simple k-form o n :!Rm i s the functional on Ck (JR"),
T * (fdy1) : cp
�---+
=
= which implies that
f f (T u u 1 du 0 T1 L f f(T cp (u)) ( OX] ) jdy1 ( T o cp) o cp (
Jk
J
Jk
))
a (T o cp ) a
o cp J du , x=q;(u) a u
o
is a k-form. Linearity of the pullback promotes this to general forms - the pullback of a form is a form. It remains to check that T * (dy 1) = dT1 . For I = (i 1 , , h), distributivity of the wedge product and the definition of the exterior derivative of a function imply that •
.
•
d T1 = d T.·I J 1\
·
·
·
1\
,) - (L d · dxS] OX T.l k
S] = l
a T;,
s
1\
·
·
1\
·
I
o Ti ) (I:" dx OX Sk = !
k
s
k
Sk
, h are fixed. All terms with repeated dummy indices The indices i 1 , s 1 , , sk are zero, so the sum is really taken as (s 1 , , sk ) varies in the set •
•
•
•
•
•
•
•
•
Multivariable Calculus
324
Chapter 5
of k-tuples with no repeated entry, and then we know that (s 1 , . . . , sk ) can be sk ) = rr J for an ascending J = (j 1 , expressed uniquely as (s 1 , A) and a permutation rr . Also, dx51 1\ 1\ dxsk = sgn (rr )dx 1 . This gives •
•
•
,
•
·
·
•
•
,
·
and hence T * (dy1) = dT1 . Here we used the description of the determinant from Appendix E. (c) For 0-forms it is clear that the pullback of a product is the product of the pullbacks. T * (fg) = T *f T 'g . If the forms a = fdy1 and f3 = gdy 1 are simple then a 1\ f3 = fgdy11 and by (b),
T * (a 1\ {3)
=
T * (fg) dT11
=
T *f T g d T1
1\
dT1
=
T * a 1\ T * fJ.
Wedge distributivity and pullback linearity complete the proof of (c) . (d) If w is a form of degree 0, w = f E Q 0 (1Rm ) , then
T * (df) (x)
=
=
T*
m
of (L � dy; ) y ,
i= l
m of L: r * ( - ) T * (dy; ) oy . i= l
,
which is merely the chain rule expression for d(f o T) =
Thus,
T * dw = dT * w for 0-forms.
d(T * f),
The General Stokes' Formula
Section 1 0
325
Next consider a simple k-form w = fdyi with k � I . Using (b), the degree-zero case, and the wedge differentiation formula, we get
d(T * w) = d(T *f dT1 ) = d(T *f) 1\ dT1 + ( - 1) 0 T *f 1\ d (dT1 ) * (df) 1\ d T1 = T = T * (df 1\ dy1 ) = T * (dw) . Linearity promotes this to general k-forms and completes the proof of (d) .
D 10
The General Stokes ' Formula
1cp dw = la{cp w ,
In this section we establish the general Stokes' formula as
where w E Q k (JRn ) and ifJ E Ck + 1 (JRn ) . Then, as special cases, we reel off the standard formulas of vector calculus. Finally, we discuss antidifferentiation of forms and briefly introduce de Rham cohomology. First we verify Stokes ' formula on a cube, and then get the general case by means of the pullback. A
Definition
0 for 1 /2 < r < l , and (r) = I for r ::: I . Then define l/f : [ - 1 , 1 ]11 --+ lR11 by
a
{v
v
l/f ( ) =
+
0
a
a
a (lvl) ( � l v l - v)
if
v :F 0
if v = 0.
In fact, this formula defines l/f on all of 1R11 and l/f carries the sphere Sr of radius r to the sphere of radius p (r ) = r
+ a (r ) ( 1 - r ) ,
sending each radial line to itself. Set cp = l/f oK where K scales /11 to [ - 1 , 1]11 by the affine map u I--* = (2u t - 1 . . . . . 2u11 - 1 ) . Then (a) cp is smooth because, away from u 0 = ( 1 /2, . . . , 1 / 2) , it is the com posite of smooth functions. and in a neighborhood of u 0 it is identi cally equal to K (u ) . (b) Since 0 .:::; p (r) .:::; 1 for all r , l/f sends lR 11 to B . Since p (r ) = I for all r ::: 1 , cp sends a /11 to a B . (c) It i s left as Exercise 65 to show that the Jacobian of l/f for r = 1 t I is p ' (r) p (r)11- j r11- • Thus, the Jacobian acpjau is always non negative, and is identically equal to 211 on the ball of radius 1 /4 at u 0 , so its integral on I" is positive.
v
vi
Step 4. Consider an ---
(n - 1 )-form a . If fJ : r- t
--+ lR11 is an (n - I )-cell
whose image lies in a B then
i a - io/3 a i T*a . =
The (n - I ) -dimensional faces of cp : I" --+ B lie in a B . Thus
{ a laq;
(23)
=
{ a. laq; T*
Step 5. Now we get the contradiction. Consider the specific (n - I )-form
Note that da = dxt
1\
·
·
·
1\
dxn is n-dimensional volume and
1 da = f tp
Jn
a
cp
au
du
>
0.
Appendix A
Perorations of Dieudonne
In fact the integral is the volume of
i da 0 such that if S : JR.n -+ JR.m and li S - T i l < 8 , then S has rank ::: k. (b) Give a specific example in which the rank of S can be greater than the rank of T, no matter how small 8 is. (c) Give examples of linear transformations of rank k for each k where 0 ::::: k ::::: min{n , m } . Let S c M. (a) Define the characteristic function X s : M -+ JR. . (b) If M is a metric space, show that X s (x ) is discontinuous at x if and only if x is a boundary point of S. On page 302 there is a definition of Z c JR.2 being a zero set that involves open rectangles. (a) Show that the definition is unaffected if we require that the rectangles covering Z are squares. (b) What if we permit the squares or rectangles to be non-open? (c) What if we use discs or other shapes instead of squares and rectangles? Assume that S c IR2 is bounded. (a) Prove that if S is Riemann measurable then so are its interior and closure. (b) Suppose that the interior and closure of S are Riemann measur able and l int(S) I = JSJ . Prove that S is Riemann measurable. (c) Show that some open bounded subsets of JR.2 are not Riemann measurable. •
* *42.
43 .
44.
45 .
*46.
•
.
.
•
Exercises
355
* 47 In the derivation of Fubini's Theorem on page 304, it is observed that for all y E [c, \ Y, where Y is a zero set, the lower and upper integrals with respect to agree, F (y) F (y) . One might think that the values of F and F on Y have no effect on their integrals. Not so. Consider the function defined on the unit square [0, 1] x [0, 1],
d]
x
if y is irrational
1
j(x, y)
1
=
-
1
q
-
x is irrational if y is rational and x = is rational
if y is rational and
p fq
and written in lowest terms.
(a) Show that f is Riemann integrable, and its integral is 1 . (b) Observe that if Y is the zero set Q n [0. 1 ], then for each y fl. Y ,
11 f(x, y) dx
exists and equals 1 . (c) Observe that if for each arbitrary manner some
y E
Y
we choose in a completely
h (y ) E [F(y) , F(y)]
and set
H(x) = { hF(y)(y) = F(y)
if y fj. Y if y E Y
then the integral of H exists and equals 1 , but if we take g 0 for all y E Y then the integral of
G (x)
= F(y) = { gF(y) (y ) = 0
(x) =
if y fl. y if y E Y
does not exist. * * *48. Is there a criterion to decide which redefinitions of the Riemann integral on the zero set Y of Exercise 47 are harmless and which are not? 49. Using the Fundamental Theorem of Calculus, give a direct proof of Green· s Formulas
- Jl fy dxdy = 1R fdx
and
!L gx dx dy = 1R g dy.
356
Chapter 5
Multivariable Calculus
R is a square in the plane and
f, 8 :
IR2
---+
lR are smooth. (Assume
that the edges of the square are parallel to the coordinate axes.) 50. Draw a staircase curve Sn that approximates the diagonal D. = { (x , y) E IR2 : 0 s x = y S 1 } to within a tolerance 1 / n . (Sn consists of both treads and risers.) Suppose that f, 8 : IR2 ---+ lR are smooth. (a) Why does length Sn fr length D. ? (b) Despite (a), prove fsn f dx ---+ JD. f dx and fsn 8 dy ---+ JD. 8 dy. (c) Repeat (b) with D. replaced by the graph of a smooth function 8 : [a , b] ---+ R (d) If C is a smooth simple closed curve in the plane, show that it is the union of finitely many arcs Ct , each of which is the graph of a smooth function y = 8 (x) or x = 8(y), and the arcs Ce meet only at common endpoints. (e) Infer that if (Sn ) is a sequence of staircase curves that converges to C then fsn f dx + 8dy ---+ fc fdx + 8dy. (f) Use (e) and Exercise 49 to give a proof of Green's Formulas on a general region D c IR2 bounded by a smooth simple closed curve C, that relies on approximatingt C, say from the inside, by staircase curves Sn which bound regions Rn composed of many small squares. (You may imagine that R1 C R2 C . . . and that Rn ---+ D.) 5 1 . A region R in the plane is of type 1 if there are smooth functions 8 1 : [a , b] ---+ IR , 82 : [a , b] ---+ lR such that 81 (x) S 8z (x } and R =
{ (x , y) : a S x S b and 8t (x)
S
y S 8z (x)} .
R is of type 2 if the roles of x and y can be reversed, and it is simple if it is of both type 1 and type 2. (a) Give an example of a region that is type 1 but not type 2. (b) Give an example of a region that is neither type 1 nor type 2. (c) Is every simple region starlike? convex? (d) If a convex region is bounded by a smooth simple closed curve, is it simple? (e) Give an example of a region that divides into three simple sub regions but not into two.
t This staircase approximation proof generalizes to regions that are bounded by fractal, non differentiable curves such as the von Koch snowflake. As Jenny Harrison has shown, it also generalizes to higher dimensions.
Exercises
* (f) If a region is bounded by a smooth simple closed curve
357
C, it
need not divide into a finite number of simple subregions. Find an example. (g) Infer that the standard proof of Green's Formulas for simple regions (as, for example, in J. Stewart's Calculus) does not im mediately carry over to the general planar region R with smooth boundary, i.e., cutting R into simple regions can fail. *(h) Show that if the curve C in (f) is analytic, then no such example exists. [Hint: C is analytic if it is locally the graph of a function defined by a convergent power series. A nonconstant analytic function has the property that for each x , there is some derivative of f which is nonzero, f (r ) (x) f= 0. 1 **52. The 2-cell q; : r -+ Bn constructed in Step 3 of the proof of Brouwer's Theorem is smooth but not one-to-one. (a) Construct a homeomorphism h : / 2 -+ B 2 where / 2 is the closed unit square and B 2 is the closed unit disc. (b) In addition make h in (a) be of class C 1 (on the closed square) and be a diffeomorphism from the interior of / 2 onto the interior of B 2 • (The derivative of a diffeomorphism is everywhere non singular.) (c) Why can h not be a diffeomorphism from ! 2 onto B 2 ? (d) Improve class C 1 in (b) to class c oo . **53. If K, L C ]Rn and if there is a homeomorphism h : K � L that extends to H : U -+ V such that U, V C JR;.n are open, H is a homeomorphism, and H , H - 1 are of class cr , 1 :S r _:s oo, then we say that K and L are ambiently cr -diffeomorphic. (a) In the plane, prove that the closed unit square is ambiently dif feomorphic to a general rectangle; to a general parallelogram. (b) If K, L are ambiently diffeomorphic polygons in the plane, prove that K and L have the same number of vertices. (Do not count vertices at which the interior angle is 1 80 degrees.) (c) Prove that the closed square and closed disc are not ambiently diffeomorphic. (d) If K is a convex polygon that is ambiently diffeomorphic to a polygon L, prove that L is convex. (e) Is the converse to (b) true or false? What about in the convex case?
Multivariable Calculus
358 (f)
Chapter 5
The closed disc is tiled by five ambiently diffeomorphic copies of the unit square as shown in Figure 1 2 1 . Prove that it cannot be tiled by fewer.
Figure 121 Five diffeomorphs of the square tile the disc.
54. 55. 56.
57. 58. 59.
(g) Generalize to dimension n :=::: 3 and show that the n -ball can be tiled by 2n + 1 diffeomorphs of the n -cube Can it be done with fewer? (h) Show that a triangle can be tiled by three diffeomorphs of the square. Infer that any surface that can be tiled by diffeomorphs of the triangle can also be tiled by diffeomorphs of the square. What happens in higher dimensions? Choose at random I, J , two triples of integers between 1 and 9. Check that dx1 1\ dxJ dxu . Show that d : Q k � Qk + 1 is a linear vector space homomorphism. Prove that the pullback acts linearly on forms, and that it is natural with respect to composition in the sense that ( T o S) * = S* o T * . See how succinct a proof you can give. True or false: if w is a k-form and k is odd, then w 1\ w = 0. What if k is even and :=::: 2? Does there exist a continuous mapping from the circle to itself that has no fixed point? What about the 2-torus? The 2-sphere? Show that a smooth map T : U � V induces a linear map of cohomology groups H k (V) � H k (U) defined by T * : [w J f--+ [T * w ] .
=
Here, [wJ denotes the equivalence class of w E z k ( V) in H k ( V). The question amounts to showing that the pullback of a closed form w i s
359
Exercises
closed and that its cohomology class depends only on the cohomology class of w. t 60. Prove that diffeomorphic open sets have isomorphic cohomology groups. 6 1 . Show that the 1 -form defined on JR.2 \ { (0, 0) } by w
-y =dx + - dy r2 r2 X
is closed but not exact. Why do you think that this L -form is often referred to as d() and why is the name problematic? 62. Show that the 2-form defined on the spherical shell by
is closed but not exact. 63 . If w is closed, is the same true of f w? What if w is exact? 64. Is the wedge product of closed forms closed? Of exact forms exact? What about the product of a closed form and an exact form? Does this give a ring structure to the cohomology classes? 65 . Prove that the n -cell l/r : [- 1 , l ] n ---+ B n in the proof of the Brouwer Fixed Point Theorem has Jacobian p '(r) p ( r ) n -l fr n - l for r I v i as claimed on page 336. **66. The Hairy Ball Theorem states that any continuous vector field X in JR.3 that is everywhere tangent to the 2-sphere S is zero at some point of S. Here is an outline of a proof for you to fill in. (If you imagine the vector field as hair combed on a sphere, there must be a cowlick somewhere.) (a) Show that the Hairy Ball Theorem is equivalent to a fixed point assertion: every continuous map of S to itself that is sufficiently dose to the identity map S ---+ S has a fixed point. (This is not needed below, but it is interesting.) (b) If a continuous vector field on S has no zero on or inside a small simple closed curve C c S. show that the net angular turning of X along C as judged by an observer who takes a tour of C in the
=
t A fancier way to present the proof of the Brouwer Fixed Point Theorem goes like this: As always, the question reduces to showing that there is no smooth retraction T of the n -ball to its boundary. Such a T would give a cohomology map T * : H k (iJB) ---. H k (B) where the cohomology groups of iJ B are those of its spherical shell neighborhood. The map T * is seen to be a cohomology group isomorphism because T o inclusiona s = inclusion a s and inclusion 0 8 * = identity. But when k = n - I :0: I the cohomology groups are non-isomorphic: they are computed to be H n - 1 (iJB) = JR and H n - 1 (B) = 0.
360
Multivariable Calculus
Chapter 5
counterclockwise direction is - 2rr . (The observer walks along C in the counterclockwise direction when S is viewed from the outside, and he measures the angle that X makes with respect to his own tangent vector as he walks along C. By convention, clockwise angular variation is negative.) Show also that the net turning is + 2rr if the observer walks along C in the clockwise direction. (c) If C, is a continuous family of simple closed curves on S, a :S t :S b, and if X never equals zero at points of C, , show that the net angular turning of X along C, is independent of t . (This is a case of a previous exercise stating that a continuous integer valued function of t is constant.) (d) Imagine the following continuous family of simple closed curves C1 • For t = 0, Co is the arctic circle. For 0 :S t :S 1 /2, the latitude of C1 decreases while its circumference increases as it oozes downward, becomes the equator, and then grows smaller until it becomes the antarctic circle when t = 1 /2. For 1 /2 :S t :S 1 , C1 maintains its size and shape, but its new cen ter, the South Pole, slides up the Greenwich Meridian until at t = 1 , C, regains its original arctic position. See Figure 1 22. Its orientation has reversed. Orient the arctic circle Co positively and choose an orientation on each C, that depends continuously on t . To reach a contradiction, suppose that X has no zero on S. (i) Why is the total angular turning of X along Co equal to -2rr ?. (ii) Why is it +2rr on C1 ? (iii) Why is this a contradiction to (c) unless X has a zero some where? (iv) Conclude that you have proved the Hairy Ball Theorem
36 1
Exercises
) Figure 122 A deformation of the arctic circle that reverses its orientation
6
Lebesgue Theory
This chapter presents a geometric theory of Lebesgue measure and integra tion. In calculus you certainly learned that the integral is the area under the curve. With a good definition of area, that is the point of view we advance here. Deriving the basic theory of Lebesgue integration then becomes a matter of inspecting the right picture.
1
Outer measure
How should you measure the length of a subset of the line? If the set to be measured is simple, so is the answer: the length of the interval ( a , b) is b - a . But what i s the length of the set of rational numbers? of the Cantor set? As is often the case in analysis we proceed by inequalities and limits. In fact one might distinguish the fields of algebra and analysis solely according to their use of equalities versus inequalities.
Definition The length of an interval I = (a , b) is b - a . It is denoted I l l The Lebesgue outer measure of a set A c lR is
m * A = inf{ L I h i : { ld is a covering of A by open intervals } . k
Tacitly we assume that the covering is countable; the series L k I h I is its total length. (Recall that ..countable" means etther finite or denumerable. )
Lebesgue Theory
364
Chapter 6
The outer measure of A is the infimum of the total lengths of all possible coverings { h} of A by open intervals. If every series L h diverges, then by definition, m * A = oo. Outer measure is defined for every A C JR. I t measures A from the outside as do calipers. A dual approach measures A from the inside. It is called inner measure, is denoted m * A, and is discussed in Section 3 . Three properties o f outer measure (the axioms of outer measure) are easy to check. (a) The outer measure of the empty set is 0, m * IZJ = 0. (b) If A C B then m * A � m * B .
kI I
00
(c) If A = U� 1 A n then m * A �
L: m * A n . n=l
(b) and (c) are called monotonicity and countable subadditivity of outer measure. (a) is obvious. Every interval covers the empty set. (b) is obvious. Every covering of B is also a covering of A. (c) uses the E j2n trick. Given E > 0 there exists for each n a covering { hn : k E N } of A n such that 00
L I hn I k=l
k
The collection U n 00
: k, n E N} covers A and 00
L l hn I = L L l hn I k.n
E < m * A n + -n . 2
00
�
n=l k=l
L n=l
(m * A n +
;J = L n=l 00
m * A n + E.
Thus the infimum of the total lengths of coverings of A by open intervals is � L n m * A n + E , and since E > 0 is arbitrary, the infimum is � L n m * A n , which is what (c) asserts. Next, suppose you have a set A in the plane and you want to measure its area. Here is the natural way to do it.
Definition The area of a rectangle = (a , b) x (c, d) is = (b - a) · (d - c), and the (planar) outer measure of A c JR2 is the infimum of the total area of countable coverings of A by open rectangles
R
m * A = inf {
I I
Lk I Rk l : {Rk}
IRI Rk
covers
A}.
If need be, we decorate and m * with subscripts " 1 " and "2" to distinguish the linear and planar quantities.
Outer measure
Section 1
365
The outer measure axioms - monotonicity, countable subadditivity, and the outer measure of the empty set being zero - are true for planar outer measure too. See also Exercise 5 . A consequence o f subadditivity concerns zero sets, sets that have zero outer measure. l Proposition
The countable union of zero sets is a zero set.
Proof If m * Zn
= 0 for all n E N and Z = U Zn then by (c), n
Two other properties enjoyed by outer measure are left for you as Exercises 1 and 2: (d) Rigid translation of a set has no effect on its outer measure. (e) Dilation of a set dilates its outer measure correspondingly. The next theorem states a property of outer measure that seems obvious. See also Exercises 3, 4.
2 Theorem The linear outer measure of a closed interval is its length; the planar outer measure of a closed rectangle is its area. E >
= [a, b] be the interval. For any 0, the single open interval (a - E , b + E ) covers I, so m * I :::: I I I + 2E . Since E is arbitrary, m * I :::: I I I .
Proof Let I
To check the reverse inequality, let { ld be a countable covering of I by open intervals. Compactness implies that for some finite N, I c U�= l h . Form a partition a Xo < · · · < b
=
Xn =
such that all the endpoints ak . bk of the intervals h (ak . bk ) that appear in I occur as partition points. Each partition interval xd is contained wholly inside some covering interval h . which implies
=
Xi = (Xi - l ,
N
IXi l :::: 'L: I Xi n /tl . k= l
I I
For one of the terms in the sum is X i itself. Also, for each of I restricts to a partition of the interval I Ik . and thus
n
n
II n h i = L IXi n hl . i=l
k, the partition
Lebesgue Theory
366
This gives
n
n
i=l
i =l
Chapter 6
N
N
I I I = L IXd ::: I: ( I: IX; n h l ) = I: ( I: IX; n h l ) N
N
k =l
n
k =l
•= 1
oo
= I: 11 n h l ::: "L: I �tl ::: I: l �t l . k=l
k =1
k= 1
which shows that I I ::; m * and completes the proof in dimension one. The case of a rectangle R [a , b] x L c, dl is similar. As with intervals, it is enough to show that any countable covering of R by open rectangles Rk (at . bk ) x (ct . dk ) has I R I ::; L I Rk l · There is an N such that R 1 , , RN cover R. Partition the sides of R as
I
I =
=
.
a
= Xo
0 there i s a covering I of X by open intervals h whose total length is
L l h l :::=: m * X + E. k
Split any interval h that contains 0 into its positive and negative open subintervals It. This gives a new covering p of X with the same total length as I. Each interval in p lies wholly in the positive or negative half axis, so p splits into disjoint coverings f/ of x±. Thus
m*X
�
m * (X n H) + m * (X n He) ::::: L I l l + L I l l = L l h l k
�
m * X + E.
Since E > 0 is arbitrary this gives measurability of H. The planar case i s similar. The y-axis has planar outer measure zero since for any E > 0 it is covered by the countable set of open rectangles ( -E/4n , E/4n ) X ( -2n , zn ) whose total area is 4E . By Proposition 7 it is then fair to exclude the y-axis from a test set X c JR2 , and the rest of the reasoning is the same as in the linear case. D
Section
3
Regularity
373
Proof of Theorem 6 Consider ffi. By Proposition 9 the half lines [0, oo) and ( - oo , 0) are measurable. Measurability is unaffected by translation, so [a , oo) and ( -oo, b) are measurable. Measurability is also unaffected by the exclusion of zero sets. Thus (a , oo) is measurable and so is
(a , b) = (a , oo ) n ( - oo , b) . Consider ffi. 2 . The translates of half planes are measurable and it is fair to exclude zero sets. Thus the vertical strip (a , b) x ffi. is measurable. Exchang ing the roles of the x-and y-axis shows that the horizontal strip JR. x (c, d) is also measurable, so their intersection (a , b) x (c, d) is measurable. Every open set in JR. is the countable union of open intervals, and every open set in the plane is the countable union of open rectangles. Since M is a rr -algebra, every open set is measurable, and since the complement of a D measurable set is measurable, every closed set is also measurable. 10 Corollary The Lebesgue measure of an interval is its length and the Lebesgue measure of a rectangle is its area. Proof By Theorem 6, the interval and rectangle are measurable, so their
measures equal their outer measures, which we know to be their length and area. D Sets that are slightly more general than open sets and closed sets arise naturally. A countable intersection of open sets is called a G� -set and a countable union of closed sets is an Fu -set. ("8" stands for the German word durschnitt and "rr " stands for "sum.") By deMorgan's Laws, the complement of a G0-set is an Fa -set and conversely. Since the rr-algebra of measurable sets contains the open sets and the closed sets, it also contains the G 0-sets and the Fer -sets. 1 1 Theorem Lebesgue measure is regular in the sense that each measur able set E can be sandwiched between an Fa -set and a G�-set, F C E C G, such that m (G \ F) = 0. Proof Assume first that E
c
JR. is bounded. There is a large closed interval
I that contains E . Measurability implies that m l = m E + m (l \ E) .
There are decreasing sequences of open sets U, and V, :J ( I \ E), m U, --+ m E, and m V, --+ m (l \
V, such that U, :J E, E) as --+ oo. The n
Lebesgue Theory
374
complements of E and
Kn = I \
Chapter 6
Vn form an increasmg sequence of closed subsets
mKn = m l - m Vn ---+ m l - m (l \ E) = m E . Thus F = U Kn is an Fa -set in E with m F = m E . Similarly there i s a G 8 -set G that contains E and has m G = m E. Because all the measures are finite, the equality m F = m E = m G implies that m ( G \ F) = 0. The same n reasoning applies to bounded subsets of the plane or !R . The unbounded case is left as Exerci se 9. D 12 Corollary Modulo zero sets, Lebesgue measurable sets are Fa -sets and/or G8 -sets. Proof
E = F U Z = G \ Z' for the zero sets Z = E \ F and Z' = G \ E .
n
0
The outer measure of a set A c IR or A c !R is the infimum of the measure of open sets that contain it. Dually, we define the inner measure of A to be the supremum of the measure of closed sets containeei in it. We denote the inner measure of A as m * A . Clearly m * A ::=: m * A and m * measures A from the inside. Also, m * is monotone: A c B implies
m * A :::: m * B .
Equivalently, m * A i s the measure of the smallest measurable set H � A while m * A is the measure of the largest measurable set N c A . (H being "smallest" means that for any other measurable H' � A, H \ H' is a zero set. Similarly, if N' is any measurable subset of A then N' \ N is a zero set. It is easy to see that H and N exist and are unique up to zero sets. They are called the hull and kernel of A . See Exercise 8 . )
!Rn ) is Lebesgue measurable if and only if its inner and outer measures are equal, m * A = m * A. Further, if B is a bounded measurable set that contains A, then A is measurable if and only if it divides B cleanly. 1 3 Theorem A bounded set A C IR (or A
C
A is measurable then it is both its own largest measurable subset and the smallest measurable subset that contains it, so m * A = m * A. O n the other hand, i f m * A = m * A, then A is sandwiched between
Proof If
measurable sets with equal finite measure, so it differs from each by a zero set and is therefore measurable. Finally suppose that B � A with B bounded and measurable. Then
m * A + m * (B \ A) = m B .
Section
3
375
Regularity
See Exercise 8.
If
A divides B cleanly. then m * A + m*( B \ A)
= mB. A
Finiteness of all the measures implies that m * A = m* and A i s measur able. The converse is obvious because a measurable set divides every test D set cleanly. Remark Lebesgue took Theorem 1 3 as his definition of measurability. He said that a bounded set is measurable if its inner and outer measures are equal, and an unbounded set is measurable if it is a countable union of bounded measurable sets. In contrast, the current definition which uses cleanness and test sets is due to Caratheodory. It is easier to use (there are fewer complemems to consider), unboundedness has no effect on it, and it generalizes more easily to "abstract measure spaces."
Regularity of Lebesgue measure has a number of uses. Here is one. See also Exercises 32 and 3 3 .
The Cartesian product ofmeasurable sets is measurable, and the measure of the product is the product of the measures. 1 4 Theorem
Proof By convention,
we claim that A
mz(A
x
x
0 oo = 0 = oo · 0. Given measurable A , B ·
JR. ,
B is measurable with respect to planar measure and
B) = mtA m t B .
A
C
·
A.
Case 1 . or B i s a zero set; say it i s Given E > 0 and an interval (c. d) there is a covering of A by open intervals ( h } with L k I < Ej(d - c) . Thus A x (c, d) is covered by rectangles h x (c, d) whose total area is < E, so X (c. d) is a zero set. S ince X JR. = u:: l A X ( - n, n), it is a zero set, and since x B C x JR., x B is also a zero set. Zero sets are measurable. which completes the proof of the theorem in this case. UAi and B = U B where Case 2. and B are open sets. Then and B are open intervals, and A x B is the disjoint countable union of open rectangles. It is therefore measurable and by countable additivity,
A
Ai
j
A
A
A
hi
A
A
A=
j
ij ij = ( L I A i i ) ( L i Bj i ) = m t A · m t B . i j
A
and B are bounded G0-sets. Then there are nested decreas Case 3 . and Vn _j,. B with m 1 U1 < oo and ing sequences of open sets, Un -1--
A
376
Chapter 6
Lebesgue Theory
Thus Un X Vn ..!- A X B as n � 00 Being a Gil -set, A X B is measurable. Downward measure continuity gives m 1 Un ---+ m 1 A, m 1 Vn � m 1 B , and by Case 2
m 1 v1
0 is arbitrary, m * (TJE) ::::; mE. The same applies to R \
Then
E,
and
so
m * (T1 E) + m* (Tt ( R \ E)) ::::; m E + m ( R \ E) = I R I = m(Tt R ) ::::; m * ( TJ E) + m * (Tt R \ Tt E). This gives equality m (T1 R) = m * (T1 E) + m * (T1 R \ T1 E), which means that Tt E divides the bounded, measurable set T1 R cleanly. Theorem 1 3 implies that Tt E is measurable. Subtraction of I R I = m E + m (R \ E) m(Tt R) = m(TtE) + m(Tt (R \ E) )
Chapter 6
Lebesgue Theory
380
gives
0 = (m E - m ( Tt E) ) + (m (R \ E) - m (Tt (R \ E) ) ) . Both terms are :::: 0, so they are zero, which completes the proof of (4) when E is bounded. Case 4. E is unbounded. Break E into countably many bounded, mea surable, disjoint pieces and apply Case 3 piece by piece. This completes D the proof of (4 ), of (g), and of the theorem. Remark The standard proof oflinearity of the Lebesgue integral is outlined in Exercise 28. 1t is no easier than this undergraph proof, and undetgraphs at least give you a picture as guidance. Definition The completed undergraph is the undergraph together with the graph itself,
Uf
= Uf
U
{ (x , f (x)) : x E
JR}.
The definitions of measurability and integral are unaffected if we replace the undergraph with the completed undergraph. 17 Theorem
[0, oo) be given, and set fn (x) = ( 1 + 1 /n ) f (x) for U Then f is the decreasing intersection nu fn together with the
Proof Let f : lR ---*
n E N. x-axis. Assume that U f is measurable. Then U fn is measurable, the intersection nu fn is measurable, and the x-axis is measurable (it is a zero set) so f) f is measurable. Also m (Uf) = m (U f) . For if m (U f) = oo, then Uf => U f implies that m (Uf) = oo, while if m (U f) < oo, then the result follows from downward measure continuity (Theorem 5) and the fact that the x -axis is a zero set. The converse is checked similarly. Set g n (x ) = ( 1 - 1 /n ) f (x ) and express U f as the increasing union of Ugn , modulo a zero set on the x -axis. Measurability of Uf implies measurability of Ugn implies measurability D of U f and by upward measure continuity m (U f) = m (Uf). Definition Let fn : lR ---* [0, oo ) be a sequence of functions. Its lower envelope and upper envelope sequences are
f (x)
:::....n
= inf{ fk (x)
:
k :::: n } and fn
f ...n increases and fn decreases as Clearly, :::. sandwich the original sequence.
n
= sup{ fk (x ) : k :::: -
n} .
oo, and the envelopes
381
Lebesgue integrals
Section 4
1 8 Theorem If the functions fn are measurable, then so are lts envelopes. Proof The undergraph of the maximum of two functions is the union of
their undergraphs, while the minimum is the intersection. Keeping track of strict inequality versus nonstrict inequality. we get the formulas
D
and measurability follows from Theorem 1 7 . 19 Corollary
The pointwise limit of measurable functions is measurable.
Proof If fn (x)
� f (x) for every x (or almost every x) as n � oo then the same is true of the envelope functions. Then =--n f t f implies that U=--n f t Uf so f is measurable. D
Suppose that fn lR � [0, oo) is a sequence of measurable functions converging pointwise to the limit function f. If there exists a function g lR � [0, oo) whose integral is finite and which is an upper bound for all the functions fm 0 ::S fn (x) ::S g (x), then J fn � J f. 20 Lebesgue Dominated Convergence Theorem
:
:
Proof Easy. See Figure 1 26. The upper and lower envelopes converge monotonically to f , and since U /n c Ug, all have finite measure. This makes both downward and upward measure continuity valid, so we see that the measures of Uf and U .fn both converge to m (Uf) . Between these "'-II measures is sandwiched m (U fn), and so it also converges to m (U f) as n � oo. D
.. .. . . .. . . . . .
.
,'
.. .. . . .. ..
.. . .. .. •
R
..
t.
t.
I
.
I
I
t.
t.
.
�
.
:
.
.
.
.. �
t. # t. ...
..
. .
.
.. . ..
.
.
. #II ,.,
Figure 126 Lebesgue Dominated Convergence.
..
##
.
.
.
.. ..
.
.. .. . ..
382
Lebesgue Theory
Chapter 6
Remark If a dominator g with finite integral fails to exist then the assertion
fails. For example the sequence of "steeple functions," shown in Figure 85 on page 204, have integral n and converge at all x to the zero function as
n --+ oo.
Suppose that fn : lR --+ [0, oo) is a sequence of measurable functions and f : lR [0, oo) is their lim inf, f (x) = lim inf{ fk {x) : k � n} . Then J f _:::: lim inf J fn · n---+ oo
21 Fatou's Lemma
--+
n---+ oo
Proof The assertion is really more about lim inf's than integrals. The as
sumption is that f is the limit of the lower envelope functions c:... f .n . Since ( x ) , we have f (x) _::: : f n c:....n The Lebesgue Monotone Convergence Theorem implies that J c:... f .n converges upward to J .f, so the integrals J fn have a lim inf that is no smaller than f f . D Remark The inequality can be strict, as is shown by the steeple functions
Up to now we have assumed the integrand f is nonnegative. If f takes both positive and negative values we define if f (x) � if f (x ) < Then f± � of f ,
0 and f
= f+ -
0 0
J_ (x) =
l�
f (x )
i f f(x) < if f(x) �
0 0.
f It is easy to see that the total undergraph _ .
{(x , y) E lR2 : y < f (x) }
is measurable if and only if f± are measurable. See Exercise 3 1 . If f± are integrable we say that f is integrable and define its integral as
The set of measurable functions f : lR --+ lR is a vector space, the set of integrable functions is a subspace, and the integral is a linear map from the latter into JR. 22 Proposition
The proof is left to the reader as Exercise 2 1 .
Section 5
5
383
Lebesgue integrals as limits
Lebesgue integrals as limits
The Riemann integral is the limit of Riemann sums. There are analogous "Lebesgue sums'' of which the Lebesgue integral is the limit. Let f : lR ---+ [0. oo) be given. take a partition Y : 0 = Yo < Y1 < Yz . on the y-axis, and set .
X;
{x E lR : Yi 1
=
-
::S
f (x )
.
< yd .
(We require that y; ---+ oo as i ---+ oo.) Assume that X; i s measurable and define the lower and upper Lebesgue sums as lA f, Y) =
L Yi -1 · m X;
L (f, Y )
i=l
=
00
L Yi · mX; . i=l
These sums represent the measure of "Lebesgue rectangles" X; x [0. Yi - l ) and X; x [0, y; ) that sandwich the undergraph. Under the measurability conditions explained below. they converge to the Lebesgue integral as the mesh of Y tends to zero,
ff
l:_ (f, Y ) t
and
L (f. Y ) -!-
f
f.
Upshot Lebesgue sums are like Riemann sums, and Lebesgue integration is like Riemann integration, except that Lebesgue partitions the value axis and takes limits, while Riemann does the same on the domain axis.
The assumption that the sets X; are measurable needs to be put in context. Definition The function
a E JR, fpre la . oo) = follows that the sets
X;
f
{x
=
: lR ---+ lR
is pre-image measurable if for each
E lR : a ::::: f (x ) } is Lebesgue measurable. (It fpre[y;_ 1 , y; ) are measurable.)
This is the standard definition for measurability of a function. We will show that it is equivalent to the geometric definition that the undergraph is measurable, and then discuss when the Lebesgue sums converge to the Lebesgue integral. First we show that there is nothing special about using closed rays [a , oo ) . 23 Proposition Thefollowing are equivalent conditionsforpre-image mea surability of f : lR ---+ R (a) The pre-image of every closed ray [a . oo) is measurable. (b) The pre-image of every open ray (a , oo) is measurable.
3 84
(c) (d) (e) (f) (g) (h)
Lebesgue Theory
Chapter 6
The pre-image of every closed ray ( -oo. a] is measurable. The pre-image of every open ray ( -oo, a) is measurable. The pre-image of every half-open interval La , b) is measurable. The pre-image of every open interval (a , b) is measurable. The pre-image of every half-open interval (a , b] is measurable. The pre-image of every closed interval [a , b] is measurable.
=> (h) =} (a). Since the pre-image of a union, an intersection. or a complement is the union. intersection, or complement of the pre-image, these implications are checked by taking pre-images of the following.
Proof We show that (a) =} (b) =}
(a) =} (b) (b) =} (c) (c) =} (d) (d) =} (e) (e) =} (f) (f) =} (g) (g) =} (h) (h) =} (a)
·
·
·
(a , oo) = U[a + ljn, oo) . (-oo, a]= (a , oo)C. (-oo, a)= U 0 is arbitrary, m * (Ea ) = 0. The Ea are monotone increasing zero sets as a t I . Let ting a = 1 1 /f, f = 1 , 2, . . . , we see that the union of all the Ea with a < 1 is also a zero set, Z. Points x E E \ Z have the property that as B ,!. x, the lim inf o f the concentration of E in B i s ::::: a for all a < 1 ; since the concentration is always _:::: 1 , this means that the limit of the concentration exists and equals 1 . Hence E \ Z is contained in E 1 and almost every point D of E is a density point of E . -
8
Lebesgue's Fundamental Theorem of Calculus
In this section we write the integral of f over a set E as JE f(x) dm . In dimension one we write it as JE f (t) dt , or as Jt f (t) dt when E = (a, {3 ) . Definition The average o f an integrable function
measurable set E
c
f : !Rn ---* IR over a
IRn with positive measure is
1 f (x) dm
fe
=
1 -- r f (x) dm .
mE )E
35 Theorem If f : !Rn ---* IR is Lebesgue integrable, then for almost every p E ]Rn lim
where B ,!.
p means
1 f (x) dm
8{-ph
=
f (p) ,
that B is a ball that contains p and shrinks down to p .
Section
8
Lebesgue's Fundamental Theorem of Calculus
397
E is a measurable set. The average of f on B is the concentration of E in B. By the Lebesgue Density Theorem, for almost every p E :!Rn this concentration converges to X E (p) . Case 2. f is integrable and nonnegative. Fix any a > 0 and consider the
Proof Case 1 . f = X E where
1h1 f(x) dm
set
A = {p E :!Rn : lim sup B {,p
l
- f (p) > a } .
It suffices to show that for each a > 0, A = A (a) is a zero set. Let E > 0 be given. We claim that m * A < E . In Section 5 we showed that for any E > 0 there is a Lebesgue lower sum function
0 there is a function 8 : [a , b] ---+ (0, oo) and -
I L f (tk ) �Xk - I I < n
E
k=I
for all Riemann sums with �Xk < 8 ( tk ) , k = 1 , . . . , n . (McLeod refers to the function 8 as a gauge and to the intermediate points tk as tags.) (a) Verify that if we require the gauge 8 (t) to be continuous then the Denjoy integral reduces to the Riemann integral. (b) Verify that the function
f (x) =
{ )x 1 00
if 0 < x ::::: 1 if x = 0
has Denjoy integral 2. [Hint: Construct gauges 8 (t) such that 8 (0) > 0 but lim 8 (t) = 0 . ] t �o+
424
Lebesgue Theory
Chapter 6
(c) Generalize (b) to include all functions defined on [a . b] for which the improper Riemann integral is finite. (d) Infer from (c) and Exercise 37 that some functions are Denjoy integrable but not Lebesgue integrable. (e) Read McLeod's book to verify that (i) Every nonnegative Denjoy integrable function is Lebesgue integrable, and the integrals are equal. (ii) Every Lebesgue integrable function is Denjoy integrable, and the integrals are equal. Infer that the difference between Lebesgue and Denjoy corre sponds to the difference between absolutely and conditionally convergent series : if f is Lebesgue integrable, so is I f I , but this is not true for Denjoy integrals . 3 9 . Let T : IR2 --+ IR2 b e a rotation and let E c IR2 b e measurable. Show that m (T E) = m E by completing the following outline. (a) The planar measure of a disc is determined solely by its radius. (b) A rectangle R is the union of a zero set and countably many disjoint discs. (c) Under T, discs are sent to discs of equal radius, and a zero set is sent to a zero set. (d) Infer from (a)-( c) that T R is measurable and m (T R) = I R I . (e) Infer from (d) that T E is measurable and m ( T E) = m E . [Hint: Regularity.] (f) Generalize to !Rn . Combined with translation invariance, this exercise shows that Lebesgue measure is invariant under all rigid motions of IRn . 40. If T : IRn --+ IRn is a general linear transformation, use Exercise 39 and the polar form of T explained in Appendix D of Chapter 5 to show that if E c IRn is measurable then m (T E) = l det T l m E . 4 1 . The balanced density of a measurable set E at x is the limit, if exists, of the concentration of E in B where B is a ball centered at x that shrinks down to x . Write 8balanced (x , E) to indicate the balanced density, and if it is 1 , refer to x as a balanced density point. (a) Why is it immediate from the Lebesgue Density Theorem that almost every point of E is a balanced density point? (b) Given a E [0, 1 ] , construct an example of a measurable set E C IR that contains a point x with 8balanced (x . E) = a . (c) Given a E [0, 1 ] , construct an example of a measurable set E c IR that contains a point x with 8 (x , E) = a .
Exercises
425
* *(d) I s there a single set that contains points of both types of density for all a E [0, 1 ] ? 42. Suppose that P c lR has the property that for every interval (a , b) c IR, m t P n (a . b)) 1 b-a 2 (a) Prove that P is nonmeasurable. [Hint: This is a one-liner.] (b) Is there anything special about 1 /2? **43 . Assume that the (unbalanced) density of E exists at every point of R not merely at almost all of them. Prove that up to a zero set, E = IR, or E = 0. ( This is a kind of measure theoretic connectedness. Topological connectedness of lR is useful in the proof.) Is this al so true in IRn ? 44. Prove that any positive measure subset of lR contains a positive mea sure Cantor set. Is the same true in IRn ? *45 . As indicated in Appendix D, U c IR n is K -quasi-round if it can be sandwiched between balls B c U c B ' such that diam B' _::::: K diam B . (a) Prove that in the plane, squares and equilateral triangles are (uniformly) quasi-round. (The same K works for all of them.) (b) What about isosceles triangles? (c) Formulate a Vitali Covering Lemma for a Vitali covering V of A c IR2 by uniformly quasi-round sets instead of discs. (d) Prove it. (e) Generalize to JRn . (f) Consider the alternate definition of K -quasi-roundness of a measurable V C IRn as diam(V) n -
--
mV
--
-
< - K.
What is the relation between the two definitions, and is the Vitali Covering Lemma true for coverings by uniformly quasi-round sets under the second definition? [Hint: Review the proof of the Vitali Covering Lemma.] *46. Construct a Jordan curve (homeomorphic copy of the circle { (x . y) : x 2 + y 2 = 1 } ) in IR2 that has positive planar measure. [Hint: Given a Cantor set in the plane. is there a Jordan curve that contains it? Is there a Cantor set in the plane with positive planar measure?] 47 . [Speculative] Density seems to be a first order concept. To say that the density of E at x is 1 means that the concentration of E in a ball
426
Lebesgue Theory
B containing x tends to 1 as B .J,
x.
Chapter 6
That is,
m (B) - m (E n B) ------� o. mB
48.
49.
50.
*5 1 . *52.
But how fast can we hope it tends to 0? We could call x a double density point if the ratio still tends to 0 when we square the denomi nator. Interior points of E are double density points. Are such points common or scarce in a measurable set? What about balanced density points? Let E c ffi.n be measurable, and let x be a point of a E, the boundary of E . (That is, x lies in both the closure of E and the closure of Ec . ) (a) Is it true that if the density 8 = 8 (x , E) exists then 0 < 8 < l ? Proof or counter-example. (b) Is it true that if 8 = 8 (x , E) exists and 0 < 8 < l then x lies in a E ? Proof or counter-example. (c) What about balanced density? Choose a pair of derivates other than the right max and left min. If f is monotone write out a proof that these derivates are equal almost everywhere. Construct a monotone function f : [0, l ] � JR. whose discontinuity set is exactly the set Q n [0, 1 ], or prove that such a function does not exist. Construct a strictly monotone function whose derivative is equal to zero almost everywhere. In Section 9 the total variation of a function f : [a , b] � ffi is defined as the supremum of all sums I:7= 1 l b. i f l where P partitions [a, hl into subintervals [xi-1 · xd and b.d = f (xi ) - f (xi-d· As sume that the total variation of f is finite (i.e., f is of bounded variation) and define
r; = sup{ L l b. d l } p
k
P; = sup{ L b.d : b.d N: =
p -
k
�
0}
i�f{ L b.d : b.d � 0} k
where P ranges through all partitions of [a, x j . Prove that (a) f is bounded. (b) r; , P; , N; are monotone nondecreasing functions of x .
427
Exercises
(c) r; = P; + N; . (d) f (x) = f (a) + P; N: . **53. Assume that f : [a , b] --+ � has bounded variation. The Banach indicatrix is the function -
(a) Prove that Ny < oo for almost every y. (b) Prove that y 1--+ Ny i s measurable. (c ) Prove that
r:
=
ld Ny dy
where c ::S min f and max f ::S d. *54. A sequence of measurable functions fn : [a , b] --+ � converges nearly uniformly to f as n --+ oo if for each E > 0 there is an E-set S C [a , bl (that is, m S < E) such that fn (x) --+ f(x) uniformly as x varies in sc and n --+ oo . (a) Contrast nearly uniform convergence with almost uniform convergence. which means uniform convergence on the com plement of a zero set. (b) Egoroff's Theorem states that almost everywhere convergence on [a , b] implies nearly uniform convergence. Prove it as fol lows. (i) Let fn : [a , b] --+ � be a sequence of measurable functions that converges almost everywhere to a function f. and set X (k ,
l) = {x E
[a ,
b] : Vn
�
k , l fn (X ) - f (x > l < l jl } .
Show that for each fixed £ , U k X (k , £) equals [a , b] modulo a zero set. (ii) Given E > 0 show that there exists a sequence k 1 < k2 < . . such that for Xe = X (ke , e ) we have m (X D < E/2£ . (iii) lnfer that X = n e xe has m (Xc) < E and that fn converges uniformly to f on X. (Avoid re-using the letter E in your proof.) (c) Your proof in (b) is also valid for functions defined on any bounded subset of Euclidean space, is it not? (d) Why is Egoroff's Theorem false for functions defined on un bounded domains such as �? *55. Show that nearly uniform convergence is transitive in the following sense. Assume that fn converges nearly uniformly to f as n --+ oo, .
428
Lebesgue Theory
Chapter 6
and that for each fixed n there is a sequence fn k wluch converges nearly uniformly to fn as k ---+ oo. (All the functions are measurable and defined on [a , b ] . ) (a) Show that there is a sequence k(n ) ---+ oo a s n ---+ oo such that fn , k(n ) converges nearly uniformly to f as n ----+ oo . In symbols .
nulim nulim fn.k = f n ---+ oo k----* 00
=}
nulim fn.k( n) = f. n -+oo
(b) Why does (a) remain true when almost everywhere convergence replaces nearly uniform convergence? [Hint: The answer is one word.] (c) Is (a) true when IR replaces [a , b l ? (d) Is (b) true when IR replaces [a , b ] ? 56. Consider the continuous functions
fn , k (x) = (cos (:rr n !x ) ) k for k, n E N and x E R (a) Show that for each x E IR, lim lim fn , k (X ) = X Q (X ) , n---> OO k -HXJ the characteristic function of the rationals. (b) Infer from Exercise 23 in Chapter 3 that there can not exist a sequence fn , k(n) converging everywhere as n ---+ oo . (c) Interpret (b) to say that everywhere convergence can not replace almost everywhere convergence or nearly uniform convergence in Exercise 55. *57. Lusin's Theorem states that a measurable function f : [a , b] ---+ IR is nearly uniformly continuous in the sense that for any E > 0 there is an E -set S c [a , b 1 such that the restriction of f to sc is uniformly continuous. Prove Lusin's Theorem as follows. (a) Show that the characteristic function of an open interval is the nearly uniform limit of continuous functions. (b) Infer from (a) that the same is true of the characteristic function of a measurable set. [Hint: Regularity and Exercise 5 5 . ] (c) Infer from (b) that the same i s true for a simple function. (d) Use Egoroff's Theorem and Exercises 28, 55 to infer from (c) that the same is true for a nonnegative measurable function: it is the nearly uniform limit of a sequence of continuous functions.
Exercises
429
(e) Given a measurable f : [a , b] ---+ lR and given E > 0 infer from (d) that there exists a sequence of continuous functions fn : [a , b] ---+ lR and an open E -set U C [a , b] such that fn ---+ f uniformly on uc as n � oo . ( f) Why does (e) imply that f i s nearly uniformly continuous? 58. At what stage, if any, does your reasoning in Exercise 57 make es sential use of one-dimensionality? Explain. 59. Let f : lR ---+ JR. be measurable. (a) Give an example of an f for which the conclusion of Lusin's Theorem is false. (b) Formulate the definition of nearly continuous and prove that f is nearly continuous. (c) Generalize to JR.n . *60. Let E be a measurable subset of the line having positive Lebesque measure. (a) Prove Steinhaus' Theorem: E meets its t-translate for all suffi ciently small t. [Hint: density points.J (b) Formulate and prove the corresponding result in higher dimen sions. (c) Prove that despite the fact that the standard Cantor set has mea sure zero, it meets each of its t-translates for I t I � 1 .
Index
1 -form, 3 1 4
c r equivalent, 29 1 c r norm, 285 Fa -set. 1 89 G0-set, 1 89 M-test, 207 , 2 1 1 , 24 1 , 285 a-Holder, 1 86, 253 8-dense, 253 E -chain, 1 23 E -principle, 2 1 E , 8-condition, 55 E /2n -argument, 1 64 a -compact, 250, 257 k-cell, 3 1 5 k-chain, 325 k-form, 3 1 5 p-adic, 1 30 p-series, 1 82 rth derivative, 1 47 t -advance map, 234 x1-area, 3 1 5 absolute continuity, 398 absolute convergence of a series, 1 8 1 absolute property, 82 abuse of notation, 7 accumulation points, 69
address string, 97 adheres, 5 9 advance map, 234 aleph null, 30 algebraic number, 48 almost everywhere, 1 63 almost uniform convergence, 427 alternating harmonic series, 1 84 alternating series. 1 84 ambient homeomorphism, I 05 arnbiently diffeomorphic, 357 analogy, 1 0 analytic function, 1 48 , 235 Analyticity Theorem, 237 antiderivative, 1 73 Anti derivative Theorem, 1 7 3 Antoine's Necklace. 1 07 arc, 1 22 Archimedean property, 20 area, 306, 364 argument by contradiction, 8 Arzela-Ascoli Theorems, 2 1 4, 2 1 6 ascending k-tuple index, 3 1 7 ascending presentation, 3 1 7 associativity, 1 4 average ( of a function), 396 average derivative, 278
432 B aire class I , 1 89 B aire's Theorem, 243 baker's transformation, 1 1 8 B anach Contraction Principle, 228 B anach indicatrix, 427 B anach space, 285 basic k-form, 3 1 5 Bernstein polynomial, 2 1 8 bijection, 30 bilinear, 275 block test. 1 97 B olzano-Weierstrass Theorem, 77 Borel's Lemma, 256 boundary of a k-cell, 325 boundary of a set, 66, 1 27 bounded function, 249 bounded linear transformation, 268 bounded metric, 1 24 bounded sequence, 49 bounded variation, 406, 426 box, 24 Brouwer's Fixed Point Theorem, 228, 334 bump function, 1 88 Burkill, J .C., 376 Cantor function, 1 75 Cantor piece, I 03 Cantor set, fat, 98, 1 93 Cantor set, middle quarters, 1 92 Cantor set, middle-thirds, 95 Cantor space, 1 03 Cantor surjection theorem, 99 cardinality, 28 Cartesian product. 2 1 Cauchy completion, 1 1 2 Cauchy condition, 1 8, 73 Cauchy Convergence Criteria, 1 9, 1 80 Cauchy product of series, 1 99 Cauchy sequence in a metric space, 73 Cauchy-Binet Formula, 323. 343 Cauchy-Riemann Equations, 340 Cauchy-Schwarz Inequality, 22 Cavalieri 's Principle, 305, 3 3 8 center o f a starI ike set, 1 2 1 chai n connected, 252 chain rule, 274 chain-connected, 1 23
Index Change of Variables Formula, 306 characteristic function. 1 60 Chebyshev Lemma, 40 1 class, 2 class C 0 , C 1 , c r . coc , c w , 1 47 clopen, 60 closed form, 329 closed neighborhood, 94 closed set, 59 closed set condition, 65 closure of a set, 66 cluster point, 69, 1 26 co-Cauchy, 1 09 coarse partition, 1 55 cohomology classes, 3 3 3 common refinement, 1 5 8 commutative diagram, 29 1 compact (in the covering sense), 8 8 compact (sequentially), 76 comparability of metrics, 7 1 comparable norms, 346 comparison test, 1 80 complement of a set, 4 1 complete, 1 3 , 1 8 completeness of a metric space, 74 Completion Theorem, I 08 complex analytic, 238 complex derivative. 339 complex differentiable, 239 complex linear map, 339 composite, 30 compound word (as an address), I 00 concentration, 395 condensation point, 69, 1 26 condition number, 34 1 conditional convergence of a series, 181 cone over a set. 1 25 connected, 83 connected component, 1 36 conorm, 345 constant rank, 290 content, 4 1 9 continuity, 36, 55, 65 continuously differentiable, 1 47 Continuum Hypothesis, 30 contraction. 228 convergence of a sequence, 1 7, 54
433
Index convergence with respect to an order relation, 1 9 1 convex, 25 convex combination, 26, 45 convex function, 46 convex hull, 1 05 countable, 30 countable additivity, 364, 368 countable base of a topology, 1 28 covering, 8 8 covering compact, 88 covering, Vitali or fine, 392 critical point, 1 93 critical value, 193 cupcake theorem, 1 34 curl of a vector field, 329 cut, 1 1 Darboux continuous , 1 44 Darhoux integral, 1 56 de Rham cohomology group, 3 3 3 decimal expansion, 42 decreasing sequence, 1 1 7 Dedekind cuts, 1 1 deMorgan' s Law, 4 1 Denj oy integral, 423 dense, 98 density, 395 density point, 395 density, balanced, 395, 424 denumerable, 30 derivate, 403 derivative, 1 39, 27 1 derivative growth rate. 235 determinant, 343 Devil ' s ski slope, 1 77 Devil's staircase function, 1 75 diagonal, 43 diameter, 79 diffeomorphism, 1 52. 289 difference of sets, 2 differentiable, 1 4 1 , 27 1 differential forms, 3 1 3 dipole, 326 directional derivative. 348 disconnected, 83 discontinuity of the first kind, 1 94 discontinuity of the second kind, 1 94 discrete metric, 52
disjoint sets, 2 distance from a point to a set, 1 1 5 distance vector, 1 24 divergence of a series, 1 79 divergence of a vector field. 328 domain of a function, 28 domination, 1 80 dot product, 22 double density point. 426 double integral, 390 dyadic, 387 dyadic filtration lemma. I 00 dyadic number, 44 dyadic ruler function, 19 3 efficient covering, 392 Egoroff's Theorem, 427 embeds, 8 1 empty set, 2 engulf, 257 envelopes, upper and lower, 380 equal cardinality, 30 equicontinuity, 2 1 3 equivalence relation, classes, 3 error factor, 275 Euclidean distance, 24 Euler characteristic, 45 Euler's product formula, 200 exact form, 329 excludes, 94 exponential growth rate, 1 83 extends to (of a function), 1 1 8 exterior derivative, 3 2 1 F sigma set, 373 fat Cantor set, 98, 1 93 Fatou ' s Lemma, 382 field, 1 5 fine partition, 1 55 finite additivity, 369 finite cardinality. 30 finite intersection property, 1 20 finite subadditivity, 4 1 9 finite word, 48 first orthant, 24 fixed point. 43. 228. 334 flow, 233 flux of a vector field, 329 Frechet derivative, 272
434 front inclusion k -cell, 326 Fubini 's Theorem, 304 function, 28 function algebra, 223 functional, 3 1 4 Fundamental Theorem of Calculus, 171 Fundamental Theorem of Continu ous Functions, 39 G delta set, 373 gap interval, 1 03 gauge, 423 general linear group, 35 1 generic, 243 geometric series. 1 80 gradient vector field, 298 grand intersection, 1 20 graph, 43 grid, 300 growing steeple, 204
Hahn-Mazurkiewicz Theorem, 1 32 Hairy Ball Theorem, 359 harmonic series, 1 80, 1 84 Hausdorff metric, 1 3 3 , 1 99, 252 Hawaiian earring, 53, 1 22 Heine- Borel Theorem, 77 Heine- Borel Theorem in C 0 ([a , b] , JR) , 217 Henstock, 423 higher order differentiability, 1 47 Hilbert cube, 1 3 1 Holder condition. 1 86, 253 holomorphic, 239 homeomorphism, 57 hull, 374, 4 1 7 identity map, 3 0 Identity Theorem, 256 image set of a function. 28 implicit function, 286 Implicit Function Theorem, 286 improper Riemann integral, 1 7 8 inclusion cell, 3 1 8 increasing sequence, 1 1 7 indefinite Lebesgue integral, 398 indicator function, 1 60 infimum, 1 7
Index infinite (cardinality) , 30 infinite product, 1 98 infinitely differentiable, 1 47 inheritance, 52, 67 Inheritance Principle. 68 initial condition, 230 injection, 29 inner measure, 364, 374 inner product, 27 inner product space, 27 integer lattice. 24 integers, I integral test, 1 8 1 integrally equivalent functions, 194 integration by parts, 1 78 integration by substitution, 1 77 integration operators, 330 interior of a set, 66, 1 27 intermediate value property, 38, 1 44 Intermediate Value Theorem, 38, 83, 84 interval, 1 9 intrinsic property. 82 Inverse Function Theorem, 1 52, 289 inverse image, 64 irrational numbers, 1 9 isometry, 1 1 6 iterate (of a function), 407 iterated integral, 390 Jacobian, 306 Jordan content, 306, 4 1 9 Jordan Curve Theorem, 1 3 2 j ump discontinuity, 46, 1 94 kernel, 374, 4 1 8 Kurzweil, 423 L' Hospital's Rule, 1 43 Lagrange multiplier, 296 lattice. 24 least upper bound property, 1 3 Lebesgue Density Theorem, 395 Lebesgue Dominated Convergence Theorem, 3 8 1 Lebesgue integrable, 377 Lebesgue integral. 376 Lebesgue measure, 367 Lebesgue number, 89, 1 66
435
Index Lebesgue outer measure. 363 Lebesgue sums, 383 Lebesgue's Fundamental Theorem of Calculus, 396 Lebesgue's Last Therorem, 40 1 Lebesgue's Main Theorem, 399 left limit, 46 Leibniz rule, 274 length, 363 length of a vector, 22 less than or equal, 1 3 letters in an address, 1 00 limit, 1 7 , 54, 5 8 limit infimum, 49 limit point, 5 8 limit set, 6 1 limit supremum, 49 linear transformation, 267 Lipschitz condition, 1 42, 1 70, 1 95, 230 Littlewood's princinples. 394 locally path-connected, 1 32 logarithm function, 1 74 lower bound, 43 lower integral, 1 5 6 lower semicontinuity, 420 lower sum. 1 56 Lusin's Theorem, 428 magnitude, 1 6 magnitude of a vector, 22 Manhattan metric, 72. 1 24 maximum stretch. 268 McLeod, 423 meager, 243 mean value property, 1 4 1 Mean Value Theorem, 1 4 1 , 277 , 278 measurability (of a function), 376 measurability, pre-image, 383 measurable, 367 measurable set, 367 measure continuity, 370 Mertens ' Theorem, 1 99 mesh of a partition. 1 55 metaphor, 9 metric, metric space, 5 1 middle quarters Cantor set, 1 92 middle-thirds Cantor set, 95 minimum stretch, 345
modulus of continuity, 253 monotone, 1 1 7 Monotone Convergence Theorem, 377 monotone function. 4 7 monotone sequence, 1 1 7 monotonicity of outer measure, 364 Moore- Kline Theorem, l 03 most (in the Baire sense), 243 name of a form, 3 1 4 natural numbers, 1 nearly, 392, 4 1 2 nearly continuous, 4 1 2, 429 nearly uniform continuity, 428 nearly uniform convergence. 4 1 2, 427 neighborhood. 62 nested sequence, 78 net x -variation, 3 1 4 norm, 27 normed space. 268 nowhere dense. 98 nowhere differentiable functions. 240 one-to-one, 29 onto, 29 open covering, 88 open map. 1 1 8 open set condition, 65 operator norm, 268 orbit, 1 1 8, 407 ordered field, 1 5 ordinate set, 377 orthant. 24 orthogonal linear map, 34 1 oscillating discontinuity, 1 94 oscillation, 1 65 outer measure, 363 outer measure axi oms, 364 outer measure, abstract, 367 parallelogram law, 50 partial derivative, 272 partition, 1 04 partition of an index set, 353 partition pair. 1 54 path, 86 path-connected, 86, 1 3 2 Peano curve, 1 02 Peano space, 1 3 2
436 perfect, 93 periodic, 407 permutation, 342 Picard's Theorem, 23 1 piece, 99 piecewise continuous function, 1 60 piecewise linear function, 245 planar outer measure, 364 Poincare's Lemma, 330 pointwise convergence, 20 1 pointwise equicontinuity, 2 1 3 , ?49 pointwise limit, 20 I polar form, 3 4 1 positive definite symmetric linear map, 34 1 positive definiteness of a metric, 52 power series, 1 85 , 2 1 1 power set, 4 7 pre-image, 64 prime number, 6 product matrix, 270 propagation, 2 1 6 proper subset, 82 pullback, 322 pushforward, 322 quasi-round, 4 1 2 Rademacher's Theorem, 406 radius of convergence, 1 8 5 range o f a function, 2 8 rank, 290 Rank Theorem, 292 Ratio Mean Value Theorem, 1 43 ratio test, 1 83 rational cut 1 2 rational numbers, 2 rational ruler function, 1 6 1 real number, 1 2 rear inclusion k-cell, 326 rearrangement of a sequence, 1 1 8 rearrangement of a series , 198 reduces to (of a covering), 8 8 refinement o f a partition, 1 5 8 Refinement Principle, 1 5 8 regularity (ofLebesgue measure), 373 regularity hierarchy, 1 48, 208 relative measure, 395
Index representative of an equivalence class. 3
residual, 243 retraction, 334 Riemann s- function, 200 Riemann integrable, 1 5 5 , 300 Riemann integral, 1 55 Riemann measurable, 306 Riemann sum, 1 54, 300 Riemann 's Integrability Criterion, 1 60 Riemann-Lebesgue Theorem, 1 63 , 302 right interval, 403 right limit, 46 right slope, 403 root test, 1 83 sawtooth function, 240 Schroeder-Bernstein Theorem, 34 scraps, 88 second derivative, 280 second-differentiable, 280 semicontinuity, 420 separable metric space, 1 28 separates points, 223 separation (disconnectedness), 83 sequence , 1 7, 52 sequentially compact, 76 sigma algebra, 368 sign of a permutation, 342 signed commutativity, 3 1 6 simple k-form, 3 1 5 simple function, 386, 42 1 simple region, 356 simply connected, 3 3 0 singleton set, 2 slice, 387 sliding secant method, 1 45 slope over an interval, 40 1 smooth, 14 7, 284 somewhere dense, 98 space filling, 1 02 staircase approximation, 356 starlike, 1 2 1 , 332 steeple functions, 204 Steinhaus' Theorem, 429 step function, 1 60 Stokes ' Theorem, 326, 327 Stone-Weierstrass Theorem, 223 strictly decreasing sequence, 1 1 7
437
Index strictly increasing sequence, 1 1 7 subadditivity, finite, 4 1 9 subcovering, 8 8 subfield, 1 5 sublinear, 27 1 subsequence, 54 subspace, 52 sup norm, 205 supersolving, 225 supremum. 1 7 surjection, 29 symmetric difference, 2 symmetrization of a bilinear map, 350 symmetry of a metric, 52 tag, 423 tail of a series , 1 80 tame, 1 06 target of a function, 28 taxicab metric, 72, 1 24 Taylor polynomial, 1 49 Taylor series, 1 5 1 , 235 Taylor's Theorem, 1 50, 238 telescoping sum, 207 term by term differentiation, 2 1 0 term b y term integration, 208 thick, 243 thin, 243 tiling a manifold with cells, 327 topological equivalence, 66 topological property, 64 topological space, 60 topologist's sine circle, 53, 1 23 topologist' s sine curve, 87 topology, 60 total derivative, 272 total length. 98, 363 total length of a covering by intervals, 1 63 total variation, 406, 426 totally bounded, 92 totally disconnected, 96 trajectory, 230 transitive relation, 1 9 1 transitivity, 1 5 translation, 1 5 translation (by a function), 378
trefoil knot, 105, 1 34 Triangle Inequality, 1 6 triangle inequality, 23, 24, 52 trichotomy, 1 5 trigonometric polynomial, 227 truncate an address string, 97 type 1 region, type 2 region, 356 ultrametric, 1 29 uncountable, 30 undergraph, 1 54, 376 undergraph, completed, 380 undergraph, total, 382, 42 1 uniform cr convergence, 284 uniform continuity, 48, 82, 1 1 8 uniform convergence, 20 I uniform equicontinuity, 249 uniform limit, 202 uniformly cr Cauchy, 284 unit ball, cube, sphere, 24 universal compact metric space, 99 upper bound, 1 3 upper integral, 1 56 upper semicontinuity, 1 36, 264, 420 upper sum, 1 56 utility problem, 1 32 vanishing at a point, 223 variation, 406 Vitali covering, 392 volume multiplier, 306 weak contraction, 254 wedge product, 3 1 8 Weierstrass M -test, 207 Weierstrass Approximation Theorem 217 weighting signed area, 3 1 5 wild, 1 06 word (as an address), 1 00 Zeno's maze, 5 3 Zeno's staircase function, 1 6 1 zero locus, 256, 286 zero set, 98, 1 63 , 302, 365 zero-th derivative, 147
