Points converge to a limit if they get physically closer and closer to it. What about a sequence of functions? When do functions converge to a limit function? What should it mean that they get closer and closer to a limit function? The simplest idea is that a sequence of functions In converges to a limit function I if for each x, the values In (x) converge to I (x) as n -+ 00. This is called pointwise convergence: a sequence of functions In : [a, b] -+ ~ converges pointwise to a limit function I : [a, b] -+ ~
if for each x E [a, b],
lim In(x) = I(x). n-HX)
The function I is the pointwise limit of the sequence (fn) and we write
In -+ I or lim In = J. n->oo
Note that the limit refers to n -+ 00, not to x -+ 00. The same definition applies to functions from one metric space to another.
The requirement of uniform convergence is stronger: the sequence of functions In : [a, b] -+ ~ converges uniformly to the limit function I: [a, b] -+ ~ if for each E > 0 there is an N such that for all n :::: Nand
The function I is the uniform limit of the sequence Un) and we write
In ~ I or unif limln = f. n-'>oo
Your intuition about uniform convergence is crucial. Draw a tube V of vertical radius E around the graph of I. For n large, the graph of In lies wholly in V. See Figure 83. Absorb this picture!
b
Figure 83 The graph of In is contained in the E-tube around the graph of I.
It is clear that uniform convergence implies pointwise convergence. The difference between the two definitions is apparent in the following standard example.
Example Define In : (0, 1) -+ lR. by In (x) = xn. For each x E (0, 1) it is clear that In (x) -+ 0. The functions converge pointwise to the zero function as n -+ 00. They do not converge uniformly: for take E = 1/10. The point Xn = ::/1/2 is sent by In to 1/2 and thus not all points x satisfy (1) when n is large. The graph of In fails to lie in the E -tube V. See Figure 84.
The lesson to draw is that pointwise convergence of a sequence of functions is frequently too weak a concept. Gravitating toward uniform convergence we ask the natural question:
Which properties 01 junctions are preserved under uniform convergence?
The answers are found in Theorem 1, Exercise 4, Theorem 6, and Theorem 9. Uniform limits preserve continuity, uniform continuity, integrability, and - with an additional hypothesis - differentiability.
Section 1 Uniform Convergence and CO[a, b) 203
o
Figure 84 Non-uniform, pointwise convergence.
1 Theorem If fn =::; f and each fn is continuous at Xo, then f is continuous at Xo. In other words, the uniform limit of continuous functions is continuous.
Proof For simplicity, assume that the functions have domain [a, b] and target R (See also Section 8 and Exercise 2.) Let E > 0 and Xo E [a, b] be given. There is an N such that for all n ::: N and all x E [a, b],
E Ifn(x) - f(x)1 < 3·
The function fN is continuous at Xo and so there is a 8 > 0 such that Ix - xol < 8 implies
If Ix - Xo I < 8 then
If(x) - f(xo) I ::: If(x) - fN(X)1 + IfN(X) - fN(XO) I + IfN(XO) - f(xo) I E E E
< - + - + - = E. - 3 3 3
Thus f is continuous at Xo E [a, b]. D
Without uniform convergence, the theorem fails. For example, define fn : [0, 1] -+ lR as before, fn(x) = xn. Then fn(x) converges pointwise to the function
I(x) = I~ ifO:Sx<l
if x = 1.
204 Function Spaces Chapter 4
The function I is not continuous and the convergence is not uniform. What about the converse? If the limit and the functions are continuous, does pointwise convergence imply uniform convergence? The answer is "no," as is shown by xn on (0, 1). But what if the functions have a compact domain of definition, [a, b]? The answer is still "no."
Example John Kelley refers to this as the growing steeple,
{
n2X
In(x) = ~n - n2x
if 0 < x < 1 - -n
if 1 < x < ~ n - - n
if~<x<1. n - -
See Figure 85.
o
Figure 85 The sequence of functions converges pointwise to the zero function, but not uniformly.
Then limn---+oo In (x) = 0 for each x, and In converges pointwise to the function I = O. Even if the functions have compact domain of definition, and are uniformly bounded and uniformly continuous, pointwise convergence does not imply uniform convergence. For an example, just multiply the growing steeple functions by 1/ n.
The natural way to view uniform convergence is in a function space. Let Cb = Cb([a, b], JR) denote the set of all bounded functions [a, b] ~ R
Section 1 Uniform Convergence and COra, b] 205
The elements of Cb are functions I, g, etc. Each is bounded. Define the sup norm on Cb as
IIIII = sup{l/(x)1 : x E [a, b]}.
The sup norm satisfies the norm axioms discussed in Chapter 1, page 27.
11/11 :::: 0 and IIIII = 0 if and only if 1= O. Ilcfll = Iclll/ll III + gil:::: IIIII + Ilgli.
As we observed in Chapter 2, any norm defines a metric. In the case at hand,
dU, g) = sup{l/(x) - g(x)1 : x E [a, b]}
is the corresponding metric on Cb . See Figure 86. To distinguish the norm II I II = sup I I (x) I from other norms on C b we sometimes write II I II sup for the sup norm.
graph/,
d(f,g)
Figure 86 The sup-norm of I and the sup-distance between the functions I and g.
The thing to remember is that Cb is a metric space whose elements are functions. Ponder this.
2 Theorem Convergence with respect to the sup-metric d is equivalent to uniform convergence.
Proof If dUn, f) ---+ 0 then sup{llnx - Ixl In :::::t I, and conversely.
3 Theorem Cb is a complete metric space.
x E [a, b]} ---+ 0, so o
206 Function Spaces Chapter 4
Proof Let Un) be a Cauchy sequence in Cb. Foreachindividualxo E [a, b] the values In (xo) form a Cauchy sequence in JR since
Thus, for each x E [a, b], lim In(x)
n-HX)
exists. Define this limit to be I (x). It is clear that In converges pointwise to I. In fact, the convergence is uniform. For let E > ° be given. There exists N such that m, n :::: N imply
Also, for each x E [a, b] there exists an m = m(x) :::: N such that
E Ilm(x) - l(x)1 < "2.
If n :::: N and x E [a, b] then
I/n(x) - l(x)1 ::: I/n(x) - 1m (x) (x) I + I Im(x) (x) - l(x)1 E E
< - + - = E. 2 2
Hence In ~ I· The function I is bounded. For IN is bounded and for all x, IIN(X) - l(x)1 < E. Thus I E Cb. By Theorem 2, uniform convergence implies d-convergence, dUn' f) -+ 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 CO = CO([a, b], JR) denote the set of continuous functions [a, b] -+
R Each I E CO belongs to Cb since a continuous function defined on a compact domain is bounded. That is, CO c Cb .
4 Corollary CO is a closed subset OICb. It is a complete metric space.
Proof Theorem 1 implies that a limit in Cb of a sequence of functions in CO lies in Co. That is, CO is closed in Cb . A closed subset of a complete space is complete. D
Section 1 Uniform Convergence and CO[a, b] 207
Just as it is reasonable to discuss the convergence of a sequence of functions we can also discuss the convergence of a series of functions, L fk. Merely consider the nth partial sum
n
Fn(x) = L fk(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
00
F(x) = 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 l!k (x) I converges, then the series L fk converges absolutely.
5 Weierstrass M-test If L Mk is a convergent series of constants and if fk E Cb satisfies II fk II ::::: Mk for all k, then L fk converges uniformly and absolutely.
Proof If n > m then the partial sums of the series of absolute values telescope as
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 Cb, and by Theorem 3 it converges uniformly. 0
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 lb fn(x) dx = lb uniflimfn(x) dx. n--+oo a a n--+oo
208 Function Spaces Chapter 4
In other words, R is a closed subset of Cb and the integral functional f f-+ f: f (x) dx is a continuous map from R to R 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 -+ 00. By the Riemann-Lebesgue Theorem, fn is bounded and there is a zero set Zn such that fn is continuous at each x E [a, b] \ Zn. Theorem 1 implies that f is continuous at each x E [a, b]\UZn, while Theorem 3 implies that f is bounded. Since UZn is a zero set, the Riemann-Lebesgue Theorem implies that fER. Finally
lib f(x)dx -ib fn(x)dxl = lib f(x) - fn(X)dxl
:s ib If(x) - fn(x)1 dx :s dU, fn)(b - a) -+ 0
as n -+ 00. Hence the integral of the limit is the limit of the integrals. 0
7 Corollary If fn E Rand fn =t f then the indefinite integrals converge uniformly,
Proof As above,
liX f(t) dt -lX fn(t) dtl :s dUn, f)(x - a) :s dUn, f)(b - a) -+ 0
when n -+ 00. o
8 Term by Term Integration Theorem A uniformly convergent series of integrable functions L A can be integrated term by term in the sense that
bOO 00 b 1 !;fk(X)dX = !; 1 A(x)dx.
Section 1 Uniform Convergence and CO [a, b] 209
Proof The sequence of partial sums Fn converges uniformly to L fk. Each Fn belongs to R since it is the finite sum of members of R. According to Theorem 6,
This shows that the series L 1: fk(X) dx converges to f: L fk(X) dx. 0
9 Theorem The uniform limit of a sequence of differentiable functions is differentiable provided that the sequence of derivatives also converges uniformly.
Proof We suppose that fn : [a, b] -+ lR is differentiable for each nand that fn ~ f as n -+ 00. Also we assume that f~ ~ 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
fn(x) fn(a) + l x
f~(t)dt ~ f(a) + l x
g(t)dt.
Since fn ~ f we see that f(x) = f(a) + f: get) dt and, again by the fundamental theorem of calculus, f' = g.
In the general case the proof is harder. Fix some x E [a, b] and define
I fn(t) - fn(x)
t-x f~(x)
I f(t) - f(x)
¢(t) = t-x g(x)
if t i= x
ift = x
if t i= x
ift = x.
Each function ¢n is continuous since ¢n(t) converges to f~(x) as t -+ x. Also it is clear that ¢n converges pointwise to ¢ as n -+ 00. We claim the convergence is uniform. For any m, n the Mean Value Theorem applied to the function fm - fn gives
¢m(t) - ¢n(t) = (fm(t) - fn(t)) - (fm(x) - fn(x)) = 1,' (e) - 1,'(e) t-x m n
210 Function Spaces Chapter 4
for some e between t and x. Since f~ ~ g the difference f/n - f~ tends uniformly to 0 as m, n --+ 00. Thus (CPn) is Cauchy in Co. Since CO is complete, CPn converges unifonnly to a limit function 1/1, and 1/1 is continuous. As already remarked, the pointwise limit of CPn is cP, and so 1/1 = cp. Continuity of 1/1 = cP implies that g(x) = f'(x). D
10 Term by Term Differentiation Theorem A uniformly convergent series of differentiable functions can be differentiated term by term, provided that the derivative series converges uniformly,
Proof Apply Theorem 9 to the sequence of partial sums. D
Note that Theorem 9 fails if we forget to assume the derivatives converge. For example, consider the sequence of functions fn : [-1, 1] --+ lR defined by
~ fn(x) = yr +;;.
Figure 87 The unifonn limit of differentiable functions need not be differentiable.
See Figure 87. The functions converge unifonnly to f(x) = lxi, a nondifferentiable function. The derivatives converge pointwise but not unifonnly. Worse examples are easy to imagine. In fact, a sequence of everywhere differentiable functions can converge unifonnly to a nowhere differentiable function. See Sections 4 and 7. It is one of the miracles of the complex numbers that a unifonn limit of complex differentiable functions is complex differentiable, and automatically the sequence of derivatives converges unifonnly to a limit. Real and complex analysis diverge radically on this point.
Section 2 Power Series 211
2 Power Series
As another application of the Weierstrass M -test we say a little more about the power series L Ckxk. A power series is a special type of series of functions, the functions being constant multiples of powers of x. As explained in Section 3 of Chapter 3, its radius of convergence is
1 R=---
lim sup ,lfCk. k-+oo
Its interval of convergence is (-R, R). If x E (-R, R), the series converges and defines a function f (x) = L Ckxk, while if x fj. [-R, R), the series diverges. More is true on compact subintervals of ( - R, R).
11 Theorem If r < R, then the power series converges uniformly and absolutely on the interval [-r, r].
Proof Choose f3, r < f3 < R. For all large k, ~ < 1/ f3 since f3 < R. Thus, if Ix I :::: r then
These are terms in a convergent geometric series and according to the M -test L Ck xk converges uniformly when x E [-r, r]. 0
12 Theorem A power series can be integrated and differentiated term by term on its interval of convergence.
For f(x) = LCkXk and Ixl < R this means
00 r f(t)dt = f ~xk+l 10 k=O k + 1
and f'(x) = L kCkXk-l. k=l
\
Proof The radius of convergence of the integral series is determined by the exponential growth rate of its coefficients,
~ ()
l/k . Ck-l. (k-ll/k 1
hmsup k I-I = hmsup (ICk_lI1/(k-ll) -k-+oo k k-+oo k
Since (k - 1) / k --+ 1 and k- 1/ k --+ 1 as k --+ 00, we see that the integral series has the same radius of convergence R as the original series. According to Theorem 8, term by term integration is valid when the series converges
212 Function Spaces Chapter 4
uniformly, and by Theorem 11, the integral series does converge uniformly on any interval [-r,r] c (-R,R).
A similar calculation for the derivative series shows that its radius of convergence too is R. Term by term differentiation is valid provided the series and the derivative series converge uniformly. Since the radius of convergence of the derivative series is R, the derivative series does converge uniformly on any [-r, r] C (-R, R). 0
13 Theorem Analytic functions are smooth, C'" C Coo.
Proof An analytic function f is defined by a convergent power series. According to Theorem 12, the derivative of f is given by a convergent power series with the same radius of convergence, so repeated differentiation is valid, and we see that f is indeed smooth. 0
The general smooth function is not analytic, as is shown by the example
1-1IX
e(x) = ~ if x > ° ifx:::O
on page 149. 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
expx = L~ k=O k!
• 00 (_l)k x2k+1
sm x = L (2k + I)! k=O
00 (_I)kx 2k
cosx = t; (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 convergence R = 00. Theorem 12 justifies term by term differentiation, yielding the usual formulas,
exp'(x) = expx sin'(x) = cosx cos'(x) = - sinx.
The logarithm has already been defined as the indefinite integral It 1/ t d t . We claim that if Ix I < 1 then log(l + x) is given as the power series
00 (_l)k+l xk log(l +x) = L k
k=l
Section 3 Compactness and Equicontinuity in CU 213
To check this, we merely note that its derivative is the sum of a geometric series,
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 CO
The Heine-Borel theorem states that a closed and bounded set in JR.m is compact. On the other hand, closed and bounded sets in CO are rarely compact. Consider for example the closed unit ball
B = {f E CO([O, 1], JR.) : Ilfll ~ I}.
To see that B is not compact we look again at the sequence fn (x) = xn. It lies in B. Does it have a subsequence that converges (with respect to the metric d of Co) to a limit in Co? No. For if fnk converges to f in CO then f(x) = lim fnk(X). Thus f(x) = 0 if x < I and f(1) = 1, but this
k-+oo
function f does not belong to Co. The cause of the problem is the fact that CO 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 CO 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 Un) in CO is equicontinuous if
V E > 0 38 > 0 such that
Is - tl < 8 and n EN=} Ifn(s) - fn(t)1 < E.
The functions fn are equally continuous. The 8 depends on E but it does not depend on n. Roughly speaking, the graphs of all the fn are similar. For total clarity, the concept might better be labeled uniform equicontinuity, in contrast to pointwise equicontinuity, which requires
V E > 0 and V x E [a, b] 38 > 0 such that
Ix - tl < 8 and n EN=} Ifn(x) - fn(t) I < E.
214 Function Spaces Chapter 4
The definitions work equally well for sets of functions, not only sequences of functions. The set £ c CO is equicontinuous if
YE > 0 38 > 0 such that
Is - tl < 8 and f E £ ::::} If(s) - f(t)1 < E.
The crucial point is that 8 does not depend on the particular f E £. It is valid for all f E £ simultaneously. To picture equicontinuity of a family £, 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 offunctions in CO([a, b], JR) has a uniformly convergent subsequence.
Section 3 Compactness and Equicontinuity in CO 215
Think of this as a compactness result. If (fn) is the sequence of equicontinuous functions, the theorem amounts to asserting that the closure of the set Un : n E N} is compact. Any compact metric space serves just as well as [a, b], and the target space JR can also be more general. See Section 8.
Proof [a, b] has a countable dense subset D = {d1, d2 , ••• }. For instance we could take D = Q n [a, b]. Boundedness of (fn) means that for some constant M, all x E [a, b], and all n EN, Ifn(x)1 :::: M. Thus (fn(dd) is a bounded sequence of real numbers. Bolzano-Weierstrass implies that some subsequence of it converges to a limit in JR, say
fl,k(dd -+ Yl as k -+ 00.
The subsequence fl,k evaluated at the point d2 is also a bounded sequence in JR, and there exists a sub-subsequence h,k such that h,k (d2 ) converges to a limit in JR, say h,k(d2) -+ Y2 as k -+ 00. The sub-subsequence evaluated at d l still converges to Yl. Continuing in this way gives a nested family of subsequences fm,k such that
(fm,k) is a subsequence of (fm-l,k)
j :::: m ::::} fm,k(d j ) -+ Yj as k -+ 00.
Choose k(m) :::: m large enough that if j :::: m and k(m) :::: k then
The superdiagonal subsequence gm (x) = fm,k(m) (x) converges to a limit at each point XED. We claim that gm (x) also converges at the other points x E [a, b] and that the convergence is uniform. It suffices to show that (gm) is a Cauchy sequence in Co.
Let E > 0 be given. Equicontinuity gives a 8 > 0 such that for all s,t E [a,b],
Is-tl<8 ::::}
Choose J large enough that every x E [a, b] lies in the 8-neighborhood of some dj with j :::: J. Since D is dense and [a, b] is compact, this is possible. See Exercise 19. Since {d1, ••• , dJ } is a finite set and gm (d j ) converges for each dj , there is an N such that for all £, m :::: N and all j :::: J,
216 Function Spaces Chapter 4
If.e, m ::: N and x E [a, b], choose d j with Idj - xl < 8 and j :::: 1. Then
Igm(x) - ge(x)1 :::: Igm(x) - gm(dj ) I + Igm(d j ) - ge(dj ) I + Ige(dj ) - ge(x)1 E E E
< - + - + - = E. - 3 3 3
Hence (gm) is Cauchy in Co, it converges in Co, and the proof is complete. D
Part of the preceding development can be isolated as the
15 Arzela-Ascoli Propagation Theorem Pointwise convergence of an equicontinuous sequence of functions on a dense subset of the domain propagates to uniform convergence on the whole domain.
Proof This is the E 13 part of the proof. D
The example cited over and over again in the equicontinuity world is the following:
16 Corollary Assume that fn : [a, b] ---+ lR is a sequence of differentiable functions whose derivatives are uniformly bounded. If for some Xo, fn (xo) is bounded as n ---+ 00, then the sequence Un) 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 Nand x E [a, b]. Equicontinuity of Un) follows from the Mean Value Theorem:
Is - tl < 8 => Ifn(s) - fn(t) I = If~(e)lls - tl :::: M8
for some e between sand t. Thus, given E > 0, the choice 8 = E/(M + 1) shows that Un) is equicontinuous.
Let C be a bound for Ifn(xo)l, valid for all n EN. Then
Ifn(x)1 :::: Ifn(x) - fn(xo) I + Ifn(xo)1 :::: Mix - xol + C
::::Mlb-al+C
shows thatthe sequence Un) is bounded in Co. The Arzela-Ascoli theorem then supplies the uniformly convergent subsequence. D
Two other consequences of the same type are fundamental theorems in the fields of ordinary differential equations and complex variables.
Section 4 Uniform Approximation in CO 217
(a) A sequence of solutions to a continuous ordinary differential equation in JR.m 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.
17 Heine-Borel Theorem in a function space A subset E C CO is compact if and only if it is closed, bounded, and equicontinuous.
Proof Assume that [; is compact. By Theorem 2.56, it is closed and totally bounded: given E > 0 there is a finite covering of E by neighborhoods in CO that have radius E/3, say ~j3(fk)' with k = 1, ... , n. Each fk is uniformly continuous so there is a 8 > 0 such that
Is - tl < 8 =}
If fEE then for some k, f E NE/ 3(fk), and Is - tl < 8 implies
If(s) - f(t)1 ::: If(s) - fk(s)1 + Ifk(s) - fkCt) I + Ifk(t) - f(t)1 E E E
<-+-+-=E 333
Thus E is equicontinuous. Conversely, assume that [; is closed, bounded, and equicontinuous. If (fn)
is a sequence in [; then by the Arzela-Ascoli theorem, some subsequence (fnk) converges uniformly to a limit. The limit lies in E since E is closed. Thus E is compact. 0
4 Uniform Approximation in CO
Given a continuous but nondifferentiable function f, we often want to make it smoother by a small perturbation. We want to approximate f in CO 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 The set of polynomials is dense in CO([a, b], JR.).
Density means that for each f E CO and each E > 0 there is a polynomial function p(x) such that for all x E [a, b],
If(x) - p(x)1 < E.
There are several proofs of this theorem, and although they appear quite
218 Function Spaces 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, b] is [0, 1]. We do so.
Proof #1 For each n E N, consider the sum
Pn(x) = t (n)Ckxk(l - x)n-k, k=O k
where Ck = f(k/n) and G) is the binomial coefficientn!/ 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 --+ 00. The proof relies on two formulas about how the functions
rk(x) = (~)Xk(l - xt-k
shown in Figure 90 behave. They are n
(2)
n
(3) I:(k - nx)2rk (x) = nx(l - x). k=O
In terms of the functions rk we write
Figure 90 The seven basic Bernstein polynomials of degree six,
(6) k 6-k k x (I-x) ,k=0, ... ,6.
Section 4 Uniform Approximation in CO 219
n n
Pn(x) = I>krk(X) I(x) = L I(x)rk(x). k=O k=O
Then we divide the sum Pn - 1= L)Ck - f)rk into the terms where kin is near x, and other terms where kin is far from x. More precisely, given E > 0 we use uniform continuity of I On [0, 1] to find 8 > 0 such that It - sl < 8 implies II(t) - I(s)1 < E/2. Then we set
Kl={kE{0, ... ,n}:I~-xl<8} and K2={0, ... n}\K1•
This gives
n
The factors ick - I (x) I in the first sum are < E 12 since Ck = I (k In) and kin 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
Note that the bottom term in (5) and the bottom two terms in (6) are O. Multiplying (5) by x and (6) by x 2 and then setting y = I - x in both equations gives
Proof #2 Let f E CO([O, 1], JR) be given and let g(x) = f(x) - (mx + b) where
f(l) - f(O) m=
1 and b = f(O).
Then g E CO and g (0) = 0 = g (1). If we can approximate g arbitrarily well by polynomials, then the same is true of f since mx + b is a polynomial. In other words it is no loss of generality to assume that f(O) = f(1) = 0 in the first place. Also, we extend f to all of JR by defining f (x) = 0 for all x E JR \ [0, 1]. Then we consider a function
f3n(t) = bn(1 - t 2 )n - 1 :::: t :::: 1,
where the constant bn is chosen so that tl f3n(t) dt = 1. As shown in Figure 91, f3n is a kind of polynomial bump function. Set
Pn(x) = L: f(x + t)f3n(t) dt.
This is a weighted average of the values of f using the weight function f3n. We claim that Pn is a polynomial and Pn(x) =+ f(x) as n -+ 00.
1.5,----,---,---,---,----,---,---,---,----,--,
0.5
Figure 91 The graph of the function f36(t) = 1.467(1 - t2)6.
To check that Pn is a polynomial we use a change of variables, u = x + t. Then
lx+1 11 Pn(x) = f(u)f3n(U - x) du = f(u)f3n(X - u) du
x-I ° since f = 0 outside of [0, 1]. The function f3n (x - u) is a polynomial in
222 Function Spaces Chapter 4
x whose coefficients are polynomials in u. The powers of x pull out past the integral and we are left with these powers of x multiplied by numbers, the integrals of the polynomials in u times I (u). In other words, by merely inspecting the last formula, it becomes clear that Pn (x) is a polynomial mx.
To check that Pn =:; I as n ---+ 00, we need to estimate fin (t). We claim that if 8 > 0 then
(10) fin (t) =:; 0 as n ---+ 00 and 8 ::::: I t I ::::: 1.
This is clear from Figure 91. Proceeding more rigorously, we have
Since 1/ e = lim (1 - 1/ n)n, we see that for some constant c and all n, n--->oo
See also Exercise 29. Hence if 8 ::::: It I ::::: 1 then
due to the fact that In tends to 00 more slowly than (1- 82 )-n as n ---+ 00.
This proves (10). From (10) we deduce that Pn =:; I, as follows: Let E > 0 be given. Uniform continuity of I gives 8 > 0 such that It I < 8
implies II(x + t) - l(x)1 < E/2. Since fin has integral 1 on [-1, 1] we have
The first integral is < E /2, while the second is at most 2M ~tl:=::8 fin (t) dt. By (10), the second integral is < E/2 when n is large. Thus Pn =:; I as claimed. 0
Section 4 Uniform Approximation in CO 223
Next we see how to extend this result to functions defined on a compact metric space M instead of merely on an interval. A subset A of CO M = CO(M, JR) is a function algebra if it is closed under addition, scalar multiplication, and function multiplication. That is, if f, g E A and c is a constant then f + g, c f, and f . g belong to A. For example, the set of polynomials is a function algebra. The function algebra vanishes at a point p if f (p) = 0 for all f E A. For example, the function algebra of all polynomials with zero constant term vanishes at x = O. The function algebra separates points if for each pair of distinct points PI, P2 E M there is a function f E A such that f(PI) f. f(P2). For example, the function algebra of all trigonometric polynomials separates points of [0, 2][) and vanishes nowhere.
19 Stone-Weierstrass Theorem If M is a compact metric space and A is a function algebra in COM that vanishes nowhere and separates points then A is dense in CO M.
Although the Weierstrass approximation theorem is a special case of the Stone-Weierstrass theorem, the proof of the latter does not stand on its own; it depends crucially on the former. We also need two lemmas.
20 Lemma If A vanishes nowhere and separates points then, given distinct points PI, P2 E M, and given constants CI, C2, there exists afunction f E A such that f(PI) = CI and f(P2) = C2.
Proof Choose g], g2 E A that satisfy gl (PI) f. 0 f. g2(P2). Then g = g? + g~ belongs to A and g(PI) f. 0 f. g(P2). Let h E A separate PI, P2, and consider the matrix
By construction a, c f. 0 and b f. d. Hence det H = acd - abc ac(d - b) f. 0, H has rank 2, and the linear equations
a~ + ab1] = CI
c~ + cd1] = C2
have a solution (~, 1]). Then f = ~g + 1]gh belongs to A and f(Pl) = c], f(P2) = C2· 0
21 Lemma The closure of afunction algebra in CO Mis afunction algebra.
Proof Clear enough. o
224 Function Spaces Chapter 4
Proof of the Stone-Weierstrass Theorem. Let A be a function algebra in CO M that vanishes nowhere and separates points. We must show that A is dense in COM: given F E COM and E > 0 we must find G E A such that for all x E M,
(11) F(x) - E < G(x) < F(x) + E.
First we observe that
(12)
where A denotes the closure of A in CO M. Let E > 0 be given. According to the Weierstrass approximation theorem, there exists a polynomial p(y) such that
(13) E
sup{lp(y) - Iyll : Iyl :s Ilfll} < 2:
After all, Iyl is a continuous function defined on the interval [- IIJII , II JII]. The constant term of p(y) is at most E/2 since Ip(O) - 1011 < E/2. Let q(y) = p(y) - p(O). Then q(y) is a polynomial with zero constant term and (13) becomes
(14) Iq(y) - Iyll < E.
Write q(y) = alY + a2y2 + ... + anyn and
g = ad +a2f2 + ... +anr·
Lemma 21 states that A is an algebra, so g E A.t Besides, if x E M and y = f(x) then
Ig(x) - If(x)1I = Iq(y) - Iyll < E.
Hence If I E A = A as claimed in (12). Next we observe that if t, g belong to A, then max(j, g) and min(j, g)
also belong to A. For
(f ) - f + g + It - g I max ,g -2 2
min(j, g) = f + g _ If - gl. 2 2
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 + a] f + ... + an fn belongs to A.
Section 4 Uniform Approximation in CO 225
Repetition shows that the maximum and minimum of any finite number of functions in A also belongs to A.
Now we return to (11). Let FE COM 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 EM. According to Lemma 20 we can find a function in A with given values at p, q, say H pq 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
(15) x E Uq =} 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 (11). See Figure 92.
q
Figure 92 In a neighborhood of q, Hpq supersolves (11) in the sense of (15).
Compactness of M implies that finitely many of these neighborhoods Uq
cover M, say Uq\ ' ••. , Uqn • Define
Then Gp E A and
226 Function Spaces Chapter 4
(16) Gp(p) = F(p) and F(x) - E < Gp(x)
for all x E M. See Figure 93.
Figure 93 G p is the maximum of Hpqi' i = 1, ... , n.
Continuity implies that each p has a neighborhood Vp such that
(17) x E Vp :::} 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 (11) everywhere.
Section 4 Uniform Approximation in CO 227
Figure 95 The graph of G lies in the E-tube around the graph of F.
M, say Vp1 , •.. , Vpm , Set
G(x) = min(Gpl (x), ... , G pm (x».
We know that G E A and (16), (17) imply (11). See Figure 95. 0
22 Corollary Any 2rr -periodic continuous function of x E JR can be uniformly approximated by a trigonometric polynomial
n n
T(x) = Lakcoskx + Lbksinkx. k=O k=O
Proof Think of [0, 2rr) parameterizing the circle Sl by x t-+ (cosx, sin x).
The circle is compact, and 2rr -periodic continuous functions on JR become continuous functions on Sl. The trigonometric polynomials on Sl form an algebra T c CO Sl that vanishes nowhere and separates points. The Stone-Weierstrass theorem implies that T is dense in CO Sl. 0
Here is a typical application of the Stone-Weierstrass Theorem: Consider a continuous vector field F : D. --+ JR2 where D. 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
n
P(x, y) = L CijXiyj
i.j=O
where the Cij are constants. They form a function algebra A in CO (D., JR) that
228 Function Spaces Chapter 4
separates points and vanishes nowhere. By the Stone-Weierstrass Theorem, A is dense in Co, so we can approximate the components of F = (FI, F2) by polynomials
FI ~ P F2 ~ 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, including 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 EM,
d(f(x), fey)) ~ kd(x, y).
23 Banach Contraction Principle Suppose that f : M -+ M is a contraction and the metric space M is complete. Then f has a unique fixed point p andfor any x E M, rex) = f 0 f 0 ... 0 f(x) -+ pas n -+ 00.
Brouwer Fixed Point Theorem Suppose that f : Bm -+ Bm is continuous where Bm is the closed unit ball in IRm. Then f has a fixed point p E Bm.
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 dynamical! See Figure 96. Choose any Xo E M and define Xn = r(xo). We claim that for all n E N
::: d(p, xn) + knd(xo, Xl) + kd(xn, p) --+ 0 as n --+ 00.
Since d(p, f(p)) is independent ofn, 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 pointxo E Mandchooserosolargethatf(Mro(xo» C Mro(xo).LetBo = Mro(xo) and Bn = r(Bn-d. The diameter of Bn is at most kn diam(Bo), and this tends to 0 as n ---+ 00. The sets Bn nest downward as n ---+ 00 and f sends Bn inside Bn+1• Since M is complete, this implies that nBn is a single point, say p, and f(p) = p. 0
Proof of Brouwer's Theorem in dimension one The closed unit I-ball is the interval [-1, 1] in lR. If f : [-1, 1] ---+ [-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 g(p) = O. That is, f(p) = p. 0
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 jRm.
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) yl(t) = F(Y(t)) and YeO) = p.
See Figure 97. In this notation we think of 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
IF(x) - F(y)1 ~ L Ix - YI.
Here, I I refers to the Euclidean length of a vector. F, x, y are all vectors
Section 5 Contractions and ODE's 231
··· ... r .. I •• &f.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.
24 Picard's Theorem Given p E U there exists an F -trajectory Y (t) in U through p. This means that Y : (a, b) ---+ U solves (20). Locally, Y is unique.
To prove Picard's Theorem it is convenient to re-express (20) as an integral 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) I: f(x) dx is approximated by Riemann sums R = L f(tk)/).Xk. (b) Continuous functions are integrable. (c) If f'(x) exists and is continuous then 1: f'ex) dx = feb) - f(a).
(d) II: f(x) dxl ::: M(b - a) where M = sup If(x)l·
The Riemann sum R in (a) has a = Xo ::: ... ::: Xk-l ::: tk ::: Xk ::: ... ::: Xn = b and all the /).Xk = Xk - Xk-l are small.
Given a vector-valued function of a real variable
f(x) = (fl(X), ... , fm(x)),
a ::: x ::: b, we define its integral componentwise as the vector of integrals
232 Function Spaces Chapter 4
Corresponding to (a) - (d) we have (a') I: f(x) dx is approximated by R = (Rl, ... , Rm), with Rj a Rie-
mann sum for h. (b') Continuous vector-valued functions are integrable. (c') If f'(x) exists and is continuous, then 1: f'(x) dx = feb) - f(a).
(d') II: f(x) dxl ::: M(b - a) where M = sup If(x)l· (a'), (b'), (c') are clear enough. To check (d') we write
R = L Rjej = L L h(tk)tukej j k
L L h(tk)ej/:),.xk = L f(tk)/:),.Xk k j k
where el, ... , em is the standard vector basis for ]Rm. Thus,
IRI ::: L If(tk)1 /:),.Xk ::: L M /:),.Xk = M(b - a). k k
By (a'), R approximates the integral, which implies (d'). (Note that a weaker inequality with M replaced by ,JrnM follows immediately from (d). This weaker inequality would suffice for most of what we do - but it is inelegant.)
Now consider the following integral version of (20),
(21) yet) = P + lot F(Y(s)) ds.
A solution of (21) is by definition any continuous curve Y : (a, b) -+ U for which (21) holds identically in t E (a, b). By (b') any solution of (21) is automatically differentiable and its derivative is F(Y(t)). That is, any solution of (21) solves (20). The converse is also clear, so solving (20) is equivalent to solving (21).
Proof of Picard's Theorem Since F is continuous, there exists a compact neighborhood N = Nr(p) and a constant M such that IF(x)1 ::: M for all x EN. Choose I' > 0 such that
(22) I'M ::: r and TL < 1.
Consider the set e of all continuous functions Y [ - T, T] -+ N. With respect to the metric
d(Y, a) = sup{IY(t) - a(t)1 : t E [-I', T]}
Section 5 Contractions and ODE's
the set e is a complete metric space. Given Y E e, define
<I>(Y)(t) = P + 11 F(Y(s»ds.
233
Solving (21) is the same as finding Y such that <I> (Y) = Y. That is, we seek a fixed point of <1>.
We just need to show that <I> is a contraction of e. Does <I> send e into itself? Given Y E e we see that <I>(Y)(t) is a continuous (in fact differentiable) vector-valued function of t and that by (22),
I <I> (Y)(t) - pi = 111 F(Y(S»dSI ::: rM::: r.
Therefore, <I> does send e into itself. <I> contracts e because
d(<I>(Y), <1>(0')) = s~p 111 F(Y(s» - F(O'(s» dsl
::: r sup IF(Y(s» - F(O'(s»1 s
::: r sup L IY (s) - O'(s)1 ::: r Ld(Y, a) s
and r L < 1 by (22). Therefore <I> has a fixed point Y, and <I> (Y) = Y implies that Y (t) solves (21), which implies that Y is differentiable and solves (20).
Any other solution a (t) of (20) defined on the interval [ - r, r] also solves (21) and is a fixed point of <1>, <I> (a) = O'. Since a contraction mapping has a unique fixed point, Y = a, which is what local uniqueness means. 0
The F -trajectories define a flow in the following way: To avoid the possibility 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]Rm. Then trajectories can be defined for all time t E R Let Y (t, p) denote the trajectory through p. Imagine all points P E ]Rm moving in unison along their trajectories as t increases. They are leaves on a river, motes in a breeze. The point PI = Y (t1, p) at which P arrives after time t1 moves according to Y (t, PI). Before P arrives at PI, however, PI has already gone elsewhere. This is expressed by the flow equation
Y(t, PI) = Y(t + f1, 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
234
Y
'" ./
/
/ JI
/ P /
Function Spaces
./ __ e---~
, /'" Pl
~/
/ /
/
",/ P2
Chapter 4
Figure 98 The time needed to flow from from P to P2 is the sum of the times needed to flow from P to PI and from PI to P2·
is locally unique. It is revealing to rewrite the flow equation with different notation. Setting CPt (p) = Y (t, p) gives
CPt+s(p) = CPt(CPs(p)) for all t, S E R
CPt is called the t-advance map. It specifies where each point moves after time t. See Figure 99. The flow equation states that t t-+ CPt is a group
Figure 99 The t-advance map shows how a set A flows to a set CPt (A).
homomorphism from lR into the group of motions of lRm. In fact each CPt is a homeomorphism oflRm onto itself and its inverse is CP-t. For CP-t 0 CPt = CPo and CPo is the time-zero map where nothing moves at all, CPo = identity map.
Section 6* Analytic Functions 235
6* Analytic Functions
Recall from Chapter 3 that a function f : (a, b) -+ lR 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 Ckhk such that for all x + h near x,
00
f(x + h) = LCkhk•
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 rth derivative grows as r -+ 00.
Let f : (a, b) -+ lR be smooth. The Taylor series for f at x E (a, b) is
~ f(k) (x) hk. ~ k! k=O
Let I = [x - a, x + a] be a subinterval of (a, b), a > 0, and denote by Mr the maximum of I f(r) (t) I for tEl. The derivative growth rate of f on I is
. ~Mr a = hmsup -. r--->oo r!
Clearly, j I f(r) (x) I / r! :::: .{j Mr / r!, so the radius of convergence
1 R = -----;::.===
r--->oo
of the Taylor series at x satisfies
1 -:::: R. a
r!
In particular, if a is finite the radius of convergence of the Taylor series is positive.
25 Theorem If aa < 1, then the Taylor series converges uniformly to f on the interval I.
236 Function Spaces Chapter 4
Proof Choose8 > o such that (a+8)(j < l.TheTaylorremainderfonnula from Chapter 3, applied to the (r - l)st order remainder, gives
r-l f(k) f(r) f(x + h) _ " (x hk = (8) hr
~ k! r! k=O
for some 8 between x and x + h. Thus, for r large
Since (a + 8)(j < 1, the Taylor series converges unifonnly to f(x + h)
001. D
26 Theorem If f is expressed as a convergent power series f (x + h) = L ckhk with radius of convergence R > (j, then f has bounded derivative growth rate on I.
The proof of Theorem 26 uses two estimates about the growth rate of factorials. If you know Stirling's fonnula they are easy, but we prove them directly.
(23) t;r lim r -=e
r---+oo r!
Taking logarithms, applying the integral test, and ignoring tenns that tend to zero as r --+ 00 gives
1 1 -(logrr -logr!) = logr - -(logr + log(r - 1) + ... + log 1) r r
l1 r
1 I
r '" log r - - log x dx = log r - - (x log x - x)
r 1 r 1
1 = 1--,
r
which tends to 1 as r --+ 00. This proves (23).
Section 6* Analytic Functions
To prove (24) we write A = e-fL for J-L > 0, and reason similarly:
f (:)A k = f k(k - l)(k - 2;! ... (k - r + 1) e-kfL
According to (23) the rth root of this quantity tends to e1- fL / min(l, J-L) as r -+ 00, completing the proof of (24).
Proof of Theorem 26 By assumption the power series L qhk has radius of convergence R and a < R. Since 1/ R is the lim sup of ~ as k -+ 00,
there is a number A < 1 such that for all large k, I Ckak I ::: A k. Differentiating the series term by term with Ih I ::: a gives
00
k=r
for r large. Thus,
According to (24),
f¥iM r 1 a = lim sup -::: - lim sup r
r---+oo r! a r---+oo
and f has bounded derivative growth rate on I. D
From Theorems 25 and 26 we deduce the main result of this section.
27 Analyticity Theorem A smooth function is analytic if and only if it has locally bounded derivative growth rate.
238 Function Spaces Chapter 4
Proof Assume that I (a, b) ---+ lR is smooth and has locally bounded derivative growth rate. Then x E (a, b) has a neighborhood N on which the derivative growth rate a is finite. Choose a > 0 such that I = [x - a, x + a] C Nand aa < 1. We infer from Theorem 25 that the Taylor series for I at x converges uniformly to I on I. Hence I is analytic.
Conversely, assume that I is analytic and let x E (a, b) be given. There is a power series L Ckhk that converges to I(x + h) for all h in some interval (-R, R), R > O. Choose 0',0 < a < R. We infer from Theorem 26 that I has bounded derivative growth rate on I. 0
28 Corollary A smooth function is analytic ifits derivatives are uniformly bounded.
An example of such a function is I (x) = sin x.
Proof If I I(r) (e) I :::: M for all rand e then the derivative growth rate of I is bounded, in fact a = 0 and R = 00. 0
29 Taylor's Theorem If I (x) = L Ckxk and the power series has radius 01 convergence R, then I is analytic on ( - R, R).
Proof The function I is smooth, and by Theorem 26 it has bounded derivative growth rate on I C (-R, R). Hence it is analytic. 0
Taylor's Theorem states that not only can I be expanded as a convergent power series at x = 0, but also at any other point Xo E (-R, R). Other proofs of Taylor's theorem rely more heavily on series manipulations and Mertens' theorem.
The concept of analyticity extends immediately to complex functions. A function I : D ---+ C is complex analytic if D is an open subset of C and for each zED there is a power series
such that for all z + ~ near Z,
00
I(z + n = LCk~k. k=O
The coefficients Ck are complex and so is the variable ~ . Convergence occurs on a disc of radius R. This lets us define eZ
, log z, sin z, cos z for the complex
Section 6* Analytic Functions 239
number z by setting
00 (_l)k+! Zk
log(l + z) = L k when Izl < 1 k=!
00 (_1)kZ2k cos Z = '""' ----
. ~ (2k)!
It is enlightening and reassuring to derive formulas such as
eie = cose + i sine
directly from these definitions. (Just plug in z = iB and use the equations i 2 = -1, i 3 = -i, i4 = 1, etc.) A key formula to check is eZ+w = eZew .
One proof involves a manipulation of product series, a second merely uses analyticity. Another formula is 10g(eZ
) = z. There are many natural results about real analytic functions that can be
proved by direct power series means; e.g., the sum, product, reciprocal, composite, and inverse function of analytic functions are analytic. Direct proofs, like those for the Analyticity Theorem above, involve major series manipulations. The use of complex variables leads to greatly simplified proofs of these real variable theorems, thanks to the following fact.
Real analyticity propagates to complex analyticity and complex analyticity is equivalent to complex differentiability. t
For it is relatively easy to check that the composition, etc., of complex differentiable functions is complex differentiable.
The analyticity concept extends even beyond C. You may already have seen such an extension when you studied the vector linear ODE
x' = Ax
in calculus. A is a given m x m matrix and the unknown solution x = x(t) is a vector function of t, on which an initial condition x (0) = Xo is usually imposed. A vector ODE is equivalent to m coupled, scalar, linear ODE's. The solution x(t) can be expressed as
x(t) = etAxo
t A function f : D ~ IC is complex differentiable or holomorphic if D is an open subset of IC and for each zED, the limit of
/';f f(z + /';Z) - f(z) /';Z /';z
exists as /';z ~ 0 in IC. The limit, if it exists. is a complex number.
240 Function Spaces Chapter 4
where
1 1 00 t k
etA = lim (/ + tA + _(tA)2 + ... + -(tAt) = ~ _Ak. n--+oo 2! n! L k!
k=O
I is the m x m identity matrix. View this series as a power series with kth coefficient t k
/ k! and variable A. (A is a matrix variable!) The limit exists in the space of all m x m matrices, and its product with the constant vector Xo does indeed give a vector function of t that solves the original linear ODE.
The previous series defines the exponential of a matrix as eA = L A k / k!. You might ask yourself - is there such a thing as the logarithm of a matrix? A function that assigns to a matrix its matrix logarithm? A power series that expresses the matrix logarithm? What about other analytic functions? Is there such a thing as the sine of a matrix? What about inverting a matrix? Is there a power series that expresses matrix inversion? Are formulas such as log A 2 = 2 log A true? These questions are explored in nonlinear functional analysis.
A terminological point on which to insist is that the word "analytic" be defined as "locally power series expressible." In the complex case, some mathematicians define complex analyticity as complex differentiability, and although complex differentiability turns out to be equivalent to local expressibility as a complex power series, this is a very special feature of C. In fact it is responsible for every distinction between real and complex analysis. For cross-theory consistency, then, one should use the word "analytic" to mean local power series expressible, and use "differentiable" to mean differentiable. Why confound the two ideas?
7* Nowhere Differentiable Continuous Functions
Although many continuous functions, such as lxi, :yx, and x sin(l/x) fail to be differentiable at a few points, it is quite surprising that there can exist a function which is everywhere continuous but nowhere differentiable.
30 Theorem There exists a continuous function f : lR -+ lR that has a derivative at no point whatsoever.
Proof The construction is due to Weierstrass. The letters k, m, n denote integers. Start with a sawtooth function (To : lR -+ lR defined as
0'0 is periodic with period 2; if t = x + 2m then ao(t) compressed sawtooth function
ak(x) = (~r 0'0 (4kx)
241
ao(x). The
has period rrk = 2/4k. 1ft = x +mrrk thenak(t) = ak(x). See Figure 100.
0"2(X)
Figure 100 The graphs of the sawtooth function and two compressed sawtooth functions.
According to the M -test, the series L ak (x) converges uniformly to a limit f, and
CXl
f(x) = L ak(x) k=O
is continuous. We claim that f is nowhere differentiable. Fix an arbitrary point x, and set On = 1/2 . 4n. We will show that
242 Function Spaces Chapter 4
!:If f(x ± on) - f(x)
!:lx On
does not converge to a limit as On -+ 0, and thus that f' (x) does not exist. The quotient is
!:If = f (jk(X ± on) - (jk(X) .
!:lx k=O On
There are three types of term in the series, k > n, k = n, and k < n. If k > n then (jk(X ± on) - (jk(X) = 0- for On is an integer multiple of the period of (jk,
o = _1_ = 4k-(n+l) . ~ = 4k-(n+1) . rr . n 2. 4n 4k k
Thus the infinite series expression for !:lfl !:lx reduces to a sum of n + 1 terms
The function (jn is monotone on either [x - On, x] or [x, x + on], since it is monotone on intervals of length 4 -n and the contiguous interval [x - On, x, X + On] at x is of length 4-n. The slope of (jn is ±3n. Thus, either
or
The terms with k < n are crudely estimated from the slope of (jk being ±3k :
Thus
I !:If I n n-l n 3
n - 1 1 n
!:lx ~ 3 - (3 + ... + 1) = 3 - 3-=1 = 2(3 + 1),
which tends to 00 as On -+ 0, so f' (x) does not exist. D
By simply writing down a sawtooth series as above, Weierstrass showed that there exists a nowhere differentiable continuous function. Yet more amazing is the fact that most continuous functions (in a reasonable sense
defined below) are nowhere differentiable. If you could pick a continuous function at random, it would be nowhere differentiable.
Recall that the set D C M is dense in M if D meets every non-empty open subset W of M, D n W =1= 0. The intersection oftwo dense sets need not be dense; it can be empty, as is the case with Q and QC in R On the other hand if U, V are open dense sets in M then U n V is open dense in M. For if W is any non-empty open subset of M then U n W is a non-empty open subset of M, and by denseness of V, V meets un W; i.e., un V n W is non-empty and U n V meets W.
Moral Open dense sets do a good job of being dense.
The countable intersection G = n G n of open dense sets is called a thick (or residualt ) subset of M, due to the following result, which we will apply in the complete metric space CO([a, b], JR). Extending our vocabulary in a natural way we say that the complement of a thick set is thin (or meager). A subset H of M is thin if and only if it is a countable union of nowhere dense closed sets, H = UHn. Clearly, thickness and thinness are topological properties. A thin set is the topological analog of a zero set (a set whose outer measure is zero).
31 Baire's Theorem Every thick subset of a complete metric space M is dense in M. A non-empty, complete metric space is not thin: if M is the union of countably many closed sets, at least one has non-empty interior.
If all points in a thick subset of M satisfy some condition then the condition is said to be generic, we also say that most points of M obey the condition. As a consequence ofBaire's theorem and a modification of Weierstrass' construction we will prove
32 Theorem The generic f E CO = CO([a, b], JR) is differentiable at no point of [a, b], nor is it monotone on any subinterval of [a, b].
Using Lebesgue's monotone differentiation theorem (monotonicity implies differentiability almost everywhere), the second assertion follows from the first, but below we give a direct proof.
Before getting into the proofs of Baire's theorem and Theorem 32, we further discuss thickness, thinness, and genericity. The empty set is always thin and the full space M is always thick in itself. A single open dense subset is thick and a single closed nowhere dense subset is thin. JR \ Z is a thick subset of JR and the Cantor set is a thin subset of R Likewise JR is a
t "Residual" is an unfortunate choice of words. It connotes smallness. when it should connote just the opposite.
244 Function Spaces Chapter 4
thin subset of]R2. The generic point of]R does not lie in the Cantor set. The generic point of]R2 does not lie on the x-axis. Although ]R \ Z is a thick subset of]R it is not a thick subset of ]R2. The set Q is a thin subset of R It is the countable union of its points, each of which is a closed nowhere dense set. QC is a thick subset of R The generic real number is irrational. In the same vein:
(a) The generic square matrix has determinant =F O. (b) The generic linear transformation]Rm -+ ]Rm is an isomorphism. (c) The generic linear transformation]Rm -+ ]Rm-k is onto. (d) The generic linear transformation]Rm -+ ]Rm+k is one-to-one. (e) The generic pair of lines in ]R3 are skew (nonparallel and disjoint). (f) The generic plane in ]R3 meets the three coordinate axes in three
distinct points. (g) The generic nth degree polynomial has n distinct roots. In an incomplete metric space such as Q, thickness and thinness have no
bite: every subset of Q, even the empty set, is thick in Q.
Proof of Baire's Theorem If M = 0, the proof is trivial, so we assume M =F 0. Let G = nGn be a thick subset of M; each Gn being open dense in M. Let Po E M and E > 0 be given. Choose a sequence of points Pn E M and radii rn > 0 such that rn < lin and
M2rl (PI) c ME(po)
M 2r2 (P2) c M'J (PI) n G I
M 2rn (Pn+l) C Mrn(Pn) n G I ··· n Gn.
See Figure 10 1. Then
The diameters of these sets tend to 0 as n -+ 00. Thus (Pn) is a Cauchy sequence and it converges to some P EM, by completeness. The point P belongs to each set Mrn(Pn) and therefore it belongs to each Gn. Thus pEG n ME (Po) and G is dense in M.
To check that M is not thin, we take complements. Suppose that M = UKn and Kn is closed. If each Kn has empty interior, then each Gn = K~ is open-dense, and
Figure 101 The closed neighborhoods Mrn (Pn) nest down to a point.
33 Corollary No subset of a complete non-empty metric space is both thick and thin.
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 is dense in M, so M is empty. D
To prove Theorem 32 we use two lemmas.
34 Lemma The set P L of piecewise linear functions is dense in Co.
Proof If ¢ : [a, b] -+ ]R is continuous and its graph consists of finitely many line segments in ]R2 then ¢ is piecewise linear. Let f E CO and E > 0 be given. Since [a, b] is compact, f is uniformly continuous, and there is 8 > 0 such that It - sl < 8 implies If(t) - f(s)1 < E. Choose n> (b-a)/8andpartition [a,b] intonequalsubintervals Ii = [Xi-I,X;], each of length < 8. Let ¢ : [a, b] -+ ]R be the piecewise linear function whose graph consists of the segments joining the points (Xi -1, f (Xi -I)) and (Xi, f (Xi)) on the graph of f. See Figure 102.
The value of ¢(t) for t E Ii lies between f(Xi-l) and f(Xi). Both these numbers differfrom f(t) by less than E. Hence, for all t E [a, b],
246 Function Spaces Chapter 4
Figure 102 The piecewise linear function ¢ approximates the continuous function f.
If(t) - ¢(t)1 < E.
In other words, d (f, ¢) < E and P L is dense in Co. o
3S Lemma If ¢ E P Land E > 0 are given then there exists a sawtooth function a such that lIa II .:::: E, a has period.:::: E, and
. 1 rmn{lslopeal} > max{lslope¢l} +-.
E
Proof Let () = max{lslope¢l} and choose c large. The compressed sawtooth a(x) = Eao(cx) has lIa II = E, period 7r = llc, and slope s = ±EC.
When c is large, 7r .:::: E, and lsi> () + liE. 0
Proof of Theorem 32 For n E N define
Rn = {f E CO : Y x E [a, b - ~] 3h > 0 such that I D./ I > n}
Ln = {f E CO : Yx E [a + ~,b] 3h < 0 such that I D./ I > n}
1 Gn = {f E CO : f restricted to any interval oflength - is non-monotone},
n
where D.f = f(x + h) - f(x). We claim that each of these sets is open dense in Co.
For denseness it is enough to prove that the closure of each contains P L, since by Lemma 34 the closure of P L is Co. Let ¢ E P Land E > 0 be given. According to Lemma 35, there is a sawtooth function a such that IIail ::::: E, a has period < lin, and
min{lslopeal} > max{lslope¢l} + n.
Consider the piecewise linear function f = ¢ + a. Its slopes are dominated by those of a, and so they alternate in sign with period < 1 12n. At any x E [a, b - lin] there is a rightward slope either> nor < -no Thus fERn. Similarly f E Ln. Any interval I of length lin 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 G n. Since d(j, ¢) = E is arbitrarily small this shows that Rn, L n, and G n are dense in Co.
Next suppose that fERn is given. For each x E [a, b - lin] there is an h = h(x) > 0 such that
Since f is continuous, there is a neighborhood Tx of x in [a, b] and a constant v = v > 0 such that this same h yields
If(t+h)-f(t)1
h >n+v
for all t E Tx. Since [a, b - lin] is compact, finitely many of these neighborhoods Tx cover it, say TXl ' ... , TXm. Continuity of f implies that for all t E T Xi'
(25) I f(t + hi) - f(t) I
hi ::: n + vi,
where hi = h(xd and Vi = V(Xi). These m inequalities for points t in the m sets TXi remain nearly valid if f is replaced by a function g with d (j, g)
small enough; (25) becomes
(26) I g(t + h~~ - g(t) I > n
which means that g E Rn and Rn is open in Co. Similarly Ln is open in Co.
248 Function Spaces Chapter 4
Checking that G n is open is easier. If (fk) is a sequence of functions in G~ and fk ~ f then we must show that f E G~. Each fk is monotone on some interval h oflength lin. There is a subsequence of these intervals that converges to a limit interval I. Its length is 1 I n and by uniform convergence, f is monotone on I. Hence G~ is closed and Gn is open. Each set Rn , Ln ,
G n is open dense in Co. Finally, if f belongs to the thick set
then for each x E [a, b] there is a sequence h n 1= 0 such that
I f(x +hn ) - f(x) I
> n. hn
The numerator of this fraction is atmost2l1fll, so hn -+ 0 as n -+ 00. 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 lin when n is large enough, f is non-monotone on J. D
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, Differentiation 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 JR. with a complete metric space Y. Let F denote the set of all functions f : X -+ Y. Recall from Exercise 2.84 that the metric d y on Y gives rise to a bounded metric
, dy(y, y') p(y,y) = 1 +dy(y,y')'
where y, y' E Y. Note that p < 1. Convergence and Cauchyness with respect to p and d y are equivalent. Thus completeness of Y with respect to dy implies completeness with respect to p. In the same way we give F the metric
d(f ) dy(f(x), g(x))
, g = sup XEX 1 + dy(f(x), g(x))
Section 8* Spaces of Unbounded Functions 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) < 00; i.e., d(f, c) < 1. Unbounded functions have d (f, c) = 1.
36 Theorem In the space F equipped with the metric d, (a) Uniform convergence of (fn) is equivalent to d-convergence. (b) Completeness ofY implies completeness of F. (c) The set Fb of bounded functions is closed in F. ( d) The set CO (X, Y) of continuous functions is closed in F.
Proof (a) f = uniflim fn means that dy (fn (x), f (x)) =4 0, which means n--+oo
that d (fn, f) -+ O. (b) If (fn) is Cauchy in F and Y is complete then, just as in Section 1,
f (x) = lim fn (x) exists for each x EX. Cauchyness with respect to the n--+oo
metric d implies uniform convergence and thus d(fn, f) -+ O. (c) If fn E Fb andd(fn, f) -+ 0 then sUPx dy (fn (x), f(x)) -+ O. Since
fn is bounded, so is f. (d) The proof that CO is closed in F is the same as in Section 1. D
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(f(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 I n we have points xn, tn E X and functions fn E [; with dx (xn , tn) < lin and dy(fn(xn), fn(tn)) :::: E. By compactness of X we may assume that Xn -+ XO. Then tn -+ XO, which leads to a contradiction of pointwise equicontinuity at XO. D
38 Theorem If the sequence of functions fn : X -+ Y is uniformly equicontinuous, X is compact, andfor each x E X, (fn(x)) lies in a compact subset of Y, then (fn) has a uniformly convergent subsequence.
Proof Being compact, X has a countable dense subset D. Then the proof ofthe ArzelaAscoli Theorem in Section 3 becomes a proof of Theorem 38.
D
250 Function Spaces Chapter 4
The space X is 0' -compact if it is a countable union of compact sets, X = UXi . For example Z, Q, ffi. and ffi.m are a -compact, while any uncountable set equipped with the discrete metric is not a -compact.
39 Theorem If X is a -compact and if Un) is a sequence of pointwise equicontinuous functions such that for each x E X, Un (x)) lies in a compact subset ofY, then (fn) has a subsequence that converges uniformly to a limit on each compact subset of X.
Proof Express X as UXi with Xi compact. By Theorem 37 Un Ix) is uniformly equicontinuous and by Theorem 38 there is a subsequence An that converges uniformly on Xl, and it has a sub-subsequence /2,n that converges uniformly on X2, and so on. A diagonal subsequence (gm) converges uniformly on each Xi. Thus (gm) converges pointwise. If A C X is compact, then (gm IA) is uniformly equicontinuous and pointwise convergent. By the proof of the Arzela Ascoli propagation theorem, (gm IA) converges uniformly. 0
40 Corollary If Un) is a sequence of pointwise equicontinuous functions ffi. --+ ffi., and for some Xo E ffi., Un (xo)) is bounded then Un) has a subsequence that converges uniformly on every compact subset offfi..
Proof Let [a, b] be any interval containing Xo. By Theorem 37, the restrictions of fn to [a, b] are uniformly equicontinuous, and there is a 8 > 0 such that if t, s E [a, b] then It - sl < 8 implies that Ifn(t) - fn(s)1 < 1. Each point x E [a, b] can be reached in::: N steps oflength < 8, starting at xo, if N > (b - a)j8. Thus Ifn(x)1 ::: Ifn(xo)1 + N, and Un (x)) is bounded for each x ERA bounded subset of ffi. has compact closure and Theorem 39 gives the corollary. 0
Exercises 251
Exercises
In these exercises, CO = CO([a, b], ffi.) is the space of continuous real valued functions defined on the closed interval [a, b]. It is equipped with the sup norm, IIfll = sup{lf(x)1 : x E [a, b]}.
1. Let M, N be metric spaces. (a) Formulate the concepts of pointwise convergence and uniform
convergence for sequences of functions fn : M -+ N. (b) For which metric spaces are the concepts equivalent?
2. Suppose that fn ~ f where f and fn are functions from the metric space M to the metric space N. (Assume nothing about the metric spaces such as compactness, completeness, etc.) If each fn is continuous prove that f is continuous. [Hint: Review the proof of Theorem 1.]
3. Let fn : [a, b] -+ ffi. be a sequence of piecewise continuous functions, each of which is continuous at the point Xo E [a, b]. Assume that fn ~ f as n -+ 00.
(a) Prove that f is continuous at Xo. [Hint Review the proof of Theorem 1.]
(b) Prove or disprove that f is piecewise continuous. 4. (a) If fn : ffi. -+ ffi. is uniformly continuous for each n E Nand
if fn ~ f as n -+ 00, prove or disprove that f is uniformly continuous.
(b) What happens for functions from one metric space to another instead of ffi. to ffi.?
5. Suppose that fn : [a, b] -+ ffi. and fn ~ f as n -+ 00. Which of the following discontinuity properties (see Exercise 37 in Chapter 3) of the functions fn carries over to the limit function? (Prove or give a counter-example. )
(a) No discontinuities. (b) At most ten discontinuities. (c) At least ten discontinuities. (d) Finitely many discontinuities. (e) Countably many discontinuities, all of jump type. (f) No jump discontinuities. (g) No oscillating discontinuities.
**6. (a) Prove that CO and ffi. have equal cardinality. [Clearly there are at least as many functions as there are real numbers, for CO includes the constant functions. The issue is to show that there are no more continuous functions than there are real numbers.]
252 Function Spaces Chapter 4
(b) Is the same true if we replace [a, b] with ~ or a separable metric space?
(c) In the same vein, prove that the collection T of open subsets of ~ and lR itself have equal cardinality.
7. Consider a sequence of functions fn in Co. The graph G n of fn is a compact subset of lR2 .
(a) Prove that Un) converges uniformly as n -+ 00 if and only if the sequence (G n) in K (lR2) converges to the graph of a function f E Co. (The space K was discussed in Exercise 2.124.)
(b) Formulate equicontinuity in terms of graphs. 8. Is the sequence of functions fn : lR -+ lR defined by
equicontinuous? Prove or disprove. 9. If f : lR -+ lR is continuous and the sequence fn(x) = f(nx) is
equicontinuous, what can be said about f? 10. Give an example to show that a sequence of functions may be uni
formly continuous, pointwise equicontinuous, but not uniformly equicontinuous, when their domain M is noncompact.
11. If every sequence of pointwise equicontinuous functions M -+ lR is uniformly equicontinuous, does this imply that M is compact?
12. Prove that if [; C CO(M, N) is equicontinuous then so is its closure. 13. Suppose that Un) is a sequence of functions lR -+ lR and for each
compact subset K C ~, the restricted sequence Un I K) is pointwise bounded and pointwise equicontinuous.
(a) Does it follow that there is a subsequence of Un) that converges pointwise to a continuous limit function lR -+ lR?
(b) What about uniform convergence? 14. Recall from Exercise 78 in Chapter 2 that a metric space M is chain
connected if for each E > 0 and each p, q E M there is a chain P = Po, ... , Pn = q in M such that
d(Pk-l, pd < E for 1::: k ::: n.
A family F of functions f : M -+ lR is bounded at P E M if the set {f(p) : f E F} is a bounded in R Show that M is chain connected if and only if pointwise boundedness of an equicontinuous family at one point of M implies pointwise boundedness at every point of M.
Exercises 253
15. A continuous, strictly increasing function fl, : (0, (0) -+ (0, (0)
is a modulus of continuity if fl,(s) -+ 0 as s -+ O. A function I : [a, b] -+ JR has modulus of continuity fl, if I/(s) - l(t)1 :::::
fl,(ls - tl). (a) Prove that a function is uniformly continuous if and only if it
has a modulus of continuity. (b) Prove that a family of functions is equicontinuous if and only
if its members have a common modulus of continuity. 16. Consider the modulus of continuity fl,(s) = Ls where L is a positive
constant. (a) What is the relation between Cf1 and the set of Lipschitz func-
tions with Lipschitz constant::::: L? (b) Replace [a, b] with JR and answer the same question. (c) Replace [a, b] with N and answer the same question. (d) Formulate and prove a generalization of (a).
17. Consider a modulus of continuity fl,(s) = Hsa where 0 < a ::::: 1 and 0 < H < 00. A function with this modulus of continuity is said to be a-HOlder, with a-HOlder constant H. See also Exercise 2 in Chapter 3.
(a) Prove that the set ca(H) of all continuous functions defined on [a, b] which are a-Holder and have a-Holder constant::::: His
equicontinuous. (b) Replace [a, b] with (a, b). Is the same thing true? (c) Replace [a, b] with JR. Is it true? (d) What about Q? (e) What about N?
18. Suppose that Un) is an equicontinuous sequence in CO and p E [a, b] is given.
(a) If In (p) is a bounded sequence of real numbers, prove that Un) is uniformly bounded.
(b) Reformulate the Arzela-Ascoli Theorem with a weaker boundedness hypothesis.
(c) Can [a, b] be replaced with (a, b)?, Ql?, JR?, N ? (d) What is the correct generalization?
19. If M is compact and A is dense in M, prove that for any 8 > 0 there is a finite subset {aI, ... , ad C A which is cS-dense in M in the sense that each x E M lies within distance 8 of at least one of the points aj,j = 1, ... ,k.
254 Function Spaces Chapter 4
20. Give an example of a sequence of smooth equicontinuous functions In : [a, b] -+ ffi. whose derivatives are unbounded.
*21. Given constants a, f3 > 0 define
for x > O. (a) For which pairs a, f3 is Ia,{3 uniformly continuous? (b) For which sets of (a, f3) in (0, (0)2 is the family equicontinu
ous? 22. Suppose that E C CO is equicontinuous and bounded.
(a) Prove that sup{f(x) : lEE} is a continuous function of x. (b) Show that (a) fails without equicontinuity. (c) Show that this continuous sup-property does not imply equicon
tinuity. (d) Assume that the continuous sup-property is true for each subset
FeE. Is E equicontinuous? Give a proof or counter-example. 23. Let M be a compact metric space, and let (in) be a sequence of
isometries in : M -+ M. (a) Prove that there exists a subsequence ink that converges to an
isometry i as k -+ 00.
(b) Does the inverse isometry i~ 1 converge to i-I? (Proof or counterexample.)
(c) Infer that the group of orthogonal 3 x 3 matrices is compact. [Hint: Each orthogonal 3 x 3 matrix defines an isometry of the unit 2-sphere to itself.]
(d) How about the group of m x m orthogonal matrices? 24. Suppose that I : M -+ M is a contraction, but M is not necessarily
complete. (a) Prove that I is uniformly continuous. (b) Why ~oe~ (a) i~ly that L extends uniquely to a continuous
ma£. I : M -+ M, where M is the completion of M? (c) Is f a contraction?
25. Give an example of a contraction of an incomplete metric space that has no fixed point.
26. Suppose that f : M -+ M and for all x, y EM, if x =1= y then d (f (x), f (y)) < d (x, y). Such an f is a weak contraction.
(a) Is a weak contraction a contraction? (Proof or counter-example.) (b) If M is compact is a weak contraction a contraction? (Proof or
counter-example.)
Exercises 255
(c) If M is compact, prove that a weak contraction has a unique fixed point.
27. Suppose that f : JR ---+ JR is differentiable and its derivative satisfies 1f'(x)1 < 1 for all x E JR.
(a) Is f a contraction? (b) A weak one? (c) Does it have a fixed point?
28. Give an example to show that the fixed point in Brouwer's Theorem need not be unique.
29. On page 222 it is shown that if bn J~1 (1 - t2)n dt = I then for some constant c, and for all n E N, bn ::: c..fii. What is the best (i.e., smallest) value of c that you can prove works? (A calculator might be useful here.)
30. Let M be a compact metric space, and let CLip be the set of continuous functions f : M ---+ JR that obey a Lipschitz condition: for some L and all p, q, E M,
If(p) - f(q)1 ::: Ld(p, q).
*(a) Prove that CLip is dense in CaCM, JR). [Hint: Stone-Weierstrass.] ***(b) If M = [a, b] and JR is replaced by some other complete metric
space, is the result true or false? ***(c) If Mis a general compact metric space and Y is a complete metric
space, is CLip(M, y) dense in CaCM, Y)? (Would M equal to the Cantor set make a good test case?)
31. Consider the ODE x' = x on JR. Show that its solution with initial condition Xo is t t-+ et Xo. Interpret et+s = et eS in terms of the flow property.
32. Consider the ODE y' = 2M where y E R (a) Show that there are many solutions to this ODE, all with the
same initial condition yeO) = O. Not only does y(t) = 0 solve the ODE, but also y(t) = t 2 does for t ::: O.
(b) Find and graph other solutions such as yet) = 0 for t ::: c and yet) = (t - c)2 for t ::: c > O.
(c) Does the existence of these non-unique solutions to the ODE contradict Picard's Theorem? Explain.
*(d) Find all solutions with initial condition yeO) = O. 33. Consider the ODE x' = x 2 on JR. 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?
256 Function Spaces Chapter 4
34. Suppose that the ODE x' = f (x) on JR is bounded, I f (x) I .:::: 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 I f (x) I .:::: C Ixl + K for all x.
(c) Repeat (a) and (b) with JRm in place ofR (d) Prove that if f : JRm -+ JRm is uniformly continuous then the
condition stated in (b) is true. Deduce that solutions of uniformly continuous ODE's defined on JRm do not escape to infinity in finite time.
**35. (a) Prove Borel's Lemma: given any sequence whatsoever of real numbers (ar ) there is a smooth function f : JR -+ JR such that f(r) (0) = ar . [Hint: Try f = L fh (x )akxk / k! where f3k is a well chosen bump function.]
(b) Infer that there are many Taylor series with radius of convergence R = O.
(c) Construct a smooth function whose Taylor series at every x has radius of convergence R = O. [Hint: Try L f3k(x)e(x + qk) where {ql, Q2, ... } = Q.]
*36. Suppose that T C (a, b) clusters at some point of (a, b) and that f, g : (a, b) -+ JR are analytic. Assume that for all t E T, f(t) = get).
(a) Prove that f = g everywhere in (a, b). (b) What if f and g are only COO? (c) What if T is an infinite set but its only cluster points are a and
b? **(d) FindanecessaryandsufficientconditionforasubsetZ C (a, b)
to be the zero locus of an analytic function f defined on (a, b), Z = {x E (a, b) : f(x) = OJ. [Hint: 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 fq(x) = d(q, x) - d(p, x).
(a) Prove that fq is a bounded, continuous function of x E M, and that the map q f-+ fq sends M isometrically onto a subset Mo of CO(M, JR).
Exercises 257
(b) Since CO(M, JR.) is complete, infer that an isometric copy of M is dense in a complete metric space, the closure of M 0, 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 (T-compact if it is the countable union of compact subsets, M = UMi .
(a) Why is it equivalent to require that M is the monotone union of compact subsets,
M=~Mi'
i.e., MI C M2 C ... ? (b) Prove that a a -compact metric space is separable. (c) Prove that Z, Q, R JR.m are a-compact.
*(d) Prove that CO is not a-compact. [Hint: Think Baire.]
*(e) IfM = l!J int(Mi)andeachM;iscompact,Mis(T*-compact. Prove that M is a * -compact if and only if it is separable and locally compact. Infer that Z, Rand JR.m are a* -compact, but Q is not.
(f) Assume that M is a* -compact, M = l!J int(M;) with each M; compact. Prove that this monotone union engulfs all compacts in M, in the sense that if A C M is compact, then for some i, ACM;.
(g) If M = l!J M; and each M; is compact show by example that this engulfing property may fail, even when M itself is compact.
**(h) Prove or disprove that a complete a-compact metric space is a * -compact.
39. (a) Give an example of a function f : [0, 1] x [0, 1] -+ JR. such that for each fixed x, y f-+ f(x, y) is a continuous function of y, and for each fixed y, x f-+ f (x, y) is a continuous function of x, but f is not continuous.
(b) Suppose in addition that the set of functions
£ = {x f-+ f(x, y) : y E [0, I]}
is equicontinuous. Prove that f is continuous. 40. Prove that JR. can not be expressed as the countable union of Cantor
sets. 41. What is the joke in the following picture?
258 Function Spaces Chapter 4
e
y' = 1l(Y)
More Pre-lim Problems
1. Let I and In,n EN, be functions fromlR to R Assume that In (Xn) ---+ I (x) as n ---+ 00 and Xn ---+ x. Show that I is continuous. [Hint: Equicontinuity is irrelevant since the functions In are not assumed to be continuous.]
2. Suppose that In E CO and for each x E [a, b],
h (x) ::: h (x) ::: ... ,
and lim In (x) = O. Is the sequence equicontinuous? Give a proof n-+oo
or counter-example. [Hint: Does In (x) converge uniformly to 0, or does it not?]
3. Let E be the set of all functions u : [0, 1] ---+ lR such that u(O) = 0 and u satisfies a Lipschitz condition with Lipschitz constant 1. Define ¢ : E ---+ lR according to the formula
¢(u) = 11 (u(x)2 - u(x)) dx.
Prove that there exists a function u E E at which ¢(u) attains an absolute maximum.
4. Let (gn) be a sequence of twice differentiable functions defined on [0, 1], and assume that for all n, gn (0) = g~ (0). Suppose also that for all n E N and all x E [0, 1], Ig~(x)1 ~ 1. Prove that there is a subsequence of (gn) converging uniformly on [0, 1].
Exercises 259
5. Let (Pn ) be a sequence of real polynomials of degree::::: 10. Suppose thatPn(x)convergespointwisetoOasn -+ 00 and x E [0, 1]. Prove that Pn (x) converges uniformly to 0.
6. Let (an) be a sequence of nonzero real numbers. Prove that the sequence of functions
has a subsequence converging to a continuous function. 7. Suppose that f : lR -+ lR is differentiable, f(O) = 0, and j'(x) >
f(x) for all x E R Prove that f(x) > ° for all x > 0. 8. Suppose that f : [a, b] -+ lR 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) -+ lR be a uniformly continuous function where [0, 1) is the half open interval. Prove that there is a unique continuous map g : [0, 1] -+ lR such that g(x) = hex) for all x E [0, 1).
10. Assume that f : lR -+ lR is uniformly continuous. Prove that there are constants A, B such that If(x)1 ::::: A + B Ixl for all x E R
11. Suppose that f (x) is defined on [-1, 1] and that its third derivative exists and is continuous. (That is, f is of class C 3 .) Prove that the senes
00
L (n(f(lln) - f(-Iln)) - 2f'(0)) n=O
converges. 12. Let A c lRm 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. 13. Let f : [0, 1] -+ lR be continuously differentiable, with f(O) = 0.
Prove that
II f112 ::::: 101
(f' (x))2 dx
where Ilfll = sup{lf(t)1 : ° ::::: t ::::: I}. 14. Let fn : lR -+ lR be differentiable functions, n 1, 2, . .. with
fn(O) = ° and If~(x)1 ::::: 2 for all n, x. Suppose that
lim fn(x) = g(x) n-+oo
for all x. Prove that g is continuous.
260 Function Spaces 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.
16. Let k ~ ° be an integer and define a sequence of maps In : lR -+ lR as
Xk
In (x) = --z-+ x n n = 1, 2, . .. . For which values of k does the sequence converge uniformly on lR? On every bounded subset of lR?
17. Let I : [0, 1] -+ lR be Riemann integrable over [b, 1] for every b
such that ° < b ::::; 1. (a) If I is bounded, prove that I is Riemann integrable over [0, 1]. (b) What if I is not bounded?
18. (a) Let S and T be connected subsets of the plane lR2 having a point in common. Prove that S U T is connected.
(b) Let {Sa} be a family of connected subsets of lR2 all containing the origin. Prove that USa is connected.
19. Let I : lR -+ lR be continuous. Suppose that lR contains a countably infinite set S such that
iq
I(x)dx = ° if p and q are not in S. Prove that I is identically zero.
20. Let I : lR -+ lR satisfy I (x) ::::; I (y) for x ::::; y. Prove that the set where I is not continuous is finite or countably infinite.
21. Let (gn) be a sequence of Riemann integrable functions from [0, 1] into lR such that Ign(x)1 ::::; 1 for all n, x. Define
Gn(x) = foX gn(t) dt.
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 I : [0, 1] -+ lR and for any
E > ° there is a function of the form
n
g(x) = LCkxk
k=O
for some n EN, and Ig(x) - l(x)1 < E for all x in [0,1]. 24. Give an example of a function I : lR -+ lR having all three of the
following properties:
Exercises
(a) f(x)=Oforallx<Oandx>2. (b) f'(1) = 1, (c) f has derivatives of all orders.
261
25. (a) Give an example of a differentiable function f : IR ---+ IR whose derivative is not continuous.
(b) Let f be as in (a). If 1'(0) < 2 < 1'0) prove that f'ex) = 2 for some x E [0, 1].
26. Let U c IRm be an open set. Suppose that the map h : U ---+ IRm is a homeomorphism from U onto IRm which is uniformly continuous. Prove that U = IRm.
27. Let Un) be a sequence of continuous maps [0, 1] ---+ IR such that
11 Un(y»2 dy ~ 5
for all n. Define gn : [0, 1] ---+ IR by
gn(x) = 11 Jx + y fn(Y) dy
(a) Find a constant K ::: ° such that Ign (x) I ~ K for all n. (b) Prove that a subsequence of the sequence (gn) converges uni
formly. 28. Consider the following properties of a map f : IRm ---+ R
(a) f is continuous. (b) The graph of f is connected in IRm x R
Prove or disprove the implications (a) =>- (b), (b) =>- (a). 29. Let (Pn ) be a sequence of real polynomials of degree ~ 10. Suppose
that
lim Pn(x) = ° n--->oo
for all x E [0, 1]. Prove that Pn (x) ~ 0, ° ~ x ~ 1. What can you say about Pn(x) for 4 ~ x ~ 5?
30. Give an example of a subset ofIR having uncountably many connected components. Can such a subset be open? Closed? Does your answer change if IR2 replaces IR?
31. For each (a, b, c) E IR3 consider the series
Determine the values of a, b, and c for which the series converges absolutely, converges conditionally, diverges.
262 Function Spaces Chapter 4
32. Let X be a compact metric space and f : X -+ X an isometry. (That is, d(f(x), f(y» = d(x, y) for all x, y EX.) Prove that f(X) = x.
33. Prove or disprove: Q is the countable intersection of open subsets ofR
34. Let f : lR -+ lR be continuous and i: If(x)1 dx < 00.
Show that there is a sequence (xn ) in lR such that Xn -+ 00,
xnf(xn) -+ 0, and xnf( -xn) -+ 0 as n -+ 00.
35. Let f : [0, 1] -+ lR be a continuous function. Evaluate the following limits (with proof)
(a) lim t xn f(x) dx n~CXJ 10 (b) lim n t xn f (x) dx.
n~oo 10 36. Let K be an uncountable subset oflRm. Prove that there is a sequence
of distinct points in K which converges to some point of K. 37. Let (gn) be a sequence of twice differentiable functions on [0, 1] such
that for all n, gn (0) = 0 and g~ (0) = O. Suppose that Ig~ (x) I ::; 1 for all n, x. Prove that there is a subsequence of (gn) which converges uniformly on [0, 1].
38. Prove or give a counter-example: Every connected locally pathwise connected set in lRm is path wise connected.
39. Let (fn) be a sequence of continuous functions [0, 1] -+ lR such that fn (x) -+ 0 for each x E [0, 1]. Suppose that
for all n where K is a constant. Does f01 fn (x) dx converge to 0 as n -+ oo? Prove or give a counter-example.
40. Let E be a closed, bounded, and non-empty subset of lRm and let f: E -+ Ebeafunctionsatisfying If(x) - f(y)1 < Ix - yl for all x, y E E, x =1= y. Prove that there is one and only one point Xo E E such that f(xo) = Xo.
41. Let f : [0, 2:rr] -+ lR be a continuous function such that
127r f(x) sin(nx) dx = 0
for all integers n ::: 1. Prove that f is identically zero.
Exercises 263
42. Let E be the set of all real valued functions u : [0, 1] ---+ ~ satisfying u(O) = 0 and \u(x) - u(y)\ .:s \x - y\ for all x, y E [0,1]. Prove that the function
¢(u) = 11 «u(x)2 - u(x» dx
achieves its maximum value at some element of E. 43. Let!l, 12, ... be continuous real valued functions on [0, 1] such that
for each x E [0, 1], /J (x) ::: hex) ::: .... Assume that for each x, fn (x) converges to 0 as n ---+ 00. Does fn converge uniformly to O? Give a proof or counter-example.
44. Let f : [0, (0) ---+ [0,00) be a monotonically decreasing function with 100
f(x) dx < 00.
Prove that lim xf(x) = O. x---+oo
45. Suppose that F : ~m ---+ ~m is continuous and satisfies
\F(x) - F(y)\ ::: A \x - y\
for all x, y E ~m and some constant A > O. Prove that F is one-toone, 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 : ~ ---+ ~ which sends open sets to open sets must be monotonic.
48. Let f : [0,00) ---+ ~ be uniformly continuous and assume that
lim rb
f(x) dx b---+oo Jo
exists (as a finite limit). Prove that lim f(x) = o. x---+oo
49. Prove or supply a counter-example: If f and g are continuously dif-ferentiable functions defined on the interval 0 < x < 1 which satisfy the conditions
lim f(x) = 0 = lim g(x) x---+o x---+o
and lim f(x) = c x---+o g(x)
and if g and g' never vanish, then lim f' (x) = c. (This is a converse x---+o g'(x)
of L'Hospital's rule.)
264 Function Spaces Chapter 4
50. Prove or provide a counter-example: If the function f from JR to JR has both a left and a right limit at each point of JR, then the set of discontinuities is at most countable.
51. Prove or supply a counter-example: If f is a non-decreasing real valued function on [0, 1] then there is a sequence fn, n = 1,2, ... of continuous functions on [0, 1] such that for each x in [0, 1], lim fn(x) = f(x).
n-+oo
52. Show that if f is a homeomorphism of [0, 1] onto itself then there is a sequence of polynomials Pn (x), n = 1, 2, ... , such that Pn --+ f uniformly on [0, 1] and each Pn is a homeomorphism of [0, 1] onto itself. [Hint: First assume that f is C 1
.]
53. Let f be a C2 function on the real line. Assume that f is bounded with bounded second derivative. Let A = sUPx !f(x)! and B = sUPx I ff! (x) I. Prove that
sup If'(x)1 ~ 2J AB. x
54. Let f be continuous on JR and let
1 n-l ( k) fn (x) = - L f x + - .
n k=O n
Prove that fn (x) converges uniformly to a limit on every finite interval [a,b].
55. Let f be a real valued continuous function on the compact interval [a, b]. Given E > 0, show that there is a polynomial p such that
for all x E [a, b].
pea) = f(a),
p'ea) = 0, and
!p(x) - f(x)! < E,
56. A function f : [0, 1] --+ JR is said to be upper semicontinuous if, givenx E [0, 1] and E > 0, there exists a 8 > ° such that Iy - xl < 8 implies that f (y) < f (x) + E. Prove that an upper semicontinuous function on [0, 1] is bounded above and attains its maximum value at some point p E [0, 1].
57. Let f and fn be functions from JR to R Assume that fn (xn) --+ f (x) as n --+ 00 whenever Xn --+ x. Prove that f is continuous. (Note: the functions fn are not assumed to be continuous.)
Exercises 265
58. Let f(x), 0 .::: x .::: 1, be a continuous real function with continuous derivative f' (x). Let M be the supremum of I f' (x) I, 0 .::: x < 1. Prove: for n = 1,2, ...
1 n-l (k) 11 M -2::f - - f(x)dx':::
2n n k=O n 0
59. Let K be a compact subset of jRm and let (B j) be a sequence of open balls which cover K. Prove that there is an E > 0 such that each E -ball centered at a point of K is contained in at least one of the balls Bj .
60. Let f be a continuous real-valued function on [0, 00) such that
lim (f(X) + r f(t) dt) x---+oo 10
exists (and is finite). Prove that limx->oo f(x) = o. 61. A standard theorem asserts that a continuous real-valued function
on a compact set is bounded. Prove the converse: if K is a subset of jRm and if every continuous real-valued function defined on K is bounded, then K is compact.
62. Let F be a uniformly bounded equicontinuous family of real valued functions defined on the metric space X. Prove that the function
g(x) = sup{f(x) : f E F}
is continuous. 63. Suppose that Un) is a sequence of nondecreasing functions which
map the unit interval into itself. Suppose that lim fn (x) = f (x) n->oo
pointwise and that f is a continuous function. Prove that fn (x) -+ f(x) uniformly as n -+ 00. Note that the functions fn are not necessarily continuous.
64. Does there exist a continuous real-valued function f(x), 0.::: x .::: 1, such that
101
xf(x) dx = 1 and
for all n = 0,2,3,4,5, ... ? Give a proof or counter-example. 65. Let f be a continuous, strictly increasing function from [0,00) onto
[0,00) and let g = f- I (the inverse, not the reciprocal). Prove that
loa f(x) dx + lob g(y) dy :::: ab
266 Function Spaces Chapter 4
for all positive numbers a, b, and determine the condition for equality. 66. Let I be a function [0,1] --+ lR whose graph {(x, I(x)) : x E [0, I]}
is a closed subset of the unit square. Prove that I is continuous. 67. Let (an) be a sequence of positive numbers such that L an converges.
Prove that there exists a sequence of numbers Cn --+ 00 as n --+ 00
such that L cnan converges. 68. Let I (x, y) be a continuous real valued function defined on the unit
square [0, 1] x [0, 1]. Prove that g(x) = max{f(x, y) : y E [0, I]} is continuous.
69. Let the function I from [0, 1] to [0, 1] have the following properties. It is of class C1
, 1(0) = ° = 1(1), and f' is nonincreasing (i.e., I is concave). Prove that the arc-length of the graph of I does not exceed 3.
70. Let A be the set of all positive integers that do not contain the digit 9 in their decimal expansions. Prove that
I L - < 00. a
aEA
That is, A defines a convergent sub-series of the harmonic series.
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