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7 and Series of Functions

HW7: Page 165: 1, 2, 3, 4, 6, 7, 8, 9, 11, 12, 13, 16, 18, 19, 20, 22 7.1 DISCUSSION OF MAIN PROBLEM

7.1 Convergence Pointwise: Suppose f , n =1, 2, 3,..., is a of functions { n} defined on a E, and suppose that the sequence of fn(x) converges for every x E. We can then define a f by { } 2 f(x)= lim fn(x),xE. n !1 2 We say that f is the of f , or f converges to f pointwise on E. { n} { n} Similarly, if f (x) converges for every x E, and if we define n 2 P 1 f(x)= f (x),xE, n 2 n=1 X we say that the function f is the sum of the series fn. Problems: The main problem which arises is to determineP whether important prop- • erties of functions are preserved under the limit operations above. For instance, if the functions fn are continuous, or di↵erentiable, or integrable, is the same true of the limit function f? What are the relations between fn0 and f 0, say, or between the of fn and that of f? To say that f is continuous at a limit point x means lim f(t)=f(x). t x ! 81 82 7 SEQUENCES AND SERIES OF FUNCTIONS

Hence, to ask whether the of continuous functions is continuous is the same as to ask whether

lim lim fn(t)= lim lim fn(t), t x n n t x ! !1 !1 ! i.e., whether the orders in which limit processes are carried out can be inter-changed: on the left, we first let n ,thent x; on the right side, t x first, then n . !1 ! ! !1 7.2 Example For m =1, 2, 3,..., n =1, 2, 3,...,let

m s = . m,n m + n Then, for each fixed n, lim sm,n =1, m !1 so that lim lim sm,n =1. n m !1 !1 On the other hand, for every fixed m,

lim sm,n =0, n !1 so that lim lim sm,n =0. m n !1 !1 7.3 Example For real x,let

x2 f (x)= ,n=1, 2, 3,.... n (1 + x2)n

and consider 1 1 x2 f(x)= f (x)= . n (1 + x2)n n=0 n=0 X X 2 Since fn(0) = 0, we have f(0) = 0. For x = 0, the last series converges to 1 + x . Hence 6 0,x=0, f(x)= 1+x2 x =0. ⇢ 6 This example shows that a of continuous functions may have a discontinuous sum.

7.5 Example For real x,let

sin nx f (x)= ,n=1, 2, 3,..., n pn 7.2 83

and f(x)= lim fn(x)=0 n !1 Then f 0(x) = 0, and fn0 (x)=pn cos nx,

so that f 0 does not converge to f 0. { n} 7.6 Example For 0 x 1, let   f (x)=nx(1 x2)n,n=1, 2, 3,.... n For 0

lim fn(x)=0, 0 x 1. n !1   It is easy to calculate 1 n f (x)dx = . n 2n +2 Z0 Thus, 1 1 1 lim fn(x)dx = =0= lim fn(x) dx. n 0 2 6 0 n !1 Z Z h !1 i

7.2 UNIFORM CONVERGENCE

7.7 Uniform Convergence: A sequence of functions fn converges uniformly on E to a function f if for every ✏>0, there is an integer{N }such that n N implies f (x) f(x) <✏ | n | for all x E. 2 Aseries f (x) converges uniformly on E if the sequence s of partial sums, n { n} P n sn(x)= fi(x), i=1 X converges uniformly on E.

It is clear that if fn converges uniformly to f on E,then fn converges pointwise • to f on E. { } { }

7.8 Theorem (Cauchy Criterion) The sequence of functions fn , defined on E, converges uniformly on E if and only if for every ✏>0 there{ exists} an integer N such that m, n N implies f (x) f (x) <✏ | n m | 84 7 SEQUENCES AND SERIES OF FUNCTIONS

for all x E. 2 Proof Suppose the Cauchy condition holds. By Theorem 3.11, the sequence fn(x) converges for each fixed x to a limit which we may call f(x). Hence the sequence{ } fn converges to f on E. Let ✏>0 be given. There is an integer N such that m,{ n} N implies f (x) f (x) <✏/2. | n m | If we let m in the inequality, by Theorem 3.19, we have !1 f (x) f(x) ✏/2 <✏, n N, | n | for all x E.Thus f converges uniformly to f. 2 { n} Conversely, suppose fn converges uniformly to f on E. Then, for any ✏>0, there is an integer N such{ that} n N implies f (x) f(x) <✏/2 | n | for all x E. Hence, if n, m N,wehave 2 f (x) f (x) f (x) f(x) + f (x) f(x) <✏ | n m || n | | m | for all x E. 2 7.9 Theorem Suppose f converges to f on E.Put { n}

Mn =sup fn(x) f(x) . x E | | 2 Then f converges uniformly to f on E if and only if M 0 as n . { n} n ! !1 Proof The proof is straightforward.

7.10 Theorem (Weierstrass M-Test) Suppose fn is a sequence of functions defined on E satisfying { } f (x) M ,n=1, 2, 3,..., | n | n for all x E.Then f converges uniformly on E if M converges. 2 n n Proof If Mn converges,P then, for any ✏>0, thereP is an integer N such that m n N implies P m Mi <✏. i=n X Hence, if m n N, m m

fi(x) Mi <✏,  i=n i=n X X for all x E. It follows from Theorem 7.8 that f converges uniformly to f on n E. 2 { } 7.3 UNIFORM CONVERGENCE AND CONTINUITY 85

7.3 UNIFORM CONVERGENCE AND CONTINUITY

7.11 Theorem Suppose fn f uniformly on a set E in a . Let x be a limit point of E, and suppose! that

lim fn(t)=An,n=1, 2, 3,.... t x ! Then A converges, and { n} lim f(t)= lim An. t x n ! !1 In other words, lim lim fn(t)= lim lim fn(t). t x n n t x ! !1 !1 !

Proof Let ✏>0 be given. Since fn converges uniformly on E, there exists an integer N such that n, m N implies{ } f (t) f (t) <✏, | n m | for all t E. Letting t x,wehave 2 ! A A ✏, | n m|

form n, m N. Hence, An is a , and therefore converges, say to A. { } From the inequality

f(t) A f(t) f (t) + f (t) A + A A , | || n | | n n| | n |

we now give the estimates for the terms on the right hand side. In fact, since fn f uniformly, we can choose n suciently large such that !

f(t) f (t) <✏/3 | n | for all t E, and such that 2 A A <✏/3. | n | For this large n, by a condition in the theorem, we can choose a neighborhood V of x such that f (t) A <✏/3 | n n| if t V E, t = x. Thus, we know that for t V E, t = x, 2 \ 6 2 \ 6 f(t) A <✏. | |

That is, lim f(t)=A =limAn. t x n ! !1

7.12 Theorem If fn is a sequence of continuous functions on E, and if fn f uniformly on E,then{ } f is continuous. ! 86 7 SEQUENCES AND SERIES OF FUNCTIONS

Proof By Theorem 7.11, for every t E,wehave 2

lim f(t)=lim lim fn(t)= lim lim fn(t)= lim fn(x)=f(x). t x t x n n t x n ! ! !1 !1 ! !1 Thus, f is continuous on E

7.13 Theorem Suppose K is compact, and (a) f is a sequence of continuous functions on K, { n} (b) f converges pointwise to a f on K, { n} (c) f (x) f (x) for all x K, n =1, 2, 3,.... n n+1 2 Then f f uniformly on K. n ! Proof Put gn = fn f.Thengn is continuous, gn 0, and gn gn+1.Wehaveto prove that g 0 uniformly on K. ! n ! Let ✏>0 be given. Write

K = x K : g (x) ✏ ,n=1, 2, 3,.... n { 2 n } Since g is continuous, by Theorem 4.8, K K is closed for each n. By Theorem n n ⇢ 2.35, Kn is compact. Since gn gn+1, we know that Kn Kn+1.Fixx K,since g (x) 0, we see that x K if n is suciently large. Hence x K 2. In other n ! 62 n 62 n words, Kn is empty. Hence KN is empty for some N, by Theorem 2.36. It follows T that 0 gn(x) <✏for all x K and all n N. This proves the theorem. T 2 1 Example Consider f (x)= on (0, 1). It is clear that f is continuous • n nx +1 n on (0, 1), fn(x) 0 on (0, 1), and fn fn+1. However, fn does not converge uniformly to 0 on! (0, 1). In fact, for ✏ =1 /2 > 0, no matter{ how} large n is, we can always find a point x (0, 1) such that 2 1 ✏. nx +1 7.14 Metric Space C (X): If X is a metric space, we define C (X) to be the set of all complex-valued, continuous, bounded functions with domain X. For each f C (X), we define its supremum norm: 2 f =sup f(x) . || || x X | | 2

In fact, is a norm defined on C (X). Since f C (X) has to be bounded, f < ||.If · || f = 0 only if f(x)=0foreveryx 2X, that is, only if f = 0. If || || 1 || || 2 c is a complex , then cf =supx X cf(x) = c f .Ifh = f + g,then h(x) f(x) + g(x) f +|| g|| for all x2 |X,whichimplies| | ||| || f + g f + g . | || | | ||| || || || 2 || || || || || || Thus, together with the supremum norm , C (X) is a metric space. || · || 7.4 UNIFORM CONVERGENCE AND INTEGRATION 87

Asequence fn converges to f with respect to the metric of C (X) if and only if • f f uniformly{ } on X. n ! 7.15 Theorem The above metric makes C (X) into a . Proof Let f be a Cauchy sequence in C (X). For any ✏>0, there exists an { n} integer N such that fn fm <✏if n, m N.Since fn(x) fm(x) fn fm for all x X, by Theorem|| || 7.8, there is a function f| with domain|X ||to which|| f converges2 uniformly. Since f is continuous for every n, by Theorem 7.12, f { n} n is continuous. Moreover, since there is an n such that f(x) fn(x) < 1 for all x X, we know that f(x) f (x) f(x) + f (x) < 1+| f which| implies that 2 | || n | | n | || n|| f(x) is bounded. Thus, f C (X). Since f f uniformly on X, we know that 2 n ! f f 0 as n . Therefore, C (X) is a complete metric space. || n || ! !1

7.4 UNIFORM CONVERGENCE AND INTEGRATION

7.16 Theorem Let ↵ be monotonically increasing on [a, b]. Suppose f f uniformly n ! on [a, b]. If f R(↵) on [a, b] for n =1, 2, 3,...,thenf R(↵) on [a, b], and n 2 2 b b f d↵ =lim fn d↵. n Za !1 Za In other words, b b lim fn d↵ =lim fn d↵. a n n a Z h !1 i !1 Z Proof We only need to prove the theorem for real functions. Put

✏n =sup fn(x) f(x) . x [a,b] | | 2 Then f ✏ f f + ✏. n n   n These inequalities give

b b (f ✏ )d↵ L(f,↵) U(f,↵) (f + ✏ )d↵. n n    n n Za Za This sequence of inequalities gives

0 U(f,↵) L(f,↵) 2✏ [↵(b) ↵(a)],   n which implies f R(↵) on [a, b], since ✏ 0 by the hypothesis. 2 n ! From the following inequality

b b fn d↵ f d↵ ✏n[↵(b) ↵(a)], a a  Z Z 88 7 SEQUENCES AND SERIES OF FUNCTIONS

we let n and obtain the desired limit. !1 Corollary:Iff R(↵) on [a, b] and • n 2

1 fn(x)=f(x) n=1 X uniformly on [a, b]. Then

b b 1 f d↵ = fn d↵. a n=1 a Z X Z

7.5 UNIFORM CONVERGENCE AND DIFFERENTIATION

7.17 Theorem Suppose f is a sequence of di↵erentiable functions on [a, b] such that { n} fn(x0) converges for some x0 [a, b]. If fn0 converges uniformly on [a, b], then {f converges} uniformly on [a, b2], and { } { n}

f 0(x)= lim fn0 (x). n !1

Proof To show that fn converges uniformly on [a, b], we consider the following estimates: for any x { [a,} b], 2 f (x) f (x) f (x) f (x) f (x )+f (x ) + f (x ) f (x ) | n m ||n m n 0 m 0 | | n 0 m 0 | f 0 (t) f 0 (t) x x + f (x ) f (x ) |n m |·| 0| | n 0 m 0 |

where t is a number between x and x0. Let ✏>0 be given. Choose N such that n, m N imply f (x ) f (x ) <✏/2, | n 0 m 0 | and ✏ f 0 (t) f 0 (t) < , | n m | 2(b a) for all t [a, b]. Hence, for n, m N, and x [a, b], 2 2 ✏ f (x) f (x) < x x + ✏/2 ✏. | n m | 2(b a) ·| 0|  Thus, f converges uniformly on [a, b]. { n} To prove the desired limit in the theorem, we let f be the limit of fn . For a fixed x [a, b], put { } 2 f (t) f (x) f(t) f(x) (t)= n n ,(t)= , n t x t x 7.6 EQUICONTINUOUS FAMILIES OF FUNCTIONS 89

where t [a, b], t = x. It is clear that 2 6

lim n(t)=fn0 (x),n=1, 2, 3,.... t x ! By Theorem 7.11, if we can show that uniformly on [a, b] x ,then n ! \{ }

lim (t)= lim fn0 (x), t x n ! !1

which is the desired limit since f 0(x)=lim(t). t x ! To show that n converges uniformly to on [a, b] x , we have the following estimates: by the{ Mean} Value Theorem, there exists t˜ \{between} t and x, such that

f (t) f (x) f (t) f (x) (t) (t) = n n m m | n m | t x t x [f (t) f (t)] [f (x) f (x)] = n m n m t x ✏ = f 0 (t˜) f 0 (t˜) < , n m 2(b a) if n, m N, t = x. Hence converges uniformly, for t = x.Since f converges 6 { n} 6 { n} to f, we know that n(t) (t) pointwise for t = x.Thus n converges uniformly to , for t = x. ! 6 { } 6 7.18 There exists a real continuous function on the real line which is nowhere di↵eren- tiable. The details are skipped.

7.6 EQUICONTINUOUS FAMILIES OF FUNCTIONS

Problem: we know that every bounded sequence of complex numbers contains a • convergent subsequence, and the question arises whether something similar is true for sequences of functions. To make the question more precise, we shall define two kinds of boundedness.

7.19 Pointwise Bounded, Uniformly Bounded: Let fn be a sequence of functions defined on E. { }

fn is pointwise bounded on E, if for a fixed point, there exists a finite-valued function{ } defined on E such that

f (x) <(x),n=1, 2, 3,.... | n |

f is uniformly bounded on E,ifthereexistsanumberM such that { n} f (x)