7 Sequences 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 sequence of functions { n} defined on a set E, and suppose that the sequence of numbers fn(x) converges for every x E. We can then define a function f by { } 2 f(x)= lim fn(x),xE. n !1 2 We say that f is the limit 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 integrals 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 limit of a sequence 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 convergent series 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 UNIFORM CONVERGENCE 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 <x 1, it is clear that lim fn(x)=0. n !1 For x = 0, fn(0) = 0. Hence 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 metric space. 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 Cauchy sequence, 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 sufficiently 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 continuous function 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 sufficiently 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.