Chapter Sequences and Series of Functions Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 9.2, Slide Section 9.2 Applications of Uniform Convergence Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 9.2, Slide Theorem 9.2.1 Let ( fn) be a sequence of continuous functions defined on a set S and suppose that ( fn) converges uniformly on S to a function Proof: f : S → Then f is continuous on S The argument is based on the inequality f ( x ) − f ( c ) ≤ f ( x ) − f n ( x ) + f n ( x ) − f n ( c ) + f n (c ) − f ( c ) The idea is to make the first and last terms small by using the uniform convergence of ( fn ) and choosing n sufficiently large Once an n is selected, the continuity of fn will enable us to make the middle term small by requiring x to be close to c To see this, let c ∈ S and let ε > Then there exists N ∈ such that n ≥ N implies that | fn(x) – f (x) | < ε /3 for all x ∈ S Since c ∈ S, we also have | fn(c) – f (c) In particular, we have | fN (x) – f (x) | < ε /3 whenever n ≥ N | < ε /3 and | Since fN (c) – f (c) | < ε /3 fN is continuous at c, there exists a δ > such that | fN (x) – fN Thus for all x ∈ S with | x – c | < δ (c) | < ε /3 whenever | x – c | < δ and x ∈ S we have | f (x) − f (c) | ≤ | f (x) − fN (x) | + | fN (x) − fN (c) | + | fN (c) − f (c) | < ε ε ε + + = ε 3 Hence f is continuous at c, and since c is any point in S, we conclude that f is continuous on S ♦ Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 9.2, Slide When we apply Theorem to series of functions, we get the following corollary Corollary 9.2.2 Let ∑ S Then ∞ f of functions defined on a set S Suppose that each fn is continuous on S and that the series converges uniformly to a function f on be a series n=0 n ∞ f = ∑ n =is0continuous f n on S Proof: sisn continuous, = ∑ k = 0Theorem f k 9.2.1 implies that n Since each partial sum f , the limit of the partial sums, is also continuous ♦ One useful application of Theorem 9.2.1 is in showing that a given sequence does not converge uniformly Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 9.2, Slide For example, let fn(x) = x n for x ∈ [0, but the limit function Then each 1] fn is continuous on [0, 1], 0, if ≤ x < 1, f ( x) =is not continuous at x = 1, if x = 1, Thus the convergence cannot be uniform on [0, y 1] (1, f (1)) f 1 f 2 f 3 ) f Copyright © 2013, 2005, 2001 Pearson Education, Inc x Section 9.2, Slide Theorem 9.2.4 Let ( fn) be a sequence of continuous functions defined on an interval [a, b ] and suppose that ( fn) converges uniformly on [a, b ] to a function f Then b f n ( x) dx n →∞ ∫a lim = b ∫a f ( x) dx “The limit of the integrals is the integral of the limit.” Proof: Theorem 9.2.1 implies that f is continuous on [a, Given any ε b], so the functions f, fn , and f are all integrable on [a, f n − b] by Theorem 7.2.2 > 0, since ( fn) converges uniformly to f on [a, b], there exists a natural number N such that for all x ∈ [a, b ] and all n ≥ N f n ( x) − f ( x) < ε , b−a (We may assume that a ≠ b, for otherwise the integrals are all zero and the result is trivial.) Thus whenever b ∫a n ≥ N, we have b f n ( x) dx − ∫ f ( x ) dx = a It follows that b ∫a [ f n ( x) − f ( x)] dx ≤ b ∫a f n ( x) − f ( x) dx ≤ b ∫a ε dx = ε b−a lim n →∞ ∫ ab f n ♦( x) dx = ∫ ab f ( x) dx Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 9.2, Slide Corollary 9.2.5 Let Σ∞n =be0 afseries n of functions defined on an interval [a, b] Suppose that each fn is continuous on [a, b] and that the series converges uniformly to a function f on [a, b ] Then ∫ab f ( x) dx = Σ∞n = ∫ab f n ( x) dx “The integral of the sum is the sum of the integrals.” Proof: Let sn = Σ nk be = 0thef knth partial sum of the series so that (s on [a, b] n) converges uniformly to f Then we have b ∫a f ( x) dx = lim ∫ b sn ( x ) n→∞ a b n = lim ∫ k∑= n →∞ a dx f k ( x) dx n b = lim ∑ ∫ f k ( x) dx a n →∞ k =0 = ∞ Theorem 9.2.4 Definition of sn The integral of a finite sum is the sum of the integrals b ∑ ∫a f n ( x) dx ♦ Definition of an infinite series n=0 Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 9.2, Slide Example 9.2.6 Σ ∞ ( −t ) n and suppose that n = 00 < r < Consider the geometric series = 1+ t When t ∈ [−r, r], then the series converges to 1/(1 + t), so In fact, since | n − t | ≤r Σ n and n r converges when < r< 1, the Weierstrass M -test (9.1.11) shows that the convergence is uniform on any interval [−r, ∞ ∑ ( −1)n t n n =0 r], where < r < Thus, if x ∈ (− 1, 1), we can integrate the series term by term from to x: ∫ x dt = 1+ t ∞ ∑ (−1)n n=0 ∫ But from calculus we know that t dt = ∫ x n ∞ x n+1 ∑ (−1) n + n=0 n x dt = ln (1 + x) 1+ t So for any x in ( –1, 1) we have x n +1 x x3 x ln (1 + x) = ∑ (−1) = x− + − +L n + n=0 ∞ n In Section 9.3 we extend this to include x = 1: ln = − 1 + − +L The next theorem deals with the derivatives of a convergent sequence of functions Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 9.2, Slide Theorem 9.2.7 Suppose that ( fn) converges to f on an interval [a, b] Suppose also that each exists and is continuous on [a, b], and the sequence converges ( f n′) for each x ∈ [a, b] uniformly on [a, b] Then f n′ lim n →∞ f n′ ( x) = f ′( x) Note that it is the sequence of derivatives that must converge uniformly, not the original functions Corollary 9.2.8 Let Σn∞=be0afseries n of functions that converges to a function f on an interval [a, b] Suppose that, for each n, uniformly convergent on [a, b] Then f ′(x) = Σn∞= f n′ for all x ∈ [a, b] exists and is continuous on [a, b] and that the series of derivatives f n′ Σn∞= f n′( x) One of the surprising results in real analysis is that there exists a continuous function that does not have a derivative at any point This is our next theorem Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 9.2, Slide is Theorem 9.2.9 There exists a continuous function defined on that is nowhere differentiable Proof: Define g (x) = | x | for x ∈ [ −1, 1] and extend the definition of g to all of by requiring that g (x + 2) = g (x) for all x It is not difficult to show (in the text) that g is continuous on ( ) n g n ( x) = g (4n x) Then, g0(x) = g(x), g1 ( x) = 34 g (4 x), g ( x) = 16 g (16 x), etc For each integer n ≥ 0, let ( ) In general, each g n+1 Now define f on oscillates four times as fast, but only three-fourths as high, as g n f ( x) = by ∞ ( ) ∑ g n ( x) = n=0 ∞ ∑( ) n=0 g (4 n x ) g2(x) g1(x) g0(x) = g(x) −2 −1 Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 9.2, Slide 10 f ( x) = ∞ ∑ n=0 g n ( x) = ∞ ∑( ) n=0 g (4 n x ) n For all x we have ≤ g (x) ≤ 1, so that ≤ gn(x) ≤ (3/4) Thus the Weierstrass M -test implies that the series converges uniformly to Since each g is continuous on n f on , Corollary 9.2.2 implies that f is continuous on The proof (in the text) that f is nowhere differentiable is tedious, but we can see from the first few partial sums what is happening As n → ∞, we get “corners” at more and more points In the limit function, f , there is a corner at every point and it is nowhere differentiable g0(x) + g1(x) + g2(x) g0(x) + g1(x) g0(x) −2 −1 Copyright © 2013, 2005, 2001 Pearson Education, Inc g2(x) g1(x) Section 9.2, Slide 11 ... idea is to make the first and last terms small by using the uniform convergence of ( fn ) and choosing n sufficiently large Once an n is selected, the continuity of fn will enable us to make... selected, the continuity of fn will enable us to make the middle term small by requiring x to be close to c To see this, let c ∈ S and let ε > Then there exists N ∈ such that n ≥ N implies that | fn(x)... nowhere differentiable Proof: Define g (x) = | x | for x ∈ [ −1, 1] and extend the definition of g to all of by requiring that g (x + 2) = g (x) for all x It is not difficult to show (in the text)