Sequences and Series of Functions Chapter 9 Sequences and Series of Functions

Section 9.2 Applications of Uniform Convergence

Proof: The argument is based on the inequality 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 f : S  . Then f is continuous on S. Proof: The argument is based on the inequality 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  > 0. Then there exists N  such that n  N implies that | fn(x) – f (x) | <  /3 for all x  S. Since c  S, we also have | fn(c) – f (c) | <  /3 whenever n  N. In particular, we have | fN (x) – f (x) | <  /3 and | fN (c) – f (c) | <  /3. Since fN  is continuous at c, there exists a  > 0 such that | fN (x) – fN (c) | <  /3 whenever | x – c | <  and x  S. Thus for all x  S with | x – c | <  we have | f (x)  f (c) |  | f (x)  fN (x) | + | fN (x)  fN (c) | + | fN (c)  f (c) | Hence f is continuous at c, and since c is any point in S, we conclude that f is continuous on S. 

When we apply Theorem 1 to series of functions, we get the following corollary. Let be a series 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 S. Then is continuous on S. Proof: Since each partial sum is continuous, Theorem 9.2.1 implies that 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.

For example, let fn(x) = x n for x  [0, 1]. Then each fn is continuous on [0, 1], but the limit function is not continuous at x = 1. Thus the convergence cannot be uniform on [0, 1]. f (1, f (1)) x y 1 f 1 f 2 f 3 )

“The limit of the integrals is the integral of the limit.” 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 “The limit of the integrals is the integral of the limit.” Proof: Theorem 9.2.1 implies that f is continuous on [a, b], so the functions f, fn , and fn  f are all integrable on [a, b] by Theorem 7.2.2. Given any  > 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. (We may assume that a  b, for otherwise the integrals are all zero and the result is trivial.) Thus whenever n  N, we have It follows that 

“The integral of the sum is the sum of the integrals.” Corollary 9.2.5 Let be a series 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 “The integral of the sum is the sum of the integrals.” Proof: Let be the nth partial sum of the series so that (sn) converges uniformly to f on [a, b]. Then we have Theorem 9.2.4 Definition of sn The integral of a finite sum is the sum of the integrals. Definition of an infinite series

Example 9.2.6 Consider the geometric series and suppose that 0 < r < 1. When t  [r, r], then the series converges to 1/(1 + t), so In fact, since |  t n |  r n and  r n converges when 0 < r < 1, the Weierstrass M -test (9.1.11) shows that the convergence is uniform on any interval [r, r], where 0 < r < 1. Thus, if x  ( 1, 1), we can integrate the series term by term from 0 to x: But from calculus we know that So for any x in ( –1, 1) we have In Section 9.3 we extend this to include x = 1: The next theorem deals with the derivatives of a convergent sequence of functions.

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 uniformly on [a, b]. Then for each x  [a, b]. Note that it is the sequence of derivatives that must converge uniformly, not the original functions. Corollary 9.2.8 Let be a series of functions that converges to a function f on an interval [a, b]. Suppose that, for each n, exists and is continuous on [a, b] and that the series of derivatives is uniformly convergent on [a, b]. Then f (x) = for all x  [a, b]. 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.

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 . For each integer n  0, let Then, g0(x) = g(x), In general, each gn + 1 oscillates four times as fast, but only three-fourths as high, as gn. Now define f on by 1 1 3 2 2 4 g2(x) g1(x) g0(x) = g(x)

For all x we have 0  g (x)  1, so that 0  gn(x)  (3/4)n. Thus the Weierstrass M -test implies that the series converges uniformly to f on . Since each gn is continuous 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. 1 1 3 2 2 4 g0(x) + g1(x) + g2(x) g0(x) + g1(x) g0(x) g2(x) g1(x)