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Unformatted text preview: Uniform convergence and its consequences The following issue is central in mathematics: On some domain D , we have a sequence of functions { f n } . This means that we really have an uncountable set of ordinary sequences, since for each x D , we have the sequence { f n ( x ) } . 1 Pointwise convergence Definition: Suppose that a sequence of functions f n : D R is given and that, for each x D , lim n f n ( x ) exists . We write f ( x ) for this limit, and say that the sequence of functions { f n } converges pointwise to f on the domain D . Example 1: On (0 , ), let f n ( x ) = nx 1 nx . Then lim n f n ( x ) = lim n 1 1 / ( nx ) 1 = 1 , x > , so f n converges pointwise to the constant function f ( x ) = 1. Example 2: On ( 1 , 1), let f n ( x ) = 1 + x + x 2 + + x n . Then, as we know, f n converges pointwise to the limit 1 / (1 x ). Of course, in this case, we write the limit as the sum of an infinite series: X n =0 x n = 1 1 x , for  x  < 1 . The issue at hand is the behavior of the limit function f . For instance, suppose each of the functions f n is continuous; is it true that f is continuous? More generally, which properties of the sequence { f n } survive the limiting process? As another example, suppose that X n =0 a n x n converges to some f ( x ) on D. Here the sequence of functions is the sequence of partial sums of the series: f n ( x ) = n X k =0 a k x k . 1 Each member f n of the sequence is a polynomial; it can be differentiated, for instance: f n ( x ) = n X k =1 ka k x k 1 . What can we say about the sequence { f n } . Does it converge? If so, does it converge to f , the derivative of the original limit? For the geometric series above, pointwise convergence of the sequence of derivatives is equivalent to the convergence of the infinite series X n =1 nx n 1 , which can be demonstrated using the ratio test: b n +1 b n = n + 1 n  x   x  as n , so the sequence of derivatives of the geometric series converges on the same interval as the original series. But whats the limit? We suspect that X n =1 nx n 1 = 1 (1 x ) 2 , but if true, this must be shown separately  it doesnt follow from the above. Exercise: (1) Use Taylors theorem to show that the above series converges to 1 / (1 x ) 2 . That is, show that the remainder term in Taylors theorem goes to 0 as n provided that  x  < 1. Naturally, things do not always work out the way we imagine. Here are some important counterexamples to bear in mind: 1. f n ( x ) = x n on the interval [0 , 1]. Here, for 0 x < 1, we have x n 0. On the other hand, for x = 1 , x n 1. So here is a case of a sequence of continuous functions converging pointwise on a closed interval to a limit function which is not continuous....
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This note was uploaded on 04/12/2010 for the course MAT MAT 157 taught by Professor Lam during the Spring '10 term at Canada College.
 Spring '10
 Lam
 Math

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