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Unformatted text preview: Sequences of Functions Uniform convergence 9.1 Assume that f n f uniformly on S and that each f n is bounded on S. Prove that { f n } is uniformly bounded on S. Proof : Since f n f uniformly on S, then given = 1 , there exists a positive integer n such that as n n , we have  f n ( x ) f ( x )  1 for all x S. (*) Hence, f ( x ) is bounded on S by the following  f ( x )   f n ( x )  + 1 M ( n ) + 1 for all x S. (**) where  f n ( x )  M ( n ) for all x S. Let  f 1 ( x )  M (1) ,...,  f n 1 ( x )  M ( n 1) for all x S, then by (*) and (**),  f n ( x )  1 +  f ( x )  M ( n ) + 2 for all n n . So,  f n ( x )  M for all x S and for all n where M = max ( M (1) ,...,M ( n 1) ,M ( n ) + 2) . Remark : (1) In the proof, we also shows that the limit function f is bounded on S. (2) There is another proof. We give it as a reference. Proof : Since Since f n f uniformly on S, then given = 1 , there exists a positive integer n such that as n n , we have  f n ( x ) f n + k ( x )  1 for all x S and k = 1 , 2 ,... So, for all x S, and k = 1 , 2 ,...  f n + k ( x )  1 +  f n ( x )  M ( n ) + 1 (*) where  f n ( x )  M ( n ) for all x S. Let  f 1 ( x )  M (1) ,...,  f n 1 ( x )  M ( n 1) for all x S, then by (*),  f n ( x )  M for all x S and for all n 1 where M = max ( M (1) ,...,M ( n 1) ,M ( n ) + 1) . 9.2 Define two sequences { f n } and { g n } as follows: f n ( x ) = x 1 + 1 n if x R, n = 1 , 2 ,..., g n ( x ) = 1 n if x = 0 or if x is irrational, b + 1 n if x is rational, say x = a b , b > . Let h n ( x ) = f n ( x ) g n ( x ) . (a) Prove that both { f n } and { g n } converges uniformly on every bounded interval. Proof : Note that it is clear that lim n f n ( x ) = f ( x ) = x, for all x R and lim n g n ( x ) = g ( x ) = 0 if x = 0 or if x is irrational, b if x is ratonal, say x = a b ,b > . In addition, in order to show that { f n } and { g n } converges uniformly on every bounded interval, it suffices to consider the case of any compact interval [ M,M ] , M > . Given > , there exists a positive integer N such that as n N, we have M n < and 1 n < . Hence, for this , we have as n N  f n ( x ) f ( x )  = x n M n < for all x [ M,M ] and  g n ( x ) g ( x )  1 n < for all x [ M,M ] . That is, we have proved that { f n } and { g n } converges uniformly on every bounded interval. Remark : In the proof, we use the easy result directly from definition of uniform convergence as follows. If f n f uniformly on S, then f n f uniformly on T for every subset T of S....
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 Fall '09
 Frade

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