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Unformatted text preview: Solutions homework 9. (1) Let f n ( x ) = x 2 n 1+ x 2 n . Prove that f ( x ) = lim n f n ( x ) exists for all x R . Does ( f n ) converge uniformly to f ? Solution: There are 3 cases:  x  < 1,  x  = 1, and  x  > 1. In case  x  < 1, then x 2 n 0 as n , which implies that f n ( x ) 0 as n for  x  < 1. In case  x  = 1, then x 2 n = 1 for all n , so f n ( x ) 1 2 when  x  = 1. In case  x  > 1, then divide the numerator and denominator by x 2 n to see that f n ( x ) 1 when  x  > 1. This shows that f ( x ) = lim n f n ( x ) exists for all x , but f is not continuous at the points x = 1, which implies that the convergence is not uniform (as each f n is continuous and uniform limits of continuous functions are continuous). (2) Define f n : [0 , 1] [0 , 1] by f n ( x ) = x n (1 x ). Prove that f n converges uniformly to 0....
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 Spring '07
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