exam1-sol - Modern Analysis, Exam 1 Possible solutions 1....

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Unformatted text preview: Modern Analysis, Exam 1 Possible solutions 1. (a) Suppose that f n does not converge uniformly to f on K . Then there is > 0 such that for all n N , there exists m n N and a point, call it x m n K , such that | f m n ( x m n )- f ( x m n ) | . Clearly we may assume that { m n } is an increasing sequence in n . By compactness of K , we may find a convergent subsequence { x p n } (say x p n x as n ). Then | f p n ( x p n )- f ( x p n ) | | f p n ( x p n )- f ( x ) | | {z } (I) + | f ( x )- f ( x p n ) | | {z } (II) . By assumption, (I) 0 as n . Since f is continuous, it follows that (II) as n . This contradicts the fact that | f m n ( x m n )- f ( x m n ) | for every n N . (b) Consider f n ( x ) = 0 if x 6 = n and f n ( n ) = 1. Clearly f n 0 pointwise but not uniformly ( f n ( n ) = 1). Now suppose that x n is a convergent sequence in R . In particular, x n is bounded. Therefore f n ( x n ) = 0 for all sufficiently large indices n . 2. (a) Since f ( n ) = 0 for every n Z , each Ces` aro mean, N f ( x ), is 0 on the nose. Since f is continuous at x , it follows that N f ( x ) f ( x ) as N (since N f ( x ) is given by a convolution with a good kernel). This implies that f ( x ) = 0. (b) Since | f ( n ) exp( inx ) | = | f ( n ) | and | f ( n ) | converges, it follows (by the Weier- strass M-test) that S N f ( X ) = N n =- N f ( n ) exp( inx ) converges uniformly to a con- tinuous function g ( x ). Note that g ( n ) = f ( n ): g ( n ) = 1 2 R - g ( x ) exp(- inx )...
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This note was uploaded on 10/18/2011 for the course MATH S4062Q taught by Professor Staff during the Summer '11 term at Columbia.

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exam1-sol - Modern Analysis, Exam 1 Possible solutions 1....

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