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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 ε > 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 )  ≥ ε. 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 )  ≥ ε 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 ˆ f ( n ) = 0 for every n ∈ Z , each Ces` aro mean, σ N f ( x ), is 0 on the nose. Since f is continuous at x , it follows that σ N f ( x ) → f ( x ) as N → ∞ (since σ N f ( x ) is given by a convolution with a good kernel). This implies that f ( x ) = 0. (b) Since  ˆ f ( n ) exp( inx )  =  ˆ f ( n )  and ∑  ˆ f ( n )  converges, it follows (by the Weier strass Mtest) that S N f ( X ) = ∑ N n = N ˆ f ( n ) exp( inx ) converges uniformly to a con tinuous function g ( x ). Note that ˆ g ( n ) = ˆ f ( n ): ˆ g ( n ) = 1 2 π R π π g ( x ) exp( inx )...
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 Summer '11
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 Metric space, Compact space, FN, Baire Category Theorem

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