This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 Mtest) 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 )...
View
Full
Document
This note was uploaded on 10/18/2011 for the course MATH S4062Q taught by Professor Staff during the Summer '11 term at Columbia.
 Summer '11
 Staff

Click to edit the document details