This preview shows pages 1–3. 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: 245B, Winter 2009, Assignment 3: Notes and selected model answers (Model solutions follow question numbers in bold .) Folland Chapter 5 19. [Note that for part b. youll need to show that any open ball containing X contains a sequence with no convergent subsequence. If the radius of such a ball is > , you can obtain such a sequence by scaling the sequence found in part a. by / 2 .] 20. Let K be the scalar field of X ( R or C ). Since M is finitedimensional we can choose for it a finite basis, say e 1 , e 2 , . . . , e m . Now define f j M * by setting f j ( e i ) := i,j for j m and extending by linearity. As linear functionals on a finitedimensional normed space these f j are all continuous. Consequently the HahnBanach Theorem gives us for each j m an extension g j X * with g j  M = f j . Finally, let N := T j m ker g j ; we show that this works. On the one hand, if x M N , then in particular x = 1 e 1 + + m e m for some 1 , 2 ,..., m K . Now since g j  M = f j and we also have x ker g j for every j m , we have 0 = g j ( x ) = f j ( x ) = j for each j m : that is, x = 0 . On the other hand, for any x X we can write x = m j =1 g j ( x ) e j + ( x m j =1 g j ( x ) e j ) , and now m j =1 g j ( x ) e j is manifestly in M while for any ` m we have g ` x m X j =1 g j ( x ) e j = g ` ( x ) m X j =1 g j ( x ) g ` ( e j ) = g ` ( x ) g ` ( x ) = 0 , so the remainder term x m j =1 g j ( x ) e j lies in N , and thus X = M + N . 56. [The main idea here appears in the proof that ( E ) = E : the point is that if this fails, then we can find some y ( E ) \ E , and now by a GramSchmidtlike 1 procedure we can adjust y so that it is actually orthogonal to E and nonzero; hence 6 = y E ( E ) , which leads to a contradiction.] 62. a. Fix > . By Follands Theorem 2.10 b. there is a sequence ( n ) n of simple functions on [0 , 1] such that  1   2  ...  f  and n f pointwise. Since  n f  2 (  n  +  f  ) 2 4  f  2 and  n f  2 pointwise, by the Dominated Convergence Theorem we have R  n f  2 2 for some sufficiently large n ....
View Full
Document
 Spring '10
 tao/analysis

Click to edit the document details