245B_W09_hwk3

245B_W09_hwk3 - 245B, Winter 2009, Assignment 3: Notes and...

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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 finite-dimensional 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 finite-dimensional normed space these f j are all continuous. Consequently the Hahn-Banach 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 Gram-Schmidt-like 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 ....
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245B_W09_hwk3 - 245B, Winter 2009, Assignment 3: Notes and...

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