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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. you’ll need to show that any open ball containing ∈ X contains a sequence with no convergent subsequence. If the radius of such a ball is ε > , you can obtain such a sequence by scaling the sequence found in part a. by ε/ 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 ) := δ 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 = λ 1 e 1 + ··· + λ m e m for some λ 1 ,λ 2 ,...,λ 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 ) = λ 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 ε > . By Folland’s Theorem 2.10 b. there is a sequence ( φ n ) n of simple functions on [0 , 1] such that ≤  φ 1  ≤  φ 2  ≤ ... ≤  f  and φ n → f pointwise. Since  φ n f  2 ≤ (  φ n  +  f  ) 2 ≤ 4  f  2 and  φ n f  2 → pointwise, by the Dominated Convergence Theorem we have R  φ n f  2 ≤ ε 2 for some sufficiently large n ....
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 Spring '10
 tao/analysis
 measure, Indicator function, Lebesgue integration, ﬁnite positive measure

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