hw2 - Math 245C Homework 2 Brett Hemenway April 29, 2007...

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Unformatted text preview: Math 245C Homework 2 Brett Hemenway April 29, 2007 Folland Chapter 5 6. Let X be a finited dimensional vector space, and e 1 ,...,e n a basis for X . Define k ∑ n 1 a j e j k 1 = ∑ n 1 | a j | . (a) We have n X 1 a j e j + n X 1 b j e j 1 = n X 1 | a j + b j | ≤ n X 1 | a j | + n X 1 | b j | = n X 1 a j e j 1 + n X 1 b j e j 1 and λ n X 1 a j e j 1 = n X 1 | λa j | = | λ | n X 1 | a j | and if k ∑ n 1 a j e j k 1 = 0, then ∑ n 1 | a j | = 0, so each a j = 0, which means ∑ n 1 a j e j = 0. So k · k 1 is a norm. 1 (b) Let φ : K n → X ( a 1 ,...,a n ) 7→ n X 1 a j e j If, as the norm on K n we take k ( a 1 ,...,a n ) k = ∑ n 1 | a j | (see remark on page 153), we have that k ( a 1 ,...,a n ) k = n X 1 | a j | = k φ (( a 1 ,...,a n )) k 1 So φ is clearly continuous. In fact φ is Lipschitz continuous with constant 1. (c) Let E = { x ∈ X : k x k 1 = 1 } . Then if S = { ( a 1 ,...,a n ) : k ( a 1 ,...,a n ) k = 1 } ⊂ K n , we have that S is compact in K n , and E = φ ( S ), so E is compact in X . (d) Let k · k be a norm on X . Let M = max( k e 1 k ,..., k e n k ). Then n X 1 a j e j ≤ n X 1 k a j e j k ≤ M n X 1 | a j | = M n X 1 a j e j 1 Thus the function f : ( X, k · k 1 ) → ( X, k · k ) is continuous. In particular, the set E defined in part (c), is compact in ( X, k · k ), since k · k is continuous on ( X, k · k ), we have that k · k takes a minimum value on E , so we let m = inf x ∈ E k x k > 0. From above, we have for any x ∈ X , we have x k x k 1 ∈ E , so m ≤ x k x k 1 ⇒ m k x k 1 ≤ k x k So we have that m k x k 1 ≤ k x k ≤ M k x k 1 So any two norms on X are equivalent. 8. Let ( X, M ) be a measure space and M ( X ) the space of complex mea- sures on ( X, M ). Following the hint, we consider an absolutely con- vergent series, and show that it converges. Suppose ∑ k μ n k < ∞ . We 2 wish to show that μ = ∑ μ n is a measure on X . For any E ∈ M , we have X | μ n | ( E ) ≤ X | μ n | ( X ) = X k μ n k < ∞ , then since the C is complete, by Theorem 5.1, ∑ μ n ( E ) converges to a complex number. Now, μ ( ∅ ) = X μ n ( ∅ ) = 0 . Let E m be a sequence of disjoint sets in M . Then μ [ m E m ! = X n μ n [ m E m ! = X n X m μ n ( E m ) = X m X n μ n ( E m ) = X m μ ( E m ) . Where we are justified in changing the order of summation by Fubini’s Theorem because X n X m | μ n ( E m ) | ≤ X n X m k μ n k < ∞ Thus μ is a complex measure, so by Theorem 5.1 M ( X ) is complete. 11. If 0 ≤ α ≤ 1, the let Λ α ([0 , 1]) be the space of H¨ older continuous functions, i.e. f ∈ Λ α ([0 , 1]) iff k f k Λ α < ∞ , where k f k Λ α = | f (0) | + sup x 6 = y | f ( x )- f ( y ) | | x- y | α (a) To show that k · k Λ α defines a norm k f + g k Λ α = | f (0) + g (0) | + sup x 6 = y | f ( x ) + g ( x )- f ( y )- g ( y ) | | x- y | α ≤ | f (0) | + | g (0) | + sup x 6 = y | f ( x )- f ( y ) | + | g ( x )- g ( y ) | | x- y | α ≤ | f (0) | + | g (0) | + sup...
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This note was uploaded on 04/06/2010 for the course MATH various taught by Professor Tao/analysis during the Spring '10 term at UCLA.

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hw2 - Math 245C Homework 2 Brett Hemenway April 29, 2007...

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