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

This preview shows pages 1–4. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## 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.

### Page1 / 13

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online