hw5 - Math 245A Homework 5 Brett Hemenway November 16, 2005...

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Unformatted text preview: Math 245A Homework 5 Brett Hemenway November 16, 2005 21. Suppose f n , f ∈ L 1 , and f n → f a.e. Recall || f n | - | f || ≤ | f n- f | which means ≤ Z | f n | - Z | f | = Z ( | f n | - | f | ) ≤ Z || f n | - | f || ≤ Z | f n- f | So if R | f n- f | → 0, then R | f n | - R | f | → 0, so R | f n | → R | f | . We also have | f n- f | ≤ | f n | + | f | So if R | f n | → R | f | , we have R ( | f n | + | f | ) → R 2 | f | . Then, using problem 20, we have Z | f n- f | → Z | f- f | = 0 Because | f n- f | → 0 in L 1 . This gives R | f n- f | → 0 iff R | f n | → R | f | . 23. Let f : [ a, b ] → R be a bounded function and H ( x ) = lim δ → sup | y- x |≤ δ f ( y ) h ( x ) = lim δ → inf | y- x |≤ δ f ( y ) 1 a. If f is continuous at x , then for any > 0, there exists a δ , such that | x- y | ≤ δ implies | f ( x )- f ( y ) | < , so f ( x )- ≤ inf | y- x |≤ δ f ( y ) ≤ sup | y- x |≤ δ f ( y ) ≤ f ( x ) + Which means H ( x ) = lim δ → sup | y- x |≤ δ f ( y ) = lim δ → inf | y- x |≤ δ f ( y ) = h ( x ) Conversely, if H ( x ) = h ( x ), then for any > 0, there exists a δ such that sup | y- x |≤ δ f ( y )- inf | y- x |≤ δ f ( y ) < so we have | f ( x )- f ( y ) | < whenever | y- x | ≤ δ b. If { P k } is sequence of ever finer partitions, and G p , g p are defined as before, G P = n X j =1 M j χ ( t j- 1 ,t j ] g P = n X j =1 m j χ ( t j- 1 ,t j ] with G = lim G P , g = lim g P . We have, G P k ( x ) ≥ H ( x ) for all x outside of P k g P k ( x ) ≤ h ( x ) for all x outside of P k because for any x ∈ ( t j- 1 , t j ) we can find a δ such that ( x- δ, x + δ ) ⊂ ( t j- 1 , t j ). Thus for x 6∈ S ∞ k =1 P k we have G P k ( x ) ≥ H ( x ) g P k ( x ) ≤ h ( x ) So outside of S ∞ k =1 P k G ( x ) ≥ H ( x ) g ( x ) ≤ h ( x ) 2 Now, let H δ ( x ) = sup | y- x |≤ δ f ( y ) h δ ( x ) = inf | y- x |≤ δ f ( y ) Then choose a k such that every interval ( t j- 1 , t j ) in P k has t j- t j- 1 < δ , then G P k ( x ) ≤ H δ ( x ) g P k ( x ) ≥ h δ ( x ) holds for all k > k , so G ( x ) ≤ H δ ( x ) g ( x ) ≥ h δ ( x ) for any δ , so G ( x ) ≤ H ( x ) g ( x ) ≥ h ( x ) So we have that G ( x ) = H ( x ) outside of S ∞ k...
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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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hw5 - Math 245A Homework 5 Brett Hemenway November 16, 2005...

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