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 DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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...
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.
 Spring '10
 tao/analysis
 Math

Click to edit the document details