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Unformatted text preview: Math 245C Homework 3 Brett Hemenway May 16, 2007 Folland Chapter 6 7. Let f L p L , so f L q for q > p . Proposition 6.10 says k f k q k f k p q p k f k 1 p q . Letting q , we have lim q k f k q k f k . To show the opposite inequality, by the definition of k f k , there exists a set E such that E = { x :  f ( x )  > (1 ) k f k } , such that 0 < ( E ) < . Then k f k q = Z  f  q 1 q Z E  f  q 1 q (1 ) k f k ( E ) 1 q . Letting q , we have lim q k f k q (1 ) k f k . Since was arbitrary, we conclude that lim q k f k q k f k . 1 8. (a) Jensons Inequality says e R log  f  q Z e log  f  q = Z  f  q . Taking q th roots of both sides, we have e R log  f  k f k q . Then taking log of both sides, we have Z log  f  log k f k q . (b) Notice that 1 + log x x for all x > 0, with equality only when x = 1. Letting x = k f k q q = R  f  q , we have 1 + log k f k q q k f k q q log k f k q q k f k q q 1 log k f k q R  f  q 1 q . For any x , lim q x q 1 q = log x, This means lim q  f  q 1 q = log  f  , which means that we can assume  f  q 1 q is bounded for small q , thus by Dominated Convergence lim q R  f  q 1 q = lim q Z  f  q 1 q = Z lim q  f  q 1 q = Z log  f  . (c) Combining parts (a) and (b), we have Z log  f  log k f k q Z  f  q 1 q . Taking exponents, we have e R log  f  k f k q e R  f  q 1 q . 2 Finally, letting q 0, and using part (b) and the continuity of e x , we have e R log  f  lim q k f k q e R log  f  10. Suppose 1 p < , and f n ,f L p and f n f almost everywhere. If k f n f p k 0, then k f n k p k f k p  k f n f k p so k f n k p k f k p . Suppose instead k f n k p k f k p . Now  f n f  p 2 p (  f n  p +  f  p ) and we have Z 2 p (  f n  p +  f  p ) d < since f n ,f L p , and lim n Z 2 p (  f n  p +  f  p ) d = 2 p +1 Z  f  p d since k f n k p k f k p . Thus by Problem 20 in Section 2.3, we have lim n Z  f n f  p d = Z lim n  f n f  p d = 0 because f n f almost everywhere. 11. Let f be measurable, on X , and define R f to be the set of all z C such that { x :  f ( x ) z  < } has positive measure for all . (a) Suppose z R c f , then there exists an > 0 such that ( { x :  f ( x ) z  < } ) = 0. Now, for any z B ( z , ), we have  z z  < , thus B ( z,  z z  ) is ball of positive radius, contained in B ( z , ) by the triangle inequality. Then { x :  f ( x ) z  <  z z } { x :  f ( x ) z  < } , so ( { x :  f ( x ) z  <  z z } ) = 0, which means z R c f ....
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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.
 Spring '10
 tao/analysis
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