# hw3 - Math 245C Homework 3 Brett Hemenway May 16, 2007...

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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 &gt; 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 ) | &gt; (1- ) k f k } , such that 0 &lt; ( E ) &lt; . 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 &gt; 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 &lt; , 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 &lt; 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 | &lt; } has positive measure for all . (a) Suppose z R c f , then there exists an &gt; 0 such that ( { x : | f ( x )- z | &lt; } ) = 0. Now, for any z B ( z , ), we have | z- z | &lt; , thus B ( z,- | z- z | ) is ball of positive radius, contained in B ( z , ) by the triangle inequality. Then { x : | f ( x )- z | &lt;- | z- z |} { x : | f ( x )- z | &lt; } , so ( { x : | f ( x )- z | &lt;- | 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.

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hw3 - Math 245C Homework 3 Brett Hemenway May 16, 2007...

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