HW6_solutions - Homework 6 solutions 1 Problem 2 page 119...

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Homework 6: solutions 1. Problem 2, page 119. We have | f ( x ) | ≤ k f k except possibly on a set of measure zero. Then k f k p p = Z 1 0 | f | p Z 1 0 k f k p = k f k p , thus k f k p ≤ k f k . On the other hand, k f k = inf { M : m { x [0 , 1] : | f ( x ) | > M } = 0 } , and therefore for any ² > 0 such that k f k - ² > 0, the measure of the set E ² = { x [0 , 1] : | f ( x ) | > k f k - ² } is non-zero. Thus, k f k p p = Z 1 0 | f | p > Z E ² ( k f k - ² ) p = ( k f k - ² ) p m ( E ² ) . Therefore, k f k - ² = ( k f k - ² ) lim p →∞ ( mE ² ) 1 /p lim inf p →∞ k f k p lim sup p →∞ k f k p ≤ k f k . Since ² is arbitrary, we obtain that lim inf p →∞ k f k p = lim sup p →∞ k f k p = k f k . 2. Prove H¨older’s inequality for p = 1 , q = . Show that equality holds if and only if | g ( x ) | = k g k at almost every point x such that f ( x ) 6 = 0. If f L 1 [0 , 1] , g L [0 , 1] then | g ( x ) | ≤ k g k except on a set of measure zero. Thus, Z 1 0 | fg | ≤ Z 1 0 | f | · k g k = k f k 1 k g k . Assume now that | g ( x ) | = k g k at almost every point x such that f ( x ) 6 = 0. Then Z 1 0 | fg | = Z { x : f ( x ) 6 =0 } | f |k g k = k f k 1 k g k . Conversely, assume that Z 1 0 | fg | = k f k 1 k g k . If there is a subset E of the set { x [0 , 1] : f ( x ) 6 = 0 } of positive measure where | g ( x ) | < k g k then Z 1 0 | fg | = Z { x : f ( x ) 6 =0 } | fg | < Z E | f |k g k = k f k 1 k g k , which contradicts to the assumption that R 1 0 | fg | = k f k 1 k g k . Therefore, a subset of the set { x [0 , 1] : f ( x ) 6 = 0 } where
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