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Unformatted text preview: 1. H¨older’s Inequality and Minkowski’s inequality. We fix p, q ∈ [1 , ∞ ] such that 1 p + 1 q = 1 . Definition 1.1. Suppose f : R n → R is Lebesgue measurable. We let  f  p = Z  f ( x )  p dx ¶ 1 /p if p < ∞ and we let  f  ∞ = sup { t ∈ (0 , ∞ ) : L n ( { f  > t } ) > } . Proposition 1.1. Suppose f : R n → R is Lebesgue measurable and c ∈ R . Then  cf  p =  c  f  p . Proof. Exercise for the reader. / Theorem 1.1. (H¨older’s Inequality.) Suppose f, g : R n → R are Lebesgue measurable. Then  fg  1 ≤  f  p  g  q . Proof. Exercise for the reader. Here are some hints. Treat the case p = ∞ or q = ∞ by a straightforward argument. When p < ∞ and q < ∞ first reduce to the case  f  p = 1 and  g  q = 1 by making use of the previous Proposition; then apply the inequality a 1 /p b 1 /q ≤ 1 p a + 1 q b for a, b ∈ (0 , ∞ ). / Theorem 1.2. Minkowski’s Inequality. Suppose f, g : R n → R are Lebesgue measurable. Then  f + g  p ≤  f  p +  g  p . Proof. Exercise for the reader. Here are some hints. The cases p = 1 and p = ∞ follow from the triangle inequality. In case 1 < p < ∞ apply H¨older’s Inequality to  f + g  p ≤  f + g  p 1 (  f  +  g  ). / 1.1. An extension of H¨older’s Inequality. Suppose p, q, r ∈ [0 , ∞ ] and 1 p + 1 q = 1 r . Theorem 1.3. Suppose f, g : R n → R are Lebesgue measurable. Then  fg  r ≤  f  p  g  q ....
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This note was uploaded on 01/14/2011 for the course ECE 210a taught by Professor Chandrasekara during the Fall '08 term at UCSB.
 Fall '08
 Chandrasekara

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