Prob 6 - ans 1

Prob 6 - ans 1 - 1. Holders Inequality and Minkowskis...

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Unformatted text preview: 1. Holders Inequality and Minkowskis 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. (Holders 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. Minkowskis 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 Holders Inequality to | f + g | p | f + g | p- 1 ( | f | + | g | ). / 1.1. An extension of Holders 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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Prob 6 - ans 1 - 1. Holders Inequality and Minkowskis...

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