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Math Help - Use Hölder's inequality

  1. #1
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    Use Hölder's inequality

    Let (X, \mathcal{B}, \mu) be a measure space. Use Hölder's inequality to prove that if \frac{1}{p}+\frac{1}{q} = \frac{1}{r} and r \geq 1 then for every f \in L^p(X, \mathcal{B}, \mu), g \in L^q(X, \mathcal{B}, \mu), and fg \in L^r(X, \mathcal{B}, \mu) with || fg ||_r \leq || f ||_p \cdot || g ||_q.

    I figured this out already.
    Last edited by Erdos32212; December 13th 2009 at 01:39 PM.
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  2. #2
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    Quote Originally Posted by Erdos32212 View Post
    Let (X, \mathcal{B}, \mu) be a measure space. Use Hölder's inequality to prove that if \frac{1}{p}+\frac{1}{q} = \frac{1}{r} and r \geq 1 then for every f \in L^p(X, \mathcal{B}, \mu), g \in L^q(X, \mathcal{B}, \mu), and fg \in L^r(X, \mathcal{B}, \mu) with || fg ||_r \leq || f ||_p \cdot || g ||_q.


    I don't see how to prove this right now. I know that this looks a lot like Young's inequality which follows from Hölder's inequality, so I was thinking to adopt a similar proof of that theorem. Right now, I just need a few hints on how to proceed. Thank you.
    Simply apply Hölder's inequality with the correct p and q... Write down what you want to prove ( (\int (f g)^r)^{1/r}\leq \cdots), rewrite it like Holder's inequality ( \int (f^r) (g^r) \leq \cdots) in order to guess the exponents, and its shouldn't be a problem.
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