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Thread: Holder's inequality

  1. #1
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    Holder's inequality

    With s>1 and r>s. How can I show that $\displaystyle L^s \supset L^r$, for expectation with respect to a probability measure?
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  2. #2
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    Hello,

    Show what you've done. Basically, you just have to apply Holder's inequality to $\displaystyle E[X^s]$, with $\displaystyle X^r$ as f and 1 as g. If X is in $\displaystyle L^r$ then $\displaystyle E[X^r]<\infty$, that's how it'll work.
    Give it a try.
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  3. #3
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    Holder inequality: $\displaystyle E|XY|\le (E|X|^p)^{1/p}(E|Y|^q)^{1/q}$
    Let q=r/s
    with r.v. $\displaystyle |Y|^s \text{ and } 1_\Omega $
    Then $\displaystyle E|Y|^s\le (E|Y|^r)^{s/r}<\infty$?

    Not very sure though
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  4. #4
    Moo
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    This is exactly it !

    Now you have to talk a little :

    Let's take $\displaystyle Y\in L^r$. Then $\displaystyle E[|Y|^r]<\infty$. - your calculations - . Thus $\displaystyle Y\in L^s$, which implies that $\displaystyle L^r\subset L^s$ : any element of the first belongs to the second.
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