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Math Help - L^p norm

  1. #1
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    L^p norm

    (X, sigma, measure) is a finite measure space. Show that as p--> infty, lim |f|_p = |f|_infty, where f lies in L^infty.

    I know that
    |f|_infty= ess sup |f|= smallest number k st measure ({x:|f||>k})=0

    I would note that given e>0, measure({x in X: |f(x)|> |f|_infty - e} ) > 0.


    But from here, I am kind of lost.
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  2. #2
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    Quote Originally Posted by math8 View Post
    (X, sigma, measure) is a finite measure space. Show that as p--> infty, lim |f|_p = |f|_infty, where f lies in L^infty.

    I know that
    |f|_infty= ess sup |f|= smallest number k st measure ({x:|f||>k})=0

    I would note that given e>0, measure({x in X: |f(x)|> |f|_infty - e} ) > 0.


    But from here, I am kind of lost.
    Since |f(x)|\leqslant\|f\|_\infty for almost all x, it follows that \textstyle\|f\|_p = \left(\int|f|^p\right)^{1/p} \leqslant \left(\int\|f\|_\infty^p\right)^{1/p} = \left(\|f\|_\infty^p\mu(X)\right)^{1/p} \to \|f\|_\infty as p→∞. Hence \limsup_{p\to\infty}\|f\|_p\leqslant \|f\|_\infty.

    Let m<\|f\|_\infty, and let S_m = \{x\in X:|f(x)|\geqslant m\}. Then \mu(S_m)>0, and \textstyle\|f\|_p \geqslant \left(\int_{S_m}\!\!m^p\right)^{1/p} = \left(m^p\mu(S_m)\right)^{1/p} \to m as p→∞. Since this holds for all m<\|f\|_\infty, it follows that \liminf_{p\to\infty}\|f\|_p\geqslant \|f\|_\infty.
    Last edited by Opalg; March 14th 2009 at 04:02 AM.
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