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Math Help - Help with bounded multiplication operator proof

  1. #1
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    Help with bounded multiplication operator proof

    Hi guys, i need to prove that the operator Tf = x*f(x) maps from L^p(-2,2) into L^p(-2,2) where:
    L^p(-2,2) = \lbrace f: [-2,2] \rightarrow \mathbb{C}:\, \int\limits_{-2}^2 |f(x)|^p\,\text{d}x < \infty \rangle
    I think that Holders inequality would be useful so i use it like this
    \int\limits_{-2}^2 |f(x)x|^{p-1}|f(x)x|\,\text{d}x \leq \int\limits_{-2}^2 \left(|f(x)x|^{p-1}\left[\int\limits_{-2}^2|f(x)|^p\, dx\right]^{1/p}\left[\int\limits_{-2}^2|x|^q\right]^{1/q}\right)dx =  ||f(x)||_p \int\limits_{-2}^2 \left(|f(x)x|^{p-1} \int\limits_{-2}^2 |x|^q\right]^{1/q}\right) dx<br />
    By doing that until the integrand of the order p equals 1 i obtain with the relation  1/p+1/q = 1:
    16 ||f(x)||_p^p 2^{p/(p-1)}(2p-1)^{1-p}(p-1)^{p-1}
    This expression converges when p goes to infinity so can i conclude that it maps as wished?
    But i cant assume that the norm^P is finite can i?
    Is there a smarter way(correct)
    Im really lost, fought this problem for almost 6 hours.
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  2. #2
    Super Member
    Joined
    Apr 2009
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    México
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    Re: Help with bounded multiplication operator proof

    Since you know Hölder's inequality the argument is easy, and in fact a more general result can be obtained in the same way: Let g\in L^\infty(-2,2) then the mapping f\mapsto gf is a bounded linear operator L^p(-2,2)\to L^p(-2,2) for any 1\leq p\leq \infty. To see this (the case p=\infty is trivial, so assume it's finite) just note that Hölders inequality (with exponents 1,\infty associated to |f|^p, |g| respectively) give \| gf\|_p \leq \| g\|_{\infty}\| f\|_p
    Last edited by Jose27; April 17th 2012 at 09:48 PM.
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