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Show an operator on L^2(0,\infty) is bounded

  1. Nov 12, 2011 #1
    1. The problem statement, all variables and given/known data

    Show that the operator on [tex] L^2(0,\infty)[/tex] defined by [tex]g \rightarrow f(x)= \int_{0}^{\infty} e^{-xy}g(y)dy[/tex] is bounded.

    2. Relevant equations

    Operator norm: [tex]||T|| = \sup_{||g||_{L^2}=1}||Tg||_{L^2}[/tex]



    3. The attempt at a solution

    I tried to get a handle on [tex]f(x)= \int_{0}^{\infty} e^{-xy}g(y)dy[/tex], by first fixing x and applying Holder's inequality. I got that for every x, [tex]f(x) \leq \frac{1}{2x}[/tex] but this didn't really get me much since [tex]\int_{0}^{\infty}(\frac{1}{2x})^2dx[/tex] doesn't converge...

    What I need is that [tex]\int_{0}^{\infty} | \int_{0}^{\infty}e^{-xy}g(y)dy|^2dx [/tex] is finite. any inequalities/tricks i'm not thinking of? I'm hoping I'm just not realizing what theorem/inequality I need to use....I really appreciate any help you all can offer
     
  2. jcsd
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