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Math Help - Norm Of The Convolution Operator

  1. #1
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    Norm Of The Convolution Operator

    Hi All;
    I need the solution of the following question:
    Let L^1(I) be the space of Lebesgue integrable Functions on a measurable subset I of \Re, with the standard norm
    \|x\|_{L^1(I)}=\int_{I}\!|x|\,dt .
    Let \kappa \in L^1(I) be a given function. Then for any function  x\in L^1({\Re}^+), the integral (Kx)(t)=\int_0^\infty\!\kappa(t-s)x(s)\, ds exists for almost t\in{\Re}^+ and (Kx)(t)\in L^1({\Re}^+) .
    The question is how it follows that \|K(x)\|_{L^1({\Re}^+)}\leq\|K\|_{L^1({\Re}^+)}\|x  \|_{L^1({\Re}^+)} .

    Also, how it follow that \|K(x)(t)\|_{L^1(D)}\leq\|K\|_{D}\|x(t)\|_{L^1(D)}.
    \|K\|_{D} is the norm of K:L^1(D) \longrightarrow L^1(D)
    \|K(x)(t)\|_{L^1(D)}=\int_{D}\!{\int_c^\infty\!\ka  ppa(t-s)x(s)\,ds}\,dt .

      D\subseteq{\Re}^+
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  2. #2
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    Quote Originally Posted by raed View Post
    Hi All;
    I need the solution of the following question:
    Let L^1(I) be the space of Lebesgue integrable Functions on a measurable subset I of \Re, with the standard norm
    \|x\|_{L^1(I)}=\int_{I}\!|x|\,dt .
    Let \kappa \in L^1(I) be a given function. Then for any function  x\in L^1({\Re}^+), the integral (Kx)(t)=\int_0^\infty\!\kappa(t-s)x(s)\, ds exists for almost t\in{\Re}^+ and (Kx)(t)\in L^1({\Re}^+) .
    The question is how it follows that \|K(x)\|_{L^1({\Re}^+)}\leq\|K\|_{L^1({\Re}^+)}\|x  \|_{L^1({\Re}^+)} .

    Also, how it follow that \|K(x)(t)\|_{L^1(D)}\leq\|K\|_{D}\|x(t)\|_{L^1(D)}.
    \|K\|_{D} is the norm of K:L^1(D) \longrightarrow L^1(D)
    \|K(x)(t)\|_{L^1(D)}=\int_{D}\!{\int_c^\infty\!\ka  ppa(t-s)x(s)\,ds}\,dt .

      D\subseteq{\Re}^+
    As far as I know, it takes a fair bit of machinery to prove such results, including Fubini's and Hölder's theorems and the duality between L^1 and L^\infty. To simplify things, I'll just indicate how to do it when all the functions are defined on the whole of \mathbb{R}. You can then modify the proof to deal with domains that are subsets of \mathbb{R}.

    So suppose that x\in L^1(\mathbb{R}), \kappa\in L^1(\mathbb{R}) and (Kx)(t)=\int_{\mathbb{R}}\!\kappa(t-s)x(s)\, ds . Denote by _s\kappa the function t\mapsto\kappa(s-t). Then _s\kappa\in L^1(\mathbb{R}) and \|_s\kappa\|_1 = \|\kappa\|_1.

    Let h\in L^\infty(\mathbb{R}). Then
    \begin{aligned}\int_{\mathbb{R}}\!|(Kx)(t)h(t)|\,d  t &\leqslant \int_{\mathbb{R}}\!  \int_{\mathbb{R}}\!|\kappa(t-s)x(s)h(t)|\,dsdt \\&= \int_{\mathbb{R}}\!  \int_{\mathbb{R}}\!|\kappa(t-s)x(s)h(t)|\,dtds \quad\text{(Fubini)} \\ &\leqslant \int_{\mathbb{R}}\! |x(s)|\|_s\kappa\|_1\|h\|_\infty \quad\text{(H\"{o}lder, applied to the t-integral)} \\ &= \|\kappa\|_1\|x\|_1\|h\|_\infty .\end{aligned}

    Since that holds for all h\in L^\infty(\mathbb{R}), it follows from duality that Kx\in L^1(\mathbb{R}), with \|Kx\|_1\leqslant \|\kappa\|_1\|x\|_1.
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