Results 1 to 2 of 2

Thread: Norm Of The Convolution Operator

  1. #1
    Member
    Joined
    Sep 2009
    Posts
    173

    Norm Of The Convolution Operator

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

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

    $\displaystyle D\subseteq{\Re}^+$
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    10
    Quote Originally Posted by raed View Post
    Hi All;
    I need the solution of the following question:
    Let $\displaystyle L^1(I)$ be the space of Lebesgue integrable Functions on a measurable subset $\displaystyle I $of $\displaystyle \Re$, with the standard norm
    $\displaystyle \|x\|_{L^1(I)}=\int_{I}\!|x|\,dt .$
    Let $\displaystyle \kappa \in L^1(I)$ be a given function. Then for any function $\displaystyle x\in L^1({\Re}^+)$, the integral $\displaystyle (Kx)(t)=\int_0^\infty\!\kappa(t-s)x(s)\, ds $ exists for almost $\displaystyle t\in{\Re}^+$ and $\displaystyle (Kx)(t)\in L^1({\Re}^+)$ .
    The question is how it follows that $\displaystyle \|K(x)\|_{L^1({\Re}^+)}\leq\|K\|_{L^1({\Re}^+)}\|x \|_{L^1({\Re}^+)}$ .

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

    $\displaystyle D\subseteq{\Re}^+$
    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 $\displaystyle L^1$ and $\displaystyle L^\infty$. To simplify things, I'll just indicate how to do it when all the functions are defined on the whole of $\displaystyle \mathbb{R}$. You can then modify the proof to deal with domains that are subsets of $\displaystyle \mathbb{R}$.

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

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

    Since that holds for all $\displaystyle h\in L^\infty(\mathbb{R})$, it follows from duality that $\displaystyle Kx\in L^1(\mathbb{R})$, with $\displaystyle \|Kx\|_1\leqslant \|\kappa\|_1\|x\|_1.$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Convolution of Functions on Lp norm
    Posted in the Advanced Applied Math Forum
    Replies: 0
    Last Post: Sep 23rd 2010, 07:17 PM
  2. convolution of exp(-a*norm(x)^2) and exp(-b*norm(x)^2) ?
    Posted in the Advanced Math Topics Forum
    Replies: 1
    Last Post: Aug 14th 2010, 11:30 AM
  3. Norm of an Operator
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: Apr 22nd 2010, 08:58 AM
  4. Operator norm
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Apr 16th 2010, 07:32 PM
  5. Convolution operator
    Posted in the Advanced Applied Math Forum
    Replies: 3
    Last Post: Sep 10th 2007, 09:03 AM

Search Tags


/mathhelpforum @mathhelpforum