# Thread: Norm Of The Convolution Operator

1. ## 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 $K:L^1(D) \longrightarrow L^1(D)$
$\|K(x)(t)\|_{L^1(D)}=\int_{D}\!{\int_c^\infty\!\ka ppa(t-s)x(s)\,ds}\,dt .$

$D\subseteq{\Re}^+$

2. Originally Posted by raed
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 $\|Kx\|_1\leqslant \|\kappa\|_1\|x\|_1.$