1
vote
1answer
21 views

Convolution operator positive definite?

Let $\mu$ be a compactly supported Borel probability measure on $\mathbb{R}^n$. Consider the convolution operator $T: L^2(\mathbb{R}^n) \rightarrow L^2(\mathbb{R}^n)$ defined by $$ Tf = f \ast \mu $$ ...
1
vote
1answer
42 views

proof that$ L^1 (G)$ is a subspace of $M(G)$

Let G be a locally compact group, and let $M(G)$ be the space of complex Radon measures on G. I identify the function f with the measure $f(x) \rm dx$ . but How do I prove this inclusion?؟ . .
2
votes
1answer
149 views

$g, f, \hat {f} \in L^{1}(\mathbb R)\cap L^{2}(\mathbb R) \cap C_{0}(\mathbb R) \implies \widehat{(fg)}= \hat{f} \ast \hat{g} ? $

Let $f, g\in L^{1}(\mathbb R)$ and it Fourier transform of $f$, $\hat{f} (y) = \int _ {\mathbb R} f(x) e^{-2\pi i x \cdot y} dx, \ (y\in \mathbb R)$ and the convolution of $f $ and $g$; $f\ast g ...
1
vote
1answer
44 views

Interesting equation in L^1

Consider $L^{1}(T) = \{ f : R \rightarrow C \text{ with period 1 and } \int_{0}^{1} |f (x)| \ dx < \infty\}$. For $f,g \in L^{1}(T)$ the convolution is given by $(f * g)(x)= ...
8
votes
1answer
498 views

How to show convolution of an $L^p$ function and a Schwartz function is a Schwartz function

We have the Schwartz space $\mathcal{S}$ of $C^\infty(\mathbb{R^n})$ functions $h$ such that $(1+|x|^m)|\partial^\alpha h(x)|$ is bounded for all $m \in \mathbb{N_0}$ and all multi-indices $\alpha$. ...
3
votes
1answer
275 views

Applications of Young's convolution inequality

Recall that the convolution of two functions is given by $$f*g(y)=\int f(x)g(y-x)dx.$$ The well known inequality known as Young's inequality, say that $$\|f*g\|_r\leq\|f\|_p\cdot\|g\|_q $$ provided ...
2
votes
1answer
654 views

Young's inequality for discrete convolution

Young's inequality for convolution of functions states that for $f\in L^p(\mathbb{R}^d)$ and $g\in L^q(\mathbb{R}^d)$ we have $$\|f\star g\|_r\le\|f\|_p\|g\|_q$$ for $p$, $q$, $r$ satisfying ...
3
votes
1answer
103 views

Convolution on noncommutative group algebras

If $G$ is a non-Abelian locally compact group, and $f$ is in $L^1{(G)}$ and $u$ is in $L^{\infty}(G)$, and $f\ast u=0$ can it be concluded that $u\ast f=0$?
2
votes
1answer
318 views

Convolution on group with measure

I was wondering about the generalization of the concept of convolution from the familiar one on real spaces and how many properties still remain. For convolution on Lebesgue-integrable real-valued ...