$f*g$ is uniformly continuous

Question

Let $f,g \in L^1( \mathbb{R}^2)$, assume that $f$ is uniformly continuous in $\mathbb{R}^2$. Is this true $f*g$ is uniformly continuous in $\mathbb{R^2}$?

Generally if $g \in L^p(\mathbb{R^2})$ then $f*g$ is uniformly continuos?

Answer 1

$\vert f*g(x)-f*g(y)\vert=\vert\int f(x-t)-f(y-t)g(t) dt\vert \leq \int \vert g(t)\vert dt. M(\vert x-y \vert)$,

where $M(h)=\sup _{\vert k \vert \leq h, z\in \bf R} \vert f(z)- f(z+k)\vert$.

If $f$ is uniformly continuous $\lim _{h\to 0 }M(h)=0$, and $f*g$ is uniformly continuous. This proof works if $f$ is $L^1$ or $L^{\infty}$. For $L^p$, $1<p<\infty$, write $f$ as a sum of a $L^1$ and a $L^{\infty}$ function, $f= f. 1_{\vert f\vert \geq 1}+f. 1_{\vert f\vert < 1}$

Answer 2

We need to be a little careful here, since for general $f,g \in L^1,$ $f*g(x)$ is only defined for a.e. $x.$ However, we are also given that $f$ is uniformly continuous. Such an $f$ has the property that $f(x)\to 0$ as $|x|\to \infty.$ That implies $f\in L^\infty.$ That's good news, since then $f*g(x)$ is defined for every $x.$

It's now easy enough to show $f*g$ is uniformly continuous. But actually more is true: $f*g(x)\to 0$ as $|x|\to \infty.$ Proof: Think of $f_x(t) = f(x-t).$ For each fixed $t,$ $f_x(t) \to 0$ as $|x|\to \infty.$ We also have $|f_x(t)g(t)| \le \|f\|_\infty |g(t)|.$ The dominated convergence theorem then proves the result.