Is the convolution of two continuous functions continuous?

Question

The title is the question:

Is it true, that the convolution of two continuous functions is continuous again?

Answer 1

Unqualified continuity on a set of infinite measure does little to control the behavior of integrals. Here is an example of what can happen when the functions are not assumed to be in $L^1$.

Let $f(x) = \sum_{n\in\mathbb Z} \max(0,1-|x-10n|)$. This is a continuous function on $\mathbb{R}$. Its graph consists of triangles with base $[10n-1,10n+1]$. So, when $x \in [2,8]$, the convolution $$ (f*f)(x) = \int_{\mathbb{R}} f(t)f(x-t)\,dx $$ is equal to $0$. However, for $x\in (-2,2)$ the integral above diverges to $+\infty$. So one can say that $f*f$ is not continuous in this case.

If you assume integrability, e.g., $f$ is continuous and bounded while $g$ is continuous and integrable, then the result is true. But in this case, the assumption of continuity is redundant: the convolution of $L^1$ function and $L^\infty$ function is continuous; and more generally, the convolution of two functions from conjugate Lebesgue spaces $L^p$ and $L^{p/(p-1)}$ is continuous.