I am just learning about convolutions in a measure theory course. We are given that the convolution of two functions $f$ and $g$ is defined as:
$$f*g(t) = \int f(u)g(t-u) \,du$$
Now, I would like to show that, if $f$ and $g$ are continuous with compact support, then the convolution $f*g$ is also continuous. I can't seem to wrap my head around how to show this.
Best Answer
$g$ is uniformly continuous. Take $\delta > 0$ so that $|g(x) - g(y)| < \epsilon$ whenever $|x-y|<\delta$. Then if $|s-t| < \delta$, $$|f*g(t) - f*g(s)| \le \int |f(u)| |g(u-t) - g(u-s)|\; du \le \epsilon \int |f(u)|\; du$$