Let $g\in L^1(\mathbb{R}^n)$ and $f:\mathbb{R}^n\to\mathbb{R}$ be bounded and continuous. Why is the convolution integral $$f*g:\mathbb{R}^n\to\mathbb{R}\;,\;\;\;\int f(\tau)g(x-\tau)\text{ d}\tau$$ also bounded and continuous?
*Proof*$\;\;\;$ It holds $$|(f*g)(x)|\leq\int |f(\tau)g(x-\tau)|\text{ d}\tau =\int |f(x-t)g(t)|\text{ dt}\le \sup_{x\in \mathbb{R}^n}|f(x)|\int |g(t)|\text{ dt}$$ Since $f$ is bounded and $g$ is integrable the supremum and the integral exists. This should be enough for at least the boundedness of $f*g$. While I would like to be sure that this is correct, I would also like to know how we show the continuity.
I would think continuity is not too bad, for fix and $x$ consider take an
$|f*g(x)-f*g(y)| = |\int f(x-t)g(t)-f(y-t)g(t)\text{ dt}| \leq \int |(f(x-t)-f(y-t))||g(t)|\text{ dt}$
Consider this:
Choose an $R>0$ large enough such that $\int_{\mathbb{R}^n/B(0,R)} |g| < \frac{\epsilon}{2K}$, where $B(0,R)$ is a ball of radius $R$ centred at the origin. Since $|f|<K$ for some $K>0$,
$|f*g(x)-f*g(y)|\leq \int_{B_R} |(f(x-t)-f(y-t))||g(t)|\text{ dt} + \int_{\mathbb{R}^n/B(0,R)} |(f(x-t)-f(y-t))||g(t)|\text{ dt} \leq \sup_{x\in B(0,R)}|(f(x-t)-f(y-t))|\int_{B(0,R)} |g| +\frac{\epsilon}{2K} 2K$
but $f$ is continuous, hence uniformly continuous on $B(0,R)$, so can choose $\delta>0$ such that $|x-y|<\delta$, then $\sup_{x\in B(0,R)}|(f(x-t)-f(y-t))|<\frac{\epsilon}{\int_{B(0,R)} |g|}$