I need some help with this exercise:
If $f\in L_p(\mathbb{R}^n)$ and $g\in L_q(\mathbb{R}^n)$, where $\frac{1}{p}+\frac{1}{q}=1$,
Is the convolution $f\ast g(x)=\displaystyle\int_{\mathbb{R}^n}f(x-y)g(y)\;dy$ a continuous function?
Thank you very much in advance.
The continuity here is even uniform. Indeed we know by the invariance of the Lebesgue measure that $\|f(\cdot+h)-f(\cdot)\|_{L^p(\Bbb R^n)}\to 0$ has $|h|\to0$ for all $f\in L^p(\Bbb R^n)$.
Thus by Holder inequality for all $h\in \Bbb R^n$, we have $$\sup_{x\in \Bbb R^n}[g*f(\cdot+h)-g*f(\cdot)](x)= \sup_{x\in \Bbb R^n}g*\big[f(\cdot+h)-f(\cdot)\big](x)\\ \leq \|f(\cdot+h)-f(\cdot)\|_{L^p(\Bbb R^n)}\|g\|_{L^q(\Bbb R^n)}\to0,~~|h|\to0. $$ Thus we have the uniform continuity.
Yes, it's true. By an approximation argument and Hölder's inequality, we can see that $f$ is the uniform limit of linear combinations of functions of the form $\chi_A*\chi_B$, where $A$ and $B$ are Borel sets of finite measure. By outer regularity of Lebesgue measure, it is enough to show that $x\mapsto \chi_A*\chi_O$ is continuous for each open set $O$ of finite measure. This can be done by dominated convergence.