Continuity of convolution $f\ast g$ for $f\in C_0(\mathbb{R})$ and $g\in L^1(\mathbb{R})$.

Question

If $f\in C_0(\mathbb{R})$ (functions of compact support) and $g\in L^1(\mathbb{R})$ show that the convolution function $$f\ast g(x)=\int_\mathbb{R}f(t)g(x-t)$$ is continuos.

I have tried to solve this exercise with the following ideas but I have not had success

$f\in C_0(\mathbb{R})$ $\Rightarrow$ $f$ is uniform continuous but i not see do with the function $g\in L^1(\mathbb{R})$.

Also I tried to use that $C_0(\mathbb{R})$ is dense in $L^1(\mathbb{R})$, and, so exist a $(a_n)\subset C_0(\mathbb{R})$ so that $a_n\to g$

Answer 1

First, note $C_0(\mathbb{R})$ is the closure of $C_c(\mathbb{R})$ in the uniform metric and $C_c(\mathbb{R})$ is dense in $L^1(\mathbb{R})$. Then choose sequences $\{f_n\} \subseteq C_c(\mathbb{R})$ and $\{g_n\} \subseteq C_c(\mathbb{R})$ such that $\| f_n - f\|_u \to 0$ and $\|g_n - g\|_1 \to 0$. Then we should be able to prove $f_n * g_n \in C_c(\mathbb{R})$. By Holder's inequality, $$\begin{split} |f_n*g_n(x)-f*g(x)|&\le |f_n*g_n(x)-f*g_n(x)|+|f*g_n(x)-f*g(x)| \\ &\le \|f_n - f\|_{\infty} \|g_n\|_1 + \|f\|_1 \|g_n - g\|_{\infty} \to 0 \end{split}$$ So $f*g \in C_0(\mathbb{R})$.

Answer 2

It's actually true for any continuous $f$ bounded on $\mathbb R.$ Let $M= \sup_{\mathbb R} |f|.$ Rewrite the convolution as

$$\int_{\mathbb R} f(x-t)g(t)\,dt.$$

Fix any $x$ and let $x_n \to x.$ By the continuity of $f,$ $f(x_n-t) \to f(x-t)$ pointwise on $\mathbb R.$ Since $|f(x_n-t)g(t)| \le M|g(t)|$ on $\mathbb R$ for all $n$ and $t,$ the dominated converge theorem shows

$$\int_{\mathbb R} f(x_n-t)g(t)\,dt \to \int_{\mathbb R} f(x-t)g(t)\,dt.$$

Thus the convolution is continuous at $x$ as desired, and we're done.