I'm trying to prove Proposition 4.19. in Brezis's Functional Analysis, i.e.,
Theorem Let $f \in \mathcal C_c (\mathbb R^n)$ and $g \in L_{\text{loc}}^1 (\mathbb R^n)$. Then $(f*g) (x)$ is well-defined for all $x \in \mathbb R^n$ and $(f * g) \in \mathcal C (\mathbb R^n)$.
Could you have a check on my below attempt?
Proof Let $S$ be the compact support of $f$.
- Fix $x \in \mathbb R^n$. Then $$ \begin{align} (f*g) (x) &= \int_{\mathbb R^n} f(x-y) g(y) \ \mathrm d y \\ &= \int_{\mathbb R^n} f(x-y) g(y) 1_{x-S} (y) \ \mathrm d y. \end{align} $$
We have $f(x-\cdot)$ belongs to $L^{\infty} (\mathbb R^n)$, and $g 1_{x-S}$ belongs to $L^1 (\mathbb R^n)$. By Hölder's inequality, $f(x-\cdot) g 1_{x-S}$ belongs to $L^1$. Hence $(f*g) (x)$ is well-defined.
- Let $x, x_n \in \mathbb R^n$ such that $x_n \to x$. Let $K$ be the closure of $\{x, x_1, \ldots, x_n, \ldots\}$. Then $K$ is compact. Notice that $$ \begin{align} (f*g) (x_n) &= \int_{\mathbb R^n} f(x_n-y) g(y) 1_{K-S} (y), \\ (f*g) (x) &= \int_{\mathbb R^n} f(x-y) g(y) 1_{K-S} (y). \end{align} $$
Because $f$ is continuous, $f(x_n-y) g(y) 1_{K-S} (y) \to f(x-y) g(y) 1_{K-S} (y)$ for all $y \in \mathbb R^n$. On the other hand, $$ |f(x_n-y) g(y) 1_{K-S} (y)| \le \|f\|_{\infty} |g(y)| 1_{K-S} (y)| \quad \forall n, \forall y \in \mathbb R^n. $$
It follows from $K-S$ is compact that $g 1_{K-S}$ is integrable. The claim then follows from dominated convergence theorem.
I think your proof is correct, although the second part could be done simpler.
Let $K={\rm supp}\,f.$ The function $f$ is uniformly continuous. Hence for $\varepsilon>0$ there is $1\ge\delta>$ such that $$\|u-v\|\le \delta \implies |f(u)-f(v)|<\varepsilon $$ Fix $x$ and let $U_\delta=\{x'\, :\,\|x'-x\|\le \delta\}.$ Then for $x'\in U_\delta$ we obtain $$|(f*g)(x)-(f*g)(x')| \\ \le \int\limits_{\mathbb{R}^n}|f(x-s)-f(x'-s)|\,|g(s)|\,ds\\ \le \varepsilon\int\limits_{U_\delta-K}|g(s)|\,ds \le \varepsilon\int\limits_{U_1-K}|g(s)|\,ds$$