Theorem: Let $f \in C_c^0(\mathbb{R}^n)$ and $g \in L_\text{loc}^1(\mathbb{R}^n)$ then the convolution $$(f*g)(x) = \int_{R^n} f(x-y)g(y)dy$$ is continuous on $\mathbb{R}^n$.
Let $x_n \rightarrow x$ and let K a fixed compact set in $\mathbb{R}^n$ such that $x_n – supp(f) \subset K$
In proving the theorem we make the explicit observation that $f$, beeing continuous over a compact set, is uniformly continuous by Heine-Borel theorem. Hence, let $\epsilon_n > 0$ and consider $\delta > 0$ such that $|x_n – x| < \delta$, by uniform continuity:
$$
|f(x_n – y) – f(x – y)| < \epsilon_n \chi_K(y) \quad \forall n \in \mathbb{N} \quad \forall y \in \mathbb{R}^n
$$
with $\epsilon_n \rightarrow 0$. Here $\chi_K(y)$ denotes the indicator function over K.
Hence we conclude by observing that, because $g \in \mathbb{L}_{loc}^1(\mathbb{R}^n)$
$$
| (f*g)(x_n) – (f*g)(x) | \leq \int_{\mathbb{R}^n} |g(y)| |f(x_n – y) – f(x – y)| dy \leq \epsilon_n \int_{K} |g(y)| dy \stackrel{n \rightarrow \infty}{\rightarrow} 0
$$
That is, the convolution is sequentially continuous hence continuous.
In this argument I don't get the point in using uniform continuity of $f$. Where is the advantage? Why is it so crucial to get the thesis? Why we can't conclude by just using the (sequential) continuity of $f$?
Best Answer
Uniform continuity in this case says that given $\epsilon>0$, there exists $\delta>0$ such that $\textit{for every}$ $x_n-y, x-y \in \mathbb{R}^n$ satisfying $|(x_n-y)-(x-y)|=|x_n-x| < \delta$, we must have $|f(x_n-y)-f(x-y)|<\epsilon$. Continuity alone makes us fix one of the points (i.e. would make us fix either $x_n-y$ or $x-y$), and then says that there exists a $\delta$ such that for all other points in the domain, we have an analogous implication. But in this context, we are integrating over $y$, so both $x_n-y$ and $x-y$ are changing. So, we need a $\delta$ that will work for all pairs of points at once.