As I understand it, the convolution
$$(f \ast g)(x) = \int_{\mathbb{R}^n} f(y) g(x-y)\,\mathrm dy$$
is defined wherever the above integral is defined, i.e., finite. This is true whether $f$ and $g$ are functions, test functions and distributions, etc.
My question is this: If $f$ and $g$ are both locally integrable, can we guarantee that their convolution always exists?
It seems like too good of a result to be true, but my argument is this:
For any compact set $K$, we have that $f, g \in L^1(K)$. Then by Minkowski's inequality,
$$\|f \ast g\|_{L^1(K)} \le \|f\|_{L^1(K)}\|g\|_{L^1(K)} < \infty.$$
Then we have that for each compact set $K$, $|(f\ast g)(x)| < \infty$ for almost every $x \in K$. If we let $N = \{x \in \mathbb{R}^n: |(f\ast g)(x)|< \infty\}$ and $B_n = \overline{B_n(0)}$, we have that
$$N = N \cap (\bigcup_{n=1}^\infty B_n) = \bigcup_{n=1}^\infty (N \cap B_n),$$
thus
$$|N| \leq \sum_{n=1}^\infty |N\cap B_n| = 0,$$
since $|(f\ast g)(x)| < \infty$ for almost every $x \in B_n$. Then the convolution is finite almost everywhere, i.e. the integral that defines it is finite and the convolution exists. Is this a valid argument, and if not what is the flaw?
Best Answer
So you are right to be skeptical. Take for instance $f(x) = x^2 = g(x)$. These are locally integrable functions but do not have finite convolution anywhere.
I think I know what the flaw in your reasoning is. The inequality $\|f * g\|_{L^1(K)} \le \|f\|_{L^1(K)} \|g\|_{L^1(K)}$ isn't actually working with the "full" convolution $f*g$. It should be:
$$\|f*g\|_{L^1(K)} = \int_K\bigg| \int_{\mathbb{R}} f(y)g(x-y)\,dy\bigg|\,dx \le \int_K \int_{\mathbb{R}} |f(y)g(x-y)|\,dy\,dx.$$
Allowing for infinities we can interchange integration order since everything is positive here. (Tonelli works for $+\infty$.) This gives us
$$\int_{\mathbb{R}} |f(y)|\int_{K}|g(x-y)|\,dx\,dy = \int_{\mathbb{R}}|f(y)|\int_{K+y}|g(x)|\,dx\,dy.$$
Note that neither integral need be bounded here. This is not in general bounded by $\|f\|_{L^1(K)} \|g\|_{L^1(K)}$. The culprit is that you dropped the fact that the convolution is really over all of $\mathbb{R}$ and instead thought of it as over $K$.
Note that there are still issues with working with $L^1(K)$ and convolutions over $\mathbb{R}$ in general anyway because the support of a convolution grows. It's tricky business that necessitates care.