Convolution of a compactly supported $f \in L^1(\mathbb{R}^n)$ with a function $g \in L^1_{\text{loc}}(\mathbb{R}^n)$

convolutionintegrationlp-spacesreal-analysis

Suppose $f$ and $g$ are two $L^1_{\text{loc}}(\mathbb{R}^n)$-functions. Their convolution is well-defined under many conditions, three of which are:

  1. When $f$ is compactly supported and continuous, and $g$ is just $L^1_{\text{loc}}(\mathbb{R}^n)$.
  2. When $f$ and $g$ are in $L^1(\mathbb{R}^n).$
  3. When $f$ and $g$ are in $L^2(\mathbb{R}^n).$

There are many more. However, I'm wondering whether the following is true:

If $f,g \in L^1_{\text{loc}}(\mathbb{R}^n)$ are such that $f$ is compactly supported (thus also in $L^1(\mathbb{R}^n)$), then $f * g$ is well-defined almost everywhere. Also, do I then have $f * g \in L^1_{\text{loc}}(\mathbb{R}^n)$?

The proof of $1.$ doesn't go through: it's based on the fact that $f$ is bounded, and I don't have that if $f$ is not continuous. The proof of $2.$ that I've found goes along the following lines: let $H(x,y) := f(x-y)g(y).$ Then $$\int_{\mathbb{R}^{2n}} |H(x,y)|dxdy = \int_{\mathbb{R}^n} \left( \int_{\mathbb{R}^n} |f(x-y)|dx \right) |g(y)| dy = ||f||_1 \cdot ||g||_1, $$so by Fubini, the integral $\int_{\mathbb{R}^n} f(x-y)g(y)dy$ exists and is integrable for a.e. $y \in \mathbb{R}^n.$ However, in my case, $H$ is not necessarily integrable.

It's hard to think of a counterexample, because we'd need $f$ to be so discontinuous as to ruin the local integrability of $g$ on a set of positive measure! I can easily do it at one point, (for example $f(y) = g(y) = \frac{1}{\sqrt{y}}$ at $x = 0$), or even at a countable set of points (think of infinitely many "volcanoes" of the above type converging to zero), but doing it for a large enough set of points seems very hard to me.

Best Answer

(Hoping I didn't make a mistake here..) If $f$ has compact support and is locally integrable, then $f$ is actually integrable (take compact $K$ such that $\operatorname{supp}(f)\subseteq K$) which will make $f*g$ locally integrable. Let $K$ be compact, then

$$\int_K |(f*g)(x)|\,dx = \int_K \bigg|\int_{\operatorname{supp}(f)}f(y)g(x-y)\,dy\bigg|\,dx \le \int_K \int_{\operatorname{supp}(f)}|f(y)||g(x-y)|\,dy\,dx.$$

By Fubini-Tonelli, we can interchange these even if it diverges to infinity, so

$$\int_K |(f*g)(x)|\,dx \le \int_{\operatorname{supp}(f)}|f(y)|\int_K |g(x-y)|\,dx\,dy.$$

The $x$ integral is finite since $g$ is locally integrable and $K+y$ is also a compact set. The $y$ integral is finite since $f$ is actually just straight up integrable.

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