Support of a convolution

Remarks

If $(f * g)(x) = \int f(y) g(x - y) \, dy \ne 0$, then $f(y) g(x - y) \ne 0$ for some $y$. So $y \in \operatorname{supp} f$ and $x - y \in \operatorname{supp} g$, and $x = y + (x - y)$ lies in the sumset $\operatorname{supp} f + \operatorname{supp} g$. This sumset is the image of the compact set $\operatorname{supp} f \times \operatorname{supp} g$ under the continuous map $(a, b) \mapsto a + b$, so it is compact and hence closed, and it contains the closure of $\set{f * g \ne 0}$.

Example: if $f$ and $g$ are continuous, positive on $(0, 1)$ and zero elsewhere, then $f * g$ is positive exactly on $(0, 2)$, and $\operatorname{supp}(f * g) = [0, 2] = [0, 1] + [0, 1]$.

The inclusion can be strict, because of cancellation in the integral. Let $\varphi$ be continuous, positive on $(0, 1)$ and zero elsewhere, and put $\psi = \varphi * \varphi$, supported on $[0, 2]$. For $f(x) = \varphi(x) - \varphi(x - 2)$ and $g(x) = \varphi(x) + \varphi(x - 2)$ the cross terms cancel and $f * g(x) = \psi(x) - \psi(x - 4)$. So $\operatorname{supp}(f * g) = [0, 2] \cup [4, 6]$, while $\operatorname{supp} f + \operatorname{supp} g = [0, 6]$. For integrable functions the same statement holds with essential supports, in the form $\operatorname{supp}(f * g) \subseteq \overline{\operatorname{supp} f + \operatorname{supp} g}$.

Going deeper

For a student

This is why convolution with a mollifier supported in a ball of radius $\varepsilon$ moves the support of a function by at most $\varepsilon$. Together with smoothing by convolution it shows that smooth functions with compact support are dense in $L^p$ for $1 \le p < \infty$. On the line, the Titchmarsh convolution theorem gives an equality for the convex hulls of the supports.