The convolution of two compactly supported functions is also compactly supported

376 Views Asked by At

If $f,g \in C_{c}(\mathbb{R})$, then we know that $\overline{\text{supp}(f) + \text{supp}(g)}$ is compact and in general, $\text{supp}(f \ast g) \subset \overline{\text{supp}(f) + \text{supp}(g)}$. It seems to me that this amounts to showing that $\text{supp}(f \ast g)$ is closed.

I thought to consider a convergent sequence $\left\{x_{n}\right\}$ in $\text{supp}(f \ast g)$ and show that its limit is also in $\text{supp}(f \ast g)$, but I'm stuck.

1

There are 1 best solutions below

1
On BEST ANSWER

In most definitions, the support of a function $f$ is defined as the closure of all points, where $f\neq 0$. In this way, the support is always a closed set. In fact to the contrary, for $f$ continuous, the set $\{f\neq 0\}$ is always open. The important property here is the boundedness, which you already mentioned in your first paragraph.