functional analysis - Convolutions Support.

It is known that:

$$
\operatorname{supp}(u *v) \subset \operatorname{supp}(u) + \operatorname{supp}(v)
$$
Where:
$$
\operatorname{supp}(u) = \overline{\{x \in \mathbb{R}^n: u(x) \neq 0\}}
$$
And:
$$
(u*v)(x)=\int_\limits{\mathbb{R}^n} u(x-y)v(y) dy

$$
I need an example of two functions $u,v$ such that $u$ has compact support, but $u*v$ has no compact support. Does anyone know any examples of this?

Answer

Let $u$ be any non-negative function with compact support which has the value $1$ on some open set. Let $v(x)=e^{-x^{2}}$. Then $(u*v)(x) >0$ for all $x$. So $u*v$ does not have compact support.

For an explicit example let $u(x)=1$ for $0 \leq x \leq 1$, $0$ for $x \geq 1+\frac 1 n$as well as for $x \leq -\frac 1 n$, $u(x)=1+nx$ for $-\frac 1 n \leq x \leq 0$ and $u(x)=1-n(x-1)$ for $1 \leq x \leq 1+\frac 1 n$.

There is no such example where both $u$ and $v$ have compact support. If $u$ has support $K$ and $v$ has support $H$ then $u*v$ vanishes on the complement of $K+H$. Since sum of two compact sets is compact it follows that $u*v$ has compact support.