Convolutions Support.

Question

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 1

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.

Answer 2

Consider any function of compact support $u$ (with non empty interior) such that $$ u \ge 1\quad \forall x\in \operatorname{int}(\operatorname{supp}(u)) $$ where $\operatorname{int}$ is the interior operation. Then by choosing $$ v(x)= \begin{cases} u(x) & x\in \operatorname{supp}(u)\\ e^{-\Vert x\Vert^2}& x\in\Bbb R^n\setminus \operatorname{supp}(u) \end{cases} $$ we have the required example, since the convolution is such that $$ \begin{split} u\ast v(x)&=\int\limits_{\Bbb R^n}u(x-y)v(y)\mathrm{d}y\\ &= \int\limits_{\operatorname{supp}(u)}u(x-y)u(y)\mathrm{d}y+\int\limits_{\Bbb R^n\setminus\operatorname{supp}(u)}\!\!u(x-y)e^{-\Vert y\Vert^2}\mathrm{d}y\\ &\ge \int\limits_{\operatorname{supp}(u)}\mathrm{d}y+ \int\limits_{\Bbb R^n\setminus\operatorname{supp}(u)}\!\!e^{-\Vert y\Vert^2}\mathrm{d}y\\ &=\operatorname{\mu}(\operatorname{supp}(u))+\!\!\!\int\limits_{\Bbb R^n\setminus\operatorname{supp}(u)}\!\!e^{-\Vert y\Vert^2}\mathrm{d}y>0\quad\forall x\in \Bbb R^n \end{split} $$ where $\mu(A)$ is the Lebesgue measure of the set $A$.