Given the following definitions, taken from Wikipedia:
Support of a function. When $X$ is a topological space and $f : X \to \mathbb C$ is a continuous function, the support of $f$ is defined topologically as the closure of the subset of $X$ where $f$ is non-zero.
${\displaystyle \operatorname {supp} (f):={\overline {\{x\in X\,|\,f(x)\neq 0\}}}={\overline {f^{-1}\left(\left\{0\right\}^{c}\right)}}.}$
Support of a distribution. Suppose that $u$ is a distribution and that $U$ is an open set in Euclidean space such that, for all test functions $f$ such that ${\rm supp}\ f \subseteq U$, ${\displaystyle u(f)=0}$. Then $u$ is said to vanish on $U$.
Now, if $u$ vanishes on an arbitrary family $U_{\alpha}$ of open sets, then for any test function $f$ supported in $\bigcup \limits_\alpha U_{\alpha }$, (…) $u(f)=0$ as well.
Hence we can define the support of $u$ as the complement of the largest open set on which $u$ vanishes.
Question. For a given function $g(x)$ and a distribution $u(f)$, assuming $g, f \in {\cal D}(\mathbb R^n)$, how can we define the support of the product $g(x)\ u(f)$? ${\rm supp}\ g \ \cap {\rm supp} \ u $?
Best Answer
Say $U$ is the largest open set in which $u$ vanishes; let $O$ be the complement of the support of $g$, so $O=(g^{-1}(0))^o$.
The main step is
Proof: Say $\phi\in C^\infty_c(U\cup O)$. Say $\psi_1,\psi_2$ is a partition of unity: $0\le\psi_j\le 1$, $supp(\psi_1)\subset U$, $supp(\psi_2)\subset O$, and $\psi_1+\psi_2=1$. Let $\phi_j=g\psi_j$. Then $\phi_1$ is supported in $U$, so $\phi_1g=0$, hence by definition $gu(\phi)=u(\phi_1g)=0$. Similarly $gu(\phi_2)=0$, since $g\phi_2$ is supported in $O$ and $u$ vanishes in $O$. So $gu(\phi)=0+0=0$.
So now you just have to figure out why that's the largest open set in which $gu$ vanishes, and then see what that is qed. having just determined that we didn't learn the definition of $gu$ before asking the question I'll be stopping here.