Showing that if $fT=0$, then $\text{supp }T\subset\{x:f(x)=0\}$

Question

Let $T$ be a distribution and $f$ be a smooth function. If $fT=0$, then for test functions $\phi$ we have

$$\langle fT,\phi\rangle=\langle T,f\phi\rangle=0.$$

We know that $f\phi$ is a test function, so that its support is contained in a compact set. By definition, the support of $T$ is the smallest closed set $E$ such that $\langle T,f\phi\rangle$ depends only on the values of $f\phi$ and all its derivatives on $E$.

We also know that $\{x:f(x)=0\}$ is closed. However, I don't know if we can deduce from this that $E\subset\{x:f(x)=0\}$?

Answer 1

The more civilized definition of $\operatorname{supp} T$ is that it's the complement of the largest open set $U$ such that $\langle T,\phi\rangle =0$ for any test function $\phi$ with $\operatorname{supp} \phi\subset U$.

With this definition, the proof is easy: let $\phi$ be a test function with support contained in $\{x:f(x)\ne 0\}$. Then $\phi/f$ is also a test function. Hence, $$\langle T, \phi\rangle = \langle fT, \phi/T\rangle = 0$$