Vtyjkyuk

Show that if $F$ is a distribution, such that $|xi|^2F=0$, then support (F) is subset of $0$.

My approach: Recall that, the support of a distribution $F$ is the complement of the union of all open sets $Uin mathbbR^n$ such that $F(varphi) = 0$ (for $varphiin mathcalD_K$ with compact $Ksubset U$).

So, if $F$ is a distribution such that $|xi|^2F=0$, then $$|xi|^2F(varphi)=int_mathbbR^n^2f(x)varphi(x)dx =0$$

Wlog, we assume that $Fneq 0$ and the support of $varphi$ is not in $0$. Then, the unique way that the above integral will be $0$, is that in $xi=0$ we have $f(xi)=0$.

In your proof you are assuming that $F$ is a regular distribution, i.e. given by a function. This is not necessarily true.

Hint: If $phi varphiin mathcalD_K$ is such that $operatornamesupp(f) subset mathbb R^d backslash 0$ then show that there exists some $psi in mathcalD$ such that
$$varphi=|xi|^2 psi$$

Then
$$F(varphi)=|xi|^2F(psi)$$