# Show that if $xT=xS$, then $T=S+c\delta$

Let $T$ and $S$ denote distributions such that their action on some test function $\phi$ is denoted by $\langle T, \phi \rangle$ and $\langle S, \phi\rangle$.

I want to show that if $xT=xS$, then $T=S+c\delta$, where $\delta$ denotes the Dirac delta distribution and $c$ is some constant.

I know that $x\delta=0$, which looks like it may be useful at some point in this question.

How do I begin?

Hint $xS=xT$ then $x(T-S)=0$.
Hint 2 Show that if $xu=0$ then $\mbox{supp}(u) \subset \{ 0\}$.
Hint 3 Conclude that $u$ has finite order, and hence $$u=\sum_{k=0^n}c_kD^k$$
Hint 4 Show that $xu=0$ implies $c_1=..c_n=0$.