I have to solve the equation $(x-a)T=0$ , T is a distribution. By definition : $(x-a)\int T(x)\varnothing (x)=0$
I know if I pose $X=x-a$ I find $XT(X)=0$ and $T(X)=\delta(X)$.
But I stuck to find the solution with the definition of the distribution.
Think for any help.
To solve $yS=0$ for distributions, $S=c\delta$ for any scalar $c$ is the general solution. It is a solution, as $(yS)(\varphi(y))=c\delta(y\varphi(y))=0$. And every solution is of this form, as the support of $S$ is at most $\{0\}$, so $S$ is a linear combination of derivatives of $\delta$'s.
Now you pose $x-a=y$ and have the general solution $T=c\delta_a$, where $\delta_a$ is the evaluation functional at the point $a$.