I'm not sure if this calculation is correct and if I interpret it correctly (from old exam).
Show that $ (x-a)^2 \delta ''_a = 2 \delta _a $.
We have for distributions $f$ and test functions $\varphi$ that $\langle\delta_a,\varphi\rangle = \varphi(a)$ and $\langle f', \varphi\rangle = -\langle f, \varphi'\rangle$.
My calculations are as follows:
Let $y = x-a$ and note that $y'=1$, then it follows that
$$\begin{align} \langle y^2\delta_a'',\varphi\rangle & = \langle \delta_a'',y^2\varphi\rangle \\ & =-\langle\delta_a',(y^2\varphi)'\rangle \\ & = -\langle\delta_a',2 y\varphi+y^2\varphi' \rangle
\\ & =\langle\delta_a,(2 y\varphi+y^2\varphi')'\rangle \\ & = \langle\delta_a,2\varphi+2y\varphi' +2y\varphi'+y^2\varphi''\rangle \\ & =2\varphi(a) \\ & =2\langle\delta_a,\varphi\rangle \end{align}$$
Any comments on the calculations?
So I guess that this shows that given the function $y=x-a$, a new distribution can be formed by $y^2 \delta_a''$ which maps test functions $\varphi$ to $2\varphi(a)$, and the distribution $y^2 \delta_a''$ is equal to $2\delta_a$ since they map test functions to the same value.
Best Answer
Since $y'=1$, $$ \begin{align} \left\langle\delta_a,(2\phi y+\phi'y^2)'\right\rangle &=\left\langle\delta_a,2\phi+2y\phi'+\phi''y^2+2y\phi'\right\rangle\\ &=\left\langle\delta_a,2\phi\right\rangle \end{align} $$ since $\langle\delta_a,y\psi(y)\rangle=0$ for any $\psi$.
So, yes, your answer seems correct.