# show that $\delta''(\frac13x^3 +x) = \delta''(x) - 2\delta(x)$

I tried to go about about this using the definition : let $$\phi$$ be a test function

\begin{align}\langle \delta''(\frac13x^3 +x) ,\phi\rangle &= \langle \delta''(\frac13x^3 +x) ,\frac{x^2+1}{x^2+1}\phi\rangle = \langle (x^2+1)\delta''(\frac13x^3 +x) ,\frac{1}{x^2+1}\phi\rangle \\ & = \langle (\delta'(\frac13x^3 +x))' ,\frac{1}{x^2+1}\phi\rangle = - \langle \delta'(\frac13x^3 +x) ,\frac{1}{x^2+1}\phi' - \frac{2x}{(x^2+1)^2}\phi\rangle \\ & = - \langle (\delta(\frac13x^3 +x))' ,\frac{1}{(x^2+1)^2}\phi' - \frac{2x}{(x^2+1)^3}\phi\rangle \\ & = \langle \delta (\frac13x^3 +x) ,\frac{1}{(x^2+1)^2}\phi'' - \frac{2x-4x(x^2+1)^2}{(x^2+1)^3}\phi' - \frac{2(x^2+1)^3-6x(x^2+1)^2}{(x^2+1)^3}\phi\rangle \end{align}

now using that $$\langle \delta(F(x)) ,\phi\rangle = \sum_{x_i \in \ker F} \frac{\phi(x_i)}{|F'(x_i)|}$$

$$F'(x) = x^2+1,\,x_i =0$$

then

$$\langle \delta''(\frac13x^3 +x) ,\phi\rangle = \phi''(0)-2\phi(0) = \langle \delta''(x) -2 \delta(x) ,\phi\rangle$$

question number 1 : did I do anything not permitted ?

question number 2 : is there some some other ways to solve this ? maybe with less computations ?

• Your calculation looks good. I cannot see anything that is not permitted. I have solved it in another way, but it doesn't really lead to less computations. – md2perpe Jan 28 '19 at 17:36