During a lecture, we were told that for two distributions $S,T \in \mathcal{D'}(\mathbb{R})$ the convolution $S* T$ is defined if either $S$ or $T$ had a compact support, or, if both of them were tempered distributions.
With that said, the Heaviside distribution does not appear to have a compact support, nor is it a tempered distribution. Meaning that $H * H$, where $H$ denotes the Heaviside functional, is not well defined.
Yet, our professor asked us to prove that the second derivative of $H * H$ is equal to Dirac's delta.
$$
(H * H)'' = \delta_0
$$
How is this possible?
Best Answer
Clearly $H$ is locally integrable. It constitutes, therefore, a Tempered Distribution and for any $\phi\in C_C^\infty$, we have
$$\langle H,\phi\rangle =\int_0^\infty \phi(x)\,dx$$
The convolution $H*H$ is also a distribution such that for any $\phi\in C_C^\infty$ we have
$$\begin{align} \langle H*H,\phi\rangle &= \langle H, H*(d_{-1} \phi)\rangle\\\\ &=\int_{-\infty}^\infty H(x) \int_{-\infty}^\infty H(y)\phi(x+y)\,dy\,dx\\\\ &=\int_0^\infty \int_0^\infty \phi(x+y)\,dy\,dx\\\\ &=\int_0^\infty \int_x^\infty \phi(y)\,dy\,dx\\\\ &=\int_0^\infty \phi(y)\int_0^y\,dx\,dy\\\\ &=\int_0^\infty y\phi(y)\,dy\\\\ &=\int_{-\infty}^\infty yH(y)\phi(y)\,dy\tag1 \end{align}$$
Therefore, we see from $(1)$ that in distribution
$$(H*H)(x)=xH(x)\tag2$$.
Finally, using $(2)$ we see that the second distributional derivative $(H*H)''$ of $H*H$ satisfies
$$\begin{align} \langle (H*H)'', \phi\rangle&=\langle H*H, \phi''\rangle\\\\ &=\int_{-\infty}^\infty xH(x) \phi''(x)\,dx\\\\ &=\int_0^\infty x\phi''(x)\,dx\\\\ &=\phi(0) \end{align}$$
so that in distribution we have
$$(H*H)''(x)=\delta(x)$$
as was to be shown!
Note that we could have derived $(2)$ directly by writing
$$\begin{align} \int_{-\infty}^\infty H(x')H(x-x')\,dx'&=\int_0^{xH(x)} \,dx'\\\\ &=xH(x) \end{align}$$