# Why is the square of Dirac delta function not a distribution?

This is the exercise from Strauss's "Partial Differential Equations", chapter 12.1.

Show that no sense can be made of the square $[\delta(x)]^2$ as a distribution.

To show this, I found a function $S(x,t)= \frac {1}{\sqrt{4\pi t}}e^{-x^2/4t}$ which converges weakly to $\delta(x)$ as $t$ goes to zero.

If $[\delta(x)]^2$ had a meaning, then $S(x,t)^2$ must converge to that as $t$ goes to zero, but intuitively, I guessed that the integral $\int_{-\infty}^{\infty}S(x,t)^2\phi(x)dx$ goes to zero as $t$ approaches to zero when $t>0$, but how can I prove this rigorously in the range of the book containing this exercise? Is there any other way of approach?

• See this mathoverflow post. – Mark Apr 6 '17 at 20:40
• A distribution $T$ is a map $C^\infty_c(\Bbb R)\to\Bbb C$. What sense can you give to $T^2$, given that the codomain of $T$ does not contain its domain? – TZakrevskiy Apr 10 '17 at 13:12

If $T$ is a distribution and $\phi \in C^\infty_c, \int_\mathbb{R} \phi(x)dx = 1, \phi_n(x) = n \phi(nx)$ then $T_n = T \ast \phi_n$ converges to $T$ in the sense of distributions.
We want to know if $S_n(x) = T_n(x)^2$ converges to some distribution.
When $T = \delta$ it is a distribution of order $0$ so instead of $\phi \in C^\infty_c$ we can take $\phi(x) = 1_{x \in [1/2,1/2]}$ so that $S_n(x) = n^2 \, 1_{x \in [1/(2n),1/(2n)]}$ and clearly it diverges as $n \to \infty$ in the sense of distributions.
Using the same kind of argument, you get that for any $\phi\in C^\infty_c, \int_\mathbb{R} \phi(x)dx = 1$ :
$(\delta \ast \phi_n)^2= \phi_n^2$ diverges in the sense of distributions and $\delta^2$ isn't well-defined.