What is the Product of $\delta$ functions with itself? was already asked some time ago. In a comment the OP states:
I want to create $\delta(t)$ such that the product with itself is also $\delta(t)$ and the innerproduct $\int_{-\infty}^\infty \delta(t)\cdot\delta(t) dt = 1$ and retain all other properties of the usual $\delta(t)$.
I feel that I'm in a comparable situation: I'd like to have a function $\Delta(x)$, such that
- $\displaystyle\int_{-\infty}^\infty\Delta(x)^2 f(x) dx=\int_{-\infty}^\infty\delta(x)f(x)dx$ and
- $\displaystyle\int_{-\infty}^\infty\Delta(x-t)\Delta(x+t)f(x)dx=0$ for $t\neq 0$.
Combining these gives: $ \int_{-\infty}^\infty\Delta(x-t)\Delta(x+t)f(x)dx=\delta_{t0}\int_{-\infty}^\infty\delta(x)f(x)dx, $ with $\delta_{xy}$ being the Kronecker Delta. I think that following could work:
Let $\Delta_a(x\pm t)=\frac{1}{a \sqrt{\pi}} \mathrm{e}^{-(x\pm t)^2/\color{red}{2}a^2}$ (little related to the so called nascent delta function). The $\color{red}{2}$ in the denominator inspired the title and I hope it was not misleading you, sorry if so.
When I now, for example, set $f(x)=1$ and calculate $ \lim_{a\to 0}\int_{-\infty}^\infty\Delta_a(x-t)\Delta_a(x+t)dx, $ I get exactly what I need. So my question is: Is this OK, and if not can I save it somehow?