# Does $\sqrt{\delta^2}$ make more sense than $\delta^2$?

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

1. $\displaystyle\int_{-\infty}^\infty\Delta(x)^2 f(x) dx=\int_{-\infty}^\infty\delta(x)f(x)dx$ and
2. $\displaystyle\int_{-\infty}^\infty\Delta(x-t)\Delta(x+t)f(x)dx=0$ 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?

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Have you thought about defining $\Delta(x) = \lim_{h\rightarrow 0} 1_{[-h,h]}(x)/\sqrt{2h}$? (where $1_A(x)$ is the indicator function on $A$) – Chris Taylor May 23 '12 at 13:45
@ChrisTaylor Indeed, I have. Would this be a better choice? – draks ... May 23 '12 at 13:46
I have no idea, but it makes the integrals easier. – Chris Taylor May 23 '12 at 13:48
@draks: Are you saying you define for $\varphi \in D$: $$\Delta(\varphi) := \lim_{a\to 0}\int_{-\infty}^\infty\int_{-\infty}^\infty\Delta_a(x-t)\Delta_a(x+t) \varphi(t) dt dx,$$ – Vobo May 24 '12 at 7:27
@Vobo is integration over $t$ needed, since the only contribution will be for $t=0$. Then it boils down to $\int_{-\infty}^\infty\delta(x)f(x)dx=f(0)$. But maybe I didn't get you right. My idea was just to get an analogy for $\delta(x)$ the has some multiplicative properities, thet $\delta$ lacks, see the linked question and discussions therein. – draks ... May 24 '12 at 8:23