# Dirac Delta definition in non-standard analysis?

What is the definition of Dirac Delta in non-standard analysis?

I would define it either as a standard distribution with $\sigma=\epsilon$ or maximum equal to $\omega$. Which is the correct answer?

• They mention an article here, which might be worth pursuing. – Omnomnomnom Oct 16 '14 at 23:55

Just like in standard analysis where there are lots of ways to represent the Dirac delta distribution as a limit of a sequence, in nonstandard analysis there are lots of ways to represent the Dirac delta as a nonstandard function.

The simplest is probably to pick an infinite $H$ and set

$$d(x) = \begin{cases} 0 & |x| > \frac{1}{2H} \\ H & |x| \leq \frac{1}{2H} \end{cases}$$

Convolving $d$ with any standard continuous function gives

\begin{align}\int_{-\infty}^{\infty} d(x) f(x) \, dx &= H \int_{-\frac{1}{2H}}^{\frac{1}{2H}} f(x) \, dx \\&= H f(\epsilon) \int_{-\frac{1}{2H}}^{\frac{1}{2H}} 1 \, dx \\&= H f(\epsilon) \frac{1}{H} \\&= f(\epsilon) \\&\approx f(0) \end{align}

where I've used the mean value theorem to go from line one to line two. (one could instead use $|f(x) - f(0)| < e$ for $|x| < \frac{1}{2H}$ if they liked)

• @Anixx: Everything the dirac delta distribution is follows from the formula for convolving with a test function. Some representations might make things more convenient, of course (e.g. the distributional derivative being represented by the derivative of a representative). Incidentally, the dirac delta distribution isn't differentiable either, unless you restrict to test functions that are differentiable at zero. (I'm not sure if you need differentiable eveywhere to make sense of differentiating a distribution) – user14972 Oct 17 '14 at 12:08
• @Anixx: I don't mean to imply it's a bad representation; functions of that shape are particularly useful in the context of Schwartz distributions due to their nice properties with respect to the Fourier transform, so in such contexts it is surely a useful choice of representation. – user14972 Oct 17 '14 at 12:28
• The upshot would seem to be: what defines the Dirac delta are its properties under integration. So we should hardly be surprised if there are (infinitely) many representations in NS analysis any more than in the standard case. But it also doesn't exclude some reps being more useful than others. – Semiclassical Oct 17 '14 at 12:39
• @Anixx: Ack, I did mean $d(x-\epsilon) - d(x)$. The value of an integrand (when it's an ordinary function) at individual points doesn't matter; such differences go away when you take the integral. Sliding the nonstandard part left or right doesn't matter either, although if you really cared about symmetry you could use something like $(d(x - \epsilon) - d(x + \epsilon)) / (2\epsilon)$ – user14972 Oct 17 '14 at 14:36
• Incidentally, there's no reason why the representative of $\delta$ has to have a maximum at zero -- or even have a large value there. Another perfectly good representative is $$\delta(x) = \begin{cases} H & x \in [3/H, 4/H] \\ 0 & x < 3/H \vee x > 4/H \end{cases}$$ If we're in a situation where we get to pick the representative, we can pick what we like, but knowing counterexamples like this is rather important if we are faced with a problem where we are merely given an otherwise unknown function that represents the $\delta$: we can't assume it resembles our favorite choices at all! – user14972 Oct 17 '14 at 14:39

$$\delta(x)=\frac{\omega e^{-(\omega x)^2}}{\sqrt{\pi }}$$
$$\theta(x)=\frac12\operatorname{erf}(\omega x)+1/2$$
$$\delta(0)=\frac{\omega}{\sqrt{\pi}}$$
• I can certainly believe that those work (and they're evidently consistent with $\theta'(x)=\delta(x)$). But to call that the answer would seem to suppose that no other formulations are possible, which surprises me given the multitude of nascent delta functions encountered in practice. Is the situation different in non-standard analysis? – Semiclassical Oct 17 '14 at 2:51
• @Semiclassical this is what is non-standard analysis. And also $\delta(0)=\frac{\omega}{\sqrt{\pi}}$ – Anixx Oct 17 '14 at 2:56