If Dirac Delta is not a function, then what is this?

Question

Is$$f(x)=\lim_{a\to\infty}\frac{e^{-(x/a)^2}}{a\sqrt{\pi}}$$ a function or not? If not, why not?

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Sorry if the question is too beginner-ish

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Well, you are right, I should have written $a\to 0$

Answer 1

What you wrote is $f(x) = 0$.

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Now, since you talk about dirac deltas, you probably meant to take the limit as $a \to 0^+$.

In that case, the limit depends on fine details of what you actually mean by the notation.

If you mean this as a limit of functions — i.e. the sort of thing you learn in introductory calculus — then it's pretty clear that $f(x) = 0$ whenever $x \neq 0$, and depending on your feelings about extended real numbers, either $f(0) = +\infty$ or $f(0)$ is undefined.

Either way, the limit has nothing to do with dirac deltas.

Now, if you instead mean this as a limit of distributions (which is pretty much always meant in a context where dirac deltas are involved, despite the fact it's pretty rare to let someone in on this secret who isn't explicitly studying the mathematical foundations of this sort of thing), the limit is indeed a dirac delta.

For the sake of precision, I will add brackets around a function when I meant to think of it as a distribution rather than a function. The way limits of distributions work is that

$$ \lim_{a \to 0^+} \left[ \frac{e^{-(x/a)^2}}{a\sqrt{\pi}} \right] $$

is the distribution satisfying, for any test function $f(x)$,

$$ \int_{-\infty}^{\infty} \left( \lim_{a \to 0^+} \left[ \frac{e^{-(x/a)^2}}{a\sqrt{\pi}} \right] \right) f(x) \mathrm{d}x = \lim_{a \to 0^+} \int_{-\infty}^{\infty} \frac{e^{-(x/a)^2}}{a\sqrt{\pi}} f(x) \mathrm{d}x $$

It's not too difficult to show that, for any test function,

$$\lim_{a \to 0^+} \int_{-\infty}^{\infty} \frac{e^{-(x/a)^2}}{a\sqrt{\pi}} f(x) \mathrm{d}x = f(0) $$

and consequently

$$\lim_{a \to 0^+} \left[ \frac{e^{-(x/a)^2}}{a\sqrt{\pi}} \right] = \delta(x) $$

Now, one thing I want to strongly emphasize is that it's also similarly easy to show

$$ \int_{-\infty}^{\infty} \left( \lim_{a \to 0^+} \frac{e^{-(x/a)^2}}{a\sqrt{\pi}} \right) f(x) \mathrm{d}x = 0$$

In particular, you cannot equate the two limits:

$$ \lim_{a \to 0^+} \left[ \frac{e^{-(x/a)^2}}{a\sqrt{\pi}} \right] \neq \lim_{a \to 0^+} \frac{e^{-(x/a)^2}}{a\sqrt{\pi}} $$

Answer 2

What you've written down is an expression. It defines the function as $f(x)=0$ for $x\in\mathbb R - \{0\}$ and the limit is $\infty$ at $x=0$ (this is assuming, as a commenter suggested, that you meant to write the limit $a\to 0^+$. The $+$ is important or it would be undefined rather than infinite.) You can look at this as a real-valued function that is only defined on $\mathbb R-\{0\}$ (the same way you would typically look at the expression $f(x) = 1/x$) or you can look at it as a function with domain $\mathbb R$ whose range is the extended reals. These two (predominantly the first one) are what I would consider the literal interpretation of what you wrote down (again, provided you change the limit to $a\to 0^+$.)

However, what you wrote down is related to the Dirac delta, and in a fuzzy sense can be interpreted as such. But note that a Dirac delta function is not merely a function that takes the value zero for $x\ne 0$ and $\infty $ when $x=0.$ There's that matter of the integral being $1$ as well, which is clearly not the case here (the integral is undefined). So your expression is not the Dirac delta. As you have been correctly informed, the delta function is a distribution, not a function.

The issue here is an order of limits. You may have heard an expression like "the delta function lives under the integral." In other words they don't aren't functions as stand-alone objects but you can integrate functions against them (as you could integrate a function against another function). In your expression, you took the limit too early. What you want to do is put the bulk of your expression under an integral but keep the limit outside the integral like so: $$ \lim_{a\to 0^+} \int_{-\infty}^\infty\phi(x) \frac{e^{-x^2/a^2}}{a\sqrt{\pi}}dx$$ where $\phi(x)$ is some nicely behaved function (called a test function in the terminology of distributions). Then we can do a substitution to turn this into $$ \lim_{a\to 0^+} \int_{-\infty}^\infty\phi(ax) \frac{e^{-x^2}}{\sqrt{\pi}}dx.$$ Now we assume the test function is nice enough so that we can pass the limit inside the integral to get $$ \phi(0)\int_{-\infty}^\infty \frac{e^{-x^2}}{\sqrt{\pi}}dx = \phi(0).$$

So we see that when we modify things so the limit starts outside the integral we get the fundamental property of the delta function that $$\int \phi(x) \delta(x) dx = \phi(0).$$

This is a nice heuristic way of looking at things, but the whole business of putting the limit outside the integral sign is a bit cumbersome for a formal theory. What's done instead is that we consider "integrating against the delta function" to be an operation we can do on test functions. More specifically, it's considered as a linear map from the space of test functions to the real numbers (If you've had linear algebra, it's a dual vector.) What this map does is simple: it takes $\phi$ and returns $\phi(0).$

Note that it is "integrating against the delta function" that is the linear map, not the delta function itself. If you've seen dual spaces before, recall that the linear maps $\mathbb R^n\to \mathbb R$ are a vector space in $n$ dimensions and we can think each such map as being represented by a vector $w$, where the corresponding linear map takes $v\to \langle w,v\rangle.$ The delta function is analogous to $w$.