The Dirac delta "function" is often introduced as a limit of normal distributions $$\delta_a(x)=\frac{1}{a\sqrt{\pi}}e^{-\frac{x^2}{a^2}}\text{ as }a\to0^+.$$ Obviously, this sequence of functions converges to $0$ when $x\neq0$ and diverges otherwise. As far as I know, what is literally meant is that $$\int_{\mathbb{R}}f(x)\delta(x)dx\text{ is defined as }\lim_{a\to0^+}\int_{\mathbb{R}}f(x)\frac{1}{a\sqrt{\pi}}e^{-\frac{x^2}{a^2}}dx\text{, when $f$ is well behaved.}$$ I've read the wikipedia page on distributions and it was helpful. My question is this:

The Dirac delta acts on well behaved functions as the evaluation at zero map (and can be tweaked to evaluate at any point). I understand the intuitive motivation for using the (abuse of) notation "$\int_{\mathbb{R}}f(x)\delta(x) dx$" to represent a functional. Are there specific instances in mathematical analysis (broadly understood) where one needs to use the definition as "$\lim_{a\to0^+}\int_{\mathbb{R}}f(x)\frac{1}{a\sqrt{\pi}}e^{-\frac{x^2}{a^2}}dx$", and thinking of the Dirac delta as an evaluation map is insufficient?

  • 3
    $\begingroup$ $\displaystyle \int_{-\infty}^\infty f(x)\delta'(x)\,dx = -f'(0)$. How does that identity follow from simply saying $\delta$ is an evaluation map? ${}\qquad{}$ $\endgroup$ – Michael Hardy Oct 26 '14 at 5:40
  • 1
    $\begingroup$ @MarianoSuárez-Alvarez Ok, so the distributional definition is defined on the space of test functions. The integral definition makes sense for the space of Riemann integrable functions. These spaces of functions do intersect non-trivially. Call whichever you want "the" Dirac delta, but the question is still well formed as stated. $\endgroup$ – J. David Taylor Oct 26 '14 at 5:41
  • 3
    $\begingroup$ @MichaelHardy I suspect that it follows from taking the distributional derivative. $\endgroup$ – J. David Taylor Oct 26 '14 at 5:43
  • 2
    $\begingroup$ @MichaelHardy, that follows immediately from the definition of derivatives of distributions, using the fact that the delta «function» is the evaluation. Any textbook treating the subject explains this. $\endgroup$ – Mariano Suárez-Álvarez Oct 26 '14 at 5:44
  • 1
    $\begingroup$ In particular, that integral is not an integral. $\endgroup$ – Mariano Suárez-Álvarez Oct 26 '14 at 5:45

The delta function, as a generalised function, is the Fourier transform of the generalised function $1$,

$$ \int_\mathbb{R} 1 e^{-2\pi i kx} dx = \delta(k) \qquad \int_\mathbb{R} \delta(k) e^{2\pi i xk} dk =1 ,$$

this intuitively captures the fact that a wave localized in space is not localized in frequency and vice versa. I would have a hard time gaining that insight by thinking of the delta function as an evaluation map on a function space.

I agree that naively the notation suggests that these identities are impossible. You will notice that I was careful to say "the generalised function 1" not "the real number 1"; the distinction makes all the difference in the world.

The generalised funciton $1$ is defined by the sequence of functions $\exp(-x^2/n^2)$. By definition the Fourier transform of a generalized function, $f(x)$, is the generalized function $g(k)$ defined by the sequence,

$$ g_n(k) = \int_\mathbb{R} e^{-2\pi i k x } f_n(x) dx $$

For the generalised function $1$ we have,

$$ g_n(k) = \int_\mathbb{R} e^{-2\pi i k x } e^{-x^2/n^2} dx $$

Evaluating this integral would produce a sequence of functions that are equivalent to the sequence defining the delta function.

  • $\begingroup$ The second equation matches perfectly with the evaluation at zero map definition. $\int_{\mathbb{R}}1e^{2\pi ikx}dx$ doesn't converge if $k=0$. As such, how do you interpret it in a way that is not equivalent to one of the two definitions I gave in my question? While the evaluation at zero map does not capture the intuition you have, I find the linear-algebra-geometric interpretation of harmonic analysis to be quite handy. $\endgroup$ – J. David Taylor Oct 26 '14 at 6:08
  • $\begingroup$ I mean $\int_{\mathbb{R}}1e^{-2\pi ikx}dx$. $\endgroup$ – J. David Taylor Oct 26 '14 at 6:09
  • 1
    $\begingroup$ I don't disagree that thinking of the delta function as an evaluation map is sufficient. I think that having multiple (rigorous) interpretations of the delta function can be helpful. I think this is similar to the idea of a tangent space in differential geometry, I know at least 4 equivalent definitions of the tangent space of a point $p$ in a manifold, but they all evoke completely different images of whats going on (for me at least). $\endgroup$ – Spencer Oct 26 '14 at 6:28
  • $\begingroup$ Point taken, and it is not immediately obvious to me how to interpret the first transform outside of what you stated (I'm racking my brain). But, do I correctly understand that you are not arguing that there are contexts to prefer the sequence of functions definition of the Dirac delta? $\endgroup$ – J. David Taylor Oct 26 '14 at 6:33
  • $\begingroup$ I'm not sure how far the theory of generalised functions has been developed. I know that in elementary situations like this one it is a good minimalist rigorous theory of delta functions. There may be contexts in which it hasn't even been developed since the theory of distributions is already there. The point I'm making here is that when they are equivalent there may be reasons of personal preference to use one over another since different formalisms can provide different insights. Obviously I can't make a sweeping statement that one is always better than another unless there is an objective- $\endgroup$ – Spencer Oct 26 '14 at 6:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.