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I understand that, in $L^2$ Hilbert space, functions do not exist pointwise, so if function $\phi(x) \in L^2$, we cannot speak of a specific value $\phi (0)$, at $x=0$, and so we cannot say $\int_{- \infty}^{\infty} \phi(x) \delta (x) dx = \phi (0)$.

But if we look at the space of functions, $\psi(x)$ (with $x \in R$), that both exist pointwise (so that $\psi(x)$ has a specifically defined value for any $-\infty \leq x \leq \infty$) and are square integrable from $-\infty$ to $\infty$, then does it make sense to let $\delta (x)$ act on those such that we can say: $\int_{- \infty}^{\infty} \psi(x) \delta (x) dx = \psi (0)$? I know that this simple definition of $\psi(x)$ does not guarantee the existence of the derivative of the delta function in this space but, that not withstanding, I think that $\int_{- \infty}^{\infty} \psi(x) \delta (x) dx = \psi (0)$ would still be true for those $\psi(x)$, especially if we define $$\int_{- \infty}^{\infty} \psi(x) \delta (x) dx \equiv \lim_{n\to\infty} \int_{- \infty}^{\infty} \psi(x) \delta_n (x) dx$$ with something like $\delta_n (x) \equiv \sqrt{\frac{n}{\pi}} \exp (-nx^2)$

What I am getting at is that it seems to me that the space of test functions for $\delta (x)$ need not be something as restrictive as Schwartz (S) or Kantorovich (K) space.

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  • $\begingroup$ The Schwartz space is the largest space such that : the distributions with compact support and their Fourier transform are continuous when acting on it. If you restrict to distributions of order $0$ as the $\delta$ and forget about the Fourier tansform, then you can consider a larger space, for example the space of piecewise continuous functions $\endgroup$ – reuns Apr 18 '17 at 2:17
  • $\begingroup$ @ user1952009. Thanks. But $FT(\delta(x-t) ) =\int_{-\infty}^{\infty} \delta(x-t) exp(-ipx) dx=exp(-ipt)$. When $exp(-ipt)$ is a functional acting on some function space, it can still do so in a continuous manner without that function space having to be as restrictive as Schwartz space, I think. $\endgroup$ – David Apr 18 '17 at 18:18
  • $\begingroup$ For instance, if the test function space is that of absolutely integrable functions then, I believe, FTs are continuous in that space. Therefore, by the definition of FT of a distribution, FTs of distributions that act on the space of absolutely integrable functions are also continuous. Please let me know if I am missing something. $\endgroup$ – David Apr 19 '17 at 0:42
  • $\begingroup$ I don't see what you mean. If you want to keep the FT, then look at $L^1 \cap FT[L^1]$ which is a space of continuous and $L^1$ functions whose FT is continuous and $L^1$. Hence both $\langle \delta,\varphi \rangle$ and $\langle 1,\varphi \rangle$ are well-defined for $\varphi$ in that space. $\endgroup$ – reuns Apr 19 '17 at 0:49
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Your example fails; consider $\psi$ given by

$$ \psi(x) = \begin{cases} 1 & x = 0 \\ 0 & x \neq 0 \end{cases} $$

$\psi$ is square integrable,

$$ \psi(0) \neq 0 = \lim_{n \to \infty} \int_{-\infty}^{\infty} \psi(x) \delta_n(x) \, \mathrm{d} x $$

so in the dual space to square-integrable functions, $\delta \neq \lim_{n \to \infty} \delta_n$.


The most reasonable general space to find delta distributions is, in my opinion, in the dual space to the space of continuous functions, if you simply want them to exist.


But you should pay attention to the fact that, usually, the point isn't simply to define delta distributions — instead one wants to define a space that:

  • Contains the objects of interest
  • Admits all of the sorts of calculations we want in our application
  • Has good analytical properties

The second point is very important. For example, in physics, Fourier transforms are very important, which makes the space of tempered distributions (on Schwartz functions) a popular one to work within.

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