Is there some sense in which one could write any distribution as a sum of this sort?
$$A(x,y)=\sum_{n=0}^{\infty}a_n(x)i^n\frac{\partial^n}{\partial x^n}\delta (x-y)$$
Provided that the rhs acting on a test function is convergent for all $x$.
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3$\begingroup$ This question would get a more sympathetic reception on MathStackExchange. $\endgroup$– paul garrettAug 29, 2019 at 17:21
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1$\begingroup$ A closely related question: MO154814 $\endgroup$– Igor KhavkineAug 30, 2019 at 10:47
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2 Answers
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Seconding @Victor Ivrii's good answer, with a few more points:
First, as Victor noted, a (properly) infinite sum of the sort written has convergence problems. This is already essentially visible if we just ignore the $y$-variable. Then we're asking whether an infinite sum of derivatives of $\delta$ (all just at $0$) can be a distribution. It is not at all obvious that this is not possible, I think. But there are at least two (covertly equivalent) proofs. First, we can recall E. Borel's theorem that an arbitrary sequence of complex numbers can be the Taylor-Maclaurin coefficients at $0$ of a smooth function. Thus, in a purported infinite linear combination, no matter how rapidly we make the coefficients decay, there is a smooth function whose Taylor-Maclaurin coefficients grow faster, and the sum (implied by evaluating the infinite sum on such a function) does not converge. Another way is to use Taylor-Maclaurin expansions with error term to classify distributions supported at $0$: finite linear combinations of derivatives of $\delta$.
On another hand, a similar argument does show that all distributions on $\mathbb R^{m+n}$ supported on $\mathbb R^m\times \{0\}$ are (finite!) sums of transverse derivatives, followed by evaluation on the smaller $\mathbb R^m$. So, for example, the finite version of your expression does give all distributions on $\mathbb R^2$ supported on the diagonal.
As in my comment, apart from inescapable issues of convergence (that is, it's not about failure to converge "pointwise", but failure to converge even in the weak dual topology, etc.), differentiation never increases support, so a-priori your expressions only give distributions supported on the diagonal.
Using the extended sense of Fourier transform is fine, of course, if one is careful, but/and will not truly produce something with larger support, etc.
answered Aug 29, 2019 at 21:43
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$\begingroup$ So in other words, the "distribution" $\exp (-(x-y)^2/a^2)$ acting on the test function $\phi (y)=y^2$ gives a finite and smooth function. But if the test function were say $\phi(y)\sim\exp (y^4)$, the result would not make sense since the integral does not converge. If one wants to define distributions acting on a set functions, they better be well behaved for all test functions in that set. Is that roughly what you are saying? $\endgroup$– fewfew4Aug 29, 2019 at 23:00
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$\begingroup$ You need to stop juggling words and think (and read). One test function does not make much sense. You define what "test functions" are and get the space of distributions. If it is $\mathscr{D}:=C_0^\infty $ you get $\mathscr{D}'$, if it is $\mathscr{E}:=C^\infty $ you get $\mathscr{E}'$, if it is $\mathscr{S} $ you get $\mathscr{S}'$. Etc $\endgroup$ Aug 30, 2019 at 9:19
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$\begingroup$ Alright, I'm sorry for being imprecise. So the singular term "test function" denotes an entire set of functions? I was trying to show how that gaussian example could not be a distribution on the set of smooth test functions. Was my argument imprecise? Or did I just use the wrong language? $\endgroup$– fewfew4Aug 30, 2019 at 17:34
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The "sum of this sort" is not a distribution unless sum is really finite. And in the latter case $A$ is supported on the diagonal $\{(x,y)\colon x=y\}$. So the answer is "No"
Actually, there are distributions of the infinite order, f.e. \begin{equation} \sum _{n\ge 0} \delta ^{(n)} (x-n). \end{equation} but these $\delta$-functions are located at points tending to the border of the domain (which here is $+\infty$)
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$\begingroup$ $\exp{(-a^2\partial^2)}\delta (x-y)\propto \exp{(x-y)^2/4a^2}$ is an example of the sort of sum mentioned that is not supported on the diagonal. $\endgroup$– fewfew4Aug 29, 2019 at 17:22
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1$\begingroup$ @LucashWindowWasher, assumming that your $\partial$ means derivative: any differentiation of a distribution cannot increase support, quite generally. Can you clarify your comment? $\endgroup$ Aug 29, 2019 at 17:24
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$\begingroup$ @paul garrett yes, that is what I am referring to. I simply expressed the delta function as $\delta (x-y)=\int_{-\infty}^{\infty}\frac{dp}{2\pi}\exp(ip(x-y))$ and performed the $p$ integral. $\endgroup$– fewfew4Aug 29, 2019 at 17:42
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$\begingroup$ First $\exp(-a^2\partial ^2)$ is really bad, since it corresponds to a heat equation with negative time. You meant probably $\exp(a^2\partial ^2)$. This is not a differential operator, so it does not contradicts, and it is definitely not $\sum_{n=0}^\infty \frac{1}{n!} (a\partial)^{2n}$ since this series converges only on some (but not all!) analytic functions . $\endgroup$ Aug 29, 2019 at 21:54
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1$\begingroup$ Yes, it converges on these functions, as I wrote "on some analytic functions". However, if you consider the remainder it will be very large for $|p|\ge \sqrt{N}$ $\endgroup$ Aug 30, 2019 at 0:32