# Is square of Delta function defined somewhere?

Hello, every one. I am wondering whether any one knows that whether the square of Dirac Delta function is defined some where?

In the beginning, this question might look strange. But by restricting the space of the test functions, I think it is still possible. For example, in order to make sense of $\delta_0^2$, we can think that it is the limit of $\frac{e^{-x^2/t}}{2\pi t}$ as $t\rightarrow 0_+$. Now choose the test function $f(x)=x^2$. It is clear that $$\int_{-\infty}^{\infty} x^2 \frac{e^{-x^2/t}}{2\pi t} d x = \frac{1}{2\sqrt{\pi t}} \int_{-\infty}^{\infty} x^2 \frac{e^{-x^2/t}}{\sqrt{\pi t}} d x = \frac{1}{2\sqrt{\pi t}} \cdot \frac{t}{2} = \frac{\sqrt{t}}{4\sqrt{\pi}}\;.$$ Then let $t$ tend to $0$, we get $<\delta_0^2,f>=0$ in this case. So we can restrict, for example, all test functions tend to 0 at the speed no less than $x^2$.

I don't want to invent the whole stuff if it already exists. Otherwise, I might take care of the every details. Thank you in advance for any hints.

EDIT: Here are some references that I found to be useful. 1: Mikusiński, J. On the square of the Dirac delta-distribution. (Russian summary) Bull. Acad. Polon. Sci. Sér. Sci. Math. Astronom. Phys. 14 1966 511–513. 44.40 (46.40)

2:Ta Ngoc Tri, The Colombeau theory of generalized functions Master thesis, 2005

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There is the <a href="en.wikipedia.org/wiki/…; Colombeau algebra </a>. Though I have no idea what applications it has. – Rbega Dec 2 '10 at 17:29
I am no expert but my understanding is that the Dirac delta function is not a function but is a distribution. Furthermore multiplication of distributions is not defined and this is the cause of much frustration in quantum field theory. – Bruce Westbury Dec 2 '10 at 17:35
@Bruce the multiplication of distributions is sometimes defined. If you are interested you should look up wavefront sets, one good reference is Chapter 5 of math.mit.edu/~rbm/18.157-F09/18.157-F09.html, see e.g. Prop 5.12. For example, this would make rigorous the notion that the product of a delta function at $0$ and a delta function at $x\neq 0$ should be $0$. However, you are right that the product of two delta functions is not well defined as a distribution. However, I think that @Anand is asking about restricting the domain of distributions to allow it to exist. – Otis Chodosh Dec 2 '10 at 17:47
@Anad: I think you have to clarify what kind of object you want your square of $\delta$ to be. E.g. restricting the test funtions does not give you something like $\delta^2$ as long as you are looking for a linear functional (which you probably don't). – Dirk Dec 2 '10 at 18:58
@RBega, @Bruce: I'm reading through Colombeau's book "New Generalized Functions and Multiplication of Distributions" right now. Colombeau's stated motivation is to make sense of the distributional products that show up in quantum field theory; in fact, the first chapter of his book is a quick introduction to quantum field theory. – Vectornaut Apr 12 '12 at 21:37

When L. Schwartz "invented" distributions (actually, he only invented the mathematical theory as a part of functional analysis, because distributions were already used by physicists), he proved incidentally that it is impossible to define a product in such a way that distributions form an algebra with acceptable topological properties. What is possible is to define the product of distributions when their wave front sets do not meet. This is why $fT$ makes sense if $T$ is a distribution and $f$ is $C^\infty$, for instance, because the front set of $f$ is void. But you can also multiply that way genuine distributions; for instance in $\mathbb R^2$, $\delta_{x=0}=\delta_{x_1=0}\delta_{x_2=0}$.

J.-F. Colombeau invented in the 70's an algebra of generalized functions, which has something to do with distributions. But each distribution has infinitely many representatives in the algebra, and you have to play with the equality and a "weak equality" (or "association"). I don't know of an example where this tool solved an open problem. In Colombeau's algebra, the square of $\delta_0$ makes sense, but is highly non unique.

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@Prof. Denis Serre, Thank you very much. Your explanation is very clear. Especially the case $\delta_{x=0}=\delta_{x_1=0}\delta_{x_2=0}$ in $R^2$. I took this results for granted before. Now I know the reason. By the way, do you have a good reference on the topic of Wave front set? I encountered this topic in Hormander's books. His books seem too difficult. Thank you very much! :-) – Anand Dec 3 '10 at 14:08

$\delta_0$ vanishes identically on the space of test functions you've defined. So it's not surprising that its square is well-defined: $0\cdot 0 = 0$.

I suspect you'll have a much harder time defining $\delta_0^2$ on test functions which don't vanish at $0$.

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My intention is just to make $\delta_0^2$ rigorous by restricting the test functions to vanish at 0. :-) – Anand Dec 2 '10 at 17:58
Yes, but you've thrown the baby out with the bathwater. – userN Dec 2 '10 at 17:59
I think that you can restrict to functions which approach their value at $0$ like $x^2$, instead of functions which are zero at $0$, e.g. $x^2 + 3$. I don't think this leads to anything interesting, but @A.J.'s objection might not be fatal. – Otis Chodosh Dec 2 '10 at 18:08
@Otis, Thanks for your comments. Since I meet this problem in my research, I need something rigorous, even it is not very interesting. :-) – Anand Dec 2 '10 at 18:11
@Anand I think that you should edit your question to describe the properties of \delta_0^1 you would like. A.J.'s argument basically shows that if you want to define $\delta^2 (f)$ to be $\lim_{t\to 0} \int_{-\infty}^\infty f e^{-x^2/t}/\sqrt{2\pi t}$ for some class of $f$, then basically the only thing you can do is to define it to be zero for functions which vanish faster than $x^2$. Perhaps you should explain why you would like to define $\delta_0^2$ and what properties you would like from such a definition. Otherwise, I doubt that anyone can say anything more useful than A.J's answer. – Otis Chodosh Dec 2 '10 at 18:19

There are whole theories in microlocal analysis that deal with the issues here, I believe. Some heuristics are that the "singular support" of a distribution controls what it can be multiplied by in a naive sense (distributions with a disjoint singular support). So squaring the delta function is the first bad case - whatever the singular support means, it must be the set containing 0 for the delta function. Need more heuristics.

One insight is that one dimension may be too few to show the real picture. "Microlocal" tends to mean localising in (co)tangential directions, and one dimension offers only two. Hyperfunctions in the case of one dimension make something of this by considering the real line as the boundary of the upper half complex plane. I.e up is not the same as down. Boundary values of functions holomorphic in the upper half plane have a candidate for the delta function analogue: take 1/z. No problem squaring that. More of a problem saying what this analogy means that is worth anything. Mikio Sato did that. Now I shall be quiet, because this is probably already wrong enough.

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@Charles, thank you very much for your comments. It is very interesting. But it is too hard for me to understand. I wish I could understand something about Microlocal analysis one day. :-) – Anand Dec 2 '10 at 21:38
The process of squaring holomorphic functions on half-planes does not yield hyperfunctions that behave like squares, even when restricting to the regular case. – S. Carnahan Dec 3 '10 at 4:38

The extent to which multiplication of distributions is defined was examined by Richards & Youn and some of the results are in their short and fairly elementary joint book on distributions. One can multiply something fairly exotic like the third derivative of the delta function by a very well-behaved function; that much everybody knows. But I think they had a result that as one factor becomes progressively less well-behaved the other must become more well-behaved in order to make multiplication possible. I don't recall the details. But I'm pretty sure theirs is not the last word on the subject.

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Yes. As you said that as one factor becomes progressively less well-behaved, the other should become more well-behaved. The assumption here might be that you consider the same space of test functions. In current case, both factor becomes ill-behaved (both of them are $\delta_0$). So we might overcome this difficulty by restricting the test functions, requiring that they should vanish at 0 at certain speed. Thanks. :-) – Anand Dec 2 '10 at 18:05
By the way, could you please give me some more details about the reference of Richards & Youn? I learnt this topic by Roudin's book, I am not an expert at this domain. Thank you very much! :-) – Anand Dec 2 '10 at 18:13
Google "Richards Youn Distributions" and it pops up in google books. – B R Dec 2 '10 at 18:44
The Theory of Distributions: A Nontechnical Introduction by J. Ian Richards and Heekyung Youn, published in 1990 by Cambridge University Press. "Nontechnical" appears to mean that the reader is not assumed to know functional analysis, topology, or measure theory. But everything is rigorously proved. Including the fact that the "tempered" distributions---those that don't grow too fast at $\pm\infty$---are closed under the Fourier transform. – Michael Hardy Dec 2 '10 at 20:51
Thanks BR and Michael Hardy. :-) – Anand Dec 2 '10 at 21:31

I'd like to point out that several of the concepts mentioned here are explained on the nLab:

while several are missing and the parts on microlocal analysis and hyperfunctions could use some help .

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@Tim van Beek. Thank you very much for the reference. It looks like what I am looking for...:-) – Anand Dec 3 '10 at 14:03

The theory of distributions and operations on them are generally only useful in so far as they extend the operations on smooth functions. If you look in Hörmander, there is a criterion in terms of wavefront sets which is very useful (mentioned by others), and you'll also notice that the wavefront sets of $\delta$ and $\delta$ collide. The reason you can't square the delta-function is that when you approximate it by smooth functions, there is no unique limit. If you wanted to restrict to a smaller space of test functions, you would clearly have to consider test functions which vanish at the origin in some way. But do you have a particular purpose in mind for this question?

EDIT: Sorry -- this was supposed to be a comment, not an answer.

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@Phil Isett, I am doing stochastic heat equation. If the initial data is a Delta distribution, then the second moments of the solution is well defined for all $t>0$, but for time $t=0$, the second moment will be formally certain Delta square. This is why I encounter this problem. Thanks you for your comment. :-) – Anand Jul 21 '11 at 9:10

Denis Serre's answer is just perfect. Let me add a couple of examples of distributions that can be squared:

(1) With $H$ the Heaviside function, define $Log(x+i0)=\ln(\vert x\vert)+i\pi H(-x)$ and $$T_1=\frac{1}{x+i0}=\frac{d}{dx}(Log(x+i0))=pv\frac 1{x}-i\pi \delta_0(x).$$ It is easy to see that $WF T_1=[0]\times (0,+\infty),$ so that $WF T_1+WF T_1$ does not meet 0. Then there is no difficulty to define $T^2$ say as $$\langle T^2,\phi\rangle=\lim_{\epsilon\rightarrow 0_+}\int\frac{\phi(x) dx}{(x+i\epsilon)^2}.$$

(2) Let us consider a smooth hypersurface $\Sigma$ of $\mathbf R^d$ defined by the equation $f(x)=0$ with a smooth $f$ such that $df\not=0$ at $f=0$ and let $\delta_\Sigma$ be the Euclidean measure on $\Sigma$. Then $$T_2=pv\frac{1}{f}-i\delta_\Sigma$$ can be squared. The reason is the same than for the previous example, since $WF T_2$ is the positive conormal of $\Sigma$. A point $(x,\xi)\in WF T_2$ iff $$x\in \Sigma\quad \xi =\lambda df(x) \text{ with \lambda >0}.$$ Then of course, if $(x,\xi_j)$, $j=1,2$ are both in $WF T_2$ then $$\xi_1+\xi_2\not=0.$$

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I've seen the idea of it used in image processing for denoising; the total variation energy

$E_{TV}(f) = \displaystyle\int\left(|\nabla f| +(f-u)^2\right)$

is generally used instead of Tikhonov regularization

$E_{Tikhonov}(f) = \displaystyle\int\left(|\nabla f|^2 +(f-u)^2\right)$

as the latter never has a discontinuous solution (since the integral would be infinite).

I don't remember how rigorously this idea was developed - "Mathematical Problems in Image Processing" by Aubert and Kornprobst was the textbook I used at the time, but there are probably some more recent references in the field.

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Check out the papers by Accardi and Boukas

[added by S. Carnahan: The relevant part of their first ArXiv paper is that they regularize powers of $\delta$, not by setting $\delta^n(x) = c_n \delta(x)$ for some real $c_n$ (which they find to work poorly for their purposes), but by using two variables and setting $\delta(t-s)^n = \delta(s)\delta(t-s)$ for all $n \geq 2$. This definition allows them to define certain representations of Lie algebras by Fock space methods. As far as I can tell, this does not yield a workable definition of $\delta^2$ for analysis on the real line.]

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I was more or less ready to delete this non-answer, but I might as well summarize what I found by Googling. – S. Carnahan Aug 18 '13 at 16:46