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I am wondering whether anyone knows if the square of Dirac Delta function is defined somewhere.

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 $\langle\delta_0^2,f\rangle=0$ 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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    $\begingroup$ There is the <a href="en.wikipedia.org/wiki/…"> Colombeau algebra </a>. Though I have no idea what applications it has. $\endgroup$
    – Rbega
    Commented Dec 2, 2010 at 17:29
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    $\begingroup$ 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. $\endgroup$ Commented Dec 2, 2010 at 17:35
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    $\begingroup$ @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. $\endgroup$ Commented Dec 2, 2010 at 17:47
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    $\begingroup$ @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). $\endgroup$
    – Dirk
    Commented Dec 2, 2010 at 18:58
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    $\begingroup$ @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. $\endgroup$
    – Vectornaut
    Commented Apr 12, 2012 at 21:37

10 Answers 10

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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 $\mathbb R^2$, $$\delta_{x=0}=\delta_{x_1=0}\delta_{x_2=0}.\tag{1}\label{1}$$

J.-F. Colombeau invented in the '70s 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.

Edit (May 2020). I'd like to share the following generalization of identity \eqref{1} above, which I found in developing my theory of Divergence-free positive symmetric tensor. In $\mathbb{R}^d$, consider the one-dimensional Lebesgue measure $\mathcal{L}_j$ along the $j$-th axis, for $1\le j\le d$. Then $$(\mathcal{L}_1\dotsm\mathcal{L}_d)^{\frac1{d-1}}=\delta_{x=0}.$$ There are a lot of reasons why this equality makes sense and is valid. For instance, if you approach $\mathcal{L}_j$ by $(2\epsilon)^{1-d}dx\rvert_{K_j(\epsilon)}$ where $K_j(t)=(-\epsilon,\epsilon)^{d-1}\times {\mathbb R}\vec e_j$, then the left-hand side equals $(2\epsilon)^{-d}dx\rvert_{(-\epsilon,\epsilon)^d}$, which approaches the Dirac at the origin. There is an analogous identity when the orthogonal axes are replaced by an arbitrary list of $d$ axes; then the right-hand sides is $C\delta$, where the constant $C$ is computed by solving a case of Minkowski's Problem.

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    $\begingroup$ @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! :-) $\endgroup$
    – Anand
    Commented Dec 3, 2010 at 14:08
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    $\begingroup$ For an introduction to the wave front set see arxiv.org/abs/1404.1778 $\endgroup$ Commented Apr 10, 2018 at 13:59
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$\delta_0$ 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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  • $\begingroup$ My intention is just to make $\delta_0^2$ rigorous by restricting the test functions to vanish at 0. :-) $\endgroup$
    – Anand
    Commented Dec 2, 2010 at 17:58
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    $\begingroup$ Yes, but you've thrown the baby out with the bathwater. $\endgroup$
    – user1504
    Commented Dec 2, 2010 at 17:59
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    $\begingroup$ 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. $\endgroup$ Commented Dec 2, 2010 at 18:08
  • $\begingroup$ @Otis, Thanks for your comments. Since I meet this problem in my research, I need something rigorous, even it is not very interesting. :-) $\endgroup$
    – Anand
    Commented Dec 2, 2010 at 18:11
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    $\begingroup$ @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. $\endgroup$ Commented Dec 2, 2010 at 18:19
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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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  • $\begingroup$ @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. :-) $\endgroup$
    – Anand
    Commented Dec 2, 2010 at 21:38
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    $\begingroup$ The process of squaring holomorphic functions on half-planes does not yield hyperfunctions that behave like squares, even when restricting to the regular case. $\endgroup$
    – S. Carnahan
    Commented Dec 3, 2010 at 4:38
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I'd like to point out that several of the concepts mentioned here are explained on the nLab:

multiplication of distributions,

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

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  • $\begingroup$ @Tim van Beek. Thank you very much for the reference. It looks like what I am looking for...:-) $\endgroup$
    – Anand
    Commented Dec 3, 2010 at 14:03
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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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  • $\begingroup$ 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. :-) $\endgroup$
    – Anand
    Commented Dec 2, 2010 at 18:05
  • $\begingroup$ 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! :-) $\endgroup$
    – Anand
    Commented Dec 2, 2010 at 18:13
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    $\begingroup$ Google "Richards Youn Distributions" and it pops up in google books. $\endgroup$
    – B R
    Commented Dec 2, 2010 at 18:44
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    $\begingroup$ 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. $\endgroup$ Commented Dec 2, 2010 at 20:51
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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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  • $\begingroup$ @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. :-) $\endgroup$
    – Anand
    Commented Jul 21, 2011 at 9:10
  • $\begingroup$ @Anand: The easy way out for your purposes is to consider the second moment (or higher) as a distribution on space time. Trying to restrict to sharp times (i.e., feeding it a "test function" $\delta(t)f(x)$ with $f\in\mathcal{D}({\rm space})$) is what causes the problem. $\endgroup$ Commented Apr 10, 2018 at 13:57
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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 $\operatorname{Log}(x+i0)=\ln(\vert x\vert)+i\pi H(-x) $ and $$T_1=\frac{1}{x+i0}=\frac{d}{dx}(\operatorname{Log}(x+i0)) = \operatorname{pv}\frac 1{x}-i\pi \delta_0(x).$$ It is easy to see that $ \DeclareMathOperator\WF{WF}\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 $$ \langle T^2,\phi\rangle=\lim_{\varepsilon\rightarrow 0_+} \int\frac{\varphi(x) \, dx}{(x+i\varepsilon)^2}. $$

  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 $df\ne0$ 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 as 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 $$ \xi_1+\xi_2\ne0. $$

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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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    $\begingroup$ I was more or less ready to delete this non-answer, but I might as well summarize what I found by Googling. $\endgroup$
    – S. Carnahan
    Commented Aug 18, 2013 at 16:46
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Looks like nobody mentioned the work A contribution to the theory of generalized functions by Egorov, which proposes an interesting approach to generalize the concept of distribution in a possibly simpler (yet more general) way than the one by Colombeau.

A nice introductory short-essay on the same argument is in the Appendix (again by Egorov) of the book: Demidov, A. S. (2001). Generalized functions in mathematical physics: main ideas and concepts (Vol. 237, Nova Publishers).

The author first introduces the concept of generalized real (or complex) numbers as equivalence classes of eventually coinciding sequences: $\{c_k\}\equiv \{z_k\}$ iff they coincide for all $k$ large enough. Then he defines a generalized function as an equivalence class of sequences of $C^\infty$ functions defined on a domain $\Omega\subset \mathbb{R}^n$, where $\{f_k\}\equiv \{g_k\}$ iff they eventually coincide on every $K\Subset \Omega$. Clearly the values of generalized functions at a point make sense only as generalized real numbers.

Then the usual Dirac $\delta$ can be recovered as the equivalence class containing $\{k\phi(kx)\}$, where $\phi\in C^\infty_0(\Omega)$ such that $\int\phi(x)dx=1$. The product of two generalized functions is naturally defined through the equivalence class containing the product of (any) two representatives of the factors. Notice that $x\delta(x)\ne 0$, contrary to what happens in the "usual" distribution theory, and that is essential by a well-known no-go theorem by Schwartz:

Let $A$ be an associative algebra, in which a derivation operator (i.e., a linear operator $D : A \to A$ such that $D(f\cdot g) = f\cdot D(g)+D(f)\cdot g$) is defined. Suppose that the space $C(\mathbb{R})$ of continuous functions on the real line is a subalgebra in $A$, and $D$ coincides with the usual derivation operator on the set of continuously differentiable functions, and the function, which is identically equal to 1, is the unit of the algebra $A$. Then $A$ cannot contain an element $\delta$ such that $x\cdot\delta(x) = 0$.

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

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

is generally used instead of Tikhonov regularization

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

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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