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Is seems to me that it makes sense to presume some relations between derivatives of Dirac delta functions and its powers. I wonder, whether someone proposed a similar theory?

Particularly, it could make sense that

$$\frac{\pi^3 \delta(0) ^3}{6}+\frac{\pi \delta(0) }{12}=-\pi \delta''(0)$$

In such formalism, the more "inner" peaks of the distribution (when we approximate it with a smooth function) would be represented by higher powers of $\pi\delta(0)$.

I wonder whether similar relations ever arised as a formal method or in any other form.

UPDATE

This paper seemingly makes an attempt, but their results are quite different. Particularly, following their paper

$$\delta''(t)=16\pi^2\delta^3(t)$$

which is different from the above expression. To my view, their approach has many weaknessess, particularly, in my view $\delta''(0)$ should have the opposite sign than $\delta^3(0)$ (which is the case if we approximate delta function with a smooth curve).

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    $\begingroup$ What formalism do you have in mind in which it would be possible to multiply a distribution by itself? (or 2 different distributions with overlapping wave front sets for that matter). $\endgroup$
    – Saal Hardali
    Commented Apr 10, 2018 at 9:11
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    $\begingroup$ @Saal Hardali I encountered such relations as entirely formal links, and they seem to work for my purpose. Yet, definitely I do not know how they should be defined from the point of view of distribution theory. Each even derivative is related to a polynomial of the delta functions, the polynomials are similar to the Bernoulli polynomials (but not exactly equal). $\endgroup$
    – Anixx
    Commented Apr 10, 2018 at 9:14
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    $\begingroup$ How exactly did you come up with that particular equation? And why do you only state it for $0$ in the argument? $\endgroup$
    – Tobias Fritz
    Commented Apr 10, 2018 at 9:28
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    $\begingroup$ Dividing the equation by $\pi$ gives $\zeta(2)\delta(0)^3 + \zeta(-1)\delta(0) = -\delta''(0)$. $\endgroup$
    – Tobias Fritz
    Commented Apr 10, 2018 at 9:29
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    $\begingroup$ @Tobias Fritz I was considering divergent integrals. This particular value represents $\int_0^\infty x^2\,dx$ (the fact this integral is equal to the right hand side can be formally derived from the Fourier transform, the left-hand side is more complicated, I was manually assessing the difference between the divergent series and the divergent integral, the divergent series can be expressed with Bernoulli polynomials of $\pi \delta(0)$. $\endgroup$
    – Anixx
    Commented Apr 10, 2018 at 9:32

2 Answers 2

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There is a serious difficulty with the notion of products for distributions ; as a matter of fact Laurent Schwartz, one of the creator of Distribution Theory, wrote an article expressing the impossibility of multiplying distributions http://sites.mathdoc.fr/OCLS/pdf/OCLS_1954__21__1_0.pdf

On the other hand some simple observations are in order: just think that you want to define $\delta_0^2$, the square of the Dirac mass at 0: a natural idea would be to approximate the Dirac mass by $\rho(\frac{x}\epsilon)\frac1{\epsilon}$ with $\rho$ smooth compactly supported with integral 1 and take a look at $$ \rho(\frac{x}\epsilon)^2\frac1{\epsilon^2}. $$ Considering now a test function $\phi$ say smooth and compactly supported, we may check $$ \int\rho(\frac{x}\epsilon)^2\frac1{\epsilon^2}\phi(x) dx= \int\rho(t)^2\frac1{\epsilon}\phi(\epsilon t) dt\sim\frac1{\epsilon} \phi(0)\int \rho(t)^2 dt$$ if $\phi(0)\not=0$. Nonetheless there is no finite limit if $\phi(0)\not=0$, but even the factor of $1/\epsilon$ depends on $\rho$, that is on the approximation process.

A more subtle remark is due to Lars Hörmander: you may define the so-called wave-front-set of a distribution and say that if $T_1, T_2$ are distributions such that $$ (x,\xi_1)\in WF T_1,\quad (x,\xi_2)\in WF T_2\Longrightarrow \xi_1+\xi_2\not=0, $$ then $T_1T_2$ makes sense. In particular you may consider $$ \left(\frac{1}{x+i0}\right)^2, \quad \text{but not }\quad \delta_0^2, $$ meaning that $ \lim_{\epsilon\rightarrow 0_+}\int \frac{\phi(x)}{(x+i\epsilon)^2} dx $ makes sense for any test function $\phi$. Somehow the notion of wave-front-set is providing the mathematical obstruction to a general rule of multiplication for distributions.

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  • $\begingroup$ But is not $$\frac1{i 0^-}=i\delta(0)$$?... $\endgroup$
    – Anixx
    Commented May 6, 2018 at 21:04
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For what it's worth, a formula relating derivatives and powers of Dirac delta functions was discussed in Products and compositions with the Dirac delta function:

The formula is credited to 1974 and 1977 papers by Fisher and Tewari in the journal "Mathematics Student", which I have not found accessible online.

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  • $\begingroup$ This is VERY interesting! But what is the second term? What does the $C$ and $\delta$ mean there? $\endgroup$
    – Anixx
    Commented Apr 10, 2018 at 14:43
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    $\begingroup$ quite frankly I don't know and I hoped to learn more from the source papers in Math.Student, which I have not been able to locate; (recent issues of this journal are online, but not older issues) $\endgroup$
    – Carlo Beenakker
    Commented Apr 10, 2018 at 14:55
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    $\begingroup$ I am trying to insert m=0, r=1 but fail to see how this holds... $\endgroup$
    – Anixx
    Commented Apr 10, 2018 at 15:02
  • $\begingroup$ Then I get $\delta'(x)=\delta(x)$. Does not hold seemingly. $\endgroup$
    – Anixx
    Commented Apr 10, 2018 at 15:06
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    $\begingroup$ Maybe in the first term it is meant to be derivative as well, and not power?... $\endgroup$
    – Anixx
    Commented Apr 10, 2018 at 15:10

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