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The following equation may be meaningful, but how can we make it well-defined $$\delta(x-a)\cdot\delta(x-b)=0$$ Question: How do we defined this equation? Or more broadly define product between generalized functions with certain restrictions. Whether this definition satisfies the product rule [$D(fg)=Df\cdot g+f\cdot Dg$].

Added: Maybe theory of hyperfunction can explain this, which I am not familiar with. Should its product satisfy the product rule?

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    $\begingroup$ This matter is discussed here. $\endgroup$
    – Iosif Pinelis
    Commented Aug 9, 2023 at 14:23
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    $\begingroup$ The book on generalized functions by Richards and Youn goes into things like this. $\endgroup$
    – Michael Hardy
    Commented Aug 9, 2023 at 18:59
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    $\begingroup$ Maybe this answer can interest you: mathoverflow.net/a/445123/167834 $\endgroup$
    – Alessandro Della Corte
    Commented Aug 10, 2023 at 20:28
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    $\begingroup$ Here I will say somewhat more: Anybody knows how to multiply a generalized function by a test function. One is a playboy and the other is a monk. More prosaically, one is wild and the other well behaved. Richards and Youn say that if you have a function less well behaved than a test function, you can still multiply it by a distribution as long as the distribution is somewhat more well behaved than some distributions are. As the space of well behaved functions grows including progessively less well behaved functions, the space of distributions by which they can be multiplied shrinks. $\endgroup$
    – Michael Hardy
    Commented Aug 10, 2023 at 20:37

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You can, of course, if you wish, consult the highly sophisticated treatise of Hörmander to verify that your result holds (for distinct values of $a$ and $b$—there is no serious text which claims this for the case $a=b$). Or you can refer to work which precedes this by decades and is completely elementary, in order to verify it as follows, without the use of functional analysis (to simplify the notation, we assume that $a=0$ and $b=1$). Then it is clear that the product exists and is equal to $0$ on each of the intervals $]-\infty,\frac 2 3[$ and $]\frac 1 3,\infty[$. We now piece these two distributions together (trivial case of “recollement des morceuses”) to get the zero distribution on the line.

We remark that the very elementary theory which justifies these simple manipulations is easily available online (reference below), where you will also find explicitly the following generalisations—just as easily proved:

  1. if $f$ is a distribution on the line which equals a smooth function on a neighbourhood of $0$, then $f\delta_0$ and indeed $f\delta_0^{(n)}$ exist and are equal to zero if $f$ vanishes there.

  2. If $a$ and $b$ are distinct, then $\delta^{(n)}(x-a)\delta^{(m)}(x-b)=0$.

The reference is to the site https://jss100.campus.ciencias.ulisboa.pt of the late portuguese mathematician J. Sebastião e Silva where you will find his text “Theory of Distributions” (the above results are in the fourth chapter “Multiplication and Change of Variables”) under the "Textos Didáticos".

In response to the above request for an explicit formula for the product of a $C^n$ function and a distribution of order $n$, I can’t comment there but it is:

$$fg=\sum_{k=0}^{n} \binom n k D^{n-k}(f^{(k)}G) $$ where $f$ is $C^n$ and $g$ is a distribution of order $n$ of the form $D^n G $ with continuous $G$ ($D$ is the distributional derivative).

This is 4.1.2 in the reference given here. The most used version, where $g=\delta^{(n)}$, is 4.2.3.

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