# Can anyone provide a proof for

## Main Question or Discussion Point

Can anyone give a proof of why the product of 2 distributions can't be defined ?? :Confused:

In fact i believe (at least it should be) that if $$f_{n} (x)$$ and $$g_{n} (x)$$ are a succesion of function for $$n \rightarrow \infty$$ then the product of the 2 sucessions should be equal to the product of the 2 distributions..

hence $$f_{n} (x) \rightarrow d(x)$$ and $$g_{n} (x) \rightarrow e(x)$$ where d(x) and e(x) are 2 distributions then :

$$f_{n} (x) g_{n} (x) \rightarrow d(x)e(x)$$ ???

I have read about 'MOllifiers' and several methods for generalizing the distribution theory to include product of distributions, also couldn't the product be always defined as a 'sum' (in fact the sum of 2 distributions is defined) since:

$$a X b = a+a+a+a+a+a+a+a+.....$$ (the sum has 'b' terms)

or $$log(a X b )=log(a) +log(b)$$ :Grumpy:

Can anyone give a proof of why the product of 2 distributions can't be defined ?? :Confused:
It can be defined, but only in certain special cases and in a very special way.

Take the delta distribution defined by $\delta[f]:=f(0)$, where $f$ is in Schwartz Space/ a test function. How would you define $\delta^2$?

you can define $$\delta (x) \delta (x) = \delta ^{2} (x)$$ in the form.

$$\delta ^{2} (x) \sim \frac{ sin ^{2} (Nx)}{\pi ^{2} x^{2}}$$

as N-->oo (N big) , do i get the 'Field medal' for it ?? :Bigrin:

or $$log(a X b )=log(a) +log(b)$$ :Grumpy:
let $$log(a )=c'$$, and $$log( b )=c"$$, let's say that both logs have a base "d"( i do not know how to write it )

then by definition we have from the first

d^c'=a, and d^c"=b

if we multiply side by side we get

ab=(d^c')(d^c")=d^(c'+c") so agani by definition we have
log_d(ab)=c'+c", , i guess you can see the rest?

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Stingray
$$f_{n} (x) g_{n} (x) \rightarrow d(x)e(x)$$ ???
This doesn't work. Say that two sequences $f_n(x)$ and $g_n(x)$ both converge to the Dirac distribution. Depending on what these functions are, it's possible to obtain $f_n(x) g_n(x) \rightarrow 0$, among many other results. The limit of the product depends on more than the limits of the individual sequences. It's not unique.
$$a X b = a+a+a+a+a+a+a+a+.....$$ (the sum has 'b' terms)