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Can anyone provide a proof for

  1. Apr 2, 2007 #1


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    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 [tex] f_{n} (x) [/tex] and [tex] g_{n} (x) [/tex] are a succesion of function for [tex] n \rightarrow \infty [/tex] then the product of the 2 sucessions should be equal to the product of the 2 distributions..

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

    [tex] f_{n} (x) g_{n} (x) \rightarrow d(x)e(x) [/tex] ???

    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:

    [tex] a X b = a+a+a+a+a+a+a+a+..... [/tex] (the sum has 'b' terms)

    or [tex] log(a X b )=log(a) +log(b) [/tex] :Grumpy:
  2. jcsd
  3. Apr 2, 2007 #2
    It can be defined, but only in certain special cases and in a very special way.

    Take the delta distribution defined by [itex]\delta[f]:=f(0)[/itex], where [itex]f[/itex] is in Schwartz Space/ a test function. How would you define [itex]\delta^2[/itex]?
  4. Apr 3, 2007 #3


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    you can define [tex] \delta (x) \delta (x) = \delta ^{2} (x) [/tex] in the form.

    [tex] \delta ^{2} (x) \sim \frac{ sin ^{2} (Nx)}{\pi ^{2} x^{2}} [/tex]

    as N-->oo (N big) , do i get the 'Field medal' for it ?? :Bigrin:
  5. Apr 5, 2007 #4
    let [tex] log(a )=c'[/tex], and [tex] log( b )=c"[/tex], 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?
    Last edited: Apr 5, 2007
  6. Apr 5, 2007 #5


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    This doesn't work. Say that two sequences [itex]f_n(x)[/itex] and [itex]g_n(x)[/itex] both converge to the Dirac distribution. Depending on what these functions are, it's possible to obtain [itex]f_n(x) g_n(x) \rightarrow 0[/itex], among many other results. The limit of the product depends on more than the limits of the individual sequences. It's not unique.

    That only makes sense if b is an integer. But multiplication of distributions by integers is already defined (as is multiplication by arbitrary complex numbers).
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