How we can compute the convolution product $$\Big(\sum_{n=0}^{+\infty} \delta_n^{(n)}\Big) \star \Big(\sum_{n=0}^{+\infty} \delta_n\Big)$$ where $\delta$ is Dirac distribution? Thank's for the help

  • $\begingroup$ $\delta_n$ is the translated Dirac distribution, $\delta_n[\varphi] = \varphi(n)$? $\endgroup$ May 4 '14 at 13:26
  • $\begingroup$ Yes, $<\delta_n, \varphi> = \varphi(n)$. $\endgroup$
    – gram
    May 4 '14 at 13:34
  • 1
    $\begingroup$ Okay. You need to argue why the convolution exists in the first place (not all distributions can be convolved with each other). Then argue why it is $$\sum_{k=0}^{+\infty} \left(\delta_k \star \sum_{n=0}^{+\infty} \delta_n^{(n)}\right).$$ You probably know what $\delta_n \star T$ is for a distribution $T$? $\endgroup$ May 4 '14 at 13:56
  • $\begingroup$ My problem is that these distribtions haven't a compact $\supp$, also, we know that for any distribution $T$, $T \star \delta_0=T$. Byt here, i don't know how we can compute. $\endgroup$
    – gram
    May 4 '14 at 15:17

Formal computation. For every test function, $\varphi\star \delta_n $ is the shifted function $t\mapsto \varphi(t-n) $; the shift is $n$ units to the right. Let's accept that the same holds for distribution. Then $$\delta_k \star\left(\sum_{n=0}^{+\infty} \delta_n^{(n)}\right) = \sum_{n=0}^{+\infty} \delta_{n+k}^{(n)}$$ And formal summation over $k$ yields $$\left(\sum_{k=0}^\infty \delta_k \right)\star \left(\sum_{n=0}^{+\infty} \delta_n^{(n)} \right) = \sum_{k=0}^\infty \sum_{n=0}^{+\infty} \delta_{n+k}^{(n)}$$ The end result can be rewritten in a way that shows we indeed have a distribution, by letting $m=n+k$: $$\sum_{m=0}^\infty \sum_{n=0}^{m} \delta_{m}^{(n)} \tag{1}$$ The main point here is that on every compact subset only finitely many terms of (1) are present.

If you want to make the above rigorous, take a (compactly supported) test function $\varphi$, evaluate (1) against it (it's a finite sum). On the other hand, the evaluation of convolution $f\star g$ amounts to applying the product distribution $f(x)g(y)$ to the function $f(x+y)$ in the plane. This gives the same result.

As an aside: if the convolution was instead $$\Big(\sum_{n=0}^{+\infty} \delta_n^{(n)}\Big) \star \Big(\sum_{n=0}^{+\infty} \delta_{-n}\Big)$$ the formula would not meaningul (at least I don't see a way to define this convolution meaningfully).


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