2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ $\delta_n$ is the translated Dirac distribution, $\delta_n[\varphi] = \varphi(n)$? $\endgroup$ – Daniel Fischer 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 $T$? $\endgroup$ – Daniel Fischer 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
1
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.