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I was wondering if there were some general results about repeated convolution products. So far, I have only dealt with single convolutions (for instance for constructing dense subspaces in $L^p$, or when working with Fourier series or Fourier transforms), but I was wondering, given a function $f$ such that $||f||_{L^1} \leq 1$, then $|| f \star f||_{L^1} \leq ||f||_{L^1} ^2 \leq 1$, and so on. Being that the $L^1$ norm is always bounded, all the functions obtained by repeatedely convoluting $f$ with itself should be well defined, at least for almost every $x$. The questions is: what can I say about this succession? Also, is this problem somehow "useful" (I mean if it is of some relevance in some scientific applied field, like for example the one-time convolution that lets us write solutions of physical problems, like heat) or is it just a useless mathematical object that we pure mathematicians love so much?

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Of course there's no such thing as an infinite convolution - if your question is to make any sense we have to recast it as a question about limits of finite convolutions. And then we have to ask in what sense the limit exists.

We never get anything interesting as a limit in $L^1$ norm:

Suppose $f\in L^1$ and let $f_n=f*f*\dots*f$. If $g\in L^1$ and $||f_n-g||_1\to0$ then $g=0$.

Proof: The Fourier transforms satisfy $(\hat f)^n\to\hat g$ pointwise. Hence $\hat g(\xi)=0$ or $1$ for every $\xi$; since $g\in C_0$ we have $\hat g=0$, hence $g=0$.

You see similarly that the norm limit simply cannot exist unless $|\hat f|<1$ everywhere.

It is possible to get interesting things from convolutions of measures, if we take the limit in the weak* sense. For example, let $$\mu_n=\frac12(\delta_{1/2^n}+\delta_{-1/2^n})$$and $$\nu_n=\mu_1*\dots*\mu_n.$$Then $\nu_n$ converges weak* to $\frac12$ times Lebesgue meaure restricted to $[-1,1]$.

Exercise Take the Fourier transform and derive an interesting infinite product for $\sin(\xi)$.

Edit: Come to think of it, what I say above is misleading, giving the impression that there's a crucial difference between weak convergence and norm convergence here. Of course that difference is going to make a big difference, but in fact the real reason for the difference bewteen the two cases presented above is that in one case we're talking about the convolution of $f$ with itself, while in the other we're talking about the convolution of infinitely many different measures: We can have $||f_1*\dots*f_n-g||_1\to0$ in nontrivial ways, while on the other hand we cannot have $\mu*\dots*\mu\to\nu$ weakly except in certain trivial cases (I'll leave it to you to work out what the trivial possibilities are, looking at the Fourier transform, and assuming that our flavor of weak convergence implies pointwise convergence for the transform.)

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  • $\begingroup$ Yeah thanks, I wrote it in a hurry and made a huge mistake. I edited the post though. Thanks for the examples! $\endgroup$ – tommy1996q Feb 25 '18 at 22:54
  • $\begingroup$ Also, what do you mean by weak convergence? My definition involves scalar products, but we are in $L^1$ so we don't have any $\endgroup$ – tommy1996q Feb 25 '18 at 23:05
  • $\begingroup$ The Banach-space dual of $C_0(\Bbb R^n)$ is the space of complex measures on $\Bbb R^n$; accordingly if $\mu_n$ is a sequence of complex measures then $\mu_n\to\mu$ in the weak* topology if $\int fd\mu_n\to\int f\,d\mu$ for every $f\in C_0$. If all the $\mu_n$ have support in a fixed compact set then $\mu_n\to\mu$ in this sense implies that $\hat\mu_n\to\hat\mu$. There are other sorts of weak convergence, defined by saying $\int f\,d\mu_n\to\int f\,d\mu$ for other classes of functions $f$; one popular version is to consider bounded continuous $f$, which also implies that $\hat\mu_n\to\hat\mu$. $\endgroup$ – David C. Ullrich Feb 25 '18 at 23:12
  • $\begingroup$ Oh I see what you mean now. Indeed I was thinking about a different thing. I actually don't know much about measures in general (always used Lebesgue so far), but I think I got the hang of it more or less $\endgroup$ – tommy1996q Feb 25 '18 at 23:17

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