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Suppose you have two continuous functions $f,g: \mathbb{R}\to\mathbb{R}$; is the product $f'g$ as a distribution, at least locally? I am interested in a local result, actually, so you can as well assume that $f$ and $g$ are compactly supported.

Obivously, $f'$ is, as $f\in L^1_{\mathrm{loc}}$.

Moreover, I am aware of the following particular cases:

  1. if $f\in W^{1,1}(I)$, then $f'g$ is well defined on $I$
  2. if $f'\in L^p(I)$ and $g\in L^q(I)$ for $p^{-1}+q^{-1}=1$, then $f'g$ is defined on $I$;
  3. if $g\in W^{1,1}(I)$, then again $f'g$ is well defined on $I$

I am interested in a general result or in a counterexample. Thank you!

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up vote 1 down vote accepted

Edited after questioner's clarifications: The analogous question for functions on a circle is somewhat easier to answer decisively, since all distributions are tempered (and have Fourier series expansions), and since the index for Fourier series is discrete. Further, the local question on the line or on $\mathbb R^n$ reduces to the analogous question on products $\mathbb T^n$ of circles, since we can smoothly truncate.

The $L^2$ Levi-Sobolev spaces $H^s$ on the circle [or products of circles, similarly] can be characterized/defined as distributions $u$ with Fourier coefficients $\hat{u}(n)$ satisfying $\sum_n (1+n^2)^s\cdot |\hat{u}(n)|^2<\infty$. A distribution (provably) lies in some $H^s$, so has Fourier coefficients of polynomial growth. [This is true on $\mathbb T^n$.]

The product of two Fourier series, one in $H^s$ the other in $H^t$, certainly converges for $s+t\ge 0$, and converse, I think, by Banach-Steinhaus/uniform-boundedness. [Similarly on $\mathbb T^n$.]

While (Levi-Sobolev imbedding) $H^{1/2+\epsilon}\subset C^o$, and while Fourier series of functions in $H^{1/2+\epsilon}$ converge uniformly pointwise to the functions, Baire category (and Riemann-Weierstrass counterexamples) show that the converse fails significantly. [Edit:] The traditional Baire category argument to prove that typical continuous functions' Fourier series fail to converge to them pointwise (hence, note, are in no $H^{1/2+\epsilon}$ for any $\epsilon>0$), as in 2012-13/05b_banach_fourier.pdf does not produce an explicit function, but is perhaps more compelling/persuasive. I hesitate somewhat to try to recap the Riemann-Weierstrass explicit versions, unless "explicit example" is essential to your/the-questioner's context.

Thus, in terms of the Levi-Sobolev spaces, for $f$ continuous but not in $H^{1/2}$ [or not in $H^{{\rm dim}/2}$, for $\mathbb T^n$] there should exist many $g\in C^o$ such that the apparent Fourier series product for $f'\cdot g$ does not make sense because the expression for the $0th$ coefficient diverges (beyond repair).

(If this kind of answer is interesting/useful, I can amplify it and worry about the details.) [Edit:] a bit added, per request...

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As I am mainly interested in the question locally, I could think of $f$ and $g$ as compactly supported functions (thinking about it, I shall edit the question accordingly!). So I think they could be seen as functions on $S^1$ (don't they?). I have more or less followed your argument and I'm quite persuaded, I just can't figure out exactly how to produce a counterexample. Actually, I didn't really put myself on it ... anyway, if you already have a clear idea on how it should be, it would be helpful. And, obviously, thanks for the answer! – wisefool Dec 11 '12 at 21:56
Just a random thought: I know that there are some funny ways of defining products of distributions (the wavefront set, the Colombeau algebra and so on), but I do not know much more about them ... Would it maybe be possible to define the product I asked about in one of those contexts? – wisefool Dec 13 '12 at 18:30
Yes, in two or more dimensions, hypotheses on non-intersecting wavefront sets can allow legitimate products of distributions, basically for sensible physical reasons. Similar things can be reflected in a local/Fourier series discussion, too, by deliberately avoiding blow-up of Fourier coefficients of products. As a trivial example, $\sum_{m,n} {\sqrt{|m|}\over 1+n^2}\cdot e^{imx+iny}$ and $\sum_{m,n} {\sqrt{|n|}\over 1+m^2}\cdot e^{imx+iny}$ can be multiplied (I think). – paul garrett Dec 13 '12 at 19:17

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