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Let $\mathcal S (\mathbb R)$ denote the space of Schwartz functions on $\mathbb R$ and $\mathcal S^* (\mathbb R)$ denote the dual space of Schwartz (a.k.a tempered) distributions. We consider $\mathcal S (\mathbb R)$ as a Frechet space and $\mathcal S^* (\mathbb R)$ as a direct limit of Banach spaces.

Let $c:\mathcal S (\mathbb R) \otimes \mathcal S^* (\mathbb R) \to \mathcal S^* (\mathbb R)$ be the convolution map. Let $\hat c:\mathcal S (\mathbb R) \hat\otimes \mathcal S^* (\mathbb R) \to \mathcal S^* (\mathbb R)$ be its extention to the completed tensor product. We have an argument that "proves" the following contradictory facts:

  1. $\mathrm{Im} (c)=\mathrm{Im} (\hat c)$
  2. $$\mathrm{Im} (c)=(f \in C^\infty(\mathbb R)|\exists \text{ a polinomial }p \text{ s.t. } \forall n\in \mathbb N \text{ the function } \frac{f^{(n)}}{p} \text{ is bounded} )$$
  3. $$\mathrm{Im} (\hat c)=(f \in C^\infty(\mathbb R)|\forall n\in \mathbb N, \exists \text{ a polinomial }p \text{ s.t. } \text{ the function } \frac{f^{(n)}}{p} \text{ is bounded} )$$
  4. $\mathcal T_u(\mathbb R) \subsetneq \mathcal T(\mathbb R)$, were $\mathcal T_u(\mathbb R)$ is the r.h.s of (1) and $\mathcal T(\mathbb R)$ is the r.h.s of (2).

What of those statments are true and what are wrong? Do you have references for any of them?

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  • $\begingroup$ Would you mind to show the argument? $\endgroup$
    – Dirk
    Commented Jul 9, 2012 at 13:08
  • $\begingroup$ Things in 4 are not defined. $\endgroup$
    – Marc Palm
    Commented Jul 9, 2012 at 13:11
  • $\begingroup$ To Dirk, it is rather long, we now write it in details and if we will not find the mistake I will upload it. $\endgroup$
    – Rami
    Commented Jul 9, 2012 at 13:23
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    $\begingroup$ The last sentence was not clear: I want to say that I think that 1 is wrong, since the Fourier transforms $\phi \ast \delta$ is the product of the Fourier transforms. Certainly not every distribution is a product of a Schwartz function and distributions, since this would give the distribution rather restrictive growth conditions probably similar to $T_u$ and $T$. So $\widehat{c}$ is surjective, but $c$ is not. $\endgroup$
    – Marc Palm
    Commented Jul 11, 2012 at 15:36
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    $\begingroup$ Note that the space you call $T(\mathbb{R})$ is more commonly called $\mathcal{O}_M$. It has a canonical locally convex topology w.r.t. which convolution is a separately continuous bilinear map $\mathcal{S}\times\mathcal{S}'\to\mathcal{O}_M$. By some abstract argument (both $\mathcal{S}$ and $\mathcal{S}'$ are barrelled) it follows that this map is even hypocontinuous (meaning it is continuous on $\mathcal{S}\times B$ and $C\times\mathcal{S}'$ for all bounded subsets $B$,$C$)... $\endgroup$
    – Johannes Hahn
    Commented Sep 28, 2014 at 21:31

2 Answers 2

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A bit late and far from answering all your questions but perhaps still worth mentioning:

A proof of the inclusion ${\rm Im}(c)\subseteq$ RHS of item 2) can be found in "Topological Vector Spaces and Distributions" by John Horváth. See in particular Proposition 4.11.7 on page 420.

BTW, as Johannes Hahn said the standard notation of for RHS of item 3) is $\mathcal{O}_M$. That for the RHS of item 2) is $\mathcal{O}_C$.

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The answer to the question in the title is: {Fourier $(\phi u)$} $_{\phi\in \mathscr S, u\in \mathscr S'}$.

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