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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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Would you mind to show the argument? –  Dirk Jul 9 '12 at 13:08
    
Things in 4 are not defined. –  plusepsilon.de Jul 9 '12 at 13:11
    
To Dirk, it is rather long, we now write it in details and if we will not find the mistake I will upload it. –  Rami Jul 9 '12 at 13:23
    
To Mrc Plm, Do you mean $\mathcal T_u(\mathbb R)$ and $\mathcal T(\mathbb R)$? As I said: $$T_u(\mathbb R)=(f \in C^\infty(\mathbb R)|\exists \text{ a polinomial }p \text{ s.t. } \forall n\in \mathbb N \text{ the function } \frac{f^{(n)}}{p} \text{ is bounded} )$$ $$T(\mathbb R)=(f \in C^\infty(\mathbb R)|\forall n\in \mathbb N, \exists \text{ a polinomial }p \text{ s.t. } \text{ the function } \frac{f^{(n)}}{p} \text{ is bounded} )$$ –  Rami Jul 9 '12 at 13:28
    
I do not believe Im$(c)=$Im$(\hat{c})$. I have not checked but the statements 2. and 3. look plausible. –  Jochen Wengenroth Jul 9 '12 at 14:57

1 Answer 1

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

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