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:

- $\mathrm{Im} (c)=\mathrm{Im} (\hat c)$
- $$\mathrm{Im} (c)=(f \in C^\infty(\mathbb R)|\exists \text{ a polinomial }p \text{ s.t. } \forall n\in \mathbb N \text{ the function } \frac{f^{(n)}}{p} \text{ is bounded} )$$
- $$\mathrm{Im} (\hat c)=(f \in C^\infty(\mathbb R)|\forall n\in \mathbb N, \exists \text{ a polinomial }p \text{ s.t. } \text{ the function } \frac{f^{(n)}}{p} \text{ is bounded} )$$
- $\mathcal T_u(\mathbb R) \subsetneq \mathcal T(\mathbb R)$, 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?

7more comments