Timeline for What functions can be obtained as a convolution of a Schwartz function and a tempered distribution?
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 23, 2018 at 17:54 | answer | added | Abdelmalek Abdesselam | timeline score: 3 | |
| Sep 28, 2014 at 21:34 | comment | added | Johannes Hahn | ... I'm not sure since I do not know topological tensor products well-enough but I think the latter implies that $\ast:\mathcal{S}\otimes\mathcal{S}'\to\mathcal{O}_M$ is continuous. | |
| Sep 28, 2014 at 21:31 | comment | added | Johannes Hahn | 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 $B$,$C$)... | |
| Jul 16, 2012 at 7:20 | comment | added | Marc Palm | Okay, I suspected already that convergence in the completed tensor-product topology isn't the same as convergence in the range. | |
| Jul 13, 2012 at 10:41 | comment | added | Rami | To Mrc Plm (follow up comment). One such example is related to the Dixmier-Malliavin Theorem that you have mentioned: Let $c':C^{\infty}_c \otimes C^{\infty}_c \to C^{\infty}$ be the convolution map and $\hat c':C^{\infty}_c \hat \otimes C^{\infty}_c \to C^{\infty}$ be its extention to the completion. Then $Im(c)=Im(\hat c)=C^{\infty}_c \neq C^{\infty}$. I believe that our situation is similar but I cant figure out what the image is in our case. | |
| Jul 13, 2012 at 10:39 | comment | added | Rami | To Mrc Plm. I do not expect $c$ or $\hat c$ to have a closed image. I understand that both maps have dense image. I do not think that $\hat c$ is surjective. Note that although the source of $\hat c$ is a complete space, it does not mean that the image of $\hat c$ is closed. There are many examples of dense image maps between complete spaces (see follow up comment). | |
| Jul 11, 2012 at 15:36 | comment | added | Marc Palm | 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. | |
| Jul 9, 2012 at 18:36 | comment | added | Marc Palm | I would have a look at MR0517765: This says that $$C_c^\infty(\mathbb{R}) \hat{\ast} C_c^\infty(\mathbb{R}) = C_c^\infty(\mathbb{R}).$$ So both maps have dense image. I am not sure why you expect that $c$ has closed image, why should every distribution be the product of a distribution and a Schwartz function, simply because you can use Fourier transform. | |
| Jul 9, 2012 at 15:41 | answer | added | Bazin | timeline score: 2 | |
| Jul 9, 2012 at 14:57 | comment | added | Jochen Wengenroth | I do not believe Im$(c)=$Im$(\hat{c})$. I have not checked but the statements 2. and 3. look plausible. | |
| Jul 9, 2012 at 13:28 | comment | added | Rami | To Mrc Plm, Do you mean $\mathcal T_u(\mathbb R)$ and $\mathcal T(\mathbb R)$? As I said: $$T_u(\mathbb R)=(f \in C^\infty(\mathbb R)|\exists \text{ a polinomial }p \text{ s.t. } \forall n\in \mathbb N \text{ the function } \frac{f^{(n)}}{p} \text{ is bounded} )$$ $$T(\mathbb R)=(f \in C^\infty(\mathbb R)|\forall n\in \mathbb N, \exists \text{ a polinomial }p \text{ s.t. } \text{ the function } \frac{f^{(n)}}{p} \text{ is bounded} )$$ | |
| Jul 9, 2012 at 13:23 | comment | added | Rami | To Dirk, it is rather long, we now write it in details and if we will not find the mistake I will upload it. | |
| Jul 9, 2012 at 13:11 | comment | added | Marc Palm | Things in 4 are not defined. | |
| Jul 9, 2012 at 13:08 | comment | added | Dirk | Would you mind to show the argument? | |
| Jul 9, 2012 at 11:45 | history | asked | Rami | CC BY-SA 3.0 |