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Background: Related question

I am trying to prove, that the countably-normed spaces $C^\infty_0(B_1)$ on the open unit ball (i.e. function and all derivatives vanish at the boundary) in $\mathbb{R}^n$ and the Schwartz Space $\mathscr{S}(\mathbb{R}^n)$ are isomorphic. Here, $\mathscr{S}$ carries the usual Schwartz topology and $C^\infty_0(B_1)$ the locally convex topology induced by the family $\|\varphi\|_i=\sup_{x\in B_1}\max_{|\alpha|\leq i}|D^\alpha\varphi|$.

My ansatz is to use, that the open unit ball and whole $\mathbb{R}^n$ are diffeomorphic, e.g. via the diffeomorphism $$ h:x\rightarrow \frac{x}{\sqrt{1+|x|^2}} $$ therefore my guess for the isomorphism is $$ I:C^\infty_0(B_1)\rightarrow \mathscr{S}(\mathbb{R}^n), f\rightarrow f\circ h $$ However, I am unable to prove, that $I$ and $I^{-1}$ are continuous with respect to the locally convex topology.

Concretely, I have to show that for every pair $p,q\in \mathbb{N}$ there exist $r\in \mathbb{N},C>0$ with $$ \sup_{x\in \mathbb{R}^n}\max_{\alpha\leq q} (1+|x|^2)^p|D^\alpha (\varphi \circ h) | \leq C \sup_{x\in B_1}\max_{\alpha\leq r} |D^\alpha \varphi | $$ for all $\varphi\in C_0^\infty(B_1)$, similar for $I^{-1}$. The crux about the continuity has to be, that the vanishing at the boundary of $B_1$ for some function $\varphi$ somehow also implies the vanishing at infinity for $I(\varphi)$, however, I am not able to formulate a clean proof for this.

Does anybody have an idea or a reference? I found a similar statement in Treves (Topological Vector Spaces) Theorem 51.4, unfortunately, the proof of the theorem was "left as exercise", and I am obviously too dumb to solve it.

I would greatly appreciate any help!

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  • $\begingroup$ Maybe you can use the rather technical arguments in 3.21 of www2.math.uni-wuppertal.de/~vogt/vorlesungen/fs.pdf $\endgroup$ – Vobo Aug 30 '14 at 14:04
  • $\begingroup$ Thank you, I already stumbled over that script. However, I am seeking for an elementary proof. My feeling is, that this has to be possible. $\endgroup$ – Daniel Aug 31 '14 at 21:35
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    $\begingroup$ Your I is not well-defined. It maps $C_0^\infty(B_1)$ to $C_0^\infty(\mathbb{R}^n)$ which is a proper super space of $\mathcal{S}(\mathbb{R}^n)$. Just because all derivatives vanish at infinity the function does not have to be Schwartz. Simple counter example $\frac{1}{1+\|x\|^2}$. $\endgroup$ – Johannes Hahn Sep 2 '14 at 16:00
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    $\begingroup$ Have you tried using a different diffeomorphism between $B(0,1)$ and $\mathbb{R}^n$, like: $$x\to -x\,\log(1-|x|)$$ ? $\endgroup$ – Jack D'Aurizio Sep 3 '14 at 19:41
  • $\begingroup$ @JohannesHahn : It seems like I have interchanged $h$ and $h^{-1}$, my bad. If I map some function from $B_1$ to a function in $\mathscr{S}$, $h$ should of course take the value back. The mapping how I defined it has also values in $\mathscr{S}$, which do not have compact support, e.g. take the image of $\exp(-\frac{1}{1-x^2})\in C_0 ^\infty(B_1)$. $\endgroup$ – Daniel Sep 4 '14 at 22:03

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