1

If $x \in \mathbb{R}^{d}$ or $\mathbb{Z}^{d}$, I will denote $||x|| := \sqrt{x_{1}^{2}+\cdots + x_{d}^{2}}$. Let: $$s(\mathbb{Z}^{d}) := \{\varphi \in \mathbb{R}^{\mathbb{Z}^{d}}: ||\varphi||_{k} := \operatorname{sup}_{x\in \mathbb{Z}^{d}}||x||^{k}|\varphi(x)| < \infty \quad \forall k \in \mathbb{N}\}$$ and $$\mathscr{S}(\mathbb{R}^{d}) := \{f \in C^{\infty}(\mathbb{R}^{d};\mathbb{C}): \operatorname{sup}_{x\in \mathbb{R}^{d}}||x||^{k}|\partial^{\alpha}f(x)| < \infty \quad \forall k \in \mathbb{R}, \alpha \in \overbrace{\mathbb{N}\times \cdots \times \mathbb{N}}^{\text{$d$ times}}\}$$

Both $s(\mathbb{Z}^{d})$ and $\mathscr{S}(\mathbb{R}^{d})$ are locally convex spaces, with family of seminorms given, respectively, by $||\cdot||_{k}$ and $||\cdot||_{k,\alpha}$.

I know for a fact that $s(\mathbb{R}^{d})$ and $\mathscr{S}(\mathbb{R}^{d})$ are homeomorphic and I once saw a proof of this fact but I no longer remember where I saw it. I tried using Hermite polynomials, but actually it led me nowhere so far; if $d=1$, I guess we could use raising and lowering operators, but I don't know how to address the problem in the general case. Any references or hints are really welcomed!

EDIT: As pointed out in the comments, the word homeomorphic should be changed to isomorphic as topological vector spaces.

JustWannaKnow
  • 4,075
  • I doubt that's true; seems like it "should" be a space of periodic functions on $\Bbb R^d$. – David C. Ullrich Nov 15 '21 at 21:44
  • @DavidC.Ullrich: It is true, but it not well known. – Abdelmalek Abdesselam Nov 16 '21 at 13:55
  • @IamWill: The word "homeomorphic" is wrong. It should be isomorphic as topological vector spaces. In other words, the homeomorphism should also be a linear map. – Abdelmalek Abdesselam Nov 16 '21 at 13:56
  • @AbdelmalekAbdesselam I'm going to fix it. Thanks for pointing it out! – JustWannaKnow Nov 16 '21 at 13:57
  • for a reference see the comment I put here https://math.stackexchange.com/questions/2268519/is-the-schwartz-function-space-separable/3105629#3105629 – Abdelmalek Abdesselam Nov 16 '21 at 13:58
  • @AbdelmalekAbdesselam thanks for the reference. I knew this paper from Barry Simon, but the problem is that all the sequence spaces used in all proofs I know (Simon's included) are indexed by $\mathbb{N}$ rather than $\mathbb{Z}^{d}$, and the latter seems more useful when applying to, say, physics problems. I always accepted the fact that these two sequence spaces are actually the same thing, but I have never proved it myself. It seems obvious, because you can pass from $\mathbb{Z}^{d}$ to $\mathbb{N}$ trivially, but the norms involved seem tricky to me. – JustWannaKnow Nov 16 '21 at 14:04
  • This is why I actually asked another question right before posting this one: https://math.stackexchange.com/questions/4307628/how-can-i-convert-norms-using-the-bijection-between-mathbbn-and-mathbb It is concerned with the relation between these two sequence spaces. I thought if I could solve the latter, I could use it to prove the former. – JustWannaKnow Nov 16 '21 at 14:05

0 Answers0