Topological isomorphism (homeomorphism) between $C^{\infty}(S^1)$ and $s(\mathbb{Z})$

The space $s(\mathbb{Z})$ of rapidly decreasing sequences is defined as follows: $$s(\mathbb{Z})= \{x = (x_n) \in \mathbb{K}^{\mathbb{Z}}:\|x\|_k = \sum_{n \in \mathbb{Z}}|x_n||n^k| < \infty \;\forall k \in \mathbb{Z_{\geq0}} \} $$ The topology on $s(\mathbb{Z})$ is determined by the family $\{\|\;.\|_k:k \in \mathbb{Z_{\geq 0}}\}$ of norms.I know how to topologize the space $C^{\infty}[0,1]$, by analogy how to topologize $C^{\infty}(S^1)$? I suppose, that topology on $C^{\infty}(S^1)$ is given by the family of norms: $$\{\|\;.\|_n: n \in \mathbb{Z_{\geq 0}} \}$$ Where $$\|f\|_{n} = max_{t \in S^1}|f^{(n)}(t)|$$ And the main question is, how to prove that $C^{\infty}(S^1)$ topologically isomorphic (homeomorphic) to $s(\mathbb{Z})$? I think this homeomorphism is given bu the following function: $$H: C^{\infty}(S^1) \longrightarrow s(\mathbb{Z})$$ $$f \longrightarrow c = (c_n)$$ Where $c_n$ is $n$-th Fourier coeffitient of $f$ by the system $e^{in\varphi}$, where $n \in \mathbb{Z}$, $\varphi \in [0, 2\pi]$. Is it true?And if it's so, how to prove that it's a topological isomorphism?

I think the continuity of $H^{-1}$ is quite simple: let $f(t)=\sum_{n=-\infty}^\infty c_ne^{int}$, then $$|f^{(k)}(t)|\leq \sum_{n=-\infty}^{\infty}|c_n (in)^ke^{int}|=\sum_{n=-\infty}^{\infty}|c_n||n^k|\quad \forall t\in S^1\mbox{ and } \forall k\in\mathbb Z_{\geq0},$$ which implies that $||f||_k\leq ||c||_k$, for all nonnegative integer $k$.

And since $f=H^{-1}(c)$, we have that $H^{-1}$ is bounded, hence continuous.

For the continuity of $H$, I have no idea how to prove it either. I've tried using Parseval's theorem, but it only gives $$||c||_{\ell^2}=\sqrt\frac{1}{\pi}||f||_{L^2}\leq||f||_{L^\infty}$$

It would be nice if someone can complete this proof.

To prove the continuity of $H$, we can use the following lemma and theorem:

Lemma Let $f:X\rightarrow Y$ be a linear map of topological vector spaces where $Y$ is locally convex. Then $f$ is continuous iff for each continuous seminorm $q$ on $Y$, there is a continuous seminorm $p$ of $X$ such that $q\circ f\leq p$.

Theorem Let $P$ be a countable family of seminorms of $X$ which defines its topology. And write

• $P:=\{p_n:n\in\mathbb Z_{\geq0}\}$,
• $P_1:=\{\sum_{n\in I}p_n:p_n\in P,I\subset\mathbb Z_{\geq0},card(I)<\infty\} $,
• $P_2:=\{\sum_{k=0}^n p_k:p_k\in P, n\in\mathbb Z_{\geq0}\}$,
• $P_3:=\{\sup_{k=0}^np_k:p_k\in P, n\in\mathbb Z_{\geq0}\}$.

Then $P,P_1,P_2,P_3$ induces the same topology on $X$.

The lemma implies the theorem. If both are true, then by defining $$\rho_n^p(x):=\sup_{k=0}^n(\sum_{m=-\infty}^\infty|m^kx_m|^p)^{(\frac{1}{p})}\quad\forall p\in [1,\infty),\,n\in\mathbb{Z}_{\geq 0},\,x\in S(\mathbb{Z})$$ $$\rho_n^{\infty}(x):=\sup_{k=0}^n\sup_{m=-\infty}^{\infty}|m^kx_m|\quad\forall n\in\mathbb{Z}_{\geq 0},\,x\in S(\mathbb{Z})$$ and the corresponding seminorms $\sigma_n^p,\sigma_n^{\infty}$, respectively, on $C^{\infty}(S^1)$ , we have that

$$\rho_n^{\infty}(c)\leq\rho_n^2(c)=\sqrt\frac{1}{\pi}\sigma_n^2(f)\leq\sigma_n^{\infty}(f)$$ and $$\rho_n^1(x)\leq(1+\frac{\pi^2}{3})\rho_{n+2}^{\infty}(x)\tag{*}$$

Notice that $\rho_n^1(x)=\sup_{k=0}^n||x||_k$, which, by theorem, defines the same topology as the given one, similar to $\sigma_n^{\infty}(f)=\sup_{k=0}^n||f||_k$. Thus $$\rho_n^1(H(f))\leq(1+\frac{\pi^2}{3})\sigma_{n+2}^{\infty}(f)$$ Therefore $H$ is bounded, hence continous.

The inequality $(*)$ is also quite simple: since $$|n^kx_n|\leq\sup_{n=-\infty}^{\infty}|n^kx_n|\leq\sup_{\ell=0}^k\sup_{n=-\infty}^{\infty}|n^\ell x_n|=\rho_k^\infty(x)$$ and especially, $$|n^kx_n|\leq\frac{\rho_{k+2}^\infty(x)}{|n|^2},\,\mbox{ for }n\neq0$$ thus for $k\neq0$, $$\sum_{n=-\infty}^\infty|n^kx_n|\leq|0^kx_0|+2\sum_{n=1}^\infty\frac{\rho_{k+2}^\infty(x)}{n^2}=(2\sum_{n=1}^\infty\frac{1}{n^2})\rho_{k+2}^\infty(x)=\frac{\pi^2}{3}\rho_{k+2}^\infty(x)$$ and for $k=0$, $$\sum_{n=-\infty}^\infty|n^kx_n|\leq|x_0|+2\sum_{n=1}^\infty\frac{\rho_{k+2}^\infty(x)}{n^2}\leq\rho_{k+2}^\infty(x)+\frac{\pi^2}{3}\rho_{k+2}^\infty(x)=(1+\frac{\pi^2}{3})\rho_{k+2}^\infty(x)$$ thus $$\rho_n^1(x)\leq(1+\frac{\pi^2}{3})\rho_{n+2}^\infty(x),\,\forall n\in\mathbb{Z}_{\geq 0}.$$