Notations:

$ H_{per}^s(0,2\pi):= \{f \in L^2(0,2\pi): \sum_{n \in \mathbb{Z}} (1+k^2)^s \vert \hat f(n) \vert^2 < +\infty \} $,

$ H_{per}^{\infty}(0,2\pi):= \bigcap_{s>0} H_{per}^s(0,2\pi) $,

$ C^{\infty}_{per}[0,2\pi] :=\{ u \in C^{\infty} [0,2\pi]: u^{(k)}(0) = u^{(k)}(2\pi), \ \forall k \in \mathbb{N}\} $.

We define

$\mathcal{U}_{per} := \{ u \in H_{per}^{\infty} (0,2\pi): \lim_{s\to +\infty} \Vert u \Vert^{\frac{1}{s}}_{H_{per}^s} \textrm{ exists} \}$.

The question is

Question: $C^{\infty}_{per}[0,2\pi] \subseteq \mathcal{U}_{per} $ ?

Notice that $ \mathcal{U}_{per} \neq \emptyset $, $ u:= a_N \cos N t \in \mathcal{U}_{per} $.

Personal thoughts:

I find a subtle place. For $ f \in C^{\infty}_{per}[0,2\pi] $, its Fourier series could be termwise differentiated in any finite times. Then

$ \lim_{n \to \infty} n^k \hat f (n) \ \textrm{exists} \ \forall k \in \mathbb{N} $.

Now, can it yields that $ \lim_{s\to +\infty} \Vert u \Vert^{\frac{1}{s}}_{H_{per}^s} $ exists? That is,

$ \lim_{k \to \infty} (\sum_{n \in \mathbb{Z}} (1+n^2)^k \hat f (n) )^{\frac{1}{k}} \textrm{exists}$ ?

Now we can observe from above formula that decay rate of polynomial of any finite order for Fourier coefficients $ \hat f(n) $ is not enough. Whether can we develop some kind of decay rate in a transcendent sense?

Thanks in advance!