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Notations:

$ H_{per}^s(0,2\pi):= \{f \in L^2(0,2\pi): \sum_{n \in \mathbb{Z}} (1+n^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 1: How can we find a precise characterization for $ \mathcal{U}_{per} $?, or a weaker version, for a appropriate subset of $ \mathcal{U}_{per} $?

The first guess strike us is

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

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

Thoughts:

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} $.

This provides decay rate of polynomial of abitrary finite order, which is not enough to validate

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

However, it inspires us to assume a more rapid decay rate. For this sake, we define

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

Now we notice that $ f \in H_{a}^s(0,2\pi) $ would possess Fourier coefficients of expotential decay rate. Would it induce that

$ H_{a}^s(0,2\pi) \subset \mathcal{U}_{per} $?

If any researcher could provide information on the $ H_{a}^s(0,2\pi) $?

Thanks in advance!

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    $\begingroup$ Your first definition is that of the periodic Sobolev space, not of the non-periodic one. The two coincide only for $s<1/2$, so it is not a good definition to let $s\to\infty$. I think you should clarify definitions before going further. $\endgroup$ Commented Apr 3, 2019 at 14:49
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    $\begingroup$ In the case of $C^\infty_{per}$, do you really not need the derivatives to match? $\endgroup$ Commented Apr 3, 2019 at 14:49
  • $\begingroup$ @ Jean Duchon It is really kind of you to specify the long-standing error in my basic knowledge for Sobolev spaces. Thank you, deeply! $\endgroup$
    – Yidong Luo
    Commented Apr 3, 2019 at 16:07
  • $\begingroup$ @ Willie Wong I think it needs. I will modify it in a while. $\endgroup$
    – Yidong Luo
    Commented Apr 3, 2019 at 16:13

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