Let $A:=\{ u \in H^k(0,2\pi): u^{(j)}(0)=u^{(j)}(2\pi) \mbox{ for } j=0,1,\ldots, k-1\}$, where $H^{k}(0,2\pi)\subseteq L^2(0, 2 \pi)$ is the Sobolev space of order $k$ on $(0, 2 \pi)$. Can we say that $u \in A$ iff 
$$\sum\_{n=-\infty}^\infty (1+n^2)^k |\hat{u}(n)|^2<\infty?$$
In the above series $\hat{u}(n)$ are the Fourier coefficients of $u$. We think that the answer to this question is affirmative because maybe we can identify $A$ with the Sobolev space of the torus $H^k(\mathbb{T})$, and use [this result](https://en.wikipedia.org/wiki/Sobolev_space#The_case_p_=_2). But we don't know how to show that there is an isomorphism between $A$ and $H^k(\mathbb{T})$.


Do you know any reference for a characterization of $A$ with Fourier series?


Thank you for any help you can provide us.



Characterization of a subset of the Sobolev space $H^k(0,2\pi)$ in terms of Fourier series
This is indeed true, and let me illustrate this in the special case $k=1$.


We are given a function $u\in H^1(0, 2\pi)$ such that $u(0)=u(2\pi)$. First we note that this boundary condition indeed makes sense because by Sobolev embedding $u$ is continuous. We also know that $u$ has a (at least distributional) derivative $v\in L^2(0, 2\pi)$. After identifying 0 and $2\pi$ and making the domain a circle, I claim that the distributional derivative of $u$ is still $v$. Indeed, the only difference from before is that we are now testing $u$ against functions that are smooth on the circle, i.e., functions on $[0, 2\pi]$ whose derivatives at the two endpoints agree to all orders. Then it is easy to see that when testing against $u$, we perform an integration by parts and get two boundary terms that cancels each other, and hence get the right distributional derivative. Thus we can naturally identify $A$ and $H^k(\mathbb R/2\pi\mathbb Z)$, and the rest follows easily.


