I have the following question: Let $(u_n)_n \in W^{s,1}(0,\pi)$, $s\in (0,1)$, be a uniformly bounded sequence with the $||.||_{W^{s,1}(0,\pi)}$ norm (with Gagliardo semi-norm). From the compact embedding $W^{s,1}(0,\pi)\subset\subset L^1(0,\pi)$ there exists a subsequence $(u_{n_k})$ that strongly converges to $u\in L^1(0,\pi)$. Is the limit function $u$ in $W^{s,1}(0,\pi)$? I can't find any reference that proves or disproves the reflexivity of this space (although I think it is not).
Any help will be appreciated! Thank you in advance.
EDIT: Taking a look at the book "Theory of functions spaces" (Hans Triebel, 1983), page 178, the Besov space $B^{s,1}$ is not reflexive, and since the space $B^{s,1}$ coincides with the space $W^{s,1}$, I cannot use reflexivity to prove the above problem (am I saying it wrong...? I know very little to nothing about Besov spaces).
So, as I edited earlier in the question, the space $W^{s,1}(0,\pi)$ is not reflexive. Fortunately, there is another way to prove the above statement, and it uses Fatou's lemma.
Because of the compact embedding, we have strong convergence in $L^1(0,\pi)$. Hence, there exists a subsequence $u_{n_k}$ converging with the $L^1$ norm to a function $u\in L^1(0,\pi)$ and, in particular, converging almost everywhere in $(0,\pi)$. Thus, the sequence $\frac{|u_{n_k}(x)-u_{n_k}(y)|}{|x-y|^{1+s}}$ will also converge almost everywhere to the function $\frac{|u(x)-u(y)|}{|x-y|^{1+s}}$. We have only left to apply Fatou's lemma on $(0,\pi)\times(0,\pi)$ to obtain \begin{equation*} \int_0^\pi\int_0^\pi \frac{|u(x)-u(y)|}{|x-y|^{1+s}} dydx \leq \lim_{n\to\infty} \inf \int_0^\pi\int_0^\pi \frac{|u_{n_k}(x)-u_{n_k}(y)|}{|x-y|^{1+s}} dydx \leq M, \end{equation*} because of the uniform bound for $u_n$.