Let $1\leq p\leq +\infty$, $0<s<1$ and $\Omega\subseteq \mathbb{R}^n$ an open set. The fractional Sobolev space $W^{s,p}(\Omega)$ is defined to be
$$ W^{s,p}(\Omega) = \left\{ u\in L^p(\Omega) : \frac{|u(x)-u(y)|}{|x-y|^{\frac{n}{p} + s}} \in L^p(\Omega\times\Omega) \right\} $$
equipped with the norm
$$ \|u\|_{W^{s,p}(\Omega)} = \left( \int_\Omega |u|^p \; dx + \int_\Omega\int_\Omega \frac{|u(x)-u(y)|^p}{|x-y|^{n+ sp}} \; dx dy \right)^{1/p}. $$
It is a well known result that this is a Banach space, but every reference I read says that this is true, without giving a proof.
Adam's Sobolev Spaces uses techniques from interpolation theory to prove this result, but I'm no familiar with the theory, so I'm asking for an "elementary proof", that is, as usual, proving that every Cauchy sequence has a limit in the space. Here is my "attempt":
Let $(u_n)$ be a Cauchy sequence in $W^{1,p}(\Omega)$, then $(u_n)$ is a Cauchy sequence in $L^p(\Omega)$, so there exists $u\in L^p(\Omega)$ such that $u_n\to u$ in $L^p(\Omega)$. I would like to prove that $u\in W^{s,p}(\Omega)$ and that $u_n\to u$ in $W^{s,p}(\Omega)$. But I don't know how to proceed here.
Thanks for the help!
