# Is fractional order Sobolev spaces reflexive?

Let $0<s<1$, we define $$W^{s,p}(\Omega):=\left\{u\in L^p(\Omega),\,\frac{|u(x)-u(y)|}{|x-y|^{\frac{N}{p}+s}}\in L^p(\Omega\times\Omega)\right\}$$ with norm $$\|u\|:=\left(\int_{\Omega} |u|^p\,dx+\int_{\Omega\times\Omega}\frac{|u(x)-u(y)|^p}{|x-y|^{N+sp}}\,dx\,dy\right)^{\frac{1}{p}}.$$ Is $W^{s,p}(\Omega)$ uniformly convex and reflexive when $1<p<\infty$, as usual Sobolev space? How to prove? What happen for general fractional order $s$?

Define $T:W^{s,p}(\Omega)\to L^p(\Omega)\times L^p(\Omega\times\Omega)$ by $$T(u)=\left(u,\frac{|u(x)-u(y)|}{|x-y|^{N/p+s}}\right),$$
where $W^{s,p}(\Omega)$ is equipped with the norm given by you and $L^p(\Omega)\times L^p(\Omega\times\Omega)$ is equipped with the norm $$\|(u,v)\|=\left(\int_\Omega |u|^p+\int_{\Omega\times\Omega}|v(x,y)|^p\right)^{1/p}.$$
Note that $$\|T(u)\|=\left(\int_\Omega |u|^p+\int_{\Omega\times\Omega}\frac{|u(x)-u(y)|^p}{|x-y|^{N+sp}}\right)^{1/p}=\|u\|_{s,p},$$
which implies that $T$ is an isometry. Can you conclude?
Remark: If $s>2$, the idea is similar.