Let $x_1\in \mathbb{R}^n$, $n\geq 3$, $\Omega=\mathbb{R}^n\backslash B_1(x_1)$, define $D_{\Omega}$ by taking the closure of $C_{c}^{\infty}(\overline{\Omega})$ under the norm \begin{align*} \|u\|_{D_{\Omega}}:=\Big(\int_{\Omega}|\nabla u|^2\Big)^{1/2}+\Big(\int_{\Omega}|u|^{2n/(n-2)}\Big)^{(n-2)/2n}. \end{align*} My question is whether it is true for the following Sobolev trace inequality $$ \int_{\Omega}|\nabla \Phi|^2 \leq C(n,\Omega) \|\varphi\|_{H^{\frac{1}{2}}(\partial B_1(x_1))}^2 $$ for any $\Phi\in D_{\Omega}$ satisfying $$\Phi|_{\partial B_1(x_1)}=\varphi.$$ where $H^{\frac{1}{2}}$ is the fractional space.
The paper [Hitchhiker’s guide to the fractional Sobolev spaces] records that: for any $p \in(1,+\infty)$, assume that the open set $\Omega \subseteq \mathbb{R}^n$ is sufficiently smooth, then the space of traces $T u$ on $\partial \Omega$ of $u$ in $W^{1, p}(\Omega)$ is characterized by $\|T u\|_{W^{1-\frac{1}{p}, p}(\partial \Omega)}<+\infty$). Moreover, the trace operator $T$ is surjective from $W^{1, p}(\Omega)$. I guess it has a similar result for the above inequality. Thanks for any help.
