Showing a linear map has a unique extension from a Sobolev space to an Lp space This is related to another question I asked: Directional derivative in a Sobolev-like inequality
Suppose that $u \in C_{0}^{\infty}(\Omega)$. Show that the linear map $u \to u(x,0) \in C^{\infty}(\mathbb{R}^{n})$ has a unique continuous extension as a map from $W^{1,p}_{0}(\Omega)$ to $L^{p}(\partial \Omega_{+})$, where $\partial \Omega_{+} \equiv \{x;(x,0)\in\Omega\}$.
I know that you can define extensions from one Sobolev space to another. Is the reason why we can do it here from a Sobolev space to an $L^{p}$ space because $L^{p}$ spaces are contained inside Sobolev spaces? Also, I was thinking about using the BLT Theorem, but do I know that the map from $u$ to $u(x,0)$ is bounded?
I'm very confused, and not as well-versed in functional analysis as I should be, so any assistance/hints you could give me for this proof would be much appreciated!
 A: $\Omega$ is bounded, so there is a K such that $\Omega \subset \mathbb{R}^n\times (-\infty,K)$. Then it's basically John's hint from the other question: For $u \in C_0^\infty(\Omega)$, we have
$$u(x,0) = - \int_0^K \frac{\partial u}{\partial x_{n+1}}(x,t)\,dt.$$
Hölder's inequality gives
$$\lvert u(x,0)\rvert \leqslant \left(\int_0^K \left\lvert \frac{\partial u}{\partial x_{n+1}}(x,t)\right\rvert^p\,dt\right)^{1/p}\cdot \left(\int_0^K 1^{p/(p-1)}\,dt\right)^{(p-1)/p},$$
and from that we obtain
$$\int_{\partial \Omega_+} \lvert u(x,0)\rvert^p\,dx \leqslant K^{p-1} \int_{\partial\Omega_+} \int_0^K \left\lvert \frac{\partial u}{\partial x_{n+1}}(x,t)\right\rvert^p\,dt\,dx \leqslant K^{p-1} \left\lVert \frac{\partial u}{\partial x_{n+1}}\right\rVert_{L^p(\Omega)}^p,$$
or, writing $\rho(u)(x) = u(x,0)$,
$$\lVert \rho(u)\rVert_{L^p(\partial \Omega_+)} \leqslant K^{1-\frac1p}\left\lVert \frac{\partial u}{\partial x_{n+1}}\right\rVert_{L^p(\Omega)} \leqslant K^{1-\frac1p}\lVert u\rVert_{W_0^{1,p}(\Omega)}.\tag{1}$$
The estimate $(1)$ shows that
$$\rho \colon C_0^{\infty}(\Omega) \to L^p(\partial\Omega_+);\quad \rho(u)(x) = u(x,0)$$
is continuous if we endow $C_0^\infty(\Omega)$ with the subspace topology induced by $W_0^{1,p}(\Omega)$. By the denseness of $C_0^\infty(\Omega)$ in $W_0^{1,p}(\Omega)$, there is a unique continuous linear $\tilde{\rho}\colon W_0^{1,p}(\Omega) \to L^p(\partial\Omega_+)$ that extends $\rho$.
