Is $W_0^{1,p}(\Omega)$ complemented in $W^{1,p}(\Omega)$? Let $\Omega\subset\mathbb{R}^N$ be a bounded domain and $p\in (1,\infty)$. It is know that there exists a unique bounded surjective linear map $T: W^{1,p}(\Omega)\to W^{1-1/p,p}(\partial\Omega)$ with the property that $$Tu=u_{|\partial\Omega},\ \forall\ u\in C^\infty(\overline{\Omega})$$
It is also know that there exists a bounded linear map $\ell : W^{1-1/p,p}(\partial\Omega) \to W^{1,p}(\Omega)$ which is a right inverse for $T$. This map is sometimes called the lift map (see Necas Theorem 5.7). In proving the existence of $\ell$, Necas explicit constructs the map $\ell$.
I want to avoid this construction and use an argument from Functional Analysis, to wit, Theorem 2.12 of Brezis. To use it, I have to prove that $W_0^{1,p}(\Omega)$ is complemented in $W^{1,p}(\Omega)$, however I am stuck here. How would I go about this?
 A: Partial answer: let's consider the simplest case: let $p=2$. Fix a $g\in H^{1/2}(\partial \Omega)$. For any $v \in H^1_0(\Omega)$, if there is some $w\in H^1(\Omega)$, the trace of $w$ is $g$, then 
$$
w\in H^1_0(\Omega)^{\perp}\iff 
\langle  w,v\rangle_{H^1}:=\int_{\Omega} (\nabla w \cdot \nabla v + wv) = 0 \quad \text{for all } v \in H^1_0(\Omega).
$$
This is equivalent to say that $w$ is the weak solution to 
$$
\begin{cases}
-\Delta w + w = 0 &\text{in } \Omega,
\\
w = g\in H^{1/2}(\partial \Omega) &\text{on }\partial \Omega.
\end{cases}\tag{1}
$$
This roughly tells us that the orthogonal complement of $H^1_0(\Omega)$ should be functions like $w$.
Such $w$ is constructed similar to an extension from a $g\in H^{1/2}(\partial \Omega)$. Due to problem (1)'s well-posedness, we can define a linear operator  based on (1):
$$L: H^{1/2}(\partial \Omega)\to H^1(\Omega),\; g\mapsto Lg:= w.$$
We also have $\|Lg\|_{H^1(\Omega)} \leq c\|g\|_{H^{1/2}(\partial \Omega)}$. This result says the boundedness of $L$. The argument can be found in the two additional lemmas in the updates below.
Also notice that if $g=0$, then $w=0$, this implies the intersection is $\{0\}$. Hence for any $u \in H^1(\Omega)$, let the trace operator be $T$
$$
u = \underbrace{L(Tu)}_{\in (H^1_0(\Omega))^{\perp}} + \underbrace{\big(u - L(Tu)\big)}_{\in H^1_0(\Omega)}.
$$
This argument can be easily extended to $p\geq 2$ case for $W^{1,p}\subset W^{1,2}$.

Some updates:

Lemma 1: Suppose $\Omega$ has whatever regularity is needed. Let $u\in H^1(\Omega)$ be the weak solution of$$
\begin{cases}
-\Delta u + u = f &\text{in } \Omega,
\\
u = g\in H^{1/2}(\partial \Omega) &\text{on }\partial \Omega.
\end{cases} 
$$
  There is a $v\in H^1(\Omega)$ such that $u-v \in H^1_0(\Omega)$, then
  $$
\|u\|_{H^1(\Omega)}\leq C\Big(\|f\|_{H^{-1}(\Omega)} + \|v\|_{H^{1}(\Omega)}\Big).\tag{2}
$$

Proof: Use $u-v\in H^1_0$ as the test function:
$$
\int_{\Omega} \nabla u\cdot\nabla(u-v) + \int_{\Omega}u(u-v) = \int_{\Omega}f(u-v),
$$
which is
$$
\|u\|_{H^1(\Omega)} := \int_{\Omega} |\nabla u|^2 +\int_{\Omega}u^2 
= \int_{\Omega}f(u-v) + \int_{\Omega}\nabla u\cdot\nabla v + \int_{\Omega}uv
\\
\leq \frac{1}{\epsilon} \|f\|_{H^{-1}(\Omega)} 
\epsilon \|u-v\|_{H^1(\Omega)} + \frac{1}{2}\int_{\Omega}
\big(|\nabla u|^2+|\nabla v|^2\big) + \frac{1}{2}\int_{\Omega}(u^2+v^2).
$$
Then use Minkowski inequality again on $\|u-v\|_{H^1(\Omega)}$, choose $\epsilon$ small enought such that the $\|u\|_{H^1(\Omega)}$ term can be absorbed by the left. Also notice the choice of $\epsilon$ is irrelevant of $v$. The estimate in (2) will follow. 

Lemma 2: The weak solution $w$ of (2) satisfies the estimate $\|w\|_{H^1(\Omega)} \leq c\|g\|_{H^{1/2}(\partial \Omega)}$.

Proof: Choose any $v\in H^1_g(\Omega):=\{v\in H^1(\Omega): Tv = g\in H^{1/2}(\partial \Omega)\}$, i.e., the traces coincide. We have $w-v\in H^1_0$, hence by (2), we have
$$
\|w\|_{H^1(\Omega)} \leq C\|v\|_{H^1(\Omega)} \quad\text{for all }v\in H^1_g(\Omega). \\
\implies c \|w\|_{H^1(\Omega)} \leq \inf_{v\in H^1_g(\Omega)}\|v\|_{H^1(\Omega)} 
=: \|g\|_{H^{1/2}(\partial \Omega)}.
$$
The last step is just using an implicit way of introducing the half norm (even though proving this quotient like norm is equivalent to the usual fractional norm on boundary still needs a constructive proof......).
