Assumption on bounded open in trace theorem- Sobolev Space I have a problem in the proof of the theorem $$T(u)=0\implies u\in W_0^{1,p}(U).$$
Assume $U\subset \Bbb R^n$ bounded open and $\partial U$ is $C^1$. Suppose $u\in W^{1,p}(U)$.
My question: Why can I assume that  $U=\Bbb R^n\cap \{x_n>0\}$ with the support of $u$ in $\overline{\Bbb R^n \cap \{x_n>0\}}$?
My attempt: the fact that $\partial U$ is $C^1$ implies by definition that for every $x_1\in \partial U$ exist $r>0$ and a function $C^1$ $\gamma:\Bbb R^{n-1}\rightarrow \Bbb R$ such that $$U\cap B(x_1,r)=\{ x\in B(x_1,r)| \gamma(x_1,...,x_{n-1}>x_n \}.$$ In this way i can define the two functions
$$\left\{ 
\begin{array}{c}
y_i=x_i=:\phi_i(x) \hspace{0.9 cm} (i=1,...,n-1) \\ 
y_n=x_n-\gamma(x_1,...,x_{n-1})=:\phi _n(x)
\end{array}\right. $$ and   $$\left\{ \begin{array}{c}
    x_i=y_i=:\Psi_i(y) \hspace{0.9 cm} (i=1,...,n-1) \\ 
 x_n=y_n+\gamma(y_1,...,y_{n-1})=:\psi _n(y)
\end{array}\right.$$
Now i define $\tilde{u} = u\circ \psi (y)$. In this way i have $T(\tilde {u})=0$ in $L^p(\partial \psi (U\cap B(x_1,r))$. The fact that $\partial U$ is compact implies exist $x_1,...,x_n$ such that $$\partial U \subset (U\cap B(x_1)) \cup \cdots \cap (U\cup B(x_n)).$$ Consider $V_0 \subset \subset U$ such that $$U\subset V_0 \cup (U\cap B(x_1)) \cup \cdots \cap (U\cup B(x_n)).$$ Now I want to use unity partition to define  a general $\tilde{u}$ but i don't know how define the function associated with the open $V_0$. How can i do? How can i construite the general function $\tilde{u}$?
 A: By choosing the balls appropriately, you can assume that $\overline{U}$ is contained in $V_0\cup\bigcup_{i=1}^mB(x_i,r_i/2)$. Consider a partition of unity $\varphi_0$, $\ldots$, $\varphi_m$ subordinated to $V_0$, $B(x_1,r_1/2)$, $\ldots$, $B(x_n,r_m/2)$.
Since  $\tilde u=0$ on $y_n=0$, if you extend $\tilde u=0$ to be zero for $y_n<0$, the extended function is still a Sobolev function (just use integration by parts to see this) in $\psi(B(x_i,r_i))$. Using the fact that translation is continuous in $L^p$, you can define $\tilde{v}_i(y)=\tilde u(y_1,\ldots,y_{n-1},y_n-\delta_i)$, where $\delta_i$ is so small that $$\|\tilde{u}-\tilde{v}_i\|_{W^{1,p}(\psi(U\cap B(x_i))}<\frac{\varepsilon}{m}.$$ The function $\tilde v_1$ is zero in a neighborhood of the boundary $y_n=0$.
Define $$v=\varphi_0u+\sum_{i=0}^m\varphi_i(\tilde v_i\circ\psi_i^{-1}).$$
Since $\varphi_0$ is compactly supported in $U$, you have that $v$ has compact support contained in $U$. Then you can mollify. I skipped a lot of details, but this is the basic idea to approximate $u$ with smooth functions with compact support.
What Jose27 is saying is correct. In each ball you have a different change of variables, so constructing a global change of variables seems very difficult and possibly impossible in most cases.
When people write "without loss of generality, we can assume that $U$ is the half-space", they usually just mean that they will give a "local proof" in a small ball.
A: I apologize in advance that I can't write a detailed answer due to insufficient time, and am going to try to briefly explain the main ideas.

*

*First of all, we can assume that $u\in C^\infty(U)$ by the Meyers-Serrin theorem, or even $u\in C^\infty(\bar U)$ because the boundary is $C^1$ (in fact the so-called segment property is enough for this).


*Then as mentioned in the other answer, we use a partition of unity to split the function into little pieces. You can flatten the boundary if you want, but this is not strictly necessary.


*Now the main point of the problem (which I think was missed by the other answer) is that we can't assume $u$ becomes $0$ as we approach $\partial U$. In other words, when we extend $u$ by $0$ outside $U$, the resulting function will in general have discontinuity across $\partial U$. To deal with this, we approximate $u$ by a smooth function supported compactly in $U$, and need to make use of the fact that the trace of $u$ vanishes. The approximation can be done through multiplication by a smooth cut-off function.


*The details of how to make use of the condition $T(u)=0$ depends (in a non-essential way) on the exact definition of trace you are following, but essentially what $T(u)=0$ means is the following. Imagine that you peel off a thin layer (like skinning a potato) from $U$, and call the resulting boundary surface $\Sigma$. Now imagine $\Sigma$ is moving outward slowly and approaching the former boundary $\partial U$. As $\Sigma$ approaches $\partial U$, the $L^p$-norm of $u$ on $\Sigma$ should go to $0$:
$$
\big\|u|_{\Sigma}\big\|_{L^p(\Sigma)}\to0.
$$


*I am sorry I need to run now but look at the 1-dimensional case first. It will make things clearer.
