Extension and trace operators for Sobolev spaces Given that $\Omega \subset \mathbb{R}^{n}$ is an open, convex, Lipschitz bounded set. Let $O \subset \Omega$ be open bounded set then consider. $$u_{m} \rightharpoonup^{*} u \text{ in } W^{1,\infty}_{o}(O)$$ Assume that we can find $v_{m}$ such that $$v_{m} \rightharpoonup^{*} u \text{ in } W^{1,\infty}(O)$$$$ v_{m} = u \text{ on } \partial O$$ Assume that we apply the linear continuous operator $$P: W^{1,\infty}(O) \rightarrow W^{1,\infty}(\Omega)$$ We then extend $u$ from $O$ to $\Omega$ in such a way that the extension $\bar{u} \in W^{1,\infty}(\Omega)$ and similarly since $v_{m} = u$ on $\partial O$ we define ${\bar{v}}_{m} = v_{m}$ in $O$ and ${\bar{v}}_{m} = \bar{u}$ in $\Omega - O$.
We then have $${\bar{v}}_{m} \rightharpoonup^{*} \bar{u} \text{ in } W^{1,\infty}_{o}(\Omega)$$


*

*In Lawrence's book "Partial Differential Equations" the trace operator isn't defined for $p = \infty$. If the trace operator isn't defined how would you describe how '$v_{m} = u$ on $\partial O$' is defined?

*Can we immediately state that $v_{m} \in W^{1,\infty}_{o}(O)$ since $v_{m} = u$ on $\partial O$ and $u \in W^{1,\infty}_{o}(O)$?

*Do we get ${\bar{v}}_{m} \rightharpoonup^{*} \bar{u}$ in $W^{1,\infty}_{o}(\Omega)$ since we have that from the extension operator it follows that $P_{2}: W^{1,\infty}_{o}(O) \rightarrow W^{1,\infty}_{o}(\Omega)$ is also a continuous linear extension operator? If this is true, how would you go about showing this?
Thanks for any assistance
 A: I'm not sure about question 1. Interesting question though, since I note also that the trace is not defined for $p = \infty$ in two of Evans books (Partial Differential Equations and Measure Theory and Fine Properties of Functions) as well as Brezis book (Function Analysis, Sobolev Spaces and Partial Differential Equations). 
For question 2: I would say yes. Since if the trace is defined for $p = \infty$, then by definition it follows that $u \in W^{1,p}_{o}(\Omega)$ iff $Tu = o$ on $\partial \Omega$. Where $T$ is the bounded linear trace operator $T: W^{1,p}(\Omega) \rightarrow L^{p}(\partial \Omega)$.
For question 3: I don't think you have to apply the extension theorem. You could just extend ${\bar{v}}_{m}$ and $\bar{u}$ by zero on $\Omega\setminus \bar{O}$. If we assume that $O$ is also lipschitz then we note that the $n$-dimensional measure of the boundary of $O$ is zero. So it does not affect our Lebesgue integrals. We can show ${\bar{v}}_{m} \rightharpoonup^{*} \bar{u}\text{   } \text{  in } W^{1,\infty}_{o}(\Omega)$ as follows:
Assuming we have the following extension:
$$\begin{align}
\bar{u} = \begin{cases} u~~~ \text{ in } O \\ 0~~~~ \text{ in } \Omega\setminus O  
\end{cases}
\end{align}
$$
 and
$${\bar{v}}_{m} = \begin{cases} v_{m}~~ \text{  in  } O \\ \bar{u}~~~~~ \text{   in  }  \Omega \setminus O.
\end{cases}
$$ 
Take any $\phi \in L^{1}(\Omega)$, this yields $$\lim\limits_{m \rightarrow \infty}  \int_{\Omega}({\bar{v}}_{m} - \bar{u})\phi dx = \int_{O}(v_{m}-u)\phi dx + \int_{\Omega \setminus O}(u-0)\phi dx.$$  
The first integral converges to $0$ since you assumed that $v_{m} \rightharpoonup^{*} u$ in $W^{1,\infty}(O)$. The second integral is $\int_{\Omega \setminus O} 0 dx = 0$. This gives ${\bar{v}}_{m} \rightharpoonup^{*} \bar{u}$ in $L^{\infty}(\Omega)$. Similary, you can show that $\nabla {\bar{v}}_{m} \rightharpoonup^{*} \nabla \bar{u}$ in $L^{\infty}(\Omega; \mathbb{R}^{n})$. Conclusion ${\bar{v}}_{m} \rightharpoonup^{*} \bar{u}$.  $\square$
A: Expanding the comment by user127096.
The space $W^{1,\infty}({\cal{O}})$ is equal to the space of bounded locally Lipschitz functions. This means for every element $u\in W^{1,\infty}({\cal{O}})$ it is possible to find a function $v$ that is Lipschitz continuous by just rearranging a set of measure zero.
The trace then can be taken as you take the trace of continuous functions by pointwise convergence.
