Extension and trace operators for Sobolev spaces

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.

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 ${\bar{v}}_{m} \rightharpoonup^{*} \bar{u}$. $\square$