Usually, for $1\leq p<\infty$, we define $W_0^{1,p}(\Omega)$, where $\Omega$ is open bounded smooth boundary, by taking the closure of $C_c^\infty(\Omega)$ under $W^{1,p}$ norm. However, we don't define $W_0^{1,\infty}(\Omega)$ in the same way since we will end up with $C_0^1(\Omega)$, which in particular does not contain piece-wise linear function.
Hence, we say $W_0^{1,\infty}(\Omega)$ usually by the following, (at least when I study the quasi-convex functions), $$ W_0^{1,\infty}(\Omega):=\{u\in W^{1,\infty}(\Omega),\,\,T[u]=0\}$$ where $T$ is the trace operator. Hence in this definition $W_0^{1,\infty}$ will contain piece-wise linear functions. Good.
Today I saw another definition of $W_0^{1,\infty}$ by define $$ E(\Omega):=\{\text{The weak star $W^{1,\infty}$ closure of }C_c^\infty(\Omega)\} $$
I think the space $E(\Omega)$ will contain all Lipschitz functions for sure and I even think $$E(\Omega)=W^{1,\infty}_0(\Omega)$$ What I can do so far is the function belongs to $E$ will at least have $0$ trace and hence $E$ contains $W_0^{1,\infty}(\Omega)$. But I can not show the converse. Can anybody help me about this? Thank you!
Update: Here I worked out one possible answer.
Let's first claim that $E(\Omega)\subset W_0^{1,\infty}(\Omega)$. Suppose $(u_n)\subset C_c^\infty(\Omega)$ and $u_n\to u$ weakly in $W^{1,\infty}$, we show that $u\in W_0^{1,\infty}(\Omega)$. Since $\Omega$ is open bounded with nice boundary, we have $u_n\to u$ strongly in $L^\infty$ and hence we conclude that $u\in C^0(\bar\Omega)$ with $T[u]=0$. Next, by lower semi-continuous of weak star convergence, we have $$ \|\nabla u\|_{L^\infty(\Omega)}\leq \liminf \|\nabla u_n\|_{L^\infty(\Omega)}<\infty $$ Hence we have $u\in W_0^{1,\infty}(\Omega)$ which conclude my first claim.
Now we work on the converse.
Take $u\in W_0^{1,\infty}(\Omega)$, our job is to find a sequence $(u_n)\subset C_c^\infty(\Omega)$ such that $u_n\to u$ weakly star in $W^{1,\infty}$. To do so, fix such $u\in W_0^{1,\infty}(\Omega)$ and in view of my old post here, we could have $(v_n)\subset C^\infty(\Omega)$ such that $v_n\to u$ in $L^\infty$ strong and $\nabla v_n\to \nabla u$ weakly star in $L^\infty$. Now we need to chopper $(v_n)$ to obtain $(u_n)$ as we want.
To do so, let's define $\Omega_\epsilon:=\{x\in\Omega,\,\,\text{dist}(x,\partial\Omega)>\epsilon\}$ and we define $\phi_\epsilon\in C_c^\infty(\Omega)$ such that $\phi\equiv 1$ on $\Omega_\epsilon$ and $\phi\equiv 0$ on $\Omega\setminus\Omega_{\epsilon/2}$. We then define $$ u_{n,\epsilon}:=\phi_\epsilon v_n $$
We first observe that $$ \lim_{\epsilon\to 0}\lim_{n\to\infty}\|u_{n,\epsilon}-u\|_{L^\infty(\Omega)}=0\tag 1 $$
Secondly, we have $$ \|\nabla (u_{n,\epsilon})\|=\|\nabla\phi_\epsilon v_n+\phi_\epsilon\nabla v_n\| \leq \frac{1}{n\epsilon}+\|\nabla v_n\|$$
Here I am going to choose $\epsilon$ and $n$ so that $1/\epsilon n$ will remain finite and hence we have $\|\nabla (u_{n,\epsilon})\|_{L^\infty}$ is uniformly bounded.
Write $u_{n,\epsilon}=u_{\epsilon_n}=u_n$ such that $u_n\to u$ in $L^\infty$ and $\nabla u$ is $L^\infty$ bounded. Hence we have $\nabla u_n\to v$ weakly star in $L^\infty$. Finally, for any $\varphi\in C_c\infty(\Omega)$, we have $$\int_\Omega \nabla u_n \varphi\,dx=-\int_\Omega u_n\nabla \varphi\to \int_\Omega u \nabla \varphi dx $$ Moreover $$ \int_\Omega \nabla u_n \varphi\,dx\to \int_\Omega v\varphi\,dx $$ hence we have $v=\nabla u$ and we done.
Together with $(1)$, we have $u_{n,\epsilon}\to u$ weakly star in $W^{1,\infty}$ for properly choosing $n$ and $\epsilon$. Hence, we done.
