Let $\Omega$ be a bounded domain in $\Bbb R^m$ with smooth enough boundary so that $W^{1,\infty}(\Omega)=\text{Lip}(\Omega)$.
Let $(u_n)$ be a sequence in $W^{1,\infty}(\Omega)$. What does it mean for the sequence to converge weak*-ly in $W^{1,\infty}(\Omega)$?
I don't know what the pre-dual of $W^{1,\infty}(\Omega)$ is. I know that for $p'=p/(p-1)$, $$ W_0^{k,p}(\Omega)^* = W^{-k,p'}(\Omega) $$ but I don't know if it makes sense to talk about $W_0^{-1,1}(\Omega)$ or what it is, if it exists at all.
I would guess that $u_n\overset{*}{\rightharpoonup} u$ in $W^{1,\infty}$means that $$ \int_\Omega u_n f + \nabla u_n\cdot \mathbf g \ dx \to \int_\Omega u f + \nabla u\cdot \mathbf g \ dx $$ for all $f\in L^1(\Omega)$ and all $\mathbf g\in L^1(\Omega;\Bbb R^m)$. Even if this is true I still want to know the name of this space and how we know that its dual is indeed $W^{1,\infty}(\Omega)$.