I understand the meaning of $u_n$ converges to $u$ weak star (it means that $u_n\in E^*$ and $(u_n,x)_{E^*,E} \to (u,x)_{E^*,E}$ for all $x\in E$) but I've some trouble for identifying a space $E$ such that $E^* = W^{1,\infty}(\Omega)$ (where $\Omega$ is an open subset of $\mathbb{R}^n$). I know that we can identify $L^\infty(\Omega)$ to $L^1(\Omega)^*$ but what happens for $W^{1,\infty}(\Omega)$ ?
It seems that we can identify the predual of $W^{1,\infty}(\Omega)$ as $W^{1,1}(\Omega)$ as Terence Tao says here but where can I find a reference of this fact?
You may view $W^{1,\infty}(\Omega)$ as a subspace of $L^\infty(\Omega)^{n+1}$ (via $u \mapsto (u,\nabla u$), which can be identified with the dual of $L^1(\Omega)^{n+1}$.
Thus, for any bounded sequence $u_n$, there is a subsequence (without relabeling), such that $u_n \stackrel{*}{\rightharpoonup} u$ and $\partial_i u_n \stackrel{*}{\rightharpoonup} v_i$ in $L^1(\Omega)^*$. For any $\varphi \in C_0^\infty(\Omega) \subset L^1(\Omega)$ you have \begin{equation*} \int_\Omega u_n \, \partial_i \varphi \, \mathrm{d}x = -\int_\Omega \partial_i u_n \, \varphi \, \mathrm{d}x \end{equation*} by definition of the weak derivative. Passing to the limit $n \to \infty$ yields \begin{equation*} \int_\Omega u \, \partial_i \varphi \, \mathrm{d}x = -\int_\Omega v_i \, \varphi \, \mathrm{d}x. \end{equation*} Hence, $u \in W^{1,\infty}(\Omega)$ and $\partial_i u = v_i$.