Let $\Omega \subset \mathbb{R}^n$ be open and bounded.

What does weak-* convergence for a sequence of functions $\{f_k\}_{k \in \mathbb{N}}$ in $W^{1,\infty}(\Omega)$ mean? It seems to me that there is no clear identification in literature of the pre-dual of $W^{1,\infty}$ hence no clear understanding of what weak star convergence effectively means computationally.

I am actually not interested in the characterization of it, but in the question highlighted in bold below.

So, suppose that $f_k \rightharpoonup^* f$ in $W^{1,\infty}(\Omega)$. Since we know that a set is weak* compact iff it is weak* closed and bounded in norm, in particular $\{f_k\}_{k \in \mathbb{N}}$ is bounded in $W^{1.\infty}(\Omega)$. For the same reason, considering now only $L^\infty$, and eventually taking a subsequence, there exist a $\bar f \in W^{1,\infty}(\Omega)$ such that $$ f_k \rightharpoonup^*\bar f \text{ in }L^\infty(\Omega) \quad \text{and} \quad \nabla f_k \rightharpoonup^* \nabla \bar f \text{ in }L^\infty(\Omega).$$

$\mathbf{\text{Question:}}$ is $f = \bar f?$

How could one prove that? It looks doable but not having any information on the pre-dual of $W^{1,\infty}$ or its canonical pairing lets me with no ideas.

Let $\Omega \subset \mathbb{R}^n$ be open and bounded.

What does weak-* convergence for a sequence of functions $\{f_k\}_{k \in \mathbb{N}}$ in $W^{1,\infty}(\Omega)$ mean? It seems to me that there is no clear identification in literature of the pre-dual of $W^{1,\infty}$ hence no clear understanding of what weak star convergence effectively means computationally.

I am actually not interested in the characterization of it, but in the question highlighted in bold below.

So, suppose that $f_k \rightharpoonup^* f$ in $W^{1,\infty}(\Omega)$. Since we know that a set is weak* compact iff it is weak* closed and bounded in norm, in particular $\{f_k\}_{k \in \mathbb{N}}$ is bounded in $W^{1.\infty}(\Omega)$. For the same reason, considering now only $L^\infty$, and eventually taking a subsequence, there exists a $\bar f \in W^{1,\infty}(\Omega)$ such that $$ f_k \rightharpoonup^*\bar f \text{ in }L^\infty(\Omega) \quad \text{and} \quad \nabla f_k \rightharpoonup^* \nabla \bar f \text{ in }L^\infty(\Omega).$$

Question: Is $f = \bar f$?

How could one prove that? It looks doable but not having any information on the pre-dual of $W^{1,\infty}$ or its canonical pairing leaves me with no ideas.