Does weak-* convergence in $W^{1,\infty}$ imply weak-* convergence in $L^\infty$?

Question

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.

Answer 1

I was wondering about the same question and was surprised that I could not find an answer to this anywhere. The answer to your question is yes.

Let $(X, \|\cdot\|)$ be a normed vector space and $Y \subset X$ a closed subspace. Then $(X / Y)^*$ is canonically isomorphic to $Y^\perp$, where $Y^\perp = \{\varphi \in X^* : \ker(\varphi) \subset Y \}$ is the annihilator of $Y$. The isomorphism is given by \begin{align} &\Phi : Y^\perp \rightarrow (X / Y)^*, \\ &\Phi(\varphi)([x]) = \varphi(x). \end{align}

Hence, if a subset $F \subset X^*$ satisfies $F = Y^\perp$ for some $Y$, then $$ (X/ Y)^* \cong F, $$ i.e. $F$ has a predual. Given $F$, a natural candidate for $Y$ is $F^\perp = \{x \in X : x \in \ker(\varphi) \text{ for all } \varphi \in F \}$, the pre-annihilator of $F$. However, in general it is not true that $$ F = (F^\perp)^\perp \tag{1}\label{1}. $$ Fortunately, $W^{1, \infty}(\Omega)$ satisfies this property for a suitable choice of $X$. For the moment, assume $(1)$ holds. Given a sequence $(\varphi_k) \subset F$, we compare weak$^*$ convergence in $X^*$ and $F$. Since $(X / F^\perp)^* \cong F$, $\varphi_k \overset{\ast}\rightharpoonup \varphi$ in $F$ means \begin{align} \Phi(\varphi_k) \overset{\ast}\rightharpoonup \Phi(\varphi) \text{ in } (X / F^\perp)^* \iff &\Phi(\varphi_k)([x]) \to \Phi(\varphi)([x]) \quad \forall [x] \in X / F^\perp \\ \iff &\varphi_k(x) \to \varphi(x) \quad \forall x \in X. \end{align} But this is the definition of weak$^*$ convergence in $X^*$, so we are done.

It remains to show that $W^{1, \infty}(\Omega)$ satisfies $(1)$. Let $X = L^1(\Omega, \mathbb{R}^{n + 1})$ and $F = W^{1, \infty}(\Omega)$. We think of $F$ as a subset of $X^* = L^\infty(\Omega; \mathbb{R}^{n + 1})$ by identifying $u \in W^{1, \infty}(\Omega)$ with $(u, D_1u, \dots, D_nu) \in L^\infty(\Omega, \mathbb{R}^{n + 1})$. By definition, $F^\perp$ consists of all $(f_0, \dots, f_n) \in L^1(\Omega, \mathbb{R}^{n + 1})$ such that $$ \int_\Omega u f_0 + \sum_{j = 1}^n D_ju f_j \, dx = 0 \quad \forall u \in W^{1, \infty}(\Omega).$$ Suppose $v \in (F^\perp)^\perp \subset L^\infty(\Omega, \mathbb{R}^{n + 1})$, i.e. $$ \int_\Omega \sum_{j = 0}^n v_j f_j \, dx = 0 \quad \forall f \in F^\perp.$$ We prove that $v \in F$ by showing that it is weakly differentiable. For $\eta \in C_c^\infty(\Omega; \mathbb{R}^n)$ we define a function $f \in L^1(\Omega; \mathbb{R}^{n + 1})$ by \begin{align} f_0 &= \text{div}\, \eta, \\ f_j &= -\eta_j \quad j = 1, \dots, n. \end{align} Observe that $f \in F^\perp$. This follows simply from the definition of the weak derivative. Hence, $$ 0 = \int_\Omega \sum_{j = 0}^n v_j f_j \, dx = \sum_{j = 1}^n \int_\Omega v_0 D_j\eta_j + v_j \eta_j \, dx \implies v_j = D_jv_0 \quad j = 1, \dots, n.$$