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

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.

• Note: MathJax supports bold on its own; please use double stars **double stars** rather than $\mathbf{\text{math mode}}$ $\mathbf{\text{math mode}}$. I have edited accordingly. Commented Oct 24, 2021 at 18:13
• Please see the book "Lipschitz Algebras" by Nik Weaver, especially Chapter 3 about the predual. Commented Oct 26, 2021 at 13:48
• What is the predual of $W^{1,\infty}$? In other words, what does weak$^*$ mean, actually? Commented Nov 2, 2021 at 19:06

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, $$0 = \int_\Omega \sum_{j = 0}^n v_j f_j \, dx = \sum_{j = 1}^n \int_\Omega v_0 D_j\eta_j + v_j \eta_j \, dx \implies v_j = D_jv_0 \quad j = 1, \dots, n.$$
• Notation: I would distinguish, for $A\subset X$ and $B\subset X^*$, between annihilator $A^\perp\subset X^*$ and pre-annihilator $B_\perp:=B^\perp\cap X\subset X$ (where we regard $X$ as canonically embedded in its bidual $X^{**}$) Commented Dec 9, 2023 at 17:23