1
$\begingroup$

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.

$\endgroup$
3
  • 3
    $\begingroup$ 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. $\endgroup$
    – LSpice
    Oct 24, 2021 at 18:13
  • $\begingroup$ Please see the book "Lipschitz Algebras" by Nik Weaver, especially Chapter 3 about the predual. $\endgroup$
    – Onur Oktay
    Oct 26, 2021 at 13:48
  • $\begingroup$ What is the predual of $W^{1,\infty}$? In other words, what does weak$^*$ mean, actually? $\endgroup$ Nov 2, 2021 at 19:06

1 Answer 1

1
$\begingroup$

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, $$ 0 = \int_\Omega \sum_{j = 0}^n v_j f_j \, dx = \sum_{j = 1}^n \int_\Omega v_0 D_j\eta_j + v_j \eta_j \, dx \implies v_j = D_jv_0 \quad j = 1, \dots, n.$$

$\endgroup$
1
  • $\begingroup$ 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 $X^{**}$) $\endgroup$ Dec 9, 2023 at 17:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.