Are lattice operations in a Lipschitz space sequentially continuous in the weak* topology?

Question

This is a follow-up on this (answered) question on math.SE, but involves a different topology. I think this time it is more appropriate for MO. I will repeat the background from the question cited above.

Denote by $Lip_0(X)$ the set of all Lipschitz functions on a metric space $X$ vanishing at some base point $e \in X$. The norm in $Lip_0$ is defined as follows $$ \|f\|_{Lip_0} := Lip(f), $$ where $Lip(f)$ denotes the Lipschitz constant. With pointwise operations $f \vee g := \max\{f,g\}$ and $f \wedge g := \min\{f,g\}$ the space $Lip_0$ becomes a Lipschitz lattice, in which the following condition holds $$ \|f \vee g\|_{Lip_0} \leq \max\{\|f\|_{Lip_0},\|g\|_{Lip_0}\}. $$ The Banach lattice condition $|f| \leq |g| \implies \|f\| \leq \|g\|$, however, fails. (Nik Weaver. Lipschitz Algebras, 2nd ed.)

For a large class of metric spaces $X$, the space $Lip_0(X)$ has a unique predual, which is called the Arens-Eels space or the Lipschitz-free space, depending on the community. It can be seen as the completion of the space of Radon measures with zero mean $\mathcal M_0(X)$ with respect to the dual Lipschitz norm $$ \|\mu\|_{Lip*} := \sup\{\langle \mu,f \rangle \colon \|f\|_{Lip_0} \leq 1\}. $$ What is added by this completion are limits as $d(x,y) \to 0$ of linear combinations of the so-called elementary molecules $$ m_{xy} := \frac{1}{d(x,y)}(\delta_x - \delta_y), $$ where $d(x,y)$ is the distance between $x,y \in X$ and $\delta_x, \delta_y$ are delta-functions placed at $x,y$. (Nik Weaver. Lipschitz Algebras, 2nd ed.)

As pointed out in the answer to the question I cited above, lattice operations $f_+ := f \vee 0$, $f_- := (-f) \vee 0$ and $|f| := f \vee (-f)$ are not continuous in the $Lip_0$ norm, i.e. $$ \|f_n - f\|_{Lip_0} \to 0 \quad \text{does not imply} \quad \|(f_n)_+ - f_+\|_{Lip_0} \to 0. $$

Question. Are operations $f_+ := f \vee 0$, $f_- := (-f) \vee 0$ and $|f| := f \vee (-f)$ sequentially continuous in the weak* topology, i.e. does $$ f_n \rightharpoonup^* f \implies (f_n)_+ \rightharpoonup^* f_+ $$ hold?

Any help will be much appreciated.

Answer 1

Yes. If $f_n \to f$ weak* then the sequence $(f_n)$ must be bounded in ${\rm Lip}_0(X)$ (Banach-Steinhaus), and for bounded nets weak* convergence is the same as pointwise convergence. So $f_n \to f$ boundedly pointwise, which easily implies the same of the positive parts.

Let me also correct a couple of inaccuracies in your post: first, we do know that ${\rm Lip}_0(X)$ has a unique predual for a large class of spaces $X$ (if it has finite diameter, or if it is convex), but this is not known for all $X$. Second, what is added in the completion is limits of linear combinations of elementary molecules.