This is a follow-up on [this (answered) question on math.SE](https://math.stackexchange.com/questions/3802588/are-lattice-operations-continuous-in-the-lipschitz-norm), 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.



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


