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

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.

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.