# Are lattice operations continuous in the Lipschitz norm?

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 fololows $$\|f\|_{Lip_0} := Lip(f),$$ 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.)

Question. Are operations $$f_+ := f \vee 0$$, $$f_- := (-f) \vee 0$$ and $$|f| := f \vee (-f)$$ continuous in the $$Lip_0$$ norm, i.e. does, e.g., $$\|f_+ - g_+\|_{Lip_0} \leq C\|f - g\|_{Lip_0}$$ hold?

I have searched a lot for either a proof or a counterexample, but couldn't find anything. Any help will be appreciated.

An easy way to see this is to consider the space of Lipschitz continuous functions on $$[0,1]$$ (without base point, and the norm being the supremum of the infinity norm and the Lipschitz constant).
Set $$f_\varepsilon(x) := x - \varepsilon$$ for all $$x \in [0,1]$$. Then $$f_\varepsilon \to f_0$$ as $$\varepsilon \to 0$$, but convergence of the absolute values does not hold.
If you insist on using a base point, add the point $$-1$$ and set all functions to $$0$$ there.
• @Yury: Thank you for the correction! I actually intended to write $f_0$ instead of $0$ :-). Corrected. Wunderfull question, by the way - for several years I have been wondering now and then whether the lattice operations in a lattice ordered Banach space with non-normal cone are automatically continuous; but I never thought thoroughly enough about it - until I read your question. Aug 28, 2020 at 15:51