Are lattice operations continuous in the Lipschitz norm?

Question

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.

Answer 1

No, the lattice operations are not continuous, in general.

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.