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.
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.