Norm inequality for Banach Lattices

Question

Let $X$ be a Banach lattice, $\varepsilon>0$ and $x,y\in S_X$.

If $\||x|\pm y\|\le1+\varepsilon$, then is it true that $\||x|+|y|\|\le1+f(\varepsilon)$?

Where $f$ is some real function satisfying $\lim_{x\rightarrow 0}f(x)=0$.

I'm stuck in answering this question. So far I noticed that this statements holds if $X=C(K)$ and in general $$|x|+|y|=|x|+(y\vee-y)=(|x|+y)\vee(|x|-y).$$ But even if $|x|+|y|$ is the $\sup$ of two elements of which we can control the norm, this doesn't imply that a similar control holds for $\||x|+|y|\|$. Is there any obvious counterexample I'm missing?

Answer 1

This seems correct in AM-spaces (such as $C(K)$) but I think it fails in AL-spaces: Consider the Banach lattice $\mathbb{R}^2$ endowed with the $l^1$-norm. For $x=(1/2,1/2)$ and $y=(1/2,-1/2)$ we have $\|x\|=\|y\|=1$. As $x=|x|$ we get $$ \||x| \pm y \| =1 ~~ \text{hence} ~~ \forall \varepsilon > 0: \||x| \pm y \| \le 1+\varepsilon, $$ but $\||x|+|y|\| = 2$.