Suppose I have a lattice $X$ with a weak Fatou norm, meaning that there exists some constant $M$ (clearly $\geq 1$) such that for all $x\geq 0$ and for all increasing nets $x_\alpha$ with $x_\alpha \uparrow x$, we have $ \| x \| \leq M \| x_\alpha \| $. Under what conditions do there exist disjoint elements $0 \leq y ,z \leq x$ where both $y$ and $z$ have relatively "large" norms? That is, is there some $K > 0$ such that if $\|x \| = 1$, I can ensure that for some positive, disjoint $y$ and $z$ bounded by $x$ we have both $\|y\| \geq K$ and $\|z\| \geq K$? I've tried looking through the main literature but can't find anything specific to this.
Fatouness is important here, since it can affect the size of the smallest norm. For example, If $X$ is the sublattice of $c$ of convergent sequences $(a_1,a_2,...)$ where $M \lim_n a_n =a_1$. Then if you have the sequence $(1, 1/M,1/M,...)$, the most $K$ can be is $1/M$, since any element $x$ containing $a_1 > 0$ must also contain cofinitely many $a_n$'s, leaving $y$ with only finite support. On the other hand, a lattice without such a norm might not guarantee any such positive $K$ (see for example, a $c_0$ sum of the lattices above with increasing $M\in \mathbb N$).
