Weakly null sequence in Banach lattices

Question

Let $(x_n)_n$ be a positive, disjoint, weakly null sequence in a Banach lattice $E$. If $(y_n)_n$ is a sequence such that $0\leq y_n\leq x_n$ for every $n\in \mathbb{N}$, we can garantee that $y_n$ is weakly null?

Remarks:

"Weakly null" means that $x_n$ converges to zero in the weak topology.

"disjoint" means that $\inf\{|x_n|,|x_m|\}=0 ~ \forall n\neq m$.

"positive" means that $x_n\geq 0$ for all $n\in\mathbb{N}$.

Thanks for any explanations.

Answer 1

If $E^*$ allows for a decomposition of functionals into positive and negative parts then this can be proven. This is true at least if $E$ is a Hilbert lattice or equal to some $L^p$ with the natural ordering ($p\in [1,+\infty)$).

Let $f\in E^*$. Then there are $f^+,f^-$ with $f=f^++f^-$ with $f^+\ge0$, $f^-\le0$. Then it follows $0\le f^+(y_n) \le f^+(x_n)$ and $0\le -f^-(y_n) \le -f^-(x_n)$. This implies $y_n\rightharpoonup0$.