Weakly null sequence in Banach lattices

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.

• I am not sure if I understand disjoint correctly. Wouldnt your definition be the same as $\inf_n \{|x_n|\}=0$? – supinf Mar 27 at 13:03
• No. Disjoint is stronger. For example, the sequence $(x_n)_n:=(1,1,0,0,0,\cdots)$ in $\mathbb{R}$ haves $\inf_n|x_n|=0$, but $\inf\{|x_1|,|x_2|\}=1\neq 0$. – L26 Mar 27 at 13:24
• Then your notation is incorrect. $\inf_{n\neq m}$ means you minimize over all $n,m$. Maybe you want to say is that $\inf \{|x_n|,|x_m|\}=0\forall n\neq m$? – supinf Mar 27 at 13:58
• It's true. My notation is incorrect. Thanks! – L26 Mar 27 at 15:10

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 $$y_n\rightharpoonup0$$.
• Thanks! I think that your explanation holds for every Banach lattice, because the norm dual of a Banach lattice $E$ coincide with its order dual. And the order dual is a Dedekind complete Riesz space, so it is possible write $f\in E'$ as $f=f^++f^-$. – L26 Mar 27 at 15:26