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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.

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  • $\begingroup$ I am not sure if I understand disjoint correctly. Wouldnt your definition be the same as $\inf_n \{|x_n|\}=0$? $\endgroup$ – supinf Mar 27 at 13:03
  • $\begingroup$ 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 $\inf\{|x_1|,|x_2|\}=1\neq 0$. $\endgroup$ – L26 Mar 27 at 13:24
  • $\begingroup$ Then your notation is incorrect. $\inf_{n\neq m}$ means you minimize over all $n,m$. Maybe you want to say is that $\inf \{|x_n|,|x_m|\}=0\forall n\neq m$? $\endgroup$ – supinf Mar 27 at 13:58
  • $\begingroup$ It's true. My notation is incorrect. Thanks! $\endgroup$ – L26 Mar 27 at 15:10
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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 $y_n\rightharpoonup0$.

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  • $\begingroup$ what is an example where this assumption is not true? $\endgroup$ – supinf Mar 27 at 14:31
  • $\begingroup$ 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 $f=f^++f^-$. $\endgroup$ – L26 Mar 27 at 15:26

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