$x'_n\overset{w^\ast}{\to}0$ and $x_n\overset{w}{\to}0$ implies $\sup_m |x_m'|(|x_n|)\to 0$?

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If $x_n'$ is weak-$\ast$-ly null and $x_n$ is weakly null in a Banach lattice, do we always have

$$\sup_m |x_m'|(|x_n|)\to 0\,?$$

Thank you!

Background:

Definition weak(-$\ast$) topology:

On the dual $E'$ of a Banach lattice $E$ we define two topologies:

  • $(x_n')$ in $E'$ converges to $x'$ in $E'$ weak-$\ast$-ly if and only if $$\forall x\in E:x_n'(x)\to x'(x)\,,$$
  • and it converges weakly if and only if $$\forall x''\in E'':x''(x_n')\to x''(x)\,.$$

Definition absolute value:

The absolute value $|x|$ of an element of a Banach lattice is defined by $|x|=x\vee -x$.