We say that a sequence $(x\_{n})\_{n}$ in a Banach space $X$ is unconditionally $p$-summable ($1\leq p<\infty$) if 
$$\sup\_{x^{\*}\in B\_{X^{\*}}}(\sum\_{n=m}^{\infty}|\langle x^{\*},x\_{n}\rangle|^{p})^{\frac{1}{p}}\rightarrow 0\quad (m\rightarrow \infty).$$ We denote the set of all unconditionally $p$-summable sequences on $X$ by $l^{u}\_{p}(X)$. Define a norm 
$$\|(x\_{n})\_{n}\|\_{p}^{w}=\sup\{(\sum\_{n=1}^{\infty}|\langle x^{\*},x\_{n}\rangle|^{p})^{\frac{1}{p}}:x^{\*}\in B\_{X^{\*}}\}, (x\_{n})\_{n}\in l^{u}\_{p}(X).$$


The following is my question: Let $(x\_{n})\_{n}\in l^{u}\_{p}(X)$ and let $\epsilon>0$. Are there $(\widetilde{x\_{n}})\_{n}\in l^{u}\_{p}(X)$ and $(\lambda\_{n})\_{n}\in c\_{0}$ such that $\|(\lambda\_{n})\_{n}\|=1$, $x\_{n}=\lambda\_{n}\widetilde{x\_{n}}$ ($n=1,2,...$) and $\|(\widetilde{x\_{n}})\_{n}\|\_{p}^{w}<\|(x\_{n})\_{n}\|\_{p}^{w}+\epsilon$? 



A question on unconditionally $p$-summable sequences
Yes. Choose $n\_1<n\_2<\dots$ s.t. for all $m$
$$
\sup\_{x^{\*}\in B\_{X^{\*}}}\sum\_{k=n\_m}^\infty |\langle x^\*, x\_k \rangle|^p < \epsilon^p/4^m.
$$
Set $\lambda\_k = 1$ for $k<n\_1$ and $\lambda\_k = 2^{-k}$ for $n\_m\le k < n\_{m+1}$.


