A question on unconditionally $p$-summable sequences

Question

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 $\|(\widetilde{x_{n}})_{n}\|_{p}^{w}<\|(x_{n})_{n}\|_{p}^{w}+\epsilon$?

Answer 1

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 $n_m\le k < n_{m+1}$.