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Bessaga-Pelczynski Selection Principle states that if $(x_{n})_{n}$ is a basis for a Banach space $X$, then every normalized weakly null sequence $(y_{n})_{n}$ in $X$ admits a subsequence that is equivalent to a block basic sequence with respect to $(x_{n})_{n}$. I am wondering whether the sequence $(y_{n})_{n}$, for every $\epsilon>0$, admits a subsequence sequence $(y_{k_{n}})_{n}$ that is $(1+\epsilon)$-equivalent to a block basic sequence with respect to $(x_{n})_{n}$. Thank you!

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If $(x_n)_{n=1}^\infty$ is normalized then yes, that is true. It is even true if we replace the normalization criterion with $\lim\|y_n\|=1$. The proof is in Albiac-Kalton, although it is not formally stated there. For a reference, see Theorem 1.1 here. (Note that we also require $(x_n)_{n=1}^\infty$ to satisfy $\lim\|x_n\|=1$, even though that is not stated in the reference. But I assume you are using a normalized $(x_n)_{n=1}^\infty$, so this is not an issue.)

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  • $\begingroup$ Good! Ben. I shall read your paper. $\endgroup$ – Dongyang Chen Jul 16 '16 at 16:18

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