1. Suppose $(x_n)$ and $(y_n)$ are two basic sequences in a separable Banach space $X$ such that $\overline{span}\{(x_n), (y_n)\}=X$. Can we always pass to subsequences $(x_{n_k})$ and $(y_{n_k})$ such that $\overline{span}\{(x_{n_k}), (y_{n_k})\}\neq X$?

I had the impression that this must be "obviously true", and perhaps still is embarrassingly easy, but I just can't see the argument.

  1. A somewhat related question, to which I do not know the answer either: Can any separable Banach space be written as the closed span of finitely many basic sequences?

1 Answer 1


I would suggest to get a positive answer to the second question as follows: Let $\{x_i\}$ be a basic sequence in $X$ (existing by Mazur's result) and let $\{z_i\}$ be a rapidly converging to zero sequence in $X$ with dense linear span. Let $\{y_i\}$ be given by $y_i=x_i+z_i$. The sequence $\{y_i\}$ is basic by the result of Krein-Milman-Rutman. The linear span of the union is dense because it contains $\{z_i\}$.

The answer to the first question is negative. It is done by using the construction for the second question, and in addition picking $\{z_i\}$ in such a way that each of it subsequences has a dense linear span in $X$. Such constructions are known, one of them: Consider a sequence $\{u_k\}$ of normalized vectors with dense linear span and let $z_i=\sum_{k=1}^\infty\left(\tau_i\right)^ku_k$, where $\{\tau_i\}$ is a rapdly converging to $0$ sequence of real numbers. Any subsequence of $\{z_i\}$ has a dense linear span because otherwise there is $x^* \in X^*$, $x^*\ne 0$, such that the analytic (in $\lambda$) function $\sum_{k=1}^\infty\lambda^kx^*(u_k)$ would be a nonzero function with infinitely many zeros in $[0,1]$.

  • $\begingroup$ Is the answer the same if the two sequences are orthonormal sequences in a separable Hilbert space? $\endgroup$
    – Theo
    Apr 27, 2018 at 1:29

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