Let $X$ be a closed subspace of $\ell^2$ over $\mathbb{C}$

Let $\{x_m\}_{m \in \mathbb{N}}$ a basis for $X$ equivalent to canonical basis $\{e_m\}_{m \in \mathbb{N}}$ for $\ell^2$

I Would like to know if it is true that: $$ \bigcap_{p=1}^\infty \overline{ \operatorname{span} } \{x_m\}_{m \geq p} = \{0\} $$



Let $(x_n)$ be a basis for $X$ and $(y_n)$ be a basis for $Y$. We say that $(x_n)$ and $(y_n)$ are equivalent if the convergence of $\sum a_n x_n$ is equivalent to that of $\sum a_n y_n$ (Sequence and series in Banach spaces. Joseph Diestel)

  • $\begingroup$ The inclusion $X\subset \ell^2$ seems to be irrelevant. You are just asking about a basis in a Hilbert space $X$ which is equivalent (in the sense you describe) to the standard o.n. basis of $\ell^2$. But isn't such a basis just a Riesz basis, if I have understood the set-up correctly? $\endgroup$ – Yemon Choi Feb 4 at 20:57
  • 4
    $\begingroup$ @kp9r4d: no, it isn't. $\endgroup$ – Nik Weaver Feb 4 at 21:12
  • $\begingroup$ @YemonChoi I don't know if it is a Riesz basis, I know just that $(x_n)$ is equivalent to $(e_n)$ $\endgroup$ – Matey Math Feb 4 at 21:40

I assume "basis" means "Schauder basis"? If so, then you don't need any additional assumption to conclude that this intersection is zero. $X$ can be any Banach space with a Schauder basis. For any $v \in X$ we can uniquely write $v = \sum a_nx_n$; define $P_n(v) = a_n$. This is a bounded linear functional, as explained on the Wikipedia page for Schauder bases. So if $v \in {\rm span}\{x_m\}_{m \geq p}$ then $P_n(v) = 0$ for all $n < p$, and by continuity the same is true for $v \in \overline{\rm span}\{x_m\}_{m \geq p}$. Thus any $v \in \bigcap_{p = 1}^\infty \overline{\rm span}\{x_m\}_{m \geq p}$ satisfies $P_n(v) =0$ for all $n$, and therefore $v = 0$.

  • $\begingroup$ thanks for your answer $\endgroup$ – Matey Math Feb 5 at 0:13
  • $\begingroup$ You are welcome! $\endgroup$ – Nik Weaver Feb 5 at 1:30

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