Intersection of Hilbert spaces with basis equivalent to canonical basis for $\ell^2$

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: $$\bigcap_{p=1}^\infty \overline{ \operatorname{span} } \{x_m\}_{m \geq p} = \{0\}$$

Thanks.

P.S.

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)

• 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? – Yemon Choi Feb 4 at 20:57
• @kp9r4d: no, it isn't. – Nik Weaver Feb 4 at 21:12
• @YemonChoi I don't know if it is a Riesz basis, I know just that $(x_n)$ is equivalent to $(e_n)$ – 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 $$v = 0$$.