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Let $\{v_n\}_{n \in \mathbb{N}}$ be a Schauder basis of $V$ subspace of $\ell^2$ over $\mathbb{C}$ and $\forall m \in \mathbb{N}$ let $V_m = \overline{\operatorname{span}} \{v_n\}_{n \geq m}$

Let $\{u_n\}_{n \in \mathbb{N}}$ be a Schauder basis of $U$ subspace of $\ell^2$ over $\mathbb{C}$ and $\forall m \in \mathbb{N}$ let $U_m = \overline{\operatorname{span}} \{u_n\}_{n \geq m}$

Under the hypothesis that $\forall m \in \mathbb{N}: V_m + U_m$ is closed, is it true that:

$$ \bigcap_{m=1}^\infty \left( V_m + U_m \right) = \{0\} $$

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In general it is not true: $V_m$ and $U_m$ could even be transverse for all $m$, giving $$ \bigcap_{m=1}^\infty \left( V_m + U_m \right) = \ell_2. $$

Let $\kappa:\mathbb{N}\to\mathbb{N}$ be a map such that every $p\in\mathbb{N}$ has fiber $\kappa^{-1}(p)$ of infinite cardinality.

Let $\{v_n\}_{n\ge1}$ be the orthonormal basis of $\ell_2$ and let $K:\ell_2\to\ell_2$ be the bounded linear operator defined by $Kv_n= 2^{-n}v_{\kappa(n)}$ for all $n\ge1$. Then $$\|K\|\le\|K\|_{HS}=\Big(\sum_{n\ge1}\|Kv_n\|^2\Big)^{1/2}=\Big(\sum_{n\ge1}4^{-n}\Big)^{1/2}=3^{-1/2}<1$$ Therefore $I+K$ is invertible and $u_n:=(I+K)v_n$ for $n\ge1$ defines a Schauder basis.

However, by the hypothesis on $\kappa$, for any $1\le p<m$ there exists $n\ge m$ such that $\kappa(n)=p$, so $u_n=v_n+2^{-n}v_p\in U_m$, and $v_p=2^nu_n-2^nv_n\in U_m+V_m$, hence $U_m+V_m\supset \operatorname{span}(v_1,\dots,v_{m-1})+V_m=\ell_2.$

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  • $\begingroup$ (Here $\mathbb{N}=\{1,2,3,\dots\}$) $\endgroup$ – Pietro Majer Apr 17 '19 at 22:11
  • $\begingroup$ Grazie @PietroMajer for your answer $\endgroup$ – Matey Math Apr 18 '19 at 11:08

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