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Let $H$ a Hilbert space and $(a_n)_{n \in \mathbb{N}}$ a complete orthonormal system, i.e. a basis. Do you know a example of a orthonormal system $(b_n)_{n\in \mathbb{N}}$ satisfying $\infty > \sum_{k \in \mathbb{N}} \lVert a_k-b_k \lVert \geq 1$, which is not complete?

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Sketch of the proof

0)Since $\sum_{k\in\mathbb{N}}\Vert a_k-b_k\Vert<+\infty$, then $\sum_{k\in\mathbb{N}}\Vert a_k-b_k\Vert^2<+\infty$. And we reduce to the case considered in the "A Hilbert Space problem book" by P.R. Halmos. This problem no 7 in Russian edition.

1) Assume that there exist $b_0\neq 0$, orthogonal to all $b_k$. Then for all $N$ the system ${b_0,\ldots,b_N}$ is linearly independent.

2) Consider $n$ such that $\sum_{k>n}\Vert a_k-b_k\Vert^2<1$.

3) Projections $\{c_0,\ldots,c_n\}$ of $b_0,\ldots,b_n$ on $\mathrm{span}\{a_1,\ldots,a_n\}$ are linearly dependent, i.e. $\sum_{k=0}^n \alpha_k c_k=0$ for some coefficients $\{\alpha_0,\ldots,\alpha_n\}$, that are not zero simultaneously.

4) Now prove that vector $d=\sum_{k=0}^n\alpha_k b_k$ is orthogonal to $a_1,\ldots,a_n$ and then prove that $$ \Vert d\Vert^2\leq\sum\limits_{k>n}\Vert d\Vert^2\Vert a_k-b_k\Vert^2. $$

6) This implies $d=0$. Hence $\{b_0,\ldots,b_n\}$ is linearly dependent. Contradiction, so for all $b_0$ orthogonal to $b_1,\ldots,b_n$ we have $b_0=0$. This means completeness of $b_1,\ldots,b_n$.

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