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It is known (and easy) to prove that if $T: H\longrightarrow H $ is compact, where $H$ is a Hilbert space, then for any orthonormal basis $ e_n $ we have $||Te_n||\longrightarrow 0$.

My question is the following: Let $P$ be a orthogonal projection on $H$. Let $ e_n $ be a fixed orthonormal basis such that $$||Pe_n||\longrightarrow 0$$ Does this guarantee that $P$ is compact, i.e. finite rank?

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No. Consider $H = (\sum_n \ell_2^{2^n})_2$ with the unit vector basis. In $\ell_2^{2^n}$, let $x_n$ be the sum of the unit vector basis. For the subspace take the closed linear span of $(x_n)$.

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No. Let the basis be $e_{1,1}, e_{2,1}, e_{2,2},\ldots e_{n,1},\ldots,e_{n,n},\ldots$, in this order, i.e., split it into groups of growing size. Then the projector to the subspace generated by $e_{1,1},\frac12(e_{2,1}+e_{2,2}),\ldots,\frac1n(e_{n,1}+\ldots+e_{n,n}),\ldots$ is a counterexample.

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