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Let $e_{n} = (0, 0, \ldots, 0, 1, 0, \ldots)$ where $1$ is in the $n$th position. Then $\{e_{n}\}$ is an orthonormal basis for the Hilbert space $\ell^{2}(\mathbb{N})$. Does there exists a linear operator $T: \ell^{2} \rightarrow \ell^{2}$ which is not continuous but also satisfies $\sum_{n, m = 1}^{\infty}|\langle Te_{n}, e_{m} \rangle|^{2} < \infty$?

user160671
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  • I recommend looking for an example where $Te_n=0$ for all $n$, but $T\neq 0$. – Jonas Meyer Jul 24 '14 at 20:44
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    Different question, but containing an answer to this one: http://math.stackexchange.com/questions/87415/boundedness-of-operator-on-hilbert-space?lq=1 – Jonas Meyer Jul 24 '14 at 20:47
  • I think I saw somewhere that the existence of an unbounded everywhere defined operator on a separable Hilbert space is equivalent to some form of the axiom of choice. – Jonas Dahlbæk Jul 24 '14 at 21:33
  • In the linked answer $T$ is not defined on all of $l_2$. I suspect that OP just meant densely defined, but if it is really supposed to be defined on all of $l_2$ and be unbounded then I doubt that examples constructed from Hamel bases (using axiom of choice) can be made to have that double sum converge. – Conifold Jul 24 '14 at 21:34
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    @Conifold We can take Hamel basis containing all $e_n$, and define $T$ to be zero on those vectors. –  Jul 25 '14 at 02:08

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