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Can someone please help me with the following problem? I have some of my work below but I am not sure if I attacked the problem wisely. Thank you for your time and consideration.

Suppose that a bounded linear operator $A$ on the Hilbert space $\mathscr{H}$ has the property that $ \sum_{n \in \mathbb{N}} \left|\,\langle \,A \varphi_n\, , \, \varphi_n \, \rangle\,\right| $ is finite for each orthonormal basis $ \left(\varphi_n \right)_{n\in \mathbb{N}}$ of $\mathscr{H}$. Prove $ A \in \mathscr{I}_1$.

$\textit{Proof.}$ As we know, a Hilbert space $\mathscr{H}$ is separable, so we can choose a countable collection of elements $F = \{hk\}$ so that $A \in \mathscr{I}_1$ in $\mathscr{H}$ so that linear combinations of elements in $F$ are dense in $\mathscr{H}.$ Finitely many elements $\mathscr{I}_1, \mathscr{I}_2, \dots, \mathscr{I}_n$ are said to be linearly independent if $$a_1\mathscr{I}_1 + a_2\mathscr{I}_2 + \dots + a_n\mathscr{I}_1 = 0$$ then $a_1 = a_2 = \dots = a_n = 0.$ From this, we can say that a countable family of elements is linearly independent, and $\sum_{n \in \mathbb{N}} \left|\,\langle \,A \varphi_n\, , \, \varphi_n \, \rangle\,\right| $ is finite for each orthonormal basis $\left(\varphi_n \right)_{n\in \mathbb{N}}$ of $\mathscr{H}.$ Hence, we conclude that $A \in \mathscr{I}_1.$

rudinsimons12
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    It may help to include the definition of $\mathscr{I}_1$ that you are working from as well as what tools you have available. (If it is what I think it is, it is possible in some settings to take this property as a definition of $\mathscr{I}_1$, and someone familiar with that as the definition may be confused about how to help.) The link https://math.stackexchange.com/questions/3049581/self-adjoint-operators-and-trace-class-property may be of interest. – leslie townes Nov 02 '21 at 18:29
  • Of course there are non-separable Hilbert space. Look like you are using a different definition. – Arctic Char Nov 02 '21 at 22:17

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