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Let $H$ be a separable Hilbert space and let $T\in B(H)$ be compact (I don't know whether this is relevant to the question), such that $\displaystyle \sum_{j=1}^\infty\langle T\xi_j,\eta_j\rangle$ converges for any choice of orthonormal bases $\{\xi_j\}$, $\{\eta_j\}$. Does this imply that $T$ is trace-class?

I think it is, but I couldn't really write a proof.

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up vote 3 down vote accepted

If $T$ is not trace class, for any orthonormal basis $\xi_j$ of your Hilbert space $H$, $\sum_j \langle |T| \xi_j, \xi_j \rangle$ diverges. In particular, there are infinitely many $\xi_j$ such that $\langle |T| \xi_j, \xi_j \rangle > 0$. By the polar decomposition, there is a partial isometry $V$ such that $T = V |T|$, where $|T| = (T^* T)^{1/2}$. This is an isometry of closed subspaces $A$ to $B$, where $B$ contains $\text{Ran}(T)$ and $A$ contains $\text{Ran}(|T|)$. Since $|T|$ is self-adjoint, $|T|v = 0$ for any $v$ orthogonal to $A$. So start with an orthonormal basis $\alpha_j$ of $A$. Corresponding to this is $\beta_j = V \alpha_j$, an orthonormal basis of $B$. We have $$\sum_j \langle T \alpha_j, \beta_j \rangle = \sum_j \langle |T| \alpha_j, V^* \beta_j \rangle = \sum_j \langle |T| \alpha_j, \alpha_j \rangle = \infty$$ The only trouble is that we might not be able to simultaneously extend both $\alpha_j$ and $\beta_j$ to orthogonal bases of the whole space, because one of $A$ and $B$ might have finite codimension while the other has infinite codimension. To fix this problem, split the index set into two infinite subsets $K$ and $L$ such that we still have $\sum_{j \in K} \langle |T| \alpha_j, \alpha_j \rangle = \infty$. Since the closed spans of $\{\alpha_j: j \in K\}$ and $\{\beta_j: j \in K\}$ both have infinite codimension, we can extend both of these to orthonormal bases $\xi_j$ and $\eta_j$.

Note that $$\sum_j \left| \langle T \xi_j, \eta_j \rangle \right| \ge \sum_{j \in K} \langle |T| \alpha_j, \alpha_j \rangle = \infty$$ so $\sum_j \langle T \xi_j, \eta_j \rangle$ does not converge absolutely. It might converge conditionally, but we can always rearrange a conditionally convergent series to be divergent.

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Thanks a lot! $ $ – Martin Argerami Oct 7 '12 at 13:53

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