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Let $H$ be a separable Hilbert space and let $T\in B(H)$, 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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2 Answers 2

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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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Robert Isreal's answer is correct, but at least to my way of thinking, not as clear as it would be possible. So, let me provide a proof which targets the hidden details:

(Note that $H$ doesn't need to be separable. So, I'll remove this assumption.)

  • $T$ is a bounded linear operator on $H$ $\Rightarrow$ There is a unique partial isometry $U:H\to H$ with $$\ker U=\ker T\tag1$$ and $$T=U|T|\tag2$$
  • $(\ker U)^\perp$ is a closed subspace of $H$ $\Rightarrow$ $(\ker U)^\perp$ admits an orthonormal basis $(e_i)_{i\in I}$
  • $U$ is an isometric isomorphism between $(\ker U)^\perp$ and $\operatorname{im}U$ $\Rightarrow$ $$f_i:=Ue_i\;\;\;\text{for }i\in I$$ is an orthonormal basis of the closed subspace $$\operatorname{im}U=(\ker U^\ast)^\perp\tag3$$ of $H$ (the only crucial thing is that $(f_i)_{i\in I}$ is an orthonormal basis of a closed subspace of $H$)
  • Now, $$U^\ast\left.U\right|_{(\ker U)^\perp}=\operatorname{id}_{(\ker U)^\perp}\tag4$$ and hence $$\sum_{i\in I}\langle Te_i,f_i\rangle_H=\sum_{i\in I}\langle U|T|e_i,Ue_i\rangle_H=\sum_{i\in I}\langle|T|e_i,U^\ast Ue_i\rangle_H=\sum_{i\in I}\langle|T|e_i,e_i\rangle_H\tag5$$ by $(2)$
  • $H=(\ker U)^\perp\oplus\ker U$ $\Rightarrow$ $(e_i)_{i\in I}$ can be supplemented to an orthonormal basis $(\tilde e_j)_{j\in J}$ of $H$ by elements of $\ker U$
  • $H=(\ker U^\ast)^\perp\oplus\ker U^\ast$ and $(3)$ $\Rightarrow$ $(f_i)_{i\in I}$ can be supplemented to an orthonormal basis $(\tilde f_k)_{k\in K}$ of $H$
  • By a simple renumbering (and insertion of zeros, if necessary), we may assume $J=K$
  • $(1)$ $\Rightarrow$ $$\ker U=\ker T=\ker|T|\tag6$$ and hence $$\langle T\tilde e_j,\tilde f_j\rangle_H=\langle U|T|\tilde e_j,\tilde f_j\rangle_H=0\;\;\;\text{for all }j\in J\text{ with }\tilde e_j\not\in(e_i)_{i\in I}\tag7$$
  • Thus, $$\sum_{j\in J}\langle T\tilde e_j,\tilde f_j\rangle_H=\sum_{i\in I}\langle Te_i,f_i\rangle_H=\sum_{i\in I}\langle|T|e_i,e_i\rangle_H=\sum_{j\in J}\langle|T|\tilde e_j,\tilde e_j\rangle_H\tag8$$ by $(5)$
  • Suppose $T$ is not nuclear $\Rightarrow$ $$\sum_{j\in J}\langle|T|\tilde e_j,\tilde e_j\rangle_H=\infty\tag9$$
  • $|T|$ is nonnegative $\Rightarrow$ $$\langle Te_i,f_i\rangle_H=\langle |T|e_i,e_i\rangle_H\ge0$$ and hence $\sum_{j\in J}\langle T\tilde e_j,\tilde f_j\rangle_H=\sum_{i\in I}\langle Te_i,f_i\rangle_H$ cannot converge by $(8)$ and $(9)$
  • Thus, we have found an orthonormal bases of $H$ for which your assumption is not satisfied
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