16
$\begingroup$

I have a compact selfadjoint operator $T$ on a separable Hilbert space. For some fixed orthonormal basis, the operator's diagonal is in $\ell^1(\mathbb{N})$.

Can we conclude that $T$ is trace class?

$\endgroup$
  • 1
    $\begingroup$ By diagonal, do you mean $\langle Te_n,e_n\rangle$, where $\{e_n\}$ is the fixed orthonormal basis? $\endgroup$ – Davide Giraudo Jul 24 '12 at 21:55
  • 1
    $\begingroup$ May be I'm mistaken somwhere, but what is wrong with the following proof. Since $T$ is self adjoint and compact we can say that $T(x)=\sum_n\lambda_n\langle x,e_n\rangle e_n$. So $|\mathrm{Tr}(T)|=|\sum_n\langle T e_n,e_n\rangle|=|\sum_n\lambda_n|\leq\Vert\lambda\Vert_1<\infty$ $\endgroup$ – Norbert Jul 24 '12 at 21:57
  • 2
    $\begingroup$ @Davide: yes, including the sum you didn't type. $\endgroup$ – Martin Argerami Jul 24 '12 at 22:01
  • 3
    $\begingroup$ @Norbert: you are assumming that $T$ is diagonal in the given basis, which is not the case. $\endgroup$ – Martin Argerami Jul 24 '12 at 22:02
18
$\begingroup$

No, we cannot conclude that the operator is trace class.

For example, let a Hilbert space have orthonormal basis $e_1,f_2,e_2,f_2,e_3,f_3,\ldots$, and $T$ interchanges $e_i,f_i$, while multiplying both by a positive real $\lambda_i$. That is, in these coordinates, the matrix of $T$ is a list of diagonal blocks, with the $i$-th diagonal block being anti-diagonal $\lambda_i,\lambda_i$.

For $\lambda_i\rightarrow 0$, the operator is compact, almost from the definition.

All the diagonal entries are $0$.

The operator is self-adjoint because the matrix is symmetric real.

However, the operator is not trace class unless $\sum_i |\lambda_i|<\infty$, which easily fails for many sequences of positive reals $\lambda_i\rightarrow 0$.

Edit: It is noteworthy that the analogous characterization (I pointedly don't say "definition") of "Hilbert-Schmidt" does not depend on choice of basis. Thus, "defining" trace-class as composition of two Hilbert-Schmidt operators is sometimes usefully more intrinsic, less basis/coordinate-dependent.

$\endgroup$
  • 2
    $\begingroup$ Nice example. Thanks! $\endgroup$ – Martin Argerami Jul 25 '12 at 1:45
  • 2
    $\begingroup$ @Martin: You probably know this, but just in case: If your operator happens to be positive (not only self-adjoint), then your desired conclusion does hold, i.e. the trace is summable independently of the orthonormal basis, see Corollary 3.4.4 on page 117 of Pedersen's Analysis Now. $\endgroup$ – t.b. Jul 25 '12 at 2:19
  • 3
    $\begingroup$ Yes, I have wished so hard for my operator to be positive... $\endgroup$ – Martin Argerami Jul 25 '12 at 2:21
2
$\begingroup$

Disclaimer: Non-Compact Operators!

Given the Hilbert space $\ell^2(\mathbb{N})$.

Consider sum of shifts:* $$A_\pm:\ell^2(\mathbb{N})\to\ell^2(\mathbb{N}):\quad A_\pm:=R\pm L$$

They have finite trace: $$\sum_n\langle A_\pm\delta_n,\delta_n\rangle=\sum_n0=0$$

But for the shifts: $$\sum_n\langle|R|\delta_n,\delta_n\rangle=\sum_n1=\infty$$ $$\sum_n\langle|L|\delta_n,\delta_n\rangle=\sum_n1=\infty$$ Thus for the sum: $$\operatorname{Tr}A_\pm<\infty\implies\operatorname{Tr}A_\mp<\infty$$ Concluding counterexample.

*Shifts: Right & Left

$\endgroup$
  • $\begingroup$ Note that no operator in this example is compact. $\endgroup$ – Martin Argerami Jan 12 '17 at 12:34
  • $\begingroup$ True that they're not compact! $\endgroup$ – C-Star-W-Star Jan 12 '17 at 23:58
  • $\begingroup$ Do you think I should, to avoid confusion for future readers, better remove this example? $\endgroup$ – C-Star-W-Star Jan 12 '17 at 23:59
  • $\begingroup$ I would leave them, but adding a disclaimer that they are not compact. $\endgroup$ – Martin Argerami Jan 13 '17 at 0:17
  • $\begingroup$ Good idea with the disclaimer, thank you! :) Done! $\endgroup$ – C-Star-W-Star Jan 13 '17 at 11:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.