# Selfadjoint compact operator with finite trace

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?

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By diagonal, do you mean $\langle Te_n,e_n\rangle$, where $\{e_n\}$ is the fixed orthonormal basis? – Davide Giraudo Jul 24 '12 at 21:55
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 $|\mathrm{Tr}(T)|=|\sum_n\langle T e_n,e_n\rangle|=|\sum_n\lambda_n|\leq\Vert\lambda\Vert_1<\infty$ – Norbert Jul 24 '12 at 21:57
@Davide: yes, including the sum you didn't type. – Martin Argerami Jul 24 '12 at 22:01
@Norbert: you are assumming that $T$ is diagonal in the given basis, which is not the case. – Martin Argerami Jul 24 '12 at 22:02

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.

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Nice example. Thanks! – Martin Argerami Jul 25 '12 at 1:45
@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. – t.b. Jul 25 '12 at 2:19
Yes, I have wished so hard for my operator to be positive... – Martin Argerami Jul 25 '12 at 2:21

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.