Take the 2-minute tour ×  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?  share|improve this
question 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 add comment  1 Answer  up vote 9 down vote accepted  No,
we cannot conclude that the operator is trace