Is a bounded operator with finite trace trace class?

Question

Let $\mathcal{H}$ be a separable Hilbert space, $A\in\mathcal{B}(\mathcal{H})$ a bounded linear Operator and assume we have an orthonormal basis $(x_n)_{n=1}^\infty$. If $A$ is trace-class, then $\sum_{n\in\mathbb{N}}\langle x_n,Ax_n\rangle$ is finite. But what about the converse, i.e. if we know that $$\sum_{n\in\mathbb{N}}\langle x_n,Ax_n\rangle<\infty,$$ can we deduce that $A$ is trace class? If not, what can be said if $A$ is known to be positive, i.e. $A\ge 0$?

Answer 1

Let me recall that: $A\in B(\mathcal H)$ is trace-class iff if for some (and hence all) orthonormal bases $e_k$ of $\mathcal H$ the sum $$Tr(A)=\sum_k\langle (A^*A)^{1/2}e_k,e_k\rangle<\infty.$$

The answer to the first question is no. Take e.g. $Ae_k=\lambda_ke_k,\ k\in\mathbb N$ with $\lambda_k=\dfrac{(-1)^k}k.$ Then the alternating series $$\sum_k\langle Ae_k,e_k\rangle=\sum_k\dfrac{(-1)^k}k<\infty$$ but $$\sum_k\langle (A^*A)^{1/2}e_k,e_k\rangle=\sum_k\dfrac1k=+\infty.$$

If $A$ is a positive operator, then $A=(A^*A)^{1/2}$ and we have $$Tr(A)=\sum_k\langle (A^*A)^{1/2}e_k,e_k\rangle=\sum_k\langle Ae_k,e_k\rangle<\infty,$$ so that $A$ is trace class.

Answer 2

The assertion is false as:

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

Consider the right shift: $$A:\ell^2(\mathbb{N})\to\ell^2(\mathbb{N}):\quad A:=R$$

Then it has finite trace: $$\sum_n\langle A\delta_n,\delta_n\rangle=\sum_n0=0$$

But it is not trace class: $$\sum_n\langle|A|\delta_n,\delta_n\rangle=\sum_n1=\infty$$

Concluding counterexample.