Is a bounded operator with finite trace trace class? 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. $A\ge 0$?
 A: 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.
A: 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.
