Edit: This question didn't get any answers after a bounty, so I cross-posted it here on Math Overflow.
Let $H$ be a separable Hilbert space and $T \in L(H)$ a trace-class operator. It is well known that the trace of $T$ is given by $\operatorname{tr}(T) = \sum_n \langle Te_n, e_n\rangle$, where $\{e_n\}_{n\in \mathbb N}$ is an orthonormal Hilbert basis. The quantity is independent of the chosen orthonormal basis and even independent of the order of summation.
In finite dimensions, however, we can represent $T$ as a matrix using any basis, not necessarily orthonormal, and summing the diagonal yields the trace. Thus, one might expect a similar result to hold for arbitrary trace-class operators.
Suppose $\{f_n\}_{n \in \mathbb N}$ is an ordered Schauder basis for $H$, and let $\pi_n$ denote the projection onto the $n\text{th}$ basis element. If $T$ is trace class, do we have $\operatorname{tr}(T) = \sum_n \pi_n(Tf_n)$? If so, is the sum always independent of order? If the suggested formula fails in general, what if we restrict to unconditional or symmetric Schauder bases?
