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I'm cross-posting this question from Math.SE, as it didn't get much attention there.

Let $H$ be a separable Hilbert space and $T \in L(H)$ a trace-class operator. It is well known that the trace of $T$, given by $\operatorname{tr}(T) = \sum_n \langle Te_n, e_n\rangle$, for some orthonormal Hilbert basis $\{e_n\}_{n\in \mathbb N}$, 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 linear functional on $H$ that selects out the coefficient of $f_n$, e.g., $\pi_1(\alpha_1f_1 + \alpha_2 f_2 + \cdots) = \alpha_1$. 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?

I'm cross-posting this question from Math.SE, as it didn't get much attention there.

Let $H$ be a separable Hilbert space and $T \in L(H)$ a trace-class operator. It is well known that the trace of $T$, given by $\operatorname{tr}(T) = \sum_n \langle Te_n, e_n\rangle$ for some orthonormal Hilbert basis $\{e_n\}_{n\in \mathbb N}$, 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 linear functional on $H$ that selects out the coefficient of $f_n$, e.g., $\pi_1(\alpha_1f_1 + \alpha_2 f_2 + \cdots) = \alpha_1$. 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?

I'm cross-posting this question from Math.SE, as it didn't get much attention there.

Let $H$ be a separable Hilbert space and $T \in L(H)$ a trace-class operator. It is well known that the trace of $T$, given by $\operatorname{tr}(T) = \sum_n \langle Te_n e_n\rangle$, for some orthonormal Hilbert basis $\{e_n\}_{n\in \mathbb N}$, 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 linear functional on $H$ that selects out the coefficient of $f_n$, e.g., $\pi_1(\alpha_1f_1 + \alpha_2 f_2 + \cdots) = \alpha_1$. 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?

I'm cross-posting this question from Math.SE, as it didn't get much attention there.

Let $H$ be a separable Hilbert space and $T \in L(H)$ a trace-class operator. It is well known that the trace of $T$, given by $\operatorname{tr}(T) = \sum_n \langle Te_n, e_n\rangle$, for some orthonormal Hilbert basis $\{e_n\}_{n\in \mathbb N}$, 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 linear functional on $H$ that selects out the coefficient of $f_n$, e.g., $\pi_1(\alpha_1f_1 + \alpha_2 f_2 + \cdots) = \alpha_1$. 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?

I'm cross-posting this question from Math.SE, as it didn't get much attention there.

Let $H$ be a separable Hilbert space and $T \in L(H)$ a trace-class operator. It is well known that the trace of $T$, given by $\operatorname{tr}(T) = \sum_n \langle Te_n e_n\rangle$, for some orthonormal Hilbert basis $\{e_n\}_{n\in \mathbb N}$, 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?

I'm cross-posting this question from Math.SE, as it didn't get much attention there.

Let $H$ be a separable Hilbert space and $T \in L(H)$ a trace-class operator. It is well known that the trace of $T$, given by $\operatorname{tr}(T) = \sum_n \langle Te_n e_n\rangle$, for some orthonormal Hilbert basis $\{e_n\}_{n\in \mathbb N}$, 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 linear functional on $H$ that selects out the coefficient of $f_n$, e.g., $\pi_1(\alpha_1f_1 + \alpha_2 f_2 + \cdots) = \alpha_1$. 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?