This is a variation of the problem questioned some time ago.

For a complex Hilbert space $H$ let $T: H \rightarrow H$ be a bounded operator. We call that $T$ is a trace-class operator if the following sum $$\sum_{i}{\langle |T|e_{i}, e_{i} \rangle} < \infty$$ converges, where $|T| = (T T^{*})^{\frac{1}{2}}$ is the absolute value of the operator.

Assume that $$\sum_{i}{\langle Te_{i}, e_{i} \rangle}$$ converges for any basis in the space $H$. How to prove that if the aformentioned property holds then the operator is a trace class operator?

The progress on the problem is the following: Given an arbitrary bounded operator $T: H \rightarrow H$, one can use the following decomposition $$T = \big( \frac{T + T^{*}}{2} \big) ^{*} + i \big( \frac{T^{*} - T}{2i} \big) ^{*}$$

The latter line gives the decomposition $$T = A + i B$$ where $A, B$ are normal operators.

For the normal operators we can apply the spectral theorem that proposes that $T$ is unitary equivalent to $$(UT U^{-1})(f(x)) = g(x) f(x)$$ where $$U: H \rightarrow L^{2}(X, \mu)$$ $$g \in L^{\infty}(X, \mu)$$

Though this decomposition classify the operator in a broad sence, i see no direct way to conclude the statement. Are the any hints that may extend the previous argument? If not, are there any ways to conclude the statement?

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    $\begingroup$ As phrased right now, your are asking how to show that if $\sum\langle|T|e_j,e_j\rangle<\infty$, then $T$ is trace-class; and that's exactly the definition of "trace-class". $\endgroup$ Dec 22, 2018 at 16:31
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    $\begingroup$ @MartinArgerami I've made a mistake while typing, now fixed. We should ask that how to prove that the operator is a trace class, provided that $\sum_{i}{\langle T e_{i}, e_{i} \rangle} < \infty$ $\endgroup$ Dec 22, 2018 at 16:35
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    $\begingroup$ It's not immediately obvious to me that the answer to my question answers yours. $\endgroup$ Dec 25, 2018 at 2:53
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    $\begingroup$ @MartinArgerami Yes, you are right. This answer provides relevant information (math.stackexchange.com/questions/2036398/…) $\endgroup$ Jan 5, 2019 at 0:37
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    $\begingroup$ hyperkahler: that answer is very wrong. $\endgroup$ Jan 5, 2019 at 3:10

1 Answer 1


The series is required to converge for any basis, and so it has to survive reordering, and thus absolute convergence.

Since the real and imaginary parts of $T$ will satisfy the hypothesis and linear combinations of trace-class are trace-class, we may assume that $T$ is selfadjoint. We may also assume that $T$ is compact; because if $T$ is not compact, there exists $\lambda\in\sigma(T)\setminus\{0\}$ and $\delta>0$ such that $\lambda-\delta>0$ and the spectral projection $E_T(\lambda-\delta,\lambda+\delta)$ is infinite, and so an orthonormal basis of its range, extended to an orthonormal basis of $H$, provides $\{f_j\}$ such that $\sum_n|\langle Te_n,e_n\rangle|=\infty$.

Knowing that $T$ is compact, by the Spectral Theorem, we know that $$\tag1T=\sum_j\lambda_jP_j,$$ where $\lambda_j\in\mathbb R\setminus\{0\}$ for all $j$, and the projections $P_j$ are rank-one and pairwise orthogonal.

If $T$ is not trace-class, then $$ \operatorname{Tr}(T)=\sum_j|\lambda_j|=\infty. $$ If we write $\lambda_j^+$ for the positive eigenvalues and $\lambda_j^-$ for the negative ones, at least one of $\sum_j\lambda_j^+$ and $\sum_j\lambda_j^-$ diverges. Then, with $\{e_j\}$ the orthonormal basis given by $(1)$ (i.e., $P_j=\langle\cdot,e_j\rangle\,e_j$), we have that $$ \sum_j\langle Te_j,e_j\rangle $$ cannot converge absolutely.


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