# Self adjoint operators and trace class property

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?

• 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". Dec 22, 2018 at 16:31
• @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 $\sum_{i}{\langle T e_{i}, e_{i} \rangle} < \infty$ Dec 22, 2018 at 16:35
• It's not immediately obvious to me that the answer to my question answers yours. Dec 25, 2018 at 2:53
• @MartinArgerami Yes, you are right. This answer provides relevant information (math.stackexchange.com/questions/2036398/…) Jan 5, 2019 at 0:37
• hyperkahler: that answer is very wrong. Jan 5, 2019 at 3:10

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.