# 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. $A\ge 0$?

• The problem is that the series $\sum_{n\in\mathbb{N}}\langle x_n,Ax_n\rangle<\infty$ might not be absolutely convergent. For example, if $A$ is a multiplication operator (i.e. $A(x_n)=\lambda_n x_n$), where $\sum_{i=1}^\infty \lambda_i$ is convergent but not absolutely convergent. Feb 3, 2015 at 9:01
• So if the ONB $(x_n)_{n\in\mathbb{N}}$ is fixed, is it true that $A$ is trace class, if and only if the series $\sum_{n\in\mathbb{N}}\langle x_n,Ax_n\rangle$ is absolutely convergent? Feb 3, 2015 at 9:10
• @DavideGiraudo: The series might be absolutely convergent so also summable but still the operator may fail to be trace class. Jul 25, 2015 at 10:58
• @RobertRauch: No, that is not true: Absolute convergence does not imply trace class! Jul 25, 2015 at 11:00

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.

• very good example.
– R.N
Sep 3, 2015 at 14:34
• @RaziehNoori: thank you :) Sep 3, 2015 at 14:35
• @RaziehNoori: By the way if you're interested I have a couple of open questions on my MSE page: Profile Sep 3, 2015 at 14:54
• I am reading your page and question. my phd thesis is about $\frak{E}_{p}(I)$ it is very related to trace class.
– R.N
Sep 3, 2015 at 14:59
• – R.N
Sep 4, 2015 at 6:17

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.

• That example lacks summability. So it will fail to converge for some choices of ONB's. Jul 22, 2015 at 16:12
• Nothing wrong. But your counterexample is not very comprising as though the series converges the sum doesn't. That will force the series to diverge for some ONB. Note the difference between sums $\sum_{n\in\mathbb{N}}a_n:=\{\sum_{n\in N}a_n\}_{N\subseteq\mathbb{N}:\#N<\infty}$ and series $\sum_{n=1}^\infty a_n:=\{\sum_{n=1}^Na_n\}_{N\in\mathbb{N}}$. (See also Trace and Summability.) Jul 23, 2015 at 16:04
• See there's an important difference: There are ordered sums, a.k.a. series, and unordered sums, a.k.a. sums. However ONB's come as unordered so there's no preferable way of ordering. The most natural then will be by inclusion on finite subsets. That is the net of finite sums or in short sum. This will be evident when considering uncountable ONB's on nonseparable Hilbert spaces that are most of all. But it will be also notable on countable ONB's as there's no preferable way of running through the naturals: Forwards, backwards or cross-cross?! Jul 24, 2015 at 13:41
• Now to the examples: The one you gave exists as series but not as sum. But according to the previous comment that is not what is desired on summation over ONB's. Instead the example I gave is summable. Jul 24, 2015 at 13:48
• Simply said: Series are the wrong concept here. Jul 24, 2015 at 14:05