Let $H$ be a complex Hilbert space. For an operator $T: H \rightarrow H$ fix an orthonormal basis and define the "absolute value" of $T$ as $$|T| = (TT^{*})^{\frac{1}{2}}$$

We say that the operator is a trace-class operator, if $\text{tr}(|A|) < \infty$, i.e. $$\sum{\langle |A|e_{i}, e_{i} \rangle} < \infty$$ for some $\{ e_{i} \}$ (and thus any) basis in $H$

Here $\langle \cdot, \cdot \rangle$ stands for the inner product on $H$).

What i know is that there exists a chain of inclusions: $$ \{ \text{finite rank operators} \} \subset \{ \text{trace class} \} \subset \{ \text{compact} \}$$

Is there is an example of an operator $S: H \rightarrow H$ such that for some basis $\{ e_{i} \}$ in $H$ the following sum $$ \sum{\langle Se_{i}, e_{i} \rangle} < \infty$$ converges, but the operator is not a trace-class?

I can provide the following "counterexample": Let $H = L^{2}([0, 1])$ with basis $ \{ e^{i n x} \} $ and consider the operator $T$ that maps $$ e^{i n x} \mapsto \frac{1}{n^{3}} \frac{d}{dx}(e^{i n x})$$

In fact it is a composition of operators $L^{2} \rightarrow l^{2} \rightarrow l^{2}$, where the first arrow stands for the derivative operator and the second one multiply each element $a_{n}$ of the sequence by $\frac{1}{n^{3}}$. One can show that this operator is not compact (is it true? it should, since the derivative operator is not compact and the second arrow does not help to make the image of a non precompact set precompact).

I claim that $\sum{\langle T (e^{inx}), e^{inx} \rangle} < \infty$, since the series $\sum{\frac{1}{n^{2}}} < \infty$, but the operator is not trace-class, since otherwise it would be compact.

Do the mentioned reasoning fail to be correct? If yes, how one can fix the details in order to provide a correct counterexample? Are there any others examples that met the requirements?


2 Answers 2


Consider the right unilateral shift $S : \ell^2 \to \ell^2$. We have $Se_n = e_{n+1}$ so $\sum_{n=1}^\infty \langle Se_n, e_n\rangle = 0$.

On the other hand, we have $S^*S = I$ so $|S| = I$. It follows $$\sum_{n=1}^\infty \langle |S|e_n, e_n\rangle = \sum_{n=1}^\infty \langle e_n, e_n\rangle = \infty$$

so $S$ is not trace class.

  • $\begingroup$ Great, thank you! $\endgroup$ Dec 14, 2018 at 22:38

Your operator is compact. If you represent it on $\ell^2(\mathbb N)$, your $T$ acts on the canonical basis by $$ Te_n=\tfrac{i}{n^2}e_{n-1}. $$ And this is compact, because it is a limit of finite-rank operators. Namely, if $$ T_kx=\sum_{|n|\leq k}\tfrac{i}{n^2}x_ne_{n-1}, $$ Then $\|T_n-T\|\leq1/n^2$.

In fact, your $T$ is also trace class. You have that $T^*$ is given by $T^*e_n=-\tfrac{i}{n^2}e_{n+1}$. Then $$T^*Te_n=\tfrac1{n^4}e_n.$$ So $|T|$ is the operator given by $$|T|e_n=\tfrac1{n^2}e_n,$$ and $$\operatorname{Tr}(|T|)=\sum_n\frac1{n^2}<\infty.$$

Using the above ideas you can play with the sequence you use in defining $T$, as in $$ T_ae_n=a_n\,e_{n-1}, $$ where $a\in\ell^\infty(\mathbb N)$. This $T_a$ will always have $\operatorname{Tr}(T)=0$, and $$|T|e_n=|a_n|\,e_n.$$ So,

  • if $\sum_n|a_n|<\infty$, you will have that $T$ is trace-class;

  • if $\lim_n|a_n|=0$, you will have that $T$ is compact;

  • if $\limsup_n|a_n| >0$, then $T$ is not compact.


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