• $U=(U,\langle\;\cdot\;,\;\cdot\;\rangle)$ be a separable Hilbert space and $\left\|\;\cdot\;\right\|$ be the norm induced by $\langle\;\cdot\;,\;\cdot\;\rangle$
  • $Q$ be a bounded, linear, nonnegative and symmetric operator on $U$ with finite trace

I've read that we can conclude that $Q$ is Hilbert-Schmidt, but I'm quite sure that this is wrong.

By the Hilbert-Schmidt theorem, there is an orthonormal basis $(e_n)_{n\in\mathbb N}$ of $U$ with $$Qe_n=\lambda_ne_n\;\;\;\text{for all }n\in\mathbb N\tag 1$$ for some $(\lambda_n)_{n\in\mathbb N}\subseteq[0,\infty)$. Since $Q$ has finite trace, $$\sum_{n\in\mathbb N}\lambda_n\stackrel{(1)}=\sum_{n\in\mathbb N}\langle Qe_n,e_n\rangle<\infty\tag 2\;,$$ but the square of the Hilbert-Schmidt norm of $Q$ is equal to $$\sum_{n\in\mathbb N}\left\|Qe_n\right\|^2\stackrel{(1)}=\sum_{n\in\mathbb N}\lambda_n^2\tag 3\;.$$

Maybe we can show that $\sum_{n\in\mathbb N}\lambda_n^2<\infty$, but I don't think so.

Am I missing something or does someone know a counter-example?


It is generally true that if $M := \sum |x_n|<\infty $, then $\sum |x_n|^2 <\infty $, simply because of $|x_n|\leq M $, so that $|x_n|^2\leq M |x_n|$.

Now apply this to $x_n =\lambda_n $.


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