If $(\lambda_n)_{n\in\mathbb N}$ is the eigenvalue sequence of a trace-class operator, can we show that $\sum_{n\in\mathbb N}\sqrt{\lambda_n}<\infty$?

Question

Let

• $U$ be a $\mathbb R$-Hilbert space
• $Q$ be a bounded, linear, nonnegative and self-adjoint operator on $U$ with finite trace
• $(e^n)_{n\in\mathbb N}$ be an orthonormal basis of $U$ with $$Qe^n=\lambda_ne^n\;\;\;\text{for all }n\in\mathbb N\tag1$$ for some $(\lambda_n)_{n\in\mathbb N}\subseteq[0,\infty)$ with $$\lambda_n\ge\lambda_{n+1}\;\;\;\text{for all }n\in\mathbb N\tag2$$ and either $$\exists n_0\in\mathbb N:\forall n\ge n_0:\lambda_n=\lambda_{n_0}\tag3$$ or $$\forall\varepsilon>0:\exists n_0\in\mathbb N:\forall n\ge n_0:\lambda_n<\varepsilon\tag4$$

By definition, $$\sum_{n\in\mathbb N}\lambda_n=\sum_{n\in\mathbb N}\langle Qe^n,e^n\rangle_U\stackrel{\text{def}}=\operatorname{tr}Q<\infty\tag5\;.$$ Can we show that $$S_N:=\sum_{n=1}^N\sqrt{\lambda_n}\;\;\;\text{for }N\in\mathbb N$$ is convergent too?

Answer 1

No, for the same reason that if a sequence $\lambda_1, \lambda_2, \ldots$ is summable, it doesn't mean that $\sqrt{\lambda_1}, \sqrt{\lambda_2}, \ldots$ is. Example: $\lambda_n = 1/n^2$. Nothing in the definition of trace-class and other properties you mention precludes this from happening: take the diagonal $Q = \text{diag}(1, 1/2^2, 1/3^2, \ldots)$ on the sequence space $U = \ell_2(\mathbb{N})$.