# Show that an operator is nonnegative

Let $(\lambda_{mn})_{(m,\:n)\in\mathbb N^2}\subseteq[0,\infty)$ with $$\lambda_{mn}^2\le\mu_m\mu_n\;\;\;\text{for all }(m,n)\in\mathbb N^2\tag1$$ for some $(\mu_n)_{n\in\mathbb N}\subseteq[0,\infty)$ such that $\sum_{n\in\mathbb N}\mu_n$ exists in $\mathbb R$. Let $(e_n)_{n\in\mathbb N}$ be an orthonormal basis of a separable $\mathbb R$-Hilbert space $H$ and $$e_m\otimes e_n:=\langle\;\cdot\;,e_m\rangle_He_n\;\;\;\text{for }(m,n)\in\mathbb N^2\;.$$ Let $\mathfrak L_1(H)$ denote the space of nuclear operators on $H$.

I want to show that $$A:=\sum_{(m,\:n)\in\mathbb N^2}\lambda_{mn}e_m\otimes e_n$$ is nonnegative, i.e. $$\langle Ax,x\rangle_H\ge0\tag2\;\;\;\text{for all }x\in H\;.$$

Let $\operatorname{HS}(H)$ denote the space of Hilbert-Schmidt operators on $H$. Note that $(e_m\otimes e_n)_{(m,\:n)\in\mathbb N^2}$ is an orthonormal basis of $\operatorname{HS}(H)$ and hence $A$ exists in $\operatorname{HS}(H)$ if and only if $\sum_{(m,\:n)\in\mathbb N^2}\left\|\lambda_{mn}e_m\otimes e_n\right\|_{\operatorname{HS}(H)}^2$ exists in $\mathbb R$. Now, $$\begin{split}\sum_{(m,\:n)\in\mathbb N^2}\left\|\lambda_{mn}e_m\otimes e_n\right\|_{\operatorname{HS}(H)}^2&=\sum_{(m,\:n)\in\mathbb N^2}\lambda_{mn}^2\underbrace{\left\|e_m\otimes e_n\right\|_{\operatorname{HS}(H)}^2}_{=\:1}\\&\le\sum_{(m,\:n)\in\mathbb N^2}\mu_m\mu_n\\&=\left(\sum_{m\in\mathbb N}\mu_m\right)\sum_{n\in\mathbb N}\mu_n\end{split}\tag3$$ and hence $A$ exists in $\operatorname{HS}(H)$. Note that $A$ is self-adjoint. Now, I've read that $A$ is nonnegative, but $$\langle Ax,x\rangle_H=\sum_{(m,\:n)\in\mathbb N^2}\lambda_{mn}\langle x,e_m\rangle_H\langle x,e_n\rangle_H\;\;\;\text{for all }x\in H\tag4$$ and I don't see the nonnegativity from $(4)$.

It is not true. Take $$\lambda_{11}=\lambda_{22}=1,\ \ \ \lambda_{21}=\lambda_{12}=2,\ \ \ \lambda_{mn}=0\ \ \text{ if } \max\{m,n\}\geq3.$$Put $$\mu_4=\mu_2=4,\ \ \ \mu_n=0\ \text{ if } n\geq3$$ And let $$x=e_1-e_2.$$Then $$\langle Ax,x\rangle =-2.$$