Let $H$ be a complex Hilbert space and $x \in T(H)$ be a positive trace class operator. I want to show that $$\text{Tr}(x-): B(H) \to \mathbb{C}: y \mapsto \text{Tr}(xy)$$ is a positive functional.
Attempt:
Let $y \geq 0$ a positive operator in $B(H)$. I need to show that $$\text{Tr}(xy) \geq 0$$
Of course, we can write $x = a^* a, y =b^ *b$ for some operators $a,b$. And then we need to show that
$$\text{Tr}(a^* a b^*b) \geq 0$$
It seems natural to try to use the cyclicity of the trace: $$\text{Tr}(a^*ab^*b) = \text{Tr}(ba^* ab^*) = \text{Tr}((ab^*)^*(ab^*))$$
and then we can conclude since the trace of a positive operator is positive.
However, I only know that $\text{Tr}(uv) = \text{Tr}(vu)$ when either $u$ or $v$ is trace class, so what I need for the above to work is that $a$ or $b$ is trace class?
Since $x$ is trace-class, so is $xy$. And $xy^{1/2}$. As you know that you can commute inside the trace if one of the two operators is trace-class, $$ \operatorname{Tr}(xy)=\operatorname{Tr}(xy^{1/2}y^{1/2})=\operatorname{Tr}(y^{1/2}xy^{1/2})\geq0, $$ as this last operator is trace-class and positive.