I am trying to understand the proof of Proposition 8.3.16 (***) in Higson and Roe's Analytic K-Homology book. They rely on Exercise 8.8.8 (which is the only argument I don't fully understand). It goes as follows:
Let $A$ be a $C^*$-algebra, $\rho:A\rightarrow \mathbb{B}(H)$ a representation of $A$ on a (separable, multigraded) Hilbert space $H$. Define $$ \mathcal{D}_\rho(A)=\{T\in\mathbb{B}(H)\;:\;[T,\rho(a)]\sim0,\;\;\;\forall a\in A\},$$ and $$\mathcal{D}_{\rho}(A//A)=\{T\in \mathcal{D}_{\rho}(A)\;:\;T\rho(a)\sim0\sim\rho(a)T,\;\;\;\forall a\in A\}, $$ where $T\sim S$ if and only if $T-S\in\mathbb{K}(H)$.
Exercise 8.8.8: Suposse $P\in\mathcal{D}_{\rho}(A)$ is self-adjoint and that $\rho(a)P\rho(a^*)$ is positive modulo compact operators for every $a\in A$. Prove that $P$ is positive modulo $\mathcal{D}_{\rho}(A//A)$.
The book provides a hint: since $P$ is self-adjoint, we can write $P=P_+-P_-$, where $P_{\pm}$ are positive and $P_+P_-=P_-P+=0$. Following this line of reasoning, the idea would be to show that $P_-\in\mathcal{D}_{\rho}(A//A)$. I have tried computing $$ \langle \rho(a)P_-x,\rho(a)P_-x\rangle,$$ for some $x\in H$, with the hope that I could prove this vanishes modulo compacts, but to no avail.
(***)- Disclaimer: the reason why I also explicted this result is because in that setting one might use that the representation $\rho$ is non-degenerate (Lemma 8.3.8), meaning that $\overline{\rho(A)}H$ is dense in $H$, although Exercise 8.8.8 appears to be true in the more general picture.
$$\rho(a^*) P_- P_- \rho(a)= P_-\ \rho(a^*) P_- \rho(a) + [\rho(a^*), P_-]\rho(a)$$ The first term is zero module compacts, the second is also compact. This means that $(P_-\rho(a))^* (P_-\rho(a))$ is compact, so $P_-\rho(a)$ is so as well. Do the same trick with compact commutators to get that $\rho(a)P_-$ is compact. So $P_-$ is in $D(A//A)$.