Let $A:H \rightarrow H$ be a bounded self-adjoint linear operator on Hilbert $H$ and let $R(z)=(A-z)^{-1}$ be its resolvent where $z\in \mathbb{C}\backslash \mathbb{R}$. Let $P$ be a rank-2 projection on $H$. Then how do you prove that $PR(z)P$ is invertible as a linear operator on the range of $P$?
The statement is clear if $P$ is a rank-1 projection. However, it becomes much more difficult for rank-2. Moreover, can this statement be generalized to rank-$n$?
I think I found a nice proof, but hopefully others can check for errors.
(1). Let $x\in H$ be nonzero and $y=R(z)x$. Then \begin{align} \langle x, R(z) x \rangle &= \langle (A-z)y, y\rangle \\ &= \langle y, Ay \rangle -\bar{z} ||y||^2 \end{align} Since $\Im{z}\ne0$, we see that $\Im{\langle x, R(z) x \rangle}/ \Im{z} >0$. This corresponds to $P$ being a rank-1 projection onto the span of $x$.
(2). Also notice that $R(\bar{z}) = R(z)^*$ since $A$ is self-adjoint. Hence, \begin{align} \Im{R(z)} &= \frac{R(z)-R(\bar{z})}{2i} \\ &= \frac{z-\bar{z}}{2i} R(\bar{z})R(z) \\ &= \Im{z} R(\bar{z}) R(z) \\ \Im{R(z)}/\Im{z} &= R(\bar{z}) R(z) > 0 \end{align} where we have striclty positivity since $R(z)$ is invertible.
(3). Now let $\underline{R}(z)=PR(z)P$ for finite-rank projection $P$ so that we can regard $\underline{R}(z)$ as a normal linear operator on the range of $P$. Following the same logic as before, we see that $$ \Im{\underline{R}(z)}/\Im{z} = \underline{R}(\bar{z}) \underline{R}(z) \ge 0 $$ Suppose that we do not have strict inequality. Then there exists nonzero $x\in \operatorname{range}{(P)}$ such that \begin{align} PR(z)Px &= 0 \\ \langle x, PR(z)Px \rangle &=0 \\ \langle x, R(z) x \rangle &=0 \end{align} This contradicts our previous result and thus we have $\Im{\underline{R}(z)}/\Im{z}>0$. Since $\underline{R}(z)$ is normal, it's then clear that it is invertible on the range of $P$.