Identity involving the resolvent of an operator. $\mathcal{H}$ is a complex separable Hilbert space, $D \subseteq \mathcal{H}$ is a dense subspace.
$L : D \rightarrow \mathcal{H}$ is a densely defined, symmetric and closed operator. $L$ is not assumed to be bounded.
Given $z \in \mathbb{C}$, let $R_z$ denote $(L-z)^{-1}$, whenever it exists.
How can I prove the following Proposition?
Proposition: Suppose that $z \in \mathbb{C}$ is such that $\mathrm{Im}(z) \neq 0$, and suppose also that both $R_z, R_{\bar{z}}$ are everywhere defined. Show that $R_z^* = R_{\bar{z}}$.
Here, $^*$ denotes the adjoint and $\bar{z}$ is the complex conjugate of $z$.
I should say that I tried every trick I could think of, to no avail. I tried starting from expressions like $\langle (R_z^* - R_{\bar{z}})v, v \rangle$ and variations of that, but I always get nowhere.
 A: Whether or not $z$ is real, if $R_{z}$ and $R_{\overline{z}}$ are assumed to be defined on all of $\mathcal{H}$, then these operators are closed because the graphs of $L-z I$, $L-\overline{z}I$ are closed and the graphs of $R_{z}$, $R_{\overline{z}}$ are their respective transposes in $\mathcal{H}\times\mathcal{H}$ and, hence, closed; so $R_{z}$, $R_{\overline{z}}$ are closed operators on a Hilbert Space $\mathcal{H}$, which forces them to be bounded by the closed graph theorem. This is a very general fact which has nothing to do with $L$ being symmetric.
Now assume $R_{z}$ and $R_{\overline{z}}$ exist and are defined on all of $\mathcal{H}$ for some $z$. It follows that $R_{z}$, $R_{\overline{z}}$ are bounded (and thus have bounded adjoints) and $(L-zI)R_{z}=I=(L-\overline{z}I)R_{\overline{z}}$. Because $L$ is symmetric, one has $(La,b)=(a,Lb)$ for all $a,b\in\mathcal{D}(L)$, which gives
$$
\begin{align}
     (R_{z}x,y) & =(R_{z}x,(L-\overline{z}I)R_{\overline{z}}y) \\
                & =((L-zI)R_{z}x,R_{\overline{z}}y)\\
                & =(x,R_{\overline{z}}y),\;\;\; x,y\in\mathcal{H}.
\end{align}
$$
Therefore $R_{z}^{\star}=R_{\overline{z}}$.
