adjoints: if $\{h,h'\} \in \sK \times \sH$, then $$\label{hilbert17}
 \{h,h'\} \in T^{*} \quad \Leftrightarrow \quad \{h,h'\} \in (T_{\s})^{\times}.$$ The identity shows the difference between $(T_{\s})^{*}$ and $(T_\op)^\times$.

Let $T$ be a closed linear relation from a Hilbert space $\sH$ to a Hilbert space $\sK$ and let $B \in \mathbf{B}(\sK)$ be selfadjoint. In order to study the linear relation $$T^{*}(I+iB)T,$$ decompose the Hilbert space $\sK$ as in and decompose the selfadjoint operator $B \in \mathbf{B}(\sK)$ accordingly: $$\label{Bdec1}
 B=\begin{pmatrix} B_{11}& B_{12}\\ B_{12}^*& B_{22}\end{pmatrix}: \,
 \begin{pmatrix} \cdom T^* \\ \mul T \end{pmatrix} \to
 \begin{pmatrix} \cdom T^* \\ \mul T \end{pmatrix}.$$ Here the operators $B_{11} \in \mathbf{B}(\cdom T^*)$ and $B_{22} \in \mathbf{B}(\mul T)$ are selfadjoint, while $B_{12} \in \mathbf{B}(\mul T, \cdom T^*)$ and $B_{12}^* \in \mathbf{B}(\cdom T^*,\mul T)$.

By means of the decomposition the following auxiliary operators will be introduced. First, define the operator $C_0 \in \mathbf{B}(\cdom T^*)$ by $$\label{c0}
C_0=I+B_{12}(I+B_{22}^2)^{-1}B_{12}^*.$$ Observe that $C_0 \geq I$ and that $(C_0)^{-1}$ belongs to $\mathbf{B}(\cdom T^*)$ and is a nonnegative operator. Next, define the operator $C \in \mathbf{B}(\cdom T^*)$ by $$\label{c}
 C=C_0^{-\half}\left[B_{11}-B_{12}(I+B_{22}^2)^{-\half}B_{22}
 (I+B_{22}^2)^{-\half}B_{12}^*\right]C_0^{-\half},$$ which is clearly selfadjoint.

\[redux\] Let $T$ be a closed linear relation from a Hilbert space $\sH$ to a Hilbert space $\sK$, let $T_\op$ 