\section{The Cayley Transform}
\label{section:cayley-transform}

\begin{definition}[Cayley Transform]
\label{definition:cayley-transform-bounded}
    Let $H$ be a complex Hilbert space and $T \in B(H)$ with $-i \not\in \sigma_A(H)$, then $U(T) = (T - i)(T + i)^{-1}$ is the \textbf{Cayley transform} of $T$. 
\end{definition}

\begin{theorem}
\label{theorem:cayley-sa-uni}
    Let $H$ be a complex Hilbert space and $U_1$ be the set of unitary operators on $H$ with $1$ not in their spectrum, then the Cayley transform $T \mapsto (T - i)(T + i)^{-1}$ is a strong-operator continuous bijection between $B(H)_{sa}$ and $U_1$. 
\end{theorem}
\begin{proof}[Proof, {{\cite[Theorem 19.3]{Zhu}}}. ]
    Let $T \in B(H)$ be self-adjoint. By \autoref{proposition:self-adjoint-spectrum}, $\sigma_{B(H)}(T) \subset \real$. By the \hyperref[Spectral Mapping Theorem]{theorem:spectral-mapping-continuous}, 
    \[
    \sigma_{B(H)}[(T-i)(T+i)^{-1}] \subset \bracsn{(t - i)/(t + i)|t \in \real} \subset \partial B_\complex(0, 1) \setminus \bracsn{1}
    \]
    
    Hence $(T - i)(T+i)^{-1}$ is a well-defined unitary element of $B(H)$ whose spectrum does not contain $1$. 

    Since the mapping $t \mapsto -i(t + 1)/(t - 1)$ is the inverse of $t \mapsto (t - i)/(t + i)$ on $\partial B_\complex(0, 1)$, the \hyperref[Spectral Mapping Theorem]{theorem:spectral-mapping-continuous} implies that $T \mapsto -i(T + I)(T - I)^{-1}$ is the inverse of the Cayley transform on $U_1$. 

    For any self-adjoint elements $S, T \in B(H)$,
    \begin{align*}
        U(S) - U(T) &= (S + i)^{-1}(S - i) - (T - i)(T+i)^{-1} \\
        &= (S + i)^{-1}[(S - i)(T + i)  - (S + i)(T - i)](T + i)^{-1} \\
        &= 2i(S + i)^{-1}(S - T)(T + i)^{-1}
    \end{align*}

    so for any $x \in H$, 
    \[
    \normn{[U(S) - U(T)]x}_H \le 2\normn{(S + i)^{-1}}_{B(H)} \cdot \normn{(S - T)(T+i)^{-1}x}_H
    \]

    By the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, $\normn{(S+i)^{-1}}_{B(H)} \le 1$. Therefore the Cayley transform is strong-operator continuous. 
\end{proof}

\begin{corollary}
\label{corollary:functional-calculus-c0-self-adjoint}
    Let $H$ be a complex Hilbert space and $f \in C_0(\real; \complex)$, then the mapping $T \mapsto f(T)$ is strong-operator continuous on $B(H)_{sa}$. 
\end{corollary}
\begin{proof}
    Let
    \[
    g: \partial B_\complex(0, 1) \to \complex \quad z \mapsto \begin{cases}
        f(-i(z+1)/(z-1)) &z \ne 1 \\
        0 &z = 1
    \end{cases}
    \]

    then since $f \in C_0(\real; \complex)$, $g \in C(\partial B_\complex(0, 1); \complex)$. For each $T \in B(H)_{sa}$, $f(T) = g(U(T))$. By \autoref{proposition:bh-adjoint-strong-continuous}, the mapping $U \mapsto g(U)$ is strong-operator continuous on the set of unitary operators on $H$. By \autoref{theorem:cayley-sa-uni}, $T \mapsto f(T)$ is the composition of two strong-operator continuous mappings. 
\end{proof}