37.2 The Cayley Transform
Definition 37.2.1 (Cayley Transform).label 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 Cayley transform of $T$.
Theorem 37.2.2.label 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 $U_{1}$.
Proof, [Theorem 19.3, Zhu93]. Let $T \in B(H)$ be self-adjoint. By Proposition 36.4.6, $\sigma_{B(H)}(T) \subset \real$. By the Spectral Mapping Theorem,
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 Spectral Mapping Theorem 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 $S, T \in B(H)$,
so for any $x \in H$,
By the continuous functional calculus, $\normn{(S+i)^{-1}}_{B(H)}\le 1$. Therefore the Cayley transform is strong-operator continuous.$\square$
Corollary 37.2.3.label 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 $B(H)_{sa}$.
Proof. Let
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 Proposition 37.1.5, the mapping $U \mapsto g(U)$ is strong-operator continuous on the set of unitary operators on $H$. By Theorem 37.2.2, $T \mapsto f(T)$ is the composition of two strong-operator continuous mappings.$\square$
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