Theorem 41.1.4 (Sz.-Nagy’s Dilation Theorem).label Let $H$ be a complex Hilbert space and $T \in B(H)$ with $\norm{T}_{B(H)}\le 1$, then there exists a unitary power dilation of $T$.
Proof, [Theorem 1.1, Po02]. Using Lemma 41.1.3 it is sufficient to show the following statements:
- (i)
For any $T \in B(H)$ with $\norm{T}_{B(H)}\le 1$, there exists an isometric power dilation of $T$.
- (ii)
For any isometry $V \in B(H)$, there exists a unitary power dilation of $V$.
(i): Since $\norm{T}_{B(H)}\le 1$, $1 - T^{*}T \ge 0$, and $(1 - T^{*}T)^{1/2}$ is well-defined. Let $D_{T} = (1 - T^{*}T)^{1/2}$, and
then for each $x \in H$,
Thus for each $x = \seq{x_n}\in l^{2}(H)$,
So $V$ is an isometry. Moreover, $[V^{n}(x, 0, \cdots)]_{1} = T^{n}x$ for all $n \in \natz$. Therefore under the identification that $H = \bracsn{(x, 0, \cdots)|x \in H}\subset l^{2}(\natp; H)$, $V$ is an isometric power dilation of $T$.
(ii): Let $P = I_{H} - VV^{*}$ be the projection onto $V(H)^{\perp}$, and $U \in B(H^{2})$ be defined by
then
so $U$ is unitary. Moreover, for each $n \in \natp$,
Under the identification that $H = \bracsn{(x, 0)|x \in H}\subset H^{2}$, $QU^{n}|_{H}= V^{n}$ for all $n \in \natp$. Therefore $U$ is a unitary power dilation of $V$.$\square$
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