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\documentclass{article} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{amsthm} \newtheorem{theorem}{Theorem} \newtheorem{lemma}{Lemma} \begin{document} \begin{theorem} Any irreducible strongly continuous one-parameter semigroup of patial isometries $S_t$, $t\ge0$, from $H_0$ is unitarily equivalent to the following semigroup $S_t^{(a)}$ in $L_2([0,a], dx)$ with some $a\ge0$: \[ (S_t^{(a)}f)(x) = \chi_{[t,a]}(x)\, f(x-t). \] \end{theorem} \begin{proof} We will proof this in several stages. 1. Any strongly continuous one-parameter semigroup of partial isometries is centered. 2. We show that $H= H_0 \oplus H_+ \oplus H_- \oplus H_1$. 3. We only need to describe the structute in $H_0$. Decompose $H_0$ wrt $S_tS_t^*$ 4. Decompose by ``length'' $a$. \begin{lemma} Let $H=H_0$. Consider $H_t = \cup_sS_s^*Q_t H$, $Q_t = S_tS_t^*$. These subspases possess the following properties: (i) $H_t\subset H_s$, $s <t$, (ii) any $H_t$ is invariant wrt $S_s$, $S_s^*$. (iii) $\cap _t H_t = 0$, $\cup_t H_t = H$, \end{lemma} \begin{proof} Indeed, (i) follows from the fact that $Q_t$ is a decreasing family of projections. To show (ii), first notice that by the construction $H_t$ is invariant with respect to $S_s^*$, $s \ge0$. For any vector of the form $S_s^*Q_tf \in H_t$ we have, if $\lambda\ge s$, \begin{align*} S_\lambda S_s^* Q_t f &= S_{\lambda-s} Q_sQ_tf = Q_\lambda S_{\lambda - s} Q_tf = Q_\lambda Q_{t+\lambda -s} S_{\lambda - s}f \\ & = Q_{t+\lambda - t}(Q_\lambda S_{\lambda - s}f) \in H_t, \end{align*} since $Q_{t+\lambda - s}H \subset Q_t H \subset H_t$. For $\lambda <s$ we have \[ S_\lambda S_s^* Q_tf = Q_\lambda S_{s-\lambda}^* Q_tf= S_{s-\lambda}^* Q_s Q_t f = S_{s-\lambda}^* Q_t(Q_sf) \in H_t, \] therefore, $H_t$ is an invariant subspace for $S_s$, $S_s^*$, $s\ge0$. Now we show (iii). For $t=0$ we have $H_t=H$, thus the union is the whole $H$. We need to show that $\cap_t H_t = 0$. First notice that $Q_\lambda H_t \subset H_t$, $\lambda$, $t>0$. Indeed, this follows from the relations $Q_\lambda S_s^*Q_t = S_s^* Q_{s+\lambda} Q_t = S_s^* Q_t Q_{s+\lambda}$ for all $\lambda$, $t$, $s$. Introduce subspaces $H_{t,\lambda} = \cup_{s \ge \lambda} S_s^*Q_t H$, $H_{t,\lambda}^0 = Q_{t-\lambda} H_t$, $H_{t,\lambda}^1 = H_t \ominus H_{t,\lambda}^0 = (I-Q_{t-\lambda}) H_t$. For $s\le \lambda$ we have \[ Q_{t-\lambda} S_s^* Q_t=S_s^* Q_{t-(\lambda -s)} Q_t = S_s^*Q_t, \] therefore, $\cup_{0<s<\lambda} S_s^* Q_t H \subset H_{t,\lambda}^0$, and \[ H_{t,\lambda}^1 \subset \cup_{s\ge \lambda} S_s^* Q_t H = H_{t,\lambda}. \] On the other hand, for $\lambda <s$ we have \[ P_\lambda S_s = S_\lambda^* S_\lambda S_\lambda^*S_{s-\lambda} = S_\lambda^*, \] which implies that $P_\lambda=I$ in $H_{t,\lambda}$, and, therefore, in $H_{t,\lambda }^1$. The space $\cap_t H_t$ is invariant with respect to $S_\lambda$, $S_\lambda^*$, $\lambda \ge0$. For any $f \in \cap _t H_t$ and any fixed $\lambda$ we have $Q_{t-\lambda} f \to 0$, $t\to \infty$, therefore, $P_\lambda f =f$. But since $P_\lambda \to0$, $t\to\infty$ strongly, this implies $f=0$. \end{proof} The lemma above implies that in the case of irreducible one-parameter semigroup of partial isometries there exists $a>0$ such that $H_t = H$, $t\le a$, and $H_t=0$, $t>a$. We will show that to each $a>0$ there corresponds exactly one, up to a unitary equivalence, irreducible one-parameter semigroup of partial isometries. \end{proof} \end{document} %%% Local Variables: %%% mode: latex %%% TeX-master: t %%% End: