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\documentclass{article}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amsthm}
\newtheorem{theorem}{Theorem}

\begin{document}

\begin{theorem}
Let $S_t$ be a strongly continuous semigroup of partial isometries in
a Hilbert space $H$. Then $H$ can be decomposed into a direct sum of
invariant with respect to $S_t$, $S_t^*$, $t\ge0$, subspaces,
$H=H_0\oplus H_+\oplus H_- \oplus H_1$ such that:

in $H_1$ the semigroup $S_t$, $t\ge0$ is unitary, i.e., all operators
$S_t$ are unitary\textup;

in $H_+$ $S_t$ is a semigroup of completely non-unitary isometries\textup;

in $H_-$ $S_t^*$ is a semigroup of completely non-unitary
isometries\textup;

in $H_0$ $S_t\to 0$, $S_{t}^* \to 0$, $t\to \infty$ strongly.
\end{theorem}

\begin{proof}
Indeed,
\end{proof}

\begin{theorem}
Let $S_t$, $t\ge0$, be a strongly continuous semigroup of completely
non-isometric partial isometries. Then it is unitarily equivalent to 
the direct integral
\[
S_t =\int _{a>0}^\oplus S_t^{(a)}\otimes I_a\,d\mu
\]
where $S_t$ is the following semigroup in $H_a = L_2([0,a],dx)$
\[
(S_tf)(x) = \chi_{[t,a]}(x)\, f(x-t)
\]
\end{theorem}

\begin{proof}
Indeed,
\end{proof}


\end{document}

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