and, after changing variables,

|\widehat h(s-t)|^2\,ds = \int_{-b}^{b}
|\widehat h(t)|^2\,ds\,.

Since $b$ is arbitrary, we get $\widehat h= 0$ a.e. on $\R$. Hence, $h=0$ a.e. which gives $Y_0 = 0$ and so $Y_{+}$ is defined uniquely by $X$.

To complete the proof, it remains to check that $\alpha_\tau  = 1$ and $G(\tau) > 0$ for every $\tau \in \R_+$ in the case when $\Hh$ is diagonal. Then,  $\gamma(\tau) = 1$ as well by (<ref>). We have $\Rr_{\Hh}(\tau) = 0$, $\tau \in \R_+$, for any diagonal Hamiltonian $\Hh$, see Lemma 2.2 in [5]. Then $G(\tau)$ is a diagonal matrix with positive entries for every $\tau \in \R_+$, in particular, $G(\tau) > 0$. Suppose for a moment that $\det\Hh = 1$ almost everywhere on $\R_+$. Then, formula $(42)$ in [8] for $z = i$ together with the relation $\Rr_{\Hh} = 0$ says that $\frac{\partial}{\partial \tau}(e^{-\tau}\Et_\tau(i)) = -\K'_{\Hh}(\tau)e^{-\tau}\Et_\tau(i)$. Since $\Et_0(i) > 0$, this shows that for such Hamiltonians $\Hh$ we have $\Et_\tau(i) > 0$, $\tau \in \R_+$. That implies $\alpha_\tau = 1$ for all $\tau \in \R_+$. Now let $\Hh \in \szcs$ be a diagonal Hamiltonian such that $\det\Hh > 0$ almost everywhere on $\R_+$. Define the new Hamiltonian $\widetilde{\Hh}: \tau \mapsto (\det\Hh(L_\tau))^{-1/2}\Hh(L_\tau)$, $\det\widetilde{\Hh} = 1$ on $\R_+$. The function $\tau \mapsto \Theta(L_{\tau},z)$ then solves Cauchy problem (<ref>) for $\widetilde{\Hh}$. The previous reasoning shows  that for the corresponding coefficient $\alpha_\tau$ we have $\widetilde{\alpha}_\tau = 1$, $\tau \in \R_+$. But $\widetilde{\alpha}_\tau = \alpha_{L_\tau}$, and we see that $\alpha_\tau = 1$ for all diagonal Hamiltonians $\Hh \in \szcs$ such that $\det\Hh > 0$ almost everywhere on $\R_+$. Then, the general case follows via an approximation argument by considering Hamiltonians of the form $\Hh_\eps = \eps\idm + \Hh$, and letting $\eps \to 0$.

§.§ Scattering and wave operators
The  following theorem answers the question: does the asymptotics $Y_{\pm}$ of a state $X$ under the evolution $U_t$ determine the state $X$ itself?  It also strengthens Theorem <ref>.

Let $\Hh \in \szcs$, let $\Di_{\Hh}$ be the corresponding self-adjoint operator (<ref>) on $H$, and let $\mu = w\,dx + \mus$ be the main spectral measure for $\Hh$. Then, the strong wave operators

\begin{equation}\label{eq35}
W_{\pm} = \lim_{t\to\pm\infty}U_{t}^{-1}\widetilde U^{0}_{\gamma, t}
\end{equation}

are correctly defined (the limits are understood in the strong operator topology) and unitary as operators from $L^2(\Hh_0)$ onto the absolutely continuous subspace $H_{\bf ac}$ of $\Di_{\Hh}$. Moreover, if $X\in H$, $Y_{X,\pm}$ are defined by (<ref>) and $P_{\bf ac}$ denotes the orthogonal projection in $H$ onto the absolutely continuous subspace of $\Di_{\Hh}$, then we have $W_{\pm}Y_{X,\pm} = P_{\bf ac}X$, so that $Y_{X,\pm}$ determine $P_{\bf ac}X$ uniquely and we have $Y_{X_1,\pm} = Y_{X_2,\pm}$ in Theorem <ref> if and only if $P_{\bf ac}X_1 = P_{\bf ac}X_2$.
The scattering operator

S = W_+^{-1}W_{-}, \qquad S: Y_{X,-} \mapsto Y_{X,+},

is a unitary operator on $L^2(\Hh_0)$, and its spectral representation takes the form

\begin{equation}\label{eq5}
\W_{\Hh_0}S\W_{\Hh_0}^{-1} f_0 = \frac{\ov D_{\mu}}{D_{\mu}}f_0, \qquad f_0 \in L^2(\R),
\end{equation}

where $D_{\mu}$ is the Szegő function of $\mu$. In particular, the operator $S$ does not depend on the choice of the phase function $\gamma$ in Theorem <ref>.

Let us first prove that the limit  in (<ref>) exists as $t \to+\infty$.  The argument for $t\to -\infty$ is similar.  In fact, we claim that for an arbitrary $X$, we have

\begin{equation}\label{sa13}
\lim_{t\to+\infty}U_{t}^{-1}\widetilde U^{0}_{\gamma, t} Y_{X, +}=P_{\bf ac} X\,.
\end{equation}

Indeed, denoting $\Delta_{b,t} = [L_{|t|-b}, L_{|t|+b}]$ for some positive numbers $b$, we have

\begin{align*}
U_{t}^{-1}\widetilde U^{0}_{\gamma,t}Y_{X, +}
U_{t}^{-1}\chi_{\Delta_{b,t}}\widetilde U^{0}_{\gamma,t} Y_{X, +} + U_{t}^{-1}\chi_{\R_+\setminus\Delta_{b,t}}\widetilde U^{0}_{\gamma,t} Y_{X, +}\\
U_{t}^{-1} \chi_{\Delta_{b,t}} U_t X + U_{t}^{-1}(\chi_{\Delta_{b,t}}\widetilde U^{0}_{\gamma,t}Y_{X, +} - \chi_{\Delta_{b,t}}U_t X) + U_{t}^{-1}\chi_{\R_+\setminus\Delta_{b,t}}\widetilde U^{0}_{\gamma,t} Y_{X, +}.
\end{align*}

To get our claim, it is enough to prove that

$(a)$ $
\lim_{b\to+\infty}\limsup_{t \to +\infty}\|\chi_{\R_+\setminus\Delta_{b,t}}\widetilde U^{0}_{\gamma,t} Y_{X, +}\|_{L^2(\Hh)}=0,$

$(b)$$\lim_{t \to +\infty}\|\chi_{\Delta_{b,t}}\widetilde U^{0}_{\gamma,t} Y_{X, +} - \chi_{\Delta_{b,t}}U_t X\|_{L^2(\Hh)} = 0$ for every $b \ge 0$,

$(c)$ $\lim_{b\to+\infty}\limsup_{t \to +\infty}\|U_{t}^{-1} \chi_{\Delta_{b,t}} U_t X - P_{\bf ac}X\|_{L^2(\Hh)} = 0$.

Clearly, $(b)$ is just a restatement of Theorem <ref>. To check $(a)$, observe that

\begin{align*}
\|\chi_{\R\setminus\Delta_{b,t}}\widetilde U^{0}_{\gamma,t}Y_{X, +}\|_{L^2(\Hh)}
&=\int_{\R_+\setminus\Delta_{b,t}}\sqrt{\det\Hh(\tau)}\|U^{0}_{t}Y_{X, +}(T(\tau))\|_{\C^2}^{2}\,d\tau\\
&=\int_{\R_+\setminus [|t|-b,|t|+b]}\|U^{0}_{t}Y_{X, +}(\tau)\|_{\C^2}^{2}\,d\tau.
\end{align*}

Now, (<ref>) yields (a).  It remains to prove $(c)$. First, notice that

\[
\|U_{t}^{-1} \chi_{\Delta_{b,t}} U_t X - P_{\bf ac}X\|^2_{L^2(\Hh)}=\|\chi_{\Delta_{b,t}} (U_t X - U_tP_{\bf ac}X)\|_{L^2(\Hh)}^2+\| \chi_{\R_+\backslash \Delta_{b,t}}  U_tP_{\bf ac}X\|_{L^2(\Hh)}^2\,.
\]

By Theorem <ref>, we have $Y_{X,+}=Y_{P_{\bf ac}X,+}$ and therefore
\lim_{t\to +\infty}\|\chi_{\Delta_{b,t}} (U_t X - U_tP_{\bf ac}X)\|_{L^2(\Hh)}=0\,.

\[
\| \chi_{\R_+\backslash \Delta_{b,t}}  U_tP_{\bf ac}X\|^2_{L^2(\Hh)}=\| U_tP_{\bf ac}X\|^2_{L^2(\Hh)}-\| \chi_{ \Delta_{b,t}}  U_tP_{\bf ac}X\|^2_{L^2(\Hh)}=\|P_{\bf ac}X\|^2_{L^2(\Hh)}-\| \chi_{ \Delta_{b,t}}  U_tP_{\bf ac}X\|^2_{L^2(\Hh)}
\]

and, by Theorem <ref>,

\begin{eqnarray*}
\limsup_{t\to +\infty}(\|P_{\bf ac}X\|^2_{L^2(\Hh)}-\| \chi_{ \Delta_{b,t}}  U_tP_{\bf ac}X\|^2_{L^2(\Hh)})=\|P_{\bf ac}X\|^2_{L^2(\Hh)}-\liminf_{t\to +\infty}\| \chi_{ \Delta_{b,t}}  \widetilde U^{0}_{\gamma,t}Y_{P_{\bf ac}X, +}   \|^2_{L^2(\Hh)}\\
=\|P_{\bf ac}X\|^2_{L^2(\Hh)}-\liminf_{t\to +\infty}(\|   \widetilde U^{0}_{\gamma,t}Y_{P_{\bf ac}X, +}   \|^2_{L^2(\Hh)}-\| \chi_{ \R_+\backslash\Delta_{b,t}}  \widetilde U^{0}_{\gamma,t}Y_{P_{\bf ac}X, +}   \|^2_{L^2(\Hh)})\,.
\end{eqnarray*}

\begin{equation}\label{rab1}
\|  \widetilde U^{0}_{\gamma,t} Y_{P_{\bf ac}X, +}   \|^2_{L^2(\Hh)}=\|P_{\bf ac}X\|^2_{L^2(\Hh)}
\end{equation}

by (<ref>) and
\[
\lim_{b\to+\infty}\limsup_{t \to +\infty}\|\chi_{\R_+\setminus\Delta_{b,t}}\widetilde U^{0}_{\gamma,t}Y_{P_{\bf ac}X, +}\|_{L^2(\Hh)}=0\]
by (a), we get (c).

Then, (<ref>) implies, in particular, that the map $\mathcal{Y}_+: X\mapsto Y_{X,+}$ is the unitary bijection from $H_{\bf ac}$ onto $L^2(\Hh_0)$. So, $\lim_{t\to+\infty}U_{t}^{-1}\widetilde U^{0}_{\gamma,t} Y$ exists for every $Y\in L^2(\Hh_0)$ and $\lim_{t\to+\infty}U_{t}^{-1}\widetilde U^{0}_{\gamma,t} Y=\mathcal{Y}_+^{(-1)}Y$.

Summarizing, we have proved that the strong wave operator $W_+$ in (<ref>) exists and $W_+ Y_{X, +} = P_{\bf ac} X$ for every $X$,  where $Y_{X, +}$ is defined as in Theorem <ref>. Analogously, one can check the existence of the wave operator $W_{-}$ and prove the formula $W_- Y_{X, -} = P_{\bf ac} X$. All other assertions of the theorem are simple consequences of these two facts.

The following corollary implies, in particular, that if Szegő measure $\mu$ is purely a.c., then every $X\in H$ propagates and the global $L^2(\Hh)$ asymptotics holds for $U_tX$. The reader can compare it to Theorem <ref> which establishes the asymptotics over the finite interval.

Let $\Hh$ be a Hamiltonian of class $\szcs$. Then there exists a phase function $\gamma$ such that the following assertion holds. For every $X \in H_{\bf ac}$, there are uniquely defined  $Y_{X,\pm} \in L^2(\Hh_0)$ such that

\begin{equation}\label{eq000}
\lim_{t \to \pm\infty}\|U_t X - \widetilde U^{0}_{\gamma,t} Y_{X,\pm}\|_{L^2(\Hh)} =0,
\end{equation}

Moreover, if $\Hh$ is diagonal, then one can take $\gamma = 1$ on $\R_+$.

Indeed, if $X \in H_{\bf ac}$, then
$X = P_{\bf ac}X$, and (<ref>) is equivalent to

\lim_{t \to \pm\infty}\|P_{\bf ac}X - U_t^{-1} \widetilde U^{0}_{\gamma,t} Y_{X,\pm}\|_{L^2(\Hh)} =0,

that holds by Theorem <ref>.

§.§ Dynamical classification of spectral types
Our analysis allows to detect the spectral types of $\Di_\Hh$ by observing the long-time dynamics of $U_t$.

Suppose $X\in H$ is given. Denote the orthogonal projections to absolutely continuous, singular continuous, and pure point subspaces of $\Di_{\Hh}$ by $P_{\bf ac}, P_{\bf sc},$ and $P_{\bf pp}$, respectively.
Our next result gives the dynamical characterization of whether $X$ has nontrivial projections to any of these subspaces.

Let $\Hh \in \szcs$. Then, for every $X \in H$ we have

\begin{align}
\lim_{b \to +\infty}\lim_{T \to +\infty}\frac{1}{T}\int_{0}^{T}\|U_{t}X\|_{L^2(\Hh, [0, L_b])}^{2}\,dt
&= \|P_{\bf pp}X\|_{L^2(\Hh)}^{2}, \label{eq83bis}\\
\lim_{b \to +\infty}\lim_{T \to +\infty}\frac{1}{T}\int_{0}^{T}\|U_t X\|_{L^2(\Hh, [L_{b}, L_{t-b}])}^{2}\,dt &= \|P_{\bf sc} X\|_{L^2(\Hh)}^{2},\label{eq89bis}\\
\lim_{b \to +\infty}\lim_{t \to +\infty}\|U_t X\|_{L^2(\Hh, [L_{t-b}, L_{t+b}])}^{2} &= \|P_{\bf ac} X\|_{L^2(\Hh)}^{2}, \label{eq82bis}\\
\lim_{b \to +\infty}\lim_{t \to +\infty}\|U_t X\|_{L^2(\Hh, [L_{t+b}, +\infty))} & = 0. \label{eq91bis}
\end{align}

The analogous statements holds when $t\to +\infty$ is replaced by $t\to -\infty$.

We start with proving (<ref>). Given $X$ and $\epsilon>0$, we can find $X_\epsilon\in H$ such that $\|X-X_\epsilon\|_{L^2(\Hh)}\le \epsilon$ and $X_\epsilon$ has compact support. From Theorem <ref>, we get

\[
\lim_{b \to +\infty}\limsup_{t \to +\infty}\|U_t X_\epsilon\|_{L^2(\Hh, [L_{b+t}, +\infty))}= 0\,.
\]

\begin{align*}
\limsup_{b \to +\infty}\limsup_{t \to +\infty}\|U_t X\|_{L^2(\Hh, [L_{b+t}, +\infty))}
\le &\limsup_{b \to +\infty}\limsup_{t \to +\infty}\|U_t (X-X_\epsilon)\|_{L^2(\Hh, [L_{b+t}, +\infty))}\\
&+\limsup_{b \to +\infty}\limsup_{t \to +\infty}\|U_t X_\epsilon\|_{L^2(\Hh, [L_{b+t}, +\infty))}\le \epsilon
\end{align*}

because $\|U_t(X-X_\epsilon)\|_{L^2(\Hh)}=\|X-X_\epsilon\|_{L^2(\Hh)}\le \epsilon$. Since $\epsilon$ is arbitrary, we get (<ref>).

Formula (<ref>) is immediate from Theorem <ref>.

To prove (<ref>), we apply Lemma <ref> from Appendix. Take $\Lambda>0$ and $b\in \R_+ \setminus \bigcup_{I \in \mathfrak{I}(\Hh)} I.$ Let $P_{[-\Lambda,\Lambda]}$ be the orthogonal projection associated with the spectral decomposition of $\Di_{\Hh}$. We claim that the operator
\chi_{[0,b]}P_{[-\Lambda,\Lambda]}
$ is compact in $H$. Indeed, this follows from the formula

\[
\chi_{[0,b]}(\tau)(P_{[-\Lambda,\Lambda]}X)(\tau)=\frac{\chi_{[0,b]}(\tau)}{\sqrt\pi}\int_{[-\Lambda,\Lambda]}\Theta(\tau,x)(\W_{\Hh}X)(x)\,d\mu(x)
\]

and the fact that the set $\{\int_{[-\Lambda,\Lambda]}\Theta(\tau,x)(\W_{\Hh}X)d\mu: \|X\|_{H}\le 1\}$ is precompact in $C[0,b]$ by Arzela-Ascoli theorem. Hence, by Lemma <ref>  applied to Hilbert space $H$, operator $\Di_{\Hh}$, and  $A=\chi_{[0,b]}$, one has

\[
\lim_{T\to\infty}\frac 1T\int_0^T \|\chi_{[0,b]}U_tP_{[-\Lambda,\Lambda]}X\|_{L^2(\Hh)}^2dt=\sum_{j}\|\chi_{[0,b]}P_{\{E_j\}}P_{[-\Lambda,\Lambda]}X\|^2_{L^2(\Hh)}\,,
\]

where $P_{\{E_j\}}$ is orthogonal projection corresponding to eigenvalue $E_j$ of $\Di_{\Hh}$ and the sum is done over all eigenvalues.
Taking $b$ to infinity (see Corollary 2 in [62]), we have

\begin{equation}\label{des6}
\lim_{b\to \infty}\lim_{T\to\infty}\frac 1T \int_0^T \|\chi_{[0,b]}U_t P_{[-\Lambda,\Lambda]}X\|_{L^2(\Hh)}^2dt=\|P_{\bf pp}P_{[-\Lambda,\Lambda]}X\|^2_{L^2(\Hh)}\,,
\end{equation}

for every $X$. Now, taking $\Lambda\to \infty$, we get (<ref>).

We are left with showing (<ref>). Fix any $X$ and $b>0$. Then,

\[
\|U_tX\|^2_{L^2(\Hh)}=\|U_tX\|^2_{L^2(\Hh,[0,L_b])}+\|U_tX\|^2_{L^2(\Hh, [L_b,L_{t-b}])}+\|U_tX\|^2_{L^2(\Hh,[L_{t-b},L_{t+b}])}+\|U_tX\|^2_{L^2(\Hh,[L_{t+b},\infty))}\,.
\]

We also have

\[
\|U_tX\|^2_{L^2(\Hh)}=\|X\|^2_{L^2(\Hh)}=\|P_{\bf ac}X\|^2_{L^2(\Hh)}+\|P_{\bf sc}X\|^2_{L^2(\Hh)}+\|P_{\bf pp}X\|^2_{L^2(\Hh)}\,.
\]

Subtracting one identity from the other and taking the Cesaro mean, we get

\[
\lim_{b\to\infty}\lim_{T\to\infty}\left(\frac{1}{T}\int_0^T \|U_tX\|^2_{L^2(\Hh, [L_b,L_{t-b}])}dt-\|P_{\bf sc}X\|^2_{L^2(\Hh)}\right)=0\,,
\]

as follows from already established (<ref>), (<ref>), and (<ref>). The arguments for $t\to -\infty$ are identical.

We will need the following technical lemma later in the text.
Suppose $\Hh$ is a proper Hamiltonian,  $r\in \R_+ \setminus \bigcup_{I \in \mathfrak{I}(\Hh)} I$, $X\in H$ is compactly supported on $[0,r]$, and $(X, \left(\begin{smallmatrix} 1\\ 0\end{smallmatrix}\right))_{L^2(\Hh)}\neq 0$. If $Y$ is defined by

\begin{eqnarray}
Y(\tau)=0,\quad \tau>r,   \nonumber \\
Y(\tau)=J\int_\tau^r \Hh(s)X(s)ds, \quad \tau<r,   \label{des1}
\end{eqnarray}

then $X$ and $P_HY$ satisfy $
|(\W_{\Hh}X)(x)|^2+|(\W_{\Hh}P_{H}Y)(x)|^2>0$ for all $x\in \R$. Such $X$ and $Y$ exist.

First, observe that

\[
\langle  Y(0), \left(\begin{smallmatrix} 0\\ 1\end{smallmatrix}\right)\rangle_{\C^2}=  \langle  J\int_0^r \Hh(s)X(s)ds, \left(\begin{smallmatrix} 0\\ 1\end{smallmatrix}\right)\rangle_{\C^2}=(X, \left(\begin{smallmatrix} 1\\ 0\end{smallmatrix}\right))_{L^2(\Hh)}\neq 0.
\]

Second, notice that $\Theta(\tau,0)=\left(\begin{smallmatrix} 1\\ 0\end{smallmatrix}\right)$ and so $(\W_{\Hh}X)(0)\neq 0$ given assumptions of the lemma.
Suppose $x$ is such that $x\neq 0$ and $(\W_{\Hh}X)(x)=0$. Observe that, by Lemma <ref>, $\widetilde \W_{\Hh}Y=\W_{\Hh}P_HY$. Then,

\begin{align*}
\sqrt{\pi} \cdot \widetilde\W_{\Hh}Y
&=\int_0^r \langle \Hh Y,\Theta(\tau,x)\rangle_{\C^2} d\tau=-x^{-1}\int_0^r  \langle J Y,\Theta'(\tau,x)\rangle_{\C^2} d\tau\\
x^{-1}\left(  -\langle  Y(0), \left(\begin{smallmatrix} 0\\ 1\end{smallmatrix}\right)\rangle_{\C^2}+\int_0^r  \langle J Y',\Theta(\tau,x)\rangle_{\C^2} d\tau \right)\\
&=x^{-1}\left(   -\langle  Y(0), \left(\begin{smallmatrix} 0\\ 1\end{smallmatrix}\right)\rangle_{\C^2}+\int_0^r  \langle \Hh(\tau) X(\tau),\Theta(\tau,x)\rangle_{\C^2} d\tau \right)\\
&=x^{-1}\left(-\langle  Y(0), \left(\begin{smallmatrix} 0\\ 1\end{smallmatrix}\right)\rangle_{\C^2}+\sqrt{\pi} (\W_{\Hh}X)(x) \right)
\neq 0,
\end{align*}

where we used (<ref>) and our other assumptions. Finally, since $\Hh$ is proper, we can always find $X$ that satisfies all conditions and define $Y$ accordingly. For example,

\[
X=\left(\begin{smallmatrix} 1\\ 0\end{smallmatrix}\right)\cdot \chi_{[0,r]}, \quad Y=\left(\begin{smallmatrix} -\int_\tau^r h(s)ds\\ \int_\tau^r h_1(s)ds\end{smallmatrix}\right)\cdot \chi_{[0,r]}
\]

is one possible choice.

Theorem <ref> gives a dynamical description of spectral types for each element $X$ but it does not tell how to detect presence of pure point, singular continuous, and absolutely continuous spectral types for $\Di_{\Hh}$ itself. We will address it in the next theorem. Recall that $\dlt$ is defined as $\dlt = \bigl[L_{|t| -b}, L_{|t| + b}\bigr]$ for $b>0$.

Let $\Hh$ be a Hamiltonian of class $\szcs$. Then, the following holds true.

$(A)$ If the singular spectrum of $\Hh$ is empty, then

\begin{equation}\label{eq0001}
\lim_{b\to +\infty}\lim_{t \to +\infty}\|U_t X\|_{L^2(\Hh,\dlt)} =\|X\|_{L^2(\Hh)}
\end{equation}

for every $X\in H$.

$(B)$ If there is some compactly supported $X\in H$ for which

\begin{equation}\label{lrt1}
\lim_{b \to +\infty}\lim_{T \to +\infty}\frac{1}{T}\int_{0}^{T}\|U_t X\|_{L^2(\Hh, [L_{b}, L_{t-b}])}^{2}\,dt=0,\end{equation}

then $\Di_{\Hh}$ has no singular continuous spectrum.

$(C)$Let vectors $X$ and $Y$ be defined as in Lemma <ref>.  If

\begin{align}
\lim_{b\to +\infty}\liminf_{t \to +\infty}\|U_t X\|_{L^2(\Hh,\dlt)} &=\|X\|_{L^2(\Hh)}, \label{eq0001e} \\
\lim_{b\to +\infty}\liminf_{t \to +\infty}\|U_t P_{H}Y\|_{L^2(\Hh,\dlt)} &=\|Y\|_{L^2(\Hh)}, \label{eq0002e}
\end{align}

then the singular spectrum of $\Di_{\Hh}$ is empty.

$(D)$ Let vectors $X$ and $Y$ be defined as in Lemma <ref>. If

\begin{align}
\lim_{b \to +\infty}\lim_{T \to +\infty}\frac{1}{T}\int_{0}^{T}\|U_{t}X\|_{L^2(\Hh, [0, L_b])}^{2}\,dt&=0, \label{des5}\\
\lim_{b \to +\infty}\lim_{T \to +\infty}\frac{1}{T}\int_{0}^{T}\|U_{t}Y\|_{L^2(\Hh, [0, L_b])}^{2}\,dt&=0, \label{des5bis}
\end{align}

then $\Di_{\Hh}$ has no bound states.

We get the same conclusions if  the limits $t\to +\infty$ are replaced by $t\to -\infty$.

Suppose the singular spectrum is empty, then $X=P_{\bf ac} X$ and our claim follows from (<ref>).

Then, suppose compactly supported  $X$ is such that (<ref>) holds. Recalling Theorem <ref>, consider $f=\W_{\Hh} X$.
Represent the measure $\mu=\mu_{\bf ac}+\mu_{\bf s}$ as a sum of absolutely continuous and singular components and further write $\mu_{\bf s}=\mu_{\bf sc}+\mu_{\bf pp}$ as a sum of singular continuous and pure point parts. Then,
(<ref>) gives

\[
\int_\R |f|^2\,d\mu_{\bf sc}=0\,.
\]

On the other hand, $f$ is an entire function that can have only countably many zeroes in $\C$. Therefore, $|f|>0$ a.e. with respect to $\mu_{\bf sc}$ and so $\mu_{\bf sc}=0$.

To show $(C)$, we only need to prove that (<ref>), (<ref>) imply that the spectrum of $\Hh$ is purely absolutely continuous.  If $f:=\W_{\Hh}X$ and $g:=\W_{\Hh}P_H Y$, then (<ref>), (<ref>) give

\[
\int_{\R} (|f|^2+|g|^2)d\mu_{\bf s}=0.
\]

That, however, contradicts Lemma <ref> unless $\mu_{\bf s}(\R)=0$.

Finally, to get $(D)$, we notice that (<ref>), (<ref>) and (<ref>) give $P_{\bf pp}X=P_{\bf pp}Y=0$ which can be rewritten as

\[
\int_{\R} (|f|^2+|g|^2)d\mu_{\bf pp}=0,
\]

where, again, $f=\W_{\Hh}X$ and $g=\W_{\Hh}P_H Y$. Since $f$ and $g$ are entire functions that have no common zeroes by Lemma <ref>, we get $\mu_{\bf pp}=0$.

The arguments for $t\to -\infty$ are identical.

§ KREIN STRINGS

In this section, we introduce  Krein strings, explain the connection with diagonal canonical systems, and use this connection to translate some results obtained in the previous section to this new setting. We will be using extensively two papers [5]  and [35].

§.§ Krein strings

Let $0 < L \le \infty$ and $M$ be a non-decreasing right-continuous function on $(-\infty,L)$, satisfying $M(\xi)=0, \xi<0$. The Lebesgue–Stieltjes measure $\mf$ on $[0,L)$ is defined by $\mf[0,\xi]=M(\xi)$. We write its decomposition into the absolutely continuous and singular parts as $\mf=\mf_{\bf ac}+\mf_{\bf s}=\rho(\xi)d\xi+\mf_{\bf s}$. Recall that in our notation $M(L-)=\lim_{\xi\to L, \xi<L}M(\xi)$ and
that the $[M,L]$ pair is called proper if $M$ and $L$ satisfy the following conditions

\begin{align}
&0<M(\xi)<M(L-), \quad \forall\xi\in (0,L).\label{sa15}
\end{align}

In this paper, we will work with proper $[M,L]$ pairs only.  Some authors (e.g., [35]) define strings with left-continuous $M$ while others (e.g., [25]) work with right-continuous $M$.
In [34] and [35], the authors considered two functions $\phi(\xi,z)$ and $\psi(\xi,z)$ that satisfy

\[
\phi(\xi,z)=1-z\int_{[0,\xi]}(\xi-s)\phi(s,z)\,d\mf (s), \quad \xi\in [0,L)\,,
\]

\[
\psi(\xi,z)=\xi-z\int_{[0,\xi]}(\xi-s)\psi(s,z)\,d\mf (s), \quad \xi\in [0,L)\,.
\]

It is customary to extend $\phi$ and $\psi$ to $(-\infty,0)$ by $\phi(\xi,z)=1$ and $\psi(\xi,z)=\xi$ where $\xi<0$.
These functions are uniquely determined by the string $[M,L]$ and they define the principal Titchmarsh-Weyl function $q$ of $[M,L]$ by

\[
q(z)=\lim_{\xi\to L} \frac{\psi(\xi,z)}{\phi(\xi,z)}, \quad z\in \mathbb{C}\backslash [0,\infty),
\]

see formula (2.21) in [35] or Theorem 10.1 in [34]. This function $q$ has the unique integral representation

\[
q(z)=\int_{\R_+} \frac{d\sigma(\lambda)}{\lambda-z}\,,
\]

where  $\sigma$, the main (or orthogonal) spectral measure of the string $[M,L]$, is a measure on $\R_+$ satisfying condition

\[
\int_{\R_+}\frac{d\sigma(\lambda)}{1+\lambda}<\infty\,.
\]

Similarly to (<ref>), we can define the generalized Fourier transform associated with the string $[M,L]$:

\begin{equation}\label{wts}
\Us: v \mapsto \int_{0}^{L}v(\xi)\phi(\xi,z)d\mf, \qquad z \in \C,
\end{equation}

starting with functions  $v\in L^2(\mf)$ that have compact support in $[0,L)$.  It is known (see Section 10 in [34]) that $\Us$ can be extended to the unitary operator from $L^2(\mf)$ onto $L^2(\sigma)$.
The inverse map is given by (see formula (2.25) in [35])

\[
v=\Us^{-1}(\Us v)=\int_0^\infty \phi(\xi,\lambda)(\Us v)(\lambda)d\sigma\,,
\]

where the last integral can be first defined on $L^2(\sigma)$ functions with compact support and then extended to all of $L^2(\sigma)$. The theory of Krein strings goes back to works by M. Krein [45] and Feller [27].

§.§ Connection between Krein strings and canonical systems with diagonal Hamiltonians

Suppose  $[M,L]$ is a proper string.
Consider the increasing function  $N: \xi \mapsto \xi+M(\xi)$ on $[0, L)$ and let $\nf$ denote the corresponding measure, $\nf[0,\xi]=N(\xi)$ for $\xi \ge 0$. Condition  (<ref>) is equivalent to $N(L-)=+\infty$. Define also the function $N^{(-1)}$ on $\R_+$ by $N^{(-1)}: y \mapsto  \inf \{ \xi\ge 0: N(\xi) \ge y \}$. The set under the last infimum is non-empty for every $y \ge 0$ because of the assumptions we made on $M$ and $L$. Using the fact that $N$ is strictly increasing, one can show that $N^{(-1)}$ is continuous on $\R_+$, and we have $N^{(-1)}(N(\xi)) = \xi$ for every $\xi \in [0, L)$. Recall that  $\rho$ is the density of the absolutely continuous part of $\mf$, so that $\mf=\rho d\xi+\mf_s$.  Define two functions on $\R_+$:

\begin{equation}
\begin{cases}
0, & \mbox{if } N^{(-1)}(\tau)\in \Omega_{\bf s}(\mf),\\
\frac{1}{1+\rho (N^{(-1)}(\tau))}, & \mbox{otherwise},
\end{cases}
\label{sd1}
\end{equation}

\begin{equation}
\begin{cases}
1, & \mbox{if } N^{(-1)}(\tau)\in \Omega_{\bf s}(\mf),\\
\frac{\rho(N^{(-1)}(\tau))}{1+\rho(N^{(-1)}(\tau))}, & \mbox{otherwise}.
\end{cases}
\label{sd2}
\end{equation}

Given a Hamiltonian $\Hh=\diag(h_1,h_2)$, define $\Hh_{\sigma_1}=\sigma_1\Hh\sigma_1=\diag(h_2,h_1)$, where $\sigma_1=\left(\begin{smallmatrix}0 & 1\\ 1  & 0\end{smallmatrix}\right)$.
If $\Hh_{\sigma_1}$ is proper and   $\trace \Hh=1$, we let

\begin{equation}\label{lk3}
\xi(\tau)=\int_0^\tau h_1(s)ds, \quad L=\int_0^\infty h_1(s)ds, \quad
M(\xi)=\int_0^{\sup\{x: \xi(x)=\xi\}} h_2(s)ds,\quad \xi<L\,.
\end{equation}

Formulas (<ref>), (<ref>), and (<ref>) establish the bijection $[M ,L] \mapsto \diag(h_2,h_1)$ between proper $[M, L]$ pairs  and  proper Hamiltonians $\Hh_{\sigma_1}= \diag(h_2, h_1)$ with unit trace.

This correspondence is explained in Section 4 of [35] where the function $M$ is taken left-continuous. That difference in normalization makes an effect only at those $\xi^*$ where $M$ has a jump. These points correspond to those indivisible intervals where $\Hh=\diag(0,1)$. Moreover, $N(\xi^*)$ is the right end of each such indivisible interval.

It is only left to comment  that
the assumption that the left end of the string is heavy is equivalent to condition that the Hamiltonian $\Hh_{\sigma_1} = \diag(h_2, h_1)$ is not equal to $\excl$ on $[0,\varepsilon]$ with some  $\varepsilon>0$. Moreover,
making assumption that
$ L+M(L-)=+\infty$ and the right end is heavy is equivalent to saying that $\Hh$ is not equal to either $\excl$ or $\left(\begin{smallmatrix} 1&0\\0&0\end{smallmatrix}\right)$ on $(\tau_0,\infty)$ for some $\tau_0>0$. Indeed, $L+M(L-)=+\infty$ gives $N(L-)=+\infty$ and $h_1$, defined by (<ref>), can not vanish on $(\tau_0,\infty)$ for some $\tau_0>0$ since otherwise we would have $(N^{(-1)}(\tau_0),L)\subseteq \Omega_{\bf s}(\mf)$. Similarly, $L+M(L-)=+\infty$ and the assumption that the right end is heavy, along with the formula (<ref>), does not allow  $h_2$ to vanish  on $(\tau_0,\infty)$ for some $\tau_0>0$. The converse statements follow similarly if (<ref>) are used: since $h_1$ does not vanish on $(\tau_0,\infty)$ for any $\tau_0$, we have $L+M(L-)=+\infty$. Then, since $h_2$ does not vanish on  $(\tau_0,\infty)$, the right end of the string is heavy.

We will need the following two results  from [35].

Let $\xi=N^{(-1)}(\tau)$.
If $\left(\begin{smallmatrix}\Theta^+(\tau,z)&\Phi^+(\tau,z)\\\Theta^-(\tau,z)&\Phi^-(\tau,z)
\end{smallmatrix}\right)$ is a monodromy matrix for the canonical system with Hamiltonian $\diag(h_1,h_2)$, then

\[
\left(\begin{smallmatrix}\Theta^+(\tau,z)&\Phi^+(\tau,z)\\\Theta^-(\tau,z)&\Phi^-(\tau,z)
\end{smallmatrix}\right)=\left(\begin{smallmatrix}  \psi'(\xi-,z^2) &   -z^{-1}\phi'(\xi-,z^2) \\ -z\psi(\xi,z^2)&\phi(\xi,z^2)
\end{smallmatrix}\right)\,.
\]

Similarly, if $\left(\begin{smallmatrix}\Theta_{\sigma_1}^+(\tau,z)&\Phi_{\sigma_1}^+(\tau,z)\\\Theta_{\sigma_1}^-(\tau,z)&\Phi_{\sigma_1}^-(\tau,z)
\end{smallmatrix}\right)$ is a monodromy matrix for the canonical system with Hamiltonian $\diag(h_2,h_1)$, then

\[
\left(\begin{smallmatrix}\Theta_{\sigma_1}^+(\tau,z)&\Phi_{\sigma_1}^+(\tau,z)\\\Theta_{\sigma_1}^-(\tau,z)&\Phi_{\sigma_1}^-(\tau,z)
\end{smallmatrix}\right)=\left(\begin{smallmatrix}\Phi^-(\tau,z)&-\Theta^-(\tau,z)\\-\Phi^+(\tau,z)&\Theta^+(\tau,z)
\end{smallmatrix}\right)=\left(\begin{smallmatrix} \phi(\xi,z^2) &z\psi(\xi,z^2) \\  z^{-1}\phi'(\xi-,z^2) &\psi'(\xi-,z^2)
\end{smallmatrix}\right)\,.
\]

This follows from Lemma 4.1 in [35] after the matrix $W_d$, which is introduced in that paper, is rewritten in terms of $\left(\begin{smallmatrix}\Theta^+(\tau,z)&\Phi^+(\tau,z)\\\Theta^-(\tau,z)&\Phi^-(\tau,z)
\end{smallmatrix}\right)$.

For the proof of the following lemma, check Theorem 4.2 in [35].

Suppose $[M,L]\mapsto \diag(h_2,h_1)$ is the bijection given by (<ref>) and (<ref>), $q$ is the Titchmarsh-Weyl function for the string given by $[M,L]$, and $m,m_{\sigma_1}$ are the Titchmarsh-Weyl functions for $\diag(h_1,h_2)$ and  $\diag(h_2,h_1)$, respectively. Then, we have

\begin{equation}\label{sdk1}
zq(z^2)=m_{\sigma_1}(z)=-m^{-1}(z), \quad z\in \mathbb{C}^+\,.
\end{equation}
Consequently, the spectral measure $\mu_{\sigma_1}$ of Hamiltonian $\diag(h_2,h_1)$ satisfies

\begin{equation}\label{sa20}
\mu_{\sigma_1}([E_1,E_2])=0.5\pi \sigma([E_1^2,E_2^2]), \quad \mu_{\sigma_1}(\{0\})=\pi\sigma(\{0\})
\end{equation}

for all $0<E_1<E_2$.

Since the measure $\mu_{\sigma_1}$ is even, the last identities define it uniquely.   Also, our integral representation (<ref>) connecting the spectral measure $\mu$ and $m$ is different from the one used  in formula (2.1) in [35] by a factor $\pi$. Hence we have $1/\sqrt{\pi}$ in (<ref>) and an extra factor $\pi$ in (<ref>).

If $\Hh=\diag(h_1,h_2)$ is any proper diagonal  Hamiltonian, then

\[
\Theta^+(\tau,z)=\Theta^+(\tau,-z),\, \Theta^-(\tau,z)=-\Theta^-(\tau,-z),
\Phi^+(\tau,z)=-\Phi^+(\tau,-z),\, \Phi^-(\tau,z)=\Phi^-(\tau,-z)\,.
\]

For such $\Hh$, $\W_{\Hh}$ establishes a unitary equivalence between the subspace $\{X=(X_1,X_2)^t\in H, X_2=0\}$ and the subspace of all even functions on $L^2(\mu)$. Similarly, $\W_{\Hh}$ establishes a unitary equivalence between the subspace of $\{X=(X_1,X_2)^t\in H, X_1=0\}$ and all odd functions on $L^2(\mu)$

The first claim follows directly from the equation

\[
\end{smallmatrix}\right)=z\left(\begin{smallmatrix}h_1(\tau)&0\\0&h_2(\tau)
\end{smallmatrix}\right)\left(\begin{smallmatrix}\Theta^+(\tau,z)&\Phi^+(\tau,z)\\\Theta^-(\tau,z)&\Phi^-(\tau,z)
\end{smallmatrix}\right), \quad \left(\begin{smallmatrix}\Theta^+(0,z)&\Phi^+(0,z)\\\Theta^-(0,z)&\Phi^-(0,z)
\end{smallmatrix}\right)=\left(\begin{smallmatrix}1&0\\0&1
\end{smallmatrix}\right)
\]

if we multiply it from left and from the right by $\left(\begin{smallmatrix}1&0\\0&-1
\end{smallmatrix}\right)$. The other statements follow if we recall that $\mu$ is even measure.

Given (<ref>) and the unitary map $\Us$, one can define the Krein string as an operator $\cal{S}_M$ by

\[
\cal{S}_Mv=\Us^{(-1)}\Bigl(\lambda (\Us v)(\lambda)\Bigr)
\]

where $\dom \cal{S}_M:=\{v\in L^2(\mf): \lambda (\Us v)(\lambda)\in L^2(\sigma)$}. Clearly, this is a self-adjoint operator in $L^2(\mf)$. Given that definition of $\cal{S}_M$, we can obtain its characterization  through the associated canonical system.

The map $\Upsilon: v(\xi)\in L^2(\mf)\mapsto X(\tau)=(v(N^{(-1)}(\tau)),0)^t$ is a unitary map onto the subspace $\{X=(X_1,X_2)^t\in H, X_2=0\}$ of the space $L^2(\Hh_{\sigma_1})$ introduced in (<ref>) with $\Hh_{\sigma_1} = \diag(h_2, h_1)$. Moreover,

\begin{equation}\label{lk6}
\Upsilon^{-1}\Di^2_{\Hh_{\sigma_1}}\Upsilon=\cal{S}_M\,.
\end{equation}

The first claim that the map is unitary follows from (<ref>) and the change of variables in the Lebesgue-Stieltjes integral (see formula (6.3) in [5] or formulas (4.35)–(4.36) in [35])

\[
\int_0^L |v(\xi)|^2d\mf=\int_0^\infty |v(N^{(-1)}(\tau)|^2h_2(\tau)d\tau\,.
\]

Clearly, the function $N(\xi)$ might have jumps and the variables $\xi$ and $\tau$ are not in bijection. However, if $N$ has a jump at a point $\xi^*$, then the  function $v(N^{(-1)}(\tau))$ is equal to a constant on the corresponding indivisible interval in variable $\tau$ which is mapped to a point $\xi^*$.

To show (<ref>), we need to prove $\Di^2_{\Hh_{\sigma_1}}\Upsilon=\Upsilon\cal{S}_M$. To show this identity, it is convenient to work on the spectral side of both $\cal{S}_M$ and $\Di_{\Hh_{\sigma_1}}$. Suppose $\sigma$ is the main spectral measure of $[M,L]$ and function $g$ satisfies $g\in L^2(\sigma)$ and $\lambda g\in L^2(\sigma)$. By our definition, this is equivalent to $\cal{U}_M^{-1}g\in \dom \cal{S}_M$. For $\cal{U}_M^{-1}g$, we can write

\[
\]

and the integral converges in $L^2(\mf)$ sense. Then,

\begin{eqnarray*}
(\Upsilon (\cal{U}_M^{-1}g))(\tau)=\left(\int_{\R_+}g(\lambda)\phi(N^{(-1)}(\tau),\lambda)d\sigma, 0\right)^t\stackrel{\eqref{sa20}}{=}\left(\frac 1\pi\int_{\R}g(x^2)\Theta^+_{\sigma_1}(\tau,x)d\mu_{\sigma_1}, 0\right)^t=\\=\W^{-1}_{\Hh_{\sigma_1}}(\pi^{-0.5}g(x^2))\,,
\end{eqnarray*}

where the last identity follows from the fact that $g(x^2)$ is even, measure $\mu_{\sigma_1}$ is even, and $\Theta_{\sigma_1}^-(\tau,x)$ is odd in $x$. The last formula shows that

\begin{equation}\label{lkj}
(\W_{\Hh_{\sigma_1}}\Upsilon\,\, \cal{U}_M^{-1}g(\lambda))(x)=\pi^{-0.5}g(x^2)\,.
\end{equation}

Notice that, thanks to (<ref>), the map $g(\lambda)\mapsto \pi^{-0.5}g(\lambda^2)$ is a unitary map of $L^2(\sigma)$ onto to the set of even functions in $L^2(\mu_{\sigma_1})$.
Moreover, if $g$, $\lambda g\in L^2(\sigma)$, and  $v=\cal{U}_M^{-1}g$,  then

\begin{equation}\label{des20}
\W_{\Hh_{\sigma_1}}\Di^2_{\Hh_{\sigma_1}}\Upsilon v=(\W_{\Hh_{\sigma_1}}\Di^2_{\Hh_{\sigma_1}}\W^{-1}_{\Hh_{\sigma_1}})\W_{\Hh_{\sigma_1}}\Upsilon v\stackrel{\eqref{lkj}}{=}x^2\cdot \frac{g(x^2)}{\sqrt\pi}\,.
\end{equation}

On the other hand, (<ref>) also yields

\begin{equation}\label{lkj_bis}
\W_{\Hh_{\sigma_1}}\Upsilon\,\, \cal{S}_Mv=x^2\cdot \frac{g(x^2)}{\sqrt\pi}\,.
\end{equation}

The comparison of (<ref>) and (<ref>) now gives
\Di^2_{\Hh_{\sigma_1}}\Upsilon v=\Upsilon\,\, \cal{S}_Mv
for every $v\in \dom \cal{S}_M$ and the proof is finished.

Formula (<ref>) provides an expression of $\cal{S}_Mv$ solely in  terms of $M$ and $v$  (see, e.g.,  p. 151 in Chapter 5 of [25], for a similar representation). In particular, that gives

\begin{equation}\label{dom1}
v\in \dom \cal{S}_M \Rightarrow  v(\xi)=C-\int_0^\xi \left(\int_{[0,\eta]}hd\mf\right)d\eta\,,\, C\in \C,\, h\in L^2(\mf)\,.
\end{equation}

§.§ Wave equation for Krein strings

The vibration of the proper string with parameters $[M,L]$  is governed by the following formal hyperbolic Cauchy problem:

\begin{equation}\label{e1}
u_{tt}+\cal{S}_Mu=0, \quad u(\xi,0)=u_0(\xi), \quad u_t(\xi,0)=0, \quad u'(0-,t)=0\,,
\end{equation}

where $u_0(\xi)$ is the initial displacement of the string, its initial velocity is equal to zero, and the Neumann boundary condition $u'(0-,t)=0$ indicates that its left end is “loose”. In this paper, we will only study solutions to (<ref>) given by the formula

\begin{equation}\label{e44}
u(\xi,t)=\cos (t\sqrt{\cal{S}_M})u_0,  \quad u_0\in L^2(\mf),
\end{equation}

where $\cos (t\sqrt{\cal{S}_M})$ is defined via Spectral Theorem:

\[
\cos (t\sqrt{\cal{S}_M})u_0=\cal{U}_M^{-1}\Bigl(\cos (t\sqrt \lambda)(\cal{U}_M u_0)(\lambda)\Bigr).
\]

Function $u$ is understood as element in $C(\R,L^2(\mf))\cap L^\infty(\R,L^2(\mf))$ in the standard mixed-norm notation. It is even in $t$ so we can consider only positive $t$. If $u_0$ is real, then $u$ is real as well. Hence, since $u$ is a linear operator of $u_0$, we can assume without loss of generality that $u_0$ is real.

In this paper, we do not address the question in what sense $u$ satisfies (<ref>) or questions of uniqueness and existence of solutions in classes of higher regularity (see, e.g., [13] where similar issues were studied in the context of wave equation with fractal Laplacian). We only notice that $u_0\in {\rm dom }\,\cal{S}_M$ implies $u\in {\rm dom }\,\cal{S}_M$ and (<ref>) yields that functions in class $\dom \cal{S}_M$ are continuous and have right and left derivatives at each $\xi\in (0,L)$ and the right derivative at zero.  Condition $u_0\in {\rm dom }\,\cal{S}_M$ ensures that identities
u_{tt}+\cal{S}_Mu=0, \,   u(\xi,0)=u_0(\xi),\,$ and $u_t(\xi,0)=0
$  hold as $L^2(\mf)$ functions and the Neumann boundary condition $u'(0-,t)=0$ can be interpreted in a standard way $u'(0,t)=0$ provided that $\mf(\{0\})=0$ (if $\mf(\{0\})>0$, then  $u'(0,t)\neq 0$ in general).
Moreover, an assumption $u_0\in {\rm dom }\,\cal{S}_M$ guarantees that the  energy is finite since

\begin{align*}
E(t)=\frac 12\int_{[0,L)} (u_t^2+(\cal{S}_Mu)u)d\mf=\frac 12\int_0^\infty \lambda(\cos^2(t\sqrt \lambda)+\sin^2(t\sqrt \lambda)) \cdot(\cal{U}_M u_0)^2d\sigma=\\
=\frac 12\|\sqrt{\cal{S}_M}u_0\|_{L^2(\mf)}^2<\infty\,.
\end{align*}

Thus, one can argue that $u_0\in {\rm dom}\, \cal{S}_M$ gives rise to a solution that can be considered physical.

In the following lemma, we show that $u(\xi,t)$ can be written via the evolution operator for the corresponding canonical system.  That will allow us to use the results of the previous section.

Let $u$ be defined by (<ref>) and $U_t$ be an evolution for the canonical system with the Hamiltonian  $\diag(h_2,h_1)$ in which $h_1$ and $h_2$ are obtained by formulas (<ref>) and (<ref>). Then,

\begin{equation}\label{nw1}
\left(\begin{smallmatrix}u(\xi,t)\\0\end{smallmatrix}\right)=\frac 12\Bigl( (U_tX)(\tau)+ (U_{-t}X)(\tau) \Bigr)\,,
\end{equation}

where $X(\tau)=\left(\begin{smallmatrix}u_0(N^{(-1)}(\tau)\\0\end{smallmatrix}\right), \,\xi=N^{(-1)}(\tau)$.

Taking $g=\cos(\sqrt\lambda t)g_0,\, g_0\in L^2(\sigma)$ in (<ref>), we have $(\W_{\Hh_{\sigma_1}}\Upsilon\,\, \cal{U}_M^{-1}\cos(\sqrt{\lambda} t)g_0(\lambda))(x)=\pi^{-0.5}\cos(xt)g_0(x^2)$, which gives

\[
0.5\Upsilon^{-1}(e^{it\Di_{\Hh_{\sigma_1}}}+e^{-it\Di_{\Hh_{\sigma_1}}})\Upsilon=\Upsilon^{-1}\cos(t\Di_{\Hh_{\sigma_1}})\Upsilon=\cos (t\sqrt{\cal{S}_M})
\]

since $g_0$ is arbitrary.

\[
T^{(S)}(\xi)=\int_0^\xi \sqrt{\rho(s)}\,ds, \qquad L^{(S)}_{\eta}=\inf\{\xi: T^{(S)}(\xi)=\eta\},
\]

for $\xi, \eta \in \R_+$. In [5] (see proof of Theorem 2), we showed that

\begin{equation}\label{sa30}
T^{(S)}(\xi)=\int_0^\tau \sqrt{h_1(\alpha)h_2(\alpha)}d\alpha=T(\tau), \quad \xi=N^{(-1)}(\tau)\,,
\end{equation}

where $T$ is defined by (<ref>) for the Hamiltonian $\diag(h_2,h_1)$. Recall that the front of $u(\xi,t)$ has been denoted by $\sff_t$ in the Introduction. Now, we can prove Theorem <ref>, which gives the formula for $\sff_t$.

Proof of Theorem <ref>.
That is immediate from Lemma <ref>, formula (<ref>), and Theorem <ref>.

If the number $L_{t+a}$ in (<ref>) is an end-point of indivisible interval, it must be its left endpoint by definition. Hence, Theorem <ref> and the formula for  operator $\Upsilon$ show that $\|u(\xi,t)\|_{L^2(\mf,\{\sff_t\})}=0$, i.e., the wavefront as a point never carries a positive $L^2(\mf)$ norm of solution.

Recal that the measure
$\sigma = \upsilon\, m + \sigma_{\bf s}$ on $\R_+$ with the density $\upsilon$ and the singular part $\sigma_{\bf s}$  belongs to the Szegő class $\szp$ if $(x+1)^{-1} \in L^1(\sigma)$ and

\int_{\R_+}\frac{\log \upsilon(x)}{\sqrt x(x+1)}\,dx > -\infty.

A simple change of variables shows that $\sigma\in \szp$ if and only if $\mu_{\sigma_1}\in \sz$, where $\mu_{\sigma_1}$ is taken from Lemma <ref>. In Introduction, the class of strings for which the spectral measure is Szegő was called $\szks$ and it was characterized in Theorem <ref>. The following result gives its dynamical description and has Theorem <ref> as a corollary.

Suppose $[M,L]$ is a proper string. If there is  $u_0 \in L^2(\mf)$, a  compactly supported function, such that

\begin{equation}\label{eqs14_cot}
\limsup_{t \to +\infty} \|u\|_{L^2(\mf, \dlto)} > 0, \qquad
\dlto := \bigl[L^{(S)}_{t+ a - \ell}, L^{(S)}_{t + a}\bigr],
\end{equation}

then $[M,L]\in \szks$. Here,  $a:=T^{(S)}(\sff_0)$.

Conversely, if $[M,L]\in \szks$, then

\begin{equation}\label{eqs147}
\liminf_{t \to +\infty} \|u\|_{L^2(\mf, \dlto)} > 0
\end{equation}

for every compactly supported  $u_0\in L^2(\mf)$ not equal to zero identically and for every $\ell>0$. Again, $a:=T^{(S)}(\sff_0)$.

That follows from Corollary <ref> and Lemma <ref>.

Combined with the Theorem <ref> below, we conclude that (<ref>) can be strengthened to

\[
\lim_{t \to +\infty} \|u\|_{L^2(\mf, \dlto)}> 0
\]

for every $\ell>0$.

When $[M,L]=[d\xi,\infty]$, we have $\phi(\xi,z)=\cos(\xi\sqrt z)$, the spectral measure is $(\pi\sqrt \lambda)^{-1}d\lambda$,

\[
(\cal{U}^{0}u_0)(\lambda)=\int_0^\infty u_0(\xi)\cos(\xi\sqrt \lambda)d\xi\,,
\]

and the solution  to (<ref>) is given by (see (<ref>))

\begin{equation}\label{e2}
V_t^0u_0=\frac{u_0(\xi + t) + u_0(\xi - t)}{2}=0.5u_0(\xi-t)+o(1)\,,
\end{equation}

where $u_0$ is extended to $\R$ as even function and $o(1)$ is in $L^2(\R_+)$. Hence the evolution is equivalent to translation when $t\to +\infty$.

Recall, that  $\Omega_{\bf ac}(\mf)$ and $\Omega_{\bf s}(\mf)$ are sets that satisfy (<ref>).
As in the case of the canonical systems, we need to introduce the modified dynamics. Given $y\in L^2(\R)$, we let

\begin{equation}\label{eq12s}
\widetilde V_{t}^0 y: \xi \mapsto \frac{1}{2}\chi_{\Omega_{\bf ac}(\mf)}\cdot \rho^{-\frac 14}(\xi)\cdot y(T^{(S)}(\xi)-t), \qquad \xi \in {[0, L)} \,.
\end{equation}

For the modified evolution, one can use change of variables to get $\|\widetilde V_{t}^0 y\|_{L^2(\mf)}\le \|y\|_{L^2(\R)}$.
If $\sigma\in \szp$, then its Szegő function is defined by

\[
{D}_\sigma^{(S)}(z):=\frac{{D}_{\mu_{\sigma_1}}(\sqrt z)}{\sqrt\pi \sqrt[4] z},
\]

where the measure $\mu_{\sigma_1}$, given by (<ref>), is the spectral measure of Hamiltonian $\Hh_{\sigma_1}$. Notice that $|D^{(S)}_\sigma(\lambda)|^2=\sigma'(\lambda)$ for a.e. $\lambda>0$.
In the case when $[M,L]\in \szks$, we can obtain the asymptotics of $u$ near its wavefront. The following result implies Theorem <ref> from the Introduction.

Suppose $[M,L]\in \szks$.  Then, there is a map $u_0 \mapsto G_{u_0}$ from $L^2(\mf)$ to $L^2(\R)$, such that for every fixed positive $a$ we have

\begin{eqnarray}\label{e110}
\lim_{t\to +\infty}\|u-\widetilde V_{t}^0 G_{u_0}\|_{L^2(\mf,[L^{(S)}_{t-a}, L^{(S)}_{t + a}])}= 0.
\end{eqnarray}

The function $G_{u_0}$ can be computed by the formula

\begin{equation}\label{vopr1}
G_{u_0}(\eta)=\frac{1}{\sqrt\pi}\int_{\Omega_{\bf ac}(\sigma)}\frac{\Re(D_\sigma^{(S)}(\alpha) e^{i\eta\sqrt{\alpha}}) g(\alpha)}{ \sqrt[4]{\alpha}}\, d\alpha,\quad  g:=\cal{U}_Mu_0.
\end{equation}

Moreover, $\|G_{u_0}\|_{L^2(\R)}^2=2\|g\cdot \chi_{\Omega_{\bf ac}(\sigma)}\|^2_{L^2(\sigma)}$ and, if $u_0\in H_{\bf ac}(\cal{S}_M)$, then

\begin{eqnarray}\label{e111}
\lim_{t\to +\infty}\|u-\widetilde V_{t}^0 G_{u_0}\|_{L^2(\mf)}= 0.
\end{eqnarray}

Fix $a> 0$ and choose $b > a$.
Define $\Hh_{\sigma_1}$ by formulas (<ref>), (<ref>) and note that  $\Hh_{\sigma_1}\in \szcs$. Set $X=\Upsilon u_0$, and let, as above, $\xi = N^{(-1)}(\tau)$ for $\tau \ge 0$. Formula (<ref>) gives

\begin{equation}\label{eq2}
\begin{pmatrix}u(\xi,t)\\0\end{pmatrix}=\frac 12\Bigl( (U_tX)(\tau)+ (U_{-t}X)(\tau) \Bigr), \qquad \tau \ge 0.
\end{equation}

Using Theorem <ref>, we obtain

\begin{equation}\label{eq3}
\frac 12\Bigl( (U_tX)(\tau)+ (U_{-t}X)(\tau) \Bigr) = \frac{1}{2}\Hh_{\sigma_1,\mathfrak{n}}^{-1/2}(\tau)(U_t^0Y_{X,+}+U_{-t}^0Y_{X,-})(T(\tau)) + o(1), \qquad t \to +\infty,
\end{equation}

with $o(1)$ in $L^2(\Hh_{\sigma_1},\dlt)$, where

\Hh_{\sigma_1,\mathfrak{n}}^{-1/2}(\tau) = \begin{cases}\begin{pmatrix}
\left(\frac{h_1(\tau)}{h_2(\tau)}\right)^{\frac 14} & 0 \\ 0 & \left(\frac{h_2(\tau)}{h_1(\tau)}\right)^{\frac 14}
\end{pmatrix}, &\tau:\; h_1(\tau)h_2(\tau) > 0,\\
0, & \tau:\; h_1(\tau)h_2(\tau) = 0,
\end{cases}

is the matrix from (<ref>), and

\W_{\Hh_{0}}Y_{X, +} = f \ov{D_{\mu_{\sigma_1}}}\cdot \chi_{\Omega_{\bf ac}(\mu_{\sigma_1})}, \qquad
\W_{\Hh_{0}}Y_{X, -} = f D_{\mu_{\sigma_1}}\cdot \chi_{\Omega_{\bf ac}(\mu_{\sigma_1})},
\qquad

Applying (<ref>), we obtain

\frac{1}{2}(U_t^0Y_{X,+}+U_{-t}^0Y_{X,-})(\tau)=
\frac{1}{2\sqrt 2} \left( \widehat h_+(\tau-t)   \left( \begin{matrix}1 \\ -i \end{matrix}\right) +\widecheck h_-(\tau-t)   \left( \begin{matrix}1 \\ {i} \end{matrix}\right)\right)+o(1),

as $t \to +\infty$, with $o(1)$ in $L^2(\Hh_0)$ and

\begin{align*}
\widehat h_+(\eta)
&=\frac{1}{\sqrt{2\pi}}\int_{\R} f(x)\overline{D}_{\mu_{\sigma_1}}(x)e^{-ix\eta}\cdot \chi_{\Omega_{\bf ac}(\mu_{\sigma_1})}\,dx,\\
\widecheck h_-(\eta)
&=\frac{1}{\sqrt{2\pi}}\int_{\R} f(x)D_{\mu_{\sigma_1}}(x)e^{ix\eta}\cdot \chi_{\Omega_{\bf ac}(\mu_{\sigma_1})}\,dx.
\end{align*}

In other words, we have

\begin{align*}
\frac{1}{2}(U_t^0Y_{X,+}+U_{-t}^0Y_{X,-})(\tau)=
\left(\begin{matrix}A(\tau-t)\\0\end{matrix}\right)+o(1) \qquad t \to +\infty,
\end{align*}

where $o(1)$ is in $L^2(\Hh_0)$ and

\[
A(\eta)=\frac{1}{2\sqrt\pi}\int_{\R} f(x)\Re (D_{\mu_{\sigma_1}}(x)e^{ix\eta})\cdot \chi_{\Omega_{\bf ac}(\mu_{\sigma_1})}dx, \qquad \eta \in \R.
\]

We now see from (<ref>) and (<ref>) that

\begin{equation}\label{eq04}
\begin{pmatrix}
u(\xi, t)\\
\end{pmatrix}
= \Hh_{\sigma_1,\mathfrak{n}}^{-1/2}(\tau)\left(\begin{matrix}A(T(\tau)-t)\\0\end{matrix}\right) + \Hh_{\sigma_1,\mathfrak{n}}^{-1/2}(\tau) R_{1,t}(T(\tau)) + R_{2,t}(\tau),
\end{equation}

where $\|R_{1,t}\|_{L^2(\Hh_0)} + \|R_{2,t}\|_{L^2(\Hh, \dlt)} \to 0$ as $t \to +\infty$. Note that considering $\xi\in \Omega_{\bf ac}(\mf)$ for which $\rho(\xi)>0$ is the same as considering those $\tau \ge 0$ for which $\det \Hh_{\sigma_1}(\tau)=h_1(\tau)h_2(\tau)>0$. Moreover, for such $\xi$ and $\tau$ we have

\rho^{-\frac 14}(\xi)=\left(\frac{h_1(\tau)}{h_2(\tau)}\right)^{\frac 14}, \qquad T^{(S)}(\xi) = T(\tau).

So, one can rewrite relation (<ref>) in the form

\begin{pmatrix}
u(\xi, t) - \chi_{\Omega_{\bf ac}(\mf)}(\xi) \rho^{-\frac 14}(\xi) A(T^{(S)}(\xi) - t)\\
\end{pmatrix}
= \Hh_{\sigma_1,\mathfrak{n}}^{-1/2}(\tau) R_{1,t}(T(\tau)) + R_{2,t}(\tau),

or in the form

\Upsilon \bigl(u(\cdot, t) - \widetilde V_{t}^0 y\bigr)(\tau) = \Hh_{\sigma_1,\mathfrak{n}}^{-1/2}(\tau) R_{1,t}(T(\tau)) + R_{2,t}(\tau),

with $y = 2A$. Since $b > a$, the mapping $\Upsilon$ sends $L^2(\mf, [L^{(S)}_{t-a}, L^{(S)}_{t + a}])$ into a subset of $L^2(\Hh, \dlt)$. Noting that

\|\Hh_{\sigma_1,\mathfrak{n}}^{-1/2} R_{1,t}(T(\cdot)) + R_{2,t}\|_{L^2(\Hh, \dlt)} \le \|R_{1,t}\|_{L^2(\Hh_0)} + \|R_{2,t}\|_{L^2(\Hh, \dlt)} \to 0, \qquad t \to +\infty,

we see that (<ref>) holds with $G_{u_0} = y = 2A$. If we put $g=\cal{U}_Mu_0$, then
$g$ and $f$ are related to each other by $f(x)=g(x^2)/\sqrt\pi$ according to (<ref>).
Recall also that ${D}_\sigma^{(S)}(z)=\frac{{D}_{\mu_{\sigma_1}}(\sqrt z)}{\sqrt\pi \sqrt[4] z}$. By construction, the function $f$ is even, and
$D_{\mu_{\sigma_1}}(-x) = \overline{D_{\mu_{\sigma_1}}(x)}$ almost everywhere on $\R$ since $\mu_{\sigma_1}$ is even.
After changing variables $x^2=\alpha$, we obtain

\begin{align*}
&=\frac{1}{\sqrt\pi}\int_{\R} f(x)\Re (D_{\mu_{\sigma_1}}(x)e^{ix\eta})\cdot \chi_{\Omega_{\bf ac}(\mu_{\sigma_1})}(x)\,dx\\
&=\frac{2}{\sqrt\pi}\int_{\R_+} f(x)\Re (D_{\mu_{\sigma_1}}(x)e^{ix\eta})\cdot \chi_{\Omega_{\bf ac}(\mu_{\sigma_1})}(x)\,dx\\
&=\frac{1}{\sqrt\pi}\int_{\R_+} \frac{f(\sqrt{\alpha})\Re (D_{\mu_{\sigma_1}}(\sqrt{\alpha})e^{i\sqrt{\alpha}\eta})\cdot \chi_{\Omega_{\bf ac}(\sigma)}(\alpha)}{\sqrt{\alpha}}\,d\alpha\\
&=\frac{1}{\sqrt{\pi}}\int_{\Omega_{\bf ac}(\sigma)}\frac{ g(\alpha)\Re (D_{\sigma}(\alpha)e^{i\sqrt{\alpha}\eta})}{\sqrt[4]{\alpha}}\,d\alpha.\\
\end{align*}

We also have
$\|G_{u_0}\|_{L^2(\R)}^2= 4 \|A\|_{L^2(\R)}^2 = 2 \|\chi_{\Omega_{\bf ac}(\mu_{\sigma_1})} f\|_{L^2(\mu_{\sigma_1})}^{2} = 2 \|P_{\bf ac} X\|_{L^2(\Hh)}^{2} = 2 \|P_{\bf ac} u_0\|_{L^2(\mf)}^{2} = 2\|g\chi_{\Omega_{\bf ac}(\sigma)}\|^2_{L^2(\sigma)}$, where in the second identity we have used the fact that

A(\eta)=\frac{1}{2\sqrt\pi}\int_{\R} f(x)\Re (D_{\mu_{\sigma_1}}(x)e^{ix\eta})\,dx = \frac{1}{2\sqrt\pi}\int_{\R} f(x)D_{\mu_{\sigma_1}}(x)e^{ix\eta}\,dx,

and the integrals are understood in the $L^2(\R)$-sense. Finally, Corollary <ref> implies (<ref>).

§.§ Examples
In this subsection, we explain how the general results can be applied to two examples considered in the Introduction.

1. Strings for which $\rho=1$.
Consider the case when $L=\infty$ and $\rho=1$. For the associated measure $\mf$, we get

\begin{equation}\label{e0}
d\mf=d\xi+d\mf_{\bf s}\,,
\end{equation}

where $\mf_{\bf s}$ is any singular measure.  If $\mf_{\bf s}=0$, then the solution $u$ is given by (<ref>).  The models described by our choice of $M$ are numerous, e.g., think about the beads with masses $\{m_j\}$ placed at points $\{\xi_j\}, \xi_0<\xi_1<\ldots$ connected by the string with uniform density equal to one.

One might want to know how the presence of “impurities” encoded by $\mf_{\bf s}$ changes the character of wave propagation. The general results from the previous subsection can be reformulated as follows. From Theorem <ref>, we immediately get

If $u_0$ has compact support and $\|u_0\|_{L^2(\mf)}>0$, then $\sff_t=\sff_0+t$.

Clearly, the front propagates with the same linear speed regardless of the nature of $\mf_{\bf s}$. For $M$ that satisfy $\rho=1$, it was established (see [5]), that

\begin{equation}\label{s_a2}
\mf_{\bf s}(\R_+)=0 \Longleftrightarrow \sigma\in \szp\,.
\end{equation}

In the next two statements, we describe how the dynamics of $u$ depends on $\mf_{\bf s}$.

If $\|u_0\|_{L^2(\mf)}>0, \sff_0<\infty$, and
$\mf_{\bf s}(\R_+)=\infty$, then

\[
\lim_{t\to\infty}\|u\|_{L^2(\mf,[\sff_t-a,\sff_t])}=0
\]

for every fixed $a>0$. Conversely, if there is $u_0$ that satisfies  $\|u_0\|_{L^2(\mf)}>0, \sff_0<\infty$, and

\[
\limsup_{t\to\infty}   \|u\|_{L^2(\mf,[\sff_t-a,\sff_t])}>0
\]

for some fixed $a>0$, then $\mf_{\bf s}(\R_+)<\infty$.

That follows from the Theorem <ref> and (<ref>).

The next result shows that
the condition $\mf_{\bf s}(\R_+)<\infty$ guarantees that part of the wave propagates like a traveling wave in (<ref>).  In that theorem, $P_{\bf ac}$ denotes the orthogonal projection to $H_{\bf as}(\cal{S}_M)$.

In the case when $\mf_{\bf s}(\R_+) < \infty$,  we have

\begin{align}
\lim_{t\to+\infty}\|u(\xi,t)\|_{L^2(\mf_{\bf s}, [t-a,t+a])} = 0,\label{first1}\\
\lim_{t\to+\infty}\|u(\xi,t)-G_{u_0}(\xi-t)\|_{L^2[t-a,t+a]}= 0,\nonumber
\end{align}

for some  $G_{u_0} \in L^2(\R)$ and all $ a>0$.
Moreover, $\|G_{u_0}\|_{L^2(\R)}>0$ if and only if $P_{\bf ac}u_0\neq 0$. If $u_0$ is not identically equal to zero and has compact support, then $P_{\bf ac}u_0\neq 0$.

These results are contained in Theorem <ref>, Theorem <ref>, and (<ref>).

The statements made in Example <ref> in Introduction now follow.

2. Strings made of two types of material. For another example, let us consider a string with $\mf_{\bf s}(\R_+)=0$ whose density $\rho$ takes two positive values: $a$ and $b$. We interpret such strings as those made from two types of material. In Example <ref> of Introduction, we claimed

Suppose $a\neq b$. We have $\sigma\in \szp$ if and only if either $|\{\xi: \rho(\xi)=a\}|<\infty$ or $|\{\xi: \rho(\xi)=b\}|<\infty$ .

We will apply the Theorem <ref> with properly chosen $\{\eta_n\}$. Define $\eta_n$ by

\[
\eta_n=\int_0^n\sqrt{\rho(\xi)}d\xi\,.
\]

Thus, $\xi_n=n, n=0,1,\ldots$. Since $\rho$ takes values $a$ and $b$, we always have condition

\[
0<C_1<\eta_{n+1}-\eta_n<C_2, \quad n=0,1,2,\ldots
\]

satisfied. For each $n \ge 0$, we have

\begin{equation}\label{eq68}
\rho(\tau)
= \begin{cases}
a, & \tau \in E_n,\\
b, & \tau \in F_n,
\end{cases}
\end{equation}

where $|E_n| = \delta_n, |F_n| = 1 - \delta_n$, and $E_n\subseteq [n,n+1), F_n\subseteq [n,n+1)$. Then,

\begin{align*}
&= 2a(\delta_n + \delta_{n+1}) + 2b(2-\delta_n - \delta_{n+1})\\
&= 4b + 2(a-b)(\delta_n + \delta_{n+1})
\end{align*}

\begin{align*}
\left(\int_{n}^{n+2}\sqrt{\rho(\tau)}\,d\tau\right)^2
(2\sqrt{b} +
(\sqrt{a} - \sqrt{b})(\delta_n + \delta_{n+1}))^2\\
&=4b + 4\sqrt{b}(\sqrt{a} - \sqrt{b})(\delta_n + \delta_{n+1}) +
(\sqrt{a} - \sqrt{b})^2(\delta_n + \delta_{n+1})^2.
\end{align*}

If we denote

A_n = 2\int_{n}^{n+2}\rho(\tau)\,d\tau - \left(\int_{n}^{n+2}\sqrt{\rho(\tau)}\,d\tau\right)^2, \qquad n \ge 0,

\begin{align*}
A_n&=2(a-b)(\delta_n + \delta_{n+1}) - (4\sqrt{b}(\sqrt{a} - \sqrt{b})(\delta_n + \delta_{n+1}) +
(\sqrt{a} - \sqrt{b})^2(\delta_n + \delta_{n+1})^2)\\
&=(2(a-b)- 4\sqrt{b}(\sqrt{a} - \sqrt{b}) -
(\sqrt{a} - \sqrt{b})^2(\delta_n + \delta_{n+1}))(\delta_n + \delta_{n+1})\\
&=(2(\sqrt{a}+\sqrt{b}) - 4\sqrt{b} -
(\sqrt{a} - \sqrt{b})(\delta_n + \delta_{n+1}))(\delta_n + \delta_{n+1})(\sqrt{a} - \sqrt{b})\\
&=(2(\sqrt{a}-\sqrt{b}) -
(\sqrt{a} - \sqrt{b})(\delta_n + \delta_{n+1}))(\delta_n + \delta_{n+1})(\sqrt{a} - \sqrt{b})\\
&=(\sqrt{a} - \sqrt{b})^2(2-\delta_n - \delta_{n+1})(\delta_n + \delta_{n+1}).
\end{align*}

Then, the string satisfies conditions in the left-hand side of (<ref>) if and only if either $a = b$ or

\sum_{n=0}^\infty (2-\delta_n - \delta_{n+1})(\delta_n + \delta_{n+1}) < \infty.

The last statement implies the claim in the proposition.

§ DIRAC OPERATORS

We start this section by recalling the definition of the one-dimensional Dirac operator. Then, we make the connection to canonical systems and explain how the results from the second section can be applied to prove theorems stated in Section <ref>.

§.§ Dirac operators

Recall that the one-dimensional Dirac operator $\Dd_Q$ on $\R_+$ is defined by

\begin{equation}\label{do}
\Dd_Q : Z \mapsto JZ' + Q Z, \qquad J = \jm, \qquad Q = \left(\begin{smallmatrix}q_1 & q_2\\ q_2  & -q_1\end{smallmatrix}\right).
\end{equation}

Here the functions $q_1, q_2$ are real and belong to $L^1_{\loc}(\R_+)$.
The “free” Dirac operator with potential $Q = 0$ will be denoted by $\Dd_0$.  The domain of $\Dd_Q$ is given by

\dom\Dd_{Q} = \left\{Z \in L^2(\C^2):\;
\begin{aligned}
&Z \mbox{ is locally absolutely continuous on  }\R_+,\\
&JZ' + QZ  \in L^2(\C^2),\\
&\langle Z(0), \zo\rangle_{\C^2} = 0.
\end{aligned}
\right\}.

With this domain, the operator $\Dd_Q$ is a densely defined self-adjoint operator on $L^2(\C^2)$, see Section 8.6 in [53] or [11] for recent developments. Let $\Psi$ denote the generalized eigenvector of $\Dd_Q$:

\begin{equation}\label{cp}
J\tfrac{\partial}{\partial \tau}\Psi(\tau, z) + Q \Psi(\tau, z) = z \Psi(\tau,z), \quad \Psi(0, z) = \oz, \qquad \tau \ge 0, \quad z \in \C.
\end{equation}

Then there exists a unique Borel measure $\mu_D$ on $\R_+$ such that the generalized Fourier transform

\begin{equation}\label{ft}
\F_Q: Z \mapsto \frac{1}{\sqrt{\pi}}\int_{\R_+}\langle Z(\tau), \Psi(\tau, \bar z)\rangle_{\C^2}\,d\tau, \qquad z \in \C,
\end{equation}

densely defined on functions with compact support, can be extended to a uitary operator from $L^2(\C^2)$ to $L^2(\mu_D)$. This measure is called the main spectral measure of $\Dd_{Q}$. For example, the Lebesgue measure on $\R$ is the main spectral measure for the free Dirac operator $\Dd_0$.

§.§ The reduction of Dirac operator to a canonical system and the Szegő condition

The following result is well-known, see, e.g., Section 2.4 in [8].

Let $Q \in L^1_{\loc}(\R_+)$ be as in (<ref>), and let the matrix-valued function $N_0$ be the solution of the Cauchy problem

\begin{equation}\label{s_a3}
JN_0'(\tau) + Q(\tau)N_0(\tau) = 0, \quad \tau \ge 0, \quad N_0(0) = \idm.
\end{equation}

Denote by $\Di_{\Hh}$ the self-adjoint operator on $H$ corresponding to the canonical system generated by the Hamiltonian $\Hh = N_0^*N_0$. Then, the main spectral measures of the operators $\Dd_{Q}: L^2(\C^2) \to L^2(\C^2)$ and $\Di_{\Hh}: H \to H$  coincide. In particular, the operators $\Dd_{Q}$ and $\Di_{\Hh}$ are unitary equivalent and the unitary equivalence is given by the operator  $V: X\mapsto N_0 X$, which is a unitary map from $H$ to $L^2(\C^2)$. Moreover, we have $\W_{\Hh} = \F_{Q}V$.

An important property of the Hamiltonian $\Hh = N_0^*N_0$ in the previous lemma is that it has unit determinant everywhere on $\R$. Indeed, the Wronskian in problem (<ref>) is constant so $\det N_0(\tau) = \det N_0(0) = 1$. Hence,

\begin{equation}\label{eq59}
\det\Hh(\tau) = 1, \quad T(\tau) = \tau, \qquad L_\eta=\eta \qquad \tau,\eta \in \R_+,
\end{equation}

for the corresponding functions $T$ and $L$ introduced for  $\Hh$ in (<ref>).

The spectral measures $\mu_D$ of the Dirac operators define a proper subset of all Poisson-finite measures on the real line that generate the canonical systems as discussed in the second section. Some of them belong to the Szegő class. Thanks to the characterization (<ref>), we have the following proposition (see Corollary 1.4 in [6]):

The condition $N_0^*N_0\in \szcs$ is necessary and sufficient for the spectral measure $\mu_D$ of Dirac operator to satisfy $\mu_D\in \sz$.

Checking that $N^*_0N_0\in \szcs$ is not always easy. However, in many cases, the application of our proposition is straightforward.

Suppose that the potential $Q$ with entries in $L^1_{\loc}(\R_+)$ has the form

\begin{equation}\label{eq60}
Q = \begin{pmatrix}q & 0\\ 0 & -q\end{pmatrix} \quad \mbox{or}\quad Q = \begin{pmatrix}0 & q\\ q & 0\end{pmatrix}.
\end{equation}

Then, for the spectral measure $\mu_D$ of the corresponding Dirac operator $\Dd_Q$ we have

\begin{equation}\label{eq61}
\mu_D\in \sz\quad  \Longleftrightarrow \quad N_0^*N_0\in \szcs\quad \Longleftrightarrow\quad  \sum_{n \ge 0}\left(\int_{n}^{n+2}h(\tau)\,d\tau\int_{n}^{n+2}\frac{d\tau}{h(\tau)} - 4\right) < \infty,
\end{equation}

where $h(\tau) = e^{2\int_{0}^{\tau}q(s)\,ds}$, $\tau \ge n$.

The first equivalence has already been discussed and we need to show the second one.  For the potentials $Q$ of the form (<ref>), define $g(\tau)=\int_{0}^{\tau}q(s)\,ds$. Then, solving the problem (<ref>) to find $N_0$ is easy.  That gives

N_0(\tau) = \begin{pmatrix}\cosh g(\tau) & \sinh g(\tau)\\\sinh g(\tau) & \cosh g(\tau)\end{pmatrix} \quad \mbox{or} \quad
N_0(\tau) = \begin{pmatrix}e^{-g(\tau)} & 0\\0 & e^{g(\tau)}\end{pmatrix},

respectively. Therefore, for $\Hh=N_0^*N_0$, we have

\Hh(\tau) = \begin{pmatrix}\cosh 2g(\tau) & \sinh 2g(\tau)\\\sinh 2g(\tau) & \cosh 2g(\tau)\end{pmatrix} \quad
\quad \mbox{or} \quad
\Hh(\tau) = \begin{pmatrix}e^{-2g(\tau)} & 0\\0 & e^{2g(\tau)}\end{pmatrix}.

In both cases $\det \Hh = 1$ on $\R_+$, and the straightforward calculation yields

\det\int_{n}^{n+2}\Hh(\tau)\,d\tau = \int_{n}^{n+2}e^{2g(\tau)}\,d\tau\int_{n}^{n+2}e^{-2g(\tau)}\,d\tau = \int_{n}^{n+2}h(\tau)\,d\tau\int_{n}^{n+2}\frac{1}{h(\tau)}\,d\tau.

So, $\Hh \in \szcs$ if and only if (<ref>) holds. The result follows.

Propositions <ref> and <ref> imply Theorem <ref>.

§.§ The evolution for Dirac equation and Møller wave (modified wave) operators.

The self-adjoint operator $\Dd_Q$ defines a unitary evolution $e^{it\Dd_Q}$. Lemma <ref> above gives the connection between $e^{it\Dd_Q}$ and evolution $e^{it\Di_{\Hh}}$ for canonical systems.
If $\Hh = N_0^*N_0$, then
for all $t\in \R$.

For every $Z$ in $L^2(\C^2) = L^2(\Hh_0)$ we define

\sff[Z]=\inf\{\tau \ge 0:\; Z(s) = 0 \; \mbox{for almost every }s > \tau\}.

Notice that all elements of $N_0$ are real-valued so $Z$ has real components if and only if $X=V^{-1}Z$ has real components.

For every real $Z \in L^2(\C^2)$ and every $t\in \R$, we have $\sff[e^{it\Dd_Q}Z] = |t|+\sff[Z]$.

This follows from Theorem <ref> and formula (<ref>).

For an arbitrary $Z \in L^2(\C^2)$, we can write each of its components as a sum of real and imaginary parts. Then, Proposition <ref> gives
$\sff[e^{it\Dd_Q}Z] \le |t|+\sff[Z]$ for all $t \in \R$.

We will also need the following proposition.
Let $\mu$ be the spectral measure of the Dirac operator $\Dd_Q$. Suppose $\mu\notin \sz$. Then, for every $Z\in L^2(\C^2)$, we have

\[
\lim_{t\to\pm\infty} \|e^{it\Dd}Z\|_{L^2(\C^2, [|t|-b,|t|+b])}=0
\]

for every $b>0$.

That follows from Corollary <ref> and formula (<ref>).

Proof of Theorem <ref>. Take a non-zero function $Z \in L^2(\C^2)$. Set

Z_+=\lim_{t \to +\infty} e^{-it\Dd_Q}M_{\gamma} e^{it\Dd_0}Z,

and notice that $\|Z_+\|_{L^2(\C^2)}=\|Z\|_{L^2(\C^2)}>0$.
This yields

\begin{equation}\label{s_a7}
\lim_{t \to +\infty} \| M_{\gamma}e^{it\Dd_0}Z-e^{it\Dd_Q}Z_+\|_{L^2(\C^2)}=0.
\end{equation}

Formula (<ref>) for $Z = \begin{pmatrix}z_1 \\ z_2\end{pmatrix}$ can be recast as

\begin{equation}\label{eq32bisbis}
e^{it\Dd_{0}}\begin{pmatrix}z_1 \\ z_2\end{pmatrix}
\frac{1}{2}
\begin{pmatrix}
z_1(\tau-t) + z_1(\tau+t)\\
-i(z_1(\tau-t) - z_1(\tau+t))
\end{pmatrix}
\frac{1}{2}
\begin{pmatrix}
i(z_2(\tau-t) - z_2(\tau+t))\\
z_2(\tau-t) + z_2(\tau+t)
\end{pmatrix}, \qquad \tau \in \R_+,
\end{equation}

where $z_1 \in L^2(\R_+)$  is extended to the whole real line $\R$ as an even function and $z_2$ is extended as odd function.
That gives

$$\liminf_{t\to +\infty}\|e^{it\Dd_0}Z_+\|_{L^2(\C^2,[t-b,t+b])}>0$$
for large enough $b$. From (<ref>), one gets $\liminf_{t\to +\infty}\|e^{it\Dd_Q}Z_+\|_{L^2(\C^2,[t-b,t+b])}>0$. Now we have $\mu\in \sz$ by Proposition <ref>. The case when $t\to -\infty$ can be handled similarly.

Notice that we have used the existence of $\lim_{t \to +\infty} e^{-it\Dd_Q}M_{\gamma} e^{it\Dd_0}Z$ for just one nonzero element $Z \in L^2(\C^2)$ in the proof of Theorem <ref>.

Proof of Theorem <ref>. Assume that the main spectral measure $\mu_D$ of $\Dd_{Q}$ is in the Szegő class. By Lemma <ref>, $\mu_D$ coincides with the spectral measure of the Hamiltonian $\Hh = N_0^* N_0$ generated by the solution of equation $J N_0' + QN_0 = 0$, $N_0(0) = \idm$. Taking into account (<ref>), Theorem <ref> tells us that for some function $\gamma_0: \R_+ \to \T$ and for every $Z\in L^2(\Hh_0)=L^2(\C^2)$, there exist the limits

\lim_{t\to +\infty}e^{-it\Di_{\Hh}}M_{\gamma_0}\Hh^{-1/2}e^{it\Di_{0}}Z, \quad \lim_{t\to -\infty}e^{-it\Di_{\Hh}}M_{\overline{\gamma}_0}\Hh^{-1/2}e^{it\Di_{0}}Z

in the norm of $L^2(\Hh)$.  Since  $\Di_0 = \Dd_0$, that implies existence of the limit

\lim_{t\to +\infty}Ve^{-it\Di_{\Hh}}M_{\gamma_0}\Hh^{-1/2}e^{it\Di_{0}}Z
\lim_{t\to +\infty}e^{-it\Dd_{Q}}VM_{\gamma_0}\Hh^{-1/2}e^{it\Dd_{0}}Z

in $L^2(\C^2)$.
Note that $(N_0(\tau)\Hh^{-1/2}(\tau))^* (N_0(\tau)\Hh^{-1/2}(\tau))$ is the identity matrix for each $\tau \in \R_+$. Since
\[
\det (N_0(\tau)\Hh^{-1/2}(\tau))=1,\]
it follows that the operator $Z \mapsto V\Hh^{-1/2}Z$ on $L^2(\C^2)$ coincides with the multiplication operator by a $2 \times 2$ matrix-valued function of the form

\Sigma_{\phi(\tau)} = \begin{pmatrix}\cos\phi(\tau) & \sin\phi(\tau) \\ -\sin\phi(\tau) & \cos\phi(\tau)\end{pmatrix}, \qquad \phi(\tau) \in [0, 2\pi).

We have

\Sigma_{\phi}\begin{pmatrix}1 \\ -i\end{pmatrix}
= \begin{pmatrix}\cos\phi - i\sin\phi \\ -\sin\phi - i\cos\phi\end{pmatrix}
e^{-i\phi}\begin{pmatrix}1 \\ -i\end{pmatrix}.

Formula (<ref>) shows that when $t \to +\infty$, for every $Z\in L^2(\C^2)$ we have

e^{it\Dd_{0}}\begin{pmatrix}z_1 \\ z_2\end{pmatrix}
\frac{z_1(\tau-t)}{2}
\begin{pmatrix}
\end{pmatrix}
\frac{iz_2(\tau-t)}{2}
\begin{pmatrix}
\end{pmatrix} + o(1),

where $o(1)$ is with respect to $L^2(\C^2)$-norm. Therefore, we have

$$e^{-it\Dd_{Q}}VM_{\gamma_0}\Hh^{-1/2}e^{it\Dd_{0}}Z = e^{-it\Dd_{Q}}M_{\gamma_0e^{-i\phi}}e^{it\Dd_{0}}Z + o(1), \qquad t \to +\infty$$
and  the limit

W_+(\Dd_{Q}, \Dd_0, \gamma)Z=\lim_{t\to+\infty}e^{-it\Dd_{Q}}M_{\gamma}e^{it\Dd_{0}}Z

exists in $L^2(\C^2)$ for all $Z \in L^2(\C^2)$ if we take $\gamma =\gamma_0 e^{-i\phi}$. The existence of the wave operator $W_{+}(\Dd_Q, \Dd_0, \gamma)$ follows. Arguing similarly, one can prove the existence of $W_{-}(\Dd_Q, \Dd_0, \gamma)$ with the modification to the dynamics given by  $M_{\overline{\gamma}}$. Moreover, the proof shows that

\Ran W_{\pm}(\Dd_Q, \Dd_0,\gamma) = V(\Ran W(\Di_{\Hh}, \Di_{0},\gamma)) = \F_{Q}^{-1}\W_{\Hh}(H_{\bf ac}(\Dd_{\Hh})) =  \F_{Q}^{-1}(L^2(\mu_{\bf ac})) = H_{\bf ac}(\Dd_{Q}),

where $\mu_{\bf ac} = w\,dx$ is the absolutely continuous part of the measure $\mu$ and $\mu=\mu_D$. In other words, the wave operators $ W_{\pm}(\Dd_Q, \Dd_0,\gamma)$ are complete. It is also clear from the proof that our construction gives $\gamma = 1$ in the case where $Q$ is anti-diagonal ($q_1 = 0$).

Proof of Theorem <ref>. Given Lemma <ref>, Theorem <ref> is a direct consequence of Theorem <ref> and Theorem <ref>.

Proof of Theorem <ref>. Given Lemma <ref>,  Theorem <ref> follows from  Theorem <ref>.

§.§ Wiegner-von Neumann potentials

In this subsection, we prove Theorem <ref>. Let us recall its statement for convenience.

Suppose that a potential $Q_{\alpha, \beta}$ with entries in $L^1_{\loc}(\R_+)$ has the form

\begin{equation}\label{eq60bis}
Q_{\alpha,\beta}= \begin{pmatrix}q & 0\\ 0 & -q\end{pmatrix}, \quad \mbox{or}\quad Q_{\alpha, \beta} = \begin{pmatrix}0 & q\\ q & 0\end{pmatrix}, \qquad q(\tau) = \frac{\sin \tau^\alpha}{\tau^\beta}, \qquad \tau \ge \tau_0,
\end{equation}

for some $\alpha, \beta \in \R$ and $\tau_0 >0$. Then,

\begin{equation}
\mu\in \sz \quad \Longleftrightarrow\quad  W_\pm(\Dd_Q, \Dd_0, \gamma) {\quad\rm exist\,}\quad\Longleftrightarrow\quad
(\alpha, \beta) \in A_1 \cup A_2 \cup A_3
\end{equation}

\begin{align*}
A_1 &= \{(\alpha,\beta): \;\; \alpha \le 0,\;\; \beta-\alpha > 1/2\}\\
A_2 &= \{(\alpha,\beta): \;\; \alpha \in (0,1),\;\; \beta > 1/2\} \\
A_3 &= \{(\alpha,\beta): \;\; \alpha \ge 1, \;\;\alpha+\beta > 3/2\}.
\end{align*}

Note that the set $A_1 \cup A_2 \cup A_3$ in Theorem <ref> is open. It is shown on Figure <ref>.

A sequence of real numbers $\{x_n\}$ is uniformly distributed $\mmod 1$ if for every interval $[a,b] \subseteq [0,1]$ we have

\lim_{n \to \infty}\frac{\sharp \{1 \le k \le n: \;\; x_k \,\mmod 1 \in [a,b] \}}{n} = b-a,

where $x\, \mmod 1$ denotes the fractional part of $x$. We will use the following result.
[Theorem 3.5 in [48], p. 29]
Let $k$ be a positive integer, and let $f$ be a function defined for $x \ge 1$, which is $k$ times differentiable for $x \ge x_0$. If $f^{(k)}$ tends monotonically to zero as $x \to \infty$ and $\lim_{x \to \infty} x |f^{(k)}(x)| = \infty$, then the sequence $\{f(n)\}_{n \ge 0}$ is uniformly distributed $\mmod 1$.

In particular, given  $\alpha \in (0, 1)$, the sequence $\{\frac{n^\alpha}{2\pi}\}_{n \ge 0}$ is uniformly distributed $\mmod 1$. Indeed, one can use the previous theorem with $f = \frac{x^\alpha}{2\pi}$. Let us notice that $\{\frac{n^\alpha}{2\pi}\}_{n \ge 0}$ is uniformly distributed $\mmod 1$ for any $\alpha > 0$, see [48] for more details.

We also need M. Korey's estimate from [43]. Recall that we use notation $\langle f \rangle_{I} = \frac{1}{|I|}\int_{I}f(x)\,dx$.

Suppose $I=[a,b]$ and  measurable function $f : I \to \R$ satisfies

\begin{equation}\label{des11}
\langle e^{f}\rangle_{I} \cdot \langle e^{-f}\rangle_I = 1 + \eps, \qquad \eps \in [0, 1].
\end{equation}

\begin{equation}\label{eq63}
\langle \left|f(x) - \langle f\rangle_I\right|\rangle_I\le c\sqrt{\eps},
\end{equation}

for a universal constant $c$.

In [43], formula (3.7), Korey shows that

\langle e^{f}\rangle_{I} \cdot  \exp\left(-\langle f\rangle_I\right) = 1 + \eps', \qquad \eps' \in [0, 1]

\langle |f-m_I(f)|\rangle\le C\sqrt{\eps'}\,,
where $m_I(f)$ denotes a median of $f$ over $I$. By Jensen's inequality,
\exp\left(-\langle f\rangle_I\right)\le  \langle e^{-f}\rangle_I\,.
Therefore, (<ref>) implies

\[
\langle |f-m_I(f)|\rangle\le C\sqrt\eps\,,
\]

Now, for every $c' \in \R$, we have

\begin{align*}
\langle \left|f(x) - \langle f\rangle_I\right|\rangle_I\le \langle \left|f(x) - c'\right|\rangle_I+|c'-\langle f\rangle_I|
\le 2\langle \left|f(x) - c'\right|\rangle_I\,.
\end{align*}

Taking $c'=m_I(f)$ finishes our proof.

Given real-valued $q\in L^1_{\rm loc}(\R_+)$, denote  $g_n(\tau) = 2\int_{n}^{\tau}q(s)\,ds$ and $\widetilde g_n = g_n-\langle g_n\rangle_{[n,n+2]}$.

If   $\lim_{n \to +\infty} \|g_n\|_{L^\infty[n,n+2]} = 0$, then

\begin{equation}\label{des16}
\sum_{n \ge 0}\left(\int_{n}^{n+2}h_n(\tau)\,d\tau\int_{n}^{n+2}\frac{1}{h_n(\tau)}\,d\tau - 4\right) < \infty\quad  \Longleftrightarrow\quad
\sum_{n \ge 0}\int_{n}^{n+2} {\widetilde g_n}^2\,d\tau < +\infty\,,
\end{equation}

where $h_n(\tau) = e^{g_n(\tau)}$, $\tau \ge n$.

Notice that

\[
\int_{n}^{n+2} {\widetilde g_n}^2\,d\tau\lesssim \int_{n}^{n+2}  g_n^2\,d\tau
\]

by Cauchy-Schwarz inequality. Hence, to establish convergence of the sum in the left-hand side of the proposition, it is enough to check that $\|g_n\|_{L^2[n,n+2]}\in \ell^2(\mathbb{N})$.

Notice that

\[\langle h_n(\tau)\rangle_{[n,n+2]} \cdot \langle h^{-1}_n(\tau)\rangle_{[n,n+2]}
=\langle e^{\widetilde g_n}\rangle_{[n,n+2]} \cdot \langle e^{-\widetilde g_n}\rangle_{[n,n+2]}\,.
\]

Then, since $\lim_{n\to\infty}\|\widetilde g_n\|_{L^\infty[n,n+2]}=0$, we use Taylor expansion to get

\begin{align*}
\int_{n}^{n+2} e^{\widetilde g_n}\,d\tau
&= 2 + \int_{n}^{n+2}\widetilde g_n(\tau)\,d\tau + \frac{1}{2}\int_{n}^{n+2}\widetilde g^2_n(\tau)\,d\tau + O\left(\int_{n}^{n+2}|\widetilde g_n(\tau)|^3\,d\tau\right), \\
\int_{n}^{n+2}   e^{-\widetilde g_n} \,d\tau
&= 2 - \int_{n}^{n+2}\widetilde g_n(\tau)\,d\tau + \frac{1}{2}\int_{n}^{n+2}\widetilde g^2_n(\tau)\,d\tau + O\left(\int_{n}^{n+2}|\widetilde g_n(\tau)|^3\,d\tau\right),
\end{align*}

as $n \to +\infty$. It follows that

\int_{n}^{n+2}  h_n(\tau)d\tau  \,d\tau \int_{n}^{n+2}\frac{1}{h_n(\tau)}\,d\tau =
4 + 2 \int_{n}^{n+2}\widetilde g^2_n(x)\,dx + o\left(\int_{n}^{n+2}\widetilde g^2_n(x)\,dx\right)

which proves the required claim.

In the proof below, many arguments will be based on the following formula. For an integer $n \ge 1$, $a,b\in \R, a\neq 0$, and $d= (a+b-1)/a$, we have

\begin{align*}
a\int_{n}^{x}\frac{\sin \tau^a}{\tau^b}\,d\tau
&= \int_{n^a}^{x^a}\frac{\sin y}{y^{d}}\,dy = \left. -\frac{\cos y}{y^{d}}\right|_{n^a}^{x^a} -d
\int_{n^a}^{x^a}\frac{\cos y}{y^{d+1}} \,dy\\
&= \left. -\frac{\cos y}{y^{d}}\right|_{n^a}^{x^a} -d \left. \frac{\sin y}{y^{d +1}}\right|_{n^a}^{x^a} -
\int_{n^a}^{x^a}\frac{\sin y}{y^{d+2}} \,dy
\end{align*}

\begin{align}\label{des17_pic}
a\int_{n}^{x}\frac{\cos \tau^a}{\tau^b}\,d\tau
&= \int_{n^a}^{x^a}\frac{\cos y}{y^{d}}\,dy = \left. \frac{\sin y}{y^{d}}\right|_{n^a}^{x^a} +d
\int_{n^a}^{x^a}\frac{\sin y}{y^{d+1}} \,dy\\
&= \left. \frac{\sin y}{y^{d}}\right|_{n^a}^{x^a} -d \left. \frac{\cos y}{y^{d +1}}\right|_{n^a}^{x^a} {-}
\int_{n^a}^{x^a}\frac{\cos y}{y^{d+2}} \,dy.\nonumber
\end{align}
A simple iteration implies the bounds for the remainders

\[
\left|\int_{n^a}^{x^a}\frac{\sin y}{y^{d+2}} \,dy\right|\lesssim n^{-a(d+2)}+x^{-a(d+2)}, \qquad \left|\int_{n^a}^{x^a}\frac{\cos y}{y^{d+2}} \,dy\right|\lesssim n^{-a(d+2)}+x^{-a(d+2)}, \quad x>n\,.
\]

Proof of Theorem <ref>.  Recall Proposition <ref>, Theorems <ref> and <ref>. To prove our result, we only need to establish for which parameters $\alpha$ and $
\beta$ the condition

\begin{equation}\label{help}
\sum_{n \ge 0}\left(\int_{n}^{n+2}h_n(\tau)\,d\tau\int_{n}^{n+2}\frac{1}{h_n(\tau)}\,d\tau - 4\right) < \infty
\end{equation}

is satisfied, where $h_n$ is defined as in Proposition <ref>.  For an integer $n \ge \max(1, \tau_0)$ and $\gamma = (\alpha+\beta-1)/\alpha$, we have

\begin{align*}
\alpha\int_{n}^{x}\frac{\sin \tau^\alpha}{\tau^\beta}\,d\tau
&= \int_{n^\alpha}^{x^\alpha}\frac{\sin y}{y^{\gamma}}\,dy =\left. -\frac{\cos y}{y^{\gamma}}\right|_{n^\alpha}^{x^\alpha} -\gamma \left. \frac{\sin y}{y^{\gamma +1}}\right|_{n^\alpha}^{x^\alpha} -
\gamma(\gamma+1)
\int_{n^\alpha}^{x^\alpha}\frac{\sin y}{y^{\gamma+2}} \,dy.
\end{align*}

Let us consider several cases.

Case $\alpha \ge 1$, $\alpha + \beta > 1$. In this case $\gamma > 0$ and we are in the setting of Proposition <ref>. Note that

\sup_{x\in [n,n+2]}\left|
-\gamma \left. \frac{\sin y}{y^{\gamma +1}}\right|_{n^\alpha}^{x^\alpha} -\gamma(\gamma+1) \int_{n^\alpha}^{x^\alpha}\frac{\sin y}{y^{\gamma+2}} \,dy \right|= O\left(\frac{1}{n^{\alpha(\gamma+1)}}\right)\in \ell^2(\mathbb{N})\,.

Thus, we only need to control the sum of dispersions

\begin{equation}\label{des18}
\sum_{n\ge 1}\Bigl\langle  \left|\frac{\cos x^\alpha}{x^{\alpha\gamma}}-   \Bigl\langle\frac{\cos x^\alpha}{x^{\alpha\gamma}}\Bigr\rangle_{[n,n+2]} \right|^2\Bigr\rangle_{[n,n+2]}=\sum_{n\ge 1}\left(\Bigl\langle  \left|\frac{\cos x^\alpha}{x^{\alpha\gamma}}\right|^2\Bigr\rangle_{[n,n+2]}-  \Bigl\langle  \frac{\cos x^\alpha}{x^{\alpha\gamma}}\Bigr\rangle_{[n,n+2]}^2  \right)\,.
\end{equation}

In what follows, we will see that the second term in the parenthesis  is negligible.
Set $\eta = \alpha\gamma + \alpha - 1$. We have $\eta \ge \gamma > 0$.
We apply (<ref>) in which $d=\eta/\alpha$ to get

\[
\int_{n}^{n+2}\frac{\cos (x^\alpha)}{x^{\alpha\gamma}}dx=\frac{1}{\alpha}\left(\frac{\sin(n+2)^\alpha}{(n+2)^{\eta}}-\frac{\sin n^\alpha}{n^{\eta}}\right)+O(n^{-\eta-\alpha})
\]

and $n^{-\eta-\alpha}\in \ell^2(\mathbb{N})$. So, the question reduces to the convergence of the series

\begin{equation}\label{eq62}
\sum_{n \ge 1}\left(2\alpha^2\int_{n}^{n+2}\frac{\cos^2 x^\alpha}{x^{2\alpha\gamma}}\,dx -  \left(\frac{\sin(n+2)^\alpha}{(n+2)^{\eta}}-\frac{\sin n^\alpha}{n^{\eta}}\right)^2  \right)\,.
\end{equation}

Notice first that  $\alpha\ge 1, \alpha+\beta>3/2$ implies $2\alpha\gamma>1, 2\eta>1$. We get convergence in that situation and (<ref>) holds.

We claim that for $\alpha>1, 1<\alpha+\beta\le 3/2$, the series diverges and (<ref>) does not holds.
Indeed, in that case, one has

\[
\int_{n}^{n+2}\frac{\cos^2 x^\alpha}{x^{2\alpha\gamma}}\,dx\sim n^{-2\alpha\gamma}, \quad n^{-2\eta}=o(n^{-2\alpha\gamma}), \,\, n\to\infty
\]

and we get the claim because $2\alpha\gamma\le 1$.

It is only left to consider $\alpha=1$ and $0<\beta\le 1/2$. In that case, $\gamma=\beta$, and

\[
\frac{\cos x}{x^\gamma}-\Bigl\langle \frac{\cos x}{x^\gamma}\Bigr\rangle_{[n,n+2]}= n^{-\beta}(\cos x-\langle \cos x\rangle_{[n,n+2]})+O(n^{-\beta-1})\,.
\]

Since $\langle (\cos x-\langle \cos x\rangle_{[n,n+2]})^2\rangle_{[n,n+2]}\sim 1$ and $\beta\le 1/2$, the series (<ref>) diverges.

To summarize, if $\alpha \ge 1$ and $\alpha + \beta > 1$, then (<ref>) holds if and only if $\alpha + \beta > 3/2$.

Case $\alpha \ge 1$,  $\alpha +\beta \le 1$. We are going to show that (<ref>) fails for this range of parameters. Assume that (<ref>) holds where, again,
$h_n = \exp\left(2\int_{n}^{x}\frac{\sin \tau^\alpha}{\tau^\beta}\,d\tau\right)$.
Then, estimate (<ref>) gives

\begin{equation}\label{eq64}
\sum_{n \ge 1}\left(\int_{n}^{n+2}\left|\int_{n}^{x}\frac{\sin \tau^\alpha}{\tau^\beta}\,d\tau - \frac{1}{2}\int_{n}^{n+2}\int_{n}^{y}\frac{\sin s^\alpha}{s^\beta}\,ds\,dy \right|\,dx\right)^2 < \infty.
\end{equation}

Recall that $\gamma = (\alpha+\beta-1)/\alpha\le 0$  and

\begin{align*}
\alpha\int_{n}^{x}\frac{\sin \tau^\alpha}{\tau^\beta}\,dy
&= \int_{n^\alpha}^{x^\alpha}y^{-\gamma}\sin y\,d\tau
= \left. - y^{-\gamma}\cos y \right|_{n^\alpha}^{x^\alpha}
\int_{n^\alpha}^{x^\alpha}\frac{\cos y}{y^{\gamma+1}} \,dy \\
&= n^{-\alpha\gamma}\cos n^{\alpha} - x^{-\alpha\gamma}\cos x^{\alpha} + O(n^{-\alpha(\gamma + 1)}).
\end{align*}

Therefore, we have

\begin{align*}
\frac{1}{2}\int_{n}^{n+2}\alpha\int_{n}^{y}\frac{\sin s^\alpha}{s^\beta}\,ds dy
&= n^{-\alpha\gamma}\cos n^{\alpha} - \frac{1}{2}\int_{n}^{n+2} y^{-\alpha\gamma}\cos y^{\alpha}\,dy + O(n^{-\alpha(\gamma + 1)}),\\
&= n^{-\alpha\gamma}\cos n^{\alpha} - \frac{1}{2\alpha}\int_{n^{\alpha}}^{(n+2)^{\alpha}} \eta^{-\gamma - 1 + 1/\alpha}\cos \eta\,d\eta + O(n^{-\alpha(\gamma + 1)}),\\
&= n^{-\alpha\gamma}\cos n^{\alpha}
+ O(n^{-\alpha(\gamma + 1) + 1}).
\end{align*}

Hence, uniformly in $x \in [n, n+2]$,

\alpha\left|\int_{n}^{x}\frac{\sin \tau^\alpha}{\tau^\beta}\,d\tau - \frac{1}{2}\int_{n}^{n+2}\int_{n}^{y}\frac{\sin s^\alpha}{s^\beta}\,ds\,dy \right| =
x^{-\alpha\gamma}|\cos x^{\alpha}| + O(n^{-\alpha(\gamma + 1) + 1}).

\int_{n}^{n+2} x^{-\alpha\gamma}|\cos x^{\alpha}|\,dx + O(n^{-\alpha(\gamma + 1) + 1}) \ge n^{-\gamma\alpha}\left(\int_{n}^{n+2} |\cos x^{\alpha}|\,dx + O(n^{-\alpha + 1})\right)

\begin{equation}
\inf_{n}\int_{n}^{n+2} |\cos x^{\alpha}|\,dx \ge C>0
\end{equation}

for every $\alpha \ge 1$, the terms in series (<ref>) can be estimated from below by $c n^{-2\gamma\alpha} \ge c > 0$ for large $n$. So, (<ref>) does not holds if $\alpha \ge 1$ and $\alpha +\beta \le 1$.

Case $\alpha \in (0,1)$, $\beta \le 1/2$. We are going to show that (<ref>) fails for this range of parameters. Assume that (<ref>) holds and $h_n = \exp\left(2\int_{n}^{x}\frac{\sin \tau^\alpha}{\tau^\beta}\,d\tau\right)$.
Then, the estimate (<ref>) gives

\sum_{n \ge 1}\left(\int_{n}^{n+2}\left|\int_{n}^{x}\frac{\sin \tau^\alpha}{\tau^\beta}\,d\tau - \frac{1}{2}\int_{n}^{n+2}\int_{n}^{y}\frac{\sin s^\alpha}{s^\beta}\,ds\,dy \right|\,dx\right)^2 < \infty.

Applying the mean value theorem, we see that

\sum_{n \ge 1}\left(\int_{n}^{n+2}\left|\int_{n}^{x}\frac{\sin \tau^\alpha}{\tau^\beta}\,d\tau - \int_{n}^{x_n}\frac{\sin s^\alpha}{s^\beta}\,ds\right|\,dx\right)^2 < \infty

for some points $x_n \in [n, n+2]$. This yields

\sum_{n \ge 1}\left(\int_{n}^{n+2}\left|\int_{x_n}^{x}\frac{\sin \tau^\alpha}{\tau^\beta}\,d\tau\right|\,dx\right)^2 < \infty,

where we interpret $\int_{b}^{a} f$ as $-\int_{a}^{b} f$ if $b>a$. For each $n \ge 0$, one can choose an interval $\Delta_n \subseteq [n, n+2]$ of length $1/4$ such that $\dist(x_n, \Delta_n) \ge 1/4$. Then, again by the mean value theorem, there exist points $\widetilde x_n \in \Delta_n$ such that

\begin{equation}\label{s_a20}
\frac{1}{16} \sum_{n \ge 1} \left|\int_{x_n}^{\widetilde x_n}\frac{\sin \tau^\alpha}{\tau^\beta}\,d\tau\right|^2  = \sum_{n \ge 1}\left(\int_{\Delta_n}\left|\int_{x_n}^{x}\frac{\sin \tau^\alpha}{\tau^\beta}\,d\tau\right|\,dx\right)^2 \le \sum_{n \ge 1}\left(\int_{n}^{n+2}\left|\int_{x_n}^{x}\frac{\sin \tau^\alpha}{\tau^\beta}\,d\tau\right|\,dx\right)^2 < \infty.
\end{equation}

Since $\sin \tau^{\alpha} - \sin n^\alpha = O(n^{\alpha - 1})$ for $\tau \in [n, n+2]$, we have

\left|\int_{x_n}^{\widetilde x_n}\frac{\sin \tau^\alpha}{\tau^\beta}\,d\tau\right| \ge |\sin n^\alpha| \cdot I_n - O( n^{\alpha-1}I_n), \qquad I_n = \left|\int_{x_n}^{\widetilde x_n}\frac{d\tau}{\tau^\beta}\,\right|.

Given that $\beta\le 1/2$, we have
I_n\sim n^{-\beta}
because $x_n$ and $\widetilde x_n$ are at least $1/4$ distance apart.
The sequence $\frac{1}{2\pi}n^{\alpha}$ is uniformly distributed mod $1$. Therefore, there is $N_0$ so that  for every $N>N_0$,
we will have a bound

\[
|\sin n^\alpha|>0.01
\]

for at least $N/2$ integer numbers $n\in [N,2N]$. Therefore, since $\alpha\in (0,1)$,

\[
\sum_{n=N}^{2N} \Bigl||\sin n^\alpha| \cdot I_n - O( n^{\alpha-1}I_n)\Bigr|^2\gtrsim N^{1-2\beta}\,.
\]

Since $\beta\le 1/2$, $\lim_{N\to\infty}N^{1-2\beta}$ does not converge to zero when $N\to\infty$ and we have a contradiction with (<ref>). Therefore, (<ref>) fails.

Case $\alpha \in (0,1)$, $\beta > 1/2$. Since $\beta$ is positive, we are in the setting of Proposition <ref>. We have

\begin{align*}
\sum_{n \ge 1}\int_{n}^{n+2}\left(\int_{n}^{x}\frac{\sin \tau^\alpha} {\tau^\beta}\,d\tau\right)^2\,dx
\sum_{n \ge 1}\int_{n}^{n+2}\left(\int_{n}^{x}\frac{d\tau}{\tau^\beta}\,d\tau\right)^2\,dx \\
2\sum_{n \ge 1}\left(\int_{n}^{n+2}\frac{d\tau}{\tau^\beta}\,d\tau\right)^2\,dx \le
8\sum_{n \ge 1} \frac{1}{n^{2\beta}} <
\infty.
\end{align*}

Hence, (<ref>) is true for this range of parameters by Remark <ref>.

Case $\alpha \le 0$, $\beta-\alpha > 1/2$. For this range of parameters, we have

\sum_{n \ge 1}\int_{n}^{n+2}\left(\int_{n}^{x}\frac{\sin \tau^\alpha} {\tau^\beta}\,d\tau\right)^2\,dx \lesssim   \sum_{n \ge 1}\frac{1}{n^{2(\beta-\alpha)}}

due to the fact that $\sin y$ is comparable to $y$ when $y \in [0,1]$. Using Remark <ref>, we conclude that (<ref>) holds  if  $\beta-\alpha > 1/2$.

Case $\alpha \le 0$, $\beta-\alpha \le 1/2$. We claim that (<ref>) fails in that situation. Arguing as in the case when $\alpha \in (0,1)$ and $\beta \le 1/2$, we assume that (<ref>) is true and obtain

\begin{equation}\label{eq66}
\sum_{n \ge 1}\left(\int_{n}^{n+2}\left|\int_{x_n}^{x}\frac{\sin \tau^\alpha}{\tau^\beta}\,d\tau\right|\,dx\right)^2 < \infty,
\end{equation}

for some points $x_n \in [n, n+2]$. For each $n \ge 0$, one can choose an interval $\Delta_n \subseteq [n, n+2]$ of length $1/4$ such that $\dist(x_n, \Delta_n) \ge 1/4$. Then, by the mean value theorem, we have

\frac{1}{4}\left|\int_{x_n}^{\widetilde x_n}\frac{\sin \tau^\alpha}{\tau^\beta}\,d\tau\right|  = \int_{\Delta_n}\left|\int_{x_n}^{x}\frac{\sin \tau^\alpha}{\tau^\beta}\,d\tau\right|dx \le \int_{n}^{n+2}\left|\int_{x_n}^{x}\frac{\sin \tau^\alpha}{\tau^\beta}\,d\tau\right|dx,

for some points $\widetilde x_n \in \Delta_n$. Since $\sin \tau^{\alpha} \ge 2\tau^{\alpha}/\pi$ for $\tau \in [n, n+2]$ and $n \ge 1$, one gets

n^{\alpha-\beta} \lesssim \int_{x_n}^{\widetilde x_n}\tau^{\alpha-\beta}\,d\tau \le   \left|\int_{x_n}^{\widetilde x_n}\frac{\sin \tau^\alpha}{\tau^\beta}\,d\tau\right|, \qquad n \ge 1,

which leads to a contradiction with (<ref>).

§ APPENDIX

In this Appendix, we collect some proofs and  auxiliary results.

1.  Proof of Proposition <ref>. To get the statement, one can modify the proofs in [6]. Alternatively, one can argue as follows.  First, we claim that (<ref>) implies  (<ref>)  for $\alpha_n=\lambda n$ with any $\lambda>0$. Indeed, if $m_\lambda(z)$ and $\mu_\lambda$ denote Weyl-Titchmarsh function and spectral measure, respectively, of canonical system with Hamiltonian $\Hh(\lambda \tau)$ (see formula (1-5) in [6]), then $m_\lambda(z)=m_1(\lambda^{-1} z)$ as easily follows from (1-5) in [6]. Now, it is enough to observe that  $\mu_\lambda\in \sz \Longleftrightarrow \mu_1\in \sz$.

Secondly, we claim that, given intervals $I^-\subseteq I, |I|=1$, and $\epsilon\in (0,1]$, the following implication holds

\begin{equation}\label{fix2}
\det \int_I \Hh(\tau) d\tau- 1=\epsilon \Rightarrow\det \int_{I^-} \Hh(\tau) d\tau- |I^-|^2\lesssim \epsilon,
\end{equation}

for every non-negative Hamiltonian $\Hh$ that satisfies $\det \Hh=1$ a.e. on $I$. Indeed, denote $A=\int_{I^-} \Hh d\tau,  B=\int_{I^+} \Hh d\tau$ where $I=I^-\cup I^+$. Then, we get (see, e.g., (A-1) in [6]):

\[
\det A\ge |I^-|^2, \quad \det B\ge |I^+|^2
\]

\[
\det (A+B)= 1+\epsilon\,.
\]

Minkowski inequality for determinants yields

\[
\det(A+B)\ge (\sqrt{\det A}+\sqrt{\det B})^2\,.
\]

Denote $\sqrt{\det A}=x, \sqrt{\det B}=y$ to get

\[
x+y\le (1+\epsilon)^{\frac 12}, \quad|I^-|+|I^+|=1,\, |I^-|\le x, \quad |I^+|\le y\,.
\]

That implies (draw the corresponding domains on the plane), that $|I^-|\le x\le |I^-|+C\epsilon$. Taking the square of the last bound yields the estimate on the right-hand side of (<ref>).

Now, if $\det \Hh=1$ a.e., the statement in  (<ref>) holds by combining these two claims. Indeed, suppose $\mu\in \szp$ and we are given sequence $\{\alpha_n\}$. Then, there is $\lambda$ such that every interval $[\alpha_n, \alpha_{n+2}]$ is inside one of the intervals $[\lambda l, \lambda (l+2)]$ or $[\lambda (l-1),\lambda (l+1)]$ for some $l$. Since the sum in (<ref>) converges for $\{\alpha_l\}=\{\lambda l\}$, we can apply (<ref>) (with dilated and translated interval $I$) to get condition in the left-hand side of (<ref>) satisfied for $\{\alpha_n\}$. Conversely, if the sum in (<ref>) converges for some $\{\alpha_n\}$, then there is some $\lambda$ such that each interval $[\lambda n, \lambda (n+2)]$ is covered by either $[\alpha_l, \alpha_{l+2}]$ or $[\alpha_{l-1}, \alpha_{l+1}]$ for some $l$. Thus, applying (<ref>) again, we get that the sum in (<ref>) converges with $\{\alpha_n\}=\{\lambda n\}$ and $\mu\in \szp$. The case of general $\Hh$ follows by making the change of variables in $\tau$ and using approximation argument.

2.   We will  show that the free Dirac evolution is, in fact, equivalent to the shift operator on the real line.  To this end, we perform two elementary unitary transformations

\[
\widehat \Dd_0=-Z^{-1}\Dd_0 Z= \left(\begin{smallmatrix} -i\partial_\tau & 0\\ 0 & i\partial_\tau\end{smallmatrix}\right), \quad Z=\frac{1}{\sqrt 2} \left(\begin{smallmatrix} i & -1\\ 1 & -i\end{smallmatrix}\right)\,.
\]

Operator $\widehat \Dd_0$, taken with suitable boundary condition at $0$: $f_1(0)=if_2(0)$, is self-adjoint on
the same Hilbert space $\left(\begin{smallmatrix} f_1\\f_2\end{smallmatrix}\right)\in L^2(\C^2)$. If one further maps

\[
\left(\begin{smallmatrix} f_1\\ f_2\end{smallmatrix}\right)\mapsto g(x)=\begin{cases} f_1(x), &x>0,\\ if_2(-x), &x<0,\end{cases}
\]

then, $\Dd_0$ becomes unitarily equivalent to $-i\partial_x$ on $L^2(\mathbb{R})$ with $e^{t\partial_x}g=g(x+t)$, which is the standard shift operator.

3. We have the following result.

If entire function $f$ has bounded type both in $\C_+$ and  $\C_-$, then its exponential type can be computed by the formula

\begin{equation}\label{eq21bis}
\type f = \limsup_{y \to +\infty}\frac{\log\max(|f(iy)|, |f(-iy)|)}{y}.
\end{equation}

Let us apply Theorem 2 in Lecture 16 of [52]. It says that for every entire function $f$ of bounded type in $\C_+$ and $\C_-$ we have

\begin{align*}
\log |f(z)| &= \sigma_+ y  + o(|z|),  \qquad y \ge 0,\\
\log |f(z)| &= \sigma_- y  + o(|z|),  \qquad y \le 0,
\end{align*}

outside of a set of disks $\{z \in \C: |z - a_j| < r_j\}$ of finite view (the latter means that $\sum_j \frac{r_j}{|a_j|} < \infty$). Here $\sigma_{\pm}  \in \R$ and $y = \Im z$. Take $\eps > 0$ and denote $\sigma = \max(\sigma_{+}, -\sigma_-)$. By the maximum principle for subharmonic functions, we have

\log |f(z)| \le (\sigma + \eps)|y| + o(|z|)

everywhere in $\C$ as $z \to \infty$. Therefore, we have $\type f \le \sigma$. On the other hand, the set of disks of finite view cannot fill any half-axis, hence

\sigma_+ = \limsup_{y \to +\infty}\frac{\log|f(iy)|}{y}, \qquad -\sigma_- = \limsup_{y \to -\infty}\frac{\log|f(iy)|}{|y|},

which proves the statement.

4. The following result in the linear algebra has been used in the main text.

For every real $2\times 2$ matrix $A$ with non-negative determinant,
there is a rotation matrix $\Sigma_\phi$ of the form

\Sigma_\phi = \begin{pmatrix}\cos\phi & \sin\phi \\ -\sin\phi & \cos\phi\end{pmatrix}, \qquad \phi \in [0, 2\pi),

such that $\Sigma_\phi A \ge 0$.

This is immediate from the proof of the polar decomposition given in  [28], p. 276.

5. In the main text, we used the following variation of a result by Robinson [62], which is based on ideas dating back to Ruelle's work [64].
Suppose $H$ is a Hilbert space, $\Di$ is a densely defined self-adjoint operator, and $P_{\Delta}$ denotes the orthogonal projector for $\Di$ relative to a set $\Delta\subseteq \R$. If $A$ is a bounded  operator on $H$ and $AP_{[-\Lambda,\Lambda]}$ is compact for some $\Lambda\ge 0$, then

\[
\lim_{T\to +\infty}\frac 1T\int_0^T \|Ae^{it\Di}P_{[-\Lambda,\Lambda]}\psi\|^2dt=\sum_{j}\|AP_{E_j}P_{[-\Lambda,\Lambda]}\psi\|^2
\]

for every $\psi\in H$, where $P_{ E_j}$ denotes the orthogonal projection on the eigenspace that corresponds to eigenvalue $E_j$ and the sum is over all eigenvalues $\{E_j\}$ of $\Di$.

The proof of this result repeats the proof of Theorem 2 in [62].

$'$ $'$ $'$

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Original part about dynamical characterization

As a corollary of Theorem <ref>, we obtain the following dynamical characterization of the absence of the singular continuous spectrum.

Let $\Hh \in \szcs$. Then the singular continuous spectrum of $\Di_{\Hh}$ is empty if and only if we have

\lim_{b \to +\infty}\limsup_{t \to +\infty}\frac{1}{T}\int_{0}^{T}\|U_t X\|_{L^2(\Hh, [L_{b}, L_{t-b}])}^{2}\,dt = 0

for some (and then for every) compactly supported nonzero element $X \in H$.

Take a compactly supported nonzero element $X \in H$. Since $\W_{\Hh} X$ is an entire function, we have $P_{cs}X = 0$ if and only if $H_{cs} = \{0\}$. It remains to use Theorem <ref>.

The proof of Theorem <ref> is based on Theorem <ref> and the following well-known fact.

[Wiener theorem]
Let $\nu \ge 0$ be a finite Borel measure on the real line. Then we have

\sum_{x \in \R:\; \nu\{x\} \neq 0}\nu(\{x\})^2 = \lim_{T \to +\infty}\frac{1}{2T}\int_{-T}^{T}|\hat\nu(y)|^2\,dy.

In particular, for every signed Borel measure $\nu$ on the real line with finite variation and without point masses we have

\lim_{T \to +\infty}\frac{1}{2T}\int_{-T}^{T}|\hat\nu(y)|^2\,dy = 0.

Let $\Hh$ be a proper Hamiltonian with the spectral measure $\mu$, and let $f = \W_{\Hh}X$ for some $X \in \dom\Di_{\Hh}$. Then the integral

\begin{equation}\label{eq84}
\widetilde X(\tau) = \frac{1}{\sqrt{\pi}}\int_{\R} f(x)\Theta(\tau, x)\,d\mu(x), \qquad \tau \in \R_+,
\end{equation}

converges absolutely  for every $\tau$ and defines a continuous function on $\R_+$.  Moreover, we have $\widetilde X \in H$ and $\widetilde X = X$ in $H$. For every $r \ge 0$, we have $\sup_{X} \max_{\tau \in [0,r]}\|\widetilde X(\tau)\|_{\C^2} < \infty$, where the supremum is taken over all $X$ such that $\|X\|_{L^2(\Hh)} + \|\Di_{\Hh}X\|_{L^2(\Hh)} \le C$.

Let us first show that the integral in (<ref>) converges absolutely and defines a continuous function $\widetilde X$ on $\R_+$. Assume that $\Hh$ has no indivisible intervals. We can estimate

\|\widetilde X(\tau)\|_{\C^2}^2 \le \frac{1}{\pi}\int_{\R}|f(x)|^2(x^2 + 1)\,d\mu(x)\int_{\R}\frac{\|\Theta(\tau, x)\|_{\C^2}^2}{1+x^2}\,d\mu(x).

Since $X \in \dom \Di_\Hh$, the first integral above is finite. We have $\|\Theta(\tau, x)\|_{\C^2}^{2} = |E_{\tau}(z)|^2$ where $E_{\tau}$ is the Hermite-Biehler function generated by $\Hh$ on $[0, \tau]$. Substituting any zero of $E_\tau$ into (<ref>), we see that

\begin{equation}\label{eq85}
\int_{\R}\frac{\|\Theta(\tau, x)\|_{\C^2}^2}{1+x^2}\,d\mu(x) = \int_{\R}\frac{|E_{\tau}(x)|^2}{1+ x^2}\,d\mu(x) < \infty.
\end{equation}

In case $E_\tau$ has no zeroes in $\C$, we have $E_\tau = ce^{-iT(\tau)z}$, $c \in \C$, and (<ref>) holds as well. Thus, the integral in (<ref>) converges absolutely for every $\tau \in \R_+$. Moreover, for every $\tau \in \R_+$ and small enough $\eps>0$ the same argument along with the continuity of the mapping $\tau \mapsto E_{\tau}(z)$ and Rouche's theorem gives

\int_{\R}\frac{\sup_{|\tau-\tau'|<\eps}\|\Theta(\tau', x)\|_{\C^2}^2}{1+x^2}\,d\mu(x) = \int_{\R}\frac{\sup_{|\tau-\tau'|<\eps}|E_{\tau'}(x)|^2}{1+ x^2}\,d\mu(x) < \infty.

By a compactness argument, $\max_{\tau \in [0,r]}\|\widetilde X(\tau)\|_{\C^2}$ is controllable by $\|(i+x)f\|_{L^2(\mu)} = \|(i+\Di_{\Hh})X\|_{L^2(\Hh)}$ and a certain quantity that depends only on $r$.
We now see from the Lebesgue dominated convergence theorem and the estimate

\|\widetilde X(\tau_1) - \widetilde X(\tau_2)\|_{\C^2}^2 \le \frac{1}{\pi}\int_{\R}|f(x)|^2(x^2 + 1)\,d\mu(x)\int_{\R}\frac{\|\Theta(\tau_1, x) - \Theta(\tau_2, x)\|_{\C^2}^2}{1+x^2}\,d\mu(x),

that $\widetilde X$ is continuous. The adaptation of the argument to the case where $\Hh$ has indivisible intervals is straightforward: it is is enough to note that $\langle Je, e\rangle_{\C^2} = 0$, $e \in \C^2$, hence

\begin{equation}\label{eq87}
\Theta(\tau, z) = \Theta(a, z) - z\langle\Theta, e_I\rangle_{\C^2}
\int_{a}^{\tau}\kappa_{I}(s)\,ds \cdot Je_{I}, \qquad \Hh(\tau) = \kappa_{I}(\tau) \langle \cdot, e_I\rangle_{\C^2}e_{I}, \qquad \|e_I\|_{\C^2}  = 1,
\end{equation}

on each indivisible interval $I = (a,b)$ of $\Hh$.

Now fix  $r\in \R_+ \setminus \bigcup_{I \in \mathfrak{I}(\Hh)} I$ (in other words, $r$ is not an inner point of an indivisible interval and the mapping $X \mapsto \chi_{[0,r]}X$ acts in $H$). Since $\Hh(\tau)\Theta(\tau, x) = \Hh(\tau)\Theta(a, x)$ is constant in $\tau$ on each indivisible interval $I = (a,b)$ of $\Hh$ (see (<ref>)), the same is true for $\Hh\widetilde X$. In particular, we have $\chi_{[0,r]}\widetilde X \in H$. Let us show that $\chi_{[0,r]}\widetilde X = \chi_{[0,r]} X$ in $H$. Then $\widetilde X \in H$ and $\widetilde X = X$ in $H$. For every $\lambda \in \C$, we have

\begin{align}
\W_{\Hh}(\chi_{[0,r]} X)(\lambda)
&=\frac{1}{\sqrt{\pi}}\int_{0}^{r}\langle\Hh(\tau)X(\tau), \Theta(\tau,  \bar\lambda)\rangle_{\C^2}\,d\tau,\notag\\
&=\frac{1}{\sqrt{\pi}}(X, \chi_{[0,r]}\Theta(\cdot, \bar \lambda))_{L^2(\Hh)}\notag\\
&=(f, k_{\mu, r, \lambda})_{L^2(\mu)},\notag
\end{align}

\begin{equation}\notag
k_{\mu, r, \lambda}(x) =
\frac{1}{\sqrt{\pi}}\W_{\Hh}(\chi_{[0,r]}\Theta(\cdot, \bar \lambda)) = \frac{1}{\pi}\int_{0}^{r}\langle\Hh(\tau)\Theta(\tau,\bar\lambda), \Theta(\tau,x)\rangle_{\C^2}\,d\tau, \qquad x \in \R.
\end{equation}

Taking into account complex conjugation in the definition of the inner product in $\C^2$, we obtain

\W_{\Hh}(\chi_{[0,r]} X)(\lambda) = (f, k_{\mu,r, \lambda})_{L^2(\mu)} =
\frac{1}{\sqrt{\pi}}\int_{0}^{r}\langle\Hh(\tau)\widetilde X(\tau),\Theta(\tau,\bar\lambda) \rangle_{\C^2}\,d\tau = \W_{\Hh}(\chi_{[0,r]} \widetilde X)(\lambda).

It follows that
$\chi_{[0, r]} X = \chi_{[0, r]} \widetilde X(\tau)$ in $H$. The lemma follows.

Let $\Hh \in \szcs$, and let $H_{cs}$ be the singular continuous subspace of $\Di_{\Hh}$. For every $X \in H_{\bf sc}$ and every $r > 0$ we have

\frac{1}{T}\int_{0}^{T}\|U_t X\|_{L^2(\Hh, [0, r])}\,dt \to 0, \qquad T \to +\infty.

Since $U_t$ is unitary on $H$, we can assume that $X \in \dom\Di_{\Hh} \cap H_{\bf sc}$ without loss of generality. Monotonicity of $\|U_t X\|_{L^2(\Hh, [0, r])}$ in $r$ allows us to assume that $r\in \R_+ \setminus \bigcup_{I \in \mathfrak{I}(\Hh)} I.$ Set

\begin{equation}\label{eq86}
\widetilde X_t(\tau) = \frac{1}{\sqrt{\pi}}\int_{\R} e^{itx}f(x)\Theta(\tau, x)\,d\mu(x), \qquad \tau \in \R_+.
\end{equation}

Then $\widetilde X_t = U_t X$ in $H$ by the Spectral Theorem and Lemma <ref>.
For every $2 \times 2$ matrix $A \ge 0$ we have

(A x, x)_{\C^2} \le \trace A \cdot \|x\|^2.

\|U_t X\|_{L^2(\Hh, [0, r])} \le \int_{0}^{r}\trace\Hh(\tau)\|\widetilde X_t(\tau)\|_{\C^2}^{2}\,d\tau, \qquad r \ge 0.

By Lemma <ref>, for each $t \in \R$ the function $\widetilde X_t(\tau)$ is continuous in $\tau$ on $[0,r]$ and its maximum there depends only on $r$ and the norm $\|U_tX\|_{L^2(\Hh)} + \|\Di_{\Hh} U_tX\|_{L^2(\Hh)} = \|X\|_{L^2(\Hh)} + \|\Di_{\Hh} X\|_{L^2(\Hh)}$. It follows that

\sup_{t \ge 0, \; \tau \in [0,r]}\|\widetilde X_t(\tau)\|_{\C^2}^{2} < \infty.

Therefore, the claim will follow from the Lebesgue dominated convergence theorem if we check

\begin{equation}\label{eq65}
\frac{1}{T}\int_{0}^{T}\|\widetilde X_t(\tau)\|_{\C^2}^{2}\,dt \to 0, \qquad T \to +\infty,
\end{equation}

for every $\tau \in [0, r]$. But (<ref>) is just a consequence of the Wiener theorem (see Theorem <ref>) and (<ref>) if we take into account that measures $\nu_{\pm} = |\Theta^{\pm}(\tau, x)f(x)|\,d\mu(x)$ are finite and do not have point masses (the latter follows from  our assumption $f \in H_{cs} \cap \dom \Di_{\Hh}$).

Let $\Hh \in \szcs$. For every $X \in H_{\bf pp}$ we have

\lim_{r\to \infty}\liminf_{t \to +\infty}\|U_t X\|_{L^2(\Hh, [0, r])} = \lim_{r\to \infty}\limsup_{t \to +\infty}\|U_t X\|_{L^2(\Hh, [0, r])} = \|X\|_{L^2(\Hh)}.

Consider a decomposition of $X$ of the form

\begin{equation}\label{eq81}
X = \sum_{\lambda \in \Sigma_{pp}(\mu)} c_{\lambda}X_{\lambda}
\end{equation}

where $\{X_{\lambda}\}_{\lambda \in \Sigma_{pp}(\mu)}$ is the complete (in $H_{\bf pp}$) system of orthonormal eigenvectors of $\Di_{\Hh}$. Since $U_t$ is unitary, we can assume that the sum in (<ref>) contains only a finite number of summands (that is, $\lambda \in S$ for a finite set $S \subset \R$). We have

U_{t}X = \sum_{\lambda \in S} c_{\lambda}e^{it\lambda}X_{\lambda}

by the definition of $U_t$. Then

\begin{align*}
\|U_t X\|_{L^2(\Hh)}&= \lim_{r\to \infty}\liminf_{t \to +\infty}\|U_t X\|_{L^2(\Hh)} \\
&\le \lim_{r\to \infty}\liminf_{t \to +\infty}\|U_t X\|_{L^2(\Hh, [0, r])} + \sum_{\lambda \in S}|c_\lambda| \cdot \lim_{r\to \infty} \max_{\lambda \in S}\|X_\lambda\|_{L^2(\Hh, [r, \infty))}\\
&\le \lim_{r\to \infty}\limsup_{t \to +\infty}\|U_t X\|_{L^2(\Hh, [0, r])}\le \|X\|_{L^2(\Hh)}.
\end{align*}

Now the statement follows from the fact that $\|X\|_{L^2(\Hh)} = \|U_t X\|_{L^2(\Hh)}$.
Let $\Hh \in \szcs$. For every $X \in H$, we have

\begin{equation}
\lim_{b \to +\infty}\liminf_{t \to +\infty}\|U_t X\|_{L^2(\Hh, \dlt)} = \lim_{b \to +\infty}\limsup_{t \to +\infty}\|U_t X\|_{L^2(\Hh, \dlt)}
= \|P_{\bf ac} X\|_{L^2(\Hh)},\label{eq82}
\end{equation}

We also have

\begin{equation}
\lim_{b \to +\infty}\limsup_{t \to +\infty}\|U_t X\|_{L^2(\Hh, [L_{t+b}, \infty))} = 0.\label{eq91}
\end{equation}

Note that $\|U_t X\|_{L^2(\Hh, \dlt)} = \|U_{-t}\chi_{\dlt}U_t X\|_{L^2(\Hh)}$ for every $b > 0$. Therefore, (<ref>) follows from assertion $(c)$ in the proof of Theorem <ref>. Relation (<ref>) holds for every compactly supported $X \in H$ by Theorem <ref>. Therefore, it holds for all $X \in H$ by continuity.
Let $\Hh \in \szcs$. For every $X \in H$ we have

\begin{align}
\lim_{b \to +\infty}\limsup_{t \to +\infty}\frac{1}{T}\int_{0}^{T}\|U_{t}X\|_{L^2(\Hh, [0, L_b])}^{2}
&= \|P_{\bf pp}X\|_{L^2(\Hh)}^{2}. \label{eq83}\\
\lim_{b \to +\infty}\liminf_{t \to +\infty}\frac{1}{T}\int_{0}^{T}\|U_{t}X\|_{L^2(\Hh, [0, L_b])}^{2}
&= \|P_{\bf pp}X\|_{L^2(\Hh)}^{2}. \label{eq88}
\end{align}

For every $b > 0$ we have

\limsup_{t \to +\infty}\frac{1}{T}\int_{0}^{T}(\|U_{t}P_{\bf ac}X\|_{\Hh, [0, L_b]}^{2} + \|U_{t}P_{\bf sc}X\|_{\Hh, [0, L_b]}^{2}) \,dt = 0

by (<ref>) and Lemma <ref>. Using Lemma <ref> and the fact that $\limsup_{t \to +\infty}\frac{1}{T}\int_{0}^{T}g(s)\,ds \le \limsup_{t \to +\infty}g(t)$ for any function $g$, we get

\lim_{b \to +\infty}\limsup_{t \to +\infty}\frac{1}{T}\int_{0}^{T}\|U_{t}X\|_{L^2(\Hh, [0, L_b])}^{2} \le \|P_{\bf pp}X\|_{L^2(\Hh)}^{2}.

Similarly, from Lemma <ref> and the fact that $\liminf_{t \to +\infty}\frac{1}{T}\int_{0}^{T}g(s)\,ds \ge \liminf_{t \to +\infty}g(t)$ for any function $g$, we get the opposite inequality

\lim_{b \to +\infty}\liminf_{t \to +\infty}\frac{1}{T}\int_{0}^{T}\|U_{t}X\|_{L^2(\Hh, [0, L_b])}^{2} \ge \|P_{\bf pp}X\|_{L^2(\Hh)}^{2}.

Combining the last two estimates, we obtain (<ref>), (<ref>).
Let $\Hh \in \szcs$. For every $X \in H$ we have

\begin{align}
\lim_{b \to +\infty}\limsup_{t \to +\infty}\frac{1}{T}\int_{0}^{T}\|U_t X\|_{L^2(\Hh, [L_{b}, L_{t-b}])}^{2}\,dt &= \|P_{\bf sc} X\|_{L^2(\Hh)}^{2},\label{eq89}\\
\lim_{b \to +\infty}\liminf_{t \to +\infty}\frac{1}{T}\int_{0}^{T}\|U_t X\|_{L^2(\Hh, [L_{b}, L_{t-b}])}^{2}\,dt &= \|P_{\bf sc} X\|_{L^2(\Hh)}^{2}\label{eq90}.
\end{align}

Since $U_t$ is unitary, we have

\| X\|_{L^2(\Hh)}^{2} = \|U_{t}X\|_{\Hh, L^2([0, L_b])}^{2} +
\|U_{t}X\|_{L^2(\Hh,[L_b, L_{t-b}])}^{2} +
\|U_{t}X\|_{L^2(\Hh,\dlt)}^{2} + \|U_{t}X\|_{L^2(\Hh,[L_{t+b}, +\infty))}^{2}.

From here and Lemmas <ref>, <ref>, we get

\|X\|_{L^2(\Hh)}^{2} \le \|P_{\bf pp}X\|_{L^2(\Hh)}^{2} +
\lim_{b \to +\infty}\liminf_{t \to +\infty}\frac{1}{T}\int_{0}^{T}\|U_{t}X\|_{\Hh, L^2([L_b, L_{t-b}])}^{2} + \|P_{\bf ac}X\|_{L^2(\Hh)}^{2}.

Lemma <ref> also gives

\lim_{b \to +\infty}\limsup_{t \to +\infty}\frac{1}{T}\int_{0}^{T}(\|U_{t}P_{\bf ac}X\|_{\Hh, L^2([L_b, L_{t-b}])}^{2} + \|U_{t}P_{\bf pp}X\|_{\Hh, L^2([L_b, L_{t-b}])}^{2})\,dt = 0.

\begin{align*}
\|X\|_{L^2(\Hh)}^{2}
\|P_{\bf pp}X\|_{L^2(\Hh)}^{2} + \lim_{b \to +\infty}\liminf_{t \to +\infty}\frac{1}{T}\int_{0}^{T}\|U_{t}X\|_{\Hh, L^2([L_b, L_{t-b}])}^{2} + \|P_{\bf ac}X\|_{L^2(\Hh)}^{2} \\
&\le \|P_{\bf pp}X\|_{L^2(\Hh)}^{2} + \lim_{b \to +\infty}\limsup_{t \to +\infty}\frac{1}{T}\int_{0}^{T}\|U_{t}P_{\bf sc}X\|_{\Hh, L^2([L_b, L_{t-b}])}^{2} + \|P_{\bf ac}X\|_{L^2(\Hh)}^{2} \\
&\le \|P_{\bf pp}X\|_{L^2(\Hh)}^{2} + \|P_{\bf sc}X\|_{L^2(\Hh)}^{2} + \|P_{\bf ac}X\|_{L^2(\Hh)}^{2}  = \|X\|_{L^2(\Hh)}^{2}.
\end{align*}

This proves the statement.

Proof of Theorem <ref>. Assertion (<ref>) is (<ref>). Assertion (<ref>) is (<ref>). Assertion (<ref>)  is (<ref>). Assertion (<ref>)  is (<ref>).

*********************  Korey

We also need M. Korey's estimate from [43]. Recall that we use notation $\langle f \rangle_{I} = \frac{1}{|I|}\int_{I}f(x)\,dx$.

[Formula $(3.7)$ in [43]]
Suppose $I=[a,b]$ and  measurable function $f : I \to \R$ satisfies

\langle e^{f}\rangle_{I} \cdot \langle e^{-f}\rangle_I = 1 + \eps, \qquad \eps \in [0, 1].

\begin{equation}\label{eq63}
\langle \left|f(x) - \langle f\rangle_I\right|\rangle_I\le c\sqrt{\eps},
\end{equation}

for a universal constant $c$.

It must be mentioned that Korey used a different constant in (<ref>)  – the median value of $f$ on the interval $[a,b]$. This, however, does not change the estimate. Indeed, for each $C \in \R$, we have

\begin{align*}
\langle \left|f(x) - \langle f\rangle_I\right|\rangle_I\le \langle \left|f(x) - C\right|\rangle_I+|C-\langle f\rangle_I|
\le 2\langle \left|f(x) - C\right|\rangle_I
\end{align*}

so if $c/2$ is a constant in (<ref>) for $C$, then $c$ works in (<ref>) for $\langle f\rangle_I$.