q_{2},$ so that for fixed $p$, the Lorentz space extends as $q$ grows. Finally, we note that the $L_{pp}$-space coincides with the usual $L_{p}$ since $\|f\|_{pp} = \|f\|_{p}.$ We will also need the following well-known interpolation theorem, so-called generalized Marcinkiewicz theorem. We refer to \cite{Stein} or \cite{BeL} for a proof. The operator $B$ is called subadditive if it satisfies $|B(f_{1}+f_{2})(k)| \leq |Bf_{1}(k)| +|Bf_{2}(k)|.$ We say that $B$ is of restricted weak type $(r,p)$ if its domain $D(B)$ contains all finite linear combinations of characteristic functions of sets of finite measure and all truncations of its members and satisfies $\|Tf\|_{p\infty} \leq C \|f\|_{r1}$ for all $f \in D \cap L_{r1}.$ \\ \noindent \bf Theorem. \it Suppose that $T$ is a subadditive operator of restricted weak types $(r_{j}, p_{j}),$ $j=1,2$ with $r_{0}2$ and $p$ such that $q^{-1}+p^{-1}=1,$ we have the following estimate for the maximal function: \begin{equation} \|Mf\|_{q} \leq C_{q} \|f\|_{pq} \;\: for \;\: every \;\: f \in L_{pq} \end{equation} {\rm (} hence, in particular, $\|Mf\|_{q} \leq C_{q}\|f\|_{p}${\rm )}. As a consequence, the integral \[ \int^{N}_{0} A(k,x)f(x)\,dx \] converges as $N \rightarrow \infty$ for almost every value of $k$ if $f \in L_{pq}$ {\rm (} and in particular if $f \in L_{p},$ $1 \leq p < 2${\rm )}. \\ \noindent \it Remark. \rm The result we prove here is suited for the applications we make in this paper. In fact, similar results hold in greater generality for more general integral operators with natural defintions of the maximal function. The proof is more involved and we plan to devote a separate publication \cite{Kis1} to this problem. \noindent \bf Proof. \rm First we remark that from the estimate on the maximal function the a.e. convergence for operator $T$ in classes $L_{pq},$ $p<2,$ follows in a standart way (see, e.g., \cite{Gar}). We sketch here this simple argument for the sake of completness. Indeed, suppose that there exists a function $f \in L_{pq}$ with $p$ and $q$ as in the theorem and a set $S$ of positive measure where the integral defining $Tf(k)$ diverges. Then we can find $\epsilon >0$ and a set $S_{\epsilon} \subset S,$ such that $m(S_{\epsilon})>0,$ and for every $k \in S_{\epsilon}$ and every positive number $N_{1}$ there exists a larger number $N_{2}$ such that $|\int_{N_{1}}^{N_{2}}A(k,x)f(x)\,dx|> \epsilon.$ Hence, for every $N_{1},$ we have $\|Mf \chi (N_{1}, \infty) \| _{q} \geq \epsilon m(S_{\epsilon})^{\frac{1}{q}}.$ On the other hand, clearly $\|f\chi (N_{1}, \infty) \|_{pq} \stackrel{N_{1} \rightarrow \infty}{\longrightarrow} 0.$ This contradicts the bound (21). We now come to a proof of the first assertion. The first step is the decompostion of the support of the function $f$ into dyadic pieces and estimates on the certain auxilliary maximal functions. A similar idea was used already by Paley \cite{Pal} in hi s proof of a.e. convergence for the series of orhtogonal functions. Let $f$ be a measurable bounded function of compact support and choose $n$ so that $2^{n-1} \leq m(\rm{supp}(f)) \leq 2^{n}.$ Let the measurable set $E$ be the support of the function $f:$ $E= \{ x| \;\, |f(x)|>0 \}.$ For every integer $m 0.$ Fix now any $q>2$ and let $q'$ satisfy $q>q'>2.$ Under the conditions of the theorem, we have that $\|Tf\|_{\infty}\leq C_{1}\|f\|_{1}$ for all $f \in L_{1}$ and $\|Tf\|_{2} \leq C_{2}\|f\|$ is satisfied for all measurable bounded functions of compact support. By interpolation, we have \[ \|Tf\|_{q'} \leq C_{q'} \|f\|_{p'q'}, \,\,\, (p')^{-1}+(q')^{-1}=1. \] Noting that \[ \|f\|^{*}_{p'q'} = \left( \frac{q'}{p'} \int\limits^{\infty}_{0} |f^{*}(t)|^{q'} t^{\frac{q'}{p'}-1}\,dt\right) ^{\frac{1}{q'}}, \] and using the equivalence of $\|\cdot \|^{*}$ and $\|\cdot\|,$ we see that in particular \[ \|M_{m,l}f(k)\|^{q'}_{q'}= \|T(f \chi (E_{m,l}))(k) \|_{q'}^{q'} \leq C_{q'}^{q'}\left( \frac{q'}{p'} \int\limits_{0}^{\infty} | (f\chi (E_{m,l}))^{*} |^{q'}t^{\frac{q'}{p'}-1} \,dt \right)\leq \] \[ \leq C_{q'}^{q'}2^{m(\frac{q'}{p'}-1)}\frac{q'}{p'} \| f\chi (E_{m,l})(t)\|^{q'}_{q'}.\] Summing over $l$ and using (23), we obtain \[ \|M_{m}f\|_{q'}^{q'} \leq C_{q'}^{q'}\frac{q'}{p'} 2^{m(\frac{q'}{p'}-1)}\|f\|_{q'}^{q'}. \] By (22), we have that \begin{equation} \| Mf \|_{q'} \leq C_{q'} \left( \frac{q'}{p'} \right)^{\frac{1}{q'}} \| f \|_{q'} \sum\limits_{m=-\infty}^{n} 2^{m (\frac{1}{p'}-\frac{1}{q'})}= B_{q'} 2^{n(\frac{1}{p'}-\frac{1}{q'})}\|f\|_{q'}. \end{equation} Now we note that in a particular case when $f$ is a characteristic function of a set, $f=\chi (E),$ (24) means \begin{equation} \| M\chi (E) \|_{q'} \leq B_{q'} 2^{n(\frac{1}{p'}-\frac{1}{q'})}2^{\frac{n}{q'}}= B_{q'} 2^{\frac{n}{p'}} \leq 2^{\frac{1}{p'}}B_{q'} \| \chi (E) \|_{p'1}. \end{equation} It is easy to check that the operator $Mf,$ defined originally on the measurable bounded functions of compact support, is a sublinear operator. It is well-known fact that from the inequality (25) for sublinear operator it follows that $\|Mf\|_{q'} \leq C\|\chi (E) \|^{*}_{p',1}$ holds for all finite combinations of simple functions (see \cite{Stein}) and hence by simple limiting argument for all measurable bounded functions of compact support. Interpolating with an obvious relation $\|Mf\|_{\infty} \leq A \|f\|_{1},$ we obtain that $\|Mf\|_{\tilde{q}} \leq C_{\tilde{q}}\|f\|_{\tilde{p} \tilde{q}}$ for every $\tilde{p},$ $\tilde{q}$ such that $q'>\tilde{q}>2$ and $\tilde{p}^{-1}+\tilde{q}^{-1}=1$ and for every function $f$ bounded and of compact support. In particular, this relation holds for the value of $q$ we fixed in the beginning of the proof (and hence for every $q>2$). It is straightforward to see that this inequality is then extended to all functions $f \in L_{pq}.$ $\Box$ \begin{center} \large \bf 3. Proofs of criteria and applications. \end{center} \normalsize Now we prove all theorems formulated in the first section. First we give proofs of the general criteria. \\ \noindent \bf Proof of Theorem 1.7. \rm By assumption, the operator $T_{1i}$ defined by \[ (T_{1i}f)(\lambda) = \frac{\chi (S_{i})}{\Im (\theta \overline{\theta}')} \int\limits_{0}^{\infty} \overline{\theta}(x, \lambda)^{2} f(x)\,dx, \] satisfies an $L_{2}-L_{2}$ bound on the bounded functions of compact support. We may also assume that on every set $S_{i}$ we have the Wronskian $W[\theta, \overline{\theta}]$ bounded away from zero by some constant $c_{i},$ or else subdivide the $S_{i}$ so that it holds true. Since the functions $\theta (x, \lambda)$ are uniforml y bounded when $\lambda \in S,$ we also have an obvious $L_{1}-L_{\infty}$ bound for $T_{1i}.$ Hence, by Theorem 2.1, the integral \begin{equation} \int\limits_{0}^{x} \overline{\theta}(y, \lambda)^{2} f(y)\,dy \end{equation} converges as $x \rightarrow \infty$ for a.e.~$\lambda \in S_{i}$ for every $f \in L_{p},$ $1\leq p <2.$ By the assumption on the decay of $V(x),$ the function $x^{\frac{1}{4}}V(x)$ belongs to $L_{2(1-\epsilon)}.$ Therefore, we find that \[ q(x, \lambda) = \int\limits_{x}^{\infty} \overline{\theta}(y, \lambda)^{2} V(y)\,dy = \int\limits_{x}^{\infty} \overline{\theta}(y, \lambda)^{2} (V(y) y^{\frac{1}{4}})y^{-\frac{1}{4}}\,dy= \] \[ = x^{-\frac{1}{4}}\int\limits_{0}^{x} \overline{\theta}(y, \lambda)^{2} (y^{\frac{1}{4}}V(y))\,dy + \frac{1}{4} \int\limits_{x}^{\infty} y^{-\frac{5}{4}}\int\limits_{0}^{y} \overline{\theta}(t, \lambda)^{2} (t^{\frac{1}{4}}V(t))\,dt. \] We conclude that for all $\lambda \in S_{i}$ such that the integral (26) converges, and hence for a.e.~$\lambda \in S_{i},$ the function $q$ satisfies $q(x, \lambda)=O(x^{-\frac{1}{4}})$ as $x \rightarrow \infty.$ Since this holds for any $i,$ we have that this estimate is also true for a.e.~$\lambda \in S.$ This implies that $q(x, \lambda)V(x) \in L_{1}$ and allows for a.e.~$\lambda \in S$ to find the asymptotics of solutions of the perturbed Schr\"odinger equation. Transforming back via (7) and (9) we find that for a.e.~$\lambda \in S,$ there exist two solutions $\phi(x, \lambda),$ $\overline{\phi}(x, \lambda)$ of the generalized eigenfunction equation (4) with the following asymptotics: \[ \phi (x, \lambda) = \theta (x, \lambda) \exp \left( - \frac{2}{\Im (\theta \overline{\theta}')} \int\limits^{x}_{0} V(t)|\theta(t, \lambda)|^{2}\,dt \right)\left(1+ O(x^{-\epsilon})\right), \] \[ \phi' (x, \lambda) = \theta' (x, \lambda) \exp \left( - \frac{2}{\Im (\theta \overline{\theta}')} \int\limits^{x}_{0} V(t)|\theta(t, \lambda)|^{2}\,dt \right)\left(1+ O(x^{-\epsilon})\right). \] Clearly the solutions $\phi,$ $\overline{\phi}$ are linearly independent, since the Wronskian $W[\phi, \overline{\phi}] = W[\theta, \overline{\theta}] \ne 0.$ This concludes the proof, given Lemma 1.5. $\Box$ \\ \noindent \bf Proof of Theorem 1.8. \rm Similarly to the previous proof, we infer that under the assumption of the theorem, for every $f \in L_{p},$ $1 \leq p <2,$ the integral \[ \int\limits_{0}^{x} \overline{\theta}(y, \lambda)^{2} \exp \left(-\frac{1}{\Im (\theta \overline{\theta}')} \int\limits_{0}^{y} V(t) |\theta(t, \lambda)|^{2}\,dt \right) f(y)\,dy \] converges for a.e.~$\lambda \in S.$ As before, integrating by parts, we find that if $|V(x)| \leq C(1+x)^{-\frac{2}{3}-\epsilon},$ then the function $q_{1}(x, \lambda)$ given by (12) satisfies \[ q_{1}(x, \lambda) = O(x^{-\frac{1}{6}}) \] for a.e.~$\lambda \in S.$ Therefore, for a.e.~$\lambda \in S$ we also have $|q_{1}(x, \lambda)^{2}V(x)| \leq C(1+x)^{-1-\epsilon}.$ This allows us to find the asymptotics of the solutions of the system (13) and then of the original Schr\"odinger equation. As in the previous proof, the Wronskian argument shows linear independence of the solution with asymptotics (18) and its complex conjugate. $\Box$ \\ The proofs of the whole line analogs of the criteria, Theorems 1.9 and 1.10, follow in the same way given Lemma 1.7. \\ Now we come to a final goal of this paper - concrete applications to the preservation of the absolutely continuous spectrum of the Schr\"odinger operators. We first discuss the free case: $U(x)=0.$ We remark that the criterion given by Theorem 1.7 applies trivially since the operator $T_{1}$ in question is just a rescaled Fourier transform and hence satisfies the $L_{2}-L_{2}$ estimate. This gives the stability of the absolutely continuous spectrum of the free Schr\"odinger operators under perturbations by all potentials $V$ satisfying $V(x) \leq C(1+x)^{-\frac{3}{4}-\epsilon}.$ This has been proven in \cite{Kis} using a more direct method rather than Theorem 1.7, which is possible becuase the integral operator is just a Fourier transform in this case. To prove Theorem 1.1, we would like to apply the criterion of Theorem 1.8. This leads to the consideration of the operator $T_{2}$ given by \begin{equation} (T_{2}f)(\lambda) = \frac{i\chi (a,b) (\lambda)}{2\sqrt{\lambda}} \int\limits_{0}^{\infty} \exp \left(-2i\sqrt{\lambda} x +\frac{i}{\sqrt{\lambda}}\int\limits_{0}^{x}V(t)\,dt\right)f(x)\,dx, \end{equation} where $(a,b) \subset (0, \infty)$ and $a > 0,$ $b < \infty.$ We seek to show that the operator $T_{2}$ satisfies the $L_{2}-L_{2}$ bound for every choice of $a,$ $b$ (although the value of the constant in the estimate may of course depend on this choice). Theorem 1.1 will then follow immediately from Theorem 1.8 given that $(a,b)$ is an arbitrary proper subinterval of $(0, \infty).$ For the proof of the $L_{2}-L_{2}$ estimate it suffices to assume that the potential $V$ satisfies $|V(x)| \leq C(1+x)^{-\frac{1}{2}-\epsilon},$ with some $\epsilon >0.$ Let $2\sqrt{\lambda}=k.$ It is clear that it is sufficient to show the $L_{2}-L_{2}$ for the operator $T_{2}'$ given by \begin{equation} (T_{2}'f)(k)= \chi (a,b) \int\limits_{0}^{\infty} \exp \left(-ik x +\frac{2i}{k}\int\limits_{0}^{x}V(t)\,dt\right)f(x)\,dx, \end{equation} The operator $T'_{2}$ looks like a pseudodifferential operator (restricted to the interval $(a,b)$) with a symbol \[ a(k,x) = \exp \left(\frac{i}{k}\int\limits_{0}^{x} V(t)\,dt \right). \] We remind that a symbol $a(k,x)$ belongs to an exotic class $S_{\rho, \sigma}$ if $a(k,x)$ is an infinitely differentiable function satisfying \begin{equation} |\partial^{n}_{k}\partial^{m}_{x}a(k,x)| \leq C_{mn} (1+|x|)^{\sigma n -\rho m} \end{equation} for every $m,$ $n.$ For the symbol classes $S_{\rho, \sigma},$ $1>\rho \geq \sigma \geq 0,$ the $L_{2}-L_{2}$ estimate is well-known (see, e.g., \cite{Stein1}). However, for our purpose, although we may without loss of generality assume that $V \in C^{\infty}$ (absorbing all lack of smoothness into short range correction which is easy to treat), there is no hope in general that an estimate like (29) holds for all integer $m,$ $n.$ Already taking the second derivative in $x,$ we should derivate $V,$ while under our assumptions we have absolutely no control over its derivative. However, Coifman and Meyer \cite{CoMe} have studied the question what is the minimal number of derivative estimates in (29) one has to ask for in order to have an $L_{2}-L_{2}$ bound. In particular, from their results it follows (Theorem 7 on page 30) that it suffices to check (29) for $m,$ $n=0,1$ for some $1> \rho \geq \sigma \geq 0$ in order to ensure an $L_{2}-L_{2}$ bound on $T.$ It is straightforward to check that for our symbol we have these estimates for every $V$ satisfying $|V(x)| \leq C(1+x)^{-\frac{1}{2}-\epsilon}$ (and hence in particular for every $V$ satisfying $|V(x)| \leq C(1+x)^{-\frac{2}{3}-\epsilon}):$ \[ |\partial_{x} a(k,x)| \leq C(1+x)^{-\frac{1}{2} -\epsilon}; \] \[ |\partial_{k} a(k,x)| \leq \frac{2C}{(1-2\epsilon)a} (1+x)^{\frac{1}{2}-\epsilon}; \] \[| \partial_{x}\partial_{k} a(k,x)| \leq C_{11} (1+x)^{-2\epsilon}. \] In particular, $a(k,x) \in ``S_{\frac{1}{2}, \frac{1}{2}}"$ (and even $``S_{\frac{1}{2}+\epsilon, \frac{1}{2}-\epsilon}"$) where quotations mean the reduced number of conditions on the derivatives, i.e. $m$ and $n$ are not greater than $1$ in (29). Hence we have a theorem: \\ \noindent \bf Theorem 3.1. \it The operator $T_{2},$ given by \rm (27), \it satisfies the $L_{2}-L_{2}$ bound $\|T_{2}f\|_{2} \leq C_{2}\|f\|_{2}$ if $V(x)$ verifies $|V(x)| \leq C(1+x)^{-\frac{1}{2}-\epsilon}.$ Also, the bound $\|T_{2}f\|_{\infty} \leq C_{1}\|f\|_{1}$ holds trivially. \\ \rm This theorem together with criterion given by Theorem 1.8 implies Theorem 1.1. \rm We also sketch an alternative proof of Theorem 3.1 which uses only an $L_{2}-L_{2}$ bound for the usual ``exotic" symbol class $S_{\frac{1}{2}, \frac{1}{2}}$ with (29) true for any number of derivatives. For this we need the following lemma: \\ \noindent \bf Lemma 3.2. \it Let $V(x)$ satisfy $|V(x)| \leq C(1+x)^{-\frac{1}{2} -\epsilon};$ then we can represent a function $V(x)$ as a sum $V(x)=V_{1}(x)+ V_{2}(x),$ where $V_{1}(x)$ satisfies \[ |V_{1}^{(m)}(x)|\leq C_{m}(1+x)^{-\frac{1}{2}(m+1)-\epsilon} \] for every integer $m \geq 0,$ and $V_{2}(x)$ is conditionally integrable: $\int_{0}^{x}V(t)\,dt$ converges as $x \rightarrow \infty.$ \\ \bf Proof. \rm Define an increasing sequence $\{a_{n} \}_{n=1}^{\infty}$ by the conditions $a_{0} =1,$ $a_{n}-a_{n-1}=a_{n-1}^{\frac{1}{2}}.$ Let $\xi$ be a $C^{\infty}$ function such that $\xi$ vanishes on the interval $(-\delta, \delta),$ $\delta$ small positive number, and $\xi=1$ outside $(-2\delta, 2\delta).$ Let us define $V_{1}(x)$ by \[ V_{1}(x) = \sum\limits_{n=1}^{\infty}C_{n} \chi (a_{n}, a_{n+1}) \xi \left(\frac{x-a_{n}}{a_{n}^{\frac{1}{2}}}\right)\xi \left( \frac{x-a_{n+1}}{a_{n+1}^{\frac{1}{2}}} \right). \] We choose each $C_{n}$ by a condition that $\int_{a_{n}}^{a_{n+1}} (V-V_{1})(t)\,dt=0$ for every $n.$ It is easy to check that $V_{1} \in C^{\infty}$ and for $x \in (a_{n}, a_{n+1})$ we have \[ |V_{1}^{(m)}(x)| \leq \sup_{x\in(a_{n},a_{n+1})} |V(x)| C_{\xi} a_{n}^{-\frac{1}{2}m},\] where $C_{\xi}$ depends only on $L_{\infty}$ norms of the derivatives of $\xi$ up to the $m$-th order. It is easy to see that $\frac{a_{n+1}}{a_{n}} \stackrel{n \rightarrow \infty}{\longrightarrow}0$ and hence we obtain $|V_{1}^{(m)}(x)| \leq C_{m} (1+x)^{-\frac{1}{2}(m+1) -\epsilon}.$ On the other hand, $\int^{a_{n}}_{0} V_{2}(t)\,dt =0$ for every $n$ and therefore it is easy to see that $V_{2}$ is conditionally integrable and in fact $|\int^{\infty}_{x} V_{2}(t)\,dt|\leq Cx^{-\epsilon}.$ $\Box$ Now let us write the symbol $a(k,x)$ as follows: \[ a(k,x) = i\chi (a,b) (k)\exp \left(\frac{i}{2k}\int\limits_{0}^{x}V_{1}(t) \,dt \right) \exp \left( \frac{i}{k} \int\limits_{0}^{x}V_{2}(t) \,dt \right). \] The first two factors constitute a symbol from the $S_{\frac{1}{2}, \frac{1}{2}}$ class by inspection. Denote this symbol by $a_{1}(k,x).$ For every bounded function $f$ of compact support a pseudodifferential operator $T_{1},$ associated with the symbol $a_{1}(k,x),$ satisfies $\|T_{1}f\|_{2} \leq C_{1}\|f\|_{2}.$ Since $V_{2}$ is a conditionally integrable function, there exists a constant $C_{2}$ such that $|\int_{0}^{x}V_{2}(t)\,dt| \leq C_{2}$ for every $x.$ Write the action of the operator $T$ as \[ Tf(k)= T_{1} \left( \sum\limits^{\infty}_{j=1} \frac{1}{j!} \left( \frac{i}{k} \int\limits_{0}^{x} V_{2}(t)\,dt \right) ^{j} f(x) \right) = \] \[ = \sum\limits_{j=1}^{\infty} \frac{1}{j!}\left(\frac{i}{k}\right)^{j} T_{1} \left( \left( \int\limits_{0}^{x}V(t)\,dt\right)^{j}f(x)\right), \] where the change of the orders of the action of $T_{1}$ and summation is justified by the absolute convergence of the series. Hence, for every bounded function of compact support, we have \[ \| Tf\|_{2} \leq C_{1}\sum\limits_{j=1}^{\infty}\frac{1}{j!a^{j}}C_{2}^{j} \|f\|_{2} \leq C_{1} \exp (\frac{C_{2}}{a}) \|f\|_{2}. \,\,\,\, \Box \] We now consider slowly decaying perturbations of Schr\"odinger operators with periodic potentials. Let $U(x)$ be a periodic, piecewise continuous function of period $T.$ It is a well-known fact (see, e.g., \cite{ReSi}) that the spectrum of the operator \[ \tilde{H}_{U}= -\frac{d^{2}}{dx^{2}}+U(x) \] acting on $L^{2}(-\infty, \infty)$ is purely absolutely continuous of multiplicity two and consisits of bands $[a_{n}, b_{n}],$ n=1,..., where $a_{n} 0,$ defined on the semi-axis with some boundary condition at zero. It is easy to see that the absolutely continuous spectrum of this operator is of multiplicity one and coincides as a set with the absolutely continuous spectrum of the corresponding whole-axis operator. This follows, for example, from the existence of the Bloch solutions, which we will discuss shortly. To prove the stabilty of the absolutely continuous spectrum of periodic Schr\"odinger operators under a new class of slowly decaying perturbations, we would like to apply Theorem 1.8. For this we need to establish an $L_{2}-L_{2}$ bound for an appropriate operator $T_{2}.$ Rather detailed knowledge of the properties of the solutions $\theta (x, \lambda),$ which we choose to be the Bloch functions, is important to achieve this goal. Let us recall the basic facts about the spectrum and the eigenfunctions of the one-dimensional Schr\"odinger operators with periodic potentials; for the missing proofs we refer to \cite{ReSi}. For every band $[a_{n}, b_{n}]$ there exists a real analytic function $\gamma (\lambda),$ which is called quasimomentum, such that $\gamma (\lambda)$ changes monotonically on $(a_{n}, b_{n})$ from $0$ to $\pi$ if $n$ is odd , and from $\pi$ to $0$ if $n$ is even. The derivative $\gamma'(\lambda)$ might only vanish at the points $a_{n}$ or $b_{n}$ and in this case respectively $b_{n-1}=a_{n}$ or $b_{n}=a_{n+1},$ i.e. there is no gap between the bands. For every energy $\lambda \in (a_{n}, b_{n})$ there exists a solution $\theta (x, \lambda),$ which is called a Bloch function, such that \[ \theta (x+T, \lambda) = \exp (i\gamma(\lambda)) \theta (x, \lambda) \] and \[ \theta '(x+T, \lambda) = \exp (i\gamma(\lambda))\theta'(x, \lambda). \] The following lemma holds: \\ \noindent \bf Lemma 3.3. \it For every $\lambda \in (a_{n}, b_{n}),$ the solutions $\theta (x, \lambda)$ and $\overline{\theta}(x, \lambda)$ are linearly independent. \\ \noindent \bf Proof. \rm Indeed, suppose that $\overline{\theta}(x, \lambda)= c\theta(x, \lambda);$ then we must have \[ \overline{\theta}(x+T, \lambda)= \exp (-i\gamma(\lambda))\overline{\theta}(x, \lambda)\] and \[ \overline{\theta} (x+T, \lambda)= c\theta(x+T, \lambda)=c \exp (i\gamma(\lambda))\theta (x, \lambda). \] Together this implies $\sin \gamma (\lambda)=0,$ which is not possible when $\lambda \in (a_{n}, b_{n}).$ $\Box$ \\ \noindent Hence, the Wronskian $W[\theta, \overline{\theta}]= \Im(\theta\overline{\theta}')\neq 0$ when $\lambda \in (a_{n}, b_{n}).$ Next, we remind that the function $\theta (x, \lambda),$ normalized by a condition $\|\theta(x, \lambda) \|_{L_{2}(0,T)} =1,$ is real analytic in $\lambda$ as a function in $L_{2}(0,T)$ when $\lambda$ belongs to $(a_{n}, b_{n}).$ Moreover, we have \\ \bf Lemma 3.4. \it The solution $\theta (x, \lambda)$ is analytic in $\lambda$ when $\lambda \in (a_{n}, b_{n})$ for every fixed $x \in [0,T].$ Moreover, the functions $\theta (x, \lambda),$ $\partial_{x} \theta (x, \lambda),$ $\partial_{\lambda}\theta (x, \lambda)$ and $\partial^{2}_{x \lambda}\theta (x, \lambda)$ are continuous functions in every rectangle $[a'_{n}, b'_{n}] \times [0,T],$ where $a_{n}From the discussion of the properties of Bloch functions it follows that the Wronskian $W[\theta, \overline{\theta}]= \Im (\theta \overline{\theta}')$ is bounded away from zero on $[a'_{n}, b'_{n}].$ Indeed, the Wronskian is continuous (and, in fact, real analytic) inside each band and vanishes only at $a_{n}$ or $b_{n}$ by Lemma 3.3. Let us denote \[ \omega_{n} =\inf_{\lambda \in (a'_{n}, b'_{n})} \Im (\theta \overline{\theta}'). \] Also, the module of the derivative of the quasimomentum, $|\gamma'(\lambda)|,$ is bounded away from zero on $(a'_{n}, b'_{n}).$ Let \[ \eta_{n} = \inf_{\lambda \in (a'_{n}, b'_{n})} |\theta'(\lambda)|. \] Next, we note that the function \[ \sigma (x, \lambda) = \left( \exp (-i\gamma(\lambda) \frac{x}{T}) \theta (x, \lambda) \right)^{2} \] is a periodic function with period $T.$ Let us consider the Fourier series for $\sigma (x, \lambda):$ \[ \sigma (x, \lambda) = \sum\limits_{j} \exp \left( 2\pi ij\frac{x}{T}\right) \hat{\sigma}_{j} (\lambda). \] By $\hat{f}_{j}$ or $\hat{f}(k)$ we denote the Fourier transform of the function $f$ in the discrete and continuous case respectively. >From the properties of $\theta (x, \lambda),$ it follows that $\partial_{x} \sigma (x, \lambda)$ is a continuous function on $[a'_{n}, b'_{n}] \times [ 0,T];$ let us denote \[ \sigma_{n} = \sup_{[a'_{n}, b'_{n}] \times [0,T]} |\partial_{x} \sigma (x, \lambda)|. \] Now note that \[ \|T_{1}f\|_{L_{2}(a'_{n}, b'_{n})} \leq \frac{1}{2\omega_{n}}\|\chi(a'_{n}, b'_{n}) \int\limits_{0}^{\infty} \exp (2i \gamma (\lambda) \frac{x}{T}) \sigma (x, \lambda) f(x)\,dx \|_{L_{2}(a'_{n}, b'_{n})} = \] \[ = \frac{1}{2\omega_{n}}\| \int\limits_{0}^{\infty} \exp (2i \gamma (\lambda) \frac{x}{T}) \sum\limits_{j} \left( \exp (2\pi ij\frac{x}{T} \hat{\sigma}_{j}(\lambda\right) f(x)\,dx \|_{L_{2}(a'_{n}, b'_{n})}. \] Since $|\partial\sigma (x, \lambda)| \leq \sigma_{n}$ for all $x \in [0,T]$ uniformly in $\lambda \in (a'_{n}, b'_{n}),$ it is a standart fact that the Fourier series (in $x$) for $\sigma (x, \lambda)$ converges absolutely. In fact, even for Lipshitz-continuous with power $\alpha > \frac{1}{2}$ function $f(x)$ one has $\sum_{n} |\hat{f}(n)| \leq C\|f\|_{\Lambda_{\alpha}},$ see, for example, \cite{Kat}. Hence, we can change the order of summation and integration in the previous formula. We have \[ \|T_{1}f\|_{L_{2}(a'_{n}, b'_{n})} \leq \frac{1}{2\omega_{n}}\| \sum\limits_{j} \hat{\sigma}_{j}(\lambda)\int\limits_{0}^{\infty} \exp \left( 2i \frac{x}{T}) (\gamma (\lambda) + j\pi) \right) f(x)\,dx \|_{L_{2}(a'_{n}, b'_{n})} \leq \] \[ \frac{1}{2\omega_{n}}\left(\, \int\limits_{a'_{n}}^{b'_{n}} \left( \sum_{j} |\hat{f} \left( \frac{2(\gamma(\lambda)+j\pi)}{T} \right) \hat{\sigma}_{j}(\lambda) | \right)^{2} \, d\lambda \right)^{\frac{1}{2}} \leq \] \[ \leq \frac{1}{2\omega_{n}} \left( \int\limits_{a'_{n}}^{b'{n}} \left( \sum\limits_{j} |\hat{\sigma}_{j}|^{2}(\lambda)\right)\left( \sum\limits _{j} \left|\hat{f}\left( \frac{2(\gamma(\lambda) +j\pi)}{T}\right)\right|^{2} \right) \,d\lambda \right)^{\frac{1}{2}} \leq \] \[ \leq \frac{\|\sigma^{2}(x,\lambda)\|_{L_{2}(0,T)}T^{\frac{1}{2}}}{2\omega_{n}\eta_{n}^{\frac{1}{2}}} \left(\sum\limits_{j} \left|\,\int\limits_{j\pi + \gamma (a'_{n})}^{j\pi +\gamma (b'_{n})} |\hat{f}(y)|^{2} \,dy\right| \right)^{\frac{1}{2}}. \] To obtain the last inequlity we changed the orders of summation and integration and introduced for each $j$ a new variable $y = \frac{2(\gamma(\lambda)+j\pi)}{T}.$ We also note that from Lemma 3.4 it follows that $\sup_{\lambda \in [a_{n}',b_{n}']}\|\sigma^{2}(x, \lambda)\|_{L_{2}(0,T)} \leq C_{n} <\infty.$ Hence, the last expression we obtained is estimated by \[ \frac{C_{n}T^{\frac{1}{2}}}{2\omega_{n} \eta_{n}^{\frac{1}{2}}} \|\hat{f}\|_{2} \] since the function $\gamma$ maps the interval $(a'_{n}, b'_{n})$ into the interval $(0, \pi).$ Therefore, we get the desired bound \[ \|T_{1}f\|_{L_{2}(a'_{n}, b'_{n})} \leq C \|f\|_{2}. \] Now note that we can write the action of $T_{2}$ in a way similar to that of $T_{1}:$ \begin{equation} (T_{2}f)(\lambda) = \frac{\chi (a_{n}', b_{n}')}{W [\theta, \overline{\theta}]} \sum\limits_{j} \hat{\sigma}_{j}(\lambda) \int\limits_{0}^{\infty} \exp \left(2i \frac{x}{T}(\gamma (\lambda) +j\pi)\right) a(\lambda, x) f(x)\,dx, \end{equation} where \[ a(\lambda, x) = \exp \left( \frac{1}{W[\theta, \overline{\theta}]}\int\limits_{0}^{x} |\sigma(t, \lambda)|^{2}V(t)\,dt \right). \] Proceeding with the estimation of the $L_{2}$ norm of the right-hand side of (30) exactly as we did it before, we find that to establish the $L_{2}-L_{2}$ bound, it is sufficient to show that it holds for an operator $\tilde{T},$ defined by \[ (\tilde{T}f)(y) = \int\limits_{0}^{\infty} \exp (iyx) \tilde{a}(y,x)\,dx, \] with a ``symbol" $\tilde{a}(y,x)$ defined by \[ \tilde{a}(y,x)= \chi (2\gamma (a_{n}'), 2\gamma (b_{n}')) \times \] \[ \times \exp\left(\frac{1}{W[ \theta (x,\gamma^{-1}(yT/2)), \overline{\theta}(x,\gamma^{-1}(yT/2)]} \int\limits_{0}^{x}|\theta(t, \gamma^{-1}(yT/2))|^{2}V(t)\,dt\right) \] if $y \in [0, \frac{2\pi}{T}],$ and periodic in $y:$ $\tilde{a}(y+\frac{2\pi}{T},x) = \tilde{a}(y,x).$ The operator $\tilde{T}$ replaces the Fourier transform which appeared in the estimate of $T_{1}.$ We note that the discontinuity in $y$ due to the presence of characteristic functions is artificail. We can always replace the characteristic functions by smooth functions of compact support equal to $1$ when $y \in (2\gamma(a_{n}', \gamma(b_{n}'))$ and vanishing outside $(0, \pi).$ Form the $L_{2}-L_{2}$ bound for such operator would follow the bound for the original one. Now it is straightforward to check, using Lemma 3.4 and properties of the quasimomentum $\gamma (\lambda),$ that we have \[ |\partial_{x}^{\alpha}\partial_{y}^{\beta} \tilde{a}(y, x)| \leq C_{\alpha \beta}(1+x)^{(-\frac{1}{2}-\epsilon)\alpha + (\frac{1}{2}-\epsilon)\beta} \] for all $\alpha,$ $\beta$ taking values in $\{0,1\}.$ Hence, the "symbol'' $\tilde{a}(\lambda, x)$ belongs to the "$S_{\frac{1}{2},\frac{1}{2}}$'' class with the reduced number of conditions on derivatives. By the Coifman-Meyer criterion \cite{CoMe} it follows that the operator $\tilde{T}$ satisfies an $L_{2}-L_{2}$ bound and therefore this bound also holds for $T_{2}.$ $\Box$ \\ To prove the whole-axis analog of Theorem 1.2, Theorem 1.4, we apply the whole axis criterion formulated in Theorem 1.10. The needed $L_{2}-L_{2}$ bounds are obtained similarly to the semi-axis case. As a final remark we note that the results parallel to those we show here also hold for Jacobi matrices case. The role of key Theorem 2.1 is played by its discrete analog ( which in particular follows from considerations in \cite{Kis1}). We plan to further develop this theme in a subsequent publication. \begin{center} \bf Acknowledgment. \end{center} \normalsize I would like to thank Prof. B.~Simon for stimulating discussions and valuable comments. 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