Content-Type: multipart/mixed; boundary="-------------0007210912660" This is a multi-part message in MIME format. ---------------0007210912660 Content-Type: text/plain; name="00-299.keywords" Content-Transfer-Encoding: 7bit Content-Disposition: attachment; filename="00-299.keywords" Sturm-Liouville operators, Dirac systems, spectral analysis ---------------0007210912660 Content-Type: application/x-tex; name="Art7-15.tex" Content-Transfer-Encoding: 7bit Content-Disposition: inline; filename="Art7-15.tex" %% This document created by Scientific Word (R) Version 2.0 %% Starting shell: mathart1 \documentstyle[12pt]{article} %%%%%%%%%%%%%%%%%%%%%%%%%%%% \setlength{\textwidth}{7in} \setlength{\textheight}{9in} \setlength{\topmargin}{-.1in} \oddsidemargin=-1cm \language=1 \begin{document} \title{On the application of one M.G.Krein's result to the spectral analysis of Sturm-Liouville operators. } \date{April, 2000 } \author{S.A.Denisov \thanks{ Moscow State University. Mathematical Subject Classification 34B24, 34L20. The work was partially supported by RFBR Grant N00-15-96104.}} \maketitle \begin{abstract} Discovered by M.G.Krein analogy between polynomials orthogonal on the unit circle and generalized eigenfunctions of certain differential systems is used to obtain some new results in spectral analysis of Sturm-Liouville operators. \end{abstract} {\bf Section A.} In this section we remind some results obtained by M.G.Krein in his famous article \cite{Kr}. In this paper author develops the ''theory of polinomials, orthogonal on the positive half-line''. And this polinoms are constructed from exponents rather than from the powers of independent variable. It's well known that there are many ways to construct the system of orthogonal polinomials on the unit circle. One of them is to start from the moments matrix. This way was chosen by M.G.Krein to obtain his results for positive half-line. Let's assume that $H(t)=\overline{H(-t)}$ - function summable on each segment $(-r,r)$. {\bf Proposition.} {\it If for any continuous }$\varphi (t)${\it \ the following inequaliy holds } \begin{equation} \int\limits_0^r\left| \varphi (s)\right| ^2ds+\int\limits_0^r\int\limits_0^rH(t-s)\varphi (t)\overline{\varphi (s)}% dtds\geq 0 \label{asr1} \end{equation} {\it for each }$r>0${\it , so, and in this case only, there exists the non-decreasing function }$\sigma (\lambda )${\it \ }$(\lambda \in R,\sigma (0)=0,\sigma (\lambda -0)=\sigma (\lambda )${\it \ }$)${\it , such that } \begin{eqnarray*} \int\limits_{-\infty }^\infty \frac{d\sigma (\lambda )}{1+\lambda ^2} &<&\infty \\ \int\limits_0^t(t-s)H(s)ds &=&\int\limits_{-\infty }^\infty (1+\frac{% i\lambda t}{1+\lambda ^2}-e^{i\lambda t})\frac{d\sigma (\lambda )}{\lambda ^2% }+(i\gamma -\frac{sign(t)}2)t \\ && \end{eqnarray*} {\it \ } {\it where }$\gamma ${\it \ is real constant.} If in addition we presume that the equality in (\ref{asr1}) is possible for $% \varphi =0$ only then the Hermit kernel $H(t-s),\ (0\leq t,s\leq r)$ has Hermit resolvent $\Gamma _r(s,t)=\overline{\Gamma _r(s,t)}$ that satisfies the relation \[ \Gamma _r(t,s)+\int\limits_0^rH(t-u)\Gamma _r(u,s)du=H(t-s)\ (0\leq s,t\leq r) \] The continuous analogues of polinomials orthogonal on the unit circle are defined by the formula \begin{eqnarray*} P(r,\lambda ) &=&e^{i\lambda r}(1-\int\limits_0^r\Gamma _r(s,0)e^{-i\lambda s}ds), \\ P_{*}(r,\lambda ) &=&1-\int\limits_0^r\Gamma _r(0,s)e^{i\lambda s}ds,\ r\geq 0. \end{eqnarray*} Using the well known properties of resolvents we obtain the following system \begin{equation} \begin{array}{ccc} \frac{dP(r,\lambda )}{dr} & = & i\lambda P(r,\lambda )-\overline{A(r)}% P_{*}(r,\lambda ), \\ \frac{dP_{*}(r,\lambda )}{dr} & = & -A(r)P(r,\lambda ), \end{array} \label{sys3} \end{equation} where $A(r)=\Gamma _r(0,r).$ {\bf Proposition.} {\it For each finite }$f(x)\in L^2(R^{+})${\it \ we have the following equality} \[ \left\| f\right\| _2^2=\int\limits_{-\infty }^\infty \left| F_P(\lambda )\right| ^2d\sigma (\lambda ),{\rm where\, } F_P(\lambda )=\int\limits_0^\infty f(r)P(r,\lambda )dr. \] Consequently we have the isometric mapping $U_P$ from $L^2(R^{+})$ into $% L^2(\sigma ,R).$ {\bf Theorem.} {\it The mapping }$U_P${\it \ is unitary if and only if the following integral diverges \\(equals to }$-\infty ${\it \ )} \begin{equation} \int\limits_{-\infty }^\infty \frac{\ln \sigma ^{^{\prime }}(\lambda )}{% 1+\lambda ^2}d\lambda . \label{int2} \end{equation} {\bf Theorem.} {\it \ The following statements are equivalent} {\it (1) The integral {\rm (\ref{int2})} is finite.} {\it (2) At least for some }$\lambda $, $\Im \lambda >0$ {\it the integral } \begin{equation} \int\limits_0^\infty \left| P(r,\lambda )\right| ^2dr \label{int3} \end{equation} {\it converges.} {\it (3) At least for some }$\lambda ${\it \ (}$\Im \lambda >0${\it \ ) the function }$P_{*}(r,\lambda )${\it \ is bounded.} {\it (4) On any compact set in the open upper half-plane integral {\rm (\ref {int3})} converges uniformly. That is equivalent to the existence of uniform limit }$\Pi (\lambda )=\lim_{r\rightarrow \infty }P_{*}(r,\lambda ).${\it \ } It's easy to verify that in cases $A(r)\in L^1(R^{+}),\ A(r)\in L^2(R^{+})$ the conditions (1)-(4) are satisfied. What is more, in the first case measure $\sigma $ is continuously differentiable with certain estimates for its derivative. Consider $E(r,\lambda )=e^{-i\lambda r}P(2r,\lambda )=\Phi (r,\lambda )+i\Psi (r,\lambda ).$ Let $E(-r,\lambda )=\overline{E(r,\lambda )}=\Phi (r,\lambda )-i\Psi (r,\lambda ).$ From (\ref{sys3}) we infer that \begin{eqnarray*} \frac{d\Phi }{dr} &=&-\lambda \Psi -a(r)\Phi +b(r)\Psi ,\ \Phi (0,\lambda )=1; \\ \frac{d\Psi }{dr} &=&\lambda \Phi +b(r)\Phi +a(r)\Psi ,\ \Psi (0,\lambda )=0. \end{eqnarray*} where $a(r)=2\Re A(2r),\ b(r)=2\Im A(2r).$ {\bf Proposition.} {\it The mapping }$U_E:f(r)\rightarrow F_E(\lambda )=\int\limits_{-\infty }^\infty f(r)E(r,\lambda )dr,${\it \ defined on the finite functions }$f(r)\in L^2(R),${\it \ generates the unitary operator from }$L^2(R)${\it \ onto }$L^2(\sigma ,R)$. The trivial case $a=b=0$ yields $E(r,\lambda )=e^{i\lambda r},\ \Psi (r,\lambda )=\sin (r\lambda ),\ \Phi (r,\lambda )=\cos (r\lambda ).$ In case when $H(t)$ is real, the function $\sigma (\lambda )$ is odd. Consequently $b(r)=0.$ Assuming that $H(t)$ is absolutely continuous, we have that $\Phi $ and $\Psi $ are solutions of the equations \begin{equation} \begin{array}{ccc} \Psi ^{^{\prime \prime }}-q\Psi +\lambda ^2\Psi =0, & \Psi (0)=0, & \Psi ^{^{\prime }}(0)=\lambda ; \\ \Phi ^{^{\prime \prime }}-q_1\Phi +\lambda ^2\Phi =0, & \Phi (0)=1, & \Phi ^{^{\prime }}(0)+a(0)\Phi (0)=0, \end{array} \label{bis} \end{equation} where $q_1(x)=a^2(x)-a^{^{\prime }}(x)$ and $q(x)=a^2(x)+a^{^{\prime }}(x).$% \vspace{1cm} {\bf Section B.} Consider the Sturm-Liouville operator on the half-line with Dirichlet boundary condition at zero \begin{equation} l(u)=-u^{^{\prime \prime }}+qu,\ u(0)=0. \label{e1} \end{equation} Let's assume that real-valued $q(x)$ admits the following representation $% q(x)=a^2(x)+a^{^{\prime }}(x)$ where $a(x)$ is absolutely continuous function on the half-line. That means that $a(x)$ is the solution of the Ricatti equation. Consider also the corresponding differential Dirac-type system (see \cite{LS} or \cite{Atk}) \begin{equation} \left\{ \begin{array}{cc} \Phi ^{^{\prime }}(x,\lambda )= & -\lambda \Psi (x,\lambda )-a(x)\Phi (x,\lambda ) \\ \Psi ^{^{\prime }}(x,\lambda )= & \lambda \Phi (x,\lambda )+a(x)\Psi (x,\lambda ) \end{array} \right. , \label{s1} \end{equation} where $\Phi (0,\lambda )=1,\Psi (0,\lambda )=0.$ From the result stated in Section A it follows that the spectral measure $% \rho (\lambda )$ of problem (\ref{e1}) is connected with the spectral measure $\hat{\sigma} (\lambda )$ of system (\ref{s1}) \footnote{% See the definition of spectral measure for the differential system in \cite {Atk}. Here the function $\hat{\sigma} (\lambda )$ is connected with $\sigma (\lambda)$ from the section A by the relation: $\hat\sigma(\lambda )=2\sigma(\lambda ).$} by the following relation \begin{equation} \rho (t)=2\int\limits_0^{\sqrt{t}}\alpha ^2d\hat{\sigma} (\alpha ). \label{m1} \end{equation} From this and results stated in Section A we can infer one very simple but significant corollary {\bf Corollary. } {\it If }$q(x)${\it \ is real-valued function such that } {\it \begin{equation} \sup\limits_{x\in R}\int\limits_x^{x+1}\left| q(s)\right| ^2ds<\infty , \end{equation} } {\it the improper integral }$W(x)=\int\limits_x^\infty q(s)ds${\it \ exists and satisfies the condition $W(x)\in L^2(R^{+}),$ then the absolutely continuous part of spectrum of operator $H_h$, generated by differential expression $l(u)=-u^{^{\prime \prime }}+qu$ and boundary condition$\ u(0)=hu^{^{\prime }}(0),\ (h\in R\cup \infty \ )$ fills the whole positive half-line.} Proof. Let's consider (\ref{s1}) with $a(x)=-W(x).$ The spectral measure of system (\ref{s1}) with chosen $a(x)$ has the needed property. Consequently the Sturm-Liouville operator with potential $q^{*}(x)=a^{^{\prime }}+a^2$ and Dirichlet boundary condition also has the a.c. component which fills the whole positive half-line. But initial potential $q(x)$ differs from $q^*(x)$ by $L^1(R^{+})$ term only. Consequently, Kuroda's theorem \cite{Kur} guarantees that operator $H_h$ has the needed property for $h=0$. But it means that this statement is true for any $h$ since the essential support of a.c. component doesn't depend on $h.$ That follows, for example, from the subordinate solutions theory.$\Box$ In the next theorems of this section we will show how Krein's results will help to establish asymptotics for generalized eigenfunctions and analyze the spectrum of some Sturm-Liouville operators. {\bf Theorem 1. } {\it If }$q(x)${\it \ is real-valued function such that } {\it \begin{equation} \sup\limits_{x\in R}\int\limits_x^{x+1} \min\{0,q(s)\}ds>-\infty, \label{aus} \end{equation} } {\it the improper integral }$W(x)=\int\limits_x^\infty q(s)ds${\it \ exists and satisfies the condition} $\left| W(x)\right| \leq \frac \gamma {x+1},\ (00$ so that $\left( \int\limits_0^xu_1^2(s,\lambda )ds\right) \left( \int\limits_0^xu_2^2(s,\lambda )ds\right) ^{-\zeta }\rightarrow \infty $ for any two linearly independent solutions $u_1,u_2$ of equation from (\ref{e1}). The refined subordinacy theory \cite{Rem1}, \cite{JL} yields that there is $% \eta (\lambda )>0$ so that $\left( D_\eta \rho \right) (\lambda ^2)=% \overline{\lim }_{\varepsilon \rightarrow 0}$ $\frac{\rho (\lambda ^2-\varepsilon ,\lambda ^2+\varepsilon )} {\left( 2\varepsilon \right) ^\eta }=0.$ Consequently $\rho $ gives zero weight to every $\Omega \subset R^{+}$ with $\dim \Omega =0.$ On the other hand we will prove that $u_1,u_2$ might be unbounded at the infinity only on the $\lambda $ set with the zero Hausdorff dimension. Consider $Q(x,\lambda ):\ Q^{^{\prime }}=-Ae^{-i\lambda x}\overline{Q(x)},\ Q(0)=1.$ So \begin{eqnarray*} Q(x) &=&1-\int\limits_0^xA(s)e^{-i\lambda s}\overline{Q(s)}% ds=1+\int\limits_0^x\left( \int\limits_s^\infty A(\tau )e^{-i\lambda \tau }d\tau \right) ^{^{\prime }}\overline{Q(s)}ds= \\ && \end{eqnarray*} \begin{equation} \left. 1+\left( \int\limits_s^\infty A(\tau )e^{-i\lambda \tau }d\tau \right) \overline{Q(s)}\right|_{s=0}^{s=x}-\int\limits_0^x\left( \int\limits_s^\infty A(\tau )e^{-i\lambda \tau }d\tau \right) \overline{% Q^{^{\prime }}(s)}ds, \label{der} \end{equation} where $\lambda $ is such that integral $\int\limits_0^\infty A(\tau )e^{-i\lambda \tau }d\tau $ converges. The last term in (\ref{der}) can be rewritten as $\int\limits_0^x\left( \int\limits_s^\infty A(\tau )e^{-i\lambda \tau }d\tau \right) A(s)e^{i\lambda s}Q^{^{}}(s)ds.$ Finally we get \[ Q(x)=J(x)-\int\limits_0^x\left( \int\limits_s^\infty A(\tau )e^{-i\lambda \tau }d\tau \right) A(s)e^{i\lambda s}{Q^{^{}}(s)}ds, \] where $J(x)=C(\lambda)+o(1)\overline{Q(x)}$. From argument used in \cite {Rem2} (Theorem 1.3, 1.4 ) it follows that the set $\Xi $ of $\lambda $ for which $\left( \int\limits_s^\infty A(\tau )e^{i\lambda \tau }d\tau \right) A(s)e^{-i\lambda x}\notin L^1(R^{+})$ has zero Hausdorff dimension. But one can easily verify that this fact leads to the boundedness of $Q(x)$ at the infinity for $\lambda \notin \Xi .$ From formula (\ref{red}) it follows that generalized eigenfunctions $u(x,\lambda )$ which correspond to the Dirichlet boundary condition at zero are bounded at the infinity for $\lambda \notin \Xi .$ At the same way we can prove that the linear independent solution $% v(x,\lambda )$ is bounded at the infinity if $\lambda \notin \Upsilon $ $% (\dim \Upsilon =0\ ).$ Consequently results obtained by Stolz \cite{St} guarantee that the support of singular measure has the zero Hausdorff dimension. Meanwhile as we have shown above spectral measure gives the zero weight to any set of zero Hausdorff measure. So the singular spectrum is absent.$\Box$ {\bf Remark 1.} We could have got rid of the condition (\ref{aus}) and prove the absence of positive eigenvalues by making use of Hardy inequality for the equation $Q(x)=\int\limits_x^\infty A(s)e^{-i\lambda s}\overline{Q(s)}ds$ which follows from $Q^{^{\prime }}=-Ae^{-i\lambda x}\overline{Q}$ and $% Q(\infty )=0 $. Really, it would mean the absence of nonzero eigenvalues for differential system (\ref{s1}). By (\ref{m1}) we would have the absence of positive eigenvalues for (\ref{e1}). {\bf Remark 2.} Conditions of Theorem 1 are often fulfilled for potentials that oscillate at the infinity (see \cite{Behn1}, \cite{Behn2} and bibliography there ). We used method different from common ones such as modified Pr\"ufer transform, $I+Q$ asymptotic integration and so on. {\bf Remark 3.} If we consider potentials which satisfy the estimate $\left| q_\varepsilon (x)\right| \leq \frac \varepsilon {x+1}$, then the point spectrum may occur on $[0,\frac{4\varepsilon ^2}{\pi ^2}]$ (see \cite{vNW}, \cite{Rem2} ) for any $\varepsilon>0 $. The situation for potentials considered in Theorem 1 is different. Really, the von Neumann -Wigner example $q_{vNW}=8\frac{\sin 2x}{x} + O(\frac{1}{x^2})$ shows that the condition \begin{equation} \left| \int\limits_x^\infty q_\gamma (s)ds\right| \leq \frac \gamma {x+1} \label{ppp} \end{equation} doesn't guarantee the absence of the positive eigenvalues for $\gamma $ large enough. Meanwhile, for small $\gamma \ (0< \gamma <1/4)$ the singular spectrum disappears on the whole positive half-line. \footnote{% It's interesing to find out is the constant $1/4$ optimal or not.} {\bf Remark 4.} From the proof of the Theorem 1 we can infer that the asymptotics for generalized eigenfunctions may not be true on the set of zero Hausdorff dimension only. And this statement holds even without the constraint (\ref{aus}). In \cite{RS} the following simple statement is proved {\sl If }$q(x)${\sl \ is continuous function which admits the representation } $q=\frac{\partial W}{\partial x},\ W\in L^1(1,\infty )${\sl \ then equation% } $-\varphi ^{^{\prime \prime }}+q\varphi =k^2\varphi ${\sl \ doesn't have nontrivial square integrable solutions for }$k\neq 0.$ {\sl What is more, if }$\left| W(x)\right| \leq C|x|^{-1-\varepsilon }${\sl \ then for any $k\neq 0 $ there exists the solution }$\varphi (x,k)${\sl \ of the same equation which has the following asymptotic} $\left| \varphi (x,k)-e^{-ixk}\right| \leq C(k) \left| x\right| ^{-\varepsilon }{\ ,x\geq 1,}${\sl where } $% |C(k)|0.$ We improved this result in power scale in some extent using method from \cite {Rem2}. {\bf Theorem 2. } {\it If }$q(x)${\it \ is real-valued function such that the improper integral }$W(x)=\int\limits_x^\infty q(s)ds${\it \ exists and satisfies the condition} $W(x)\in L^p(R^{+})\cap L^2(R^{+}),\ (10.$ {\bf Remark 2. }In paper \cite{Den} it was considered the dependence of absence of singular component on certain interval on the local smoothness of Fourier transform. {\bf Remark 3. }From the corollary and method used in theorem 3 it follows that if $q(x)${\it \ is such that (A) and (B) are satisfied, }$\widehat{q(\omega )% }=\widehat{q(0)}+\widehat{\varphi (\omega )}${\it \ where }$\left| \widehat{% \varphi (\left| \omega \right| )}\right| \leq C|\omega |^{1/2+\varepsilon }\ ${\it \ in the vicinity of zero for some positive }$\varepsilon ${\it \ and }% $\frac{\widehat{\varphi (\left| \omega \right| )}}{|\omega |+1}\in L^2(R),$% {\it \ then the essential support of spectral measure of operator is }$% R^{+}. $\vspace{1cm} {\bf Section C.} In this section we will discuss the dependence of spectral measure $\sigma (\lambda )$ on the coefficient $A(x)$ of system (\ref{sys3}). In fact function $A(x)$ plays the role of sequence $a_n$ for polynomials orthogonal on the unit circle (see \cite{Ger} ). We will see that for the Dirac-type systems the situation is not so simple. The basic reason is the possible oscillation of $A(x)$. The following Lemma is true {\bf Lemma. }If measurable bounded function $A(x)$ is such that \[ \int\limits_x^\infty e^{-s}A(s)ds=\overline{o}(e^{-x}),\ A(x)e^x\int\limits_x^\infty A(s)e^{-s}ds\in L_1(R^{+}) \] then conditions (1)-(4) from section A are satisfied. Proof. Consider system (\ref{sys3}) with $\lambda =i$. If $P=e^{-x}Q$ then we have \begin{equation} \begin{array}{ccc} Q^{^{\prime }} &=&-Ae^xP_{*} \\ P_{*}^{^{\prime }} &=&-Ae^{-x}Q \end{array} \label{ff1} \end{equation} Consequently \begin{eqnarray*} P_{*}(x,i) &=&1-\int\limits_0^xA(s)e^{-s}ds+\int\limits_0^xA(s)e^sP_{*}(s,i)\int% \limits_s^\infty A(\xi )e^{-\xi }d\xi ds-\int\limits_x^\infty A(\xi )e^{-\xi }d\xi \int\limits_0^xA(s)e^sP_{*}(s,i)ds \\ && \end{eqnarray*} And now it suffices to use the standard argument. Let $M_n=\max_{x\in [0,n]}|P_{*}(x,i)|=|P_{*}(x_n,i)|$ So $M_n\leq 1+C+M_n\int\limits_0^\infty |A(s)|e^s\left| \int\limits_s^\infty A(\xi )e^{-\xi }d\xi \right| ds+\overline{o}(1)M_n$ If the whole integral in the last formula is less then $1$ then $M_n$ is bounded. Otherwise we should start to solve the equations (\ref{ff1}) not from zero but from some other point $x_0$ for which this condition is satisfied. {\bf Example. }$A(x)=(x^2+1)^{-\alpha }\sin (x^\beta )$ where $\alpha ,\beta >0.$ One can easily verify that conditions of Lemma are satisfied if $2\alpha +\beta /2>1$. Meanwhile $A(x)\in L_2(R^{+})$ if and only if $\alpha >1/4$. Nevertheless for nonpositive $A(x)$ with bounded derivative the condition $% A(x)\in L^2(R^{+})$ is necessary for (1)-(4) from Section A to be true. {\bf Proposition. }If one of the conditions (1)-(4) is true, $A(x)\leq 0$ and $A^{^{\prime }}(x)$ is bounded then $A(x)$ is from $L^2(R^{+}).$ Proof. Really, since $A(x)\leq 0$ both $P$ and $Q$ are not less then 1. Consequently if one of (1)-(4) holds then $P_{*}(x,i)$ is bounded and as it follows from (\ref{ff1}) $\int\limits_0^x|A(s)|e^s\int\limits_s^x\left| A(\xi )\right| e^{-\xi }d\xi ds$ is bounded as well. But we have the inequality \[ \int\limits_0^x|A(s)|e^s\int\limits_s^x\left| A(\xi )\right| e^{-\xi }d\xi ds\geq e^{-1}\int\limits_0^{x-1}|A(s)|\int\limits_s^{s+1}\left| A(\xi )\right| d\xi ds\geq C\int\limits_0^{x-1}\left| A(s)\right| ^2ds. \] which concludes the proof of proposition. The latter inequality follows from the boundedness of $A^{^{\prime }}(x).$ $\Box$ Function $A(x)$ from the example above with $\alpha =1/4,\ 1<\beta \leq 3/2\ $ satisfies the conditions of Lemma (consequently (1)-(4) holds ), has bounded derivative but is not from $L^2(R^{+}).$ The explanation is that this function is not nonpositive. We would like to conclude the paper with two open problems the first of which is much more difficult then the second one. {\bf Open problems}. 1. Prove that Theorem 2 holds for $W\in L^2(R^{+})$. 2. Prove that the presence of a.c. component on the half-line pertains to those potentials which Fourier transform is from $L^{2}$ near the zero. Specifically, if $q$ admits the Fourier transform $\widehat{q}$ such that $% \widehat{q}\in L^{2}_{loc}(R)$ and $\frac{\widehat{q}}{|\omega |+1}$$\in L^2(R), $ then the a.c. part of the spectrum fills the whole positive half-line. This conjecture seems reasonable at least with some additional constraints since we can represent $\widehat{q}=\widehat{q_1}+\widehat{q_{2}}$. Where the $\widehat{q_1}$ is localized near the zero and is from $L^2$ so the methods of paper \cite {Den} works. The other function $\widehat{q_{2}}$ is such that Theorem 3 can be applied.% \vspace{1cm} Acknowledgment. Author is grateful to C.Remling for attention to this work. e-mail address: saden@cs.msu.su \begin{thebibliography}{99} \bibitem{Kr} M.G.Krein Continuous analogues of propositions on polynomials orthogonal on the unit circle, Dokl. Akad. Nauk SSSR, {\bf 105}, 637-640, (1955). \bibitem{LS} B.M.Levitan, I.S.Sargsjan Introduction to the Spectral Theory, Trans. Math. Monographs, 39, Amer. Math. 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