Content-Type: multipart/mixed; boundary="-------------0107090249488" This is a multi-part message in MIME format. ---------------0107090249488 Content-Type: text/plain; name="01-249.keywords" Content-Transfer-Encoding: 7bit Content-Disposition: attachment; filename="01-249.keywords" Sturm-Liouville operators, square summable potentials, singular continuous spectrum. ---------------0107090249488 Content-Type: application/x-tex; name="New1.tex" Content-Transfer-Encoding: 7bit Content-Disposition: inline; filename="New1.tex" \documentstyle[12pt]{article} %%%%%%%%%%%%%%%%%%%%%%%%%%%% \setlength{\textwidth}{7in} \setlength{\textheight}{9in} \setlength{\topmargin}{-.1in} \oddsidemargin=-1cm \language=1 \begin{document} \title{On the coexistence of absolutely continuous and singular continuous components of the spectral measure for some Sturm-Liouville operators with square summable potential. } \date{{\small {\it Moscow State University, Faculty of Computational Mathematics and Cybernetics, Vorob'evy Gory, Moscow, Russia, 119899. e-mail address: saden@cs.msu.su }}} \author{S.A.Denisov} \maketitle \begin{abstract} We prove that there exist some Sturm-Liouville operators with square summable potentials such that the singular continuous component of the spectral measure lies on the positive half-line. \end{abstract} In \cite{Sim}, B.Simon states 15 open problems, one of which says: does $% q(x) $ exist so that $|q(x)|0$, and operator on the positive half-line $Hu=-u^{\prime \prime }+q(x)u$ has singular continuous spectrum for some boundary condition at zero? In the current paper, we construct Sturm-Liouville operator with smooth, square summable potential such that the singular continuous spectrum exists. Our technique is based on the relation between some Sturm-Liouville operators and so-called Krein systems. The problem considered in this article has its analogs in the theory of polynomials orthogonal on the unit circle and in the theory of Krein systems. For polynomials, the answer is well known: Szeg\"o condition on the measure $\int\limits_0^{2\pi }\ln \sigma ^{\prime }(\lambda )d\lambda >-\infty $ is equivalent to $a_n\in \ell ^2$, where $a_n$ are Geronimus coefficients \cite{Ger}. That means even for $\ell ^2$ Geronimus coefficients, the singular continuous component of the measure can coexist with the absolutely continuous one. For the Krein systems, the criterion for ''Szeg\"o condition'' was given in terms of coefficient $A(x)$, provided that it is uniformly square summable \cite{Den1}. The Krein system is continuous analog of difference equations that determine polynomials orthogonal on the unit circle \cite{Atk, Ger}. Let us recall some basic facts from the theory of Krein systems \cite{Kr, Sakh}. Consider $% H(t)=\overline{H(-t)}$ - function summable on each segment $(-r,r)$. {\bf Proposition. (M.G.Krein, \cite{Kr} ) } {\it If for any continuous }$% \varphi (t)${\it \ the following inequality 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 , then, 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 $\Gamma _r(s,t)=\overline{\Gamma _r(t,s)}$ that satisfies the relation \begin{equation} \Gamma _r(t,s)+\int\limits_0^rH(t-u)\Gamma _r(u,s)du=H(t-s),\ (0\leq s,t\leq r). \label{selina} \end{equation} The continuous analogs of polynomials orthogonal on the unit circle are defined by the formulas \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, which is called Krein system \begin{equation} \left\{ \begin{array}{ccc} \frac{dP(x,\lambda )}{dx} & = & i\lambda P(x,\lambda )-\overline{A(x)}% P_{*}(x,\lambda ),\ P(0,\lambda )=1; \\ \frac{dP_{*}(x,\lambda )}{dx} & = & -A(x)P(x,\lambda ),\ P_{*}(0,\lambda )=1, \end{array} \right. \label{sys} \end{equation} where $A(x)=\Gamma_x(0,x)$. On the other hand, system (\ref{sys}) itself defines measure $\sigma$, etc. \cite{Ryb}. Coefficient $A(x)$ is continuous analog of Geronimus parameters $a_n$. These parameters enter the system of difference equations that determine polynomials orthogonal on the unit circle. Consider $A$ to be real valued coefficient. From \cite{Kr}, we know that $e^{-i\lambda x}P(2x,\lambda )=\Phi (x,\lambda )+i\Psi (x,\lambda )$, where $\Phi $ and $\Psi $ are generalized eigenfunctions of equations \[ \Psi^{\prime \prime }-q(x) \Psi +\lambda^2\Psi =0, \Psi(0,\lambda)=0,\, \Psi^{\prime}(0,\lambda)=\lambda, \] \[ \Phi^{\prime \prime }-q_1(x) \Phi +\lambda^2\Phi =0, \Phi(0,\lambda)=1,\, \Phi^{\prime}(0,\lambda)+a(0)\Phi(0,\lambda)=0, \] where $q=a^2+a^{\prime },\,q_1=a^2-a^{\prime },$ and $a(x)=2A(2x)$. It is known \cite{Den} that the spectral measure $\varsigma $ of operator $% Hu=-u^{\prime \prime }+qu,\,u(0)=0$ is related to the spectral measure $% \sigma (t)$ of system (\ref{sys}) by the formula \begin{equation} \varsigma (t)=4\int\limits_0^{\sqrt{t}}\alpha ^2d\sigma (\alpha ). \label{liza} \end{equation} Assume that $\sigma \ $satisfies the following condition $t^2\ln \left( 2\pi \sigma ^{^{\prime }}(t)\right) \in L^1(R).$ {\bf Lemma. } {\it If $A(x)$ is such that } {\it \begin{equation} A(0)=A^{\prime }(0)=0 \label{int} \end{equation} } {\it and $A^{(j)}(x)\in L^1(R^{+}),\ j=0,\ldots 3$, then } {\it \begin{equation} -8A^{^{\prime \prime }}(0)-\int\limits_0^\infty q^2dx=\frac 4\pi \int\limits_{-\infty }^\infty t^2\ln \left( 2\pi \sigma ^{\prime }(t)\right) dt. \label{main} \end{equation} } {\bf Proof}. The idea is quite simple. It is known \cite{Kr, Sakh}, that \begin{equation} \Pi (\lambda )=\exp \left\{ \frac 1{2\pi i}\int\limits_{-\infty }^\infty \frac{1+t\lambda }{\lambda -t}\frac{\ln \left( 2\pi \sigma ^{\prime }(t)\right) }{1+t^2}dt\right\} \label{gus} \end{equation} where $\Pi(\lambda)=\lim_{x\to\infty} P_* (x,\lambda)$ for $\Im \lambda >0$, provided that this limit exists. We let $\lambda =id,d\to +\infty .$ Look at the asymptotics of the integral in the right hand side of (\ref{gus}). It is as follows \begin{eqnarray*} \frac 1{2\pi i}\int\limits_{-\infty }^\infty \frac{1+idt}{id-t}\frac{\ln \left( 2\pi \sigma ^{\prime }(t)\right) }{1+t^2}dt &=&-\frac 1{2\pi }\int\limits_{-\infty }^\infty \ln \left( 2\pi \sigma ^{\prime }(t)\right) \left[ \frac 1d-\frac{t^2}{d^3}+\frac{t^4}{d^3(t^2+d^2)}\right] dt= \\ &&-\frac 1{2\pi d}\int\limits_{-\infty }^\infty \ln \left( 2\pi \sigma ^{\prime }(t)\right) dt+\frac 1{2\pi d^3}\int\limits_{-\infty }^\infty t^2\ln \left( 2\pi \sigma ^{\prime }(t)\right) dt+\overline{o}(d^{-3}). \end{eqnarray*} To get $\overline{o}(d^{-3}),$ we assumed that $t^2 \ln \left( 2\pi \sigma ^{\prime }(t)\right)\in L^1(R) $. This assumption will be true throughout the paper. Consequently, \begin{eqnarray*} \Pi (id) &=&1-\frac 1{2\pi d}\int\limits_{-\infty }^\infty \ln \left( 2\pi \sigma ^{\prime }(t)\right) dt+\frac 1{2\pi d^3}\int\limits_{-\infty }^\infty t^2\ln \left( 2\pi \sigma ^{\prime }(t)\right) dt+ \\ &&\frac 1{8\pi ^2d^2}\left[ \int\limits_{-\infty }^\infty \ln \left( 2\pi \sigma ^{\prime }(t)\right) dt\right] ^2-\frac 1{48\pi ^3d^3}\left[ \int\limits_{-\infty }^\infty \ln \left( 2\pi \sigma ^{\prime }(t)\right) dt\right] ^3+\overline{o}(d^{-3}). \end{eqnarray*} On the other hand, we can compute the asymptotics of $\Pi (id)$ using equations (\ref{sys}) directly. Indeed, we have the system \begin{eqnarray*} P^{^{\prime }} &=&-dP-AP_{*},\ P(0)=1; \\ P_{*}^{^{\prime }} &=&-AP,\ P_{*}(0)=1. \end{eqnarray*} Introduce the function $Q=e^{dx}P.\ $So, \begin{eqnarray*} Q^{^{\prime }} &=&-Ae^{dx}P_{*},\ Q(0)=1; \\ P_{*}^{^{\prime }} &=&-Ae^{-dx}Q,\ P_{*}(0)=1. \end{eqnarray*} For $P_{*},$\ we have integral equation \begin{equation} P_{*}(x,id)=1-\int\limits_0^xA(s)e^{-ds}ds+\int% \limits_0^xP_{*}(s,id)A(s)e^{ds}\int\limits_s^xA(\tau )e^{-d\tau }d\tau ds. \label{iren} \end{equation} The Gronwall Lemma yields $P_{*}(x,id)\rightarrow 1\ $as $d\rightarrow \infty \ $uniformly in $x\in R^{+}.$ Let us iterate (\ref{iren}). We have \begin{eqnarray*} P_{*}(x,id) &=&1-\int\limits_0^xA(s)e^{-ds}ds+\int\limits_0^x\left[ A(s)e^{ds}\int\limits_s^xA(\tau )e^{-d\tau }d\tau \right] \left[ 1-\int\limits_0^sA(s_1)e^{-ds_1}ds_1\right] ds+ \\ &&\int\limits_0^x\left[ A(s)e^{ds}\int\limits_s^xA(\tau )e^{-d\tau }d\tau \right] \left[ \int\limits_0^sP_{*}(s_1,id)A(s_1)e^{ds_1}\int\limits_{s_1}^sA(\tau _1)e^{-d\tau _1}d\tau _1ds_1\right] ds. \end{eqnarray*} Consequently, integrating by parts and using conditions $A(0)=A^{^{\prime }}(0)=0,$ we have \[ \Pi (id)=1+\frac 1d\int\limits_0^\infty A^2(s)ds+\frac 1{2d^2}\left[ \int\limits_0^\infty A^2(s)ds\right] ^2-\frac 1{d^3}A^{^{\prime \prime }}(0)-\frac 1{d^3}\int\limits_0^\infty A^{^{\prime }}(s)^2ds- \] \[ \frac 1{d^3}\int\limits_0^\infty A^4(s)ds+\frac 1{6d^3}\left[ \int\limits_0^\infty A^2(s)ds\right] ^3+\overline{o}(d^{-3}). \] Comparing the coefficients for $d^{-1}$ and $d^{-3}$, we get \begin{eqnarray*} -\frac 1{2\pi }\int\limits_{-\infty }^\infty \ln \left( 2\pi \sigma ^{\prime }(t)\right) dt &=&\int\limits_0^\infty A^2(s)ds, \\ -\frac 1{2\pi }\int\limits_{-\infty }^\infty t^2\ln \left( 2\pi \sigma ^{\prime }(t)\right) dt &=&A^{^{\prime \prime }}(0)+\int\limits_0^\infty A^{^{\prime }}(s)^2ds+\int\limits_0^\infty A^4(s)ds. \end{eqnarray*} Recall that $a(x)=2A(2x),\ q=a^{^{\prime }}+a^2.\ $Therefore, we have \begin{eqnarray*} -\frac 1\pi \int\limits_{-\infty }^\infty \ln \left( 2\pi \sigma ^{\prime }(t)\right) dt &=&\int\limits_0^\infty q(s)ds, \\ -8A^{^{\prime \prime }}(0)-\frac 4\pi \int\limits_{-\infty }^\infty t^2\ln \left( 2\pi \sigma ^{\prime }(t)\right) dt &=&\int\limits_0^\infty q^2(s)ds. \end{eqnarray*} $\Box $ The further construction is as follows. Consider odd function $\sigma (t)=\left( 2\pi \right) ^{-1}t$ on $|t|\leq 1$ and $|t|\geq 2$. On the segment $[1,2]$ it has absolutely continuous component, that satisfies ''local Szeg\"o condition'', and singular continuous component. Construct the corresponding potential $q(x)$ using the Krein approach \cite{Kr}. So, formally, using lemma, we may hope that $q\in L^2(R^{+})$. But the problem is that we obtained (\ref{main}) using smoothness of $A(x)$ and conditions (% \ref{int}). Now we will concentrate on making these arguments precise. {\bf Theorem.} {\it There exists smooth, square summable function $q$, such that the spectrum of operator $Hu=-u^{\prime \prime }+qu,\,u(0)=0$ contains the singular continuous component.} {\bf Proof}. Let $\sigma (t)=\left( 2\pi \right) ^{-1}+\varepsilon (t-1)+\sigma _{sc}(t)$ on the interval $[1,2].$ Here $\varepsilon $ is some positive constant smaller than $\left( 2\pi \right) ^{-1}.$ $\sigma _{sc}(t)$ can be chosen as standard Cantor ladder such that $\sigma _{sc}(1)=0,\ \sigma _{sc}(2)=\left( 2\pi \right) ^{-1}-\varepsilon .$ Together with $% \sigma (t),$ we consider its infinitely smooth approximations $\sigma _n(t).$ To be more specific, we choose $\sigma _n(t)$ in such a way that \[ \sigma _n(t)=\left( 2\pi \right) ^{-1}t \quad {\rm for } \quad |t|>5/2 \quad {\rm and}\,\quad |t|<1/2, \] \[ \sigma _n^{^{\prime }}(t)>\varepsilon /2, \] and \[ \left\| \sigma _n(t)-\sigma (t)\right\| _{C(R)}\rightarrow 0\,\quad {\rm as}% \,\quad n\rightarrow \infty \] The existence of these approximations is obvious. For each $\sigma _n(t)$ and $\sigma (t),$ we can construct the corresponding accelerants $H_n(x)=$ $\int\limits_{-\infty }^\infty \cos (xt)d(\sigma _n(t)-\left( 2\pi \right) ^{-1}t)=$ $2\int\limits_{1/2}^{5/2}\cos (xt)(\sigma _n^{^{\prime }}(t)-\left( 2\pi \right) ^{-1})dt,$ $% H(x)=\int\limits_{-\infty }^\infty \cos (xt)d(\sigma (t)-\left( 2\pi \right) ^{-1}t).$ $H_n(x)$ is from Schwartz space and $H_n(0)=0$. It is due to the infinite smoothness of $\sigma _n$ and equality $\sigma _n(t)=\left( 2\pi \right) ^{-1}t$ for $|t|>5/2.$ Consider $\rho _n(t)=1+2\int\limits_0^\infty H_n(x)\cos \left( xt\right) dx=$ $1+2\pi (\sigma _n^{^{\prime }}(t)-\left( 2\pi \right) ^{-1})=2\pi \sigma _n^{^{\prime }}(t)>\pi \varepsilon $ for all $t.$ Then, recover $A_n(x)$ and $A(x)$ that are related to $\sigma _{n}$ and $\sigma $. To do that we should solve the following integral equations first \begin{eqnarray*} \Gamma _r^{(n)}(x)+\int\limits_0^rH_n(x-s)\Gamma _r^{(n)}(s)ds &=&H_n(x),\ 0\pi \varepsilon ,$ operator $% \Theta ^{(n)}$ is invertible in the classes $L^1(R^{+})$ and $L^\infty (R^{+})$ \cite{Kr1}. In addition, for $\Gamma _{}^{(n)}(x)$, we have an estimate \begin{equation} \left| \Gamma _{}^{(n)}(x)\right| \leq \frac{C_{n,j}}{\left( 1+x\right) ^j}% ,\ j=1,2,\ldots \ . \label{kat} \end{equation} Indeed, for $\widehat{\Gamma }_{}^{(n)}(\omega )=\int\limits_0^\infty \Gamma _{}^{(n)}(x)e^{i\omega x}dx$, we have the formula \begin{equation} \widehat{\Gamma }_{}^{(n)}=\frac{\xi ^{(n)}}{\xi ^{(n)}+1} \label{meri} \end{equation} where $\xi ^{(n)}(\omega )=\exp \left[ \int\limits_0^\infty \gamma ^{(n)}(x)e^{i\omega x}dx\right] -1,$ and $\gamma ^{(n)}(x)$ is defined by the relation $\ln \rho _n(\omega )=\int\limits_{-\infty }^\infty \gamma ^{(n)}(x)e^{i\omega x}dx.$ Function $\gamma ^{(n)}(x)$ is from Schwartz space. Therefore, $\xi ^{(n)}(\omega )$ has derivatives of any order summable on the whole line and $\xi ^{(n)}(\pm \infty )$. From (\ref{meri}), we infer that all derivatives of $\widehat{\Gamma }_{}^{(n)}(\omega )$ are summable on $R$ and $\widehat{\Gamma }_{}^{(n)}(\pm \infty )=0$. $\Gamma _{}^{(n)}(x)$ is bounded because $\Theta ^{(n)}$ is invertible in $L^\infty (R).$ Therefore, we have (\ref{kat}). Consider $\Theta ^{(n)}$ as the operator acting in $L^\infty (R^+)$. It is invertible \cite{Kr1}. Consider also the function \[ \delta _r^{(n)}(x)=\left\{ \begin{tabular}{ll} $\Gamma _{}^{(n)}(x)-\Gamma _r^{(n)}(x),$ & $0r.$% \end{tabular} \right. \] On the interval $[0,r]$, we have \begin{equation} \delta _r^{(n)}(x)+\int\limits_0^rH_n(x-s)\delta _r^{(n)}(s)ds=\delta _r^{(n)}(x)+\int\limits_0^\infty H_n(x-s)\delta _r^{(n)}(s)ds=-\int\limits_r^\infty H_n(x-s)\Gamma _{}^{(n)}(s)ds. \label{jenny} \end{equation} $||\cdot ||_{L^2[0,r]}$-norm of the right hand side is less then $\frac{% C_{n,j}}{(1+r)^j}$ for all integer $j.$ Because \\$\inf spec\left[ \int\limits_0^rH_n(x-s)f(s)ds,f\in L^2(0,r)\right]$ $\geq \inf spec\left[ \int\limits_0^\infty H_n(x-s)f(s)ds,f\in L^2(R^{+})\right] >-1$, we have $% \left\| \delta _r^{(n)}(x)\right\| _{L^2(0,r)}\leq \frac{C_{n,j}}{1+r^j}.$ We can also write (\ref{jenny}) as $\Theta ^{(n)}\delta _r^{(n)}(x)=-\int\limits_r^\infty H_n(x-s)\Gamma _{}^{(n)}(s)ds$ on $[0,r].$ On the interval $x\in [r,\infty ),$ we have \[ \Theta ^{(n)}\delta _r^{(n)}(x)=\int\limits_0^rH_n(x-s)\delta _r^{(n)}(s)ds. \] $\left\| \int\limits_r^\infty H_n(x-s)\Gamma _{}^{(n)}(s)ds\right\| _{L^\infty (0,r)}\leq \frac{C_{n,j}}{\left( 1+r\right) ^j}$, and $% \int\limits_0^r\left| H_n(x-s)\right| \left| \delta _r^{(n)}(s)\right| ds\leq ||H_n||_{L^2(R)}||\delta _r^{(n)}||_{L^2[0,r]}\leq \frac{C_{n,j}}{% \left( 1+r\right) ^j}.$ Consequently, $\left\| \delta _r^{(n)}(x)\right\| _\infty \leq \frac{C_{n,j}}{\left( 1+r\right) ^j}.$ $\Gamma _r^{(n)}(x)$ is continuous in $x$ and $r\ $on $0\leq x\leq r<+\infty .$ Therefore, $% A^{(n)}(r)=\Gamma _r^{(n)}(r)$ is continuous, and $\left| A^{(n)}(r)\right| \leq \frac{C_{n,j}}{\left( 1+r\right) ^j}.$ So, $A^{(n)}(r)\in L^1(R^{+}).$ Notice that $H_n(0)=0.$ Therefore, $A^{(n)}(0)=0$. To see that $\frac{% dA^{(n)}(r)}{dr}$, $\frac{d^2A^{(n)}(r)}{dr^2}$, $\frac{d^3A^{(n)}(r)}{dr^3}% \in L^1(R^{+}),$ it is enough to differentiate equality \[ A^{(n)}(r)+\int\limits_0^r\Gamma _r^{(n)}(s)H_n(r-s)ds=H_n(r) \] successively three times. Then, one should use estimates on $\delta _r^{(n)}(x),\ \Gamma _{}^{(n)}(x),\ H_n(r)$ and its derivatives. For example, using identity $\frac{\partial \Gamma _r(x)}{\partial r}=-\Gamma _r(r)\Gamma _r(r-x)$ \cite{Kr}, we get \begin{equation} \frac{dA^{(n)}(r)}{dr}+\int\limits_0^r\Gamma _r^{(n)}(s)H_n^{^{\prime }}(r-s)ds-A(r)\int\limits_0^rH_n(s)\Gamma _r^{(n)}(s)ds=\frac{dH_n(r)}{dr}. \label{helga} \end{equation} Trivial analysis shows that $\frac{dA^{(n)}(r)}{dr}\in L^1(R^{+}).$ For example, to prove that $\int\limits_0^r\Gamma _r^{(n)}(s)H_n^{^{\prime }}(r-s)ds\in L^1(R^{+}),$ one can write this integral in the following way \[ \int\limits_0^r\Gamma _{}^{(n)}(s)H_n^{^{\prime }}(r-s)ds-\int\limits_0^r\delta _r^{(n)}(s)H_n^{^{\prime }}(r-s)ds. \] From the Young inequality for convolutions, we see that the first term is summable on $R^{+}.$ For the second one, we can use an estimate on$\ \delta _r^{(n)}(s)$ to get \[ \left| \int\limits_0^r\delta _r^{(n)}(s)H_n^{^{\prime }}(r-s)ds\right| \leq \frac{C_{n,j}}{\left( 1+r\right) ^j}. \] In the same way, one can prove that $\frac{d^2A^{(n)}(r)}{dr^2}$, $\frac{% d^3A^{(n)}(r)}{dr^3}\in L^1(R^{+}).$ Because $H_n(r)$ is even, $\frac{dH_n}{% dr}(0)=0$. Therefore, $\frac{dA^{(n)}}{dr}(0)=0.$ Thus, Lemma 1 is applicable. Notice that $\frac{d^2A^{(n)}}{dr^2}(0)=\frac{d^2H_n}{dr^2}(0).$ Consequently, $\left| \frac{d^2A^{(n)}}{dr^2}(0)\right| $ is bounded in $n$. Let $q_n,q$ be potentials corresponding to coefficients $A_n$, $A$ respectively. Notice that $H(x)$ is infinitely smooth, therefore, $A(x)$ and $q(x)$ are infinitely smooth as well. It follows from the Fredholm formulas for resolvent of the integral equation. From Lemma 1, we see that $||q_n||_2$ is bounded in $n$. It is because we have the lower bound from zero for $% \sigma _n^{^{\prime }}(t)$. Our goal is to prove that $q$ is from $L^2(R^{+}).$ Consider the difference\\% $H_n(x)-H(x)=$ $2\int\limits_{1/2}^{5/2}\cos (tx)$ $d\left[ \sigma _n(t)-\sigma (t)\right] $. Integrating by parts, we have \[ H_n(x)-H(x)=2\int\limits_{1/2}^{5/2}x\sin (tx)\left[ \sigma _n(t)-\sigma (t)\right] dt. \] $\left\| \sigma _n(t)-\sigma (t)\right\| _{C(R)}\rightarrow 0$ as $% n\rightarrow \infty $ so, $H_n(x)\rightarrow H(x)$ uniformly on each finite segment. In the same way, one can show that $H_n^{^{\prime }}(x)\rightarrow H^{^{\prime }}(x)$ on each finite segment. Fredholm formulas for the resolvent of integral equation implies that $\Gamma _r^{(n)}(s,x)\rightarrow \Gamma _r(s,x)$ uniformly in $0\leq x,s\leq r\leq M$ for each fixed $M$ as $% n\rightarrow \infty $. Consequently, $A_n\rightarrow A$ uniformly on each finite segment. From (\ref{helga}), we infer that $A_n^{^{\prime }}\rightarrow A^{^{\prime }}$ (and so $q_n\rightarrow q$ ) uniformly on each finite interval as well. Boundedness of $||q_n||_2$ leads to inclusion $q\in L^2(R^{+}).$ Therefore, due to (\ref{liza}), we constructed potential $q\in L^2(R^{+})$, such that the corresponding Sturm-Liouville operator with Dirichlet boundary condition at zero has singular continuous component that coexists with absolutely continuous one. {\bf Remark.} The construction of potential, that admits the estimate suggested by B.Simon, is one step further in this direction. It requires very detailed study of the integral equation (\ref{selina}) that determines the resolvent kernel for each $r>0$. We should also mention that though potential $q$ from the Theorem is smooth and square summable, generally speaking, it might not be decaying at the infinity. To prove that decaying potentials from $L^2(R^{+})$ class can admit coexistence of s.c. and a.c. components of the spectrum, one should use some identities that involve first derivative of $q$ also. Then, using more or less the same technique, it might be possible to prove that properly chosen measure leads to potential from $W^{1,2}(R^{+})$ class. That means $q(x)$ decays at the infinity. \begin{center} {\bf Acknowledgment.} \end{center} Author is grateful to Courant Institute of Mathematical Sciences, New York University, where most of this work was done. \begin{thebibliography}{99} \bibitem{Atk} F.V. Atkinson Discrete and continuous boundary value problems. Academic Press, New York, 1964. \bibitem{Den} S.A. Denisov On the application of some M.G.Krein's results to the spectral analysis of Sturm-Liouville operators, (to appear in the J. Math. Anal. Appl. ). \bibitem{Den1} S.A. Denisov To the spectral theory of Krein systems (to appear in Intergal Equations Operator Theory ). \bibitem{Ger} Ya.L. Geronimus polynomials orthogonal on the segment and unit circle (Russian ). Fizmatgiz, Moscow, 1958. \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{Kr1} M.G. Krein Integral equations on the half-line with a kernel depending on the difference of the arguments (Russian ), Uspehi Mat. Nauk (in Russian ), {\bf 13}, N5 (83), 3-120, (1958). \bibitem{Kr2} M.G. Krein Theory and applications of Volterra operators in Hilbert space. Translations of Mathematical Monographs, {\bf 24}, AMS, Providence, R.I., 1970. \bibitem{Kr3} M.G. Krein On determination of the potential of a particle from its $S$--function (Russian ), Dokl. Akad. Nauk SSSR, {\bf 105}, 433-436, (1955). \bibitem{Ryb} A.M. Rybalko On the theory of continual analogues of orthogonal polynomials, Teor. Funktsii, Funktsional. Anal. i Prilozhen., {\bf 3}, 42-60, (1966). \bibitem{Sakh} L.A. Sakhnovich Spectral theory of a class of canonical systems, Funct. Anal. Appl., {\bf 34}, N2, 119-129, (2000). \bibitem{Sim} B. Simon Schr\"odinger operator in the 21-st century, (preprint ). \end{thebibliography} \end{document} ---------------0107090249488--