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Sturm-Liouville operators, square summable potentials, 
 singular continuous spectrum.
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\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 $|q(x)|<C/(1+x)^{1/2+\varepsilon }$, where $%
C,\varepsilon >0$, 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<x<r, \\
\Gamma _r^{}(x)+\int\limits_0^rH(x-s)\Gamma _r(s)ds &=&H(x),\ 0<x<r,
\end{eqnarray*}
where $\Gamma _r(x)=\Gamma _r(0,x)$, $\Gamma _r^{(n)}(x)=\Gamma _r^{(n)}(0,x)
$. From the Krein results \cite{Kr, Kr3, Kr2}, we infer that these two
equations are solvable and $\Gamma _r^{(n)}(x),\Gamma _r^{}(x)-$ continuous
in $x$ and $r.$ Let $A_n(x)=\Gamma _x^{(n)}(x),\ A(x)=\Gamma _x(x).$ We want
to show that Lemma 1 is applicable to systems with coefficients $A_n(x).$
Together with $\Gamma _r^{(n)}(x),$ consider $\Gamma _{}^{(n)}(x)$ which is
the solution of the following equation 
\[
\Gamma _{}^{(n)}(x)+\int\limits_0^\infty H_n(x-s)\Gamma
_{}^{(n)}(s)ds=H_n(x),\ 0<x<\infty . 
\]

It is so-called Wiener-Hopf equation. Consider operator $\Theta ^{(n)}:$ $%
\Theta ^{(n)}f=f(x)+\int\limits_0^\infty H_n(x-s)f(s)ds.$ Because even
function $H_n(x)\in L^1(R)$ and $\rho _n(t)>\pi \varepsilon ,$ 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 
\[
\delta _r^{(n)}(x)=\left\{ 
\begin{tabular}{ll}
$\Gamma _{}^{(n)}(x)-\Gamma _r^{(n)}(x),$ & $0<x<r,$ \\ 
$0,$ & $x>r.$%
\end{tabular}
\right. 
\]

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.

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