%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % % % % % % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \input amstex \documentstyle{amsppt} \magnification=1200 \hcorrection{.25in} \topmatter \title The Krein spectral shift and \hbox{rank one} perturbations of spectra \endtitle \author A. Poltoratski\thanks Research at MSRI is supported in part by NSF grant DMS-9022140 \endthanks\endauthor \address MSRI, 1000 Centennial Drive, Berkeley, California 94720\endaddress \email agp\@msri.org \endemail \abstract We use recent results on the boundary behavior of Cauchy integrals to study the Krein spectral shift of a rank one perturbation problem for self-adjoint operators. As an application, we prove that all self-adjoint rank one perturbations of a self-adjoint operator are pure point if and only if the spectrum of the operator is countable. We also study pairs of pure point operators unitarily equivalent up to a rank one perturbation and give various examples of rank one perturbations of singular spectra. \endabstract \subjclass 47A55, 30E20\endsubjclass \keywords The Krein spectral shift, rank one perturbations, self-adjoint operators, Clark's measures \endkeywords \endtopmatter \document \tolerance=1000 \font\twelvecmmi=cmmi12 \mathchardef\lesha="7F1F \def\bigchi{\textfont15=\twelvecmmi\lesha} \pageno1 \define\nec{\underset{ z @>>\ngtr> x }\to {-\kern -5pt -\kern -5pt-\kern -5pt-\kern -5pt\longrightarrow}} \define\Clos{\operatorname{Clos}} \define\supp{\operatorname{supp}} \define\im{\operatorname{Im}} \define\re{\operatorname{Re}} \define\dist{\operatorname{dist}} \define\C{\Cal} \define\a{\Cal A} \define\Cb{\Cal B}\define\Cc{\Cal C} \heading 1. Introduction. \endheading In this paper we use well-known connections between one-dimensional perturbation theory and complex analysis. We apply recent results on the boundary behavior of Cauchy integrals to study the spectral properties of rank one perturbations of self-adjoint operators. Section 2 discusses the structure of a family of the spectral measures of rank one perturbations of a self-adjoint operator. We establish the relations between such families and the characteristic function of the operator. Section 3 is devoted to the properties of the Krein spectral shift of a rank one perturbation problem. We study spectral shifts in terms of the boundary behavior of the resolvent function. In Section 4 we prove that all self-adjoint rank one perturbations of a self-adjoint operator are pure point if and only if the spectrum of the operator is countable. Section 5 analyzes relative spectral properties of two self-adjoint pure point operators unitarily equivalent up to a rank one perturbation. In Section 6 we give an example of ``the absence of mixed spectra''. We construct a family $A+\lambda(\cdot,\varphi)\varphi$ of rank one perturbations of a self-adjoint operator such that the operators corresponding to the coupling constants $\lambda\in [0;1] $ are singular continuous and the operators corresponding to the coupling constants $\lambda\in \Bbb R\setminus [0;1]$ are pure point. \remark{\bf Acknowledgments} I would like to thank Nikolai Makarov for asking some of the questions discussed in this paper. I am also grateful to Aleksei Aleksandrov, Dmitri Jakubovich and Fedor Nazarov for useful discussions and comments, and to the Mathematical Sciences Research Institute for hospitality while this work was completed. \endremark \heading 2. Families $\Cal M_\varphi$ and rank one perturbations of self-adjoint operators. \endheading We will denote by $\Cal M( \hat\Bbb R)$ the space of Borel complex measures on $ \hat\Bbb R=\Bbb R\cup{\infty}$ with the norm $$||\mu||=\frac 1\pi\int\limits_{-\infty}^\infty \frac {d|\mu(t)|}{1+t^2}+|\mu|(\infty).$$ We will also use the notation $\Cal M_+( \hat\Bbb R)$ for the subset consisting of nonnegative measures. Let $\phi$ be a nonconstant analytic function in the upper half plane $\Bbb C_+$ such that $|\phi|\leq 1$. Then for any $\alpha$ from the unit circle $\Bbb T$ function $\frac{\alpha+\phi}{\alpha-\phi}$ has positive real part in $\Bbb C_+$ . Thus there exists a measure $\mu_\alpha\in \Cal M_+( \hat\Bbb R)$ such that its Poisson integral satisfies $$ (\Cal P\mu_\alpha)(x+iy)= \frac 1\pi \int\limits_{-\infty}^{\infty} \frac{y d\mu_\alpha(t)}{(x-t)^2+y^2} +y\mu_\alpha(\infty) =\re \frac{\alpha+\phi}{\alpha-\phi} .\tag 2.1$$ We will denote by $\Cal M_\phi( \hat\Bbb R)$ the family of such measures $\{\mu_\alpha\}_{\alpha\in\Bbb T}\ $ corresponding to function $\phi$. If $\phi$ is defined in the unit disk one can replace the Poisson integral in (2.1) by the Poisson integral in the disk and consider the analogous family $\Cal M_\phi(\Bbb T)$ consisting of positive measures on the unit circle $\Bbb T$. Families $\Cal M_\phi $ possess many interesting properties, see [A1], [A2], [C], [P1], [P2] and [Sa]. As was shown by Clark in [C], if $\phi$ is an inner function in the unit disk $\Bbb D$ then $\Cal M_\phi(\Bbb T)$ is the system of the spectral measures of all unitary rank one perturbations of the model contraction with the characteristic function $\phi$. By adopting the argument of Clark to the case of the upper half-plane, one can make a similar connection between rank one perturbations of self-adjoint operators and families $\Cal M_\phi( \hat\Bbb R)$. Here we present a different way to establish this connection. If $A_0$ is a bounded cyclic self-adjoint operator and $\varphi$ is its cyclic vector, then we can consider the family of rank one perturbations $$A_\lambda=A_0+ \lambda(\cdot , \varphi)\varphi , $$ $\lambda\in\Bbb R . $ Let $\nu_\lambda$ be the spectral measure of $\varphi$ for $A_\lambda$. Since $A_\lambda$ is bounded, $\supp\nu_\lambda\subset\Bbb R$ is compact. Thus the Cauchy transform of $\nu_\lambda$ $$\Cal K\nu_\lambda(z)=\frac 1\pi\int\limits_{-\infty}^{\infty}\frac {d\nu_ \lambda(t)}{t-z}=\frac 1\pi((A_\lambda-z )^{-1}\varphi , \varphi)\tag 2.2$$ converges for any $z\in\Bbb C_+$. The relation for the resolvents$$ (A_0 - z)^{-1} -(A_\lambda-z)^{-1} =\left[\lambda(\cdot , \varphi)\left((A_\lambda -z)^{-1}\varphi\right)\right] (A_0-z)^{-1},$$ for any $z\in\Bbb C_+$, gives us $$ \Cal K\nu_\lambda(z)=\frac {\Cal K\nu_0(z)}{1+\pi\lambda \Cal K\nu_0(z)},\tag 2.3 $$ see [A]. Since $\nu_0$ is a nonnegative Borel measure on $ {\Bbb R}$, $-i\Cal K\nu_0$ is an analytic function with a nonnegative real part in $\Bbb C_+$: $$\re -i\Cal K\nu_0=\Cal P\nu_0.$$ Thus $$-i\Cal K\nu_0=\frac{1+\phi}{1-\phi}\tag 2.4$$ for some bounded analytic function $\phi,\ |\phi|\leq1$ in $\Bbb C_+$. Hence by (2.3) $$ (\Cal P\nu_\lambda)(x+iy)=\re -i\Cal K\nu_\lambda(x+iy)=\re \frac{\frac{1-\phi}{1+\phi}} {1+i\pi\lambda\frac{1-\phi}{1+\phi}} = c\re\frac{\beta-\phi}{\beta+\phi} \tag2.5$$ where $\beta=\frac{1+i\pi\lambda}{1-i\pi\lambda} $, $c=\frac1{1+\pi^2\lambda^2} $. Thus $$ \Cal P\nu_\lambda=c\Cal P\mu_\beta$$ where $\{\mu_\beta\}_{\beta\in\Bbb T}=\Cal M_\phi( \hat\Bbb R)$. Conversely, if $\phi$ is an analytic function in $\Bbb C_+$ with $|\phi|\leq 1$ we can consider the operator $A_\mu$ of multiplication by $ z$ in $L^2(\mu)$ where $\mu=\mu_1\in\Cal M_\phi( \hat\Bbb R)$. Reversing the above argument we can show that for any $\alpha\in\Bbb T$ the measure $\mu_\alpha\in\Cal M_\phi$ is a spectral measure for the rank one perturbation $A+\lambda(\cdot,1)1$ where $\lambda=\frac i\pi\frac {1-\alpha}{1+\alpha}$. Therefore, each family $\Cal M_\phi( \hat\Bbb R)$ is a family of spectral measures of rank one perturbations of some self-adjoint operator. Each measure from $\Cal M_+(\hat\Bbb R)$ belongs to a family $\Cal M_\phi$ for some $\phi$ ($\phi$ can be found from (2.1) uniquely up to a M\"obius transform). One can also say that two given measures $\mu$ and $\nu$ from $\Cal M(\hat\Bbb R)$ belong to the same family $\Cal M_\phi$ iff there exist operator $A$ and its rank one perturbation $A_\lambda=A+\lambda(\cdot,\varphi)\varphi$ such that $\mu$ and $\nu$ are the spectral measures of $\varphi$ for $A$ and $A_\lambda$. Indeed, as we proved above, such $A$ and $A_\lambda$ exist iff $c_1\mu,c_2\nu\in\Cal M_\psi$ for some function $\psi$ and positive constants $c_1$ and $c_2$. Suppose $c_1\mu=\mu_\alpha, c_2\nu=\mu_\beta$ for some $\mu_\alpha,\mu_\beta\in\Cal M_\psi$. Then $\mu,\nu\in\Cal M_\phi$ for $\phi=b\circ\psi$ where $b$ is the M\"obius transform of the unit disk $\Bbb D$ such that $|b'(\alpha)|=1/c_1, |b'(\beta)|=1/c_2$, see [A2]. Further relations between the resolvent functions of the rank one perturbations of a self-adjoint operator are discussed in [A], [D], [M-P], [S-W], [S] and [R-J-L-S]. \remark{\bf Remark} The same argument works for the families of unitary operators (in particular for those studied by Clark [C]). Let $\Cal U_1$ be a unitary cyclic operator. Let probability measure $\mu_1\in\Cal M_+(\Bbb T)$ be the spectral measure of some cyclic vector $v$ for $\Cal U_1$. Then we can consider the family of unitary rank one perturbations of $\Cal U_1$ : $$\Cal U_\alpha= \Cal U_1+(\alpha-1)(\cdot , \Cal U_1^{-1}v)v, $$ $\alpha\in\Bbb T$. For the resolvents we have $$(\Cal U_1-z)^{-1}-(\Cal U_\alpha-z)^{-1}= (\Cal U_1-z)^{-1}[\alpha(\cdot , \Cal U_1^{-1}v)v](\Cal U_\alpha-z)^{-1} .$$ Denote by $\mu_\alpha$ the spectral measure of $v/||v||$ for $U_\alpha$. Let $ K\mu_\alpha$ and $ P\mu_\alpha$ be the standard Cauchy and Poisson integrals of measure $\mu_\alpha$ in the unit disk $\Bbb D$. Since $ K\mu_\alpha=((\Cal U_\alpha-z)^{-1}v , v)$ for $z\in\Bbb D$, we have that $$ K\mu_\alpha=\frac{\alpha K\mu_\alpha} {1+ (\alpha-1) K\mu_1} .\tag 2.6 $$ If we consider an analytic function $\theta$ in $\Bbb D$ such that $ |\theta|\leq1, \theta(0)=0$ and $$ P\mu_1=\re\frac{1+\theta}{1-\theta}$$ in $\Bbb D$, then $$ K\mu_1=\frac1{1-\theta}$$ and by (2.6) $$ P\mu_\alpha=2\re K\mu_\alpha-1= 2\re\left[\frac{\frac\alpha{1-\theta}}{1+\frac{\alpha-1}{1-\theta}}-\frac12 \right]= \re \frac{\alpha+\theta}{\alpha-\theta}.$$ Thus $\{\mu_\alpha\}_{\alpha\in\Bbb T} =\Cal M_\theta (\Bbb T)$. \endremark As we mentioned before, this result was obtained in [C] for rank one unitary perturbations of the model contraction $T_\theta=\Cal S P_\theta $ , where $\theta$ is an inner function, $\Cal S:f\mapsto zf$ is a shift operator in the Hardy space $ H^2$ and $ P_\theta$ is the orthogonal projector from $ H^2$ onto the model space $\theta^*( H^2)= H^2\circleddash\theta H^2$. Any singular cyclic unitary operator can be represented as a rank one perturbation of $T_\theta$ for some inner $\theta$. Conversely, if the operator $\Cal U_1$ from the last remark is singular, then $\ \Cal U_1-(\cdot , \Cal U_1^{-1}v)v $ is a $C_0$ completely nonunitary contraction with the characteristic function $\theta$. \heading 3. The Krein spectral shift. \endheading In this section we discuss the notion of the Krein spectral shift in the case of rank one perturbations. For basic results and definitions in this area we refer to [K], [A-D], [M-P] and [S]. Let $A$ be a bounded self-adjoint operator, $\varphi $ its cyclic vector. Let $\mu\in\Cal M_+(\hat\Bbb R)$ be the spectral measure of $\varphi$ for $A$. Since $\mu$ is a nonnegative measure, its Cauchy transform has nonnegative imaginary part. Therefore for each $\lambda\in\Bbb R$ the function $1+\pi\lambda\Cal K$ admits the following representation: $$1+\pi\lambda\Cal K\mu=\exp[\Cal Ku]\tag 3.1$$ fore some real valued function $u\in L^\infty(\Bbb R), || u||\leq\pi$ with compact support. If $\lambda\geq 0$ then function $u$ can be chosen to satisfy $$u=\arg(1+\pi\lambda\Cal K\mu)\tag 3.2$$ where $\arg$ stands for the principal branch of argument taking values in $(-\pi;\pi]$ (we assume that $\Cal K\mu$ is defined a. e. on $\Bbb R$ by its angular boundary values). In this case we will have $0\leq u\leq \pi$. Similarly if $\lambda<0$ then $u$ can be chosen to satisfy (3.2) with $\arg$ taking values in $[-\pi;\pi)$. In this case we will have $0\leq u\leq -\pi$. \definition{Definition} Such $u$ is called the Krein spectral shift of the perturbation problem $\left(A\mapsto A+\lambda(\cdot,\varphi)\varphi\right)$. \enddefinition The Krein spectral shift is usually defined in more general settings of compact perturbations $\left(A\mapsto A+K\right)$, see [K] and [M-P]. When $K$ is one-dimensional the general definition can be reduced to the one given above, see [M-P]. Let $\nu$ be the spectral measure of $ \varphi$ for $A+\lambda(\cdot,\varphi)\varphi$. Even though measure $\nu$ does not appear in (3.1), it is uniquely determined by $\mu$ and $\lambda$ and satisfies a similar equation. \proclaim{Lemma 3.1} Measures $\mu$, $\nu$ and the Krein spectral shift $u$ satisfy $$1+\pi\lambda\Cal K\mu=\exp[\Cal K(u)]=\left[1-\pi\lambda\Cal K\nu\right]^{-1}.\tag3.3$$ \endproclaim \demo{Proof} Apply formulas (3.1) and (2.3) with $\nu_0=\mu$ and $\nu_\lambda=\nu$. $\blacktriangle$ \enddemo Since in this paper we will mostly deal with the spectral measures of operators, it will be more convenient for us to call $u$ the phase shift of the pair of measures $(\mu,\nu)$ for the coupling constant $\lambda$. If $\lambda=1$ we will call $u$ the phase shift of the pair $(\mu;\nu)$. One can use formula (3.3) as an alternative definition of the phase shift. Note that in this definition $u$ determines $\mu$ and $\nu$ uniquely. This reflects the fact that among the spectral measures of $A$ and $A+\lambda(\cdot,\varphi)\varphi$ we always choose those corresponding to the vector $\varphi$. Any function $u\in L^\infty(\Bbb R), 0\leq u \leq\pi$ with compact support is the Krein spectral shift of some perturbation problem (the phase shift of some pair of measures). Indeed, for any $\lambda>0$ one can consider a nonnegative measure $\mu\in\Cal M(\hat\Bbb R)$ with compact support satisfying (3.1). Then $u$ is the Krein spectral shift of the perturbation problem $\left(A_\mu\mapsto A_\mu+\lambda(\cdot,1)1\right)$. Similarly each $u\in L^\infty(\Bbb R), 0\geq u \geq-\pi$ with compact support is also the Krein spectral shift of some perturbation problem (with $\lambda<0$). As was shown in the previous Section, there exist operator $A$ and its cyclic vector $\varphi$ such that $\mu$ and $\nu$ are spectral measures for $\varphi$ of $A$ and $A+(\cdot,\varphi)\varphi$ iff both measures belong to the same family $\Cal M_\phi$ for some function $\phi$ analytic in $\Bbb C_+$. One can show that if we take $\mu_\alpha,\mu_\beta\in\Cal M_\phi$ $\alpha=e^{i\theta_1} ,\beta=e^{i\theta_2}, 0\leq \theta_1<\theta_2<2\pi$ where $\phi$ is inner then the phase shift $u$ for the pair $(\mu_\alpha,\mu_\beta)$ can be defined as $$u=\pi\chi_{\{x\in\Bbb R|\phi(x)=e^{i\theta}, \theta_1<\theta<\theta_2\}}$$ where $\chi_E$ denotes the indicator function of the set $E$ (we again assume that $\phi$ is defined a. e. on $\Bbb T$ by its angular boundary values). As before, let $u$ be the phase shift of the pair of measures $(\mu,\nu)$ and $\phi$ be a characteristic function: $\mu,\nu\in\Cal M_\phi$. Then $u$ has the following properties. \proclaim{Lemma 3.2} Let $K\subset \Bbb R$ be a measurable set. Then the following conditions are equivalent i) the restrictions of $\mu$ and $\nu$ on $K$ are singular, ii) $|u|$ is equal to $0$ or $\pi$ a. e. on $K$, iii) $|\phi|=1$ a. e. on $K$. \endproclaim \demo{Proof} For i)$\Leftrightarrow $ii) see [M-P] or [S]; for i)$\Leftrightarrow$iii) see [C].$\blacktriangle$ \enddemo We will denote by $Qu$ the conjugate Poisson integral of $u$: $$Qu(x+iy)=-\re \Cal Ku(x+iy)=\frac1\pi\int\limits_{-\infty}^{\infty} \frac{x-t}{(x-t)^2+y^2}u(t)dt.$$ It is well-known that if $v\in L^\infty (\Bbb R)$ with a compact support then the difference $$ Qv(x+iy)- \int\limits_{\Bbb R\setminus (x-y;x+y)}\frac {v(t)dt} {x-t}\tag 3.4$$ is $O(1)$ as $x+iy\underset\ngtr\to\rightarrow x_0$ for any $x_0\in\Bbb R$ and $o(1)$ for any Lebesgue point $x_0$ of $v$, see [G]. Let $E\subset\Bbb R$ be a measurable set, $\mu\in \Cal M(\Bbb R)$. We will denote $$p.\ v.\ \int\limits_Ed\mu(x)= \lim_{\epsilon\to 0}\int\limits_{E\setminus (-\epsilon;\epsilon)}d\mu(x).$$ We will denote by $\mu^s$ the singular component of measure $\mu$. The definition of the phase shift and relation (3.4) imply the following \proclaim{Lemma 3.3} i) For $\mu^s$-a.e $x$ $$p.\ v.\ \ \int\limits_\Bbb R \frac{u(x+t)dt}{t}= \infty.\tag 3.5$$ For $\nu^s$-a.e $x$ $$p.\ v.\ \ \int\limits_\Bbb R \frac{u(x+t)dx}{t}=- \infty.\tag 3.6$$ ii) There exist $\alpha,\beta\in\Bbb T$ such that for $\mu^s$-a.e $x$ $$\phi(z) \nec \alpha;$$ For $\nu^s$-a.e $x$ $$\phi(z) \nec \beta.$$ \endproclaim \demo{Proof}Follows from formulas (2.1) and (3.3). For i) see [S]; for ii) see [C].$\blacktriangle$ \enddemo \example{\bf Example 3.4} Let $x,y\in\Bbb R, x-\infty).$$ We will write that $$p.\ v.\ \int\limits_E f(x)dx<\infty \ \ (>-\infty)$$ if $$\liminf_{\epsilon\to 0}\int\limits_{E\setminus (-\epsilon;\epsilon)}d\mu(x)<\infty\ \ \ (\limsup>-\infty).$$ \proclaim{Corollary 3.6} Let $\sigma,\gamma\in\Cal M_+{\Bbb R})$ be singular measures with compact supports. Define holomorphic function $F$ in $\Bbb C_+$ as $$F=\frac{1+\Cal K\sigma}{1+\Cal K\gamma}.$$ Let $K$ be a measurable subset of $\Bbb R$. Then $|F|<\infty$ $\sigma$-a. e. on $K$ iff the restriction of $\sigma$ on $K$ is absolutely continuous with respect to $\gamma$. \endproclaim \demo{Proof} Let $\sigma=f\gamma+\eta$ where $f\geq 0, f\in L^1(\gamma),\eta\perp\gamma$. Then $$F=\frac{ 1+\Cal K\sigma}{1+\Cal K\gamma}=\frac{1+\Cal K\sigma}{1+\Cal K(\sigma+\gamma)}\times \frac{1+\Cal K(\sigma+\gamma)}{1+\Cal K\gamma}.\tag 3.7$$ By Theorem 3.5 the first fraction has finite nonzero limits $|\sigma|$-a. e. Since $$ \frac{1+\Cal K(\sigma+\gamma)}{1+\Cal K\gamma}=\left[\frac{1+\Cal K\gamma}{1+\Cal K((1+f)\gamma+\eta)}\right]^{-1},$$ by Theorem 3.5 the second fraction in (3.7) has finite limits $|\gamma|$-a. e. and tends to infinity $\eta$-a. e. Thus $|F|<\infty$ $\sigma$-a. e. on $K$ iff $\eta(K)=0$. $\blacktriangle$ \enddemo \proclaim{Lemma 3.7} Let $u_1$ and $u_2$ be the phase shifts of the pairs of measures $(\mu_1,\nu_1)$ and $(\mu_2,\nu_2)$ respectively. Let $K\subset\Bbb R$ be a measurable set. Then i) $$p.\ v.\ \int\limits_\Bbb R (u_1(t+x)-u_2(t+x))\frac{dt}{t}<\infty$$ for $\mu_1$-a. e. $x\in K$ iff the restriction of $\mu_1^s$ on $K$ is absolutely continuous with respect to $\mu_2$; $$p.\ v.\ \int\limits_\Bbb R (u_1(t+x)-u_2(t+x))\frac{dt}{t}>-\infty$$ for $\nu_1$-a. e. $x\in K$ iff the restriction of $\nu_1^s$ on $K$ is absolutely continuous with respect to $\nu_2$; ii) if $\mu_2^s$-a. e. $x\in K$ is a Lebesque point of $u_1-u_2$ and $$p.\ v.\ \int\limits_\Bbb R (u_1(t+x)-u_2(t+x))\frac{dt}{t}= f(x)<\infty$$ for $\mu_2^s$-a. e. $x\in K$ then the restriction of $\mu_1^s$ on $K$ is equal to the restriction of $e^{f}\mu_2^s$ on $K$; if $\nu_2^s$-a. e. $x\in K$ is a Lebesque point of $u_1-u_2$ and $$p.\ v.\ \int\limits_\Bbb R (u_1(t+x)-u_2(t+x))\frac{dt}{t}= f(x)>-\infty$$ for $\nu_2^s$-a. e. $x\in K$ then the restriction of $\nu_1^s$ on $K$ is equal to the restriction of $e^{-f}\nu_2^s$ on $K$. \endproclaim \demo{Proof} Notice that by the defining formula (3.3) $$\exp\left[\Cal K(u_1-u_2)\right]=\frac{1+\Cal K\mu_1}{1+\Cal K\mu_2} =\frac{1+\Cal K\nu_2}{1+\Cal K\nu_1}.$$ Now Corollary 3.6 and formula (3.4) imply part $i)$. Since the difference (3.4) is $o(1)$ at Lebesgue points of $u_1-u_2$, part $ii)$ follows from part $i)$ and Theorem 3.5.$\blacktriangle$ \enddemo \proclaim{Lemma 3.8 [M-P]} Measure $\mu$ has a point mass at $x$ iff $$\int\limits_{x-1}^{x+1} (\pi\chi_{(x;x+1)}-u(y))\frac {dy}{y-x}<\infty.\tag3.8$$ Measure $\nu$ has a point mass at $x$ iff $$\int\limits_{x-1}^{x+1} (\pi\chi_{(x-1;x)}-u(y))\frac {dy}{x-y}<\infty.\tag3.9$$ \endproclaim \demo{Proof} As was shown in Example 3.5 measure $\mu_0$ corresponding to the phase shift $u_1=\pi\chi_{(x;x+1)}$ has a point mass at $x$. It is left to apply statement i) of Lemma 3.7 to $u_1, u_2=u$, and $K=\{x\}$. The statement can also be proved directly, see [M-P]. $\blacktriangle$\enddemo \remark{\bf Remark} In terms of the characteristic function $\phi$ a point mass of $\mu_\alpha\in\Cal M_\phi$ can be described in the following way, see [C], [A-C] or [Sa]. For any $x\in\Bbb R$ $\mu_\alpha(x)>0$ iff $\phi(z)\to \alpha $ as $z \underset \ngtr \to \longrightarrow x$ and $\phi$ has a finite nontangential derivative $\phi'(x)$ at $x$. In terms of other measures from the family $\Cal M_\phi$, $\mu_1(x)>0$ iff $\Cal K\mu_{-1}(z)\to 0$ as $z\underset \ngtr \to \longrightarrow x$ and $(x-y)^{-1}\in L^2(\mu_{-1})$, see [C], [S-W] or [S]. \endremark \heading 4. On the stability of the absence of continuous spectra \endheading Numerous examples given in [A-D], [D], [S] and [R-J-L-S] show that for a general self-adjoint operator the property of not having a continuous part is not stable under rank one perturbations. For instance [D] contains an example of a self-adjoint operator $A$ and its cyclic vector $\varphi$ such that $A$ is pure point but $A+\lambda(\cdot,\varphi)\varphi$ are singular continuous for all real $\lambda\neq 0$. Here we prove that in order to have $only$ pure point rank one perturbations the operator must have a very ``thin'' spectrum. We will denote the spectrum of $A$ by $\sigma(A)$. \proclaim{ Theorem 4.1} Let $A$ be a self-adjoint operator. Then the following two conditions are equivalent: 1) All self-adjoint rank one perturbations of $A$ are pure point, 2) $\sigma (A)$ is countable. \endproclaim \remark{\bf Remark} We assume that $A$ itself is included in the set of all its rank one perturbations. In 1) ``all'' can not be replaced with ``almost all'' in any reasonable sense: it is not difficult to show that if $\mu$ is a standard singular Cantor measure on $[0;1]$ then $A_\mu+\lambda(\cdot,\varphi)\varphi$ is pure point for a.e. $\lambda\in\Bbb R$ for any cyclic vector $\varphi$. \endremark To prove Theorem 4.1 we will need the following two Lemmas. \proclaim{Lemma 4.2} Let $F\subset\Bbb R, \ |F|=0$ be a closed set, $F=\Bbb R\setminus\bigcup_{n=1}^{\infty} I_n$ where $I_n=(x_n;y_n)$ are disjoint open intervals. Then one can choose two disjoint sets of positive integers $L$ and $M$ such that for any $y\in F\setminus \{x_1,x_2,x_3,...\}$ $$\int\limits_ {(y;y+1)\cap \bigcup\limits_{n\in L}I_n}\frac{dx}{x-y} =\int\limits_ {(y;y+1)\cap \bigcup\limits_{n\in M}I_n}\frac{dx}{x-y}=\infty.\tag4.1$$ and for any $y\in F\setminus \{y_1,y_2,y_3,...\}$ $$\int\limits_ {(y-1;y)\cap \bigcup\limits_{n\in L}I_n}\frac{dx}{y-x} =\int\limits_ {(y-1;y)\cap \bigcup\limits_{n\in M}I_n}\frac{dx}{y-x}=\infty.\tag4.2$$ \endproclaim \demo{Proof} Since $|F|=0$ we can choose $\{I_{n^{1}_l}\}_{l=1}^\infty$ in such a way that $$\int\limits_ {(y;y+1)\cap \bigcup\limits_{l=1}^\infty I_{n^1_l}}\frac 1{x-y}dx>1$$ for any $y\in F\setminus \{x_1,x_2,x_3,...\},$ $$\int\limits_ {(y-1;y)\cap \bigcup\limits_{l=1}^\infty I_{n^1_l}}\frac 1{y-x}dx>1$$ for any $y\in F\setminus \{y_1,y_2,y_3,...\}$ and the set of the endpoints of the intervals $I_{n_l^1}$ has no cluster points except $\pm\infty$. Indeed, for each $n\in\Bbb N$ we can choose a finite number of intervals covering more than a half (in measure) of the interval $[n;n+1]$ . Then we can put $\{I_{n^1_l}\}$ to be the set of all chosen intervals for all $n\in\Bbb N$. Since the set of the endpoints of $I_{n^1_l}$ has no cluster points in $\Bbb R$, the set $(0;1)\setminus \bigcup_{l=1}^\infty I_{n^1_l}$ is a union of disjoint closed intervals. Therefore in a similar way we can choose intervals $I_{n^2_l} \subset \Bbb R\setminus \bigcup_{l=1}^\infty I_{n^1_l}$ such that $$\int\limits_ {(y;y+1)\cap \bigcup\limits_{l=1}^\infty I_{n^2_l}}\frac 1{x-y}dx>1$$ for any $y\in F\setminus \{x_1,x_2,x_3,...\},$ $$\int\limits_ {(y-1;y)\cap \bigcup\limits_{l=1}^\infty I_{n^2_l}}\frac 1{y-x}dx>1$$ for any $y\in F\setminus \{y_1,y_2,y_3,...\}$ and the set of the endpoints of the intervals $I_l^2$ has no cluster points except possibly the ends of the intervals $I_l^1$ and $\pm\infty$. Proceeding in this way we will obtain $ \{I_{n^{k}_l}\}_{l=1}^\infty$ for $k=1,2,3...$. After that we can put $L=\{n^{2l-1}_k \}_{l,k\in\Bbb N}$ and $M=\{n^{2l}_k\}_{l,k\in\Bbb N}$. $\blacktriangle$\enddemo \proclaim{Lemma 4.3}Let $F\subset (0;1)$ be a closed set, $|F|=0$, $F=(0;1)\setminus\bigcup I_n$ where $I_n=(x_n;y_n)$ are disjoint open intervals. Let $\{z_n\}_{n>0}$ be a sequence of points such that $z_n\in I_n$ and $y_n-z_n<|I_n|^2$. Let $M$ be as in the previous Lemma. Then there exists $N\subset M$ such that for all $y\in F$ i) if $y=y_n$ for some $n\in\Bbb N$ then $$\int\limits_{(y;y+1)\cap\bigcup\limits_{n\in N}I_n}\frac {dx}{x-y}=+\infty.\tag4.3$$ ii) if $y\neq x_n,y_n$ for any $n$ and $E=\bigcup\limits_{n\in N}I_n\cup\bigcup\limits_{n\in\Bbb N}[z_n;y_n]$, then $$p.\ v.\ \int\limits_{(-1;1)}\chi_E(y+x)\frac{dx}{x}<+\infty.\tag4.4$$ %i.e. there exists a sequence $\{\epsilon_n\}_{n>0}$ such that $\epsilon_n\to 0$ and %$$\liminf_{n\to\infty}\int\limits_{E\setminus(y-\epsilon_n;y+\epsilon_n)}\frac{dx}{x-y}<\infty.\tag4.4$$ \endproclaim \demo{\bf Proof} WLOG we can assume that the intervals $I_n$ are enumerated in such a way that $|I_1|\geq|I_2|\geq...$. Let $\delta_l $ be a small positive constant (the exact choice of $\delta_l$ will be made later). For each $l\in\Bbb N$ choose a sequence of intervals $\{I_{n_k^l}\}_{k>0}$ such that for any $k\in\Bbb N$ $\ \ n_k^l\in M$, $I_{n^l_k}\in (y_l;y_l+\delta_l)$, $\ \ y_{n^1_1}>x_{n^1_1}>y_{n^1_{2}}>x_{n^1_2}>...$ and $$\int\limits_{(y_l;y_l+1)\cap\bigcup\limits _{k>0}I_k^l}\frac{dx}{x-y_l}=\infty.$$ Put $N=\{n^l_k |\ l,k\in\Bbb N\}$. Then condition $i)$ is satisfied. We will show that if constants $\delta_l$ are chosen small enough then $N$ also satisfies $ii)$. Denote $J_l=\bigcup\limits_{k>0}I_{n^l_k}$. Choose constants $\delta_l$ to satisfy both $$\delta_l<|I_l|^2\tag4.5$$ and $$\delta_l<\frac1{4}(y_l-z_l)\dist(y_l;(y_l;\infty)\cap\bigcup\limits_{k0$. Consider the sets of integers $M'=\{l|x_l>y\},\ M''= \{l|\ y_l+\delta_ly\}$. Then $\Bbb N=M'\cup M''\cup M'''$. Put $$J'=\bigcup\limits_{l\in M'}J_l,\ \ J''=\bigcup\limits_{l\in M''}J_l \ \ {\text { and}}\ \ J'''=\bigcup\limits_ {l\in M'''}J_l.$$ Then, since $J_l\subset (y_l;y_l+\delta_l)$, $$\left|\int\limits_{J'}\frac{dx}{x-y}\right|<\infty$$ by (4.5). Also obviously $$\int\limits_{J''}\frac{dx}{x-y}<0<\infty$$ since $J''\subset(y-1;y)$. Let $\{n_k\}_{k\in\Bbb N}$ be some enumeraton of $M'''$ such that $ n_1\dist(y_{n_k};(y_{n_k};\infty)\cap\bigcup\limits_{l $$$$>\frac34\dist(y_{n_k};(y_{n_k};\infty)\cap\bigcup\limits_{l3\frac{\delta_{n_k}}{(y_{n_k}-z_{n_k})}. $$ Since $|\bigcup\limits_{l \left|\int\limits_{(y;\infty)\cap\bigcup\limits_{l \left|\int\limits_{(y;\infty)\cap \left[\bigcup\limits_{n\in\Bbb N}(z_n;y_n)\cup\bigcup\limits_{l0\right\}\setminus (y-\epsilon_k;y+\epsilon_k)=(y;\infty)\cap \left[\bigcup\limits_{n\in\Bbb N}(z_n;y_n)\cup\bigcup\limits_{l0$ both sets $ (x -\epsilon;x)\cap \a$ and $ (x;x +\epsilon)\cap \a$ are nonempty. Let $I_1=(x_1;y_1),I_2=(x_2;y_2),...$ be disjoint open intervals such that $F=(0;1)\setminus\bigcup I_n$. We always can assume that $F$ contains no isolated points i.e. $x_i\neq y_j$ for any $i,j$. First, using Lemmas 4.2 and 4.3 we will construct a phase shift $u_0$ of a pair $(\mu_0;\nu_0)$ such that $\mu_0=\sum c_n\delta_{a_n}$ for some $\ a_n\in\a, c_n>0$ and $\nu_0$ is continuous with $\supp\nu_0\subset F$. By Lemma 4.2 we can choose two sets of integers $L$ and $M$ satisfying (4.1) and (4.2). For each $n\in\Bbb N$ choose $z_n\in I_n\cap\Cal A$ such that $y_n-z_n<|I_n|^2$. By Lemma 4.3 we can choose $N\subset M$ satisfying (4.3) and (4.4). Denote $O=\Bbb N\setminus N$. For each $ k\in O$ put $a_k=z_k$. For each $ k\in N$ choose $a_k\in I_k\cap\a$ such that $a_k-x_k<|I_k|^2$. After that put $u_0=\pi$ on $\bigcup(a_k;y_k)$ and $u_0=0$ elsewhere. Then $u_0$ has compact support. Consider the pair of measures $(\mu_0;\nu_0)$ such that $u_0$ is its phase shift. By Lemma 3.2 $\mu_0$ and $\nu_0$ are singular. To prove that $\mu_0=\sum a_n\delta_{a_n}$ we need to show that $\mu_0(F)=0$. To show that $\mu_0(F)=0$ let us notice that if $y\in F$ then $$0\geq\int\limits_{(y-1;y)\cap\bigcup\limits_{k\in N}(x_k;a_k)}\frac{dx}{x-y}\geq -C_1>-\infty$$ because $a_k-x_k<|I_k|^2$ for any $k\in N$. Thus $$\int\limits_{\{u_0=\pi\}}\frac{dx}{x-y}<\int\limits_{\bigcup\limits_{k\in\Bbb N} (z_k;y_k)\cup\bigcup\limits_{k\in N}I_k}\frac{dx}{x-y}+C_1. \tag4.9$$ If $y\neq y_i,x_i$ for any $i\in\Bbb N$ then (4.9) and Lemma 4.3 imply that there exist positive $\epsilon_1,\epsilon_2,...$ such that $\epsilon_n\to 0$ as $n\to\infty$ and $$\int\limits_{\{u_0=\pi\}\setminus(y-\epsilon_n;y+\epsilon_n)}\frac{dx}{x-y}0$. Then $$\int\limits_{(y;y+1)\cap\bigcup\limits_{k\in N}I_k}\frac{dx}{x-y}=\infty$$ because $N$ satisfies (4.3). Since for $k\in N$ we have $a_k-x_k<(y_k-x_k)^2$, this implies that $$\int\limits_{(y;y+1)\cap\bigcup\limits_{k\in N}(a_k;y_k)}\frac{dx}{x-y}=\infty.$$ Since on $(a_k;y_k)$ $u_0=\pi$, condition (3.9) is not satisfied. Thus $\nu_0(y)=0$. If $y\in F$ and $y\neq y_i$ for any $i$, then $$\int\limits_{(y-1;y)\cap\bigcup\limits_{k\in O}I_k}\frac{dx}{y-x}=\infty$$ by (4.2) because $M\subset O$. Since for $k\in O$ we have $y_k-a_k<(y_k-x_k)^2$, this implies that $$\int\limits_{(y-1;y)\cap\bigcup\limits_{k\in O}(x_k;a_k)}\frac{dx}{y-x}=\infty.$$ Again, since $u_0=0$ on $(x_k;a_k)$, condition (3.9) does not hold and $\nu_0(y)=0$. Hence, $\nu_0$ is continuous. Our next goal is to transform $u_0$ into the phase shift $u$ of a pair $(\mu;\nu)$ such that $\mu$ is a spectral measure of $A$ and $\nu$ has a nontrivial continuous part (note that $\mu_0$ does not have point masses at some points of $\a$, therefore it is not a spectral measure of $A$). Let $\{b_n\}_{n=1}^\infty$ be some enumeration of the set $\a\setminus \{a_n\}_{n=1}^\infty$. For each $b_n$ let us choose $c_n$ such that if $b_n\in I_k$ then $c_n\in I_k$, $$\ \ |b_n-c_n|<\frac{\dist(b_n;(\Bbb R\setminus I_k) \cup\{b_1,b_2,...,b_{n-1}\})}{2^{n+1}}\tag4.10$$ and for each $n\in\Bbb N$ the sets $\Cb_n=\{a_i\}_{i=1}^{\infty}\cup \{b_1,b_2,...,b_n\}$ and $\Cc_n=\{c_1,c_2,...,c_n\}\cup F$ are well-mixed. For each $k\in\Bbb N$ define the function $u_k$ in the following way: 1) $|u_k|=0$ or $\pi$ everywhere on $\Bbb R$; 2) $u_k$ is continuous everywhere except $ \Cb_k\cup\Cc_k$; 3) $u_k$ jumps from $0$ to $\pi$ at each point of $\Cb_k$ and from $\pi$ to $0$ at each point of $\{c_1,c_2,...,c_k\}$. Let $u_k$ be the phase shift of a pair $(\mu_k;\nu_k)$. Then $$\mu_k=\sum _{n=1}^\infty\alpha^k_n\delta_{a_n} +\sum_{n=1}^k \beta^k_n\delta_{b_n}$$ for some positive constants $\alpha^k_i$ and $\beta^k_i$ and $$\nu_k=\sum_{n=1}^k\gamma^k_n \delta_{c_n}+f_k\nu_0$$ for some positive constants $\gamma^k_1$ and some positive function $f_k\in L^1(\nu_0)$. Condition (4.10) implies that the sequence $\{u_k\}$ converges in measure to some function $u$. Since $\supp u_k\subset [-1;2]$ for any $k\in\Bbb N$, $\ \ u $ has compact support. Let $u$ be the phase shift of a pair $(\mu;\nu)$. Then $\mu_k\to\mu$ and $\nu_k\to\nu$ in the $*$-weak topology. Since by (4.10) $$\int\limits_{\Bbb R}\left|\frac {u_k(x)-u_{k+1}(x)}{x-y}dx\right|<\frac 1{2^{k+1}}\tag 4.11$$ for any $y\in\Cb_k$, by part ii) of Lemma 3.7 $$\mu_{k+1}=g_k\mu_k+\beta^{k+1}_{k+1}\delta_{b_{k+1}}$$ where $g_k\in L^\infty( \mu_k)$, $$1-\frac1{2^k}i} g_k(b_i)=\beta_i<\infty.$$ Thus $A$ is unitarily equivalent to $ A_\mu$. Also, since $\nu=\eta+\sigma$ where $\eta$ is some positive measure and $\sigma$ is a $*$-weak limit of the sequence $\{f_k\nu_0\}$, (4.13) implies that $\sigma=f\nu_0$ for some $f\in L^1(\nu_0)$, $$01}f_k(x)=f(x)<3f_1(x)$$ $\nu_0$-a. e. Thus $A_\mu+(\cdot,1)1$ has a nontrivial continuous part. $\blacktriangle$\enddemo \proclaim{Corollary 4.4} Let $A$ be a self-adjoint operator. Then all trace class perturbations of $A$ are pure point iff $\sigma(A)$ is countable. \endproclaim \demo{Proof} The ``if'' part follows from the stability of the essential spectrum under trace class perturbations. The ``only if'' part follows from Theorem 4.1. The statement also follows from the Theorem of Carey and Pincus [C-P] on the equivalence modulo the trace class. $\blacktriangle$ \enddemo \heading 5. The problem of two spectra \endheading \define\rank{\operatorname {rank}} \definition{Definition} We will say that two operators $A$ and $B$ are equivalent up to a rank one perturbation if there exist operators $A'$ and $B'$ acting in the same space such that $A$ is unitarily equivalent to $A'$, $B$ is unitarily equivalent to $B'$ and $\rank (A'-B')=1$. \enddefinition In this section we will give a partial answer to the following question. \remark{The problem of two spectra} Let $\mu$ and $\nu$ be two finite Borel measures on $ {\Bbb R}$. When do there exist two cyclic self-adjoint operators $A$ and $B$ equivalent up to a rank one perturbation such that $\mu$ and $\nu$ are spectral measures of $A$ and $B$ respectively? \endremark \definition{Definition} If such $A$ and $B$ exist, we will say that $\mu$ and $\nu$ solve the problem of two spectra (PTS). \enddefinition We will say that $\mu$ is equivalent to $\nu$ and write $\mu\sim\nu$ if there exists $f\in L^1(\mu)$ such that $f>0\ \ \ \mu$-a. e. and $f\mu=\nu$. We will say that pairs of measures $(\mu,\nu)$ and $(\mu',\nu')$ are equivalent, $(\mu,\nu)\sim(\mu',\nu')$, if $\mu\sim\mu'$ and $\nu\sim\nu'$. In terms of the families $M_\phi$, we can say that $\mu$ and $\nu$ solve PTS if and only if there exists $\phi$ such that $(\mu,\nu)\sim(\mu',\nu')$ for some $\mu',\nu'\in\Cal M_\phi$. In terms of the phase shift, $\mu$ and $\nu$ solve PTS if and only if there is an equivalent pair possessing a phase shift i. e. satisfying (3.3) for some real function $u\in L^\infty(\Bbb R), \ ||u||_\infty\leq\pi$ and $\lambda\in\Bbb R$. In this section we will discuss the case of pure point measures $\mu$ and $\nu$. The first result in this direction is the following theorem, proved by Gelfand and Levitan. \proclaim{Theorem 5.1} Let $\Cal A=\{a_n\}_{n=1}^\infty$ and $\Cal B=\{b_n\}_{n=1}^\infty$ be two disjoint sequences of real numbers, $\lim_{n\to\infty}a_n=\lim_{n\to\infty}b_n=c\in\Bbb R$ . There exist unitarily equivalent up to a rank one perturbation cyclic self-adjoint operators $A$ and $B$ such that $\sigma(A)=\Clos\Cal A$ and $\sigma(B)=\Clos\Cal B$ if and only if the sets $\{a_1,a_2,...\}$ and $\{b_1,b_2,...\}$ are well-mixed. If such $A$ and $B$ exist then they are unique up to a unitary equivalence. \endproclaim \demo{Proof} %To prove this result we can use the notion of the Krein spectral shift. Define function $u$ on $\Bbb R$ to satisfy the following conditions: 1) $u$ is continuous and $|u|$ is equal to $\pi$ or 0 everywhere on $\Bbb R\setminus (\{a_n\}\cup\{b_n\})$, 2) at the points $\{a_n\}$ $u$ jumps from $0$ to $\pi$, 3) at the points $\{b_n\}$ $u$ jumps from $\pi$ to $0$. WLOG $\inf_{n\in\Bbb N}a_n\leq\inf_{n\in\Bbb N}$ and $\sup_{n\in\Bbb N}a_n\leq \sup_{n\in\Bbb N}b_n$. Then $u$ has compact support. As we discussed before, there are unique measures $\mu,\nu\in\Cal M_+(\Bbb R)$ such that $u$, $\mu$ and $\nu$ satisfy formula (3.3) with $\lambda=1$. Put $A=A_\mu$ and $B=A_\nu$. Then $B$ is unitarily equivalent to $ A+(\cdot,1)1$. By Lemma 3.3 i) $\supp\mu\subset\Clos\Cal A$ and $\supp\nu\subset\Clos\Cal B$. Also by Lemma 3.8 the measures $\mu$ and $\nu$ have point masses at points $a_i$ and $b_i$ respectively, therefore $\supp\mu=\Clos\Cal A$ and $\supp\nu=\Clos\Cal B$. Thus the operators $A$ and $B$ satisfy the condition of the Theorem. Since the sequences are well-mixed and have only one cluster point there exists a unique function $u$ satisfying properties 1)-3). Therefore $A$ and $B$ are unique up to a unitary equivalence. $\blacktriangle$\enddemo \remark{\bf Remark} If $\mu=\sum \alpha_n\delta_{a_n}$ and $\nu=\sum \beta_n\delta_{b_n}$, where $\alpha_i,\beta_i>0$, solve PTS then the sets $\Cal A=\{a_1,a_2,...\}$ and $\Cal B=\{b_1,b_2,...\}$ are well-mixed. To prove it one can, for instance, notice that if $\Cal A$ and $\Cal B$ are not well-mixed then the function $u$ satisfying properties 1)-3) from the above proof does not exist. Note that Theorem 5.1 provides us with operators such that $\mu$ and $\nu$ are only {\it absolutely continuous } with respect to spectral measures of $A$ and $A+(\cdot,\phi)\phi$ but not necessarily {\it equivalent} to them. Therefore it does not imply that $\mu$ and $\nu$ solve PTS. As shown in Example 5.2 below, the condition that $\Cal A$ and $\Cal B$ are well-mixed is not sufficient for $\mu$ and $\nu$ to solve PTS. \endremark \proclaim{Example 5.2} Put $ a_n=(-1)^n/2^n$ for $n=1,2,3,...$, $b_1=-1,\ b_n=a_{n-1}+(-1)^n/4^n$ for $n=2,3,...$. Then the sets $\{a_1,a_2,...\}$ and $\{b_1,b_2,...\}$ are well-mixed. Let $\mu$ and $\nu$ be defined as in the last remark. Suppose that $\mu$ and $\nu$ solve PTS. Then there exist a pair $(\mu,\nu)\sim(\mu',\nu')$ possessing a phase shift $u$. Such $u$ must be continuous on $\Bbb R\setminus (\{a_n\}\cup\{b_n\})$, jump from $0$ to $\pi$ at any $a_n$ and jump from $\pi$ to $0$ at any $b_n$. Thus $ u=0$ on $(1/2^{2n}-1/4^{2n+1};1/2^{2n})$ and on $(-1/2^{2n-1}+1/4^{2n};-1/2^{2n+1})$ for $n=1,2,...$ and $u=\pi$ on the rest of $\Bbb R$. But then Lemma 3.8 implies that $\mu(0)>0$. Since $0\neq a_n$ for any $n\in\Bbb N$ we have a contradiction. \endproclaim Hence in Theorem 5.1 one can not prescribe arbitrarily whether or not spectral measures of $A$ and $B$ have a point mass at $c$. If the sequences $\{a_1,a_2,...\}$ and $\{b_1,b_2,...\}$ are finite, disjoint and well-mixed then $\mu$ and $\nu$ obviously solve PTS. More interesting example is provided by the following theorem. \proclaim{Theorem 5.3 ([A3])} Let $\Cal A$ and $\Cal B$ be two disjoint countable sets on the unit circle $\Bbb T$. Suppose that ${\Clos\Cal A}=\Clos{\Cal B}=\Bbb T$. Let $\{a_n\}_{n=1}^\infty$ and $\{b_n\}_{n=1}^\infty$ be some enumerations of $\Cal A$ and $\Cal B$ respectively. Then there exist sequences of positive real numbers $\{\alpha_n\}$ and $\{\beta_n\}$ such that the measures $\sum \alpha_n\delta_{a_n}$ and $\sum \beta_n\delta_{b_n}$ belong to the same family $M_\varphi$ \endproclaim In terms of operator theory Theorem 5.3 means that any two cyclic pure point self-adjoint operators whose spectrum is equal to $\Bbb R$ are unitarily equivalent up a rank one perturbation. Note that the condition ${\Clos\Cal A}=\Clos{\Cal B}=\Bbb T$ automatically implies that $\Cal A$ and $\Cal B$ are well-mixed. However the following example shows that in general the condition $\Clos{\Cal A}=\Clos{\Cal B}$ does not imply that that the corresponding measures solve PTS even if $\Cal A$ and $\Cal B$ are well-mixed. \proclaim{Example 5.4} Let $C$ be the standard Cantor null set on the unit interval $[0;1]$, $C=[0;1]\setminus \bigcup I_n$ where $I_n=(x_n;y_n)$ are disjoint open intervals. Let $\Cal A$ and $\Cal B$ be two disjoint well-mixed countable sets of points of $C$ such that $$ 0, x_1, x_2,..\in \Cal A, 1\in\Cal B.\tag 5.1$$ Then $\Clos{\Cal A}=\Clos{\Cal B}=C$. Let $\{a_n\}_{n=1}^\infty$ and $\{b_n\}_{n=1}^\infty$ be some enumerations of $\Cal A$ and $\Cal B$ respectively. Define $\mu=\sum \frac 1{2^n}\delta_{a_n}$ and $\nu=\sum \frac 1{2^n}\delta_{b_n}$. Suppose that $\mu$ and $\nu$ solve PTS. Then there exists the phase shift $u$ of the pair $(\mu';\nu')$ where $(\mu,\nu)\sim(\mu',\nu')$. Function $u$ must be constant on each $I_n$. Condition (5.1) implies that $u=0$ a. e. on $\Bbb R$. This means that $\mu'$ and $\nu'$ are trivial and we obtain a contradiction. \endproclaim It turns out that this example is, in a certain sense, typical: the following Theorem shows that such $\mu$ and $\nu$ solve PTS for any two well-mixed dense in $K$ sequences if and only if the set $K$ is not too ``porous''. \proclaim{Theorem 5.5} Let $K\subset\Bbb R$ be a closed set. Denote by $I_1,I_2,...$ the disjoint open intervals $I_k=(x_k;y_k)$ such that $K=\Bbb R\setminus\bigcup I_n$. Then the following two conditions are equivalent: i)Any two cyclic self-adjoint pure point operators $A$ and $B$ such that the sets of their eigenvalues are well-mixed and $\sigma(A)=\sigma(B)=K$ are unitarily equivalent up to a rank one perturbation. ii) If $y\in K\setminus\{x_1,y_1,x_2,y_2,...\}$ then $$\int\limits_{\bigcup\limits_{n\in \Bbb N} I_n} \frac{dx}{|y-x|}<\infty;\tag 5.2$$ if $y=x_k$ or $y=y_k$ for some $k\in \Bbb N$ then $$\int\limits_{\bigcup\limits_{n\in \Bbb N, n\neq k} I_n} \frac{dx}{|y-x|}<\infty.\tag 5.3$$ \endproclaim \example{\bf Example 5.6} Condition ii) fails for $K=C$ for any standard Cantor set $C$ (even if $|C|>0$). Condition ii) obviously holds if $\partial K$ is a finite set. We will show that there exist more complicated examples of such $K$, namely that for any nowhere dense set $L\subset\Bbb R$ there exists a set $K$ such that $K$ satisfies ii) and $\partial K\supset L$. Denote by $J_i=(x_i;y_i)$ the disjoint open intervals such that $\Clos L=\Bbb R\setminus\bigcup J_n$ . On each $J_i$ choose a sequence of intervals $\{\Delta_i^k\}_{-\infty}^\infty, \Delta_i^k=(x_i^k;y_i^k)$ such that $x_i^k\to x_i$ as $k\to-\infty$, $x_i^k\to y_i$ as $k\to\infty$ and $$\int\limits_{\cup_{k}\Delta_i^k}\frac{dx}{|x-y|}<\frac1{2^i}$$ for all $y\in\Bbb R\setminus J_i$. Then we can put $K=\Bbb R\setminus \bigcup_{i,k}\Delta_i^k$. Similarly, one can show that for any closed $L\subset \Bbb R$ there exists a closed $K\supset L$ satisfying ii) and such that $\partial K\supset\partial L$. \endexample \demo{Proof of Theorem 5.5} ii)$\Rightarrow$i) WLOG $K\subset(0;1)$. Let $\Cal A$ and $\Cal B$ be the sets of eigenvalues of $A$ and $B$ respectively. Let $\{a_n\}_{n=1}^\infty$ and $\{b_n\}_{n=1}^\infty$ be some enumerations of $\Cal A$ and $\Cal B$ respectively. a) Let us first consider the case when the endpoints of the intervals $I_k$ belong to neither $\Cal A$ nor $\Cal B$. Let us find reenumerations $\{a_{n_k}\}$ and $\{b_{m_k}\}$ of the sequences $\Cal A=\{a_n\}$ and $\Cal B=\{b_m\}$ such that for any $k\in\Bbb N$ 1) the sets $\{a_{n_1},...,a_{n_{k}}\}$ and $\{b_{m_1}, ...,b_{m_{k}}\}$ are well-mixed and 2) the distance $|a_{n_k}-b_{m_k}|$ is at least $2^{k+1}$ times less than the distance between the $a_{n_k}$ and any of the points $a_{n_1},...,a_{n_{k-1}},b_{m_1}, ...,b_{m_{k-1}}$. To find such reenumerations we can act in the following way. WLOG $a_1>b_1$. Choose $n_1=1$ and $m_1=1$. After that for $k=2,3,...$ consequently do the following. Choose a point $x$ in $\Cal A\cup\Cal B\setminus \{a_{n_1},...,a_{n_{k-1}},b_{m_1}, ...,b_{m_{k-1}}\}$ with minimal index in the initial enumeration. Suppose $x\in \Cal A$. Then put $a_{n_k}=x$ and choose $b_{m_k}\in\Cal B$ to satisfy conditions 1) and 2) above. Similarly, if $x\in\Cal B$ then put $b_{m_k}=x$ and choose $a_{n_k}\in\Cal A$ to satisfy conditions 1) and 2) above. Note that it is always possible to choose $b_{m_{k}}$ or $a_{n_{k}}$ to satisfy 2) because $a_{n_k},b_{m_{k}}\in K\setminus \{x_1,y_1,x_2,y_2,...\}$. Now for each $k$ define a function $u_k$ on $\Bbb R$ in such a way that $|u_k|\in\{0,\pi\}$ on $\Bbb R$, $u_k$ is continuous everywhere except $a_{n_1},a_{n_2},...a_{n_k},b_{n_1},b_{n_2},...,b_{n_k}$ , $u_k$ jumps from $0$ to $\pi$ at each $a_{n_i}, 0b_1$, $u_k$ has a compact support. Therefore $u_k$ is the phase shift of a pair of measures $(\mu_k;\nu_k)$. It is easy to see that $\mu_k=\sum_{i=1}^k \alpha^k_i\delta_{a_{n_i}}$ and $\nu_k=\sum_{i=1}^k \beta^k_i\delta_{b_{m_i}}$ for some positive constants $\alpha^k_1,...,\alpha^k_k,\beta^k_1,...,\beta^k_k$. Condition 2) implies that for any $l>k$ $|\{u_k\neq u_l\}|\leq \sum_{i=k}^l 1/2^i$. Thus the sequence $\{u_k\}$ converges in measure to some function $u$. That implies that the sequences of measures $\{\mu_k\}$ and $\{\nu_k\}$ converge in the $*$-week topology to some measures $\mu$ and $\nu$ (respectively) such that $u$ is the phase shift of the pair $(\mu;\nu)$. To complete the proof we need to show that $\mu=\sum \alpha_n\delta_{a_n}$ and $\nu=\sum \beta_n\delta_{b_n}$ for some positive constants $\alpha_i$ and $\beta_i$. Since $$||\mu_k||= \im \exp \left[\Cal Ku_k(i)\right] \to \im \exp \left[\Cal Ku(i)\right],$$ we have that $$||\mu_k||-||\mu_{l}||\to 0$$ as $k,l\to\infty$. Thus $$ \sum_{i=k}^l\alpha^{i}_{l}\leq\left|\ ||\mu_k||-||\mu_l||\ \right|+\sum_{i=1}^k|\alpha^i_k-\alpha^i_l| \leq\left|||\mu_k||-||\mu_l||\right|+k/2^{k-1}\to 0\tag5.4$$ as $k,l\to\infty$. But condition 2) implies that $$\int\limits_{\{u_k\neq u_{k+1}\}}\frac{dx}{|x-a_i|}<\frac1{2^k}$$ for any $k$ and any $i\leq k$. Hence, by part ii) of Lemma 3.7 $$|\alpha^k_i/\alpha^{k+1}_i-1|<1/2^{k-1}\tag5.5$$ for any $k$ and any $ i\leq k$. Since $\supp u_k\subset (0;1)$, $\alpha^{k+1}_i\leq ||\mu_k||\leq 1$. Hence by (5.5) $$|\alpha_i^k-\alpha_i^{k+1}|<1/2^{k-1}.\tag5.6$$ Together (5.4) and (5.6) imply that $$||\mu_k-\mu_l||\to 0$$ as $k,l\to\infty$. Put $\alpha_i=\lim_{k\to\infty}\alpha_i^k$, and $\mu'=\sum_{i=1}^\infty\alpha_i\delta_{a_{n_i}}$. Since the sequence $\{\mu_k\}$ converges in norm, it must converge to $\mu'$. But since $\mu$ is the $*$-weak limit of $\{\mu_k\}$, $\mu=\mu'=\sum_{i=1}^\infty\alpha_i\delta_{a_{n_i}}$. It is left to notice that (5.5) implies that $$\alpha_i=\alpha_i^i\prod_{l>i}\frac {\alpha_i^l}{\alpha_i^{l+1}}>0$$ for each $i$. In the same way we can prove that $\nu=\sum \beta_i\delta_{b_i}$ for some positive $\beta_i$. b) In the general case, denote $N=\{n| \partial I_n\cap K\neq\emptyset\}$. Consider disjoint countable sets $\Cal A'=\Cal A\cup\Cal C$ and $\Cal B'=\Cal B\cup\Cal D$ where $\Cal C=\{c_n\}\subset[\cup_{n\in N}I_n]\cap (0;1)$ and $\Cal D=\{d_n\}\subset [\cup_{n\in N}I_n]\cap (0;1)$ satisfy: 1) $\Cal C,\Cal D$ are dense in $[\cup_{n\in N}I_n]\cap (0;1)$ and 2) if for some $n\in N$ one of the endpoints $x$ of the interval $I_n$ belongs to $\Cal A\ (\Cal B)$ but the other endpoint $y$ of $I_n$ is not in $\Cal A\cap\Cal B$ then $y\in \Cal A'\ (\Cal B')$ (i. e. both $x$ and $y$ are in the same set $\Cal A'$ or $\Cal B'$). (Notice that since $\Clos\Cal A=\Clos\Cal B=K$ and $\Cal A\cap\Cal B=\emptyset$, $K$ does not have isolated points. Hence $x_k\neq y_l$ for any $k, l \in \Bbb N$.) Then the set $\Cal A'\cup\Cal B'$ contains all the endpoints of the intervals $I_n,\ n\in\Bbb N$. Then $K'=K\cup\bigcup_{n\in N}I_n,\ \Cal A'$ and $\Cal B'$ satisfy the conditions of part a) and we can construct a phase shift $v$ corresponding to measures $\mu'=\sum\alpha_i\delta_{a_i}+\sum\sigma_i\delta_{c_i}$ and $\nu'=\sum\beta_i\delta_{b_i}+\sum\eta_i\delta_{d_i}$ for some positive $\alpha_i,\sigma_i,\beta_i,\eta_i$. Define $N_\pi=\{n| x_n\in\Cal A\ or\ y_n\in\Cal B\}$ and $N_0=\{n| x_n\in\Cal B\ or\ y_n\in\Cal A\}$. Then $N_\pi\cap N_0=\emptyset$ because the sets $\Cal A$ and $\Cal B$ are well-mixed and $N=N_\pi\cup N_0$. Let function $u$ be such that $u=v$ outside $\cup_{n\in N}I_n$, $u=\pi$ on each $I_n, n\in N_\pi$ and $u=0$ on each $I_n, n\in N_0$. Let $\mu$ and $\nu$ be the measures such that $u$ is the phase shift of the pair $(\mu;\nu)$. Then ii) together with Lemma 3.7 imply that the restriction of $\mu$ on $K\setminus\{x_1,y_1,x_2,y_2,...\}$ is absolutely continuous with respect to $\mu'$. Also, obviously $\mu(\cup_{n\in N} I_n)=0$ because $u$ is constant on each $I_n, \ n\in N$. It is left to check the endpoints of the intervals $I_n, n\in N$. If for some $n\in N_\pi$ $x_n\in \Cal A$, then $\mu'$ has a point mass at $x_n$. Thus $v$ satisfies the condition (3.8) at the point $y=x_n$. Since $u=\pi$ on $I_n$, by (3.8) and ii) we have $$\int\limits_{\{u\neq v\}}\frac{dx}{|x-x_n|}<\infty.$$ Hence by (3.8) $\mu$ also has a point mass at $x_n$. If for some $n\in N_\pi$ $x_n$ belongs to neither $\Cal A$ nor $\Cal B$ then $\nu'$ has a point mass at $x_n$. Thus $v$ satisfies (3.9) at $y=x_n$. Since $u=\pi$ on $I_n$, by (3.8) and ii) we have that $$\int\limits_{\{u\neq \pi\}}\frac{dx}{|x-x_n|}<\infty.$$ Thus by (3.8) and (3.9) $\mu(x_n)=0$ and $\nu(x_n)=0$. Other endpoints of $I_n,\ n\in N$ can be checked in the same way. Thus $\mu=\sum \alpha_n\delta_{a_n}$ for some positive $\alpha_n$. Similarly we can show that $\nu=\sum \beta_n\delta_{b_n}$ for some positive $\beta_n$. i)$\Rightarrow$ii) Suppose condition ii) is not satisfied at some $y\in K$. Let $\Cal A$ and $\Cal B$ be disjoint, dense in $K$ and well-mixed sequences such that $y\in\Cal B$ and $y_i\in\Cal A$ for all $i$ such that $y_i\neq\infty$. Then the phase shift $u$ of the pair $(\mu;\nu)$ is equal to $0$ on $\cup I_n$. Thus condition (3.9) is not satisfied at $y$ and $\nu(y)=0$. $\blacktriangle$ \enddemo Here is one more way to avoid the situation of Example 5.4. \proclaim{Theorem 5.6} Let $K\subset\Bbb R$ be a closed set. Denote by $I_1=(x_1;y_1),I_2=(x_2;y_2),...$ disjoint open intervals such that $K=\Bbb R\setminus\bigcup I_n$. Let $A$ and $B$ be two self-adjoint cyclic pure point operators, $\Cal A$ and $\Cal B$ be the sets of all eigenvalues of $A$ and $B$ respectively. Suppose that $\sigma (A)=\sigma(B)=K$ and $\Cal A\cap\{x_1,y_1,x_2,y_2,...\}= \Cal B\cap\{x_1,y_1,x_2,y_2,...\}=\emptyset$. Then $A$ and $B$ are equivalent up to a rank one perturbation. \endproclaim \demo{Proof} Notice that in part a) of the implication ii)$\Rightarrow$i) in the previous proof we did not use condition ii). $\blacktriangle$ \enddemo In the rest of this Section we are going to discuss the following question. \remark{\bf Remark} Let $A$ be a singular self-adjoint cyclic operator, $\varphi$ and $\psi$ its noncollinear cyclic vectors ($\varphi\neq c\psi$). Is it possible that operators $A^\varphi=A+(\cdot,\varphi)\varphi$ and $A^\psi=A+(\cdot,\psi)\psi$ are unitarily equivalent? If $A$ is a finite rank operator, then the answer is negative. Indeed, denote by $\mu$ and $\nu$ the spectral measures of $\varphi$ for $A$ and $A^\varphi$ respectively; denote by $\mu'$ and $\nu'$ the spectral measures of $\psi$ for $A$ and $A^\psi$ respectively. If $\mu$ and $\nu$ are linear combinations of point masses at points $a_1,a_2,..,a_n$ and $b_1,b_2,..,b_n$ respectively, $a_10$. At the same time we again can prove that $u'$ is the phase shift of the pair $(\sum \alpha_i'\delta_{a_i},\sum \beta_i'\delta_{b_i})$ for some positive $\alpha_i',\beta_i'$. Therefore $u$ and $u'$ are different phase shifts of equivalent pairs of measures. In terms of the families $M_\varphi$, we can say that if $(\mu,\nu)\in\Cal M_\varphi$ and $(\mu',\nu')\in\Cal M_\phi$ are equivalent pairs of measures and $\varphi$ is a finite Blacshke product, then $\varphi$ is equal to $b\circ\phi$ for some M\"obius transform $b$. However, if $\varphi$ is more complicated then $\varphi$ and $\phi$ can be completely different functions. In fact, modifying the construction from the proof of Theorem 5.5 one can show that there exist infinitely many different phase shifts such that the corresponding pairs of measures are equivalent to $(\sum \alpha_i\delta_{a_i},\sum \beta_i\delta_{b_i})$. It means that if $A$ and $B$ are two pure point operators whose eigenvalues are dense in the same set $K\subset\Bbb R$ satisfying condition ii) of Theorem 5.5, then there exist infinitely many different (corresponding to noncollinear vectors) rank one perturbations of $A$ unitarily equivalent to $B$. However, it is absolutely unclear if such situation is possible in the case when $A$ or $B$ has a nontrivial singular continuous part. \endremark \heading 6. One example of the absence of the mixed spectrum \endheading Let as usual $A$ be a cyclic self-adjoint operator, $\phi$ its cyclic vector, $A_\lambda=A+\lambda(\cdot,\phi)\phi$ $\ \ \lambda\in \Bbb R$. Let $\mu_\lambda$ be the spectral measure of $\phi$ for $A_\lambda$. Let us denote by $\Pi$ and $\Sigma$ the sets of $\lambda$ for which $\mu_\lambda$ has nontrivial pure point and nontrivial singular continuous part on $[0;1]$ respectively. Then the set $\Pi\cap\Sigma$ will consist of those $\lambda$ for which the corresponding measures are ``mixed'' on the interval $[0;1]$. The paper [D] of Donoghue gives examples in which $\Pi=\{0\},\Sigma=\Bbb R\setminus \{0\}$ and $\Sigma=\{0\},\Pi=\Bbb R\setminus \{0\}$. Therefore the set $\Pi\cap\Sigma$ can be empty when both $\Pi$ and $\Sigma$ are nonempty. One of the natural questions which arise from the recent results on rank one perturbations (see [R-J-L-S]) is {\it whether the set $\Pi\cap\Sigma$ can be empty (or almost empty) when the sets $\Pi$ and $\Sigma$ are sufficiently big (topologically or in measure)}. The following example gives a partial answer to this question. \example{Example 6.1} We will show that there exist a self-adjoint cyclic operator $A$ such that for some cyclic vector $\phi$ the operators $A+\lambda(\cdot,\phi)\phi$ are singular continuous on $[0;1]$ for all $0\geq\lambda\geq 1$ and pure point for all other $\lambda\in\Bbb R$. We will first construct the Krein spectral shift $u$. Let $\{a_n\}$ be a sequence of real numbers $00\tag6.1$$ and $$1-\prod_{k=n}^\infty (1-a_k)\geq \frac1n.\tag6.2$$ Consider the Cantor set $C$ corresponding to the sequence $\{a_n\}$: let $$C_0=I_0^0=[0;1], \ \ C_1=I_1^1\cup I_2^1,...,\ \ C_n=I_1^n\cup... \cup I_{2^n}^n,...$$ where $$I_{2k}^n\cup I_{2k-1}^n=I_k^{n-1}\setminus \Delta_k^{n-1}$$ and $\Delta_k^n$ is the open interval placed in the center of the interval $I_k^n$ and such that $|\Delta_k^n|=a_n|I_k^n|$ and let $C=\bigcap_{n=0}^\infty C_n$. The Cantor set $C$ has the following properties: $$|C|=c\tag6.3$$ and $$\frac{|I_n^k\cap C|}{|I_n^k|}= \prod_n^\infty(1-a_n) \geq\frac1n\geq\frac{\ln 2}{-\ln |I_n^k|}\tag 6.4$$ for any $n,k\in\Bbb N$. Define $u=\pi$ on $C$ and $u=0$ elsewhere on $\Bbb R$. Denote $U(z)=\Cal Ku(z)$ for $z\in\Bbb C_+$. \proclaim{Claim} $U$ has a finite nontangential derivative $U'(x)=\lim_{ z \underset \ngtr \to \longrightarrow x} \frac {U(x)-U(z)}{x-z}$ at a point $x\in\Bbb R$ if and only if $x\not\in C$. \endproclaim \demo{Proof} Since $u$ is locally constant on $\Bbb R\setminus C$, $U'$ obviously exists there. Let intervals $I^n_k$ be the same as in the construction of $C$ above. Let $x\in C$. Let $\{I_{n_k}^k\}_{k=1}^\infty$ be the sequence of intervals containing $x$. Denote by $x_k$ the middle of the interval $I_{n_k}^k$ and put $y_k=|x-x_k|$. Then (6.4) imply that $$\left|\pi -\Cal Pu(x_n+iy_n)\right|\to 0$$ (since $\prod_n^\infty(1-a_n)\to 0$) but $$\left|\pi -\Cal Pu(x_n+iy_n)\right|\geq\frac d{|\ln y_n|}$$ for some $d>0$. Thus $U(x)-U(z)\neq O(x-z) $ as $z\underset\ngtr\to\rightarrow x$ and $U'(x)$ does not exist.$\blacktriangle$ \enddemo Consider a real measure $\nu_0$ such that $u$ is the phase shift of a pair $(\nu_0,\nu)$ for some measure $\nu$. Put $A=A_\nu, A_\lambda=A+\lambda(\cdot,1)1$. Let $\nu_\lambda$ be the spectral measure of $1$ for $A_\lambda$ (then $\nu=\nu_1$). By Lemma 3.2 $\nu$ is singular. Therefore all $\nu_\lambda$ are singular. Hence $$|\Cal K\nu_\lambda|\nec\infty$$ for $\nu_\lambda$-a. e. $x$. By formula (2.3) this means that $$1+\pi\Cal K\nu_0=\exp U(z)\nec 1-\frac1{\lambda}$$ for $\nu_\lambda$-a. e. $x$. Thus by the definition of the phase shift for all $\lambda\in (0;1)$ $$\arg(1+\pi\Cal K\nu_0(z))=\Cal Pu(z)\nec \pi$$ for $\nu_\lambda$-a. e. $x$. Thus all $\nu_\lambda, \lambda\in (0;1)$ are concentrated on the set $C$. Since $U'$ does not exist on $C$ and $$\Cal K\nu_0=\exp U-1 \nec 1-1/\lambda\neq 0$$ $\nu_\lambda$-a. e., $(\Cal K\nu_0)'$ does not exist on $C$ for all $\lambda\in (0;1)$. Hence the characteristic function $\phi$ defined in $\Bbb C_+$ by formula (2.4) does not have a nontangential derivative on $C$. Thus by the Remark after Lemma 3.8 all $\nu_\lambda, \lambda\in (0;\infty)$ are singular continuous. Similarly all $\nu_\lambda$ for $ \lambda\in \Bbb R\setminus [0;1]$ are concentrated on $\Bbb R\setminus C$. Since $U'$ exists everywhere on $\Bbb R\setminus C$, $\ \mu_\lambda$ for $\lambda\in \Bbb R\setminus [0;1]$ are pure point. To prove that $\nu_0$ is continuous, let us notice that since by Lemma 3.3 $$p.\ v.\ \int\limits_\Bbb R \frac{u(t+x)dt}{t}=\infty$$ for $\nu_0$-a. e. $x$, $\nu_0$ is concentrated on $C$. Let $y\in C$ and let $I^k_{n_k}$ be the sequence of intervals from the construction of $C$ containing $y$. Then (6.4) implies that condition (3.8) is not satisfied at $y$. Thus $\nu_0$ can not have a point mass at $x$. Hence $\nu_0$ is continuous. Similarly, $\nu_1$ is continuous. \endexample \remark{\bf Remark} It is still unclear if there exist examples such that the sets $\Pi$ and $\Sigma$ are big (topologically or in measure) but $\Pi\cap\Sigma=\emptyset$ when $\sigma(A)$ contains the interval $[0;1]$. Modifying the above example, we can obtain a dense in $[0;1]$ set $C'$ by inserting smaller Cantor sets into each complimentary interval $I_n^k$, then inserting smaller Cantor sets into each new complimentary interval and so on. If the size of these Cantor sets decreases to $0$ fast enough then replacing $C$ in the above example with $C'$ we will obtain an example of $A$ and $\phi$ such that $\sigma(A)=[0;1]$, $A+\lambda(\cdot,\phi)\phi$ is continuous on $[0;1]$ {\it for all} $\lambda\in [0;1]$ and pure point {\it for almost all} $\lambda\in \Bbb R\setminus [0;1]$. 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