Content-Type: multipart/mixed; boundary="-------------0409231513143" This is a multi-part message in MIME format. ---------------0409231513143 Content-Type: text/plain; name="04-304.keywords" Content-Transfer-Encoding: 7bit Content-Disposition: attachment; filename="04-304.keywords" orthogonal polynomials, sum rules, szego theorem ---------------0409231513143 Content-Type: application/x-tex; name="sz_2004.TEX" Content-Transfer-Encoding: 7bit Content-Disposition: inline; filename="sz_2004.TEX" \documentclass[reqno,centertags, 12pt]{amsart} \usepackage{amsmath,amsthm,amscd,amssymb} \usepackage{latexsym} %\usepackage{showkeys} \sloppy %%%%%%%%%%%%% fonts/sets %%%%%%%%%%%%%%%%%%%%%%% \newcommand{\bbN}{{\mathbb{N}}} \newcommand{\bbR}{{\mathbb{R}}} \newcommand{\bbD}{{\mathbb{D}}} \newcommand{\bbP}{{\mathbb{P}}} \newcommand{\bbZ}{{\mathbb{Z}}} \newcommand{\bbC}{{\mathbb{C}}} \newcommand{\bbQ}{{\mathbb{Q}}} \newcommand{\calH}{{\mathcal H}} \newcommand{\calI}{{\mathcal I}} %%%%%%%%%%%%%%%%%% abbreviations %%%%%%%%%%%%%%%%%%%%%%%% \newcommand{\dott}{\,\cdot\,} \newcommand{\no}{\nonumber} \newcommand{\lb}{\label} \newcommand{\f}{\frac} \newcommand{\ul}{\underline} \newcommand{\ol}{\overline} \newcommand{\ti}{\tilde } \newcommand{\wti}{\widetilde } \newcommand{\Oh}{O} \newcommand{\oh}{o} \newcommand{\marginlabel}[1]{\mbox{}\marginpar{\raggedleft\hspace{0pt}#1}} \newcommand{\tr}{\text{\rm{Tr}}} \newcommand{\dist}{\text{\rm{dist}}} \newcommand{\loc}{\text{\rm{loc}}} \newcommand{\spec}{\text{\rm{spec}}} \newcommand{\rank}{\text{\rm{rank}}} \newcommand{\ran}{\text{\rm{ran}}} \newcommand{\dom}{\text{\rm{dom}}} \newcommand{\ess}{\text{\rm{ess}}} \newcommand{\ac}{\text{\rm{ac}}} \newcommand{\singc}{\text{\rm{sc}}} \newcommand{\s}{\text{\rm{s}}} \newcommand{\sing}{\text{\rm{sing}}} \newcommand{\pp}{\text{\rm{pp}}} \newcommand{\supp}{\text{\rm{supp}}} \newcommand{\AC}{\text{\rm{AC}}} \newcommand{\bi}{\bibitem} \newcommand{\hatt}{\widehat} \newcommand{\beq}{\begin{equation}} \newcommand{\eeq}{\end{equation}} \newcommand{\ba}{\begin{align}} \newcommand{\ea}{\end{align}} \newcommand{\veps}{\varepsilon} %\newcommand{\Ima}{\operatorname{Im}} %\newcommand{\Real}{\operatorname{Re}} %\newcommand{\diam}{\operatorname{diam}} % use \hat in subscripts % and upperlimits of int. % % Rowan's unspaced list % \newcounter{smalllist} \newenvironment{SL}{\begin{list}{{\rm\roman{smalllist})}}{% \setlength{\topsep}{0mm}\setlength{\parsep}{0mm}\setlength{\itemsep}{0mm}% \setlength{\labelwidth}{2em}\setlength{\leftmargin}{2em}\usecounter{smalllist}% }}{\end{list}} %%%%%%%%%%%%%%%%%%%%%% renewed commands %%%%%%%%%%%%%%% %\renewcommand{\Re}{\text{\rm Re}} %\renewcommand{\Im}{\text{\rm Im}} %%%%%%%%%%%%%%%%%%%%%% operators %%%%%%%%%%%%%%%%%%%%%% \DeclareMathOperator{\Real}{Re} \DeclareMathOperator{\Ima}{Im} \DeclareMathOperator{\diam}{diam} \DeclareMathOperator*{\slim}{s-lim} \DeclareMathOperator*{\wlim}{w-lim} \DeclareMathOperator*{\simlim}{\sim} \DeclareMathOperator*{\eqlim}{=} \DeclareMathOperator*{\arrow}{\rightarrow} \allowdisplaybreaks \numberwithin{equation}{section} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%% end of definitions %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \newtheorem{theorem}{Theorem}[section] \newtheorem*{t0}{Theorem} \newtheorem*{t1}{Theorem 1} \newtheorem*{t2}{Theorem 2} \newtheorem*{t3}{Theorem 3} \newtheorem*{c4}{Corollary 4} \newtheorem*{p2.1}{Proposition 2.1} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} %\newtheorem{hypothesis}[theorem]{Hypothesis} %\theoremstyle{hypothesis} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{xca}[theorem]{Exercise} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} % Absolute value notation \newcommand{\abs}[1]{\lvert#1\rvert} \begin{document} \title[Higher-Order Szeg\H{o} Theorems With Two Singular Points] {Higher-Order Szeg\H{o} Theorems\\ With Two Singular Points} \author[B. Simon and A. Zlato\v{s}]{Barry Simon$^{1}$ and Andrej Zlato\v{s}$^{2}$} \thanks{$^1$ Mathematics 253-37, California Institute of Technology, Pasadena, CA 91125. E-mail: bsimon@caltech.edu. Supported in part by NSF grant DMS-0140592} \thanks{$^2$ Department of Mathematics, University of Wisconsin, Madison, WI 53706. E-mail: andrej@math.wisc.edu} \date{September 16, 2004} \begin{abstract} We consider probability measures, $d\mu=w(\theta) \f{d\theta}{2\pi} +d\mu_\s$, on the unit circle, $\partial\bbD$, with Verblunsky coefficients, $\{\alpha_j\}_{j=0}^\infty$. We prove for $\theta_1\neq\theta_2$ in $[0,2\pi)$ and $(\delta\beta)_j=\beta_{j+1}$ that \[ \int [1-\cos(\theta-\theta_1)][1-\cos(\theta-\theta_2)] \log w(\theta) \, \f{d\theta}{2\pi} >-\infty \] if and only if \[ \sum_{j=0}^\infty \, \bigl|\bigl\{(\delta -e^{-i\theta_2}) (\delta -e^{-i\theta_1}) \alpha\bigr\}_j\bigr|^2 +\abs{\alpha_j}^4 <\infty \] We also prove that \[ \int (1-\cos\theta)^2 \log w(\theta)\, \f{d\theta}{2\pi} >-\infty \] if and only if \[ \sum_{j=0}^\infty \abs{\alpha_{j+2}-2\alpha_{j+1} +\alpha_j}^2 + \abs{\alpha_j}^6 <\infty \] \end{abstract} \maketitle \section{Introduction} \lb{s1} This paper is a contribution to the theory of orthogonal polynomials on the unit cirle (OPUC); see \cite{GBk,OPUC1,OPUC2,Szb} for background. Throughout, $d\mu$ will be a nontrivial probability measure on the unit circle, $\partial\bbD$, in $\bbC$, which we suppose has the form \begin{equation} \lb{1.1} d\mu = w(\theta)\, \f{d\theta}{2\pi} + d\mu_\s \end{equation} where $d\mu_\s$ is singular with respect to Lebesgue measure $d\theta$ on $\partial\bbD$. The Carath\'eodory and Schur functions, $F$ and $f$, associated to $d\mu$ are given for $z\in\bbD$ by \begin{align} F(z) &= \int \f{e^{i\theta}+z}{e^{i\theta}-z}\, d\mu(\theta) \lb{1.2} \\ &= \f{1+zf(z)}{1-zf(z)} \lb{1.3} \end{align} The Verblunsky coefficients $\{\alpha_j\}_{j=0}^\infty$ can be defined inductively by the Schur algorithm \begin{equation} \lb{1.4} f(z) = \f{\alpha_0 + zf_1 (z)}{1+z\bar\alpha_0 f_1(z)} \end{equation} which defines $\alpha_0\in \bbD$ and $f_1$. Iterating gives $\alpha_1, \alpha_2, \dots$ and $f_2, f_3, \dots$. That $\alpha_j\in\bbD$ (rather than just $\bar\bbD$) follows from the assumption that $d\mu$ is nontrivial, that is, has infinite support so $f$ is not a finite Blaschke product. Actually, \eqref{1.4} defines what are usually called Schur parameters; the Verblunsky coefficients are defined by a recursion relation on the orthogonal polynomials. The equality of these recursion coefficients and the Schur parameters of \eqref{1.4} is a theorem of Geronimus \cite{Ger44}; see \cite{OPUC1}. We will use the definition in \eqref{1.4}. The most famous result in OPUC is Szeg\H{o}'s theorem which, in Verblunsky's format \cite{V36}, says \begin{equation} \lb{1.5} \log \biggl(\, \prod_{j=0}^\infty (1-\abs{\alpha_j}^2)\biggr) = \int \log (w(\theta)) \, \f{d\theta}{2\pi} \end{equation} In this expression, both sides are nonpositive (since $\abs{\alpha_j}<1$, and Jensen's inequality implies $\int \log (w(\theta))\, \f{d\theta}{2\pi}\leq \log (\int w(\theta)\f{d\theta}{2\pi}) \leq \log (\mu(\partial\bbD))$). Moreover, \eqref{1.5} includes the statement that both sides are finite (resp.~$-\infty$) simultaneously. Thus \eqref{1.5} implies a spectral theory result. \begin{theorem}\lb{T1.1} \begin{equation} \lb{1.6} \int \log(w(\theta))\, \f{d\theta}{2\pi} > -\infty \Leftrightarrow \sum_{j=0}^\infty \, \abs{\alpha_j}^2 <\infty \end{equation} \end{theorem} This form of the theorem has caused considerable recent interest due to work of Deift-Killip \cite{DeiftK} and Killip-Simon \cite{KS} which motivated a raft of papers \cite{DenJDE,Kup1,Kup2,LNS,NPVY,Sim288,SZ,Zl2004}. In \cite[Section~2.8]{OPUC1}, Simon found a higher-order analog to \eqref{1.6} that allows $\log (w(\theta))$ to be singular at a single point: \begin{theorem}\lb{T1.2} \begin{equation} \lb{1.7} \int (1-\cos\theta) \log(w(\theta))\, \f{d\theta}{2\pi} >-\infty \Leftrightarrow \sum_{j=0}^\infty \, \abs{\alpha_{j+1} -\alpha_j}^2 + \abs{\alpha_j}^4 <\infty \end{equation} \end{theorem} This result allows a single singular point of order $1$ in $\log (w(\theta))$ at $\theta=0$. By a simple rotation argument \cite{OPUC1}, if $\cos(\theta)$ is replaced by $\cos(\theta-\theta_1)$, $\abs{\alpha_{j+1}-\alpha_j}^2$ is replaced by $\abs{\alpha_{j+1} -e^{-i\theta_1}\alpha_j}^2$. Our goal in this paper is to analyze two singularities or a single double singularity. We will prove that \begin{theorem} \lb{T1.3} For $\theta_1\neq \theta_2$, \begin{equation} \lb{1.8} \begin{split} \int (1-\cos&(\theta-\theta_1))(1-\cos(\theta-\theta_2)) \log(w(\theta)) \, \f{d\theta}{2\pi} >-\infty \\ &\Leftrightarrow \sum_{j=0}^\infty \bigl|\bigr\{ \, (\delta -e^{-i\theta_2})(\delta -e^{-i\theta_1}) \alpha\bigr\}_j\bigr|^2 + \abs{\alpha_j}^4 <\infty \end{split} \end{equation} \end{theorem} In this theorem, $\delta$ is the operator on sequences \begin{equation} \lb{1.9} (\delta\alpha)_j =\alpha_{j+1} \end{equation} We will also prove a result for $\theta_1 =\theta_2$. \begin{theorem}\lb{T1.4} \begin{equation} \lb{1.10} \int (1-\cos\theta)^2 \log (w(\theta)) \, \f{d\theta}{2\pi} >-\infty \Leftrightarrow \sum_{j=0}^\infty \, \abs{\alpha_{j+2} -2\alpha_{j+1} +\alpha_j}^2 + \abs{\alpha_j}^6 <\infty \end{equation} \end{theorem} Again, one can replace $\cos(\theta)$ by $\cos(\theta-\theta_1)$ if $\{\alpha_{j+2} -2 \alpha_{j+1} +\alpha_j\}_j$ is replaced by $\{(\delta-e^{-i\theta_1})^2\alpha\}_j$. Given the form of these theorems, it is natural to conjecture the situation for arbitrarily many singularities: \begin{conjecture} \lb{Cn1.5} For $\{\theta_k\}_{k=1}^\ell$ distinct in $[0,2\pi)$, \begin{equation} \lb{1.11} \begin{split} \int \prod_{k=1}^\ell &(1-\cos (\theta- \theta_k))^{m_k} \log (w(\theta)) \, \f{d\theta}{2\pi} >-\infty \\ &\Leftrightarrow \sum_{k=0}^\infty \biggl| \biggl\{\, \prod_{k=1}^\ell [\delta - e^{-i\theta_k}]^{m_k} \alpha \biggr\}_j\biggr|^2 + \abs{\alpha_j}^{2\max (m_k)+2}<\infty \end{split} \end{equation} \end{conjecture} Independently of our work, Denisov-Kupin \cite{DenKup2} have found conditions on the $\alpha$'s equivalent to the left side of \eqref{1.11} being finite. However, their conditions are complicated and even for the case $\sum_{k=1}^\ell m_k =2$, it is not clear they are equivalent to the ones we have in Theorems~\ref{T1.3} and \ref{T1.4} (although they must be!). In Section~\ref{s2}, we review the features we need of the relative Szeg\H{o} function which will play a critical role in our proofs, and we compute its first two Taylor coefficients. In Section~\ref{s3}, we prove Theorem~\ref{T1.3} in the special case $\theta_1=0$, $\theta_2=\pi$, and in Section~\ref{s4}, we prove Theorem~\ref{T1.4}. With these two warmups done, we turn to the general result, Theorem~\ref{T1.3}, in Section~\ref{s5}. The details of this are sufficiently messy that we do not think this direct approach is likely to yield our conjecture. \smallskip We would like to thank S.~Denisov for telling us of his work \cite{DenKup2}. \smallskip \section{The Relative Szeg\H{o} Function} \lb{s2} In Section~2.9 of \cite{OPUC1}, Simon introduced the relative Szeg\H{o} function, defined by \begin{equation} \lb{2.1} (\delta_0 D)(z) = \f{1-\bar\alpha_0 f(z)}{\rho_0} \, \f{1-zf_1(z)}{1-zf(z)} \end{equation} where \begin{equation} \lb{2.2} \rho_k = (1-\abs{\alpha_k}^2)^{1/2} \end{equation} and $f,f_1$ are given by \eqref{1.3} and \eqref{1.4}. The key property of $\delta_0 D$ we will need and the reason it was introduced is \begin{theorem}[\mbox{\cite[Theorem 2.9.3]{OPUC1}}] \lb{T2.1} Let $d\mu_1$ be the measure whose Verblunsky coefficients are $(\alpha_1, \alpha_2, \dots)$. Let $w$ be given by \eqref{1.1} and $w_1$ by \begin{equation} \lb{2.3} d\mu_1 = w_1(\theta) \, \f{d\theta}{2\pi} + d\mu_{1,\s} \end{equation} Suppose $w(\theta)\neq 0$ for a.e.~$e^{i\theta}$ in $\partial\bbD$. Then the same is true for $w_1$ and \begin{equation} \lb{2.4} (\delta_0D)(z) =\exp\biggl( \f{1}{4\pi} \int \f{e^{i\theta}+z}{e^{i\theta}-z}\, \log \biggl( \f{w(\theta)}{w_1(\theta)}\biggr)d\theta\biggr) \end{equation} \end{theorem} As in \cite{KS,Sim288,SZ}, this is the basis for step-by-step sum rules, as we will see. To prove Theorems \ref{T1.3} and \ref{T1.4}, we will need to start with computing the first three Taylor coefficients of $\log ((\delta_0 D)(z))$. \begin{theorem}\lb{T2.2} We have that \begin{equation} \lb{2.5} \log (\delta_0 D(z)) =A_0+A_1 z + A_2 z^2 +O(z^3) \end{equation} where \begin{align} A_0 &=\log \rho_0 \lb{2.6} \\ A_1 &= \alpha_0 -\alpha_1 -\bar\alpha_0\alpha_1 \lb{2.7} \\ A_2 &= \tfrac12\, \alpha_0^2 -\tfrac12\, \alpha_1^2 + \alpha_1 -\alpha_2 -\alpha_1 \abs{\alpha_0}^2 + \alpha_2 \abs{\alpha_1}^2 -\bar\alpha_0 \alpha_2 \rho_1^2 + \tfrac12\, \bar\alpha_0^2 \alpha_1^2 \lb{2.8} \end{align} \end{theorem} \begin{proof} $f_2(0) = \alpha_2$, so \[ f_1 = \f{zf_2+\alpha_1}{1+z\bar\alpha_1 f_2} = \alpha_1 + z\alpha_2 \rho_1^2 + O(z^2) \] Thus \[ f= \f{zf_1 + \alpha_0}{1+z\bar\alpha_1 f_1} = \alpha_0 + z\alpha_1 \rho_0^2 + z^2 \rho_0^2 (\alpha_2 \rho_1^2 -\bar\alpha_0 \alpha_1^2) + O(z^3) \] Plugging these into \eqref{2.1} yields the required Taylor coefficients. \end{proof} {\it Remarks.} 1. Denisov-Kupin \cite{DenKup2} do what is essentially the same calculation using the CMV matrix. \smallskip 2. \eqref{3.2} and \eqref{3.3} below show that \eqref{2.4} implies \begin{align} \int \log \biggl( \f{w(\theta)}{w_1(\theta)}\biggr) \f{d\theta}{2\pi} &= 2A_0 \lb{2.9} \\ \int \log \biggl( \f{w(\theta)}{w_1(\theta)}\biggr)e^{-im\theta}\, \f{d\theta}{2\pi} &= \begin{cases} A_m & m=1,2 \\ \bar A_{-m} & m=-1,-2 \end{cases} \lb{2.10} \end{align} \smallskip \section{Singularities at Antipodal Points} \lb{s3} As a warmup, in this section we prove the following, which is Theorem~\ref{T1.3} for $\theta_1 =0$, $\theta_2=\pi$. By the remark after Theorem \ref{T1.2} this also gives the result for any antipodal $\theta_1$ and $\theta_2$. \begin{theorem}\lb{T3.1} \begin{equation} \lb{3.1} \int (1-\cos^2 (\theta)) \log w(\theta)\, \f{d\theta}{2\pi} >-\infty \Leftrightarrow \sum_{j=0}^\infty \, \abs{\alpha_{j+2} -\alpha_j}^2 + \abs{\alpha_j}^4 <\infty \end{equation} \end{theorem} {\it Remark.} Let $\alpha_j$ be given and let $\beta_j$ be the sequence $(\alpha_0, 0, \alpha_1, 0, \alpha_2, 0, \dots)$. Then (see Example~1.6.14 of \cite{OPUC1}), $w^{(\beta)} (\theta) =\f12 w^{(\alpha)}(\f12\theta)$ and the RHS of \eqref{3.1} for $\beta =$ the RHS of \eqref{1.7} for $\alpha$. Thus \eqref{3.1} for $\beta$ is \eqref{1.7} for $\alpha$. This shows, in particular, that if a result like \eqref{3.1} holds, it must involve $\abs{\alpha_j}^4$, rather than, say, $\abs{\alpha_j}^6$. \smallskip We begin by noting that if $Q(\theta)$ is real and \begin{equation} \lb{3.2} Q(\theta) =\sum_{n=-\infty}^\infty b_n e^{in\theta} \end{equation} then \begin{equation} \lb{3.3} \int \f{e^{i\theta}+z}{e^{i\theta}-z}\, Q(\theta) \f{d\theta}{2\pi} = b_0 + 2\sum_{n=1}^\infty b_n z^n \end{equation} since $(e^{i\theta}+z)/(e^{i\theta}-z) = 1+2\sum_{n=1}^\infty z^n e^{-in\theta}$. Thus, by \eqref{2.9}, \eqref{2.10}, and \begin{equation} \lb{3.4} 1-\cos^2 (\theta) =\tfrac14\, (2-e^{2i\theta} -e^{-2i\theta}) \end{equation} we have \begin{equation} \lb{3.5} \int (1-\cos^2(\theta))\log \biggl( \f{w(\theta)}{w_1(\theta)}\biggr) \f{d\theta}{2\pi} =A_0 -\tfrac12\, \Real (A_2) \end{equation} with $A_0$ given by \eqref{2.6} and $A_2$ by \eqref{2.8}. \begin{lemma} \lb{L3.2} We have that \begin{equation} \lb{3.6} A_0 -\tfrac12\, \Real (A_2) =B_0 + C_0 + D_0 + F_0 -F_1 + G_0 -G_2 \end{equation} where \begin{align} B_j &= \tfrac12\, \bigl[ \log (1-\abs{\alpha_j}^2) + \abs{\alpha_j}^2 + \tfrac 12\abs{\alpha_j}^4 \bigr] \lb{3.7} \\ C_j &= -\tfrac14\, (1-\abs{\alpha_{j+1}}^2) \abs{\alpha_j-\alpha_{j+2}}^2 \lb{3.8} \\ D_j &= -\tfrac18\, (\abs{\alpha_{j+1}^2 + \alpha_j^2}^2 + 4\abs{\alpha_j \alpha_{j+1}}^2)\lb{3.9} \\ F_j &= -\tfrac12\, \Real ( \tfrac12\, \alpha_j^2 + \alpha_{j+1} -\alpha_{j+1} \abs{\alpha}^2) + \tfrac14\, \abs{\alpha_{j+1}}^2 \abs{\alpha_j}^2 - \tfrac18\, \abs{\alpha_j}^4 \lb{3.10} \\ G_j &= -\tfrac14\, \abs{\alpha_j}^2 \notag \end{align} \end{lemma} {\it Remark.} \eqref{3.5}/\eqref{3.6} is thus the step-by-step sum rule in the spirit of \cite{KS,Sim288,SZ}. \begin{proof} This is a straightforward but tedious calculation. The first term in $B_0$ is just $A_0$ (since $\log \rho_j =\f12 \log (1-\abs{\alpha_j}^2)$). $A_2$ is responsible for the $\Real(\cdot)$ terms in $F_0 - F_1$ and the cross-terms in $\abs{\alpha_j-\alpha_{j+2}}^2$ and $\abs{\alpha_{j+1}^2 +\alpha_j^2}^2$. The $\abs{\alpha_j}^2 + \abs{\alpha_{j+2}}^2$ term in $C_0$ is turned into $2\abs{\alpha_j}^2$ by $G_0 -G_{2}$, and then cancelled by the $\abs{\alpha_j}^2$ term in $B_0$. Similarly, the $\abs{\alpha_j}^4 + \abs{\alpha_{j+1}}^4$ in $D_0$ (after adding the $\abs{\alpha_j}^4$ terms in $F_0-F_1$) cancels the $\abs{\alpha_j}^4$ term in $B_0$. Finally, the $\abs{\alpha_{j+1}}^2 (\abs{\alpha_j}^2 + \abs{\alpha_{j+2}}^2)$ term in $C_0$ (after being turned into $2\abs{\alpha_{j+1}}^2\abs{\alpha_{j}}^2$ by the $\abs{\alpha_{j+1}}^2 \abs{\alpha_j}^2$ term in $F_0-F_1$) cancels the $4\abs{\alpha_j\alpha_{j+1}}^2$ term in $D_0$. \end{proof} By iterating \eqref{3.5}/\eqref{3.6} and noting the cancellations from the telescoping $F_j -F_{j+1}$ and $G_j -G_{j+2}$ yields \begin{equation} \lb{3.11} \begin{split} \int (1- &\cos^2 (\theta)) \log \biggl( \f{w(\theta)}{w_{2m}(\theta)}\biggr) \f{d\theta}{2\pi} \\ &= F_0 -F_{2m} + G_0 + G_1 - G_{2m} -G_{2m+1} + \sum_{j=0}^{2m-1} (B_j + C_j + D_j) \end{split} \end{equation} As a final preliminary, we need, \begin{lemma}\lb{L3.3} \begin{SL} \item[{\rm{(i)}}] $\abs{F_j} \leq \f{13}8$; $\abs{G_j} \leq \f14$ %\item[{\rm{(ii)}}] $0\geq B_j \geq - \abs{\alpha_j}^2/6$ \item[{\rm{(ii)}}] $\abs{\alpha_j} <\f12 \Rightarrow c_1\abs{\alpha_j}^6 \le -B_j \le c_2\abs{\alpha_j}^6$ for some $c_2>c_1>0$. \item[{\rm{(iii)}}] $\abs{\alpha_{j+1}}^4 + \abs{\alpha_j}^4 \leq -8D_j \leq 4 (\abs{\alpha_{j+1}}^4 +\abs{\alpha_j}^4)$ \end{SL} \end{lemma} \begin{proof} (i) follows from $\abs{\alpha_j}\leq 1$, (ii) from $-\log (1-x) = \sum_{j=1}^\infty x^j/j$, and (iii) by noting that $2\Real (\alpha_j^2 \alpha_{j+1}^2) + 2\abs{\alpha_j^2 \alpha_{j+1}^2} \geq 0$ and repeated use of $\abs{xy} \leq \f12 (|x|^2 + |y|^2)$. \end{proof} %{\it Remark.} (ii) is actually not needed! \begin{proof}[Proof of Theorem~\ref{T3.1}] We follow the strategy of \cite{KS} as modified by \cite{SZ}. Suppose first that the RHS of \eqref{3.1} holds. Let $w^{(n)}$ be the weight for the $n^{\rm th}$ Bernstein-Szeg\H{o} approximation with Verblunsky coefficients $(\alpha_0, \alpha_1, \dots, \alpha_{n-1}, 0, \dots, 0, \dots)$. By \eqref{3.11} and $(w^{(n)})_{2m}\equiv 1$ for large $m$, \[ \int (1-\cos^2 (\theta)) \log (w^{(n)}(\theta))\, \f{d\theta}{2\pi} = F_0^{(n)} + G_0^{(n)} + G_1^{(n)} + \sum_{j=0}^{n-1} (B_j^{(n)} + C_j^{(n)} + D_j^{(n)}) \] so, by Lemma~\ref{L3.3}, $\abs{\alpha_j}^6 \leq \abs{\alpha_j}^4\to 0$, and RHS of \eqref{3.1}, \begin{equation} \lb{3.12} \inf_n \biggl[ \int (1-\cos^2 (\theta)) \log (w^{(n)} (\theta)) \f{d\theta}{2\pi} \biggr] > -\infty \end{equation} Up to a constant, $\int (1-\cos^2 (\theta)) \log w(\theta)\f{d\theta}{2\pi}$ is an entropy and so upper semicontinuous \cite{KS}. Thus \eqref{3.12} implies \begin{equation} \lb{3.13} \int (1-\cos^2 (\theta)) \log w(\theta)\, \f{d\theta}{2\pi} > -\infty \end{equation} Conversely, suppose \eqref{3.13} holds. Since $\int (1-\cos^2(\theta))\log (w_{2m}(\theta)) \f{d\theta}{2\pi}$ is an entropy up to a constant, it is bounded above \cite{KS}, and so the left side of \eqref{3.11} is bounded below as $m$ varies. Since $F$ and $G$ are bounded and $B,C,D$ are negative, we conclude \[ \sum_{j=0}^\infty -(B_j +C_j +D_j) <\infty \] Since $\sum (-D_j) <\infty$, Lemma~\ref{L3.3} implies $\sum \abs{\alpha_j}^4 <\infty$. This implies $\alpha_j \to 0$, so $\sum (-C_j) <\infty$ implies $\sum \abs{\alpha_j -\alpha_{j+2}}^2 <\infty$. \end{proof} \smallskip \section{Singularity of Order $2$} \lb{s4} Our goal here is to prove Theorem~\ref{T1.4}. Since \begin{align*} (1-\cos\theta)^2 &= \tfrac14\, (2-e^{i\theta}-e^{-i\theta})^2 \\ &= \tfrac32 - e^{i\theta} -e^{-i\theta} + \tfrac14 \, e^{2i\theta} + \tfrac14\, e^{-2i\theta} \end{align*} we see, by \eqref{2.9}/\eqref{2.10} that \begin{equation} \lb{4.1} \int \log \biggl( \f{w(\theta)}{w_1(\theta)}\biggr) (1-\cos\theta)^2 \, \f{d\theta}{2\pi} = 3A_0 -2\Real (A_1) + \tfrac12\, \Real (A_2) \end{equation} with $A_0, A_1, A_2$ given by \eqref{2.6}--\eqref{2.8}. \begin{lemma}\lb{L4.1} The RHS of \eqref{4.1} $= H_0 + I_0 +J_0 + K_0 -K_1 + L_0 -L_2$ where \begin{align*} H_j &=\tfrac32\, [\log (1-\abs{\alpha_j}^2) + \abs{\alpha_j}^2] \\ I_j &= -\tfrac14\, \abs{\alpha_{j+2} -2\alpha_{j+1} +\alpha_j}^2 \\ J_j &=\tfrac14\, (\alpha_j \bar\alpha_{j+2} + \bar\alpha_j \alpha_{j+2}) \abs{\alpha_{j+1}}^2 + \tfrac18\, (\alpha_j^2 \bar\alpha_{j+1}^2 + \bar\alpha_j^2 \alpha_{j+1}) \\ K_j &= -2\Real (\alpha_j) + \tfrac14\, \Real (\alpha_j^2) \\ &\qquad +\tfrac12\, \Real (\alpha_{j+1}) - \tfrac12\, \Real (\alpha_{j+1} \abs{\alpha_j}^2) + \Real [\bar\alpha_{j+1} \alpha_j] - \abs{\alpha_j}^2 \\ L_j &= -\tfrac14\, \abs{\alpha_j}^2 \end{align*} \end{lemma} \begin{proof} The non-cross-terms in $I_0$ are \[ -\tfrac14\, (\abs{\alpha_2}^2 + 4\abs{\alpha_1}^2 + \abs{\alpha_0}^2) = -\tfrac32\, \abs{\alpha_0}^2 + (\abs{\alpha_0}^2 - \abs{\alpha_1}^2) + \tfrac14\, (\abs{\alpha_0}^2 -\abs{\alpha_2}^2) \] which cancel the $\abs{\alpha_0}^2$ term in $H_0$, the final $\abs{\alpha_j}^2$ term in $K_0-K_1$, and the $L_0-L_2$ term. The cross-terms in $I_0$ are \[ \begin{split} -\tfrac12\, \Real (\bar\alpha_2 &\alpha_0) - \Real (\bar\alpha_2 \alpha_1 + \bar\alpha_1 \alpha_0) \\ &=-\tfrac12\, \Real (\bar\alpha_2\alpha_0) + 2\Real (\bar\alpha_0\alpha_1) - \Real (\bar\alpha_0\alpha_1) + \Real (\bar\alpha_1\alpha_2) \end{split} \] The first term comes from the piece of $\tfrac 12\Real (A_2)$ (since $\bar\alpha_0\alpha_2\rho_1^2 = \bar\alpha_0 \alpha_2 (1-\abs{\alpha_1}^2)$, the second from the last term in $-2\Real (A_1)$, and the last two are cancelled by the $\Real (\bar\alpha_{j+1}\alpha_j)$ term in $K_0 -K_1$. The $\alpha_0-\alpha_1$ term in $A_0$ leads to the first term in $K_1 -K_0$. The first term in $J_0$ comes from the second half of $\bar\alpha_0\alpha_2\rho_1^2 =\bar\alpha_0\alpha_2 -\bar\alpha_0 \alpha_2 \abs{\alpha_1}^2$ (the first half in this expression gave a cross-term in $I_j$). The second term in $J_0$ is the $\f12 \bar\alpha_0^2 \alpha_1^2$ term in $A_2$. The remaining terms in $A_2$ give precisely the remaining terms in $K_0 -K_1$. \end{proof} \begin{lemma} \lb{L4.2} The RHS of \eqref{4.1} $= \ti H_0 + \ti I_0 + \ti J_0 + \ti K_0 - \ti K_1 + \ti L_0 - \ti L_2$, where \begin{align*} \ti H_j &= \tfrac32\, \bigl[\log (1-\abs{\alpha_j}^2) + \abs{\alpha_j}^2 + \tfrac 12\abs{\alpha_j}^4 \bigr] \\ \ti I_j &= I_j \\ \ti J_j &= -\tfrac14\, \abs{\alpha_{j+1}}^2 \abs{\alpha_j -\alpha_{j+2}}^2 - \tfrac18\, \abs{\alpha_{j+1}^2 -\alpha_j^2}^2 - \tfrac14\, (\abs{\alpha_{j+1}}^2 - \abs{\alpha_j}^2)^2\\ \ti K_j &= K_j -\tfrac38\, \abs{\alpha_j}^2 - \tfrac14\, \abs{\alpha_{j+1}}^2 \abs{\alpha_j}^2 \\ \ti L_j &= L_j \end{align*} \end{lemma} \begin{proof} The non-cross-terms in the last two terms in $\ti J_0$ give \[ -\tfrac38\, (\abs{\alpha_0}^4 + \abs{\alpha_1}^4) = -\tfrac34\, \abs{\alpha_0}^4 + \tfrac38 (\abs{\alpha_0}^4 -\abs{\alpha_1}^4) \] The first term cancels the $\ti H_0 - H_0$ term, and the second, the first term in $(K_0 - \ti K_0) - (K_1 -\ti K_1)$. The cross-term in $-\f14 (\abs{\alpha_{j+1}}^2 - \abs{\alpha_j}^2)^2$ and the non-cross-terms in $-\f14 \abs{\alpha_{j+1}}^2 \abs{\alpha_j -\alpha_{j+2}}^2$ combine to $-\f14 \abs{\alpha_{j+2}}^2 \abs{\alpha_{j+1}}^2 + \tfrac14 \abs{\alpha_{j+1}}^2 \abs{\alpha_j}^2$ and are cancelled by the second term in $(K_0 -\ti K_0) - (K_1 -\ti K_1)$. The cross-term in $-\f18 \abs{\alpha_{j+1}^2 -\alpha_j^2}^2$ is the second term in $J_0$ and finally, the cross-term in $-\f14 \abs{\alpha_{j+1}}^2 \abs{\alpha_j -\alpha_{j+2}}^2$ is the first term in $J_0$. \end{proof} \begin{lemma}\lb{L4.3} \begin{SL} \item[{\rm{(i)}}] $\abs{\ti K_j} \leq \f{47}{8}$; $\abs{\ti L_j} \le \f14$ %\item[{\rm{(ii)}}] $0\geq \ti H_j \geq - \abs{\alpha_j}^6/2$ \item[{\rm{(ii)}}] $\abs{\alpha_j} < \f12 \Rightarrow d_1\abs{\alpha_j}^6 \le -\ti H_j \le d_2\abs{\alpha_j}^6$ for some $d_2>d_1>0$. \item[{\rm{(iii)}}] $\ti J_j \le 0$ \item[{\rm{(iv)}}] $\sum_{j=0}^\infty (-\ti I_j) + \abs{\alpha_j}^6 <\infty \Rightarrow \sum_{j=0}^\infty \abs{\alpha_{j+1} -\alpha_j}^3 <\infty$ \item[{\rm{(v)}}] $\sum_{j=0}^\infty (-\ti I_j) + \abs{\alpha_j}^6 <\infty \Rightarrow \sum_{j=0}^\infty (-\ti J_j) <\infty$ \end{SL} \end{lemma} {\it Remark.} (iv) is essentially a discrete version of the inequality of Gagliardo \cite{Gag} and Nirenberg \cite{Nir}. \begin{proof} (i) follows from $\abs{\alpha_j} <1$, (ii) is just (ii) of Lemma~\ref{L3.3} (since $\ti H_j =3B_j$), and (iii) is trivial. To prove (iv), we let $\delta$ be given by \eqref{1.9} and let \begin{equation} \lb{4.2} \partial = \delta -1 \end{equation} so since $\delta^* =\delta^{-1}$ ($\delta$ is unitary on $\ell^2$), we have \begin{equation} \lb{4.3} \partial^* = \delta^* -1 = -\delta^{-1} \partial = -\delta^*\partial \end{equation} As a result, if $\alpha$ is a finite sequence, then \begin{align} \sum_n \, \abs{(\partial\alpha)_n}^3&= \sum_n (\partial\alpha)_n (\partial\bar\alpha)_n \abs{\partial\alpha}_n \notag \\ &=-\sum_n (\delta\alpha)_n [\partial \{(\partial\bar\alpha) \abs{\partial\alpha}\}]_n \lb{4.4} \end{align} Moreover, we have a discrete Leibnitz rule, \begin{align} \partial (fg) &= (\delta f)(\delta g)-fg \notag \\ &= (\delta f) \partial g + (\partial f)g \lb{4.5} \end{align} and since $\abs{a-b} \geq \abs{a}-\abs{b}$ by the triangle inequality, \begin{equation} \lb{4.6} \abs{\partial\abs{f}} \leq \abs{\partial f} \end{equation} which is a discrete Kato inequality. By \eqref{4.5}, \[ \partial \{(\partial \bar\alpha)\abs{\partial\alpha}\} = [\delta(\partial\bar\alpha)] \partial \abs{\partial\alpha} + (\partial^2 \bar\alpha)\abs{\partial\alpha} \] so, by \eqref{4.6}, \[ \abs{\partial \{(\partial\bar\alpha)\abs{\partial\alpha}\}} \leq \abs{\partial^2 \alpha} \, \abs{\delta(\partial\bar\alpha)} + \abs{\partial^2 \alpha} \, \abs{\partial\alpha} \] Using H\"older's inequality with $\f16 + \f12 + \f13 =1$ and \eqref{4.4}, we get \[ \|\partial\alpha\|_3^3 \leq 2 \| \alpha\|_6 \|\partial^2 \alpha\|_2 \|\partial\alpha\|_3 \] (because $\|\delta\alpha\|_p=\|\alpha\|_p$), so \begin{equation} \lb{4.7} \sum_n \, \abs{(\partial\alpha)_n}^3 \leq 2^{3/2} \biggl(\,\sum_n \, \abs{\alpha_n}^6\biggr)^{1/4} \biggl(\, \sum_n \, \abs{(\partial^2 \alpha)_n}^2\biggr)^{3/4} \end{equation} Having proven \eqref{4.7} for $\alpha$'s of finite support, we get it for any $\alpha$ with the right side finite since $\sum_n \abs{\alpha_n}^6 <\infty$ implies $\alpha_n\to 0$, which allows one to cut off $\alpha$ at $N$ and take $N\to\infty$ in \eqref{4.7}. But \eqref{4.7} implies (iv). To prove (v), we control the individual terms in $\sum (-\ti J_j)$. First, \begin{align*} \|\abs{\alpha}^2 \abs{\delta^2 \alpha -\alpha}^2\|_1 &\leq \| \, \alpha^2\|_3 \, \| \abs{\delta^2 \alpha -\alpha}^2\|_{3/2} \\ \intertext{(by H\"older's inequality with $\f13 + \f23 =1$)} &\leq 4 \|\alpha\|_6^2\, \|\partial\alpha\|_3^2 <\infty \end{align*} (by first using $\|\delta^2\alpha-\alpha\|_3\le 2\|\partial\alpha\|_3$ and then (iv)). Next, \[ \abs{\alpha_{j+1}^2 -\alpha_j^2}^2 \leq (\abs{\alpha_{j+1}} + \abs{\alpha_{j+1}})^2 \abs{\alpha_{j+1} -\alpha_j}^2 \] can be controlled as the first term was and the final term is controlled in the same way since $|\alpha_{j+1}|^2-|\alpha_j|^2 \le |\alpha_{j+1}^2-\alpha_j^2|$. \end{proof} \begin{proof}[Proof of Theorem~\ref{T1.4}] Suppose first that the right-hand side of \eqref{1.10} holds, that is, $\alpha\in\ell^6$ and $\partial^2\alpha\in\ell^2$. Iterate $n$ times \eqref{4.1}/Lemma~\ref{L4.2} for the $n^{\rm th}$ Bernstein-Szeg\H{o} approximation (with weight $w^{(n)}$) to obtain \[ \inf_n\, \biggl[ \int (1-\cos\theta)^2 \log (w^{(n)}(\theta)) \,\f{d\theta}{2\pi} \biggr] > -\infty \] since the left side is just \[ \inf_n \Big[\ti K_0^{(n)}+\ti L_0^{(n)}+\ti L_1^{(n)} +\sum_{j=0}^{n-1} (\ti H_j^{(n)} + \ti I_j^{(n)} + \ti J_j^{(n)}) \Big] \] which is finite by Lemma~\ref{L4.3} and the hypothesis. Again we have that $\int (1-\cos\theta)^2 \log w(\theta) \f{d\theta}{2\pi}$ is an entropy up to a constant and so upper semicontinuous. Thus RHS of \eqref{1.10} $\Rightarrow$ LHS of \eqref{1.10}. For the opposite direction, as in the last section, we use iterated \eqref{4.1}/Lemma~\ref{L4.2} plus the fact that $\int (1-\cos\theta)^2 \log (w_{2m}(\theta)) \f{d\theta}{2\pi}$ is bounded from above to conclude \[ \sum_{j=0}^\infty -(\ti H_j + \ti I_j + \ti J_j) <\infty \] Since each is positive, $\sum (-\ti H_j) <\infty$, which implies $\sum \abs{\alpha_j}^6 <\infty$ by (ii) of Lemma~\ref{L4.3}, and $\sum_{j=0}^\infty (-\ti I_j)<\infty$, which implies $\partial^2\alpha\in\ell^2$. \end{proof} \smallskip \section{The General Case} \lb{s5} Finally, we turn to the general case of Theorem \ref{T1.3}, and we define \begin{equation} \lb{5.1} \calI_m \equiv \int \big[1-\cos(\theta-\theta_1)\big] \big[1-\cos(\theta-\theta_2)\big] \log \biggl( \f{w(\theta)}{w_m(\theta)}\biggr) \f{d\theta}{2\pi} \end{equation} %for $\theta_2-\theta_1\neq 0,\pi$. Using \eqref{2.9} and \eqref{2.10}, we obtain \begin{equation} \lb{5.2} \calI_1 = \frac {4+e^{i(\theta_1-\theta_2)}+e^{-i(\theta_1-\theta_2)}} 4 A_0 - \Real \big[ (e^{i\theta_1}+e^{i\theta_2}) A_1 \big] + \tfrac 12 \Real \big[ e^{i(\theta_1+\theta_2)} A_2 \big] \end{equation} The situation is now somewhat more complicated than in the previous sections and it will be more convenient to work with $\calI_m$ from the start, only keeping track of the essential components of the sums (analogs of $\sum (B_j+C_j+D_j)$ and $\sum (\ti H_j+\ti I_j+\ti J_j)$ above) and ignore the ones that are always bounded and hence irrelevant for us (analogs of $F_0-F_1+G_0+G_1-G_m-G_{m+1}$ and $\ti K_0-\ti K_m+\ti L_0+\ti L_1-\ti L_m+\ti L_{m+1}$). Hence substituting \eqref{2.6}--\eqref{2.8} in \eqref{5.2} and iterating, we obtain \begin{align*} \calI_m & = C_{\alpha,m} + \frac {4+e^{i(\theta_1-\theta_2)}+e^{-i(\theta_1-\theta_2)}}4 \sum_{j=0}^{m-1} \log (1-|\alpha_j|^2) \\ & + \sum_{j=0}^{m-1} \Real \Big\{ \big( e^{i\theta_1}+e^{i\theta_2} \big) \alpha_{j+1}\bar\alpha_j - \tfrac 12 e^{i(\theta_1+\theta_2)} \big[ \alpha_{j+2}\bar\alpha_j(1-|\alpha_{j+1}|^2) -\tfrac 12 \alpha_{j+1}^2\bar\alpha_j^2 \big] \Big\} \end{align*} where \begin{align*} C_{\alpha,m} & \equiv -\Real \big[ (e^{i\theta_1}+e^{i\theta_2}) (\alpha_0-\alpha_m) \big] \\ &+ \tfrac 12 \Real \big[ e^{i(\theta_1+\theta_2)} \big(\tfrac 12 \alpha_0^2 - \tfrac 12 \alpha_m^2 + \alpha_1 - \alpha_{m+1} - \alpha_1|\alpha_0^2| + \alpha_{m+1}|\alpha_m|^2 \big) \big] \end{align*} We let \[ \beta_j\equiv \alpha_j e^{i(\theta_1+\theta_2)j/2} \] and \[ a\equiv \tfrac 12 \big( e^{i(\theta_1-\theta_2)/2}+e^{-i(\theta_1-\theta_2)/2} \big) \in(-1,1) \] We will assume $a\neq 0$ since the case when $\theta_1$ and $\theta_2$ are antipodal follows from Theorem \ref{T3.1}. With $C_{\beta,m}\equiv C_{\alpha,m}$ and all the sums taken from $0$ to $m-1$, the above becomes \begin{align} \calI_m = & C_{\beta,m} + \big(\tfrac 12 +a^2 \big) \sum \log (1-|\beta_j|^2) + a \sum \big[ \beta_{j+1}\bar\beta_j + \bar\beta_{j+1}\beta_j \big] \notag \\ & -\tfrac 14 \sum \big[ \beta_{j+2}\bar\beta_j (1-|\beta_{j+1}|^2) + \bar\beta_{j+2}\beta_j (1-|\beta_{j+1}|^2) \big] \notag \\ & + \tfrac 18 \sum \big[ \beta_{j+1}^2\bar\beta_j^2 + \bar\beta_{j+1}^2\beta_j^2 \big] \lb{5.3} \end{align} In the following manipulations with the sums, we will use $C_{\beta,m}$ as a general pool/depository of terms that will be added/left over in order to keep all the sums from $0$ to $m-1$. Its value will therefore change along the argument, but it will always depend on a few $\beta_j$'s with $j$ close to $0$ or $m$ only (i.e., it will gather all the ``irrelevant'' terms) and will always be bounded by a universal constant. \begin{lemma} \lb{L5.1} With $C_{\beta,m}$ universally bounded, we have \begin{align} \calI_m =& C_{\beta,m} + \big(\tfrac 12 +a^2 \big) \sum \big[\log (1-|\beta_j|^2) + |\beta_j|^2 + \tfrac 12 |\beta_j|^4 \big] \notag \\ & - \tfrac 14 \sum (1-|\beta_{j+1}|^2) \big|\beta_{j+2}-2a\beta_{j+1}+\beta_j \big|^2 \notag \\ & -\tfrac 14 \sum |\beta_{j+1}|^2 \big|\beta_{j+2}-2a\beta_{j+1} \big|^2 -\tfrac 14 \sum |\beta_{j+1}|^2 \big| \beta_{j}-2a\beta_{j+1} \big|^2 \notag \\ & - \tfrac 18 \sum \big|\beta_{j+1}^2-\beta_j^2 \big|^2 + \tfrac{1}2 a^2 \sum|\beta_j|^4 \lb{5.4} \end{align} with all the sums taken from $0$ to $m-1$. \end{lemma} {\it Remarks.} 1. This enables us to prove the ``$\Leftarrow$'' part of \eqref{1.8} (even if $a=0$) since \begin{equation} \lb{5.5} \big| \big\{ (\delta-e^{-i\theta_2})(\delta-e^{-i\theta_1}) \alpha \big\}_j \big| = \big|\beta_{j+2}-2a\beta_{j+1}+\beta_j \big| \end{equation} But to prove the other implication, we first need to deal with the last sum in \eqref{5.4}, which has the ``wrong'' sign. \smallskip 2. Note that we actually did not need to exclude the case $a=0$ since then the last sum in \eqref{5.4} vanishes and an examination of \eqref{5.4} shows that $\lim_{m\to\infty} \calI_m>-\infty$ if and only if the RHS of \eqref{1.8} holds. An argument from the proofs of Theorems \ref{T1.3} and \ref{T1.4} then gives the ``$\Rightarrow$'' part of \eqref{1.8}. \begin{proof} Multiplying out the terms in the second, third, and fourth sums of \eqref{5.4} and after obvious cancellations, we are left with \[ -\tfrac 14 \sum \Big[ |\beta_{j+1}|^2 \big( 4a^2 |\beta_{j+1}|^2 - \beta_{j+2}\bar\beta_j -\bar\beta_{j+2}\beta_j \big) + \big|\beta_{j+2}-2a\beta_{j+1}+\beta_j \big|^2 \Big] \] But this is just \begin{equation} \lb{5.6} -\tfrac 14\sum \big[ |\beta_{j+2}|^2 + 4a^2|\beta_{j+1}|^2 + |\beta_{j}|^2 + 4a^2 |\beta_{j+1}|^4 \big] \end{equation} plus the second and third sums in \eqref{5.3}, the latter written as \[ \tfrac 12 a \sum [ \beta_{j+2}\bar\beta_{j+1} + \bar\beta_{j+2}\beta_{j+1} +\beta_{j+1}\bar\beta_j + \bar\beta_{j+1}\beta_j ] \] (with $C_{\beta,m}$ keeping the change). Adding the fifth and sixth sums in \eqref{5.4} to \eqref{5.6} and subtracting the last sum in \eqref{5.3}, we obtain \[ -\tfrac 14 \sum (2+4a^2)|\beta_j|^2 - \tfrac 18 \sum (2+4a^2)|\beta_j|^4 \] (again replacing all $|\beta_{j+1}|$ and $|\beta_{j+2}|$ by $|\beta_{j}|$ and adding the difference to $C_{\beta,m}$). But this together with the first sum in \eqref{5.4} gives exactly the first sum in \eqref{5.3}. \end{proof} We define \[ \gamma_j\equiv \beta_{j+2}-2a\beta_{j+1}+\beta_j \] then the second, third, and fourth sums in \eqref{5.4} involve $|\gamma_j|$, $|\gamma_j-\beta_j|$ and $|\gamma_j-\beta_{j+2}|$. Using $|x-y|^2 \ge |x|^2 + |y|^2 - 2|x||y|$ for the last two, we obtain (with a new $C_{\beta,m}$) \begin{align} (-8)\calI_m \ge & C_{\beta,m} + \sum O(|\beta_j|^6) + \sum (2+2|\beta_{j+1}|^2) |\gamma_j|^2 \notag \\ & + 4 \sum |\beta_{j+1}|^2|\beta_j|^2 -4 \sum |\beta_{j+1}|^2 \big( |\beta_{j+2}| + |\beta_{j}| \big) |\gamma_j| \notag \\ & + \sum \big|\beta_{j+1}^2-\beta_j^2 \big|^2 - 4a^2 \sum|\beta_{j+1}|^4 \lb{5.7} \end{align} since \[ \log (1-|\beta_j|^2) + |\beta_j|^2 + \tfrac 12 |\beta_j|^4 = O(|\beta_j|^6) \] Next, we use $-4xy\ge -8x^2-\tfrac 12 y^2$ with $x=|\beta_{j+1}|^2 ( |\beta_{j+2}| + |\beta_{j}|)$ and $y=|\gamma_j|$ to estimate the fourth sum by $\sum O(|\beta_j|^6)-\tfrac 12\sum |\gamma_j|^2$. Also, \begin{align*} - 4a^2 \sum |\beta_{j+1}|^4 &= - \sum|\beta_{j+1}|^2 |\beta_{j+2}+\beta_j-\gamma_j|^2 \\ & \ge - \sum |\beta_{j+1}|^2 |\beta_{j+2}+\beta_j|^2 - \sum |\beta_{j+1}|^2 |\gamma_j|^2 \\ & \qquad \qquad -2 \sum |\beta_{j+1}|^2|\beta_{j+2}+\beta_j||\gamma_j| \\ & \ge C_{\beta,m} - 4\sum |\beta_{j+1}|^2|\beta_{j}|^2 - \sum |\beta_{j+1}|^2 |\gamma_j|^2 \\ & \qquad \qquad - \sum O(|\beta_j|^6) - \tfrac 14 \sum |\gamma_j|^2 \end{align*} again using $-2xy\ge -4x^2-\tfrac 14 y^2$. Plugging these into \eqref{5.7}, we have \[ (-8)\calI_m \ge C_{\beta,m} + \sum O(|\beta_j|^6) + \sum \big( \tfrac 54+|\beta_{j+1}|^2 \big) |\gamma_j|^2 + \sum \big|\beta_{j+1}^2-\beta_j^2 \big|^2 \] The last sum is just $\sum \tfrac 12 (|\beta_{j+2}^2-\beta_{j+1}^2|^2 + |\beta_{j+1}^2-\beta_j^2|^2)$ plus a piece that goes into $C_{\beta,m}$. Letting $\veps\equiv \tfrac 13\min\{2|a|, 2-2|a|\}>0$, we obtain \begin{align*} |\beta_{j+1}|^2 |\gamma_j|^2 +\tfrac 12 |\beta_{j+2}^2-\beta_{j+1}^2|^2 + \tfrac 12 |\beta_{j+1}^2-\beta_j^2|^2 \ge \tfrac 12\veps^4|\beta_{j+1}|^4 \end{align*} Indeed, if the third term is smaller than $\tfrac 12\veps^4|\beta_{j+1}|^4$, then $|\beta_{j}-\beta_{j+1}|$ or $|\beta_{j}+\beta_{j+1}|$ is less than $\veps|\beta_{j+1}|$, and similarly for the second term. But then $|\beta_{j+2}+\beta_j|/|\beta_{j+1}|\in [0,2\veps)\cup(2-2\veps,2+2\veps)$ and so $|\gamma_j|/|\beta_{j+1}|\ge \min\{2|a|-2\veps, 2-2\veps-2|a|\}\ge\veps$, meaning that the first term is at least $\veps^2 |\beta_{j+1}|^4$. So finally, \[ (-8)\calI_m \ge C_{\beta,m} + \sum O(|\beta_j|^6) + \sum |\gamma_j|^2 + \tfrac 12\veps^4\sum |\beta_{j}|^4 \] that is (by \eqref{5.5} and the definition of $\beta_j$, $\gamma_j$), \begin{equation} \lb{5.8} \calI_m \le C_{\alpha,m} + \sum O(|\alpha_j|^6) - \tfrac 18 \sum \big| \big\{ (\delta-e^{-i\theta_2})(\delta-e^{-i\theta_1}) \alpha \big\}_j \big|^2 - \tfrac 1{16}\veps^4\sum |\alpha_{j}|^4 \end{equation} \begin{proof}[Proof of Theorem \ref{T1.3}] If the RHS of \eqref{1.8} holds, then the RHS of \eqref{5.4} for the $n^{\rm th}$ Bernstein-Szeg\H o approximation (with $m\ge n$) is bounded (in $n$), and so \[ \inf_n \bigg[ \int \big[1-\cos(\theta-\theta_1)\big] \big[1-\cos(\theta-\theta_2)\big] \log (w^{(n)}(\theta)) \f{d\theta}{2\pi} \bigg] > -\infty \] By upper semicontinuity of the above integral (which is again an entropy up to a constant), we obtain the LHS of \eqref{1.8}. Conversely, assume the LHS of \eqref{1.8} holds. Then the essential support of $w$ all of $\partial\bbD$, and so by Rakhmanov's theorem \cite{Rak}, $|\alpha_j|\to 0$. Hence, starting from some $j$, we have $O(|\alpha_j|^6)\le \tfrac 1{32}\veps^4|\alpha_j|^4$ and so \begin{equation} \lb{5.9} \calI_m \le D_{\alpha,m} - \tfrac 18 \sum \big| \big\{ (\delta-e^{-i\theta_2})(\delta-e^{-i\theta_1}) \alpha \big\}_j \big|^2 - \tfrac 1{32}\veps^4\sum |\alpha_{j}|^4 \end{equation} for large $m$ and some bounded (in $m$) $D_{\alpha,m}$. As in the previous sections, $\int \big[1-\cos(\theta-\theta_1)\big] \big[1-\cos(\theta-\theta_2)\big] \log (w_m(\theta)) \f{d\theta}{2\pi}$ is bounded above, and so $\calI_m$ is bounded below by the hypothesis. \eqref{5.9} then shows that the RHS of \eqref{1.8} holds. \end{proof} \bigskip %%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{thebibliography}{100} %%%%%%%%%%%%%%%%%%%%%%%%%%%%% \smallskip \bi{DeiftK} P.A. Deift and R. 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