\documentclass[12pt,draft]{amsart} \usepackage{amsmath,amsthm} %\usepackage{mathtext} \usepackage{amsfonts} \usepackage[T2A]{fontenc} \usepackage[cp1251]{inputenc} %\usepackage[centertags]{amsmath} \usepackage{amsfonts} \newcommand*{\cal}{\mathcal} \newtheorem{theorem}{Theorem} \newtheorem{prop}{Proposition} \DeclareMathOperator{\spann}{span} \tolerance 900 \begin{document} \begin{center}{\large\bf SCHWARZ LEMMA AND OPTIMAL RECOVERY OF FUNCTIONS IN $H^2$} \end{center} \vspace{3mm} \begin{center} {\bf Konstantin Yu. Osipenko,$^1$, Michael I. Stessin$^2$} \end{center} \begin{center}\textit{$^1$ "MATI" - Russian State Technological University, Orshanskaya Str. 3, Moscow, 121552, Russia} \end{center} \begin{center} \textit{$^2$ Department of Mathematics and Statistics, University at Albany, Albany, NY 12222, USA\\ e-mail: $^{1}$kosipenko@yahoo.com, $^2$stessin@math.albany.edu }\end{center} \begin{abstract} This paper is devoted to optimal method of the reconstruction of the value of the function $f$ whose values on some sets are known. \end{abstract} {\it AMS Subject Classification: 30H05, 30E10} \medskip {\bf Keywords. Extremal problems, optimal recovery, Hardy spaces, Shwartz lemma} \bigskip Let $D \subset C^k$ be a domain, $\nu $ be a probability measure on $\overline{D}$ and $X$ be a closed subspace of $L^2(\nu).$ Consider $D_0,\dots ,D_n \subset D$ and probability measures $\mu_0,\dots , \mu_n$ on $D_0,\dots ,D_n$ respectively. We suppose that $X \subset L^2(\mu_j), j = 0, 1,\dots , n.$ We allow one of $D_j$ to coincide with $D.$ In this case we assume that $\mu_j$ coincides with $\nu.$ Write ${\cal D} = (D_0,\dots,D_n)$, $\mu = (\mu_0,\dots , \mu_n)$, $\mu = (\mu_1,\dots , \mu_n)$, $y =(y_1,\dots , y_n)$. \section{Optimal recovery problem} Given $y_1,\dots , y_n$ defined on $D_1,\dots ,D_n$ such that $$ \|f_j - y_j\|_{L^2(\mu j )} \le \delta_j,\quad j = 1,\dots , n, $$ we are to reconstruct $f$. Here $f_j$ is the restriction of $f$ to $D_j$ and $\delta_j \ge 0$, $j = 1,\dots , n$ are accuracy levels. In particular, $\delta_j = 0$ means that $f$ is known precisely on $D_j$. A recovery algorithm (method, procedure, etc.) is an operator $$ A\colon L^2(\mu_1)\times \dots \times L^2(\mu_n) \to L^2(\mu_0). $$ We consider $A(y)$, $y = (y_1,\dots , y_n),$ to be the recovered value of $f$ on $D_0.$ At this point we impose no conditions on $A.$ The maximal possible error of a method $A$ is $$e(X,{\cal D},\mu,\delta,A)=\sup_{\substack{f\in X,\ y\in L^2(\mu_1)\times\dots\times L^2(\mu_n)\\\|f_j-y_j\|_{L^2(\mu_j)}\le \delta_j, j=1,\dots,n}}\|f_0 - A(y)\|_{L^2(\mu_0)}$$ The optimal recovery error is $$ E(X,{\cal D}, \mu, \delta) = \inf_{A\colon L^2(\mu_1)\times \dots \times L^2(\mu_n)\to L^2(\mu_0)}e(X,{\cal D}, \mu, \delta, A). $$ A method ${\hat A}$ such that $$ E(X,{\cal D}, \mu, \delta) = e(X,{\cal D}, \mu, \delta, {\hat A}) $$ is called an {\it optimal recovery method.} The problem of finding an optimal recovery method (and sometimes an extremal function at which the optimal recovery error is attained) is usually referred to as {\it optimal recovery problem.} \section{Extremal problem} The optimal recovery problem is closely related to the following extremal problem. Find \begin{equation}\label{e1} \|f_0\|_{L^2(\mu 0)}\to\max,\quad \|f_j\|^2_{L^2(\mu_j)}\le\delta_j^2,\ j = 1,\dots , n,\quad f\in X. \end{equation} A special case of this extremal problem is when $D$ is the unit disk $\mathbb{D}$, $\mu_0$ and $\mu_1$ are point masses and $\mu_2$ is the normalized Lebesgue measure on the unit circle. Here the problem turns into $$ \max\{|f(a_0)|: |f(a_1)|\le \delta_1,\, \|f\|_{H^2}\le \delta_2\}, $$ which might be viewed as a version of the classical Schwarz lemma. Here we consider another variant of Scwarz Lemma. Let $a \in \mathbb{D}$ and $\Gamma$ be a circle inside of the unit disk, $\mu$ be the normalized Lebesgue measure on $\Gamma$, and $\mu > 0.$ Find \begin{equation}\label{e2} \sup \left \{\int_{\Gamma} |f|^2d\mu: f\in H^2,\, \|f\|^2_{H^2}\le 1,\, |f(a)|\le \delta \right \}. \end{equation} We will consider the case when the circle $\Gamma$ passes through the origin and its center lies on the real axis, so that $$ \Gamma = \{ z\in \mathbb{C}: |z -\rho|=\rho\},\quad0<\rho <1/2.$$ The corresponding optimal recovery problem is: {\it Reconstruct a Hardy function $f$ on the circle $\Gamma$ from its value at $a$ given with some tolerance.} There are several papers where similar problems were considered for Hardy and Bergman spaces in connection with optimal recovery in both one and several dimensional cases (see, for example, \cite{OsS1}--\cite{OsS3}). \section{Euler equation for the general problem} Let $K(z,w)$ be the reproducing kernel of $X.$ Write $$ \tilde{\mu} = -\mu_0 + \sum_{j=1}^n\lambda_j\mu_j. $$ Then $\tilde{\mu}$ is a regular measure on $D$ and every function from $X$ is square- integrable with respect to $\tilde{\mu}$. For $w \in D$ we introduce $$ d\tilde{\mu}_w(z) = K(z,w)d \tilde{\mu}(z). $$ Obviously every function from $X$ is $\tilde{\mu}_w$-integrable. We further define $$ \tau_w^{\lambda}(z) = \int_D K(z \tau ) d\tilde{\mu}_w(\tau ). $$ \begin{theorem} If $\tilde{f} \in X$ is a solution of the general extremal problem above, then there exists a non-negative vector $\widehat{\lambda} = (\widehat{\lambda}_1,\dots ,\widehat{\lambda}_n)$ such that $$ \widehat{f} = ( \spann \{\tau_{w}^{\widehat{\lambda}}, w\in D\})^{\perp}, $$ and $$ \widehat{\lambda}_j(\|f\|_{L_2(\mu_j)}-\delta_j) = 0,\, j = 1,\dots , n. $$ \end{theorem} We say that a non-negative vector $ \lambda = (\lambda_1,\dots , \lambda_n)$ belongs to the spectrum of the problem, if there exists an admissible for this problem function $f \in X$ such that \begin{align*} 1.\ & \lambda_j(\|f\|_{L^2(\mu_j )}-\mu_j) = 0.\\ 2.\ & f \in (\spann\{\tau_w^{\lambda}: w \in D\})^{\perp}. \end{align*} In this case we call $f$ a {\it spectral function.} \begin{theorem} Let $\Lambda$ be the spectrum of the problem. Then \begin{equation}\label{e3} \sup_{\substack{f\in X\\ \|f\|_{L_2(\mu_j)}\le \delta_j,\ j=1,\dots, n}} = \sup_{\lambda \in \Lambda}\sum_{j=1}^n\lambda_j\delta_j^2. \end{equation} \end{theorem} We call a spectral point $(\widehat{\lambda_1},\dots , \widehat{\lambda_n})$ {\it extremal,} if the maximum of the right-hand side of \eqref{e3} is attained at $(\widehat{\lambda_1},\dots , \widehat{\lambda_n}).$ \section{Spectrum of the Schwarz Lemma} Here we have. $$ \tau_w^{\lambda} = -{1\over \pi} \int_{\Gamma}{1\over 1-z\overline{\tau}}\cdot {1\over 1-\tau\overline{w}}\cdot {|d\tau |\over |\tau - \rho|} + $$ $$ \lambda_1{1\over 1-z\overline{a}}\cdot {1\over 1-a\overline{w}} + {\lambda_2\over 2\pi}\int_{|\tau|=1}{1\over 1-z\overline{\tau}}\cdot {1\over 1-\tau\overline{w}}|d\tau |= $$ $$ -{1\over 1-z\rho -\rho \overline{w}}+{\lambda_1\over (1-z\overline{a})(1-a\overline{w})}+{\lambda_2\over 1-z\overline{w}}. $$ By Theorem 1 every extremal function satisfies the following equation $$ {1\over 1-\rho w}f\left( {\rho\over 1-\rho w}\right) = \lambda_1{f(a)\over 1-\overline{a}w} $$ for some $\lambda_1, \lambda_2 \ge 0$ and all $w\in \mathbb{D}.$ Let $$ b = {1-\sqrt{1-4\rho^2}\over 2\rho }. $$ Then $b$ is the Denjoy-Wolff point of the following self-mapping of $\mathbb{D}$ $$ z \to {\rho \over 1-\rho z}, $$ and the disk bounded by the circle $\Gamma$ is a hyperbolic neighborhood of $b.$ Consider the following functions $$ \varphi_j(z) = {\sqrt{1-b^2}\over 1-bz} \left ( b-z\over 1-bz\right )^j,\quad j=0,1,\dots\,\,. $$ These functions form an orthonormal basis of $H^2,$ and they are eigenfunctions of the operator $$ Tf(z) = {1\over 1-\rho z}f \left ({\rho \over 1-\rho z} \right ), $$ and the corresponding eigenvalues are %\begin{equation}\label{e4} $$\alpha_j = {b^{2j}\over 1-\rho b}.$$ %\end{equation} \begin{theorem} Let $a\neq b$. $1$. If $$ \left | a- {\rho \over 1-\rho^2}\right | \ge {\rho^2\over 1-\rho^2}, $$ or $$ \delta > {\sqrt {|a|^2\rho^2 - |\rho-a|^2}\over a\rho +\overline{a}\rho - |a|^2}, $$ then the spectrum of Schwarz Lemma extremal problem consists of two parts $\Lambda = \Lambda_1\cup \Lambda_2$, where $$ \Lambda_1 = \{(0, \alpha_j) : |\varphi_j(a)| \le \delta\}, $$ $$ \Lambda_2 = \{(\lambda_1, \lambda_2) : \lambda_1, \lambda_2 > 0,\ F(\lambda_2) = \delta^{-2},\ \lambda_1 = h(\lambda_2)\}, $$ where $$ F(\lambda) = \sum_{j=0}^{\infty}{|\varphi_j(a)|^2\over (a_j-\lambda)^2}h^2(\lambda),\quad h(\lambda) = \left (\sum_{j=0}^{\infty}{|\varphi_j(a)|^2\over a_j-\lambda}\right )^{-1}. $$ $2$. If $$ \left | a- {\rho \over 1-\rho^2}\right | < {\rho^2\over 1-\rho^2}, $$ and $$ \delta \le {\sqrt {|a|^2\rho^2 - |\rho-a|^2}\over a\rho +\overline{a}\rho - |a|^2}, $$ then the spectrum includes in addition the point $$ \Lambda_3 = \left \{ \left ({a\rho +\overline{a}\rho - |a|^2\over \rho^2},0 \right )\right \}. $$ \end{theorem} \begin{theorem} Let $a = b$, $$ \Lambda_1 = \{(0, \alpha_j):\, j = 1, 2,\dots , \}, $$ $$ \Lambda_2 = \{((1-b^2)(\alpha_0-\alpha_j), \alpha_j):\, j = 1, 2,\dots , \}. $$ Then the spectrum of problem is $\Lambda = \Lambda_1\cup \Lambda_2,$ if $\delta <\dfrac1{\sqrt{1-b^2}}$, and $\Lambda = \Lambda_1\cup \Lambda_2 \cup\{(0,\alpha_0)\},$ if $\delta \ge \dfrac1{\sqrt{1-b^2}}.$ \end{theorem} It turns out that $\Lambda_2$ is the most important part of the spectrum. \begin{prop} If $a$ lies outside $\Gamma$, then $F(\lambda) \to \infty$ as $\lambda \to 0.$ \end{prop} This Proposition implies that if $a$ lies outside $\Gamma$, then $\Lambda_2$ contains only finite number of points. Now we will use Theorem 2 to describe the extremal points of the spectrum. \begin{prop} If $\delta \ge |\varphi_0(a)|,$ then $(0, \alpha_0)$ is the extremal point of the spectrum. \end{prop} \begin{prop} If $a = b$ and $\delta < 1/\sqrt {1-b^2}$, then the extremal spectral point is $$ (\widehat{\lambda}_1, \widehat{\lambda}_2) = ((1- b^2)(\alpha_0 - \alpha_1), \alpha_1). $$ \end{prop} \begin{prop} If $\delta < |\varphi_0(a)|$, then $\Lambda_1$ does not contain extremal spectral points. \end{prop} Note that the function %\begin{equation}\label{e7} $$ g(\lambda) = \sum_{j=0}^{\infty}{|\varphi_j(a)|^2 \over \alpha_j-\lambda}$$ %\end{equation} is monotone and increases from $-\infty$ to $+\infty$ when $\lambda \in (\alpha_{j+1}, \alpha_j).$ Let $\zeta_j$ be the only zero of $g$ on the interval $(\alpha_{j+1}, \alpha_j).$ \begin{prop} Let $a\neq b.$ If $\delta \le |\varphi_1(a)|,$ then the extremal spectral point $(\widehat{\lambda}_1, \widehat{\lambda}_2)$ is unique, belongs to $\Lambda_2$ and is determined by the condition $\zeta_0 < \widehat{\lambda}_2 < \alpha_0.$ \end{prop} \begin{prop} Assume that $|\varphi_1(a)| <\delta < |\varphi_0(a)|$ and $$ \gamma = \left |{b-a\over 1-ab} \right | \ge b^{2/3}, $$ then the conclusion of Proposition $5$ is valid, that is, the extremal spectral point $(\widehat{\lambda}_1, \widehat{\lambda}_2)$ is unique, belongs to $\Lambda_2$ and is determined by the condition $\zeta_0 < \widehat{\lambda}_2 < \alpha_0.$ \end{prop} \section{Optimal Recovery Method} To construct optimal recovery methods we need the following result (several results of this type may be found in \cite{MM}, \cite{MO2}, \cite{Os}). \begin{theorem} Assume that there exist $\widehat{\lambda}_j\ge 0,\, j = 1,\dots , n,$ such that the value of the extremal problem $$\|f_0\|^2_{L_2(\mu_0)}\to \max,\quad\sum_{j=1}^{\infty}\widehat{\lambda}_j\|f_j\|^2_{L_2(0,\mu_j)} \le\sum_{j=1}^{\infty}\widehat{\lambda}_j\delta_j^2,\quad f\in X, $$ is the same as in \eqref{e1}. Moreover, assume that for every $\tilde{y} = (\tilde{y}_1,\dots , \tilde{y}_n) \in Y_1 \times \cdots \times Y_n,$ where $Y_j$ are dense in $L^2(\mu_j),$ there exists $f_{\tilde{y}}$ which is a solution of the extremal problem $$ \sum_{j=1}^{\infty}\widehat{\lambda}_j\|f_j-\tilde{y}_j\|^2_{L_2(0,\mu_j)} \to \min,\quad f\in X. $$ Moreover, let $\widehat{A}\colon L^2(\mu_1)\times \cdots \times L^2(\mu_n)\to L^2(\mu_0)$ be a linear continuous operator, where the norm in $L^2(\mu_1)\times \cdots \times L^2(\mu_n)$ is defined as $$ \|y\|=\left (\sum_{j=1}^{n}\|y_j\|^2_{L^2(\mu_j)}\right )^{1/2}, $$ such that for all $\tilde{y} = (\tilde{y}_1,\dots , \tilde{y}_n) \in Y_1 \times \cdots \times Y_n,$ $$ \widehat{A}(\tilde{y})=(f_{\tilde{y}})_0. $$ Then $$ E(X,{\cal D}, \mu, \delta) = \sup_{\substack{f\in X\\ \|f_j\|_{L^2(\mu_j)}\le \delta_j,\ j=1,\dots, n}}\|f_0\|_{L^2(\mu_0)} $$ and the method $\widehat{A}(y)$ is optimal. \end{theorem} We will apply Theorem 5 to the construction of optimal recovery method for the Schwarz Lemma type problem considered above. Consider the extremal problem \begin{equation}\label{e9} \int_{\Gamma}|f|^2d\mu \to \max,\quad \widehat{\lambda}_1|f(a)|^2 + \widehat{\lambda}_2\|f\|^2_{H^2} \le \widehat{\lambda}_1\delta^2+\widehat{\lambda}_2,\quad f\in H^2, \end{equation} where as before $\mu $ is the normalized Lebesgue measure on $\Gamma$ and $(\widehat{\lambda}_1,\widehat{\lambda}_2)$ is an extremal spectral point for problem \eqref{e2}. \begin{prop} Suppose that either \noindent $1$. $a \neq b$ and $\delta \le |\varphi_1(a)|,$ or $|\varphi_1(a)| <\delta < |\varphi_0(a)|$ and $ \gamma = \left |\dfrac{b-a}{1-ab}\right | \ge b^{2/3}, $ \noindent or \noindent $2$. $a = b$ and $\delta < \varphi (b) = 1/\sqrt{1-b^2}.$ Then the values of extremal problems \eqref{e2} and \eqref{e9} are the same. \end{prop} \begin{theorem} Suppose that one of the following conditions is satisfied $1$. $\delta \ge |\varphi_0(a)|,$ $2$. $\delta \le |\varphi_1(a)|,$ $3$. $ |\varphi_1(a)|<\delta < |\varphi_0(a)|,\quad \gamma \ge b^{2/3},$ $4$. $a = b,$ \noindent and $(\widehat{\lambda}_1, \widehat{\lambda}_2)$ is the corresponding extremal spectral point. Then the error of optimal recovery is given by $$ \sqrt{\widehat{\lambda}_1\delta^2+\widehat{\lambda}_2}$$ and the method \begin{equation}\label{e10} \widehat{A}(y)(z) = {\widehat{\lambda}_1y \over \widehat{\lambda}_1+\widehat{\lambda}_2(1-|a|^2)}\cdot{1-|a|^2\over 1-\overline{a}z} \end{equation} is optimal. \end{theorem} Note that for $a = b$ the optimal method of recovery \eqref{e10} does not depend on $\delta$ and has the form $$ \widehat{A}(y)(z) = {1-|b|^2\over 1-bz}. $$ \section{Open problems} 1. It would be desirable to identify the extremal spectral point in all possible cases. We have shown that in a number of cases the extremal spectral point is the only point in $\Lambda_2$ such that $\zeta_0<\widehat{\lambda}_2 < \alpha_0.$ Our attempts to find a nontrivial-case when this point is not extremal failed. Thus, we are tempted to conjecture that the point of $\Lambda_2$ with the biggest $\lambda_2$ is always extremal. {\bf Conjecture.} {\it If $a \neq b$ and $\delta < |\varphi_0(a)|$, the point in $\Lambda_2$ such that $\zeta_0<\widehat{\lambda}_2 < \alpha_0$ is always the spectral extremal point for problem \eqref{e2}.} \medskip 2. It is natural to ask which choice of $a$ minimizes the value of problem \eqref{e2} (of course, this choice of $a$ leads to the least optimal recovery error). It follows from above discussion that the point $b$ plays a special role. {\bf Problem.} {\it Does the choice $a = b$ always lead to the least mean square optimal recovery error?} \medskip 3. Finally, if in problem \eqref{e2} we replace the constraint $|\varphi(a)|\le \delta$ with $$ {1\over 2\pi r}\int_{|z-a|=r}|f(z)|^2|d(z-a)|\le \delta, \quad 0