% Pomerenke's Conjecture % Schwarzian Derivatives of Hyp Convex functions % Second Revision for PLMS - Summer 2005 \documentclass{lms} \usepackage{amsmath,amssymb,amsfonts,graphics,epsfig,rotating,cite,rotating} \newcommand{\C}{\mathbb C} \newcommand{\D}{\mathbb D} \newcommand{\bdryD}{\partial \mathbb D} \newcommand{\R}{\mathbb R} \newcommand{\Ub}{\mathbb U} \newcommand{\Lb}{\mathbb L} \newcommand{\HH}{\mathbb H} \newcommand{\Z}{\mathbb Z} \newcommand{\K}{\mathcal K} \newcommand{\Kh}{\widetilde{\mathcal K}} \newcommand{\Dh}{\widetilde{D}} \newcommand{\Ph}{\widetilde{P}} \newcommand{\Sh}{\widetilde{S}} \newcommand{\Pt}{\widetilde{P}} \newcommand{\Tt}{\widetilde{T}} \newcommand{\tgamma}{\widetilde\gamma} \newcommand{\ft}{\widetilde f} \newcommand{\LL}{\mathcal L} \newcommand{\PP}{\mathcal P} \newcommand{\T}{\mathcal T} \newcommand{\CC}{\mathcal C} \newcommand{\hh}{\widehat h} \newcommand{\tp}{{\tau^\prime}} \newcommand{\tn}{{\tau_n}} \newcommand{\ti}{{\tau_\infty}} \newcommand{\Kp}{\mathcal K^\prime} \newcommand{\Kpt}{\mathcal \widetilde{K^\prime}} \newcommand{\Qt}{\widetilde Q} \newcommand{\ap}{\alpha^\prime} \newcommand{\tpn}{{\tau_n^\prime}} \newcommand{\tphi}{\widetilde \varphi} \newcommand{\abs}[1]{\lvert#1\rvert} \newcommand{\norm}[1]{\lvert\lvert#1\rvert\rvert} \newcommand{\Snf}{\norm{S_f}_\D} \newcommand{\PSL}{\text{PSL}} \newcommand{\z}{\zeta} \newcommand{\e}{\epsilon} \newcommand{\iG}{\int_{\gamma}} \newcommand{\G}{{\scriptscriptstyle{\Gamma}}} %\theoremstyle{plain} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{julia}{Julia Variational Formula} \newtheorem{thm}{Theorem 1.1} \newtheorem{step-down}{Step Down Lemma} %\theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{notation}[theorem]{Notation} %\theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \newtheorem{example}[theorem]{Example} \newtheorem{note}[theorem]{Note} \DeclareMathOperator{\m}{mod} \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} \DeclareMathOperator{\Arg}{Arg} \DeclareMathOperator{\Log}{Log} \DeclareMathOperator{\interior}{int} \numberwithin{equation}{section} \begin{document} \title[Deformation of a Disc Under a Hyperbolically Convex Map]{The Sharp Bound for the Deformation of a Disc under a Hyperbolically Convex Map} %[Schwarzian Derivatives of Hyperbolically Convex Functions]{A Sharp %Bound on the Schwarzian Derivatives of Hyperbolically %Convex Functions} \date{\today} \author{Roger W. Barnard, Leah Cole, Kent Pearce, and G. Brock Williams} %\keywords{Hyperbolically Convex, Schwarzian Derivative, Julia Variation} %\subjclass{30C70} %Insert `2000 Mathematics Subject Classification' numbers here! \classno{30C70} \maketitle \begin{abstract} We complete the determination of how far convex maps can deform discs in each of the three classical geometries. The euclidean case was settled by Nehari in 1976, and the spherical case by Mej\'ia and Pommerenke in 2000. We find the sharp bound on the Schwarzian derivative of a hyperbolically convex function and thus complete the hyperbolic case. This problem was first posed by Ma and Minda in a series of papers in the 1980's. Mej\'ia and Pommerenke then produced partial results and a conjecture as to the extremal function in 2000. Their function maps onto a domain bounded by two proper geodesic sides, a ``hyperbolic strip.'' Applying a generalization of the Julia variation and a critical Step Down Lemma, we show that there is an extremal function mapping onto a domain with at most two geodesic sides. We then verify using special function theory that among the remaining candidates, Mej\'ia and Pommerenke's two-sided domain is in fact extremal. This correlates nicely with the euclidean and spherically convex cases in which the extremal is known to be a map onto a two-sided ``strip.'' \end{abstract} %****************************************************************** \section{Introduction}\label{S:intro} Hyperbolic convexity is a natural generalization of euclidean convexity; a region $G$ in the Poincar\'e model $\D$ of the hyperbolic plane is \textbf{hyperbolically convex} if for any two points in $G$, the hyperbolic geodesic segment between them lies entirely in $G$. Such regions arise naturally in Teichm\"uller theory, for example, since the fundamental domains of Fuchsian groups are hyperbolically convex~\cite{aB83, oL87}. A conformal map $f:\D\to\D$ is hyperbolically convex if its range is hyperbolically convex. Hyperbolically convex functions have been extensively studied by Ma and Minda~\cite{MM94,MM99} and Mej\'ia and Pommerenke~\cite{PM00,MP00,MP01,MP02,MP00b}, as well as Beardon~\cite{aB83} and Solynin~\cite{aS99,aS98}, among others. One frequently cited open problem is to find for hyperbolically convex functions a sharp bound on the Schwarz norm \[ \norm{S_f}_\D=\sup\{\abs{S_f(z)} \eta_\D^{-2}(z):\: z\in \D\}, \] where $\eta_\D(z)=\frac{1}{1-\abs{z}^2}$ is the hyperbolic density of $\D$ and $S_f$ is the Schwarzian derivative \[ S_f = \left(\frac{f^{\prime\prime}}{f^\prime}\right)^\prime - \frac{1}{2}\left(\frac{f^{\prime\prime}}{f^\prime}\right)^2. \] The Schwarz norm of an analytic function $f$ has long been a primary tool in understanding its geometric behavior. For example, $\norm{S_f}_\D =0$ if and only if $f$ is a M\"obius transformation. Thus $\norm{S_f}_\D$ is thought of as measuring how closely the geometric behavior of $f$ resembles that of a M\"obius transformation. Since the image of $\D$ under a M\"obius transformation must be a disc, $\norm{S_f}_\D$ also measures the difference between the conformal geometry of $f(\D)$ and that of a disc. Lehto has used this idea to great effect, producing a pseudo-metric on the set of all simply connected proper subdomains of $\C$. See Lehto's book~\cite{oL87}, for example, for a detailed discussion. Nehari showed that if $f(\D)$ is convex (in the euclidean sense), then $\norm{S_f}_\D\leq 2$, with equality if and only if $f(\D)$ is an ``infinite strip'' bounded by two parallel lines~\cite{zN76}. Similarly, Mej\'ia and Pommerenke showed that the extremal domain for spherically convex functions is a ``spherical strip''~\cite{MP00b}. The problem of finding a similar bound for the Schwarz norm of hyperbolically convex functions has been intensely studied by a number of authors, including Ma, Minda, Mej\'ia, Pommerenke and Vasilev~\cite {MM94,MM99,MP00,MPV01,MP02,PM00}. Mej\'ia and Pommerenke~\cite{PM00} found partial results on the bound and conjectured that the extremal value of $\norm{S_f}_\D$ is attained by a map of the form \begin{equation} \label{E:tanh} f_\alpha (z)=\tan\left( \alpha\int_0^z \left(1-2\xi^2\cos 2\theta+\xi^4 \right)^{-1/2}\,d\xi \right), \end{equation} where $\alpha = \frac{\pi}{2K(\cos \theta)}$, and $K$ is the elliptic integral of the first kind. The range of $f_\alpha$ is a ``hyperbolic strip'' bounded by two geodesics through $\pm i\tanh \left(\frac{\pi K(\sin\theta)}{4K(\cos\theta)}\right)$ and perpendicular to the imaginary axis. See Figure~\ref{F:convex3}. \begin{figure} \begin{center}\scalebox{0.45}{ \epsfig{file=convex3} } \caption{The extremal domains for maximizing the Schwarz norm in euclidean (left), spherical (center), and hyperbolic (right) geometry.} \label{F:convex3} \end{center} \end{figure} In this paper, we prove the following theorem, thus verifying Mej\'ia and Pommerenke's conjecture and completing the classification of the extremal domains for the Schwarzian in all three of the classical geometries. \begin{thm*} The maximal value of the Schwarz norm for hyperbolically convex functions is $S_{f_\alpha}(0)$, where \[ f_\alpha(z) =\tan\left( \alpha\int_0^z \left(1-2\xi^2\cos 2\theta+\xi^4 \right)^{-1/2}\,d\xi \right), \qquad \alpha = \frac{\pi}{2K(\cos \theta)}, \] $K$ is the elliptic integral of the first kind, and $\alpha$ is chosen so that $\cos \theta$ is the unique critical point of the function \[ g(s) = 4s^2-2 +\frac{\pi^2}{2K^2({s})} \] on $(0,1)$. \end{thm*} A computer calculation produces a maximal value for the Schwarz norm for a hyperbolically convex function of approximately $2.383635$. %\begin{thm} %The maximal value of the Schwarz norm for a hyperbolically convex %function is approximately $2.383635$ and is achieved by the ``hyperbolic strip %map'' $f_\alpha$ with $\alpha\approx 0.5343598$. %\end{thm} \begin{remark} Our variational techniques are used to argue that if an extremal function exists, it must have certain properties. To guarantee the existence of an extremal function, we restrict our attention to a dense collection of compact classes (hyperbolically convex functions which map onto domains with at most a fixed number of (proper) sides) and argue that an extremal function exists for these classes. Our methods show that the map $f_\alpha$ defined in~\eqref{E:tanh} is extremal in each such class for a common value of $\alpha$ that is approximately equal to $0.5343598$. Moreover, since the classes are dense in the space of all hyperbolically convex functions, this $f_\alpha$ must be extremal for all hyperbolically convex functions. Indeed, up to disc automorphisms, $f_\alpha$ can be the only extremal function which maps onto a finite-sided hyperbolically convex polygon. However, this does not rule out the possibility, that some function onto a non-polygonal region could have the same Schwarz norm as $f_\alpha$. Thus our methods do not guarantee uniqueness, but we conjecture that the extremal is in fact unique. \end{remark} %\begin{remark} %Our arguments and the variations we construct can also be extended to the %euclidean and spherical settings. Thus we have a unified approach %to all three classical geometries. %\end{remark} In Section~\ref{S:hypercon} we develop background material on hyperbolic convexity and the Schwarzian derivative and discuss more of the history of the problem. In Section~\ref{S:prelim} we first show by an approximation argument that the problem can be reduced to studying the value of the Schwarzian at the origin for those functions which map onto finite-sided polygons. % to proving that %among all hyperbolically convex polygons with at most a fixed number %of sides, the extremal domain has at most $2$ sides. %Thus we need %only work with compact families where the existence of an extremal %function is assured and our variational methods can be applied. %The $\sup$ can be replaced by $\max$ and We then give a generalization of the Julia Variational Formula that we will use in Sections~\ref{S:twovar} and~\ref{S:effect} to develop two variations which preserve hyperbolic convexity. In Section~\ref{S:2sides} we apply these variations to show the extremal polygon has at most two proper sides. Finally, in Section~\ref{S:compute} we employ inequalities on elliptic integrals to verify that the two-sided domain conjectured by Mej\'ia and Pommerenke~\cite{MP00,MP01,PM00} is indeed extremal. We would like to thank the Phillip Brown, the referee and the editor for their very valuable suggestions. %After developing some background material on hyperbolic convexity and a %generalization of %the Julia variation in Sections~\ref{S:hypercon} and~\ref{S:julia}, %we develop two variations in Sections~\ref{S:twovar} and~\ref{S:effect} %which preserve %hyperbolic convexity. These variations %allow us to show in Section~\ref{S:2sides} that the %extremal value occurs for a function whose range has at most two %sides. \section{Hyperbolic Convexity and Schwarzians}\label{S:hypercon} \subsection{Hyperbolic Geometry} The unit disc $\D$ equipped with the metric \[ d_h (z,w)=\inf \left\{\int_\gamma \frac{1}{1-\abs{z}^2}\,\abs{dz} : \gamma \text{ is a rectifiable curve joining $z$ and $w$}\right\} \] forms a model for the hyperbolic plane~\cite{aB83}. Notice that the Poincar\'e density \[ \eta_\D(z)=\frac{1}{1-\abs{z}^2} \] goes to infinity as $z$ moves toward the boundary of the disc. Consequently, integrating $\eta_\D$ over curves near the boundary produces large values of the integral. If $z$ and $w$ do not lie on a ray through the origin, then the euclidean line segment joining them will produce a larger integral than a curve which bends away from the boundary. In fact, the infimum will be achieved by an arc of a circle perpendicular to $\partial \D$. Such curves are hyperbolic geodesics. Since disc automorphisms \[ M(z)=e^{i\theta}\frac{z-a}{1-\overline{a}z}, \] where $\theta\in[0,2\pi)$ and $a\in\D$, preserve circles orthogonal to $\partial \D$, they are precisely the isometries of $\D$. Any region $G$ conformally equivalent to $\D$ also carries a hyperbolic metric defined in the same manner using the density \[ \eta_G(z)=\frac{\abs{f'(z)}}{1-\abs{f(z)}^2}, \] where $f$ is a conformal map of $G$ onto $\D$. Notice that it doesn't matter which map $f$ is chosen as any two such maps must differ by a disc automorphism. \subsection{Convexity} The euclidean notion of convexity generalizes to hyperbolic regions in an obvious manner. \begin{definition}A region $\Omega\subset\D$ is \textbf{hyperbolically convex} if for any two points $z,w\in\Omega$, the hyperbolic geodesic segment joining $z$ and $w$ lies completely in $\Omega$. \end{definition} Notice that since the disc automorphisms are the isometries of the hyperbolic plane, the image $M(\Omega)$ of $\Omega$ under a disc automorphism $M$ is hyperbolically convex if and only if $\Omega$ is. The fundamental domains of discrete groups of disc automorphisms provide a great many useful examples of hyperbolically convex domains. See Beardon~\cite{aB83} for an extensive discussion of these regions. We will call a hyperbolically convex region $\Omega$ bounded by a finite number of either geodesic arcs lying inside $\D$ or arcs of $\partial \D$ a \textbf{hyperbolically convex polygon}. We call the bounding geodesic arcs \textbf{proper sides} and the arcs of $\partial \D$ \textbf{improper sides}. For $n\geq 0$, we let \begin{align}\notag K_n=\{&\text{hyperbolically convex polygons containing $0$}\\\notag & \text{and having at most $n$ proper sides}\} \cup\{0\}. \end{align} %Notice that $K_n$ contains \begin{definition}A conformal map $f:\D\to\Omega$ is called a \textbf{hyperbolically convex function} if its range is hyperbolically convex. We let $\mathbf {\mathit H}$ denote the class of all hyperbolically convex functions that fix the origin and let $\bf{\mathit H_{\mathit n}}$ denote the subset of functions whose range is in $K_n$. \end{definition} %A sequence of polygons in $K_n$ can converge (in the sense of Cartheodory) only to another polygon %in $K_n$. This convergence carries over to hyperbolically convex functions so that $H_n$ is %compact. On the other hand, every hyperbolically convex region is the limit of a sequence %of hyperbolically convex polygons; hence $\cup_n H_n$ is dense in $H$. \subsection{Schwarzians} Much of the geometric behavior of an analytic function is described by its Schwarzian derivative~\cite{GL2000,oL87}. \begin{definition} The \textbf{Schwarzian derivative} (or just ``Schwarzian'') of an analytic function $f$ is \[ S_f = \left( \frac{f''}{f'} \right)' - \frac12 \left( \frac{f''}{f'} \right)^2 . \]. \end{definition} \begin{proposition} The Schwarzian of an analytic function is identically $0$ if and only if it is a M\"obius transformation. Moreover, the Schwarzian satisfies the chain rule \[ S_{f\circ g}=\left(S_f \circ g\right)\left(g'\right)^2 + S_g. \] \end{proposition} Thus, if $M$ is M\"obius, then \[ S_{M\circ g} = S_g. \] and \[ S_{f\circ M}=\left(S_f \circ M\right)\left(M'\right)^2. \] Hence the Schwarzian is unchanged by post-composition with a M\"obius transformation, but pre-composition produces an extra quadratic factor. \begin{definition} Let $f$ be defined on a simply connected region $G\subsetneq\C$. The \textbf{Schwarz norm} of $f$ is given by \[ \norm{S_f}_{G} = \sup_{z \in G} \eta_G^{-2}(z)\abs{S_f (z)}. \] \end{definition} By taking into account the density of the hyperbolic metric, the Schwarz norm is completely M\"obius invariant. It is easy to show for any M\"obius $M$ that \[ \eta_{M^{-1}(G)}(z)=\eta_{G} (M(z)) \abs{M'(z)}, \] and thus \[ \frac{\abs{S_{f\circ M}(z)}}{\eta^2_{M^{-1}(G)}(z)} = \frac{\abs{S_f(M(z))}\,\abs{M'(z)}^2}{\eta^2_{G}(M(z))\,\abs{M'(z)}^2} = \frac{\abs{S_f(w)}}{\eta^2_{G}(w)}. \] where $w=M(z)$. Thus \[ \norm{S_f}_G=\norm{S_{f\circ M}}_{M^{-1}(G)} \] and \[ \norm{S_f}_G=\norm{S_{M\circ f}}_G. \] In particular, notice that $\norm{S_f}_\D$ is unchanged by disc automorphisms. \subsection{Computing the Schwarz Norm of a Function} %When computing the Schwarz norm of a function, %it can sometimes be very helpful to %change its domain. Because of the M\"obius invariance of the Schwarz norm, we can change the domain of a function so that %take our function to be defined on a region where either the the hyperbolic density or the expression of the function itself is simpler. \begin{example}\label{E:one-side} Consider the function \[ g(z)=\frac{\sqrt{z}-1}{\sqrt{z}+1} \] from the upper half plane $\HH$ onto the upper half-disc $\D^+=\{z\in\D:\:\im z>0\}$. Notice that any hyperbolically convex domain bounded by exactly one proper side can be mapped onto $\D^+$ by a disc automorphism. Moreover, $g$ could have been defined on $\D$ by precomposition with a M\"obius transformation; however, that would not change the Schwarz norm. Consequently, the Schwarz norm for any hyperbolically convex function whose image is bounded by exactly one proper side must be equal to $\norm{S_g}_\HH$. Next notice that $g$ itself is the composition of $h(z)=\sqrt{z}$ and another M\"obius transformation. Thus $\norm{S_g}_\HH=\norm{S_h}_\HH$. It is now easy to compute \[ S_h(z)=\frac{3}{8z^2}. \] The hyperbolic density on $\HH$ has an especially useful representation \[ \eta_\HH(z)=\frac{1}{2\im z}. \] Since both $\eta_\HH$ and $S_h$ are invariant under translations of the form $x + iy \rightarrow (x+a) + iy$, \[ \norm{S_h}_\HH=\sup_{z\in\HH} \eta^{-2}(z)\abs{S_h(z)} =\sup_{y>0} 4y^2\left|\frac{3}{8(iy)^2}\right| = \frac{3}{2}. \] \end{example} %M\"obius invariance also allows another useful trick for computing the %Schwarz norm. \subsection{Geometry of the Schwarzian} Since $\Snf=0$ if and only if $f$ is M\"obius, we can view $\Snf$ as measuring how close $f$ is to being a M\"obius transformation. Since any M\"obius transformation would send $\D$ to another disc or half plane, $\Snf$ also measures the amount of deformation between $f(\D)$ and a disc. This notion was formalized by Lehto~\cite{oL87} to produce a pseudometric between regions conformally equivalent to a disc. There are a number of results that show that if $\norm{S_f}_\D$ is small, then $f(\D)$ possesses disc-like properties. The two most important for our purposes are due to Nehari~\cite{zN49,zN76}. \begin{theorem} If $\Snf<2$, then $f$ is univalent and $f(\D)$ is a quasidisc. Moreover, if $f$ is univalent, then $\Snf\leq 6$. \end{theorem} \begin{theorem} If $f(\D)$ is convex (in the euclidean sense), then $\Snf\leq 2$, with equality if and only if $f(\D)$ is an infinite strip. \end{theorem} Mej\'ia and Pommerenke~\cite{MP00b} proved a similar result for spherically convex regions. \begin{theorem} If $f(\D)$ is spherically convex, then \[ \norm{S_f}_\D \leq 2(1-\sigma(f)^2), \] where \[ \sigma(f)=\max_{z\in\D} (1-\abs{z}^2)\frac{\abs{f^\prime}}{1+\abs{f}^2}. \] For a fixed value of $\sigma(f)$, this maximum value of $\Snf$ is achieved by a map of the form $f_\phi(z)=i\tanh\left(\frac{2\phi}{\pi}\Log \left(\frac{1+z}{1-z}\right)\right)$ which takes $\D$ onto a ``spherical strip,'' that is, a lune bounded by great circles through $\pm i$ and making an angle $2\phi$ with the imaginary axis. \end{theorem} Thus, convex and spherically convex regions cannot be deformed too far from being a disc in the sense of the Lehto pseudometric, and the regions with the greatest amount of deformation are strips. It has been conjectured by Mej\'ia and Pommerenke~\cite{MP00,PM00,MP01} %and by Ma and Minda~\cite{MM99} that the same must hold for hyperbolically convex regions. In particular, Mej\'ia and Pommerenke conjectured that the maximum value of $\Snf$ over all hyperbolically convex maps $f$ is achieved by \begin{equation}\label{E:f_alpha} f_\alpha(z) =\tan\left( \alpha\int_0^z \left(1-2\xi^2\cos 2\theta+\xi^4 \right)^{-1/2}\,d\xi \right), \qquad \alpha = \frac{\pi}{2K(\cos \theta)}, \end{equation} where $K$ is the elliptic integral of the first kind. The range of $f_\alpha$ is a ``hyperbolic strip'' bounded by two geodesics through $\pm i \tanh \left(\frac{\pi K(\sin\theta)}{4K(\cos\theta)}\right)$. See Figure~\ref{F:strip}. \begin{figure} \begin{center}\scalebox{0.65}{ \epsfig{file=f_alpha-onto-2-sides} } \caption{The hyperbolically convex extremal function for maximizing the Schwarz norm. The extremal domain consists of an odd symmetric polygon bounded by two proper geodesic sides - a ``hyperbolic strip.''} \label{F:strip} \end{center} \end{figure} %As our primary tool for proving their conjecture, we will %modify an extension of the Julia variation. %~\cite{BS84,BL75}. \section{Preliminary Simplifications}\label{S:prelim} \subsection{Reduction to Polygonal Domains} As a first step toward verifying Mej\'ia and Pommerenke's conjecture, we first reduce the problem to hyperbolically convex polygons. \begin{lemma} To determine the extremal value of $\norm{S_f}_\D$ over $H$, it suffices to determine the extremal value over each $H_n$. Moreover, each $H_n$ is compact. \end{lemma} \begin{proof} From its definition as a supremum, it follows that the Schwarz norm is a lower semi-continuous functional on $H$; that is, for any point $f \in H$, if $\{f_n\}$ is a sequence in $H$ which converges to $f$ (locally uniformly on compacta), then $\liminf \norm{S_{f_n}}_\D \ge \Snf$. To see this, let $\e > 0$. Then, for any compact subset $K$ of $\D$, there exists an index $N$ such that for $n \ge N$ we have $\norm{S_{f_n\vert_{\interior K}}}_\D \ge \norm{S_{f\vert_{\interior K}}}_\D - \e$. Hence, for $n \ge N$ we have $\norm{S_{f_n}}_\D\ge \norm{S_{f_n\vert_{\interior K}}}_\D \ge \norm{S_{f\vert_{\interior K}}}_\D - \e$. Hence, $\liminf \norm{S_{f_n}}_\D \ge \norm{S_{f\vert_{\interior K}}}_\D - \e$. Since $K$ was abritrary we must have, $\liminf \norm{S_{f_n}}_\D \ge \Snf - \e$. Since $\e$ was arbitrary, we have $\liminf \norm{S_{f_n}}_\D \ge \Snf$. For a fixed $n$, a sequence $D_k$ of hyperbolically convex polygonal domains with at most $n$ (proper) sides can converge in the sense of Carath{\'e}odory~\cite{cP92} only to another polygonal domain $D_0$, where $D_0$ has at most $n$ (proper) sides. This follows because if $D_0$ had more than $n$ proper sides, then (from the boundary behavior) the Carath{\'e}odory convergence~\cite{cP92} would imply, for $k$ sufficiently large, that $D_k$ would have at least $n+1$ proper sides. From Carath{\'e}odory 's Convergence Theorem, this carries over to hyperbolically convex functions in $H_n$. Thus, $H_n$ is compact. On the other hand, the boundary $\Gamma$ of a hyperbolically convex domain $D$ can be approximated by geodesics lying in $\overline{\D}$ by connecting points on $\Gamma$ sufficiently close together with geodesics lying in $\overline{\D}$. It follows that we can find a sequence of hyperbolically convex polygons that converge, in the sense of Carath\'eodory, to $D$. Hence, % (by~\cite[cP92]) the mapping functions onto these polygons converge locally uniformly to the map onto $D$. As a result, $\cup_n H_n$ is dense in $H$. Thus any extremal function in $H$ can be approximated by functions in $H_n$. As a result, it suffices to determine the supremum of $\norm{S_{f}}_\D$ in each $H_n$. \end{proof} %Consequently, we need only show that $f_\alpha$ is extremal in each $H_n$. \subsection{Reduction to $S_f(0)$} Recall that the Schwarz norm was defined by taking the supremum of $\abs{S_f(z}\left(1-\abs{z}^2\right)^2$ over all points $z\in\D$. Since $\abs{S_f(z}\left(1-\abs{z}^2\right)^2$ is invariant under disc automorphisms, we need only consider $\abs{S_f(0)}$. \begin{lemma}\label{L:sf0} For each $n\geq 2$, \[ \sup_{f\in H_n} \norm{S_f}_\D = \max_{f\in H_n} \abs{S_f(0)}. \] \end{lemma} \begin{proof} Recall that every $f\in H_n$ is conjugate by a M\"obius transformation to a map $g:\HH\to\HH$ which is hyperbolically convex in the upper half plane model. Suppose $g$ maps $\HH$ onto a curvilinear polygon with angles $\alpha_1\pi,\alpha_2\pi,\dots, \alpha_n\pi$, and the preimages of the vertices are $a_1,a_2,\dots,a_n\in\R$. Then the Schwarzian of $g$ has the following very useful representation (see Nehari~\cite{zN52}) \begin{equation} S_g(z) = \frac{1}{2}\sum_{k=1}^n \frac{1-\alpha_k^2}{(z-a_k)^2} +\sum_{k=1}^n\frac{\beta_k}{z-a_k}, \end{equation} where $\beta_1,\beta_2,\dots,\beta_n$ are real constants determined by the angles. Thus \[ \norm{S_g}_\HH = \sup_{z\in\HH} 4(\im z)^2 \left|\frac{1}{2}\sum_{k=1}^n \frac{1-\alpha_k^2}{(z-a_k)^2} +\sum_{k=1}^n\frac{\beta_k}{z-a_k}\right|. \] Notice however, that \[ \lim_{\overset{z\to a\in \R}{a\neq a_k}} 4(\im z)^2 \left|\frac{1}{2}\sum_{k=1}^n \frac{1-\alpha_k^2}{(z-a_k)^2} +\sum_{k=1}^n\frac{\beta_k}{z-a_k}\right| = 0 \] and \[ \lim_{z\to a_k} 4(\im z)^2 \left|\frac{1}{2}\sum_{k=1}^n \frac{1-\alpha_k^2}{(z-a_k)^2} +\sum_{k=1}^n\frac{\beta_k}{z-a_k}\right| = 2(1-\alpha_k^2)<2. \] Applying the relationships between $\alpha_i$ and $\beta_i$ given by Nehari~\cite{zN52}, we can similarly show \[ \lim_{z\to \infty} 4(\im z)^2 \left|\frac{1}{2}\sum_{k=1}^n \frac{1-\alpha_k^2}{(z-a_k)^2} +\sum_{k=1}^n\frac{\beta_k}{z-a_k}\right| \] is also bounded by 2. In view of the invariance of the Schwarz norm under M\"obius transformations, the same bounds must be true for mappings of $\D$ onto hyperbolic polygons in $\D$. It is not difficult~\cite{PM00} to construct mappings of $\D$ onto hyperbolic polygons with $\norm{S_f}_\D \geq 2$. Thus if $f$ maps $\D$ onto a hyperbolic polygon and $f$ is extremal for maximizing the Schwarz norm, then the supremum in the definition of the Schwarz norm must in fact be attained at some interior point. Suppose $f$ is extremal in $H_n$ for maximizing the Schwarz norm and \[ \max_{z\in\D} (1-\abs{z}^2)^2\abs{S_f(z)}= (1-\abs{z_0}^2)^2\abs{S_f(z_0)}. \] Since $(1-\abs{z}^2)^2\abs{S_f(z)}$ is invariant under M\"obius transformations, precomposing $f$ with a disc automorphism which sends zero to $z_0$ will produce a map $f_0$ which has the same Schwarz norm as $f$ but for which $\norm{S_{f_0}}= \abs{S_{f_0}(0)}$. Notice by precomposing $f$ with a disc automorphism, our new map $f_0$ need not fix $0$ and thus need not remain in $H_n$. However, by postcomposing $f_0$ with a disc automorphism which sends $f_0(0)$ to $0$ and then rotates so that $S_{f_0}(0)$ is real, we can constuct an extremal function $F\in H_n$ with \[ \norm{S_f}_\D=\norm{S_F}_\D=S_{F}(0). \] Since the functional $S_f(0)$ is continuous on $H_n$ and $H_n$ is compact, then \[ \sup_{f\in H_n} \norm{S_f}_\D = \sup_{f\in H_n} S_f(0) = \max_{f\in H_n} S_f(0). \] \end{proof} %Next we describe a variation we can apply to the class $H_n$. \section{Variational Techniques}\label{S:twovar} \subsection{The Julia Variation and Extensions}\label{S:julia} Let $\Omega$ be a region bounded by a piecewise analytic curve $\Gamma$ and $\phi(w)$ be a positive piecewise $C^1$ function on $\Gamma$, vanishing where $\Gamma$ is not analytic. Denote the outward pointing unit normal vector at each point $w$ where $\Gamma$ is smooth by $n(w)$. For $\epsilon$ near $0$, construct a new curve \[ \Gamma_\epsilon=\{w+\epsilon\phi(w)n(w):\: w\in\Gamma\} \] and let $\Omega_\epsilon$ be the new region bounded by $\Gamma_\e$. For our purposes, we may assume that $\Omega$ is simply connected. Thus if $\epsilon$ is sufficiently small and $\epsilon >0$, then $\Gamma$ is ``pushed outside'' the domain, while if $\epsilon < 0$, then $\Gamma$ is ``pushed inside'' the domain. With the above notation, we state Julia's %modification of the %Hadamard variational formula. %; see~\cite{BL75,BS84} for similar variations of %circular boundary arcs. \begin{julia*} Let $f$ be a conformal map from $\D$ onto $\Omega$ with $f(0)=0$, and suppose that $f$ has a continuous extension to $\partial\D$, which we also denote by $f$. Then for $\e$ sufficiently small, a similarly normalized conformal map $f_{\e}$ from $\D$ onto $\Omega_\epsilon$, with $f_{\e}(0) = 0$, is given by \begin{align}\label{E:julia} f_\epsilon(z) = f(z)+\frac{\epsilon z f'(z)}{2\pi}\int_{\partial \D} \frac{1+\xi z}{1-\xi z}\,d\Psi + E(\e,z), \end{align} where $d\Psi=\frac{\phi(f(\xi))}{\abs{f'(\xi)}}\,d\theta$ and $\xi=e^{i\theta}$ and $E(\e,z)$ is $o(\e)$ for $z$ on compact subsets of $\D$ and is continuously differentiable in $\epsilon$ for each fixed $z\in\D$. \end{julia*} Notice the restriction that $\phi$ vanish at the points of non-smoothness of $\partial \Omega$ was a strong one. It implies, for example, that while we could vary the (proper) sides of a hyperbolic polygon, there is a difficulty at the vertices. However, it follows from the work of the first author and J. Lewis~\cite{BL75} that such an extended version of the Julia variation is possible (except at internal cusps, ie., where two proper sides meet at an angle of measure $2\pi$) which allows $f^\prime$ to vanish at the corners. Moreover, they showed that the resulting function will agree with the Julia variational formula on compact subsets up to $o(\epsilon)$ terms. This will allow us to create variations which avoid any problems at the corners and preserve hyperbolic convexity. %and apply the Julia formula to vary the (proper) %sides of hyperbolically %convex polygons while preserving the hyperbolic convexity. \begin{figure} \begin{center}\scalebox{0.7}{ \epsfig{file=julia-out} } \caption{The variation produced by ``pushing out'' a side.} \label{F:julia-out} \end{center} \end{figure} \subsection{Two Variations}\label{SS:twovar} We next describe in detail two variations for functions in $H_n$ which preserve hyperbolic convexity. First, if $f$ maps onto a hyperbolically convex polygon $\Omega$ with a proper side $\Gamma$, we can ``push'' $\Gamma$ to a nearby geodesic $\Gamma_\epsilon$ in such a way that the varied function $f_\epsilon$ will still be hyperbolically convex. See Figure~\ref{F:julia-out}. \begin{lemma} \label{L:julia-out} Suppose $f\in H_n$ and $f$ is not constant nor the identity map. If $\Gamma=f(\gamma)$ is a proper side of $\Omega=f(\D)$, then for $\epsilon$ sufficiently small there exists a variation $f_\epsilon\in H_n$ which ``pushes'' $\Gamma$ to a nearby geodesic $\Gamma_\epsilon$, % That is, $f_\epsilon(\D)$ %s a hyperbolic polygon whose sides are the same as those of $f$, except that %$\Gamma$ has been replaced by $\Gamma_\epsilon$, where $\Gamma_\epsilon\to\Gamma$ as $\epsilon\to 0$. % is a geodesic through $m+\epsilon n(m)$ and $m$ is the euclidean midpoint % of $\Gamma$. This variation, $f_\epsilon$, agrees with the Julia Variational Formula on compact subsets up to $o(\epsilon)$ terms. \end{lemma} \begin{proof} Let $\Gamma$ be a (proper) side of the boundary of $\Omega$ which lies on a circle $\Lambda$. Without loss of generality we can assume that $\Omega$ has been rotated so that the center of $\Lambda$ lies on the positive real-axis. To define our variation first map $\Omega$ into the right half-plane under the map $h(z) = \frac{1+z}{1-z}$ and let $\Upsilon = h(\Gamma)$ and $\Delta = h(\Omega)$. Let $g = h \circ f$. We note that $\Delta = g(\D)$. Under $h$, the geodesics in $\D$ map to arcs in the right half-plane which lie on circles whose centers are on the imaginary axis. We note that $h(\Lambda)$ is a circle centered at the origin. Let the arc $\widehat{AB}$ denote $\Upsilon $ on the circle $h(\Lambda)$ and let $r_{\Lambda}$ be the radius of $h(\Lambda)$. Now, considering the case where $\e > 0$, i.e., the case where we ``push out'' a (proper) side of the domain by a euclidean distance $\e$ and thus enlarge the domain. Let the circle $\chi$ of radius $r = r_{\Lambda} - \e$ be obtained by multiplying $h(\Lambda)$ by $1- \e/r_\Lambda$ and let $\widehat{CD}$ be the arc of $\chi$ which connects the extensions of the circles defined by the images of the $\Gamma$'s adjacent geodesics under $h$, when $A$ and $B$ are in the right-half plane. If $A$ or $B$ is on the imaginary axis, then $C$ or $D$ will be the point on the imaginary axis where $\chi$ intersects the imaginary axis, resp. In this case, the interval $[A,C]$ or $[D,B]$ will be added to the boundary of our varied domain, resp. Now recall in the original Julia variation formula, the function $\phi$ must be $0$ at the end points of $\widehat{AB}$. Thus at each end of $\widehat{AB}$, we construct approximating smooth curves $\sigma_A$ and $\sigma_B$ which connect $\widehat{AB}$ to $\widehat{CD}$ such that $\sigma_A$ and $\sigma_B$ are orthogonal to $\widehat{AB}$ at the endpoints $A$ and $B$ and $\sigma_A$ and $\sigma_B$ intersect $\widehat{CD}$ tangentially at points $C'$ and $D'$, resp. They can be chosen to be within $\e^2$ of the arc they are approximating. See Figure~\ref{F:julia-half-plane}. \begin{figure} \begin{center}\scalebox{0.7}{ \epsfig{file=julia-half-plane} } \caption{The geodesics in the right half plane produced by our variation.} \label{F:julia-half-plane} \end{center} \end{figure} Now for $w \in \widehat{AB}$ let $n(w)$ be the outward unit normal to $g(\Lambda)$ and define $\phi(w) = |w^*-w|$ where $w^*$ is the nearest point on $\sigma_A \cup \widehat{C'D'} \cup \sigma_B$ along the normal $n(w)$. Let the domain $\Delta^*_{\e}$ be the variation of $\Delta$ where the arc $\widehat{AB}$ on the boundary of $\Delta$ has been replaced by $\sigma_A \cup \widehat{C'D'} \cup \sigma_B$. To obtain a variation of $\Delta$ which preserves hyperbolic convexity, let the domain $\Delta_{\e}$ be the variation of $\Delta$ where the arc $\widehat{AB}$ has been replaced by $\widehat{CD}$ and connecting intervals, if necessary. Let $g^*_{\e}$ and $g_{\e}$ be the corresponding mapping functions onto $\Delta_\e^*$ and $\Delta_\e$, resp. Let $\phi(w) = 0$ for $w \in \partial \Delta \setminus \widehat{AB}$. Then, by the Julia Variational Formula, $g^*_{\e}$ satisfies \begin{equation*} g^*_{\e}(z) = g(z)+\frac{\e z g'(z)}{2\pi}\int_{\partial \D} \frac{1+\xi z}{1-\xi z}\,d\Psi + o(\e), \end{equation*} \noindent on compact subsets, where $d\Psi=\frac{\phi(g(\xi))}{\abs{g'(\xi)}}\,d\theta$ and $\xi=e^{i\theta}$. Rewriting $g$ as $h \circ f$, we have \begin{equation*} g^*_{\e}(z) = h(f(z))+\frac{\e z h'(f(z))f'(z)}{2\pi}\int_{\partial \D} \frac{1+\xi z}{1-\xi z}\,d\Psi + o(\e), \end{equation*} \noindent on compact subsets, where $d\Psi=\frac{\phi(h(f(\xi)))} {\abs{h'f((\xi))f'(\xi)}}\,d\theta$ and $\xi=e^{i\theta}$. The function $g^*_{\e}$ was constructed so that it satisfies the requirements for representation by the Julia Variational Formula. The function $g_{\e}$ was constructed so that image, $g_{\e}(\D)$, was hyperbolically convex. The arguments used in \cite{BL75}, pp 348-356, show that $g^*_{\e} - g_{\e} = o(\e)$ on compact subsets. Then, defining $f_{\e} = h^{-1} \circ g_{\e} \equiv \frac{g_{\e}-1}{g_{\e}+1}$, a straightforward argument shows that \begin{equation*} f_\epsilon(z) = f(z)+\frac{\epsilon z f'(z)}{2\pi}\int_{\partial \D} \frac{1+\xi z}{1-\xi z}\,d\Psi + o(\e) \end{equation*} on compact subsets, where $\Psi$ is a real measure on $\partial \D$ and $\xi=e^{i\theta}$. Thus, $f_{\e}$ maps $\D$ onto a hyperbolically convex domain, i.e., $f_{\e} \in H_n$ for $\e$ sufficiently small. A similar variation can be defined for $\e < 0$ that ``pushes in'' the (proper) side $\Gamma$ to a nearby geodesic. \end{proof} %Applying this technique on a subarc of $\Gamma$ ending on an endpoint of %$\Gamma$, If $\Gamma$ intersects another side $\Gamma^*$ at $z^*$, and $z_0\in\Gamma$, then we can vary $f$ so as to replace the portion of $\Gamma$ between $z_0$ and $z^*$ with a new geodesic between $z_0$ and some $z^*_\epsilon\in\Gamma^*$. See Figure~\ref{F:kick-in}. A similar argument to the above then gives: \begin{lemma} \label{L:kick-in} Suppose $f\in H_n$ and $f$ is not constant nor the identity map, and $\Gamma=f(\gamma)$ is a proper side of $f(\D)$ meeting a side $\Gamma^*$. Then, there exists a variation $f_\epsilon\in H_{n+1}$ which adds a proper side to $f(\D)$ by pushing one end of $\Gamma$ to a nearby side $\Gamma_\epsilon$. That is, $f_\epsilon(\D)$ is a hyperbolic polygon whose sides are the same as those of $f$, except that one end of $\Gamma$ has been replaced by $\Gamma_\epsilon$ and $\Gamma^*$ has been shortened. Moreover, $f_\epsilon$ agrees with the Julia Variational Formula on compact subsets up to $o(\epsilon)$ terms. \end{lemma} \begin{figure} \begin{center}\scalebox{0.65}{ \epsfig{file=julia-kick-in} } \caption{The variation produced by pushing in one end of a side.} \label{F:kick-in} \end{center} \end{figure} \begin{remark}\label{R:end} Notice that in order to maintain hyperbolic convexity, we can only push in the ``end'' of a side, that is, a subarc that ends at a vertex of the polygon. However, we can choose this subarc to be as long, or more importantly, as short, as we wish. \end{remark} \section{The Effect of the Variations on the Schwarzian}\label{S:effect} %If the supremum in the definition of the Schwarz norm occurs at some %point $z_0$ in $\D$, we could pre-compose $f$ with a disc automorphism %sending $0$ to $z_0$. This will leave both the Schwarz norm and the %range of $f$ unchanged. %for many purposes we can assume $f$ has been normalized so that %$\Snf=S_f(0)$. Recall from Lemma~\ref{L:sf0}, we can assume our extremal function satisfies $\norm{S_f}_\D = S_f(0)$. In this case, the Schwarz norm has a simple representation in terms of power series coefficients~\cite{oL87,PM00}. \begin{proposition}\label{P:Sf0} If $f=\alpha(z + a_2 z^2 + a_3 z^3 +\dots)$, % is extremal over $H_n$ for $\norm{S_f}_\D$, then % has been normalized so that the sup in the definition of the %Schwarz norm occurs at $z=0$, then \[ S_f(0)=6(a_3 - a_2^2). \] \end{proposition} \begin{proof} Since $f'(0)=\alpha$, $f''(0)={2}\alpha a_2$, and $f'''(0)={6}\alpha a_3$, \begin{align}\notag \frac{d}{dz} \frac{f''(z)}{f'(z)}\bigg\vert_{z=0} &=\frac{f'(0)f'''(0)- (f''(0))^2}{(f'(0))^2}\\\notag &=\frac{6\alpha^2 a_3 - {4}\alpha^2 a_2^2}{\alpha^2}\\\notag &=6a_3 -4 a_2^2. \end{align} Also, \[ \frac{1}{2}\left(\frac{f''(0)}{f'(0)}\right)^2 =\frac{1}{2}\frac{4 \alpha^2 a_2^2}{\alpha^2} = 2 a_2^2. \] Consequently, \[ S_f(0)= %\eta_\D(0) \left( \left(\frac{f''}{f'} \right)' - \frac12 \left( \frac{f''}{f'} \right)^2\right) = 6(a_3-a_2^2). \] \end{proof} Now applying the representation for the variations provided by the Julia Formula, we can calculate to first order in $\epsilon$ the Schwarzian of our varied functions. \begin{lemma}\label{L:Sg} If $f(z)=\alpha(z +a_2 z^2 +a_3 z^3 + \dots)\in H_n$ %has been normalized so that $\norm{S_f}_\D=S_f(0)$ and either of the variations described in Lemma \ref{L:julia-out} and Lemma \ref{L:kick-in} are applied to $f$, then the new function $f_\epsilon(z)$ has $S_{f_\epsilon}(0)$ given by \[ 6\left( \left( a_3 +\frac{\epsilon}{2\pi} \int_\Gamma (3a_3 + 4a_2\xi + 2\xi^2) \, d\Psi \right) - \left( a_2 +\frac{\e}{2\pi} \int_\Gamma (2a_2 + 2\xi) \, d\Psi \right)^2\right) +o(\epsilon). \] \end{lemma} \begin{proof} Let $\gamma$ be the preimage in $\partial \D$ of the portion of the boundary we are varying. Expanding $\frac{1+\xi z}{1-\xi z}$ as a series, we obtain on compact subsets \begin{align}\notag f_\epsilon(z)&=f(z) + \frac{\epsilon z f^\prime(z)}{2\pi} \int_\gamma \frac{1 +\xi z}{1- \xi z} \,d\Psi + o(\epsilon) \\\notag %&= f(z) + \frac{\epsilon z f^\prime(z)}{2\pi} %\int_\gamma (1+\xi z)(1+ \xi z + \xi^2 z^2 + \xi^3 z^3 +\dots) \, d\Psi %+ o(\epsilon)\\\notag &= f(z) + \frac{\epsilon z f^\prime(z)}{2\pi} \int_\gamma (1+2 \xi z+2 \xi^2 z^2 +2 \xi^3 z^2 +\dots)\, d\Psi+o(\epsilon). \end{align} Now since $zf'(z)=\alpha(z+2a_2 z^2 + 3 a_3 z^3 +\dots)$, we have \begin{align}\notag f_\epsilon(z) = % &f(z) + \frac{\epsilon}{2\pi} % \int_\gamma \alpha(z+2a_2 z^2 + 3 a_3 z^3 +\dots) \\\notag % &(1+2 \xi z+2 \xi^2 z^2 +2 \xi^3 z^2 +\dots)\, d\Psi+o(\epsilon)\\\notag = & f(z)+\frac{\epsilon\alpha}{2\pi}\int_\gamma [z + 2(a_2 + \xi) z^2 + (3a_3 + 4a_2 \xi +2\xi^2)z^3 +\dots] \, d\Psi + o(\epsilon) \end{align} on compact subsets. Finally, gathering the powers of $z$, we arrive at \begin{align}\label{E:fepsilon} f_\epsilon(z)=\alpha\Biggl(& \biggl(1+\frac{\epsilon}{2\pi} \int_\gamma \, d\Psi\biggr) z+ \biggl(a_2+\frac{\epsilon}{2\pi} \int_\gamma 2(a_2+\xi)\, d\Psi\biggr)z^2\\\notag & + \left(a_3+\frac{\epsilon}{2\pi} \int_\gamma (3a_3+4a_2\xi+2\xi^2) \, d\Psi\right)z^3 + \dots \Biggr)+o(\epsilon) \end{align} on compact subsets. Hence by Proposition~\ref{P:Sf0}, \begin{align}\notag \frac{1}{6} S_{f_\epsilon}(0) = &\left(a_3 +\frac{\epsilon}{2\pi}\int_\gamma (3a_3+4a_2\xi+2\xi^2) \, d\Psi\right)- \\\notag &\left(a_2+ \frac{\epsilon}{2\pi}\int_\gamma 2(a_2+\xi)\,d\Psi\right)^2 +o(\epsilon). \end{align} \end{proof} Taking the derivative of the new Schwarzian with respect to $\epsilon$, we have \begin{align}\notag \frac{1}{6}\frac{\partial}{\partial \e} S_{f_\epsilon}(0) = &\frac 1{2\pi}\iG (3a_3 + 4a_2\xi + 2\xi^2) \, d\Psi \\ \notag &- 2 a_2\frac{1}{2\pi} \iG 2(a_2+\xi) \, d\Psi - \frac{2\e}{2\pi}\left(\iG 2(a_2 + z) \, d\Psi\right)^2 +\dots \\\notag & +o(1). \end{align} When $\e=0$, the derivative becomes \begin{align}\label{E:Sg} \frac{\partial}{\partial \e} S_{f_\epsilon}(0) \bigg\vert_{\e=0} %= &6\left( %\frac{1}{2\pi}\iG 3a_3 +4a_2\xi +\xi^2 -4a_2^2 - 4a_2\xi \, d\Psi\right) \\\notag &=\frac{6}{2\pi}\iG (3a_3-4a_2^2 + 2\xi^2) \, d\Psi. \end{align} \section{Reduction to at Most Two Proper Sides}\label{S:2sides} With these computations, we can now show that any function mapping onto a region with more than two disjoint proper sides cannot be extremal. We do this in two steps, first reducing the extremal domain to one having at most four proper sides, then employing our Step Down Lemma to further reduce the possibilities to at most two proper sides. \begin{lemma}\label{L:4sides} Suppose $f \in H_n$ is a hyperbolic convex function onto a region with more than four proper sides, then $f$ cannot be extremal for maximizing the Schwarz norm. \end{lemma} \begin{proof} Suppose $f\in H_n$ is extremal, that $\norm{S_f}_\D = S_f(0)$ and that $f$ maps onto a region with more than four proper sides. %In Lemma~\ref{L:sf0} we showed %that we may assume the extremal function $f$ satisfies %$\norm{S_f}_\D = S_f(0)$. We will apply our variations to show that $f$ cannot be extremal. %Recall, however, that our variations were %designed to control $\abs{S_f(0)}$, not $\norm{S_f}_\D$ itself. %the supremum in the definition of the Schwarz norm for %the extremal $f$ is acutally a %maximum, and the maximum is atained at some interior point $z_0\in\D$. %Thus we can pre-compose $f$ with a disc automorphism %which sends $0$ to $z_0$ and produce a function with the same range, %but normalized so that %\begin{equation}\label{E:normalized} %\norm{S_f}_\D=S_f(0). %\end{equation} %On the other hand, given a function onto a hyperbolically %convex $n$-gon which has been normalized %so that~\eqref{E:normalized} holds, pre- and post-composition %with disc automormhisms can be used produce a map which fixes the %origin and thus lies in $H_n$. Thus we may work with either %normalization %and in light of our variational lemmas, % we assume that %If $f$ is extremal, then Recall from equation~\eqref{E:Sg} that for either of our variations along a side $\Gamma=f(\gamma)$ \[ \frac{\partial}{\partial \epsilon} S_{f_\epsilon} (0) \bigg|_{\epsilon=0} = \frac6{2\pi} \iG \left(3a_3 - 4a_2^2 + 2\xi^2\right) d\Psi. \] By composing $f_\epsilon$ with a rotation (which will not change $S_{f_\epsilon}$), we may assume \[ \frac{\partial}{\partial \epsilon} S_{f_\epsilon} (0) \bigg|_{\epsilon=0} =\frac{6}{2\pi} \iG \re \{3a_3 - 4a_2^2 + 2\xi^2\} \,d\Psi. \] But now notice that $\K (\xi) = 3a_3 - 4a_2^2 + 2\xi^2$ wraps $\{\abs{\xi}=1\}$ twice around the image circle $C$. If $\K(\gamma)$ lies completely in the right half plane, then we push the (proper) side $\Gamma$ out, with $\e > 0$ as in Lemma \ref{L:julia-out}. Then, since $\re \K(\gamma) > 0$ for $\gamma = f^{-1}(\Gamma)$, we will have that $\iG \re\{3a_3 - 4a_2^2 + 2\xi^2\} \, d\Psi > 0 $. Similarly, if $\K (\gamma)$ lies completely in the left half plane, $\Gamma$ can be pushed in so that $\iG \re\{3a_3 - 4a_2^2 + 2\xi^2 \} \, d\Psi < 0$. On the other hand, if $f$ is extremal, then $\frac{\partial }{\partial \epsilon} S_{f_\epsilon}(0) |_{\epsilon=0}$ must equal $0$. Otherwise, \[ \norm{S_f}_\D=S_f(0)0$. This varied function $f_\epsilon$ will thus lie in $H_5\setminus H_4$ and have a larger Schwarz norm than any function in $H_4$. But this contradicts Lemma~\ref{L:4sides}, since the extremal function in $H_5$ can have at most $4$ proper sides. As a result, the extremal domain for a map in $H_5$ can have at most $3$ proper sides. %Since $H_4\subset H_5$, %But %since the extremal value is monotonic as a function of $n$, this %would violate %the fact that the extremal domain for $H_5$ has at most 4 sides. %the extremal domain for $H_4$ cannot have 4 sides. However, if the extremal domain has exactly 3 proper sides, some endpoint must again lie in the left half-plane. As before, we may add a proper side and increase the Schwarz norm. Thus the extremal domain can have at most two proper sides. \end{proof} \section{Computing the Schwarz Norms for One- and Two-Sided Functions}\label{S:compute} Now we need only compute the Schwarz norm for the remaining possible extremal functions to complete the proof of our theorem. First, however, we need a couple of facts about the behavior of elliptic integrals~[1,~pp.~53-54]. \begin{lemma}~\label{L:matti} If $K$ and $E$ are the complete elliptic integrals of the first and second kind, respectively, then \begin{enumerate} \item For each $c\in[1/2,2]$, the function $g(x)=(\sqrt{1-x^2})^c K(x)$ is decreasing from $[0,1)$ onto $(0,\pi/2]$. \item The function \[ h(t)=\frac{E(t)-(1-t^2)K(t)}{t^2} \] is increasing from $(0,1)$ onto $(\pi/4,1)$. \end{enumerate} \end{lemma} With this lemma, we can prove our main result: \begin{thm*} The maximal value of the Schwarz norm for hyperbolically convex functions is $S_{f_\alpha}(0)$, where \[ f_\alpha(z) =\tan\left( \alpha\int_0^z \left(1-2\xi^2\cos 2\theta+\xi^4 \right)^{-1/2}\,d\xi \right), \qquad \alpha = \frac{\pi}{2K(\cos \theta)}, \] $K$ is the elliptic integral of the first kind, and $\alpha$ is chosen so that $\cos \theta$ is the unique critical point of the function \[ g(s) = 4s^2-2 +\frac{\pi^2}{2K^2(s)} \] on $(0,1)$. \end{thm*} Numerical calculation indicates the extremal value of the Schwarz norm is approximately $2.383635$ and is achieved by $S_{f_\alpha}(0)$ with $\alpha\approx 0.5343598$. \begin{proof} By Lemma~\ref{L:4sides} and the Step Down Lemma, the extremal function in $H_n$, $n>2$, cannot map onto a region with more than $2$ proper sides. Thus we need only determine whether the extremal function has $0$, $1$, or $2$ proper sides. The only hyperbolically convex functions onto a region with no proper sides are disc automorphisms which have Schwarz norm equal to $0$. Any hyperbolically convex function onto a region with exactly one proper side differs from the function $g$ of Example~\ref{E:one-side} by composition with M\"obius transformations, and hence has the same Schwarz norm, namely $3/2$. Note that Mej\'ia and Pommerenke~\cite{PM00} also explicitly computed these maps and their Schwarz norm. Next notice that for two-sided domains whose proper sides intersect inside the disc, the intersection point can be moved to the origin by a disc automorphism. Moreover, domains bounded by two proper sides which do not intersect can be mapped via a disc automorphism onto an odd symmetric domain, as in figure~\ref{F:strip}, and domains whose two proper sides intersect on the boundary of the disc can be approximated by domains whose (proper) sides do not intersect. Thus there are two types of regions bounded by exactly two proper sides that we must consider: \begin{enumerate} \item \label{I:origin} Domains in which the two proper sides intersect at the origin \item Odd symmetric domains in which the proper sides do not intersect \end{enumerate} For case~\eqref{I:origin}, consider the family of maps \[ g_\alpha(z)= \left(\frac{\sqrt{z}-1}{\sqrt{z}+1}\right)^\alpha,\quad 0<\alpha<1 \] from the upper half plane onto a sector of $\D$ with an opening of angle $\alpha\pi$ at the origin. Functions which map onto domains of the first type can be obtained from one of the maps $g_\alpha$ by composition with M\"obius transformations, thus we may limit our attention to this family. Computation and simplification reveals \[ S_{g_\alpha}(z)=\frac{3z^2-(2+4\alpha^2)z +3}{8z^2(z-1)^2}. \] Because both the Schwarzian and the hyperbolic density $\eta_\HH(z)=1/(2\im{z})$ are invariant under horizontal translation, we need only consider pure imaginary numbers when computing the Schwarz norm. Thus \begin{align}\notag \norm{S_{g_\alpha}}_\HH=&\sup_{z\in\HH} 4\left(\im z\right)^2 \left|\frac{3z^2-(2+4\alpha^2)z +3}{8z^2(z-1)^2}\right| \\\notag &=\sup_{y>0} \frac{1}{2}\frac{\abs{3y^2+(2+4\alpha^2)iy -3}}{y^2+1}. \end{align} The derivative of the right hand side with respect to $y$ is \[ \frac{-8(\alpha^2-1)(\alpha^2+2)(y^3-y)} {\sqrt{9y^4+(16\alpha^4+16\alpha^2-14)y^2 +9}(y^2+1)^2}. \] Clearly the denominator is positive for all $y>0$, $0<\alpha<1$. Moreover, for all $0<\alpha<1$ the numerator is negative for $0 -\min (1-x^2)^2 {S_f(x)}$. We will show that \begin{equation} \label{claim_min} \min (1-x^2)^2 {S_f(x)} \ge -2 \end{equation} \vspace{0.1in} \noindent over $ 0 < x < 1$ and $0<\theta<\pi/2$. Then, as Mej\'ia and Pommerenke~\cite{PM00} noted (and as we will show) that $\max (1-x^2)^2 {S_f(x)} > 2$, the claim will hold. We simplify our algebraic computations by replacing $x$ with $\sqrt{x}$ in the above expression and obtain \begin{equation} \label{E:m} (1-x)^{2} \frac{2( (\cos(2\theta)+\alpha^2) + (\cos^2(2\theta)-2\alpha^2\cos(2 \theta) -3)x + (\cos(2\theta) +\alpha^2)x^2)}{(1-2\cos(2\theta)x +x^2)^2} \end{equation} \noindent where $ 0 \le x < 1$ and $0<\theta<\pi/2$. To minimize (\ref{E:m}), we note that the coefficient of $\alpha^2$ is $1-2\cos(2\theta)x +x^2$ which is nonnegative. Hence, substituting $\alpha=0$ into (\ref{E:m}), we have that (\ref{E:m}) is bounded below by \begin{equation} \label{E:m1} (1-x)^{2} \frac{2( \cos(2\theta) + (\cos^2(2\theta) -3)x + \cos(2\theta) x^2)}{(1-2\cos(2\theta)x +x^2)^2} \end{equation} We will show that (\ref{E:m1}) is bounded below by $-2$ by showing that \begin{align}\notag &(1-x)^{2} \frac{2( \cos(2\theta) + (\cos^2(2\theta) -3)x + \cos(2\theta) x^2)}{(1-2\cos(2\theta)x +x^2)^2} + 2 \\ \notag &= \frac{p(c,x)}{(1-2\cos(2\theta)x +x^2)^2} > 0 \end{align} \noindent where $$ p(c,x) = (2x+4x^2 + 2x^3)c^2 + (2-12x+4x^2-12x^3+2x^4)c + (2-6x+16x^2-6x^3+2x^4) $$ \noindent and $c = \cos(2\theta)$ satistifies $-1 < c < 1$. It is sufficient to show that $p(c,x) > 0$ for $-1 < c < 1$ and $0 < x < 1$. We note that $p(c,x)$ is quadratic in $c$ for each fixed $x$ and the coefficent of $c^2$ is positive. Let $$ c(x) = -\frac{2-12x+4x^2-12x^3+2x^4}{2(2x+4x^2 + 2x^3)}, $$ \noindent then $c(x)$ is the location of the vertex of the quadratic $p(c,x)$ for each fixed $x$. We claim that $c(x)$ is an increasing function of $x$ on $(0,1)$. To see this, we note that $$c'(x) = \frac{n(x)}{2(x+1)(1+2x+x^2)x^2}$$ \noindent where $n(x) = -x^5-3x^4+14x^3-14x^2+3x+1$. Clearly, $n(0) = 1$ and $n(1) = 0$. A Sturm sequence argument shows that $n(x)$ has exactly one zero on $(0,1]$. Hence, we have $c'(x) > 0$ on $(0,1)$ and $c(x)$ is increasing on $(0,1)$. Let $x_0$ be the unique solution of $c(x) = -1$ for $0 < x < 1$, $x_0 \approx 0.1197$. For $0 < x < x_0$, the location $c(x)$ of the vertex of $p(c,x)$ satisfies $c(x) < -1$. Hence over $-1 < c < 1$ and $0 < x < 1$, we have $ \min p(c,x) > p(-1,x) = 8x + 16x^2 +8x^3 > 0$. For $x_0 < x < 1$, we have $-1 < c(x) < 1$ and, hence, $$ \min p(c,x) = p(c,x)|_{c=c(x)} = \frac{q(x)}{2x(1+2x+x^2)} $$ \noindent where $$ q(x) = -x^8 + 16x^7 - 44x^6 + 48x^5 - 3xx^4 +48x^3-44x^2+16x -1. $$ Clearly, $q(1) = 0$ and $q(1/2)=\frac{143}{256} > 0$. A Sturm sequence argument shows that on the interval $(x_0,1]$ that $q$ has exactly one zero. Hence, on $(0,1)$, $q(x) > 0$. Thus, $p(c,x) > 0$ for $-1 < c < 1$ and $0 < x < 1$ and, consequently, $\min (1-x^2)^2 {S_f(x)} \ge -2$. To maximize (\ref{E:m}), we substitute $\alpha=\frac{\pi}{2 K(\cos(\theta))}$, take the derivative with respect to $x$, and simplify to obtain \begin{align} \frac{(x^2-1)(\cos(2\theta) -1)} {K^2(\cos(\theta))(x^2-2\cos(2\theta)x+1)^3} &(K^2(\cos\theta)(-10x^2 \cos(2\theta) \\\notag &-6x^2 +12x\cos(2\theta)-4x\cos^2(2\theta)\\\notag &+24x-10\cos(2\theta)-6) -\pi^2(x^2-2\cos(2\theta)+1)) \end{align} Notice that the denominator of this expression is always positive. %since %\[ %x^2-2\cos(2\theta)x+1 > x^2-2x+1 =(x-1)^2 >0. %\] On the other hand, the numerator changes sign as $x$ varies. However, we will see that for fixed $\theta$, the numerator can change sign only once. To justify this claim, fix $\theta$ and let \begin{align} \notag A(x)=K^2(\cos\theta)(&-10x^2 \cos(2\theta) -6x^2 +12x\cos(2\theta)-4x\cos^2(2\theta)\\\notag & +24x-10\cos(2\theta)-6) -\pi^2(x^2-2\cos(2\theta)+1). \end{align} Notice that $(x^2-1)(\cos(2\theta)-1)>0$ for $0\leq x<1$ and $0<\theta<\pi/2$, so we need only show that $A(x)$ changes sign only once. It follows from known properties~\cite{AVV97} of $K$ that $K(\cos(\theta))>\pi/2$; thus, since the coefficient of $K(\cos(\theta))$ in \eqref{E:ktheta} is positive, \begin{align}\label{E:ktheta} A'(x) & = K^2(\cos(2\theta))(-20x\cos(2\theta) -12x+12\cos(2\theta) - 4\cos^2(2\theta) + 24) - 2\pi^2(x-\cos(2\theta))\\\notag &>\left(\frac{\pi}{2}\right)^2(-20x\cos(2\theta) -12x+12\cos(2\theta) - 4\cos^2(2\theta) + 24) - 2\pi^2(x-\cos(2\theta))\\\notag &= (-10\pi^2\cos^2\theta)x + (14\pi^2\cos^2\theta-4\pi^2\cos^4\theta ) \end{align} The right hand side of the above inequality is linear in $x$ and positive at both $x=0$ and $x=1$. Thus $A'(x)>0$ and $A(x)$ is increasing for all $0\leq x <1$. Consequently, $A(x)$ can change sign at most once. Notice \[ A(0)=-10\cos(2\theta)K^2(\cos(2\theta)) -6K^2(\cos(\theta))-\pi^2 <0 \] for all $0<\theta<\pi/2$. Thus, as a consequence of the above argument, for each fixed $\theta$, $ %\frac\partial}{\partial x} (1-x^2)^2S_{f_\alpha}(x)$ is decreasing at $x=0$ and switches from decreasing to increasing at most once on the interval $0\leq x <1$. (Recall that to establish this fact, we replaced $x$ by $\sqrt{x}$ in our calculations. We now return $x$ to its original meaning.) Clearly $(1-x^2)^2S_{f_\alpha}(x)$ equals $0$ when $x=1$. Thus even after $(1-x^2)^2S_{f_\alpha}(x)$ begins increasing, it never becomes larger than $0$. On the other hand, when $x=0$, \begin{equation}\notag (1-x^2)^2S_{f_\alpha}(x)|_{x=0} = 2\cos(2\theta)+\frac{\pi^2}{2K^2(\cos(\theta))}. \end{equation} Applying Lemma~\ref{L:matti} with $c=1/2$ %from~[1,~p.~54], we obtain \begin{align}\notag (1-x^2)^2S_{f_\alpha}(x)|_{x=0} &\geq 2\cos(2\theta)+\frac{\pi^2}{\frac{\pi^2}{2 \sqrt{1-\cos^2(2\theta)}}} \\\notag &=2(1+2\sin(\theta))(1-\sin(\theta))>0 \end{align} for all $0<\theta<\pi/2$. Thus, $\sup_{0\leq x<1}(1-x^2)^2 S_{f_\alpha}(x)$ is indeed achieved for $x=0$. See Figure~\ref{F:x-running}. \begin{figure} \begin{center}\scalebox{0.45}{ %\begin{rotate}{-90} %\begin{minipage}{\linewidth} \epsfig{file=x-running,angle=-90} } %\end{minipage} %\end{rotate} } \caption{The graph of $(1-x^2)^2{S_{f_\alpha}(x)}$ for $0\leq x<1$ and $\alpha=0.5343598$ ($\theta=0.218202$).} \label{F:x-running} \end{center} \end{figure} Now the only question is to determine for which value (or values) of $\theta$ \[ \norm{S_{f_\alpha}}_\D=S_{f_\alpha}(0) = 4\cos^2(\theta)-2+\frac{\pi^2}{2K^2(\cos(\theta))} \] is maximized. To this end, let $s=\cos \theta$ and let \[ g(s)=4s^2-2+\frac{\pi^2}{2K^2({s})}. \] Notice that \[ \frac{dg}{d\theta}=\frac{dg}{ds}\frac{ds}{d\theta}, \] and $\frac{ds}{d\theta}=-\sin(\theta)<0$ for $0<\theta<\pi/2$. Thus since $\cos(\theta)$ is decreasing for $0<\theta<\pi/2$, if $\frac{dg}{ds}$ switches signs at most once, then $\norm{S_{f_\alpha}}_\D$ will have a relative maximum for at most one value of $\theta$. Computing this derivative and simplifying, we obtain \begin{equation}\label{E:dg} \frac{dg}{ds}(s)= 8s - \frac{\pi^2}{s}\frac{ E\left(s\right)-(1-s^2)K( s))}{(1-s^2)K^3({s})}. \end{equation} For $0