\documentclass[12pt,twoside]{amsart} \usepackage{mathrsfs} \usepackage{mathtools} \usepackage{amssymb} \usepackage{verbatim} \usepackage{amsmath} \usepackage{comment} \usepackage{amsthm,thmtools,xcolor} \usepackage[colorlinks,linkcolor=blue,citecolor=blue, pdfstartview=FitH] {hyperref} \usepackage{backref} \setlength{\textwidth}{5.8in} \setlength{\oddsidemargin}{0.3in} \setlength{\evensidemargin}{0.3in}\setlength{\footskip}{0.3in} \setlength{\headsep}{0.25in} \usepackage{bm} \usepackage{a4wide} %\usepackage{babel} \usepackage[latin1]{inputenc} \usepackage[T1]{fontenc} \usepackage{times} \usepackage{hyperref} \usepackage{amssymb,latexsym} \usepackage{etoolbox} \makeatletter \patchcmd\maketitle {\uppercasenonmath\shorttitle} {} {}{} \patchcmd\maketitle {\@nx\MakeUppercase{\the\toks@}} {\the\toks@} {} {}{} \patchcmd\@settitle {\uppercasenonmath\@title} {} {}{} \patchcmd\@setauthors {\MakeUppercase{\authors}} {\authors} {}{} \usepackage{enumerate} \newcommand{\op} {\overline{\partial}} \newcommand{\dbar}{\ensuremath{\overline\partial}} \newcommand{\dbarstar}{\ensuremath{\overline\partial^*}} \newcommand{\C}{\ensuremath{\mathbb{C}}} \newcommand{\R}{\ensuremath{\mathbb{R}}} \newcommand{\D}{\ensuremath{\mathbb{D}}} \newcommand{\T}{\ensuremath{\mathbb{T}}} \newcommand{\B}{\ensuremath{\mathbb{B}}} \newcommand{\X}{\ensuremath{\mathcal{X}}} \newcommand{\h}{\ensuremath{\mathfrak{H}}} \newcommand{\s}{\ensuremath{\bm\theta}} \newcommand\ad{{\rm ad}} \usepackage{amsthm,thmtools,xcolor} \declaretheoremstyle[ headfont=\color{brown}\normalfont\bfseries, bodyfont=\color{blue}\normalfont\itshape, ]{colored} \declaretheorem[ style=colored, name=Theorem, ]{thm} \usepackage{amsthm,thmtools,xcolor} \declaretheoremstyle[ headfont=\color{brown}\normalfont\bfseries, bodyfont=\color{blue}\normalfont\itshape, ]{colored} \declaretheorem[ style=colored, name=Proposition, ]{pr} \usepackage{amsthm,thmtools,xcolor} \declaretheoremstyle[ headfont=\color{brown}\normalfont\bfseries, bodyfont=\color{blue}\normalfont\itshape, ]{colored} \declaretheorem[ style=colored, name=Lemma, ]{lm} \usepackage{amsthm,thmtools,xcolor} \declaretheoremstyle[ headfont=\color{brown}\normalfont\bfseries, bodyfont=\color{blue}\normalfont\itshape, ]{colored} \declaretheorem[ style=colored, name=Corollary, ]{co} \usepackage{amsthm,thmtools,xcolor} \declaretheoremstyle[ headfont=\color{brown}\normalfont\bfseries, bodyfont=\color{blue}\normalfont\itshape, ]{colored} \declaretheorem[ style=colored, name=Definition, ]{de} \usepackage{amsthm,thmtools,xcolor} \declaretheoremstyle[ headfont=\color{brown}\normalfont\bfseries, bodyfont=\color{blue}\normalfont\itshape, ]{colored} \declaretheorem[ style=colored, name=Remark, ]{re} \makeatletter \newcommand{\sumprime}{\if@display\sideset{}{'}\sum% \else\sum'\fi} \makeatother \begin{document} \numberwithin{equation}{section} % \thanks{Research supported by Knut and Alice Wallenberg Foundation, and the China Postdoctoral Science Foundation.} \bigskip \address{DEPARTMENT OF MATHEMATICAL SCIENCES, NORWEGIAN UNIVERSITY OF SCIENCE AND TECHNOLOGY, NO-7491 TRONDHEIM, NORWAY} \medskip \email{xu.wang@ntnu.no} \title[]{Complex analysis with potential theory} \author{Xu Wang} \date{\today} \begin{abstract} These are the course notes for the course TMA4175 [2023] "Complex Analysis" at NTNU. It is mainly based on the complex analysis book \cite{Ahlfors} of Ahlfors and will be updated after each lecture. \end{abstract} \maketitle % \section{Introduction} % Variation of Hodge star asso fibration. \tableofcontents %We shall follow the Ahlfors book in section 2-5. Around 40/60 pages 7 weeks for Chapter 1-4, 6 weeks for Chapter 5-7. \section{Basic complex analysis} \subsection{Holomorphic functions} \medskip \begin{de}\label{de:open} A set $U\subset \mathbb C$ is said to be \textbf{open} if for every $z\in U$, $$ \mathbb D_z(r) :=\{w\in \mathbb C: |w-z|0$. \end{de} \medskip \emph{Example}: The unit disk (centered at the origin) $$ \mathbb D:=\{z\in \mathbb C: |z|<1\} $$ is open. The \emph{annulus} $$ \mathbb D_{r,1}:=\{z\in \mathbb C: r<|z|<1\}, \ \ 00$ small). Since $U$ is bounded, there are only finite number of squares, say $Q_1,\cdots, Q_n$, which are contained in $U$. Then let us define $\Gamma:=\partial (Q_1\cup\cdots \cup Q_n)$ as the boundary of $Q_1\cup\cdots \cup Q_n$. \smallskip \emph{Step 3 --- show that $f(z) = \frac{1}{2\pi i }\int_{\Gamma} \frac{f(\zeta) d\zeta}{\zeta-z}, \ \ \text{for all}\ z\in \gamma$}: Assume that $z$ lies in the interior of $Q_{j_0}$ for some $1\leq j_0\leq n$, then one may deform the boundary of the square $Q_{j_0}$ to a small circle around $z$, hence Theorem \ref{th:CIH} gives $$ f(z) = \frac{1}{2\pi i }\int_{\partial Q_{j_0}} \frac{f(\zeta) d\zeta}{\zeta-z}, $$ moreover, the other squares $Q_j$, $j\neq j_0$, satisfy $\partial Q_j\sim_{U\setminus\{z\}} 0$, hence Theorem \ref{th:CIH} further gives $$ 0= \frac{1}{2\pi i }\int_{\partial Q_{j}} \frac{f(\zeta) d\zeta}{\zeta-z}, \ \ j\neq j_0, \ 1\leq j \leq n. $$ Thus \begin{equation}\label{eq:Cauchy-h2} f(z) = \frac{1}{2\pi i }\int_{\partial Q_{1} +\cdots+\partial Q_n} \frac{f(\zeta) d\zeta}{\zeta-z}, \end{equation} for all $z$ lies in the interior of $Q_{j_0}$, $1\leq j_0\leq n$. Note that each $\partial Q_{1} +\cdots+\partial Q_n$ is a sum of oriented line segments, some of them are shared by two squares, which will be canceled in the integral in \eqref{eq:Cauchy-h2}, the remaining line segments are precisely those that make up $\Gamma$. Hence \begin{equation}\label{eq:Cauchy-h3} f(z) = \frac{1}{2\pi i }\int_{\Gamma} \frac{f(\zeta) d\zeta}{\zeta-z}, \end{equation} for all $z$ lies in the interior of $Q_{j_0}$, $1\leq j_0\leq n$. But notice that both sides of \eqref{eq:Cauchy-h2} are continuous functions on the interior of $ Q_1\cup\cdots \cup Q_n$, so \eqref{eq:Cauchy-h2} actually holds for all $z$ in the interior of $ Q_1\cup\cdots \cup Q_n$, which whould contain $\gamma$ when $\delta$ is small enough. Hence \eqref{eq:Cauchy-h2} holds for all $z\in \gamma$. \smallskip \emph{Step 4 --- show that $n(\gamma, \zeta)=0 \ \text{for all}\ \zeta\in\Gamma$}: Observe that $\zeta\in \Gamma$ if and only if $\zeta \in \partial Q_j\cap \partial Q$ for some $1\leq j\leq n$ and some square $Q$ that does not fully lie in $U$, in which case we can choose $\zeta_0\in Q\setminus U$, then the line segment joining $\zeta$ and $\zeta_0$ lies in $Q$ and therefore does not meet $\gamma$, hence $n(\gamma, \zeta)= n(\gamma, \zeta_0) =0$. The proof now is complete. \end{proof} \medskip \noindent \textbf{Remark}: \emph{Note that $$ n(\gamma_0, a) = n(\gamma_1, a), \ \ \forall \ a\notin U, $$ if $\gamma_0\sim_U \gamma_1$. Hence $\gamma_0\sim_U \gamma_1$ implies that $\gamma_0 -\gamma_1$ is homologous to zero in $U$ and the above theorem generalizes Theorem \ref{th:CIH}. \textcolor{red}{We stopped here in the Jan 12th lecture}.} \subsubsection{Homology theory in the complex plane} \medskip \begin{de} An open set $\Omega\subset \mathbb C$ is called a \textbf{domain (region)} if any two points in $\Omega$ can be connected by a piecewise smooth curve in $\Omega$. \end{de} \medskip \noindent \textbf{Remark}: \emph{Our definition is different from the Ahlfors definition in page 57, Definition 4. But in any case, they are equivalent by the following theorem.} \medskip \begin{thm}[Page 56, Theorem 3]\label{th:domain} Let $\Omega$ be a nonempty open set in $\mathbb C$. Then the followings are equivalent: \begin{itemize} \item[(1)] $\Omega$ is a domain; \smallskip \item[(2)] If $\Omega=\Omega_1\cup \Omega_2$ with $\Omega_1$, $\Omega_2$ open and $\Omega_1\cap \Omega_2 =\emptyset$ then either $\Omega_1$ or $\Omega_2$ is empty; \smallskip \item[(3)] Any two points in $\Omega$ can be connected by a piecewise horizontal/vertical curve in $\Omega$. \end{itemize} \end{thm} \medskip \begin{proof} \emph{(1) implies (2)}: Otherwise, both are nonempty, so we can choose $p\in \Omega_1$, $q\in \Omega_2$ and connect $p, q$ with a piecewise smooth curve $\gamma(t)$ in $\Omega$ with $\gamma(0)=p$, $\gamma(1)=q$. Consider $$ T:=\sup\{0\leq t\leq 1: \gamma(t)\in \Omega_1\}, $$ Since $\gamma(1) \notin \Omega_1$, we know that $00$, this of course can not happen by the maximum property of $T$. So we must have $\gamma(T)\in \Omega_2$, then continuity of $\gamma$ implies that $\gamma(T-\varepsilon, T+\varepsilon) \subset \Omega_2$, this can not happen either. So we know that one of $\Omega_1$, $\Omega_2$ must be empty. \smallskip \emph{(2) implies (3)}: If $p, q$ in $\Omega$ can not be connected by a piecewise horizontal/vertical curve in $\Omega$, then one may define $\Omega_1$ to be the collection of points in $\Omega$ that connects to $p$ by a piecewise horizontal/vertical curve in $\Omega$. Then the complement, say $\Omega_2$, of $\Omega_1$ is precisely the collection of those points that can not be connected to $p$ by a piecewise horizontal/vertical curve in $\Omega$. The basic observation is that $\Omega_1$ and $\Omega_2$ are disjoint nonempty open, which contradicts (2). \smallskip \emph{(3) implies (1)}: directly from the definition. \end{proof} \medskip \begin{thm}[see page 106, Theorem 1 in the Ahlfors book] Let $p,q$ be smooth functions on a domain $\Omega$ in $\mathbb C$. The line integral $$ \int_{\gamma} p \,dx +q \,dy $$ depends only on the end points of piecewise smooth curves $\gamma$ in $\Omega$ if and only if the differential $p \,dx +q \,dy$ is \textbf{exact} in $\Omega$, i.e. there exists a smooth function $U(x,y)$ in $\Omega$ with $$ \frac{\partial U}{\partial x} =p, \ \frac{\partial U}{\partial y} =q. $$ \end{thm} \medskip \begin{proof} \emph{Sufficiency}: Assume that $p \,dx +q \,dy$ is exact, we need to show that $$ \int_{\gamma} p \,dx +q \,dy $$ depends only on the end points, say $\gamma(a), \gamma(b)$, of $\gamma : [a,b] \to \Omega$. Write $$ \gamma(t)=(x(t), y(t)), $$ then $$ \int_{\gamma} p \,dx +q \,dy = \int_{a}^b \frac{\partial U}{\partial x} (x(t), y(t)) x'(t) + \frac{\partial U}{\partial y} (x(t), y(t)) y'(t) \, dt. $$ Note that $$ \frac{\partial U}{\partial x} (x(t), y(t)) x'(t) + \frac{\partial U}{\partial y} (x(t), y(t)) y'(t) = \frac{d}{dt} U(x(t), y(t)) = \frac{d}{dt} U(\gamma(t)), $$ hence $$ \int_{\gamma} p \,dx +q \,dy = \int_{a}^b \frac{d}{dt} U(\gamma(t))\, dt = U(\gamma(b))- U(\gamma(a)) $$ depends only on the end points. \smallskip \emph{Sufficiency}: Fix $z_0\in\Omega$, then we can define $$ U(x,y)=\int_{\gamma_{z_0, x+iy}} p dx+q dy, $$ where $\gamma_{z_0, x+iy}$ is an arbitrary piecewise smooth curve joining $z_0$ and $x+iy$ in $\Omega$. By our assumption, the integral depends only on the end points, so $U$ does not depends on the choice of $\gamma_{z_0, x+iy}$. If we choose the last segment of $\gamma_{z_0, x+iy}$ horizontal (we can do this by Theorem \ref{th:domain} (3), try!), then $$ U(x,y) = \int^x p(t, y)\, dt + \text{constant} $$ gives $$\frac{\partial U}{\partial x} =p.$$ In the same way, by choosing the last segment vertical, we can show that $\frac{\partial U}{\partial y} =q$. \end{proof} \begin{co}[see page 107]\label{co:Morera} The integral $\int_\gamma f \,dz$, with smooth $f$, depends only on the end points of $\gamma$ if and only if $f$ is the derivative of a holomorphic function in $\Omega$. \end{co} \medskip \begin{proof} Assume that $\int_\gamma f \,dz$ depends only on the end points of $\gamma$, apply the above theorem to $$ f \,dz = f\, dx + if\, dy, $$ we know there exists a smooth function $U$ with \begin{equation}\label{eq:fU} \frac{\partial U}{\partial x} = f, \ \ \frac{\partial U}{\partial y} = i f, \end{equation} which gives $$ \frac{\partial U}{\partial z} = \frac12 \frac{\partial U}{\partial x} +\frac1{2i} \frac{\partial U}{\partial y} =f $$ and $$ \frac{\partial U}{\partial \bar z} = \frac12 \frac{\partial U}{\partial x} -\frac1{2i} \frac{\partial U}{\partial y} =0. $$ Hence $U$ is holomorphic and $f= U'$ by \eqref{eq:CDerivative}. One may further check that if $f=U'$ with $U$ holomorphic then \eqref{eq:fU} holds, hence the other direction is also true. \end{proof} \begin{co}\label{co:Morera1} If $f$ is a smooth function on a domain $\Omega\subset \mathbb C$ and $\int_\gamma f \,dz =0$ for all cycle $\gamma$ in $\Omega$, then $f$ is holomorphic. \end{co} \medskip \begin{proof} By Corollary \ref{co:Morera}, we know that $f$ is the derivative of a holomorphic function. Thus $f$ is holomorphic. \end{proof} \medskip \noindent \textbf{Remark}: \emph{The above statement is also true if $f$ is only assumed to be continuous, which is known as the Morera theorem, see page 122 of the Ahlfors book.} \medskip \begin{thm}[see page 144, Theorem 16 in the Ahlfors book] Let $p \,dx +q \,dy$ be a locally exact (exact in some neighborhood of each point in $\Omega$) differential in a domain $\Omega \subset \mathbb C$. Then $$ \int_\gamma p \,dx +q \,dy =0 $$ for every cycle $\gamma$ homologous to zero in $\Omega$. \end{thm} \begin{proof} The main step is to show that if $\gamma$ is homologous to zero and $p \,dx +q \,dy$ is locally exact then $$ \int_\gamma p \,dx +q \,dy = \sum_{j} n_j \int_{\partial R_j} p \,dx +q \,dy, $$ for some integers $n_j$ and rectangles $R_j$ in $\Omega$, see page 144-146 for details. \end{proof} \medskip \noindent \textbf{Remark}: \emph{It is known that $u=p \,dx +q \,dy $ is locally exact if and only if it is \textcolor{blue}{closed}, i.e. $$ du:=\left(\frac{\partial q}{\partial x} -\frac{\partial p}{\partial y}\right) dx \wedge dy=0 $$ (see the Wikipedia page for "Closed and exact differential forms ", especially the "Poincar\'e lemma" part) or equivalently $$ \frac{\partial p}{\partial y}= \frac{\partial q}{\partial x}. $$ \textcolor{red}{We stopped here in the Jan 17th lecture}.} \subsection{Simply connected domains in $\mathbb C$} \medskip \begin{de}\label{de:scd} A domain $\Omega\subset \mathbb C$ is said to be \textbf{simply connected} if $\gamma\sim_\Omega 0$ for every piecewise smooth curve $\gamma$ in $\Omega$. \end{de} \medskip \noindent \textbf{Remark}: \emph{Our definition is different from the one given by Ahlfors in page 139, where $\Omega$ is said to be simply connected if its complement with respect to the extended plane $\mathbb C\cup \{\infty\}$ is connected, or equivalently (see page 139, Theorem 14), every cycle in $\Omega$ is homologous to zero in $\Omega$.} \medskip \begin{thm}[See page 141, Corollary 1]\label{th:scd} If $f$ is holomorphic in a simply connected domain $\Omega\subset\mathbb C$ then $$ \int_\gamma f(z)\, dz=0 $$ for every piecewise smooth closed curve $\gamma$ in $\Omega$, in particular, every cycle in $\Omega$ is homologous to zero in $\Omega$. \end{thm} \medskip \noindent \textbf{Remark}: \emph{By the above theorem and Theorem 14 in page 139 of the Ahlfors book, we know that simply connected-ness in Definition \ref{de:scd} above implies the Ahlfors one in page 139, Definition 1. The Riemann mapping theorem (will be proved in section 2) further implies that they are equivalent (for a proof without using the Riemann mapping theorem, see the Wikipedia page for "Riemann mapping theorem").} \medskip \emph{Example}: Convex open sets are simply connected. The \emph{annulus} $$ \mathbb D_{r,1}:=\{z\in \mathbb C: r<|z|<1\}, \ \ 00$. Then Theorem \ref{th:CIT0} implies that \begin{equation}\label{eq:CIF0} \frac1{2\pi i}\int_{\gamma} \frac{f(\zeta)}{\zeta-z}\, d\zeta= \lim_{\varepsilon\to 0+} \frac1{2\pi i}\int_{\gamma_\varepsilon} \frac{f(\zeta)}{\zeta-z}\, d\zeta = f(z). \end{equation} In case $\gamma$ is given by $|\zeta-a|=r$ (\textcolor{blue}{always with the anti-clockwise orientation}), then \eqref{eq:CIF0} gives: \medskip \begin{thm}[Cauchy's integral formula - the circle case]\label{th:CIF} Let $f$ be a holomorphic function on a neighborhood of the disk $|\zeta-a| \leq r$. Then \begin{equation}\label{eq:CIF} \frac1{2\pi i}\int_{|\zeta-a|=r}\frac{f(\zeta)}{\zeta-z}\, d\zeta=f(z) , \ \ \ \forall \ z \ \text{with}\ |z-a|0$, if $$ f^{(h)}(a) \neq 0 \ \text{and}\ \ f^{(\nu)}(a)=0, \ \ \forall \ 0 \leq \nu 0. $$ The point $a$ is called an \textcolor{blue}{isolated singularity} of $f$. By Theorem \ref{th:Laurent} about Laurent series expansions $f$ can be written as $$ f(z) =\sum_{n=-\infty}^\infty c_n (z-a)^n $$ in $\mathbb D_{0, R}$. We shall follow Berndtsson's notes on residue calculus \cite[Definition 1]{B} and define: \medskip \begin{de}\label{de:residue} The coefficient $c_{-1}$ above is called the \textbf{residue} of $f$ at $a$. We write $$ c_{-1}:={\rm Res}_a f. $$ (One may compare with Definition 3, page 149 in the Ahlfors book). \end{de} \medskip Consider now the integral $$ \int_{|z-a|=r} f(z) \, dz, $$ where $00$, if its Laurent series reduces to $$ f(z)= \sum_{n=-h}^\infty c_n (z-a)^n, \ \ c_{-h} \neq 0. $$ In this case we shall write $h= {\rm Ord}^{P}_a f$ ($P$ for pole). A function that is holomorphic except for poles is called a \textbf{meromorphic function}. \end{de} \medskip \noindent \textbf{Remark}: \emph{It is clear that $a$ is an order $h$ pole of $f$ if and only if $(z-a)^h f(z)$ extends to a holomorphic function, say $g$, in a neighborhood of $a$ with $g(a)\neq 0$. Hence we know that a function is meromorphic if and only if it is locally a quotient of two holomorphic functions.} \medskip \begin{de}\label{de:singularity} Let $f$ be holomorphic on a punctured disk around $a$. We say the $a$ is a \textbf{removable singularity} of $f$ if its Laurent series has no negative terms ($c_{n}=0$ for all $n<0$). In case its Laurent series has infinitely many negative terms, we call $a$ an \textbf{essentially singularity} of $f$. \end{de} \medskip The following theorem of Weierstrass gives a characterization of the behavior of a function around an essential singularity. \medskip \begin{thm}[Page 129, Theorem 9, Ahlfors]\label{th:essing} A holomorphic function comes arbitrarily close to any value in every neighborhood of an essential singularity. \end{thm} \begin{proof} Otherwise, we can find a complex number $A$ and $\delta>0$ such that $|f(z)-A| >\delta$ around $a$. Then $a$ must be a removable singularity of $$ g(z):=\frac{1}{f(z)-A}, $$ (one may check that for $g$, all $c_n=0$, $n<0$), thus $g$ is holomorphic around $a$ and $$ f=\frac1g+A $$ is meromorphic near $a$. Hence $a$ can not be an essential singularity of $f$. \end{proof} \textcolor{red}{We stopped here in the Jan 26th lecture, in the next lecture, we will first recall Theorem \ref{th:residue1} and do the examples in page 155-158 of the Ahlfors book}. \subsubsection{Argument principle} \begin{thm}[Argument principle, circle version]\label{th:argument0} Let $f$ be meromorphic on a neighborhood of the disk $|z-z_0|\leq r$. Then $f$ has finite zeros, say $\{a_j\}_{1\leq j \leq N}$ and finite poles, say $\{b_k\}_{1\leq k \leq M}$ in that disk. Assume that all $a_j, b_k$ are away from the circle $|z-z_0|=r$, then \begin{equation}\label{eq:argument0} \frac{1}{2\pi i } \int_{|z-z_0|=r} \frac{f'(z)}{f(z)} \, dz =\sum_{j=1}^N {\rm Ord}_{a_j} f -\sum_{k=1}^M {\rm Ord}^P_{b_k} f. \end{equation} \end{thm} \begin{proof} By Proposition \ref{pr:uniqueness2}, we know that $f$ has only finite zeros and poles in the disk. To prove \eqref{eq:argument0}, we shall write $$ f(z)= \frac{(z-a_1)^{{\rm Ord}_{a_1} f} \cdots (z-a_N)^{{\rm Ord}_{a_N} f} }{(z-b_1)^{{\rm Ord}^P_{a_1} f} \cdots (z-b_M)^{{\rm Ord}^P_{b_M} f}} \, g, $$ where $g$ is holomorphic and $\neq 0$ in $|z-z_0|\leq r$. Forming the logarithmic derivative we obtain $$ \frac{f'(z)}{f(z)}= \sum_{j=1}^N \frac{{\rm Ord}_{a_j} f}{z-a_j} -\sum_{k=1}^M \frac{{\rm Ord}^P_{b_k} f}{z-b_k} + \frac{g'(z)}{g(z)} $$ for $z\neq a_j, b_k$, and particularly on the circle $|z-z_0|=r$, Since $g(z)\neq 0$ in the disk, we know that $g'/g$ is holomorphic around the disk, thus Theorem \ref{th:CIH} yields $$ \int_{|z-z_0|=r} \frac{g'(z)}{g(z)} \, dz =0, $$ together with $$ \int_{|z-z_0|=r} \frac{dz}{z-a_j} = \int_{|z-z_0|=r} \frac{dz}{z-b_k} = 2\pi i , $$ we obtain \eqref{eq:argument0}. \end{proof} \medskip Apply Theorem \ref{th:CIT0}, Theorem \ref{th:argument0} can be generalized to \medskip \begin{thm}[Argument principle, homotopy version]\label{th:argument1} Let $f$ be meromorphic on a domain $\Omega\subset \mathbb C$. Let $\gamma$ be a piecewise smooth closed curve enclosing zeros $\{a_j\}_{1\leq j\leq N}$ and poles $\{b_k\}_{1\leq k \leq M}$ of $f$ in $\Omega$. Assume that $f$ has no zeros and poles on $\gamma$. Then \begin{equation}\label{eq:argument1} \frac{1}{2\pi i } \int_{\gamma} \frac{f'(z)}{f(z)} \, dz =\sum_{j=1}^N {\rm Ord}_{a_j} f -\sum_{k=1}^M {\rm Ord}^P_{b_k} f. \end{equation} \end{thm} \medskip \noindent \textbf{Remark}: \emph{The function $w=f(z)$ maps $\gamma$ onto a closed curve, say $\Gamma$, in the $w$-plane and we find \begin{equation}\label{eq:arg1} \int_{\gamma} \frac{f'(z)}{f(z)} \, dz = \int_{\Gamma}\frac{dw}{w}. \end{equation} Thus the left hand side of \eqref{eq:argument1} equals $n(\Gamma, 0)$ --- the change of argument of $f(z) \in \Gamma$ as $z$ traverses $\gamma$, this explains why the above theorem is referred to as the \textcolor{blue}{argument principle}.} \subsubsection{Local description of holomorphic mappings} In case $f$ is holomorphic on a neighborhood of the disk $|z-z_0|\leq r$, applying Theorem \ref{th:argument0} to $f-a$ we obtain: \medskip \begin{pr}\label{pr:value-number} If $f(z)\neq a$ on the circle $|z-z_0|=r$ then \begin{equation}\label{eq:value-number} \frac{1}{2\pi i } \int_{|z-z_0|=r} \frac{f'(z)}{f(z)-a} \, dz = \# \{z: f(z)=a, |z-z_0|0$ is sufficiently small then $$ n=\# \{z: f(z)=a, |z-z_0|<\varepsilon\}, $$ for all $a$ in some neighborhood of $w_0$, moreover, if $\varepsilon$ is sufficiently small, then the above equality is also true without counting multiplicity. \end{thm} \begin{proof} Since the zero of $f(z)-w_0$ is isolated, it suffices to choose $\varepsilon$ so that $f(z)$ is defined and holomorphic on the disk $|z-z_0|\leq \varepsilon$ and so that $z_0$ is the only zero of $f(z)-w_0$ in this disk. For the final statement, it suffices to rake $\varepsilon$ so that $$ f'(z) \neq 0 \ \ \text{for}\ 0< |z-z_0|< \varepsilon, $$ then all zeros of $f(z)-a$ are of order one for $a$ in a small punctured neighborhood of $w_0$. \end{proof} \medskip \noindent \textbf{Remark}: \emph{By the above theorem, if $z_0$ is a finite order zero of $f(z)-w_0$ then the $f$ image of every sufficiently small disk $|z-z_0| <\varepsilon$ contains a neighborhood of $w_0$. Hence we have} \medskip \begin{co}[Page 132, Corollary 1]\label{co:open} A non-constant holomorphic function maps open sets onto open sets, \end{co} \medskip Notice that the followings are equivalent in Theorem \ref{th:local0}: \medskip \begin{itemize} \item[(1)] $n=1$;\\ \item[(2)] $f'(z_0)\neq 0$;\\ \item[(3)] $f$ is one to one near $z_0$. \end{itemize} \medskip In particular, $(3)\Rightarrow (2)$ explains the remark after Definition \ref{de:conformal}. To summarize, we have \medskip \begin{co}[Page 132, Corollary 2]\label{co:oneone} If $f$ is holomorphic near $z_0$ then $f'(z_0)\neq 0$ if and only if $f$ maps a neighborhood of $z_0$ conformally onto a neighborhood of $f(z_0)$. The inverse of a conformal mapping is also conformal. \end{co} \medskip \noindent \textbf{Remark}: \emph{In fact, there is a very precise formula for the inverse of a conformal mapping. To find that formula we need to generalize \eqref{eq:value-number} to $$ \frac{zf'(z)}{f(z)-a}, $$ which has residue $z(a) {\rm Ord}_{z(a)} (f-a) $ at a zero $z(a)$ of $f(z)-a$. If $f$ is conformally then $$ z(a) {\rm Ord}_{z(a)} (f-a) = z(a) =f^{-1}(a), $$ which proves the following theorem.} \medskip \begin{thm}\label{th:inverse} Assume that $f$ is conformal near $|z-z_0| \leq r$, then \begin{equation}\label{eq:inverse} f^{-1}(w) = \frac{1}{2\pi i } \int_{|z-z_0|=r} \frac{zf'(z)}{f(z)-w} \, dz \end{equation} holds true for $w$ in a neighborhood of $f(z_0)$. \end{thm} \medskip To count the number of zeros for a general holomorphic function, the following \textcolor{blue}{Rouch\'e's theorem} (see the corollary in page 153 of the Ahlfors book for generalizations) is often useful. \medskip \begin{thm}[Rouch\'e's theorem]\label{th:Rouche} Let $f, g$ be holomorphic near $|z-z_0| \leq r$. Assume that $$ |f-g| <|f| \ \text{on}\ |z-z_0| =r, $$ then $f, g$ have the same number (counting multiplicity) of zeros in the open disk $|z-z_0|0$ such that $f(z)\neq 0$ on $0<|z-z_0| \leq r$. Since $f_n$ converges locally uniformly to $f$, Theorem \ref{th:w} and Proposition \ref{pr:compact} imply that $f_n$ and $f_n'$ converge uniformly to $f$ and $f'$ on the compact set $|z-z_0|=r$, which gives $$ \lim_{n\to \infty} \frac{1}{2\pi i} \int_{|z-z_0|=r} \frac{f_n'(z)}{f_n(z)} \, dz = \lim_{n\to \infty} \frac{1}{2\pi i} \int_{|z-z_0|=r} \frac{f'(z)}{f(z)} \, dz, $$ by Theorem \ref{th:argument0}, the integrals in the left hand side are all zero, thus the right hand side is also zero, which gives (by Theorem \ref{th:argument0}) $f(z_0)\neq 0$. Since $z_0$ is arbitrary, the theorem follows. \end{proof} \medskip The above theorem can be used to study conformal mappings (also called \textcolor{blue}{univalent function} in page 230 of the Ahlfors book). \medskip \begin{lm}[The Hurwitz Lemma, page 231]\label{le:Hurwitz} If a sequence of conformal mappings $f_n$ converges to $f$ locally uniformly on a domain $\Omega\subset \mathbb C$, then $f$ is either a constant or conformal on $\Omega$. \end{lm} \begin{proof} It suffices to apply the Hurwitz thorem to $f_n(z)-f_n(z_0)$ and the domain $\Omega\setminus\{z_0\}$ (why it is still a domain, try!) for an arbitrarily fixed point $z_0$ in $\Omega$, \end{proof} \subsection{Normal families} \begin{de}[Definition 2, page 220]\label{de:normal} A family $\mathcal F$ of holomorphic functions on a domain $\Omega\subset \mathbb C$ is said to be \textbf{normal} if every sequence $\{f_n\}$ of functions $f_n \in \mathcal F$ contains a subsequence which converges locally uniformly on $\Omega$. \end{de} \medskip \begin{thm}[Montel's theorem]\label{th:montel} If $|f| \leq 1$ for all $f\in \mathcal F$ then $\mathcal F$ is normal. \end{thm} \begin{proof} Let $\{f_n\}$ be a sequence in $\mathcal F$. Denote by $\{w_m\}$ the sequence of rational points in $\Omega$. Since $\{f_n(w_1)\}$ is a bounded sequence in $\mathbb C$, the Bolzano-Weierstrass theorem gives a convergent subsequence, say $\{f_{1n}(w_1)\}$. Similarly, $\{f_{1n}(w_2)\}$ has a convergent subsequence, say $\{f_{2n}(w_2)\}$. Continue this process, we obtain a double sequence $\{f_{kn}\}$ with diagonal $f_{nn}$ satisfying \begin{equation}\label{eq:double} \lim_{n\to \infty} f_{nn}(w_k) \ \text{exists for all} \ k\geq 1. \end{equation} By Proposition \ref{pr:compact}, it suffices to check that $f_{nn}$ converges locally uniformly: for $a\in \Omega$, take $r>0$ such that $|z-a|< 2r$ lies in $\Omega$. Recall that (see \eqref{eq:taylor1}) \begin{equation}\label{eq:montel} \frac{f_{nn}(z)-f_{nn}(w_k)}{z-w_k} = \frac{1}{2\pi i } \int_{|\zeta-a|=r} \frac{f_{nn}(\zeta)}{(\zeta-w_k)( \zeta-z)} \, d\zeta, \end{equation} hence $|f_{nn}|<1$ gives $$ \frac{|f_{nn}(z)-f_{nn}(w_k)|}{|z-w_k|} \leq \frac4r, \ \ \text{for all} \ \ |z-a|< r/2, \ |w_k-a|0$, first we can take rational $w_{k}$ with $$ \frac{8|z-w_k|}{r} \leq \frac{\varepsilon}2, $$ for such $w_k$, by \eqref{eq:double}, we can further take $N$ such that $$ |f_{nn}(w_k)-f_{mm}(w_k)| \leq \frac{\varepsilon}2, \ \text{for all} \ n, m \geq N. $$ Hence \eqref{eq:montel1} gives $$ \sup_{|z-a|0$. \end{thm} \begin{proof} \emph{1. Uniqueness.} If $f_1$ and $f_2$ are two such mappings, then $S:=f_1f_2^{-1}$ defines a conformal mapping of $|w|<1$ onto itself with $S(0)=0$ and $S'(0)>0$. By Proposition \ref{pr:mobius}, we know that $S$ must be the identity mapping. Hence $f_1=f_2$. \smallskip The existence part is divided into the following steps. \smallskip \emph{2. Define $f$ as the solution of an optimization problem.} Consider the following $$ \text{optimization problem}: \ \text{Find $f\in\mathcal F$ with} \ f'(z_0)=B:=\sup\{g'(z_0): g\in \mathcal F\}, $$ where $$ \mathcal F:=\left\lbrace g: g'(z_0)>0,\, g(z_0)=0,\, \sup_{z\in \Omega}|g(z)|\leq 1 \, \text{and}\, g: \Omega \to g(\Omega) \, \text{is conformal}\right\rbrace. $$ We shall show that the solution $f$ of this optimization problem exists and fits our needs. \smallskip \emph{3. $\mathcal F$ is not empty.} We note there exists, by assumption, a point $a\notin \Omega$. Since $\Omega$ is simply connected, $h(z):=\sqrt{z-a}$ is well defined in $\Omega$ by Corollary \ref{co:branch}. Note that if $h(z_1)=\pm h(z_2)$ then $$ (h(z_1))^2 = z_1-a = z_2-a = (h(z_2))^2 $$ gives $z_1=z_2$. Hence we know that $h: \Omega \to h(\Omega)$ is conformal and \begin{equation}\label{eq:empty} h(\Omega) \cap -h(\Omega) =\emptyset. \end{equation} By Corollary \ref{co:open}, $h(\Omega)$ is open thus covers a disk $|w-h(z_0)|<\rho$, thus \eqref{eq:empty} gives $$ h(\Omega)\cap \{w\in\mathbb C: |w+h(z_0)|<\rho\}=\emptyset. $$ In other words, $|h(z)+h(z_0)|\geq \rho$ for all $z\in \Omega$. Then one map verify that the function $$ g_0(z):= \frac{\rho}{4} \,\frac{|h'(z_0)|}{|h(z_0)|^2} \cdot\frac{h(z_0)}{h'(z_0)} \cdot\frac{h(z)-h(z_0)}{h(z)+h(z_0)} $$ belongs to $\mathcal F$ (see page 230 of the Ahlfors book for details). \smallskip \emph{4. Solve the optimization problem.} A priori, the constant $B$ in our optimization problem could be infinite. In any case, one may tale $g_n \in \mathcal F$ with $g_n'(z_0)\to B$ as $n\to\infty$. By Theorem \ref{th:montel}, $\{g_n\}$ contains a subsequence, say $\{g_{n_k}\}$ which converges locally uniformly to a holomorphic function $f$ on $\Omega$. It is clear that $|f|\leq 1$ on $\Omega$, $f(z_0)=0$ and by Theorem \ref{th:w}, $f'(z_0)=B$ (this proves that $B<\infty$). Lemma \ref{le:Hurwitz} further implies that $f: \Omega \to f(\Omega)$ is conformal. Hence $f\in \mathcal F$ solves the optimization problem. \smallskip \emph{5. $f(\Omega)$ is the unit disk.} Otherwise $w_0\notin f(\Omega)$ for some $|w_0|<1$. Again since $\Omega$ is simply connected, $$ F(z)=\sqrt{\frac{f(z)-w_0}{1-\overline{w_0} f(z)}} $$ is well defined in $\Omega$ by Corollary \ref{co:branch}. Similar to $h$, we know that $F: \Omega \to F(\Omega)$ is conformal and $|F| \leq 1$. To normalize it we form $$ G(z):=\frac{|F'(z_0)|}{F'(z_0)} \cdot \frac{F(z)-F(z_0)}{1-\overline{F(z_0)} F(z)}, $$ so that $G\in\mathcal F$. After brief computation, $$ G'(z_0)=\frac{|F'(z_0)|}{1-|F(z_0)|^2} = \frac{1+|w_0|}{2\sqrt{|w_0|}} \, B >B. $$ This is a contradiction, so $f(\Omega)$ must be the whole unit disk. \end{proof} \medskip \noindent \textbf{Remark (Page 231, Ahlfors book).} \emph{At first glance, it may seem like pure luck that our computation yields $G'(z_0)>f'(z_0)$. This is not quite so, for we can write $f=T(G)$ (try to find $T$ yourself) for some holomorphic function $T$ which maps $|w|<1$ into itself with $T(0)=0$. The Schwarz lemma gives $|T'(0)|<1$, thus $$ |f'(z_0)|=|T'(0) G'(z_0)| < |G'(z_0)|. $$} Another remark is that: $\log|f|$ is usually called \textcolor{blue}{the Green function} of $\Omega$ with a pole at $z_0$, moreover the constant $B$ in the above proof satisfies $$ B= f'(z_0)=\lim_{z\to z_0}e^{\log(|f(z)|)-\log(|z-z_0|)}. $$ The right hand side is known as the \textcolor{blue}{logarithmic capacity} of $\mathbb C\setminus \Omega$ with respect to $z_0$, which is the central concept in the \textcolor{blue}{potential theory} (see \cite[Chapter 5]{Ransford}; in case $\Omega$ is not simply connected, $B$ is also an important \textcolor{blue}{conformal invariant}, see \cite{AB}). Hence the above proof of the Riemann mapping theorem directly leads us to the potential theory. In fact, the Riemann mapping theorem is equivalent to the existence of Green's function on $\Omega$ (see Theorem 4.4.11 in \cite{Ransford} for a nice Green's function proof of the Riemann mapping theorem). In the next section we shall study basic facts of the potential theory in the complex plane. \textcolor{red}{We stopped here in the Feb 9th lecture}. \section{Harmonic functions} \subsection{Definitions and basic properties} \begin{de} A real valued smooth function $u$ on a domain $\Omega\subset\mathbb C$ is said to be \textbf{harmonic} if $$ u_{z\bar z}=0, \ \ \ u_{z\bar z}:=\frac{\partial^2 u}{\partial z\partial \bar z}. $$ \end{de} \medskip \noindent \textbf{Remark.} \emph{If $f=u+iv$ is holomorphic then $$ 0=f_{z\bar z} =u_{z\bar z}+ i v_{z\bar z}, $$ since both $u_{z\bar z}$ and $v_{z\bar z}$ are real, they must vanish. Hence the real and imaginary parts of a holomorphic function are always harmonic. Later we shall prove a partial converse (see Theorem \ref{th:hh} below): a harmonic function is \textcolor{blue}{locally} the real part of a holomorphic function.} \medskip \begin{thm}\label{th:hh} Let $u$ be a harmonic function on a domain $\Omega\subset \mathbb C$. If $\Omega$ is simply connected then $u={\rm Re}\, f$ for some $f$ holomorphic on $\Omega$. Moreover $f$ is unique up to adding a constant. \end{thm} \medskip \begin{proof} \emph{Uniqueness}: If $u={\rm Re}\, f$ for some holomorphic function $f$, say $f=u+iv$, then $$ u_x+ iv_x=f_x =f'= \frac{f_y}{i}= \frac{u_y+iv_y}{i} $$ gives $ u_x =v_y, \ u_y =-v_x $ and \begin{equation}\label{eq:df} df= du+ i\, dv =du+ i\star du, \end{equation} where \begin{equation}\label{eq:cd} \star du:= -u_y\, dx + u_x \, dy \end{equation} is called the \textcolor{blue}{conjugate differential} of $du$. In particular, \eqref{eq:df} implies that $f$ is unique up to adding a constant (since $df$ is fully determined by $du$). \smallskip \emph{Existence}: \eqref{eq:df} also suggests to define $$ f(z):= u(z_0)+\int_{z_0}^z du+ i\star du, $$ where $z_0$ is a fixed point $\Omega$ and the integral is taken over any piecewise smooth curve $\gamma_{z_0, z}$ connecting $z_0, z$ in $\Omega$. Note that (try!) \begin{equation}\label{eq:cd1} du+ i\star du= (u_x-i\,u_y) \, dz \ \text{and $u_x-i\,u_y$ is holomorphic on $\Omega$}, \end{equation} by the Cauchy integral theorem --- Theorem \ref{th:scd}, we know that $f(z)$ does not depend on the choice of $\gamma_{z_0, z}$. Hence $f$ is a well defined holomorphic function on $\Omega$ and $$ df= du+ i\star du $$ gives $d({\rm Re}\, f) = du$, together with ${\rm Re}\, f(z_0) =u(z_0)$ we know that ${\rm Re}\, f = u$ on $\Omega$. Hence $f$ fits our needs. \textcolor{blue}{Remark}: The simply-connected-ness is used in Theorem \ref{th:scd}. \end{proof} \medskip \noindent \textbf{Remark [page 163, Ahlfors].} \emph{By Theorem \ref{th:Cauchy-homology}, \eqref{eq:cd1} also implies that \begin{equation}\label{eq:cd2} \int_{\gamma} \star\, du= 0 \end{equation} for all cycles $\gamma$ which are homologous to zero in $\Omega$. If $\gamma$ is smooth with equation $z=z(t)$, the direction of the tangent is determined by the angle $\alpha={\rm arg} \,z'(t)$ (i.e. $z'(t)=|z'(t)|e^{i\alpha}$) and we can write $$ dx=|dz|\, \cos \alpha, \ \ dy=|dz|\, \sin\alpha. $$ The normal which points to the right of the tangent has the direction $\beta=\alpha-\pi/2$, and thus $\cos\alpha= -\sin\beta, \ \ \sin\alpha=\cos\beta$. The expression \begin{equation}\label{eq:nd} \frac{\partial u}{\partial n} = u_x \cos\beta+ u_y\sin\beta \end{equation} is called \textcolor{blue}{the right hand normal derivative} of $u$ with respect to the curve $\gamma$. We obtain \begin{equation}\label{eq:nd1} \star \,du:= -u_y\, dx + u_x \, dy = -u_y |dz|\, \cos \alpha+ u_x |dz|\, \sin\alpha = \frac{\partial u}{\partial n}\, |dz| \ \ \text{on}\, \gamma. \end{equation} Thus \eqref{eq:cd2} is equivalent to \begin{equation}\label{eq:cd3} \int_{\gamma} \frac{\partial u}{\partial n}\, |dz|= 0, \end{equation} see page 164 of the Ahlfors book for more explanations.} \medskip The following theorem is a generalization of \eqref{eq:cd2}. \medskip \begin{thm}[See Ahlfors, page 164, Theorem 19 for another proof]\label{th:green0} If $u_1,u_2$ are harmonic on a neighborhood of a piecewise smooth bounded domain $\Omega\subset\mathbb C$ then \begin{equation}\label{eq:green0} \int_{\partial \Omega} u_1\, \star du_2 - u_2\, \star du_1 =0, \end{equation} where the orientation of $\partial \Omega$ is chosen so that $\Omega$ lies to the left. \end{thm} \medskip \begin{proof} The idea is to use \textcolor{blue}{Green's theorem} (see the proof below or Wikipedia) \begin{equation}\label{eq:Green} \int_{\partial \Omega} p\, dx+ q\, dy = \int_{\Omega} (q_x-p_y) \, dx dy. \end{equation} We may directly verify that $q_x-p_y =0$ in case $ p\, dx+ q\, dy = u_1\, \star du_2 - u_2\, \star du_1. $ \end{proof} \begin{proof}[Proof of \eqref{eq:Green}] Covering $\mathbb C$ by net of squares, it suffices to prove the case that $\Omega$ is a rectangle (Try!). Denote the upper and lower horizontal part of $\partial \Omega$ by $H_1$ and $H_2$ respectively. The $y$-coordinate on $H_1, H_2$ is a constant, we denote it by $Y_1, Y_2$. Now we have $$ \int_{\partial \Omega} p\, dx = \int_a^b p(x, Y_2)-p(x, Y_1) \, dx. $$ Hence the Newton-Lebniz formula $$ p(x, Y_2)-p(x, Y_1) = - \int_{Y_2}^{Y_1} p_y (x, y) \, dy $$ gives $$ \int_{\partial \Omega} p\, dx = -\int_a^b \left(\int_{Y_2}^{Y_1} p_y (x, y) \, dy\right) dx= -\int_{\Omega} p_y\, dx dy. $$ A similar argument gives $\int_{\partial \Omega} q\, dy = \int_{\Omega} q_x \, dx dy$. Hence \eqref{eq:Green} follows. \end{proof} \medskip \noindent \textbf{Remark.} \emph{For general smooth functions $u, v$, \eqref{eq:green0} gives \begin{equation}\label{eq:Green1} \int_{\partial \Omega} u_1\, \star du_2 - u_2\, \star du_1 = \int_{\Omega} (u_1\, \Delta u_2 - u_2\, \Delta u_1) \, dx dy \end{equation} where $$ \Delta u:= u_{xx} + u_{yy} $$ is called the \textcolor{blue}{Laplacian} of $u$. By \eqref{eq:nd1}, one may write \eqref{eq:Green1} as \begin{equation}\label{eq:Green11} \int_{\partial \Omega} \left( u_1\, \frac{\partial u_2}{\partial n}- u_2\, \frac{\partial u_1}{\partial n} \right) |dz| = \int_{\Omega} (u_1\, \Delta u_2 - u_2\, \Delta u_1) \, dx dy, \end{equation} this is known as \textcolor{blue}{Green's formula}. \textcolor{red}{We stopped here in the Feb 14th lecture}.} \medskip \emph{Exercise 1}: With $z=x+iy$, show that for smooth function $u$ we have $$ u_{z\bar z}= \frac14\left(u_{xx}+u_{yy}\right). $$ Verify that all linear functions $ax+by$ are harmonic. \medskip \emph{Exercise 2}: Check that $\log|z|$ is harmonic on $\mathbb C\setminus\{0\}$ and find a holomorphic function $f$ on a simply connected domain $\Omega\subset \mathbb C\setminus\{0\}$ such that $\log|z| ={\rm Re}\, f$. \medskip \emph{Exercise 3}: Do the Exercise 1.1, Problem 1,2,3,4 in \cite[Page 7]{Ransford}. \subsection{The mean-value property} A nice application of Theorem \ref{th:hh} is the following: \medskip \begin{thm}[Mean-value property]\label{th:MVP} Let $u$ be a function harmonic on an open neighborhood of the disk $|z-z_0| \leq r$. Then \begin{equation}\label{eq:MVP} u(z_0)=\frac1{2\pi} \int_0^{2\pi} u(z_0+r e^{i\theta})\, d\theta. \end{equation} \end{thm} \begin{proof} By Theorem \ref{th:hh}, $u$ is the real part of a holomorphic function $f$, one may check that (try!) the Cauchy integral formula for $f$ gives \eqref{eq:MVP}. \end{proof} \medskip \eqref{eq:MVP} leads directly to the \textcolor{blue}{maximum principle} for harmonic functions. \medskip \begin{thm}[Maximum Principle]\label{th:MP} Let $u$ be a harmonic function on a domain $\Omega\subset \mathbb C$. \begin{itemize} \item[(1)] If $u$ attains a maximum on $\Omega$ then $u$ is a constant; \item[(2)] Assume that $\Omega$ is bounded. If $u$ extends continuously to $\overline{\Omega}$ and $u\leq 0$ on $\partial \Omega$, then $u\leq 0$ on $\Omega$. \end{itemize} \end{thm} \begin{proof} The proof of (1) is the same as for Theorem \ref{th:max}. For the proof of (2), as $\overline{\Omega}$ is compact and $h$ is continuous there, $h$ must attain a maximum at some point $z_0\in \overline{\Omega}$, i.e. $$ h(z_0)=\sup_{\overline{\Omega}} h. $$ If $z_0\in \partial \Omega$, then $h(z_0) \leq 0$ by assumption, and so $h\leq 0$ on $\Omega$. If $z_0\in \Omega$ then (1) implies that $h$ is a constant on $\Omega$. Since $h$ is continuous on $\overline{\Omega}$, we know $h$ must be a constant on $\overline{\Omega}$. Thus in this case, our assumption also implies $h\leq 0$ on $\Omega$. \end{proof} \begin{de}[Maximum principle]\label{de:max} We say that a function $h: \Omega \to [-\infty, \infty)$ satisfies the \textbf{maximum principle} on a domain $\Omega\subset\mathbb C$ if $h$ has no maximum in $\Omega$ unless it is a constant (i.e. either $h$ is a constant or $$ h(z) < \sup_{\Omega} h, $$ for all $z\in \Omega$). \end{de} \medskip \noindent \textbf{Remark.} \emph{Later we shall study the class of \textcolor{blue}{subharmonic functions}, which can be defined directly using the maximum principle (see Definition \ref{de:sub}).} \medskip \subsubsection{The Green formula approach to the mean value property} Apply \eqref{eq:green0} to the case that $$ \Omega=\{z\in \mathbb C: r_1<|z|r_2, s \varepsilon$} \, \text{as} \ z\to e^{i\theta_0}, $$ (note that $|z|\to 1$ and $|e^{i\theta}-z| \to |e^{i\theta}-e^{\theta_0}| > C>0$ as $|\theta-\theta_0| >\varepsilon$), so the second integral also tends to zero as $z\to e^{i\theta_0}$. It follows that $P_U(z) \to 0= U(\theta_0)$ as $z\to e^{i\theta_0}$. \textcolor{red}{We stopped here in the Feb 21st lecture}. \end{proof} \subsection{Functions with the mean value property} A nice application of Schwarz's theorem is the following: \begin{thm}[Ahlfors, page 242]\label{th:har-mvp} A continuous function $f$ on a domain $U\subset \mathbb C$ is harmonic if and only if it satisfies the mean value property $$ u(z_0) =\frac1{2\pi} \int_0^{2\pi} u(z_0+re^{i\theta})\, d\theta $$ for all disk $|z-z_0|\leq r$ contained in $\Omega$. \end{thm} \begin{proof} It suffices to show that $u$ is harmonic on $|z-z_0|0\}, \ \ \sigma:=\{z\in \mathbb D: {\rm Im}\, z=0\}. $$ Suppose that $v$ is continuous on $\mathbb D^+ \cup \sigma$, harmonic on $\mathbb D^+$ and $v=0$ on $\sigma$. Then $v$ extends to a harmonic function (still denoted by $v$) on $\mathbb D$ satisfying $v(\bar z)= -v(z)$ for $z\in \mathbb D$. \textcolor{red}{We stopped here in the Feb 23rd lecture}. \end{thm} \begin{proof} See Ahlfors page 173. \end{proof} \begin{co}\label{co:ref} With the notation above. Let $v$ be the imaginary (resp. real) part of an holomorphic function $f$ on $\mathbb D^+$. Assume that $$ \lim_{z\to z_0} v(z)= 0 \ \text{for all} \ z_0\in \sigma. $$ Then $f$ extends to a holomorphic function (still denoted by $f$) on $\mathbb D$ satisfying $f( z)= \overline{f(\bar z)}$ (resp. $f(z) = -\overline{f(\bar z)}$) for $z\in \mathbb D$. \end{co} \begin{proof} See Ahlfors page 173. \end{proof} \subsection{Use of the reflection principle} The reflection principle can be used to study regularity property of the the solution of the \textcolor{blue}{Dirichlet problem}: \begin{equation}\label{eq:Dirichlet-r} \Delta u=0 \ \text{on $\Omega$ with $u|_{\partial \Omega}=\phi$} \ \ (\text{$u|_{\partial \Omega}$ means the restriction of $u$ to $\partial\Omega$}). \end{equation} We say that $u$ is a \textcolor{blue}{solution of \eqref{eq:Dirichlet-r}}, if $u$ is harmonic on $\Omega$, continuous on $\overline{\Omega}$ with $u|_{\partial \Omega}=\phi$. Theorem \ref{th:MP} implies that the solution, if it exists, must be unique. The following theorem is a deep result in PDE theory: \medskip \begin{thm}[See Theorem 9.9 in \cite{Agmon}]\label{th:regular0} If $\Omega$ is smoothly bounded and $\phi$ is smooth then the solution of \eqref{eq:Dirichlet-r} is unique and smooth up to the boundary (i.e. it extends to a smooth function on a neighborhood of $\overline{\Omega}$). \end{thm} \medskip We are not able to prove the above result using theories covered in this course. But in case $\Omega$ and $\phi$ are real analytic, we shall show that the reflection principle gives (in fact, a stronger version of) the above result. \medskip \begin{de}\label{de:real-ana} We say that $\Omega$ and $\phi$ are \textbf{real analytic} if for every $\zeta\in\partial\Omega$, there exists a conformal mapping $f$ from $\mathbb D$ onto an open neighborhood, say $V_\zeta$, of $\zeta$ such that $$ f(\mathbb D^+)=\Omega\cap V_\zeta, \ \ f(\sigma)= \partial\Omega\cap V_\zeta, $$ and $\phi(f)={\rm Re}\, h$ on $\sigma$ for some holomorphic function $h$ on $\mathbb D$. \end{de} \medskip \begin{thm}\label{th:ref1} Assume that $u$ is a solution of \eqref{eq:Dirichlet-r}. If $\Omega$ and $\phi$ are real analytic then $u$ extends to a harmonic function on a neighborhood of $\overline{\Omega}$. \end{thm} \begin{proof} By Definition \ref{de:real-ana}, we know that $\phi(f)={\rm Re}\, h$, hence the harmonic function $u(f)-{\rm Re}\, h$ on $\mathbb D^+$ vanishes on $\sigma$. By Theorem \ref{th:ref}, $u(f)-{\rm Re}\, h$ extends to a harmonic function on $\mathbb D$. Thus $u$ extends to a harmonic function on $V_\zeta$. Since harmonic functions are real analytic, we know that the extensions to overlapping $V_\zeta$, $\zeta\in \partial \Omega$, must coincide and define a harmonic function on a neighborhood of $\overline{\Omega}$. \end{proof} \medskip Similarly, Corollary \ref{co:ref} implies the following result (see Theorem 3, 4 in page 233-235 of the Ahlfors book). \medskip \begin{thm}\label{th:ref2} Let $\Omega$ be a bounded simply connected domain with real analytic boundary in $\mathbb C$. Then the Riemann mapping function $f$ which maps $\Omega$ onto the unit disk extends to a holomorphic function on a neighborhood of $\overline{\Omega}$. \end{thm} \begin{proof} Note that $|f(z)| \to 1$ when $z \to \partial \Omega$, hence $\log f$, which is well defined on $\Omega\cap V_\zeta$ (choose a smaller $V_\zeta$ if necessary, one may assume that $f$ has no zero in $\Omega\cap V_\zeta$), satisfies that $$ {\rm Re}\, \log f(z)= \log|f(z)| \to 0, \ \ \text{as $z \to \partial \Omega\cap V_\zeta$ }. $$ Thus by Corollary \ref{co:ref}, $\log f$ extends holomorphically to $V_\zeta$. Thus $f$ extends to $V_\zeta$ (hence to $\overline{\Omega}$). \end{proof} \subsection{Analytic arcs} A non-trivial example of bounded simply connected domain with real analytic boundary is the \textcolor{blue}{elliptical disk} $\Omega$ defined as the interior of the \textcolor{blue}{ellipse} \begin{equation}\label{eq:ellipse} z=\frac12 \left(\rho \,e^{it}+\rho^{-1} e^{-it}\right), \ \ 0\leq t<2\pi, \end{equation} where $\rho>1$ is a constant. Write $z=x+iy$, \eqref{eq:ellipse} gives $$ x= \frac12 \left(\rho \,\cos t+\rho^{-1} \cos t\right), \ \ y= \frac12 \left(\rho \,\sin t -\rho^{-1} \sin t\right). $$ Hence $\partial \Omega$ is given by $$ \frac{x^2}{a^2} +\frac{y^2}{b^2}=1, \ \ \ a:=\frac{\rho+\rho^{-1}}{2}, \ \ b:=\frac{\rho-\rho^{-1}}{2} $$ and $$ \Omega=\left\lbrace x+iy \in \mathbb C: \frac{x^2}{a^2} +\frac{y^2}{b^2}<1\right\rbrace. $$ Since $\Omega$ is convex, we know that $\Omega$ is simply connected. We shall show that $\Omega$ is real analytic in the sense of Definition \ref{de:real-ana}. In fact, from \eqref{eq:ellipse}, we know that $\partial \Omega = g(\mathbb R)$ for \begin{equation}\label{eq:ellipse1} g(w)= \frac12 \left(\rho \,e^{iw}+\rho^{-1} e^{-iw}\right), \end{equation} the main observation is that $$ g'(w) \neq 0, \ \text{for all}\ w\in \mathbb R. $$ Hence for every $w_0 \in\mathbb R$, $g$ is conformal on $$ |w-w_0|<\varepsilon $$ for some $\varepsilon$ (one may take $\varepsilon =\pi$ by the exercise below). Now we know that the conformal mapping $f$ defined by $$ f(\zeta):=g(w_0+\varepsilon \zeta) $$ maps $\mathbb D$ onto a neighborhood, say $V_\zeta$, of $\zeta=f(0)$. One may check that for small $\varepsilon$, $$ f(\mathbb D^+)=\Omega\cap V_\zeta, \ \ f(\sigma)= \partial\Omega\cap V_\zeta, $$ hence we know that $\Omega$ has real analytic boundary. \medskip \noindent \textbf{Remark.} \emph{\eqref{eq:ellipse} is an example of analytic arc defined in page 234 in the Ahlfors book, by a similar discussion, we know that $\Omega$ is real analytic if and only if its boundary consists of free one-side regular analytic arcs (see page 234, Ahlfors). } \medskip \emph{Exercise}: Show that $g$ defined in \eqref{eq:ellipse1} is conformal on $|{\rm Re}\, (w-w_0)| <\pi$ for every $w_0\in \mathbb R$. \textcolor{red}{We stopped here in the Feb 28th lecture}. \section{Explicit Riemann mappings} By Theorem \ref{th:ref2}, we know that the Riemann mapping function for the elliptical disk extends holomorphically to the boundary. In this section, we shall show how to find a precise formula for the Riemann mapping function of an elliptical disk. The main idea is to use the Bergman kernel of the elliptical disk. \subsection{Bergman kernel and the Riemann mapping function} Let $\Omega$ be a domain in $\mathbb C$. Denote by $\mathcal O(\Omega)$ the space of holomorphic functions on $\Omega$ and $$ H:=\left\lbrace f\in\mathcal O(\Omega): ||f||^2:=\int_{\Omega} |f|^2\, dxdy<\infty \right\rbrace $$ the space of $L^2$-holomorphic functions on $\Omega$. It is clear that $H$ is a $\mathbb C$-linear space with the following $L^2$ inner product $$ (f, g):=\int_{\Omega} f\bar g \, dxdy, \ \ f,g\in H. $$ The first observation is the following: \medskip \begin{thm}\label{th:complete} $H$ is a Hilbert space with a countable orthonormal basis $\{f_j\}_{j\geq 1}$. \end{thm} \begin{proof} Recall that Hilbert space is a complete inner product space, where "complete" means that every Cauchy sequence converges. Let $\{g_n\}$ be a Cauchy sequence in $H$. By the mean value property below, we have $$ |g_n(a) -g_m(a)| = \big|\frac1{\pi r^2} \int_{|z-a|0$ for some $a\in \Omega$. Then \begin{equation}\label{eq:Berg-R} f(z)=\int_a^z \sqrt{\frac{\pi}{K(a,a)}} K(w,a)\, dw \end{equation} for all $z\in \Omega$, where $K$ denotes the Bergman kernel on $\Omega$. \end{co} \begin{proof} By \eqref{eq:Berg-c} and \eqref{eq:K5}, we have $$ \frac1\pi f'(z) f'(a)=K(z, a). $$ hence $f'(a)= \sqrt{\pi K(a,a)}$ and $$ f'(z)= \sqrt{\frac{\pi}{K(a,a)}} K(z,a). $$ Thus the theorem follows by integrating this identity along any curves connecting $a$ and $z$ (since $\Omega$ is simply connected, the integral does not depend on the choice of such curves). \end{proof} \eqref{eq:Berg-R} suggests to compute the Riemann mapping function using the Bergman kernel. \textcolor{red}{We stopped here in the 2nd March lecture}. \subsection{Test exam 1} $\ $ \medskip \textbf{Exercise 1.} Compute $$ \int_{|z|=2} \frac{e^z\, dz}{(z-1)^2 \sin z}. $$ \emph{Answer}: The poles of $\frac{e^z\, dz}{(z-1)^2 \sin z}$ in $|z|<2$ is $z=0,1$, so $$ \int_{|z|=2} \frac{e^z\, dz}{(z-1)^2 \sin z} = 2\pi i \left(1+ \left(\frac{e^z}{\sin z}\right)'(1)\right) = 2\pi i \left(1+\frac{e(\sin 1-\cos 1)}{\sin^2 1}\right). $$ \medskip \textbf{Exercise 2.} Compute $$ \int_{0}^{2\pi} \frac{\cos t\, dt}{a+b\cos t}, \ \ a>b>0. $$ \emph{Answer}: $2\cos t =e^{it} + e^{-it}$, put $z=e^{it}$, we get $dt=\frac{dz}{iz}$, hence $$ \int_{0}^{2\pi} \frac{\cos t\, dt}{a+b\cos t} = \int_{|z|=1} \frac{z+z^{-1}}{2a +b (z+z^{-1})} \, \frac{dz}{iz}. $$ Then one may apply the Reidue theorem, the computation is similar to the example in page 155 of the Ahlfors book, so we will not repeat here. The solution is $$ \frac{2\pi}{b} \left( 1 -\frac a{\sqrt{a^2-b^2}} \right). $$ \medskip \textbf{Exercise 3.} Compute $$ \int_0^\infty \frac{dx}{x^4+1}. $$ \emph{Answer}: This is a special case of Example 2 in page 156 of the Ahlfors book, the answer is $$ \pi i \sum_{y>0} {\rm Res} \frac{1}{z^4+1}. $$ Since $$ z^4+1 = (z-e^{\pi i /4})(z-e^{3\pi i /4}) (z-e^{5\pi i /4}) (z-e^{7\pi i /4}), $$ we get $$ \pi i \sum_{y>0} {\rm Res} \frac{1}{z^4+1} = \pi i\, {\rm Res}_{z=e^{\pi i /4}} \frac{1}{z^4+1} + \pi i \,{\rm Res}_{z=e^{3\pi i /4}} \frac{1}{z^4+1} = \frac\pi {2\sqrt 2}. $$ \medskip \textbf{Exercise 4.} Use the argument principle to find the number of solutions to $z^3+z^2+z+4=0$ in the right half plane. \emph{Answer}: By the argument principle (see the Remark in page 18), the solution is $$ \frac{\text{change of argument of $z^3+z^2+z+4$ on $C$}}{2\pi}, $$ where $C=[iR, -iR] +C_R$, $C_R$ is the half circle from $-iR$ to $iR$ in the right half plane, when $R$ is large, the change of argument of $z^3+z^2+z+4$ on $C_R$ is close to $3\pi $; on the other hand, for $z= it$, $t$ goes from $R$ to $-R$, we have $$ z^3+z^2+z+4 = 4-t^2 + it(1-t^2) $$ so the argument changes approximately from close to $-\pi/2$ (at $z=iR$) to $0$ (at $z=i$) then to close to $\pi/2$ at $z=-iR$. Hence the change of argument of $z^3+z^2+z+4$ on $C$ is $4\pi$, and the solution is $2$. \medskip \textbf{Exercise 5.} Let $u(z+iy)=e^x\sin y$, show that $u$ is harmonic on $\mathbb C$ and find holomorphic $f$ on $\mathbb C$ such that $u={\rm Re}\, f$ \emph{Answer}: A direct computation gives $u_{xx}+u_{yy}=0$ (you should write down the details if it is a real exam) and one may take $$ f(z)=-i e^z. $$ \medskip \textbf{Exercise 6.} Compute $$ \int_0^{2\pi} \log|e^{i\theta}-a| \, d\theta $$ \emph{Answer}: If $|a|>1$ then $\log|z-a|$ is harmonic near $|z|\leq 1$ so the mean value property gives $$ \int_0^{2\pi} \log|e^{i\theta}-a| \, d\theta =2\pi \log |a|, $$ If $|a|<1$ then $\log|1-az|$ is harmonic near $|z|\leq 1$, hence $$ \int_0^{2\pi} \log|e^{i\theta}-a| \, d\theta = \int_0^{2\pi} \log|1-ae^{i\theta}| \, d\theta=0. $$ When $|a|=1$, we also get $\int_0^{2\pi} \log|e^{i\theta}-a| \, d\theta=0$ by taking limit. The solution is $\max\{0, 2\pi\log|a|\}$. \medskip \textbf{Exercise 7.} Find the Bergman kernel for $1<|z|<2$. \emph{Answer}: Every holomorphic function $f$ has a Laurent series expansion on $1<|z|<2$: $$ f(z)=\sum_{n\in\mathbb Z} c_n z^n. $$ One may check that $\{z^n\}$ is orthogonal with respect to the $L^2$ inner product, so $$ K(z, a)= \sum_{n=-\infty}^\infty \frac{z^n \bar a^n}{||z_n||^2}, $$ where $$ ||z_n||^2 = \int_{1<|z|<2} |z^n|^2 \, dxdy = \pi \frac{4^{n+1} -1}{n+1}, \ \ n\neq -1, \ \ ||z^{-1}||^2 = 2\pi \log 2. $$ \medskip \textbf{Exercise 8.} Define $u$ such that $u(z)-\log |z-\frac12|$ is harmonic on $|z|<1$, continuous on $|z|\leq 1$, and $u=0$ on $|z|=1$. Show that such $u$ is unique. \emph{Answer}: If $u_1, u_2$ satisfies the assumptions, then $u_1-u_2$ is harmonic on $|z|<1$, continuous on $|z|\leq 1$, and vanishes on $|z|=1$, so the maximum principle for $u_1-u_2$ and $u_2-u_1$ gives that $u_1=u_2$. Thus $u$ is unique. In fact $$ u= \log|\frac{z-\frac12}{1-\frac12 z}| $$ \medskip \textbf{Exercise 9.} Let $f$ be a holomorphic function on $|z|<1$. Assume that ${\rm Re}\,f =3$. Show that $f$ is a constant. \emph{Answer}: Otherwise the image of $f$ would be an open set (see Corollary \ref{co:open}), so we get a contradiction. \medskip \textbf{Exercise 10.} Use the Harnack inequality to prove that bounded harmonic functions on $\mathbb C$ are constants. \emph{Answer}: If $v$ is bounded and harmonic then $v+C\geq 0$. Hence the Harnack inequality \ref{eq:harnack1} applies to $u:=v+C$, which gives $$ \frac{\rho-r}{\rho+r} u(0) \leq u(z) \leq \frac{\rho+r}{\rho-r} u(0) $$ for all $|z|1$ is a constant. Write $z=x+iy$, \eqref{eq:ellipse-new} gives $$ x= \frac12 \left(\rho +\rho^{-1} \right) \cos t, \ \ y= \frac12 \left(\rho -\rho^{-1}\right) \sin t. $$ Since $\cos^2 t +\sin^2 t =1$, we know that the coordinate $(x,y)$ in the ellipse satisfies $$ \frac{x^2}{a^2} +\frac{y^2}{b^2}=1, \ \ \ a:=\frac{\rho+\rho^{-1}}{2}, \ \ b:=\frac{\rho-\rho^{-1}}{2}. $$ Hence we have \begin{equation}\label{eq:ellip-new} \Omega=\left\lbrace x+iy \in \mathbb C: \frac{x^2}{a^2} +\frac{y^2}{b^2}<1\right\rbrace. \end{equation} In order to find the Bergman kernel of $\Omega$, we need to find an orthonormal basis of the space $H$ of $L^2$-holomorphic functions on $\Omega$. It is clear that all polynomials lies in $H$. Applying the \textcolor{blue}{Gram-Schmidt process} (see the Wikipedia page) to $\{z^n\}_{n\geq 0}$, we obtain an orthonormal system $\{p_n\}_{n\geq 0}$ for $H$ (i.e. $||p_n||=1$ and $(p_n, p_m)=0$ for $n\neq m$), where each $p_n$ is a degree $n$ polynomial. To prove that $\{p_n\}$ is a basis of $H$, we have to show that \textcolor{blue}{polynomials are dense in $H$}. \medskip \begin{thm} Polynomials are dense in $H$ for the elliptical disk. \end{thm} \begin{proof} Fix $h\in H$, put $$ h_r(z):=rh(rz), \ \ 00$, there exist $01$. Hence the Cauchy integral formula gives $$ g(z)=\frac1{2\pi i }\int_{\partial (s\Omega)} \frac{g(\zeta)}{\zeta-z}\, d\zeta. $$ Since the above integral is the limit (in fact, uniform limit for $z\in\Omega$) of finite sums $$ \sum_{j=1}^N \lambda_j \frac{g(\zeta_j)}{\zeta_j-z}, \ \ \zeta_j\in \partial (s\Omega), $$ it suffices to prove the theorem for $g(z)=(\zeta-z)^{-1}$ for $\zeta\in \partial (s\Omega)$. Fix an open disk $D \subset \mathbb C \setminus \{\zeta\}$ with $\overline\Omega \subset D$, then it suffices to take $P_n$ to be the first $n$-terms of the Taylor series of $g$ on $D$. \end{proof} In order to find the orthonormal polynomials $p_n$, we need the following \textcolor{blue}{complex form of the Green formula}. \medskip \begin{thm}\label{th:gree1} Let $\Omega$ be bounded domain in $\mathbb C$ with smooth boundary. Then \begin{equation}\label{eq:c-green} \int_{\partial \Omega} f\, dz = \int_{\Omega} 2i\, f_{\bar z} \, dxdy \end{equation} for every $f$ smooth on a neighborhood of $\overline{\Omega}$. \end{thm} \begin{proof} Since $f\, dz= fdx+ if dy$, the usual greee formula \eqref{eq:green0} gives $$ \int_{\partial \Omega} f\, dz = \int_{\Omega} \left(if_x-f_y\right) \, dxdy. $$ Hence \eqref{eq:c-green} follows from $if_x-f_y= 2i\, f_{\bar z}$. \end{proof} A direct consequence of \eqref{eq:c-green} is \begin{equation}\label{eq:c-green0} \int_{\Omega} 2i\, f' \overline{g'} \, dxdy = \int_{\partial \Omega} f'\bar g\, dz, \end{equation} for all functions $f,g$ holomorphic on a neighborhood of $\overline{\Omega}$. This formula reduces our orthonormal polynomial problem to $\partial\Omega$, which is much easier since in our case $\partial \Omega$ is the conformal image of the circle $|w|=\rho$ under the \textcolor{blue}{Joukowski map} (a conformal map used in the airfoil design, see Wikipedia) $$ j: w\mapsto z=j(w)= \frac12\left(w+\frac1w\right). $$ As a standard Fourier series result, we know that $\{z^n\}_{n\geq 0}$ define a natural orthogonal basis for the $L^2$ space of the circle $|w|=\rho$. The transform of $z^n$ to $\partial\Omega$ under the Joukowski map is usually called the \textcolor{blue}{Chebyshev polynomial}. \medskip \begin{de}\label{de:ch1} The degree $n$ polynomial $T_n$ defined by $T_n(j(w))=j(w^n)$ is called the \textbf{Chebyshev polynomial of the first kind}. \end{de} \medskip \noindent \textbf{Remark.} \emph{Note that $$ T_0=1, \ \ \ T_1(z)=z; $$ moreover $$ T_{n+1}(z)= \frac12\left(w^{n+1}+\frac1{w^{n+1}}\right) = \left(w+\frac1w\right) \frac12\left(w^{n}+\frac1{w^{n}}\right)- \frac12\left(w^{n-1}+\frac1{w^{n-1}}\right) $$ gives \begin{equation}\label{eq:ch11} T_{n+1}(z)=2zT_n(z)-T_{n-1}(z). \end{equation} In this way, one may get a precise formula for all $T_n$.} \medskip \begin{lm}\label{lm:ch} Let $\Omega$ be the elliptical disk defined in \eqref{eq:ellip-new}. Then we have $$ (T'_{m}, T'_n)= \begin{cases} 0 & m \neq n;\\ \frac{n\pi}{4} \left(\rho^{2n} -\rho^{-2n}\right) & n=m. \end{cases} $$ \end{lm} \begin{proof} By \eqref{eq:c-green0}, we have $$ (T'_{m}, T'_n)=\frac1{2i} \int_{\partial \Omega} T_m' \overline{T_n} \,dz. $$ Under the change of variable $z=j(w)$ we have $$ dz= j'(w) dw = \frac12(1-w^{-2}) dw, $$ and $$ \frac{d j(w^m)}{dw}= \frac{d T_m(j(w))}{dw}= T_m'(j(w))j'(w)= \, \frac12(1-w^{-2}) T_m'(j(w)) $$ Hence $$ T_m'(j(w)) =\frac{m (w^{m-1}-w^{-m-1})}{1-w^{-2}} $$ we have \begin{align*} (T'_{m}, T'_n) & =\frac1{2i} \int_{|w|=\rho} \frac{m (w^{m-1}-w^{-m-1})}{1-w^{-2}} \,\overline{\frac{(w^n+w^{-n})}{2}}\, \frac12(1-w^{-2}) \,dw \\ & =\frac1{8i} \int_{|w|=\rho} m (w^{m-1}-w^{-m-1})(\bar w^n+\bar w^{-n})\,dw \\ & =\frac1{8i} \int_{|w|=\rho} m (w^{m-1}-w^{-m-1})\left( (\rho^2/w)^n+(\rho^2/w)^{-n}\right)\,dw. \end{align*} Hence our lemma follows from the Residue theorem. \end{proof} \medskip \begin{de}\label{de:ch2} The degree $n$ polynomial $U_n$ defined by $$ U_n:= \frac{T'_{n+1}}{n+1} $$ is called the \textbf{Chebyshev polynomial of the second kind}. \end{de} \medskip \noindent \textbf{Remark.} \emph{By Lemma \ref{lm:ch}, we have $$ (U_{m}, U_n)= \begin{cases} 0 & m \neq n;\\ \frac{\pi}{4(n+1)} \left(\rho^{2n+2} -\rho^{-2n-2}\right) & n=m. \end{cases} $$ Hence $p_n=U_n/||U_n||$ is the orthonormal polynomial that we need, and we have the following Bergman kernel formula for $\Omega$ $$ K(z, a) = \frac4\pi \sum_{n=0}^\infty \frac{U_n(z)\overline{U_n(a)}(n+1)}{\rho^{2n+2} -\rho^{-2n-2}}. $$ Together with \eqref{eq:Berg-R}, we get the Riemann mapping function for the elliptical disk.} \textcolor{red}{We stopped here in the 7th March lecture}. \subsection{Extremal property of the Chebyshev polynomial and capacity of a compact set} The original definition of $T_n$ is to use the following \textcolor{blue}{extremal property}. \medskip \begin{thm}\label{th:extremal} For $n\geq1$, $t_n:=2^{1-n}T^n$ satisfies \begin{equation}\label{eq:extremal} t_n \in \mathcal M_n, \ \ \max_{x\in [-1,1]} |t_n(x)| = 2^{1-n}= \inf_{q\in \mathcal M_n} \max_{x\in [-1,1]} |q(x)|, \end{equation} where $$ \mathcal M_n:=\{z^n+ a_1 z^{n-1} +\cdots a_n: a_j \in \mathbb C, \ 1\leq j\leq n\} $$ denotes the space of all degree $n$ monic polynomials on $\mathbb C$. \end{thm} \begin{proof} \eqref{eq:ch11} gives $t_n\in \mathcal M_n$. Since $j(e^{it})=\cos t$, we have $ T_n(\cos t) =\cos nt. $ Thus $$ \max_{x\in [-1,1]} |t_n(x)| = 2^{1-n}, \ \ t_n\left(x_k\right) =(-1)^k 2^{1-n}, \ \ x_k:= \cos \frac{k\pi}{n}, \ \ 0\leq k \leq n. $$ Let us use proof by contradiction. Assume that $\max_{[-1,1]} |q| < 2^{1-n}$ for some $q \in \mathcal M_n$. Let us write $ q = {\rm Re}\, q + i \, {\rm Im}\, q, $ where ${\rm Re}\, q$ and ${\rm Im}\, q$ are polynomials with real coefficients, then $$ t_n(x_0) = 2^{1-n} > {\rm Re}\, q(x_0), \ \ t_n(x_1) =- 2^{1-n} < {\rm Re}\, q(x_1), \ \ t_n(x_2) = 2^{1-n} > {\rm Re}\, q(x_2) , \ \cdots. $$ Hence the mean value theorem for zeros implies that $t_n-{\rm Re}\, q$ has at least $n$ zeros in $[-1,1]$. Since the degree of $t_n-{\rm Re}\, q$ is no bigger than $n-1$, this can happen only when $t_n-{\rm Re}\, q=0$, hence $\max_{[-1,1]} | {\rm Re}\, q| = \max_{[-1,1]} |t_n| =2^{1-n}$ and we have $\max_{[-1,1]} |q| \geq 2^{1-n}$, which contradicts the assumption $\max_{[-1,1]} |q| < 2^{1-n}$. \end{proof} \medskip One may replace $[-1,1]$ by a general compact set $K \subset \mathbb C$, then we have the following result. \medskip \begin{thm}[Exercise 4 in page 159 of \cite{Ransford}]\label{th:extremal1} Let $K$ be a compact set in $\mathbb C$. Put \begin{equation}\label{eq:extremal1} m_n(K):=\inf_{q\in \mathcal M_n} ||q||_K, \ \ ||q||_K:=\max_K |q|, \end{equation} then there exists $Q \in \mathcal M_n$ such that $||Q||_K=m_n(K)$. \end{thm} \begin{proof} If the number of points in $K$ is $\leq n$ then we can write $K=\{z_1, \cdots, z_k\}$, $k\leq n$. Take $$ Q(z) =(z-z_1)\cdots (z-z_k) z^{n-k}, $$ we know that $||Q||_K=0=m_n(K)$. If the number of points in $K$ is $\geq n$ then we can fix $n$ distinct points $z_1, \cdots, z_n$ in $K$. Note that for $$ q(z)= z^n+ a_1 z^{n-1} +\cdots +a_n, $$ we have $$ \begin{pmatrix} q(z_1)-z_1^n \\ \vdots \\ q(z_n)-z_1^n \end{pmatrix} = V \begin{pmatrix} a_1 \\ \vdots \\ a_n, \end{pmatrix} $$ where \begin{equation}\label{eq:Vandermonde} V:=\begin{pmatrix} z_1^{n-1} & z_1^{n-2} & \cdots & z_1 & 1 \\ z_2^{n-1} & z_2^{n-2} & \cdots & z_2 & 1 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ z_n^{n-1} & z_n^{n-2} & \cdots & z_n & 1 \end{pmatrix} \end{equation} is known as the \textcolor{blue}{Vandermonde matrix}. Since $$ \det V= \Pi_{1\leq j0$ with $|z-z_0|0$ then our assumption implies that $v(z_0)=M$ for some $z_0\in \Omega$, hence $\{v=M\}=\Omega$ and $M\leq 0$, we get a contradiction. \end{proof} \medskip \begin{pr}\label{pr:sub00} If $v_1, v_2$ are subharmonic, then $$ c_1v_1+c_2 v_2, \ \ \max\{v_1,v_2\}, $$ are also subharmonic, where $c_1\geq 0, \,c_2 \geq 0$ are constants. \end{pr} \medskip \begin{pr}[page 247, Ahlfors]\label{pr:psub} Let $v$ be a subharmonic function on a domain $\Omega\subset \mathbb C$. Let $D$ be an open disk with $\overline D \subset \Omega$. Then there exists a unique subharmonic function $v_D$ on $\Omega$ such that \begin{equation}\label{eq:psub} \text{$v_D$ is harmonic on $D$ and $v_D=v$ on $\Omega\setminus D$}. \end{equation} (We call $v_D$ the \textbf{Poisson Modification} of $v$ on $D$). \textcolor{red}{We stopped here in the 9th March lecture}. \end{pr} \begin{proof} By Schwarz's theorem, there is an unique continuous function, say $P_v$, on $\bar D$ such that $P_v=v$ on $\partial D$ and $P_v$ is harmonic function on $D$. Hence it suffices to check that \begin{equation}\label{eq:psub1} v_D= \begin{cases} P_v & \text{on}\ D \\ v & \text{outside} \ D \end{cases} \end{equation} is subharmonic. Note that $v-v_D=0$ on $\partial D$, Theorem \ref{th:MP-sub} implies that $v\leq v_D$ on $D$. Hence $v\leq v_D$ on $\Omega$. Now we have $$ v_D(z_0)= v(z_0) \leq \frac1{2\pi} \int_{0}^{2\pi} v(z_0+r e^{i\theta})\, d\theta \leq \frac1{2\pi} \int_{0}^{2\pi} v_D(z_0+r e^{i\theta})\, d\theta $$ for every $z_0\in \partial D$ and small $r>0$. Since $v_D$ obviously satisfies the submean inequality for small discs outside $\partial D$, we know that $v_D$ is subharmonic everywhere on $\Omega$. \end{proof} \medskip \noindent \textbf{Remark}: \emph{Assume that $D$ is given by $|z-a|0$ on $\overline{\Omega}\setminus\{\zeta_0\}$}. $$ $\Omega$ is said to be \textbf{regular} if every boundary point of $\Omega$ possesses a barrier. \end{de} \medskip \emph{Exercise}: Show that $0<|z|<1$ is not regular. \medskip \begin{thm}[See Ahlfors, page 250]\label{th:Dirichlet} Let $\Omega$ be a bounded domain in $\mathbb C$. The Dirichlet problem is solvable for $\Omega$ if and only if $\Omega$ is regular. \end{thm} \begin{proof} \emph{Proof of $\Rightarrow$.} If the Dirichlet problem is solvable for $\Omega$ then for every $\zeta_0\in\partial\Omega$, the function $\phi$ defined by $$ \phi(\zeta):=|\zeta-\zeta_0|^2, \ \ \ \zeta \in\partial\Omega, $$ extends to a continuous function, say $\omega$ on $\bar\Omega$, harmonic on $\Omega$. The maximum principle, Theorem \ref{th:MP}, for $-\omega$ implies that $\omega$ is positive on $\Omega$. Thus $\omega$ is a barrier at $\zeta_0$. Since $\zeta_0$ is arbitrary, we know that $\Omega$ is regular. \medskip \emph{Proof of $\Leftarrow$.} Assume that $\Omega$ is regular, it suffices to show that for every continuous function $\phi$ on $\partial\Omega$, the Perron envelope $u_\phi$ satisfies that \begin{equation}\label{eq:Dirichlet0} \lim_{z\to \zeta_0} u_\phi(z) = \phi(\zeta_0), \end{equation} for every $\zeta_0\in \partial\Omega$. Since $\Omega$ is bounded, we know that $\phi$ is bounded, so we can take $M>0$ such that $|\phi|\leq M$ on $\partial \Omega$. For every $\varepsilon>0$, there exists a small disk $D$ around $\zeta_0$ such that $$ |\phi(\zeta)-\phi(\zeta_0)|<\varepsilon, $$ for $\zeta\in D\cap \bar\Omega$. Let $\omega$ be a barrier at $\zeta_0$, we have $$ \omega_0:=\inf_{\bar\Omega\setminus D} \omega >0. $$ Consider $$ W(z):=\phi(\zeta_0)+\varepsilon +\frac{\omega(z)}{\omega_0} (M-\phi(\zeta_0)). $$ For $\zeta\in D\cap\partial \Omega$, we have $W(\zeta)\geq \phi(\zeta_0)+\varepsilon > \phi(\zeta)$; for $\zeta\in \partial\Omega\setminus D$ we obtain $$ W(\zeta) \geq \phi(\zeta_0)+\varepsilon + M-\phi(\zeta_0) =M+\varepsilon >\phi(\zeta). $$ By the maximum principle any function $v\in \mathcal B_\phi$ must hence satisfy $v< W$. Hence $u_\phi\leq W$ and we have \begin{equation}\label{eq:Dirichlet} \limsup_{z\to \zeta_0} u_\phi(z) \leq W(\zeta_0)= \phi(\zeta_0)+\varepsilon. \end{equation} For the lower limit, we consider $$ V(z):=\phi(\zeta_0)-\varepsilon -\frac{\omega(z)}{\omega_0} (M+\phi(\zeta_0)). $$ One may verify that $V\in \mathcal B_\phi$, hence $u_\phi \geq V$ gives \begin{equation}\label{eq:Dirichlet1} \liminf_{z\to \zeta_0} u_\phi(z) \geq V(\zeta_0)= \phi(\zeta_0)-\varepsilon. \end{equation} Since $\varepsilon$ is arbitrary, \eqref{eq:Dirichlet} and \eqref{eq:Dirichlet1} together give \eqref{eq:Dirichlet0}. \end{proof} \medskip It remains to formulate geometry conditions which imply the existence of a barrier. To begin with the simplest case, suppose that $\bar\Omega$ is contained in the half space ${\rm Im}\, z>0$, except for the $\zeta_0=0$ which lies in $\partial \Omega$. Then $w(z):={\rm Im} \,z$ is a barrier at $\zeta_0$. More generally, suppose that $\zeta_0$ is the end point of a line segment, say $[\zeta_0, \zeta_1]$, all of whose point, except $\zeta_0$, lie in $\mathbb C\setminus \bar\Omega$. By a linear change of coordinate, let us assume that $\zeta_0=0, \zeta_1=1$, we know that there is a conformal mapping (see the picture below (should be $\frac{z}{1-z}=\zeta$)) $$ \eta(z):= \sqrt{\frac{z}{1-z}}, $$ \includegraphics[scale=0.1]{2} \medskip which maps $\mathbb C\setminus [0,1]$ onto ${\rm Im}\, \eta>0$. Hence $$ \omega(z):= {\rm Im}\, \eta(z)= {\rm Im}\, \sqrt{\frac{z}{1-z}} $$ defines a barrier at $\zeta_0$. To summarize, we have: \medskip \begin{thm}\label{th:D-line} The Dirichlet problem can be solved for any bounded domain $\Omega\subset \mathbb C$ such that each boundary point is the end point of a line segment whose other points lie in $\mathbb C\setminus \bar\Omega$. In particular, any bounded domain with continuous boundary (i.e. the boundary is locally the graph of a continuous function) is regular. \end{thm} \medskip For simply connected domains, we have the following result (see Theorem 4.2.1 in \cite{Ransford} for the proof). \textcolor{red}{We stopped here in the 16th March lecture.} \medskip \begin{thm}\label{th:D-scd} Every simply connected bounded domain $\Omega\subset \mathbb C$ is regular. \end{thm} \section{Potential theory in the complex plane} \subsection{Green's functions as envelopes} \subsubsection{Green's functions for regular bounded domains} Let $\Omega$ be a bounded domain in $\mathbb C$. Fix $w\in\Omega$, assume that $\Omega$ is regular, then by Theorem \ref{th:Dirichlet}, there exists a \begin{equation}\label{eq:Green0} \text{harmonic function}\ u(z) \ \text{on $\Omega$, continuous on $\bar\Omega$ and $u(\zeta)=-\log|\zeta-w|$ for $z\in \partial\Omega$}. \end{equation} \begin{de}[Definition of Green's function for regular bounded domains]\label{de:Green} We call $$ G_\Omega(z, w):=u(z)+\log|z-w|, \ \ z\in \bar\Omega, \ w\in \Omega, $$ the \textbf{Green function with a pole at $w\in\Omega$}. \end{de} \medskip The maximum principle implies the following result: \medskip \begin{pr}\label{pr:Green} Let $\Omega$ be a bounded regular domain in $\mathbb C$. Then $G_\Omega(\cdot, w)$ is the unique function such that $G_\Omega(z, w)=0$ for $z\in \partial\Omega$, $G_\Omega(z, w)-\log|z-w|$ (as a function of $z$) is harmonic on $\Omega$ and continuous on $\bar\Omega$. \end{pr} \medskip The above proposition and Theorem \ref{th:D-scd} together imply: \medskip \begin{pr}\label{pr:D-Green} Let $\Omega$ be a bounded simply connected domain in $\mathbb C$. Then $G_\Omega(\cdot, w)=\log|f_w(z)|$, where $f_w$ is the Riemann mapping from $\Omega$ to the unit disk such that $f_w(w)=0$ and $f_w'(w)>0$. \end{pr} \medskip This proposition suggests a \emph{Green function proof} of the Riemann mapping theorem, see the proof of Theorem 4.4.11 in \cite{Ransford} for details. In case $\Omega$ is the unit disk $\mathbb D$, we know that (try, use Proposition \ref{pr:mobius}) $$ f_w(z) =\frac{z-w}{1-\bar w z}, $$ hence we get \begin{equation}\label{eq:Green-disk} G_{\mathbb D} (z, w)= \log \big| \frac{z-w}{1-\bar w z} \big|. \end{equation} \subsubsection{Green's function as an envelope} By the proof of Theorem \ref{th:Dirichlet}, we know that the function $u$ in \eqref{eq:Green0} satisfies that $$ u= \sup\{v: v \,\text{is subharmonic on $\Omega$ with}\, v^*(\zeta) \leq -\log|\zeta-w| \ \text{for $\zeta\in\partial\Omega$}\} $$ hence we get the following result. \medskip \begin{pr}\label{pr:E-Green} Let $\Omega$ be a bounded regular domain in $\mathbb C$. Then \begin{equation}\label{eq:E-Green} G_\Omega(\cdot, w)= \sup\{\psi\leq 0: \psi(z)-\log|z-w| \,\text{is subharmonic for $z\in\Omega$}\}, \end{equation} for every fixed $w\in\Omega$. \end{pr} \medskip Not that the envelope \eqref{eq:E-Green} is also well defined (and harmonic on $\Omega$ by Theorem \ref{th:Dirichlet}) for non-regular $\Omega$. Hence, one can use it to define Green's function for general bounded domains. \medskip \begin{de}[Definition of Green's function for general bounded domains]\label{de:Green-G} Let $\Omega$ be a bounded domain in $\mathbb C$. We call $G_\Omega$ defined in \eqref{eq:E-Green} the \textbf{Green function with a pole at $w\in\Omega$}. \end{de} \subsection{Poisson kernels and harmonic measures} Assume that $\Omega$ has real analytic boundary, then the reflection principle (see Theorem \ref{th:ref1}) implies that the Green function $G_\Omega(\cdot, w)$ extends to a harmonic function on a neighborhood of $\bar\Omega$. Thus one may apply Green's formula to $$ \int_\Omega G_{\Omega}(z, w) \Delta u(z)\, dxdy, $$ where $u$ is an arbitrary smooth function on a neighborhood of $\bar\Omega$. Then we obtain the following generalization of \eqref{eq:Green12}. \medskip \begin{thm}[See Theorem 4.5.1 in \cite{Ransford} for generalizations] \label{th:GP} Let $\Omega$ be a bounded domain in $\mathbb C$ with real analytic boundary. Then \begin{equation}\label{eq:GP} u(w) = \frac1{2\pi} \int_{\Omega} G_{\Omega}(z, w) \Delta u(z) \, dxdy + \frac1{2\pi} \int_{\partial \Omega} \frac{\partial G_\Omega(z, w)}{\partial n_z}\, u(z)\, |dz|, \ \ w\in \Omega, \end{equation} for all function $u$ smooth on a neighborhood of $\bar\Omega$, where $n_z$ denotes the \textbf{outward unit normal vector} at $z\in \partial\Omega$. in particular, if $u$ is harmonic on $\Omega$ then \begin{equation}\label{eq:GP1} u(w) = \frac1{2\pi} \int_{\partial \Omega} \frac{\partial G_\Omega(z, w)}{\partial n_z}\, u(z)\, |dz|, \ \ w\in \Omega. \end{equation} \end{thm} \begin{proof} Put $\Omega_\varepsilon:= \Omega\setminus \{|z-w|<\varepsilon\}$. Since $G_\Omega(z, w)$ is harmonic for $z\in \Omega_\varepsilon$, the Green's formula \eqref{eq:Green11} gives \begin{equation}\label{eq:GP111} \int_{\partial \Omega_\varepsilon} \left( G_{\Omega}(z, w) \, \frac{\partial u}{\partial n_z}- u \, \frac{\partial G_{\Omega}(z, w)}{\partial n_z} \right) |dz| = \int_{\Omega_\varepsilon} G_{\Omega}(z, w) \Delta u(z) \, dx dy, \end{equation} Since $ G_{\Omega}(z, w)=0$ for $z=\partial\Omega$ and $\partial \Omega_\varepsilon = \partial\Omega-\{|z-w|<\varepsilon\}$, we have (try!) \begin{equation}\label{eq:GP112} \lim_{\varepsilon \to 0}\int_{\partial \Omega_\varepsilon} \left( G_{\Omega}(z, w) \, \frac{\partial u}{\partial n}- u \, \frac{\partial G_{\Omega}(z, w)}{\partial n} \right) |dz| = 2\pi u(w) - \int_{\partial \Omega} \frac{\partial G_\Omega(z, w)}{\partial n_z}\, u(z)\, |dz|. \end{equation} Hence \eqref{eq:GP} follows. \end{proof} \medskip \noindent \textbf{Remark.} \emph{By the maximum principle, if $u\geq 0$ on $\partial\Omega$ and harmonic on $\Omega$ then $u\geq 0$ on $\Omega$, thus \eqref{eq:GP1} implies that $p(z, w)|dz|$, with \begin{equation}\label{eq:GP2} p(z, w):= \frac1{2\pi} \frac{\partial G_\Omega(z, w)}{\partial n_z}, \end{equation} defines a \textcolor{blue}{probability measure} on $\partial \Omega$. Since its integral \eqref{eq:GP1} gives the value of harmonic functions, we call $p(z, w)|dz|$ the \textcolor{blue}{harmonic measure} on $\partial \Omega$ with respect to $w\in \Omega$.} \medskip \begin{de}\label{de:GP} Let $\Omega$ be a bounded domain in $\mathbb C$ with real analytic boundary. We call $p(z, w)$, $z\in \partial\Omega$, $w\in\Omega$, defined in \eqref{eq:GP2} the \textbf{Poisson kernel} of $\Omega$. The corresponding probability measure $p(z, w)|dz|$ is called the \textbf{harmonic measure} on $\partial \Omega$ with respect to $w\in \Omega$. \textcolor{red}{We stopped here in the 21st March lecture}. \end{de} \subsection{Green's function with a pole at infinity and equilibrium measure} Let $\Omega$ be a bounded domain in $\mathbb C$ with Green's function $G_\Omega$ defined in \eqref{eq:E-Green}. Note that the following mapping $$ z\mapsto \zeta:=\frac1{z-w} $$ maps $w$ to $\infty$ and $\Omega$ to a domain, say $\Omega'$, around $\infty$. Let us define $G_{\Omega'}$ such that $$ G_{\Omega'}(\zeta)= G_\Omega(z, w), $$ or equivalently (since $z=w+\zeta^{-1}$) \begin{equation}\label{eq:G-infty0} G_{\Omega'}(\zeta)= G_\Omega\left( w+ \frac1\zeta, w\right) . \end{equation} \medskip \begin{pr}\label{pr:G-infty0} Assume that $\Omega$ has real analytic boundary. Then $G_{\Omega'}$ is the unique function on $\Omega'$ such that \smallskip (1) $G_{\Omega'}\leq 0$ on $\Omega'$; \smallskip (2) $G_{\Omega'}$ extends to a harmonic function near $\overline{\Omega'}$ and $G_{\Omega'}(\zeta)=0$ for $\zeta\in \partial \Omega'$; \smallskip (3) $G_{\Omega'}(z^{-1})-\log|z|$ extends to a harmonic function near $z=0$. \end{pr} \begin{proof} (1) is obviuous. For (2), since $\Omega$ has real analytic boundary, by the reflection principle (see Theorem \ref{th:ref1}), we know that $G_\Omega(\cdot, w)$ extends to a harmonic function on a neighborhood of $\bar\Omega$ and $G_\Omega(z, w)=0$ for $z\in\partial\Omega$ (since $\Omega$ is regular), thus (2) follows. For (3), note that $$ G_{\Omega'}(z^{-1})-\log|z| = G_\Omega\left( w+z, w\right) - \log|z|, $$ thus (3) follows from the fact that $G_\Omega\left( z, w\right) - \log|z-w|$ is harmonic near $z=w$. For the uniqueness, one may pull back $G_{\Omega'}$ to $\Omega$ and then apply the maximum principle. \end{proof} \medskip \noindent \textbf{Remark}. \emph{In general, let $K$ be a compact set in $\mathbb C$ such that the unbounded component, say $\Omega_K$, of $\mathbb C\setminus K$ has analytic boundary. Then $\Omega_K=\Omega'$ for some bounded domain $\Omega$ with analytic boundary, in this case, we shall write $G_{\Omega'}$ as $G_{\Omega_K}$ and the above proposition gives:} \medskip \begin{pr}\label{pr:G-infty} Let $K$ be a compact set in $\mathbb C$. Denote by $\Omega_K$ the unbounded connected component of $\mathbb C\setminus K$. Assume that $\Omega_K$ has real analytic boundary. Then there is a unique function $G_{\Omega_K}$ on $\Omega_K$ such that \smallskip (1) $G_{\Omega_K}\leq 0$ on $\Omega$; \smallskip (2) $G_{\Omega_K}$ extends to a harmonic function near $\overline{\Omega_K}$ and $G_{\Omega_K}(\zeta)=0$ for $\zeta\in \partial \Omega_K$; \smallskip (3) $G_{\Omega_K}(z^{-1})-\log|z|$ extends to a harmonic function near $z=0$. \end{pr} \medskip \begin{de}\label{de:G-infty} Let $K$ be a compact set in $\mathbb C$. Assume that $\Omega_K$ has real analytic boundary. Then we call $G_{\Omega_K}$ in Proposition \ref{pr:G-infty} the \textbf{Green function of $\Omega_K$ with a pole at $\infty$}. The following limit \begin{equation}\label{eq:Robin-infty} \gamma:= \lim_{z\to 0} \left(G_{\Omega_K}(z^{-1})-\log|z|\right) \end{equation} is called the \textbf{Robin constant} of $K$. \end{de} \medskip We shall use the following result in the next subsection. \medskip \begin{lm}\label{lm:G-infty} Let $K$ be a compact set in $\mathbb C$. Assume that $\Omega_K$ has real analytic boundary. Put \begin{equation}\label{eq:GP-infty} p(\zeta,\infty):= \frac1{2\pi} \frac{\partial G_{\Omega_K}(\zeta)}{\partial n_\zeta}, \end{equation} where outward unit normal vector at $\zeta\in \partial\Omega_K$. Then \begin{equation}\label{eq:mu_K} d\mu_K(\zeta):= p(\zeta,\infty) |d\zeta|, \ \ \zeta\in \partial\Omega_K, \end{equation} defines a probability measure $\mu_K$ on $K$ supported on $\partial \Omega_K$ satisfying \begin{equation}\label{eq:mu_K1} p_{\mu_K}(z):=\int_{K} \log|z-\zeta| \,d\mu_K(\zeta)= \begin{cases} \gamma-G_{\Omega_K}(z) & z\in \Omega_K \\ \gamma & z\in \mathbb C\setminus \Omega_K. \end{cases} \end{equation} \end{lm} \begin{proof} Similar to \eqref{eq:GP}, we have \begin{equation}\label{eq:GP-infty0} u(\infty) = \frac1{2\pi} \int_{\Omega_K} G_{\Omega_K}(\zeta) \Delta u(\zeta) \, dxdy + \frac1{2\pi} \int_{\partial \Omega_K} \frac{\partial G_{\Omega_K}(\zeta)}{\partial n_\zeta}\, u(\zeta)\, |d\zeta|, \ \ w\in \Omega, \end{equation} for all function $u$ smooth on a neighborhood of $\bar\Omega_K$ such that $u(1/z)$ extends to a smooth function near $z=0$, here $u(\infty):=\lim_{z\to 0} u(1/z)$. Thus we know that $\mu_K$ is a probability measure supported on $\partial \Omega_K$ in $K$. Hence it suffices to prove \eqref{eq:mu_K1}. By the Fubini theorem and the Fatou theroem (see the next subsection), $$ p_{\mu_K}(z)= \frac1{2\pi} \int_{\partial \Omega_K} \log|z-\zeta|\, \frac{\partial G_{\Omega_K}(\zeta)}{\partial n_\zeta}\, |d\zeta| $$ satisfies the submean inequality, is upper semi-continuous on $\mathbb C$ and harmonic outside $\partial \Omega_K$. For fixed $z\in \mathbb C\setminus \overline{\Omega_K}$, apply \eqref{eq:GP-infty0} to $$ u(\zeta):= \log|z-\zeta| + G_{\Omega_K}(\zeta), $$ we get $$ p_{\mu_K}(z) =u(\infty) = \gamma. $$ Since $\partial \Omega_K$ is smooth and $G_{\Omega_K}$ is smooth near $\partial \Omega_K$, we know that $p_{\mu_K}$ is continuous near $\partial\Omega_K$, thus $p_{\mu_K}(z) =\gamma$ also for $z\in \partial\Omega_K$. Now it remains to show that $p_{\mu_K} = \gamma-G_{\Omega_K}$ on $\Omega_K$, we already know that they are equal on $\partial \Omega_K$, note that $$ \lim_{z\to \infty} (p_{\mu_K}(z)- \log|z|) = 0 = \lim_{z\to \infty} (\gamma-G_{\Omega_K}(z)-\log|z|), $$ so they must be equal on $\Omega_K$ by the maximum principle. \end{proof} \medskip \begin{de}\label{de:equilibrium} Let $K$ be a compact set in $\mathbb C$. Assume that $\Omega_K$ has real analytic boundary. Then we call $\mu_K$ defined in \eqref{eq:mu_K} the \textbf{equilibrium measure} on $K$. We also call the function $p_{\mu_K}$ in \eqref{eq:mu_K1} the \textbf{equilibrium potential} of $K$. \end{de} \medskip In the later sections, we shall follow \cite[section 2.2]{A1} to study the extremal property of $\mu_K$ in the space of Borel measures on $K$ and prove that $e^\gamma$ equals the logarithmic capacity $c(K)$ of $K$. \textcolor{red}{We stopped here in the 23rd March lecture}. \subsection{A short summary and test exam 2} \subsubsection{A short summary} $\ $ \medskip 1. The first main result is the Cauchy integral theorem (see Theorem \ref{th:CIT0}), which implies Theorem \ref{th:scd}, Theorem \ref{th:residue1} and Theorem \ref{th:argument1}, etc. \medskip For example, Theorem \ref{th:CIT0}) directly gives (try!) $$ \int_{|z|+|z|^9=3} z^{20} +e^z \, dz =0 $$ and can be used to prove Corollary \ref{co:branch}, which is used in the proof of the Riemann mapping theorem. Theorem \ref{th:residue1} can be used to compute integrals (see page 154-161 in the Ahlfors book, especially the exercise in page 161) $$ \int_0^{2\pi} \frac{dt}{a+b\cos t} = \frac{2\pi}{\sqrt{a^2-b^2}}, \ \ a>b\geq 0; \ \ \int_{-\infty}^\infty \frac{e^{-ix}\, dx}{1+x^2} =\frac\pi e. $$ Theorem \ref{th:argument1} can be used to prove that $$ z+e^{-z} =\lambda, \ \ \lambda>1 \ \text{is a constant}, $$ has exactly one solution in the right half plane and prove Corollary \ref{co:open}, Corollary \ref{co:oneone}, Theorem \ref{th:inverse} and Theorem \ref{th:Rouche} etc. \medskip 2. The second main result is the Riemann mapping theorem, Theorem \ref{th:Riemann}. We know that the Riemann mapping function is directly related to the Bergman kernel (see Corollary \ref{co:Berg-c} and the Green function of a simply connected domain (see Proposition \ref{pr:Green}). In order to study the regularity property of the Riemann mapping function (see Theorem \ref{th:ref2}), we introduce the theory of harmonic functions. We prove the mean value property, Theorem \ref{th:MVP}, for harmonic functions, and obtain the Poisson formula, Theorem \ref{th:Poisson}, using the mean value property and the Mobius transform. Then we prove the crucial Schwarz's theroem, Theorem \ref{th:Schwarz}, for the Poisson integral and the Harnack inequality, Theorem \ref{th:harnack1}, for positive harmonic functions. Applications include the Harnack principle --- Theorem \ref{th:hp} (which is used in the proof of Theorem \ref{th:Dir-har}) and the reflection principle ---Theorem \ref{th:ref} (which implies Theorem \ref{th:ref2}). In order to compute the Bergman kernel of the elliptical disk, we introduce the Chebyshev polynomial and define the capacity of a compact set (see Definition \ref{de:capacity}). \medskip 3. The third main result is the solution of the \emph{Dirichlet Problem}, Theorem \ref{th:Dirichlet}, which is used to define the crucial Green's function for a regular domain. Then we use Green's function to define the Poisson kernel, harmonic measure, Robin constant, equilibrium measure and the equilibrium potential. In the later sections we shall prove the extremal property (see Theorem \ref{th:extremal-p}) of the equilibrium measure and introduce a "weighted" version of the Green function (called the Hele-Shaw envelope, see section 6.8-6.9). \textcolor{red}{We stopped here in the 28th March lecture}. \subsubsection{Test exam 2 (NOT a normal exam, just a collection of related exercises)} $\ $ \medskip \textbf{Exercise 1.} Compute $$ \int_{|z|=5} \frac{ dz}{(z^2-1)^2}, \ \ \int_{|z|=1} \frac{f(z)}{z^2}\, dz $$ where $f$ is a given holomorphic function on $|z|<2$. \medskip \textbf{Exercise 2.} Show that $\log|z^3+z+10|$ is harmonic on $|z|< 1$ and find a holomorphic $f$ with $$ {\rm Re}\, f = \log|z^3+z+10| $$ on $|z|<1$. \medskip \textbf{Exercise 3.} Compute $$ \int_{|z|=1} \frac{|dz|}{|z-3|^2}, \ \ \int_0^\infty \frac{\cos x}{x^2+1}\, dx, $$ and show that $$ \int_{|z|=1} P(z)\, d\bar z =-2\pi iP'(0), $$ where $P$ is a polynomial on $\mathbb C$. \medskip \textbf{Exercise 4.} Show that $$ \int_0^{2\pi} \log|e^{i\theta}+3|\, d\theta= 2\pi \, \log 3 $$ and $$ \log2 \leq \frac{1}{2\pi i} \int_{|z|=20} \log|z^9+8z^7 +2|\, \frac{dz}{z}. $$ \medskip \textbf{Exercise 5.} Find a conformal mapping from $\mathbb H:=\{z\in \mathbb C: {\rm Im}\, z>0\}$ onto the unit disk, compute the Bergman kernel of the unit disk and then use it to get the Green's function of $\mathbb H$ with a pole at $i$ and the Bergman kernel of $\mathbb H$. \medskip \textbf{Exercise 6.} Let $f$ be a conformal mapping from a bounded simply connected domain $\Omega$ onto the unit disk such that $f(a)=0$ for some $a\in\Omega$. \smallskip a) Find Green's function of $\Omega$ with a pole at $a$; \smallskip b) Assume that $\Omega$ has real analytic boundary, find the equilibrium measure and the capacity of $\mathbb C\setminus \Omega'$, where $$ \Omega':=\{(z-a)^{-1}: z\in \Omega\setminus \{a\}\}. $$ \medskip \textbf{Exercise 7.} Solve Exercise 2 in page 31. \medskip \textbf{Exercise 8.} Let $u$ be a positive subharmonic function on a domain $\Omega\subset \mathbb C$. Assume that the disk $|z|\leq \rho$ lies in $\Omega$. Show that $$ u(z) \leq \frac1{2\pi}\, \frac{\rho+r}{\rho-r} \int_0^{2\pi} u(\rho e^{i\theta})\, d\theta $$ for all $z$ with $|z|=r<\rho$. You might use the maximum principle, Schwarz's theorem and the proof of the Harnack inequality. Another proof is to use a generalization of Exercise 7. \subsection{A short course on Borel measures} \subsubsection{Riesz representation theorem} We shall mainly follow the Ransford book \cite{Ransford} in this part. Let $X$ be a topological space. A function $ \phi: X\to \mathbb R $ is said to be \textcolor{blue}{continuous} if $$ \phi^{-1}(a,b):=\{x\in X: a<\phi(x)t\}\, dt. \end{equation} The right hand side is the integral of a decreasing function. \medskip \begin{thm}[Fatou's lemma]\label{th:Fatou} If $f_j \geq 0$ are Borel then (see \cite[Page 18]{LL} for the proof) $$ \liminf_{j\to\infty} \int_X f_j \, d\mu \geq \int_{X} \liminf_{j\to\infty} f_j \, d\mu. $$ \end{thm} \medskip \begin{de}\label{de:com-int} A complex function $f: X\to \mathbb C$ is said to be \textbf{Borel} if both its real and imaginary parts are Borel. A complex Borel function $f$ is said to be \textbf{integrable} if $$ \int_X |f| \, d\mu<\infty. $$ The \textbf{$L^p$ space}, $1\leq p< \infty$, is defined by (see \cite[Chapter 7]{Ax}) $$ L^p(X, \mu):=\left\lbrace \text{complex Borel}\, f: \int_X |f|^p\, d\mu<\infty\right\rbrace/ \sim, $$ where $\sim$ means we identify functions which are equal outside a $\mu$-measure zero set. \end{de} \medskip \noindent \textbf{Remark}: \emph{It is known that each $L^p(X, \mu)$ is a complex Banach space (i.e. complete complex normed space, see \cite[page 67]{Rudin} and \cite[page 52]{LL}) and $C_c(X)$ is dense in $L^p(X, \mu)$ (see \cite[Page 69]{Rudin}). People often identify $f$ with its equivalent class in $L^p(X, \mu)$. In this way, $f=g$ means they are equal outside a $\mu$-measure zero set, and $f_j \to f$ pointwise on $X$ means $f_j(x) \to f(x)$ for all $x$ outside a $\mu$-measure zero set.} \medskip \begin{thm}[Monotone convergence]\label{th:Mon} If $f_j \in L^1(X, \mu)$ are real and $f_1 \leq f_2 \leq \cdots$, then $$ \lim_{j\to\infty} \int_X f_j \, d\mu = \int_{X} \lim_{j\to\infty} f_j \, d\mu, $$ see \cite[page 17]{LL} for the proof. \end{thm} \medskip \begin{thm}[Dominated convergence]\label{th:Dom} If $f_j \in L^1(X, \mu)$ pointwise converge to $f$ and there exists $G\in L^1(X, \mu)$ with $|f_j|\leq G$ for all $j$, then $f\in L^1(X, \mu)$ and $$ \lim_{j\to\infty} \int_X f_j \, d\mu = \int_{X} \lim_{j\to\infty} f \, d\mu, $$ see \cite[page 19]{LL} for the proof. \end{thm} \subsubsection{Complex Borel measures} \begin{de}\label{de:cm} A \textbf{complex Borel measure} is a function $$ \mu: \mathcal B \to \mathbb C $$ such that $\mu(\emptyset)=0$ and $$ \sum_{j=1}^\infty |\mu(A_j)| <\infty, \ \ \sum_{j=1}^\infty \mu(A_j) = \mu(\cup_{j=1}^\infty A_j) $$ for all disjoint set $A_j\in \mathcal B$. \end{de} \medskip \noindent \textbf{Remark}: \emph{If $\mu$ is a complex measure, then its \textcolor{blue}{variation} $|\mu|$ defined by $$ |\mu|(A):= \sup \sum_{j} |\mu(A_j)|, $$ where the supremum runs over all sequences of disjoint Borel sets $A_j$ whose union is $A$, is a Borel measure. We also know that its \textcolor{blue}{total variation} $||\mu||:=|\mu|(X)<\infty. $ We have the following \textcolor{blue}{Riesz Representation theorem for complex Borel measures} (see Theorem 6.19 in page 130 in \cite{Rudin} for the statement and the proof, we do not need "regular" because all finite Borel measures on a metric space is regular, see Theorem A.2.2 in \cite{Ransford}.)} \medskip \begin{thm}\label{th:Riesz-C} Let $X$ be a metric space with compact exhaustion. Let $$ \Lambda : C_c(X, \mathbb C) \to \mathbb C $$ be a $\mathbb C$-linear mapping. If $$ ||\Lambda||:=\sup \left\lbrace|\Lambda(\phi)|: \phi \in C_c(X, \mathbb C), \ \sup_X |\phi|=1 \right\rbrace<\infty, $$ then there exists a unique complex Borel measure $\mu$ on $X$ such that $$ \Lambda(\phi) =\int_X \phi \, d\mu, \ \ \ \forall \ \phi\in C_c(X). $$ Moreover, we have $||\mu||=||\Lambda||$. \end{thm} \subsubsection{Fubini theorem} Let $X_1, X_2$ be metric spaces with compact exhaustion. Let $\mu_j$ be Borel measures on $X_j$ such that $\mu_j(K_j)<\infty$ for every compact $K_j \subset X_j$, $j=1,2$. The \textcolor{blue}{Borel $\sigma$-algebra $\mathcal B_{X_1\times X_2}$} on $X_1\times X_2$ equals the smallest $\sigma$-algebra, say $\mathcal B_{X_1} \times \mathcal B_{X_2}$, containing $A_1\times A_2$ for all $A_1\in \mathcal B_{X_1}$ and $A_2\in \mathcal B_{X_2}$. It is known that (see page 11 in \cite{LL} for the uniqueness and page 23 in \cite{LL} for the existence) there exists a unique measure $\mu$ on $\mathcal B_{X_1\times X_2}$ such that $$ \mu(A_1 \times A_2) =\mu_1(A_1) \times \mu_2(A_2), \ \ \forall \ A_1\in \mathcal B_{X_1}, \, A_2\in \mathcal B_{X_2}. $$ We write \textcolor{blue}{$\mu:=\mu_1\times \mu_2$}. We have the following \textcolor{blue}{Fubini's theorem} (see \cite[page 25]{LL}). \medskip \begin{thm}\label{th:Fubini} With the notation above, if $f\geq 0$ is Borel on $X_1\times X_2$ then \begin{equation}\label{eq:Fubini} \int_{X_1\times X_2} f\, d(\mu_1\times \mu_2)= \int_{X_1} \left( \int_{X_2} f\, d\mu_2\right) \, d\mu_1 = \int_{X_2} \left( \int_{X_1} f\, d\mu_1\right) \, d\mu_2. \end{equation} If $f$ is complex valued then \eqref{eq:Fubini} holds if one assumes in addition that $$ \int_{X_1\times X_2} |f|\, d(\mu_1\times \mu_2)<\infty, $$ (see \cite[page 25]{LL} for the proof and generalizations). \end{thm} \subsection{General subharmonic functions and extremal property of the equilibrium measure} \begin{de}[See \cite{Ransford}, page 28] Let $\Omega$ be a domain in $\mathbb C$. We say that a function $$ v: \Omega \to [-\infty, \infty) $$ is \textbf{upper semicontinuous (usc)} if $$ \limsup_{z\to z_0} v(z) \leq v(z_0) $$ for every $z_0\in \Omega$. An usc function $v$ on $\Omega$ is said to be \textbf{subharmonic} if it satisfies the \textbf{local submean inequality}, i.e. for every $z_0\in \Omega$, there exists $r_0>0$ with $|z-z_0|-\infty$.} \medskip \noindent \textbf{Remark 2}. \emph{An interesting class of non-continuous subharmonic functions are $\log|f|$ (see Theorem \cite[Theorem 2.2]{Ransford}), where $f$ is holomorphic.} \medskip \noindent \textbf{Remark 3}. \emph{One may check that Theorem \ref{th:MP-sub}, Proposition \ref{pr:psub}, Proposition \ref{pr:sub00} and Theorem \ref{th:GSI} also apply to non-continuous subharmonic functions.} \begin{comment} \medskip \begin{pr}\label{pr:loc} Let $\Omega$ be a domain in $\mathbb C$. If $v$ is subharmonic in a neighborhood of each point $z\in \Omega$, then it is subharmonic in $\Omega$. \end{pr} \medskip \begin{proof} Otherwise, assume that $g:=v-u$ does not satisfies the maximum principle in a domain $\Omega'\subset \Omega$ for some harmonic function $u$ in $\Omega'$, i.e. there exists $z_0\in \Omega'$ such that $$ \sup_{\Omega'} g =g(z_0), $$ but $g$ is not a constant in $\Omega'$ (i.e. $g(z_1)0$ such that $$ g(z) \leq g(z_T) -\varepsilon $$ for $z$ in a neighborhood $U$ of $\hat z$. Take $$ \phi(z):=g(z_T)- \chi(z), $$ where $\chi\in C_c(U)$ such that $0 \leq \chi \leq \varepsilon$ and $\chi(z_T) =\varepsilon$. We know that $\phi \geq g$ is continuous in $|z-z_T| \leq r$, but $$ g(z_T) > \int_0^{2\pi} \phi(z_T+re^{i\theta})\, d\theta. $$ Hence $g$ does not satisfy \eqref{eq:submean}, so $v$ can not also. We get a contradiction. \medskip \emph{Subharmonicity implies \eqref{eq:submean}}. Let us consider the Poisson integral $P_\phi$ for continuous $\phi \geq v$ in $|z-z_0| \leq r$. Now $v-P_\phi =v-\phi \leq 0$ on the boundary of that disk, so the maximum principle implies that $v-P_\phi \leq 0$ on the whole disk. In particular, we get $$ v(z_0) \leq P_\phi (z_0) = \frac1{2\pi}\int_0^{2\pi} \phi(z_0+re^{i\theta})\, d\theta, $$ take the infimum over $\phi$ we get \eqref{eq:submean}. \end{proof} \end{comment} \medskip In potential theory, we shall study a class of subharmonic functions called "potentials" (generalization of \eqref{eq:mu_K1}). \medskip \begin{de}\label{de:potential} Let $\mu$ be a Borel probability measure on a compact set $K\subset \mathbb C$. Its \textbf{potential} is the function $p_\mu: \mathbb C\to [-\infty, \infty)$ defined by \begin{equation}\label{eq:potential} p_\mu(z):= \int_K \log|z-w| \, d\mu(w). \end{equation} \end{de} \medskip \noindent \textbf{Remark.} \emph{The potential $p_\mu$ is a natural generalization of $\frac1n \log|P_n|$, where $P_n$ is a monic degree $n$ polynomial with zeros lie in $K$. In fact, if we write $$ P_n(z)=(z-z_1)\cdots(z-z_n), \ \ \ \ z_j \in K, \ \ \ 1\leq j\leq n, $$ then $$ \frac1n \log|P_n| = p_\mu, \ \text{for} \ \mu:=\frac{\delta_{z_1}+\cdots+\delta_{z_n}}{n}, $$ where $\delta_{z_j}: \phi \mapsto \phi(z_j)$ denotes the Delta measure at $z_j$.} \medskip \begin{pr}\label{pr:potential} Let $\mu$ be a Borel probability measure on a compact set $K\subset \mathbb C$. Then $p_\mu$ is subharmonic on $\mathbb C$, harmonic on $\mathbb C\setminus {\rm supp}\, \mu$, \begin{equation}\label{eq:potential00} \lim_{z\to 0} p_\mu(z^{-1})+\log|z| =0 \end{equation} and $p_\mu(z^{-1})+\log|z|$ extends to a harmonic function near $z=0$. \end{pr} \begin{proof} \emph{Step 1: $p_\mu$ is usc on $\mathbb C$ and continuous on $\mathbb C\setminus {\rm supp}\, \mu$}. First, let us show that $p_\mu$ is usc on $\mathbb C$, i.e. $$ \limsup_{z\to z_0} p_\mu(z) \leq \int_K \limsup_{z\to z_0} \log|z-w| \, d\mu(w) = p_\mu(z_0), $$ But this is a direct consequence of Fatou's lemma (for $w\in K$, $z$ close to $z_0$, apply Fatou to $C-\log|z-w|$, which is positive when $C$ is big enough). Similarly, apply Fatou to $C+\log|z-w|$, we get that $p_\mu$ is lsc on $\mathbb C\setminus {\rm supp}\, \mu$. \smallskip \emph{Step 2: $p_\mu$ satisfies the submean inequality on $\mathbb C$ and mean value property on $\mathbb C\setminus {\rm supp}\, \mu$}. Follows from the Fubini theorem, subharmonicity of $z\mapsto \log|z-w|$ and harmonicity of $z\mapsto \log|z-w|$ for $z\neq w$. \medskip Step 1 and 2 imply that $p_\mu$ is subharmonic on $\mathbb C$, harmonic on $\mathbb C\setminus {\rm supp}\, \mu$. The remaining part follows from $$ p_\mu(z^{-1})+\log|z| = \int_K \log|1-zw| \, d\mu(w). $$ (Try! Apply step 1 and 2 to this integral). \end{proof} \medskip \begin{de}\label{de:energy} Let $\mu, \nu$ be a Borel probability measures on a compact set $K\subset \mathbb C$. We shall define \begin{equation}\label{eq:energy} V_\mu:= \inf_{z\in \mathbb C} p_\mu(z), \ \ W_\mu:= \sup_{z\in K} p_\mu(z), \ \ I(\mu, \nu):=\int_K p_\mu\, d\nu. \end{equation} We call $I(\mu):= I(\mu, \mu)$ the \textbf{energy} of $\mu$. \end{de} \medskip \noindent \textbf{Remark.} \emph{Since $p_\mu$ is harmonic on $\mathbb C\setminus K$ and $p_{\mu}(z) \to \infty$ as $z \to \infty$, one may apply the maximum principle to conclude that $ V_\mu = \inf_{z\in K} p_\mu(z)$.} \medskip The main theorem in this section is the following extremal property of $\mu_K$ in \eqref{eq:mu_K}. \medskip \begin{thm}\label{th:extremal-p} Let $K$ be a compact set in $\mathbb C$. Assume that $\mathbb C\setminus K$ has real analytic boundary. Then \begin{equation}\label{eq:extremal-P} \gamma= V_{\mu_K} =W_{\mu_K}= I(\mu_K) = \sup_\mu I(\mu) =\sup_{\mu} V_\mu= \inf_{\mu} W_\mu, \end{equation} where the supremum and infimum are taken over all Borel probability measures on $K$. \end{thm} \begin{proof} $\gamma= V_{\mu_K} = W_{\mu_K}= I(\mu_K)$ follows directly from $p_{\mu_K} =\gamma$ on $K$ by \eqref{eq:mu_K1}. It remains to do the following two steps. \smallskip \emph{Step 1: $\gamma= \sup_{\mu} V_\mu = \inf_{\mu} W_\mu$}. By Fubini's theorem and \eqref{eq:mu_K1}, we have \begin{equation}\label{eq:Ahlfors-miss} \int_{K\times K} \log|z-w| \, d\mu(z) d\mu_K(w) = \int_K p_\mu \,d\mu_K = \int_K p_{\mu_K} \, d\mu = \gamma, \end{equation} for any Borel probability measure $\mu$ on $K$. Hence $$ V_\mu \leq \int_K p_\mu \,d\mu_K =\gamma \leq W_\mu $$ together with $\gamma= V_{\mu_K} =W_{\mu_K}$, we obtain $\gamma= \sup_{\mu} V_\mu= \inf_{\mu} W_\mu$. \smallskip \emph{Step 2: $\gamma= \sup_{\mu} I(\mu) $}. It suffices to show $I(\mu) \leq \gamma$ for any Borel probability measure $\mu$ on $K$. Note that \eqref{eq:mu_K1} gives $$ I(\mu, \mu_K)= \int_K p_{\mu_K} \, d\mu =\gamma = I(\mu_K), $$ hence $$ I(\mu) = I(\mu-\mu_K) -I(\mu_K) + 2I(\mu, \mu_K) = \gamma + I(\mu-\mu_K), $$ and step 2 follows from the lemma below. \textcolor{blue}{Remark (optional). \emph{A simple proof of $I(\mu-\mu_K) \leq 0$ using the language of currents: write $u=p_\mu-p_{\mu_K}$note that $dd^c (p_\mu) =dd^c(\log|z| \star d\mu) =d\mu$ (where we choose $d^c$ such that $dd^c\log|z| =\delta_0$) implies that $$ I(\mu-\mu_K) = \int_{\mathbb C} u \,dd^c u = \lim_{R\to \infty} \left( \int_{|z|=R} u \, d^c u - \int_{|z|0, $$ where $D_j$ (no intersection) are small open disks in $|z-z_0|\leq r$, we have \begin{equation}\label{eq:potential-disk} \gamma= \log r, \ \ p_{\mu_K}(z) =\max\{\log r, \log |z|\}, \end{equation} and $\mu_K = \frac{|dz|}{2\pi r}$ (we call it the normalized Haar measure the circle $|z-z_0|=r$). \end{pr} \begin{proof} Follows directly from $$ G_{\Omega_K}(z) = G_{\{|z-z_0|>r\}} (z)= -\log|\frac{z-z_0}{r}| $$ and Lemma \ref{lm:G-infty}. \end{proof} \subsection{Robin constant, capacity and transfinite diameter} We shall follow page 23-24 in \cite{A1} and page 153-154 (Fekete--Szeg\"o theorem) in \cite{Ransford} to prove that $\gamma =\log c(K)$. \medskip \begin{de}\label{de:n-diameter} Let $K$ be a compact set in $\mathbb C$. The \textbf{order $n$ diameter} ($n\geq 2$) of $K$ is defines as $$ d_n:=\sup \left\lbrace \prod_{1\leq j 0\ \ \ \text{on} \ \mathbb C. \end{equation} We shall introduce the following definition. \medskip \begin{de} A function $\psi$ on an open set $U\subset \mathbb C$ is said to be \textbf{$\phi$-subharmonic} if $\phi+\psi$ is subharmonic on $U$. The space of $\phi$-subharmonic functions on $U$ is denoted by ${\rm Sh} (U, \phi)$. For $\psi \in {\rm Sh} (U, \phi)$, the \textbf{Lelong number} of $\psi$ at $z_0\in U$ is defined as $$ \nu_{z_0} (\psi):= \sup\{c\geq 0: \psi \leq c\log |z-z_0|+ O(1)\}, $$ where the inequality is to be understood as meaning there is a constant $C$ such that $$\psi(z) \leq c\log |z-z_0|+C$$ for $z$ near $z_0$. \end{de} \medskip \noindent \textbf{Remark.} \emph{We observe the supremum is actually attained, so if $v_{z_0}(\psi) =t$ then $\psi \leq t\log |z-z_0|+ O(1)$. To see this, let $B$ be a small disk centered at $z_0$, for every $c0 $$ on $B\setminus\{z_0\}$ and define $\rho_\varepsilon(0) =-\infty$. Since $\rho$ is bounded above near $z_0$, we know that $\rho_\varepsilon(z) \to -\infty$ as $z \to z_0$. Thus $\rho_\varepsilon+\phi$ is usc at $z_0$ and satisfies the sub-mean inequality for small disks centered at $z_0$. Hence $\rho_\varepsilon \in {\rm Sh} (B, \phi)$, which implies that the usc regularization, say $\tilde \rho$, of $\lim_{\varepsilon \to 0} \rho_\varepsilon $ lies in ${\rm Sh} (B, \phi)$. Note that $\tilde \rho$ is an extension of $\rho$. The proof is complete. \end{proof} \medskip The Lelong number is closely related to the Laplacian operator $\Delta$ by the following lemma. \medskip \begin{lm} For smooth function $\chi$ with compact support in $\mathbb C$, \begin{equation}\label{eq:lelong} \chi(z_0) = \int_{ \mathbb C} \frac{\log|z-z_0| \Delta \chi}{2\pi}\, dxdy. \end{equation} \end{lm} \medskip \begin{proof} By a simple translation, it is enough to prove the $z_0=0$ case. Note that $$ \lim_{\varepsilon \to 0} \int_{|z| <\varepsilon} \log |z|^2 dxdy = \lim_{\varepsilon \to 0} 2\pi \int_0^\varepsilon (\log r^2) \, rdr = \lim_{\varepsilon \to 0} (\pi \varepsilon^2 \log \varepsilon^2 -\pi \varepsilon^2)=0, $$ we have (note that $\chi$ has compact support) $$ \int_{\mathbb C} \frac{\log|z|^2 \Delta \chi}{4\pi}\, dxdy = \lim_{\varepsilon \to 0} \int_{\varepsilon<|z| R$. By Green's formula, we have $$ \int_{\varepsilon<|z| 0$ then $$ \psi_t(z) = \rho_t(z):= \begin{cases} t (\log |z|^2-\log t) +t-|z|^2 & \text{if} \ |z|^2 0$ then $\rho_t \leq \psi_t \leq 0$; \smallskip \item[(3)] If $t> 0$ then $\psi_t +\phi$ is subharmonic on $\mathbb C$; \smallskip \item[(4)] If $t> 0$ then $\psi_t +\phi -2\log|z|$ is harmonic on $\Omega_t$; \smallskip \item[(5)] Any first order derivatives of $\psi_t$ are Lipschitz continuous on $\mathbb C\setminus \{0\}$; \smallskip \item[(6)] The Lebesgue measure of the boundary of $\Omega_t$ is zero; \smallskip \item[(7)] If $t> 0$ then $$\frac{ \Delta \psi}{4\pi} = t\delta_{0}-\chi_{\Omega_t} \frac{ \Delta \phi}{4\pi} $$ in the sense of distribution on $\mathbb C$, where $$ \chi_{\Omega_t}(z) := \begin{cases} 1 & \text{if} \ z\in \Omega_t \\ 0 & \text{if} \ z\notin \Omega_t \end{cases} $$ is the characteristic function of $\Omega_t$; \smallskip \item[(8)] If $t> 0$ then $\int_{\Omega_t} \frac{ \Delta \phi}{4\pi}\, dxdy =t $; \item[(9)] If $t> 0$ then $\Omega_t$ is a domain with $0\in\Omega_t \subset \{|z|^2