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\begin{document}
\title{Assignment 9: Harmonic functions, and spectral theory}

\date{Assigned Wed 05/30, Due Wed 06/06.}
\maketitle

\begin{questions}
\item Let $A \in B(X, X)$. Define $R_A: \C - \sigma(A) \to B(X,X)$ by $R_A(\lambda) = (A - \lambda I)\inv$. Show that $R_A$ is strongly holomorphic. \noteb{I never carried out the details completely in class.}

\item
\begin{parts}
\item Let $A \in B(X, X)$ be such that $A^n = 0$ for some $n \in \N$. Show that $I + A$ is invertible. \hintb{What is $\sigma(A)$?}
\item (Unrelated) If $\rho(A) < 1$, then show that $(A^n) \to 0$. Is the converse true? Could $\norm{A} \geq 1$?
\end{parts}

\item
\begin{parts}
\item Say $f: \Omega \to \C$ is holomorphic, and $u = \Re(f)$, $v = \Im(f)$ are the real and imaginary parts respectively. Show that $u$ and $v$ are harmonic.
\uplevel{The converse is \textit{almost} true. We deal with this next. We define the \textit{Hilbert transform} $H:\Lpspace{2} \to \Lpspace{2}$ by $Hf = \hat f(0) - i\sum_1^\infty (\hat f(n) e_n - \hat f(-n) e_{-n})$, where $e_n(x) = e^{2 \pi i n x}$. There are a few amazing facts related to the Hilbert transform: The first is that Hilbert transform can be expressed as a convolution with $\pi \cot \pi x$! Now inspection of this function shows immediately that it is not in any $\Lpspace{p}$, so you can't apply Young's inequality to show that this is bounded. Amazing fact 2: For all $1 < p < \infty$, $H \in B(\Lpspace{p}, \Lpspace{p})$. This is false when $p = \infty$, and a `weak' version is true when $p = 1$ (dejavu?). The proof when $p \neq 2$ is quite deep, and is the foundations of modern harmonic analysis. When $p = 2$, it's quite easy \smiley.}
\item Show that $H \in B(\Lpspace{2}, \Lpspace{2})$.
\item Say $u:D \to \C$ is harmonic, and $u$ converges (in the sense of 3(f) from last homework) to some $\Lpspace{2}$ function $F$ on the boundary of $D$. Let $G = H(F)$, and $v$ be the unique harmonic function $v:D \to \C$ such that $v$ converges (in $\Lpspace{2}$) to $G$ on the boundary of $D$. Show that $f = u + i v$ is holomorphic! Further if $u$ is real valued, then show that $v$ is also real valued. \noteb{Amazing fact 3: The boundary values of the real and imaginary part of a holomorphic function are related by the Hilbert transform.}
\sitems Let $u(z) = \ln \abs{z}$. Show that $u:D - \{0\} \to \R$ is harmonic, but there does not exist $f:D - \{0\} \to \C$ such that $f$ is holomorphic and the real part of $f$ equals $u$.
\end{parts}

\item This is a continuation of Problem 3 on your previous homework:
\begin{parts}
\item Let $\Omega \subset \C$ be an open subset such that the boundary $\del \Omega$ can be parametrized by a $C^1$ curve. Suppose $u, v:\Omega \to \C$ are $C^2$ on the interior of $\Omega$, continuous on $\bar{\Omega}$, and have a continuous normal derivative on the boundary $\del \Omega$. Show that $\int_\Omega (u \lap v - v \lap u) = \int_{\del\Omega} (u \frac{\partial v}{\partial n} - v \frac{\partial u}{\partial n})$. \noteb{$\frac{\partial u}{\partial n}$ denotes the outward normal derivative of $u$.}

\uplevel{\noindent Given $f\in \Lpspace{2}(D)$, we're going to define $\lap\inv f$ to be a function such that $\lap u = f$, and $u|_{S^1} = 0$. The intuition is as follows: Suppose $G:D \times D \to \R$ is such that $G(y,x) = G(x,y) = 0$ for all $x \in D$, $y \in S^1$, and $\int u(x) \lap_x G(x, y) dx = u(y)$. (Note $x,y \in D$. Thus if $x = (x_1, x_2)$, by $\lap_x G(x,y)$ we mean $\frac{\del^2}{\del x_1^2} G(x,y) + \frac{\del^2}{\del x_2^2} G(x,y)$.) Now the previous subpart immediately shows $u(y) = \int_D G(x,y) f(x) dx$. Thus given $f$, we can define $\lap\inv f$ by $\lap\inv f(y) = \int_D G(x,y) f(x) dx$, once we find the function $G$. The function $G$ is called the Greens function which we construct below.}

\item Let $N:\R^2 - \{0\} \to \R$ be defined by $N(x) = \frac{1}{2\pi}\ln(\abs{x})$. Suppose $u \in C^2(\R^2)$ is such that $\lim_{\abs{x}\to\infty} \abs{u(x)} = 0$. Show that $\int_{\R^2} u \lap N = \int_{\R^2} N \lap u = u(0)$. Note: Both functions under the integral `blow up' at $0$. Thus by $\int_{\R^2} (\cdots)$ we mean $\lim_{\epsilon\to 0} \int_{\R^2-B_\epsilon} (\cdots)$. \hintb{First check $\lap N = 0$. Next pick $\epsilon > 0$, and show that $\int_{\R^2 - B_\epsilon} N \lap u = \int_{\del B_\epsilon} (N \frac{\del u}{\del n} - u\frac{\del N}{\del n})$. Now, show that the first term converges $0$ as $\epsilon \to 0$, and the second term converges to $u(0)$. $N$ is called the \textit{Newton potential}.}

\uplevel{\noindent We define $h: \bar D \times \bar D \to \R$ as follows: For any $y \in D$, let $h(x, y)$ be the unique continuous function so that $\lap_x h = 0$ in $D$, and for any $x \in S^1$, $h(x,y) = N(x-y)$. From your previous homework, you know $h(re^{i\theta}, y) = \int P_r(\phi) N(e^{i(\theta - \phi)} - y) \, d\phi$.

Let $\delta = \{(x,x) \st x \in \bar{D} \}$, and define $G:\bar D \times \bar D - \delta \to \R$ by $G(x,y) = N(x-y) - h(x,y)$.

Given $f:\bar D \to \R$ continuous, define $u:\bar{D} \to \R$ by $u(y) = \int_D G(x,y) f(x) dx$. Ideally we would like to show that $\lap u = f$, and $u|_{S^1} = 0$. We do this in a somewhat indirect manner:}
\item If $v \in C^2(D)$ is such that $v|_{S^1} = 0$. Show that $\int_D u \lap v = \int_D f v$.
\item Show that $u|_{S^1} = 0$.
\item If $u \in C^2(D)$, show that $\lap u = f$.

\uplevel{If $f \in \Lpspace{2}(D)$, we define $\lap\inv f$ by $\lap\inv f(x) = \int_D G(x,y) f(y) dy$. Assignment 7, problem 5 almost immediately shows that $\lap\inv \in B(\Lpspace{2}, \Lpspace{2})$ is a compact operator.}
\item Show that $\lap\inv \in B(\Lpspace{2}, \Lpspace{2})$ is Hermitian.
\uplevel{There are of course infinitely many consequences of the fact that $\lap\inv$ is a compact Hermitian operator (e.g. most of Physics). We end with one `cute' application (the last subpart).}
\item Suppose $u \in C^2(D)$ is such that $\lap u = \lambda u$, with $u|_{S^1} = 0$. Show that $u$ is radial (i.e. $u(z) = u(\abs{z})$). \hintb{First check that for any $\theta \in \R$, $v(z) = u(e^{i\theta} z)$ satisfies $\lap v = \lambda v$. Then use the fact that the eigenspaces of $\lap\inv$ must be finite dimensional.}
\item Suppose all $u \in L^2(D)$ such that $\lap u = \lambda u$, are in fact $C^2$ functions such that $u|_{S^1} = 0$, and are differentiable enough so that $\lap\inv \lap u = \lap \lap \inv u = u$. Compute all eigenvalues (with multiplicities) of $\lap\inv$. \noteb{Showing that the eigenfunctions are $C^2$ standard Sobolev tricks, but it involves the continuous Fourier transform which we didn't do in class.}
\item Given $u_0 \in \Lpspace{2}$, show that there exists $u:\bar D \times [0,\infty)$ such that
\begin{enumerate}
\item $u(x, 0) = u_0(x)$ for all $x \in \bar{D}$.
\item $u(x, t) = 0$ for all $x \in S^1$.
\item $\del_t u - \lap u = 0$ on the set $D \times (0, \infty)$.
\end{enumerate}
\hintb{Use the spectral theorem to write $u(x, t) = \sum c_i(t) f_i(x)$, where $(f_i)$ is some orthogonal complete sequence of eigenfunctions of $\lap\inv$. Your result about the eigenvalues of $\lap$ will help you justify term by term differentiation.}
\end{parts}
\end{questions}

\section{Power series representations of holomorphic functions}%{{{
Since I couldn't figure out up one detail of this in class, here's the proof:
\begin{theorem}
Let $f:\Omega \to \C$ be holomorphic. Suppose $z_0 \in \Omega$, and $\bar B_R(z_0) \subset \Omega$. Then there exists a sequence $(a_n)$ such that $f(z) = \sum a_n (z-z_0)^n$ for all $z \in B_R(z_0)$.
\end{theorem}
\begin{proof}
Here's a simpler version of the proof I was doing in class: Assume $z_0 = 0$, and let $\Gamma = \del B_R(0)$. Then by the Cauchy integral formula,
$$f(z) = \frac{1}{2\pi i} \oint_\Gamma f(\zeta) \,\frac{d\zeta}{\zeta - z}.$$
Note that when $\abs{z} < \abs{\zeta}$, we know
$$\frac{1}{\zeta - z} = \frac{1/\zeta}{1 - z/\zeta} = \sum \frac{z^n}{\zeta^{n+1}}$$
Note also that for fixed $z$, this series converges uniformly (in $\zeta$) on $\Gamma$. Thus term by term integration is legal, and hence
$$f(z) = \frac{1}{2\pi i} \sum \oint_\Gamma \frac{f(\zeta)z^n}{\zeta^{n+1}}\,d\zeta.$$
so for $a_n = \frac{1}{2\pi i} \oint \frac{f(\zeta)}{\zeta^{n+1}} \,d\zeta$, we have $f(z) = \sum a_n z^n$ for all $\abs{z} < R$.
\end{proof}%}}}
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