\documentclass{amsart}
\usepackage{calculus}
\usepackage[margin=3cm]{geometry}
\thispagestyle{empty}

\begin{document}
\title{Assignment 8: Spectral theory}

\date{Assigned Wed 05/23. Due Wed 05/30.}
\maketitle

\begin{questions}
\item
\begin{parts}
\item Let $K\in B(X, X)$ be compact Hermitian. Show that $\im(I + K)$ is closed, and $X/\im(I+K)$ is finite dimensional. Further show $\dim\ker(I + K) = \dim(X/\im(I+K))$. \noteb{This is still true if $K$ is not Hermitian, but of course false if $K$ is not compact.}
\item Let $K\in B(X, X)$ be compact Hermitian. Show that $\sigma(K) - \{0\}$ consists of only eigenvalues. \noteb{This is still true if $K$ is not Hermitian, but of course false if $K$ is not compact.}
\item Let $K\in B(X,X)$ be compact. Show that if $X$ is infinite dimensional $0 \in \sigma(K)$.
\item Let $k:[0,1]\times[0,1] \to \C$ be a continuous function such that $k(x,y) = \bar k(y,x)$. Then for any $\lambda \in \C - \{0\}$ show that either for all $g \in \Lpspace{2}([0,1])$ the (functional) equation $\int k(x,y) f(y) dy - \lambda f(x) = g(x)$ has a unique solution, or the (functional) equation $\int k(x,y) f(y) - \lambda f(x) = 0$ has a nonzero solution. \noteb{Note that in the equations above, both the left and right hand sides are elements of $\Lpspace{2}$, and it does not make sense to talk about the values of these functions for any particular $x$. Thus when we say the (functional) equation $f(x) = g(x)$ holds, we mean that $f$ and $g$ are equal as elements of $\Lpspace{2}$. This result is true even without the assumption that $k(x,y) = \bar k(y,x)$, and is called the \textit{Fredholm Alternative}. }
\end{parts}

\item Let $\Omega \subset \C$ be open. Recall $f:\Omega \to \C$ is \textit{holomorphic} if for all $z \in \Omega$, $\lim_{\zeta \to z} \frac{f(\zeta) - f(z)}{\zeta - z}$ exists. From standard complex analysis, we recall that holomorphic functions can be represented by power series. Namely for any $z \in \Omega$, there exists a sequence $(c_n)_n$ of complex numbers such that $f(\zeta) = \sum c_n (\zeta - z)^n$, and this series has radius of convergence $\rho = \inf\{ \abs{z-z'} \st z' \not\in \Omega \}$.
\begin{parts}
\item Let $A \in B(X,X)$, and $f:\C \to \C$ be holomorphic. Show that $f(A) \in B(X,X)$, and $f(\sigma(A)) \subset \sigma(f(A))$. \noteb{We define $f(A) \in B(X, X)$ as follows: Since $f$ is holomorphic on all of $\C$, we write $f(z) = \sum c_iz^i$ for all $z \in \C$. We know that this series converges absolutely for all $z \in \C$, and thus we define $f(A) \in B(X, X)$ by $f(A) = \sum c_i A^i$.}
\uplevel{For those familiar with a little more complex analysis, the above can be extended to meromorphic functions.}
\item Let $f:\C \to \C$ be meromorphic, such that all poles of $f$ are outside $\sigma(A)$. Show that $\sigma(f(A)) = f(\sigma(A))$. \noteb{The definition of $f(A)$ is possibly less transparent in this situation: Let $\{\mu_k\}$ be the (discrete) set of poles of $f$. Then we know that there exists natural numbers $n_k$, and holomorphic entire functions $g, g_k$ such that $f(z) = g(z) + \sum \frac{1}{(z - \mu_k)^{n_k}} g_k(z)$. Now $g_k(A)$ is defined as in the previous subpart. Now since $\mu_k \not \in \sigma(A)$, $\frac{1}{A - \mu_k I}$ can be defined as $(A-\mu_k)\inv$. Thus we define $f(A) = g(A) + \sum ((A - \mu_k I)\inv)^{n_k} g_k(A)$. }
\item If $n \in \Z$, show that $\sigma(A^n) = \{ \lambda^n \st \lambda \in \sigma(A) \}$. \noteb{If $n < 0$, you can assume $A$ is invertible. If you couldn't figure out the last subpart, do this one explicitly. If you got the last subpart, this should be one line.}
\end{parts}

\item In the last homework we saw $\lap \inv$ (defined for \textit{periodic} functions) was compact. This is a general principle, valid in a more general setting. We prove this here for functions defined on the unit disk in $\C$, which is generally the first step towards proving it for general domains.

Let $D = \{z \in \C \st \abs{z} < 1 \}$, and $S^1 = \{z \in \C \st \abs{z} = 1 \}$. We define $C^2(D)$ to be the set of all functions which are twice differentiable on $D$, and continuous on $\bar{D}$. We say $u \in C^2$ is \textit{harmonic} if $\lap u = \partial^2_x u + \partial^2_y u = 0$. Our first goal is to show that given $f:S^1 \to \C$, find a harmonic function $u:\bar D \to \C$ such that $u|_{S^1} = f$.
\begin{parts}
\item Let $n \in \N$. Show that $u(z) = z^n$ and $v(z) = \bar z^n$ are Harmonic.
\uplevel{Given $f:\R \to \C$ periodic, we can identify it with a function $F:S^1 \to \C$, by defining $F(e^{i\theta}) = f(\theta)$. Thus for $F:S^1 \to \C$, we define $\hat F(n) = \frac{1}{\sqrt{2\pi}} \hat f(n) = \frac{1}{2\pi} \int_0^{2\pi} f(\theta) e^{-in\theta} = \frac{1}{2\pi i} \oint_{S^1} F(z) z^{-n}\, \frac{dz}{z}$.}
\item If $F \in L^2(S^1)$, show that $F(z) = \sum \hat{F}(n) z^n$.
\uplevel{To extend $F:S^1\to \C$ to a harmonic function $u:\bar{D} \to \C$, one would be tempted to write $u(z) = \sum \hat{F}(n) z^n$. But unfortunately, this causes problems at the origin when $n < 0$. Thus we realize that when $\abs{z} = 1$, $\bar{z} = z\inv$, and hence $F(z) = \hat F(0) + \sum_1^\infty (\hat F(n) z^n + \hat F(-n) \bar z^n)$. This expression yields itself to a natural harmonic extension inside $D$. }
\item Let $F:S^1 \to \C$ be continuous. Define $u:D \to \C$ by $u(z) = \hat F(0) + \sum_1^\infty \hat F(n) z^n + \hat F(-n) \bar{z}^{n}$. Show that $u$ is harmonic.
\item When $r < 1$, show that $u(re^{i\theta}) = \int_0^{2\pi} f(\phi) P_r(\theta - \phi) \, d\phi$, where $P_r(\phi) = \frac{1}{2\pi} \frac{1 - r^2}{1 + r^2 -2r\cos\phi}$. $P_r$ is called the Poisson Kernel. We abuse notation and define $f * P:D \to \C$ by $f * P(r e^{i\theta}) = u(r e^{i\theta}) = \int_0^{2\pi} f(\phi) P_r( \theta - \phi) \,d\phi$.
\item Show that $P_r$ is an approximate identity. Namely, show that for $r \in [0,1)$, $\int P_r(\theta) \,d\theta = 1$, $P_r(\theta) \geq 0$, and for any $\epsilon > 0$, $\dlim_{r \to 1^-} \int_{\abs{\theta} > \epsilon} P_r(\theta) \, d\theta = 0$.
\item Let $F \in \Lpspace{p}(S^1)$. Show that $(u_r) \to F$. Explicitly, show that $\dlim_{r \to 1^-} \int \abs{u(re^{i\theta}) - F(e^{i\theta})}^p \,d\theta = 0$. \noteb{Thus you can think of $u:\bar{D} \to \C$ to be a harmonic function such that $u|_{S^1} = F$ in the $\Lpspace{p}$ sense.}
\item If $F:S^1 \to \C$ is continuous, show that $u:\bar{D} \to \C$ is continuous, (even at the boundary of $D$), and $u|_{S^1} = F$.
\item If $\hat{F} \in \lpspace{\infty}$, show that $\lap u = 0$.
\item \textit{(Uniqueness)} If $F \in \Lpspace{p}$, and $u, v$ satisfy $\lap u = \lap v = 0$ inside $D$, and $u|_{S^1} = v|_{S^1} = F$ (in the sense of part (f)), then show $u = v$. \noteb{This is a little tricky with no hint.}
%\uplevel{If $F:S^1 \to \C$, we define $u_F$ to be the Harmonic function $u:D\to \C$ (constructed above) such that $u|_{S^1} = F$. We now attempt to invert the Laplacian on $D$.}
%\item Let $g:\bar{D} \to \C$ be continuous, with $g|_{S^1} = 0$. Define $u_g:\C \to \C$ by $u_g(x) = \int_\C g(y) \frac{1}{2\pi} \ln \abs{x - y}$.
\end{parts}
\end{questions}

On your next problem set, we will use problem 3 above to construct the inverse of the Laplacian on the unit disk. We will use our explicit construction to show that the inverse is a compact Hermitian operator, and then apply the spectral theorem to show that the eigenvalues of the Laplacian on the disk are a discrete set of negative reals, which diverge to infinity. This (with some physics) will explain why if you build a circular drum, frequencies it can produce are discrete, with a (strictly) positive lowest frequency the drum can produce. In this case it turns out that audible frequencies are multiples of this lowest frequency.
\end{document}