
\documentclass{article}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\usepackage{amsmath}

\setcounter{MaxMatrixCols}{10}
%TCIDATA{OutputFilter=LATEX.DLL}
%TCIDATA{Version=5.00.0.2606}
%TCIDATA{<META NAME="SaveForMode" CONTENT="1">}
%TCIDATA{BibliographyScheme=Manual}
%TCIDATA{Created=Sunday, November 13, 2005 09:50:03}
%TCIDATA{LastRevised=Sunday, January 08, 2006 09:29:40}
%TCIDATA{<META NAME="GraphicsSave" CONTENT="32">}
%TCIDATA{<META NAME="DocumentShell" CONTENT="Standard LaTeX\Blank - Standard LaTeX Article">}
%TCIDATA{CSTFile=40 LaTeX article.cst}

\newtheorem{theorem}{Theorem}
\newtheorem{acknowledgement}[theorem]{Acknowledgement}
\newtheorem{algorithm}[theorem]{Algorithm}
\newtheorem{axiom}[theorem]{Axiom}
\newtheorem{case}[theorem]{Case}
\newtheorem{claim}[theorem]{Claim}
\newtheorem{conclusion}[theorem]{Conclusion}
\newtheorem{condition}[theorem]{Condition}
\newtheorem{conjecture}[theorem]{Conjecture}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{criterion}[theorem]{Criterion}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{exercise}[theorem]{Exercise}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{notation}[theorem]{Notation}
\newtheorem{problem}[theorem]{Problem}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{solution}[theorem]{Solution}
\newtheorem{summary}[theorem]{Summary}
\newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}}
\input{tcilatex}

\begin{document}


\begin{center}
Inverse Problems
\end{center}

\bigskip

\noindent Fredholm Integral of the first kind%
\begin{equation*}
g(x,y)=\tint\limits_{0}^{1}\tint\limits_{0}^{1} k(x-x^{\prime
},y-y^{\prime})f(x^{\prime },y^{\prime }) dx^{\prime }dy^{\prime }=Kf(x,y)
\end{equation*}

\noindent Fourier Transform

\begin{align*}
\widehat{g} &= \widehat{K}\widehat{f} \\
\widehat{f} &= \widehat{K}^{-1}\widehat{g}
\end{align*}

\noindent So%
\begin{equation*}
f=\mathcal{F}^{-1}\left\{ \frac{\mathcal{F}(g)}{F(K)}\right\}
\end{equation*}

\noindent example: atmospheric turbulence%
\begin{equation*}
k(x,y)=Ce^{-\frac{x^{2}+y^{2}}{2\gamma ^{2}}}
\end{equation*}

\noindent telescope

\begin{equation*}
k(x,y)=\left\vert \mathcal{F}^{-1} \left\{ Ae^{i\varphi }\right\}
\right\vert^2
\end{equation*}

\noindent A=aperture \qquad $\varphi$ is the phase

\bigskip

\noindent Given $k$ and $g$ find $x$

\noindent Problems:

\begin{itemize}
\item $K$ is ill=conditioned

\item Noise
\end{itemize}

\bigskip

\begin{definition}
Well posed: 
\begin{equation*}
(\ast) \qquad Kf=g \qquad L: H_1 \rightarrow H_2
\end{equation*}

\begin{enumerate}
\item for each $g\in H_{2}$ there exists $f\in H_{1}$ where $(\ast )$ holds

\item f is unique

\item Stability

\noindent If $K f_{\ast} = g_{\ast}$

\noindent \quad $K f = g$

\noindent and $g\rightarrow g_{\ast }$ then $f\rightarrow f_{\ast }$
\end{enumerate}
\end{definition}

\bigskip

\begin{center}
Tikhonov Regularization
\end{center}

\bigskip

\begin{itemize}
\item replace $(\ast$ by least square minimization. This is equivalent to 
\begin{equation*}
K^{\ast} K f = K^{\ast} g
\end{equation*}

\item Regularize 
\begin{equation*}
\left( K^{\ast }K+\alpha I\right) f=K^{\ast }g
\end{equation*}%
\noindent So%
\begin{equation*}
f_{\alpha }=\left( K^{\ast }K+\alpha I\right) ^{-1}K^{\ast }g
\end{equation*}

\item iterate regularization
\end{itemize}

\noindent More generally we consider%
\begin{align*}
f_{\alpha } &=\arg \min ||Kf-g||^{2}+\underbrace{\alpha ||Lf||^{2}} \\
&\qquad \qquad \qquad \qquad \text{penalty function}
\end{align*}

\noindent Even more general%
\begin{equation*}
f_{\alpha }=\arg \min ||Kf-g||^{2}+\rho \left( f\right)
\end{equation*}
So the penalty function is given by 
\begin{equation*}
\rho(f)
\end{equation*}

\noindent For example 
\begin{equation*}
\rho (f)=||Lf||^{2}\qquad \text{in some norm}
\end{equation*}%
\noindent e.g. 
\begin{equation*}
\rho (f)=\frac{1}{2}\tiint \sum_{i=1}^{d}\left( \frac{\partial f}{\partial
x_{i}}\right) ^{2}dx_{i}
\end{equation*}

\noindent This penalizes non-smooth functions. Integrate by parts%
\begin{equation*}
\int \frac{\partial f}{\partial x}\frac{\partial f}{\partial x}dx=-\int f%
\frac{\partial ^{2}f}{\partial x^{2}}dx+\text{boundary terms}
\end{equation*}

\noindent Let $Lf=-\Delta f$ . Then%
\begin{equation*}
\rho (f)=\frac{1}{2}\tiint f\,Lf\,dx
\end{equation*}

\end{document}
