Lecture17 - \documentclass[12pt,letterpaper]cfw_article

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\documentclass[12pt,letterpaper]{article} \ \usepackage{amsmath} \usepackage{amssymb} \usepackage{latexsym} \usepackage{array} \usepackage{enumerate} \usepackage{amsthm} \usepackage{amscd} %\usepackage{verbatim} % \newtheorem{lemma}{Lemma} \newtheorem{theorem}{Theorem} \newtheorem{definition}{Definition} \newtheorem*{remark}{Remark} \ \newcommand{\bC}{{\mathbb C}} \newcommand{\bH}{{\mathbb H}} \newcommand{\bR}{{\mathbb R}} \newcommand{\ra}{\rightarrow} \newcommand{\Ra}{\Rightarrow} \ \newcommand{\ramp}{\mapsto} \newcommand{\myU}{$U$} \newcommand{\CartProd}{X_{1}\times X_{2}} \ \date{} \ \begin{document} \ \title{\bf Lecture 17} \author{} \maketitle \ \noindent {\bf Example}: (Rudin, Chp. 9) Define $f:\bR \ra \bR$ by \[ f(t) = \left\{ \begin{array}{rl} 0 &t = 0 \end{array} \right. . \] Then $f$ is differentiable and $f'(0) = 1$, yet there is no open interval about $0$ where $f$ is increasing. The lack of continuity of $f'$ at $0$ allows $f$ to be decreasing on arbitrarily small subintervals about $0$. The Inverse Function Theorem {\bf fails} because $f$ is not injective near $0$. \newline \ \noindent {\bf Example}: (Diedonn\'{e}) Define $H: \bR^2 \ra \bR^2$ by \[H(x,y) = (x,g(x,y))\] Where, \[ g(x,y) = \left\{ \begin{array}{rl} \dfrac{y(y-x^2)}{x^2} & y^2 \ge y \ge 0 \\
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\end{array} \right. \] \newline Then in {\bf fact} \begin{itemize} \item $H$ is differentiable on $\bR^2$ \item $H'_{(0,0)} = I$ \item $H$ is not injective on any open set about $(0,0)$. \end{itemize}
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This note was uploaded on 11/09/2011 for the course MAT 6932 taught by Professor Staff during the Spring '10 term at University of Florida.

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Lecture17 - \documentclass[12pt,letterpaper]cfw_article

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