Lecture12

# Lecture12 - \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 12} \author{} \maketitle \ \begin{definition} Let $U$ be open in $X$. When $F:U \ra Y$ is differentiable at all points $a \in U$, its {\bf derivative} is the map $F' : U \ra L(X,Y) : a \ramp F'_{a}$ and we call $F$ {\bf continuously differentiable} when $F'$ is continuous. \end{definition} \ \begin{theorem} Suppose $U$ contained in $\CartProd$; then $F:U\ra Y$ is {\bf continuously differentiable} if and only if $\partial_{1}F$ and $\partial_{2}F$ are defined and continuous at each point of $U$. \end{theorem} \ \begin{proof} "Scribal Exercise" \newline For any $a = (a_1,a_2) \in U$, let $||a|| = \max\left\{||a_1||,||a_2|| \right\}$.\newline $\left(\Rightarrow\right)$ \newline First, suppose that $F$ is continuously differentiable. This means that $F'_a$ is defined for all $a \in U$ and that $F'$ is continuous at $a$. By the last theorem of Lecture 9'' $\partial_1 F_a$ and $\partial_2 F_a$ exist at every $a \in U$ and $F'_a(h_1,h_2) = \partial_1 F_a(h_1) + \partial_2 F_a(h_2) .$ Specifically, \begin{align*} Now consider the maps defined below: \begin{align*} \alpha_1 : X_1 & \ra X_1\times X_2 : h_1 \ramp (h_1,0) \\

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\alpha_2 : X_2 & \ra X_1\times X_2 : h_2 \ramp (0,h_2). \end{align*} $\alpha_1$ and $\alpha_2$ are linear and \begin{align*} \left\| \alpha_1(h_1) \right\| & = \left\| (h_1,0) \right\| = \left\| h_1 \right\| \\ \left\| \alpha_2(h_2) \right\| & = \left\| (0,h_2) \right\| = \left\| h_2 \right\| . \end{align*} It follows that $\left\| \alpha_1 \right\| = 1 = \left\| \alpha_2 \right\|$. We can see that \begin{align*} Since $F'$ is continuous at each $a$ let $\varepsilon > 0$ and choose $\delta$ such that if $b \in U$ with $||a - b|| < \delta$ then $\left\| F'_a - F'_b \right\| < \varepsilon,$ thus for $i = 1$ or $2$ \begin{align*} \left\| \partial_i F_a - \partial_i F_b \right\| & = \left\| F'_a \circ \alpha_i - F'_b \circ \alpha_i \right\| \\ & = \left\| \left( F'_a - F'_b \right) \circ \alpha_i \right\| \\ & \le \left\| F'_a - F'_b \right\| \left\| \alpha_i \right\| \\ & = \left\| F'_a - F'_b \right\|
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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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Lecture12 - \documentclass[12pt,letterpaper]cfw_article

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