Lecture08

# Lecture08 -...

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Unformatted text preview: \documentclass[12pt,letterpaper]{article} \ \usepackage{amsmath} \usepackage{amssymb} \usepackage{latexsym} \usepackage{array} \usepackage{enumerate} \usepackage{amsthm} \usepackage{amscd} \usepackage{amsfonts} \ \newtheorem{lemma}{Lemma} \newtheorem{theorem}{Theorem} \newtheorem{definition}{Definition} \ \newcommand{\bC}{{\mathbb C}} \newcommand{\bH}{{\mathbb H}} \newcommand{\bR}{{\mathbb R}} \newcommand{\ra}{\rightarrow} \newcommand{\Ra}{\Rightarrow} \ \date{} \ \begin{document} \ \title{\bf Lecture 8} \author{} \maketitle \ \noindent \textbf{Example:} Recall that if $F:U \ra Y$ and $G:U \ra Z$ are differentiable at $a \in U$, then so is $H=(F,G):U \ra Y \times Z$ and, further, $H^{'}_a(h)=(F^{'}_a(h),G^{'}_a(h)).$ Conversely, if $H$ is differentiable at $a$ then so are $F$ and $G$. \\ \\ \noindent [Let $P:Y \times Z \ra Y$ be first-factor projection: $P(y,z)=y$. This is linear and bounded (if $Y \times Z$ has the norm given by, say, $\| (y,z) \| = \| y \| \vee \| z \|$ then $\| P(y,z) \| \le \|(y,z)\|$); Thus, $P \in L(Y \times Z, Y)$ and so $P$ is differentiable, with $P^{'}_{(b,c)} = P$ for $b \in Y , c \in Z$. Now $F=P \circ H$ is differentiable at $a$ by the chain rule and $\begin{array}{rcl} F^{'}_a(h) &=& P^{'}_{(F(a),G(a))} \circ H^{'}_a(h) \\ &=& P(H^{'}_a(h)) \end{array}$ A similar argument applies to $G$.] \\ A \noindent \textbf{Consequence:} a map $F: \bR^m \ra \bR^n$ is differentiable at $a \in \bR^m$ iff its components are all differentiable at $a$....
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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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Lecture08 -...

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