Lecture03 - \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} \ \newtheorem{lemma}{Lemma} \newtheorem{theorem}{Theorem} \newtheorem{definition}{Definition} \newtheorem*{remark}{Remark} \renewcommand{\leq}{\leqslant} \renewcommand{\geq}{\geqslant} \renewcommand{\epsilon}{\varepsilon} %\renewcommand{\implies}{\Rightarrow} \newcommand{\field}[1]{\ensuremath{\mathbb #1}} \newcommand{\abs}[1]{\ensuremath{\left\vert#1\right\vert}} \newcommand{\norm}[1]{\ensuremath{\left\Vert#1\right\Vert}} \newcommand{\set}[1]{\ensuremath{\left\{ #1 \right\}}} \newcommand{\mat}[1]{\ensuremath{\begin{bmatrix}#1\end{bmatrix}}} \newcommand{\R}{\field{R}} \newcommand{\C}{\field{C}} \ \date{} \ \begin{document} \title{\bf Lecture 3} \author{} \maketitle Recall that if \[A=\set{\norm{T(x)} : x\in X \text{ and } \norm{x} \leq 1 }\] and \[B=\set{M : x\in X \implies \norm{T(x)} \leq M\norm{x}},\] we claim that \[\sup A = \inf B.\] We have already shown that $\sup A \leq \inf B$, so all that remains is to prove that $\sup A \geq \inf B$. \begin{proof} Suppose $\sup A \not\in B$. Then there is an $x \in X$ such that \[\norm{T(x)} > (\ sup A)\norm{x};\] so $x\neq 0$ and \[\norm{T\left(\frac{x}{\norm{x}}\right)} > \sup A.\] But $\norm{T\left(x/\norm{x}\right)} \in A$, a contradiction; hence $\sup A \in B$ and thus $\sup A \geq \inf B$. \end{proof} \begin{definition} The common value $\sup A = \inf B$ in the preceding theorem is denoted $\norm{T}$ (the `\textbf{operator norm}' of $T$). \end{definition} Notice that $x\in X \implies \norm{T(x)} \leq \norm{T}\norm{x}$ (because $\sup A \in B$). \begin{theorem}
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$L(X,Y)$ is a vector space upon which the operator norm is a norm. \end{theorem} \begin{proof} Note that addition and scalar multiplication are pointwise: \begin{align*} \end{align*}
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Lecture03 - \documentclass[12pt,letterpaper]cfw_article

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