Lecture05 - \documentclass[12pt,letterpaper]{article}...

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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} \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} \newcommand{\norm}[1]{\| #1 \|} \newcommand{\skipline}{\vspace{\baselineskip}} \date{} \title{\bf Lecture 5} \author{} \begin{document} \maketitle \begin{center} (Continuing from last time...) \end{center} \skipline Conversely, suppose the bilinear map $$ S: X \times Y \ra Z $$ is (separately) continuous. For $x \in X$ we can define $T_x : Y \ra Z$ by $y \in Y \Ra T_x(y) = S(x, y)$. Then $T_x$ is continuous and linear, and in fact $T_x \in L(Y,Z)$, and $$ T: X \ra L(Y,Z) ; x \ra T_x, $$ is linear. If we knew that $T$ were bounded (that is, that $x \in X$ implies that $\norm{T_x} \leq M \norm{x}$) then $$ \norm{S(x,y)} = \norm{T_x(y)} \leq \norm{T_x} \norm{y} \leq M \norm{x} \norm{y}. $$ So $S$ is bounded as a bilinear map. As we know, $T$ is bounded when $X$ is finite- dimensional. More generally, $T$ is bounded when $X$ is complete; this follows as an application of the `uniform boundedness principle' (or Banach-Steinhaus theorem, for those in the know). \begin{theorem} If $S: X \times Y \ra Z$ is a bounded bilinear map, then $S$ is differentiable at...
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Lecture05 - \documentclass[12pt,letterpaper]{article}...

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