Lecture02 - \documentclass[12pt,letterpaper]cfw_article \

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\documentclass[12pt,letterpaper]{article} \usepackage{amsmath,amssymb,latexsym,array,enumerate,amsthm,amscd} \ \usepackage[all]{xy} \ \newtheorem{theorem}{Theorem} \newtheorem{definition}{Definition} \ \newcommand{\bR}{\mathbb{R}} \newcommand{\ra}{\rightarrow} \newcommand{\Ra}{\Rightarrow} \newcommand{\nv}{\|v\|} \ \date{} \author{} \ \title{\bf Lecture 2} \ \begin{document} \ \maketitle \ \noindent {\bf Exercise}: If $\mathrm{dim}V > 1$ then `the' $\ell^P$ norm arises from an I.P. precisely when $p=2$. p \begin{proof} ($\Ra$) Since $\mathrm{dim} V > 1$, there are at least two vectors $v_1$, $v_2$ in the basis used to define the norm. If the $\ell^p$ nrom comes from an I.P. then the parallelogram law must hold: p \begin{displaymath} \|v_2\|^2_{p})\ar@{=}[d]\\ \end{displaymath} So $2\cdot 2^{\frac{1}{p}}=2\cdot 2$ and hence $\frac{2}{p}=1$ or $p=2$. S ($\Leftarrow$) This was seen before. \end{proof} \ \noindent {\bf Exercise}: If $V$ is infinite-dimensional, then it is not complete for the $\ell^{\infty}$ norm associated to a basis. $ \begin{proof} Since the dimension of $V$ is infinite, we can find a countable (sub)collection $(v_m)_{m=1}^{\infty}$ of basis vectors used to define the norm. Define a new sequence $(w_m)_{m=1}^{\infty}$ by setting $w_m=\sum_{k=1}^m{\frac{1}{k}v_{k}}$. Then for $n>m$ we have: w \[ \|w_n-w_m\|_{\infty}=\|\frac{1}{n}v_{n}+\frac{1}{n-1}v_{n-1}+\cdots+\frac{1} {m+1}v_{m+1 }\|_{\infty} \]
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Thus $d_{\infty}(w_n,w_m)=\mathrm{max}\{\frac{1}{n},\frac{1}{n-1},\ldots,\frac{1}
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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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Lecture02 - \documentclass[12pt,letterpaper]cfw_article \

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