EveryFiniteDimensionalNormedVectorSpaceIsABanachSpace 2005-01-09 14:50:03 2005-02-18 12:43:59 Theorem normed vector space % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{amsthm} \usepackage{mathrsfs} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions % % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \newcommand{\sR}[0]{\mathbb{R}} \newcommand{\sC}[0]{\mathbb{C}} \newcommand{\sN}[0]{\mathbb{N}} \newcommand{\sZ}[0]{\mathbb{Z}} \usepackage{bbm} \newcommand{\Z}{\mathbbmss{Z}} \newcommand{\C}{\mathbbmss{C}} \newcommand{\R}{\mathbbmss{R}} \newcommand{\Q}{\mathbbmss{Q}} \newcommand*{\norm}[1]{\lVert #1 \rVert} \newcommand*{\abs}[1]{| #1 |} \newtheorem{thm}{Theorem} \newtheorem{defn}{Definition} \newtheorem{prop}{Proposition} \newtheorem{lemma}{Lemma} \newtheorem{cor}{Corollary} \begin{thm} Every finite dimensional normed vector space is a Banach space. \end{thm} \emph{Proof.} Suppose $(V,\Vert\cdot\Vert)$ is the normed vector space, and $(e_i)_{i=1}^N$ is a basis for $V$. For $x=\sum_{j=1}^N \lambda_j e_j$, we can then define $$ \Vert x \Vert' = \sqrt{\sum_{j=1}^N |\lambda_j|^2} $$ whence $\Vert\cdot\Vert'\colon V\to \sR$ is a norm for $V$. Since \PMlinkname{all norms on a finite dimensional vector space are equivalent}{ProofThatAllNormsOnFiniteVectorSpaceAreEquivalent}, there is a constant $C>0$ such that $$ \frac{1}{C} \Vert x \Vert' \le \Vert x \Vert \le C \Vert x \Vert', \quad x\in V. $$ To prove that $V$ is a Banach space, let $x_1,x_2,\ldots$ be a Cauchy sequence in $(V,\Vert\cdot \Vert)$. That is, for all $\varepsilon>0$ there is an $M\ge 1$ such that $$ \Vert x_j-x_k \Vert <\varepsilon, \ \ \mbox{for all} j,k\ge M. $$ Let us write each $x_k$ in this sequence in the basis $(e_j)$ as $x_k=\sum_{j=1}^N \lambda_{k,j} e_j$ for some constants $\lambda_{k,j}\in \C$. For $k,l\ge 1$ we then have \begin{eqnarray*} \Vert x_k-x_l\Vert &\ge& \frac{1}{C} \Vert x_k-x_l \Vert' \\ &\ge& \frac{1}{C} \sqrt{\sum_{j=1}^N |\lambda_{k,j}-\lambda_{l,j}|^2} \\ &\ge& \frac{1}{C} |\lambda_{k,j}-\lambda_{l,j}| \end{eqnarray*} for all $j=1,\ldots, N$. It follows that $(\lambda_{k,1})_{k=1}^\infty, \ldots, (\lambda_{k,N})_{k=1}^\infty$ are Cauchy sequences in $\C$. As $\C$ is complete, these converge to some complex numbers $\lambda_1, \ldots, \lambda_N$. Let $x=\sum_{j=1}^N \lambda_j e_j$. For each $k=1,2,\ldots$, we then have \begin{eqnarray*} \Vert x-x_k\Vert &\le& C \Vert x-x_k\Vert' \\ &\le& C \sqrt{\sum_{j=1}^N |\lambda_{j}-\lambda_{k,j}|^2}. \end{eqnarray*} By taking $k\to \infty$ it follows that $(x_j)$ converges to $x\in V$. $\Box$