一、怎么建立和编译文档
二、编写
% Finite Dimensional Normed Linear Linear Space.tex
\documentclass{ctexart}
\usepackage{amsmath}
\usepackage{amssymb}
\title{Finite Dimensional Normed Linear Linear Space}
\author{杨书惠}
\begin{document}
\section{Theorem}
Let $(X,\Vert \cdot \Vert)$ be a finite-dimensional normed linear space with basis ${ x_1 , x_2 ,\dots , x_n}$ .Then there is a constant $m>0$ such that for every choice of scalars $ \alpha_1 , \alpha_2 , \dots ,\alpha_n$ ,we have
\[
m\sum_{j=1}^n \vert \alpha_j \vert \le \lVert \sum_{j=1}^n \alpha_j x_j \rVert .
\]
\\\ \textbf{Proof.}
If $\sum_{j=1}^n \vert \alpha_j \vert =0 $ ,then $\alpha_j=0$ for all $j=1,2, \dots ,n$ and the inequality holds for any $m>0$.
\\\ Assume that $\sum_{j=1}^n \lvert \alpha_j \rvert \neq 0$ .We shall prove the result for a set of scalars $\lbrace \alpha_1 ,\alpha_2 ,\cdot , \alpha_n \rbrace$ that satisfy the condition $\sum_{j=1}^n \lvert \alpha_j \rvert=1$ . Let
\[
A=\lbrace (\alpha_1 , \alpha_2 , \dots , \alpha_n)\in \mathbb{F}_n \vert \sum_{j=1}^n \lvert\alpha_j\rvert=1\rbrace .
\]
Since $A$ is a closed and bounded subset of $\mathbb{F}_n$ ,it is compact. Define $f:A\to\mathbb{R}$ by
\[
f(\alpha_1,\alpha_2,\dots,\alpha_n)=\lVert \sum_{j=1}^n\alpha_jx_j \rVert .
\]
\end{document}
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