norm on $ℝ^n$

hbghlyj

Analysis and Geometry第4页脚注3

Giving a norm $\|\cdot\|$ on $\mathbb{R}^n$ is equivalent to giving the set $B_{\|\cdot\|}=\{v \in V:\|v\| \leq 1\}$ of vectors in its closed unit ball. Such a set $B_{\|\cdot\|}$ must be closed and bounded (both with respect to the Euclidean metric), convex, and preserved by the map $x \mapsto-x$, but otherwise can be arbitrary.

在$\mathbb{R}^n$任意取一个关于0对称的有界凸闭集作为$\{v \in V:\|v\| \leq 1\}$可以导出一个范数.

其它的条件较容易验证,下面证明$\|\cdot\|$满足三角不等式:
$\forall a,b\in\mathbb{R}^n,$ 取$\lambda={\|a\|\over\|a\|+\|b\|},$ 根据凸性$\left\|\lambda\frac{a}{\|a\|}+(1-\lambda)\frac{b}{\|b\|}\right\|\le1,$ 即$\|a+b\|\leqslant\|a\|+\|b\|$.

Czhang271828

上面完全没把定义式讲明白啊. 在一般的赋范线性空间上给定关于原点对称的有界凸集 $V$, 定义范数作
\[
\|x\|_V:=\inf_{a\in \mathbb R_+}\{a^{-1}x\in V\}.
\]