一直以来都不理解向量无穷范数如何从p范数得来,最近正看到极限,借此推导一遍。
1- p-范数
若 $x=left[x_{1}, x_{2}, cdots, x_{n} ight]^{T}$,那么 $$| x|_{p}=(|x_{1}|^{p}+|x_{2}|^{p}+cdots+|x_{n}|^{p})^{frac{1}{p}}$$
当 $p$ 取 $1, 2, infty$ 时, 分别得到:
$1$-范数: $|x|_{1}=|x_{1}|+|x_{2}|+cdots+|x_{n}|$
$2$-范数: $|x|_{2}=(|x_{1}|^{2}+|x_{2}|^{2}+cdots+|x_{n}|^{2})^{frac{1}{2}}$
$infty$-范数: $|x|_{infty}=max(|x_{1}|, |x_{2}|, cdots, |x_{n}|)$
2- $infty$-范数的推导
证明:
由定义$| x|_{p}=(|x_{1}|^{p}+|x_{2}|^{p}+cdots+|x_{n}|^{p})^{frac{1}{p}}$,记 $x_{max}=max(|x_{1}|, |x_{2}|, cdots, |x_{n}|)$
[egin{eqnarray}limlimits_{p oinfty}|x|_{p}&=&limlimits_{p oinfty}x_{max}cdotBig(ig(frac{|x_{1}|}{x_{max}}ig)^{p}+ig(frac{|x_{2}|}{x_{max}}ig)^{p}+cdots+ig(frac{|x_{n}|}{x_{max}}ig)^{p}Big)^{frac{1}{p}} onumber\&=&x_{max}cdotlimlimits_{p oinfty}Big(ig(frac{|x_{1}|}{x_{max}}ig)^{p}+ig(frac{|x_{2}|}{x_{max}}ig)^{p}+cdots+ig(frac{|x_{n}|}{x_{max}}ig)^{p}Big)^{frac{1}{p}} onumberend{eqnarray}]
因 $1leqsumlimits_{i}ig(frac{|x_{i}|}{x_{max}}ig)^{p}leq n$,故由 $limlimits_{n oinfty}n^{frac{1}{n}}=1$ 及 夹逼原理,有$$limlimits_{p oinfty}Big(ig(frac{|x_{1}|}{x_{max}}ig)^{p}+ig(frac{|x_{2}|}{x_{max}}ig)^{p}+cdots+ig(frac{|x_{n}|}{x_{max}}ig)^{p}Big)^{frac{1}{p}}=1$$
从而 $limlimits_{p oinfty}|x|_{p}=x_{max}$.