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时间:2014-06-25 13:12:36      阅读:263      评论:0      收藏:0      [点我收藏+]

标签:line   c   5   rac   mat   re   

证明:令$d = \mathop {inf}\limits_{y \in M} \left\| {x - y} \right\|$,由下确界的定义知,存在${x_n} \in M$,使得\[\mathop {\lim }\limits_{n \to \infty } \left\| {{x_n} - x} \right\| = d\]

   下面我们证明$\left\{ {{x_n}} \right\}$是基本列.由平行四边形公式知\[{\left\| {{x_m} - x} \right\|^2} + {\left\| {{x_n} - x} \right\|^2} = 2\left( {{{\left\| {\frac{{{x_m} + {x_n}}}{2} - x} \right\|}^2} + {{\left\| {\frac{{{x_m} - {x_n}}}{2}} \right\|}^2}} \right)\]由于$\frac{{{x_m} + {x_n}}}{2} \in M$,则$\left\| {\frac{{{x_m} + {x_n}}}{2} - x} \right\| \ge d$,从而可知\[0 \le \frac{1}{2}{\left\| {{x_m} - {x_n}} \right\|^2} \le {\left\| {{x_m} - x} \right\|^2} + {\left\| {{x_n} - x} \right\|^2} - 2{d^2}\]令$n,m \to \infty $,则$\left\| {{x_m} - {x_n}} \right\| \to 0$,所以$\left\{ {{x_n}} \right\}$是基本列

又由于$M$为$\bf{Hilbert}$空间$X$的闭子空间,则存在${x_0} \in M$,使得${x_n} \to {x_0}$,此时\[\left\| {x - {x_0}} \right\| = \mathop {\lim }\limits_{n \to \infty } \left\| {x - {x_n}} \right\| = d\]

   下面我们证明$x - {x_0} \bot M$.由线性子空间的定义知,对任意的$z \in M,z \ne 0$,以及$\lambda  \in K$,有${x_0} + \lambda z \in M$,于是\[\left\| {x - \left( {{x_0} + \lambda z} \right)} \right\| \ge d\]所以有\[{\left\| {\left( {x - {x_0}} \right) - \lambda z} \right\|^2} = {\left\| {x - {x_0}} \right\|^2} - 2{\mathop{\rm Re}\nolimits} \left( {\overline \lambda  \left( {x - {x_0}} \right),z} \right) + {\left| \lambda  \right|^2}{\left\| z \right\|^2} \ge {d^2}\]令$\lambda  = \frac{{\left( {x - {x_0},z} \right)}}{{{{\left\| z \right\|}^2}}}$,则有

\[{\left\| {x - {x_0}} \right\|^2} - 2\frac{{{{\left| {\left( {x - {x_0},z} \right)} \right|}^2}}}{{{{\left\| z \right\|}^2}}} + \frac{{{{\left| {\left( {x - {x_0},z} \right)} \right|}^2}}}{{{{\left\| z \right\|}^2}}} = {\left\| {x - {x_0}} \right\|^2} - \frac{{{{\left| {\left( {x - {x_0},z} \right)} \right|}^2}}}{{{{\left\| z \right\|}^2}}} \ge {d^2}\]由$\left\| {x - {x_0}} \right\| = d$可知,$\left( {x - {x_0},z} \right) = 0$,即$x - {x_0} \bot M$

   下面我们证明唯一性.假设还存在${x_0}^\prime  \in M$,${x_1}^\prime  \in {M^ \bot }$,使得\[x = {x_0}^\prime  + {x_1}^\prime \]则${x_0} - {x_0}^\prime  \in M,{x_1} - {x_1}^\prime  \in {M^ \bot }$,从而由\[{x_1} - {x_1}^\prime  = \left( {x - {x_0}} \right) - \left( {x - {x_0}^\prime } \right) = {x_0}^\prime  - {x_0} \in M\]可知${x_1}^\prime  = {x_1},{x_0}^\prime  = {x_0}$

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265656555

标签:line   c   5   rac   mat   re   

原文地址:http://www.cnblogs.com/ly758241/p/3806789.html

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