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${x_1}^\prime = {x_1},{x_0}^\prime = {x_0}$
原文地址：http://www.cnblogs.com/ly758241/p/3806789.html
