3655522

证明：考虑由$Z$与$x_0$张成的子空间$Z_1$，由于${x_0} \ne Z$，则$Z_1$中任一元素$x$可唯一地表示为\[x = x' + t{x_0},x' \in Z\]令\[g\left( x \right) = td = td\left( {{x_0},Z} \right)\]则$g$是$Z_1$上的线性泛函，且$g\left( {{x_0}} \right) = d$，当$x \in Z$时，$g(x)=0$

下证$g$是$Z_1$上的有界泛函，事实上，由于\[\left| {g\left( x \right)} \right| = \left| t \right|d = \left| t \right|d\left( {{x_0},Z} \right) \le \left| t \right|\left\| {{x_0} + \frac{{x'}}{t}} \right\| = \left\| {x' + t{x_0}} \right\| = \left\| x \right\|\]所以$g$是$Z_1$上的有界泛函，且$\left\| g \right\| \le 1$

另一方面，由$d=d\left( {{x_0},Z} \right) $知，存在$Z$中的点列${x_n}^\prime $，使得\[\mathop {\lim }\limits_{n \to \infty } \left\| {{x_n}^\prime - {x_0}} \right\| = d\]又由于\[d = g\left( {{x_0}} \right) = g\left( {{x_0} - {x_n}^\prime } \right) \le \left\| g \right\|\left\| {{x_0} - {x_n}^\prime } \right\|\]令$n \to \infty $，则$\left\| g \right\| \ge 1$，所以有$\left\| g \right\| =1$，最后由$\bf{Hahn-Banach}$泛函延拓定理，则命题得证