证明：考虑由$Z$与$x_0$张成的子空间$Z_1$，由于${x_0} e Z$，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[gleft( x ight) = td = tdleft( {{x_0},Z} ight)]则$g$是$Z_1$上的线性泛函，且$gleft( {{x_0}} ight) = d$，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| {gleft( x ight)} ight| = left| t ight|d = left| t ight|dleft( {{x_0},Z} ight) le left| t ight|left| {{x_0} + frac{{x'}}{t}} ight| = left| {x' + t{x_0}} ight| = left| x ight|]所以$g$是$Z_1$上的有界泛函，且$left| g ight| le 1$
另一方面，由$d=dleft( {{x_0},Z} ight) $知，存在$Z$中的点列${x_n}^prime $，使得[mathop {lim }limits_{n o infty } left| {{x_n}^prime - {x_0}} ight| = d]又由于[d = gleft( {{x_0}} ight) = gleft( {{x_0} - {x_n}^prime } ight) le left| g ight|left| {{x_0} - {x_n}^prime } ight|]令$n o infty $，则$left| g ight| ge 1$，所以有$left| g ight| =1$，最后由$f{Hahn-Banach}$泛函延拓定理，则命题得证