证明：必要性.若存在$X$上的连续线性泛函$f$满足条件，则egin{align*}left| {sumlimits_{upsilon = 1}^k {{t_upsilon }{alpha _upsilon }} }
ight| &= left| {sumlimits_{upsilon = 1}^k {{t_upsilon }fleft( {{x_upsilon }}
ight)} }
ight| = left| {fleft( {sumlimits_{upsilon = 1}^k {{t_upsilon }{x_upsilon }} }
ight)}
ight|\
&le left| f
ight| cdot left| {sumlimits_{upsilon = 1}^k {{t_upsilon }{x_upsilon }} }
ight| le Mleft| {sumlimits_{upsilon = 1}^k {{t_upsilon }{x_upsilon }} }
ight|
end{align*}
充分性.若对任意数，有$left| {sumlimits_{upsilon = 1}^k {{t_upsilon }{alpha _upsilon }} } ight| le Mleft| {sumlimits_{upsilon = 1}^k {{t_upsilon }{x_upsilon }} } ight|$
令$M = spanleft{ {{x_1},{x_2}, cdots ,{x_k}} ight}$，则$M$为赋范线性空间$X$的子空间.对任意$sumlimits_{upsilon = 1}^k {{t_upsilon }{x_upsilon }} in M$，定义$M$上的线性泛函$f_0$：[{f_0}left( {sumlimits_{upsilon = 1}^k {{t_upsilon }{x_upsilon }} } ight) = sumlimits_{upsilon = 1}^k {{t_upsilon }{alpha _upsilon }} ]由于[left| {{f_0}left( {sumlimits_{upsilon = 1}^k {{t_upsilon }{x_upsilon }} } ight)} ight| = left| {sumlimits_{upsilon = 1}^k {{t_upsilon }{alpha _upsilon }} } ight| le Mleft| {sumlimits_{upsilon = 1}^k {{t_upsilon }{x_upsilon }} } ight|]则$left| {{f_0}} ight| le M$，即$f_0$为$M$上的连续线性泛函，于是由$f{Hahn-Banach}$泛函延拓定理知，存在$X$上的连续线性泛函$f$，使得当$x in M$时，有$fleft( x ight) = {f_0}left( x ight)$，且${left| f ight|_X} = {left| {{f_0}} ight|_M}$，所以充分性得证