证明：我们易知，当$f=0$时，取$y=0$即可，因此只需证明$f e 0$时定理成立
若$f$为$X$上的非零连续线性泛函，则$M = left{ {x|fleft( x ight) = 0} ight}$为$X$的闭真子空间，从而由投影定理知，存在$u in Xackslash M$，以及存在${u_0} in M,z in {M^ ot }$，使得$z = u - {u_0}$，于是$z in {M^ ot }$且$z e 0$
由于$M cap {M^ ot } = left{ 0 ight}$，则$fleft( z ight) e 0$.对于任意的$xin X$，由$fleft( {x - frac{{fleft( x ight)}}{{fleft( z ight)}}z} ight) = 0$可知[x - frac{{fleft( x ight)}}{{fleft( z ight)}}z in M]故[left( {x - frac{{fleft( x ight)}}{{fleft( z ight)}}z,z} ight) = 0]所以有[fleft( x ight) = frac{{fleft( z ight)}}{{{{left| z ight|}^2}}}left( {x,z} ight) = left( {x,frac{{overline {fleft( z ight)} }}{{{{left| z ight|}^2}}}z} ight)]令$y = frac{{overline {fleft( z ight)} }}{{{{left| z ight|}^2}}}z$，则对任意的$xin X$，有$fleft( x ight) = left( {x,y} ight)$
假设还存在$y' in X$，使得$fleft( x ight) = left( {x,y'} ight)$对任意的$xin X$成立，则对于$y - y' in X$，有[fleft( {y - y'} ight) = left( {y - y',y} ight) = left( {y - y',y'} ight) = 0]因此有[{left| {y - y'} ight|^2} = left( {y - y',y - y'} ight) = 0]即${y = y'}$，所以$fleft( x ight) = left( {x,y} ight)$的表示是唯一的，并且此时有$left| f ight| = left| y ight|$