Separation theorem. Let XX be a normed space over K{R,C}\mathbb K\in\{\mathbb R,\mathbb C\}, let YXY\subseteq X be a , and suppose x0Xx_0\in X has positive distance

d(x0,Y):=infyYx0y=d>0.d(x_0,Y):=\inf_{y\in Y}\|x_0-y\|=d>0.

Then there exists a f:XKf:X\to\mathbb K such that

fY=0,f(x0)=1,f=1d.f|_Y=0,\qquad f(x_0)=1,\qquad \lVert f\rVert=\frac1d.
Proof idea

On Yspan{x0}Y\oplus\operatorname{span}\{x_0\}, define f0(y+λx0)=λf_0(y+\lambda x_0)=\lambda. The distance assumption gives f0=1/d\lVert f_0\rVert=1/d, and extends f0f_0 to XX without increasing its norm.