Let XX be a and let z0X{0}z_0\in X\setminus\{0\}.

Corollary: There exists a f:XKf:X\to\mathbb{K} such that

f=1andf(z0)=z0.\|f\|=1\quad\text{and}\quad f(z_0)=\|z_0\|.

This is a direct consequence of by applying it to Y={0}Y=\{0\} and x0=z0/z0x_0=z_0/\|z_0\|.