Let XX be a and let z0∈X∖{0}z_0\in X\setminus\{0\}.

Corollary: There exists a f:X→Kf: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/∥z0∥x_0=z_0/\|z_0\|.