Theorem 4 (Krein, Krasnoselskii, Milman)

Let E be a normed space, and let F and G be vector subspaces of E. Suppose that the dimension of G is finite and strictly less than that of F. There exists an element of F of norm 1 whose distance to G is equal to 1.

It is enough to treat the case where the field K is equal to $\mathbf{R}$. Let $n$ be the dimension of G. Replacing F by a vector subspace of F of dimension $n+ 1$ containing G, we are reduced to the case where dim(F) $=n+ 1$. Let us reason by contradiction, supposing that the conclusion of the theorem is not satisfied. Let S be the unit sphere of F. For every $x\in S$, the distance from $x$ to G is then strictly less than $\|x\|= 1$, and one can choose an element $y(x)$ of G such that $\|x-y(x)\|<1$. Let $V_x$ denote the set of elements $z$ of S such that $\|z-y(x)\|<1$; this is an open neighbourhood of $x$ in S. There exists a continuous locally finite partition of unity $(\varphi_x)_{x\in S}$ on S subordinate to the covering $(V_x)_{x\in S}$ of S (TG, IX, p. 46, prop. 3 and p. 51, prop. 6). Let $f: S\rightarrow G$ be the continuous mapping given by

for every $z\in S$. Let $z\in S$. One has $\varphi_x(z)\geqslant 0$ for every $x\in S$, and there exists $x\in S$ such that $\varphi_x(z)>0$, since $\sum_{x\in S}\varphi_x(z) = 1$, hence

This property of $f$ contradicts Theorem 3.