Theorem 2 (Leray-Schauder)

Let X be a nonempty compact topological space, homeomorphic to a convex subset of a locally convex space. Every continuous mapping of X into X has a fixed point.

One may suppose that X is a convex subset of a locally convex space E. Let N be the closure of $\{0\}$ in E and $\pi : E\rightarrow E/N$ the canonical surjection. Since X is a separated space, the restriction of $\pi$ to X is injective; since X is compact, it therefore defines a homeomorphism of X onto a nonempty compact convex subset of the separated locally convex space $E/N$. This allows us to suppose that the space E is separated.

Let $f: X\rightarrow X$ be a continuous mapping. The mapping $h: X\rightarrow E$ defined by $h(x) =f(x)-x$ is continuous; since X is compact, the image $h(X)$ is closed. It is therefore enough to prove that 0 adheres to $h$(X), that is to say that for every convex open neighbourhood U of 0 in E, there exists a point $b$ of X such that $f(b)-b\in U$.

Let U be a convex neighbourhood of 0 in E. For every $a\in X$, let $V_a$ denote the set of elements $x$ of X such that $f(x)-f(a)\in U$. It is open in X and contains $a$. Since X is compact and nonempty, there exists a nonempty finite subset A of X such that the sets $V_a$, for $a\in A$, cover X. Let $(\varphi_a)_{a\in A}$ be a continuous partition of unity subordinate to the covering $(V_a)_{a\in A}$ (TG, IX, p. 43, prop. 1 and p. 47, th. 3). Let F denote the vector subspace of E generated by $f(A)$. For every $x\in F\cap X$, put

This defines a continuous mapping $g: F\cap X\rightarrow E$. Its image is contained in the convex envelope of $f$(A), hence in $F\cap X$. Since the set $F\cap X$ is a nonempty compact convex subset of a finite-dimensional vector space, the mapping $g$ has a fixed point $b\in F\cap X$ by Theorem 1. We have

For every $a\in$ A such that $\varphi_a(b)\not = 0$, one has $b\in V_a$ so that $f(b)-f(a)\in U$. Since U is convex, it follows that $f(b)-b\in U$, which completes the proof.