PlanetMath: proof of Brouwer fixed point theorem

The $n$ -dimensional simplex $\mathcal{S}_n$ is the following subset of $\mathbb{R}^{n+1}$ $$ \left\{(\alpha_1,\alpha_2,\ldots,\alpha_{n+1}) \, \Big| \, \sum_{i=1}^{n+1}\alpha_i=1, \quad \alpha_i\geq0 \quad \forall i=1,\ldots,n+1\right\} $$

Given an element $x=\sum_i\alpha_i e_i\in\mathcal{S}_n$ we denote $[x]_i=\alpha_i$ (i.e., the $i$ -th barycentric coordinate). We also denote $F(x)=\{\, i\, |\, [x]_i \neq 0\}$ . An $I$ -face of $\mathcal{S}_n$ is the subset $\{\, x\, |\, F(x)\subseteq I \}$ .

As was noted in the statement of the theorem, the 'shape' is unimportant. Therefore, we will prove the following variant of the theorem using the KKM lemma.

Proof. Clearly, $\sum_{i=1}^n [y]_i = 1$ for any $y\in\mathcal{S}_n$ and $L=f(L)$ if and only if $[L]_i = [f(L)]_i$ for all $i=1,2,\ldots,n+1$ . For each $i=1,2,\ldots, n+1$ we define the following subset $C_i$ of $\mathcal{S}_n$ : $$ C_i = \left\{x\in\mathcal{S}_n \,\Big|\, [x]_i\geq[f(x)]_i\right\} $$ We claim that if $x$ is in some $I$ -face of $\mathcal{S}_n$ ($I\subseteq\{1,2,\ldots,n+1\}$ ) then there is an index $i\in I$ such that $x \in C_i$ . Indeed, if $x$ is in some $I$ -face then $F(v) \subseteq I$ . Thus, if $[x]_i \neq 0$ then $i\in I$ . This shows that $$ \sum_{i\in I} [x]_i = 1 $$ Assuming by contradiction that $x\not\in C_i$ for all $i\in I$ implies that $[x]_i < [f(x)]_i$ for all $i\in I$ . But this leads to a contradiction as the following inequality shows: $$ 1 = \sum_{i\in I} [x]_i < \sum_{i\in I} [f(x)]_i \leq \sum_{i=1}^n [f(x)]_i = 1 $$ This dicussion establishes that each $I$ -face is contained in the union $\cup_{i\in I} C_i$ . In addition, the subsets $C_i$ are all closed. Therefore, we have shown that the hypothesis of the KKM Lemma holds.

By the KKM lemma there is a point $L$ that is in every $C_i$ for $i=1,2,\ldots,n+1$ . We claim that $L$ is a fixed point of $f$ . Indeed, $[L]_i\geq[f(L)]_i\geq0$ for all $i=1,2,\ldots,n+1$ and thus: $$ 1 = [L]_1 + [L]_2 + \cdots + [L]_{n+1} \geq [f(L)]_1 + [f(L)]_2 + \cdots + [f(L)]_{n+1} = 1 $$ Therefore, $[L]_i=[f(L)]_i$ for all $i=1,2,\ldots,n+1$ which implies that $L=f(L)$ . $\qedsymbol$