proof of Brouwer fixed point theorem

The $n$-dimensional simplex ${𝒮}_n$ is the followingsubset of ${ℝ}^{n+1}$

$$\left\{({\alpha }_1,{\alpha }_2,\mathrm{\dots },{\alpha }_{n+1})\right|\sum _{i=1}^{n+1}{\alpha }_i=1,{\alpha }_i\ge 0\mathit{\hspace{1em}}\forall i=1,\mathrm{\dots },n+1\}$$

Given an element $x={\sum }_i{\alpha }_i{e}_i\in {𝒮}_n$ we denote${[x]}_i={\alpha }_i$ (i.e., the $i$-th barycentric coordinate). Wealso denote $F(x)=\{i|{[x]}_i\ne 0\}$. An $I$-face of${𝒮}_n$ is the subset $\{x|F(x)\subseteq I\}$.

As was noted in the statement of the theorem, the ’shape’ isunimportant. Therefore, we will prove thefollowing variant of the theorem using the KKM lemma.

Theorem 1 (Brouwer’s Fixed Point Theorem).

Let $f\mathrm{:}{\mathrm{S}}_n\mathrm{\to }{\mathrm{S}}_n$ be a continuous function.Then, $f$ has a fixed point, namely, there is an$L\mathrm{\in }{\mathrm{S}}_n$ such that $L\mathrm{=}f\mathit{}\mathrm{(}L\mathrm{)}$.

Proof.

Clearly, ${\sum }_{i=1}^n{[y]}_i=1$ for any $y\in {𝒮}_n$ and$L=f(L)$ if and only if ${[L]}_i={[f(L)]}_i$ for all$i=1,2,\mathrm{\dots },n+1$. For each $i=1,2,\mathrm{\dots },n+1$ we define thefollowing subset $C_i$ of ${𝒮}_n$:

$$C_i=\left\{x\in {𝒮}_n\right|{[x]}_i\ge {[f(x)]}_i\}$$

We claim that if $x$ is in some $I$-face of ${𝒮}_n$($I\subseteq \{1,2,\mathrm{\dots },n+1\}$) 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 ${[x]}_i\ne 0$ then $i\in I$. Thisshows that

$$\sum _{i\in I}{[x]}_i=1$$

Assuming by contradiction that $x\notin C_i$ for all $i\in I$implies that for all $i\in I$. But this leadsto a contradiction as the following inequality shows:

This dicussion establishes that each $I$-face is contained in theunion ${\cup }_{i\in I}{C}_i$. In addition, the subsets $C_i$ are allclosed. Therefore, we have shown that the hypothesis of the KKMLemma holds.

By the KKM lemma there is a point $L$ that is in every $C_i$ for$i=1,2,\mathrm{\dots },n+1$. We claim that $L$ is a fixed point of $f$.Indeed, ${[L]}_i\ge {[f(L)]}_i\ge 0$ for all $i=1,2,\mathrm{\dots },n+1$ andthus:

$$1={[L]}_1+{[L]}_2+\mathrm{\cdots }+{[L]}_{n+1}\ge {[f(L)]}_1+{[f(L)]}_2+\mathrm{\cdots }+{[f(L)]}_{n+1}=1$$

Therefore, ${[L]}_i={[f(L)]}_i$ for all $i=1,2,\mathrm{\dots },n+1$ whichimplies that $L=f(L)$.∎