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}\geq 0\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.

Theorem 1 (Brouwer’s Fixed Point Theorem).

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

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, $[L]_{i}\geq[f(L)]_{i}\geq 0$ 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 $L=f(L)$ . ∎

Title	proof of Brouwer fixed point theorem
Canonical name	ProofOfBrouwerFixedPointTheorem1
Date of creation	2013-03-22 18:13:24
Last modified on	2013-03-22 18:13:24
Owner	uriw (288)
Last modified by	uriw (288)
Numerical id	4
Author	uriw (288)
Entry type	Proof
Classification	msc 47H10
Classification	msc 54H25
Classification	msc 55M20