# proof of Brouwer fixed point theorem

The $n$-dimensional simplex ${\mathcal{S}}_{n}$ is the following subset of ${\mathbb{R}}^{n+1}$

$$\left\{({\alpha}_{1},{\alpha}_{2},\mathrm{\dots},{\alpha}_{n+1})\right|\sum _{i=1}^{n+1}{\alpha}_{i}=1,{\alpha}_{i}\ge 0\mathit{\hspace{1em}}\forall i=1,\mathrm{\dots},n+1\}$$ |

Given an element $x={\sum}_{i}{\alpha}_{i}{e}_{i}\in {\mathcal{S}}_{n}$ we denote
${[x]}_{i}={\alpha}_{i}$ (i.e., the $i$-th barycentric coordinate^{}). We
also denote $F(x)=\{i|{[x]}_{i}\ne 0\}$. 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 $f\mathrm{:}{\mathrm{S}}_{n}\mathrm{\to}{\mathrm{S}}_{n}$ 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 $y\in {\mathcal{S}}_{n}$ 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 the following subset ${C}_{i}$ of ${\mathcal{S}}_{n}$:

$${C}_{i}=\left\{x\in {\mathcal{S}}_{n}\right|{[x]}_{i}\ge {[f(x)]}_{i}\}$$ |

We claim that if $x$ is in some $I$-face of ${\mathcal{S}}_{n}$ ($I\subseteq \{1,2,\mathrm{\dots},n+1\}$) 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 $i\in I$. This shows 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 leads
to a contradiction as the following inequality^{} shows:

$$ |

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 $i=1,2,\mathrm{\dots},n+1$. We claim that $L$ is a fixed point of $f$. Indeed, ${[L]}_{i}\ge {[f(L)]}_{i}\ge 0$ for all $i=1,2,\mathrm{\dots},n+1$ and thus:

$$1={[L]}_{1}+{[L]}_{2}+\mathrm{\cdots}+{[L]}_{n+1}\ge {[f(L)]}_{1}+{[f(L)]}_{2}+\mathrm{\cdots}+{[f(L)]}_{n+1}=1$$ |

Therefore, ${[L]}_{i}={[f(L)]}_{i}$ for all $i=1,2,\mathrm{\dots},n+1$ 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 |