# proof of Schauder fixed point theorem

The idea of the proof is to reduce to the finite dimensional case where we can apply the Brouwer fixed point theorem.

Given $\epsilon>0$ notice that the family of open sets $\{B_{\epsilon}(x)\colon x\in K\}$ is an open covering of $K$. Being $K$ compact there exists a finite subcover, i.e. there exists $n$ points $x_{1},\ldots,x_{n}$ of $K$ such that the balls $B_{\epsilon}(x_{i})$ cover the whole set $K$.

Define the functions $g_{1},\ldots,g_{n}$ by

 $g_{i}(x):=\begin{cases}\epsilon-\|x-x_{i}\|,&\;\;\textrm{if \|x-x_{i}\|\leq% \epsilon}\\ 0,&\;\;\textrm{if \|x-x_{i}\|\geq\epsilon}\end{cases}$

It is clear that each $g_{i}$ is continuous, $g_{i}(x)\geq 0$ and $\sum_{i=1}^{n}g_{i}(x)>0$ for every $x\in K$.

Thus we can define a function in $K$ by

 $g(x):=\frac{\sum_{i=1}^{n}g_{i}(x)x_{i}}{\sum_{i=1}^{n}g_{i}(x)}$

The above function $g$ is a continuous function from $K$ to the convex hull $K_{0}$ of $x_{1},\ldots,x_{n}$. Moreover one can easily prove the following

 $\|g(x)-x\|\leq\epsilon\quad\;\;\forall_{x\in K}$

Now, define the function $B:=g\circ f$. The restriction $\tilde{B}$ of $B$ to $K_{0}$ provides a continuous function $K_{0}\longrightarrow K_{0}$.

Since $K_{0}$ is compact convex subset of a finite dimensional vector space, we can apply the Brouwer fixed point theorem to assure the existence of $z\in K_{0}$ such that

 $B(z)=\tilde{B}(z)=z$

Therefore $g(f(z))=z$ and we have the inequality

 $\|f(z)-z\|=\|f(z)-g(f(z))\|\leq\epsilon$

Summarizing, for each $\epsilon>0$ there exists $z=z(\epsilon)\in K$ such that $\|f(z)-z\|\leq\epsilon$. Then

 $\forall_{m\in\mathbb{N}}\quad\;\;\exists_{z_{m}\in K}\quad\;\;\|f(z_{m})-z_{m}% \|\leq\frac{1}{m}$

As $f(z_{m})$ is in the compact space $K$, there is a subsequence $z_{m_{k}}$ such that $f(z_{m_{k}})\longrightarrow x_{0}$, for some $x_{0}\in K$.

We then have

 $\displaystyle\|z_{m_{k}}-x_{0}\|$ $\displaystyle=$ $\displaystyle\|z_{m_{k}}-f(z_{m_{k}})+f(z_{m_{k}})-x_{0}\|$ $\displaystyle\leq$ $\displaystyle\|f(z_{m_{k}})-z_{m_{k}}\|+\|f(z_{m_{k}})-x_{0}\|$ $\displaystyle\leq$ $\displaystyle\frac{1}{m_{k}}+\|f(z_{m_{k}})-x_{0}\|\longrightarrow 0$

which means that $z_{m_{k}}\longrightarrow x_{0}$.

As $f$ is continuous we have $f(z_{m_{k}})\longrightarrow f(x_{0})$. Both limits of $f(z_{m_{k}})$ must coincide, so we conclude that

 $f(x_{0})=x_{0}$

i.e. $f$ has a fixed point. $\square$

Title proof of Schauder fixed point theorem ProofOfSchauderFixedPointTheorem 2013-03-22 13:45:22 2013-03-22 13:45:22 asteroid (17536) asteroid (17536) 10 asteroid (17536) Proof msc 47H10 msc 46T99 msc 46T20 msc 46B50 msc 54H25