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. $\\square$