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 $ϵ>0$ notice that the family of open sets $\{B_{ϵ}(x):x\in K\}$ is an open covering of $K$. Being $K$ compact there exists a finite subcover, i.e. there exists $n$ points $x_1,\mathrm{\dots },x_n$ of $K$ such that the balls $B_{ϵ}(x_i)$ cover the whole set $K$.

Define the functions $g_1,\mathrm{\dots },g_n$ by

It is clear that each $g_i$ is continuous, $g_i(x)\ge 0$ 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 $x_1,\mathrm{\dots },x_n$. Moreover one can easily prove the following

$$\parallel g(x)-x\parallel \le ϵ\mathit{\hspace{1em}\hspace{0.5em}}{\forall }_{x\in K}$$

Now, define the function $B:=g\circ f$. The restriction $\tilde{B}$ of $B$ to $K_0$ provides a continuous function $K_0⟶K_0$.

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

$$\parallel f(z)-z\parallel =\parallel f(z)-g(f(z))\parallel \le ϵ$$

Summarizing, for each $ϵ>0$ there exists $z=z(ϵ)\in K$ such that $\parallel f(z)-z\parallel \le ϵ$. Then

$${\forall }_{m\in \mathbb{N}}\mathit{\hspace{1em}\hspace{0.5em}}{\exists }_{z_m\in K}\mathit{\hspace{1em}\hspace{0.5em}}\parallel f(z_m)-z_m\parallel \le \frac{1}{m}$$

As $f(z_m)$ is in the compact space $K$, there is a subsequence $z_{m_k}$ such that $f(z_{m_k})⟶x_0$, for some $x_0\in K$.

We then have

$\parallel z_{m_k}-x_0\parallel$ $=$ $\parallel z_{m_k}-f(z_{m_k})+f(z_{m_k})-x_0\parallel$

$\le$ $\parallel f(z_{m_k})-z_{m_k}\parallel +\parallel f(z_{m_k})-x_0\parallel$

$\le$ ${\displaystyle \frac{1}{m_k}}+\parallel f(z_{m_k})-x_0\parallel ⟶0$

which means that $z_{m_k}⟶x_0$.

As $f$ is continuous we have $f(z_{m_k})⟶f(x_0)$. 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. $\mathrm{\square }$

Title	proof of Schauder fixed point theorem
Canonical name	ProofOfSchauderFixedPointTheorem
Date of creation	2013-03-22 13:45:22
Last modified on	2013-03-22 13:45:22
Owner	asteroid (17536)
Last modified by	asteroid (17536)
Numerical id	10
Author	asteroid (17536)
Entry type	Proof
Classification	msc 47H10
Classification	msc 46T99
Classification	msc 46T20
Classification	msc 46B50
Classification	msc 54H25