# Clarification on the difference between Brouwer Fixed Point Theorem and Schauder Fixed point theorem

From Zeidler's Applied Functional Analysis

Brouwer

The continuous operator $A:M \to M$ has a fixed point provided $M$ is compact, convex, nonempty set in a finite dimensional normed space over $\mathbb{K}$

Schauder

The compact operator $A:M \to M$ has a fixed point provided $M$ is a bounded, closed, convex, nonempty subset of a Banach space $\mathbb{X}$ over $\mathbb{K}$

Claim: if $\dim(\mathbb{X}) < \infty$ then Schauder = Brouwer

Just to clarify:

Why is the operator $A$ assumed to be compact for Schauder but merely continuous for Brouwer? What does finite/infinite dimension have to do with this assumption?

In $\ell_2(\mathbb{N})$ consider the operator $T(x)=(\sqrt{1 - \| x\|^2},x_1, x_2, \dots)$ defined for $\|x\| \leq 1$, where $x=(x_1, x_2, \dots)$ and $\|x\|^2= \sum_{i=1}^{\infty} |x_i|^2$. $T$ is continuous, in fact
\begin{align} \|T(x) - T(y)\|^2 & = \left| \sqrt{1 - \|x\|^2} - \sqrt{1 - \|y\|^2}\right|^2 + \|x - y\|^2 \leq \\ & \leq \left| \|x\|^2 - \|y\|^2\right| + \|x - y\|^2 \leq \\ &\leq ( \|x\| + \|y\|)\|x-y\| + \|x-y\|^2 \leq \\ &\leq 2\|x-y\| + \|x-y\|^2. \end{align} Moreover $T$ maps the closed unit ball to its boundary because $\|T(x)\|^2= 1 - \|x\|^2 + \|x\|^2=1$. $T$ does not have fixed points, by contradiction, if we had $T(x)=x$, we would have $\|x\|=1$, but also $(0,x_1,x_2, \dots) = (x_1,x_2, \dots)$, that is $x_i=0$ for every $i$. In that case we would have $\|x\|=0 \neq 1$, which is a contradiction.
The continuous operator $A:M \to M$ has a fixed point provided $M$ is bounded, closed, convex, nonempty set in a finite dimensional normed space over $\mathbb{K}$.