Brouwer fixed point theorem states that if $f: B^n \rightarrow B^n$ is continuous map from unit ball $B^n \subset \mathbb{R}^n$ into itself, then there is some $x \in B^n$ such that $f(x) = x$.
This result fails in infinite dimensional spaces and im trying to find continuous (or even lipschitz) map $f: B_{\ell_{\infty}} \rightarrow B_{\ell_{\infty}}$, where $B_{\ell_{\infty}}$ is unit ball in real Banach space $\ell_{\infty}$, without fixed points. I had troubles in finding such example so I started from smaller space $c_0$ (of all real sequences $(x_n)$ such that $\lim x_n = 0$). There is simple isometry $f: B_{c_0} \rightarrow B_{c_0}$ without fixed point given by $f(x) = (1,x_1,x_2,x_3,\ldots)$. We can pass to bigger space $c$ of all convergent sequences and consider $f: B_c \rightarrow B_c$ but now $f$ has fixed point $(1,1,1,\ldots)$. Little modification gives $\hat{f}(x) = (1,-1,x_1,x_2,x_3,\ldots)$ and we get continuous map $\hat{f}:B_{c} \rightarrow B_{c}$ without fixed point. Im trying to extend these ideas to space $\ell_{\infty}$ but im stuck. Any hints and comments will be appreciated.
Fixed Point Theorems – Counterexample to Brouwer in ?? Space
banach-spacesfixed-point-theorems
Best Answer
Consider $f:B_{l_\infty} \to B_{l_\infty}$ defined as $$ f(x)=f((x_k))=(1-\|x\|,\sqrt{|x_1|}, \sqrt{|x_2|},\sqrt{|x_3|}, \dots). $$ Assume $y=f(y)$ for some $y=(y_k) \in B_{l_\infty}$. Then $$ y_1=1-\|y\|, ~ y_2=\sqrt{|y_1|}, ~ y_3=\sqrt{|y_2|}=|y_1|^{1/4}, ~ y_4=\sqrt{|y_3|}=|y_1|^{1/8}, \dots $$ hence $y_k=|y_1|^{1/2^{k-1}}$ $(k \ge 2)$.
case $\|y\|=1$: Then $(y_k)=(0)$, so $\|y\|=0$, a contradiction.
case $\|y\|<1$: Then $y_1 > 0$ and $y_k \to 1$ as $(k \to \infty)$. Hence $\|y\|=1$, a contradiction.
Clearly $f$ is continous, but not Lipschitz-continous.