Counterexample to Brouwer Fixed Point Theorem in $\ell_{\infty}$ space 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.
 A: 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.
A: Thank you for your attention and answers. I tried to modify Gerd's example by replacing map $t \mapsto \sqrt{|t|}$, which is acting on coordinates, with a nicer function. I think it worked very well.
Fix $k > 1$ and consider map $f: B_{\ell_{\infty}} \rightarrow B_{\ell_{\infty}}$ given by
$$ f(x) = ( 1 - \|x\|, \varphi(k x_1), \varphi(k x_2), \varphi(k x_3), \ldots),$$
where $\varphi(t) = \min\{t,1\}$. Notice that for any $t, s \in \mathbb{R}$ we have $|\varphi(t) - \varphi(s)| \leq |s - t|$, which means that $\varphi$ satisfies Lipschitz condition with constant $1$. From this it easily follows that $f$ is $k$-lipschitzian.
Now I claim that $f$ has no fixed points. Suppose $x = f(x)$ for some $x \in B_{\ell_{\infty}}$. Then
\begin{equation} 
 (1) \quad x_1 = 1 - \|x\|, \quad x_2 = \varphi(kx_1), \quad x_3 = \varphi(kx_2), \quad x_4 = \varphi(kx_3), \quad \ldots
\end{equation}
Lets consider three cases.
If $\|x\| = 1$, then it follows from (1) that $x_i = 0$ for every $i \in \mathbb{N}$, but this is impossible.
Suppose $\|x\| \leq 1 - \frac{1}{k} < 1$. It follows that $k(1-\|x\|) \geq 1$ and consequently $x_2 = \varphi(kx_1) = \varphi(k(1-\|x\|)) = 1$. Thus  $\|x\| = 1$ contrary to the assumption.
For the last case assume $1 - \frac{1}{k} \leq \|x\| < 1$. This time we have $k(1-\|x\|) \leq 1$ and consequently $x_2 = \varphi(k(1-\|x\|)) = k(1-\|x\|)$. Now $x_3 = \varphi(kx_2) = \varphi(k^2(1-\|x\|))$ and this value is either $1$ or $k^2(1-\|x\|)$. However there is $n \in \mathbb{N}$ such that $k^n(1-\|x\|) > 1$ and at some point we'll reach $x_{n+1} = 1$. Again $\|x\| = 1$ contradicting our assumption.
A: An example of a Lipschitz map that invariates the unit ball and does not have a fixed point
$$f(x_1, x_2, x_3, \ldots) = (\sup_{i\ge 1} (1-|x_i|), x_1^2, x_2^2, x_3^2, \ldots)$$
Assume that $(x_1, x_2, x_3, \ldots)$ is a fixed point. Then we have $x_{i+1} = x_i^2$ for all $i\ge 1$, and  $x_1 = \sup_{i \ge 1}(1-|x_i|)$. But if $|x_1|< 1$ then $x_n \to 0$ so $\sup_{i\ge 1} (1-|x_i|) = 1$, while if $|x_1|=1$, then all $|x_i|=1$, so $\sup(1-|x_i|) = 0$, contradiction.
Note that $f$ is Lipschitz with constant $2$. We could choose  $|x_i|^k$ instead of $x_i^2$ ($k>1$)  that is
$$f(x) = (\|\,  1- |x| \, \|, |x|^k)$$
and get a $k$-Lipschitz map.
$\bf{Added:}$
This is a comment on the methods of @Gerd and @triple backflip:
Consider a map $\phi\colon [-1,1]\to [-1,1]$, continuous such that
$\phi(0) = 0$, $\phi(1) = 1$, and $\phi(x) > x$ on $(0,1)$.
Define now the map $f(x)$ on $\ell_{\infty}$,
$$f(x_0, x_1, \ldots) = (1-\|x\|, \phi(x_0), \phi_(x_1), \ldots)$$
Assume that $x$ is a fixed point of $f$. Then we have $x_n = \phi^n(x_0)$ for all $n\ge 0$, and also $x_0= 1-\|x\|$.
Now,  notice that the sequence $x_n$ is constant $0$ if $x_0=0$, and converges to $1$ if $x_0>0$. Therefore, we cannot have $x_0 = 1-\|x\|$, and so we do not have a fixed point in the unit ball.
Notice also that for $x = (\alpha, \phi(\alpha), \phi^2(\alpha), \ldots)$, where $\alpha \in (0,1]$, we have $\|x\|=1$ and so
$$x - f(x) = \alpha$$
which can be made as small as desired.
Now, we can take $\phi$ a $k$ Lipschitz function ($k>1$) of the form
$$\phi(x) = x + \epsilon \cdot x(1-x^2)$$
for $\epsilon>0$ small.
A: I think the following example should work.
Let $f:B_{l_{\infty}}\rightarrow B_{l_{\infty}}$ such that $f(x_1, x_2, ...) = (y_1, y_2, ...)$, with
$$
y_1 =
\begin{cases}
1\;\;\;\;\;\text{if $\|(x_1,x_2,...)\|_{\infty} < 1$} \\
0\;\;\;\;\;\text{if $\|(x_1,x_2,...)\|_{\infty} = 1$}
\end{cases}
$$
and $y_n = x_{n-1}$ for $n>1$. It is obvious that is well defined.
Now I prove this function is continuous. Let $x=(x_1, x_2, ...)$, $x'=(x'_1, x'_2, ...)$ be elements of $B_{l_{\infty}}$ and let $f(x)=(y_1, y_2, ...)$ and $f(x')=(y'_1, y'_2, ...)$. Fix $0<\varepsilon<1$ and suppose $\|f(x)-f(x')\|_{\infty} < \varepsilon$.
Since $\varepsilon < 1$, we have then by definition that $\|x\|_{\infty}=1$ iff $\|x'\|_{\infty}=1$ (otherwise, we would have $\|f(x)-f(x')\|_{\infty}=1$). But then, we have $|x_n-x'_n|<\varepsilon$ for each $n$ (by definition of $f$) and so $\|x-x'\|_{\infty}\leq \varepsilon$. So $f$ is continuous.
Finally, suppose by contradiction that $x$ is a point fixed by $f$. By induction and the definition of $f$ it follows that $x_n=x_1$ for each $n$. Now, if $\|x\|_{\infty}=1$ we have $x = f(x)=(0, 0, 0, ...)$ and this is not possible; so $\|x\|_{\infty}<1$, but then $x=f(x)=(1, 1, 1, ...)$ and this is not possible too. So we get a contradiction.
