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I was curious about counter-examples for Brouwer fixed point theorem in infinite dimension spaces.

I already know an easy counterexample in $l^p$ spaces, see this question. I also know one in $l^\infty$.

I wanted to know if there are some counter-examples for spaces of functions, for instance in $L^p(\Omega)$ for $\Omega$ a bounded domain or in $L^p(\mathbb{R}^n)$.

I guess that for $p=2$, one can enjoy the Hilbert structure and use Fourier or use a mapping to $l^2$.

Thanks for your help !

Velobos
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    Maybe you can try to construct something along the following lines. Embed $i:\ell^p \rightarrow L^p(\Omega)$, try to complement $i(\ell^p)$. If that works, the projection $p:L^p(\Omega)\rightarrow i(\ell^p)$ is continuous. Furthermore, denote by $F$ the pushforward of $\ell^p\rightarrow \ell^p$ to $i(\ell^p)\rightarrow i(\ell^p)$. Then the map $f=F\circ p$ might give you a counterexample. Namely, if the map had a fixed point, then this fixed point would live in $i(\ell^p)$, however, then $F$ would have a fixed point, which is a contradiction. – Severin Schraven Jan 10 '24 at 17:07
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    Let $\Omega \subseteq \mathbb{R}^n$ a bounded domain such that $B_2(0)\subset \Omega$. Then one could study the map $T: L^p(\Omega) \rightarrow L^p(\Omega)$ given by $$ Tf(x)=\begin{cases} \left(\frac{1-\Vert f \Vert_p^p}{\vert A_0\vert}\right)^{1/p},& x\in A_0, \ \left( \frac{1}{\vert A_n \vert} \int_{A_{n+1}} \vert f \vert^p \right)^{1/p},& x\in A_n \text{ for } n\in \mathbb{N}. \end{cases} $$ Here $A_0=\Omega \setminus B_1(0)$ and $A_n=\overline{B_{2^{-n}}(0)}\setminus B_{2^{-(n+1)}}(0)$. We have $\Vert Tf\Vert_p=1$ and it admits no fixed point. Would need to show that the map is cont. – Severin Schraven Jan 10 '24 at 17:59
  • What is your counterexample for $\ell^\infty$? Maybe one can also adapt it for $L^\infty$. – Severin Schraven Jan 10 '24 at 22:12
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    See this question for $l^\infty$ : https://math.stackexchange.com/questions/4560006/counterexample-to-brouwer-fixed-point-theorem-in-ell-infty-space – Velobos Jan 11 '24 at 08:40

1 Answers1

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In Counterexamples to Brouwer's fixed point theorem for the closed unit ball in Banach space the counterexample in $\ell^p$ for $1\leq p <\infty$ looks like

$$ T: \ell^p \rightarrow \ell^p, (x_1, x_2, \dots) \mapsto (\vert 1-\Vert x \Vert_p^p\vert^{1/p}, x_1, x_2, \dots ). $$

The point is then that $\Vert Tx \Vert_p =1$ for $\Vert x\Vert_p\leq 1$ and hence for a fixed point on the unit ball the first entry vanishes. However, for a fixed point also holds $x_n=x_{n+1}$ and thus the only possible fixed point is $x=(0,0, \dots)$ which cannot be as $\Vert x \Vert_p=1$. Therefore, $T$ does not admit any fixed point.

We try to mimic this construction in $L^p(\Omega)$. For this we start with $\Omega \subseteq \mathbb{R}^k$ having positive Lebesgue measure (I'll denote by $\vert S \vert$ the Lebesgue measure of a set $S$) and further assume that there exists a partition $(A_n)_{n\in \mathbb{N}_0}$ such that $0<\vert A_n \vert<\infty$ (in the end we will show that this last assumption holds always true for $\vert \Omega \vert>0$).

Then we define for $1\leq p <\infty$ \begin{align*} T: L^p(\Omega) \rightarrow L^p(\Omega), Tf(x)=\begin{cases} \left\vert \frac{1-\Vert f \Vert_{L^p(\Omega)}}{\vert A_0\vert} \right\vert^{1/p},& x\in A_0,\\ \left( \frac{1}{\vert A_n \vert} \int_{A_{n-1}} \vert f(y)\vert^p dy \right)^{1/p},& x\in A_n \text{ for } n\in \mathbb{N}. \end{cases} \end{align*} To get a "sequence-like" behaviour we have made our function constant on the partition, which allows us to essentially run the same argument as in $\ell^p$.

One readily checks that for $\Vert f \Vert_{L^p(\mathbb{\Omega}}\leq 1$ we have $\Vert Tf\Vert_{L^p(\Omega)}=1$. Thus, $T$ can be restricted to a well-defined map from the unit ball of $L^p(\Omega)$ to itself.

Assume that there exists a function $\Vert f \Vert_{L^p(\Omega)}\leq 1$ which is a fixed point of $T$. Then we have $\Vert f \Vert_{L^p(\Omega)} = \Vert Tf \Vert_{L^p(\Omega)}=1$ and $f$ is constant on $A_n$. We have that $f$ on $A_0$ is equal to $$ \frac{1-\Vert f \Vert_{L^p(\Omega)}}{\vert A_0\vert} = 0. $$ Now inductively we show that $f=0$ on $A_n$. Indeed, if $f$ vanishes identically on $A_m$, then $f$ is equal to $$ \left( \frac{1}{\vert A_{m+1} \vert} \int_{A_{m}} \vert f(y)\vert^p dy \right)^{1/p} =0 $$ on $A_{m+1}$. Thus, we would get $f\equiv 0$. However, this contradict $\Vert f \Vert_{L^p(\Omega)}=1$.

Hence, we are left to show that $T$ is a continuous map. Let $f,g\in L^p(\Omega)$. Then we compute

\begin{align*} \Vert Tf-Tg\Vert_{L^p(\Omega)}^p &= \int_{\Omega} \vert Tf(x)-Tg(x) \vert^p dx \\ &=\left\vert (1-\Vert f\Vert_{L^p(\Omega)}^p)^{1/p}- (1-\Vert g\Vert_{L^p(\Omega)}^p)^{1/p}\right\vert^p \\ &\quad + \sum_{n=1}^\infty \left( \Vert f \Vert_{L^p(A_{n-1})}- \Vert g \Vert_{L^p(A_{n-1})}\right)^p \\ &\leq \left\vert (1-\Vert f\Vert_{L^p(\Omega)}^p)^{1/p}- (1-\Vert g\Vert_{L^p(\Omega)}^p)^{1/p}\right\vert^p \\ &\quad +\sum_{n=1}^\infty \Vert f-g\Vert_{L^p(\Omega)}^p \\ &\leq\left\vert (1-\Vert f\Vert_{L^p(\Omega)}^p)^{1/p}- (1-\Vert g\Vert_{L^p(\Omega)}^p)^{1/p}\right\vert^p + \Vert f-g\Vert_{L^p(\Omega)}^p. \end{align*} Using the fact that $\mathbb{R}\rightarrow \mathbb{R}, x\mapsto \vert 1-\vert x \vert\vert^{1/p}$ is continuous, we see that $T$ is continuous.

Furthermore, the map $$T_\varepsilon f(x)=\begin{cases} \left\vert \frac{\varepsilon-\Vert f \Vert_{L^p(\Omega)}}{\vert A_0\vert} \right\vert^{1/p},& x\in A_0,\\ \left( \frac{1}{\vert A_n \vert} \int_{A_{n-1}} \vert f(y)\vert^p dy \right)^{1/p},& x\in A_n \text{ for } n\in \mathbb{N}. \end{cases}$$ (for $0<\varepsilon\leq 1$) would yield a counterexample if we wanted to replace the unit ball in $L^p(\Omega)$ by the ball of radius $\varepsilon$.

Finally, we note that for any $\Omega\subseteq \mathbb{R}^k$ of positive measure admits a partition $(A_n)_{n\in \mathbb{N}_0}$ such that $0<\vert A_n \vert < \infty$. Indeed, it is enough to show the existence of such a partition for $\Omega$ of finite measure (write $\Omega = \bigcup_{a\in \mathbb{Z}^k} \Omega \cap\prod_{j=1}^k [a_j, a_{j}+1]$, find a partition of the intersections, relabel and potentially alter on zero sets). In the finite measure case we can show our claim by induction. Thus, it boils down to show the statement that every set of finite, positive measure can be written as the union of two sets of finite, positive measure. This can be proved by bisecting (or if you want to have them to have equal measure, one can use Every set $E \subseteq \mathbb{R}$ of positive measure is the disjoint union of two sets $E = B \cup C$ such that $\mu(B) = \frac12\mu(E) = \mu(C)$).

Added: A counterexample for $\ell^\infty$ was provided in Counterexample to Brouwer Fixed Point Theorem in $\ell_{\infty}$ space. They used the map $$T:\ell^\infty \rightarrow \ell^\infty, x=(x_1, x_2, \dots) \mapsto (1-\Vert x \Vert_\infty, \sqrt{\vert x_1\vert}, \sqrt{\vert x_2\vert}, \dots).$$ If $x$ is a fixed point of $T$ with $\Vert x\Vert_\infty \leq 1$, then $x_1=1-\Vert x\Vert_\infty \geq 0$ and $0\leq \sqrt{\vert x_n\vert}=x_{n+1}$. Thus, $x_n=x_1^{1/(2^{n-1})}$. If $\Vert x\Vert_\infty<1$ (i.e. $x_0>0$), then $\lim_{n\rightarrow \infty} x_n=1$ and hence yielding the contradiction $\Vert x\Vert_\infty=1$. On the other hand if $\Vert x\Vert_\infty =1$, then $x_1=0$ and hence $x=0$, which is again a contradiction. Thus, $T$ does not admit any fixed point on the closed unit ball.

We mimic this approach for $L^\infty(\Omega)$ with $\Omega\subseteq \mathbb{R}^k$ with positive measure. We pick again a partition $(A_n)_{n\in \mathbb{N}_0}$ with $0<\vert A_n\vert<\infty$ and define

$$ T: L^\infty(\Omega)\rightarrow L^\infty(\Omega), Tf(x)=\begin{cases} \vert 1-\Vert f\Vert_{L^\infty(\Omega)}\vert,& x\in A_0,\\ \sqrt{\Vert f\Vert_{L^\infty(A_{n-1})}},& x\in A_n \text{ for } n\in \mathbb{N}. \end{cases}$$ It is easy to check that this yields a counterexample for the closed unit ball in $L^\infty(\Omega)$.

Severin Schraven
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