Let’s recall Brouwer fixed-point theorem.

Theorem (Brouwer): Every continuous function from a convex compact subset $K$ of a Euclidean space to $K$ itself has a fixed point.

We here describe an example of a homeomorphism of the unit ball of a Hilbert space having no fixed point. Let $E$ be a separable Hilbert space with $(e_n)_{n \in \mathbb{Z}}$ as a Hilbert basis. $B$ and $S$ are respectively $E$ closed unit ball and unit sphere.

There is a unique linear map $u : E \to E$ for which $u(e_n)=e_{n+1}$ for all $n \in \mathbb{Z}$. For $x = \sum_{n \in \mathbb{Z}} \xi_n e_n \in E$ we have $u(x)= \sum_{n \in \mathbb{Z}} \xi_n e_{n+1}$. $u$ is isometric as $$\Vert u(x) \Vert^2 = \sum_{n \in \mathbb{Z}} \vert \xi_n \vert^2 = \Vert x \Vert^2$$ hence one-to-one. $u$ is also onto as for $x = \sum_{n \in \mathbb{Z}} \xi_n e_n \in E$, $\sum_{n \in \mathbb{Z}} \xi_n e_{n-1} \in E$ is an inverse image of $x$. Finally $u$ is an homeomorphism.

We now define $f : E \to E$ by $$f(x)=\frac{1 – \Vert x \Vert}{2}e_0+u(x)$$

• For $\Vert x \Vert = 1$, $\Vert f(x) \Vert = \Vert u(x) \Vert = \Vert x \Vert = 1$, therefore $f$ is an homeomorphism of the unit sphere $S$.
• For $\Vert x \Vert \le 1$ $$\Vert f(x) \Vert \le \frac{1 – \Vert x \Vert}{2} + \Vert u(x) \Vert \le \frac{1 + \Vert x \Vert}{2} \le 1$$ hence $f(B) \subset B$.

We now prove that $f$ is bijective from $B$ to $B$ and we find the inverse mapping.

First we have $f(0)=\frac{e_0}{2}$ and conversely, $f(x)=\frac{e_0}{2}$ implies $\Vert x \Vert = 0$ hence $x = 0$. For $x \in B$ non zero, $0$, $x$ and $x / \Vert x \Vert$ are aligned. We notice that $(\lambda, 1-\lambda)=(\Vert x \Vert, 1- \Vert x \Vert)$ are the barycentric coordinates of $x$ on the segment $[x/\Vert x \Vert,0]$ as $$x=\lambda \frac{x}{\Vert x \Vert} + (1-\lambda)0$$ One can verify that the following equality holds $$f(x)=\lambda f(\frac{x}{\Vert x \Vert})+(1-\lambda)f(0)$$ which means that $e_0/2$, $f(x)$ and $f(x/\Vert x \Vert)$ are also aligned and that $(\lambda,1-\lambda)$ are the barycentric coordinates of $f(x)$ on the segment $[f(x/\Vert x) \Vert,e_0/2]$

Consequently, to find the inverse image under $f$ of a point $x \in B$, $x \neq e_0/2$ one has to proceed in the following way:

1. Find the intersection $z = \varphi(y)$ (we’ll prove that $z$ is uniquely defined) of the ray $D$ having $e_0/2$ as origin and passing through $y$.
2. Find the barycentric coordinates $(\lambda,1-\lambda)$ of $y$ as a point of the segment $[z,e_0/2]$.
3. $x=f^{-1}(y)$ is then defined by $$\lambda u^{-1}(\frac{z}{\Vert z \Vert})$$

The equation of the ray $D$ is $$y=\lambda z + (1-\lambda)\frac{e_0}{2}$$ where $\lambda \in [0,+\infty)$. Denoting $\mu = 1/\lambda$ we get $z = e_0/2+\mu(y-e_0/2)$ and $z$ belongs to $S$ if and only if $$\Vert y-e_0/2 \Vert^2 \mu^2 + \Re \langle e_0/2,y-e_0/2 \rangle \mu – 3/4 = 0$$ One can verify that this quadratic equation has a unique solution $\mu = \mu(y) > 0$ as its value is strictly negative for $\mu=0$. Considering the formulaes used to solve quadratic equations, we conclude that the function $\mu(y) : B \setminus \{e_0/2\} \to (0,+\infty)$ is continuous. Denoting $\lambda(y)=1/\mu(y)$ we notice that $\lambda(y)$ tends to $0$ as $y$ tends to $e_0/2$. Hence, $\lambda$ can be extended to a continuous function on the full ball $B$.

For $y \in B \setminus \{e_0/2\}$ $x=f^{-1}(y)=\lambda(y) u^{-1}(z/\Vert z \Vert)$ with $z=e_0/2+\mu(y)(y-e_0/2)$. This proves that the map $f^{-1}$ is continuous on $B \setminus \{e_0/2\}$ hence on $B$ as $\Vert f^{-1}(y) \Vert \le \vert \lambda(y) \vert$. Finally $f$ is an homeomorphism from $B$ to $B$.

$\left.f\right|_B$ has no fixed point

If $x = \sum_{n \in \mathbb{Z}} \xi_n e_n = f(x)$ was a fixed point, we would have $\xi_n=\xi_{n-1}$ for all $n \neq 0$ hence $\xi_n=\xi_0$ for all $n \ge 0$ and $\xi_{-n}=\xi_{-1}$ for all $n \le -1$ which implies $\xi_n = 0$ for all $n \in \mathbb{Z}$ and $x=0$. In contradiction with $f(0)=e_0/2$.

This counterexample is from an exercise of Claude Wagschal French book Topologie et analyse fonctionnelle.