# A homeomorphism of the unit ball having no fixed point

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.