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.

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