The Fixpoint Theorem of Brouwer

Norbert A'Campo

The following proof of the Fixpoint Theorem of Brouwer was outlined as classroom exercise at the University of Montpellier in the early sixties.

Theorem 1 (Brouwer)   Let $D$ be the unit Euclidean ball in ${\bf R}^n$ and let $f:D \to D$ be a continuous mapping. Then there exists $x \in D$ with $f(x)=x$.

a. Reduction to the $C^1$-differentiable case. Let $r: {\bf R}^n \to {\bf R}$ be the function $r(x)=r(x_1,x_2, \dots ,x_n)=\sqrt{x_1^2+x_2^2+ \dots +x_n^2}$. Let $\phi:{\bf R}^n \to {\bf R}$ given by: $\phi(x)=a_n(1/4r(x)^4-1/2r(x)^2+1/4)$ on $D$ and equal to $0$ in the complement of $D$, where the constant $a_n$ is chosen such that the integral of $\phi$ over ${\bf R}^n$ equals $1$. Let $f:D \to D$ be continuous. Let $F:{\bf R}^n \to D$ be the extension of $f$ to ${\bf R}^n$, for which we have $F(x)=f(x/{\vert\vert x\vert\vert)}$ on ${\bf R}^n \setminus D$. For $k\in {\bf N}, k\not=0,$ put $f_k(x):=\int_{{\bf R}^n}k^n\phi(ky)F(x-y)dy$. Show that the restriction of $f_k$ to $D$ maps $D$ into $D$. Show that the mappings $F_k$ are continuously differentiable and approximate in the topology of uniform convergence the mapping $F$. Show that if there exists a continuous mapping $f:D \to D$ without fixpoints, then there will also exist a continuously differential map without fixpoints. It follows, that it suffices to proof the Brouwer Fixpoint Theorem only for continuously differentiable mappings.

b. Proof for $C^1$-differentiable mappings. The proof is by contradiction. Assume, that the continuously differential mapping $f:D \to D$ has no fixpoints. Let $g:D \to \partial{D}$ the mapping, such that for every point $x \in D$ the points $f(x),x,g(x)$ are in that order on a line of ${\bf R}^n$. The mapping $g$ is also continuously differentiable and satisfies $g(x)=x$ for $x\in \partial{D}$. We write $g(x)=(g_1(x),g_2(x), \dots ,g_n(x))$ and have for $x\in \partial{D}$ the identities: $g_i(x_1,x_2, \dots ,x_n)=x_i, i=1 \dots n$. Observe $dg_1 \wedge dg_2 \wedge \dots \wedge dg_n=0$ since $g_1^2+g_2^2+ \dots +g_n^2=1$ holds. The following is a contradiction:

$$0\not=\int_D dx_1 \wedge dx_2 \wedge \dots \wedge dx_n=$$

$$\int_{\partial{D}} x_1 \wedge dx_2 \wedge \dots \wedge dx_n=$$

$$\int_{\partial{D}} g_1 \wedge dg_2 \wedge \dots \wedge dg_n=$$

$$\int_{D} dg_1 \wedge dg_2 \wedge \dots \wedge dg_n=$$

$$\int_{D} 0=0$$