Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 2 / 11F (Topics in Analysis)

The planar Brouwer fixed-point theorem states that every continuous map from a nonempty compact convex subset of ${\mathbb{R}}^2$ to itself has a fixed point. In particular it applies to a closed disk or a closed triangle. Convexity and compactness are essential hypotheses in this formulation.

Let $v_I,v_J,v_K$ be the vertices opposite the corresponding sides, and let ${\lambda }_I,{\lambda }_J,{\lambda }_K$ be their barycentric coordinates. We prove the equivalence by negating each assertion.

Suppose the closed covering sets have empty triple intersection. For $x\in T$ put $d_I(x)=\mathrm{dist}(x,A)$, and define $d_J,d_K$ similarly. At least one distance is positive, since $x$ cannot belong to all three sets, and at least one is zero, since they cover $T$. Consequentlyis continuous and lies in the boundary: one barycentric coordinate vanishes. On $I$ its first coefficient is zero because $I\subseteq A$, so $f(I)\subseteq I$, and likewise for the other sides. This contradicts assertion (1), proving (1) implies (2).

Conversely, suppose such a boundary-valued map exists. Define closed subsets of ${\mathbb{R}}^2$ byThey are closed because $T$ is compact and the coordinate functions are continuous. They cover $T$ because $f$ is boundary-valued; the side-preserving condition gives $I\subseteq A$, $J\subseteq B$, $K\subseteq C$. Their triple intersection is empty because barycentric coordinates sum to one. Thus failure of (1) implies failure of (2), completing the equivalence.

In fact both assertions hold. If the map in (1) existed, let $R$ rotate the equilateral triangle through $120^{\circ }$. Brouwer fixed-point theorem applied to $R\circ f:T\to T$ gives a fixed point $x$ on the boundary. On any side $I$ containing $x$, side preservation puts $x$ also on $R(I)$, so it is their common vertex. A side-preserving map fixes every vertex because that vertex lies on two sides. Therefore $x=Rf(x)=Rx$, impossible for a noncentral vertex. This also proves the covering-intersection assertion through the established equivalence.

On the unit circle the two equations reduce to the same conditionsince $y^2=1-x^2$. Define $H(x,y)=x^2(f(x,y{)}^2+g(x,y{)}^2)-f(x,y{)}^2$. At either point $(\pm 1,0)$ it is strictly positive, and at either point $(0,\pm 1)$ it is strictly negative. On each of the four closed quarter-circle arcs, continuity and the intermediate value theorem give a zero in its interior. These four interiors are disjoint, so there are at least four distinct solutions, one in each open quadrant. No solution here lies on an axis because $f,g$ are strictly positive.