Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 4 / 12G (Topics in Analysis)

The Brouwer fixed-point theorem says that every continuous self-map of the closed unit disc has a fixed point.

A union of two disjoint closed discs does not have the fixed-point property. Define a map that is constant from the first component to a point in the second and constant from the second component to a point in the first. It is continuous because the components are disjoint and has no fixed point.

The open interval does not have the fixed-point property. The continuous self-maphas its only possible fixed point at the excluded endpoint $(0,0)$.

The closed interval does have the fixed-point property. If $f(x,0)=(F(x),0)$, then $F(0)-0\ge 0$ and $F(1)-1\le 0$. The intermediate value theorem supplies $x\in [0,1]$ with $F(x)-x=0$.

The annulus does not have the fixed-point property. The antipodal mapis a continuous self-map of the annulus and has no fixed point there.

Let $T:{\mathbb{R}}^2\to D$ be the nearest-point projection,It is continuous. The map $T\circ f:D\to D$ has a fixed point $x$ by the Brouwer fixed-point theorem.

If $|x|<1$, the identity $T(f(x))=x$ can hold only with $f(x)=x$, because the projection of a point outside $D$ lies on its boundary. If $|x|=1$, the hypothesis gives $|f(x)|\le 1$, so $T(f(x))=f(x)$ and again $f(x)=x$. Thus $f$ has a fixed point.

The set $E={\Delta }_2\times {\Delta }_2$ is a compact convex square. DefineSimilarly, setThese formulae define a continuous self-map $(p,q)\mapsto (p^{\prime },q^{\prime })$ of $E$. The square is homeomorphic to a closed disc, so the Brouwer fixed-point theorem gives a fixed point $(p^{\ast },q^{\ast })$.

Let $U=u_1+u_2$ at that fixed point. From $p^{\prime }=p$,If $U>0$, every $i$ with $p_i>0$ has $A(e_i,q)-A(p,q)>0$. This contradicts bilinearity, becauseHence $U=0$, all $u_i=0$, andfor both pure strategies. By linearity this holds for every mixed strategy $p$. The same argument with $v_j$ givesfor every $q$. Therefore

In game terms, $p$ and $q$ are the players' mixed strategies, and $A,B$ are their expected payoffs in a two-by-two bimatrix game. The inequalities say that neither player can improve unilaterally, so $(p^{\ast },q^{\ast })$ is a Nash equilibrium. This is the Brouwer proof of Nash equilibrium for a two-by-two game.