Past exam of the mathematics course of the University of Cambridge / 2001 / ia / Paper 1 / 10D (Analysis I)

To prove the intermediate value theorem in this setting, let $A=\{t\in [a,b]:f(t)<v\}$ and $c=\sup A$. The endpoint inequalities and continuity give an interval immediately to the right of $a$ contained in $A$ and an interval immediately to the left of $b$ disjoint from $A$. Thus $a<c<b$. Choose points of $A$ tending to $c$; continuity gives $f(c)\le v$. If $f(c)<v$, continuity would place some point larger than $c$ in $A$, contradicting the supremum. Hence $\boxed{f(c)=v}$ at an interior point.

For a continuous self-map of $[a,b]$, define $g(t)=f(t)-t$. The range condition implies $g(a)\ge 0$ and $g(b)\le 0$. If either is zero, its endpoint is a fixed point. Otherwise apply the just-proved intermediate value theorem to $-g$ with target zero to obtain $d\in (a,b)$ and $g(d)=0$. ThereforeThis proves the fixed-point property of a closed interval without requiring $f$ itself to be monotone.

The fixed-point property fails on an open interval. The continuous functionmaps $(a,b)$ into itself, because $a<t<f(t)<b$. Its only possible fixed point satisfies $t=(t+b)/2$, hence $t=b$, which is outside the domain. It is therefore a counterexample even with the self-map requirement retained.

The fixed-point property also fails on the real line. The continuous function $\boxed{f(t)=t+1}$ maps $\mathbb{R}$ into itself but has no fixed point, since $t+1=t$ is impossible.