Past exam of the mathematics course of the University of Cambridge / 2011 / ia / Paper 1 / 10E (Analysis I)

The intermediate value theorem states that if $f:[a,b]\to \mathbb{R}$ is continuous, then every real value between $f(a)$ and $f(b)$ is attained at some $c\in [a,b]$. Here is a proof using completeness of the real numbers. Endpoint values are immediate, so suppose $f(a)<y<f(b)$; reversing the sign of $f$ handles the other ordering. LetThe set is nonempty and bounded above. Continuity at $a$ puts points to the right of $a$ in $S$, while continuity at $b$ excludes an entire interval to the left of $b$, so $a<c<b$. If $f(c)<y$, continuity puts some point larger than $c$ in $S$, contradicting its supremum. If $f(c)>y$, continuity excludes $S$ in a neighborhood of $c$, contradicting the existence of points of $S$ arbitrarily close to its supremum from below. Therefore $f(c)=y$, proving the theorem.

For the fixed-point assertion, put $g(x)=f(x)-x$. This is continuous, with $g(0)=f(0)\ge 0$ and $g(1)=f(1)-1\le 0$. If either endpoint value is zero it already gives a fixed point; otherwise the intermediate value theorem gives an interior zero. HenceThe proof needs both continuity and the closed-interval endpoint information; the following counterexamples separate those hypotheses.

No. The continuous function $f(x)=(x+1)/2$ maps $(0,1)$ into $(0,1)$, but $f(x)-x=(1-x)/2>0$ throughout the domain. Its only possible fixed point is the excluded endpoint one.

No. The continuous function $f(x)=x+1$ is a self-map of $\mathbb{R}$, but the fixed point equation would require $x+1=x$, which is impossible.

No. Define $f(x)=1$ for $0\le x<1$ and $f(1)=0$. This self-map of $[0,1]$ is discontinuous at one. Every point smaller than one is sent to a different point, and one is sent to zero, so it has no fixed point.

Yes: a tangential zero and a crossing zero can give exactly two fixed points. One explicit continuous piecewise linear example isAdjacent formulas agree at each junction. Their endpoint values lie in $[0,1]$, so linear interpolation proves that the whole image lies in $[0,1]$; also $f(0)=1$ and $f(1)=0$. On the four pieces, $f(x)-x$ is respectively $1-3x$, $3x/2-1/2$, $1-3x/2$, and $2-3x$. Their only zeros on their respective intervals are $1/3$ and $2/3$. Thus the complete fixed point set is $\boxed{\{1/3,2/3\}}$. Continuity forces at least one zero, but does not force every zero to change sign.