Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics II / 6

The mean value theorem states that if $f:[a,b]\to \mathbb{R}$ is continuous on $[a,b]$ and differentiable on $(a,b)$, where $a<b$, there is $c\in (a,b)$ such thatNo continuity assumption on the derivative is needed.

The constancy theorem states that a differentiable function on an interval whose derivative vanishes everywhere is constant. At any included endpoints, continuity there is understood.

Indeed, take arbitrary $x<y$ in the interval. The mean value theorem applies on $[x,y]$ and gives $f(y)-f(x)=f^{\prime }(c)(y-x)=0$. Every pair of values is equal, proving the assertion. The interval hypothesis ensures that the whole segment $[x,y]$ lies in the domain.

The derivative of $h(x)=f(x)-\sin x$ vanishes. By the constancy theorem on the interval $I$,This need not hold on a disconnected domain. On $I=(-\infty ,0)\cup (0,\infty )$, define $f(x)=\sin x$ for $x<0$ and $f(x)=\sin x+1$ for $x>0$. The function is differentiable throughout $I$ with $f^{\prime }(x)=\cos x$, but no single $C$ works on both components.

Suppose distinct fixed points $x<y$ existed. Then $h(t)=f(t)-t$ satisfies $h(x)=h(y)=0$. Rolle's theorem supplies $c\in (x,y)$ with $h^{\prime }(c)=0$, hence $f^{\prime }(c)=1$, contrary to the hypothesis. Thus there is at most one fixed point. The argument does not require $f^{\prime }$ to be continuous.

For every real $t$, $f(t)-t=1/(1+e^t)>0$, so there is no fixed point. Direct differentiation givesWriting $u=e^t>0$, the inequality $(1+u{)}^2\ge 4u$ showsIn particular $0<f^{\prime }(t)<1$, but $f^{\prime }(t)\to 1$ as $t\to \pm \infty$. There is no common contraction constant smaller than one. Strict pointwise bounds on the derivative therefore do not suffice for the contraction mapping theorem.

By the mean value theorem, $|f(x)-f(y)|\le A|x-y|$ for all real $x,y$. Thus $f$ is a contraction mapping on the complete metric space $\mathbb{R}$. Starting from arbitrary $x_1$, define $x_{n+1}=f(x_n)$. Induction gives $|x_{n+1}-x_n|\le A^{n-1}|x_2-x_1|$. For $m>n$, the triangle inequality and a geometric series yieldHence $(x_n)$ is a Cauchy sequence and completeness of the real numbers supplies a limit $x$. The Lipschitz continuity of $f$ implies $f(x)=\lim f(x_n)=\lim x_{n+1}=x$. Thus the limit is a fixed point, independently of the initial value. If $y$ were another, $|x-y|\le A|x-y|$ would force $x=y$. This proves the contraction mapping theorem in this setting, with error estimate