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

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