Past exam of the fixed point methods for nonlinear partial differential equations course of the University of Oxford / 2023 / 1 / a

The Banach fixed-point theorem states that a contraction mapping $T:X\to X$ of a nonempty complete metric space $(X,d)$ has a unique fixed point, and iterations of a map $x_{n+1}=T(x_n)$ converge to it from every $x_0\in X$. Precisely, assume a single constant $q\in [0,1)$ satisfies $d(Tx,Ty)\le qd(x,y)$ for every $x,y\in X$.

Induction gives $d(x_{n+1},x_n)\le q^{n}d(x_1,x_0)$. The triangle inequality and the geometric series therefore give, for $m>n$,Thus $(x_n)$ is a Cauchy sequence. By completeness it converges to some $x_{\ast }\in X$. The contraction mapping is a continuous map, so $T{x}_{\ast }={\lim }_{n}T{x}_n={\lim }_n{x}_{n+1}=x_{\ast }$. If $y_{\ast }$ is another fixed point, then $d(x_{\ast },y_{\ast })\le qd(x_{\ast },y_{\ast })$, forcing $x_{\ast }=y_{\ast }$. Letting $m\to \infty$ in the estimate also gives the contraction iteration error bound:If $q=0$, $T$ is constant and the sequence reaches its unique fixed point after the first step; this also removes any ambiguity about $0^0$ in the displayed estimate at $n=0$.

The native paper prints $n\ge 1$ while specifying $x_0$ alone. This does not determine $x_1$. We use the intended convention $x_{n+1}=\sin x_n$ for $n\ge 0$; even with an independently chosen real $x_1$, the same proof gives the requested convergence.

For the sine fixed-point iteration, after one step $x_1\in [-1,1]$. On this interval, the sine has the same sign as its argument, andFor example, $t-\sin t={\int }_0^t(1-\cos s)ds>0$, since the integrand is positive for $0<s\le 1$. By oddness of the sine, $a_n=|x_n|$ consequently satisfies $a_{n+1}=\sin a_n\le a_n$ for $n\ge 1$. This monotone sequence is bounded below by zero, so it has a limit $a\in [0,1]$. The continuity of the sine gives $a=\sin a$, and the strict inequality above forces $a=0$.

Hence $x_n\to 0$, and $0$ is a fixed point of the sine. It is the only real fixed point: every such point has absolute value at most one, where the preceding strict inequality applies. ThusThe Banach fixed-point theorem does not directly apply to the sine on $[-1,1]$: the ratios $|\sin t-\sin 0|/|t-0|$ tend to one as $t\to 0$, so no uniform contraction mapping constant smaller than one exists there.