 Solution

A contraction mapping satisfies $d(Tx,Ty)\le qd(x,y)$ for all $x,y$, with one fixed $0\le q<1$. The contraction mapping theorem states that such a map on a nonempty complete metric space has a unique fixed point and that every sequence of iterates converges to it.

Choose $x_0$ and put $x_{n+1}=T{x}_n$. Iteration gives $d(x_{n+1},x_n)\le q^{n}d(x_1,x_0)$, so for $m>n$,Thus the iterates form a Cauchy sequence and converge by completeness to $x_{\ast }$. The Lipschitz bound gives continuity of $T$, hence $T{x}_{\ast }=x_{\ast }$. Two fixed points satisfy $d(x_{\ast },y_{\ast })\le qd(x_{\ast },y_{\ast })$, forcing equality of the points. Passing $m\to \infty$ also gives the usual geometric error estimate. For $q=0$, convergence occurs after one application.

If $f^k$ is a contraction mapping, let $x_{\ast }$ be its unique fixed point. The commutation identity $f^{k}f=f{f}^k$ gives $f^k(f{x}_{\ast })=f{x}_{\ast }$, so uniqueness implies $f{x}_{\ast }=x_{\ast }$. Any fixed point of $f$ is a fixed point of $f^k$, giving uniqueness, without even assuming $f$ continuous. On the complete discrete metric space $\{0,1,2\}$, the map $f(0)=f(1)=0$, $f(2)=1$ has constant second iterate but is not a contraction: $d(f2,f0)=d(2,0)=1$.

For the specified function, $x+\sin x\ge 0$ on $[0,\infty )$ because it starts at zero and has derivative $1+\cos x\ge 0$. Therefore $f$ maps the complete half-line into itself. Its derivative satisfiesThe mean value theorem therefore makes it a contraction of constant at most $2/3$. Hence