= 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 $x_{n+1}=Tx_n$. Iteration gives $d(x_{n+1},x_n)\le q^nd(x_1,x_0)$, so for $m>n$,
$$
d(x_m,x_n)\le\sum_{j=n}^{m-1}q^jd(x_1,x_0)\le\frac{q^n}{1-q}d(x_1,x_0).
$$
Thus the iterates form a <Cauchy sequence> and converge by completeness to $x_*$. The Lipschitz bound gives continuity of $T$, hence $Tx_*=x_*$. Two fixed points satisfy $d(x_*,y_*)\le qd(x_*,y_*)$, 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 $x_*$ be its unique fixed point. The commutation identity $f^kf=ff^k$ gives $f^k(fx_*)=fx_*$, so uniqueness implies $fx_*=x_*$. 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, $x+\sin x\ge0$ on $[0,\infty)$ because it starts at zero and has derivative $1+\cos x\ge0$. Therefore $f$ maps the complete half-line into itself. Its derivative satisfies
$$
-\frac13\le f'(x)=\frac13\left(1+\cos x-\frac1{(x+1)^2}\right)\le\frac23.
$$
The <mean value theorem> therefore makes it a contraction of constant at most $2/3$. Hence
$$
\boxed{f\text{ has exactly one fixed point on }[0,\infty).}
$$
Back to article page
