= Solution
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 $d(Tx,Ty)\leq qd(x,y)$ for every $x,y\in X$.
Induction gives $d(x_{n+1},x_n)\leq q^n d(x_1,x_0)$. The <triangle inequality> and the <geometric series> therefore give, for $m>n$,
$$
d(x_m,x_n)\leq\sum_{j=n}^{m-1}q^j d(x_1,x_0)
\leq\frac{q^n}{1-q}d(x_1,x_0).
$$
Thus $(x_n)$ is a <Cauchy sequence>. By <completeness> it converges to some $x_*\in X$. The <contraction mapping> is a <continuous map>, so $Tx_*=\lim_n Tx_n=\lim_n x_{n+1}=x_*$. If $y_*$ is another <fixed point>, then $d(x_*,y_*)\leq qd(x_*,y_*)$, forcing $x_*=y_*$. Letting $m\to\infty$ in the estimate also gives the <contraction iteration error bound>:
$$
\boxed{Tx_*=x_*,\qquad d(x_n,x_*)\leq\frac{q^n}{1-q}d(x_1,x_0).}
$$
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 $n=0$.
Back to article page
