Contraction mapping theorem

A contraction of a nonempty complete metric space has one fixed point, and every orbit converges to it geometrically.

For contraction mappings $f,g$, compare their fixed points using $f$ on both arguments, then bound the change from $f$ to $g$ by the uniform metric. The resulting inequality gives the displayed estimate. Continuity at $f$ only needs its own contraction constant $q_f<1$, even if nearby maps have constants approaching one.

A metric space is complete if and only if every contraction mapping of every nonempty closed subset into itself has a fixed point. For the converse, a nonconvergent Cauchy sequence contains a rapidly approaching subsequence whose point set is closed and whose shift is a contraction without a fixed point. Requiring the property only on the whole space is a different, weaker condition.

A contraction mapping $T$ on a metric space satisfies $d(Tx,Ty)\le qd(x,y)$ for every $x,y$ and some constant $0\le q<1$.

Suppose $T_{\lambda }$ are contractions of one complete metric space with a common constant $k<1$, and $T_{\lambda }x$ depends continuously on $\lambda$ for each fixed $x$. Their fixed points $x_{\lambda }$ depend continuously on $\lambda$, because for a fixed parameter ${\lambda }_0$,

If some iterate $T^n$ is a contraction on a complete metric space, then $T$ itself has exactly one fixed point: uniqueness for $T^n$ forces $T$ to fix its fixed point.

For the Newton map $g(x)=x-f(x)/f^{\prime }(x)$ near a simple root $r$,Bounds $|f^{\prime }|\ge \delta$ and $|f^{\prime \prime }|\le M$ make $|g^{\prime }|<1$ on a sufficiently small closed interval around $r$. The contraction mapping theorem then gives the unique local fixed point.