# Contraction mapping

(Redirected from Subcontraction map)

In mathematics, a contraction mapping, or contraction or contractor, on a metric space (M, d) is a function f from M to itself, with the property that there is some nonnegative real number ${\displaystyle 0\leq k<1}$ such that for all x and y in M,

${\displaystyle d(f(x),f(y))\leq k\,d(x,y).}$

The smallest such value of k is called the Lipschitz constant of f. Contractive maps are sometimes called Lipschitzian maps. If the above condition is instead satisfied for k ≤ 1, then the mapping is said to be a non-expansive map.

More generally, the idea of a contractive mapping can be defined for maps between metric spaces. Thus, if (M, d) and (N, d') are two metric spaces, then ${\displaystyle f:M\rightarrow N}$ is a contractive mapping if there is a constant ${\displaystyle 0\leq k<1}$ such that

${\displaystyle d'(f(x),f(y))\leq k\,d(x,y)}$

for all x and y in M.

Every contraction mapping is Lipschitz continuous and hence uniformly continuous (for a Lipschitz continuous function, the constant k is no longer necessarily less than 1).

A contraction mapping has at most one fixed point. Moreover, the Banach fixed-point theorem states that every contraction mapping on a non-empty complete metric space has a unique fixed point, and that for any x in M the iterated function sequence x, f (x), f (f (x)), f (f (f (x))), ... converges to the fixed point. This concept is very useful for iterated function systems where contraction mappings are often used. Banach's fixed-point theorem is also applied in proving the existence of solutions of ordinary differential equations, and is used in one proof of the inverse function theorem.[1]

Contraction mappings play an important role in dynamic programming problems.[2][3]

## Firmly non-expansive mapping

A non-expansive mapping with ${\displaystyle k=1}$  can be strengthened to a firmly non-expansive mapping in a Hilbert space ${\displaystyle {\mathcal {H}}}$  if the following holds for all x and y in ${\displaystyle {\mathcal {H}}}$ :

${\displaystyle \|f(x)-f(y)\|^{2}\leq \,\langle x-y,f(x)-f(y)\rangle .}$

where

${\displaystyle d(x,y)=\|x-y\|}$ .

This is a special case of ${\displaystyle \alpha }$  averaged nonexpansive operators with ${\displaystyle \alpha =1/2}$ .[4] A firmly non-expansive mapping is always non-expansive, via the Cauchy–Schwarz inequality.

The class of firmly non-expansive maps is closed under convex combinations, but not compositions.[5] This class includes proximal mappings of proper, convex, lower-semicontinuous functions, hence it also includes orthogonal projections onto non-empty closed convex sets. The class of firmly nonexpansive operators is equal to the set of resolvents of maximally monotone operators[6]. Surprisingly, while iterating non-expansive maps has no guarantee to find a fixed point (e.g. multiplication by -1), firm non-expansiveness is sufficient to guarantee global convergence to a fixed point, provided a fixed point exists. More precisely, if ${\displaystyle {\text{Fix}}f:=\{x\in {\mathcal {H}}\ |\ f(x)=x\}\neq \varnothing }$ , then for any initial point ${\displaystyle x_{0}\in {\mathcal {H}}}$ , iterating

${\displaystyle (\forall n\in \mathbb {N} )\quad x_{n+1}=f(x_{n})}$

yields convergence to a fixed point ${\displaystyle x_{n}\to z\in {\text{Fix}}f}$ . This convergence might be weak in an infinite-dimensional setting.[5]

## Subcontraction map

A subcontraction map or subcontractor is a map f on a metric space (M, d) such that

${\displaystyle d(f(x),f(y))\leq d(x,y);}$
${\displaystyle d(f(f(x)),f(x))

If the image of a subcontractor f is compact, then f has a fixed point.[7]

## Locally convex spaces

In a locally convex space (E, P) with topology given by a set P of seminorms, one can define for any p ∈ P a p-contraction as a map f such that there is some kp < 1 such that p(f(x) − f(y))kp p(xy). If f is a p-contraction for all p ∈ P and (E, P) is sequentially complete, then f has a fixed point, given as limit of any sequence xn+1 = f(xn), and if (E, P) is Hausdorff, then the fixed point is unique.[8]