Definition. Let ( X , d ) {\displaystyle (X,d)} be a metric space. Then a map T : X → X {\displaystyle T:X\to X} is called a contraction mapping on X if there exists q ∈ [ 0 , 1 ) {\displaystyle q\in [0,1)} such that
for all x , y ∈ X . {\displaystyle x,y\in X.}
Banach fixed-point theorem. Let ( X , d ) {\displaystyle (X,d)} be a non-empty complete metric space with a contraction mapping T : X → X . {\displaystyle T:X\to X.} Then T admits a unique fixed-point x ∗ {\displaystyle x^{*}} in X (i.e. T ( x ∗ ) = x ∗ {\displaystyle T(x^{*})=x^{*}} ). Furthermore, x ∗ {\displaystyle x^{*}} can be found as follows: start with an arbitrary element x 0 ∈ X {\displaystyle x_{0}\in X} and define a sequence ( x n ) n ∈ N {\displaystyle (x_{n})_{n\in \mathbb {N} }} by x n = T ( x n − 1 ) {\displaystyle x_{n}=T(x_{n-1})} for n ≥ 1. {\displaystyle n\geq 1.} Then lim n → ∞ x n = x ∗ {\displaystyle \lim _{n\to \infty }x_{n}=x^{*}} .
Remark 1. The following inequalities are equivalent and describe the speed of convergence:
Any such value of q is called a Lipschitz constant for T {\displaystyle T} , and the smallest one is sometimes called "the best Lipschitz constant" of T {\displaystyle T} .
Remark 2. d ( T ( x ) , T ( y ) ) < d ( x , y ) {\displaystyle d(T(x),T(y))<d(x,y)} for all x ≠ y {\displaystyle x\neq y} is in general not enough to ensure the existence of a fixed point, as is shown by the map
which lacks a fixed point. However, if X {\displaystyle X} is compact, then this weaker assumption does imply the existence and uniqueness of a fixed point, that can be easily found as a minimizer of d ( x , T ( x ) ) {\displaystyle d(x,T(x))} , indeed, a minimizer exists by compactness, and has to be a fixed point of T . {\displaystyle T.} It then easily follows that the fixed point is the limit of any sequence of iterations of T . {\displaystyle T.}
Remark 3. When using the theorem in practice, the most difficult part is typically to define X {\displaystyle X} properly so that T ( X ) ⊆ X . {\displaystyle T(X)\subseteq X.}
Let x 0 ∈ X {\displaystyle x_{0}\in X} be arbitrary and define a sequence ( x n ) n ∈ N {\displaystyle (x_{n})_{n\in \mathbb {N} }} by setting x n = T ( x n − 1 ) {\displaystyle x_{n}=T(x_{n-1})} . We first note that for all n ∈ N , {\displaystyle n\in \mathbb {N} ,} we have the inequality
This follows by induction on n {\displaystyle n} , using the fact that T {\displaystyle T} is a contraction mapping. Then we can show that ( x n ) n ∈ N {\displaystyle (x_{n})_{n\in \mathbb {N} }} is a Cauchy sequence. In particular, let m , n ∈ N {\displaystyle m,n\in \mathbb {N} } such that m > n {\displaystyle m>n} :
Let ε > 0 {\displaystyle \varepsilon >0} be arbitrary. Since q ∈ [ 0 , 1 ) {\displaystyle q\in [0,1)} , we can find a large N ∈ N {\displaystyle N\in \mathbb {N} } so that
Therefore, by choosing m {\displaystyle m} and n {\displaystyle n} greater than N {\displaystyle N} we may write:
This proves that the sequence ( x n ) n ∈ N {\displaystyle (x_{n})_{n\in \mathbb {N} }} is Cauchy. By completeness of ( X , d ) {\displaystyle (X,d)} , the sequence has a limit x ∗ ∈ X . {\displaystyle x^{*}\in X.} Furthermore, x ∗ {\displaystyle x^{*}} must be a fixed point of T {\displaystyle T} :
As a contraction mapping, T {\displaystyle T} is continuous, so bringing the limit inside T {\displaystyle T} was justified. Lastly, T {\displaystyle T} cannot have more than one fixed point in ( X , d ) {\displaystyle (X,d)} , since any pair of distinct fixed points p 1 {\displaystyle p_{1}} and p 2 {\displaystyle p_{2}} would contradict the contraction of T {\displaystyle T} :
Several converses of the Banach contraction principle exist. The following is due to Czesław Bessaga, from 1959:
Let f : X → X be a map of an abstract set such that each iterate fn has a unique fixed point. Let q ∈ ( 0 , 1 ) , {\displaystyle q\in (0,1),} then there exists a complete metric on X such that f is contractive, and q is the contraction constant.
Indeed, very weak assumptions suffice to obtain such a kind of converse. For example if f : X → X {\displaystyle f:X\to X} is a map on a T1 topological space with a unique fixed point a, such that for each x ∈ X {\displaystyle x\in X} we have fn(x) → a, then there already exists a metric on X with respect to which f satisfies the conditions of the Banach contraction principle with contraction constant 1/2.8 In this case the metric is in fact an ultrametric.
There are a number of generalizations (some of which are immediate corollaries).9
Let T : X → X be a map on a complete non-empty metric space. Then, for example, some generalizations of the Banach fixed-point theorem are:
In applications, the existence and uniqueness of a fixed point often can be shown directly with the standard Banach fixed point theorem, by a suitable choice of the metric that makes the map T a contraction. Indeed, the above result by Bessaga strongly suggests to look for such a metric. See also the article on fixed point theorems in infinite-dimensional spaces for generalizations.
In a non-empty compact metric space, any function T {\displaystyle T} satisfying d ( T ( x ) , T ( y ) ) < d ( x , y ) {\displaystyle d(T(x),T(y))<d(x,y)} for all distinct x , y {\displaystyle x,y} , has a unique fixed point. The proof is simpler than the Banach theorem, because the function d ( T ( x ) , x ) {\displaystyle d(T(x),x)} is continuous, and therefore assumes a minimum, which is easily shown to be zero.
A different class of generalizations arise from suitable generalizations of the notion of metric space, e.g. by weakening the defining axioms for the notion of metric.10 Some of these have applications, e.g., in the theory of programming semantics in theoretical computer science.11
An application of the Banach fixed-point theorem and fixed-point iteration can be used to quickly obtain an approximation of π with high accuracy. Consider the function f ( x ) = sin ( x ) + x {\displaystyle f(x)=\sin(x)+x} . It can be verified that π is a fixed point of f, and that f maps the interval [ 3 π / 4 , 5 π / 4 ] {\displaystyle \left[3\pi /4,5\pi /4\right]} to itself. Moreover, f ′ ( x ) = 1 + cos ( x ) {\displaystyle f'(x)=1+\cos(x)} , and it can be verified that
on this interval. Therefore, by an application of the mean value theorem, f has a Lipschitz constant less than 1 (namely 1 − 1 / 2 {\displaystyle 1-1/{\sqrt {2}}} ). Applying the Banach fixed-point theorem shows that the fixed point π is the unique fixed point on the interval, allowing for fixed-point iteration to be used.
For example, the value 3 may be chosen to start the fixed-point iteration, as 3 π / 4 ≤ 3 ≤ 5 π / 4 {\displaystyle 3\pi /4\leq 3\leq 5\pi /4} . The Banach fixed-point theorem may be used to conclude that
Applying f to 3 only three times already yields an expansion of π accurate to 33 digits:
This article incorporates material from Banach fixed point theorem on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.
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