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# Banach fixed-point theorem

In mathematics, the Banach fixed-point theorem (also known as the contraction mapping theorem or contraction mapping principle) is an important tool in the theory of metric spaces; it guarantees the existence and uniqueness of fixed points of certain self-maps of metric spaces, and provides a constructive method to find those fixed points. The theorem is named after Stefan Banach (1892–1945), and was first stated by him in 1922.

## The theorem

Let (X, d) be a non-empty complete metric space. Let T : XX be a contraction mapping on X, i.e.: there is a nonnegative real number q < 1 such that $d(T(x),T(y)) \le q\cdot d(x,y)$

for all x, y in X. Then the map T admits one and only one fixed-point x* in X (this means T(x*) = x*). Furthermore, this fixed point can be found as follows: start with an arbitrary element x0 in X and define an iterative sequence by xn = T(xn−1) for n = 1, 2, 3, ... This sequence converges, and its limit is x*. The following inequality describes the speed of convergence: $d(x^*, x_n) \leq \frac{q^n}{1-q} d(x_1,x_0).$

Equivalently, $d(x^*, x_{n+1}) \leq \frac{q}{1-q} d(x_{n+1},x_n)$

and $d(x^*, x_{n+1}) \leq q d(x^*,x_n).$

Any such value of q is called a Lipschitz constant for T, and the smallest one is sometimes called "the best Lipschitz constant" of T.

Note that the requirement d(T(x)T(y)) < d(xy) for all unequal x and y is in general not enough to ensure the existence of a fixed point, as is shown by the map T : [1,∞) → [1,∞) with T(x) = x + 1/x, which lacks a fixed point. However, if the metric space 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(xT(x)) : indeed, a minimizer exists by compactness, and has to be a fixed point of T. It then easily follows that the fixed point is the limit of any sequence of iterations of T.

When using the theorem in practice, the most difficult part is typically to define X properly so that T actually maps elements from X to X, i.e. that T(x) is always an element of X.

## Proof

Choose any $x_0 \in (X, d)$. For each $n \in \{1, 2, \ldots\}$, define $x_n = T(x_{n-1})\,\!$. We claim that for all $n \in \{1, 2, \dots\}$, the following is true: $d(x_{n+1}, x_n) \leq q^n d(x_1, x_0). \,$

To show this, we will proceed using induction. The above statement is true for the case $n = 1\,\!$, for $d(x_{1+1}, x_1) = d(x_2, x_1) = d(T(x_1), T(x_0)) \leq qd(x_1, x_0). \,$

Suppose the above statement holds for some $k \in \{1, 2, \ldots\}$. Then we have \begin{align} d(x_{(k + 1) + 1}, x_{k + 1}) & = d(x_{k + 2}, x_{k + 1}) \\[6pt] & = d(T(x_{k + 1}), T(x_k)) \\[6pt] & \leq q d(x_{k + 1}, x_k) \\[6pt] & \leq q \cdot q^kd(x_1, x_0) \\[6pt] & = q^{k + 1}d(x_1, x_0). \end{align}

The inductive assumption is used going from line three to line four. By the principle of mathematical induction, for all $n \in \{1, 2, \ldots\}$, the above claim is true.

Let $\epsilon > 0\,\!$. Since $0 \leq q < 1$, we can find a large $N \in \{1, 2, \ldots\}$ so that $q^N < \frac{\epsilon(1-q)}{d(x_1, x_0)}.$

Using the claim above, we have that for any $m\,\!$, $n \in \{0, 1, \ldots\}$ with $m > n \geq N$, \begin{align} d\left(x_m, x_n\right) & \leq d(x_m, x_{m-1}) + d(x_{m-1}, x_{m-2}) + \cdots + d(x_{n+1}, x_n) \\[6pt] & \leq q^{m-1}d(x_1, x_0) + q^{m-2}d(x_1, x_0) + \cdots + q^nd(x_1, x_0) \\[6pt] & = d(x_1, x_0)q^n \cdot \sum_{k=0}^{m-n-1} q^k \\[6pt] & < d(x_1, x_0)q^n \cdot \sum_{k=0}^\infty q^k \\[6pt] & = d(x_1, x_0)q^n \frac{1}{1-q} \\[6pt] & = q^n \frac{d(x_1, x_0)}{1-q} \\[6pt] & < \frac{\epsilon(1-q)}{d(x_1, x_0)}\cdot\frac{d(x_1, x_0)}{1-q} \\[6pt] & = \epsilon. \end{align}

The inequality in line one follows from repeated applications of the triangle inequality; the series in line four is a geometric series with $0 \leq q < 1$ and hence it converges. The above shows that $\{x_n\}_{n\geq 0}$ is a Cauchy sequence in $(X, d)\,\!$ and hence convergent by completeness. So let $x^* = \lim_{n\to\infty} x_n$. We make two claims: (1) $x^*\,\!$ is a fixed point of $T\,\!$. That is, $T(x^*) = x^*\,\!$; (2) $x^*\,\!$ is the only fixed point of $T\,\!$ in $(X, d)\,\!$.

To see (1), we take the limit of both sides of the recurrence $x_n = T(x_{n-1})$, $\lim_{n\to\infty} x_n = \lim_{n\to\infty} T(x_{n-1})$

Since T is a contraction mapping, it is continuous, so we may take the limit inside: $\lim_{n\to\infty} x_n = T\left(\lim_{n\to\infty} x_{n-1} \right)$. Thus, $x^* = T(x^*)\,\!$.

To show (2), we suppose that $y\,\!$ also satisfies $T(y) = y\,\!$. Then $0 \leq d(x^*, y) = d(T(x^*), T(y)) \leq q d(x^*, y).$

Remembering that $0 \leq q < 1$, the above implies that $0 \leq (1-q) d(x^*, y) \leq 0$, which shows that $d(x^*, y) = 0\,\!$, whence by positive definiteness, $x^* = y\,\!$ and the proof is complete.

The above is essentially Banach's original proof. What follows is a considerably simpler proof that appeared recently in the Journal of Fixed Point Theory and its Application (see reference).

By the triangle inequality, for any x and y , in X, $d(x,y) \le d(x,T(x)) + d(T(x),T(y)) + d(T(y),y) \le d(x,T(x)) + q d(x,y) + d(T(y),y)$

so subtracting the middle term on the right from both sides and dividing by the positive quantity (1-q) we get the Fundamental Contraction Inequality": $d(x,y) \le \frac {1} {1-q} (d(T(x), x) + d(T(y),y)),$

and we note that if x and y are both fixed points then this implies that d(x,y) = 0, so x =y, proving that $T$ has at most one fixed point. Now define the mapping $T^n$ by composing $T$ with itself n times and note by induction that it satisfies a Lipschitz condition with constant $q^n$. It remains to show that for any $x _0 \in X$, the sequence $T^n (x_0)$ is Cauchy and so converges to a point $x^*$ of X, which as noted above is clearly a fixed point of $T$. If in the Fundamental Inequality we replace x and y by $T^n(x_0)$ and $T^m(x_0)$, we find that $d(T^n(x_0), T^m(x_0)) \le \frac {1} {1-q} (d(T^n(T(x_0)) ,T^n(x_0)) + d(T^m(T(x_0)), T^m(x_0)))$

or $d(T^n(x_0), T^m(x_0)) \le \frac {q^n + q^m} {1-q} d(T(x_0) ,x_0)$

and since q<1 , the right hand side of this latter inequality converges to zero as n and m tend to infinity, proving that $T^n(x_0)$ is Cauchy. Note also that letting m tend to infinity in the latter inequality gives the inequality $d(T^n(x_0), x^*) \le \frac {q^n} {1-q} d(T(x_0) ,x_0)$ derived in the first proof that gives the rate at which $T^n(x_0)$ converges to $x^*$.

## Applications

• A standard application is the proof of the Picard–Lindelöf theorem about the existence and uniqueness of solutions to certain ordinary differential equations. The sought solution of the differential equation is expressed as a fixed point of a suitable integral operator which transforms continuous functions into continuous functions. The Banach fixed-point theorem is then used to show that this integral operator has a unique fixed point.
• One consequence of the Banach fixed-point theorem is that small Lipschitz perturbation of the identity are bi-lipschitz homeomorphisms. Let $\Omega$ be an open set of a Banach space $E$; let $I:\Omega\to E$ denote the identity (inclusion) map and let $g:\Omega\to E$ be a Lipschitz map of constant k<1. Then (i) $\Omega':=(I+g)(\Omega)$ is an open subset of $E$ :precisely, for any $x\in\Omega$ such that $B(x,r)\subset\Omega$ one has $\textstyle B\left((I+g)(x),\, r(1-k)\right)\subset\Omega'$ ; (ii) $I+g:\Omega\to\Omega'$

is a bi-lipschitz homeomorphism; precisely, $(I+g)^{-1}$ is still of the form $I+h:\Omega'\to\Omega$,

with $h$ a Lipschitz map of constant $k/(1-k).$

A direct consequence of this result yields the proof of the inverse function theorem.

## Converses

Several converses of the Banach contraction principle exist. The following is due to Czesław Bessaga, from 1959:

Let $f\colon X \to X$ be a map of an abstract set such that each iterate ƒn has a unique fixed point. Let q be a real number, 0 < q < 1. Then there exists a complete metric on X such that ƒ is contractive, and q is the contraction constant.

Indeed, very weak assumptions suffice to obtain such a kind of converse. E.g. if $f\colon X \to X$ is a map on a T1 topological space (i.e., a T1 space) with a unique fixed point a, such that for each $x\in X$ we have that ƒn(x) converges to 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 $\frac{1}{2}$. In this case the metric is in fact an ultrametric.

## Generalizations

There are a number of generalizations as immediate corollaries, which are of some interest for the sake of applications. Let $T:X\to X$ be a map on a complete non-empty metric space.

• Assume that some iterate $T^n$ of T is a contraction. Then T has a unique fixed point.
• Assume that for all $x$ and $y$ in $X$, $\sum_n d(T^n(x),T^n(y))<\infty.$ Then T has a unique fixed point.

However, in most applications the existence and unicity of a fixed point 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.

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. Some of these have applications, e.g., in the theory of programming semantics in theoretical computer science.

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