Banach fixed point theorem

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 : X → X be a contraction mapping on X, i.e: there is a nonnegative real number q < 1 such that
 * $$d(Tx,Ty) \le q\cdot d(x,y)$$

for all x, y in X. Then the map T admits one and only one fixed point x* in X (this means Tx* = 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 = Txn−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
 * $$d(x^*, x_{n+1}) \leq q d(x_n,x^*).$$

The smallest such value of q is sometimes called the Lipschitz constant.

Note that the requirement d(Tx, Ty) < d(x, y) 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,&infin;) &rarr; [1,&infin;) with T(x) = x + 1/x, which lacks a fixed point. However, if the space X is compact, then this weaker assumption does imply all the statements of the theorem.

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 Tx is always an element of X.

Proof
Choose any $$x_0 \in (X, d)$$. For each $$n \in \{1, 2, \ldots\}$$, define $$x_n = Tx_{n-1}\,\!$$. 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


 * $$d(x_{1+1}, x_1) = d(x_2, x_1) = d(Tx_1, Tx_0) \leq qd(x_1, x_0).$$

Suppose the above statement holds for some $$k \in \{1, 2, \ldots\}$$. Then we have




 * $$d(x_{(k + 1) + 1}, x_{k + 1})\,\!$$
 * $$= d(x_{k + 2}, x_{k + 1})\,\!$$
 * $$= d(Tx_{k + 1}, Tx_k)\,\!$$
 * $$\leq q d(x_{k + 1}, x_k)$$
 * $$\leq q \cdot q^kd(x_1, x_0)$$
 * $$= q^{k + 1}d(x_1, x_0).\,\!$$
 * }
 * $$\leq q d(x_{k + 1}, x_k)$$
 * $$\leq q \cdot q^kd(x_1, x_0)$$
 * $$= q^{k + 1}d(x_1, x_0).\,\!$$
 * }
 * $$\leq q \cdot q^kd(x_1, x_0)$$
 * $$= q^{k + 1}d(x_1, x_0).\,\!$$
 * }
 * $$= q^{k + 1}d(x_1, x_0).\,\!$$
 * }

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 $$m > n \geq N$$,




 * $$d\left(x_m, x_n\right)$$
 * $$\leq d(x_m, x_{m-1}) + d(x_{m-1}, x_{m-2}) + \cdots + d(x_{n+1}, x_n)$$
 * $$\leq q^{m-1}d(x_1, x_0) + q^{m-2}d(x_1, x_0) + \cdots + q^nd(x_1, x_0)$$
 * $$= d(x_1, x_0)q^n \cdot \sum_{k=0}^{m-n-1} q^k$$
 * $$< d(x_1, x_0)q^n \cdot \sum_{k=0}^\infty q^k$$
 * $$= d(x_1, x_0)q^n \frac{1}{1-q}$$
 * $$= q^n \frac{d(x_1, x_0)}{1-q}$$
 * $$< \frac{\epsilon(1-q)}{d(x_1, x_0)}\cdot\frac{d(x_1, x_0)}{1-q}$$
 * $$= \epsilon.\,\!$$
 * }
 * $$< d(x_1, x_0)q^n \cdot \sum_{k=0}^\infty q^k$$
 * $$= d(x_1, x_0)q^n \frac{1}{1-q}$$
 * $$= q^n \frac{d(x_1, x_0)}{1-q}$$
 * $$< \frac{\epsilon(1-q)}{d(x_1, x_0)}\cdot\frac{d(x_1, x_0)}{1-q}$$
 * $$= \epsilon.\,\!$$
 * }
 * $$= q^n \frac{d(x_1, x_0)}{1-q}$$
 * $$< \frac{\epsilon(1-q)}{d(x_1, x_0)}\cdot\frac{d(x_1, x_0)}{1-q}$$
 * $$= \epsilon.\,\!$$
 * }
 * $$< \frac{\epsilon(1-q)}{d(x_1, x_0)}\cdot\frac{d(x_1, x_0)}{1-q}$$
 * $$= \epsilon.\,\!$$
 * }
 * $$= \epsilon.\,\!$$
 * }

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, $$Tx^* = x^*\,\!$$; (2) $$x^*\,\!$$ is the only fixed point of $$T\,\!$$ in $$(X, d)\,\!$$.

To see (1), we note that for any $$n \in \{0, 1, \ldots\}$$,


 * $$0 \leq d(x_{n+1}, Tx^*) = d(Tx_n, Tx^*) \leq q d(x_n, x^*).$$

Since $$qd(x_n, x^*) \to 0$$ as $$n \to \infty$$, the squeeze theorem shows that $$\lim_{n\to\infty} d(x_{n+1}, Tx^*) = 0$$. This shows that $$x_n \to Tx^*$$ as $$n \to \infty$$. But $$x_n \to x^*$$ as $$n \to \infty$$, and limits are unique; hence it must be the case that $$x^* = Tx^*\,\!$$.

To show (2), we suppose that $$y\,\!$$ also satisfies $$Ty = y\,\!$$. Then


 * $$0 \leq d(x^*, y) = d(Tx^*, Ty) \leq q d(x^*, y).$$

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.

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.

Another application is 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 &fnof;n has a unique fixed point. Let q be a real number, 0 &lt; q &lt; 1. Then there exists a complete metric on X such that &fnof; is contractive, and q is the contraction constant.

Generalizations
See the article on fixed point theorems in infinite-dimensional spaces for generalizations.

Limerick
The Banach fixed point theorem can be remembered by the following tongue-in-cheek limerick:

If M ' s a complete metric space,

And non-empty, it's always the case,

If f ' s a contraction,

Then under its action,

Exactly one point stays in place!