Banach Fixed-Point Theorem

Theorem
Let $\left({M, d}\right)$ be a complete metric space.

Let $f: M \to M$ be a contraction.

That is, there exists $q \in \left[{0 \,.\,.\, 1}\right)$ such that for all $x, y \in M$:


 * $d \left({f \left({x}\right), f \left({y}\right)}\right) \le q d \left({x, y}\right)$

Then there exists a unique fixed point of $f$.

Uniqueness
Suppose $f$ has two fixed points $p_1, p_2 \in M$.

Then:

which is only possible if $d \left({f \left({p_1}\right), f \left({p_2}\right)}\right) = 0$ since $q < 1$.

But since $d$ is a metric, this means that $f \left({p_1}\right) = f \left({p_2}\right)$ and so $p_1 = p_2$.

Existence
We will find a fixed point by selecting an arbitrary member of $M$ and repeatedly taking the image under $f$.

Take any $a_0 \in M$ and define recursively:


 * $a_{n+1} = f \left({a_n}\right)$

for $n \in \N$.

Then by assumption:


 * $d \left({a_{n+2}, a_{n+1} }\right) \le q d \left({a_{n+1}, a_n}\right)$

Therefore, for all $k, n \in \N$, we have:

from which it follows that $d \left({a_{n+1}, a_n}\right) \le q^n d \left({a_1, a_0}\right)$ for all $n \in \N$.

So for any $n > m$:

This last quantity can be made arbitrarily small for all sufficiently large choices of $m$, so the sequence $\left\langle{a_n}\right\rangle_{n \mathop \in \N}$ is Cauchy.

Since $M$ is complete metric space:


 * $\displaystyle a := \lim_{n \mathop \to \infty} a_n \in M$

Finally:

and the quantity on the right converges to zero since $a = \displaystyle \lim_{n \mathop \to \infty} a_n$.

Thus:
 * $a = f \left({a}\right)$

Also known as
Also known as the Contraction Mapping Theorem.