Banach Fixed-Point Theorem

Theorem
Suppose $\left({M, d}\right)$ is a complete metric space, and suppose $f: M \to M$ is a contraction.

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


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

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

But since $d$ is a metric, this means that $f(p_1) = f(p_2)$ 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:


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

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 $(a_n)_{n \in \N}$ is Cauchy.

Since $M$ is complete metric space:


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

Finally:


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

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

Thus $a = f(a)$.

Also known as
Also known as the Contraction Mapping Theorem.