Banach Fixed-Point Theorem

Theorem
Suppose $(M, d)$ is a complete metric space, and suppose $f:M\to M$ satisfies the condition
 * $d(f(x), f(y)) \leq qd(x, y)$

for some $q\in [0, 1)$ and each $x, y\in M$. Then there exists a unique fixed point of $f$.

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

which is only possible if $d(f(p_1), f(p_2)) = 0$ 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 find a fixed point by selecting an arbitrary member of $M$ and repeatedly taking the image under $f$.

Take any $a\in M$ and define $a_0 = a\ $, $a_{n+1} = f(a_n)\ $ for $n\in\N$. Then by assumption, $d(a_{n+2}, a_{n+1}) \leq qd(a_{n+1}, a_n)$.

Therefore

for each $n, k\in\N$, from which it follows that $d(a_{n+1}, a_n) \leq q^nd(a_1, a_0)$.

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.

Therefore, by assumption that the metric space is complete, $\exists a\in M : d(a, a_n)\to 0$.

Finally:
 * $d(a, f(a)) \leq d(a, a_{n+1}) + d(a_{n+1}, f(a)) \leq d(a, a_{n+1}) + qd(a, a_n) \to 0 + q\cdot 0 = 0$

So $a = f(a)\ $.