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, $$d(a_{n+2}, a_{n+1}) \leq d(a_{n+1}, a_n)$$.

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 $$a = f(a)\ $$.