Definition. Let $(X,d)$ be a metric space. $T:X\to X$ is called a contraction mapping on $X$, if it is Lipschitz continuous with constant $L\in [0,1)$.

Theorem (Banach fixed-point theorem). Let $(X,d)$ be a non-empty, complete metric space, $f:X\to X$ a contraction mapping with Lipschitz constant $L\in [0,1)$. Then $f$ admits an unique fixed point $x^{\star }$ in $X$. (i.e. $f\left(x^{\star }\right)=x^{\star }$)

Proof. First, we prove existence. Let $a_0\in X$ be an arbitrary point, $\forall n\in N^{+}$ let $a_n=f\left(a_{n-1}\right)$. We may assume $\exists L\in [0,1):d\left(f\right(a_n),f(a_{n-1}\left)\right)\le Ld(a_n,a_{n-1})$. Thus, $d(a_{n+1},a_n)\le Ld(a_n,a_{n-1})\le L^{2}d(a_{n-1},a_{n-2})\le \cdots \le L^{n}d(a_1,a_0)$.

It follows that $\forall n,k\in N$ such that $n<k$, we have $d(a_k,a_n)\le d(a_k,a_{k-1})+\cdots +d(a_{n+1},a_n)\le L^{k-1}d(a_1,a_0)+\cdots +L^{n}d(a_1,a_0)\le \frac{L^n}{1-L}d(a_1,a_0)$, therefore $\forall \epsilon \in R^{+}:\exists N:N<n<k:d(a_n,a_k)<\epsilon$.

Then, $a_n$ is Cauchy. We may assume $(X,d)$ is complete, therefore $a_n$ is convergent. $\exists !a=\underset{n\to \infty }{\lim }{a}_n$. Since $f$ is Lipschitz, it is continuous, therefore $\underset{n\to \infty }{\lim }f\left(a_n\right)=f\left(\underset{n\to \infty }{\lim }{a}_n\right)⟺\underset{n\to \infty }{\lim }{a}_{n+1}=f\left(\underset{n\to \infty }{\lim }{a}_n\right)⟺a=f\left(a\right)$.

Now, we prove uniqueness. Let $a_0,b_0\in X$ and $a_{n+1}=f\left(a_n\right),b_{n+1}=f\left(b_n\right)$. The sequences $a_n,b_n$ converge to fixed points $a,b$ respectively, as shown above. Since $f$ is Lipschitz, $d\left(f\right(a),f(b\left)\right)\le Ld(a,b)⟺d(a,b)\le Ld(a,b)$. If $d(a,b)>0$, then $\frac{d(a,b)}{d(a,b)}\le L⟺1\le L$, which is a contradiction, therefore $d(a,b)=0$, hence $a=b$. $\square$