**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. $$f(x^\star) = x^\star$$)

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

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

Then, $${a_n}$$ is Cauchy. We may assume $$(X,d)$$ is complete, therefore $${a_n}$$ is convergent. $$\exists ! a = \lim_{n \to \infty} a_n$$. Since $$f$$ is Lipschitz, it is continuous, therefore $$\lim_{n \to \infty} f(a_n) = f(\lim_{n \to \infty} a_n) \iff \lim_{n \to \infty} a_{n+1} = f(\lim_{n \to \infty} a_n) \iff a = f(a)$$.

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