If \((X, d)\) is a metric space, recall that a map \(f: X \to X\) is called a contraction if there is a number \(\alpha < 1\) such that
\[ d(f(x), f(y)) \leq \alpha \, d(x, y) \]
for all \(x, y \in X\). Show that if \(f\) is a contraction of a complete metric space, then there is a unique point \(x \in X\) such that \(f(x) = x\). Compare E-70TGS.
Solution
Choose \(x_0\in X\) and define \[ x_{n+1}=f(x_n). \] If \(c=d(x_1,x_0)\), repeated use of the contraction inequality gives \[ d(x_{n+1},x_n)\le \alpha^n c. \] Hence for \(m>n\), \[ d(x_m,x_n) \le\sum_{k=n}^{m-1}d(x_{k+1},x_k) \le c\sum_{k=n}^{m-1}\alpha^k \le \frac{c\alpha^n}{1-\alpha}. \] The right side tends to \(0\), so \((x_n)\) is Cauchy. Completeness gives \(x_n\to x\in X\).
The contraction inequality implies that \(f\) is Lipschitz, hence continuous. Therefore \[ f(x)=f\left(\lim_nx_n\right) =\lim_nf(x_n) =\lim_nx_{n+1} =x. \] Thus a fixed point exists.
If \(x\) and \(y\) are fixed points, then \[ d(x,y)=d(f(x),f(y))\le\alpha d(x,y). \] Since \(\alpha<1\), this forces \(d(x,y)=0\), hence \(x=y\). The fixed point is unique.