Topology

Induction, Geometric series, Cauchy sequence.

For $\displaystyle n\ge1$, we have $\displaystyle d(x_{n+1},x_n)=d(f(x_{n}),f(x_{n-1}))\le \alpha\ d(x_{n},x_{n-1})$. Thus by induction, $\displaystyle d(x_{n+1},x_n)\le \alpha^n\ d(x_{1},x_{0})$. Hence, for $\displaystyle m> n\ge1$, by the triangle inequality, $$\begin{aligned} d(x_{m},x_n) & \le \sum_{k=n}^{m-1}\alpha^k\ d(x_{1},x_{0})\\ & \le \frac{\alpha^n}{1-\alpha} d(x_{1},x_{0}). \\ \end{aligned}$$ Since $\displaystyle \lim_{n\to\infty}\alpha^n =0$, we deduce that $\displaystyle (x_n)$ is a Cauchy sequence in the complete space $\displaystyle X$, so $\displaystyle (x_n)$ converges to $\displaystyle p\in X$. Since $\displaystyle f$ is continuous, $\displaystyle \lim_{n\to\infty}f(x_n)=f(p)$. But we also have $\displaystyle \lim_{n\to\infty}x_{n+1}=p$. Thus $\displaystyle f(p)=p$. Assume that $\displaystyle f(p')=p'$. From $\displaystyle d(f(p'),f(p))\le \alpha\ d(p',p)$, we obtain $\displaystyle (1-\alpha)d(p',p)\le 0$ and then $\displaystyle d(p',p)=0$ since $\displaystyle 1-\alpha>0$. So the fixed point $\displaystyle p$ is unique. Finally, taking the limit in $\displaystyle d(x_{m},x_n)$ as $\displaystyle m\to\infty$, we obtain $$d(p,x_n)\le C \alpha^n, \quad C>0,$$ i.e. the iterating sequence, starting from any point $\displaystyle x_0\in X$, converges with exponential speed to the unique fixed point of $\displaystyle f$.