show that $\{x_n\}$ is a cauchy sequence Let $f:\Bbb R \to \Bbb R$ be differentiable on $\Bbb R$ and $|f'(x)| \leq \frac {1}2$ for all $x\in \Bbb R$. Let $a \in \Bbb R$ and define $x_1 = a, x_{n+1} = f(x_n)$ for all $n \in \Bbb N$. Show that $\{x_n\}$ is a Cauchy sequence. And show that $\exists x \in \Bbb R$ such that $f(x) = x$.

By the mean value theorem, for every $x,y\in \Bbb R$, there exists $z\in[x,y]$ such that
$$\frac{|f(x)-f(y)|}{|x-y|}=|f'(z)|\leq\frac{1}{2}.$$
It follows that $|f(x)-f(y)|\leq \frac{1}{2}|x-y|$ for every $x,y\in \Bbb R$. Thus, $f$ is a strict contraction and by the Banach contraction principle, there exists a unique $x\in \Bbb R$ such that for every $x_0=a\in \Bbb R$, the sequence $x_{n+1}=f(x_n)$, $n\in \Bbb N$ converge to $x$. 
