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 $f(x) = x$.

$f$ is differentiable  on $\mathbb R.$ so, we can use MVT at any interval.
$\forall n\in \mathbb N$
$$|x_{n+1}-x_n|=|f(x_n)-f(x_{n-1})|=$$
$$|x_n-x_{n-1}|f'(c|\leq \frac{1}{2}|x_n-x_{n-1}|$$
.
.
$$\leq \frac{1}{2^n}|x_1-x_0|.$$
From here, by triangular  inequality, we have
$\forall \; n>p\geq 0$
$$|x_{n+p}-x_n|\leq \frac{ |x_1-x_0|}{2^n}\sum_{k=0}^{p-1}\frac{1}{2^k}$$
$$=2\frac{|x_1-x_0|}{2^n}(1-\frac{1}{2^p})\leq \frac{|x_1-x_0|}{2^{n-1}}.$$
Now, you conclude.
