Prove $f: \mathbb{R}\rightarrow\mathbb{R}, x_{n+1}=f(x_n) $ with $|f'(x)|< \theta < 1$ converges for $n \rightarrow \infty$ and it has a fixed point.

Question

Show that

$f: \mathbb{R} \rightarrow \mathbb{R}$

$x_{n+1} :=f(x_n)$ with arbitrary $x_0 \in \mathbb{R}$

with $|f'(x)|< \theta < 1$

converges such that $\lim_{n\to\infty} x_n=x^*$ and $f(x^*)=x^*$.

So far I have shown that $f$ is Lipschitz-continuous and that it follows that

$|x_n-x_{n-1}|\leq \theta^{n-1} |x_1-x_0|$, which for $n \to \infty$ approaches $0$.

How can I show that $x_n$ is a Cauchy sequence, i.e:

$\forall \varepsilon>0 \quad \exists N\in\mathbb{N} \quad \forall m,n \ge N \colon \quad \left|x_m-x_n \right|<\varepsilon $

I know that fixed point theorems use what I found out, however I am afraid I cannot use them.

Answer 1

$|x_n - x_m| \le |x_n - x_{n+1}| + |x_{n+1}-x_{n+2}| + \ldots + |x_{m-1} - x_m| \le (\theta^n + \theta^{n+1} + \ldots + \theta^{m-1})|x_1 - x_0| < (\theta^n + \theta^{n+1} + \ldots)|x_1 - x_0| = \frac{\theta^n}{1-\theta}|x_1-x_0|$

The first inequality is triangle inequality, the second is by what you showed, the third is because we added more positive terms, and the fourth is by convergence of geometric series with ratio less than $1$. Now, you see that it's Cauchy, so can you take it from here?