Show that $(x_n)_{n=0}^{\infty}$ converges against a fixpoint of $f$

Question

Given:

Let $f: [0, 1] \rightarrow [0,1]$ be a Lipschitz-continuous function with Lipschitz-constant $L \ge 0$, $g: [0,1] \rightarrow \mathbb{R}$ be a function with $$g(x):= \frac{1}{1+L}f(x)+\frac{L}{1+L}x$$ and $(x_n)_{n=0}^{\infty}$ be a recursively defined sequence for a $x_0 \in [0,1]$ with $x_{n+1}:=g(x_n) \space (n \in \mathbb{N}_0)$.

Already shown:

1. $g$ is monotonically increasing,
2. $x_n \in [0,1]$ for all $n \in \mathbb{N}_0$ and
3. for $x_0 \le x_1$, $(x_n)_{n=0}^{\infty}$ is monotonically increasing and for $x_0 \ge x_1$, $(x_n)_{n=0}^{\infty}$ is monotonically decreasing.

What's left: show that $(x_n)_{n=0}^{\infty}$ converges against a fixpoint of $f$.

Can someone help at this, as I don't know how to show it?

Thanks in advance!

Answer 1

From 2) and 3), $x_n$ converges to some $x_\infty\in[0,1]$. By continuity of $g$, $g(x_\infty)=x_\infty$ and hence $f(x_\infty)=x_\infty$.