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

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 $$x_{n+1}:=g(x_n) \space (n \in \mathbb{N}_0)$$.

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?

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 $$f(x_\infty)=x_\infty$$.