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!


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.