Lipschitz function in Real Analysis

X0 isn't a it's alpha α

That explains something. Imagine my misreading it.

Okay, why is ##U_n## bounded below by a? By zero yes.

pasmith
Homework Helper
From part three we can see that ##g## is strictly decreasing so that means ##(U_{n})## is monotonic decreasing and bounded below by a>0 so therefore it converges to a number greater than 0.

No. In the case that $x_0 = \alpha = s$ you have $U_n = 0$ for all $n$.

It is enough that $U_n$ is bounded below by zero, and you need to show your working in support of the conclusion that $U_n$ is monotonic decreasing.

To show that ##(U_n)## is monotonic decreasing I thought of the following: ##U_{n+1}-U_{n}=\frac{1}{k}(|f(x_{n})-f(s)|-|f(x_{n-1})-f(s)|)##, but I don't think I can remove the absolute value signs because we don't know if the difference is positive or not.

You want to use what you learned about g being monotonic in working through why U is.