Let $1<a<e^{1/e}$, and define $f(x)=a^x$
(b) Show that if $x_1$ is any point in the interval $(1,e)$ and $p$ is a fixed point of $f(x)$ in the interval $(1,e)$, then
$$|f(x_1) - p | <|x_1 - p |$$
(c) If $x_1$ is as in (b), define a sequence $\{x_n\}_{n=1}^{\infty}$ as follows:
Let $x_2 = f(x_1),\:x_3=f(x_2),\ldots x_{n+1}=f(x_n), \ldots$
Show that the sequence $\{x_n\}$ converges to $p$
Solutions(?):
(b)
From the mean value theorem it follows that there exists some $x_2$ between $x_1$ and $p$ where $x_1 \neq p$ such
that
$f'(x_2)=|\frac{f(x_1)-p}{x_1-p}|$ then it follows:
$$|\frac{f(x_1)-p}{x_1-p}|=f'(x_2)<|\frac{\max\{f(e)=a^e:a\in(1,e^{1/e})\}-\min\{f(1)=a^1:a\in (1,e^{1/e})\}}{e-1}|<|\frac{e-1}{e-1}|=1$$
From this it follows that $|\frac{f(x_1)-p}{x_1-p}|<1 \Rightarrow |f(x_1) - p | <|x_1 - p | $
(c) I feel like it would have something to do with the Banach fixed-point but that would mean I did something wrong with (b) since it needs to be Lipschitz continuous with its constant being less than $1$ but I feel I did something wrong in (b) that makes (c) impossible or it is some constraint I have left unaccounted for.
