$|f '(x)| \le c$ for some $c<1$ implies a unique $a$ such that $f(a) = a$

Question

Let $f\colon I \to I$ be differentiable in the closed interval $I$ s.t. $|f'(x)|\le c$ for some $c<1$.

1. Show that there exists a single $a$ in $I$ s.t. $f(a) = a$

2. Given $n\in\mathbb{N}$, define $f_n = f \circ f\circ\dots\circ f$ (composed $n$ times). Show that as $n$ approaches infinity $f_n(x)$ approaches $a$ for all $x\in I$.

In 1. I've been able to prove that there can't be more than one, but I couldn't yet prove there must be one.

In 2. I've said that for every $x \ne a$, $f(x) \ne x$, which makes it so the sucessive composite functions go around $I$, and when they eventually get to $a$, they stay there. Is it enough?

Answer 1

$I=[u,v]$ Suppose that $(a)=a$, $f(b)=b, b\neq a$, you have $|f(b)-f(a)|=|f'(d)||b-a|$ this implies that $|b-a|<b-a|$ impossible. Thus $a$ is unique if it exists.

$g(x)=f(x)-x)$, $g(u)\geq 0, g(v)\leq 0$, IVT implies that there exists $a$ such that $g(a)=0$ and $f(a)=a$.

$|f^{n+1}(x)-f^n(x)|<c|f^{n}(x)-f^{n-1}(x)|<c^n|f(x)-x|$. This implies that $n>m$, $|f^n(x)-f^m(x)|\leq |f^n(x)-f^{n-1}(x)|+..+|f^{m+1}(x)-f^m(x)|\leq |f(x)-x|(c^{n-1}+...+c^m)=|f(x)-x|c^m(1-c^{n-m})/(1-c)$, so it is a Cauchy sequence contained in $I$ so it converges in $I$, write $y=lim_nf^n(x)$, $f(y)=f(lim_nf^n(x))=lim_nf^{n+1}(x)$ since $f$ is continue and $f(y)=y$. We deduce $y=a$ since $a$ is the unique fixed point.