Suppose that $f:(a,b) \to (a,b)$ has a fixed point $p$ in $(a, b)$ and that $f$ is differentiable at $p$. Furthermore, assume that $|\,f'(p)|<1$.
Question: How do I prove that $p$ is an attracting fixed point for $f$.
I know that $p$ is an attracting fixed point for $f$ if there exists a $0<\delta<1$ such that $|\,f(x) - p|<|x - p|$ whenever $|x - p|<\delta, x \neq p$. I've tried to think of what the relationship of the derivative of $f$ with the fixed point $p$ would be, but I'm kind of stuck..
Thanks in advance!
We are told that $$ f'(p) = \lim_{x\to p}\frac{f(x) - f(p)}{x-p} $$ exists, and also told that $-1<f'(p)<1$. By using the definition of limit, this means that there is a $\delta>0$ such that for any $x\in (p-\delta, p+\delta)$ (in other words, any $x$ with $|x-p|<\delta$) we have $$ -1<\frac{f(x) - f(p)}{x-p}<1 \iff\left|\frac{f(x) - f(p)}{x-p}\right|<1 $$ The defining property of attacting fixed point follows immediately from this.