How do I prove the following statement :
$$ \forall k \geqslant 1, \ |f^k(x)-p| \leqslant a^k|x-p|$$
inductively? $p$ is a stable fixed point of $f$ and $f'(p) \leqslant a < 1$. $f$ is a continuous and differentiable function.
Let $k = 1$. Then :
$$ |f(x)-p| \leqslant a|x-p| $$
The mean value theorem says there exists a range $e$ so that the slope of the line, that goes through $p$ and any $x$ in $[p+e,p-e]$ and $x \neq p$ is equal or less than $a$.
Assume $k = n$> Then :
$$|f^n(x)-p| \leqslant a^n|x-p|$$
Let $k = n + 1$
$$|f^n+1(x)-p| \leqslant a^n+1|x-p| $$
$$RHS = a^n+1|x-p| = a*a^n|x-p| $$
$a|f^n(x)-p|$ is less or equal by the assumption and this is where I get stuck. I've been trying to modify the last expression to all directions to be equal to LHS or larger without really getting anywhere.
Use that $p=f(p)$.\begin{align} |f^{n+1}(x)-p|&=|f(f^n(x))-f(p)|\\ &\le a\,|f^n(x)-p|\\ &\le a^{n+1}|x-p| \end{align}