Let $f \in C^2([a,b])$ be a function with the following properties:
(i) $f(a)f(b) <0 $
(ii) $f'(x) \neq 0$ $\forall x$
(iii) $\frac{|f(a)|} {|f'(a)|} <(b-a)$ and $\frac{|f(b)|} {|f'(b)|} <(b-a)$
(iv) $ f''(x) \leq 0$ $\forall x$
Then $h(x) := x-\frac{f(x)} {f'(x)} $ is a contraction.
What I did is:
Suppose wlog $f(a)<0<f(b)$ and so $f'(x)>0$. We have $|h'(x)| \leq... \leq L<1$ I don't know how to do the middle part, maybe we have $\frac{|f(x)|} {|f'(x)|} <(b-a)$ $\forall x$? (not sure)
Curiosity: thie statement will prove a theorem of converge for the Newton method