Given a function $F(x)$ defined on $[a,b]$ such that:
- $f$ $\in$ $C^2([a,b])$
- $F(a)F(b)$<0
- $F'(x) \neq 0$, $\forall$ $x$ $\in$ $[a,b]$
- $F''(x)$ $\geq$ $0$ or $F''(x)$ $\leq$ $0$, $\forall$ $x$ $\in$ $[a,b]$
- $(*)$$\frac{|F(c)|}{|F'(c)|}$ $<$ $b-a$, where $c$ is the endpoint of $[a,b]$ where $|F'(x)|$ reaches the minimum value among them $$\\$$ Then $f$ converges to the only zero in $[a,b]$ for any initial approximation $\mathrm{p_{0}}$ $\in$ $[a,b]$.
I'm having trouble in showing that statement $(*)$ guarantees that any initial approximation $\in$ $[a,b]$ will converge to the only root of $F$ in $[a,b]$. I could write $G(x)=x-\frac{F(x)}{F'(x)}$ and if I could guarantee that it's a contraction, i.e., $G([a,b])=[a,b]$ and that $\exists$ $\ $ $0<\lambda < 1$: $\forall x,y \in$$[a,b]$, $|G(x)-G(y)|$ $\leq$ $\lambda$$|x-y|$ , then by the Fixed-Point Theorem, there is an unique fixed point $p$ in $[a,b]$ and $G(p)=p \iff p- \frac{F(p)}{F'(p)}=p \iff \frac{F(p)}{F'(p)}=0 \iff F(p)=0$. I know that $G'(x)=\frac{F(x)F''(x)}{(F'(x))^2}$ and it could also be shown that $\exists$ $\ $ $0<\lambda < 1$: |$G'(x)$| $\leq$ $\lambda$, $\forall$ $x$ $\in$ $[a,b]$ leading to the same conclusion above, but I've not found a way to show it. Any suggestions?