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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?

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  • $\begingroup$ You can prove it by using the convergence of a monotone bounded sequence. $\endgroup$ Oct 25, 2023 at 21:11
  • $\begingroup$ So the last statement assures that every element of the sequence will belong to the interval $[a,b]$? $\endgroup$
    – J P
    Oct 25, 2023 at 21:14
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    $\begingroup$ Yes. That's the purpose of it. $\endgroup$ Oct 25, 2023 at 21:20
  • $\begingroup$ Have a look at Darboux theorem numdam.org/article/NAM_1869_2_8__17_0.pdf $\endgroup$ Oct 26, 2023 at 9:07

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