Can someone help me with this question?
Let $f:\mathbb{R}\to\mathbb{R}$ twice continuously differentiable. Further let be $\zeta\in\mathbb{R}$ such that $f(\zeta)=0$ and $f’(\zeta)\neq0.$ Show that there is $\delta\in\mathbb{R}^{+}$ such that the Banach fixed-point theorem is applicable on $g:I\to\mathbb{R}$ with $g(x):= x-\frac{f(x)}{f’(x)}$ and $I:=[\zeta-\delta,\zeta+\delta].$
Then show that it follows that the Newtonian method for $f$ for all $x_0 \in I$ converges to $\zeta.$
Thanks a lot!
$g'(x)=1-\frac{f'(x)^2-f(x)f''(x)}{f'(x)^2}$, thus $g'(\xi)=0$ and since $g'$ is continuous, there exists $\delta>0$ such that $|g'(x)|\leqslant\frac{1}{2}$ for all $x\in I:=[\xi-\delta,\xi+\delta]$. Thus for all $(x,y)\in I^2$, we have $|g(x)-g(y)|\leqslant\frac{1}{2}|x-y|$, you can use the Banach fixed-point theorem because $g$ is a contraction on $I$. Now let $(x_n)_{n\geqslant 0}$ defined by $x_0\in I$ and $x_{n+1}=g(x_n)$ (this is Newton's method), Banach fixed-point theorem states that $\lim\limits_{n\rightarrow +\infty}x_n=x^*$ where $g(x^*)=x^*$, this means that $f(x^*)=0$.