Let $\phi \in C^1([a,b]), x^* \in (a,b)$ such that $\phi(x^*)=x^*$ and $|\phi^{'}(x^*)| > 1$.
Show that: $\exists \delta > 0 : 0 < |x^* - x_0| < \delta \implies |x^* -x_0| < |x^*-x_1|$ (fixed point iteration method associated to function $\phi$ diverges).
What I've tried: We know that $|\phi^{'}(x^*)| > 1 \implies \lim\limits_{x \to x^*}|\frac{\phi(x)-\phi(x^*)}{x-x^*}| > 1 \implies \lim\limits_{x \to x^*}|\frac{\phi(x)-x^*}{x-x^*}| > 1 \implies $ when $x$ gets closer to $x^*$ we have that $|\phi(x) - x^*| > |x-x^*|$. If we have that $(x_n)_{n \geq 0}$ converges to $x^*$, then we have that for when $n$ is greater: $|\phi(x_n)-x^*| > |x_n - x^*| \implies |x_{n+1} - x^*| > |x_n - x^*| \implies $ fixed point iteration method associated to function $\phi$ diverges.
Is my proof ok? I didn't take care of $\exists \delta > 0 : 0 < |x^* - x_0| < \delta$. A idea for a proof using this statement? Thanks!
$\textbf{Edit:}$
I think I managed to integrate that part. If $\delta \to 0$, then $x_0$ is close to $x^*$, so by replacing $x_0 \leftrightarrow x_n, x_1 \leftrightarrow x_{n+1}$ is above result we get that $|x_1 - x^*| > |x_0 - x^*|$. It's ok now?