Proof of non-convergence of fixed point iteration Suppose we have a function $f$ which has a fixed point $\alpha$. Moreover, $|f'(\alpha)|>1$.
I aim to prove that the fixed point iteration does not converge locally to $\alpha$ (only sequences eventually constant to $\alpha$ converge). Is the following proof correct and can be perfected?
Take $\varepsilon>0$ sufficiently small so that
$$|f'(\alpha)|-\varepsilon>1.$$
Since $f$ is differentiable at $\alpha$ there exists some $\delta>0$ such that
$$|f(x)-\alpha - f'(\alpha)(x-\alpha)|\leq\varepsilon |x-\alpha|,$$
for every $|x-\alpha|<\delta$. Therefore,
\begin{align*}|f(x)-\alpha|&\geq |f'(\alpha)(x-\alpha)|-|f(x)-\alpha-f'(\alpha)(x-\alpha)|\\
&\geq (|f'(\alpha)-\varepsilon)|x-\alpha|\\
&>|x-\alpha|,\end{align*}
whenever $0<|x-\alpha|<\delta$.
Now, take a point $x_0\in\left]\alpha-\delta,\alpha+\delta\right[$
and construct the iteration sequence $\{x_n\}:=\{f^n(x_0)\}$. Suppose that $x_n\neq \alpha$ for every $n\in\mathbb{N}$. For the sake of contradiction, suppose that $x_n$ tends to $\alpha$. Thus, there exists some $n_0\in\mathbb{N}$ such that for every $n>n_0$ we have that $|x_n-\alpha|<\delta$. Then,
$$|x_n-\alpha|=|f(x_{n-1})-\alpha|>|x_{n-1}-\alpha|>\cdots>|x_{n_0}-\alpha|.$$
Taking limits we deduce that
$$\lim_n |x_n-\alpha|\geq |x_{n_0}-\alpha|>0.$$
This is a contradiction.
 A: The main idea of your proof is very good. I could nitpick about a few things:

*

*"Taking limits we deduce that..." - it is nowhere said that $\lim_n |x_n-\alpha|$ exists. (However, all the values $|x_n-\alpha|$ being at least as big as $|x_{n_0}-\alpha|>0$ is enough to conclude that $x_n$ does not converge to $\alpha$.)

*The point $x_0$ may not need to be in $(\alpha-\delta, \alpha+\delta)$, all we need is that, for some $n_0$, assuming convergence to $\alpha$, all elements of the sequence, starting from $x_{n_0}$, will be in $(\alpha-\delta, \alpha+\delta)$.


As for the alternative proof, consider the sequence $x_n$ given by $x_{n+1}=f(x_n)$ and assume it converges to $\alpha$ but $x_n\ne \alpha$ for all $n$. Then:
$$\lim_{n\to\infty}\frac{x_{n+1}-\alpha}{x_n-\alpha}=\lim_{n\to\infty}\frac{f(x_n)-f(\alpha)}{x_n-\alpha}=f'(\alpha)$$
so, taking moduli (absolute values) and recalling that the limit of the modulus is the modulus of the limit:
$$\lim_{n\to\infty}\frac{|x_{n+1}-\alpha|}{|x_n-\alpha|}=\lim_{n\to\infty}\left\lvert\frac{x_{n+1}-\alpha}{x_n-\alpha}\right\rvert=|f'(\alpha)|>1$$
which implies that, for some $n_0$, and all $n\ge n_0$ we also have $\frac{|x_{n+1}-\alpha|}{|x_n-\alpha|}>1$ i.e. $|x_{n+1}-\alpha|>|x_n-\alpha|$ which then implies (as in your proof) that $|x_n-\alpha|>|x_{n_0}-\alpha|>0$ for $n>n_0$, so $x_n$ cannot converge to $\alpha$.
