I am analyzing the iterated sequence of the function $\lambda \sin( \pi x)$ for $x, \lambda \in [0,1]$, where $x_n=f(x_{n-1})$ for a paper I am writing. I know that all fixed points of this function are either globally stable or unstable but I am not sure of how to prove this. Is there a theorem that guarantees this, or a way to prove this?
That is, is there a way to show that any cycle or fixed point is globally convergent for this iterated sequence?
Thanks for your help
edit: I'm sorry, I wrote this when I was exhausted and forgot to include the main point of this. I want to show that each fixed point is globally stable or unstable. I am considering the |f'|<1 for stable and the opposite for unstable.
I don't believe this statement is correct. A look at a few graphs of $f_{\lambda}(x)=\lambda \sin(\pi x)$ for a few choices of $\lambda$, together with the line $y=x$ should convince you otherwise. For example:
In this image, the red graphs correspond to $\lambda=1/2$ and $\lambda=1$. The points of intersection with the line $y=x$ are the fixed points. For $\lambda=1/2$, the slope at the fixed point is zero, while for $\lambda=1$, the slope at the fixed point is clearly less than $-1$. Thus, there must be a value of $\lambda$ where the slope at the fixed point is exactly equal to one. Furthermore, this value of $\lambda$, together with the value of the fixed point $x$, must satisfy \begin{align} \lambda \sin(\pi x) &= x \\ \lambda \pi \cos(\pi x) &= -1. \end{align} You can use a numerical equation solver to estimate the value of $\lambda$ to be approximately $0.71996$ and the associated fixed point to be approximately $0.64577$.
On the other hand, if you iterate the function $f(x)=\lambda \sin(\pi x)$ for this value of $\lambda$, you will find convergence to the fixed point. The convergence will be very slow, as is typical at a neutral fixed point. Here's an illustration of the first 1000 iterates from the critical value of $1/2$ using a cobweb plot: