We wish to solve the equation $f_a(x) = 0$, where $a ∈ R$ is a parameter and $f_a(x) = (1−a)x+ax^3$. We will study the fixed point iteration $x^{k+1} = φ(x^k)$, where the function $φ(x) = ax(1 − x^2)$.
(i) Find the positive values of $a$ such that the fixed point iteration converges to the root $α_1 = 0$ for any initial guess $x^0 ∈ [0, 1]$.
(ii) Find a condition on $a$ under which a second zero $α_2 > 0$ exists in the interval [0, 1] and determine for which $a$ the fixed point iteration can approximate $α_2$.
(iii) For what value of $a$ can the fixed point iteration approximate $α_2$ with second order convergence?
I tried taking the limit for part (i) using the proof of Lemma: Let $x^k$ be a fixed point iteration. Suppose that $x^k \to \alpha$ as $k \to \infty$, and that $\phi$ is continuous in the neighbourhood of $\alpha$. Then $\alpha=\phi(\alpha)$.
Where the proof stated $\alpha = \lim_{k \to \infty} x^{k+1} = \lim_{k \to \infty} \phi(x^{k}) = \phi(\lim_{k \to \infty} x^{k}) = \phi(\alpha)$. I keep getting $a=0$ which is evidently wrong. Not sure where to go from part (ii) and (iii) either.
I believe the fixpoint iteration is with respect to $x$, not $a$. In order to find the fixpoints, you can do as you mentioned, which leads to the following equation: $$\alpha = a\alpha(1-\alpha^2).$$ Here, either $\alpha = 0$ (which would be one possible fixpoint) or $\alpha \neq 0$, in which case we can divide the $\alpha$ away to obtain $$1 = a (1-\alpha^2).$$ Solving this for $\alpha$ yields $$\alpha = \pm \sqrt{1-\frac{1}{a}}$$ which should already give you a criterion to see when other fixpoints can appear.
Concerning the convergence of the iteration, I would use the Banach fixpoint theorem, i.e. show that the function has Lipschitz constant (that's just the maximum absolute value of the derivative) strictly smaller than 1 over some neatly-chosen domain, and that it sends this domain to itself; I tried this directly on $x \in [0,1]$ but this is not good enough, so you might need to look at the maximum value of $x_1 \in \phi([0,1]) = [0, m]$, and work with $[0,m]$ instead, since $m$ should be smaller than 1. It may also be that there is a better way to do this, depending on what you have seen in your lecture.