(Multiple options could be correct!)
Q. Let $f:[0,1]\to[0,1]$ be a twice continuously differentiable function with a unique fixed-point $f(x_*)=x_*$. For a given $x_0\in [0,1],$ consider the iteration $x_{n+1}=f(x_n)$ for $n\ge 0.$ If $L=\max_{x\in[0,1]}|f'(x)|$, then which of the following are true?
(a) If $L< 1,$ then $(x_n)$ converges to $x_*.$
(b) $(x_n)$ converges to $x_*$ provided $L\ge 1.$
(c) The error $e_n=x_n-x_*$ satisfies $|e_{n+1}|<L|e_n|.$
(d) If $f'(x_*)=0,$ then $|e_{n+1}|<C|e_n|^2$ for some $C>0.$
My attempt:
I know that when a numerical equation $\phi(x)=0$ is expressed as $f(x)=x,$ the real roots can be found by the process of fixed-point iteration. Furthermore, the condition of convergence requires that $L$ (as written in the question!) be less than $1$ throughout the interval $(x_0,x_*).$ So, option (a) is correct. This implies option (b) is false. In addition, we have: $$f(x_{i-1})-f(x_*)=x_i-x_*,\,i=1,2,\ldots,n,(n+1),\ldots.$$
Also, LMVT gives us: $$f(x_i)-f(x_*)=(x_i-x_*)f'(\xi_i),\,x_i\le \xi_i \le x_*,\,i= 0,1,\ldots,n,(n+1),\ldots.$$
Therefore, $$x_i-x_*=(x_{i-1}-x_*)f'(\xi_{i-1}),\,i=1,2,\ldots,n,(n+1),\ldots.$$
That is, $$e_i=e_{i-1}f'(\xi_{i-1}),\,i=1,2,\ldots,n,(n+1),\ldots.\tag1$$
For $i=n+1,$ Eq. $(1)$ becomes: $$e_{n+1}=e_n f'(\xi_{i-1})\\ \Rightarrow |e_{n+1}|=|e_n|\cdot|f'(\xi_{i-1})|\le L|e_n|.$$
This suggests that option (c) is correct. Am I right? If yes, is there an easier way to confirm this?
Now, $f'(x_*)=0\Rightarrow x_i=x_*,$ or $e_i=0,$ according to Eq. $(1).$ On the surface, option (d) appears false. But, I am not sure! Please give some insights...
EDIT. Changed the title, modified the tags, and fixed some typos.
Assume $f''$ is continuous (or bounded) on $[0,1].$ By the Taylor formula for $n=2$ we get $$f(x)=f(x^*)+f'(x^*)(x-x^*)+{1\over 2}f''(\xi)(x-x^*)^2$$ for a point $\xi$ between $x^*$ and $x.$ If $f'(x^*)=0$ the middle term vanishes and we get $$|f(x)-x^*|=|f(x)-f(x^*)|\le M(x-x^*)^2,\qquad M={1\over 2}\sup_{0\le t\le 1}|f''(t)|$$ Thus $$|e_{n+1}|=|x_{n+1}-x^*|=|f(x_{n})-x^*|\le M(x_n-x^*)^2=Me_n^2$$ Remark I consider the strict inequalities in $(c)$ and $(d)$ misprints. Let $f(x)=0$ for all $0\le x\le 1.$ The point $x^*=0$ is the only fixed point. For any $x_0$ we get $x_n=0$ for $n\ge 1.$ Thus $e_n=0$ for $n\ge 1,$ and we cannot have strict inequalities in (c) and (d).