A fixed-point rearrangement is useless unless it contracts on an invariant interval

Let $I=[1/2,1]$ and let $f(x)=x^2+x-1$. A root of $f$ in $I$ is a fixed point of any $g$ satisfying $g(x)=x\iff f(x)=0$ at that point. Consider the three candidate rearrangements \begin{align*} g_A(x)&=1-x^2,\\ g_B(x)&=\frac1{1+x},\\ g_C(x)&=\sqrt{1-x}, \end{align*} each of which is algebraically equivalent to $f(x)=0$ wherever it is defined and the original equation holds. 1. State the contraction-mapping theorem on a closed real interval: if $g:I\to I$ satisfies $|g(x)-g(y)|\le k|x-y|$ for some $k\in[0,1)$ and all $x,y\in I$, then $g$ has a unique fixed point $L\in I$, the Picard iterates $x_{n+1}=g(x_n)$ converge to $L$ from every $x_0\in I$, and \[ |x_n-L|\le\frac{k^n}{1-k}|x_1-x_0|. \] Prove uniqueness and the displayed a priori bound from the geometric estimate $|x_{j+1}-x_j|\le k|x_j-x_{j-1}|$. If $g$ is differentiable, record the mean-value test $|g'|\le k$ on $I$. 2. For each of $g_A$, $g_B$, and $g_C$, decide whether $g(I)\subseteq I$ and whether $\sup_I|g'|<1$ (where the derivative exists). Select the unique candidate that is a contraction self-map of $I$, and exhibit an explicit Lipschitz constant $k<1$. 3. Starting from $x_0=1$, compute the Picard iterates $x_1,\ldots,x_4$ of the successful map as exact rationals. Identify the sequence with consecutive Fibonacci ratios and evaluate the exact limit $L$. 4. Apply the a priori bound of part 1 at $n=4$ with your $k$. Compare it with the true error $|x_4-L|$. Explain why an algebraically correct rearrangement can still be an illegal iteration on $I$, using $g_A$ as the witness. Do not replace the argument by a Newton basin analysis. Do not declare a map usable on $I$ merely because it shares the correct fixed-point equation.

Uniqueness is immediate from $(1-k)|x-y|\le 0$. The a priori bound is the tail of a geometric series of increments.