Let $f:\mathbb{R}^n \rightarrow \mathbb{R}^n$ be a non-expansive map, i.e. $$\|f(x) - f(y)\| \leq \|x - y\|$$ for all $x,y\in\mathbb{R}^n$. Further, assume $f$ has at least one fixed-point $x^\star$. Define another map $g:[0,1]\times\mathbb{R}^n \rightarrow \mathbb{R}^n$ as \begin{eqnarray} g(\alpha, x) = \alpha f(x) + \gamma\cdot(1-\alpha) x \end{eqnarray} for some $\gamma \in (0,1)$. I see that $g$ is a simple homotopy map which deforms $\gamma Id$ into $f$, where $Id$ is the identity map, as we sweep $\alpha$ from $0$ to $1$. Also, $g$ has the convenient property that, for all $x,y\in\mathbb{R}^n$,
$$\|g(\alpha,x) - g(\alpha,y)\| \leq (1-\alpha + \alpha\gamma) \|x-y\|$$
where $(1-\alpha + \alpha\gamma) < 1$ and so we know that iterating $g(\alpha,x)$ for fixed $\alpha$ yields a fixed-point (contraction mapping principle).
My question, however, is this: Under what conditions can a continuation scheme such as what follows be guaranteed to find a fixed-point of $f$:
Initialization: select $\alpha^0 = 0.9$, $x^0$ at random, and a sufficiently small step size $\Delta_{\alpha}>0$.(I chose 0.9 for $\alpha^0$ arbitrarily since any $\alpha<1$ will give a fixed-point.)
Computation at stage $k$:
- find the fixed-point $x^k$ of $g$ by iterating $g(\alpha^k, \cdot)$ starting from $x^{k-1}$
- set $\alpha^{k+1} = \alpha^k + \Delta_{\alpha}$
Disclaimer: I'm just beginning to learn continuation methods, so insight is really what I'm looking for here.