Let $f:[0,1]\to [0,1]$ and $g:[0,1]\to [0,1]$ be two continuous functions, each having a unique fixed point $x_f$ and $x_g$. Assume $\Vert f-g\Vert_\infty<\epsilon$. Is it possible to say something about $\vert x_f-x_g\vert$, maybe under some additional hypothesis on $f$ and $g$.
The problem I'm interseted in is more of the following form: $f_n$ is a sequence of functions (each of these functions have a unique fixed point $x_f^n$) uniformly converging to $g$, in the sense that $\Vert f_n-g\Vert_\infty\leq \epsilon(n)$, with $\epsilon(n)\to 0$.
I would like to prove that $x_f^n$ converges to $x_g$ as $n\to\infty$.
I'll assume $g$ is continuous.
If $x_g$ is the unique fixed point of $g$ on $[0,1]$, then for any $\delta > 0$ we have $\epsilon = \inf \{|g(x)-x|: x \in [0,1], |x - x_g| \ge \delta\} > 0$.
If $f_n \to g$ uniformly, there is $N$ such that $|f_n(x) - g(x)| < \epsilon$ for all $n > N$ and $x \in [0,1]$, and then if $|x - x_g| \ge \delta$ we'd have $|f_n(x) - x| \ge |g(x) - x| - |f_n(x) - g(x)| > 0$, so $x$ is not a fixed point of $f_n$. Thus if $f_n$ is to have a fixed point $x_{f_n}$, we must have $|x_{f_n} - x_g| < \delta$. This shows $x_{f_n} \to x_g$ as $n \to \infty$.