I am trying to check if the following property holds for fixed points: Suppose:
$ f(x)= x $ is given, with solution $x = \theta \gt 0 $
I would like to show :
$ \forall \epsilon \in (0,1), \forall x : 1/\epsilon > |x -\theta| > \epsilon => \inf_x |f(x) - x | > 0 $.
I am not sure if this property holds for the particular case. I would be grateful is someone can give me any idea on the topic.
The best I was able to come up with was that due to the fact that the function is continuous it does not change sign within some vicinity of $\theta$. Because $f(x) = x $ has a solution for $x=\theta$ then we can find $ |f(x) -x|<\epsilon $, when $ x \to \theta $, so $ x-\epsilon<f(x) $ and since $\epsilon$, can be chosen arbitrary small and $x>0$ than $ 0<x-\epsilon<f(x)$ hold, for a small $\epsilon$ vicinity of $\theta$.
Thanks,
If $f$ is continuous and the equation $f(x)=x$ has a unique solution $x=\theta$, then the statement is true: $$\inf\{ |f(x) - x |: \epsilon \le |x-\theta|\le \epsilon^{-1} \} > 0,\quad \text{for all } \ 0<\epsilon<1$$ Indeed, since the set $\{ x : \epsilon \le |x-\theta|\le \epsilon^{-1} \} $ is compact, the continuous function $|f(x)-x|$ attains its minimum on this set. The minimum is positive since $f(x)\ne x$ for $x\ne \theta$.