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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,

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    $\begingroup$ What if $f(x_n)=x_n$ for a sequence of values of $x_n$ converging to $\theta$? $\endgroup$ Apr 23, 2013 at 13:02
  • $\begingroup$ Hi, I just made an update of my question, perhaps this is what you are proposing? Actually I have continuous values. But the idea is quite the same. What bothers me is the fact that the annuls I am looking at becomes very large as $\epsilon \to 0 $. $\endgroup$ Apr 23, 2013 at 13:32
  • $\begingroup$ Stefan Hansen made an effort to make your problem look nicer, and then you put it back to looking like crap. Why? $\endgroup$ Apr 23, 2013 at 13:38
  • $\begingroup$ I just updated it, I am sorry if I did something wrong. $\endgroup$ Apr 23, 2013 at 13:40
  • $\begingroup$ Anyway, are you not aware that it is possible for a continuous function to vanish at, say, all and only the points $1,1/2,1/3,1/4,\dots$ and zero? If not, try to come up with an example. $\endgroup$ Apr 23, 2013 at 13:40

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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 $x\ne \theta$.

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