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I have a question about fixed points If I have one function $f$ (that is not monotonous!) I would like to demostrate that this function hasn't fixed points. I need to find a funciton $f$ for which this is valid. My problem is that I can not find a valid example of $f$, but mainly I don't know how to prove that this function hasn't any fixed points. Can someone help me?

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  • $\begingroup$ Think of $f(x)=x^2+1$. $\endgroup$ – PierreCarre Feb 20 at 20:30
  • $\begingroup$ Just a side remark: 'monotonous' is not the word. It is 'monotonic'. $\endgroup$ – Kavi Rama Murthy Feb 20 at 23:33
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There are functions that are not monotonic, but have fixed points. For example, the function $f(x)=x^2$ is not monotonic, yet it has a fixed point at $x=0$.

A function that is not monotonic that has no fixed points would be $g(x) = x^2+1$.

You can prove that this function has no fixed points by definition. Namely, to get its fixed points you need to solve $$x^2+1=x$$ This is an ordinary quadratic equation, which has no solutions, and thus there are no fixed points.


To show that $g$ is not monotonic, we will show that $g(x+1)-g(x)$ can sometimes be positive and it can sometimes be negative.

We have $$g(x+1)-g(x) = (x+1)^2+1 - (x^2+1) = x^2+2x+1-x^2-1=2x$$ Depending on the sign of $x$ this can be positive or negative.

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  • $\begingroup$ Hi! Tanks fot the help. I forgot something: for all x in the sample (real valued) x <= f(x) $\endgroup$ – claudioz Feb 20 at 20:16
  • $\begingroup$ @claudioz You can verify that $x\leq x^2+1$. $\endgroup$ – Haris Gusic Feb 20 at 20:17
  • $\begingroup$ sorry but I do not understand how it can be not monotonous. I see that for all x in the domain such as x < x+1, the result of the function g is g(x) < g(x+1) $\endgroup$ – claudioz Feb 20 at 20:27
  • $\begingroup$ @claudioz For example, $g(-1) > g(0)$. $\endgroup$ – Robert Israel Feb 20 at 20:31
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Are we assuming real valued? Use $f(x)=x^2+2$, its trivially nonmonotonous, and to test for fixed points show $x^2+2=x$ has no real valued solutions.

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  • $\begingroup$ Hi! Tanks fot the help. I forgot something: for all x in the sample (real valued) x <= f(x) $\endgroup$ – claudioz Feb 20 at 20:15
  • $\begingroup$ That is true already. If $x \leq f(x)$, then $f(x)-x \geq 0$, and if you are dealing with monotone functions and fixed points, you should be able to solve a quadratic inequality. $\endgroup$ – siegehalver Feb 20 at 20:19
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If I have one function 𝑓 (that is not monotonous!) I would like to demostrate that this function hasn't fixed points.

$$ f(x) = x + 3 + 2 \sin x $$ Since $$ f(x) \geq x+1, $$ we have $$ f(x) \neq x $$

However, $$ f'(x) = 1 + 2 \cos x$$ is sometimes positive and sometimes negative

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