How to show that one not monotonous f doesn't have fixpoints?

Question

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?

Answer 1

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.

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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.

Answer 2

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.

Answer 3

If I have one function (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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