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If I have $f(x)\leq f(f(x))$ for all real $x$, can I deduce $x\leq f(x)$?

Thank you.

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You can deduce $x \le f(x)$ for all $x$ in the image of $f$ (i.e., all those numbers $x$ which actually occur as the function value at some $y$, namely $x = f(y)$). You cannot deduce anything about other $x$. – ShreevatsaR Nov 4 '12 at 18:07

11 Answers 11

up vote 51 down vote accepted

No. Suppose $f$ is a constant function.....

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If $f(x)$ is a constant function then $x\le f(x)$, so it is not a counterexample. ?????? Or was this perhaps meant as a hint, that the OP should try to find a function for which $x\le f(x)$ does not hold, but $f(f(x))$ is constant? – Martin Sleziak Nov 4 '12 at 6:57
@Martin: If $f(x)$ is a constant, what constant could it be that is greater than all $x$? – Rahul Nov 4 '12 at 7:02
@MartinSleziak: If $f(x)$ is a constant function, then there is a unique $x_0$ such that $f(x_0)=x_0$. Then for all $x>x_0$ we have $x>x_0=f(x_0)=f(x)$. – Cameron Buie Nov 4 '12 at 7:03
@Rahul, Cameron: You are correct, sorry for my blunder. – Martin Sleziak Nov 4 '12 at 7:04

For a slightly non-trivial example, $f(x) = - \vert x \vert$. We then have that $$f(f(x)) = - \vert f(x) \vert = -\vert-\vert x \vert \vert = - \vert x \vert = f(x)$$ However, $x \geq f(x)$ for all $x \in \mathbb{R}$ and in fact $x > f(x)$ for all $x \in \mathbb{R}^+$.

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Let $f(x)=1$. Then $f(f(x))\leq f(1)=1$. However, $2\geq f(2)=1$ so that the conclusion does not follow.

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You example makes sense. But if I substitute $a=f(x)$, $a\leq f(a)$, which leads us $x\leq f(x)$, this seems make sense too... So what's wrong? – ᴊ ᴀ s ᴏ ɴ Nov 4 '12 at 7:04
@jasoncube when you substitute $a = f(x)$ you imply that $a \in \mathrm{ Range}(f)$ so you have this inequality for those $a$ you cannot deduce it holds for every $x$ – clark Nov 4 '12 at 7:07
@jasoncube: Note that for the function $f(x) = 1$ (the constant function which takes value $1$ for all $x$), if you substitute $a = f(x)$, then necessarily $a = 1$. So although $a \le f(a)$ is true (for $a = 1$), for any other $x > 1$, it is not true that $x \le f(x)$, because $x > 1$ but $f(x) = 1$. – ShreevatsaR Nov 4 '12 at 18:04

Let $f$ map $-1$ to $-2$ and all other values to $0$. Then $f(-1)<-1$ but $f(f(x))\geq f(x)$ for all $x$.

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For an example showing that even strict inequality $f(x)<f(f(x))$ for all $x$ does not imply $x\leq f(x)$ for all $x$, consider $$ f: x\mapsto \begin{cases} -\frac{|x|}2 &\text{if }x\neq0\\ \\-1&\text{if }x=0\\ \end{cases} $$ The discontinuity at $x=0$ is inevitable, since a continuous function $\mathbf R\to\mathbf R$ that has both points $x$ where $x<f(x)$ and other points where $x>f(x)$ must have some fixed points with $x=f(x)$.

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Suppose $f(x)=1$.

$\left\{ \begin{array}{l} f(x)=f(f(x))=1 \\ 1\leq1 \end{array} \right. \therefore f(x) \leq f(f(x)) $ for all real $x$.

If $x>1$, $f(x)<x \therefore x\not\leq f(x)$.


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  • YES if $f$ is monotonically increasing in the smallest interval containing $x$ and $f(x)$
  • NO if $f$ is monotonically decreasing in the smallest interval containing $x$ and $f(x)$
  • If $f = k$ is a constant function, then it is TRUE for all $x\le k$ and FALSE for all $x>k$
  • NOT NECESSARILY TRUE in any other case.
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If $f$ satisfies the above conditions, then:

  1. For all $x \in Im(f)$ we have $x \leq f(x)$. In particular, if $f(x)$ is onto then, the problem is true. To see this, let $y$ be so that $x=f(y)$, then $f(y) \leq f(f(y)) \Rightarrow x \leq f(x)$.

  2. There exists non onto functions which satiesfy the above condition for which, $x \leq f(x)$ if and only if $x \in Im(f)$. For example $f(x)=\frac{x-|x|}{2}$.

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No, when the function which is a decreasing function did not meet the result. Such as $f(x)=-x$,only when $x\geq 0$, $f(x)\leq f(f(x))$; but at the same time, $x\geq f(x)$.

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For some basic information about writing math at this site see e.g. here, here, here and here. – Julian Kuelshammer Nov 4 '12 at 11:56

If $f(x)<f(f(x))$ for all $x$ and $f$ is continuous then $x<f(x)$ for all $x$:

Define a continuous function $g$ by $g(x)=x-f(x)$. Now $g(f(0))=f(0)-f(f(0))<0$, so $g$ is below the $x$-axis at some point. Since $g(x)=0$ would imply $x=f(x)$ and hence $f(x)=f(f(x))$, we get $g(x)\ne 0$ for all $x$. Thus, $g(x)<0$ for every $x$.

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Well.. here is a sufficient condition:

If $f$ is invertible and monotonically increasing then $f^{-1}$ exits and is also monotonically increasing and therefore $f(x)\leq f(f(x)) \implies f^{-1}(f(x)) \leq f^{-1}(f(f(x))) \implies x \leq f(x)$.

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Actually, as mentioned in my answer, onto is enough ;) – N. S. Nov 4 '12 at 18:04
@N.S. You're right! :) – Amatya Nov 4 '12 at 20:37

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