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What could be some non-trivial examples of positive valued continuous functions $f: \mathbb{R_{+}}\to \mathbb{R_{+}}$ so that: $\forall x, y \in \mathbb{R_{+}}, x<y$, $f(x) \geq \frac{1}{2}f(y)$?

Of course, the constant and monotone non-increasing functions satisfy this property trivially. Also we note that locally, when $x, y$ are close enough, this property is valid, since for close enough $x,y$, $f(x)$ is approximately equal to $f(y)$, hence $f(x) \geq \frac{1}{2}f(y)$. I'm looking for other examples, or rather a class of examples, or a way to construct the examples globally. See below.

For a non-constant example, one can define: $f: (0,\frac{\mu}{\lambda})\to \mathbb{R_{+}}$ by: $f(x)= \lambda x + \mu$, and this will satisfy the desired property. But $f$ won't satisfy the desired property when $x > \frac{\mu}{\lambda}$, so the definition of $f$ needs to be modified on $[\frac{\mu}{\lambda}, \infty)$.

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    $\begingroup$ Does $\mathbb R^+$ contain $0?$ $\endgroup$
    – zhw.
    Jan 10, 2020 at 18:56
  • $\begingroup$ @zhw. No it doesn't contain 0 :) $\endgroup$ Jan 11, 2020 at 15:01

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This is achieved if $M\le2m$ with $m:=\inf S,\,M:=\sup S$, with $S$ the set of values of $f$. So take your favourite bounded function, and linearly transform it to any $m,\,M$ of your choosing with $M\le2m$. If we first transform so $f$ ranges from $0$ to $1$ (e.g. $f=\frac{2}{\pi}\arctan x$), $f$ doesn't work, but $f+c$ does for any $c\ge1$.

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    $\begingroup$ Thanks for your answer, but I'm not sure about few points. Why "so $f(x)\in[m,\,M]$ for all $x\in\Bbb R^+$?" $f$ can take values outside its limiting values at $0$ and $\infty$. Secondly, what do you mean by "linearly transform" $f$? Is it constructing the new function $x \mapsto af(x)+b$? I'm also not sure why this is true: "if $m \geq M/2$, we're done"? Are you saying that that alone is a sufficient condition? If yes, how do we take care of the values of $f$ at $x \in (0, \infty)$? $\endgroup$ Jan 10, 2020 at 15:15
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    $\begingroup$ @Mathmath I've fixed it to not require an increasing $f$. By linear I mean changing $f$ to $af+b$. $\endgroup$
    – J.G.
    Jan 10, 2020 at 15:18

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