# Positive functions which doesn't get multiplied too fast/doesn't grow too fast

What could be some non-trivial examples of positive valued continuous functions $$f: \mathbb{R_{+}}\to \mathbb{R_{+}}$$ so that: $$\forall x, y \in \mathbb{R_{+}}, x, $$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 $$[\frac{\mu}{\lambda}, \infty)$$.

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

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 $$c\ge1$$.
• 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 $x \in (0, \infty)$? Jan 10, 2020 at 15:15
• @Mathmath I've fixed it to not require an increasing $f$. By linear I mean changing $f$ to $af+b$.