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Let $f$, $g$ and $h$ be real functions of x, $x \geq 0$. Moreover, let

$f(x) = g(x)h(x)$

Is it enough to know that both $f$ and $h$ are non decreasing in $x$, to conclude that $g$ must be monotone? My intuition says yes, but I can't seem to find a way to prove it...

Further information:

  • $ 0 \leq f(x),g(x),h(x) < \infty, \forall x$
  • $\lim_{x\rightarrow 0^+}f(x)=\lim_{x\rightarrow 0^+}h(x)=0$
  • $\lim_{x\rightarrow \infty}f(x)=\lim_{x\rightarrow \infty}h(x)=1$
  • $f,g$ and $h$ can be assumed to be continuous on $\mathbf{R}^+$.

Any help? Thanks!

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No. Let $f(x)=1-{1\over x^2+1}$ and $h(x)=1-{1\over x+1}$. Then just by graphing $f(x)\over h(x)$ we can see that it is not monotone.

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  • $\begingroup$ Thanks a lot for the counterexample! $\endgroup$ – Federico Andreis Feb 9 '15 at 13:17
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For the general case, consider $f(x)=x^2$, $h(x)=e^x$, $g(x)=x^2e^{-x}$.

With your special restrictions for the boundary, we certainly have $\lim_{x\to\infty}g(x)=1$. Now just pick $f,h$ sich that at some point $x_o$, we have $f(x_0)>h(x_0)$. Then $g(x_0)>1$ and $g$ cannot be non decreasing.

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