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Suppose I have a function $g$ on $[-1,1]$ which is increasing, concave and such that $g(-1)=0$. I set $h(t)=\frac{g(t)}{1+t}$. Is it true that $h$ is decreasing and convex?

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Take $g(x)=\cos\left(\frac{\pi x}{4}-\frac{\pi}{4}\right)$

Hence, with $h(x)=\frac{\cos\left(\frac{\pi x}{4}-\frac{\pi}{4}\right)}{x+1}$ there are problems.

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  • $\begingroup$ Are you sure? Because to me this is not a counter-example. $\endgroup$ – Babyblog Feb 5 '17 at 16:16
  • $\begingroup$ @Babyblog $g$ is an increasing and concave, which you wish. $\endgroup$ – Michael Rozenberg Feb 5 '17 at 17:21
  • $\begingroup$ But $h$ is decreasing and convex. My question is: is it true that for all such $g$ and $h$ we have $h$ convex and decreasing? $\endgroup$ – Babyblog Feb 5 '17 at 18:25
  • $\begingroup$ @Babyblog My $h$ is not convex. $h''(0)<0$ $\endgroup$ – Michael Rozenberg Feb 5 '17 at 18:34
  • $\begingroup$ Ok you're wright your $h$ is concave. Thank you. $\endgroup$ – Babyblog Feb 5 '17 at 20:56

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