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?
From concavity to convexity
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convex-analysis
1 Answers
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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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0Are you sure? Because to me this is not a counter-example. – 2017-02-05
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0@Babyblog $g$ is an increasing and concave, which you wish. – 2017-02-05
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0But $h$ is decreasing and convex. My question is: is it true that for all such $g$ and $h$ we have $h$ convex and decreasing? – 2017-02-05
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0@Babyblog My $h$ is not convex. $h''(0)<0$ – 2017-02-05
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0Ok you're wright your $h$ is concave. Thank you. – 2017-02-05