Suppose $h(x)\ge0$ is increasing and concave for all $x\ge0$. For $\Delta>0$, let $$ f(x)=\frac{h(x+\Delta)-h(x)}{\Delta}. $$ I feel that $f$ is a convex function, i.e. $$ f(tx+(1-t)y)\le t f(x)+(1-t)f(y), 0<t<1. $$ But, I don't have any good progress on proving it. Is it obvious true? Or, is there any counterexample? Any reference, help or adivce would be of great help. Thanks.
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2Do you mean for all $x \in \Bbb R$? Then your conclusion is “trivially” true because the only concave functions $h: \Bbb R \to (0, \infty)$ are the constant functions. – Martin R May 08 '22 at 13:47
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Sorry, I made a mistake. $h:\mathbb{R}\to [0,\infty)$. – user377704 May 09 '22 at 02:47
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Could you please give me more hints or advices on proving this? Thank you. – user377704 May 09 '22 at 02:53
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1@Tony Whether you include $,0,$ or not in the codomain makes no difference. The only concave functions on $\mathbb R$ bounded below are the constant functions. See Show bounded and convex function on $\mathbb R$ is constant. – dxiv May 09 '22 at 03:14
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Thanks for your response. But, $\sqrt(x) $ is nonnegative, increasing and concave on $x\ge0$. What if i change the domain? – user377704 May 09 '22 at 03:48
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1@Tony Please make up your mind on what the question is that you mean to ask, then edit it into the main post, not as a comment. The answer would likely be negative i.e. a counterexample, but no one can guess your next "what if" in advance. – dxiv May 09 '22 at 04:02
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I see. Thank you very much. – user377704 May 09 '22 at 04:05
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1Try a small $,\Delta,$ with $,h(x) = \sin x,$ for $,x \in [0, \pi/2],$ and $,h(x)=1,$ for $,x \gt \pi/2,$. – dxiv May 09 '22 at 04:11
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Thank you very much for your help! I learned a lot. – user377704 May 09 '22 at 07:20