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I just learned that a continuously differentiable function $f$ is strictly concave if $f(x+z)<f(x)+f'(x)z$ holds strictly for all $x$ and $z\neq 0$. So, this is a sufficient but not necessary condition. In other words, if $f$ is strictly concave, I cannot conclude that $f(x+z)<f(x)+f'(x)z$ holds strictly for all $x$ and $z\neq 0$. But I don't know why.

I just cannot come up an example that when a continuously differentiable $f$ is strictly concave, it can have $f(x+z)=f(x)+f'(x)z$ for some $z\neq 0$ at some x.

Could you please help me with this?

Thanks a lot!

Hi all,

I would like to thank Kavi's very patient answer. I guess I understand the theorem wrongly. And it should be a sufficient and necessary condition, so we do not need the example.

zz273
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  • I don't understand your question. A counterexample to what do you want to find? So far it looks like this: "Assume that $f$ is strictly concave. I can't show that it's not strictly concave". – Dmitry Aug 08 '20 at 22:30
  • Hi Dmitry, think you for letting me know the confusion. I refine my question. Hopefully, this is clear now. – zz273 Aug 08 '20 at 22:36
  • It doesn't hold when $f$ is not differentiable at $x$. I think $-\max(x^2, (x-2)^2)$ would be an example - strictly concave, but not differentiable at $1$. – Dmitry Aug 08 '20 at 22:45
  • Yes, you are totally right. I think I would modify the question again to focus only on continuously differentiable functions. Thanks! – zz273 Aug 08 '20 at 23:00

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Suppose equality holds with $z>0$. Let $S(y)=\frac {f(y)-f(x)} {y-x}$ for $x <y \leq x+z$. It is well known (and fairly easy to check from definition of concavity) that $S$ is monotonically decreasing. Its limiting value at $y=x$ is $f'(x)$ and we are assuming that its value at $x+z$ is also $f'(x)$. Hence $S$ is a constant on $[x,x+z]$. But that contradicts the fact that $f$ is strictly concave. A similar argument works when $z <0$. Hence the example you are looking for does not exist.

Kavi Rama Murthy
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  • Hi Kavi, thanks a lot for your answer. But I would like to find (+)=()+′() for a strictly concave function. Could you please help me with this? Thanks – zz273 Aug 08 '20 at 23:35
  • @zz273 I have proved that no such function exists. I assumed that it exists and arrived at a contradiction. – Kavi Rama Murthy Aug 08 '20 at 23:37
  • Oh! Thanks! Now I understand your answer. But what I mean is (+)=()+′() for some $z$. I guess I should modify my question a little bit to make it clear. Could you please help me with the version of "some $z$" instead of any $z$? Thanks! – zz273 Aug 08 '20 at 23:40
  • But for $f(x) = -e^x$ I cannot find some $x$ and $z$ to have (+)=()+′()? What I want to find is to have $(+)=()+′()$ for a strictly concave function. I think I just don't understand why strictly concave is not "if and only if" to (+)<()+′(). – zz273 Aug 08 '20 at 23:55
  • But ()=−^3 is not a concave or convex function in [-2,1]? – zz273 Aug 09 '20 at 00:13
  • @zz273 You cannot have equality even for some $z \neq 0$. I have added a proof. – Kavi Rama Murthy Aug 09 '20 at 00:39
  • Thank you, Kavi. Now I understand it should be an if and only if statement. – zz273 Aug 09 '20 at 02:53