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June 27th, 2017, 11:21 AM   #11
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Quote:
 Originally Posted by Dattier edit : I must think What about this inequality $\displaystyle 4(\max(g)-\min(g))\geq \min(g'')$, when $\displaystyle g \in C^2([0,1])$
No. With $g(x)=x(x-1)=\frac14(2x-1)^2-\frac14$ we have $\max{(g)}-\min{(g)}=0-(-\frac14)=\frac14$ and $g''(x)=2$.

 June 27th, 2017, 01:01 PM #12 Member   Joined: May 2017 From: France Posts: 57 Thanks: 1 edit : I must think Last edited by Dattier; June 27th, 2017 at 01:07 PM.
 June 27th, 2017, 01:23 PM #13 Member   Joined: May 2017 From: France Posts: 57 Thanks: 1 the idea $\displaystyle g(x)-\frac{\min(g'')}{2}x^2$ is convex.

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