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June 27th, 2017, 12:21 PM   #11
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Originally Posted by Dattier View Post
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 $g''(x)=2$.
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June 27th, 2017, 02:01 PM   #12
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edit : I must think

Last edited by Dattier; June 27th, 2017 at 02:07 PM.
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June 27th, 2017, 02:23 PM   #13
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the idea $\displaystyle g(x)-\frac{\min(g'')}{2}x^2$ is convex.
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