Let $f$ a $C^2[0,1]$ function that is strictly concave up: for any $0\leq x<y\leq 1$ $a\in(0,1)$ $f(ax+(1-a)y)<af(x)+(1-a)f(y)$.
How to show that $f''(x)\geq 0$ for $x\in(0,1)$? Any reference to read the proof of this result?
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The proofs I am familiar with, such as from Schramm's Introduction to Real Analysis, uses the Taylor series approximation. – Carl Morris Aug 26 '14 at 18:58
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2Maybe a start would be to show: strictly concave up (and differentiable) implies the derivative is strictly increasing. – GEdgar Aug 26 '14 at 19:00
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@GEdgar Any hints on how to show that? – Kal S. Aug 26 '14 at 22:55
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A hint is maybe the pictures here http://math.stackexchange.com/a/615161/442 – GEdgar Aug 27 '14 at 00:20