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I'm reading a paper that claims that if $\forall a,b\in \mathbb{R}^2$ s.t. $a_i>b_i$ for $i=1,2$, one has $$ f(a_1+b_2)+f(a_2+b_1)>f(a_1+a_2)+f(b_1+b_2) $$ then $f$ is strictly concave. It's not difficult to see the function is mid-point concave. But what about full concavity?

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  • $\begingroup$ I tried to construct a counterexample, but as per math.stackexchange.com/questions/1952560/… a counterexample cannot easily be expressed. However, I keep sharing your sceptism toward to validity of the claim. $\endgroup$ – LinAlg Oct 5 '16 at 1:55
  • $\begingroup$ The previous function that you posted was not a valid counterexample since it didn't respect the condition. $\endgroup$ – Lorenzo Bastianello Oct 5 '16 at 14:13
  • $\begingroup$ Is $f$ continuous? In that case it's a typical exercise that midpoint-concavity implies full concavity $\endgroup$ – Del Oct 5 '16 at 14:39
  • $\begingroup$ Continuity is not assumed $\endgroup$ – Lorenzo Bastianello Oct 5 '16 at 15:52
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Try to prove that $f$ strictly concave over I is equivalent to

$\forall (A,B,C,D)\in I^4$, such that $A<C\leq D<B$, $$\frac{f(B)-f(D)}{B-D}<\frac{f(C)-f(A)}{C-A}.$$

and put $B=b_1+b_2$, $A=a_1+a_2$, $C=min(a_2+b_1,a_1+b_2)$, $D=max(a_2+b_1,a_1+b_2)$

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  • $\begingroup$ Sorry, it is a proof of the converse assumption... $\endgroup$ – Stephane Gonzalez Oct 6 '16 at 14:38

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