0

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?

Lorenz
  • 135
  • I tried to construct a counterexample, but as per http://math.stackexchange.com/questions/1952560/characterization-of-concave-functions/1952575?noredirect=1#comment4011605_1952575 a counterexample cannot easily be expressed. However, I keep sharing your sceptism toward to validity of the claim. – LinAlg Oct 05 '16 at 01:55
  • The previous function that you posted was not a valid counterexample since it didn't respect the condition. – Lorenz Oct 05 '16 at 14:13
  • Is $f$ continuous? In that case it's a typical exercise that midpoint-concavity implies full concavity – Del Oct 05 '16 at 14:39
  • Continuity is not assumed – Lorenz Oct 05 '16 at 15:52

1 Answers1

0

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

Stephane Gonzalez
  • 21