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I want to show that a function $ψ(a_1,a_0)$, which is separable (additively decomposed) in two quasiconcave functions, is also quasiconcave (QC). I know that the sum of QC functions is not generally QC, but my numerical simulations seem to suggest that $ψ(a_1,a_0)$ is.

I have tried using the definition (i.e. trying to show that it has convex upper contour sets by plugging $ψ\big(\lambda a + (1-\lambda)b\big)\geq$... etc), but I can not prove the inequality. I also tried using the border Hessian, but i cannot sign the determinant. Any other ideas??

Below is a more detailed description of the problem:


Where $a_1 ∈[0,1]$ and $a_0 ∈[0,1]$, and

$S(a_j)=F\big(Bh(a_j)\big) \int_{a_j}^{1} \big(t*P(t)\big)dt + F\big(Bl(a_j)\big)\int_{a_j}^{1} \big(t*(1-P(t))\big)dt$

for $j∈\{1,0\}$

and the functions $F(.), Bl(.), Bh(.)$ and $P(.)$ are strictly increasing. Additionally, $Bl''>0$, and $Bh''<0$.

$F''$ and $P''$ can be positive or negative (as required to guarantee quasiconcavity --my simulations suggest that only $P''>0$ is necessary).

I would appreciate any hints --including other forums where I could post this question

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