Hello all,


Assume we have a sequence of **quasiconcave** functions (in $X$) denoted by $f\_{i,j}(X)$ for $i,j = 1,\ldots,n$. Denote by $F(X)$ the $n\times n$ matrix whose $(i,j)$ entry is the function $f\_{i,j}(X)$. 


Assuming that $F\succ0$ (positive definite for all $X$), I want to prove (or disprove) that the function $g(X)=a^TF^{-1}a$, where $a\in\mathbb{R}\_+^{n\times 1}$, is quasiconvex. 


Someone have any idea?


Thank you!




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Correction: $f\_{i,j}(X)$ are quasiconcave and not quasiconvex. Credit to Robert.



Proving that a specific function is quasiconvex
It's not true.


Consider the $2 \times 2$ matrix $$F(X) = \pmatrix{f(X) & 0\cr 0 & f(X-2)\cr}$$
where $f$ is an even function, everywhere $> 0$, and decreasing on $[0,\infty)$. Take $a = (1,1)^T$.
Then $g(X) = a^T F(X)^{-1} a = 1/f(X) + 1/f(X-2)$.
In particular $g(0) = g(2) = 1/f(0) + 1/f(2)$ while $g(1) = 2/f(1)$.  

If $2/f(1) > 1/f(0) + 1/f(2)$, $g$ is not quasiconvex. For example, we could take 
$f(0) = 4$, $f(1) = 2$, $f(2) = 1$.


