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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!

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

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    $\begingroup$ What do you mean by $F \succ 0$? Positive definite for all $X$? What about the $1 \times 1$ example $f_{1,1}(X) = 1+x^2$? $\endgroup$
    – Robert Israel
    Commented Aug 26, 2012 at 22:17
  • $\begingroup$ Hmm, this is exactly what I meant. Oops!! I meant that $f_{i,j(X)}$ are QUASICONCAVE !! Thank you for the correction! $\endgroup$
    – Josh
    Commented Aug 27, 2012 at 5:53

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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 $f(0) = 4$, $f(1) = 2$, $f(2) = 1$.

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