Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
What do you mean by $F \succ 0$? Positive definite for all $X$? What about the $1 \times 1$ example $f_{1,1}(X) = 1+x^2$? –  Robert Israel Aug 26 '12 at 22:17
Hmm, this is exactly what I meant. Oops!! I meant that $f_{i,j(X)}$ are QUASICONCAVE !! Thank you for the correction! –  Josh Aug 27 '12 at 5:53

1 Answer 1

up vote 1 down vote accepted

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

share|improve this answer
Great example. Thanks again –  Josh Aug 27 '12 at 18:28

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.