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Consider a function $f:\mathbb{R}^n_{\geq 0}\rightarrow \mathbb{R}$ that is separately convex, i.e. such that $\frac{d^2f}{dx_i^2}\geq 0$ for all $i\in \{1,\dots n\}$. Assume also that $f$ is decreasing in each of its arguments and is positively homogenous of degree 1 so that $f(tx)=tf(x)$ for all $t\geq 0$. Is $f$ quasiconvex?

According to this article, if $n=2$ then $f$ is convex (so also quasiconvex) but this result does not generalize to $n>2$. I tried to find a counterexample but all my attempts have ended up with $f$ convex.

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  • $\begingroup$ I doubt that this is true. First I would try to construct any function which is 1 homogeneous, separately convex, increasing in each variable but not jointly convex. Do you have such example? $\endgroup$
    – Paata Ivanishvili
    Commented Oct 8, 2016 at 2:45
  • $\begingroup$ I haven't been able to find a counterexample. I have also relaxed some of the assumptions on $f$ since I couldn't come up with an example that was not linear. $\endgroup$
    – user_lambda
    Commented Oct 8, 2016 at 14:58
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    $\begingroup$ this example $B(x,y,z)=\frac{xy}{z}$ is 1-homogeneous, separately convex, and increasing only in x and y variables but not in z, is not quasicovex. I understand that in general counterexample cannot be made by product functions like $B(x,y,z,..)=f_{1}(x) f_{2}(y) f_{3}(z)..$. This will never work. So joint structure should be involved. $\endgroup$
    – Paata Ivanishvili
    Commented Oct 8, 2016 at 15:20
  • $\begingroup$ Thanks, that's useful to know. I've edited the question more time (sorry about the back and forth) to impose that $f$ is decreasing in each of its arguments. The examples I've found are all convex. $\endgroup$
    – user_lambda
    Commented Oct 8, 2016 at 15:36

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