Let function $F(x,y,z)$ defined on $[0,1]\times[0,1]\times[0,1]$ be increasing in $x$ and $y$. By increasing I mean $\frac{\partial F}{\partial x}\geq0$ and $\frac{\partial F}{\partial y}\geq0$.

By quasi-concave I mean $$F(x',y',z)\geq F(x,y,z)\Rightarrow[\frac{\partial F}{\partial x},\ \frac{\partial F}{\partial y}][x'-x,\ y'-y ]^T\geq0.$$

Define $$H(x,y)=\min\limits_{z\in[x,1]}F(x,y,z).$$

My question: Is function $H(x,y)$ quasi-concave in $x$ and $y$? Why?

EDIT 1: I think $H(x,y)$ is still increasing in $x$ and $y$, because for any $x'\geq x$ and $y'\geq y$, we have $\min\limits_{z\in[x',1]}F(x',y,z)\geq \min\limits_{z\in[x,1]}F(x,y,z)$ and $\min\limits_{z\in[x,1]}F(x,y',z)\geq \min\limits_{z\in[x,1]}F(x,y,z)$. Is it right?

EDIT 2: Correction of the quasi-concavity.

  • $\begingroup$ Well. If $F$ is increasing I can also say it's quasi convex right? I believe by increasing you mean increasing in any variable if others are fixed. For e.g. increasing in $z$ if $x$ and $y$ are fixed. $\endgroup$ Jun 25, 2020 at 11:52
  • $\begingroup$ @Shiv Tavker Yes, but I'm confused about the minimized function. I want to see if function H is quasi-concave or not... $\endgroup$
    – HXW
    Jun 25, 2020 at 12:08
  • $\begingroup$ Isn't the argmin just $z^* = x$, given the way you are setting up the problem? $\endgroup$
    – user762914
    Jun 25, 2020 at 12:26
  • $\begingroup$ @Renard But function F is not necessarily increasing in z... $\endgroup$
    – HXW
    Jun 25, 2020 at 13:26
  • $\begingroup$ @Huaixin Now the First inequality in Edit 1 may not be correct. $\endgroup$ Jun 25, 2020 at 14:07

2 Answers 2


An increasing function is only necessarily quasiconcave if it is a map from $\mathbb{R}$ to $\mathbb{R}$. The function $x^2+y^2+z^2$ is strictly convex on $[0,1]^3$, and increasing for any meaningful definition of increasing.

  • $\begingroup$ True. I guess calling quasi convex is safer? $\endgroup$ Jun 25, 2020 at 12:31
  • $\begingroup$ Thanks! I corrected my formulation. $\endgroup$
    – HXW
    Jun 25, 2020 at 13:52

Since, $F(x, y, z)$ is increasing, $$\ H(x, y) = \min_{z\in[x, 1]}F(x, y, z) = F(x, y, x) $$ Can you finish it now?

  • $\begingroup$ F is increasing in x and y, but it is not necessarily increasing in z. $\endgroup$
    – HXW
    Jun 25, 2020 at 13:23
  • $\begingroup$ Oh you edited the question! $\endgroup$ Jun 25, 2020 at 14:03
  • $\begingroup$ I just corrected the formulation, sorry... $\endgroup$
    – HXW
    Jun 25, 2020 at 14:06

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