Let $f_1,f_2$ be two smooth quasiconcave functions defined on a convex subset of $\mathbb{R}^d$.

It is known that $f_1+f_2$ is not necessarily quasiconcave.

Does there always exist monotonically increasing functions $g_1,g_2$ such that the sum $g_1\circ f_1 + g_2\circ f_2$ is quasiconcave?

The motiavation to this question is economic. In economics, it is often assumed that people have quasiconcave utilities: if they are indifferent between $x$ and $y$, then they prefer $(x+y)/2$ to $x$ and to $y$, which is (approximately) the definition of quasiconcavity. Given two or more people, we would like to create a social welfare function by taking the sum of the utility functions of the agents. In general, such a sum is not guaranteed to be quasiconcave. However, a utility function is only a numeric representation of a preference-relation, so we are allowed to transform it using arbitrary monotone transformations and it will still represent the same preference. The question is: can we always transform the utility functions of the agents such that the resulting social-welfare function be quasiconcave?

There are two cases for which I know the answer.

  1. If $f_1$ and $f_2$ are concavifiable, then the answer is "yes", since we can choose $g_1,g_2$ such that $g_1\circ f_1$ and $g_2\circ f_2$ are concave, which makes their sum concave too, and concavity implies quasic

  2. If $f_1$ and $f_2$ are are only weakly quasiconcave, then the answer is "not always", as shown by the example of Taneli Huuskonen below

So the question now becomes: is this always possible to find such $g_1,g_2$ when $f_1,f_2$ are strictly quasiconcave*?

Reminder: $f$ is quasiconcave iff $\forall x,y: \forall r\in(0,1): f(r x + (1-r) y)\geq \min(f(x),f(y))$. It is strictly quasiconcave iff $\forall x,y: \forall r\in(0,1): f(r x + (1-r) y)> \min(f(x),f(y))$.

  • $\begingroup$ Are $f_i$ and $g_i$ meant to be continuous, or not? And what is the the domain of $f_i$: a convex subset of $\mathbb{R}^n$, or the whole space? $\endgroup$ Commented Feb 12, 2018 at 18:43
  • $\begingroup$ @PietroMajer to focus on the quasiconcavity aspect, let's assume that the $f_i$ are smooth, and the $g_i$ can be arbitrarily increasing functions. The domain can be any convex subset of $\mathbb{R}^n$. $\endgroup$ Commented Feb 13, 2018 at 8:40
  • $\begingroup$ @PeterHeinig OK, I did not know that. $\endgroup$ Commented Feb 15, 2018 at 18:20

2 Answers 2


There is an easy counterexample. Let $$ f_1(x)=\begin{cases} x-n,&\mbox{when $2n\le x\le 2n+1$,}\\ n+1,&\mbox{when $2n+1\le x\le 2n+2$,} \end{cases} $$ and let $f_2(x)=f_1(-x)$. Now whenever $g_1$ and $g_2$ are strictly increasing, the function $g_1\circ f_1+g_2\circ f_2$ is strictly increasing on intervals $[2n,2n+1]$ and strictly decreasing between them, hence not quasiconcave.

You can replace $f_1$ and $f_2$ with smooth functions if you wish, but I cannot see any obvious way to obtain a similar counterexample with strictly quasiconcave functions.

  • 1
    $\begingroup$ Very interesting. It seems that the main problem is the fact that the functions are constant in some range, since no monotone transformation can change this behavior. [for visualization, here are some plots of this function: mathematica.stackexchange.com/a/165674/49101] $\endgroup$ Commented Feb 12, 2018 at 8:32

If $f$ is strictly quasi-concave and $C^2$, it can be concavified on any compact set on which it does not take a maximum. You can see this by differentiation. This is enough for many applications.

  • $\begingroup$ Can you write it as comment? $\endgroup$ Commented Apr 22, 2020 at 0:07
  • $\begingroup$ I don't have enough reputation points to comment. $\endgroup$
    – Augustin
    Commented Dec 23, 2020 at 21:55

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