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](https://en.wikipedia.org/wiki/Concavification#Concavification_of_a_quasiconcave_function_by_monotone_transformation), 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 $\forall x,y: \forall r\in(0,1): f(r x + (1-r) y)> \min(f(x),f(y))$*.



Can the sum of quasiconcave functions always be made quasiconcave?
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.


