Example and Counter Example

As an example, consider the function ${\displaystyle x\mapsto f(x)=x^{2}}$ on the domain ${\displaystyle x\geq 0}$. This function is quasiconcave, but it is not concave (in fact, it is strictly convex). It can be concavified, for example, using the monotone transformation ${\displaystyle t\mapsto g(t)=t^{1/4}}$, since ${\displaystyle x\mapsto g(f(x))={\sqrt {x}}}$ is concave.

Not every concave function can be concavified in this way. A counter example was shown by Fenchel.^[2] His example is: ${\displaystyle (x,y)\mapsto f(x,y):=y+{\sqrt {x+y^{2}}}}$. Fenchel proved that this function is quasiconcave, but there is no monotone transformation ${\displaystyle g:\mathbb {R} \to \mathbb {R} }$ such that ${\displaystyle (x,y)\mapsto g(f(x,y))}$ is concave.^[3]: 7–9

Based on these examples, we define a function to be concavifiable if there exists a monotone transformation that makes it concave. The question now becomes: what quasiconcave functions are concavifiable?

Concavifiability

Yakar Kannai treats the question in depth in the context of utility functions, giving sufficient conditions under which continuous convex preferences can be represented by concave utility functions.^[4]

His results were later generalized by Connell and Rasmussen,^[3] who give necessary and sufficient conditions for concavifiability. They show that the function ${\displaystyle (x,y)\mapsto f(x,y)=e^{e^{x}}\cdot y}$ violates their conditions and thus is not concavifiable. They prove that this function is strictly quasiconcave and its gradient is non-vanishing, but it is not concavifiable.