Sufficient conditions for Schur-concavity to imply concavity.

Question

I know that any concave and symmetric function is Schur concave. However, Schur concavity does not imply concavity. Is someone aware of sufficient conditions for Schur concavity to imply concavity?

Answer 1

In principle there need not be any connection between Schur concave and concave. Indeed, every function $\phi:\mathbb R\to\mathbb R$ is Schur concave (because $x\prec y$ on $\mathbb R$ is equivalent to $x=y$, so trivially $\phi(x)=\phi(y)$). However if $\phi$ is continuous, then Schur concave implies concave (reminder that we are still in one dimension!).

This characterization extends to higher dimensions for those Schur concave functions which "do not mix different components of the input". More precisely,

Theorem. Let $I\subseteq\mathbb R$ be an interval and let $\phi:I^n\to\mathbb R$ Schur-concave be given. If there exists $g:I\to\mathbb R$ continuous such that $\phi(x)=\sum_{i=1}^n g(x_i)$ for all $x\in\mathbb R^n$, then $g$ is concave.

For $n\geq 2$ one can waive continuity of $g$ if the conclusion is modified slightly. For more on this refer to Chapter 3, result C.1.c ff. in "Inequalities: Theory of Majorization and Its Applications" (2011, 2nd ed) by Marshall et al. Beyond Schur-concave functions of the above "sum-form", however, I am not aware of any results which get one from Schur-concave back to concave.