Can a function be both Schur-concave and Schur-convex?

Question

I know that any linear and symmetric function $f:\mathbb{R}^n_+\rightarrow \mathbb{R}_+$ is both Schur-concave and Schur-convex. Let $f$ be also increasing and continuous, and suppose that $f$ is non-linear, can it still be both Schur-concave and Schur-convex?

Answer 1

Short answer: Indeed linearity is not required.

Detailed answer: Let $f$ be Schur-concave and Schur-convex (so in particular $f$ is symmetric), that is, for all vectors $x,y$ with $x\prec y$ one has $f(x)\leq f(y)\leq f(x)$, hence $f(x)=f(y)$. In particular $f$ has to be constant on each trace hyperplane $T_c:=\{x\in\mathbb R^n:\sum_{j=1}^n x_j=c\}$, $c\in\mathbb R$ because $m_c:=\frac{c}n(1,\ldots,1)^\top\prec x$ for all $x\in T_c$. In other words because all vectors can be related to each other via majorization (through the unique minimal vector $m_c$ w.r.t $\prec$) $f$ cannot distinguish between any of them.

This allows us to define an auxiliary map $\alpha:\mathbb R\to\mathbb R$ via $\alpha(c):=f(\frac{c}n(1,\ldots,1)^\top)$ to find $$ \forall_{x\in\mathbb R^n}\quad f(x)=f\big(\tfrac{\sum_{j=1}^nx_j}n(1,\ldots,1)^\top\big)=\alpha\Big(\sum_{j=1}^nx_j\Big)\,.\tag{1} $$ for all $x\in\mathbb R^n$. In other words the only information $f$ can distill from the input is the sum of its entries. And this is as far as we get; we know that $f$ has to of this form, and for all choices of $\alpha$ the function $f$ will be Schur-convex and Schur-concave (i.e. (1) is a characterization). Even adding the requirement that $f$ is continuous will not make $\alpha$ linear (only continuous), and for good reason:

Example. The map $f:\mathbb R^2\to\mathbb R$, $(x_1,x_2)\mapsto (x_1+x_2)^2$ (non-linear!) is Schur-convex and Schur-concave.