A Schur Convex function defined on $(x_1,x_2)$ is one if it satisfies the following inequality:
$$(x_1-x_2) (\frac{\partial}{\partial {x_1}}f -\frac{\partial}{\partial {x_2}}f) \ge 0, \quad \quad\text{for all } (x_1,x_2) \in \mathbb{R^2}.$$
Question:
If $f$ (all partial derivatives exist) is
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symmetric along line $x_1=x_2$,
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and $f$ is monotonic increasing (or decreasing) in each of its argument
then does it implies that f is Schur concave (or Schur convex) ?
Best Answer
Let $$f(x,y) = (\arctan(x) + \pi/2) (\arctan(y) + \pi/2)$$ Then $f$ is symmetric in $x$ and $y$, and monotonically increasing in each argument since $\arctan$ is positive and monotonically increasing. Now $$(x-y)(f_x - f_y) = (x-y) \left(\frac{\arctan(y)+\frac{\pi }{2}}{x^2+1}-\frac{\arctan(x)+\frac{\pi }{2}}{y^2+1}\right)$$ is not everywhere-nonnegative. For example, setting $x = 1$, $y = -1$ yields $-\pi/2 < 0$. Thus, $f$ is not schur convex.
Edit: neither is this quantity everywhere-nonpositive! Setting $x = 0$, $y = -2$ yields $2 \left(\frac{2 \pi }{5}-\arctan(2)\right) \approx 0.3$.