unabuenanuevasld

2022-11-21

Let $f:\mathrm{\Omega }\subseteq {\mathbb{R}}^{n}\to {\mathbb{R}}_{\ge 0}$ be a continuous differentiable function over $\mathrm{\Omega }$. Suppose that the function $f$ is concave, and fix two points $\mathbf{x}=\left({x}_{1},\dots ,{x}_{n}\right),\mathbf{y}=\left({y}_{1},\dots ,{y}_{n}\right)\in \mathrm{\Omega }$,$\mathbf{x}=\left({x}_{1},\dots ,{x}_{n}\right),\mathbf{y}=\left({y}_{1},\dots ,{y}_{n}\right)\in \mathrm{\Omega }$.
If ${x}_{i}\le {y}_{i}$ for all $i=1,\dots ,n$ and $\mathrm{\Omega }={\mathbb{R}}^{n}$, does it hold $\parallel {\mathrm{\nabla }}_{\mathbf{x}}f\parallel \ge \parallel {\mathrm{\nabla }}_{\mathbf{y}}f\parallel$?

sliceu4i

Is false. Take a concave function which is symmetric about the origin (e.g. $f=-‖x{‖}^{2}$). if $0\le {x}_{i}$ and $‖\mathrm{\nabla }f\left(\mathbf{x}\right)‖>‖\mathrm{\nabla }f\left(0\right)‖$, then due to symmetry we'd get $-{x}_{i}\le 0$ but $‖\mathrm{\nabla }f\left(-\mathbf{x}\right)‖<‖\mathrm{\nabla }f\left(0\right)‖$.