# Let f:Omega sube R^n->R_(>=0) be a continuous differentiable function over Omega. Suppose that the function f is concave, and fix two points x=(x_1,…,x_n),y=(y_1,…,y_n) in Omega. If x_i≤y_i for all i=1,…,n and Omega=R^n, does it hold ∥grad_xf∥>=∥grad_yf∥?

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 $\parallel {\mathrm{\nabla }}_{\mathbf{x}}f\parallel \ge \parallel {\mathrm{\nabla }}_{\mathbf{y}}f\parallel$?
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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 $‖\mathrm{\nabla }f\left(-\mathbf{x}\right)‖<‖\mathrm{\nabla }f\left(0\right)‖$.