Let $f:\mathbb{R}^n\rightarrow\mathbb{R}$ be a convex function. Denote $X^\*$ as the set of minimizers of $f$ and assume $X^\*$ is unbounded. Is it possible that $\|g\_x\|$ is unbounded when $d(x,X^\*)$ is bounded where $g\_x\in\partial f(x)$. Here $\partial f(x)$ denotes the set of subgradient vectors of $f$ at $x$ and $d(x,X^\*)$ denotes the distance between point $x$ and set $X^\*$.




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I know $X^\*$ is a closed convex set and I am thinking whether it is true that $\|g\_x\|$ is bounded given $d(x,X^\*)=1$. It must be true if $X^\*$ is bounded, but not sure when it is unbounded. Also, does it help if we further assume $f$ is differentiable, so that $\partial f(x)=\{\nabla f(x)\}$?




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**Edit**: I realized that what I really want to ask is whether it is possible that $\|g\_x\|$ is unbounded but $d(x,X^\*)$ is bounded. It is not equivalent to the boundedness of the ratio of $\|g\_x\|$ and $d(x,X^\*)$.



An upper bound of gradient norm for convex functions near minimizer
The function $q$ might be unbounded even in 1-neighborhood of $X^\*$; here is an example of such function $f$ on the $(x,y)$-plane.


Let $\phi(t)=|t|-t$.
Choose a sequence $x\_n\to \infty$.
For each $n$ consider function $f\_n(x,y)=\phi[-2\cdot x\_n\cdot(x-x\_n)+ (y-x\_n^2)]$.
Let $f=\max\_n\{f\_n\}$.


