# An upper bound of gradient norm for convex functions near minimizer

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 $$X^*$$.

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 $$\partial f(x)=\{\nabla f(x)\}$$?

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 $$d(x,X^*)$$.

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 $$f=\max_n\{f_n\}$$.
• So by saying "the answer is no", you mean for your example, it is possible that $d(x,X^*)=1$ but $\|g_x\|$ is unbounded? or you mean it is not possible that $q(x)$ is unbounded? If it is the second case, then you need to prove $q(x)$ is bounded for any example $f$ right? Commented Apr 16, 2023 at 13:09
• @JeanLegall, in the constructed example, $q$ is unbounded, and "yes" you can make such example smooth with unbounded $q$ in $1$-nbhd from $X^*$. (Use a smoothed version of $\phi$ and sum instead of max.) Commented Apr 16, 2023 at 13:50
Take the function $$f:\mathbb{R}^2\to\mathbb{R}$$, $$f(x,y)=y^4$$. Then $$X^*$$ is the $$x$$ axis, $$dist(\;(x,y), X^*)= |y|$$, $$\Vert \nabla f(x,y)\Vert =4|y|^3$$. $$\newcommand{\bR}{\mathbb{R}}$$ $$\newcommand{\ve}{{\varepsilon}}$$
• Thanks for your answer. What I actually want is the case when $dist(x,X^*)$ is bounded but $\nabla f(x)$ unbounded. However, for the current statement of the question, your example also works. Commented Apr 16, 2023 at 13:55