The sequence has a stationary accumulation point

Question

Let $f:\mathbb{R}^n\rightarrow \mathbb{R}$ be a smooth (continuously differentiable), convex function with a non-empty set of minimizers and $\{x^k\}$ be a sequence such that

(a) $\{x^k\}$ has an accumulation (a cluster) point, i.e., there is some $\bar x$ together with some infinite set $J\subset\mathbb{N}$ such that $x^k\overset{J}\rightarrow \bar x$.

(b) $\liminf_{k\rightarrow\infty} \left\|\nabla f(x^k)\right\|=0.$

Does it mean that $\{x^k\}$ has a stationary accumulation point, i.e. some $\bar x$ with $\nabla f(\bar x)=0?$

I know how to proceed when $\{x^k\}$ is bounded, but removing this assumption is not straightforward for me.

The proof for the bounded case is as follows: As $\liminf_{k\rightarrow\infty} \left\|\nabla f(x^k)\right\|=0,$ we find some infinite set (corresponding with a subsequence) $J$ such that $\left\|\nabla f(x^k)\right\|\overset{J}{\rightarrow}0.$ As $\{x^k\}$ is bounded, $\{x^k\}_{k\in J}$ is bounded as well, which gives us some $\bar x$ and some infinite set $I\subset J$ such that $x^k\overset{J}{\rightarrow}\bar x$. Using $\left\|\nabla f(x^k)\right\|\overset{J}{\rightarrow}0,$ $I\subset J$, $x^k\overset{J}{\rightarrow}\bar x$ together with the continuity of $\nabla f$ gives us the desired conclusion.

Thanks a lot for any help, comment, or discussion. The counterexample is also welcomed.

Answer 1

No, this is not true. Take $f(x,y)=x^2$ so that the minimizers are all points $(0,y)$ and $P^{2k}=(\frac 1k, k)$, $P^{2k+1}=(1,1)$. Then $\liminf f(P^k)=0$ but the only cluster point is $(1,1)$.