Find counterexample of gradient system with non globally convex energy function has globally asymptotic stable equilibrium point?

Question

Given $$\dot x=-\nabla f(x)$$ and suppose it has an equilibrium point $x=0$.

It is known that if $f(x)$ is globally convex, then $x=0$ is globally asymptotic stable.

I am interested its converse: can you find counter example that although if $f(x)$ is not convex everywhere (this is called globally convex right?), $x=0$ is still globally asymptotic stable?

Thanks in advance for any suggestion!

Answer 1

Try an example where $f$ is radially symmetric and increasing with distance from the origin, but is not convex.

Answer 2

Maybe the paper 1 provides an answer to this question at least in the local neighborhood of $x=0$:

First, it depends on the condition of the energy function $f(x)$.

If $f(x)$ is smooth function ($\in C^\infty$), it is possible that $x=0$ is locally stable, while $f(x)$ is not convex; And, it is possible that $f(x)$ is not convex, while $x=0$ is locally stable. The examples are provided in section 2.

If $f(x)$ is real analytical function, we have something much better: $x=0$ is a stable equilibrium point if and only if it is a local minimum of $f(x)$.

Note:

In this post, I am asking about the convexity of $f(x)$, thus there is still a gap between convexity and local minimum, which remained to be addressed in the future couple of days.

This answer only addresses locally asymptotic stability, whereas global results still need open.

Papers:

[1] On the stable equilibrium points of gradient systems

[2] Asymptotically Stable Equilibria of GradientSystems (points out error in text books)