Do Pareto critical points of a multicriteria optimization problem form an attractor of the dynamical system induced by a descent algorithm?

Question

Let $d\in\mathbb N$, $k\in\mathbb N$ and $f:\mathbb R^d\to\mathbb R^k$ be differentiable. Say that $v\in\mathbb R^d$ is a descent direction at $x\in\mathbb R^d$ if ${\rm D}f(x)v<0$ (component-wise) and that $x$ is locally Pareto critical if there is no descent direction at $x$.

A typical descent algorithm is of the form of a discrete dynamical system $$x_{n+1}=g(x_n):=x_n+t(x_n)v(x_n)\tag1,$$ where $t:\mathbb R^d\to(0,\infty)$ is a step size function and $v:\mathbb R^d\to\mathbb R^d$ is such that either $v(x)=0$ or $v(x)$ is a descent direction. Note that a point is a fixed point of $g$ if and only if it is Pareto critical.

I've read that the set of locally Pareto critical points is an attractor of the dynamical system $(1)$. How can we show this?

Answer 1

I'm not sure if this can be shown without some additional condition on $t$. Consider the trivial case $d=1$, $k=1$, $f(x)=x^2$; then $x=0$ is certainly a fixed point. But for the constant step size function $t(x)=1$ with $v(x)$ normalized to $v\in\{-1,0,1\}$, and $x_0\not\in\mathbb{Z}$, the system enters a limit cycle oscillating between two points on either side of zero. So without some condition on $t$, the set of fixed points is not necessarily an attractor, and a proof of the desired statement would have to depend on (currently unstated) properties of $t$.