Suppose I have a differential equation
$$\dot{x}(t)=f(x(t))$$
with a bounded $x(t)$, does an $f$ exist, s.t.
$$\|\dot{x}(t)\| = \|f(x(t))\| \to 0\qquad t\to \infty$$
but there is no $x^*$ s.t. $x(t)\to x^*$?
I am thinking that such a function should exist (c.f. Does a bounded function converge if its derivative tends to zero?) but I can not think of an autonomous example.
Application to Gradient Decent
Ideally I am looking for an integrable $f$ with a lower bounded integrand. I.e. there should exist a function $L \ge0$ s.t.
$$f(x) = -\nabla L(x)$$
Essentially I am trying to answer the question, whether the gradient in gradient decent can converge without the iteration sequence itself and
$$x(t_{n+1}) = x(t_{n}) - \eta\nabla L(x(t_n))$$
is basically just a discretized ODE. So answering whether $\nabla L(x(t))$ can converge to zero without $x(t)$ converging should probably be equivalent.
Where to look
If $f'(x) = \nabla^2 L(x) \ge \mu >0 $ this won't work as there are theorems which prove local convergence in that case. The second derivative must get close to zero fast around the point of interest.
For the unbounded case:
$$f(x)=\exp(-x)$$
results in
$$x(t) = \log(c+t)$$
being a solution to the ODE. $L(x)=\exp(-x)$ does the job as a loss function. As suggested in the comment it might be possible to coil up this infinite 1-dimensional path into a bounded 2-dimensional spiral. If anyone can come up with a clean/easy to understand formulation of that I would be happy :)
