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Let $f: \mathbf{R^n} \to \mathbf{R}$ be differentiable, nonconvex function. The gradient descent iteration is given by $$x^{k+1} = x^{k} - t_k \nabla f(x^k)$$ To ensure $\lim_{k \to \infty} x^k$ exists, one condition for $f$ is that the critical point of $f$ is isolated, see paper http://proceedings.mlr.press/v49/lee16.pdf. The critical point is defined as $\left\{x\in \mathbf{R}^n : \nabla f(x) = 0\right\}$. A critical point $x$ is isolated if there is a neighborhood $U$ around $x$, and $x$ is the only critical point in $U$.

This condition is new to me. How to understand this condition? What type of function satisfies the isolate critical points condition?

  • Are you looking for examples of functions that have critical points that are not isolated? What about functions that are bounded below but have no critical point? – Brian Borchers Nov 16 '22 at 03:40
  • Think of one dimensional examples, $\sin$, $\exp$ (which has no critical points). There is no easy classification of functions satisfying the isolated condition, it is a convenience for purposes of proof. – copper.hat Nov 16 '22 at 03:45
  • I want to use this condition, but I do not know how to verify it. Some good examples would help me to understand this condition. @Brian Borcher – Ronglong Fang Nov 16 '22 at 03:55
  • Yes. It looks like the isolated critical point condition is not easy to verify. @ copper.hat – Ronglong Fang Nov 16 '22 at 04:02

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