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I asked a similar question here but did not specify all the details of the function I encountered. As that question has been answered, I ask the clarified version here.

Suppose $f: U \to \mathbb R$ is a $C^{\infty}$ smooth function where $U \in \mathbb R^n$ is an open simply connected set. Further, $f$ has following property: if $U \ni \|x_n\| \to \infty$, $f(x_n) \to +\infty$; if $\{x_n\} \subseteq U$ and $ \{x_n\} \to x \in \partial U$ (boundary of $U$), $f(x_n) \to + \infty$. This is equivalent to say all the sublevel sets of $f$ are compact. Now I know if $x_0 \in U$ is a critical point (that is, $\nabla f(x_0) = 0$), then $x_0$ is a strict local minimizer (that is, the Hessian $\nabla^2 f(x_0)$ is positive definite). My question is whether these information allows us to deduce that $f$ has only one critical point. A counterexample is certainly welcome.

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  • $\begingroup$ I'm not sure how your conjecture isn't countered by such a simple function as $f(x) = x^2(x-1)^2$. It has multiple critical points/strict local minimizers, despite diverging to $+\infty$ in either direction. $\endgroup$ – Feryll Nov 14 '19 at 8:10
  • $\begingroup$ @Feryll The function in your example has a local maximum at the stationary point $x=1/2$. However, in OP, it is required that any stationary point implies local minimum. $\endgroup$ – River Li Nov 14 '19 at 8:55
  • $\begingroup$ As commenters to the sister problem indicated, what you want is Morse theory, which is explained in Milnor's book of that title. If all your critical points have full rank Hessians then (modulo other stuff I've forgotten) the number of minima plus the number of maxima minus the number of saddle points depends only on the homology of the domain. $\endgroup$ – kimchi lover Nov 14 '19 at 11:25
  • $\begingroup$ Re my comment: I'm assuming $n=2$ in the last sentence. $\endgroup$ – kimchi lover Nov 14 '19 at 14:41
  • $\begingroup$ Your title does not match the problem. $\endgroup$ – zhw. Nov 14 '19 at 16:32

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