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Suppose $U \in \mathbb R^n$ is an open simply connected set and $f: U \to \mathbb R$ is a real valued $C^{\infty}$ function. I am wondering whether the following is possible: $f$ has more than $1$ local minimizers, say $x_1, x_2 \in U$ but does not have any other saddle points or local maximizers.

I believe if $n=1$ this cannot happen but not sure whether things change in higher dimensions.

  • I think this can be answered using Morse Theory but you'll have to find an intro-level exposition to see the application. https://en.wikipedia.org/wiki/Morse_theory – Gerry Myerson Nov 14 '19 at 00:40
  • how are you defining saddle point here – eyeballfrog Nov 14 '19 at 00:41
  • Maybe http://elib.mi.sanu.ac.rs/files/journals/tm/27/tm1427.pdf is a good reference. The first paragraph on the second page seems to imply an answer. – Gerry Myerson Nov 14 '19 at 11:47
  • So, have you had a look at any Morse Theory reference? – Gerry Myerson Nov 15 '19 at 11:54
  • @GerryMyerson: Thanks for your suggestion. I read some references and it seems beyond my grasp right now. I think to this question, some counterexamples are pointed out. The function I actually encountered is described here https://math.stackexchange.com/questions/3434922/can-a-smooth-function-with-compact-sublevel-sets-only-admit-local-minimizers – MyCindy2012 Nov 15 '19 at 22:06
  • And at that question also you are being encouraged to look at Morse Theory. Maybe there's no easier way to answer your question. It happens sometimes that simple-sounding questions have difficult answers. – Gerry Myerson Nov 15 '19 at 22:11

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You can construct a function $f:\mathbb R^2\to\mathbb R$ with local minima at $(\pm1,0)$ and a saddle point at $(0,0)$, and no other critical points, such that $f(x,y)\to+\infty$ as $x^2+y^2\to\infty$. Now let $U$ equal $\mathbb R^2\setminus A$, where $A$ all points in the plane within distance $\le \epsilon$ of the positive $y$ axis, for some small $\epsilon$. The set $U$ is simply connected. But $f$ has two local minima in $U$, and no other critical points.

I don't really understand what you are asking for, so I don't know if this is a counterexample, or what. I offer it as an invitation for you to clarify your question.

kimchi lover
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  • Thanks for your example. You answered exactly what I asked here. However, my situation is more specific, to not cause confusion, I asked a new question here https://math.stackexchange.com/questions/3434922/can-a-function-which-diverges-to-infinity-at-boundary-only-admit-local-minimizer. If possible, please take a look. – MyCindy2012 Nov 14 '19 at 04:11