# Can a multivariate function only have local minimum? [duplicate]

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. en.wikipedia.org/wiki/Morse_theory – Gerry Myerson Nov 14 '19 at 0:40
• how are you defining saddle point here – eyeballfrog Nov 14 '19 at 0:41
• Maybe 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 math.stackexchange.com/questions/3434922/… – MyCindy2012 Nov 15 '19 at 22:06

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.