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Given a function $f: \mathbb{R}^n \rightarrow \mathbb{R}$ which has only one critical point and it's a local minimum, for what $n$ is it a global minimum?

For a convex function with one variable a local minimum is always global.

For functions with two variables, it's not true. There are many counterexamples:

$f(x,y) = e^{3x} + y^3-3ye^x$.

Here the only solution of $f_x=3e^{3x}-3ye^x=0$, and $f_y=3y^2-3e^x=0$ is $(0,1)$ which is a local minimum by the second derivative test.

But $f(0,-3)=-17<f(0,1)=-1$

$f(x,y)=x^2+y^2(1+x)^3$ has the same property.

What about higher dimensions?

Could you help me determine the condition on $n$ for which the only local minimum is global?

Thank you.

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  • $\begingroup$ You need to consider $f_x=0$ or $f_y=0$, not and. $\endgroup$
    – Tim Ratigan
    Commented Dec 7, 2013 at 9:17
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    $\begingroup$ Why? I thought that in critical points both derivatives have value zero. $\endgroup$
    – Roses
    Commented Dec 7, 2013 at 9:20
  • $\begingroup$ That's true, but $f_x$ can start growing while $f_y$ is still negative, for example, and then $f_y$ becomes positive later, thus going from strictly decreasing to strictly increasing without a critical point. By considering or, you find local maxima/minima when you disregard the other variable, which tells you considerably more. $\endgroup$
    – Tim Ratigan
    Commented Dec 7, 2013 at 9:26
  • $\begingroup$ Sorry, I realize my first comment might have been misconstrued, it looks somewhat forceful upon rereading. $\endgroup$
    – Tim Ratigan
    Commented Dec 7, 2013 at 9:27
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    $\begingroup$ This web page has the graph of mentioned function for anyone wanting a graphical example: math.tamu.edu/~tvogel/gallery/node16.html $\endgroup$
    – meguli
    Commented Aug 8, 2017 at 18:04

1 Answer 1

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Using your example, one can show that similar examples exists for all $n\geq 3$: Let $F :\mathbb R^{2+n} \to \mathbb R$, $n\geq 1$ be defined by

$$F(x, y, z_1, \cdots, z_n) = f(x, y) + \epsilon(z_1^2 + \cdots + z_n^2)$$

where $f$ is the first example you gave. Then $dF=0$ only at the point $(0,1,0,\cdots, 0)$ and it is a local minimum by the second derivative test. Also $F(0,-3, 0,\cdots, 0) <F(0,-1, 0\cdots, 0)$ for some small $\epsilon$.

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  • $\begingroup$ Thank you very, very much. I'm afraid I don't have enough reputation to vote up. Maybe it's a stupid question, but how small should $\epsilon$ be to make the construction work? $\endgroup$
    – Roses
    Commented Dec 7, 2013 at 10:33
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    $\begingroup$ Well, I am a bit misleading. Actually any positive $\epsilon$ would do. Because in order that $dF=0$, $z_1, \cdots, z_n$ has to be zero, and it is a local minimum as long as $\epsilon >0$. Also, $F(0, -3, 0, \cdots, 0)< F(0,1,0, \cdots, 0)$ has nothing to do with $\epsilon$. $\endgroup$
    – user99914
    Commented Dec 7, 2013 at 10:42
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    $\begingroup$ Ok, I see. The fact that $\epsilon$ is a constant is enough to show that $z_i$'s must be zero. Thank you a lot. $\endgroup$
    – Roses
    Commented Dec 7, 2013 at 10:50

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