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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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1 Answer

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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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