If a two variable smooth function has two global minima, will it necessarily have a third critical point?

Question 1

Assume that $f:\mathbb{R}^2\to\mathbb{R}$ a $C^{\infty}$ function that has exactly two minimum global points. Is it true that $f$ has always another critical point?

A standard visualization trick is to imagine a terrain of height $f(x,y)$ at the point $(x,y)$ , and then imagine an endless rain pouring with water level rising steadily on the entire plane.

Because there are only two global minima, they must both be isolated local minima also. Therefore, initially the water will collect into two small lakes around the minima.
Those two points are connected by a compact line segment $K$ . As a continuous function, $f$ attains a maximum value $M$ on the set $K$ . This means that when the water level has reached $M$ , the two lakes will have been merged.
The set $S$ of water levels $z$ such that two lakes are connected is thus non-empty and bounded from below. Therefore it has an infimum $m$ .
It is natural to think that at water height $m$ there should be a critical point. A saddle point is easy to visualize. For example the function (originally suggested in a deleted answer) $f(x,y)=x^2+y^2(1-y)^2$ has a saddle point at the midway point between the two local minima at $(0,0)$ and $(0,1)$ . But, can we prove that one always exists?

Follow-ups:

Does the answer change, if we replace $\Bbb{R}^2$ with a compact domain? What if $f$ is a $C^\infty$ function on a torus ( $S^1\times S^1$ ) or the surface of a sphere ( $S^2$ ). Ok, on a compact domain the function will have a maximum, but if we assume only isolated critical points, what else is implied by the presence of two global minima?
Similarly, what if we have local minima instead of global?
If it makes a difference you are also welcome to introduce an extra condition (like when the domain is not compact you could still assume the derivatives to be bounded – not sure that would be at all relevant, but who knows).

Question 2

With respect to the first part of your question: No, a function with two global minima does not necessarily have an additional critical point. A counterexample is
$f(x, y) = (x^2-1)^2 + (e^y - x^2)^2 \, .$
$f$ is non-negative, with global minima at $(1, 0)$ and $(-1, 0)$ .

If the gradient
$\nabla f(x, y) = \bigl( 4x(x^2-1) - 4x(e^y - x^2) \, , \, 2e^y(e^y-x^2) \bigr)$
is zero then $e^y =x^2$ and $x(x^2-1) = 0$ . $x= 0$ is not possible, so that the gradient is zero only if $x=\pm1$ and $y=0$ , that is only at the global minima.

The construction is inspired by Does $f$ have a critical point if $f(x, y) \to +\infty$ on all horizontal lines and $f(x, y) \to -\infty$ on all vertical lines?. We have $f(x, y) = g(\phi(x, y))$ where:

$g(u, v) = (u^2-1)^2 + v^2$ has two global minima, but also an additional critical point at $(0, 0)$ , and
$\phi(x, y) = ( x , e^y-x^2)$ is a diffeomorphism from the plane onto the set $\{ (u, v) \mid v > -u^2 \}$ . The image is chosen such that it contains the minima of the function $g$ , but not its critical point.

With respect to the “connected lakes” approach: The level sets
$L(z) = \{ (x, y) \mid f(x, y) \le z \}$
connect the minima $(-1, 0)$ and $(1, 0)$ exactly if $z > 1$ . The infimum of such levels is therefore $m=1$ , but $L(1)$ does not connect the minima (it does not contain the y-axis). Therefore this approach does not lead to a candidate for a critical point.

The above approach can also be used to construct a counterexample with bounded derivatives. Set $f(x, y) = g(\phi(x, y))$ with

$g(u, v) = \frac{(u^2-1)^2}{1+u^4} + \frac{v^2}{1+v^2}$ , which has two global minima at $(\pm 1, 0)$ , one critical point at $(0, 0)$ , and bounded derivatives.
$\phi(x, y) = (x, \log(1+e^y) +1 -\sqrt{1+x^2} )$ , which is a diffeomorphism from $\Bbb R^2$ with bounded derivatives onto the set $\{ (u, v) \mid v > 1- \sqrt{1+v^2} \}$ , which contains the points $(\pm 1, 0)$ but not the point $(0, 0)$ .

If a two variable smooth function has two global minima, will it necessarily have a third critical point?

Answer

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