TITLE: If a two variable smooth function has two global minima, will it necessarily have a third critical point?
QUESTION [74 upvotes]: 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).

REPLY [4 votes]: We want to explain the intuition in the OP : at the level where the lakes join, there exists a critical point.
Let $M$ be a topological space that is normal, locally path connected and path-connected (like a connected manifold).
Let $f$ be a continuous proper map $f\colon M \to \mathbb{R}$. Moreover, let $b$ a point of local minimum of $f$, and $a\ne b$ such that $f(a) \le f(b)$. Then there exists $c\ne a,b$ that is critial point of $f$.
Definition 1.: $c\in M$ is a critical point of $f$ if $c$ is a local minimum of $f$, or for every $D$ neighborhood of $c$ the set $D \cap M_{< f(c)}$ is not connected (think of a saddle point).
Definition 2.  The points $a$, $b$ of a topological space $X$ are separated in $X$ if there exists a partition of $X$ into open (close) subsets $U$, $V$, with $a \in U$, $b\in V$.
On to the proof. First, some easy things:  consider $U$ an open subset $b$ not containing $a$, such that $f(b) \le f(x)$ for all $x \in U$. Let $b \in V$ open such that $\overline V \subset U$. If for some point $y \in \partial V$ we have $f(y) = f(b)$, then $y$ is again a local minimum. Let's not allow this possibility. Therefore, we have $f(x) > f(b)$, for all $x \in \partial V$. Note that $a \in M \backslash \overline V$, and $b \in V$. Therefore, we get $a$, $b$ separated in $M_{\le f(b)}$.
Now let us show : if $a$, $b$ are separated in $M_{\le t}$, then there exists $\delta> 0$ such that $a$, $b$ separated in $M_{\le t+\delta}$. Indeed, let $C\ni a$, $D\ni b$ closed, disjoint, with union $M_{\le t}$. Consider $U$, $V$ disjoint open subsets, $C\subset U$, $D\subset V$. Note that $f^{-1}(t) \subset U \cup V$. Since $f$ is proper, there exists $\delta> 0$ such that $f^{-1}([t, t+\delta]\subset U\cup V$. We conclude that $M_{\le t+\delta} \subset U\cup V$. We get a separation of $a$, $b$ in $M_{\le t+\delta}$.
Since $M$ is path-connected, there exists $\gamma$ a path from $a$ to $b$. Let $s= \sup_{\gamma} f$. We conclude that $a$, $b$ are in the same component of $M_{\le s}$, so they are not separated in $M_{\le s}$.
Consider the set
$$\{ t \in [f(b), \infty) \ | \ a, b \textrm{ separated in } M_{\le t} \}$$
Since the $s$ above in not in the set, we conclude that the set is bounded, so it has a supremum $t^*$. Now, $t^*$ is not in the set, by the above.
Now, we have $a$, $b$ in the open subset $M_{< t^*}$. Can $a$, $b$ in the same component? If they were, we would have ( locally path-connectedness!) a path $\eta$ from $a$ to $b$ in $M_{< t^*}$. But that would mean that $a$, $b$ are not separated in some $M_{\le t^*-\epsilon}$, contradiction.
Therefore, we have $U$, $V$ disjoint open subsets, $a\in U$, $b\in V$, $U\cup V = M_{< t^*}$.
Now, consider $\overline M_{< t^*} \subset M_{\le t^*}$. If the inclusion were strict, we would have $c \in M_{= t^*}$ that is a local minimum, done.
Let's assume now that $\overline M_{< t^*} = M_{\le t^*}$, and so $\overline U \cup \overline V = M_{\le t^*}$.  The sets $\overline U$, $\overline V$ cannot be disjoint, since then we woud have a separation of $a$, $b$ in $M_{\le t^*}$. Let then $c \in \overline U \cap \overline V$. We have $f(c) = t^{*}$, and for every $D$ neighborhood of $c$, we have $D\cap M_{< f(c)} = (D\cap U) \cup (D\cap V)$. Hence, $D\cap M_{< f(c)}$ is not connected. We conclude that $c$ is a critical point.
$\bf{Note:}$ In the example of @Martin R: $f(x,y) = (x^2-1)^2 + (e^y-x^2)^2$, the level set $f^{-1}(t)$ looks like OO for $t \in (0, \frac{1}{2})$, like $\cap \cap$ for $t\in [\frac{1}{2},1]$ and like M for $t > 1$. Worth understanding.