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

differential-topologymorse-theorymultivariable-calculusreal-analysis

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).

Best Answer

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 $(0, 0)$.

Related Solutions

[Math] Unique critical point does not imply global maximum/global minimum

Your intuitive picture is probably as follows: Imaging filling water into the hollow at the origin. As the water rises, since there are points lower than the origin, you expect the water to start spilling over into the lower lying areas at some point. And that point would need to be a critical point! The way to resolve the paradox is to note that before any spilling over occurs, the lake will in fact extend off to infinity. Look at a cross section of your function graph for fixed $y$, as $y$ starts at $0$ and then grows. In other words, study the function $x\mapsto x^2+y^2(1-x)^3$ as $y^2$ grows. You will see a local minimum, starting at the origin and approaching $x=1$ from below, and a local maximum, starting at infinity and approaching $x=1$ from above. As $y^2$ grows, the value at the local minimum grows, while the value at the local maximum decreases. So the aforementioned lake will extend out to $(1,\pm\infty)$ and then start to spill over the edge (the local maximum) at infinity before it spills over anywhere else.

A picture is tremendously helpful. If you have access to maple, try this:

plots[animate](plot,
  [x^2+y^2*(1-x)^3, x = 0 .. 2, view = [0 .. 2, -1 .. 2]], y = 0 .. 10);

Or you can take a look at this static plot of $f(x,y)$. It is easy to see how $(0,0)$ is a local, but not a global, maximum:

x^2+y^2(1-x)^3

[Math] Only one critical point, local minimum but not global

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 $\epsilon$.

Best Answer

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[Math] Unique critical point does not imply global maximum/global minimum

[Math] Only one critical point, local minimum but not global

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