Assume that f:R2→R a C∞ 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)=x2+y2(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 R2 with a compact domain? What if f is a C∞ function on a torus (S1×S1) or the surface of a sphere (S2). 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).

**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)=(x2−1)2+(ey−x2)2.

f is non-negative, with global minima at (1,0) and (−1,0).

If the gradient

∇f(x,y)=(4x(x2−1)−4x(ey−x2),2ey(ey−x2))

is zero then ey=x2 and x(x2−1)=0. x=0 is not possible, so that the gradient is zero only if x=±1 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)→+∞ on all horizontal lines and f(x,y)→−∞ on all vertical lines?. We have f(x,y)=g(ϕ(x,y)) where:

- g(u,v)=(u2−1)2+v2 has two global minima, but also an additional critical point at (0,0), and
- ϕ(x,y)=(x,ey−x2) is a diffeomorphism from the plane onto the set {(u,v)∣v>−u2}. 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)∣f(x,y)≤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(ϕ(x,y)) with

- g(u,v)=(u2−1)21+u4+v21+v2, which has two global minima at (±1,0), one critical point at (0,0), and bounded derivatives.
- ϕ(x,y)=(x,log(1+ey)+1−√1+x2), which is a diffeomorphism from R2 with bounded derivatives onto the set {(u,v)∣v>1−√1+v2}, which contains the points (±1,0) but not the point (0,0).

**Attribution***Source : Link , Question Author : Jyrki Lahtonen , Answer Author : Jyrki Lahtonen*