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

Assume that f:R2R 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(1y)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)=(x21)2+(eyx2)2.
f is non-negative, with global minima at (1,0) and (1,0).

If the gradient
f(x,y)=(4x(x21)4x(eyx2),2ey(eyx2))
is zero then ey=x2 and x(x21)=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)=(u21)2+v2 has two global minima, but also an additional critical point at (0,0), and
  • ϕ(x,y)=(x,eyx2) 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.

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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)=(u21)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)+11+x2), which is a diffeomorphism from R2 with bounded derivatives onto the set {(u,v)v>11+v2}, which contains the points (±1,0) but not the point (0,0).

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

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