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

Assume that $$f:R2→Rf:\mathbb{R}^2\to\mathbb{R}$$ a $$C∞C^{\infty}$$ function that has exactly two minimum global points. Is it true that $$ff$$ has always another critical point?

A standard visualization trick is to imagine a terrain of height $$f(x,y)f(x,y)$$ at the point $$(x,y)(x,y)$$, 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 $$KK$$. As a continuous function, $$ff$$ attains a maximum value $$MM$$ on the set $$KK$$. This means that when the water level has reached $$MM$$, the two lakes will have been merged.
• The set $$SS$$ of water levels $$zz$$ such that two lakes are connected is thus non-empty and bounded from below. Therefore it has an infimum $$mm$$.
• It is natural to think that at water height $$mm$$ there should be a critical point. A saddle point is easy to visualize. For example the function (originally suggested in a deleted answer) $$f(x,y)=x2+y2(1−y)2f(x,y)=x^2+y^2(1-y)^2$$ has a saddle point at the midway point between the two local minima at $$(0,0)(0,0)$$ and $$(0,1)(0,1)$$. But, can we prove that one always exists?

Follow-ups:

• Does the answer change, if we replace $$R2\Bbb{R}^2$$ with a compact domain? What if $$ff$$ is a $$C∞C^\infty$$ function on a torus ($$S1×S1S^1\times S^1$$) or the surface of a sphere ($$S2S^2$$). 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).

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
$$f(x,y)=(x2−1)2+(ey−x2)2. f(x, y) = (x^2-1)^2 + (e^y - x^2)^2 \, .$$
$$ff$$ is non-negative, with global minima at $$(1,0)(1, 0)$$ and $$(−1,0)(-1, 0)$$.

$$∇f(x,y)=(4x(x2−1)−4x(ey−x2),2ey(ey−x2)) \nabla f(x, y) = \bigl( 4x(x^2-1) - 4x(e^y - x^2) \, , \, 2e^y(e^y-x^2) \bigr)$$
is zero then $$ey=x2e^y =x^2$$ and $$x(x2−1)=0x(x^2-1) = 0$$. $$x=0x= 0$$ is not possible, so that the gradient is zero only if $$x=±1x=\pm1$$ and $$y=0y=0$$, 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 $$f(x,y)=g(ϕ(x,y))f(x, y) = g(\phi(x, y))$$ where:

• $$g(u,v)=(u2−1)2+v2g(u, v) = (u^2-1)^2 + v^2$$ has two global minima, but also an additional critical point at $$(0,0)(0, 0)$$, and
• $$ϕ(x,y)=(x,ey−x2) \phi(x, y) = ( x , e^y-x^2)$$ is a diffeomorphism from the plane onto the set $${(u,v)∣v>−u2}\{ (u, v) \mid v > -u^2 \}$$. The image is chosen such that it contains the minima of the function $$gg$$, but not its critical point.

With respect to the “connected lakes” approach: The level sets
$$L(z)={(x,y)∣f(x,y)≤z} L(z) = \{ (x, y) \mid f(x, y) \le z \}$$
connect the minima $$(−1,0)(-1, 0)$$ and $$(1,0)(1, 0)$$ exactly if $$z>1z > 1$$. The infimum of such levels is therefore $$m=1m=1$$, but $$L(1)L(1)$$ 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 $$f(x,y)=g(ϕ(x,y))f(x, y) = g(\phi(x, y))$$ with

• $$g(u,v)=(u2−1)21+u4+v21+v2g(u, v) = \frac{(u^2-1)^2}{1+u^4} + \frac{v^2}{1+v^2}$$, which has two global minima at $$(±1,0)(\pm 1, 0)$$, one critical point at $$(0,0)(0, 0)$$, and bounded derivatives.
• $$ϕ(x,y)=(x,log(1+ey)+1−√1+x2)\phi(x, y) = (x, \log(1+e^y) +1 -\sqrt{1+x^2} )$$, which is a diffeomorphism from $$R2\Bbb R^2$$ with bounded derivatives onto the set $${(u,v)∣v>1−√1+v2}\{ (u, v) \mid v > 1- \sqrt{1+v^2} \}$$, which contains the points $$(±1,0)(\pm 1, 0)$$ but not the point $$(0,0)(0, 0)$$.
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