Optimization – Is a Unique Local Maximum a Global Maximum for a Function of Two Variables?

multivariable-calculusoptimizationpartial derivativereal-analysis

$f(x,y)$ has partial derivatives in all $\mathbb R^2$ and a unique critical point at $(x_0,y_0)$ (local maximum). Is it a global maximum?

I know that in compact sets, it isn't enough to say that if a point is the only maximum inside the set, then it's a global maximum, because in the frontier of the set it could happen that the function has a maximum greater than the inside one. But for the entire $\mathbb R^2$ there is no frontier, therefore can I admit that this unique point is a point of global maximum?

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Loose description of the geometry: imagine a flat plane and then you put a lone hill with a peak on it. Now tilt the plane a little. Now you still have one peak, but hopefully you can also see that you have introduced a saddle point. Imagine sliding the saddle point location off to infinity to make it effectively no longer there.

Basic description of the construction I saw somewhere once: you can have something whose cross sections in one direction are cubic, but whose cross-sections in the other direction is a sum of exponentials in just the right way for one of the cubic cross-sectional peaks to meet the peak of one of the exponential sum cross-sections. This can happen with no other critical points, and of course the cubic component alone is unbounded in either direction (so there is no global max or min). The direction that the exponentials taper off in aligns with the idea of the saddle point being slid away to infinity.

Consider $f(x,y)=x^3+axe^{my}+be^{ny}$ for some constants $a,b,m,n$. We have: $$\begin{align} \frac{\partial}{\partial x}f(x,y)&=3x^2+ae^{my}&\frac{\partial}{\partial y}f(x,y)&=amxe^{my}+bne^{ny} \end{align}$$

To find a critical point, the $\frac{\partial}{\partial x}$ shows us $a$ must be negative. We are just looking for one example, so we pick some numbers for the constants that make the algebra that follows simpler. With $a=-3$, $b=1$, $m=1$, $n=3$:

$$\begin{align} \frac{\partial}{\partial x}f(x,y)&=3x^2-3e^{y}&\frac{\partial}{\partial y}f(x,y)&=-3xe^{y}+3e^{3y}\\ &=3(x^2-e^y)&&=3e^y(e^{2y}-x) \end{align}$$

So at a critical point, $x=e^{2y}$ according to the latter relation.

Then using the $\frac{\partial}{\partial x}$ relation: $$e^{4y}-e^y=0\implies e^{3y}=1\implies y=0$$ which in turn says that $x=1$. So the only critical point is at $(1,0)$. Is it a local min, max, or neither?

$$\begin{align} f_{xx}(1,0)&=6&f_{xy}(1,0)&=-3&f_{yy}=6 \end{align}$$

So the Hessian determinant is $6^2-(-3)^2=27>0$, so this critical point is either a local max or a local min. OK, it turns out to be a local min, since for example $f_{xx}(1,0)$ is positive. But $f$ has no global min, since you could fix $y$ and let $x\to-\infty$ and get unbounded large negative output.

Take this $f$ and negate it to get the direct counterexample. Here is an image of $-f$ from GeoGebra. Note how there would be a saddle point, except it has been slid off the map (to infinity) to the left.

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[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 $\epsilon$.

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Related Solutions

[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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