General method of finding global maxima / minima of function of 2 variables on open sets

Question

I am interested in a method that allows to find global maxima / minima of any elementary function (continuous on their domain) of 2 variables on open sets. If there is no global maxima / minima, then find any of upper or lower bounds of the range of the function. If they don't exist also, prove that.
I am not interested in specific solutions that can be conveniently applicable to a specific example below. I'm interested in formally described method (if the method does exist), applicable to all of them. I'll be glad for any links to scientific literature. Will be also glad if the method can be generalized to functions of more that 2 variables.

Examples just for better understanding. Find global maxima and minima, or any of lower or upper bounds of

1. f(x;y) = xy on R²,
2. f(x;y) = 1/(xy) on the domain,
3. f(x;y) = x² + 2x + 3y² where x² + y² < 1
4. f(x;y) = log₂(x) + logₓ(y) + log_y(8) where x,y > 1 (global minima here is ∛81)
   
   or prove that they don't exist.

Answer 1

The steps to investigating global extrema of smooth functions on open sets are:

• Find all the local extrema.
• Examine the behavior of the function outside compact sets - can you remove a compact set such that the function is bounded (above or below)? What is the infimum/supremum? If the function is unbounded in the corresponding direction, the infimum is $-\infty$, the supremum is $+\infty$.
• If the function is unbounded in a direction, then it has no global extremum in that direction. Otherwise,
• If any of the local minimums is less than the infimum, then the lowest is the global minimum. Otherwise the infimum is global (but not a minimum). If any of the local maximums is greater than the supremum, then the highest is the global maximum, otherwise the supremum is global.

You can interpret the "behavior outside compact sets" to mean the limits along curves leaving the domain at some point, or going off to infinity in some fashion.

In your examples,

1. Clearly, $\lim_{x \to \pm \infty} f(x, 1) = \pm\infty$, so this function is unbounded in both directions.
2. Equally clearly, $\lim_{x\to 0\pm} f(x,1) = \pm\infty$, also unbounded.
3. This function extends continuously to the closure of the given domain. The behavior of $f$ outside compact sets is exactly the behavior on that boundary. So you handle this by solving the problem on the closed disk. The only difference is if the extremum occurs only on the boundary it is an infimum/supremum, not a maximum, minimum.
4. Each term is bounded below by $0$, it is not unbounded in that direction. In the other direction, if $x \to \infty$, so does the first term, and as they are bounded below, the other two terms cannot compensate, so the entire function goes to $\infty$ regardless of what $y$ does. If $x$ does not go to $\infty$, then the second term goes to $\infty$ with $y$, and if $y \to 1$, the third term goes to $\infty$ regardless of what $x$ does. Finally if $y$ does not go to $1$, the second term goes to $\infty$ as $x \to 1$. So on all the "boundaries", the function goes to $\infty$. The lowest local minimum will be the global minimum.