Mathematical methods for economic theory

Martin J. Osborne

4.3 Existence of an optimum

Let f be a function of n variables defined on the set S. The problems we consider take the form
maxxf(x) subject to x ∈ S
where x = (x1, ..., xn).

Before we start to think about how to find the solution to a problem, we need to think about whether the problem has a solution. Here are some specifications of f and S for which the problem does not have any solution.

  • f(x) = x, S = [0, ∞) (i.e. S is the set of all nonnegative real numbers). In this case, f increases without bound, and never attains a maximum.
  • f(x) = 1 − 1/x, S = [1, ∞). In this case, f converges to the value 1, but never attains this value.
  • f(x) = x, S = (0, 1). In this case, the points 0 and 1 are excluded from S (which is an open interval). As x approaches 1, the value of the function approaches 1, but this value is never attained for values of x in S, because S excludes x = 1.
  • f(x) = x if x < 1/2 and f(x) = x − 1 if x ≥ 1/2; S = [0, 1]. In this case, as x approaches 1/2 the value of the function approaches 1/2, but this value is never attained, because at x = 1/2 the function jumps down to −1/2.
The difficulties in the first two cases are that the set S is unbounded; the difficulty in the third case is that the interval S is open (does not contain its endpoints); and the difficulty in the last case is that the function f is not continuous. If S is a closed interval [ab] (where a and b are finite) and f is continuous, then none of the difficulties arise.

For functions of many variables, we need to define the concept of a bounded set.

The set S is bounded if there exists a number k such that the distance of every point in S from the origin is at most k.
A bounded set does not extend “infinitely” in any direction.
The set [−1, 100] is bounded, because the distance of any point in the set from 0 is at most 100. The set [0, ∞) is not bounded, because for any number k, the number 2k is in the set, and the distance of 2k to 0 is 2k which exceeds k.
The set {(xy): x2 + y2 ≤ 4} is bounded, because the distance of any point in the set from (0, 0) is at most 2.
The set {(xy): xy ≤ 1} is not bounded, because for any number k the point (2k, 0) is in the set, and the distance of this point from (0, 0) is 2k, which exceeds k.
We say that a set that is closed and bounded is “compact”.
A set is compact if it is closed and bounded.
The following result generalizes the observations in the examples at the top of this page.
Proposition (Extreme value theorem)  
Let f be a function of many variables defined on a set X and let S be a subset of X. If f is continuous and S is compact then the problems of maximizing and minimizing f(x) subject to x ∈ S both have solutions.
For proofs of a slightly more general result, see Rudin (1976), Theorem 4.16 (p. 89) and Apostol (1974), Theorem 4.28 (p. 83).
Note that the requirement of boundedness is on the set, not the function. Requiring that the function be bounded is not enough, as the second example on this page shows.

Note also that the result gives only a sufficient condition for a function to have a maximum. If a function is continuous and is defined on a compact set then it definitely has a maximum and a minimum. The result does not rule out the possibility that a function has a maximum and/or minimum if it is not continuous or is not defined on a compact set. (Refer to the page on logic if you are unclear on this point.)