Mathematical methods for economic theory

Martin J. Osborne
Previous section Previous page Contents Text Exercises Search Next page Next section

5.3 Conditions under which a stationary point is a global optimum

One variable

Let f be a differentiable concave function of a single variable. Then by a previous result, for every point x, no point on the graph of f lies above the tangent to f at x. Thus if f'(x) = 0 (i.e. if x is a stationary point of f), the point x is a (global) maximizer of f. Similarly, a differentiable convex function lies nowhere below any of its tangents. Combining these observations with a previous result, we have the following result.
Proposition 5.3.1
Let f be a differentiable function of a single variable defined on the interval I.
Proof

First suppose that f is concave. If f'(x) = 0 then by a previous result we have f(x') − f(x) ≤ 0 for all x' ∈ I, or f(x') ≤ f(x) for all x' ∈ I. Thus x is a (global) maximizer of f in I. If x is a (global) maximizer of f in the interior of I then f'(x) = 0 by a previous result.

Symmetric arguments apply if f is convex.

Note that we cannot reach the same conclusions for quasiconcave and quasiconvex functions. For example, the function f defined by f(x) = x3 for all x is quasiconcave (and quasiconvex) and f'(0) = 0, but 0 is not a maximizer or minimizer of f.

Now, a twice-differentiable function is concave if and only if its second derivative is nonpositive (and similarly for a convex function), so we deduce that if f is a twice-differentiable function defined on the interval I and x is in the interior of I then

  • f"(z) ≤ 0 for all z ∈ I ⇒ [x is a global maximizer of f in I if and only if f'(x) = 0]
  • f"(z) ≥ 0 for all z ∈ I ⇒ [x is a global minimizer of f in I if and only if f'(x) = 0].
Example 5.3.1
Consider the problem maxx−x2 subject to x ∈ [−1,1]. The function −x2 is concave; its unique stationary point is 0. Thus its global maximizer is 0, which is in [−1,1] and is thus the solution of the problem.
Example 5.3.2
A firm produces output using a single input. The unit price of the input is w and the unit price of output is p. The firm's output from x units of the input is √x. For what value of x is its profit maximized?

The firm's profit is p√x − wx. The derivative of this function is (1/2)px−1/2 − w, and the second derivative is −(1/4)px−3/2 ≤ 0 for all x, so the function is concave. So the global maximizer of the function is a stationary point. Hence this maximizer solves (1/2)px−1/2 − w = 0, so that x = (p/2w)2.

What happens if the production function is x2?

Many variables

A differentiable concave function of many variables always lies below, or on, its tangent plane, and a differentiable convex function always lies above, or on, its tangent plane. Thus as for functions of a single variable, every stationary point of a concave function of many variables is a maximizer, and every stationary point of a convex function of many variables is a minimizer, as the following result claims.
Proposition 5.3.2
Let f be a differentiable function of n variables defined on the convex set S.
Proof

First suppose that f is concave. If f'i(x) = 0 for i = 1, ..., n then by a previous result we have f(x') − f(x) ≤ 0 for all x' ∈ S, or f(x') ≤ f(x) for all x' ∈ S. Thus x is a (global) maximizer of f in S. If x is a (global) maximizer of f in the interior of S then f'i(x) = 0 for i = 1, ..., n by a previous result.

Symmetric arguments apply if f is convex.

As in the case of functions of a single variable, we can combine this result with a previous result characterizing twice-differentiable concave and convex functions to conclude that if f is a function with continuous partial derivatives of first and second order on a convex set S, and x is in the interior of S, then
  • H(z) is negative semidefinite for all z ∈ S ⇒ [x is a global maximizer of f in S if and only if x is a stationary point of f]
  • H(z) is positive semidefinite for all z ∈ S ⇒ [x is a global minimizer of f in S if and only if x is a stationary point of f],
where H(x) denotes the Hessian of f at x.

Be sure to notice the difference between the form of this result and that of the the result on local optima. To state briefly the results for maximizers together:

Sufficient conditions for local maximizer: if x* is a stationary point of f and the Hessian of f is negative definite at x* then x* is a local maximizer of f
Sufficient conditions for global maximizer: if x* is a stationary point of f and the Hessian of f is negative semidefinite for all values of x then x* is a global maximizer of f.
Example 5.3.3
Consider the function f(x, y) = x2 + xy + 2y2 + 3. We have
f'x(x, y)  =  2x + y
f'y(x, y)  =  x + 4y
so that the function has a single stationary point, (x, y) = (0, 0).

We have

f"xx(x, y) = 2, f"yy(x, y) = 4, and f"xy(x, y) = 1,
so the Hessian of f is
left parenthesis 2 1 right parenthesis
1 4
which is positive definite for all values of (x, y). (The matrix is independent of (x, y).) Hence f is convex (in fact, strictly convex).

Thus the global minimizer of the function is (0, 0); the minimum value of the function is 3.

Example 5.3.4
Consider the function f(x, y) = x4 + 2y2. We have
f'x(x, y)  =  4x3
f'y(x, y)  =  4y,
so that the function has a single stationary point, (x, y) = (0, 0). We have
f"xx(x, y) = 12x2, f"yy(x, y) = 4, and f"xy(x, y) = 0,
so the Hessian of f is
left parenthesis 12x2 0 right parenthesis
0 4
.
This matrix is positive semidefinite for all values of (x, y), so that (0, 0) is the unique global minimizer of the function.
Example 5.3.5
Consider the function f(x, y) = x3 + 2y2. We have
f'x(x, y)  =  3x2
f'y(x, y)  =  4y,
so that the function has a single stationary point, (x, y) = (0, 0). We have
f"xx(x, y) = 6x, f"yy(x, y) = 4, and f"xy(x, y) = 0,
so the Hessian of f is
left parenthesis 6x 0 right parenthesis
0 4
.
This matrix is neither positive semidefinite nor negative semidefinite for all values of (x, y). (For x < 0 it is not positive semidefinite and for x > 0 it is not negative semidefinite.) So f is neither convex nor concave. Thus we cannot conclude that (0, 0) is a global maximizer or that is it a global minimizer.

At (x, y) = (0, 0) the matrix is positive semidefinite, but not positive definite. Thus we cannot tell from this analysis whether (0, 0) is a local maximizer or local minimizer, or neither.

We conclude that if f has a minimizer, then this minimizer is (0, 0), and if f has a maximizer, then this maximizer is (0, 0).

In fact, f does not have either a minimizer or a maximizer: for any given value of y, f(x, y) is an arbitrarily large negative number when x is an arbitrarily large negative number, and for any given value of x, f(x, y) is an arbitrarily large positive number when y is an arbitrarily large positive number. Further, (0, 0) is neither a local maximizer nor a local minimizer: for all ε > 0 we have f(ε, 0) > 0 and f(−ε, 0) < 0.

Example 5.3.6
Consider the function f(x1, x2, x3) = x2
1 + 2x2
2 + 3x2
3 + 2x1x2 + 2x1x3, found to be strictly convex in a previous example. It has a unique stationary point, (x1, x2, x3) = (0, 0, 0). Hence its unique global minimizer is (0, 0, 0), with a value of 0.
Example 5.3.7
Consider a firm with the production function f, defined over vectors (x1, ..., xn) of inputs. Assume that f is concave. The firm's profit function is
π(x1, ..., xn) = pf(x1, ..., xn) − ∑n
j=1wjxj
where p is the price of the output of the firm and wj is the price of the jth input. The second term in the profit function is linear, hence concave, and so the function π is concave (by a previous result). So the input bundle (x*1, ..., x*n) > 0 maximizes the firm's profit if and only if it is a stationary point of π, or
pf'j(x*1, ..., x*n) = wj for j = 1, ..., n
(i.e. the value of the marginal product of each input j is equal to its price).