Math 438 ACTIVITY 22:
WHY:
Optima of Convex Functions
This activity is designed to review the ways to determine where a function is convex or concave and also to find the
absolute maxima and minima of these functions. Convex and concave functions are especially nice for optimization
purposes since they have a unique relative min and max (respectively) and hence this relative min or max is also the
absolute min or max. Although functions encountered in practice may not be convex or concave, they can
sometimes be closely approximated by convex or concave functions over regions of interest.
VOCABULARY:
Recall the following terms for a function f defined on a convex subset of Rn,
Convex Function
f is convex if f(aX + (1-a)Y) ≤ af(X) + (1-a)f(Y) for all 0 ≤ a ≤ 1
Strictly Convex Function f is strictly convex if f(aX + (1-a)Y) < af(X) + (1-a)f(Y) for all 0 ≤ a ≤ 1
Concave Function
f is concave if f(aX + (1-a)Y) ≥ af(X) + (1-a)f(Y) for all 0 ≤ a ≤ 1
Strictly Concave Function
f is strictly concave if f(aX + (1-a)Y) > af(X) + (1-a)f(Y) for all 0 ≤ a ≤ 1
Affine Function
f is affine if f(aX + (1-a)Y) = af(X) + (1-a)f(Y) for all 0 ≤ a ≤ 1
Epigraph
The epigraph of a function f, denoted epi(f), is the set {(X, y) | y ≥ f(X)}.
Hypograph
The hypograph of a function f, denoted hyp(f), is the set {(X, y) | y ≤ f(X)}.
Necessary and Sufficient Conditions for Convexity/Concavity
Functions of One Variable
f(x)>0
 strictly convex 
f(x)≥0
convex
For an interval D in R, a function f : D  R is strictly concave on D iff 
 for all x  D 
f(x)<0
 concave 
f(x)≤0
Functions of Several Variables
 convex 
 positive semidefinite 
For a convex set K in Rn , a function f:K  R is concave on K iff the Hessian is negative semidefinite at every




point of K.
LEARNING OBJECTIVES:
1.
Understand how to find stationary points for a function of several variables and determine which of its stationary points are
max and min points.
2.
Discover the role of the Hessian in showing that a function of several variables is convex.
3.
Understand how group roles enhance group learning
CRITERIA:
1.
Quality of the answers to the Critical Thinking Questions.
2.
Diligence with which each member of the team performed her role.
RESOURCES:
1.
Section 4.7: Strategic Mathematics
2.
30 minutes
PLAN
1.
Work through the model
2.
Answer the Critical Thinking Questions
3
Complete the exercise
MODEL:
We will see if there is a minimum value for the function f(x, y) = x2 + 6xy – 2y3 + 24y defined on the whole plane (and find the
minimum, if it exists):
First, we find the gradient vector and the Hessian matrix.
f(x, y) = (2x + 6y, 6x - 6y2 + 24)
2

6
H = 
.
6 12 y 18
Solving the gradient for point(s) which make it zero we get:
2x + 6y = 0 or x = -3y

6(-3y) – 6y2 + 24 = 0 so y2 + 3y – 4 = 0
y2 + 3y – 4 = (y + 4)(y - 1) = 0 so y = –4 or y = 1
When y = –4, x = 12, and when y = 1, x = –3. – so the stationary points are (12, -4) and (–3, 1)
It is easy to check that H is positive definite at (12, -4), and thus the function has a local minimum there. At (-3, 1) the
2

0
Hessian is indefinite, so this gives a saddle point. In fact, the diagonal matrix D we get when reducing H is 
.
0 12 y  36
As long as –12y – 36 > 0 , that is, wherever y < -3 , the Hessian is positive definite, so the function is convex. For the half
plane y < -3 the absolute minimum is at (12, -4) with value f(12, -4) = -112. We cannot give a nice description of the
behavior for y -3 ; we can check the function value at some points in the halfplane: At (0,5) , f(x, y) = -250 + 120 = -130
 the function has no
< -112 (so the value at (12,-4) is not an absolute min over the whole x,y-plane). Over the entire plane,
absolute minimum, but decreases to -∞.
CRITICAL THINKING QUESTIONS:
1.
How can we be sure that H is positive definite at (12, -4) [in the model]? Show how this can be verified.
2.
How do you know that the function in the model has no global minimum? What is the significance of finding a point whose
value is less than the value at the local minimum?
3.
If a function of several variables is not convex everywhere, how can you determine where it is convex? (Hint, look at the
model above).
4.
If the Hessian matrix is indefinite at a stationary point, does this mean that the point is a saddle point? Why or why not?
5.
Is the product of two convex functions always convex? (Hint: look at the function f(x,y) = xy which is the product of two
convex functions. What is its Hessian? Is the Hessian positive semidefinite?) Can you find a condition which ensures that
the product of convex functions satisfying this condition is convex?
Exercise
Let f(x1,x2,x3) = 4x12 + x22 + x32 - 2x1x2 - 2x1x3 - 2x1 + 8 with domain {X | X ≥ 0})
Find the gradient vector for this function, its Hessian matrix, its stationary point(s). Determine where the function is convex.
Classify the stationary point(s) as relative min or relative max or saddle point(s). Does this function have an absolute
minimum value (if so, where, and what is the value?)?