Chapter_4

Chapter 4 Quasiconcave and Quasiconvex Functions

1. Chapter Objectives: In this chapter we shall consider the new class of functions known as Quasiconcave and Quasiconvex functions and their relations with concave and convex functions. We shall characterize these functions and study the properties of their derivatives. We shall state prove the Kuhn Tucker theorem for this class of functions in optimization.

2. Quasiconcave and Quasiconvex Functions

Quasiconcave Function: Let U be a convex subset of Rn. A real valued function defined on U is quasiconcave if for each a in R, the set

U f (a )={x∈U∨f (x )≥a} is a convex set.

Quasiconvex Function: Let U be a convex subset of Rn. A real valued function defined on U is quasiconvex if for each a in R, the set

U f (a )={x∈U∨f (x )≤a} is a convex set.

Theorem 1: A function f :U→R is Quasiconcave on D if and only if for all x , y∈U and for all λ ϵ(0,1), f (λx+(1−λ ) y )≥min ( f ( x ) , f ( y ))

A function f :U→R is Quasiconvex on D if and only if for all x , y∈U and for all λ ϵ(0,1), f (λx+(1−λ ) y )≤max ( f (x ) , f ( y ))

Proof: Suppose f is quasiconcave.

⇒U f (a ) is aconvex set for eacha∈R

Let x , y∈U∧λ∈ (0,1 )∧f ( x )≥ f ( y) Letting f(y) = a, we observe that , x, y ∈U f (a)

U f (a ) is a convex set ⇒ λx+(1− λ ) y ϵU f (a)

⇒ f (λx+ (1−λ ) y )≥a=f ( y )=min {f ( x ) , f ( y ) }

Conversely, f ( λx+ (1−λ ) y )≥min {f ( x ) , f ( y ) } for all x , y in U λ ϵ(0,1), a∈R

If U f (a ) is empty or contains only one point then it is clearly convex. Let us assume that it contains at least two points x and y. Then, f ( x )≥a∧f ( y )≥a .

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⇒min {f ( x ) , f ( y )} ≥aBy hypothesis f ( λx+ (1−λ ) y )≥min {f ( x ) , f ( y ) }≥a

⇒ λx+(1− λ ) y∈U f (a ) .HenceU f (a ) is a convex set. Thus, f is quasiconcave.Similar argument proves the theorem for quasiconvex function.Note: A function f is said to be strictly quasiconcave or quasiconvex if strict inequality is used in the definition.

Theorem 2: Let f :U⊂Rn⟶ R is concave on U, it is also quasi concave on U. If f is convex on U, it is quasiconvex on U.

Proof: Suppose f is concave.

⇒ f (λx+(1− λ ) y)≥ λf (x)+(1−λ ) f ( y)

⇒ f (λx+(1− λ ) y)≥ λmin {f (x ) , f ( y ) }+ (1−λ )min {f ( x ) , f ( y )} =min {f ( x ) , f ( y )}

∴ f is quasiconcave.

Note: The converse of the result is not true.

For, let f : R ⟶ R be a non decreasing function. Then f is both quasiconcave and quasiconvex function. To see this, let x > y ⇒ x>¿ λx+(1−λ ) y> y, λ ϵ(0,1)

⇒ f (x )≥ f ( λ x+ (1−λ ) y )≥ f ( y )

⇒max {f ( x ) , f ( y ) }≥ f ( λ x+ (1−λ ) y )≥min {f ( x ) , f ( y )}

Therefore, f is both quasiconcave and quasiconvex.

Consider f = x3, f is a nondecreasing function, hence f is both quasiconcave and quasi convex. However, f is not a concave function.

Theorem 3: f :U⊂Rn⟶ R is quasiconcave on U and ϕ :R→R is a monotone non decreasing function, then the composite function ϕοf is a quasiconcave function from U to R. In particular, any monotone transformation of a concave function results in quasiconcave function.

Proof: Let x, y in U and λ ϵ(0,1)

Since f is quasiconcave , we have f (λ x+(1−λ ) y )≥min {f ( x ) , f ( y )}



ϕ is a nondecreasing function ⇒ ϕ ( f (λ x+(1−λ ) y )) ≥ϕ (min { f (x ) , f ( y ) })

= min (ϕ ( f ( x )) , ϕ (f ( y ))¿ ,

Therefore, ϕοf Quasiconcave function.

Note:

1. Quasi-concave and Quasi- convex functions are not necessarily continuous in the interior of the domains

2. Quasi-concave functions can have local maxima that are not global maxima, quasi-convex functions that have local minima that are not global minima.

3. First order conditions are not sufficient to identify even local optima under quasiconvexity.

For, let us consider the function,

f ( x )={x3 x ϵ [0,1]1 , xϵ [1,2]x3 , x>2

Clearly, f is nondecreasing function. It is both quasiconcave and quasiconvex on R.

The function is discontinuous at x = 2.

f is constant on (1, 2). Every point is a local maximum as well as local minimum.

No point in (0, 1) is either a global maximum or global minimum.

f’(0) = 0 but 0 is neither a local maximum nor a local minimum.

Theorem 4: If the production function f(K,L) is concave and f(0,0) = 0, then it has decreasing or constant return to scale. However, quasiconcave Cobb-Douglas production function can have increasing return to scale.

Proof: Suppose that s > 1.

⇒ 0 < 1/s < 1. The function f is concave implies that

f (K , L )=f ( 1s ( sK , sL )+(1−1s ) f (0,0)) ≥

1sf (sK , sL )+(1−1s ) f (0,0)



¿ 1sf (sK , sL)⇒ f (sK , sL)≤sf (K , L)

We know that if α+β>1, the Cobb-Douglas production function has an increasing return to scale. The production function is not concave but it is quasiconcave as it can be written as an monotonic transformation of a convex function. For,Kα Lβ=(K α

α+ β Lβ

α+β )α+β

Kα Lβ=(r )α+β where r=Kα

α+ β Lβ

α+ β

Kα Lβ=q (r )=(r )α+ β

When α+β>0, q(r) is a strictly increasing function.

Therefore, Kα Lβ quasiconcave function.

Note: Recall the assumption that the indifference curve is concave. Then this implies that the indifference curve is quasiconcave..

3. Derivatives of Quasiconcave and Quasiconvex functions

Theorem 5: Let f :U⊂Rn⟶ R be a C1 function and where U is convex and open. Then f is Quasiconcave on D if and only if for any x , y in U f ( y )≥ f (x)⇒Df (x)( y−x)≥0

Proof: Suppose that f is quasiconcave in U and x , y in U such that f ( y )≥ f (x).

Let t ϵ (0,1). Since f is quasiconcave , we have

f (x+t ( y−x ) )=f ( (1−t ) x+ty )≥min { f (x ) , f ( y ) }=f ( x )

⇒ f ¿¿ .

⇒ f ¿¿ Letting t →0, we get Df(x)(y-x) ≥0

Conversely, for x, y in U, f ( y )≥ f (x) we have Df (x)( y−x)≥0

Let min { f (x ) , f ( y ) }=f ( x ). We shall show that f ( (1−t ) x+ty )≥min { f ( x ) , f ( y ) }=f ( x ) , so that f becomes quasiconcave.

Let g(t) = f (x+t ( y−x ) ) ⇒ g (0 )=f ( x )≤ f ( y )=g(1)



g' (t )=Df ( (1−t ) x+ty )( y−x )≥0

This implies that g is a non decreasing function.

Therefore, g(o) = f(x) ≤ g(t) = f (x+t ( y−x ) ). Hence f is Quasiconcave.

Similar result for quasiconvex function can be established.

Second Derivative Test for QuasiConcavity and Quasiconvexity

Let f :U⊂Rn⟶ R be a C2 function and where U is convex and open. Then, if (−1 )k|C k ( x )|>0 for all k = 1,2,…n then f is quasiconcave on D. The matrix C k ( x ) is

[ 0∂ f∂ x1

⋯ ∂ f∂ xk

∂ f∂x1

∂2 f∂ x1

2

∂2 f∂ x1∂ xk

❑

∂ f∂xk

∂2 f∂ x k∂ x1

❑ ⋯ ∂2 f∂ xk

2]

4. Kuhn-Tucker Theorem with quasiconcavity and quasiconvexity

Suppose that the function g defined on U is quasiconcave and differentiable and that the derivatives of g at x* are not all zero. Suppose that the functions h1, h2,….,hm defined on U are quasiconvex and differentiable. Then the Kuhn Tucker conditions are sufficient for X* to solve the problem of

maximizing g(x) on U subject to the constraints h j (x )≤k j , j=1,2…m

The K-T conditions are:

(1) The first order conditionDg ¿(2) The Lagrange multipliers are non negative.(3) The solution x* must be feasible, that is h(x*) ≤k(4) The Complementary slackness condition are satisfied, λ¿¿

Proof: Suppose that the theorem does not hold.

⇒ There is solution x in U such that g(x) >g(x*) satisfying the constraints.

The function g is quasiconcave and differentiable and the derivatives g at x* are not all zero. Therefore we have,



⇒∑1

n

(x i¿−xi¿)∂ g¿¿¿ (1)

If h j(x*) ¿h j(x) then by CS condition λ j=0 , so ∑1

n

(x i¿−xi¿)¿¿

Suppose h j(x*)=k, then the multipliers are non negative and the constraints are convex

and hence ∑1

n

(x i¿−xi¿)¿¿

Combining the above two cases, we have, ∑1

n

(x i¿−xi¿)¿¿ (2)

From KT first order conditions, we have ∂ g¿¿

⇒∑1

n

(x i¿−xi¿)∂ g¿¿¿ (3)

The equations (1) and (2) contradicts(3). The result is true.


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