20. Homogeneous and Homothetic Functions
11/10/20
Homogeneous and homothetic functions are closely related, but are used in different ways in economics. We start with a look at homogeneity when the numerical values themselves matter. This happens with production functions. You should be familiar with the idea of returns to scale. Let’s focus on constant returns to scale. A production function exhibits constant returns to scale if, whenever we change all inputs in a given proportion, output changes by the same proportion. Rm R In equations, let f: + be a production function. It exhibits constant returns to scale if → f(tx)= tf(x) for every x Rm and t > 0. ∈ + 2 MATH METHODS 20.1 Homogeneous Functions
Rm Homogeneous Function. On + , a real-valued function is homogeneous of degree γ if
f(tx)= tγf(x) for every x Rm and t > 0. ∈ + The degree of homogeneity can be negative, and need not be an integer. Constant returns to scale functions are homogeneous of degree one. If γ > 1, homogeneous functions of degree γ have increasing returns to scale, and if 0 < γ < 1, homogeneous functions of degree γ have decreasing returns to scale. ◮ Example 20.1.1: Cobb-Douglas Production. Consider the Cobb-Douglas production R2 α β function defined on + by f(x, y)= Ax y with A,α,β > 0. Then f(tx, ty)= A(tx)α(ty)β = tα+βAxαyβ = tα+βf(x, y) showing that f is homogeneous of degree α + β. If α + β < 1, f has decreasing returns to scale, if α + β = 1, f has constant returns to scale, and if α + β > 1, f has increasing returns to scale. ◭ 20.HOMOGENEOUSANDHOMOTHETICFUNCTIONS 3 20.2 Examples of Homogeneous Functions Homogeneous functions arise in both consumer’s and producer’s optimization prob- lems. The cost, expenditure, and profit functions are homogeneous of degree one in prices. Indirect utility is homogeneous of degree zero in prices and income. Further, homogeneous production and utility functions are often used in empirical work. m γi The Cobb-Douglas functions u(x) = A Qi=1 xi with γi > 0 are homogeneous of Rm −ρ degree i γi on + . The constant elasticity of substitution function f(x) = [δx + −P 1 (1 δ)x ρ]ν/ρ for ν > 0 and ρ > 1, ρ = 0 is homogeneous of degree ν. The Leontief − 2 − 6 function u(x) = min xi is homogeneous of degree 1. Any monomial, m γi a Y xi i=1 m Rm is homogeneous of degree γ = Pi=1 γi on + . A sum of monomials of degree γ is homogeneous of degree γ, the sum of monomials of differing degrees is not homogeneous. Additional examples of homogeneous functions include:
p x, x3/2 +3x1/2y, √x + y + z · T x8 +5x4y4 + 10x3y5 +7xy7 + 13y8,Q(x)= x Ax.
The following are not homogeneous:
sin xyz, x2 + y3, 3xyz +3x2z 3xy − xy + yz + xz , exp(x2 + y2). x + z2
1/2 Most quasi-linear utility functions, such as u(x) = x1 + x2 are not homogeneous of any degree. The linear term means that they can only be homogeneous of degree one, meaning that the function can only be homogeneous if the non-linear term is also homogeneous of degree one. Non-linear cases that are homogeneous of degree one require at least three goods. One example is
x1 + (x2x3)1/2 which is homogeneous of degree one. 4 MATH METHODS 20.3 Cones and Homogeneity Homogeneous functions require us to know f(tx) when we know f(x). That means that tx dom f whenever x dom f. Sets with that property are called cones, and cones are∈ the natural domain of∈ homogeneous functions. Cone. Let V be a vector space. A set C V is a cone if for every x C and t > 0, tx C. Equivalently, C is a cone if tC C⊂for all t > 0. ∈ ∈ ⊂ Rm Rm Cones include the positive orthant + , the strictly positive orthant ++, any vector subspace of Rm, any ray in Rm and any set of non-negative linear combinations of a collection of vectors {x ,..., x }. The set {(x, y) : x, y 0, y x} is as example of the 1 i ≥ ≤ last as it can also be written {x = t1(1, 0)+ t2(1, 1) : t1,t2 0}. We can redefine homogeneity to apply to functions defined≥ on any cone. Homogeneous Function. Let C be a cone in a vector space V. A function f: C R is homogeneous of degree γ if → f(tx)= tγf(x) for every x Rm and t > 0. ∈ Restricting the domain of a homogeneous function so that it is not all of Rm allows us to expand the notation of homogeneous functions to negative degrees by avoiding division by zero. E.g., 1/ x is homogeneous of degree 1 on the cone Rm . Restricting the domain k k2 − ++ also allows us to consider f(x)/g(x) where f is homogeneous of degree γ1 and g is Rm homogeneous of degree γ2, both on ++. The quotient is homogeneous of degree γ γ on Rm . 1 − 2 ++ 20.HOMOGENEOUSANDHOMOTHETICFUNCTIONS 5 20.4 Derivatives of Homogeneous Functions When a function is homogeneous, homogeneity is inherited by the derivative, but with the degree reduced by one. Theorem 20.4.1. Let C Rm be an open cone and f: C R is C1 and homogeneous of degree γ. Then the derivative⊂ Df is homogeneous of degree→ (γ 1). − Proof. By homogeneity, f(tx)= tγf(x). Taking the x derivative of that equation, we obtain tDf(tx)= tγ Df(x). Dividing by t, we find that Df(tx)= t(γ−1) Df(x). We assumed C was open so that Df could be defined at all points of C. The converse fails. If Df is homogeneous of degree β, we cannot conclude that f is homogeneous of degree (β + 1). For example, let m = 2 and consider f(x)=1+ x1x2, which is not homogeneous of any degree. A quick calculation shows that (Dxf)(x) = (x2, x1) which is homogeneous of degree one, even though f is not homogeneous. Fortunately, addition of a constant is the main thing that goes wrong with the converse when Df is homogeneous of degree β for β = 1. 6The− case β = 1 can suffer from two other types of complications. The first involves logarithmic functions.− Suppose f(x) = b ln φ(x) where φ is homogeneous of degree one with φ > 0. Then b Df = Dφ(x), φ(x) which is homogeneous of degree minus one. The β = 1 case has a second type of complication when there is more than one variable. In −Rm with m > 1, functions can be homogeneous of degree zero without being constant. One such function is g(x)= x1/(x1 + x2). Its derivative is 1 Dg x , x , = 2 ( 2 1) (x1 + x2) − which is clearly homogeneous of degree minus one. 6 MATH METHODS 20.5 Homogeneity and Indifference Curves The fact that derivatives of homogeneous functions are also homogeneous has conse- quences for the shape of indifference curves and isoquants. It implies that the marginal rate of substitution and marginal rate of technical substitution are constant along rays through the origin. That is, they are homogeneous of degree zero. Tosee this let u be homogeneous of degree γ and consider MRSij = (∂u/∂xi) (∂u/∂xj). We calculate ∂u (tx) tγ−1 ∂u (x) ∂u (x) ∂xi ∂xi ∂xi MRSij(tx)= = = = MRSij(x). ∂u (tx) tγ−1 ∂u (x) ∂u (x) ∂xj ∂xj ∂xj
Of course, the calculation for the marginal rate of technical substitution is essentially the same. Consequences of this include that facts that income expansion paths and scale ex- pansion paths are rays through the origin whenever the original production or utility function is homogeneous. 20.HOMOGENEOUSANDHOMOTHETICFUNCTIONS 7 20.6 Euler’s Theorem The second important property of homogeneous functions is given by Euler’s Theorem. Rm R 1 Euler’s Theorem. Let f: ++ be C . Then f is homogeneous of degree γ if and only if Dxf(x) x = γf(x), that→ is m ∂f X xi (x)= γf(x). ∂xi i=1
Proof. Define ϕ(t)= f(tx). By the chain rule, dϕ/dt = Df(tx) x. If f is homogeneous of degree γ, we can also write ϕ(t) = tγf(x) so that dϕ/dt = γtγ−1f(x). Setting t = 1 we obtain (Df) x = γf(x). · Conversely, if Dxf(x) x = γf(x), dϕ 1 γf(tx) γϕ(t) = Dxf(tx) x = Dxf(tx) (tx)= = . dt t t t It follows that d(ln ϕ) 1 dϕ γ = = . dt ϕ dt t Integrating from t = 1 to t, we obtain ln ϕ(t) ln ϕ(1) = γ ln t. Taking the expo- nential yields ϕ(t) = ϕ(1)tγ. This can be rewritten− f(tx) = tγf(x), showing that f is homogeneous of degree γ. Gradients and Euler’s Theorem. Euler’s Theorem is sometimes written using the gradient, f = DfT , which is a column vector. In that case, the condition can be written ∇ x xf = γf(x). Either way, it means that ·∇ ∂f X xi = γf(x). ∂xi i 8 MATH METHODS 20.7 Wicksteed’s Theorem Suppose f is a constant returns to scale production function. If the firm is a price-taker in both input and output markets, profits are pf(x) w x where p is the output price and w the vector of factor prices. − · To maximize profit we set its derivative equal to zero.
∗ T pDf(x )= w .
This tells us that for each good i, the value of marginal profit must be equal to the factor price: ∂f p = wi. ∂xi What about profits? Euler’s Theorem provides the answer. By Euler’s Theorem,
∂f f(x)= X xi . ∂x i i
We multiply by p to obtain ∂f pf(x)= X xip . ∂x i i Then substitute the first order conditions to find
pf(x)= X xiwi = w x, · i showing that revenue equals cost and that profit is zero. If the production function was homogeneous of degree γ, 0 < γ < 1, the same argument would show profit is positive, while if γ > 1, profit would be negative. 20.HOMOGENEOUSANDHOMOTHETICFUNCTIONS 9 20.8 Homogeneity and Monotonicity We define three types of monotonicity for real-valued functions of m variables. Monotonicity. A real-valued function f defined on a subset of Rm is monotonic if f(x) f(y) whenever x y. It is strictly monotonic is f(x) > f(y) whenever x y. Finally,≥ it is strongly monotonic≥ if f(x) >f(y) whenever x > y. ≫ For functions on R, strong and strict monotonicity are the same. We saw earlier that the homogeneous of degree zero function g(x) = x1/(x1 + x2) is not monotonic. This is normal for such functions. Indeed, Euler’s Theorem can be used to show that functions that are homogeneous of degree zero cannot be monotonic when there are two or more variables. 1 Rm Theorem 20.8.1. Any function f C ( ++) for m > 1 that is homogeneous of degree zero is not monotonic. ∈ Proof. We prove this by applying Euler’s Theorem,
∂f 0= X xi . ∂xi i
If every ∂f/∂xi > 0, the right-hand side is positive. This is impossible as it must be zero, so there has to be j with ∂f/∂xj < 0,. The function f cannot be monotonic. However, it is possible to combine a logarithmic form with a homogeneous of degree zero form to get an increasing function with a derivative that is homogeneous of degree minus one. The function
x1 f(x1, x2)= + ln(x1 + x2) x1 + x2 is such a case. 10 MATH METHODS 20.9 Cardinal Functions vs. Ordinal Functions The terms cardinal and ordinal are often used when discussing utility functions. How- ever, they are not properties of the functions themselves, but of our interpretation of them. When we say a production function is cardinal, we mean that the output numbers have direct meaning. Two tons of steel and three tons of steel are different things. In contrast, with utility functions what is important is not the particular level of utility, but rather comparisons of utility levels—which is more and which is less. To that end, we say that utility functions u, v: X R are equivalent utility functions if u(x) u(y) if and only if v(x) v(y) for all x, y →X. ≥ ≥ ∈ Theorem 20.9.1. Equivalence between utility functions from X R is an equivalence relation. → Proof. The relation is reflexive. A function u: X R is equivalent to itself because u(x) u(y) if and only if u(x) u(y). Yes, that was→ pretty trivial. The≥ relation is symmetric. If≥u is equivalent to v, then v is trivially equivalent to u. Here u(x) u(y) if and only if v(x) v(y), so v(x) v(y) if and only if u(x) u(y). Finally, the≥ relation is transitive. This≥ is barely harder≥ to show. If u is equivalent≥ to v and v is equivalent to a function w then u(x) u(y) if and only if v(x) v(y) if and only if w(x) w(y), so u(x) u(y) if and only≥ if w(x) w(y), showing≥ that u is equivalent to w.≥ This proves transitivity.≥ ≥ A property of a function f: X R is ordinal if it shared by all equivalent utility functions. → 20. HOMOGENEOUSANDHOMOTHETICFUNCTIONS 11 20.10 Two Types of Equivalence You’ll notice that this treatment is slightly different from that in Simon and Blume (section 20.3) with an inequality rather than an equals sign. The reason is that if we used equality, as they do, the utility functions u(x, y) = x + y and v(x, y) = x y would be equivalent functions. However, as utility functions, they are completely− − different. For u, more is better, for v, more is worse. Our definition maintains the sense of utility. If two utility functions are equivalent in our sense, they are equivalent in the Simon and Blume sense. As the above example shows, the converse is false. Theorem 20.10.1. If u and v are equivalent utility functions on X, then u(x) = u(y) if and only if v(x)= v(y) for all x, y X. ∈ Proof. Now u(x) = u(y) if and only if both u(x) u(y) and u(y) u(x). By equivalence, that holds if and only if both v(x) v(y) and≥ v(y) v(x). The≥ last pair of statements hold if and only if v(x)= v(y). ≥ ≥ 12 MATH METHODS 20.11 Monotonic Transformations Monotonic transformations will allow us to characterize all equivalent utility functions. Monotonic Transformation. Let I be an interval in the real line. A function φ: I R is a monotonic transformation of I if φ is strictly increasing on I. Moreover, if f→is a real-valued function of m variables and φ is a monotonic transformation on its image, we say φ f: x φ f(x) ◦ 7 → is a monotonic transformation of f. If φ is differentiable with φ′ > 0, then φ is a monotonic transformation. 20. HOMOGENEOUSANDHOMOTHETICFUNCTIONS 13 20.12 Monotonic Transformations and Utility Equivalence One important result is that two utility functions are equivalent if and only if there are monotonic transformations of each into the other. Theorem 20.12.1. Let u, v: X R. Then u and v are equivalent utility functions on X if and only if there is a monotonic→ transformation φ with u = φ v. ◦ Proof. If: Suppose u = φ v. Because φ is strictly increasing, v(x) v(y) if and only ◦ ≥ if u(x)= φ v(x) φ v(y) = u(x), showing that the two functions are equivalent. ≥ Only If: Suppose u andv are equivalent. Letu ¯ ran u. There is x X with u¯ = u(x). Define φ(¯u)= v(x). ∈ ∈ The function v is well-defined.1 If we had picked any other y with u(y)=u ¯, we would have v(x) = v(y) by Theorem 20.10.1. This shows that the definition of φ(¯u) depends only onu ¯, not on our choice of x with u(x)=¯u. Now v(x) = φ u(x) for all x X. We need only show that φ is increasing. ∈ Supposeu ¯0 < u¯1. Take xi with u(xi)=u ¯i, i = 0, 1. Then u(x1) > u(x0). By the definition of equivalence, v(x1) v(x0). By Theorem 20.10.1, v(x1)= v(x0) if and only if u(x ) = u(x ). Since u(x ) =≥u(x ), v(x ) = v(x ), showing that v(x ) > v(x ). Then 0 1 0 6 1 0 6 1 1 0 φ(¯u1) = v(x1) > v(x0) = φ(¯u0). This shows that φ is increasing and completes the proof. R 2 7 x Thus on +, the functions x, x , x , e , and ln x are all monotonic transformations of f(x) = x, and so all equivalent. Similarly, xy, xy + (xy)4, log xy, and (xy)1/2 are all R2 equivalent on +. When utility is differential, one property preserved by monotonic differentiable trans- formations is the marginal rate of substitution. Suppose v(x) = φ u(x) . Then by the Chain Rule, ∂v ′ ∂u ∂u ∂ φ ∂ ∂ MRSv = xi = xi = xi = MRSu . ij ∂v φ′ ∂u ∂u ij ∂xj ∂xj ∂xj
1 “Well-defined” is a mathematics term meaning that the definition is unambiguous. 14 MATH METHODS 20.13 Homothetic Functions Homothetic functions are the ordinal equivalent of homogeneous functions. Homothetic Function. Let C be a cone. A function f: C R is homothetic if for every x, y C and t > 0, f(x) f(y) if and only if f(tx) f(ty→). ∈ ≥ ≥ One consequence of the definition of homotheticity is that f is equivalent to g defined by g(x)= f(tx). Any homogeneous utility function is also homothetic. And since homotheticity is an ordinal property, any increasing transformation of a homogeneous utility function is homothetic too. However, not all homothetic preferences have a homogeneous utility representation. Lexicographic preferences are homothetic, but cannot be represented by any utility function—homogeneous or otherwise. If we require that preferences are also mono- tonic and continuous, we can represent homothetic preferences by a homogeneous utility function. For a utility function, homotheticity means that preferences are invariant under scalar multiplication in the sense that the set of indifference curves is unchanged when all consumption bundles are multiplied by the same positive number. More precisely, preferences are invariant under homothetic transformations centered on the origin. Homothetic preferences include commonly used functional forms such as Cobb- Douglas utility and constant elasticity of substitution utility. One special type of homothetic utility is homogeneous utility, where multiplying the consumption bundle by a scalar multiplies utility by some power of that scalar. The Homothetic Representation Theorem will show that monotonic continuous and homothetic preferences can be represented by a homogeneous utility function. Homotheticity in economics is based on comparing positive scalar multiples of vec- tors. By restricting our attention to consumption sets that are cones, we ensure that scalar multiplication is always possible. Such scaling preserves the shapes of objects, including indifference surfaces. It only changes their scale. 20. HOMOGENEOUSANDHOMOTHETICFUNCTIONS 15 20.14 Homothetic and Homogeneous Functions One remarkable fact is that every continuous homogeneous utility functions is a mono- tonic transformation of a homogeneous utility function. The key part of the theorem is separated in a lemma, as it is of general use. ∗ Rm Rm Lemma 20.14.1. Suppose x ++ and u is continuous and increasing on + . If x Rm, there is a unique α ∈0 with u(αx∗)= u(x). ∈ + ≥ Rm ∗ + Proof. Let x + . Chooseα ¯ such thatα ¯x x 0 and define A = {α [0, α¯] : u(αx∗) u(x)}∈and A− = {α [0, α¯] : u(x) ≫u(α≥x∗)}. By continuity, both∈A+ and A− are≥ closed sets. Further, they∈ obey A+ A≥− [0, α¯]. Now 0 A− andα ¯ A+, so both are non-empty. Since intervals are∪ connected,⊃ A+ A−∈is non-empty.∈ By construction, u(x) = u(αx∗) whenever α (A+ A−). Transitivity∩ and the fact that u is increasing imply there is only one such α∈. ∩ 16 MATH METHODS 20.15 Homothetic Representation Theorem We are now ready to prove the Homothetic Representation Theorem. Rm R Homothetic Representation Theorem. Suppose u: + is increasing, homothetic, Rm and continuous on + . Then u(x) is a monotonic transformation→ of a homogeneous of degree one function v. Proof of Theorem. Apply Lemma 20.14.1 to x∗ = e. Then for every x Rm, there ∈ + is a α 0 with u(αe)= u(x). In that case, define v(x)= α, so that u v(x)e = u(x). ≥
b α¯e
αe b
b x
Figure 20.15.1: The position of the intersection of the indifference curve with the diagonal determines the unique α with u(x) = u(αe).
Now v(x) v(y) if and only if ≥ u(x)= u v(x)e u v(y)e = v(y). ≥ It follows that v is utility equivalent to u. We know u(x)= u v(x)e . By homotheticity of u, u(tx)= u tv(x)e . But u(tx)= u v(tx)e , so u tv(x)e = u v(tx)e . (20.15.1) so tv(x) = v(tx) because Lemma 20.14.1 shows that equation (20.15.1) has a unique solution. It follows that v is homogeneous of degree one. Since u and v are equivalent, by Theorem 20.12.1, there is a monotonic transforma- tion φ with u(x) = (φ v)(x). ◦ Although homothetic functions are related to homogeneous functions, there are differences. Nothing like Euler’s Theorem need hold for functions that are merely homothetic. To see this, consider f(x)= Pℓ αℓ ln xℓ. Then Dxf(x) x = Pℓ αℓ, which cannot be written in the desired form. 20. HOMOGENEOUSANDHOMOTHETICFUNCTIONS 17 20.16 Homotheticity and Marginal Rates of Substitution When preferences are described by a smooth utility function, we can also describe homotheticity by saying that the marginal rate of substitution is the same anywhere on a given ray through the origin. The marginal rates of substitution are constant along any ray through the origin. This means that the shape of the indifference curves are preserved under scalar multiplication. All slopes remain unchanged. Theorem 20.16.1. Let C Rm be an open cone and f: C R with Df(x) 0 on C. ⊂ ≫ Suppose f: C R is homothetic and differentiable. Then→MRSkℓ(x) = MRSkℓ(tx) for all t > 0 and x→ C. ∈ The requirement that the marginal rate of substitution exists rules out cases where we are dividing by zero. The proof is easy if we can write f = F φ, F monotonic, φ homogeneous of degree ◦ one, and both F and φ differentiable. Use the Chain Rule to compute MRSkℓ(x) = (∂φ/∂xk) (∂φ/∂xℓ). By Theorem 20.4.1, both derivatives are homogeneous of degree zero, and so is the marginal rate of substitution. Below, we prove this in a slightly more general fashion. Proof. Consider the upper contour sets U(x) = {y : f(y) f(x)}. By homotheticity, U(tx)= tU(x). Homogeneity of the inner product then shows≥ that if p supports U(x), it also supports U(tx). Since f is differentiable, p and Df(x) are proportional by Theorem 21.22.1. Thus Df(x) and Df(tx) are also proportional. The factors of proportionality cancel when constructing the marginal rates of substitution, so MRSkℓ(x) = MRSkℓ(tx). There is a converse, which we will state, but not prove. Rm R 2 Theorem 20.16.2. Suppose f: ++ is C , Df 0, and MRSkℓ is homogeneous of degree zero in x for every k and ℓ.→ Then f is homothetic.≫ Proof. A proof may be found in Lau.2 Lau uses the slightly weaker assumption that there is some j with ∂f/∂x = 0, in which case the MRS condition must be restated to j 6 work around the fact that MRSkℓ may not be defined for all pairs k and ℓ. Lau does not do this, but the replacement for the MRS condition is that for all k and ℓ, there are homogeneous of degree zero functions gkℓ such that ∂f/∂xk = gkℓ (∂f/∂xℓ). When Df 0, this is equivalent to the marginal rates of substitution being× homogeneous of degree≫ zero in x.
2 Lemma 1 in Lau, Lawrence J. (1969) Duality and the structure of utility functions J. Econ. Theory, 1, 374–396. 18 MATH METHODS
21. Concave and Quasiconcave Functions
11/12/20
NB: Problems 14, 18 and 22 from Chapter 19 and problems 1, 11, and 18 from Chapter 20 are due on Thursday, November 19. Convex, concave, and similar functions arise naturally in economics. They include the indirect utility function, cost function, expenditure function, and profit function. Moreover, concavity is usually assumed of utility as it ensures a diminishing (or at least non-increasing) marginal rate of substitution. It often applies to production. 21.CONCAVEANDQUASICONCAVEFUNCTIONS 19 21.1 Convex and Concave Functions Recall that in any vector space, the line segment between x and y is given by ℓ(x, y)= {(1 t)x + ty : 0 t 1}, and that a set S is convex if it contains ℓ(x, y) whenever x, y− S. ≤ ≤ ∈ Convex and Concave Functions. Let S be a convex set. A function f: S R is convex if for all x, y S and 0 t 1, f tx + (1 t)y • ∈ ≤ ≤ − ≤ tf(x)+(1 t)f(y→). A function−f: S R is concave if for all x, y S and 0 t 1, f tx + (1 t)y • ∈ ≤ ≤ − ≥ tf(x)+(1 t)f(→y). When the inequalities− are strict for 0 < t < 1, we say the function is strictly convex or strictly concave. One consequence is that for convex functions, every chord of the graph of the function lies on or above the graph. For concave functions, every chord lies on or below the graph.
Convex Concave
b
b
b
b
Figure 21.1.1: The left panel shows a convex function, where every chord connecting any two points of the graph lies above the graph. The right panel illustrates a concave function, and every chord connecting any two points of the graph lies below the graph.
◮ Example 21.1.2: Convex Functions. Any linear function f(x) = p x is both concave and convex, as is the generic affine function f(x) = a + p x. The· function f(x) = m 2 m 1/2 · x Pℓ=1 xℓ is convex while f(x) = Pℓ=1 xℓ is concave. The function f(x) = e is convex, while f(x) = ln x is concave. ◭ 20 MATH METHODS 21.2 Basic Properties of Concave and Convex Functions Several easily established properties of convex and concave functions are collected together without proof in Theorem 21.2.1. Theorem 21.2.1. 1. A function f is (strictly) convex if and only if f is (strictly) concave. − 2. A positive scalar multiple of a concave (convex) function is concave (convex). 3. The sum of two concave (convex) functions is concave (convex). 4. If φ is concave (convex) and weakly increasing on R and f is a concave (convex) function, then φ f is concave (convex). ◦ 5. The pointwise limit of a sequence of concave (convex) functions is concave (con- vex). 6. The infimum (supremum) of a sequence of concave (convex) functions is concave (convex). Proof. You should be able to prove these yourself. 21.CONCAVEANDQUASICONCAVEFUNCTIONS 21 21.3 Support Property
Supporting Hyperplane. We say a vector p supports a set S at x0 if either p x p x0 for every x S or p x p x for every x in S. · ≥ · ∈ · ≤ · 0 Of course, the set H = {x : p x = p x0} is a hyperplane. So supporting the set means that the set is on one side of· the hyperplane.· Moreover, they necessarily touch at x0. That means we can restate the definition in terms of half-spaces. A vector p supports + S at x0 if and only if S is contained in the one of the two half-spaces H (p, p x0) and − · H (p, p x0). Let f:·S R be a function where S Rm. The upper contour set is defined by U(y) = {x → S : f(x) f(y)}. The lower⊂ contour set is defined by L(y) = {x S : f(x) f(y)}.∈ ≥ ∈ ≤ ◮ Example 21.3.1: A Convex Upper Contour Set. Figure 21.14.2 illustrates an upper R2 R contour set for u: + given by u(x, y)= xy. →
p x · = p U(x0) x · 0 pp
b x0
Figure 21.3.2: The shaded area is the upper contour set U(x0) for u(x, y) = xy. The half- + space H (p, x0) is the hatched area when x0 = (1, 1) and p = (1, 1). The vector p supports U(x0) at x0.
◭ 22 MATH METHODS 21.4 Support Property Theorem The Support Property Theorem shows that a differentiable function f is concave if and only if Df(x ) supports the upper contour set at x for every x dom f. It also shows 0 0 0 ∈ that f is convex if and only if Df(x0) supports the lower contour set at x0 for every x dom f. 0 ∈ Support Property Theorem. Suppose f: U R is C1 where U is an open convex set U Rm. The function f is concave if and only→ if ⊂ f(y) f(x)+ Df(x)(y x) (21.4.1) ≤ − for all x, y U. The function f is convex if and only if ∈ f(y) f(x)+ Df(x)(y x) (21.4.2) ≥ − for all x, y U. ∈ Proof (Only If). First suppose f is concave and take ε with 0 < ε < 1.
f x + ε(y x) = f (1 ε)x + εy εf(y)+(1 ε)f(x). − − ≥ − We can rearrange to obtain
f x + ε(y x) f(x) ε f(y) f(x) . − − ≥ − Dividing by ε > 0 and letting ε 0 yields → Df(x)(y x) f(y) f(x). − ≥ − When f is convex, the only change to this part of the proof is that all three inequalities must be reversed.
Proof Continues... 21.CONCAVEANDQUASICONCAVEFUNCTIONS 23 21.5 Support Property, continued Proof (If). Now suppose the inequality
f(y) f(x)+ Df(x)(y x) (21.4.1) ≤ − is satisfied for all x, y U. Replace x by x′ = x + (1 α)(y x) = αx + (1 α)y where 0 α 1. Then∈ − − − ≤ ≤ ′ ′ ′ ′ f(y) f(x )+ αDf(x )(y x) f(x ) αDf(x )(x y). (21.5.3) ≤ − ≤ − − Rewrite the support equation (21.4.1) by replacing x with x′ and y with x. This yields
′ ′ f(x) f(x )+(1 α) Df(x )(x y). (21.5.4) ≤ − − Then multiply equation (21.5.3) by (1 α), equation (21.5.4) by α, and add them together. The Df(x′) terms cancel, leaving−
′ αf(x)+(1 α)f(y) f(x )= f αx + (1 α)y − ≤ − establishing concavity. The proof for the convex case is the same, but with every inequality reversed.
One consequence of the Support Property Theorem is that if f is concave, the Df(x0) supports the upper contour set U(x0) at x0. To see this, suppose f(y) f(x0). Then equation (21.4.1) implies ≥
0 f(y) f(x ) Df(x )(y x ), ≤ − 0 ≤ 0 − 0 so Df(x )y Df(x )x . This implies U(x ) H+ Df(x ), x as in Figure Fig:SupportedSet. 0 ≥ 0 0 0 ⊂ 0 0 24 MATH METHODS 21.6 The Support Property in R By the Support Property Theorem, a differentiable function is concave if and only if equation (21.4.1) holds. This important inequality can be taken as the definition of concavity for differentiable functions. When f is a function on a subset of the real line, we can use the right-hand side of equation (21.4.1) to define a line, y = f(x )+ f′(x )(x x ). This line is tangent to 0 0 − 0 the graph of f at the point x0,f(x0) , and the graph of the function is in the lower half-space that the tangent line defines. The tangent line supports the graph of f in the sense that the graph lies within one of the half-spaces defined by the tangent. The tangent line supports both the graph and subgraph of f at x0,f(x0) . The subgraph is sub f{(x, y) : f(x) y} at x ,f(x ) . ≤ 0 0 y f b f∗(p) − p
x x0 sub f
Figure 21.6.1: The tangent line at x has the equation y = f(x ) + f′(x )(x x ). Because f 0 0 0 − 0 is concave, the tangent line supports the subgraph of f. The graph is never above the tangent line and touches it at x ,f(x ) . Let p = f′(x ). The vector p = (p, 1) is perpendicular 0 0 0 − to the tangent line. Its vertical intercept is the negative of the concave conjugate function f∗(p) = px f(x ). 0 − 0 21.CONCAVEANDQUASICONCAVEFUNCTIONS 25 21.7 Support Property in Rm ′ In Figure 21.6.1 p = f (x0) is the slope of the tangent line. We now rewrite the equation of the tangent in a way that that shows it is a hyperplane in R2. Thus x (p, 1) = px0 f(x0). (21.7.5) − y − The vector p = (p, 1) is perpendicular to the tangent line. The right-hand side− of equation (21.7.5) is not zero unless tangent goes through the origin. It tells us how much the tangent line is offset from the origin. That value is called ∗ ′ the concave conjugate function and is denoted f (p)= px0 f(x0) when p = f (x0). In fact, f∗(p) is the negative of the vertical intercept of the tangent− line. The equation of the tangent line then becomes
x ∗ (p, 1) = f (p) − y and the supergradient inequality is
x ∗ (p, 1) f (p). (21.7.6) − f(x) ≥ More generally, if f is a function of m variables, we can consider its graph in Rm+1, and the picture is much the same.
x x0 (Df(x0), 1) (Df(x0), 1) − f(x) ≤ − f(x0)
where p = Df(x0). 26 MATH METHODS 21.8 Supporting Non-Differentiable Functions We can often support concave functions in the same way even if they aren’t differen- tiable. However, there may be more than one vector p that supports them in such a case.
y
b
f∗(p) − f x x0
Figure 21.8.1: As you can see, there are many lines that support the function f at x0,f(x0) . The function has slope +1 to the left of x0 and 1 to the right. Any line with a slope between − 1 and +1 that is tangent to the graph of f at x ,f(x ) will satisfy the support property. − 0 0 Supergradient and Subgradient. If f is a concave function on a convex set U Rm, and p Rm satisfies ⊂ ∈ f(y) f(x)+ p (y x), (21.8.7) ≤ · − for all x, y U, we call p a supergradient of f at x. It is a type of generalized derivative. There’s also∈ one for convex functions, called a subgradient. A vector p is a subgradient of f at x if f(y) f(x)+ p (y x), (21.8.8) ≥ · − for all x, y U. ∈ The Support Function Theorem tells us that when f is differentiable and concave (convex) Df(x0) is the only supergradient (subgradient) of f at x0. 21.CONCAVEANDQUASICONCAVEFUNCTIONS 27 21.9 Supergradients and Optimization When f is concave (convex) the first order conditions are not only necessary for an optimum, they are sufficient too. Theorem 21.9.1. Let U Rm be convex. ⊂ (a) Suppose 0 is a supergradient of f at x∗. Then x∗ is a global maximizer of f over U. In particular, this applies if f is differentiable and Df(x∗)= 0. (b) Suppose 0 is a subgradient of f at x∗. Then x∗ is a global minimizer of f over U. In particular, this applies if f is differentiable and Df(x∗)= 0. Proof. Setting p = 0 in the supergradient inequality (21.8.7), yields f(y) f(x∗) for all y U. ≤ ∈ This even works on some concave functions that aren’t differentiable, as in Figure 21.8.1 where a horizontal line (p = 0) supports the function at x0. That means x0 is a maximum, as is also clear from the graph. This result is extremely useful in economics, as we are faced with a convex feasible set U and a concave utility function we wish to maximize or convex function such as w z needing to be minimized. In either case, the first order necessary conditions become· sufficient for optimization. We can go a bit further than this for differentiable functions. Theorem 21.9.2. Let U Rm be convex. ⊂ (a) Suppose f is a concave C1 function on U and x∗ obeys Df(x∗)(y x∗) 0 for all y U. Then x∗ is a global maximizer of f on U. − ≤ ∈ ∗ ∗ ∗ (b) Suppose f is a convex C1 function on U and x obeys Df(x )(y x ) 0 for all y U. Then x∗ is a global minimizer of f on U. − ≥ ∈ Proof. For the concave case, if Df(x∗)(y x∗) 0, then f(y) f(x∗)+ Df(x∗)(y x∗) f(x∗). The convex case is similar. − ≤ ≤ − ≤ ◮ Example 21.9.3: Decreasing Functions. Suppose f is a decreasing C1 function of one variable on an interval [a,b]. Then f′(a)(x a) 0 for all x [a,b] because x a 0 and f′(a) 0. It follows that a is a maximizer− of≤f on [a,b] ∈◭ − ≥ ≤ It works in Rm too. Here’s an example for R2. ◮ Example 21.9.4: Global Minimum via Support. Let f(x, y) = x2 y2 on the set [0, 4] [0, 3] R2. Then Df = ( 2x, 2y). We will show− that −x∗ = (4, 3) is a minimizer.× Now⊂ Df(x∗) = ( 8, 6) and− Df−(x∗)(x x∗) = ( 8, 6) (x 4, y 3). Here both x 4 and y 3, so Df−(x∗−)(x x∗) 0, showing− that−x∗ is− a global· − minimum.− ◭ ≤ ≤ − ≥ 28 MATH METHODS 21.10 Hessian Convexity Tests: Necessity The support property can be used to show that the Hessian D2f(x) is negative semidef- inite when f is concave and positive semidefinite when f is convex. Theorem 21.10.1. Suppose f: U R is C2 on an open convex set U Rm. If f is concave, then the Hessian D2f(x) is→ negative semidefinite for all x U.⊂ If f is convex, then the Hessian D2f(x) is positive semidefinite for all x U. ∈ ∈ Proof. We will prove the concave case, the convex case is similar, with inequalities reversed. By the Support Property Theorem (equation 21.4.1)
f(y) f(x)+ Df(x)(y x)= f(x) Df(x)(x y) ≤ − − − reversing x and y in equation (21.4.1) yields
f(x) f(y)+ Df(y)(x y). ≤ − Adding the equations together and simplifying, we obtain
0 Df(y) Df(x) (x y). ≤ − − Now set y = x + h. We then have
Df(x + h) Df(x) h 0. − ≤ Divide by h and let h 0, obtaining hT D2f(x)h 0. In other words, D2f(x) is negative semidefinitek k fork allk →x U. ≤ ∈ 21.CONCAVEANDQUASICONCAVEFUNCTIONS 29 21.11 Hessian Convexity Tests: Necessity and Sufficiency When f is C2, we can use the Hessian matrix and Taylor’s formula to determine whether f is concave or convex. Theorem 21.11.1. Suppose f: U R is C2 on an open convex set U Rm. ⊂ 1. The function f is concave if and→ only if D2f(x) is negative semidefinite for all x U. ∈ 2. If D2f(x) is negative definite for all x U, f is strictly concave. ∈ 3. The function f is convex if and only if D2f(x) is positive semidefinite for all x U. ∈ 4. If D2f(x) is positive definite for all x U, f is strictly convex. ∈ Proof. Taylor’s formula tells us that
T f(y)= f(x)+ Df(x)(y x) + (y x) D2f(z)(y x). (21.11.9) − − − for some z on the line segment between x and y. (1) If D2f is negative semidefinite on U, this implies the support property, so f is concave by the Support Property Theorem. Conversely, if f is concave, Proposition 21.10.1 shows that D2f(x) is negative semidefinite on U. (2) If f is negative definite on U, equation (21.11.9) yields f(y)
a11 a13 a16 a31 a33 a36
a61 a63 a66
where the 2nd,4th, and 5th rows and columns have been deleted. Keep in mind that there are usually many kth-order principal minors. We can rewrite our results relating convexity and definiteness by applying the usual determinant tests to the Hessian. That yields the following theorem. Theorem 21.12.1. Suppose f: U R is C2 on an open convex set U Rm. Let H(x)= D2f(x) denote the Hessian of→f. ⊂ 1. The function f is convex if and only if every kth-order principal minor obeys H˜ (x) 0 for all x U. k ≥ ∈ 2. The function f is concave if and only if every kth-order principal minor obeys ( 1)kH˜ (x) 0 for all x U. − k ≥ ∈ 3. Suppose the leading principal minors obey Hk(x) > 0 for k = 1,...,m and all x U. Then f is strictly convex. ∈ k 4. Suppose the leading principal minors obey ( 1) Hk(x) > 0 for k =1,...,m and all x U. Then f is strictly concave. − ∈ 21.CONCAVEANDQUASICONCAVEFUNCTIONS 31 21.13 Using the Determinant Test ◮ Example 21.13.1: Cobb-Douglas Utility. Consider the Cobb-Douglas utility function R2 α 1−α on ++ defined by u(x)= x1 x2 with 0 < α < 1. The Hessian is
α−2 1−α α−1 −α α(α 1)x1 x2 α(1 α)x1 x2 H(x)= − − − − − − . α(1 α)xα 1x α (1 α)( α)xαx α 1 − 1 2 − − 1 2 α−2 1−α Clearly, det H = 0. The two first-order minors are α(α 1)x1 x2 < 0 and (1 α −α−1 −R2 R2 − α)( α)x1 x2 < 0, which means that u is concave on ++. Concavity on + then follows− from continuity. Note that u is not strictly concave because αu(0)+(1 α)u(e)=1 α = u (1 − − − α)e . ◭ ◮ Example 21.13.2: CES Utility. Another example is the constant elasticity of substitution − − utility function u(x) = [x ρ + x ρ]−1/ρ where ρ > 1, ρ = 0. The Hessian is 1 2 − 6 − − − 1+2ρ ρ ρ ( ρ ) 2 [x1 + x2 ] x2 x1x2 H(x)=(1+ ρ) 2+ρ 2+ρ − 2 . x x x1x2 x1 1 2 − The two first-order minors are both negative, while the second-order minor is 0. Thus R u is concave on ++. ◭ 32 MATH METHODS 21.14 Example: Upper and Lower Contour Sets Upper contour sets are convex for concave functions while lower contour sets are convex for convex functions. It is often the case that the lower contour set of a concave function is not convex. This happens in the figure below. ◮ Example 21.14.1: Upper and Lower Contour Sets. The left side Figure 21.14.2 illus- R2 R 1/3 2/3 trates an upper contour set for u: + given by u(x, y)= x y . The right side shows the lower contour set for the same→ function. It is not convex.
Convex Not Convex
U(y) L(y)
x b x b
Figure 21.14.2: The shaded area in the left panel is the upper contour set U(y) for u(x, y) = x1/3y2/3 and y = (1, 1). The shaded area in right panel is the lower contour set for the same function. The dashed line segment would be in L(y) if L(y) were convex. The line segment goes outside L(y), so the lower contour set is not convex.
◭ 21.CONCAVEANDQUASICONCAVEFUNCTIONS 33 21.15 Convexity of Upper and Lower Contour Sets In Example 21.14.2, the upper contour set was convex. This was no accident. Rather, it is a consequence of using a concave function. Theorem 21.15.1. Let U Rm be a convex set. ⊂ 1. If f: U R is a concave function, then the upper contour set U(y) is convex. 2. If f: U → R is a convex function, then the lower contour set L(y) is convex. Proof. Let x→, x′ U(y). Then f(x),f(x′) f(y). By concavity of f ∈ ≥ ′ ′ f (1 t)x + tx (1 t)f(x)+ tf(x ) (1 t)f(y)+ tf(y)= f(y), − ≥ − ≥ − showing that the convex combination (1 t)x + tx′ U(y) whenever x, x′ U(y). The proof of (2) is similar. − ∈ ∈ Contour Sets are Ordinal. Notice that U(x) and L(x) remain unchanged when a mono- tonic transformation is applied to f. This is because f(x) f(y) if and only if (φ f)(x) (φ f)(y) when φ is a monotonic transformation. ≥ ◦ ≥ ◦ 34 MATH METHODS 21.16 Quasiconvex and Quasiconcave Functions Quasiconvexity and quasiconcavity are the ordinal versions of convexity and concavity. They are defined in terms of the upper and lower contour sets, respectively. We just saw that upper and lower contour sets are unaffected by monotonic transformations, and hence ordinal. Quasiconvexity and Quasiconcavity. Let f: S R. 1. The function f is quasiconcave on S if→ and only if the upper contour set U(y)= {x S : f(x) f(y)} is a convex set for every y S. ∈ ≥ ∈ 2. The function f is quasiconvex on S if and only if the lower contour set L(x) = {x S : f(x) f(y)} is a convex set for every y S. ∈ ≤ ∈ From the definition, it is easy to see that f is quasiconvex if and only if f is quasi- concave. − There are other ways to characterize quasiconvexity and quasiconcavity. One is the following. Theorem 21.16.1. Let f: S R where S is a convex set. 1. The function f is quasiconcave→ if and only if for all t obeying 0 t 1, f(tx + (1 t)y) min{f(x),f(y)}. ≤ ≤ − ≥ 2. The function f is quasiconvex if and only if for all t obeying 0 t 1, f(tx + (1 t)y) max{f(x),f(y)}. ≤ ≤ − ≤ Of course, if f: S R is quasiconcave or quasiconvex on S, S must be a convex set because if x′, x′′ S→, convex combinations of x′ and x′′ are in either {x S : f(x) min[f(x′),f(x′′)]} (f∈quasiconcave) or in {x S : f(x) max[f(x′),f(x′′)]} (f quasiconvex).∈ ≥ In Theorem 21.15.1 we saw that convex∈ functions≤ are quasiconvex and concave functions are quasiconcave. The relationship is only one-way. Indeed, when f: R R, every monotone function is simultaneously quasiconvex and quasiconcave! In contrast,→ the only functions that are both convex and concave on R are the affine functions. 21.CONCAVEANDQUASICONCAVEFUNCTIONS 35 21.17 Quasiconvexity on the Real Line For f defined on a convex subset of R, we can completely characterize the quasiconcave and quasiconvex functions. Theorem 21.17.1. Let I be an interval in R. Suppose f: I R. 1. The function f is quasiconcave if and only f is either→ monotone or there exists a number x∗ such that f is weakly increasing when x x∗ and weakly decreasing when x x∗. ≤ ≥ 2. The function f is quasiconvex if and only if f is either monotone or there exists a number x∗ such that f is weakly decreasing when x x∗ and weakly increasing when x x∗. ≤ ≥ Proof. It is enough to prove the quasiconcave case as f is quasiconcave if f is quasiconvex. − We first prove sufficiency. As noted above, monotone functions are quasiconcave. We need only consider the case where x∗ exists. Let y I be given and let x and x ∈ 1 2 be arbitrary points in the upper contour set U(y) with x1 < x2. We must show that the interval [x1, x2] U(y). ∗ ⊂ If x is not in [x1, x2], then f is either always weakly decreasing or always weakly increasing on [x , x ]. Either way, for any z [x , x ], f(z) min{f(x ),f(x )} f(y), so 1 2 ∈ 1 2 ≥ 1 2 ≥ z U(y). This shows that [x1, x2] U(y). ∈ ∗ ⊂ ∗ ∗ If x [x1, x2], f is weakly increasing on [x1, x ] and weakly decreasing on [x , x2]. Again, for∈ any z [x , x ], f(z) min{f(x ),f(x )} f(y), so z U(y) and hence ∈ 1 2 ≥ 1 2 ≥ ∈ [x1, x2] U(y). In other words, f is quasiconcave. We now⊂ turn to necessity. Suppose that f is quasiconcave on I, but not monotone. ∗ We prove this part by contradiction. If x does not exist, we can find x1, x2, and x3 in I such that x < x < x and f(x ) < min{f(x ),f(x )}. If f(x ) f(x ), set y = x , 1 2 3 2 1 3 1 ≤ 3 1 otherwise take y = x3. Then the upper contour set U(y) includes x1 and x3, but not x2. It is not convex, contradicting the fact that f is quasiconcave. To sum up, quasiconcave functions defined on a real interval are either monotonic or single-peaked, while quasiconcave functions are either monotonic or single-troughed. Either the peak or trough may be an interval. 36 MATH METHODS 21.18 Strict Quasiconvexity and Quasiconcavity One consequence of Proposition 21.16.1 is that a function is quasiconcave if for every x, y with f(x) f(y) and every t, 0 t 1, we have f (1 t)x + ty f(y). We can ≥ ≤ ≤ − ≥ use this criterion for quasiconcavity to define strict quasiconcavity . Strict Quasiconcavity and Strict Quasiconvexity. Let f: S R where S is a convex set. 1. The function f is strictly quasiconcave on S if and→ only if for every x, y S with ∈ f(x) f(y) and every t with 0 < t < 1, we have f (1 t)x + ty >f(y). ≥ − 2. The function f: S R is strictly quasiconvex if and only if for every x, y S ∈ with f(x) f(y) and→ every t with 0 < t < 1, we have f (1 t)x + ty Proof. Since both xi maximize f over S, f(x1) = f(x0). If x0 = x1, define x2 = 1 1 6 2 x0 + 2 x1. By strict quasiconcavity, f(x2) > f(x0), showing that x0 is not a maximum. This contradiction shows that x0 = x1. A similar result holds for minima of a strictly quasiconvex function. 1 This has been known since de Finetti (1949). The problem of finding whether a given quasiconcave function is an increasing transformation of a concave function was posed (in terms of level sets) by Fenchel (1953). Although some results are known, it still does not have a completely satisfactory solution. 21.CONCAVEANDQUASICONCAVEFUNCTIONS 37 21.19 Support Property Theorem II There are also support properties that characterize quasiconvex and quasiconcave functions. Support Property Theorem II. Suppose f is C1 on U, an open convex subset of Rm. The function f is quasiconcave if and only if for all x, y U ∈ f(y) f(x) implies Df(x) (y x) 0. (21.19.10) ≥ − ≥ The function f is quasiconvex if and only if for all x, y U ∈ f(y) f(x) implies Df(x) (y x) 0. ≥ − ≤ then f is quasiconcave. Proof (Only if). Here f is quasiconcave. Suppose f(y) f(x). Consider (1 ε)x+εy = x + ε(y x) for 0 < ε < 1. By quasiconcavity, ≥ − − f x + ε(y x) f(x) f x + ε(y x) f(x) so − − 0. − ≥ ε ≥ Let ε 0 to obtain [Df(x)](y x) 0, which establishes the result. (Only If) → − ≥ Proof Continues... 38 MATH METHODS 21.20 Support Property Theorem II, Part II Proof (If). Now suppose that for all x, y U; f(y) f(x) implies [Df(x)](y x) 0. We must show that f is quasiconcave. ∈ ≥ − ≥ We do this by contradiction. If f is not quasiconcave, there are x0, x1 and 0 We can then write f(xt0 ) Of course, x0 xt = t(x0 x1) and x1 xt = (1 t)(x1 x0). We substitute into equation (21.20.11)− to obtain− − − − tDf(x )(x x ) 0 and (1 t)Df(x )(x x ) 0. − t 1 − 0 ≥ − t 1 − 0 ≥ Since t < 0 and (1 t) > 0, we must have − − Df(x )(x x )=0 (21.20.12) t 1 − 0 for all t I,0 < t < 1. ∈ Now 0 < f(x0) f(xt0 ) = f(xt1 ) f(xt0 ), so by the Mean Value Theorem, there is t (t ,t ) with − − 3 ∈ 1 0 0 n−1 has bordered principal minors that obey ( 1) An 0 for n = 3,...,m + 1. This provides a second-derivative test for quasiconcavity.− ≥ 40 MATH METHODS 21.22 Uniqueness of Supports When f is differentiable, we can show that the derivative is the only vector (up to scalar multiplication) that can support the upper (or lower) contour set. More precisely, we say p supports a set S at x if p x p x whenever x S. We apply this idea to the 0 · ≥ · 0 ∈ upper contour set {x : f(x) f(x0)} at x0. It is important to realize that this result does not require quasiconcavity.≥ However, if the function is not quasiconcave, the upper contour set may not have supports. Theorem 21.22.1. Suppose f is differentiable at x0 with Df(x0) = 0 and that p supports {x : f(x) f(x )} at x . Then p = αDf(x ) for some α =0. 6 ≥ 0 0 0 6 Proof. Let z = Df (p Df/ p 2)p. Then p z = 0 and Df z = Df 2 |p Df|2/ p 2. By the Cauchy-Schwartz− · k k inequality, either· p is proportional· tokDfk(and− we· are done)k k or Df z > 0. In the case where· Df z > 0, the first-order Taylor approximation f(x )+ αDf z > · 0 · f(x0) shows f(x0 + αz) >f(x0) for small α > 0. Continuity of f then yields f(x + αz εp) >f(x ) 0 − 0 for ε > 0 small. Apply Support Property II to obtain p x p (x + αz εp) · 0 ≤ · 0 − = p x ε p 2 · 0 − k k < p x . · 0 This contradiction shows Df z > 0 is impossible. Therefore p is proportional to · Df(x0). 21.CONCAVEANDQUASICONCAVEFUNCTIONS 41 21.23 Bordered Hessian Test for Quasiconcavity Since λp = Df, we can instead use the bordered Hessian with Df. The bordered principal minors are multiplied by λ2, so the signs are unchanged. This yields the following theorem. Theorem 21.23.1. Suppose f: U R is a C2 function on some open convex set U Rm with m > 1. Consider the→ bordered Hessian ⊂ 0 ∂f ... ∂f ∂x1 ∂xm ∂f ∂2f ∂2f 0 Df ∂x ∂x2 ... ∂x ∂x H = = 1 1 1 m T 2 . . . . Df D f . . .. . ∂f ∂2f ∂2f ∂ ∂ ∂ ... ∂ 2 . xm x1 xm xm n−1 1. If the bordered leading principal minors Hk obey ( 1) Hn > 0 on U for n =3,...,m +1, then f is quasiconcave on U. − 2. If all non-trivial bordered leading principal minors are negative on U, then f is quasiconvex on U. th 3. If f is quasiconcave on U, then every k order bordered principal minor H˜ k obeys ( 1)n−1H˜ 0 on U for n =3,...,m +1. − n ≥ 4. If f is quasiconvex on U, then all non-trivial bordered principal minors are non- positive on U. Proof. For the quasiconcave case, see Arrow and Enthoven (1961). Applying the result to f yields the quasiconvex case. − November 13, 2020 Copyright c 2020 by John H. Boyd III: Department of Economics, Florida International University, Miami, FL 33199