A Class of Convex Preferences Without Concave Representation ∗

Every continuous convex preference relation on R+ has a continuous utility representation U : R+ → R which is quasi-concave. It is natural to ask if the preference relation has another representation: one that is concave. The first study on this subject1 is de Finetti (1949). There he already mentions the importance of this question in utility theory. Fenchel (1953) has a deeper study of this problem. He presents necessary and sufficient conditions. His conditions are easy to check except one (i. e. condition VII on page 124–125.) Kannai (1977) deepens the study of Fenchel’s condition VII – and obtains a formula (see for example, Theorem 2.4 page 9) for the concave representation when there is one. A nice example of a preference relation that has not a concave representation appears in Arrow and Enthoven (1961) on footnote 6 page 781. The function U (x,y) = x − 1 + √ (1− x) + 4 (x+ y) is quasi-concave, strictly monotonic. The indifference curve U−1 (u) is the straight line connecting (u/2,0) and ( 0, 2+2u 4 ) . They mention that Fenchel (1953) has proved that such a function cannot have a concave representation. It is not clear however if they mean smooth representations or the general


INTRODUCTION
Every continuous convex preference relation on R l + has a continuous utility representation U : R l + → R which is quasi-concave.It is natural to ask if the preference relation has another representation: one that is concave.The first study on this subject 1 is de Finetti (1949).There he already mentions the importance of this question in utility theory.Fenchel (1953) has a deeper study of this problem.He presents necessary and sufficient conditions.His conditions are easy to check except one (i.e. condition VII on page 124-125.)Kannai (1977) deepens the study of Fenchel's condition VII -and obtains a formula (see for example, Theorem 2.4 page 9) for the concave representation when there is one.
A nice example of a preference relation that has not a concave representation appears in Arrow and Enthoven (1961) on footnote 6 page 781.The function is quasi-concave, strictly monotonic.The indifference curve U −1 (u) is the straight line connecting (u/2,0) and 0, u 2 +2u 4 . They mention that Fenchel (1953) has proved that such a function cannot have a concave representation.It is not clear however if they mean smooth representations or the general case.It is easy to check that such a function has no smooth concave representation.Another example is given in Schummer (1998).Analytically his example is U (x,y) = 2x 2−y ,0 < x,y ≤ 1.The indifference curve is the part of the straight line connecting (u,0) and (0,2) that intersect (0,1] Here I will prove a strong result in this set up.I show that if a strictly monotonic preference on R l + has affine indifference curves4 and two of these affine sets are not parallel then there is no concave representation.In other words, a preference relation with affine indifference curves that has a concave representation necessarily has a linear utility representation.That is, there exists b ∈ R l ++ such that U (x) = b • x represents the preference relation.

Let
for every x,y ∈ X and 0 < r < 1.It is easy to check that every concave function is quasi-concave as well.

Definition 1 [concavifiability]. The quasi-concave function U has a concave representation (or is concavifiable) if there is a strictly increasing function
The next lemma is a simplified version of (Fenchel, 1953, page 123 § 53).
Lemma 1 [Fenchel].Suppose f : U (X) → I is strictly increasing and continuous onto I. Let g := f −1 : Proof.Let z ,z ∈ I and 0 < r < 1.For a given > 0 there exist x ,x ∈ X such that Since is arbitrary the proof is finished.Definition 3. A set A ⊂ R l is affine if it is the translation of a vector subspace.That is A = a + V with V a vector subspace of R l .The affine set A has codimension 1 if V has codimension 1. Definition 4. The utility function U : X → R has affine indifference curves if for every u ∈ U (X) we have that U −1 (u) = H (u) ∩ X where the affine set H (u) has codimension 1.
The following lemma is central.
Lemma 2. Let I and J ⊂ (0,∞) be open intervals.Suppose g : I → J and φ : J → (0,∞) are strictly increasing functions and that g (I) = J.Suppose also that for every k > 0 the function min {g (z) ,kφ (g (z))} is convex.Then g and g • φ are convex and there is a t > 0 such that φ (u) = tu for every u ∈ J.
The proof is in the Appendix.

MAIN THEOREM
I begin defining in more detail the class of functions I use.Let b (u) ∈ R l \ {0} and τ (u) > 0 be functions defined for u ∈ (0,∞).The set is an affine set with codimension 1. Dividing by τ (u) we may suppose without loss of generality that τ (u) ≡ 1 for u > 0. And therefore Lemma 3. Suppose U : R l + → R is a strictly monotonic function such that U (0) = 0. Suppose also that + where H (u) is defined in (*).Then b i (u) is strictly decreasing for u > 0, continuous and onto (0,∞).Moreover U is quasi-concave and continuous.
The proof is in the Appendix.The theorem I want to prove is the following: Theorem 1. Suppose U : R l + → R + is a onto strictly monotonic utility function with affine indifference curves.If it has a concave representation then there is a strictly positive vector b >> 0 such that RBE Rio de Janeiro v. 64 n. 1 / p. 81-86 Jan-Mar 2010

Paulo Klinger Monteiro
The linearity of U −1 (u) implies that Fenchel's Lemma implies the concavity of − min i k i φ i (g (z)).And therefore min i k i φ i (g (z)) is convex.Fix k 1 = 1 and j = 1.Making k i → ∞ for every i = j we conclude that min {φ 1 (g (z)) ,k j φ j (g (z))} is convex for every k j > 0. Defining g = φ 1 • g we have that is convex for every k j > 0. Lemma 2 implies that φ j • φ −1 1 (u) = t j u for some t j > 0. Thus xi ti is a linear function.QED Remark 1.It is easy to generalize5 Theorem 1 to convex subsets with non-empty interior: For each interior point there is a small copy of the positive cone.Then from the theorem the utility is linear in this neighborhood.Then connectedness implies that the linear function is the same everywhere.