Revisiting the problem of integrability in utility theory

August 2, 2015 Optimization Hadjisavvas˙Penot˙final

To appear in OptimizationVol. 00, No. 00, Month 20XX, 1–15

Revisiting the problem of integrability in utility theory

Nicolas Hadjisavvasa∗ and Jean-Paul Penotb

aMathematics and Statistics Department, King Fahd University of Petroleum and

Minerals, Dhahran 31261, Kingdom of Saudi Arabia; bSorbonne Universites, UPMC

Universite Paris 6, UMR 7598 Laboratoire Jacques-Louis Lions, F-75005 PARIS France

(Received 00 Month 20XX; accepted 00 Month 20XX)

We revisit the problem of integrability in the consumer theory, focusing on the main difficul-ties. First we look for a neat and simple local existence result, and then for a global solution.Second, observing that a utility function (or indirect utility function) cannot be determineduniquely, we propose a means to get a kind of uniqueness result. Our approach is coordinate-free and can be used both in the classical case of a finite-dimensional commodity space andin the case an infinite dimensional model is adopted.

Keywords: Consumer theory; Frobenius Theorem; Integrability problem; Revealedpreferences; Utility function

AMS Subject Classification: 91B16; 91B42; 90C99

1. Introduction

We tackle a classical problem of the consumer theory: can the knowledge of thedemand map determine the utility function or the indirect utility function? Such aproblem called the problem of revealed preference or the integration problem hasattracted a great number of researchers ([1], [2], [5], [6], [7], [13], [16], [18], [25],[29], [38], [40], [41], [42], [43], [44], [45]...). Its importance stems from the fact thatutility functions are conceptual objects but cannot be observed, while demand interms of prices can be measured.

Various forms of the problem have been considered in the literature. In particular,discrete versions have supplemented the original continuous problem and formu-lations in terms of preference relations have received a great attention. The firstcontributions by Antonelli, Frobenius, Pareto, Pfaff, Slutsky and Volterra pointedout the necessity of symmetry conditions. They benefited from the theory of dif-ferential forms, a tool issued from exterior algebra and differential calculus andillustrated by Pfaff, Poincare, Cartan, Darboux among others (see [2] for an ac-count of the history of the problem). This angle of attack is expounded and enlargedin the book by Chiappori and Ekeland [6] making use of the Darboux Theorem forthe study of the aggregate demand. Another method relies on the study of differ-ential equations, a more usual tool. For that reason, P. Samuelson is often creditedfor the solution of the problem, although [40] and [41] deal with the case of aneconomy with two goods, the general case being left with an heuristic character.The use of differential equations is also prominent in [21] that used an additional

∗Corresponding author. Email: [email protected]

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boundedness assumption, and defined the utility function only on the range ofthe demand map; the structure of this set is not made explicit, so instead of thequasiconcavity of the utility one obtains some kind of convexity of the indifferencesurfaces. Recently, Crouzeix and Rapcsack [16] used ordinary differential equationsto obtain the existence of a utility function, given a demand map with values in the(strictly) positive orthant. They deal with two general problems and they assertthat the integration problem of the consumer theory can be treated similarly butthat the solution is global as ”it is no more necessary to restrict” the neighborhoodC.

In all these works, the main difficulty lies in the global construction of a utilityfunction, as underlined in [2] and [18], for instance. In [2, p. 177] S.N. Afriat writes:“the global conditions that are the only ones the economic interest recognizes”. Itis this difficulty we face here. As in [16], we focus our attention on the continuousproblem under smoothness assumptions and we first determine an indirect utilityfunction. The clear correspondence between utility functions and indirect utilityfunctions established in [20], [9], [31] then allows to construct the utility function.We recall it and adapt it for the reader’s convenience. The availability of thiscorrespondence makes a difference with the case of excess demand maps or incomecompensation functions (see [5], [21], [23], [32]).

Our main tool is a classical result of differential calculus and differential geometry,the Frobenius Theorem (see for instance [19], [28]), a result related to the famousPoincare Lemma for one-forms and equivalent to the Implicit Function Theorem([33]). We recall it in the next section in its simplest analytical form and we providea global version. The following section is devoted to a short presentation of theconsumer theory. The solution to the integration problem is presented in Section 4,in a setting that is possibly infinite dimensional. As well-known, infinite dimensionalspaces permit a modelization of economies with infinite horizon, or with uncertaintyover an infinite state space [3, 17]. That point is not crucial for our aims and thereader may suppose throughout that X = Rn, X+ = Rn+. However, the setting ofabstract normed spaces makes a clear distinction between the space of goods andthe dual spaces of prices and such a distinction is desirable.

2. Preliminaries

The main tool we shall apply in order to solve the integration problem is the classi-cal theorem of Frobenius. We shall use a special case of it, but, for the convenienceof the reader, we recall its general form, as taken from [19, Thms 10.9.4, 10.9.5]. Inthe sequel, given a normed space X with dual X∗, we denote by 〈x∗, x〉 the value ofx∗ ∈ X∗ at x ∈ X; however, if we consider x as an element of X∗∗ via the canonicalembedding, we occasionally write 〈x, x∗〉. We denote by Df the derivative of a mapf : X → Y between two normed spaces. If X := W ×W ′, D1f (resp. D2f) standsfor the partial derivative of f with respect to its first variable w ∈W (resp. secondvariable w′ ∈W ′).

Theorem 2.1 Let W and X be Banach spaces, let Z be an open subset of W ×X,and let F : Z → L(W,X) be a map of class C1 with values in the space L(W,X)of continuous linear maps from W into X. Suppose that for all (w, x) ∈ Z the

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following condition is satisfied: for all w′, w′′ ∈W(D1F (w, x)w′

)w′′ +

(D2F (w, x)F (w, x)w′

)w′′

=(D1F (w, x)w′′

)w′ +

(D2F (w, x)F (w, x)w′′

)w′. (1)

Then, for all (w, x) ∈ Z, there exist some neighborhoods U and V of w and xrespectively and a unique map s : U×U×V → X of class C1 such that s(u, u, v) = vand

D1s(w, u, v) = F (w, s(w, u, v)) for all (w, u, v) ∈ U × U × V. (2)

Uniqueness can be made stronger in the following sense: given a pair (u, v) ∈ U×Vand a C1 map su,v : U → X satisfying su,v(u) = v and Dsu,v(w) = F (w, su,v(w)),one has su,v(w) = s(w, u, v) for all w ∈ U .

When F does not depend on its second variable x the symmetry condition (1)reduces to (DF (w)w′)w′′ = (DF (w)w′′)w′, a natural condition since then su,v :=s(·, u, v) is of class C2, with Dsu,v(w) = F (w) and the second derivative of su,v isknown to be symmetric. When W = R, the so-called total differential equation (2)is just an ordinary differential equation and the condition s(u, u, v) = v representsthe initial data. In that case, L(W,X) can be identified with X (via the map` 7→ `(1)) and the symmetry condition (1) is automatically satisfied, w′, w′′ beingscalars. Another case of interest (which is the case we need) is the case X = R. Insuch a case, (2) determines a function w 7→ su,v(w) := s(w, u, v) whose derivativeis given by F.

We need a globalization of the preceding theorem. Here, we say that a subsetU of a vector space is starshaped with respect to some u ∈ U if for all u′ ∈ Uand t ∈ [0, 1] we have (1 − t)u + tu′ ∈ U. Clearly, U is convex if, and only if it isstarshaped with respect to all its points.

Theorem 2.2 Let W, X, Z, F be as in the Frobenius Theorem. Then, for everygiven (u, v) ∈ Z there exists a greatest open set Uu,v which is starshaped with respectto u and a unique map su,v : Uu,v → X of class C1 such that

Dsu,v(w) = F (w, su,v(w)) for all w ∈ Uu,v (3)

su,v(u) = v. (4)

Proof. Given (u, v) ∈ Z, let s1 : U1 → X and s2 : U2 → X be two solutionsdefined, respectively, on open sets U1 and U2 that are starshaped with respectto u. Then s1 = s2 on U1 ∩ U2. Indeed, U1 ∩ U2 is starshaped with respect tou, hence arcwise connected. The Frobenius theorem asserts that the set of pointsw ∈ U1 ∩ U2 such that s1(w) = s2(w) is open. Since by continuity it is obviouslyclosed in U1 ∩ U2, this set is U1 ∩ U2 as it is nonempty (u belongs to it).

Thus, if si : Ui → X, i ∈ I is the family of all possible solutions of the initial valueproblem (3), (4) with Ui starshaped with respect to u, we can set Uu,v =

⋃i∈IUi,

an open subset that is starshaped with respect to u, and we define s : U → X bys(w) = si(w) whenever w ∈ Ui. This map s is clearly the solution to (3), (4) withthe greatest starshaped open domain.

There is a simple case in which the greatest domain can be determined. As weshall not use it, we just state the result.

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Theorem 2.3 Let W, X, Z, F be as in the Frobenius Theorem. Suppose Z =W0×X, where W0 is an open subset of W starshaped with respect to some u ∈W0

and F is locally bounded on W0, uniformly with respect to X. Then, for all v ∈ Xthere exists a unique map su,v : Uu,v → X of class C1 satisfying (3), (4) withUu,v = W0.

3. The problem of revealed preference

Let X be a Banach space and let X+ be a closed convex cone in X. We set P = X∗,and

P+ = p ∈ P : 〈p, x〉 ≥ 0, ∀x ∈ X+,

in which case, by the bipolar theorem,

X+ = x ∈ X : 〈p, x〉 ≥ 0, ∀p ∈ P+.

We assume intP+ 6= ∅ and that P+ is pointed i.e. that −p /∈ P+ if p ∈ P+. Weunderstand X+ as the set of commodity bundles, and intP+ as the set of prices.

Given a so-called utility function u : X+ → R∪−∞, the indirect utility functionv : intP+ → R is defined by

v(p) = supu(x) : x ∈ X+, 〈p, x〉 ≤ 1. (5)

Here we normalize prices in order that the income of the consumer is 1 (replacingp with p/r if the income is r > 0). For each r ∈ R, if we set Ur := x ∈ X+ : u(x) >r, the sublevel set S≤(r) := p ∈ intP+ : v(p) ≤ r of v for level r can be writtenas

S≤(r) =⋂x∈Ur

p ∈ intP+ : 〈p, x〉 > 1,

since when p ∈ (intP+)\S≤(r) i.e. v(p) > r, one can find some x ∈ X+ satisfying〈p, x〉 ≤ 1 and u(x) > r, so that x ∈ Ur and p /∈ p ∈ intP+ : 〈p, x〉 > 1 andconversely, when p does not belong to the right-hand side of this relation thereexists some x ∈ Ur such that 〈p, x〉 ≤ 1, hence v(p) ≥ u(x) > r, so that p /∈ S≤(r).

We infer that v is nonincreasing and w∗-evenly quasiconvex in the sense thatits sublevel sets are w∗-evenly convex, i.e. the trace on intP+ of intersections ofw∗-open halfspaces. In particular, v is quasiconvex.

The Walrasian or Marshallian demand correspondence: X(·) : intP+ → 2X+ isdefined by

X(p) = x ∈ X+ : u(x) = v(p), 〈p, x〉 ≤ 1.

One says that the non satiation condition is satisfied if for any (x, p) ∈ X+ ×intP+ such that 〈p, x〉 < 1 there exists x′ ∈ X+ such that u(x′) > u(x) and〈p, x′〉 ≤ 1. This is equivalent to saying that for all x ∈ X(p), 〈p, x〉 = 1 holds. Notethat such a property is satisfied whenever u has no local maximizer on X+, inparticular when u is (strictly) increasing along rays. If the non satiation conditionholds, then it is easy to see that v takes its values in R ∪ +∞.

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The following result is known in several variants [9, 20, 31]. It will be crucial forour method. We include a proof of the version that we will use, for the reader’sconvenience.

Proposition 3.1 Let v : intP+ → R be w∗-evenly quasiconvex and nonincreasing.Then v is the indirect utility function of the quasiconcave utility function u : X+ →R ∪ −∞ given by

u(x) = infv(p) : p ∈ intP+, 〈p, x〉 ≤ 1. (6)

If moreover v has no local minimizers, then the non satiation condition is satisfied.

Proof. We have to prove that

∀p ∈ intP+, v(p) = supu(x) : x ∈ B(p) (7)

where B(p) := x ∈ X+ : 〈p, x〉 ≤ 1 is the budget set. Fixing p ∈ intP+, it is clearthat for all x ∈ B(p), u(x) ≤ v(p) holds, so that supu(x) : x ∈ B(p) ≤ v(p).Thus, to prove (7), it suffices to show that for all r < v(p) we can find somexr ∈ B(p) such that r ≤ u(xr). Setting as before

S≤(r) := p′ ∈ intP+ : v(p′) ≤ r

we note that p /∈ S≤(r). Since v is w∗-evenly quasiconvex, S≤(r) is an intersectionof w∗-open halfspaces and intP+. Hence there exists a w∗-open halfspace containingS≤(r) but not p, i.e., there exist x0 ∈ X\0, t ∈ R such that

∀p′ ∈ S≤(r),⟨p′, x0

⟩> t ≥ 〈p, x0〉 . (8)

Since v is nonincreasing, S≤(r) + P+ ⊂ S≤(r) so that (8) implies that 〈p′, x0〉 ≥0 for all p′ ∈ P+. It follows from the bipolar theorem that x0 ∈ X+. Since p ∈ intP+,we deduce 〈p, x0〉 > 0 and t > 0. Set xr = x0/t; then xr ∈ B(p) and 〈p′, xr〉 > 1whenever p′ ∈ intP+ satisfies v(p′) ≤ r. Thus, for every p′ ∈ intP+ satisfying〈p′, xr〉 ≤ 1, we have v(p′) > r so u(xr) ≥ r by the definition (6) of u. That shows(7).

Suppose v has no local minimizer. Let us show that the non satiation conditionis satisfied. If that is not the case, there exists p ∈ intP+ and x ∈ X(p) such that〈p, x〉 < 1. Then, for all p′ in a neighborhood V ⊂ intP+ of p, we have 〈p′, x〉 < 1.Then we get v(p) = u(x) ≤ v(p′) for all p′ ∈ V , i.e., p is a local minimizer, acontradiction.

In the above proposition we can replace the assumptions “v is w∗-evenly qua-siconvex” and “v has no local minima” by the stronger assumptions “v is lowersemicontinuous and quasiconvex” and “v is decreasing along rays”.

Now let us draw some consequences of smoothness properties upon the assump-tions we shall make. First, we observe that the definitions of v and X(p) showthat

v(p) = infv(p′) : p′ ∈ intP+, 〈p′, x〉 ≤ 1

whenever x ∈ X(p) since for p′ ∈ intP+, 〈p′, x〉 ≤ 1 one has x ∈ B(p′), hencev(p′) ≥ u(x) = v(p). Thus, if v is differentiable at p, if x 6= 0 and if p′ ∈ Psatisfies 〈p′, x〉 = 0 one has Dv(p)(p′) = 0, so that there exists µ ∈ R+ such that

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Dv(p) = −µx (that also follows from the Lagrange multiplier rule or the Karush-Kuhn-Tucker theorem and the fact that v is nonincreasing along rays). If moreoverDv(p) 6= 0 and the nonsatiation condition holds, one has µ = 〈Dv(p), p〉 6= 0 andx is uniquely determined in X(p) by x = (1/〈Dv(p), p〉)−1Dv(p). Hence we maywrite µ(p) and x(p) instead of µ and x and

Dv(p) = −µ(p)x(p). (9)

If v is twice differentiable, then µ(·) is differentiable since µ(p) = 〈Dv(p), p〉 andx(·) is differentiable too.

Since µ(p) 6= 0, x(p) = −(1/µ(p))Dv(p) is orthogonal at p to the indifferenceprice manifold defined by

S(p) := p′ ∈ intP+ : v(p′) = v(p).

Now for every p1, p2 ∈ x(p)⊥ := p′ ∈ P : 〈p′,x(p)〉 = 0 we obtain (note thatDv(p) ∈ X, D2v(p) ∈ L(P,X)):

D2v(p)(·) = −Dµ(p)(·)x(p)− µ(p)Dx(p)(·)⇒⟨p2, D

2v(p)p1

⟩= −Dµ(p)(p1) 〈p2,x(p)〉 − µ(p) 〈p2, Dx(p)p1〉 = −µ(p) 〈p2, Dx(p)p1〉

and likewise,⟨p1, D

2v(p)p2

⟩= −µ(p) 〈p1, Dx(p)p2〉. Since D2v(p) is symmetric, we

deduce 〈p2, Dx(p)p1〉 = 〈p1, Dx(p)p2〉 hence Dx(p) is symmetric on the subspacex(p)⊥, provided that Dv(p) 6= 0.

Since v(·) is quasiconvex, for all p ∈ intP+ D2v(p) is positive semidefinite onx(p)⊥ by [11, Thm 2.9]. Thus, as

⟨q,D2v(p)q

⟩= −µ(p) 〈q,Dx(p)q〉 for all q ∈

x(p)⊥, when Dv(p) 6= 0 we get that Dx(p) is negative semidefinite on x(p)⊥.

4. Existence of a utility function

4.1. The standing assumptions on the demand map

Let us now suppose that x : intP+ → X+ is a given map satisfying the followingproperties:

(H1) x : intP+ → X+ is of class C1;(H2) for all p ∈ intP+ one has 〈p,x(p)〉 = 1;(H3) for all p ∈ intP+ the restriction of Dx(p) to the subspace x(p)⊥ orthogonal

to x(p) is symmetric and negative semidefinite.We will show that given a map x(·) satisfying the above conditions, there exists

a utility function u : X+ → R∪−∞ whose demand correspondence X(·) satisfiesX(p) = x(p) for all p ∈ intP+.

Conditions (H2) and (H3) are familiar in the classical demand theory. However,for the sake of completeness, let us discuss them briefly. Here p is an arbitrary fixedelement of intP+. Condition (H2) is obviously the translation of the non satiationcondition. Condition (H3) is a reasonable requirement, as we saw above, and isalso usually expressed ([2], [24], [23], [32]) with the help of the Slutsky operators(p) = Dx(p)− x(p)⊗Dx(p)p ∈ L(P,X), i.e. the linear operator defined by

s(p)p′ = Dx(p)p′ −⟨p′,x(p)

⟩Dx(p)p, ∀p′ ∈ P. (10)

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Obviously s(p) = Dx(p) t(p), where t(p) is given by t(p)(p′) := p′−〈p′,x(p)〉 p.Since t(p)t(p) = t(p) and t(p)(p′) ∈ x(p)⊥, t(p) is a projector on x(p)⊥. We recallthe properties of the Slutsky operator in case (H1), (H2) and (H3) hold:

(i) s(p)p = 0 (obvious from (H2));(ii) for every p′ ∈ x(p)⊥, s(p)p′ = Dx(p)p′ (obvious from (10));(iii) for every p′ ∈ x(p)⊥, 〈p, s(p)p′〉 = 0 (since 〈p,x(p)〉 = 1 implies by differen-

tiation 〈p,Dx(p)(·)〉+ 〈·,x(p)〉 = 0 so that 〈p, s(p)p′〉 = 〈p,Dx(p)p′〉 = 0 whenever〈p′,x(p)〉 = 0).

(iv) s(p) is symmetric if, and only if the restriction of Dx(p) to x(p)⊥ is symmet-ric. In fact, for p′, p′′ ∈ P , setting β′ = 〈p′,x(p)〉 , β′′ = 〈p′′,x(p)〉 and q′ = p′−β′p,q′′ = p′′ − β′′p we have q′, q′′ ∈ x(p)⊥ and q′ = p′ when p′ ∈ x(p)⊥, so that usingsuccessively (i), (iii) and (ii), we get⟨

p′′, s(p)p′⟩

=⟨p′′, s(p)q′

⟩+⟨p′′, β′s(p)p

⟩=⟨p′′, s(p)q′

⟩=⟨q′′, s(p)q′

⟩+ β′′

⟨p, s(p)q′

⟩=⟨q′′, s(p)q′

⟩=⟨q′′, Dx(p)q′

⟩.

(v) s(p) is negative semidefinite on P if, and only if Dx(p) is negative semidefiniteon x(p)⊥. That follows from the preceding equalities in which we take p′ = p′′.

We recall that a function f : U → R of class C1 on an open convex subset U ofP is pseudoconvex if for all u, v ∈ U one has f(v) ≥ f(u) whenever Df(u)(v−u) ≥0. An operator T : U → P ∗ is pseudomonotone if for every p, p′ ∈ U one has〈Tp′, p′ − p〉 ≥ 0 whenever 〈Tp, p′ − p〉 ≥ 0. It is known that f is pseudoconvex if,and only if, Df is pseudomonotone ([11]).

The following result is known in some variants (see for instance [2], [15], [24]).We include a proof for completeness.

Proposition 4.1 If (H1), (H2) and (H3) hold, then −x(·) is pseudomonotone.

Proof. Given p, p′ ∈ intP+, set pt = p+ t(p′−p) and define g(t) = 〈p′ − p,x(pt)〉.We want to show that g(0) ≤ 0⇒ g(1) ≤ 0. That will prove pseudomonotonicity of−x(·). Assume that g(1) > 0. Then there exists t0 ∈ [0, 1) such that g(t0) = 0 andg(t) > 0, ∀t ∈ (t0, 1]. It follows that ln g(t)→ −∞ as t t0 so (ln g(t))′ = g′(t)/g(t)is not bounded from above on (t0, 1]. We know that (H2) and (H3) imply that theSlutsky operator (10) is negative semidefinite. Hence

g′(t) =⟨p′ − p,Dx(pt)(p

′ − p)⟩

=⟨p′ − p, s(pt)(p

′ − p)⟩

+⟨p′ − p,Dx(pt)pt

⟩ ⟨p′ − p,x(pt)

⟩≤⟨p′ − p,Dx(pt)pt

⟩g(t).

Hence g′(t)/g(t) is bounded from above by a bounded continuous function, a con-tradiction.

4.2. A decomposition of the spaces X and P

We will decompose the spaces X and P as follows. We choose a vector e∗ ∈intP+ and then e ∈ X+ such that 〈e∗, e〉 = 1, noting that 〈e∗, x〉 > 0 for allx ∈ X+\0. For X := Rn endowed with the Euclidean norm, we can takee = e∗ = (1/

√n, ..., 1/

√n). Let Y and Q be the subspaces e∗⊥ ⊂ X and e⊥ ⊂ P ,

respectively. Each p ∈ P has the unique decomposition p = q + re∗ with q ∈ Qand r ∈ R. Obviously, r = 〈p, e〉. If in addition p ∈ intP+, then it is known that

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〈p, e〉 > 0, hence r > 0. So P can be written as a direct sum P = Q⊕ Re∗ and wewill frequently write for simplicity p = (q, r) ∈ Q × R. Likewise, each x ∈ X hasthe unique decomposition x = y+ te with y ∈ Y and t ∈ R; here again, t = 〈e∗, x〉and t > 0 for x ∈ X+\0 since e∗ ∈ intP+. Hence again X = Y ⊕Re and we willwrite x = (y, t) ∈ Y ×R. Then for all (y, t) ∈ Y ×R and (q, r) ∈ Q×R the dualityproduct of X and P is given by 〈(q, r), (y, t)〉 = 〈q, y〉+ rt.

We can describe intP+ 6= ∅ with the help of the function r(·) defined on Q by

r(q) := inft ∈ R : q + te∗ ∈ intP+. (11)

That is an infimum over a nonempty subset of R+, since for ε > 0 such thatB(e∗, ε) ⊂ P+ one has e∗ + q/t ∈ intP+ for t > ‖q‖ /ε. Moreover r(tq) = tr(q) fort ∈ R+ and one has

q + re∗ ∈ intP+ ⇔ r > r(q).

We will also obtain an estimate for the components of x = (y, t) when y+te ∈ X+.Since e∗ ∈ intP+, we know that the set

A := x ∈ X+ : 〈e∗, x〉 = 1

is a (closed bounded) base of X+ (cf. Theorem 3.8.4 in [22]); that means that everyx ∈ X+ can be written x = tw with t ∈ R+, w ∈ A in a unique way. For everyx = y + te ∈ X+\0, x/t = x/ 〈e∗, x〉 ∈ A. Thus, y/t ∈ A− e and if c is a boundof the set A− e, then

∀x = (y, t) ∈ X+, t ≥ (1/c) ‖y‖ . (12)

We will use the decompositions X = Y ⊕ R, P = Q ⊕ R to write the demandmap x : intP+ → X+ in the form

x(p) := x(q, r) := (a(q, r), b(q, r)) = (a(p), b(p)) ∈ Y × R.

Note that b(p) = 〈e∗,x(p)〉 > 0 for every p ∈ intP+ since e∗ ∈ intP+, x(p) ∈ P+

and x(p) 6= 0 in view of Assumption (H2).

4.3. Construction of an indirect utility function v

In order to construct the utility function u, we will first construct the indirectutility function v. To this end, given p0 = (q0, r0) ∈ intP+, we will first determinethe indifference price manifold S(p0) containing p0 as the graph of a C1 functions : Q0 → R where Q0 ⊂ Q is open. Bearing in mind relation (9), we will define sso that its graph is orthogonal to x(p) at each point p = (q, s(q)), q ∈ Q0. Sincex(q, s(q)) = (a(q, s(q)), b(q, s(q))), this implies that we should seek s as a solutionof the total differential equation

Ds(q) = −a(q, s(q))

b(q, s(q)), s(q0) = r0. (13)

We will show that a solution of (13) exists; additionally, that its graph (which is tobe interpreted as an indifference price manifold) intersects the half-line ]0,+∞[ e∗

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at one point - or, equivalently, that 0 ∈ Q0. By giving to v a prescribed value on]0,+∞[ e∗, it is thus possible to define v on the graph of s; the value of v on thegraph of s will be constant and equal to the value of v at the intersection of thegraph with ]0,+∞[ e∗. Ultimately this will permit to define v on the whole intP+

since the initial data p0 := (q0, r0) is arbitrary in intP+. Uniqueness of v (up to anarbitrary scalarization function) will ensue from uniqueness of the solution s. Wewill show that this v has good properties, and defines a utility function u for whichx(·) is the demand correspondence.

We define the open convex cone K by

K = (q, r) ∈ Q× R : q + re∗ ∈ intP+ = (q, r) ∈ Q× R : r > r(q)

and the map F : K → Y by

F (q, r) := −a(q, r)

b(q, r). (14)

Proposition 4.2 For every p0 = q0 + r0e∗ ∈ intP+, equation (13) has a unique

solution in a neighborhood of p0.

Proof. The Frobenius Theorem ensures that equation (13) is locally solvable if,and only if, for every (q, r, q1, q2) ∈ K ×Q×Q one has

〈q1, D1F (q, r)q2〉+ 〈q1, 〈q2, F (q, r)〉D2F (q, r)〉

= 〈q2, D1F (q, r)q1〉+ 〈q2, 〈q1, F (q, r)〉D2F (q, r)〉

Omitting the argument p := (q, r), this relation can be written after multiplicationby b2

− b〈q1, D1aq2〉+ 〈D1b, q2〉〈q1, a〉+ 〈q2, a〉〈q1, D2a〉 − b−1〈q2, a〉〈q1, a〉D2b

= −b〈q2, D1aq1〉+ 〈D1b, q1〉〈q2, a〉+ 〈q1, a〉〈q2, D2a〉 − b−1〈q1, a〉〈q2, a〉D2b

or

− b〈q1, D1aq2〉+ 〈D1b, q2〉〈q1, a〉+ 〈q2, a〉〈q1, D2a〉

= −b〈q2, D1aq1〉+ 〈D1b, q1〉〈q2, a〉+ 〈q1, a〉〈q2, D2a〉 (15)

Now, since x(p)⊥ is the set (b(p)q′,−〈q′, a(p)〉) : q′ ∈ Q, the symmetry ofDx(p) on x(p)⊥ can be written (with the coupling of Y × R with Q× R)

〈(bq1,−〈q1, a〉), (Da(bq2,−〈q2, a〉), Db(bq2,−〈q2, a〉))〉

= 〈(bq2,−〈q2, a〉), (Da(bq1,−〈q1, a〉), Db(bq1,−〈q1, a〉))〉

or

〈bq1, bD1aq2 − 〈q2, a〉D2a〉 − 〈q1, a〉(b〈D1b, q2〉 − 〈q2, a〉D2b)

= 〈bq2, bD1aq1 − 〈q1, a〉D2a〉 − 〈q2, a〉(b〈D1b, q1〉 − 〈q1, a〉D2b).

Simplifying again by the term 〈q2, a〉〈q1, a〉D2b, dividing both sides of this relation

9


by b and rearranging terms, we get (15). Thus, equation (13) has a unique solution.

Given p0 = q0 + r0e∗ ∈ intP+, by solution of (13) we will always consider the

uniquely defined solution s with greatest starshaped domain with respect to q0.We now show:

Proposition 4.3 If s : U → R is the solution issued from (q0, r0) ∈ K withlargest domain which is starshaped with respect to q0, then 0 ∈ U .

Proof. Let us denote by T the set of t ∈ [0, 1] such that there exists some ε > 0 forwhich [tq0, q0] + εB ⊂ U, B denoting the open unit ball of Q. Since 1 ∈ T, T is anonempty interval. Let t := inf T, q := tq0. Note that, by (12), for all q ∈ U ,

‖Ds(q)‖ =‖a(q, s(q))‖b(q, s(q))

≤ c (16)

since a(p) + b(p)e ∈ X+. Consider a sequence (tk) t and set qk = tkq0 for k ∈ N.It follows from (16) and the mean value theorem that the sequence (s(qk)) is aCauchy sequence. Let us denote by s(q) the limit of this sequence. Define r(q) asin (11). Recall that for all (q, r) ∈ intP+, (a(q, r), b(q, r)) = x(q, r) ∈ X+ and(q, r(q)) ∈ P+ so 〈(a(q, r), b(q, r)), (q, r(q))〉 ≥ 0. Hence for t ∈ T ,

d

dt(s(tq0)− r(tq0)) =

d

dt(s(tq0)− tr(q0)) = 〈Ds(tq0), q0〉 − r(q0)

= −〈a(tq0, s(tq0)), tq0〉+ b(tq0, s(tq0))r(tq0)

tb(tq0, s(tq0))≤ 0.

Thus, s(tq0)− r(tq0) is nonincreasing as a function of t. By continuity, s(tq0)−r(tq0) ≥ s(q0)− r(q0) > 0 since (q0, s(q0)) ∈ intP+. Hence (q, s(q)) ∈ intP+.

Let us prove that t = 0 and T = [0, 1]. By the Frobenius Theorem, there existsa ball B(q, α) and a C1 map s : B(q, α)×B(q, α)×]t− α, t+ α[→ R such that forall (q, q′, r′) ∈ B(q, α)×B(q, α)×]t− α, t+ α[, one has

D1s(q, q′, r′) = −a(q, s(q, q′, r′))

b(q, s(q, q′, r′)), s(q′, q′, r′) = r′.

Choose t′ ∈]t, t+ α[ such that q′ := t′q0 ∈ B(q, α). Let ε be given by the definitionof T : [t′q0, q0] + εB ⊂ U. Define s1 : B(q, α) → R by s1(q) = s(q, q′, s(q′)). Thens1(q′) = s(q′) and C := B(q, α) ∩ ([q′, q0] + εB) is convex, hence connected. Sincethe set of coincidence is open by the Frobenius theorem and closed in C, s1 = son C. Taking β > 0 smaller than α and ε, we can find t1 ∈]t − α, t[ such that[t1q0, q0] + βB ⊂ U . It follows that t belongs to T . In addition, we have t = 0 sinceotherwise we can take 0 < t1 < t and arrive to a contradiction. Hence, U containssome open set of the form [0, q0] + εB for some ε > 0. In particular, 0 ∈ U.

Given (q, r) ∈ intP+, let us denote by s(·, q, r) the solution to equation (13)issued from (q0, r0) := (q, r). Since F is of class C1, s is of class C1. By Proposition4.3, s(0, q, r) is defined for all (q, r) ∈ intP+.

Given any smooth decreasing function h :]0,+∞[→ R satisfying h′(t) < 0 for allt ∈]0,+∞[ and given p := q + re∗ ∈ intP+, we set

v(p) := h(s(0, q, r)). (17)

10


Then, since s(0, 0, r) = r for all r > 0 by construction of s, one gets v(re) = h(r)for all r > 0. The function v thus defined is of class C1 and is constant on the graphof s(·, q, r) because for all q′ in the domain of s(·, q, r), setting r′ := s(q′, q, r) onehas s(q′, q′, r′) = r′ = s(q′, q, r), hence, by uniqueness, s(0, q′, r′) = s(0, q, r). Notethat the function s is uniquely defined but v is not, due to the arbitrariness in thechoice of h. However, once the values of v along ]0,+∞[ e∗ have been assigned, vis uniquely determined.

4.4. Properties of the function v

Since v is constant on the graph of s and since s(q, q, r) = r for every (q, r) ∈ K,one has

D1v(q, r) +D2v(q, r)D1s(q, q, r) = 0

or

D1v(q, r)−D2v(q, r)a(q, r)/b(q, r) = 0.

Setting µ(q, r) = −D2v(q, r)/b(q, r) it follows that (D1v(q, r), D2v(q, r)) =−µ(q, r)(a(q, r), b(q, r)) or

Dv(p) = −µ(p)x(p). (18)

Since x(·) satisfies the condition 〈p,x(p)〉 = 1, one has µ(p) = −〈p,Dv(p)〉. More-over, since for all r > 0 one has v(re∗) = h(s(0, 0, r)) = h(r), one gets

µ(re∗) = −〈re∗, Dv(re∗)〉 = − limt→0+

v(re∗ + tre∗)− v(re∗)

t

= − limt→0+

h(r + tr)− h(r)

t= −rh′(r) > 0.

Let us show more.

Lemma 4.4 For all p ∈ intP+ one has Dv(p) 6= 0, µ(p) > 0 and in fact〈e∗, Dv(p)〉 < 0.

Proof. Suppose that 〈e∗, Dv(p0)〉 = 0 for some p0 := q0 + r0e∗ ∈ intP+.

Then, since v(q, r) = h(s(0, q, r)), the chain rule yields 〈e∗, Dv(p0)〉 =h′(s(0, q0, r0))D3s(0, q0, r0) and since h′(s(0, q0, r0)) < 0, one gets D3s(0, q0, r0) =0. Now, setting S(t) := s(q0 − tq0, q0, r0), since S(·) is the solution to the ordinarydifferential equation

S′(t) = −F (q0 − tq0, S(t)).q0, S(0) = r0,

the classical result about the derivative with respect to the initial data for ordi-nary differential equations ensures that the function k given by k(t) := D3s(q0 −tq0, q0, r0) is the solution to the linear differential equation

k′(t) = −D2F (q0 − tq0, S(t))k(t) k(0) = 1,

11


where S is defined as above. Since this equation has a unique solution, if k(1) = 0,one has k(t) = 0 for all t, a contradiction with k(0) = 1. Thus 〈e∗, Dv(p0)〉 6= 0,hence by (18), µ(p0) 6= 0. Since µ(re∗) > 0 and intP+ is connected, one hasµ(p) > 0 for all p ∈ intP+.

4.5. Construction of a direct utility function

Now a direct utility function can be constructed using Proposition 3.1.

Theorem 4.5 Given a map x(·) : intP+ → X+ satisfying (H1), (H2) and (H3),there exists a quasiconcave, upper semicontinuous utility function u : X+ → Rsatisfying the non satiation condition, such that for all p ∈ intP+, the demand setX(p) associated to u is x(p). Moreover the associated indirect utility function v isreal-valued, differentiable, decreasing along rays, nonincreasing and pseudoconvex.

Proof. We define v by the construction described above. We saw that v is real-valued, (18) holds with µ(p) > 0 for all p ∈ intP+, and −x(·) is pseudomonotone.Thus Dv(·) = −µ(·)x(·) is pseudomonotone. It follows that v is pseudoconvex (seefor instance [11]). In addition, for all p, p′ ∈ intP+,⟨

p′, Dv(p)⟩

= −µ(p)⟨p′,x(p)

⟩< 0

as −µ(p) < 0 and 〈p′,x(p)〉 > 0. Thus t 7→ v(p+ tp′) is decreasing and v(p+ p′) <v(p). In particular, v is decreasing along rays. Moreover, if p ∈ intP+, p′ ∈ P+,taking a sequence (p′n) in intP+ with limit p′, we get v(p + p′) ≤ v(p) since v iscontinuous and even of class C1.

Let us define u : X+ → R ∪ +∞ by

u(x) := infv(p) : p ∈ intP+, 〈p, x〉 ≤ 1. (19)

By Proposition 3.1 or [31], [36] v is the indirect utility function associated with u,i.e., for all p ∈ intP+

v(p) = supu(x) : x ∈ X+, 〈p, x〉 ≤ 1. (20)

Given p ∈ intP+, let us show that u(x(p)) = v(p). Since 〈p,x(p)〉 = 1, we shall getthat x(p) is a maximizer of u on B(p) := x′ ∈ X+ : 〈p, x′〉 ≤ 1, i.e. x(p) ∈ X(p).

Now, for all p′ ∈ intP+ satisfying 〈p′,x(p)〉 ≤ 1 we have 〈p′ − p,x(p)〉 ≤ 0, hence⟨p′ − p,Dv(p)

⟩= −µ(p)

⟨p′ − p,x(p)

⟩≥ 0.

Since v is pseudoconvex, we get v(p′) ≥ v(p). Then the definition (19) of u showsthat u(x(p)) ≥ v(p) and since the reverse inequality holds because v satisfies (20)and x(p) ∈ B(p), we get u(x(p)) = v(p).

Since v is pseudoconvex, it is also quasiconvex. In addition it is continuous anddecreasing along rays, hence the non satiation condition holds by Proposition 3.1.Changing signs in ([35, Lemma 1]), we see that u is upper semicontinuous since vis continuous.

If x′ ∈ X(p) then v(p) = u(x′), so that by (19),

v(p) = minv(p′) : p′ ∈ intP+,⟨p′, x′

⟩≤ 1,

12


i.e., p minimizes v under the condition 〈p, x′〉 ≤ 1. Since v is smooth, there exists amultiplier λ ≥ 0 such that Dv(p) = −λx′. Here we use the fact that x′ 6= 0 by nonsatiation in view of Proposition 3.1. Thus, using the relations Dv(p) = −µ(p)x(p),Dv(p) = −λx′ and 〈p, x′〉 = 〈p,x(p)〉 = 1, we deduce x′ = x(p). Thus X(p) =x(p).

5. Examples

Let us illustrate our construction with some examples.Example. Let X be a Banach space, let X+ be a closed convex cone in X and lete ∈ X+. It is easy to check that the map x(·) defined by

x(p) :=e

〈p, e〉(p ∈ intP+) (21)

satisfies conditions (H1)-(H2)-(H3). With the preceding notation one has x(p) :=x(q, r) := (a(q, r), b(q, r)) = (0, 1/(〈p, e〉)). Thus, the solution to equation (13) iss(·, q, r) = r. Taking h(r) := 1/r for r ∈]0,+∞[, we get v(p) = 1/〈p, e〉. It can beshown that the utility function u deduced from v is given by

u(x) = supt ∈ R+ : te ≤ x x ∈ X+.

As a particular case, we consider X := Rn, X+ := Rn+ and take e := (c1, ..., cn).Then the utility function u can be given the expression

u(x1, ..., xn) := min(x1

c1, ...,

xncn

),

a nonsmooth utility function. In fact, for all p ∈ intP+ we have u(x(p)) = 1/〈p, e〉and if u(x1, ..., xn) > 1/〈p, e〉 we have xi > ci/〈p, e〉 for all i = 1, ..., n, hence〈p, x〉 > (p1c1 + ... + pncn)/〈p, e〉 = 1 and x /∈ B(p). Thus, for all x ∈ B(p) wehave u(x) ≤ 1/〈p, e〉 and v(p) = 1/〈p, e〉. For c1 = ... = cn = 1 we get the Leontieffutility function.

Another particular case is X := L1(Ω), where Ω is a probability space, P :=L∞(Ω), 〈p, x〉 :=

∫Ω p(ω)x(ω)dω, X+ := f ∈ X : f(ω) ≥ 0 a.e.. The map given

by (21) again satisfies conditions (H1)-(H2)-(H3). With the choice h(r) := 1/r againwe get v(p) = 1/〈p, e〉. If e(ω) = 1 for all ω ∈ Ω we get u(x) = ess− infωx(ω).Example. Now let us consider the case X := R2, X+ := R2

+, α, β ∈]0, 1[ withα+ β = 1, e := (1, 1), e∗ := (α, β), x(·) being given by

x(p) := (α

p1,β

p2) p := (p1, p2) ∈ R2

+.

Given p := (p1, p2) ∈ intP+, let us find a curve t 7→ (t, s(t)) along which vis constant and such that s(p1) = p2. The relation 0 = Dv(t, s(t))(1, s′(t)) =µ(t, s(t))〈(α/t, β/s(t)), (1, s′(t))〉 yields s′(t)/s(t) = −α/(βt), hence ln s(t) =

ln ct−α/β and s(t) = ct−α/β. The initial condition s(p1) = p2 yields c = pα/β1 p2.

The curve intersects the half-line R+e∗ at a point re∗ such that s(rα) = rβ;

hence, cα−α/βr−α/β = rβ, so that r1/β = cβ−1α−α/β and r = α−αβ−βpα1 pβ2 .

Using the function h : r 7→ 1/r we define v at (p1, p2) by v(p1, p2) = v(re∗) =

h(r) = ααββ/pα1 pβ2 . It corresponds to the Cobb-Douglas utility function u given

13


by u(x1, x2) = xα1xβ2 . This example can be generalized to Rn, replacing (α, β) with

(α1, ..., αn) with α1 > 0,...,αn > 0, α1 + ...+ αn = 1.

Acknowledgement

The authors are grateful to Ivar Ekeland for providing some references that weremissing in the preliminary version of the paper we disseminated and for somecomments on that version.

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Revisiting the problem of integrability in utility theory

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