LNA

ISSN 2188-8167

2021

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YUHKI HOSOYA

model, introduced by Ramsey [21] and improved by Cass [3] and Koopmans [16],can be made into a “reduced form” that was treated by Boldrin and Montrucchio [1],but the converse is not true. Thus, Boldrin-Montrucchio’s theorem does not imply
the existence of a Ramsey-Cass-Koopmans model that produces chaos. Hosoya [8]presented a method of fully characterizing the class of policy functions that canbe derived from a Ramsey-Cass-Koopmans model and calculating the objectivefunction backwards from the policy function. This is an example of an inverse
problem that requires a “form”.
A more extreme example is the integrability problem in consumer theory. The
purpose of this problem is to calculate the consumer’s preference backwards from
his/her behavior, and the “uniqueness” of the corresponding objective function isthus needed. Mas-Colell [17] gave a classical result for this uniqueness requirement,
and this result has recently been extended by Hosoya [11]. The existence anduniqueness of the objective function have been treated simultaneously by Hosoya[7, 9].
The present paper deals with a branch of consumer theory called revealed preference theory. This is a research area that asks whether a candidate of the solution function for the consumer’s behavior can actually be represented as a solutionfunction for the consumer’s optimization problem. The weak axiom of revealed
preference proposed by Samuelson [23] is a straightforward necessary condition fora candidate of a solution function to be a solution function. However, Gale [5]showed that it is not suﬃcient. Houthakker [12] developed the strong axiom of
revealed preference based on the idea of the Euler approximation of diﬀerential
equations, and Richter [22] proved that this axiom is an appropriate necessary andsuﬃcient condition for the present problem.
Meanwhile, consumer theory is a part of general equilibrium theory. In thisregard, recent general equilibrium theory has evolved in the direction of dealing
with preferences of a form that cannot be represented by some “objective function”.
This is because the “negative transitivity” of the strong order is not desirable foran axiom of the consumer’s preference. For this reason, the theory of consumer
behavior in which the preference is described by a binary relation that does notsatisfy transitivity, and the theory of equilibrium that includes it, has developed.
For example, Hildenbrand [6], Shafer and Sonnenschein [24], and Mas-Colell [18]are typical studies on this issue. The problem here is the inverse problem of that.
What kind of demand behavior can be handled by such a model?
The strong axiom of revealed preference is undesirable as an assumption, because
it is too strong and can only handle solution functions that can be expressed byan ordinary maximization of a transitive preference. Let us now turn our attention
to the weak axiom of revealed preference. The problem of expressing a candidateof a solution function satisfying this axiom by a result of maximizing some (possibly non-transitive) binary relation was studied by Kim and Richter [13] and Quah[20]. However, the orders they created are unnatural. In particular, if we take thesolution function from an ordinary maximization problem and calculate the binary
relation that appears in their research, then we obtain binary relations that are notWA AND INVERSE PROBLEMS IN CONSUMER THEORY

21transitive. In short, their research focuses only on “existence”, and “uniqueness” isalmost completely compromised.
Therefore, in this paper, we derive a natural binary relation from a candidate of
the solution function that satisﬁes the weak axiom of revealed preference and showthat this candidate is actually the solution function of the optimization problem ofthis binary relation (Theorem 2.4). In fact, under certain conditions, the binary
relation that we have derived is the “unique” corresponding binary relation thatsatisﬁes several requirements (Proposition 2.8).
In Section 2.1, we explain standard concepts and terminology used in consumer
theory. Section 2.2 is devoted to the proof of the main result and an explanation ofrelated open problems.
2. Model and Results
2.1. Preliminaries. We ﬁx n ≥ 2. Let Rn+ = {x ∈ Rn |xi ≥ 0 for all i} andRn++ = {x ∈ Rn |xi > 0 for all i}. For vectors x, y ∈ Rn , we write x ≥ y if x−y ∈ Rn+
and x  y if x − y ∈ Rn++ .
A set Ω ⊂ Rn is called the consumption set, which denotes the set of all possible
consumption plans. For a plan x ∈ Ω, xi denotes the amount of consumption of
the i-th commodity. By this interpretation, xi ≥ 0 is usually assumed. Thus,
throughout this paper, we assume that Ω ⊂ Rn+ . We also assume that Ω is nonempty
and convex, and that Ω = Ω + Rn+ : that is, x ∈ Ω and y ≥ x implies y ∈ Ω.
Let a vector p  0 denote the price system. Then, pi denotes the price of the i-thcommodity. The consumer’s money is denoted by m > 0. Therefore, the consumer’s
behavior is represented by a function f (p, m), where the domain P of f is a subset
of Rn++ × R++ . Deﬁne∆(p, m) = {x ∈ Ω|p · x ≤ m},
and P = Rn++ × R++ \ ∆−1 (∅). We call a function f : P → Ω a candidate of
demand (CoD) if p · f (p, m) ≤ m for all (p, m) ∈ P . We consider that this function
f represents the consumer’s choice behavior. If p · f (p, m) = m for all (p, m) ∈ P ,then f is said to satisfy Walras’ law. Meanwhile, if f (p, m) = f (ap, am) for all(p, m) ∈ P and a > 0, then f is said to be homogeneous of degree zero. We saythat a CoD f is income-Lipschitzian if f is locally Lipschitz in m: that is, forevery compact subset C ⊂ P , there exists L > 0 such that if (p, m1 ), (p, m2 ) ∈ C,then
kf (p, m1 ) − f (p, m2 )k ≤ L|m1 − m2 |.
Let R(f ) denote the range of f .
Next, consider a binary relation ≿ on Ω: that is, ≿⊂ Ω2 . We write x ≿ y insteadof (x, y) ∈≿. Consequently, x 6≿ y means (x, y) ∈≿.
/We also write x  y if x ≿ y
and y 6≿ x, and x ∼ y if x ≿ y and y ≿ x.
For each (p, m) ∈ P , deﬁnef ≿ (p, m) = {x ∈ ∆(p, m)|∀y ∈ ∆(p, m), y 6 x}.

22

YUHKI HOSOYAThis multi-valued function f ≿ : P ↠ Ω is called the demand function corresponds
to ≿. Note that every single-valued demand function is a CoD that is homogeneousof degree zero. We say that a CoD f corresponds to ≿, or equivalently, ≿ corresponds to f when f = f ≿ .
Suppose that u : Ω → R satisﬁes the following relationship:u(x) ≥ u(y) ⇔ x ≿ y.
Then, we say that u represents ≿, or equivalently, u is a utility function of thisrelation ≿. If u represents ≿, then we sometimes write f u instead of f ≿ , and saythat this function f u is the demand function corresponds to u. Note that, f u is thesolution function of the following optimization problem:
max u(x)(2.1)

subject to. x ∈ Ω,
p · x ≤ m.

In economics, this optimization problem is called the utility maximization problem.
We say that a binary relation ≿ on Ω is
• complete if either x ≿ y or y ≿ x for all (x, y) ∈ Ω2 ,• transitive if x ≿ y and y ≿ z imply x ≿ z,
• p-transitive if dim(span{x, y, z}) ≤ 2, x ≿ y, and y ≿ z imply x ≿ z,
• asymmetric if x ≿ y implies y 6≿ x,• monotone if x  y implies x  y,• convex if y ≿ x and 0 < t < 1 imply (1 − t)x + ty ≿ x,
• strictly convex if y 6= x, y ≿ x and 0 < t < 1 imply (1 − t)x + ty  x,• upper semi-continuous if the set U (x) = {y ∈ Ω|y ≿ x} is closed withrespect to the relative topology of Ω for all x, and• continuous if ≿ itself is closed with respect to the relative topology of Ω2 .
Note that, if u represents ≿, then ≿ must be complete and transitive. Conversely,
if ≿ is complete, transitive, and upper semi-continuous, then there exists an uppersemi-continuous function u that represents ≿. If, in addition, ≿ is continuous, thenwe can construct such u as a continuous function.1 Consequently, if Ω is closed and≿ is complete, transitive, and upper semi-continuous, then f ≿ is nonempty-valued
and homogeneous of degree zero. If, in addition, ≿ is monotone, then we can easily
show that f ≿ satisﬁes Walras’ law. Furthermore, if ≿ is strictly convex, then we canalso easily show that f ≿ is single-valued, and thus it is a CoD that is homogeneousof degree zero. In conclusion, we veriﬁed that if Ω is closed and ≿ is complete,transitive, upper semi-continuous, monotone, and strictly convex, then f ≿ is a CoD
that satisﬁes Walras’ law and homogeneity of degree zero.
Note that if ≿ is transitive, then either x ≿ y  z or x  y ≿ z implies x  z.
The proof of this fact is easy.
1See Debreu [4] or Bridges and Mehta [2]. Note that our Ω is convex, and thus it is connected.

WA AND INVERSE PROBLEMS IN CONSUMER THEORY

23

Next, we introduce axioms of revealed preference. We say that a CoD f satisﬁes
the weak axiom of revealed preference (WA) if for every (p, m), (q, w) ∈ P ,if p · f (q, w) ≤ m, then either f (p, m) = f (q, w) or q · f (p, m) > w. We also saythat f satisﬁes the strong axiom of revealed preference (SA) if for every ﬁnite
sequence (p1 , m1 ), ..., (pN , mN ), if pi ·f (pi+1 , mi+1 ) ≤ mi for every i ∈ {1, ..., N −1},then either f (pN , mN ) = f (p1 , m1 ) or pN ·f (p1 , m1 ) > mN . We can easily check thatSA implies WA. Meanwhile, Gale [5] provided an example of a CoD that satisﬁes
WA but violates SA. Therefore, WA does not imply SA.2
The interpretation of WA is as follows. Suppose that x = f (p, m), y = f (q, w)
and x 6= y. If p · y ≤ m, then the consumer chooses x though he/she can choose y,
and thus it is revealed that the consumer prefers x to y. Therefore, we can expectq · x > w: if not, then f (q, w) must not be y.
2.2. Results. The following result is known as Houthakker-Uzawa-Richter’s theorem. The proof of this proposition is found in Theorem 3.J.1 of Mas-Colell et al.
[19] or Theorem 3.2 of Hosoya [10].
Proposition 2.1. A CoD f satisﬁes SA if and only if f = f ≿ for some complete
and transitive binary relation ≿.
Using this result, we can obtain the following uniqueness result. Recall that R(f )denotes the range of f . For the rigorous proof of this corollary, see Theorem 1 of
Hosoya [11].
Corollary 2.2. Suppose that Ω = Rn+ and P = Rn++ × R++ . Let f be a continuous
and income-Lipschitzian CoD that satisﬁes Walras’ law, and R(f ) be open withrespect to the relative topology of Ω and include Rn++ . Then, f satisﬁes SA if andonly if f = f u for some upper semi-continuous real-valued function u. Moreover,
if f = f ≿ for some complete, transitive, and upper semi-continuous binary relation≿, then u represents ≿.
Remark 2.3. We can obtain u constructively as follows. First, ﬁx p̄ ∈ Rn++ . Choose
any x ∈ R(f ), and let x = f (p, m). Consider the following partial diﬀerential
equation:(2.2)∇E(q) = f (q, E(q)), E(p) = m.

We can show that there uniquely exists a concave solution E : Rn++ → R++ of thisequation (2.2). Deﬁne u(x) = E(p̄). Next, choose x ∈ Ω \ R(f ). Deﬁneu(x) = lim sup{u(y)|y ∈ R(f ), kx − yk < δ}.
δ→0Then, we can show that u satisﬁes all requirements of the above corollary.
In fact, the solution of (2.2) can be described using a diﬀerent feature. Because fsatisﬁes SA, we have that f = f ≿ for some complete and transitive binary relation≿. LetE x (q) = inf{q · y|y ≿ x}.
2For a detailed arguments, see section 4.2 of Hosoya [10].

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YUHKI HOSOYA

If ≿ is represented by v, then it is the value function of a dual problem of (2.1):
min q · y
(2.3)

subject to. y ∈ Ω,v(y) ≥ v(x).

This problem (2.3) is called the expenditure minimization problem, and thefunction E x is called the expenditure function. Lemma 1 of Hosoya [11] showedthat if x = f (p, m), then E x is a concave solution of (2.2). This result is called
Shephard’s lemma in economics.
We want to extend the above results to CoDs with WA. Before presenting thenext result, we need several preparations. First, let ▷ be any binary relation on Ω,
and P denote the set of all p-transitive binary relations that includes ▷. BecauseΩ2 ∈ P, we have that P is nonempty. It is easy to show that▷∗ = ∩▷′ ∈P ▷′ ∈ P.
We call this ▷∗ the p-transitive closure of ▷.
We write x r y if and only if there exists (p, m) ∈ P such that x = f (p, m), y ∈∆(p, m) and x 6= y. Then, WA is equivalent to the asymmetry of r . Let irp bethe p-transitive closure of r , and x ≿f y if and only if y 6irp x. We now complete
the preparation to present our main result.
Theorem 2.4. If a CoD f satisﬁes Walras’ law and WA, then ≿f is complete andf = f ≿f .
Proof. First, we introduce a lemma.
Lemma 2.5. Suppose that f is a CoD that satisﬁes Walras’ law. Then, r is
asymmetric if and only if irp is asymmetric.3
Proof of Lemma 2.5. Clearly, if irp is asymmetric, then r is asymmetric because r ⊂irp . Thus, it suﬃces to show the opposite direction.
First, suppose that r is asymmetric, and there exists a sequence x1 , ..., xk ∈ Ωsuch that dim(span{x1 , ..., xk }) ≤ 2 and xi r xi+1 for all i ∈ {1, ..., k − 1}. We willshow that xk 6r x1 . We use mathematical induction on k. If k = 2, this claim iscorrect because r is asymmetric.
Suppose that our claim holds if k ≤ k ∗ for k ∗ ≥ 2, and consider the case k =∗k + 1. Choose any sequence x1 , ..., xk such that dim(span{x1 , ..., xk }) ≤ 2 andfor all i ∈ {1, ..., k − 1}, xi r xi+1 . Then, xi 6= xi+1 and there exists (pi , mi )such that xi = f (pi , mi ) and pi · xi+1 ≤ mi . Suppose that xk r x1 . Then,xk 6= x1 and there exists (pk , mk ) such that xk = f (pk , mk ) and pk · x1 ≤ mk .
Deﬁne V = span{x1 , ..., xk }. If dim V = 1, then xi = ci x1 for ci ∈]0, 1], and
1 = c1 > c2 > ... > ck . Therefore,
pk · x1 > pk · ck x1 = pk · xk = mk ,3If Ω is Rn , then this result is just Theorem 4.1 of Hosoya [10]. However, the proof of this+theorem in that paper has a slight error.

WA AND INVERSE PROBLEMS IN CONSUMER THEORY

25

which is a contradiction. Thus, we have that dim V = 2. Let PV be the orthogonal
projection mapping from Rn onto V . By deﬁnition of the orthogonal projection
mapping, we have that PV pi · xj = pi · xj for i, j ∈ {1, ..., k}. Deﬁne qi = pi1·x1 PV pi .
Then, we have qi · x1 = 1 for all i ∈ {1, ..., k}, and thus all qi are included in theline {q ∈ V |q · x1 = 1}.
We separate our proof into three cases.
Case 1. q1 ∈ [qk−1 , qk ]. In this case, q1 = (1 − t)qk−1 + tqk for some t ∈ [0, 1].
Therefore,
q1 · (x1 − xk ) = q1 · (x1 − x2 ) + q1 · (x2 − xk )= q1 · (x1 − x2 ) + (1 − t)qk−1 · (x2 − xk ) + tqk · (x2 − xk )= q1 · (x1 − x2 ) + (1 − t)qk−1 · (x2 − xk−1 )+ (1 − t)qk−1 · (xk−1 − xk ) + tqk · (x2 − xk ) ≥ 0,
where the last inequality follows from the induction hypothesis. Therefore, we havethat x1 r xk , which contradicts the asymmetry of r .
Case 2. qk−1 ∈ [q1 , qk ]. In this case, qk−1 = (1 − t)q1 + tqk for some t ∈ [0, 1].
Therefore,(1 − t)q1 · (x1 − xk ) + tqk · (x1 − xk )= qk−1 · (x1 − xk )= qk−1 · (x1 − xk−1 ) + qk−1 · (xk−1 − xk ) > 0.
Because qk · (x1 − xk ) ≤ 0, we have (1 − t)q1 · (x1 − xk ) > 0. Therefore, we havex1 r xk , which is absurd.
Case 3. The other case. Deﬁne v = q1 − qk . The case in which v = 0 is included
in Case 1, and thus we have v 6= 0 and qi = qk + ti v for some ti ∈ R. By deﬁnition,
t1 = 1. The case tk−1 ≥ 0 is included in either Case 1 or Case 2, and thus we havetk−1 < 0. Therefore, we must have that there exists i ∈ {1, ..., k − 2} such thatti ≥ 0 and ti+1 ≤ 0, and thus, qk ∈ [qi , qi+1 ]. Then, qk = (1 − t)qi + tqi+1 for somet ∈ [0, 1], and
0 ≥ qk · (x1 − xk )= qk · (x1 − xi+1 ) + qk · (xi+1 − xk )= (1 − t)qi · (x1 − xi+1 ) + tqi+1 · (x1 − xi+1 ) + qk · (xi+1 − xk )= (1 − t)qi · (x1 − xi ) + (1 − t)qi · (xi − xi+1 )+ tqi+1 · (x1 − xi+1 ) + qk · (xi+1 − xk ) > 0,by the induction hypothesis, which is a contradiction.
Therefore, in all cases there is a contradiction. Thus, we conclude that xk 6r x1 ,and hence our claim is correct.

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YUHKI HOSOYA

Second, suppose that r is asymmetric. We will show that x irp y if and only ifeither x ∈ R(f ) and y = ax for a ∈ [0, 1[ or x is not proportional to y and there exists
a ﬁnite sequence x1 , ..., xN such that x1 = x, xN = y, span{x1 , ..., xN } = span{x, y},and xi r xi+1 for every i ∈ {1, ..., N − 1}. Let R be the set of all (x, y) such that 1)x ∈ R(f ) and y = ax for some a ∈ [0, 1[, or 2) x is not proportional to y and there
exists a ﬁnite sequence x1 , ..., xN such that x = x1 , y = xN , span{x1 , ..., xN } =span{x, y}, and xi r xi+1 for all i ∈ {1, ..., N − 1}. By Walras’ law, we have thatr ⊂ R. Choose x, y, z ∈ Ω such that (x, y), (y, z) ∈ R and dim(span{x, y, z}) ≤ 2.
If y = ax or z = by for a, b ∈ [0, 1[, then we can easily show that (x, z) ∈ R.
Suppose that x is not proportional to y and z = ax for a ≥ 0. Then, x ∈ R(f ).
If a ≥ 1, then we have that there exists a ﬁnite sequence x1 , ..., xN on span{x, y}such that x1 = x, xi r xi+1 for all i ∈ {1, ..., N − 1}, and xN r x1 , which is acontradiction. Therefore, we have a < 1, and thus (x, z) ∈ R. Finally, if each pairof x, y, z is independent, then clearly (x, z) ∈ R. Therefore, R is p-transitive, andthus irp ⊂ R. Conversely, suppose that (x, y) ∈ R. If x is proportional to y, thenx r y, and thus x irp y. If x is not proportional to y, then there exists a ﬁnite
sequence x1 , ..., xN such that x = x1 , y = xN , span{x1 , ..., xN } = span{x, y}, andxi r xi+1 for all i ∈ {1, ..., N − 1}. Because r ⊂irp , we have that xi irp xi+1 .
By p-transitivity of irp , we have that x = x1 irp xN = y, which implies thatR ⊂irp . Thus, our claim is correct.
Now, suppose that r is asymmetric, and irp is not asymmetric. Then, there
exists x, y such that x irp y and y irp x. This implies that x irp x, which
contradicts the arguments in the above paragraph. Therefore, we conclude that if
r is asymmetric, then irp is also asymmetric. This completes the proof of thislemma.
□
This lemma immediately implies that ≿f is complete. Next, suppose that x =f (p, m) and p · y ≤ m. If x = y, then clearly y 6irp x, and thus y 6f x. If x 6= y,then x r y, and by the above lemma, we have that y 6irp x, which implies thaty 6f x and x f y. Therefore, we have that x = f ≿f (p, m) as desired. This
completes the proof.
□
We can obtain the following corollary.
Corollary 2.6. Suppose that Ω = Rn++ and f = f u for some C 2 strictly quasiconcave function u such that ∇u(x)  0 for all x ∈ Ω. Then, u represents ≿f .4
Proof. First, we show that x f v implies that u(x) > u(v). Suppose that x f v.
As we have proved in the proof of Theorem 2.4, either x = av for a > 1 or there exists
x1 , ..., xN such that x1 = x, xN = v, and xi r xi+1 for all i ∈ {1, ..., N }. In theformer case, clearly u(x) > u(v). In the latter case, we have that u(xi ) > u(xi+1 ),and thus u(x) > u(v). Therefore, our claim is correct.
Second, we show the opposite direction: that is, we show that u(x) > u(v) implies
that x f v. Deﬁne p(x) = ∇u(x). By Lagrange’s multiplier rule, we have thatx = f (p(x), p(x) · x) for all x ∈ Ω.
4Note that, by the requirement of u, f satisﬁes SA and Walras’ law.

WA AND INVERSE PROBLEMS IN CONSUMER THEORY

27

Suppose that u(x) > u(v). If v is proportional to x, then x  v and thus clearlyx f v. Therefore, we assume that v is not proportional to x. Deﬁneξ(y) = (p(y) · x)v − (p(y) · v)x.
Then, p(y) · ξ(y) = 0 for all y ∈ Ω. Consider the following diﬀerential equation:ẏ(t) = ξ(y), y(0) = x.
Let y(t) be the nonextendable solution of the above equation. Hosoya [7] showedthat there exists t∗ > 0 such that y(t∗ ) is proportional to v. Becausedu(y(t)) = 0,dt
we have that u(y(t∗ )) = u(x), which implies that y(t∗ )  v. Now, for small h > 0,deﬁne
x0 = x, xi+1 = xi + hξ(xi ).
Then, the sequence (xi ) is a forward Euler approximated solution of the above
diﬀerential equation, and thus for i(h) = max{i ≥ 0|ih < t∗ }, xi(h) → y(t∗ ) as
h → 0. This implies that if h is suﬃciently small, then there exists k such thatxk  v. Now, because p(xi ) · xi+1 = p(xi ) · xi , we have that xi r xi+1 . Moreover,span{x0 , ..., xk } = span{x, v}. Therefore, we have that x irp xk and xk r v,
which implies that x f v. This completes the proof.
□
To the best of our knowledge, there is no other previous research that has derived
a binary relation with such properties in this context. For example, binary relations
deﬁned by Kim and Richter [13] and Quah [20] do not have this property. This is
the main reason why we consider ≿f is so natural.
Note that the converse relationship of Theorem 2.4 is open: that is, whether or
not there is a demand function f ≿ such that ≿ is complete but f ≿ violates WA is
unknown.
We should mention another open problem related to Theorem 2.4. First, supposethat Ω is closed. As we have argued, if f = f ≿ for some complete, transitive, uppersemi-continuous, monotone, and strictly convex binary relation, then f is a singlevalued demand function that satisﬁes Walras’ law, homogeneity of degree zero, and
SA. Conversely, suppose that f is a continuous and income-Lipschitzian CoD suchthat R(f ) is relatively open subset of Rn+ that includes Rn++ . If f satisﬁes Walras’law, homogeneity of degree zero, and SA, then by Corollary 2.2, we have thatf = f ≿ for some complete, transitive, and upper semi-continuous binary relation≿. Therefore, we obtain an “almost” necessary and suﬃcient relationship between
the existence of a good ≿ corresponds to f and SA. Moreover, such a binary relationis unique. We want to extend this result for the case in which f satisﬁes only WA.
However, there is a technical diﬃculty, and this problem is still open.
In fact, the following result was obtained by Sonnenschein [25].
Proposition 2.7. Suppose that Ω is closed, and ≿ is a complete, upper semicontinuous, and strictly convex binary relation on Ω such that for every x ∈ Ω, the28

YUHKI HOSOYA

set
P (x) = {y ∈ Ω|y  x}is convex. Then, f ≿ is a single-valued demand function deﬁned on P .
In Theorem 2.4, we proved that under several conditions, a CoD f is f ≿ for somecomplete binary relation ≿. We want to prove the reverse relationship. That is, wewant to prove that if ≿ is a complete, upper semi-continuous, and strictly convex
binary relation, then f ≿ becomes a single-valued demand function. However, thisproblem is open. In fact, we cannot determine whether or not P (x) in Proposition
2.7 is convex in this case. Because the proof of Proposition 2.7 uses the KKM
theorem, the convexity of P (x) is crucial.5
Suppose that f is surjective and ≿ is a complete, upper semi-continuous, andstrictly convex preference relation such that f = f ≿ . Choose any y ∈ P (x). Then,by strict convexity, we have that (1 − t)x + ty ∈ P (x) for every t ∈]0, 1[.6 In thiscase, is P (x) convex? In other words, we are confronted with the following problem.
PROBLEM: Suppose that C ⊂ RN , and there exists x ∈ ∂C such that if y ∈ C
and 0 < t < 1, then (1 − t)x + ty ∈ C. Is C convex?
Unfortunately, the answer to the above problem is NEGATIVE. Chaowen Yu
presented the following counterexample. Let
C =]0, 1[×{(x2 , x3 )||x3 | < |x2 |}, x = (1, 0, 0).
This C and x satisfy all requirements of PROBLEM, and C is not convex. Therefore,
our P (x) might be not convex.
However, we may be able to extend Proposition 2.7 to our cases using techniques
diﬀerent from the KKM theorem, and thus this problem is still open.
We also mention a uniqueness result on ≿. Suppose that a CoD f is continuously
diﬀerentiable, and deﬁne∂fi∂fi
sij (p, m) =(p, m)fj (p, m).
(p, m) +∂pj∂m
The n × n matrix-valued function Sf (p, m) = (sij (p, m))ni,j=1 is called the Slutsky
matrix. We say that f satisﬁes (R) if the rank of Sf (p, m) is always n − 1.
Suppose that Ω = Rn++ , and f : P → Ω is a surjective and continuously diﬀerentiable CoD that satisﬁes Walras’ law and WA. We can easily show that this f ishomogeneous of degree zero. By Proposition 1 of Hosoya [7], f satisﬁes (R) if andonly if there uniquely exists a function g : Ω → Rn++ such that the n-th coordinate ofthis function is always 1, and f (g(x), g(x) · x) = x for every x ∈ Ω. Moreover, in thiscase g is continuously diﬀerentiable. Consider the following diﬀerential equation:(2.4)ẏ(t) = (g(y(t)) · x)v − (g(y(t)) · v)x, y(0; x, v) = x.

5Probably, Komiya’s polytopal KKMS theorem is more useful for proving this theorem. See

Komiya [14].
6In fact, if ≿=≿ , R(f ) is convex, and x ∈ R(f ), then the same result holds even if ≿ is possiblyfnot strictly convex. The proof is easy.

WA AND INVERSE PROBLEMS IN CONSUMER THEORY

29

Let w = (v · x)v − (v · v)x. Deﬁne t(x, v) = inf{t ≥ 0|y(t; x, v) · w = 0}, wherey(t; x, v) is a nonextendable solution of (2.4). Letu(x, v) =ky(t(x, v); x, v)k.
kvk

Using Hosoya’s [7] results, we obtain the following result.
Proposition 2.8. Suppose that Ω = Rn++ , and f : P → Ω is a surjective andcontinuously diﬀerentiable CoD that satisﬁes Walras’ law, WA, and (R). Then,≿f = u−1 ([1, +∞[), and ≿f is the unique complete, p-transitive, and continuous
binary relation that corresponds to f .7
Proof. We mention several facts proved in Hosoya [7]. First, t(x, v) is well-deﬁned forevery (x, v) ∈ Ω2 . Second, y(t(x, v); x, v) is proportional to v, and y(t(x, v); x, v) =u(x, v)v. Third, u(x, v) ≥ 1 if and only if u(v, x) ≤ 1. Fourth, if x r v, thenu(x, v) > 1. Fifth, a 7→ u(ax, v) is increasing and u(x, av) = a−1 u(x, v). Sixth, onspan{x, v},u(z1 , u(z2 , z3 )z3 ) = u(z1 , z2 ).
Now, choose h > 0 so small, and let (xhk ) be a forward Euler approximated solutionof (2.4): that is,φ(y) = (g(y) · x)v − (g(y) · v)x,xh0 = x,xhk+1 = xhk + hφ(xhk ).
Since φ(y) ∈ span{x, v}, we have that xhk ∈ span{x, v} for all k. Because g(y)·φ(y) =0, we have that for k ≥ 1, either x = xhk orx irp xhk .
If u(x, v) > 1, then y(t(x, v); x, v)  v. Hence, for suﬃciently small h > 0, there
exists xhk such that xhk  v. This implies that xhk r v, and thus x irp v. Therefore,
we have that if u(x, v) > 1, then x irp v.
Next, suppose that x irp v. If x is proportional to v, then t(x, v) = 0 andu(x, v) = ∥x∥∥v∥ > 1. Hence, we assume without loss of generality that x is notproportional to v, and there exists a ﬁnite sequence x1 , ..., xN such that x1 = x, xN =v, xi r xi+1 for each i, and span{x1 , ..., xN } = span{x, v}. Deﬁne a0 = 1, and if
ai−1 is already deﬁned, then deﬁne ai = u(ai−1 xi , xi+1 ). Because xi r xi+1 , bymathematical induction, we have thatai = u(ai−1 xi , xi+1 ) ≥ u(xi , xi+1 ) > 1
for all i ∈ {1, ..., N − 1}. Therefore,
1 = u(x1 , x1 ) = u(x1 , aN −1 xN ) < u(x1 , xN ) = u(x, v),
7In the proof of this proposition, we frequently use knowledge of the proof of Theorem 1 in

Hosoya [7].

30

YUHKI HOSOYA

which implies that x irp v if and only if u(x, v) > 1. Because u(x, v) ≥ 1 if andonly if u(v, x) ≤ 1, we have that x ≿f v if and only if u(x, v) ≥ 1. The remainderof the proof was presented in Hosoya [7].
□Hosoya [7] also showed that, if f satisﬁes SA, then f = f uv for uv : x 7→ u(x, v),
and uv is continuously diﬀerentiable. In this case, our Proposition 2.8 implies thatuv represents ≿f , and thus we immediately obtain the following corollary. This
result also indicates a diﬀerent feature of ≿f from several known binary relations
correspond to f in this context.
Corollary 2.9. Suppose that Ω = Rn++ and f is a surjective and continuously
diﬀerentiable CoD that satisﬁes Walras’ law, SA, and (R). Then, f = f u for somecontinuously diﬀerentiable function u, and u represents ≿f . In particular, ≿f istransitive and continuous.
Note that, if Ω = Rn++ , then P = Rn++ ×R++ . This restriction could be consideredrather strong, because ∆(p, m) is not compact. In fact, Hosoya [7] showed that asimilar result holds even if P is just an open cone in Rn++ × R++ . However, thecontinuous diﬀerentiability assumption of f is strong, and thus we want to relax it.
This is another open problem.
Acknowledgement
I would like to thank Chaowen Yu for presenting a good counterexample formy conjecture. Moreover, I am grateful to the anonymous reviewer for his/her
appropriate comments.
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Manuscript received 3 March 2021
revised 19 June 2021

Y. Hosoya
742-1 Higashinakano, Hachioji-shi, Tokyo, 192-0393 Japan
E-mail address: ukki@gs.econ.keio.ac.jp