An Axiom for Concavifiable Preferences in View of Alt's Theory
aa r X i v : . [ ec on . T H ] F e b An Axiom for Concavifiable Preferences inView of Alt’s Theory
Yuhki Hosoya ∗ Faculty of Economics, Chuo University † February 16, 2021
Abstract
We present a necessary and sufficient condition for Alt’s system tobe represented by a concave utility function. This condition can beseen as an extension of Gossen’s first law, and thus has an economicinterpretation. Together with the above result, we present a rigor-ous proof of Alt’s representation theorem on a Hausdorff, separable,and path-connected topological space, and provide a necessary andsufficient condition for Alt’s utility to be continuously differentiable.
JEL codes . D11, C65, D60.
Keywords . Alt’s system, cardinal utility, Gossen’s first law, path-connectedness.
In economics, there are a number of conditions that can usually be assumedfor utility functions, such as continuity, quasi-concavity, and strict quasi-concavity. These conditions are accepted because the axioms of the prefer-ence relation that can be represented by a utility function satisfying suchconditions are easy to understand and have a natural interpretation in eco-nomics. Differentiability, on the other hand, is unlike these. The conditionsfor a preference relation that can be represented by a differentiable utilityfunction are neither easy to understand nor naturally interpretable in terms ∗ E-mail: hosoya(at)tamacc.chuo-u.ac.jp †
1f economics. Often, however, the differentiability of utility functions is ac-cepted simply because it is overwhelmingly convenient.In this paper, we treat the concavity of utility functions. Concavity is notusually considered a good assumption for utility functions. There are severalreasons for this, but the most important one is the difficulty of interpretingconcavity in economic terms. A characterization of such a preference rela-tion that can be represented by a concave utility function can be found inKannai (1977). Three axioms are presented, corresponding to the conditionof concavity without differentiation, with first-order derivatives, and withsecond-order derivatives. However, all of these axioms are too complicatedfor their meaning to be interpreted in economic terms. On the contrary, it isdifficult to determine, even mathematically, whether or not these conditionsare satisfied for a given preference relation. Because of this problem, theassumption that the utility function is concave is frowned upon, at least intheoretical research.The advantages of assuming concavity on the utility function are, how-ever, too great to ignore. First, if the utility function is concave, then La-grange’s multiplier rule can be applied to the derivation of the demand func-tion using subdifferential analysis. This advantage is not negligible in realcalculations. Second, there are some concepts that cannot be defined with-out the existence of a concave utility function. For example, Debreu’s (1976)least concave utility is one of these concepts. Kannai (1980) connected thisconcept with ALEP substitution and complementarity in equilibrium theory.Epstein and Zhang (1999) showed that the order among concave utility rep-resentations that was introduced by Debreu (1976) is related to a concept ofuncertainty aversion. Thus, there are many applications of the least concaveutility function. Third, there are several properties that emerge from thepresence of concave utility functions as the input to a Samuelson-Bergsontype social welfare function, and concavity has a non-negligible effect onwelfare analysis in applied research.Let us elaborate a little on the last point. If one extends the preference re-lation from the usual space to the space of simple lotteries, then some resultsare known about the concave representability of the Neumann-Morgensternutility function. However, the use of the Neumann-Morgenstern utility func-tion in welfare analysis has long been criticized, by Luce and Raiffa (1957)and others, as being interpretatively unnatural. Furthermore, because thebasic model of general equilibrium theory is not a stochastic model, the useof the Neumann-Morgenstern utility function is not desirable. See, for example, Rockafeller (1996) or Ioffe and Tikhomirov (1979) for detailed argu-ments.
2n this paper, we revisit the existence theorem of concave utility functionfrom Alt’s representation theory (Alt (1936)). To the best of our knowledge,this is the oldest theory to have discussed the possibility of utility represen-tation of preference relations. Although this theory is also a kind of cardinalutility theory, it has an interpretation that is more suitable for welfare anal-ysis than that of Neumann-Morgenstern utility. Furthermore, Alt’s theoryis not especially incompatible with general equilibrium theory. In these re-spects, it is worth deriving the necessary and sufficient conditions for Alt’sutility function to be concave.In this paper, we first derive a necessary and sufficient condition for theexistence of a continuous utility function that represents Alt’s system (The-orem 1). Although Alt showed such a result, the axioms he used were toonumerous and not independent, and difficult to interpret economically. Simi-lar problems can be found in more recent studies, such as Kranz et al. (1971).Seidl and Schmidt (1997) elaborated on this result, but again the numberof conditions is too large and not independent. In contrast, Shapley (1975)derived an existence theorem for Alt’s utility function with only three simpleaxioms. However, the space treated by Shapley is limited to a subset of theone-dimensional space. Our first result can be seen as an extension of Shap-ley’s result to Alt’s system on a Hausdorff, separable, and path-connectedtopological space.We then derive a necessary and sufficient condition for Alt’s utility func-tion to be concave (Theorem 2). Because this condition can be seen as a gen-eralization of Gossen’s first law, we name this condition “generalized Gossen’sfirst law”. Compared with Kannai’s axiom, this generalized Gossen’s first lawis much easier to interpret economically.Finally, because Gossen’s first law is usually described as a feature ofpartial derivatives of the utility function, we want to obtain the conditions forAlt’s utility function to be differentiable. Hence, we present a necessary andsufficient condition for Alt’s utility function to be continuously differentiableand nondegenerate (Theorem 3).Note that, generalized Gossen’s first law is stronger than Kannai’s axiomas a condition for the existence of concave utility functions. Our condition isa necessary and sufficient condition for Alt’s utility function to be concave,whereas Kannai’s axiom only guarantees the existence of a concave utilityfunction to represent a given preference relation. We believe this axiomis still useful because it has a much more natural economic interpretationthan Kannai’s axiom. This is, in our opinion, one of the most importantrequirements when discussing whether or not to assume such a property forthe utility function.In subsection 2.1, we introduce the notion of Alt’s system, and discuss3ts interpretation. In subsections 2.2 and 2.3, we present three axioms forAlt’s system, and state that they are equivalent to the existence of the cor-responding utility function. In subsections 2.4 and 2.5, we treat generalizedGossen’s first law, and show that this axiom is equivalent to the concavityof Alt’s utility function. In subsection 2.6, we examine the differentiabilityof Alt’s utility. In section 3, we discuss the relationship between this workand several related studies. Because the proofs of Theorems 1 and 3 aresomewhat lengthy, they are given in appendix.
Let X be a nonempty set. A pair ( % , ≥ ) is called an Alt’s system on X ifand only if % is a weak order on X and ≥ is a weak order on X . A function u : X → R is said to represent Alt’s system ( % , ≥ ) (or to be a utility functionof this system) if and only if the following two requirements hold. First, forevery x ∈ X and y ∈ X , u ( x ) ≥ u ( y ) ⇔ ( x, y ) ∈ % . (1)Second, for every ( x, y ) ∈ X and ( z, w ) ∈ X , u ( x ) − u ( y ) ≥ u ( z ) − u ( w ) ⇔ ( x, y, z, w ) ∈≥ . (2)The first relationship (1) says that u is a utility function that represents % . Meanwhile, the second relationship (2) says that the function ( x, y ) u ( x ) − u ( y ) represents ≥ .The interpretation of Alt’s system is as follows. First, as usual, ( x, y ) ∈ % means that x is either preferred or indifferent to y . Second, ( x, y, z, w ) ∈≥ means that the strength of the improvement from y to x is not weakerthan that from w to z . If u represents this system, then u ( x ) − u ( y ) measuresthis strength, and thus u is a sort of cardinal utility function.As usual, we write x % y instead of ( x, y ) ∈ % . Moreover, we write[ y → x ] ≥ [ w → z ] instead of ( x, y, z, w ) ∈≥ . These notations allow usto understand what is occurring. We write x ≻ y if x % y and y % x and x ∼ y if x % y and y % x . Additionally, we write [ y → x ] = [ w → z ] if[ y → x ] ≥ [ w → z ] and [ w → z ] ≥ [ y → x ], and [ y → x ] > [ w → z ] if[ y → x ] ≥ [ w → z ] and [ w → z ] = [ y → x ]. This arrow notation was originally used by Alt, and is employed throughout this paper. .2 Axioms We now present several axioms on Alt’s system. The first one is necessaryfor the system to match its interpretation appropriately.
Definition 1 . An Alt’s system ( % , ≥ ) is said to satisfy consistency if andonly if, for every x, y, z ∈ X , x % y ⇔ [ z → x ] ≥ [ z → y ] . Note that, under this consistency axiom, if there exists z ∈ X such that[ z → x ] ≥ [ z → y ], then x % y . Conversely, if x % y , then for every z ,we have that [ z → x ] ≥ [ z → y ]. This gap (between ‘there exists’ and ‘forevery’) may be a source of confusion.If there is a representation u of ( % , ≥ ), then consistency implies that u ( x ) ≥ u ( y ) ⇔ u ( x ) − u ( z ) ≥ u ( y ) − u ( z ) , and thus, consistency is trivially satisfied.The next axiom relates to the property of =. Definition 2 . An Alt’s system ( % , ≥ ) is said to satisfy the crossover axiom if and only if, for every x, y, z, w ∈ X , [ y → x ] = [ w → z ] ⇔ [ z → x ] = [ w → y ] . The relation = is symmetric, and therefore the crossover axiom also meansthat [ y → x ] = [ w → z ] ⇔ [ y → w ] = [ x → z ] . If there is a representation u of ( % , ≥ ), then the crossover axiom implies that u ( x ) − u ( y ) = u ( z ) − u ( w ) ⇔ u ( x ) − u ( z ) = u ( y ) − u ( w ) , and thus, this axiom is also trivially satisfied.We note two facts. First, for every x, y ∈ X ,[ y → x ] = [ y → x ] . If the crossover axiom holds, then this implies that[ x → x ] = [ y → y ] . The name ‘the crossover axiom’ was used in Miyake (2016). x ∼ y ⇔ [ z → x ] = [ z → y ] ⇔ [ y → x ] = [ z → z ]for every x, y, z ∈ X . That is, “ x is indifferent to y if and only if thestrength of the improvement from y to x is the same as that in the unchangedsituation.” This is also natural.To treat the third axiom, we must consider X as a topological space. Definition 3 . Suppose that X is a topological space. Then, an Alt’s system( % , ≥ ) is said to satisfy continuity if and only if ≥ is closed in X .Note that, if ( % , ≥ ) satisfies the consistency and continuity, then % = { ( x, y ) ∈ X | [ y → x ] ≥ [ x → x ] } must be closed in X .If there is a continuous representation u of ( % , ≥ ), then clearly ≥ is closed,because ≥ = { ( x, y, z, w ) ∈ X | u ( x ) + u ( w ) − u ( y ) − u ( z ) ≥ } . The next theorem is our first result.
Theorem 1 . Suppose that X is a Hausdorff, separable, and path-connectedtopological space, and ( % , ≥ ) is an Alt’s system on X . Then, there existsa continuous representation u : X → R of ( % , ≥ ) if and only if this systemsatisfies consistency, the crossover axiom, and continuity. If so, such a rep-resentation is unique up to a positive affine transform: that is, if u , u arecontinuous representations of ( % , ≥ ), then there exist a > b ∈ R suchthat u ( x ) = au ( x ) + b for every x ∈ X .The proof of this theorem is in the appendix. A topological space X is said to be path-connected if and only if for every x, y ∈ X ,there exists a continuous function f : [0 , → X such that f (0) = x, f (1) = y . Forexample, every convex set of some topological vector space is path-connected. It is well-known that every path-connected space is connected. .4 Generalized Gossen’s First Law Our main purpose is to present a necessary and sufficient condition for u tobe concave. Hence, we introduce the following axiom. Definition 4 . Suppose that X is a separable and convex subset of a Haus-dorff topological vector space. Then, an Alt’s system ( % , ≥ ) on X is said tosatisfy generalized Gossen’s first law if and only if, for every x, y ∈ X ,[ x → z ] ≥ [ z → y ]for z = ( x + y ). If, in addition,[ x → z ] > [ z → y ]whenever x = y , then ( % , ≥ ) is said to satisfy generalized strong Gossen’sfirst law .We should interpret the meaning of generalized Gossen’s first law. Sup-pose that X is a convex subset of R n , and x ∈ X and y = ( x +2 a, x , ..., x n ) ∈ X for a >
0. Then, z = ( x + a, x , ..., x n ) ∈ X . Suppose that generalizedGossen’s first law is satisfied. Then, we have that [ x → z ] ≥ [ z → y ]. If u is a continuously differentiable representation of this system, then by meanvalue theorem, u ( z ) − u ( x ) = a ∂u∂x ( z − θae ) ,u ( y ) − u ( z ) = a ∂u∂x ( z + ηae )for some θ, η ∈ [0 , e = (1 , , ..., x → z ] ≥ [ z → y ] implies that the mapping b ∂u∂x ( x + b, x , ..., x n )is nonincreasing around a . Because x and a are arbitrary, we conclude thatunder generalized Gossen’s first law, the marginal utility of x is nonincreas-ing in x . This explains why this axiom is named as such.In the definition of the generalized Gossen’s first law, z = x + ( z − x ) and y = z + ( z − x ). Therefore, this axiom can be transformed as follows: forevery v = 0 such that x + 2 v ∈ X ,[ x → ( x + v )] ≥ [( x + v ) → ( x + 2 v )] . u represents this system, then u ( x + v ) − u ( x ) ≥ u ( x + 2 v ) − u ( x + v ) , and thus, under this axiom, the marginal improvement of the utility for achange of consumption in the v direction decreases as the length of the changeincreases. The following is our second main result.
Theorem 2 . Suppose that X is a separable and convex subset of a Hausdorfftopological vector space, ( % , ≥ ) is an Alt’s system on X , and u : X → R isa continuous function that represents this system. Then, u is concave (resp.strictly concave) if and only if ( % , ≥ ) satisfies generalized Gossen’s first law(resp. generalized strong Gossen’s first law).Note that, because X is convex, it is path-connected, and thus all as-sumptions in Theorem 1 are satisfied. Proof . First, suppose that u is concave. Choose any x, y ∈ X , and define z = ( x + y ). Then,12 u ( z ) + 12 u ( z ) = u ( z ) ≥ u ( x ) + 12 u ( y ) , (3)and thus u ( z ) − u ( x ) ≥ u ( y ) − u ( z ) , as desired. If, in addition, u is strictly concave and y = x , then the inequalityin (3) is strengthened, and thus u ( z ) − u ( x ) > u ( y ) − u ( z ) , as desired.Conversely, suppose that ( % , ≥ ) satisfies generalized Gossen’s first law.We want to show that u ((1 − t ) x + ty ) ≥ (1 − t ) u ( x ) + tu ( y ) (4)for every x, y ∈ X and t ∈ [0 , x, y ∈ X , and define z = ( x + y ). By generalized Gossen’s first law,[ x → z ] ≥ [ z → y ] , u ( z ) − u ( x ) ≥ u ( y ) − u ( z ) , which implies that u ( z ) ≥ u ( x ) + 12 u ( y ) . Therefore, we have that (4) holds if t = . By mathematical induction, wecan easily show that (4) holds if t is a dyadic rational m ℓ , where ℓ ≥ ≤ m ≤ ℓ −
1. Because the set of all dyadic rationals is dense in [0 , u , (4) holds for all t ∈ [0 , % , ≥ ) satisfies generalized strong Gossen’s first law. Chooseany x, y ∈ X such that x = y . Define z = ( x + y ). Then,[ x → z ] > [ z → y ] , and thus, u ( z ) − u ( x ) > u ( y ) − u ( z ) , which implies that u ( z ) > u ( x ) + 12 u ( y ) . Therefore, we have that u ((1 − t ) x + ty ) > (1 − t ) u ( x ) + tu ( y ) (5)if t = . By mathematical induction, we can easily show that (5) holds if t is a dyadic rational m ℓ , where ℓ ≥ ≤ m ≤ ℓ −
1. Now, choose any t ∈ ]0 ,
1[ such that t is not a dyadic rational. Then, if ℓ is sufficiently large,there exists m ∈ { , ..., ℓ − } such that t = m ℓ < t < m +12 ℓ = t . In thiscase, t = t − tt − t t + t − t t − t t , and thus, by the continuity of u , we have that u ((1 − t ) x + ty ) ≥ t − tt − t u ((1 − t ) x + t y ) + t − t t − t u ((1 − t ) x + t y ) > t − tt − t [(1 − t ) u ( x ) + t u ( y )]+ t − t t − t [(1 − t ) u ( x ) + t u ( y )]= (1 − t ) u ( x ) + tu ( y ) , as desired. This completes the proof. (cid:4) .6 Result for the Differentiability of Alt’s CardinalUtility Usually, Gossen’s laws are stated in the language of differential calculus.Hence, we provide a sufficient condition for Alt’s utility to be differentiable.First, let x, y ∈ R n . We write x ≥ y if and only if x i ≥ y i for all i ∈ { , ..., n } , and x ≫ y if and only if x i > y i for all i ∈ { , ..., n } . Define R n + = { x ∈ R n | x ≥ } and R n ++ = { x ∈ R n | x ≫ } . We introduce threeaxioms. Definition 5 . Let X = R n + and ( % , ≥ ) be an Alt’s system on X . Then, thissystem is said to be monotone if x ≻ y for every x, y ∈ X such that x ≫ y . Definition 6 . Let X = R n + and ( % , ≥ ) be an Alt’s system on X . Then, thissystem is said to satisfy Debreu’s smoothness if and only if the followingset I = { ( x, y ) ∈ R n ++ | x ∼ y } is a 2 n − C manifold. Definition 7 . Let X = R n + and ( % , ≥ ) be an Alt’s system on X that satisfiesconsistency, the crossover axiom, continuity, and monotonicity. Define e =(1 , , ..., ∈ X . By Theorem 1 and the intermediate value theorem, we havethat, for every a, b > a < b , there uniquely exists f ( a, b ) > [( b − a ) e → f ( a, b ) e ] = [ f ( a, b ) e → ( b + a ) e ] . We say that this system satisfies line smoothness if and only iflim a ↓ b − f ( a, b ) a = 0for all b > Theorem 3 . Let X be either R n + or R n ++ , ( % , ≥ ) be an Alt’s system on X that satisfies consistency, the crossover axiom, continuity, generalizedGossen’s first law, and monotonicity. Let u be a representation of ( % , ≥ ).Then, u is continuously differentiable on R n ++ and Du ( x ) = 0 for every x ∈ R n ++ if and only if ( % , ≥ ) satisfies both Debreu’s smoothness and linesmoothness. The existence of such an f can also be shown using Lemma 4 in the appendix. v ( x ) = √ x x and g ( c ) = ( c − c ≤ , c − if c > . Define u ( x ) = g ( v ( x )), and suppose that ( % , ≥ ) is an Alt’s system thatcorresponds to u . Then, this system satisfies Debreu’s smoothness, because I = { ( x, y ) ∈ R n ++ | v ( x ) = v ( y ) } . However, f ( a,
1) = 1 − a , and thus,lim a ↓ − f ( a, a = 14 = 0 , which implies that this system violates line smoothness.Meanwhile, every function that is increasing and homogeneous of degreeone satisfies line smoothness, although its indifference curve is kinked. There-fore, line smoothness does not imply Debreu’s smoothness. That is, Debreu’ssmoothness is independent of line smoothness. To the best of our knowledge, Alt’s system was first argued in Alt (1936), whoproved that under several axioms (including our consistency, the crossoveraxiom, and continuity), there exists a continuous representation. He wrotehis paper in German, and it was translated in English in 1971 (Chipman et al.(1971)). Actually, his axioms include several ‘facts’ that can be shown fromthe above three axioms. For example, our Lemma 5 is one of Alt’s axiom.Similar axioms can be found in Kranz et al. (1971), where our Lemma 5 isalso treated as an axiom. Shapley (1975) used only our three axioms andderived a continuous representation. In his proof, however, X is treated as aconvex subset of the real line. Hence, our Theorem 1 is an extension of hisresult to a general path-connected topological space.Miyake (2016) considered an Alt’s system on R n ++ , and showed that,under several axioms, there exists a representation log u ( x ) such that u ishomogeneous of positive degree. Surprisingly, his result does not use thecrossover axiom. Therefore, his result is independent of our Theorem 1.We note our assumption of path-connectedness. A topological space X is said to be connected if and only if for every pair of nonempty closed sets A, B in X , either A ∪ B = X or A ∩ B = ∅ . Debreu (1954) showed that if X is a Hausdorff, separable, and connected topological space, and a weak order11 on X is closed in X , then there exists a continuous function u : X → R such that u ( x ) ≥ u ( y ) ⇔ x % y. Compared with this result, our Theorem 1 uses path-connectedness insteadof connectedness. It is known that every path-connected topological space isconnected. Meanwhile, it is also known that the following set X = { ( x − , sin x ) | x = 0 } ∪ { (0 , y ) | − ≤ y ≤ } is closed and connected, but not path-connected. Therefore, path-connectednessis stronger than connectedness. In the appendix, readers find that path-connectedness is only used in the proof of Lemma 7, and in the remainingparts of this proof, only connectedness is used. Therefore, if Lemma 7 canbe proved avoiding the use of path-connectedness, we could extend Theorem1 for a general connected space. However, at least we are not able to proveLemma 7 without the aid of path-connectedness.Debreu (1976) treated the notion of the least concave utility function.Suppose that X is a separable and convex set of a Hausdorff topologicalvector space, and % is a weak order of X that is closed in X . Let U be theset of all continuous and concave functions u such that u ( x ) ≥ u ( y ) ⇔ x % y. If u, v ∈ U , then we write u (cid:23) v if and only if there exists a concave function ϕ such that u = ϕ ◦ v . Debreu showed that if U is nonempty, then thereexists a least element in U with respect to this partial order (cid:23) . This leastelement is called the least concave utility function . He showed that theleast concave utility function is unique up to a positive affine transform,and for fixed x ∈ X , if there exists a continuously differentiable element u ∈ U at x , then the least concave utility function is also continuouslydifferentiable at x . The proof is very similar to Benveniste-Scheinkman’senvelope theorem. In this paper, we characterized the existence of such acontinuously differentiable u ∈ U in Theorem 3.Finally, we mention Alexandrov’s theorem. This theorem states that forevery convex function, its subdifferential mapping is ‘differentiable’ in thesense of set-valued differentiation almost everywhere. Applying this theo-rem, we can obtain the following result: if u is a concave and continuouslydifferentiable utility function derived from Theorem 3, then it is twice differ-entiable almost everywhere. Thus, ALEP substitution and complementarity can be defined almost everywhere. This may be useful in equilibrium theory. See Alexandrov (1939) or Howard (1998) for detailed arguments. The definition of ALEP substitution and complementarity uses second cross-derivative Proofs
A.1 Proof of a Lemma
In this subsection, we introduce a property of an Alt’s system ( % , ≥ ) on X .This system is said to satisfy second consistency if and only if for every x, y, z ∈ X , x % y ⇔ [ y → z ] ≥ [ x → z ] . We show the following lemma.
Lemma 1 . Suppose that X is a Hausdorff and connected topological spaceand ( % , ≥ ) is an Alt’s system on X that satisfies consistency, the crossoveraxiom, and continuity. Then, this system satisfies second consistency. Proof . We first show the following fact. Suppose that[ y → x ] ≥ [ w → z ] ≥ [ y → x ] . Then, there exists x ∈ X such that[ y → x ] = [ w → z ] . To show this, let U = { x ∈ X | [ y → x ] ≥ [ w → z ] } , D = { x ∈ X | [ w → z ] ≥ [ y → x ] } . Because x ∈ U and x ∈ D , we have that both U and D are nonempty. Bycontinuity, both U and D are closed. Since U ∪ D = X , by the connectednessof X , we have that there exists x ∈ U ∩ D . Then, clearly[ y → x ] = [ w → z ] . Note that, [ y → x ] ≥ [ y → x ] ≥ [ y → x ] , and thus, by consistency, we have that x % x % x .Second, suppose that x ∼ y . Then, by consistency,[ z → x ] = [ z → y ] . of the utility function. This notion was argued by von Auspitz and Lieben (1889), Edge-worth (1897), and Pareto (1906). The name ‘ALEP’ comes from the first letters of theseresearchers. y → x ] = [ z → z ] , and again by the crossover axiom,[ y → z ] = [ x → z ] . Hence, second consistency holds for every x, y, z ∈ X such that x ∼ y .Now, suppose that second consistency is violated. Then, by the abovearguments, there exists x, y, z ∈ X such that y ≻ x, [ y → z ] ≥ [ x → z ] . (6)If y ≻ z , then by consistency, the crossover axiom, and (6),[ x → x ] = [ y → y ] > [ y → z ] ≥ [ x → z ] . Therefore, we have that there exists w ∈ X such that[ x → w ] = [ y → z ] , x % w % z. By the crossover axiom, [ z → w ] = [ y → x ] . By consistency and the crossover axiom,[ z → z ] = [ y → y ] > [ y → x ] = [ z → w ] ≥ [ z → z ] , which is a contradiction.Therefore, we must have that z % y . By the transitivity of % , we havethat z ≻ x . Thus, by (6), consistency, and the crossover axiom,[ y → z ] ≥ [ x → z ] > [ x → x ] = [ y → y ] . Thus, we have that there exists w ∈ X such that[ y → w ] = [ x → z ] , z % w % y. By the crossover axiom, [ z → w ] = [ x → y ] . By the crossover axiom and consistency,[ z → z ] ≥ [ z → w ] = [ x → y ] > [ x → x ] = [ z → z ] , which is a contradiction. This completes the proof. (cid:4) .2 Proof of Theorem 1 Suppose that X is a Hausdorff, separable, and path-connected topologicalspace and ( % , ≥ ) is an Alt’s system on X . If there exists a continuousfunction u : X → R that represents ( % , ≥ ), then clearly ( % , ≥ ) satisfies con-sistency, the crossover axiom, and continuity. Therefore, it suffices to showthe opposite direction, in other words, that if ( % , ≥ ) satisfies consistency,the crossover axiom, and continuity, then there exists a continuous function u : X → R that represents ( % , ≥ ).First, we introduce several lemmas. Lemma 2 . If [ y → x ] ≥ [ w → z ] ≥ [ y → x ] , then there exists x ∈ X such that[ y → x ] = [ w → z ] . (7)Moreover, x % x % x , and if x also satisfies (7), then x ∼ x . Proof . We have already proved this result in the proof of Lemma 1, exceptfor the last claim. Hence, suppose that[ y → x ] = [ w → z ] . Then, we have that [ y → x ] = [ y → x ]by the transitivity of =, and by consistency, x ∼ x , as desired. This completes the proof. (cid:4) Lemma 3 . Suppose that[ x → y ] ≥ [ w → z ] ≥ [ x → y ] . Then, there exists x ∈ X such that[ x → y ] = [ w → z ] . (8)Moreover, x % x % x , and if x also satisfies (8), then x ∼ x . Proof . By using second consistency instead of consistency, we can show thisresult using almost the same arguments as in the proof of Lemma 2. Hence,we omit the proof. (cid:4) emma 4 . Suppose that z ≻ x . Then, there exists y ∈ X such that[ x → y ] = [ y → z ] . (9)Moreover, z ≻ y ≻ x , and if y ′ also satisfies (9), then y ∼ y ′ . Proof . Define U = { w ∈ X | [ x → w ] ≥ [ w → z ] } , D = { w ∈ X | [ w → z ] ≥ [ x → w ] } . Then, both U and D are closed. Because [ x → z ] > [ x → x ] = [ z → z ], wehave that z ∈ U and x ∈ D , and thus both U and D are nonempty. Because U ∪ D = X , by the connectedness of X , there exists y ∈ U ∩ D . Clearly,[ x → y ] = [ y → z ] . If y % z , then y ≻ x , and thus[ x → y ] ≥ [ x → z ] > [ y → z ] = [ x → y ] , which is absurd. Therefore, we have z ≻ y . Symmetrically, we can show that y ≻ x .Choose any y ′ ∈ X . If y ≻ y ′ , then by consistency and second consistency,[ y ′ → z ] > [ y → z ] = [ x → y ] > [ x → y ′ ] . If y ′ ≻ y , then again by consistency and second consistency,[ x → y ′ ] > [ x → y ] = [ y → z ] > [ y ′ → z ] . Therefore, if [ y ′ → z ] = [ x → y ′ ], then we must have y ∼ y ′ . This completesthe proof. (cid:4) Lemma 5 . Suppose that x ≻ y . If z % x , then there exists a finite sequence a , a , ..., a k ∈ X such that a = y, a = x, [ a i → a i +1 ] = [ a i +1 → a i +2 ] for all i,z % a k , [ a → a ] > [ a k → z ] . Note that in the proof of Lemmas 1-4, we only used the connectedness of X ; theseparability and path-connectedness of X were not used. However, in the proof of Lemma5, we must use Debreu’s representation theorem, and so X must be separable. If there is a representation u : X → R , then this statement says that ( k − u ( x ) − u ( y )) ≤ u ( z ) − u ( x ) < k ( u ( x ) − u ( y )) for some k ∈ N . y % w , then there exists a finite sequence a , a , ..., a − k such that a = y, a = x, [ a i → a i +1 ] = [ a i +1 → a i +2 ] for all i,a − k % w, [ a → a ] > [ w → a − k ] . Proof . We treat only the claim on z , because the claim on w can be provedsymmetrically. Suppose that such a finite sequence does not exist. Define a = y and a = x . By the negation of the claim of this lemma for k = 1, wemust have [ a → z ] ≥ [ a → a ] > [ a → a ] . By Lemma 2, there exists a ∈ X such that[ a → a ] = [ a → a ] , z % a ≻ a . We assume that a , ..., a k is already defined, and[ a → a ] = [ a i → a i +1 ]for every i ∈ { , ..., k − } . By the negation of the claim of this lemma, wemust have [ a k → z ] ≥ [ a → a ] > [ a k → a k ] . Therefore, by Lemma 2, there exists a k +1 ∈ X such that[ a k → a k +1 ] = [ a → a ] , z % a k +1 ≻ a k . Hence, by mathematical induction, we obtain an infinite sequence ( a k ) suchthat z % a k and [ a → a ] = [ a k → a k +1 ]for every k ∈ N .We will show that [ a i → a i + k ] = [ a j → a j + k ] (10)for every i, j, k . If k = 1, then it has already been proved. By the crossoveraxiom, [ a i → a j ] = [ a i +1 → a j +1 ] . By the transitivity of =, we have that, for every k ,[ a i → a j ] = [ a i + k → a j + k ] . a i → a i + k ] = [ a j → a j + k ] , and thus (10) holds.Now, by Debreu’s representation theorem, there exists a continuousfunction v : X → R such that v ( x ′ ) ≥ v ( y ′ ) ⇔ x ′ % y ′ for every x ′ , y ′ ∈ X . Let c ∗ = sup k v ( a k ). Then, we have that v ( z ) ≥ c ∗ >v ( a ), and thus c ∗ ∈ R . Because v is continuous and X is connected, v ( X )is also a connected set in R . In R , connectedness is equivalent to convexity.Therefore, v ( X ) is convex, and thus there exists z ∗ ∈ X such that v ( z ∗ ) = c ∗ .By Lemma 4, there exists w ∗ ∈ X such that [ a → w ∗ ] = [ w ∗ → z ∗ ] and z ∗ ≻ w ∗ ≻ a . Because c ∗ = v ( z ∗ ) > v ( w ∗ ) > v ( a ), there exists k such that v ( a k +1 ) > v ( w ∗ ) ≥ v ( a k ) . Then, by (10),[ a k +1 → a k +2 ] = [ a → a k +1 ] > [ a → w ∗ ] = [ w ∗ → z ∗ ] > [ a k +1 → z ∗ ] , and thus, by consistency, we have that a k +2 ≻ z ∗ . This implies that v ( a k +2 ) > c ∗ , which contradicts the definition of c ∗ . Thiscompletes the proof. (cid:4) Now, suppose that x ∼ y for every x, y ∈ X . Then, we have that [ y → x ] = [ y → z ] = [ w → z ] for every x, y, z, w ∈ X , and thus u : X → R represents this system if and only if u is a constant function. Hence, wehereafter assume that there exists x ∗ , y ∗ ∈ X such that x ∗ ≻ y ∗ .We will recursively construct (possibly finite or infinite) sequences ( a ki )for k ∈ N . First, we define a = y ∗ and a = x ∗ . Suppose that a i is defined This theorem states that if X is a Hausdorff, separable, and connected topologicalspace and % is a weak order on X that is closed in X , then there exists a continuousfunction v : X → R such that v ( x ) ≥ v ( y ) ⇔ x % y for every x, y ∈ X . For the proof of this theorem, see, for example, Debreu (1954) orBridges and Mehta (1995). Later, we construct our utility function u to satisfy u ( a ki ) = i k . i ≥
1. If there exists x ∈ X such that [ a i → x ] ≥ [ a → a ], thenby Lemma 2, there exists a i +1 ∈ X such that [ a i → a i +1 ] = [ a → a ]. Ifthere is no such x , then we do not define a i +1 . Similarly, suppose that a i isdefined for some i ≤
0. If there exists x ∈ X such that [ x → a i ] ≥ [ a → a ],then by Lemma 3, there exists a i − ∈ X such that [ a i − → a i ] = [ a → a ].If there is no such x , then we do not define a i − .Next, suppose that ( a ki ) is defined for k = 0 , ..., k ∗ and the following tworelationships hold: 1) a ki = a k +12 i and 2) [ a ki → a ki +1 ] = [ a k → a k ]. We define asequence ( a k ∗ +1 i ) as follows. First, if a k ∗ i is defined, then we define a k ∗ +12 i = a k ∗ i .Second, if both a k ∗ i , a k ∗ i +1 are defined, then by Lemma 4, there exists a k ∗ +12 i +1 ∈ X such that [ a k ∗ +12 i → a k ∗ +12 i +1 ] = [ a k ∗ +12 i +1 → a k ∗ +12 i +2 ]. Third, suppose that a k ∗ i isdefined and a k ∗ i +1 is undefined. Note that, in this case we have i ≥ k ∗ . Ifthere exists x ∈ X such that [ a k ∗ i → x ] ≥ [ a k ∗ +10 → a k ∗ +11 ], then by Lemma2, there exists a k ∗ +12 i +1 ∈ X such that [ a k ∗ +12 i → a k ∗ +12 i +1 ] = [ a k ∗ +10 → a k ∗ +11 ]. Ifsuch an x does not exist, then we do not define a k ∗ +12 i +1 . Similarly, supposethat a k ∗ i is defined and a k ∗ i − is undefined. Then, we have that i ≤
0. If thereexists x ∈ X such that [ x → a k ∗ i ] ≥ [ a k ∗ +10 → a k ∗ +11 ], then by Lemma 3, thereexists a k ∗ +12 i − ∈ X such that [ a k ∗ +12 i − → a k ∗ +12 i ] = [ a k ∗ +10 → a k ∗ +11 ]. If such an x does not exist, then we do not define a k ∗ +12 i − .By the above arguments, we can define ( a ki ) recursively, and these se-quences satisfy a ki = a k +12 i and [ a ki → a ki +1 ] = [ a k → a k ]. Let D ⊂ X be theset of all a ki . We need the following lemma. Lemma 6 . If x ≻ y , then there exists z ∈ D such that x ≻ z ≻ y . Proof . Suppose that x ≻ y and there is no z ∈ D such that x ≻ z ≻ y . Wetreat the case in which x ≻ a because the remaining case can be treatedsymmetrically.Because of our assumption, we must have y % a . By Lemma 5, thereexists a finite sequence b , b , ..., b − k such that b = x , b = y , [ b i → b i +1 ] =[ b → b ] for all i ∈ {− k, ..., − } , b − k % a and [ b → b ] > [ a → b − k ].Meanwhile, again by Lemma 5, for every m ≥
0, there exists ℓ ≥ y % a mℓ and [ a m → a m ] > [ a mℓ → y ]. If there does not exist a mℓ +1 , then by our Note that, if a k ∗ +12 i +1 is defined and a k ∗ i +1 is undefined, then there is no x ∈ X such that[ a k ∗ +12 i +1 → x ] ≥ [ a k ∗ +10 → a k ∗ +11 ]. Suppose that such an x exists. Then, we can define a k ∗ +12 i +2 . By the same arguments as in the proof of Lemma 5, we can show that (10) holdsfor ( a k ∗ +1 i ), and thus we have that[ a k ∗ → a k ∗ ] = [ a k ∗ i → a k ∗ +12 i +2 ] , which contradicts undefinedness of a k ∗ i +1 . a mi ), we have that [ a m → a m ] > [ a mℓ → x ]. If a mℓ +1 exists, then a mℓ +1 ≻ y . Because a mℓ +1 ∈ D and there is no z ∈ D such that x ≻ z ≻ y , wehave that a mℓ +1 % x , and thus[ a m → a m ] = [ a mℓ → a mℓ +1 ] ≥ [ a mℓ → x ] . Therefore, in any case we have [ a m → a m ] ≥ [ a mℓ → x ].Now, we assume that ℓ ≥ m . Because a mℓ = a m +12 ℓ , we canchoose m so large that ℓ > k . Define c = a mℓ . Then, clearly b % c % a mℓ .Note that, [ a m → a m ] ≥ [ a mℓ → x ] ≥ [ y → x ] = [ b → b ] . Meanwhile, we have that[ a mℓ − → b ] ≥ [ a mℓ − → a mℓ ] = [ a m → a m ] ≥ [ b → b ] , and thus, by Lemma 3, there exists c ∈ X such that[ c → b ] = [ a m → a m ] . If k ≥
1, then [ c → b ] ≥ [ b → b ] = [ b − → b ] , and thus, we have that b − % c % a mℓ − . Inductively, we can show that thereexists c k ∈ X such that b − k % c k % a mℓ − k . Then,[ a mℓ − k − → b − k ] ≥ [ a mℓ − k − → a mℓ − k ] ≥ [ y → x ] ≥ [ b − k → b − k ] , and thus, there exists b − k − ∈ X such that b − k − % a mℓ − k − % a and[ b − k − → b − k ] = [ y → x ] , which contradicts our definition of k .Therefore, we have that ℓ = 0 for every m . This means that there isno z ∈ D such that y % z ≻ a , and thus, there is no z ∈ D such that x ≻ z ≻ a . By Lemma 5, there exists a finite sequence b , ..., b k such that b = a , b = x, [ b i → b i +1 ] = [ b i +1 → b i +2 ] , a % b k , [ b → b ] > [ b k → a ] . Choose any m such that k + 1 < m . Then, a = a m m . Suppose that a m m − % b k . Then,[ b → b ] > [ b k → a m m ] ≥ [ a m m − → a m m ] = [ a m → a m ] , b ≻ a m ≻ b , which contradicts our initial assump-tion. Therefore, b k ≻ a m m − . Because k +1 < m , by the pigeonhole principle,there exist i, j such that b i +1 ≻ a mj +1 ≻ a mj % b i . Then,[ b → b ] = [ b i → b i +1 ] > [ b i → a mj +1 ] ≥ [ a mj → a mj +1 ] = [ a m → a m ] , and thus b ≻ a m ≻ b , which is a contradiction. This completes the proof. (cid:4) Define D ′ = { x ∈ X |∃ x ′ ∈ D s.t. x ∼ x ′ } . Let x ∈ D ′ . Then, there exist i, k such that x ∼ a ki . Define u ( x ) = i k . Clearly, u ( x ) does not depend on the choice of k . Choose any x, y, z, w ∈ D ′ .Then, there exist k and i x , i y , i z , i w such that x ∼ a ki x , y ∼ a ki y , z ∼ a ki z , w ∼ a ki w . Then, u ( x ) ≥ u ( y ) ⇔ i x ≥ i y ⇔ a ki x % a ki y ⇔ x % y. Next, suppose that u ( x ) − u ( y ) ≥ u ( z ) − u ( w ) . This means that i x − i y ≥ i z − i w . If i x − i y ≥ ≥ i z − i w , then[ y → x ] ≥ [ y → y ] = [ z → z ] ≥ [ w → z ] . If i x − i y ≥ i z − i w ≥
0, define i ∗ = min { i y , i w } . Then, both a ki ∗ + i x − i y and a ki ∗ + i z − i w are defined, and by (10), [ y → x ] = [ a ki ∗ → a ki ∗ + i x − i y ] ≥ [ a ki ∗ → a ki ∗ + i z − i w ] = [ w → z ] . By the symmetrical arguments, we can show that if 0 ≥ i x − i y ≥ i z − i w , then[ y → x ] ≥ [ w → z ]. In conclusion, we have that u ( x ) − u ( y ) ≥ u ( z ) − u ( w )implies that [ y → x ] ≥ [ w → z ]. Moreover, if u ( x ) − u ( y ) > u ( z ) − u ( w ) , Note that, we can show (10) for ( a mi ) by the same arguments as in the proof of Lemma5. i x − i y > i z − i w , and by repeating the above arguments, we have that[ y → x ] > [ w → z ] . Therefore, u represents ( % , ≥ ) on D ′ .We extend the definition of u on X . If x is not a least element of X on % , then by Lemma 6, there exists y ∈ D such that x ≻ y . Therefore, define u ( x ) = sup { u ( y ) | y ∈ D, x % y } . Conversely, if x is not a greatest element of X on % , then by Lemma 6, thereexists y ∈ D such that y ≻ x . Therefore, define u ( x ) = inf { u ( y ) | y ∈ D, y % x } . We can easily show that these two definition of u coincides at every x suchthat both can be defined.If x % y , then clearly u ( x ) ≥ u ( y ). If x ≻ y , then by Lemma 6, thereexists z, w ∈ D such that x ≻ z ≻ w ≻ y . Therefore, u ( x ) ≥ u ( z ) > u ( w ) ≥ u ( y ) , and thus u ( x ) > u ( y ). Hence, we have that x % y ⇔ u ( x ) ≥ u ( y ) . Next, we show the continuity of u . For this purpose, we should showthat both u − (] − ∞ , a ]) and u − ([ a, + ∞ [) are closed for every a ∈ R . Ifinf u < a < sup u , then u − ([ a, + ∞ [) = X \ ∪ y ∈ D : u ( y ) a { z ∈ X | z ≻ y } , and both are clearly closed. The case in which either a ≤ inf u or sup u ≤ a can be treated easily, and thus we omit the proof in such cases.To prove that u represents ( % , ≥ ), it suffices to show that[ y → x ] ≥ [ w → z ] ⇔ u ( x ) − u ( y ) ≥ u ( z ) − u ( w ) . For this, it suffices to show the following two claims:i) if u ( x ) − u ( y ) ≥ u ( z ) − u ( w ), then [ y → x ] ≥ [ w → z ], andii) if u ( x ) − u ( y ) > u ( z ) − u ( w ), then [ y → x ] > [ w → z ].22e show that i) implies ii). Suppose that i) is correct. Choose any x, y, z, w ∈ X such that u ( x ) − u ( y ) > u ( z ) − u ( w ). Note that, because X is connected, u ( X ) is also connected in R , and thus it is convex. If u ( x ) > inf u , then thereexists x ′ ∈ X such that u ( x ′ ) − u ( y ) ≥ u ( z ) − u ( w ) and x ≻ x ′ . Then,[ y → x ] > [ y → x ′ ] ≥ [ w → z ] , which implies that [ y → x ] > [ w → z ]. If u ( x ) = inf u , then w > inf u because u ( x ) + u ( w ) > u ( y ) + u ( z ). Therefore, there exists w ′ such that u ( x ) − u ( y ) ≥ u ( z ) − u ( w ′ ) and w ≻ w ′ . Then,[ y → x ] ≥ [ w ′ → z ] > [ w → z ] , which implies that [ y → x ] > [ w → z ]. Therefore, our claim is correct, andthus it suffices to show i).We need the following lemma. Lemma 7 . Suppose that x ∈ X . Then, there exist x ′ ∈ X and a sequence( x m ) on D ′ such that u ( x ) = u ( x ′ ) and x m → x ′ as m → ∞ . Moreover, if u ( x ) < sup u (resp. u ( x ) > inf u ), then we can assume that u ( x m ) > u ( x )(resp. u ( x m ) < u ( x )) for every m . Proof . We treat only the case in which u ( x ) < sup u , because the remainingcase can be treated symmetrically.By assumption, there exists ˆ x ∈ X such that u (ˆ x ) > u ( x ). Because X ispath-connected, there exists a continuous function f : [0 , → X such that f (0) = x, f (1) = ˆ x . Define t ′ = sup { t ∈ [0 , | u ( f ( t )) ≤ u ( x ) } and x ′ = f ( t ′ ).Clearly, t ′ < u ( x ′ ) = u ( x ). Let g ( t ) = u ( f ( t )). Then, we have thatif t ′ < t <
1, then g ( t ) > g ( t ′ ) = v ( x ). Therefore, by the intermediatevalue theorem, there exists t such that t ′ < t < f ( t ) ∈ D ′ . Define x = f ( t ). Next, suppose that t m and x m = f ( t m ) is defined, where t m > t ′ and x m ∈ D ′ . Because t m > t ′ , we have that t ′ < t < t m implies g ( t ) > u ( x ).Therefore, again by the intermediate value theorem, there exists t m +1 suchthat t ′ < t m +1 < t ′ + t m and x m +1 = f ( t m +1 ) ∈ D ′ . By construction, we havethat 0 < t m − t ′ < ( t − t ′ ) / m − , and thus t m → t ′ as m → ∞ . By the continuity of f , we have that x m → x ′ as m → ∞ . This completes the proof. (cid:4) In the proof of Lemma 7, we need the path-connectedness of X . In proving Theorem1, only Lemma 7 requires the path-connectedness of X , and the remaining part uses onlyconnectedness. Therefore, if there is a proof of Lemma 7 that does not use the path-connectedness of X , we can extend Theorem 1 to a general Hausdorff, separable, andconnected topological space. u ( x ) − u ( y ) ≥ u ( z ) − u ( w ) . Then, we have that u ( x ) + u ( w ) ≥ u ( y ) + u ( z )If u ( y ) = u ( z ) = inf u , then clearly[ y → x ] ≥ [ w → x ] ≥ [ w → z ] . If u ( x ) = u ( w ) = sup u , then clearly[ y → x ] ≥ [ y → z ] ≥ [ w → z ] . Therefore, without loss of generality, we can assume that either u ( x ) < sup u or u ( w ) < sup u , and either u ( y ) > inf u or u ( z ) > inf u . We treat the casein which u ( x ) < sup u and u ( y ) > inf u , because the remaining cases can betreated similarly.Because u ( x ) < sup u and u ( y ) > inf u , by Lemma 7, there exist x ′ , y ′ , z ′ , w ′ ∈ X and sequences ( x m ) , ( y m ) , ( z m ) , ( w m ) on D ′ such that u ( x m ) > u ( x ) and u ( y m ) < u ( y ) for all m , u ( x ′ ) = u ( x ) , u ( y ′ ) = u ( y ) , u ( z ′ ) = u ( z ) , u ( w ′ ) = u ( w ) , and x m → x ′ , y m → y ′ , z m → z ′ , w m → w ′ as m → ∞ . Then, for every m , u ( x m ) − u ( y m ) > u ( z ) − u ( w ) , and thus, there exists k ( m ) ≥ m such that u ( x m ) − u ( y m ) ≥ u ( z k ( m ) ) − u ( w k ( m ) ) . Because u represents ( % , ≥ ) on D ′ , we have that[ y m → x m ] ≥ [ w k ( m ) → z k ( m ) ] . Therefore, by continuity, we have that[ y → x ] = [ y ′ → x ′ ] ≥ [ w ′ → z ′ ] = [ w → z ] . This implies i). Hence, u represents ( % , ≥ ) on X .The rest of our claim is the uniqueness of such a u up to a positive affinetransform. Suppose that v : X → R is a continuous representation of ( % , ≥ ).24f x ∼ y for all x, y ∈ X , then v is a constant function, in which case theuniqueness is obvious. Hence, we assume that x ∗ , y ∗ ∈ X such that x ∗ ≻ y ∗ ,and define u : X → R as in the above proof. Then, u is a representation of( % , ≥ ) such that u ( a ki ) = i k for all a ki ∈ D . We can easily show that v ( a ki ) = v ( a ki ) − v ( a k ) + v ( y ∗ )= i ( v ( a k ) − v ( a k )) + v ( y ∗ )= i k ( v ( x ∗ ) − v ( y ∗ )) + v ( y ∗ )= ( v ( x ∗ ) − v ( y ∗ )) u ( a ki ) + v ( y ∗ ) . Because v is continuous, by Lemma 7, v ( x ) = ( v ( x ∗ ) − v ( y ∗ )) u ( x ) + v ( y ∗ ) , as desired. This completes the proof. (cid:4) A.3 Proof of Theorem 3
Suppose that u is continuously differentiable on R n ++ and Du ( x ) = 0 forevery x ∈ R n ++ . Define v ( x, y ) = u ( x ) − u ( y ). Then, I = { ( x, y ) ∈ R n ++ | x ∼ y } = v − (0) . By the preimage theorem, we have that I is a 2 n − C manifold,and thus Debreu’s smoothness holds. Next, recall that e = (1 , , ..., g ( b ) = u ( be ). Then, g is a concave, increasing, and continuouslydifferentiable function defined on R ++ , and because Du ( x ) = 0, we havethat g ′ ( b ) = Du ( be ) e >
0. Therefore, by the inverse function theorem, wehave that g − is also continuously differentiable. This implies thatlim a ↓ b − f ( a, b ) a = 0 ⇔ lim a ↓ g ( b ) − g ( f ( a, b )) a = 0 . Moreover, because g is concave, 2 g ( b ) ≥ g ( b + a ) + g ( b − a ), and thus, wehave that b ≥ f ( a, b ). This implies thatlim inf a ↓ g ( b ) − g ( f ( a, b )) a ≥ . See section 1.4 of Guillemin and Pollack (1974). ≤ lim sup a ↓ g ( b ) − g ( f ( a, b )) a = lim a ↓ g ( b ) − g ( b − a ) a − lim inf a ↓ g ( f ( a, b )) − g ( b − a ) a = lim a ↓ g ( b ) − g ( b − a ) a − lim inf a ↓ g ( b + a ) − g ( f ( a, b )) a ≤ lim a ↓ g ( b ) − g ( b − a ) a − lim inf a ↓ g ( b + a ) − g ( b ) a = g ′ ( b ) − g ′ ( b ) = 0 , which implies that line smoothness is satisfied.Conversely, suppose that ( % , ≥ ) satisfies both Debreu’s smoothness andline smoothness. First, we show that the function g ( b ) = u ( be )is continuously differentiable on R ++ . Choose any b >
0. Because g isconcave, it is locally Lipschitz, and thus there exist ε > L > | g ( c ) − g ( b ) | ≤ L | c − b | for all c ∈ [ b − ε, b + ε ]. Moreover, the left- andright-side derivatives D − g ( b ) , D + g ( b ) can be defined and are real numbers,and g is differentiable if and only if D − g ( b ) = D + g ( b ). Because 2 g ( b ) ≥ g ( b + a ) + g ( b − a ), we have that f ( b, a ) ≤ b for all a >
0, and thus bymonotonicity, we have that g ( b ) ≥ g ( f ( a, b )). Moreover,0 ≤ lim inf a ↓ g ( b ) − g ( f ( a, b )) a ≤ lim sup a ↓ g ( b ) − g ( f ( a, b )) a ≤ L lim a ↓ b − f ( a, b ) a = 0 , and thus lim a ↓ g ( b ) − g ( f ( a, b )) a = 0 . D + g ( b ) = lim a ↓ g ( b + a ) − g ( b ) a = lim a ↓ g ( b + a ) − g ( f ( a, b )) a = lim a ↓ g ( f ( a, b )) − g ( b − a ) a = lim a ↓ g ( b ) − g ( b − a ) a = D − g ( b ) , as desired. Because every differentiable and concave function is continuouslydifferentiable, we have that g is continuously differentiable.Because g is increasing and concave, we must have g ′ ( b ) > b > u is continuous and nondecreasing, for every x ∈ R n ++ , thereuniquely exists a ( x ) > u ( x ) = u ( a ( x ) e ) = g ( a ( x )). Therefore,it suffices to show that, under monotonicity and Debreu’s smoothness, a : x a ( x ) is continuously differentiable and Da ( x ) = 0 on R n ++ . We treatthis result as a lemma. Lemma 8 . Suppose that X is either R n + or R n ++ , and % is a closed weakorder on X that satisfies monotonicity and Debreu’s smoothness. Then,for every x ∈ R n ++ , there uniquely exists a ( x ) such that x ∼ a ( x ) e , where e = (1 , , ..., a : R n ++ → R ++ is continuouslydifferentiable, and Da ( x ) = 0 for every x ∈ R n ++ . Proof . It is easy to show that a ( x ) is well-defined for all x ∈ R n ++ , and thuswe omit its proof. Choose any x ∈ R n ++ , and define I x = { y ∈ R n ++ | x ∼ y } . If I is a C manifold, then this result was obtained by Debreu (1972), althoughhis proof has a gap. If I is a C ∞ manifold, then this result was rigorously proved inch.7 of Bridges and Mehta (1995). However, we can only assume that I is C , andthus these results cannot be applied. Mas-Colell (1977) noted that Moulin solved thisproblem in 1973. However, we could not obtain Moulin’s monograph, and thus this claimis unverifiable. This is why we present the proof of the smoothness of a ( x ). Note that both monotonicity and Debreu’s smoothness are requirements on % alone,and that there is no relationship between ≥ and these two requirements. See Proposition 3.C.1 of Mas-Colell, Whinston, and Green (1995).
27e first show that I x is connected. For this, it suffices to show that, for every y ∈ I x , there exists a continuous function f : [0 , → I x such that f (0) = x and f (2) = y .Define z i = min { x i , y i } >
0, and b ( t ) = ( (1 − t ) x + tz if 0 ≤ t ≤ , (2 − t ) z + (1 + t ) y if 1 ≤ t ≤ . Let M = max i {| x i − y i |} . By monotonicity and the intermediate valuetheorem, there uniquely exists c ( t ) ∈ [0 , M ] such that b ( t ) + c ( t ) e ∈ I x . Let( t k ) be the sequence in [0 ,
2] such that t k → t as k → ∞ . Suppose that c ( t k ) c ( t ). Because c ( t k ) ∈ [0 , M ], there exists a subsequence t m ( k ) suchthat c ( t m ( k ) ) → α = c ( t ) as k → ∞ . Because % is closed in X , we havethat b ( t ) + αe ∈ I x , which contradicts the uniqueness of c ( t ). Therefore,we have that c ( t ) is continuous, and f ( t ) = b ( t ) + c ( t ) e satisfies all of ourrequirements.For x ∈ R n ++ , define L x as the set of all y ∈ I x such that the function a iscontinuously differentiable around y and Da ( y ) = 0. To prove this lemma,it suffices to show that x ∈ L x for all x ∈ R n ++ . Fix x ∈ R n ++ . To show that x ∈ L x , it suffices to show that L x = I x . Because I x is connected, it sufficesto show that L x is nonempty, open, and closed with respect to the relativetopology of I x .It is clear that L x is open.Choose any y ∈ I x . Then, a ( y ) = a ( x ). Because I is a 2 n − C manifold, there exist open neighborhoods U ⊂ R n − of 0 and V ⊂ I of( y, y ), and a C diffeomorphism ϕ : U → V such that ϕ (0) = ( y, y ). Let T ( y,y ) ( I ) be the image of R n − for the linear operator Dϕ (0). This linearspace is called the tangent space of I at ( y, y ). It is known that v ∈ T ( y,y ) ( I )if and only if there exists a continuously differentiable function c : [ − ε, ε ] → I such that c (0) = ( y, y ) and ˙ c (0) = v . Because c ( t ) = ( y + te, y + te ) ∈ I , wehave that ( e, e ) ∈ T ( y,y ) ( I ). If (0 , e ) ∈ T ( y,y ) ( I ), then( − e, e ) = 2(0 , e ) − ( e, e ) ∈ T ( y,y ) ( I ) , and thus, there exists a continuously differentiable function c : [ − ε, ε ] → I such that ˙ c (0) = ( − e, e ). This means that there exists ( z, w ) ∈ I such that w ≫ y and y ≫ z . However, by monotonicity, w ≻ y ≻ z ∼ w, The statement “ ϕ is a C diffeomorphism” means that ϕ is a bijection and both ϕ and ϕ − are continuously differentiable. Note that, V is open not in R n but only in I ,and thus to understand this statement rigorously, we must define the differentiability offunctions defined on non-open sets. See section 1.1-1.2 of Guillemin and Pollack (1974). ψ ( v, a ) = ϕ ( v ) + (0 , ae ) . Then, by the above result, we have that Dψ (0 ,
0) is a regular matrix, andthus, by the inverse function theorem, there exist open neighborhoods U ′ ⊂ R n of (0 ,
0) and V ′ ⊂ R n ++ of ( y, y ) such that ψ : U ′ → V ′ is a C diffeo-morphism. Note that for ( z, w ) ∈ V ′ , z ∼ w if and only if ψ − ( z, w ) = ( v, v . Let χ ( z, w ) = ψ − ( z, w ), and h ( z, b ) = χ n ( z, y + be ). Then, h iscontinuously differentiable, and z ∼ ( y + be ) if and only if h ( z, b ) = 0.We show that D y h ( y, = 0 and ∂h∂b ( y, = 0. Because h ( y + be, b ) = 0for all b , we have that D y h ( y, e = − ∂h∂b ( y, ∂h∂b ( y, = 0.Suppose that ∂h∂b ( y,
0) = 0. This implies that Dχ n ( y, y )(0 , e ) = 0 . Meanwhile, (0 , e ) = Dψ (0 , Dχ ( y, y )(0 , e ) , and Dψ (0 , w, b ) = Dϕ (0) w + b (0 , e ) . Therefore, we have that (0 , e ) = Dϕ (0) w for some w ∈ R n − . This implies that (0 , e ) ∈ T ( y,y ) ( I ), which contradictsour previous arguments. Therefore, our claim is correct. By the implicitfunction theorem, we have that there exists an open neighborhood W ′ of y and a continuously differentiable function b : W ′ → R such that z ∼ y + b ( z ) e and Db ( z ) = 0 for every z ∈ W ′ .If y = a ( x ) e , then b ( z ) coincides with a ( z ) − a ( x ), and thus we have that a ( x ) e ∈ L x and L x is nonempty.Thus, it suffices to show that L x is closed in I x . Choose a sequence ( y m )on L x and suppose that y m → y ∈ I x as m → ∞ . Consider a function b : U → R as above. Because of monotonicity, we have that b is increasing,and thus Db ( z ) e > z ∈ W ′ . Because y m → y as m → ∞ , we havethat there exists m such that y m ∈ W ′ . Consider the following equation: b ( z ) − b ( y m + te ) = 0 . This equation holds for ( z, t ) = ( y, W ′′ ⊂ W ′ of y and a continuously29ifferentiable function t : W ′′ → R such that b ( z ) = b ( y m + t ( z ) e ), whichimplies that a ( z ) = a ( y m + t ( z ) e ) . Because y m ∈ L x , we have that a is continuously differentiable around y and Da ( y ) = 0, which implies that y ∈ L x and L x is closed. This completes theproof. (cid:4) Because of Lemma 8, we have that u satisfies all of our requirements.This completes the proof. (cid:4) Acknowledgement
We are grateful to Chaowen Yu and Shinsuke Nakamura for his helpful com-ments and suggestions.