Non-Smooth Integrability Theory

We study a method for calculating the utility function from a candidate of a demand function that is not differentiable, but is locally Lipschitz. Using this method, we obtain two new necessary and sufficient conditions for a candidate of a demand function to be a demand function. The first concerns the Slutsky matrix, and the second is the existence of a concave solution to a partial differential equation. Moreover, we show that the upper semi-continuous weak order that corresponds to the demand function is unique, and that this weak order is represented by our calculated utility function. We provide applications of these results to econometric theory. First, we show that, under several requirements, if a sequence of demand functions converges to some function with respect to the metric of compact convergence, then the limit is also a demand function. Second, the space of demand functions that have uniform Lipschitz constants on any compact set is compact under the above metric. Third, the mapping from a demand function to the calculated utility function becomes continuous. We also show a similar result on the topology of pointwise convergence.


Introduction
The question of how to measure utility had already been raised since the time of Bentham (1789). This question took on a special place in economics as Walras (1874) incorporated utility into his model and this model spread as a general equilibrium theory. Because of the difficulty of the problem, economics at one time placed the greatest emphasis on analyses in which utility need not be measured. That is the so-called axiomatic approach. The axiomatic approach does not require any particular form of utility, but assumes only 'axioms' that seem reasonable and deduces properties about the economy from them. The existence of equilibrium and the fundamental theorems of welfare economics are the typical results of this approach.
However, Debreu (1974) showed the Sonnenschein-Mantel-Debreu theorem, which demonstrated the limitation of the axiomatic approach. According to this theorem, for any continuous function on the unit simplex, we can construct an economy such that the given function coincides with the excess demand function of this economy except for some neighborhood of the boundary. The only conditions imposed on this economy are the following: the economy is a pure exchange economy and all consumers have continuous, increasing, and strictly quasi-concave utility functions, which are all natural and frequently used in the axiomatic approach. This result was taken to mean that there is almost nothing that can be said about the economy from the usual axioms used in the axiomatic approach.
An alternative to the axiomatic approach, which has therefore come into use for economic analysis, is the method of estimating utility functions from data. However, an important problem arises here. That is, the only data available to us for estimating utility functions is data on the consumer's purchase behavior. In essence, therefore, we can estimate not the utility function, but only the demand function. Because of this problem, the estimation method currently used is basically a method of parameter estimation where the demand function and the utility function is exogenously assumed to correspond one-to-one through their parameters. This method is known as calibration.
Our purpose is to construct a general theory that calculates the utility function from an arbitrary demand function without giving such a exogenous assumption. The history of such works is very old and was discussed by Antonelli (1886) long before the axiomatic approach reached its zenith. This work was succeeded by Pareto (1906), followed by Samuelson (1950) and then greatly improved by Hurwicz and Uzawa (1971) to its present form. This research area is known as integrability theory.
In this context, Hosoya (2017) gave a specific method for calculating the corresponding utility function from a given candidate of the demand function. This paper also discussed how to handle the case where the demand function contains errors. Let us elaborate on this issue. In order to obtain the demand function, we need to estimate it from the purchase behavior of the consumer, as discussed above. The problem addressed by Hosoya (2017) is the following. If the error in the demand function is small enough, is the error in the utility function calculated by our constructed method also small enough?
The problem is essentially that of continuity. In other words, it requires the property that when the demand function moves a little, the corresponding utility function also moves only a little. And the most important aspect of the continuity problem is the topology. Because Hosoya (2017) assumed that the demand function is assumed to be continuosly differentiable, the local C 1 topology was used for the space of demand functions. However, this result was problematic in many ways. The most significant problem is that there are few econometric results that discuss the space of the demand function with the local C 1 topology. This is because there is almost no currently known estimation method that allows the error to converge to 0 in the local C 1 topology as the data size increases. Therefore, even if good results are obtained in this topology, the results cannot be used in econometrics.
Therefore, a weaker topology is needed. A common use in econometric theory for this problem is the topology of compact convergence: that is, the topology whose convergence concept coincides with the uniform convergence on any compact set. However, using this topology means that the space of continuously differentiable functions is not complete. We therefore need to reproduce the results of Hosoya (2017) without assuming differentiability for the demand function. This is the research objective of this paper.
However, if we weaken the assumption of differentiability to that of continuity, we are confronted with the problem found by Mas-Colell (1977). He showed the existence of a continuous demand function such that there are two corresponding continuous utility functions representing different preference relations. Applying this to our context means that in the space of continuous demand functions, even if the error of the demand function is 0, the estimated error of the utility function may not be 0. This is highly undesirable for us.
Thus, as a compromise, we assume that the demand function is locally Lipschitz. Any locally Lipschitz function is differentiable almost everywhere by Rademacher's theorem, and thus, it can be seen that for a locally Lipschitz candidate of the demand function, we can define the Slutsky matrix almost everywhere. In this paper, we first use this result to extend Theorem 1 of Hosoya (2017). Namely, when a candidate of the demand function satisfies Walras' law and locally Lipschitz, if its Slutsky matrix is symmetric and semi-negative definite almost everywhere, then the corresponding utility function can be constructed through the calculation of a differential equations (Theorem 1). We present an example of such a calculation where this method works effectively (Example 1).
This result has several important consequences. First, using this result, we can obtain two necessary and sufficient condition for a candidate of the demand function to actually be a demand function (Corollary 1). As was the case under continuous differentiability, symmetry and negative semidefiniteness of the Slutsky matrix "almost everywhere" are one of them. More important necessary and sufficient condition is another one, which is the existence of a unique global concave solution to some partial differential equation for an arbitrary initial value condition. Because this property is robust for limit manipulation, it provides an important stepping stone for the next arguments in this paper.
Moreover, our method of constructing utility functions is upper semicontinuous if it is restricted on the range of the demand function, and we guarantee the uniqueness of the corresponding upper semi-continuous preference relation on the same space. For outside of the range of the demand function, Hosoya (2020) derives a construction method for an upper semicontinuous utility function and ensures the uniqueness of the corresponding upper semi-continuous preference relation when the range of the demand function is sufficiently wide. We slightly variate this construction method and prove that, again, if the range of the demand function is sufficiently wide, an upper semi-continuous utility function can be constructed outside the range of the demand function and the corresponding upper semicontinuous preference relation is unique (Corollary 2). Hence, Mas-Colell's (1977) counterexample vanishes in such a demand function.
With these results as a groundwork, we can finally discuss what we wanted to discuss, namely the continuity of the mapping from the demand function to the utility function. As already mentioned, the topology given to the space of demand functions is only the topology corresponding to uniform convergence on compact sets. On this space, even if a sequence of locally Lipschitz demand functions converges to some function (which is not necessarily demand function), it is not guaranteed that the limit is locally Lipschitz. However, if the limit happens to be locally Lipschitz, we can derive that the limit of a demand function is still a demand function. (Theorem 2) With this in mind, we construct a certain space of functions. It is the space of demand functions such that they satisfy Walras' law and have a Lipschitz constant uniformly on any compact set. Using Theorem 2, we can prove the completeness of this space (Corollary 3). That is, on this space, if a sequence of demand functions is a Cauchy sequence, then it converges to a demand function.
On the other hand, however, for a unique derivation of the utility function, the range of the demand function must be sufficiently wide, as discussed in the explanation of the results of Corollary 2 above. We have found an example of a sequence of demand functions that satisfy all the assumptions of Corollary 3, all of which have a sufficiently wide range, yet in the limit the range is very small (Example 2). Therefore, another additional assumption is needed for the continuity result we wanted. Namely, the range of the function in the limit must also be sufficiently wide. In addition to this, the wanted continuity proposition can be obtained when all functions satisfy an axiom called the "C axiom", which is introduced by Hosoya (2017, 2020) (Theorem 3). That is, when a sequence of demand functions converges to a demand function with respect to the topology discussed above, then the corresponding sequence of utility functions also converges to the corresponding utility function uniformly on any compact set in the positive orthant.
Note that this result does not hold at the corner. We can only derive more naive results at the corner (Corollary 4). Actually, we construct an example where the sequence of the value of the utility function do not converge to the value of the limit utility function at the corner (Example 3). This is an unfortunate result, but we have to accept it.
Finally, by strengthening the C axiom, we can strengthen the result of Corollary 3 and Theorem 3. We construct a new space of demand functions such that the C axiom is uniformly imposed on the whole functions. Then, we can show that this space is complete, and furthermore, the mapping from this space to the corresponding utility function is also continuous (Corollary 5). This is our desired result.
The results in the second half of the paper are specifically constructed with a view to discussing the consistency of estimation methods. For an estimation method of the true value x, it is said to be consistent if the estimated value x N converges in probability to x as the data size N increases. To summarise our results, we obtain the following: if the estimation method for the demand function satisfies consistency, then the estimation method for the utility function that is constructed by the given estimation method for the demand function and the computational process in Theorem 1 also satisfies consistency. We believe that this presents a new way of estimating utility functions for econometric theory. In particular, this result can be applied to any estimation method, whether parametric or non-parametric.
The structure of this paper is as follows. First, Section 2 defines several terms in consumer theory that are necessary for reading this paper. Section 3 presents the main result, whereas Subsection 3.1 discusses the construction method of the utility function, and Subsection 3.2 discusses the completeness of the space and the continuity of the calculation results. Section 4 discusses the position of the present work in the context of related research and provides a list of open problems. Section 5 is the conclusion. As many of the theorems in this paper have long proofs, all proofs are placed in Section 6.

Preliminary
Fix n ≥ 2. Throughout this paper, we use the following notations: R n + = {x ∈ R n |x i ≥ 0 for all i ∈ {1, ..., n}}, and R n ++ = {x ∈ R n |x i > 0 for all i ∈ {1, ..., n}}. The former set is called the nonnegative orthant and the latter set is called the positive orthant. We write Let Ω denote the consumption set. We assume that Ω = R n + unless otherwise stated. A set A ⊂ Ω 2 is called a binary relation on Ω.
For a binary relation A ⊂ Ω 2 , we say that it is • upper semi-continuous if for every x ∈ Ω, the set U(x) = {y ∈ Ω|(y, x) ∈ A} is closed, • upper semi-continuous on B if for every x ∈ B, the set U B (x) = {y ∈ B|(y, x) ∈ A} is closed with respect to the relative topology of B, and A binary relation on Ω is called a weak order if it is complete and transitive. For a weak order , we write x y instead of (x, y) ∈ and x y instead of (x, y) / ∈ . Moreover, we write x ≻ y if x y and y x, and x ∼ y if x y and y x.
Suppose that is a weak order on Ω. If there exists a function u : Ω → R such that x y ⇔ u(x) ≥ u(y), then we say that u represents , or u is a utility function of .
Consider a function f : R n ++ × R ++ → Ω. The following is called the budget inequality.
If the budget inequality holds for all (p, m) ∈ R n ++ × R ++ , then we call this function f a candidate of demand (CoD). If, moreover, for all (p, m) ∈ R n ++ × R ++ , then we say that this CoD f satisfies Walras' law.
Let be a weak order on Ω. For each (p, m) ∈ R n ++ × R ++ , we define We call the set-valued function f the demand relation of , and if it is single-valued, then we call this function f the demand function of . If u represents , then f u denotes f . For a CoD f , if f = f , then we say that f corresponds to and corresponds to f . Of course, if f = f u , then we say that f corresponds to u and u corresponds to f . Let f be a CoD. We say that it is income-Lipschitzian if for ev- Moreover, we say that f is locally Lipschitz if for every compact subset C ⊂ R n ++ × R ++ , there exists L > 0 such that if (p 1 , m 1 ), (p 2 , m 2 ) ∈ C, then Obviously, every locally Lipschitz CoD is income-Lipschitzian. It is well known that every continuously differentiable CoD is locally Lipschitz. Suppose that f is a CoD and f is differentiable at (p, m). Define and let S f (p, m) denote the n×n-matrix whose (i, j)-th component is s ij (p, m). This matrix-valued function S f (p, m) is called the Slutsky matrix. An alternative expression of this matrix is as follows: where f T (p, m) denotes the transpose of f (p, m). Note that, if f is locally Lipschitz, then by Rademacher's theorem, f is differentiable almost everywhere, and thus the Slutsky matrix is defined almost everywhere. We say For a CoD f , we define R(f ) as the range of f . That is, Finally, suppose that f is a CoD such that R(f ) includes R n ++ . Define for each x ∈ R n ++ . This multi-valued function is called the inverse demand correspondence of f . We say that f satisfies the C axiom if G f is compactvalued, convex-valued, and upper hemi-continuous on R n ++ .

First Result: Constructing a Reverse Calculation Method
The first result of this study is to construct a rigorous and effective method for calculating a utility function corresponds to the given CoD.
Theorem 1. Suppose that f is a locally Lipschitz CoD that satisfies Walras' law, (S), and (NSD). Fixp ≫ 0, and define u f,p (x) as follows. First, if x / ∈ R(f ), then define u f,p (x) = 0. Second, if x = f (p, m) for some (p, m), then consider the following differential equatioṅ and define u f,p (x) = c(1). Then, the following holds.
1. u f,p is well-defined, 1 and f = f u f,p .
2. u f,p is upper semi-continuous on R(f ).
3. If f = f for some weak order on Ω that is upper semi-continuous on R(f ), then for every x, y ∈ R(f ), x y ⇔ u f,p (x) ≥ u f,p (y).
As a corollary, we obtain the following result.
Corollary 1. Suppose that f is a locally Lipschitz CoD that satisfies Walras' law. Then, the following four statements are equivalent.
(i) f = f for some weak order on Ω.
(ii) f = f u f,p , where u f,p is defined in Theorem 1.
(iv) For every (p, m) ∈ R n ++ ×R ++ , the following partial differential equation has a unique concave solution.
There are a few remarks on Theorem 1 and Corollary 1. In Hosoya (2017), the same result as Theorem 1 was obtained for continuously differentiable CoDs. Hosoya (2018) showed that the same result holds for differentiable and locally Lipschitz CoDs. In these theorem, the Slutsky matrix is assumed to be symmetric and negative semi-definite at every (p, m). In contrast, our Theorem 1 only requires the symmetry and negative semi-definiteness of the Slutsky matrix at almost every (p, m). This change heavily increase the difficulty of the proof by the following two reasons. First, in (1), f ((1 − t)p + tp, c(t)) appears. However, the set A = {((1 − t)p + tp, c(t))|t ∈ [0, 1]} is a null-set with respect to the Lebesgue measure. Hence, the Slutsky matrix may be undefined in all points of A. This fact denies many techniques used in related research. Second, the classical techniques that derives such a result is constructed in Hurwicz and Uzawa (1971). However, in Hurwicz-Uzawa's proof, the weak axiom of revealed preference was first derived (lemma 5 in their paper), and then the main result was proved using the weak axiom. Because the claim of the weak axiom includes a strict inequality, this property vanishes by any limit manipulation. This indicates that the usual approximation approach cannot work well for the proof of Theorem 1.
In Hosoya (2021), similar result is obtained by using several techniques of partial differential equation. The proof of Hosoya (2021) solved the above difficulties by some purterbation techniques on partial differential equation, although this is difficult to understand. In contrast, we construct the proof of Theorem 1 based on the knowledge of ordinary differential equations. Note also that every continuously differentiable CoD is locally Lipschitz, and thus Theorem 1 is a pure extension of Hosoya's (2017) result.
We mention that (iv) of Corollary 1 is a new necessary and sufficient condition for a CoD to be a demand function of some weak order. It is well known that the strong axiom of revealed preference is necessary and sufficient for a CoD to be a demand function (Richter (1966), Mas-Colell et al. (1995)), and recently it is shown that (S) and (NSD) are necessary and sufficient for a continuously differentiable CoD with Walras' law to be a demand function (Hosoya (2017)). Our condition (iv) is a new alternative necessary and sufficient condition for a locally Lipschitz CoD with Walras' law to be a demand function. Later, we discuss that this condition is crucial for this paper.
Note that, equation (2) is deeply related to the expenditure function. For given weak order on Ω and x ∈ Ω, define This function is called the expenditure function. Actually, E x coincides with the value function of the following minimization problem: This is traditionally called the expenditure minimization problem in consumer theory. It is well known that the expenditure function is concave and continuous. Moreover, if f = f and f is continuous, and if x = f (p, m), then it satisfies (2). This result is usually called Shephard's lemma. See Lemma 1 of Hosoya (2020). Therefore, (i) implies (iv). It is obvious that (ii) implies (i), and it is easy to show that (iv) implies (iii). Finally, Theorem 1 claims that (iii) implies (ii). This is the background logic of Corollary 1.
If f = f and f is continuous and income-Lipschitzian, then we can easily show that E x (p) is the unique solution to (2), and u f,p (x) = E x (p) for all x ∈ R(f ). If R(f ) includes R n ++ and is open in R n + , then by applying a similar proof as Theorem 1 of Hosoya (2020), we obtain the following result.
Then, f = f v f,p and v f,p is upper semi-continuous. Moreover, the following holds.
1) The function f satisfies the C axiom if and only if v f,p is continuous on R n ++ .
2) If f = f for some upper semi-continuous weak order , then v f,p represents . In particular, such a must be unique.
3) f = f for some continuous weak order if and only if v f,p is continuous.
The corresponding demand function is: This function satisfies all requirements of Theorem 1 but is not continuously differentiable. Moreover, , and thus this also satisfies all requirements of Corollary 2. Setp = (1, 1), and choose If necessary, we can replace p 2 with min{2 √ p 1 m, p 2 }, and thus we can assume that p 2 2 ≤ 4p 1 m. Moreover, again if necessary, we can replace (p, m) with 1 p 2 (p, m), and thus we can assume that p 2 = 1. Let us try to solve (1) and to determine u f,p (x) and v f,p (x).
First, define and considerċ To solve (4), we have that and because p 2 =p 2 = 1, In particular, if s = 0 and c 2 (0) = m, then Second, suppose that 4c 2 (1) ≥ 1, where c 2 (0) = m. By our initial as- 1], and thus we obtain that In this case, 4m + 1−p 1 p 1 ≥ 1, and thus we have that Because Moreover, Therefore, we obtain that Third, suppose that 4c 2 (1) < 1, where c 2 (0) = m. By the same argument as above, we have that this assumption is equivalent to Note that 1 = p 1 . If 1 > p 1 , then (p 1 + t(1 − p 1 ))c 2 (t) is increasing, and thus we have that 4p 1 m < 1, which contradicts our initial assumption. Thus, we have that 1 < p 1 . We guess that Then, and hence, we obtain In particular, In conclusion, we obtain that This completes our calculation.

Second Result: Continuity of Calculation
We are now able to calculate a utility function u f,p from a CoD f . However, in real world, we can only obtain a finite data of f , and because f includes infinite data, we cannot obtain f rigorously. Hence, our CoD f must be considered as an estimated value of the true demand function. In this view, we need a continuity result: that is, we wish that if f ′ is near to f , then u f ′ ,p is also near to u f,p . If this condition is violated, then we are confronted with a methodological difficulty: that is, ensuring "consistency" for the estimated utility function becomes hard. We should explain this point in detail. Consider some estimation problem with the true value x ∈ X. Suppose that there is a given estimation method, and for some data with size N, let x N be the estimated value of x. Then, x N is a random variable on X. This estimation method is said to consistent if x N converges to x in probability as N → ∞.
Suppose that there is an estimation method for the true demand function f , and f N is an estimated value of f for some data set with size N. Suppose also that this estimation method is consistent with respect to some topology on the space of demand functions. For each f N , we can calculate the utility function u f N ,p , and thus u f N ,p can be treated as an estimated value of the "true utility function" u f,p . Our question is as follows: does u f N ,p converge to u f,p ? If not, our estimation method violates the consistency, and thus it is not useful.
Hence, the continuity of u f,p with respect to f is very important. In this regard, we first show the following result.

Theorem 2.
Suppose that (f k ) is a sequence of locally Lipschitz demand functions that satisfy Walras' law, and for every compact set C ⊂ R n ++ ×R ++ , f k converges to a CoD f uniformly on C as k → ∞. If f is locally Lipschitz, then f is also a demand function.
As a corollary, we obtain an important result.
where f, f ′ are CoDs. We can easily show that ρ is actually a metric in the space of CoDs, and a sequence (f k ) converges to f with respect to ρ if and only if for every compact set C ⊂ R n ++ × R ++ , (f k ) converges to f uniformly. Suppose that L = (L ν ) is a sequence of positive real numbers. Define F L as the set of demand functions such that f ∈ F L if and only if f satisfies Walras' law and Corollary 3. The space F L is complete with respect to the metric ρ.
By the way, for obtaining the uniqueness of the upper semi-continuous weak order corresponding to f , we need to use Corollary 2, and thus an additional assumption is needed. That is, Suppose that (f k ) is a sequence of F L that converges to f , and for every k, f k satisfies all requirements in Corollary 2. Does f also satisfy requirements in Corollary 2? Unfortunately, the following example indicates that the answer is negative.
Example 2. Consider the class of CES utility functions: where σ < 1 and σ = 0. The corresponding demand function is We assume that σ < 0. To differentiate this function, we have that for j = i and (p, m) ∈ ∆ ν , and thus, if we define L ν = ν 5 , then f σ i ∈ F L . Moreover, for every σ < 0, we have that R(f σ ) = R 2 ++ . However, it is well known that f σ converges to a function f with respect to the metric ρ, where and R(f ) = {(c, c)|c > 0}, even though R(f ρ ) = R n ++ . This fact implies that the limit manipulation in F L may shrinks the range of the demand function.
The above example shows that for our purpose, an additional assumption is needed. One of the easiest way to solve this problem is assuming that the limit CoD f satisfies all assumptions of Corollary 2. The result is as follows.
Theorem 3. Let (f k ) be a sequence on F L that converges to f . Suppose that R(f k ) includes R n ++ and f k satisfies the C axiom for all k, and f also satisfies these conditions. Then, for every compact set D ⊂ R n ++ , The definition of v f,p in Theorem 2 only depends on the values if u f,p on R n ++ . Hence, it seems to be shown that v f k ,p converges to v f,p pointwise, here v f k ,p is defined in Corollary 2. However, there are several technical difficulties, and thus we cannot obtain such a result. Instead, we can show the following result.
or not is unknown. Actually, we find the following example in which ++ , and consider the following equation By the same arguments as in the proof of Example 4 in Hosoya (2020), we can show that there exists a unique solution c * > 0 to the above equation, and if we define u h (x 1 , x 2 ) = c * , u h is C ∞ , monotone, and strictly quasi-concave.
By the same arguments as in the proof of Corollary 2, we can show that u h is upper semi-continuous on R 2 + . Moreover, again by the same arguments as in the proof of Example 4 in Hosoya (2020), we can show that f u h is also then we can easily show that u h ′ → u h with respect to the local C 2 topology on R n ++ , and by Proposition 2.7.2 of Mas-Colell (1985), we have that f h ′ → f h with respect to the metric ρ.

Corollary 5.
Suppose that (f k ) is a sequence on F L,M that converges to f with respect to the metric ρ. Then, f ∈ F L,M , and for every compact set Note that, in our Theorem 3, the C axiom is required. Corollary 2 stated that if R(f ) is relatively open in Ω, this axiom is equivalent to the continuity of v f,p on R n ++ . Actually, the requirement that R(f ) is relatively open in Ω does not used in the proof of this fact, and thus we can show by the same proof as this theorem that under assumptions of our Theorem 3, u f k ,p and u f,p are continuous on R n ++ . Actually, we use continuity of u f,p on R n ++ in the proof of Theorem 3: see Lemma 4. However, v f,p is not necessarily continuous on Ω itself even if R(f ) is open: see Example 4 of Hosoya (2020). If f is a continuously differentiable demand function such that the rank of S f (p, m) is always n − 1, then we can show that the inverse demand correspondence G f (x) is a single-valued continuously differentiable function, and thus the C axiom is automatically satisfied. For a proof, see Proposition 1 of Hosoya (2013). In this regard, if (f k ) is a sequence on F L such that every f k is continuously differentiable and the rank of . This is another sufficient condition for the limit function f of (f k ) to satisfy all requirements of Theorem 3.

Comparizon with Related Literature
The history of integrability theory begins with Antonelli (1886). This theory aims to calculate a utility function from the consumer's purchase behavior. Hurwicz (1971) classified this theory into two categories. One derives the inverse demand function from the purchase behavior and then calculates a utility function by finding a function that satisfies Lagrange's first-order conditions for the inverse demand function, which is called the indirect approach. On the other hand, the other derives the demand function from the purchase behavior, solves Shepherd's lemma as a partial differential equation to calculate the expenditure function, and then calculates the utility function from this expenditure function, which is called the direct approach. Antonelli (1886) used the indirect approach and most of the classical results including Samuelson (1950), Katzner (1970), and Debreu (1972), also used the indirect approach. On the other hand, Hurwicz and Uzawa (1971) are a classical result for the direct approach. A recent paper, Hosoya (2013), is an example of the indirect approach, while Hosoya (2017) is an example of the direct approach. Theorem 1 in this paper is categorized as a direct approach.
To understand the position of this paper in integrability theory with direct approach, let us look at the classical result due to Hurwicz-Uzawa. They showed that if a CoD is differentiable and local Lipschitz, satisfies Walras' law, (S), (NSD), and a condition called the "strong income-Lipschitzian" requirement, then this CoD is a demand function. Although they do not specify in their theorem how to derive the utility function, their utility function appears in the proof essentially coincides with our u f,p . After this paper, several studies tried to remove the "strong income-Lipschitzian" requirement, and Hosoya (2017, 2018) finally succeeded in doing so.
Let us explain the logic that allowed us to eliminate the strong income-Lipschitzian requirement. In the proof of Hurwicz-Uzawa's Theorem 2, this condition is only used for deriving the existence of the solution to the partial differential equation (2). In fact, it is well known that the necessary and sufficient condition for the existence of local solutions to (2) is (S), which was proved in Theorem 10.9.4 of Dieudonne (1969). Hurwicz-Uzawa constructed a similar proof as theorem of Nikliborc (1929) to verify the existence of a global solution to (2), in which the strong income-Lipschitzian condition and (S) were used. Then, from it and (NSD), they proved the claim that is the same as Step 4 in the proof of our Theorem 1. Once the existence of global solutions to (2) and the statement in Step 4 have been shown, we do not need to use any differentiability of f from that point onwards for proving this theorem. In contrast, Hosoya (2017) brought (NSD) into the proof of the existence of the global solution to (2) and verified that the existence of global solutions can be shown without the strong income-Lipschitzian condition. This is why we can eliminate the strong income-Lipschitzian condition from this theorem. Hosoya (2017) treated continuously differentiable CoDs, and Hosoya (2018) extended this result to differentiable and locally Lipschitz CoDs.
On the other hand, this paper removes even differentiability and assumes only local Lipschitz conditions for CoDs. This poses two difficulties in the proof. First, as already mentioned, the Lebesgue measure of the trajectory of ((1−t)p+tp, c(t)) in (1) is 0. By Rademacher's theorem, any locally Lipschitz function is differentiable almost everywhere. However, the demand candidate f may not be differentiable at any point on the above trajectory because this trajectory is a null set. And even if it is differentiable, the Slutsky matrix may not have good properties. This means that most of the tools used in integrability theory up to now cannot be used as they are. This problem is solved for perturbing the solution to the differential equation by the income. See Lemma 1 in the proof. Another difficulty arises in Step 4 of the proof of Theorem 1. This Step claims that p · y > m. The problem is that this inequality is strong. The proof using the past techniques is no longer possible, because the strong inequality is replaced by a weak inequality by the limit manipulation. This can be solved by rigorous evaluation fot the inequality of the perturbed trajectory, but this evaluation is not so straightforward. See the proof in Step 4.
Corollary 1 is derived from Theorem 1. This is one of the newest results in integrability theory. As implied by Houthakker (1950) and shown by Uzawa (1960) and Richter (1966), a necessary and sufficient condition for a CoD to be a demand function is the strong axiom of revealed preference. For this result, there is no topological condition imposed on the CoD. On the other hand, if a CoD satisfies Walras' law and is continuously differentiable, then a necessary and sufficient conditions for it to be a demand function are (S) and (NSD). This is shown in Hosoya (2017). Corollary 1 shows that the same result holds even if the CoD is not differentiable but only locally Lipschitz. The most important thing about Corollary 1, however, is that it shows another necessary and sufficient condition for a CoD to be a demand function. That is the existence of a concave global solution to the partial differential equation (2).
To illustrate the importance of this result, we begin by recalling the strong axiom of revealed preference. A CoD f satisfies the strong axiom of revealed preference if and only if for every finite sequence x 1 , ..., x k such that The problem is that there is a strong inequality in this last claim. Even if f k satisfies this condition, this strong inequality changes a weak inequality in the limit f , and thus whether f is actually a demand function becomes unknown. This implies that the strong axiom of revealed preference cannot be used for proving results such as Theorem 2.
A similar problem arises when we discuss this problem using conditions (S) and (NSD). Even if f k converges to f with respect to the metric ρ, it is uncertain that the derivative converges. Therefore, even if f k satisfies (S) and (NSD), f may violate (S) or (NSD), which fails to prove that f is a demand function. If we change the metric and use a stronger topology, we can show that (S) and (NSD) hold in the limit f . In this case, however, it becomes difficult to find results on econometric theory corresponding to such a topology. Therefore, these conditions are also undesirable.
In contrast, condition (iv) of Corollary 1 fundamentally resolves this problem. Indeed, in the proof of Theorem 2, we confirm that f satisfies condition (iv). This property (iv) is not broken by the convergence with respect to ρ, which makes such a proof possible. We can say that all the rest of our results also depend on this property (iv).
Theorem 3 requires the C axiom. This axiom was first discovered by Hosoya (2017). This axiom did not appear in Hosoya (2015), who showed a result similar to Theorem 3 in integrability theory with the indirect approach. Since G f (x) is assumed to be a single-valued, continuously differentiable function in the indirect approach, this axiom automatically holds. This is the reason why the C axiom disappears in Hosoya (2015). The C axiom is known to be equivalent to another axiom called the NLL axiom. In Theorem 2 of Hosoya (2020), it was shown that for an income-Lipschitzian demand function f that satisfies Walras' law, f = f u for some function u : Ω → R such that u is continuous on R n ++ if and only if f satisfies the C axiom. In this paper, this result is required for proving Theorem 3. See Lemma 4.

Several Open Problems
In this paper, we have attempted to produce the best results as far as possible. However, there remained several problems that we just could not solve. Here we describe a few of them, which we consider important.
First, in Corollary 1, we proved the equivalence of (i) and (iv) by assuming that f is locally Lipschitz. Can this equivalence also be proved when f is continuous and income-Lipschitzian? If it could be said, then the income-Lipschitzian requirement is sufficient for the proof of Theorem 2, so the result of Theorem 2 can be discussed on a wider space than the space of local Lipschitz CoDs. This would also strengthen Corollary 3 and Theorem 3.
Second, the question remains as to whether Corollary 4 can be strengthened. We have only shown that lim sup k→∞ v f k ,p (x) ≤ v f,p (x), and find an example such that the inequality becomes strong. However, maybe this inequality could be turned into an equality with some weak additional assumption. In particular, the equality may be guaranteed when x is an element of R(f ). If we could prove this, our result would be much better.
The third problem is related to the second problem. In Example 2, we saw an example where R(f ) shrinks at the limit. However, we do not yet know of any examples where R(f ) expands by limit operations. What we want to know is that whether R(f ) ⊂ lim sup k R(f k ) when R(f ) is an open set and f k → f . If this can be said, then the difficulty in solving the second problem will be reduced.
The fourth problem concerns whether the condition that R(f ) is an open set is necessary in the first place. In the proof of Corollary 2, we show that v f,p coincides with the utility function defined in Hosoya (2020). It is necessary for some technical reasons that R(f ) is an open set in order to guarantee that f = f v f,p . However, there is no known counterexample of f such that

R(f ) is not an open set and
Finally, there remains the task of identifying the conditions for v f,p to be continuous. Condition a. of Theorem 6 in Hurwicz and Uzawa (1971) is a frequently used condition in this context. This condition states that if p ≥ 0, p = 0 and p i = 0 for some i, then for any convergent sequence (p k ) to p on R n ++ and any (q, w) ∈ R n ++ , f (p k , v f,p k (f (q, w))) is unbounded. However, when discussing this condition, Hurwicz-Uzawa restricts the domain of the utility function to R(f ). Therefore, whether the continuity of v f,p outside R(f ) can be guaranteed by this condition is one of the open questions.

Conclusion
In this study, we obtained a calculate procedure for a utility function from a given locally Lipschitz CoD that satisfies Walras' law. Using this procedure, we found two necessary and sufficient conditions for a locally Lipschitz CoD that satisfies Walras' law to be a demand function. Moreover, under the assumption that the range of this CoD includes positive orthant and is open in the consumption space, we obtained the uniqueness result for corresponding upper semi-continuous weak order to this CoD, and derived an upper semicontinuous utility function that represents this weak order.
Using these results, we proved some kind of completeness for the space of demand functions. That is, we showed that for every sequence of demand functions that is locally Lipschitz and satisfies Walras' law, if it converges to some function with respect to topology of compact convergence, then the limit function is also a demand function. From this result, we showed that the space of demand function that has uniform Lipschiz constant on any compact set is complete under this topology.
Furthermore, we showed that if every function has sufficiently wide range and satisfies the C axiom, then our derived utility function is continuous on the demand function. Using this result, we showed that the space of demand functions that has uniform local Lipschitz constant and uniformly satisfies the C axiom is complete, and the mapping from the space of demand functions into the space of utility function is continuous.
We also provided three examples. The first exhibits how our calculating procedure of utility function works well. The second example shows that the range of the CoD may shrink by limit manipulation. The third example indicates that our continuity result never holds in the corner of the consumption space. We think that all examples are meaningful in this context.
Although there are many open problems concerning our results, we believe that the results in this paper is sufficiently good and worthwhile for applied economic research. In particular, we think that our results are a foundation to apply integrability theory into econometric theory.

Mathematical Knowledge on Lipschitz Analysis and Differential Equations
We must use the technique of the Lipschitz analysis in the proof of our theorems repeatedly. However, the Lipschitz property of the solution function (defined later) for ODEs are not so known, and thus we provide knowledge on these facts in this subsection. First, recall the definition of the locally Lipschitz function. Let f : U → R N be some function, where U ⊂ R M is open. This function is called locally Lipschitz if for every compact set C ⊂ U, there exists L > 0 such that for every x, y ∈ C, Because the following property is important, we present a proof in this subsection.
Then, f is locally Lipschitz if and only if for every x ∈ U, there exists r > 0 and L > 0 such that if y, z ∈ U, y − x ≤ r, and z − x < r, then To prove the converse relationship, we use the contraposition method. Suppose that f is not locally Lipschitz. Then, there exists a compact set C and sequences (x k ), (y k ) on C such that for all k, Because C is compact, we can assume that x k → x * , y k → y * as k → ∞.
for every k, which implies that x * = y * . Choose any r > 0 and L > 0. Then, there exists k > L such that x k , y k ∈B r (x * ), and thus the latter claim of this fact is violated. This completes the proof.
By the above fact, we have that every continuously differentiable function is locally Lipschitz. Of course, the converse is not true: consider f (x) = |x|.
The next fact is known as Rademacher's theorem. Because the proof of this fact is long, we omit the proof. Next, we explain some notion on ordinary differential equations (ODEs). First, consider the following ODE: where g : U → R N and U ⊂ R × R N is open. We call a subset I of R an interval if it is a convex set containing at least two points. We say that a function x : I → R N is a solution to (6) if and only if 1) I is an interval containing t 0 , 2) x(t 0 ) = x * , 3) x is absolutely continuous on every compact subset C ⊂ I, and 4)ẋ(t) = g(t, x(t), y) for almost every t ∈ I. Let x : I → R N and y : J → R N be two solutions. Then, we call x an extension of y if J ⊂ I and y(t) = x(t) for all t ∈ J. A solution x : I → R N is called a nonextendable solution if there is no extension except x itself. The next fact is well known, and thus we omit the proof. 3 Fact 3. Suppose that g is locally Lipschitz. Then, for every interval I including t 0 , there exists at most one solution to (6) defined on I, and if there exists a solution, it is continuously differentiable. In particular, there exists a unique nonextendable solution x : I → R N to (6), where I is open and x(t) is continuously differentiable. Moreover, for every compact set C ⊂ U, there exists t 1 , t 2 ∈ I such that if t ∈ I and either t < t 1 or t 2 < t, then (t, x(t)) / ∈ C.
Next, consider the following parametrized ODE: where h : U → R N and U ⊂ R×R N ×R M is open. We assume that h is locally Lipschitz. Fix (y, z) such that (t 0 , z, y) ∈ U. Then, (7) can be seen as (6), where g(t, x) = h(t, x, y) and x * = z. Hence, we can define a nonextendable solution x y,z : I → R N according to Fact 3. We write x(t; y, z) = x y,z (t), and call this function x : (t, y, z) → x(t; y, z) the solution function of (7). The following fact is needed but not so known, and thus we prove this fact in this paper. Proof of Fact 4. First, we introduce a lemma.
Let V ⊂ R × R M × R N be the domain of the solution function x(t; y, z). Choose any (t * , y, z) ∈ V . By Fact 3, there exists an open interval I such that t → x(t; y, z) is a nonextendable solution, and t * ∈ I. Choose t 1 , t 2 ∈ I such that t 1 < min{t * , t 0 } ≤ max{t * , t 0 } < t 2 . Consider the following differential equation: If z ′ = z and y ′ = y, then x * : t → x(t; y, z) is a solution to (8). Choose a, b > 0 so small and define By definition, Π(a, b) and Π ′ (a, b) are compact. We assume that a, b is sufficiently small that Π ′ (a, b) ⊂ U. Because h is locally Lipschitz, there exists L > 0 such that for every (t ′ . Suppose that y ′ −y ≤ b, z ′ −z ≤ b, and define t + 2 (d ′ , c * ) as the supremum of the set of all t ∈]t 0 , t 2 ] such that there exists a solutionx : [t 0 , t] → R N to (8) and (s,x(s), y ′ , z ′ ) ∈ Π(a, b) for all s ∈ [t 0 , t]. By Fact 3, we have that Therefore, by Lemma 1, We show that if Π(a, b), and thus x(t + 2 (y ′ , z ′ )) − x * (t + 2 (y ′ , z ′ )) < a, which contradicts the definition of t + 1 (y ′ , z ′ ). Therefore,x cannot be defined at t + 2 (y ′ , z ′ ). By Fact 3, there exists , which contradicts (9). Hence, if b ′ satisfies (10) and

By a symmetric argument, we have that if
and Clearly, and thus the domain V of the solution function x includes which is a neighborhood of (t * , y, z). Moreover, and thus by Fact 1, we have that the solution function is locally Lipschitz. This completes the proof.
Finally, we mention the formula of the solution to linear differential equations. Consider the following ODE: where I is an interval including t 0 and a : I → R is a bounded measurable function on I. Then, the formula of the solution to the above equation is the following: For the proof, use Theorem 1 of section 0.4 of Ioffe and Tikhomirov (1979).

Proof of Theorem 1
First, consider the following parametrized ODE: and let c(t; p, q, w) denote the solution function of (12). We introduce two lemmas.
Lemma 2. Let U = R n ++ × R ++ . Choose any (p, m) ∈ U. Suppose that W ⊂ U and the Lebesgue measure of U \ W is zero. Moreover, suppose that q ∈ R n ++ and there exists i * ∈ {1, ..., n} such that q i * = p i * , and the domain of the solution function c(t; p, q, w) of the ODE (12) includes [0, t * ] ×P 1 ×P 2 for t * > 0, where P 1 is a bounded open neighborhood of q and P 2 is a bounded open neighborhood of m, andP j denotes the closure of P j . For every (t,r, w) ⊂ R n+1 such that t ∈ [0, t * ], r ∈ P 1 for and w ∈ P 2 , define ξ(t,r, w) = ((1 − t)p + tr, c(t; p, r, w)).
Then, the Lebesgue measure of ξ −1 (U \ W ) is also zero.
Next, define We show that ξ −1 is Lipschitz on V ℓ . Define , and hence they are Lipschitz on V ℓ . Next, consider the following ODE: Let d(s; t, v, c) be the solution function of the above ODE. DefineP 1 as the closure ofP 1 is Lipschitz on this set. Therefore, where L, M > 0 are some constants, and therefore our claim is correct. Now, recall that the Lebesgue measure of U \ W is zero. Because ξ −1 is Lipschitz on V ℓ , we have that the Lebesgue measure of is zero. Therefore, the Lebesgue measure of is also zero. Clearly, the Lebesgue measure of is zero, because this set is included in {(t,r, w)|t ∈ {0, t * }}. This completes the proof of Lemma 2. Lemma 3. Choose any (p, m) ∈ R n ++ . Then, there exists a solution E : R n ++ → R ++ of the following partial differential equation if and only if the domain of the solution function of (12) includes [0, 1] × {p} × R n ++ × {m}. Moreover, in this case Proof. Suppose that a solution E : R n ++ → R ++ to (13) exists. Choose any and by the uniqueness of the solution to ODE (Fact 3), we have that d(t) ≡ c(t; p, q, m). Hence, the domain of the solution function c includes [0, 1] × {p} × R n ++ × {m}, and moreover, E(q) = d(1) = c(1; p, q, m). We show the converse is also true. Suppose that the domain of the solution function c is defined on [0, 1] × {p} × R n ++ × {m}. Define E(q) = c(1; p, q, m). We show that E(q) is a solution to (13).
First, let ∆ * be the set of all (q, w) such that f is differentiable and S f is symmetric and negative semi-definite at ((1 − t)p + tq, c(t; p, q, w)) for almost every t ∈ [0, 1], and the mapping r → c(t; p, r, w) is differentiable at r = q for almost every t ∈ [0, 1]. Suppose that (q, w) ∈ ∆ * and let e i denote the i-th unit vector. Then, for each i ∈ {1, ..., n}, 4 lim h→0 c(t; p, q + he i , w) − c(t; p, q, w) h by the dominated convergence theorem, and thus ∂c ∂q i (t; p, q, w) is defined for all t ∈ [0, 1] and absolutely continuous in t. Define the following absolutely continuous function t)p + tq, c(t; p, q, w)).
By the above evaluation and the symmetry of the Slutsky matrix, we have thaṫ where a(t) be some bounded measurable function. By the formula of the solution to linear ODEs, we have that However, we can easily check that ϕ(0) = 0, and thus ϕ(t) ≡ 0. In particular, ϕ(1) = 0 and thus ∂c ∂q i (1; p, q, w) = f i (q, c(1; p, q, w)).
Second, suppose that q n = p n and i ∈ {1, ..., n − 1}. By Lemma 2 and Fubini's theorem, there exists δ > 0 and a sequence (q k , w k ) on ∆ * such that q k → q, w k → m as k → ∞, and for every k, i ∈ {1, ..., n − 1} and almost every h ∈] − δ, δ[, (q k + he i , w k ) ∈ ∆ * . Then, for every h ∈] − δ, δ[, By the dominated convergence theorem, we have that Third, suppose that q n = p n and i ∈ {1, ..., n − 1}. Let e = (1, 1, ..., 1) and define q k = q + k −1 e. Then, q k n = p n , and thus, for every h ∈] − q i , q i [, and by the dominated convergence theorem, To summarize the above result, we obtain the following: for every q ∈ R n ++ and i ∈ {1, ..., n − 1}, Replacing the role of n to that of 1 and repeating the above arguments, we can show that (14) holds for i = n, and thus DE(r) = f (r, E(r)). This completes the proof.
We now complete the preparation to prove Theorem 1. We separate the proof of Theorem 1 into ten steps.
Proof of Step 1. We show only the former claim, because the latter claim can be shown symmetrically. Define p(t, r) = (1 − t)p + tr, and let ∆(t * ) be the set of all (r, w) such that the domain of t → c(t; p, r, w) includes [0, t * ], and for almost every t ∈ [0, t * ], f is differentiable and S f is symmetric and negative semi-definite at (p(t, r), c(t; p, r, w)). By Lemma 2 and the Fubini's theorem, there exists a sequence (q k , w k ) ∈ ∆(t * ) that converges to (q, m) as k → ∞. Define d(t) = p · x(t) and d k (t) = p · f (p(t, q k ), c(t; p, q k , w k )). Then, d k is absolutely continuous, and for almost all t ∈ [0, t * ], d k (t) = p T S f (p(t, q k ), c(t; p, q k , w k ))(q k − p).
On the other hand, to differentiate both side of the Walras' law, we obtain that (p(t, q k )) T S f (p(t, q k ), c(t; p, q k , w k ))(q k − p) = 0.
Subtracting the latter from the former, we have thaṫ and thus d k (t) is nondecreasing. Because d k (t) → d(t) for every t, we have that d(t) is also nondecreasing, and thus which completes the proof of Step 1.
Step 2. The domain of the solution function c(t; p, q, w) Proof of Step 2. Suppose not. Then, there exists p, q ∈ R n ++ and m ∈ R ++ such that c(t; p, q, m) is defined only on [0, t * [, where t * ≤ 1. Let p(t) = (1 − t)p + tq and x = f (p, m). By Fact 3, we have that the trajectory of the function (p(t), c(t; p, q, m)) is excluded from any compact set in R n ++ × R ++ as t → t * , and thus either lim sup t→t * c(t; p, q, m) = +∞ or lim inf t→t * c(t; p, q, m) = 0. By Step 1, max{p·x, q ·x} ≥ p(t)·x ≥ c(t; p, q, m) for every t ∈ [0, t * [, and thus there exists an increasing sequence (t k ) such that t k → t * and c(t k ; p, q, m) → 0 as k → ∞. Define x k = f (p(t k ), c(t k ; p, q, m)). Because p · x k ≥ m = p · x and p(t k ) · x ≥ c(t k ; p, q, m) = p(t k ) · x k , we have that q · x ≥ q · x k . Hence, x k is a sequence in the following compact set {y ∈ Ω|q · y ≤ q · x}.
Therefore, without loss of generality, we can assume that x k → x * as k → ∞. Because p · x * ≥ m, we have that x * = 0. However, which is a contradiction. This completes the proof of Step 2. which is a contradiction. This completes the proof of Step 3.
Define p(t) = (1 − t)p + tq and d(t) = p · f (p(t), c(t; p, q, m * )). We have already shown in the proof of Step 1 that d(t) is nondecreasing. Therefore, if m * > m, then as desired. Thus, hereafter we assume that m * = m. In this regard, we have that w = c(1; p, q, m), and c(1 − t; q, p, w) = c(t; p, q, m).
Define ∆ * as the set of all (r, c) such that f is differentiable and S f is symmetric and negative semi-definite at ((1 − t)p + tr, c(t; p, r, c)) for almost all t ∈ [0, 1]. By Lemma 2 and Fubini's theorem, we can show that there exists a sequence (q k , w k ) on ∆ * such that (q k , w k ) → (q, w) as k → ∞. Let 2ε = x − y , and define p k (t) = (1 − t)p + tq k and x k (t) = f (p k (t), c(t; p, q k , w k )). Then, x k (1) → y and x k (0) → x as k → ∞, and thus without loss of generality, we can assume that x k (1) − x k (0) ≥ ε for all k. By assumption, x k (t) is a Lipschitz function defined on [0, 1]. If f is differentiable at (p k (t), c(t; p, q k , w k )), then define S k t = S f (p k (t), c(t; p, q k , w k )). By our assumption, S k t can be defined and is symmetric and negative semi-definite for almost all t ∈ [0, 1]. Because S k t is symmetric and negative semi-definite, there exists a positive semi-definite matrix A k t such that S k t = −(A k t ) 2 . Moreover, the operator norm A k t is equal to S k t . 5 Because f is locally Lips-5 Because S k t is symmetric, there exists an orthogonal transform P such that chitz, there exists L > 0 such that S k t ≤ L for all k and almost all t ∈ [0, 1]. Define d k (t) = p · x k (t), and choose δ > 0 such that ε 2 > 2L 2 δ q k − p 2 for every sufficiently large k. By the same arguments as in the proof of Step 1, we can show thaṫ x k (t) = S k t (q k − p) for almost all t ∈ [0, 1]. Therefore, and thus, Letting k → ∞, we have that as desired. This completes the proof of Step 4.
Step 5. If x = y, x = f (p, m), y = f (q, w) and p · y ≤ m, then q · x > w. 6 where λ i is some eigenvalue of S k t . Because S k t is negative semi-definite, λ i ≤ 0 for every i. Hence, if we define Moreover, because the operator norm S k t (resp. A k t ) coincides with max i |λ i |, (resp. max i |λ i |,) all our claims are correct. 6 In other words, f satisfies the weak axiom of revealed preference.
Proof of Step 5. By the contrapositive of Step 4, we have that c(1; p, q, m) > w. By Step 3, m > c(1; q, p, w). By Step 4, we obtain that q · x > w, as desired. This completes the proof of Step 5.
By Steps 2 and 7, we can define u f,p (x) for all x ∈ Ω, and our definition of u f,p (x) is independent of the choice of (p, m).
Step 8. f = f u f,p .
Therefore, x = f u f,p (p, m), as desired. This completes the proof of Step 8.
Step 9. u f,p is upper semi-continuous on R(f ).

Proof of Step 9.
Suppose that x = f (p, m) and u f,p (x) < a. By Fact 4, the solution function c is continuous, and thus there exists ε > 0 such that c(1; p,p, m + ε) < a. Define y = f (p, m + ε). Then, the set is a neighborhood of x in the relative topology of R(f ), and for every z ∈ U, u f,p (z) < u f,p (y) < a. This completes the proof of Step 9.
Step 10. Suppose that f = f for some weak order , and is upper semi-continuous on R(f ). Then, for every x, y ∈ R(f ), Proof of Step 10. First, choose any x ∈ R(f ) and suppose that x = f (p, m).
By Lemma 1 of Hosoya (2020), we have that which implies that d(t) = c(t; p,p, m) for every t ∈ [0, 1]. In particular, Now, choose any ε > 0. Then, there exists w ∈ Ω such that p · w < E x (p) + ε and w x. Define z ε = f (p, E x (p) + ε). Then, z ε w, and thus z ε x. Letting ε → 0, by the upper semi-continuity of , we obtain that z x.
By the same arguments as above, we can show that E z (p) = m, and thus x z. Hence, x ∼ f (p, u f,p (x)) for all x ∈ R(f ). Now, choose any x, y ∈ R(f ). Then, as desired. This completes the proof of Step 10.
Steps 8-10 indicates that all our claims in Theorem 1 is correct. This completes the proof.

Proof of Corollary 1
It is obvious that (ii) implies (i).
Suppose that (iv) holds. Choose any (p, m) ∈ R n ++ × R ++ such that f is differentiable at (p, m). Let E be a concave solution to (2). By easy calculation, we obtain that Because E is concave, the left-hand side of the above equation is negative semi-definite. Moreover, by extended Young's theorem, 7 the left-hand side of the above equation is symmetric. Therefore, f satisfies (S) and (NSD), and (iii) holds.
Finally, our Theorem 1 says that (iii) implies (ii). This completes the proof.

Proof of Corollary 2
Define Theorems 1-2 of Hosoya (2020) stated the following facts: 1) f = f w f,p , 2) w f,p is upper semi-continuous, 3) w f,p is continuous on R n ++ if and only if f satisfies the C axiom, and 4) if is an upper semi-continuous weak order on Ω such that f = f , then for each x, y ∈ Ω, x y ⇔ w f,p (x) ≥ w f,p (y).
We first show that w f,p (x) = v f,p (x) for all x ∈ Ω.
If x ∈ R n ++ , then w f,p (x) = u f,p (x) = v f,p (x). Suppose that x / ∈ R(f ). Choose any ε > 0, y − x < ε and y ∈ R(f ). Then, there exists z ∈ R n ++ such that z ≫ y and z − x < ε. If z = f (p, m), then p · y < m, and thus u f,p (z) > u f,p (y). This indicates that sup{u f,p (y)|y ∈ R(f ), y − x < ε} = sup{u f,p (y)|y ∈ R n ++ , y − x < ε}, Suppose that x ∈ R(f )\R n ++ . Let e = (1, 1, ..., 1) and define x k = x+k −1 e. Then, x k ∈ R n ++ . It is easy to show that Because u f,p is upper semi-continuous on R(f ), we have that On the other hand, if x k = f (p k , m k ), then p k · x < m k , and thus u f,p (x k ) > u f,p (x). Therefore, we have that To combine these inequalities, we have that The rest of the claim of this corollary is 3). If v f,p is continuous, then f = f for a continuous weak order defined as Let us show the converse. Suppose that there exists a continuous weak order on Ω such that f = f . Debreu (1954) showed that there exists a continuous function u : Ω → R such that u represents . By the above arguments, v f,p also represents , and thus v f,p and u have the same order. On the other hand, in the proof of Theorem 1 of Hosoya (2020), it was shown for all x ∈ Ω, and v f,p (0) = 0. Therefore, if v f,p (x) > 0, then and if v f,p (x) = 0, then u(x) = u(0).
Let I * be the set of all t ∈ [0, 1] such that c(s) is defined and c(s) ∈ [K −1 , K] for all s ∈ [0, t], and suppose that and thus, by Lemma 1, This indicates that c k (t) → c(t) as k → ∞, and thus c(t) ∈ [w 0 , w 1 ] for all t ∈ [0, t * [. Because c(t) is a nonextendable solution to (15), by Fact 3, we have that c(t) is defined at t * . By continuity of which is a contradiction. Therefore, I * = [0, t * ]. If t * < 1, then c(t * ) ∈]K −1 , K[, and thus there exists t > t * such that t ∈ I * , which is a contradiction. Thus, t * = 1 and I * = [0, 1], which implies that c(t) is defined on [0, 1]. This completes the proof.

Proof of Corollary 3
Suppose that (f k ) is a Cauchy sequence in F L . It is easy to show that f k → f for some function f : R n ++ × R ++ → Ω, and for every compact set Clearly f satisfies Walras' law. Because ∆ ν is compact, we have that (f k ) converges to f uniformly on ∆ ν . Moreover, if (p, m), (q, w) ∈ ∆ ν , then which implies that f ∈ F L . This completes the proof.

Proof of Theorem 3
First, we show the following lemma.
Lemma 4. Suppose that f is a locally Lipschitz demand function that satisfies Walras' law, and R(f ) includes R n ++ . If f satisfies the C axiom, then u f,p is continuous on R n ++ . 8 Proof. Recall the differential equation (1): Let c(t; p, m) be the solution function. We have that u f,p (x) = c(1; p, m) if x = f (p, m). Choose any sequence (x k ) in R n ++ such that x k → x ∈ R n ++ as k → ∞, and suppose that u f,p (x k ) → u f,p (x). Taking a subsequence, we can assume that there exists ε > 0 such that |u f,p (x k ) − u f,p (x)| > ε for every k. Choose any p k ∈ G f (x k ). Again, taking a subsequence, we can assume that p k → p * ∈ G f (x). Then, which is a contradiction. Therefore, u f,p is continuous on R n ++ . This completes the proof.
Choose any compact set D ⊂ R n ++ . Let x ∈ D. We first show that there exists an open neighborhood U of x and ε > 0 such that if p ∈ G f k (y) for some y ∈ U and k, then p i ≥ ε for all i ∈ {1, ..., n}.
Suppose not. Then, there exists p ℓ ∈ R n ++ and z ℓ = f k(ℓ) (p ℓ , p ℓ · z ℓ ) such that j p ℓ j = 1, z ℓ → x and min j p ℓ j → 0 as ℓ → ∞. First, suppose that k(ℓ) = k for infinitely many ℓ. Taking a subsequence, we can assume that k(ℓ) = k for any ℓ. By the C axiom, the inverse demand correspondence G f k is compact-valued and upper hemi-continuous. Moreover, p ℓ ∈ G f k (z ℓ ) for all ℓ and z ℓ → x as ℓ → ∞. Therefore, if we choose then ε > 0 and min j p ℓ j ≥ ε for sufficiently large ℓ, which is a contradiction. Hence, we can assume that k(ℓ) is increasing. Taking a subsequence, we can assume that p ℓ → p * ∈ R n + , where j p * j = 1 and min i p * i = 0. Choose i, j ∈ {1, ..., n} such that p * i > 0 and p * j = 0.
Let C = {p ∈ R n ++ |p ∈ G f (x) for some x ∈ D}, C k = {p ∈ R n ++ |p ∈ G f k (x) for some x ∈ D}. By the compact-valuedness and upper hemi-continuity of the inverse demand functions, we have that C, C k are compact. Because of our previous arguments, there exists a compact set K ⊂ R n ++ that includes C and all C k . Define m 1 = min{p · x|p ∈ K, x ∈ D} > 0 and m 2 = max{p · x|p ∈ K, x ∈ D} > 0.
Third, it is easy to show that for any demand function f ′ and x ∈ R(f ′ ), G f ′ (x) is convex. Therefore, G f is convex-valued.
Last, suppose that G f is not upper semi-continuous at x. Then, there exists an open neighborhood U of G f (x), sequences (x ℓ ) and (p ℓ ) such that x ℓ → x as ℓ → ∞ and p ℓ ∈ G f (x ℓ ) \ U for all ℓ. Choose ν such that x ℓ ∈]ν −1 , ν[ n for all ℓ. Then, (p ℓ ) is a sequence in the compact set P ν , and thus by taking a subsequence, we can assume that p ℓ → p * ∈ P ν as ℓ → ∞. Because f is continuous, we have that p * ∈ G f (x) ⊂ U, which is a contradiction. Therefore, f satisfies the C axiom. The rest of the claim of Corollary 5 follows from Theorem 3. This completes the proof.