# Non-Smooth Integrability Theory

Yuhki Hosoya E-mail: ukki(at)gs.econ.keio.ac.jp, TEL:+81-90-5525-5142Faculty
of Economics, Chuo University. 742-1, Higashinakano, Hachioji, Tokyo, 192-0393
Japan.

###### Abstract

We study a calculation method for utility function from a candidate of a
demand function that is not differentiable but locally Lipschitz. Using this
method, we obtain two new necessary and sufficient condition for a candidate
of a demand function to be a demand function. First is the symmetry and
negative semi-definiteness of the Slutsky matrix, and second is the global
existence of a unique concave solution to some partial differential equation.
Moreover, we present under several assumptions, the upper semi-continuous weak
order that corresponds to the demand function is unique, and this weak order
is represented by our calculated utility function. We also 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 a metric corresponds to the topology of compact
convergence, then the limit function is also a demand function. Second, the
space of demand functions that have uniform Lipschitz constant on any compact
set is complete under the above metric. Third, under some additional
requirement, the mapping from a demand function into a utility function we
calculate becomes continuous. This implies that a consistent estimation method
for the demand function immediately defines a consistent estimation method for
the utility function by using our calculation method.

Keywords: Integrability Theory, Locally Lipschitz Demand Function,
Rademacher’s Theorem, Completeness of the Space of Demand Functions,
Consistency of the Estimation.

JEL codes: D11, C61, C65, C13.

##  1 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 semi-
definiteness 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 semi-
continuous 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 semi-continuous
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.

##  2 Preliminary

Fix $n\geq 2$. Throughout this paper, we use the following notations:
$\mathbb{R}^{n}_{+}=\\{x\in\mathbb{R}^{n}|x_{i}\geq 0\mbox{ for all
}i\in\\{1,...,n\\}\\}$, and
$\mathbb{R}^{n}_{++}=\\{x\in\mathbb{R}^{n}|x_{i}>0\mbox{ for all
}i\in\\{1,...,n\\}\\}$. The former set is called the nonnegative orthant and
the latter set is called the positive orthant. We write $x\geq y$ if
$x-y\in\mathbb{R}^{n}_{+}$ and $x\gg y$ if $x-y\in\mathbb{R}^{n}_{++}$.

Let $\Omega$ denote the consumption set. We assume that
$\Omega=\mathbb{R}^{n}_{+}$ unless otherwise stated. A set
$A\subset\Omega^{2}$ is called a binary relation on $\Omega$.

For a binary relation $A\subset\Omega^{2}$, we say that it is

  * •

complete if for every $(x,y)\in\Omega^{2}$, either $(x,y)\in A$ or $(y,x)\in
A$,

  * •

transitive if $(x,y)\in A$ and $(y,z)\in A$ imply $(x,z)\in A$,

  * •

upper semi-continuous if for every $x\in\Omega$, the set
$U(x)=\\{y\in\Omega|(y,x)\in A\\}$ is closed,

  * •

continuous if $A$ is closed in $\Omega^{2}$,

  * •

upper semi-continuous on $B$ if for every $x\in B$, the set $U^{B}(x)=\\{y\in
B|(y,x)\in A\\}$ is closed with respect to the relative topology of $B$, and

  * •

continuous on $B$ if $A\cap B^{2}$ is closed in $B^{2}$.

A binary relation $\succsim$ on $\Omega$ is called a weak order if it is
complete and transitive. For a weak order $\succsim$, we write $x\succsim y$
instead of $(x,y)\in\succsim$ and $x\not\succsim y$ instead of
$(x,y)\notin\succsim$. Moreover, we write $x\succ y$ if $x\succsim y$ and
$y\not\succsim x$, and $x\sim y$ if $x\succsim y$ and $y\succsim x$.

Suppose that $\succsim$ is a weak order on $\Omega$. If there exists a
function $u:\Omega\to\mathbb{R}$ such that

$x\succsim y\Leftrightarrow u(x)\geq u(y),$

then we say that $u$ represents $\succsim$, or $u$ is a utility function of
$\succsim$.

Consider a function $f:\mathbb{R}^{n}_{++}\times\mathbb{R}_{++}\to\Omega$. The
following is called the budget inequality.

$p\cdot f(p,m)\leq m.$

If the budget inequality holds for all
$(p,m)\in\mathbb{R}^{n}_{++}\times\mathbb{R}_{++}$, then we call this function
$f$ a candidate of demand (CoD). If, moreover,

$p\cdot f(p,m)=m$

for all $(p,m)\in\mathbb{R}^{n}_{++}\times\mathbb{R}_{++}$, then we say that
this CoD $f$ satisfies Walras’ law.

Let $\succsim$ be a weak order on $\Omega$. For each
$(p,m)\in\mathbb{R}^{n}_{++}\times\mathbb{R}_{++}$, we define

$\Delta(p,m)=\\{x\in\Omega|p\cdot x\in m\\},$
$f^{\succsim}(p,m)=\\{x\in\Delta(p,m)|x\succsim y\mbox{ for all
}y\in\Delta(p,m)\\}.$

We call the set-valued function $f^{\succsim}$ the demand relation of
$\succsim$, and if it is single-valued, then we call this function
$f^{\succsim}$ the demand function of $\succsim$. If $u$ represents
$\succsim$, then $f^{u}$ denotes $f^{\succsim}$. For a CoD $f$, if
$f=f^{\succsim}$, then we say that $f$ corresponds to $\succsim$ and
$\succsim$ 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 every compact
subset $C\subset\mathbb{R}^{n}_{++}\times\mathbb{R}_{++}$, there exists $L>0$
such that if $(p,m_{1}),(p,m_{2})\in C$, then

$\|f(p,m_{1})-f(p,m_{2})\|\leq L|m_{1}-m_{2}|.$

Moreover, we say that $f$ is locally Lipschitz if for every compact subset
$C\subset\mathbb{R}^{n}_{++}\times\mathbb{R}_{++}$, there exists $L>0$ such
that if $(p_{1},m_{1}),(p_{2},m_{2})\in C$, then

$\|f(p,m_{1})-f(p,m_{2})\|\leq L\|(p_{1},m_{1})-(p_{2},m_{2})\|.$

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

$s_{ij}(p,m)=\frac{\partial f_{i}}{\partial p_{j}}(p,m)+\frac{\partial
f_{i}}{\partial m}(p,m)f_{j}(p,m),$

and let $S_{f}(p,m)$ denote the $n\times n$-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:

$S_{f}(p,m)=D_{p}f(p,m)+D_{m}f(p,m)f^{T}(p,m),$

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
that $f$ satisfies (S) if $S_{f}(p,m)$ is symmetric almost everywhere, and
(NSD) if $S_{f}(p,m)$ is negative semi-definite almost everywhere.

For a CoD $f$, we define $R(f)$ as the range of $f$. That is,

$R(f)=\\{x\in\Omega|x=f(p,m)\mbox{ for some
}(p,m)\in\mathbb{R}^{n}_{++}\times\mathbb{R}_{++}\\}.$

Finally, suppose that $f$ is a CoD such that $R(f)$ includes
$\mathbb{R}^{n}_{++}$. Define

$G^{f}(x)=\left\\{p\in\mathbb{R}^{n}_{++}\left|\sum_{i}p_{i}=1,\ f(p,p\cdot
x)=x\right.\right\\}$

for each $x\in\mathbb{R}^{n}_{++}$. This multi-valued function is called the
inverse demand correspondence of $f$. We say that $f$ satisfies the C axiom if
$G^{f}$ is compact-valued, convex-valued, and upper hemi-continuous on
$\mathbb{R}^{n}_{++}$.

##  3 Main Results

###  3.1 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). Fix $\bar{p}\gg 0$, and define $u_{f,\bar{p}}(x)$ as
follows. First, if $x\notin R(f)$, then define $u_{f,\bar{p}}(x)=0$. Second,
if $x=f(p,m)$ for some $(p,m)$, then consider the following differential
equation

$\dot{c}(t)=f((1-t)p+t\bar{p},c(t))\cdot(\bar{p}-p),\ c(0)=m,$ (1)

and define $u_{f,\bar{p}}(x)=c(1)$. Then, the following holds.

  1. 1.

$u_{f,\bar{p}}$ is well-defined,111That is, $c(1)$ is always defined, and
independent of the choice of $(p,m)$. and $f=f^{u_{f,\bar{p}}}$.

  2. 2.

$u_{f,\bar{p}}$ is upper semi-continuous on $R(f)$.

  3. 3.

If $f=f^{\succsim}$ for some weak order $\succsim$ on $\Omega$ that is upper
semi-continuous on $R(f)$, then for every $x,y\in R(f)$,

$x\succsim y\Leftrightarrow u_{f,\bar{p}}(x)\geq u_{f,\bar{p}}(y).$

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.

  1. (i)

$f=f^{\succsim}$ for some weak order $\succsim$ on $\Omega$.

  2. (ii)

$f=f^{u_{f,\bar{p}}}$, where $u_{f,\bar{p}}$ is defined in Theorem 1.

  3. (iii)

$f$ satisfies (S) and (NSD).

  4. (iv)

For every $(p,m)\in\mathbb{R}^{n}_{++}\times\mathbb{R}_{++}$, the following
partial differential equation

$DE(q)=f(q,E(q)),\ E(p)=m,$ (2)

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),
$f((1-t)p+t\bar{p},c(t))$ appears. However, the set
$A=\\{((1-t)p+t\bar{p},c(t))|t\in[0,1]\\}$ 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 $\succsim$ on $\Omega$ and $x\in\Omega$, define

$E^{x}(p)=\inf\\{p\cdot y|y\succsim x\\}.$

This function is called the expenditure function. Actually, $E^{x}$ coincides
with the value function of the following minimization problem:

$\displaystyle\min~{}~{}~{}~{}~{}$ $\displaystyle~{}p\cdot y$ subject to.
$\displaystyle~{}y\in\Omega,$ $\displaystyle~{}y\succsim x.$

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 $f=f^{\succsim}$ 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 $f=f^{\succsim}$ and $f$ is continuous and income-Lipschitzian, then we can
easily show that $E^{x}(p)$ is the unique solution to (2), and
$u_{f,\bar{p}}(x)=E^{x}(\bar{p})$ for all $x\in R(f)$. If $R(f)$ includes
$\mathbb{R}^{n}_{++}$ and is open in $\mathbb{R}^{n}_{+}$, then by applying a
similar proof as Theorem 1 of Hosoya (2020), we obtain the following result.

Corollary 2. Suppose that $f$ is a locally Lipschitz CoD that satisfies
Walras’ law, (S), and (NSD). Moreover, suppose that $R(f)$ includes
$\mathbb{R}^{n}_{++}$ and is relatively open in $\Omega$. Define

$v_{f,\bar{p}}(x)=\begin{cases}u_{f,\bar{p}}(x)&\mbox{if
}x\in\mathbb{R}^{n}_{++},\\\
\inf_{\varepsilon>0}\sup\\{u_{f,\bar{p}}(y)|y\in\mathbb{R}^{n}_{++},\
\|y-x\|<\varepsilon\\}&\mbox{if }x\notin\mathbb{R}^{n}_{++}.\end{cases}$ (3)

Then, $f=f^{v_{f,\bar{p}}}$ and $v_{f,\bar{p}}$ is upper semi-continuous.
Moreover, the following holds.

  1. 1)

The function $f$ satisfies the C axiom if and only if $v_{f,\bar{p}}$ is
continuous on $\mathbb{R}^{n}_{++}$.

  2. 2)

If $f=f^{\succsim}$ for some upper semi-continuous weak order $\succsim$, then
$v_{f,\bar{p}}$ represents $\succsim$. In particular, such a $\succsim$ must
be unique.

  3. 3)

$f=f^{\succsim}$ for some continuous weak order $\succsim$ if and only if
$v_{f,\bar{p}}$ is continuous.

Example 1. Consider the utility function $u(x_{1},x_{2})=\sqrt{x_{1}}+x_{2}$.
The corresponding demand function is:

$f(p,m)=\begin{cases}\left(\frac{m}{p_{1}},0\right)&\mbox{if }p_{2}^{2}\geq
4p_{1}m,\\\
\left(\frac{p_{2}^{2}}{4p_{1}^{2}},\frac{4p_{1}m-p_{2}^{2}}{4p_{1}p_{2}}\right)&\mbox{otherwise.}\end{cases}$

This function satisfies all requirements of Theorem 1 but is not continuously
differentiable. Moreover,
$R(f)=\\{(x_{1},x_{2})\in\mathbb{R}^{2}_{+}|x_{1}>0\\}$, and thus this also
satisfies all requirements of Corollary 2. Set $\bar{p}=(1,1)$, and choose any
$x\in R(f)$. Then, $x=f^{u}(p,m)$ for some
$(p,m)\in\mathbb{R}^{2}_{++}\times\mathbb{R}_{++}$. If necessary, we can
replace $p_{2}$ with $\min\\{2\sqrt{p_{1}m},p_{2}\\}$, and thus we can assume
that $p_{2}^{2}\leq 4p_{1}m$. Moreover, again if necessary, we can replace
$(p,m)$ with $\frac{1}{p_{2}}(p,m)$, and thus we can assume that $p_{2}=1$.
Let us try to solve (1) and to determine $u_{f,\bar{p}}(x)$ and
$v_{f,\bar{p}}(x)$.

First, define

$f^{1}(q,w)=\left(\frac{w}{q_{1}},0\right),\
f^{2}(q,w)=\left(\frac{q_{2}^{2}}{4q_{1}^{2}},\frac{4q_{1}w-q_{2}^{2}}{4q_{1}q_{2}}\right),$

and consider

$\dot{c}_{i}(t)=f^{i}((1-t)p+t\bar{p},c_{i}(t))\cdot(\bar{p}-p).$ (4)

To solve (4), we have that

$c_{1}(t)=c_{1}(s)\frac{p_{1}+t(1-p_{1})}{p_{1}+s(1-p_{1})},$

and because $p_{2}=\bar{p}_{2}=1$,

$c_{2}(t)=c_{2}(s)-\frac{1}{4}\left[\frac{1}{p_{1}+t(1-p_{1})}-\frac{1}{p_{1}+s(1-p_{1})}\right].$

In particular, if $s=0$ and $c_{2}(0)=m$, then

$c_{2}(t)=m-\frac{1}{4}\left[\frac{1}{p_{1}+t(1-p_{1})}-\frac{1}{p_{1}}\right].$

Second, suppose that $4c_{2}(1)\geq 1$, where $c_{2}(0)=m$. By our initial
assumption, $4p_{1}c_{2}(0)\geq 1$. Moreover, $(p_{1}+t(1-p_{1}))c_{2}(t)$ is
monotone in $t$. Therefore, $c(t)=c_{2}(t)$ is a solution to (1) defined on
$[0,1]$, and thus we obtain that

$c(1)=c_{2}(1)=m+\frac{1-p_{1}}{4p_{1}}.$

In this case, $4m+\frac{1-p_{1}}{p_{1}}\geq 1$, and thus we have that

$4p_{1}m+1\geq 2p_{1}.$ (5)

Because

$x_{1}=\frac{1}{4p_{1}^{2}},\ x_{2}=m-\frac{1}{4p_{1}},$

we have that (5) is equivalent to

$u(x)\geq\frac{1}{2}.$

Moreover,

$c(1)=\sqrt{x_{1}}+x_{2}-\frac{1}{4}.$

Therefore, we obtain that

$u_{f,\bar{p}}(x)=u(x)-\frac{1}{4},$

if $u(x)\geq\frac{1}{2}$.

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

$u(x)<\frac{1}{2}.$

Note that $1\neq p_{1}$. 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 $c(t)=c_{2}(t)$ on
$[0,t^{*}]$, and $c(t)=c_{1}(t)$ on $[t^{*},1]$, where
$c(t^{*})=c_{1}(t^{*})=c_{2}(t^{*})$ and
$\dot{c}_{1}(t^{*})=\dot{c}_{2}(t^{*})$. Then,

$\frac{c(t^{*})(1-p_{1})}{p_{1}+t^{*}(1-p_{1})}=\dot{c}_{1}(t^{*})=\dot{c}_{2}(t^{*})=\frac{1-p_{1}}{4(p_{1}+t^{*}(1-p_{1}))^{2}},$

and thus

$c(t^{*})=c_{1}(t^{*})=c_{2}(t^{*})=\frac{1}{4(p_{1}+t^{*}(1-p_{1}))}.$

Then,

$c_{2}(t^{*})=m-\frac{1}{4}\left[\frac{1}{p_{1}+t^{*}(1-p_{1})}-\frac{1}{p_{1}}\right]=\frac{1}{4(p_{1}+t^{*}(1-p_{1}))},$

and hence, we obtain

$t^{*}=\frac{p_{1}}{1-p_{1}}\left[\frac{1-4p_{1}m}{4p_{1}m+1}\right].$

Because $1-p_{1}<0$ and $1\leq 4p_{1}m$, we have that $t^{*}\geq 0$. Moreover,
because $(p_{1}+t(1-p_{1}))c_{2}(t)$ is decreasing in $[0,1]$, $4c_{2}(1)<1$
and $4(p_{1}+t^{*}(1-p_{1}))c_{2}(t^{*})=1$, we have that $t^{*}<1$.
Therefore, $t^{*}\in[0,1]$. Hence,

$c(t)=\begin{cases}m-\frac{1}{4}\left[\frac{1}{p_{1}+t(1-p_{1})}-\frac{1}{p_{1}}\right]&\mbox{if
}t\leq t^{*},\\\
\frac{(p_{1}+t(1-p_{1}))}{4(p_{1}+t^{*}(1-p_{1}))^{2}}&\mbox{if }t\geq
t^{*}.\\\ \end{cases}$

In particular,

$c(1)=\frac{1}{4(p_{1}+t^{*}(1-p_{1}))^{2}}=\frac{16p_{1}^{2}m^{2}+8p_{1}m+1}{16p_{1}^{2}}.$

Because

$x_{1}=\frac{1}{4p_{1}^{2}},\ x_{2}=m-\frac{1}{4p_{1}},$

we have that

$c(1)=(\sqrt{x_{1}}+x_{2})^{2}=(u(x))^{2}.$

Therefore, if $u(x)<\frac{1}{2}$, then

$u_{f,\bar{p}}(x)=(u(x))^{2}.$

In conclusion, we obtain that

$u_{f,\bar{p}}(x)=\begin{cases}u(x)-\frac{1}{4}&\mbox{if }x_{1}>0,\
u(x)\geq\frac{1}{2},\\\ (u(x))^{2}&\mbox{if }x_{1}>0,\ u(x)<\frac{1}{2},\\\
0&\mbox{if }x_{1}=0.\end{cases}$

Of course, in this case,

$v_{f,\bar{p}}(x)=\begin{cases}u(x)-\frac{1}{4}&\mbox{if
}u(x)\geq\frac{1}{2},\\\ (u(x))^{2}&\mbox{if }u(x)<\frac{1}{2}.\end{cases}$

This completes our calculation.

###  3.2 Second Result: Continuity of Calculation

We are now able to calculate a utility function $u_{f,\bar{p}}$ 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
$f^{\prime}$ is near to $f$, then $u_{f^{\prime},\bar{p}}$ is also near to
$u_{f,\bar{p}}$. 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 $x\in X$. 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 $N\to\infty$.

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 $u_{f_{N},\bar{p}}$, and thus $u_{f_{N},\bar{p}}$ can be
treated as an estimated value of the “true utility function” $u_{f,\bar{p}}$.
Our question is as follows: does $u_{f_{N},\bar{p}}$ converge to
$u_{f,\bar{p}}$? If not, our estimation method violates the consistency, and
thus it is not useful.

Hence, the continuity of $u_{f,\bar{p}}$ 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
$C\subset\mathbb{R}^{n}_{++}\times\mathbb{R}_{++}$, $f^{k}$ converges to a CoD
$f$ uniformly on $C$ as $k\to\infty$. If $f$ is locally Lipschitz, then $f$ is
also a demand function.

As a corollary, we obtain an important result. Let
$\Delta_{\nu}=[\nu^{-1},\nu]^{n+1}$. Define a metric

$\rho(f,f^{\prime})=\sum_{\nu=1}^{\infty}\frac{1}{2^{\nu}}\arctan\left(\sup_{(p,m)\in\Delta_{\nu}}\|f(p,m)-f^{\prime}(p,m)\|\right)$

where $f,f^{\prime}$ are CoDs. We can easily show that $\rho$ is actually a
metric in the space of CoDs, and a sequence $(f^{k})$ converges to $f$ with
respect to $\rho$ if and only if for every compact set
$C\subset\mathbb{R}^{n}_{++}\times\mathbb{R}_{++}$, $(f^{k})$ converges to $f$
uniformly.

Suppose that $L=(L_{\nu})$ is a sequence of positive real numbers. Define
$\mathscr{F}_{L}$ as the set of demand functions such that
$f\in\mathscr{F}_{L}$ if and only if $f$ satisfies Walras’ law and

$\|f(p,m)-f(q,w)\|\leq L_{\nu}\|(p,m)-(q,w)\|$

for all $(p,m),(q,w)\in\Delta_{\nu}$.

Corollary 3. The space $\mathscr{F}_{L}$ is complete with respect to the
metric $\rho$.

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, $R(f)$ includes $\mathbb{R}^{n}_{++}$ and is
open in $\mathbb{R}^{n}_{+}$. Suppose that $(f^{k})$ is a sequence of
$\mathscr{F}_{L}$ 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:

$u^{\sigma}(x)=(x_{1}^{\sigma}+x_{2}^{\sigma})^{\frac{1}{\sigma}},$

where $\sigma<1$ and $\sigma\neq 0$. The corresponding demand function is

$f_{i}^{\sigma}(p,m)=\frac{p_{i}^{\frac{-1}{1-\sigma}}m}{p_{1}^{\frac{-\sigma}{1-\sigma}}+p_{2}^{\frac{-\sigma}{1-\sigma}}}.$

We assume that $\sigma<0$. To differentiate this function, we have that for
$j\neq i$ and $(p,m)\in\Delta_{\nu}$,

$\displaystyle\left|\frac{\partial f_{i}^{\sigma}}{\partial
p_{i}}(p,m)\right|=$
$\displaystyle~{}\frac{[(1-\sigma)p_{i}^{\frac{-2}{1-\sigma}}+p_{i}^{\frac{-2+\sigma}{1-\sigma}}p_{j}^{\frac{-\sigma}{1-\sigma}}]m}{(1-\sigma)(p_{1}^{\frac{-\sigma}{1-\sigma}}+p_{2}^{\frac{-\sigma}{1-\sigma}})^{2}}\leq\frac{\nu^{5}}{2},$
$\displaystyle\left|\frac{\partial f_{i}^{\sigma}}{\partial
p_{j}}(p,m)\right|=$ $\displaystyle~{}\frac{-\sigma
p_{i}^{\frac{-1}{1-\sigma}}p_{j}^{\frac{-1}{1-\sigma}}m}{(1-\sigma)(p_{1}^{\frac{-\sigma}{1-\sigma}}+p_{2}^{\frac{-\sigma}{1-\sigma}})^{2}}\leq\frac{\nu^{5}}{4},$
$\displaystyle\left|\frac{\partial f_{i}^{\sigma}}{\partial m}(p,m)\right|=$
$\displaystyle~{}\frac{p_{i}^{\frac{-1}{1-\sigma}}}{p_{1}^{\frac{-\sigma}{1-\sigma}}+p_{2}^{\frac{-\sigma}{1-\sigma}}}\leq\frac{\nu^{2}}{2}.$

and thus, if we define $L_{\nu}=\nu^{5}$, then
$f_{i}^{\sigma}\in\mathscr{F}_{L}$. Moreover, for every $\sigma<0$, we have
that $R(f^{\sigma})=\mathbb{R}^{2}_{++}$. However, it is well known that
$f^{\sigma}$ converges to a function $f$ with respect to the metric $\rho$,
where

$f_{1}(p,m)=f_{2}(p,m)=\frac{m}{p_{1}+p_{2}},$

and $R(f)=\\{(c,c)|c>0\\}$, even though $R(f^{\rho})=\mathbb{R}^{n}_{++}$.
This fact implies that the limit manipulation in $\mathscr{F}_{L}$ 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 $\mathscr{F}_{L}$ that converges to
$f$. Suppose that $R(f^{k})$ includes $\mathbb{R}^{n}_{++}$ and $f^{k}$
satisfies the C axiom for all $k$, and $f$ also satisfies these conditions.
Then, for every compact set $D\subset\mathbb{R}^{n}_{++}$,

$\sup_{x\in D}|u_{f^{k},\bar{p}}(x)-u_{f,\bar{p}}(x)|\to 0$

as $k\to\infty$.

The definition of $v_{f,\bar{p}}$ in Theorem 2 only depends on the values if
$u_{f,\bar{p}}$ on $\mathbb{R}^{n}_{++}$. Hence, it seems to be shown that
$v_{f^{k},\bar{p}}$ converges to $v_{f,\bar{p}}$ pointwise, here
$v_{f^{k},\bar{p}}$ 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.

Corollary 4. In addition to the assumptions of Theorem 3, suppose that $R(f)$
and all $R(f^{k})$ are relatively open in $\Omega$. Then,
$\limsup_{k\to\infty}v_{f^{k},\bar{p}}(x)\leq v_{f,\bar{p}}(x)$ for every
$x\in\Omega$.

Because $v_{f^{k},\bar{p}}(x)\geq 0$ for every $x\in\Omega$, if
$v_{f,\bar{p}}(x)=0$, then $v_{f^{k},\bar{p}}(x)$ converges to
$v_{f,\bar{p}}(x)$. However, in other case, whether $v_{f^{k},\bar{p}}(x)$
converges to $v_{f,\bar{p}}(x)$ or not is unknown. Actually, we find the
following example in which $\lim_{k\to\infty}v_{f^{k},\bar{p}}(x)\neq
v_{f,\bar{p}}(x)$.

Example 3. Suppose that $h:\mathbb{R}_{++}\to\mathbb{R}_{++}$ is $C^{\infty}$,
nondecreasing and $\lim_{c\to 0}h(c)=0,\lim_{c\to\infty}h(c)=\infty$. Choose
any $(x_{1},x_{2})\in\mathbb{R}^{2}_{++}$, and consider the following equation

$(x_{1}^{\frac{1}{1+\frac{1}{c}}}+x_{2}^{\frac{1}{1+\frac{1}{c}}})^{1+\frac{1}{c}}=h(c).$

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 $C^{\infty}$, monotone,
and strictly quasi-concave. If
$(x_{1},x_{2})\in\mathbb{R}^{2}_{+}\setminus\mathbb{R}^{2}_{++}$, then we
define

$u^{h}(x_{1},x_{2})=\inf_{\varepsilon>0}\sup\\{u^{h}(x_{1}^{\prime},x_{2}^{\prime})|(x_{1}^{\prime},x_{2}^{\prime})\in\mathbb{R}^{2}_{++},\
\|(x_{1},x_{2})-(x_{1}^{\prime},x_{2}^{\prime})\|<\varepsilon\\}.$

By the same arguments as in the proof of Corollary 2, we can show that $u^{h}$
is upper semi-continuous on $\mathbb{R}^{2}_{+}$. 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 $C^{\infty}$, and $R(f^{u^{h}})=\mathbb{R}^{2}_{++}$.
Furthermore, if $h^{\prime}\to h$ in the sense of $C^{2}$, then we can easily
show that $u^{h^{\prime}}\to u^{h}$ with respect to the local $C^{2}$ topology
on $\mathbb{R}^{n}_{++}$, and by Proposition 2.7.2 of Mas-Colell (1985), we
have that $f^{h^{\prime}}\to f^{h}$ with respect to the metric $\rho$.

Let $h:\mathbb{R}_{+}\to\mathbb{R}_{+}$ be a $C^{\infty}$ nondecreasing
function such that $h(0)=0,h(1)=h(2)=1$, $h^{\prime}(c)>0$ if $c\notin[1,2]$,
and $\lim_{c\to\infty}h(c)=+\infty$. Also, let
$\eta:\mathbb{R}_{+}\to\mathbb{R}_{+}$ be a $C^{\infty}$ function such that
$\eta(c)\equiv 0$ on $[0,1]$ and $[4,+\infty[$, $\eta(c)$ is increasing on
$[1,2]$, constant on $[2,3]$, decreasing on $[3,4]$, and
$\max_{c\in[1,4]}|\eta^{\prime}(c)|<\min_{c\in[3,4]}h^{\prime}(c)$. Define
$h^{k}(c)=h(c)+k^{-1}\eta(c)$. Then, $h^{k}\to h$ with respect to $C^{2}$
topology, and thus $f^{h^{k}}\to f^{h}$ with respect to $\rho$. On the other
hand, choose $\bar{p}=(1,1)$. Then, we can easily check that

$v_{f^{h^{k}},\bar{p}}(1,0)=\frac{1}{2},$

for all $k$, and

$v_{f^{h},\bar{p}}(1,0)=\frac{1}{\sqrt{2}},$

which implies that $\lim_{k\to\infty}v_{f^{h^{k}},\bar{p}}(1,0)\neq
v_{f^{h},\bar{p}}(1,0)$.

Finally, we can obtain another completeness result. Choose a sequence
$M=(M_{\nu})$ of positive real numbers and define $\mathscr{F}_{L,M}$ as the
set of all $f\in\mathscr{F}_{L}$ such that $R(f)$ includes
$\mathbb{R}^{n}_{++}$, $f$ satisfies the C axiom, and if
$x\in]\nu^{-1},\nu[^{n}$, then for all $p\in G^{f}(x)$ and
$i\in\\{1,...,n\\}$, $p_{i}\geq M_{\nu}$.

Corollary 5. Suppose that $(f^{k})$ is a sequence on $\mathscr{F}_{L,M}$ that
converges to $f$ with respect to the metric $\rho$. Then,
$f\in\mathscr{F}_{L,M}$, and for every compact set
$D\subset\mathbb{R}^{n}_{++}$,

$\sup_{x\in D}|u_{f^{k},\bar{p}}(x)-u_{f,\bar{p}}(x)|\to 0$

as $k\to\infty$.

Note that, in our Theorem 3, the C axiom is required. Corollary 2 stated that
if $R(f)$ is relatively open in $\Omega$, this axiom is equivalent to the
continuity of $v_{f,\bar{p}}$ on $\mathbb{R}^{n}_{++}$. Actually, the
requirement that $R(f)$ is relatively open in $\Omega$ 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, $u_{f^{k},\bar{p}}$ and
$u_{f,\bar{p}}$ are continuous on $\mathbb{R}^{n}_{++}$. Actually, we use
continuity of $u_{f,\bar{p}}$ on $\mathbb{R}^{n}_{++}$ in the proof of Theorem
3: see Lemma 4. However, $v_{f,\bar{p}}$ is not necessarily continuous on
$\Omega$ 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
$\mathscr{F}_{L}$ such that every $f^{k}$ is continuously differentiable and
the rank of $S_{f^{k}}(p,m)$ is always $n-1$, then there exists a sequence
$M=(M_{\nu})$ such that every $f^{k}$ is in $\mathscr{F}_{L,M}$ if and only if
for every compact set $C\subset\mathbb{R}^{n}_{++}$,
$\inf_{k}\min_{i}\min_{x\in C}G_{i}^{f^{k}}(x)>0$. This is another sufficient
condition for the limit function $f$ of $(f^{k})$ to satisfy all requirements
of Theorem 3.

##  4 Discussion

###  4.1 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 $u_{f,\bar{p}}$. 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
$((1-t)p+t\bar{p},c(t))$ 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 $p\cdot y>m$. 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 $x^{i}=f(p^{i},m^{i})$ and $p^{i}\cdot x^{i+1}\leq m^{i}$, $p^{k}\cdot
x^{1}>m^{k}$. 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 $\rho$, 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 $\rho$, 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 $u:\Omega\to\mathbb{R}$
such that $u$ is continuous on $\mathbb{R}^{n}_{++}$ if and only if $f$
satisfies the C axiom. In this paper, this result is required for proving
Theorem 3. See Lemma 4.

###  4.2 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 $\limsup_{k\to\infty}v_{f^{k},\bar{p}}(x)\leq
v_{f,\bar{p}}(x)$, 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 $R(f)\subset\limsup_{k}R(f^{k})$ when $R(f)$ is an open set and
$f^{k}\to f$. 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
$v_{f,\bar{p}}$ 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 $f=f^{v_{f,\bar{p}}}$. However, there is no known
counterexample of $f$ such that $R(f)$ is not an open set and $f\neq
f^{v_{f,\bar{p}}}$. Maybe $f=f^{v_{f,\bar{p}}}$ can be said even if $R(f)$ is
not an open set.

Finally, there remains the task of identifying the conditions for
$v_{f,\bar{p}}$ 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 $p\geq 0,p\neq 0$ and $p_{i}=0$ for some $i$, then for any
convergent sequence $(p^{k})$ to $p$ on $\mathbb{R}^{n}_{++}$ and any
$(q,w)\in\mathbb{R}^{n}_{++}$, $f(p^{k},v_{f,p^{k}}(f(q,w)))$ is unbounded.
However, when discussing this condition, Hurwicz-Uzawa restricts the domain of
the utility function to $R(f)$. Therefore, whether the continuity of
$v_{f,\bar{p}}$ outside $R(f)$ can be guaranteed by this condition is one of
the open questions.

##  5 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 semi-
continuous 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.

##  6 Proofs

###  6.1 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
$f:U\to\mathbb{R}^{N}$ be some function, where $U\subset\mathbb{R}^{M}$ is
open. This function is called locally Lipschitz if for every compact set
$C\subset U$, there exists $L>0$ such that for every $x,y\in C$,

$\|f(x)-f(y)\|\leq L\|x-y\|.$

Because the following property is important, we present a proof in this
subsection.

Fact 1. Let $f:U\to\mathbb{R}^{N}$, where $U\subset\mathbb{R}^{M}$ is open.
Then, $f$ is locally Lipschitz if and only if for every $x\in U$, there exists
$r>0$ and $L>0$ such that if $y,z\in U,\ \|y-x\|\leq r,$ and $\|z-x\|<r$, then

$\|f(y)-f(z)\|\leq L\|y-z\|.$

Proof of Fact 1. Suppose that $f$ is locally Lipschitz. For each $x\in U$,
there exists $r>0$ such that
$\bar{B}_{r}(x)\equiv\\{y\in\mathbb{R}^{M}|\|y-x\|\leq r\\}\subset U$, and
$\bar{B}_{r}(x)$ is compact. This implies that there exists $L>0$ such that if
$y,z\in\bar{B}_{r}(x)$, then

$\|f(y)-f(z)\|\leq L\|y-z\|.$

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 $k$,

$\|f(x^{k})-f(y^{k})\|>k\|x^{k}-y^{k}\|.$

Because $C$ is compact, we can assume that $x^{k}\to x^{*},\ y^{k}\to y^{*}$
as $k\to\infty$. Then,

$\|f(x^{*})-f(y^{*})\|\geq k\|x^{*}-y^{*}\|$

for every $k$, which implies that $x^{*}=y^{*}$. Choose any $r>0$ and $L>0$.
Then, there exists $k>L$ such that $x^{k},y^{k}\in\bar{B}_{r}(x^{*})$, and
thus the latter claim of this fact is violated. This completes the proof.
$\blacksquare$

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.222See, for example, Heinonen (2004).

Fact 2. Suppose that $f:U\to\mathbb{R}^{N}$, where $U\subset\mathbb{R}^{M}$ is
open. If $f$ is locally Lipschitz, then it is differentiable almost
everywhere.

Next, we explain some notion on ordinary differential equations (ODEs). First,
consider the following ODE:

$\dot{x}(t)=g(t,x(t)),\ x(t_{0})=x^{*},$ (6)

where $g:U\to\mathbb{R}^{N}$ and $U\subset\mathbb{R}\times\mathbb{R}^{N}$ is
open. We call a subset $I$ of $\mathbb{R}$ an interval if it is a convex set
containing at least two points. We say that a function $x:I\to\mathbb{R}^{N}$
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
$C\subset I$, and 4) $\dot{x}(t)=g(t,x(t),y)$ for almost every $t\in I$. Let
$x:I\to\mathbb{R}^{N}$ and $y:J\to\mathbb{R}^{N}$ be two solutions. Then, we
call $x$ an extension of $y$ if $J\subset I$ and $y(t)=x(t)$ for all $t\in J$.
A solution $x:I\to\mathbb{R}^{N}$ 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.333See, for example, Theorem 1.1 and Theorem 3.1 of chapter 2
in Hartman (1982).

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
$x:I\to\mathbb{R}^{N}$ to (6), where $I$ is open and $x(t)$ is continuously
differentiable. Moreover, for every compact set $C\subset U$, there exists
$t_{1},t_{2}\in I$ such that if $t\in I$ and either $t<t_{1}$ or $t_{2}<t$,
then $(t,x(t))\notin C$.

Next, consider the following parametrized ODE:

$\dot{x}(t)=h(t,x(t),y),\ x(t_{0})=z,$ (7)

where $h:U\to\mathbb{R}^{N}$ and
$U\subset\mathbb{R}\times\mathbb{R}^{N}\times\mathbb{R}^{M}$ is open. We
assume that $h$ is locally Lipschitz. Fix $(y,z)$ such that $(t_{0},z,y)\in
U$. 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 $x^{y,z}:I\to\mathbb{R}^{N}$
according to Fact 3. We write $x(t;y,z)=x^{y,z}(t)$, and call this function
$x:(t,y,z)\mapsto x(t;y,z)$ the solution function of (7). The following fact
is needed but not so known, and thus we prove this fact in this paper.

Fact 4. Under the assumption that $h$ is locally Lipschitz, the domain of the
solution function is open, and the solution function is locally Lipschitz.

Proof of Fact 4. First, we introduce a lemma.

Lemma 1 (Gronwall’s inequality). Suppose that $g:[t_{0},T]\to\mathbb{R}$ is
continuous, and

$0\leq g(t)\leq\int_{t_{0}}^{t}[Ag(s)+B]ds$

for every $t\in[t_{0},T]$, where $A>0$ and $B\in\mathbb{R}$. Then,

$g(t)\leq\frac{B}{A}(e^{A(t-t_{0})}-1).$

Proof. Omitted. See, for example, Problem 5.2.7 of Karatzas and Shreve (1991)
or Lemma 8 of Hosoya (2017).

Let $V\subset\mathbb{R}\times\mathbb{R}^{M}\times\mathbb{R}^{N}$ be the domain
of the solution function $x(t;y,z)$. Choose any $(t^{*},y,z)\in V$. By Fact 3,
there exists an open interval $I$ such that $t\mapsto x(t;y,z)$ is a
nonextendable solution, and $t^{*}\in I$. Choose $t_{1},t_{2}\in I$ such that
$t_{1}<\min\\{t^{*},t_{0}\\}\leq\max\\{t^{*},t_{0}\\}<t_{2}$. Consider the
following differential equation:

$\dot{x}(t)=h(t,x(t)+z^{\prime}-z,y^{\prime}),\ x(t_{0})=z.$ (8)

If $z^{\prime}=z$ and $y^{\prime}=y$, then $x^{*}:t\mapsto x(t;y,z)$ is a
solution to (8). Choose $a,b>0$ so small and define

$\Pi(a,b)=\\{(t,x,y^{\prime},z^{\prime})|t\in[t_{1},t_{2}],\
\|x-x^{*}(t)\|\leq a,\ \|y^{\prime}-y\|\leq b,\ \|z^{\prime}-z\|\leq b\\},$
$\Pi^{\prime}(a,b)=\\{(t,x+z^{\prime}-z,y^{\prime})|(t,x,y^{\prime},z^{\prime})\in\Pi(a,b)\\}.$

By definition, $\Pi(a,b)$ and $\Pi^{\prime}(a,b)$ are compact. We assume that
$a,b$ is sufficiently small that $\Pi^{\prime}(a,b)\subset U$. Because $h$ is
locally Lipschitz, there exists $L>0$ such that for every
$(t_{1}^{\prime},z_{1}^{\prime},y_{1}^{\prime}),(t_{2}^{\prime},z_{2}^{\prime},y_{2}^{\prime})\in\Pi^{\prime}(a,b)$,

$\|h(t_{1}^{\prime},z_{1}^{\prime},y_{1}^{\prime})-h(t_{2}^{\prime},z_{2}^{\prime},y_{2}^{\prime})\|\leq
L[|t_{1}^{\prime}-t_{2}^{\prime}|+\|z_{1}^{\prime}-z_{2}^{\prime}\|+\|y_{1}^{\prime}-y_{2}^{\prime}\|].$

Suppose that $\|y^{\prime}-y\|\leq b,\ \|z^{\prime}-z\|\leq b$, and define
$t^{+}_{2}(d^{\prime},c^{*})$ as the supremum of the set of all
$t\in]t_{0},t_{2}]$ such that there exists a solution
$\tilde{x}:[t_{0},t]\to\mathbb{R}^{N}$ to (8) and
$(s,\tilde{x}(s),y^{\prime},z^{\prime})\in\Pi(a,b)$ for all $s\in[t_{0},t]$.
By Fact 3, we have that $t^{+}_{2}(y^{\prime},z^{\prime})>t_{0}$. If
$t\in[t^{*},t^{+}_{2}(y^{\prime},z^{\prime})[$, then

$\displaystyle\|\tilde{x}(t)-x^{*}(t)\|\leq$
$\displaystyle~{}\int_{t_{0}}^{t}\|h(s,\tilde{x}(s)+z^{\prime}-z,y^{\prime})-h(s,x^{*}(s),y)\|ds$
$\displaystyle\leq$
$\displaystyle~{}\int_{t_{0}}^{t}\|h(s,\tilde{x}(s)+z^{\prime}-z,y^{\prime})-h(s,x^{*}(s)+z^{\prime}-z,y^{\prime})\|ds$
$\displaystyle~{}+\int_{t_{0}}^{t}\|h(s,x^{*}(s)+z^{\prime}-z,y^{\prime})-h(s,x^{*}(s),y)\|ds$
$\displaystyle\leq$
$\displaystyle~{}\int_{t^{*}}^{t}[L\|\tilde{x}(s)-x^{*}(s)\|+L(\|y^{\prime}-y\|+\|z^{\prime}-z\|)]ds.$

Therefore, by Lemma 1,

$\|\tilde{x}(t)-x^{*}(t)\|\leq(\|y^{\prime}-y\|+\|z^{\prime}-z\|)(e^{L(t-t_{0})}-1).$
(9)

Choose $b^{\prime}\in]0,b[$ so small that

$b^{\prime}(e^{L(t_{2}-t_{0})}-1)<a.$ (10)

We show that if $\|y^{\prime}-y\|+\|z^{\prime}-z\|\leq b^{\prime}$, then
$t^{+}_{2}(y^{\prime},z^{\prime})=t_{2}$. Suppose not. If $\tilde{x}$ is
defined at $t^{+}_{2}(y^{\prime},z^{\prime})$, then
$(t^{+}_{2}(y^{\prime},z^{\prime}),\tilde{x}(t^{+}_{2}(y^{\prime},z^{\prime})),y^{\prime},z^{\prime})\in\Pi(a,b)$,
and thus
$\|\tilde{x}(t^{+}_{2}(y^{\prime},z^{\prime}))-x^{*}(t^{+}_{2}(y^{\prime},z^{\prime}))\|<a$,
which contradicts the definition of $t^{+}_{1}(y^{\prime},z^{\prime})$.
Therefore, $\tilde{x}$ cannot be defined at
$t^{+}_{2}(y^{\prime},z^{\prime})$. By Fact 3, there exists
$t_{3}<t^{+}_{2}(y^{\prime},z^{\prime})$ such that if
$t_{3}<t<t^{+}_{2}(y^{\prime},z^{\prime})$, then
$(t,\tilde{x}(t)+z^{\prime}-z,y^{\prime})\notin\Pi^{\prime}(a,b)$, which
contradicts (9). Hence, if $b^{\prime}$ satisfies (10) and
$\|y^{\prime}-y\|+\|z^{\prime}-z\|\leq b^{\prime}$, then $\tilde{x}(t)$ is
defined on $[t_{0},t_{2}]$ and if $t\in[t_{0},t_{2}]$, then

$\|\tilde{x}(t)-x^{*}(t)\|\leq(\|y^{\prime}-y\|+\|z^{\prime}-z\|)(e^{L(t_{2}-t_{1})}-1).$

By a symmetric argument, we have that if $b^{\prime}\in]0,b[$ is so small that

$b^{\prime}(e^{L(t_{0}-t_{1})}-1)<a,$ (11)

and $\|y^{\prime}-y\|+\|z^{\prime}-z\|\leq b^{\prime}$, then (8) has a
solution $\tilde{x}$ defined on $[t_{1},t_{0}]$, and if $t\in[t_{1},t_{0}]$,

$\|\tilde{x}(t)-x(t)\|\leq(\|y^{\prime}-y\|+\|z^{\prime}-z\|)(e^{L(t_{2}-t_{1})}-1).$

Clearly,

$x(t;y^{\prime},z^{\prime})=\tilde{x}(t)+z^{\prime}-z,$

and thus the domain $V$ of the solution function $x$ includes

$[t_{1},t_{2}]\times\\{(y^{\prime},z^{\prime})|\|y^{\prime}-y\|+\|z^{\prime}-z\|\leq
b^{\prime}\\},$

which is a neighborhood of $(t^{*},y,z)$. Moreover,

$\max_{t\in[t_{0},t_{1}]}\|x(t;y^{\prime},z^{\prime})-x(t;y,z)\|\leq(\|y^{\prime}-y\|+\|z^{\prime}-z\|)(e^{L(t_{2}-t_{1})}-1),$

and thus by Fact 1, we have that the solution function is locally Lipschitz.
This completes the proof. $\blacksquare$

Finally, we mention the formula of the solution to linear differential
equations. Consider the following ODE:

$\dot{x}(t)=a(t)x(t),$

where $I$ is an interval including $t_{0}$ and $a:I\to\mathbb{R}$ is a bounded
measurable function on $I$. Then, the formula of the solution to the above
equation is the following:

$x(t)=x(t_{0})e^{\int_{t_{0}}^{t}a(s)ds}.$

For the proof, use Theorem 1 of section 0.4 of Ioffe and Tikhomirov (1979).

###  6.2 Proof of Theorem 1

First, consider the following parametrized ODE:

$\dot{c}(t)=f((1-t)p+tq,c(t))\cdot(q-p),\ c(0)=w,$ (12)

and let $c(t;p,q,w)$ denote the solution function of (12). We introduce two
lemmas.

Lemma 2. Let $U=\mathbb{R}^{n}_{++}\times\mathbb{R}_{++}$. Choose any
$(p,m)\in U$. Suppose that $W\subset U$ and the Lebesgue measure of
$U\setminus W$ is zero. Moreover, suppose that $q\in\mathbb{R}^{n}_{++}$ and
there exists $i^{*}\in\\{1,...,n\\}$ such that $q_{i^{*}}\neq p_{i^{*}}$, and
the domain of the solution function $c(t;p,q,w)$ of the ODE $(\ref{eq6.1})$
includes $[0,t^{*}]\times\bar{P}_{1}\times\bar{P}_{2}$ for $t^{*}>0$, where
$P_{1}$ is a bounded open neighborhood of $q$ and $P_{2}$ is a bounded open
neighborhood of $m$, and $\bar{P}_{j}$ denotes the closure of $P_{j}$. For
every $(t,\tilde{r},w)\subset\mathbb{R}^{n+1}$ such that $t\in[0,t^{*}]$,
$r\in P_{1}$ for

$r_{i}=\begin{cases}\tilde{r}_{i}&\mbox{if }i<i^{*},\\\ q_{i}&\mbox{if
}i=i^{*},\\\ \tilde{r}_{i-1}&\mbox{if }i>i^{*},\end{cases}$

and $w\in P_{2}$, define

$\xi(t,\tilde{r},w)=((1-t)p+tr,c(t;p,r,w)).$

Then, the Lebesgue measure of $\xi^{-1}(U\setminus W)$ is also zero.

Proof. Without loss of generality, we assume that $i^{*}=n$. Throughout the
proof of Lemma 2, we use the following notation. If $r\in\mathbb{R}^{n}$, then
$\tilde{r}=(r_{1},...,r_{n-1})\in\mathbb{R}^{n-1}$. Conversely, if
$\tilde{r}\in\mathbb{R}^{n-1}$, then $r=(r_{1},...,r_{n-1},q_{n})$.

Let $\tilde{P}_{1}=\\{\tilde{r}|r\in P_{1}\\}$. We show that $\xi$ is one-to-
one on the set $]0,t^{*}]\times\tilde{P}_{1}\times P_{2}$. Suppose that
$t_{1}\neq 0\neq t_{2}$ and
$\xi(t_{1},\tilde{r}_{1},w_{1})=\xi(t_{2},\tilde{r}_{2},w_{2})=(z,c)$. Because
$z_{n}=(1-t_{1})p_{n}+t_{1}q_{n}=(1-t_{2})p_{n}+t_{2}q_{n}$ and $p_{n}\neq
q_{n}$, we have $t_{1}=t_{2}$. Because
$z_{i}=(1-t_{1})p_{i}+t_{1}r_{1i}=(1-t_{1})p_{i}+t_{1}r_{2i}$ and $t_{1}\neq
0$, we have that $r_{1i}=r_{2i}$, and thus $\tilde{r}_{1}=\tilde{r}_{2}$.
Therefore, it suffices to show that $c(t;p,r,w)$ is increasing in $w$. Suppose
that $w_{1}<w_{2}$ and $c(t;p,r,w_{1})\geq c(t;p,r,w_{2})$. Because
$c(0;p,r,w_{1})=w_{1}<w_{2}=c(0;p,r,w_{2})$, by the intermediate value
theorem, there exists $s\in[0,t]$ such that $c(s;p,r,w_{1})=c(s;p,r,w_{2})$.
Then, by Fact 3, we have $w_{1}=c(0;p,r,w_{1})=c(0;p,r,w_{2})=w_{2}$, which is
a contradiction.

Next, define

$V^{\ell}=\xi([\ell^{-1}t^{*},t^{*}[\times\tilde{P}_{1}\times P_{2}).$

We show that $\xi^{-1}$ is Lipschitz on $V^{\ell}$. Define

$t(v)=\frac{v_{n}-p_{n}}{q_{n}-p_{n}},$
$\tilde{r}(v)=\frac{1}{t(v)}[(t(v)-1)\tilde{p}+\tilde{v}].$

Suppose that $(v_{1},c_{1}),(v_{2},c_{2})\in V^{\ell}$ and
$(v_{j},c_{j})=\xi(t_{j},\tilde{r}_{j},w_{j})$. Then, we have $t_{j}=t(v_{j})$
and $\tilde{r}_{j}=\tilde{r}(v_{j})$. Clearly, the functions $t(v)$ and
$\tilde{r}(v)$ are Lipschitz on
$\xi([\ell^{-1}t^{*},t^{*}]\times\tilde{P}_{1}\times P_{2})$, and hence they
are Lipschitz on $V^{\ell}$. Next, consider the following ODE:

$\dot{d}(s)=f((1-(s+t-t_{2}))p+(s+t-t_{2})r(v),d(s))\cdot(r(v)-p),\
d(t_{2})=c.$

Let $d(s;t,v,c)$ be the solution function of the above ODE. Define
$\hat{P}_{1}$ as the closure of $\tilde{P}_{1}$. If $(v,c)=\xi(t,\tilde{r},w)$
for some
$(t,\tilde{r},w)\in[{\ell}^{-1}t^{*},t^{*}]\times\hat{P}_{1}\times\bar{P}_{2}$,
then $d(s;t,v,c)=c(s+t-t_{2};p,r,w)$. Moreover, the set

$\\{(t,v,c)|t\in[\ell^{-1}t^{*},t^{*}],\ (v,c)=\xi(t,\tilde{r},w)\mbox{ for
some }(\tilde{r},w)\in\hat{P}_{1}\times\bar{P}_{2}\\}$

is compact, and by Fact 4, $(t,v,c)\mapsto d(t_{2}-t;t,v,c)$ is Lipschitz on
this set. Therefore,

$\displaystyle|w_{1}-w_{2}|=$
$\displaystyle~{}|d(t_{2}-t_{1};t_{1},v_{1},c_{1})-d(t_{2}-t_{2};t_{2},v_{2},c_{2})|$
$\displaystyle\leq$
$\displaystyle~{}L[|t_{1}-t_{2}|+\|(v_{1},c_{1})-(v_{2},c_{2})\|]$
$\displaystyle=$
$\displaystyle~{}L[|t(v_{1})-t(v_{2})|+\|(v_{1},c_{1})-(v_{2},c_{2})\|]$
$\displaystyle\leq$ $\displaystyle~{}L(M+1)\|(v_{1},c_{1})-(v_{2},c_{2})\|,$

where $L,M>0$ are some constants, and therefore our claim is correct.

Now, recall that the Lebesgue measure of $U\setminus W$ is zero. Because
$\xi^{-1}$ is Lipschitz on $V^{\ell}$, we have that the Lebesgue measure of

$\xi^{-1}(V^{\ell}\cap(U\setminus W))$

is zero. Therefore, the Lebesgue measure of

$\cup_{\ell}\xi^{-1}(V^{\ell}\cap(U\setminus W))$

is also zero. Clearly, the Lebesgue measure of

$\xi^{-1}(U\setminus
W)\setminus\left(\cup_{\ell}\xi^{-1}(V^{\ell}\cap(U\setminus W))\right)$

is zero, because this set is included in
$\\{(t,\tilde{r},w)|t\in\\{0,t^{*}\\}\\}$. This completes the proof of Lemma
2. $\blacksquare$

Lemma 3. Choose any $(p,m)\in\mathbb{R}^{n}_{++}$. Then, there exists a
solution $E:\mathbb{R}^{n}_{++}\to\mathbb{R}_{++}$ of the following partial
differential equation

$DE(q)=f(q,E(q)),\ E(p)=m,$ (13)

if and only if the domain of the solution function of (12) includes
$[0,1]\times\\{p\\}\times\mathbb{R}^{n}_{++}\times\\{m\\}$. Moreover, in this
case

$E(q)=c(1;p,q,m).$

Proof. Suppose that a solution $E:\mathbb{R}^{n}_{++}\to\mathbb{R}_{++}$ to
(13) exists. Choose any $q\in\mathbb{R}^{n}_{++}$. Let $d(t)=E((1-t)p+tq)$.
Then, $d(0)=E(p)=m$ and

$\dot{d}(t)=f((1-t)p+tq,d(t))\cdot(q-p),$

and by the uniqueness of the solution to ODE (Fact 3), we have that
$d(t)\equiv c(t;p,q,m)$. Hence, the domain of the solution function $c$
includes $[0,1]\times\\{p\\}\times\mathbb{R}^{n}_{++}\times\\{m\\}$, 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
$[0,1]\times\\{p\\}\times\mathbb{R}^{n}_{++}\times\\{m\\}$. Define
$E(q)=c(1;p,q,m)$. We show that $E(q)$ is a solution to (13).

First, let $\Delta^{*}$ 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 $t\in[0,1]$, and the mapping
$r\mapsto c(t;p,r,w)$ is differentiable at $r=q$ for almost every $t\in[0,1]$.
Suppose that $(q,w)\in\Delta^{*}$ and let $e_{i}$ denote the $i$-th unit
vector. Then, for each $i\in\\{1,...,n\\}$,444In this step, we frequently
abbreviate several variables for simplicity.

$\displaystyle~{}\lim_{h\to 0}\frac{c(t;p,q+he_{i},w)-c(t;p,q,w)}{h}$
$\displaystyle=$ $\displaystyle~{}\lim_{h\to 0}\frac{1}{h}\times$
$\displaystyle~{}\left[\int_{0}^{t}f((1-s)p+s(q+he_{i}),c(s;p,q+he_{i},w))\cdot(q+he_{i}-p)ds\right.$
$\displaystyle~{}-\left.\int_{0}^{t}f((1-s)p+sq,c(s;p,q,w))\cdot(q-p)ds\right]$
$\displaystyle=$
$\displaystyle~{}\int_{0}^{t}\left[f_{i}+\sum_{j=1}^{n}\left[s\frac{\partial
f_{j}}{\partial p_{i}}+\frac{\partial f_{j}}{\partial m}\frac{\partial
c}{\partial q_{i}}\right](q_{j}-p_{j})\right]ds,$

by the dominated convergence theorem, and thus $\frac{\partial c}{\partial
q_{i}}(t;p,q,w)$ is defined for all $t\in[0,1]$ and absolutely continuous in
$t$. Define the following absolutely continuous function

$\varphi(t)=\frac{\partial c}{\partial
q_{i}}(t;p,q,w)-tf_{i}((1-t)p+tq,c(t;p,q,w)).$

By the above evaluation and the symmetry of the Slutsky matrix, we have that

$\displaystyle\dot{\varphi}(t)=$
$\displaystyle~{}f_{i}+\sum_{j=1}^{n}\left[t\frac{\partial f_{j}}{\partial
p_{i}}+\frac{\partial f_{j}}{\partial m}\frac{\partial c}{\partial
q_{i}}\right](q_{j}-p_{j})-f_{i}-t\sum_{j=1}^{n}\left[\frac{\partial
f_{i}}{\partial p_{j}}+\frac{\partial f_{i}}{\partial
m}f_{j}\right](q_{j}-p_{j})$ $\displaystyle=$
$\displaystyle~{}t\sum_{j=1}^{n}\left[\frac{\partial f_{j}}{\partial
p_{i}}-\frac{\partial f_{i}}{\partial p_{j}}-\frac{\partial f_{i}}{\partial
m}f_{j}\right](q_{j}-p_{j})+\frac{\partial c}{\partial
q_{i}}\sum_{j=1}^{n}\frac{\partial f_{j}}{\partial m}(q_{j}-p_{j})$
$\displaystyle=$ $\displaystyle~{}\varphi(t)\sum_{j=1}^{n}\frac{\partial
f_{j}}{\partial m}(q_{j}-p_{j})$ $\displaystyle\equiv$
$\displaystyle~{}a(t)\varphi(t),$

where $a(t)$ be some bounded measurable function. By the formula of the
solution to linear ODEs, we have that

$\varphi(t)=\varphi(0)e^{\int_{0}^{t}a(s)ds}.$

However, we can easily check that $\varphi(0)=0$, and thus $\varphi(t)\equiv
0$. In particular, $\varphi(1)=0$ and thus

$\frac{\partial c}{\partial q_{i}}(1;p,q,w)=f_{i}(q,c(1;p,q,w)).$

Second, suppose that $q_{n}\neq p_{n}$ and $i\in\\{1,...,n-1\\}$. By Lemma 2
and Fubini’s theorem, there exists $\delta>0$ and a sequence $(q^{k},w^{k})$
on $\Delta^{*}$ such that $q^{k}\to q,w^{k}\to m$ as $k\to\infty$, and for
every $k$, $i\in\\{1,...,n-1\\}$ and almost every $h\in]-\delta,\delta[$,
$(q^{k}+he_{i},w^{k})\in\Delta^{*}$. Then, for every $h\in]-\delta,\delta[$,

$c(1;p,q^{k}+he_{i},w^{k})-c(1;p,q^{k},w^{k})=\int_{0}^{h}f_{i}(q^{k}+se_{i},c(1;p,q^{k}+se_{i},w^{k}))ds.$

By the dominated convergence theorem, we have that

$E(q+he_{i})-E(q)=\int_{0}^{h}f_{i}(q+se_{i},E(q+se_{i}))ds,$

which implies that

$\frac{\partial E}{\partial q_{i}}(q)=f_{i}(q,E(q)).$

Third, suppose that $q_{n}=p_{n}$ and $i\in\\{1,...,n-1\\}$. Let
$e=(1,1,...,1)$ and define $q^{k}=q+k^{-1}e$. Then, $q_{n}^{k}\neq p_{n}$, and
thus, for every $h\in]-q_{i},q_{i}[$,

$E(q^{k}+he_{i})-E(q^{k})=\int_{0}^{h}f_{i}(q^{k}+se_{i},E(q^{k}+se_{i}))ds,$

and by the dominated convergence theorem,

$E(q+he_{i})-E(q)=\int_{0}^{h}f_{i}(q+se_{i},E(q+se_{i}))ds,$

which implies that

$\frac{\partial E}{\partial q_{i}}(q)=f_{i}(q,g(q)).$

To summarize the above result, we obtain the following: for every
$q\in\mathbb{R}^{n}_{++}$ and $i\in\\{1,...,n-1\\}$,

$\frac{\partial E}{\partial q_{i}}(q)=f_{i}(q,E(q)).$ (14)

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. $\blacksquare$

We now complete the preparation to prove Theorem 1. We separate the proof of
Theorem 1 into ten steps.

Step 1. Suppose that $t^{*}>0$ and the domain of the solution function
$c(t;p,q,w)$ of (12) includes $[0,t^{*}]\times\\{(p,q,m)\\}$. Define
$p(t)=(1-t)p+tq$ and $x(t)=f(p(t),c(t;p,q,m))$. Then, $p\cdot x(t^{*})\geq m$
and $p(t^{*})\cdot x(0)\geq c(t^{*};p,q,m)$.

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 $\Delta(t^{*})$ be
the set of all $(r,w)$ such that the domain of $t\mapsto c(t;p,r,w)$ includes
$[0,t^{*}]$, and for almost every $t\in[0,t^{*}]$, $f$ 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
$(q^{k},w^{k})\in\Delta(t^{*})$ that converges to $(q,m)$ as $k\to\infty$.
Define $d(t)=p\cdot x(t)$ and $d^{k}(t)=p\cdot
f(p(t,q^{k}),c(t;p,q^{k},w^{k}))$. Then, $d^{k}$ is absolutely continuous, and
for almost all $t\in[0,t^{*}]$,

$\dot{d}^{k}(t)=p^{T}S_{f}(p(t,q^{k}),c(t;p,q^{k},w^{k}))(q^{k}-p).$

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 that

$\dot{d}^{k}(t)=-t(q^{k}-p)^{T}S_{f}(p(t,q^{k}),c(t;p,q^{k},w^{k}))(q^{k}-p)\geq
0,$

and thus $d^{k}(t)$ is nondecreasing. Because $d^{k}(t)\to d(t)$ for every
$t$, we have that $d(t)$ is also nondecreasing, and thus

$p\cdot x(t^{*})=d(t^{*})\geq d(0)=m,$

which completes the proof of Step 1. $\blacksquare$

Step 2. The domain of the solution function $c(t;p,q,w)$ includes
$[0,1]\times\mathbb{R}^{n}_{++}\times\mathbb{R}^{n}_{++}\times\mathbb{R}_{++}$.

Proof of Step 2. Suppose not. Then, there exists $p,q\in\mathbb{R}^{n}_{++}$
and $m\in\mathbb{R}_{++}$ such that $c(t;p,q,m)$ is defined only on
$[0,t^{*}[$, where $t^{*}\leq 1$. 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 $\mathbb{R}^{n}_{++}\times\mathbb{R}_{++}$ as $t\to
t^{*}$, and thus either $\limsup_{t\to t^{*}}c(t;p,q,m)=+\infty$ or
$\liminf_{t\to t^{*}}c(t;p,q,m)=0$. By Step 1, $\max\\{p\cdot x,q\cdot
x\\}\geq p(t)\cdot x\geq c(t;p,q,m)$ for every $t\in[0,t^{*}[$, and thus there
exists an increasing sequence $(t^{k})$ such that $t^{k}\to t^{*}$ and
$c(t^{k};p,q,m)\to 0$ as $k\to\infty$. Define
$x^{k}=f(p(t^{k}),c(t^{k};p,q,m))$. Because $p\cdot x^{k}\geq m=p\cdot x$ and
$p(t^{k})\cdot x\geq c(t^{k};p,q,m)=p(t^{k})\cdot x^{k}$, we have that $q\cdot
x\geq q\cdot x^{k}$. Hence, $x^{k}$ is a sequence in the following compact set

$\\{y\in\Omega|q\cdot y\leq q\cdot x\\}.$

Therefore, without loss of generality, we can assume that $x^{k}\to x^{*}$ as
$k\to\infty$. Because $p\cdot x^{*}\geq m$, we have that $x^{*}\neq 0$.
However,

$0<p(t^{*})\cdot x^{*}=\lim_{k\to\infty}p(t^{k})\cdot
x^{k}=\lim_{k\to\infty}c(t^{k};p,q,m)=0,$

which is a contradiction. This completes the proof of Step 2. $\blacksquare$

Step 3. For all $t\in[0,1]$, $c(1-t;p,q,m)=c(t;q,p,c(1;p,q,m))$. Moreover, if
$m>m^{\prime}$, then $c(t;p,q,m)>c(t;p,q,m^{\prime})$ for every $t\in[0,1]$.

Proof of Step 3. First,

$\frac{d}{dt}c(1-t;p,q,m)=f((1-t)q+tp,c(1-t;p,q,m))\cdot(p-q),$

and thus by the uniqueness of the solution to ODE (Fact 3), we have that

$c(1-t;p,q,m)=c(t;q,p,c(1;p,q,m)).$

Next, suppose that $c(t;p,q,m)\leq c(t;p,q,m^{\prime})$ for every $t\in[0,1]$.
By the intermediate value theorem, there exists $s\in[0,1]$ such that
$c(s;p,q,m)=c(s;p,q,m^{\prime})$. Again by the uniqueness of the solution to
ODE, we have that

$m=c(0;p,q,m)=c(0;p,q,m^{\prime})=m^{\prime},$

which is a contradiction. This completes the proof of Step 3. $\blacksquare$

Step 4. Suppose that $x\neq y,\ x=f(p,m),\ y=f(q,w)$ and $w\geq c(1;p,q,m)$.
Then, $p\cdot y>m$.

Proof. First, define $m^{*}=c(1;q,p,w)$. By Step 3, we have that
$c(t;q,p,w)=c(1-t;p,q,m^{*})$, and thus $m^{*}\geq m$. Moreover,
$w>c(1;p,q,m)$ if and only if $m^{*}>m$.

Define $p(t)=(1-t)p+tq$ and $d(t)=p\cdot f(p(t),c(t;p,q,m^{*}))$. We have
already shown in the proof of Step 1 that $d(t)$ is nondecreasing. Therefore,
if $m^{*}>m$, then

$p\cdot y=p\cdot f(q,w)=d(1)\geq d(0)=m^{*}>m,$

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 $\Delta^{*}$ 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 $t\in[0,1]$. By Lemma 2 and Fubini’s
theorem, we can show that there exists a sequence $(q^{k},w^{k})$ on
$\Delta^{*}$ such that $(q^{k},w^{k})\to(q,w)$ as $k\to\infty$. Let
$2\varepsilon=\|x-y\|$, 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, $x^{k}(1)\to y$ and
$x^{k}(0)\to x$ as $k\to\infty$, and thus without loss of generality, we can
assume that $\|x^{k}(1)-x^{k}(0)\|\geq\varepsilon$ 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
$S_{t}^{k}=S_{f}(p^{k}(t),c(t;p,q^{k},w^{k}))$. By our assumption, $S_{t}^{k}$
can be defined and is symmetric and negative semi-definite for almost all
$t\in[0,1]$. Because $S_{t}^{k}$ is symmetric and negative semi-definite,
there exists a positive semi-definite matrix $A_{t}^{k}$ such that
$S_{t}^{k}=-(A_{t}^{k})^{2}$. Moreover, the operator norm $\|A_{t}^{k}\|$ is
equal to $\sqrt{\|S_{t}^{k}\|}$.555Because $S_{t}^{k}$ is symmetric, there
exists an orthogonal transform $P$ such that
$S_{t}^{k}=P^{T}\begin{pmatrix}\lambda_{1}&0&...&0\\\ 0&\lambda_{2}&...&0\\\
\vdots&\vdots&\ddots&\vdots\\\ 0&0&...&\lambda_{n}\end{pmatrix}P,$ where
$\lambda_{i}$ is some eigenvalue of $S_{t}^{k}$. Because $S_{t}^{k}$ is
negative semi-definite, $\lambda_{i}\leq 0$ for every $i$. Hence, if we define
$A_{t}^{k}=P^{T}\begin{pmatrix}\sqrt{-\lambda_{1}}&0&...&0\\\
0&\sqrt{-\lambda_{2}}&...&0\\\ \vdots&\vdots&\ddots&\vdots\\\
0&0&...&\sqrt{-\lambda_{n}}\end{pmatrix}P,$ then $S_{t}^{k}=-(A_{t}^{k})^{2}$.
Moreover, because the operator norm $\|S_{t}^{k}\|$ (resp. $\|A_{t}^{k}\|$)
coincides with $\max_{i}|\lambda_{i}|$, (resp.
$\max_{i}\sqrt{|\lambda_{i}|}$,) all our claims are correct. Because $f$ is
locally Lipschitz, there exists $L>0$ such that $\|S_{t}^{k}\|\leq L$ for all
$k$ and almost all $t\in[0,1]$. Define $d^{k}(t)=p\cdot x^{k}(t)$, and choose
$\delta>0$ such that $\varepsilon^{2}>2L^{2}\delta\|q^{k}-p\|^{2}$ for every
sufficiently large $k$. By the same arguments as in the proof of Step 1, we
can show that

$\dot{d}^{k}(t)=-t(q^{k}-p)^{T}S_{t}^{k}(q^{k}-p)=t\|A_{t}^{k}(q^{k}-p)\|^{2},$
$\dot{x}^{k}(t)=S_{t}^{k}(q^{k}-p)$

for almost all $t\in[0,1]$. Therefore,

$\displaystyle\varepsilon^{2}\leq\|x^{k}(1)-x^{k}(0)\|^{2}=$
$\displaystyle~{}\left\|\int_{0}^{1}\dot{x}^{k}(t)dt\right\|^{2}\leq\int_{0}^{1}\|\dot{x}^{k}(t)\|^{2}dt=\int_{0}^{1}\|S_{t}^{k}(q^{k}-p)\|^{2}dt$
$\displaystyle\leq$
$\displaystyle~{}\int_{0}^{1}\|A_{t}^{k}\|^{2}\|A_{t}^{k}(q^{k}-p)\|^{2}dt\leq
L\int_{0}^{1}\|A_{t}^{k}(q^{k}-p)\|^{2}dt$ $\displaystyle=$
$\displaystyle~{}L\int_{0}^{\delta}\|A_{t}^{k}(q^{k}-p)\|^{2}dt+L\int_{\delta}^{1}\frac{1}{t}\dot{d}^{k}(t)dt$
$\displaystyle\leq$
$\displaystyle~{}L^{2}\delta\|q^{k}-p\|^{2}+\frac{L}{\delta}(d^{k}(1)-d^{k}(\delta)),$

and thus,

$\frac{\delta\varepsilon^{2}}{2L}\leq d^{k}(1)-d^{k}(\delta).$

Letting $k\to\infty$, we have that

$p\cdot y=d(1)\geq d(\delta)+\frac{\delta\varepsilon^{2}}{2L}>d(0)=m,$

as desired. This completes the proof of Step 4. $\blacksquare$

Step 5. If $x\neq y,\ x=f(p,m),\ y=f(q,w)$ and $p\cdot y\leq m$, then $q\cdot
x>w$.666In 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 $q\cdot x>w$, as desired.
This completes the proof of Step 5. $\blacksquare$

Step 6. For every $q\in\mathbb{R}^{n}_{++}$,
$c(1;q,\bar{p},c(1;p,q,m))=c(1;p,\bar{p},m)$.

Proof of Step 6. Define $p(t)=(1-t)p+tq$, $m^{*}=c(1;p,\bar{p},m)$ and
$w(t)=c(1;\bar{p},p(t),m^{*})$. By Step 3,
$c(1-t;p,\bar{p},m)=c(t;\bar{p},p,m^{*})$, and thus $w(0)=c(0;p,\bar{p},m)=m$.
Moreover, by Lemma 3 and Step 2, we have that $E(r)=c(1;\bar{p},r,m^{*})$
satisfies the following differential equation:

$DE(r)=f(r,E(r)).$

Therefore,

$\displaystyle\dot{w}(t)=$
$\displaystyle~{}\frac{d}{dt}c(1;\bar{p},(1-t)p+tq,m^{*})=\frac{d}{dt}E((1-t)p+tq)$
$\displaystyle=$ $\displaystyle~{}f((1-t)p+tq,E((1-t)p+tq))\cdot(q-p)$
$\displaystyle=$ $\displaystyle~{}f((1-t)p+tq,w(t))\cdot(q-p).$

By the uniqueness of the solution to ODE, we have that

$w(t)=c(t;p,q,m)$

for all $t\in[0,1]$. Now, define $m^{+}=c(1;p,q,m)$. Then,

$c(1;\bar{p},q,m^{*})=w(1)=m^{+}.$

By Step 3, we have $c(1-t;\bar{p},q,m^{*})=c(t;q,\bar{p},m^{+})$, and thus

$m^{*}=c(1;q,\bar{p},m^{+}),$

as desired. This completes the proof of Step 6. $\blacksquare$

Step 7. Suppose that $x=f(p,m)=f(q,w)$. Then,
$c(1;p,\bar{p},m)=c(1;q,\bar{p},w)$.

Proof of Step 7. Let $p(t)=(1-t)p+tq$ and $m(t)=(1-t)m+tw$. Suppose that
$f(p(t),m(t))=y\neq x$ for some $t\in[0,1]$. By Walras’ law, $p(t)\cdot
y=m(t)=p(t)\cdot x$, and thus $p\cdot y>m$ and $q\cdot y>w$ by Step 5.
However, this implies that $p(t)\cdot y>m(t)$, which is a contradiction.
Therefore, $f(p(t),m(t))\equiv x$, and thus

$\dot{m}(t)=w-m=x\cdot(q-p)=f(p(t),m(t))\cdot(q-p).$

By the uniqueness of the solution to ODE,

$m(t)=c(t;p,q,m),$

and thus $w=c(1;p,q,m)$. Therefore, by Step 6, we have that
$c(1;p,\bar{p},m)=c(1;q,\bar{p},w)$. This completes the proof of Step 7.
$\blacksquare$

By Steps 2 and 7, we can define $u_{f,\bar{p}}(x)$ for all $x\in\Omega$, and
our definition of $u_{f,\bar{p}}(x)$ is independent of the choice of $(p,m)$.

Step 8. $f=f^{u_{f,\bar{p}}}$.

Proof of Step 8. Suppose that $x=f(p,m),y\neq x$ and $p\cdot y\leq m$. If
$y\notin R(f)$, then $u_{f,\bar{p}}(y)=0<u_{f,\bar{p}}(x)$. If $y=f(q,w)$ for
some $(q,w)$, then by the contrapositive of Step 4, we have that
$c(1;p,q,m)>w$. By Step 3,

$c(t;q,\bar{p},c(1;p,q,m))>c(t;q,\bar{p},w)$

for every $t\in[0,1]$. By Step 6,

$u_{f,\bar{p}}(x)=c(1;p,\bar{p},m)=c(1;q,\bar{p},c(1;p,q,m))>c(1;q,\bar{p},w)=u_{f,\bar{p}}(y).$

Therefore, $x=f^{u_{f,\bar{p}}}(p,m)$, as desired. This completes the proof of
Step 8. $\blacksquare$

Step 9. $u_{f,\bar{p}}$ is upper semi-continuous on $R(f)$.

Proof of Step 9. Suppose that $x=f(p,m)$ and $u_{f,\bar{p}}(x)<a$. By Fact 4,
the solution function $c$ is continuous, and thus there exists $\varepsilon>0$
such that $c(1;p,\bar{p},m+\varepsilon)<a$. Define $y=f(p,m+\varepsilon)$.
Then, the set

$U=\\{z\in R(f)|p\cdot z<m+\varepsilon\\}$

is a neighborhood of $x$ in the relative topology of $R(f)$, and for every
$z\in U$, $u_{f,\bar{p}}(z)<u_{f,\bar{p}}(y)<a$. This completes the proof of
Step 9. $\blacksquare$

Step 10. Suppose that $f=f^{\succsim}$ for some weak order $\succsim$, and
$\succsim$ is upper semi-continuous on $R(f)$. Then, for every $x,y\in R(f)$,

$x\succsim y\Leftrightarrow u_{f,\bar{p}}(x)\geq u_{f,\bar{p}}(y).$

Proof of Step 10. First, choose any $x\in R(f)$ and suppose that $x=f(p,m)$.
Define $z=f(\bar{p},u_{f,\bar{p}}(x))$, and let

$E^{x}(q)=\inf\\{q\cdot w|w\succsim x\\}.$

By Lemma 1 of Hosoya (2020), we have that

$DE^{x}(q)=f(q,E^{x}(q)),\ E^{x}(p)=m.$

Define $d(t)=E^{x}((1-t)p+t\bar{p})$. Then,

$\dot{d}(t)=f((1-t)p+t\bar{p},d(t))\cdot(\bar{p}-p),$

which implies that $d(t)=c(t;p,\bar{p},m)$ for every $t\in[0,1]$. In
particular,

$E^{x}(\bar{p})=d(1)=c(1;p,\bar{p},m)=u_{f,\bar{p}}(x).$

Now, choose any $\varepsilon>0$. Then, there exists $w\in\Omega$ such that
$p\cdot w<E^{x}(\bar{p})+\varepsilon$ and $w\succsim x$. Define
$z^{\varepsilon}=f(\bar{p},E^{x}(\bar{p})+\varepsilon)$. Then,
$z^{\varepsilon}\succsim w$, and thus $z^{\varepsilon}\succsim x$. Letting
$\varepsilon\to 0$, by the upper semi-continuity of $\succsim$, we obtain that
$z\succsim x$.

Next, define

$E^{z}(q)=\inf\\{q\cdot w|w\succsim z\\}.$

By the same arguments as above, we can show that $E^{z}(p)=m$, and thus
$x\succsim z$. Hence, $x\sim f(\bar{p},u_{f,\bar{p}}(x))$ for all $x\in R(f)$.

Now, choose any $x,y\in R(f)$. Then,

$x\succsim y\Leftrightarrow f(\bar{p},u_{f,\bar{p}}(x))\succsim
f(\bar{p},u_{f,\bar{p}}(y))\Leftrightarrow u_{f,\bar{p}}(x)\geq
u_{f,\bar{p}}(y),$

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. $\blacksquare$

###  6.3 Proof of Corollary 1

It is obvious that (ii) implies (i).

Suppose that (i) holds, and choose any
$(p,m)\in\mathbb{R}^{n}_{++}\times\mathbb{R}_{++}$. Let $x=f(p,m)$ and define

$E^{x}(q)=\inf\\{q\cdot y|y\succsim x\\}.$

By Lemma 1 of Hosoya (2020), this $E^{x}$ is a concave solution to (2).
Suppose that $E$ is another soution of (2). For each
$q\in\mathbb{R}^{n}_{++}$, define $c_{1}(t)=E^{x}((1-t)p+tq)$ and
$c_{2}(t)=E((1-t)p+tq)$. Then,

$\dot{c}_{i}(t)=f((1-t)p+tq,c_{i}(t))\cdot(q-p),\ c_{i}(0)=m,$

and thus, by the uniqueness of the solution to ODE (Fact 3), we have that
$c_{1}\equiv c_{2}$. In particular, $E^{x}(q)=c_{1}(1)=c_{2}(1)=E(q)$, and
thus $E=E^{x}$. Therefore, the solution is unique, and (iv) holds.

Suppose that (iv) holds. Choose any
$(p,m)\in\mathbb{R}^{n}_{++}\times\mathbb{R}_{++}$ such that $f$ is
differentiable at $(p,m)$. Let $E$ be a concave solution to (2). By easy
calculation, we obtain that

$D^{2}E(p)=S_{f}(p,m).$

Because $E$ is concave, the left-hand side of the above equation is negative
semi-definite. Moreover, by extended Young’s theorem,777See Lemma 3.2 of
Hosoya (2021). 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.
$\blacksquare$

###  6.4 Proof of Corollary 2

Define

$w_{f,\bar{p}}(x)=\begin{cases}u_{f,\bar{p}}(x)&\mbox{if }x\in R(f),\\\
\inf_{\varepsilon>0}\sup\\{u_{f,\bar{p}}(y)|y\in R(f),\
\|y-x\|<\varepsilon\\}&\mbox{if }x\notin R(f).\end{cases}$

Theorems 1-2 of Hosoya (2020) stated the following facts: 1)
$f=f^{w_{f,\bar{p}}}$, 2) $w_{f,\bar{p}}$ is upper semi-continuous, 3)
$w_{f,\bar{p}}$ is continuous on $\mathbb{R}^{n}_{++}$ if and only if $f$
satisfies the C axiom, and 4) if $\succsim$ is an upper semi-continuous weak
order on $\Omega$ such that $f=f^{\succsim}$, then for each $x,y\in\Omega$,

$x\succsim y\Leftrightarrow w_{f,\bar{p}}(x)\geq w_{f,\bar{p}}(y).$

We first show that $w_{f,\bar{p}}(x)=v_{f,\bar{p}}(x)$ for all $x\in\Omega$.

If $x\in\mathbb{R}^{n}_{++}$, then
$w_{f,\bar{p}}(x)=u_{f,\bar{p}}(x)=v_{f,\bar{p}}(x)$.

Suppose that $x\notin R(f)$. Choose any $\varepsilon>0$, $\|y-x\|<\varepsilon$
and $y\in R(f)$. Then, there exists $z\in\mathbb{R}^{n}_{++}$ such that $z\gg
y$ and $\|z-x\|<\varepsilon$. If $z=f(p,m)$, then $p\cdot y<m$, and thus
$u_{f,\bar{p}}(z)>u_{f,\bar{p}}(y)$. This indicates that

$\sup\\{u_{f,\bar{p}}(y)|y\in R(f),\
\|y-x\|<\varepsilon\\}=\sup\\{u_{f,\bar{p}}(y)|y\in\mathbb{R}^{n}_{++},\
\|y-x\|<\varepsilon\\},$

and thus, $v_{f,\bar{p}}(x)=w_{f,\bar{p}}(x)$.

Suppose that $x\in R(f)\setminus\mathbb{R}^{n}_{++}$. Let $e=(1,1,...,1)$ and
define $x^{k}=x+k^{-1}e$. Then, $x^{k}\in\mathbb{R}^{n}_{++}$. It is easy to
show that

$\lim_{k\to\infty}u_{f,\bar{p}}(x^{k})=v_{f,\bar{p}}(x).$

Because $u_{f,\bar{p}}$ is upper semi-continuous on $R(f)$, we have that

$\lim_{k\to\infty}u_{f,\bar{p}}(x^{k})\leq u_{f,\bar{p}}(x).$

On the other hand, if $x^{k}=f(p^{k},m^{k})$, then $p^{k}\cdot x<m^{k}$, and
thus $u_{f,\bar{p}}(x^{k})>u_{f,\bar{p}}(x)$. Therefore, we have that

$\lim_{k\to\infty}u_{f,\bar{p}}(x^{k})\geq u_{f,\bar{p}}(x).$

To combine these inequalities, we have that

$v_{f,\bar{p}}(x)=\lim_{k\to\infty}u_{f,\bar{p}}(x^{k})=w_{f,\bar{p}}(x),$

as desired. Hence, $v_{f,\bar{p}}(x)=w_{f,\bar{p}}(x)$.

The rest of the claim of this corollary is 3). If $v_{f,\bar{p}}$ is
continuous, then $f=f^{\succsim}$ for a continuous weak order $\succsim$
defined as

$x\succsim y\Leftrightarrow v_{f,\bar{p}}(x)\geq v_{f,\bar{p}}(y).$

Let us show the converse. Suppose that there exists a continuous weak order
$\succsim$ on $\Omega$ such that $f=f^{\succsim}$. Debreu (1954) showed that
there exists a continuous function $u:\Omega\to\mathbb{R}$ such that $u$
represents $\succsim$. By the above arguments, $v_{f,\bar{p}}$ also represents
$\succsim$, and thus $v_{f,\bar{p}}$ and $u$ have the same order. On the other
hand, in the proof of Theorem 1 of Hosoya (2020), it was shown that if
$v_{f,\bar{p}}(x)>0$, then

$v_{f,\bar{p}}(x)=v_{f,\bar{p}}(f(\bar{p},v_{f,\bar{p}}(x)))$

for all $x\in\Omega$, and $v_{f,\bar{p}}(0)=0$. Therefore, if
$v_{f,\bar{p}}(x)>0$, then

$u(x)=u(f(\bar{p},v_{f,\bar{p}}(x))),$

and if $v_{f,\bar{p}}(x)=0$, then

$u(x)=u(0).$

If $v_{f,\bar{p}}(x)>0$, then for every sufficiently small $\varepsilon>0$,
there exists $\delta>0$ such that if $y\in\Omega$ and $\|y-x\|<\delta$, then

$u(f(\bar{p},v_{f,\bar{p}}(x)-\varepsilon))<u(y)<u(f(\bar{p},v_{f,\bar{p}}(x)+\varepsilon)),$

which implies that $|v_{f,\bar{p}}(y)-v_{f,\bar{p}}(x)|<\varepsilon$. If
$v_{f,\bar{p}}(x)=0$, then for every $\varepsilon>0$, there exists $\delta>0$
such that if $y\in\Omega$ and $\|y-x\|<\delta$, then

$u(0)\leq u(y)<u(f(\bar{p},v_{f,\bar{p}}(x)+\varepsilon)),$

which implies that $0\leq v_{f,\bar{p}}(y)<\varepsilon$. Therefore,
$v_{f,\bar{p}}$ is continuous. This completes the proof. $\blacksquare$

###  6.5 Proof of Theorem 2

Define

$I(t,c,g,p,q)=g((1-t)p+tq,c)\cdot(q-p),$

where $g:\mathbb{R}^{n}_{++}\times\mathbb{R}_{++}\to\mathbb{R}^{n}$ is locally
Lipschitz. Consider the following ODE:

$\dot{c}(t)=I(t,c(t),g,p,q),\ c(0)=m,$ (15)

and let $c(t;g,p,q,m)$ be the solution function. If the domain of $c$ includes
$[0,1]\times\\{f\\}\times\mathbb{R}^{n}_{++}\times\mathbb{R}^{n}_{++}\times\mathbb{R}_{++}$,
then by Corollary 1, we have that $f$ is a demand function. Hence, it suffices
to show that for every $(p,m)\in\mathbb{R}^{n}_{++}\times\mathbb{R}_{++}$ and
$q\in\mathbb{R}^{n}_{++}$, the domain of the mapping $t\mapsto c(t;f,p,q,m)$
includes $[0,1]$.

Choose any $(p,m)\in\mathbb{R}^{n}_{++}\times\mathbb{R}_{++}$ and
$q\in\mathbb{R}^{n}_{++}$, and define $p(t)=(1-t)p+tq$. For every continuous
function $g:\mathbb{R}^{n}_{++}\times\mathbb{R}_{++}\to\mathbb{R}^{n}$, define

$H_{K}(g)=\sup_{(r,c)\in[K^{-1},K]^{n+1}}\|g(r,c)\|.$

Hereafter, we abbreviate $c(t;f,p,q,m)$ as $c(t)$ and $c(t;f^{k},p,q,m)$ as
$c^{k}(t)$. Note that $c^{k}(t)$ is defined on $[0,1]$ for all $k$ according
to Corollary 1.

Because $f^{k}$ is a demand function, there exists a weak order $\succsim_{k}$
on $\Omega$ such that $f^{k}=f^{\succsim_{k}}$. Define

$E^{k}(r)=\inf\\{r\cdot y|y\succsim_{k}f^{k}(p,m)\\}.$

Then, by Shephard’s lemma (Lemma 1 of Hosoya (2020)), we have that
$c^{k}(t)=E^{k}(p(t))$. Define

$\phi(r,w)=\max\\{p\cdot y|y\in\Omega,\ r\cdot y=w\\}.$

Then, $\phi(r,0)=0$, and by Berge’s maximum theorem, $\phi$ is continuous.
Therefore, there exists $w_{0}>0$ such that if $p(t)\cdot y=w_{0}$ for some
$t\in[0,1]$, then $p\cdot y<m$, and thus $y\not\succsim^{k}f^{k}(p,m)$ for all
$k$. By definition, $E^{k}(p(t))\geq w_{0}$, and thus $c^{k}(t)\geq w_{0}$ for
all $k$ and $t\in[0,1]$. On the other hand,

$\displaystyle c^{k}(t)=$ $\displaystyle~{}E^{k}(p(t))\leq p(t)\cdot
f^{k}(p,m)$ $\displaystyle\leq$ $\displaystyle~{}(1-t)m+t\max\\{q\cdot
x|p\cdot x=m\\}$ $\displaystyle\leq$
$\displaystyle~{}\max\\{(1-s)m+s\max\\{q\cdot x|p\cdot
x=m\\}|s\in[0,1]\\}\equiv w_{1}$

for all $k$ and $t\in[0,1]$. Choose $K\in\mathbb{R}$ such that
$p,q\in[K^{-1},K]^{n}$ and $w_{0},w_{1}\in]K^{-1},K[$. Because $f$ is locally
Lipschitz, there exists $L>0$ such that if
$(p^{\prime},m^{\prime}),(q^{\prime},w^{\prime})\in[K^{-1},K]^{n+1}$, then

$\|f(p^{\prime},m^{\prime})-f(q^{\prime},w^{\prime})\|\leq
L\|(p^{\prime},m^{\prime})-(q^{\prime},w^{\prime})\|.$

Let $I^{*}$ be the set of all $t\in[0,1]$ such that $c(s)$ is defined and
$c(s)\in[K^{-1},K]$ for all $s\in[0,t]$, and suppose that $I^{*}=[0,t^{*}[$.
For any $t\in[0,t^{*}[$,

$\displaystyle|c^{k}(t)-c(t)|\leq$
$\displaystyle~{}\int_{0}^{t}\|f^{k}(p(s),c^{k}(s))-f(p(s),c(s))\|\|q-p\|ds$
$\displaystyle\leq$
$\displaystyle~{}\int_{0}^{t}[\|f^{k}(p(s),c^{k}(s))-f(p(s),c^{k}(s))\|$
$\displaystyle~{}+\|f(p(s),c^{k}(s))-f(p(s),c(s))\|]\|q-p\|ds$
$\displaystyle\leq$
$\displaystyle~{}\int_{0}^{t}[L\|q-p\||c^{k}(s)-c(s)|+H_{K}(f^{k}-f)\|q-p\|]ds,$

and thus, by Lemma 1,

$|c^{k}(t)-c(t)|\leq\frac{H_{K}(f^{k}-f)}{L}(e^{L\|q-p\|}-1).$ (16)

This indicates that $c^{k}(t)\to c(t)$ as $k\to\infty$, and thus
$c(t)\in[w_{0},w_{1}]$ for all $t\in[0,t^{*}[$. Because $c(t)$ is a
nonextendable solution to (15), by Fact 3, we have that $c(t)$ is defined at
$t^{*}$. By continuity of $c(t)$, $c(t^{*})\in[w_{0},w_{1}]\subset[K^{-1},K]$,
which is a contradiction. Therefore, $I^{*}=[0,t^{*}]$. If $t^{*}<1$, then
$c(t^{*})\in]K^{-1},K[$, and thus there exists $t>t^{*}$ such that $t\in
I^{*}$, 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.
$\blacksquare$

###  6.6 Proof of Corollary 3

Suppose that $(f^{k})$ is a Cauchy sequence in $\mathscr{F}_{L}$. It is easy
to show that $f^{k}\to f$ for some function
$f:\mathbb{R}^{n}_{++}\times\mathbb{R}_{++}\to\Omega$, and for every compact
set $C\subset\mathbb{R}^{n}_{++}\times\mathbb{R}_{++}$,

$\lim_{k\to\infty}\sup_{(p,m)\in C}\|f^{k}(p,m)-f(p,m)\|=0.$

Clearly $f$ satisfies Walras’ law. Because $\Delta_{\nu}$ is compact, we have
that $(f^{k})$ converges to $f$ uniformly on $\Delta_{\nu}$. Moreover, if
$(p,m),(q,w)\in\Delta_{\nu}$, then

$\|f(p,m)-f(q,w)\|=\lim_{k\to\infty}\|f^{k}(p,m)-f^{k}(q,w)\|\leq
L_{\nu}\|(p,m)-(q,w)\|,$

which implies that $f\in\mathscr{F}_{L}$. This completes the proof.
$\blacksquare$

###  6.7 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 $\mathbb{R}^{n}_{++}$. If $f$
satisfies the C axiom, then $u_{f,\bar{p}}$ is continuous on
$\mathbb{R}^{n}_{++}$.888Note that, in this lemma, $R(f)$ need not be
relatively open in $\Omega$, and thus Corollary 2 cannot be directly applied.

Proof. Recall the differential equation (1):

$\dot{c}(t)=f((1-t)p+t\bar{p},c(t))\cdot(\bar{p}-p),\ c(0)=m.$

Let $c(t;p,m)$ be the solution function. We have that
$u_{f,\bar{p}}(x)=c(1;p,m)$ if $x=f(p,m)$. Choose any sequence $(x^{k})$ in
$\mathbb{R}^{n}_{++}$ such that $x^{k}\to x\in\mathbb{R}^{n}_{++}$ as
$k\to\infty$, and suppose that $u_{f,\bar{p}}(x^{k})\not\to u_{f,\bar{p}}(x)$.
Taking a subsequence, we can assume that there exists $\varepsilon>0$ such
that $|u_{f,\bar{p}}(x^{k})-u_{f,\bar{p}}(x)|>\varepsilon$ for every $k$.
Choose any $p^{k}\in G^{f}(x^{k})$. Again, taking a subsequence, we can assume
that $p^{k}\to p^{*}\in G^{f}(x)$. Then,

$u_{f,\bar{p}}(x^{k})=c(1;p^{k},p^{k}\cdot x^{k})\to c(1;p^{*},p^{*}\cdot
x)=u_{f,\bar{p}}(x),$

which is a contradiction. Therefore, $u_{f,\bar{p}}$ is continuous on
$\mathbb{R}^{n}_{++}$. This completes the proof. $\blacksquare$

Choose any compact set $D\subset\mathbb{R}^{n}_{++}$. Let $x\in D$. We first
show that there exists an open neighborhood $U$ of $x$ and $\varepsilon>0$
such that if $p\in G^{f^{k}}(y)$ for some $y\in U$ and $k$, then
$p_{i}\geq\varepsilon$ for all $i\in\\{1,...,n\\}$.

Suppose not. Then, there exists $p^{\ell}\in\mathbb{R}^{n}_{++}$ and
$z^{\ell}=f^{k(\ell)}(p^{\ell},p^{\ell}\cdot z^{\ell})$ such that
$\sum_{j}p^{\ell}_{j}=1,\ z^{\ell}\to x$ and $\min_{j}p^{\ell}_{j}\to 0$ as
$\ell\to\infty$. First, suppose that $k(\ell)=k$ for infinitely many $\ell$.
Taking a subsequence, we can assume that $k(\ell)=k$ for any $\ell$. By the C
axiom, the inverse demand correspondence $G^{f^{k}}$ is compact-valued and
upper hemi-continuous. Moreover, $p^{\ell}\in G^{f^{k}}(z^{\ell})$ for all
$\ell$ and $z^{\ell}\to x$ as $\ell\to\infty$. Therefore, if we choose

$\varepsilon=\frac{1}{2}\min\\{\min_{i}p_{i}|p\in G^{f^{k}}(x)\\},$

then $\varepsilon>0$ and $\min_{j}p^{\ell}_{j}\geq\varepsilon$ for
sufficiently large $\ell$, which is a contradiction. Hence, we can assume that
$k(\ell)$ is increasing. Taking a subsequence, we can assume that $p^{\ell}\to
p^{*}\in\mathbb{R}^{n}_{+}$, where $\sum_{j}p^{*}_{j}=1$ and
$\min_{i}p^{*}_{i}=0$. Choose $i,j\in\\{1,...,n\\}$ such that $p^{*}_{i}>0$
and $p^{*}_{j}=0$.

Let $e_{j}$ denote the $j$-th unit vector. Choose a small $\delta>0$, and set
$y_{j}=x_{j}+2$, $y_{i}(\delta)=x_{i}-\delta$, and $y_{m}=x_{m}$ for every
$m\in\\{1,...,n\\}\setminus\\{i,j\\}$. Let
$y(\delta)=(y_{1},...,y_{i}(\delta),...,y_{n})$. Because $G^{f}$ is upper
semi-continuous, there exists a sequence $(\delta_{\nu})$ of positive real
numbers such that $\delta_{\nu}\to 0$ as $\nu\to 0$ and there exists
$p^{\nu}\in G^{f}(y(\delta_{\nu}))$ such that $p^{\nu}\to p^{+}\in
G^{f}(y(0))$ as $\nu\to\infty$. Because $p^{+}\cdot y(0)>p^{+}\cdot(x+e_{j})$,
we have that $p^{\nu}\cdot y(\delta_{\nu})>p^{\nu}\cdot(x+e_{j})$ for
sufficiently large $\nu$. Choose such a $\nu$, and define $q=p^{\nu}$ and
$y=y(\delta_{\nu})$. Then, $q\in G^{f}(y)$ and thus $y=f(q,q\cdot y)$. Define
$y^{k}=f^{k}(q,q\cdot y)$. Then, $y^{k}\to y$ as $k\to\infty$. By assumption,
$q\cdot z^{\ell}+q_{j}<q\cdot y$ for sufficiently large $\ell$, and thus we
have that $q\cdot z^{\ell}\leq q\cdot y^{k(\ell)}$ for sufficiently large
$\ell$. However,

$\lim_{\ell\to\infty}p^{\ell}\cdot z^{\ell}=p^{*}\cdot x>p^{*}\cdot
y=\lim_{\ell\to\infty}p^{\ell}\cdot y^{k(\ell)},$

and thus $p^{\ell}\cdot z^{\ell}\geq p^{\ell}\cdot y^{k(\ell)}$ if $\ell$ is
sufficiently large, which contradicts the weak axiom of $f^{k(\ell)}$.
Therefore, our initial claim is correct. Define $U_{x}$ as such a neighborhood
that corresponds with $x\in D$. Then $(U_{x})$ is an open covering of $D$, and
thus there exists $\varepsilon^{*}>0$ such that if $\sum_{j}p_{j}=1$ and
$f^{k}(p,p\cdot x)=x$ for some $x\in D$ and $k$, then
$p_{i}\geq\varepsilon^{*}$ for all $i\in\\{1,...,n\\}$.

Let

$C=\\{p\in\mathbb{R}^{n}_{++}|p\in G^{f}(x)\mbox{ for some }x\in D\\},$
$C_{k}=\\{p\in\mathbb{R}^{n}_{++}|p\in G^{f^{k}}(x)\mbox{ for some }x\in
D\\}.$

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 $K\subset\mathbb{R}^{n}_{++}$ that
includes $C$ and all $C_{k}$. Define $m_{1}=\min\\{p\cdot x|p\in K,x\in
D\\}>0$ and $m_{2}=\max\\{p\cdot x|p\in K,x\in D\\}>0$.

It suffices to show that $\sup_{x\in
D}|u_{f^{k},\bar{p}}(x)-u_{f,\bar{p}}(x)|\to 0$ as $k\to\infty$. Suppose not.
Then, there exist $\varepsilon>0$ and a sequence $(x^{\ell})$ in $D$ such that
$|u_{f^{k(\ell)},\bar{p}}(x^{\ell})-u_{f,\bar{p}}(x^{\ell})|\geq\varepsilon$
for all $\ell$, where $k(\ell)$ is increasing in $\ell$. Because $D$ is
compact, we can assume that $x^{\ell}\to x^{*}\in D$ as $\ell\to\infty$.
Suppose that $x^{\ell}=f^{k(\ell)}(p^{\ell},m^{\ell})$, where $p^{\ell}\in
C_{k(\ell)}$ and $m^{\ell}=p^{\ell}\cdot x^{\ell}$. Taking a subsequence, we
can assume that $p^{\ell}\to p^{*}\in K$. Define $m^{*}=p^{*}\cdot x^{*}$.
Then, $(p^{\ell},m^{\ell}),(p^{*},m^{*})\in K\times[m_{1},m_{2}]$, and thus

$\displaystyle\|f^{k(\ell)}(p^{\ell},m^{\ell})-f(p^{*},m^{*})\|\leq$
$\displaystyle~{}\|f^{k(\ell)}(p^{\ell},m^{\ell})-f(p^{\ell},m^{\ell})\|$
$\displaystyle~{}+\|f(p^{\ell},m^{\ell})-f(p^{*},m^{*})\|\to 0,$

as $\ell\to\infty$. This implies that $f(p^{*},m^{*})=x^{*}$.

Now, consider the following differential equation:

$\dot{c}(t)=I(t,c(t),g,p,m),\ c(0)=m^{*},$ (17)

where $g:\mathbb{R}^{n}_{++}\times\mathbb{R}_{++}\to\Omega$ is locally
Lipschitz and $I(t,c,g,p,m)=g((1-t)p+t\bar{p},c+m-m^{*})\cdot(\bar{p}-p)$. Let
$c(t;g,p,m)$ be the solution function of (17). We abbreviate
$c(t;f,p^{*},m^{*})$ as $c^{*}(t)$ and $c(t;f^{k(\ell)},p^{\ell},m^{\ell})$ as
$c^{\ell}(t)$. By Theorem 1, the domain of $c^{*}(t)$ and $c_{\ell}(t)$
includes $[0,1]$, $u_{f,\bar{p}}(x^{*})=c^{*}(1)$, and
$u_{f^{k(\ell)},\bar{p}}(x^{\ell})=c^{\ell}(1)+m^{\ell}-m^{*}$.

Choose $a>0$ and $b>0$ so small, and define

$\Pi=\\{(c,p,m)|\exists t\in[0,1]\mbox{ s.t. }|c^{*}(t)-c|\leq a,\
\|p-p^{*}\|+|m-m^{*}|\leq b\\},$

and $\hat{\Pi}$ as the set of all $g\in\mathscr{F}_{L}$ such that $\|g-f\|\leq
b$, where

$\|h\|=\sup_{(t,c,p,m)\in[0,1]\times\Pi}\|h((1-t)p+t\bar{p},c+m-m^{*})\|.$

Define $\tilde{\Pi}=\Pi\times\hat{\Pi}$. We assume that $a,b$ are so small
that 1) $\Pi$ is a compact set in
$\mathbb{R}_{++}\times\mathbb{R}^{n}_{++}\times\mathbb{R}_{++}$, 2)
$\min\\{c+m-m^{*}|(c,p,m)\in\Pi\\}>0$, 3) there exists $L>0$ such that if
$(t.c,p,m,g),(t,c^{\prime},p,m,g)\in[0,1]\times\tilde{\Pi}$, then

$\|g((1-t)p+t\bar{p},c+m-m^{*})-g((1-t)p+t\bar{p},c^{\prime}+m-m^{*})\|\|\bar{p}-p\|\leq
L|c-c^{\prime}|,$

and 4) there exists $B>0$ such that if $(t,c,p,m,g)\in[0,1]\times\tilde{\Pi}$,
then

$|I(t,c,g,p,m)-I(t,c,f,p^{*},m^{*})|\leq B[\|p-p^{*}\|+|m-m^{*}|+\|g-f\|].$

Now, for any sufficiently large $\ell$, $|m^{\ell}-m^{*}|\leq a$,
$\|p^{\ell}-p^{*}\|+|m^{\ell}-m^{*}|\leq b$ and $\|f^{k(\ell)}-f\|\leq b$. For
such an $\ell$, define $t_{\ell}=\sup\\{t\in[0,1]|\forall
s\in[0,t],(c_{\ell}(s),p^{\ell},m^{\ell})\in\Pi\\}$. Because
$c_{\ell}(0)=m^{*}=c^{*}(0)$, we have that $t_{\ell}$ is well-defined and
positive. If $t\in[0,t_{\ell}]$, then

$\displaystyle~{}|c_{\ell}(t)-c^{*}(t)|$ $\displaystyle\leq$
$\displaystyle~{}\int_{0}^{t}|I(\tau,c_{\ell}(\tau),f^{k(\ell)},p^{\ell},m^{\ell})-I(\tau,c^{*}(\tau),f,p^{*},m^{*})|d\tau$
$\displaystyle\leq$
$\displaystyle~{}\int_{0}^{t}|I(\tau,c_{\ell}(\tau),f^{k(\ell)},p^{\ell},m^{\ell})-I(\tau,c^{*}(\tau),f^{k(\ell)},p^{\ell},m^{\ell})|d\tau$
$\displaystyle+\int_{0}^{t}|I(\tau,c^{*}(\tau),f^{k(\ell)},p^{\ell},m^{\ell})-I(\tau,c^{*}(\tau),f,p^{*},m^{*})|d\tau$
$\displaystyle\leq$
$\displaystyle~{}\int_{0}^{t}[L|c_{\ell}(\tau)-c^{*}(\tau)|+B(\|p^{\ell}-p^{*}\|+|m^{\ell}-m^{*}|+\|f^{k(\ell)}-f\|_{C^{1}})]d\tau.$

By Lemma 1, we have for any $t\in[0,t_{\ell}]$,

$\displaystyle|c_{\ell}(t)-c^{*}(t)|\leq$
$\displaystyle~{}\frac{B(\|p^{\ell}-p^{*}\|+|m^{\ell}-m^{*}|+\|f^{k(\ell)}-f\|)}{L}(e^{L}-1)$
$\displaystyle\equiv$
$\displaystyle~{}C(\|p^{\ell}-p^{*}\|+|m^{\ell}-m^{*}|+\|f^{k(\ell)}-f\|)$

for some $C>0$. Now, choose any $b^{\prime}\in]0,b]$ with $Cb^{\prime}<a$. If
$\ell$ is sufficiently large, then
$\|p^{\ell}-p^{*}\|+|m^{\ell}-m^{*}|+\|f^{k(\ell)}-f\|_{C^{1}}\leq
b^{\prime}$. For such $\ell$, we have that $t_{\ell}=1$: if not, then
$a\leq|c_{\ell}(t_{\ell})-c^{*}(t_{\ell})|\leq Cb^{\prime}<a$, which is a
contradiction. Therefore,

$|c_{\ell}(1)-c^{*}(1)|\leq
C(\|p^{\ell}-p^{*}\|+|m^{\ell}-m^{*}|+\|f^{k(\ell)}-f\|_{C^{1}}),$

and thus if $\ell$ is sufficiently large, then

$|u_{f^{k(\ell)},\bar{p}}(x^{\ell})-u_{f,\bar{p}}(x^{*})|\leq|c_{\ell}(1)-c^{*}(1)|+|m^{\ell}-m^{*}|<\frac{\varepsilon}{2}.$

By Lemma 4, $u_{f,\bar{p}}$ is continuous at $x^{*}$, and thus
$|u_{f,\bar{p}}(x^{\ell})-u_{f,\bar{p}}(x^{*})|<\frac{\varepsilon}{2}$ if
$\ell$ is sufficiently large. Then,

$\displaystyle\varepsilon\leq$
$\displaystyle~{}|u_{f^{k(\ell)},\bar{p}}(x^{\ell})-u_{f,\bar{p}}(x^{\ell})|$
$\displaystyle\leq$
$\displaystyle~{}|u_{f^{k(\ell)},\bar{p}}(x^{\ell})-u_{f,\bar{p}}(x^{*})|+|u_{f,\bar{p}}(x^{*})-u_{f,\bar{p}}(x^{\ell})|$
$\displaystyle<$ $\displaystyle~{}\varepsilon,$

which is a contradiction. This completes the proof. $\blacksquare$

###  6.8 Proof of Corollary 4

By Theorem 3, it suffices to show that
$\limsup_{k\to\infty}v_{f^{k},\bar{p}}(x)\leq v_{f,\bar{p}}(x)$ for all
$x\in\Omega\setminus\mathbb{R}^{n}_{++}$. Choose any
$x\in\Omega\setminus\mathbb{R}^{n}_{++}$. Let $e=(1,1,...,1)$ and define
$x^{\ell}=x+\ell^{-1}e$. By the same argument as in the proof of Corollary 2,
we can show that $u_{f,\bar{p}}(x^{\ell})>v_{f,\bar{p}}(x)$ and
$\lim_{\ell\to\infty}u_{f,\bar{p}}(x^{\ell})=v_{f,\bar{p}}(x)$. The same fact
is true for $v_{f^{k},\bar{p}}$. Now, suppose that
$\limsup_{k\to\infty}v_{f^{k},\bar{p}}(x)=a>v_{f,\bar{p}}(x)$. Because
$v_{f^{k},\bar{p}}(x)<u_{f^{k},\bar{p}}(x^{1})$ and $u_{f^{k},\bar{p}}(x^{1})$
converges to $u_{f,\bar{p}}(x)$ as $k\to\infty$, we have that the sequence
$(v_{f^{k},\bar{p}}(x))$ is bounded, and thus $a<+\infty$. Taking a
subsequence, we can assume that $\lim_{k\to\infty}v_{f^{k},\bar{p}}(x)=a$. Let
$3\varepsilon=a-v_{f,\bar{p}}(x)$. Choose $\ell$ so large that
$u_{f,\bar{p}}(x^{\ell})<v_{f,\bar{p}}(x)+\varepsilon$. By Theorem 3, for
sufficiently large $k$, $v_{f^{k},\bar{p}}(x)\leq
u_{f^{k},\bar{p}}(x^{\ell})<u_{f,\bar{p}}(x^{\ell})+\varepsilon<v_{f,\bar{p}}(x)+2\varepsilon$,
which is a contradiction. This completes the proof. $\blacksquare$

###  6.9 Proof of Corollary 5

First, choose any $x\in\mathbb{R}^{n}_{++}$. Then, there exists $\nu$ such
that $x\in]\nu^{-1},\nu[^{n}$. Choose $p^{k}\in G^{f^{k}}(x)$ for each $k$.
Because $(p^{k})$ is a sequence of the compact set

$P_{\nu}=\\{p\in\mathbb{R}^{n}_{++}|\sum_{i}p_{i}=1,\ \min_{i}p_{i}\geq
M_{\nu}\\},$

there exists a subsequence $p^{\ell(k)}$ such that
$\lim_{k\to\infty}p^{\ell(k)}=p^{*}\in P_{\nu}$. Because $f^{\ell(k)}$
converges to $f$ uniformly on any compact set,

$\lim_{k\to\infty}f^{\ell(k)}(p^{\ell(k)},p^{\ell(k)}\cdot
x)=f(p^{*},p^{*}\cdot x),$

and thus $p^{*}\in G^{f}(x)$. This implies that $R(f)$ includes
$\mathbb{R}^{n}_{++}$.

Second, choose any $x\in]\nu^{-1},\nu[^{n}$, and suppose that there exists
$p\in G^{f}(x)$ such that $\min_{i}p_{i}<M_{\nu}$. Let $x^{k}=f^{k}(p,p\cdot
x)$. Then, $x^{k}\to x$ as $k\to\infty$, and thus $x^{k}\in]\nu^{-1},\nu[^{n}$
for sufficiently large $k$. Because $p\in G^{f^{k}}(x^{k})$, we have that
$f^{k}\notin\mathscr{F}_{L,M}$, which is a contradiction. Therefore, if $p\in
G^{f}(x)$, then $\min_{i}p_{i}\geq M_{\nu}$. Because $G^{f}(x)$ is obviously
closed, this implies that $G^{f}$ is compact-valued.

Third, it is easy to show that for any demand function $f^{\prime}$ and $x\in
R(f^{\prime})$, $G^{f^{\prime}}(x)$ 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 $(x^{\ell})$ and
$(p^{\ell})$ such that $x^{\ell}\to x$ as $\ell\to\infty$ and $p^{\ell}\in
G^{f}(x^{\ell})\setminus U$ for all $\ell$. Choose $\nu$ such that
$x^{\ell}\in]\nu^{-1},\nu[^{n}$ for all $\ell$. Then, $(p^{\ell})$ is a
sequence in the compact set $P_{\nu}$, and thus by taking a subsequence, we
can assume that $p^{\ell}\to p^{*}\in P_{\nu}$ as $\ell\to\infty$. Because $f$
is continuous, we have that $p^{*}\in G^{f}(x)\subset U$, 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. $\blacksquare$

## Acknowledgement

This work was supported by JSPS KAKENHI Grant Number JP21K01403.

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