# An Axiom for Concavifiable Preferences in View of Alt’s Theory

Yuhki Hosoya
Faculty of Economics, Chuo University E-mail:
hosoya(at)tamacc.chuo-u.ac.jp742-1 Higashinakano, Hachioji-shi, Tokyo,
192-0393 Japan.

###### Abstract

We present a necessary and sufficient condition for Alt’s system to be
represented by a continuous utility function. Moreover, we present a necessary
and sufficient condition for this utility function to be concave. The latter
condition can be seen as an extension of Gossen’s first law, and thus has an
economic interpretation. Together with the above results, we provide a
necessary and sufficient condition for Alt’s utility to be continuously
differentiable.

JEL codes. D11, C65, D60.

Keywords. Alt’s system, cardinal utility, Gossen’s first law, path-
connectedness.

##  1 Introduction

In economics, there are a number of conditions that can usually be assumed for
utility functions, such as continuity, quasi-concavity, and strict quasi-
concavity. These conditions are accepted because the axioms of the preference
relation that can be represented by a utility function satisfying such
conditions are easy to understand and have a natural interpretation in
economics. Differentiability, on the other hand, is unlike these. The
conditions for a preference relation that can be represented by a
differentiable utility function are neither easy to understand nor naturally
interpretable in terms of economics. Often, however, the differentiability of
utility functions is accepted simply because it is overwhelmingly convenient.

In this paper, we treat the concavity of utility functions. Concavity is not
usually considered a good assumption for utility functions. There are several
reasons for this, but the most important one is the difficulty of interpreting
concavity in economic terms. A characterization of such a preference relation
that can be represented by a concave utility function can be found in Kannai
(1977). Three axioms are presented, corresponding to the condition of
concavity without differentiation, with first-order derivatives, and with
second-order derivatives. However, all of these axioms are too complicated for
their meaning to be interpreted in economic terms. On the contrary, it is
difficult to determine, even mathematically, whether or not these conditions
are satisfied for a given preference relation. Because of this problem, the
assumption that the utility function is concave is frowned upon, at least in
theoretical research.

The advantages of assuming the concavity for the utility function are,
however, too great to ignore. First, if the utility function is concave, then
Lagrange’s multiplier rule can be applied to the derivation of the demand
function using subdifferential analysis.111See, for example, Rockafeller
(1996) or Ioffe and Tikhomirov (1979) for detailed arguments. This advantage
is not negligible in real calculations. Second, there are some concepts that
cannot be defined without the existence of a concave utility function. For
example, Debreu’s (1976) least concave utility is one of these concepts.
Kannai (1980) connected this concept with ALEP substitution and
complementarity in equilibrium theory. Epstein and Zhang (1999) showed that
the order among concave utility representations that was introduced by Debreu
(1976) is related to the concept of uncertainty aversion. Thus, there are many
applications of the least concave utility function. Third, there are several
properties that emerge from the presence of concave utility functions as the
input to a Samuelson-Bergson type social welfare function, and concavity has a
non-negligible effect on welfare analysis in applied research.

Let us elaborate a little on the last point. If one extends the preference
relation from the usual space to the space of simple lotteries, then some
results are known about the concave representability of the Neumann-
Morgenstern utility function. However, the use of the Neumann-Morgenstern
utility function in welfare analysis has long been criticized, by Luce and
Raiffa (1957) and others, as being interpretatively unnatural. Furthermore,
because the basic model of general equilibrium theory is not a stochastic
model, the use of the Neumann-Morgenstern utility function is not desirable.

In this paper, we revisit the existence theorem of concave utility function
from Alt’s representation theory (Alt, 1936). To the best of our knowledge,
this is the oldest theory to have discussed the possibility of utility
representation of preference relations. Although this theory is also a kind of
cardinal utility theory, it has an interpretation that is more suitable for
welfare analysis than that of Neumann-Morgenstern utility. Furthermore, Alt’s
theory is not especially incompatible with general equilibrium theory. In
these respects, it is worth deriving the necessary and sufficient conditions
for Alt’s utility function to be concave.

Note that the fact that Alt’s utility function is concave is more important
than just the fact that the preference is represented by some concave utility
function. Because Alt’s utility function has an interpretation that is
suitable for welfare analysis, it can be treated as an input to the Samuelson-
Bergson type social welfare function naturally. Recall that almost all typical
Samuelson-Bergson type social welfare functions, including the Bentham, Rawls,
and Nash types, are quasi-concave. Now, consider a quasi-concave social
welfare function. Suppose that there are two individuals with the same utility
function. If this utility function is concave, then bringing the states of
these two individuals closer together does not decrease welfare. If the
utility function is strictly concave, then bringing the states of these two
individuals closer together will increase welfare in many cases. Thus, if the
utility function is concave, the ethical principle that “like people should be
in like states” is justified. In contrast, if the utility function is not
concave, then it may be possible to improve welfare by putting similar people
in different situations. Thus, when discussing welfare analysis, whether the
utility function is concave or not has a decisive influence on the ethical
principle. From this viewpoint, we think that the concavity of Alt’s utility
function is important for welfare analysis.

In this paper, we first derive a necessary and sufficient condition for the
existence of a continuous utility function that represents Alt’s system
(Theorem 1). Although Alt essentially showed such a result, the axioms he used
were too numerous and not independent, and were difficult to interpret
economically. Similar problems can be found in more recent studies, such as
Kranz et al. (1971). Seidl and Schmidt (1997) elaborated on this result, but
again the number of conditions is too large and not independent. In contrast,
Shapley (1975) derived an existence theorem for Alt’s utility function with
only three simple axioms. However, the space treated by Shapley is limited to
a subset of the one-dimensional space. Our first result can be seen as an
extension of Shapley’s result to Alt’s system on a Hausdorff, separable, and
path-connected topological space.

We then derive a necessary and sufficient condition for Alt’s utility function
to be concave (Theorem 2). Because this condition can be seen as a
generalization of Gossen’s first law, we name this condition “generalized
Gossen’s first law”. Compared with Kannai’s axiom, this generalized Gossen’s
first law is much easier to interpret economically.

Finally, because Gossen’s first law is usually described as a feature of
partial derivatives of the utility function, we want to obtain the conditions
for Alt’s utility function to be differentiable. Hence, we present a necessary
and sufficient condition for Alt’s utility function to be continuously
differentiable and nondegenerate (Theorem 3). This result is also useful for
analysis of ALEP substitution and complementarity.

Note that generalized Gossen’s first law is stronger than Kannai’s axiom as a
condition for the existence of concave utility functions. Our condition is a
necessary and sufficient condition for Alt’s utility function to be concave,
whereas Kannai’s axiom only guarantees the existence of a concave utility
function to represent a given preference relation. We believe this axiom is
still useful because it has a much more natural economic interpretation than
Kannai’s axiom. This is, in our opinion, one of the most important
requirements when discussing whether or not to assume such a property for the
utility function.

In subsection 2.1, we introduce the notion of Alt’s system, and discuss its
interpretation. In subsections 2.2 and 2.3, we present three axioms for Alt’s
system, and state that they are equivalent to the existence of the
corresponding utility function. In subsections 2.4 and 2.5, we treat
generalized Gossen’s first law, and show that this axiom is equivalent to the
concavity of Alt’s utility function. In subsection 2.6, we examine the
differentiability of Alt’s utility. In section 3, we discuss the relationship
between this work and several related studies. Because the proofs of Theorems
1 and 3 are somewhat lengthy, they are given in the appendix.

##  2 Results

###  2.1 Preliminaries: Alt’s System

Let $X$ be a nonempty set. A binary relation $\geq$ on $X^{2}$ is called an
Alt’s system on $X$. A function $u:X\to\mathbb{R}$ is said to represent Alt’s
system $\geq$ (or to be a utility function of this system) if and only if the
following requirement holds: for all $(x,y,z,w)\in X^{4}$,

$u(x)-u(y)\geq u(z)-u(w)\Leftrightarrow(x,y,z,w)\in\geq.$ (1)

This relatiohship (1) says that the function $(x,y)\mapsto u(x)-u(y)$
represents $\geq$.

For an Alt’s system $\geq$, we define

$\succsim=\\{(x,y)\in X^{2}|(x,y,y,y)\in\geq\\}.$ (2)

If (1) holds for some function $u$, then

$u(x)\geq u(y)\Leftrightarrow(x,y)\in\succsim.$ (3)

This relationship (3) says that $u$ is a utility function that represents
$\succsim$.

The interpretation of Alt’s system is as follows. $(x,y,z,w)\in\geq$ means
that the strength of the improvement from $y$ to $x$ is not weaker than that
from $w$ to $z$. If $u$ represents this system, then $u(x)-u(y)$ measures this
strength, and thus $u$ is a sort of cardinal utility function. Clearly, if $u$
represents this system, then $(x,y)\in\succsim$ means that $x$ is preferred to
$y$.

As usual, we write $x\succsim y$ instead of $(x,y)\in\succsim$. Moreover, we
write $[x,y]\geq[z,w]$ instead of $(x,y,z,w)\in\geq$.222Alt originally used
the arrow notation $[y\to x]\geq[w\to z]$. On the other hand, recent related
papers (e.g. Wakker (1988), Miyake (2016), Gerasimou (2021)) usually use the
following simple notation $(x,y)\geq(z,w)$. However, the notation $(x,y)$ is
not desirable in this paper because we later treat elements of
$\mathbb{R}^{n}$, and thus this abstract pair $(x,y)$ may be mistaken for an
element of $\mathbb{R}^{2}$. Hence, we use $[x,y]$ instead of $(x,y)$. These
notations allow us to understand what is occurring. We write $x\succ y$ if
$x\succsim y$ and $y\not\succsim x$ and $x\sim y$ if $x\succsim y$ and
$y\succsim x$. Additionally, we write $[x,y]=[z,w]$ if $[x,y]\geq[z,w]$ and
$[z,w]\geq[x,y]$, and $[x,y]>[z,w]$ if $[x,y]\geq[z,w]$ and
$[z,w]\not\geq[x,y]$.

###  2.2 Axioms

We now present several axioms on Alt’s system. The first one is necessary for
the system to match its interpretation appropriately.

Definition 1. An Alt’s system $\geq$ is said to satisfy consistency if and
only if, for every $x,y,z\in X$,

$x\succsim y\Leftrightarrow[x,z]\geq[y,z].$

Note that, under this consistency axiom, if there exists $z\in X$ such that
$[x,z]\geq[y,z]$, then $x\succsim y$. Conversely, if $x\succsim y$, then for
every $z$, $[x,z]\geq[y,z]$. This gap (between ‘there exists’ and ‘for every’)
may be a source of confusion.

Under consistency, we can prove that $\succsim$ is a weak order on $X$. First,
choose any $x,y\in X$. If $x\not\succsim y$, then $[x,y]\not\geq[y,y]$, and
thus $[y,y]\geq[x,y]$. By consistency, we have that $y\succsim x$, which
implies that $\succsim$ is complete. Second, suppose that $x\succsim y$ and
$y\succsim z$. Then,

$[x,y]\geq[y,y],\ [y,z]\geq[z,z].$

Because of consistency, $[x,z]\geq[y,z]$, and thus $[x,z]\geq[z,z]$ by
transitivity of $\geq$, which implies that $x\succsim z$, and thus $\succsim$
is transitive. Therefore, if $\geq$ satisfies consistency, then $\succsim$ is
a weak order.

If there is a representation $u$ of $\geq$, then consistency implies that

$u(x)\geq u(y)\Leftrightarrow u(x)-u(z)\geq u(y)-u(z),$

and thus, consistency is trivially satisfied.

The next axiom relates to the property of $=$.

Definition 2. An Alt’s system $\geq$ is said to satisfy the crossover axiom if
and only if, for every $x,y,z,w\in X$,333The name ‘the crossover axiom’ was
used in Miyake (2016).

$[x,y]=[z,w]\Leftrightarrow[x,z]=[y,w].$

The relation $=$ is symmetric, and therefore the crossover axiom also means
that

$[x,y]=[z,w]\Leftrightarrow[w,y]=[z,x].$

If there is a representation $u$ of $\geq$, then the crossover axiom implies
that

$u(x)-u(y)=u(z)-u(w)\Leftrightarrow u(x)-u(z)=u(y)-u(w),$

and thus, this axiom is also trivially satisfied.

We note two facts. First, for every $x,y\in X$,

$[x,y]=[x,y].$

If the crossover axiom holds, then this implies that

$[x,x]=[y,y].$

That is, “the strength of the improvement under the unchanged situation is the
same at every point.” This is quite natural. Second, under consistency and the
crossover axiom,

$x\sim y\Leftrightarrow[x,z]=[y,z]\Leftrightarrow[x,y]=[z,z]$

for every $x,y,z\in X$. That is, “$x$ is indifferent to $y$ if and only if the
strength of the improvement from $y$ to $x$ is the same as that in the
unchanged situation.” This is also natural.

To treat the third axiom, we must consider $X$ as a topological space.

Definition 3. Suppose that $X$ is a topological space. Then, an Alt’s system
$\geq$ is said to satisfy continuity if and only if $\geq$ is closed in
$X^{4}$.

Note that, if $\geq$ satisfies consistency and continuity, then $\succsim$ is
also closed in $X^{2}$ by (2).

If there is a continuous representation $u$ of $\geq$, then clearly $\geq$ is
closed, because

$\geq=\\{(x,y,z,w)\in X^{4}|u(x)+u(w)-u(y)-u(z)\geq 0\\}.$

###  2.3 Result for the Existence of Alt’s Cardinal Utility

The next theorem is our first result.

Theorem 1. Suppose that $X$ is a Hausdorff, separable, and path-connected
topological space, and $\geq$ is an Alt’s system on $X$.444A topological space
$X$ is said to be path-connected if and only if for every $x,y\in X$, there
exists a continuous function $f:[0,1]\to X$ such that $f(0)=x,\ f(1)=y$. For
example, every convex set of some topological vector space is path-connected.
It is well known that every path-connected space is connected. Note that, we
do not know whether Theorem 1 still holds on a connected topological space
that is not path-connected. See section 3 and footnote 16 for more detailed
arguments. Then, there exists a continuous representation $u:X\to\mathbb{R}$
of $\geq$ if and only if this system satisfies consistency, the crossover
axiom, and continuity. If so, such a representation is unique up to a positive
affine transform: that is, if $u_{1},u_{2}$ are continuous representations of
$\geq$, then there exist $a>0$ and $b\in\mathbb{R}$ such that

$u_{2}(x)=au_{1}(x)+b$

for every $x\in X$.

The proof of this theorem is in the appendix.

###  2.4 Generalized Gossen’s First Law

Our main purpose is to present a necessary and sufficient condition for $u$ to
be concave. Hence, we introduce the following axiom.

Definition 4. Suppose that $X$ is a separable and convex subset of a Hausdorff
topological vector space. Then, an Alt’s system $\geq$ on $X$ is said to
satisfy generalized Gossen’s first law if and only if, for every $x,y\in X$,

$[z,x]\geq[y,z]$

for $z=\frac{1}{2}(x+y)$. If, in addition,

$[z,x]>[y,z]$

whenever $x\neq y$, then $\geq$ is said to satisfy generalized strong Gossen’s
first law.

We should interpret the meaning of generalized Gossen’s first law. Suppose
that $X$ is a convex subset of $\mathbb{R}^{n}$, and $x\in X$ and
$y=(x_{1}+2a,x_{2},...,x_{n})\in X$ for $a>0$. Then,
$z=(x_{1}+a,x_{2},...,x_{n})\in X$. Suppose that generalized Gossen’s first
law is satisfied. Then, $[z,x]\geq[y,z]$. If $u$ is a continuously
differentiable representation of this system, then by the mean value theorem,

$u(z)-u(x)=a\frac{\partial u}{\partial x_{1}}(z-\theta ae_{1}),$
$u(y)-u(z)=a\frac{\partial u}{\partial x_{1}}(z+\eta ae_{1})$

for some $\theta,\eta\in[0,1]$, where $e_{1}=(1,0,...,0)$. Thus, roughly
speaking, $[z,x]\geq[y,z]$ implies that the mapping

$b\mapsto\frac{\partial u}{\partial x_{1}}(x_{1}+b,x_{2},...,x_{n})$

is nonincreasing around $a$. Because $x$ and $a$ are arbitrary, we conclude
that under generalized Gossen’s first law, the marginal utility of $x_{1}$ is
nonincreasing in $x_{1}$. This explains why this axiom is named as such.

In the definition of generalized Gossen’s first law, $z=x+(z-x)$ and
$y=z+(z-x)$. Therefore, this axiom can be transformed as follows: for every
$v\neq 0$ such that $x+2v\in X$,

$[x+v,x]\geq[x+2v,x+v].$

If $u$ represents this system, then

$u(x+v)-u(x)\geq u(x+2v)-u(x+v),$

and thus, under this axiom, the marginal improvement of the utility for a
change of consumption in the $v$ direction decreases as the length of the
change increases.

###  2.5 Result for the Concavity of Alt’s Cardinal Utility

The following is our second main result.

Theorem 2. Suppose that $X$ is a separable and convex subset of a Hausdorff
topological vector space, $\geq$ is an Alt’s system on $X$, and
$u:X\to\mathbb{R}$ is a continuous function that represents this system. Then,
$u$ is concave (resp. strictly concave) if and only if $\geq$ satisfies
generalized Gossen’s first law (resp. generalized strong Gossen’s first law).

Note that, because $X$ is convex, it is path-connected, and thus all
assumptions in Theorem 1 are satisfied.

Proof. First, suppose that $u$ is concave. Choose any $x,y\in X$, and define
$z=\frac{1}{2}(x+y)$. Then,

$\frac{1}{2}u(z)+\frac{1}{2}u(z)=u(z)\geq\frac{1}{2}u(x)+\frac{1}{2}u(y),$ (4)

and thus

$u(z)-u(x)\geq u(y)-u(z),$

as desired. If, in addition, $u$ is strictly concave and $y\neq x$, then the
inequality in (4) is strengthened, and thus

$u(z)-u(x)>u(y)-u(z),$

as desired.

Conversely, suppose that $\geq$ satisfies generalized Gossen’s first law. We
want to show that

$u((1-t)x+ty)\geq(1-t)u(x)+tu(y)$ (5)

for every $x,y\in X$ and $t\in[0,1]$. Choose any $x,y\in X$, and define
$z=\frac{1}{2}(x+y)$. By generalized Gossen’s first law,

$[z,x]\geq[y,z],$

and thus,

$u(z)-u(x)\geq u(y)-u(z),$

which implies that

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

Therefore, (5) holds if $t=\frac{1}{2}$. By mathematical induction, we can
easily show that (5) holds if $t$ is a dyadic rational $\frac{m}{2^{\ell}}$,
where $\ell\geq 1$ and $1\leq m\leq 2^{\ell}-1$. Because the set of all dyadic
rationals is dense in $[0,1]$, by the continuity of $u$, (5) holds for all
$t\in[0,1]$.

Suppose that $\geq$ satisfies generalized strong Gossen’s first law. Choose
any $x,y\in X$ such that $x\neq y$. Define $z=\frac{1}{2}(x+y)$. Then,

$[z,x]>[y,z],$

and thus,

$u(z)-u(x)>u(y)-u(z),$

which implies that

$u(z)>\frac{1}{2}u(x)+\frac{1}{2}u(y).$

Therefore,

$u((1-t)x+ty)>(1-t)u(x)+tu(y)$ (6)

if $t=\frac{1}{2}$. By mathematical induction, we can easily show that (6)
holds if $t$ is a dyadic rational $\frac{m}{2^{\ell}}$, where $\ell\geq 1$ and
$1\leq m\leq 2^{\ell}-1$. Now, choose any $t\in]0,1[$ such that $t$ is not a
dyadic rational. Then, if $\ell$ is sufficiently large, there exists
$m\in\\{1,...,2^{\ell}-2\\}$ such that
$t_{0}=\frac{m}{2^{\ell}}<t<\frac{m+1}{2^{\ell}}=t_{1}$. In this case,

$t=\frac{t_{1}-t}{t_{1}-t_{0}}t_{0}+\frac{t-t_{0}}{t_{1}-t_{0}}t_{1},$

and thus, by the continuity of $u$,

$\displaystyle u((1-t)x+ty)\geq$
$\displaystyle~{}\frac{t_{1}-t}{t_{1}-t_{0}}u((1-t_{0})x+t_{0}y)+\frac{t-t_{0}}{t_{1}-t_{0}}u((1-t_{1})x+t_{1}y)$
$\displaystyle>$
$\displaystyle~{}\frac{t_{1}-t}{t_{1}-t_{0}}[(1-t_{0})u(x)+t_{0}u(y)]$
$\displaystyle~{}+\frac{t-t_{0}}{t_{1}-t_{0}}[(1-t_{1})u(x)+t_{1}u(y)]$
$\displaystyle=$ $\displaystyle~{}(1-t)u(x)+tu(y),$

as desired. This completes the proof. $\blacksquare$

###  2.6 Result for the Differentiability of Alt’s Cardinal Utility

Usually, Gossen’s laws are stated in the language of differential calculus.
Hence, we provide a sufficient condition for Alt’s utility to be
differentiable.

First, let $x,y\in\mathbb{R}^{n}$. We write $x\geq y$ if and only if
$x_{i}\geq y_{i}$ for all $i\in\\{1,...,n\\}$, and $x\gg y$ if and only if
$x_{i}>y_{i}$ for all $i\in\\{1,...,n\\}$. Define
$\mathbb{R}^{n}_{+}=\\{x\in\mathbb{R}^{n}|x\geq 0\\}$ and
$\mathbb{R}^{n}_{++}=\\{x\in\mathbb{R}^{n}|x\gg 0\\}$. We introduce three
axioms.

Definition 5. Let $X\subset\mathbb{R}^{n}$ and $\geq$ be an Alt’s system on
$X$. Then, this system is said to be monotone if $x\succ y$ for every $x,y\in
X$ such that $x\gg y$.

Definition 6. Let $X$ be either $\mathbb{R}^{n}_{+}$ or $\mathbb{R}^{n}_{++}$,
and $\geq$ be an Alt’s system on $X$. Then, this system is said to satisfy
Debreu’s smoothness if and only if the following set

$I=\\{(x,y)\in\mathbb{R}^{2n}_{++}|x\sim y\\}$

is a $2n-1$ dimensional $C^{1}$ manifold.

Definition 7. Let $X$ be either $\mathbb{R}^{n}_{+}$ or $\mathbb{R}^{n}_{++}$,
and $\geq$ be an Alt’s system on $X$ that satisfies consistency, the crossover
axiom, continuity, and monotonicity. Define $e=(1,1,...,1)\in X$. By Theorem 1
and the intermediate value theorem, for every $a,b>0$ with $a<b$, there
uniquely exists $f(a,b)>0$ such that555The existence of such an $f$ can also
be shown using Lemma 4 in the appendix.

$[f(a,b)e,(b-a)e]=[(b+a)e,f(a,b)e].$

We say that this system satisfies line smoothness if and only if

$\lim_{a\downarrow 0}\frac{b-f(a,b)}{a}=0$

for all $b>0$.

Theorem 3. Let $X$ be either $\mathbb{R}^{n}_{+}$ or $\mathbb{R}^{n}_{++}$,
and $\geq$ be an Alt’s system on $X$ that satisfies consistency, the crossover
axiom, continuity, generalized Gossen’s first law, and monotonicity. Suppose
that $u$ is a continuous representation of $\geq$. Then, $u$ is continuously
differentiable on $\mathbb{R}^{n}_{++}$ and $Du(x)\neq 0$ for every
$x\in\mathbb{R}^{n}_{++}$ if and only if $\geq$ satisfies both Debreu’s
smoothness and line smoothness.

The proof of this theorem is in the appendix. Note that Debreu’s smoothness
does not imply line smoothness. For example, let $v(x)=\sqrt{x_{1}x_{2}}$ and

$g(c)=\begin{cases}c-1&\mbox{if }c\leq 1,\\\ \frac{c-1}{2}&\mbox{if
}c>1.\end{cases}$

Define $u(x)=g(v(x))$, and suppose that $\geq$ is an Alt’s system that
corresponds to $u$. Then, this system satisfies Debreu’s smoothness, because
$I=\\{(x,y)\in\mathbb{R}^{2n}_{++}|v(x)=v(y)\\}$. However,
$f(a,1)=1-\frac{1}{4}a$, and thus,

$\lim_{a\downarrow 0}\frac{1-f(a,1)}{a}=\frac{1}{4}\neq 0,$

which implies that this system violates line smoothness.

On the other hand, every function that is increasing and homogeneous of degree
one satisfies line smoothness, although its indifference curve is kinked.
Therefore, line smoothness does not imply Debreu’s smoothness. That is,
Debreu’s smoothness is independent of line smoothness.

We should mention Alexandrov’s theorem. This theorem states that for every
convex function, its subdifferential mapping is ‘differentiable’ in the sense
of set-valued differentiation almost everywhere.666See Alexandrov (1939),
section 6 of Howard (1998), or theorem 3.12.3 of Niculescu and Persson (2018)
for detailed arguments. Applying this theorem, we can obtain the following
result: if $u$ is a concave and continuously differentiable utility function
derived from Theorem 3, then it is twice differentiable almost everywhere.
Thus, ALEP substitution and complementarity can be defined almost everywhere.

Recall the definition of ALEP substitution and complementarity. This notion
uses second cross-derivatives of the utility function. If
$\frac{\partial^{2}u}{\partial x_{i}\partial x_{j}}<0$ (resp.
$\frac{\partial^{2}u}{\partial x_{i}\partial x_{j}}>0$), then commodity $i$ is
a substitute (resp. complementary) good of commodity $j$. In other words,
commodity $i$ is a substitute good of commodity $j$ if and only if an increase
in consumption of commodity $i$ undermines the attractiveness of commodity
$j$. This notion was argued by von Auspitz and Lieben (1889), Edgeworth
(1897), and Pareto (1906), and the name ‘ALEP’ comes from the first letters of
these researchers’ surnames.

Clearly, this concept only works if the difference in the value of $u$ implies
the intensity of the improvement. Therefore, Alt’s utility theory is very
useful for this concept. Samuelson (1974) discussed in detail the definition
of ALEP complementarity using cardinal utility, and argued the relationship
between this notion and Alt’s utility. This discussion is also explained in
Kawamata (2010). The fact mentioned in the paragraph above means that by using
our Theorem 3, we can discuss the substitution-complementarity relationship in
the sense of ALEP to Alt’s utility. In this sense, Theorem 3 can be considered
as adding more value to Alt’s utility.

##  3 Remarks

To the best of our knowledge, Alt’s system was first argued by Alt (1936), who
proved that under several axioms (including our consistency and continuity,
and moreover, concatenation axiom discussed later), there exists a continuous
representation. He wrote his paper in German, and it was translated into
English in 1971 (Chipman et al. (1971)). Actually, his axioms include several
‘facts’ that can be shown from our three axioms. For example, our Lemma 5 is
one of Alt’s axioms. Similar axioms can be found in Theorem 4.2 of Kranz et
al. (1971), where our Lemmas 2 and 5 are treated as axioms. Shapley (1975)
used only our three axioms and derived a continuous representation.777Shapley
defined Alt’s system as a pair $(\succsim,\geq)$ of weak orders on $X$ and
$X^{2}$. However, using our definition of $\succsim$ (see (2)), there is no
difference between Shapley’s setup and ours. In his proof, however, $X$ is
treated as a convex subset of the real line. Hence, our Theorem 1 is an
extension of his result to a general path-connected topological space.

Theorem 5.3 of Wakker (1988) treats similar axioms to this paper and derives a
similar result to our Theorem 1 on a connected topological space. His axioms
are continuity and the following two axioms: 1) (reversal condition) If
$[x,y]\geq[z,w]$, then $[w,y]\geq[z,x]$. 2) (concatenation axiom)
$[x,y]\geq[x^{\prime},y^{\prime}]$ and $[y,z]\geq[y^{\prime},z^{\prime}]$
imply $[x,z]\geq[x^{\prime},z^{\prime}]$. First, reversal condition is
stronger than the crossover axiom, because the crossover axiom only states
that reversal condition holds for not $\geq$ but only $=$. Second, under
reversal condition and continuity, we can easily derive consistency.
Therefore, our axioms are weaker than Wakker’s axioms. Instead, we need not
only connectedness, but also path-connectedness of the space, and thus we
think that our Theorem 1 and Wakker’s result is independent.888We guess that,
under our three axioms, Lemma 7 is essentially needed to show that
concatenation axiom holds. Later, we will see that, in proving Theorem 1, only
Lemma 7 requires the path-connectedness. Therefore, we think that the reason
why Wakker’s result does not need path-connectedness is assuming concatenation
axiom directly.

Miyake (2016) considered an Alt’s system on $\mathbb{R}^{n}_{++}$, and showed
that, under several axioms, there exists a representation $\log u(x)$ such
that $u$ is homogeneous of positive degree. Surprisingly, his result does not
use the crossover axiom. Therefore, his result is independent of our Theorem
1.

Gerasimou (2021) showed that, under several axioms, our $\geq$ is represented
by a general preference intensity function $g(x,y)$. His axioms are weaker
than our axioms in Theorem 1. Instead, under our axioms, we can take
$g(x,y)=u(x)-u(y)$, which is stronger than Gerasimou’s result. Therefore, our
Theorem 1 is independent of Gerasimou’s result.

We should mention our assumption of path-connectedness. A topological space
$X$ is said to be connected if and only if, for every pair of nonempty closed
sets $A,B$ in $X$, either $A\cup B\neq X$ or $A\cap B\neq\emptyset$. Debreu
(1954) showed that if $X$ is a Hausdorff, separable, and connected topological
space, and a weak order $\succsim$ on $X$ is closed in $X^{2}$, then there
exists a continuous function $u:X\to\mathbb{R}$ such that

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

Compared with this result, our Theorem 1 uses path-connectedness instead of
connectedness. It is known that every path-connected topological space is
connected. On the other hand, it is also known that the following set

$X=\\{(x^{-1},\sin x)|x\neq 0\\}\cup\\{(0,y)|-1\leq y\leq 1\\}$

is closed and connected, but not path-connected. Hence, path-connectedness is
stronger than connectedness. In the appendix, path-connectedness is only used
in the proof of Lemma 7, and in the remaining parts of this proof, only
connectedness is used. Therefore, if Lemma 7 can be proved without the use of
path-connectedness, we could extend Theorem 1 for a general connected space.
However, at least we are not currently able to prove Lemma 7 without the aid
of path-connectedness.

Debreu (1976) treated the notion of the least concave utility function.
Suppose that $X$ is a separable and convex set of a Hausdorff topological
vector space, and $\succsim$ is a weak order of $X$ that is closed in $X^{2}$.
Let $\mathscr{U}$ be the set of all continuous and concave functions $u$ such
that

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

If $u,v\in\mathscr{U}$, then we write $u\succeq v$ if and only if there exists
a concave function $\varphi$ such that $u=\varphi\circ v$. Debreu showed that
if $\mathscr{U}$ is nonempty, then there exists a least element in
$\mathscr{U}$ with respect to this partial order $\succeq$. This least element
is called the least concave utility function. He showed that the least concave
utility function is unique up to a positive affine transform, and for fixed
$x\in X$, if there exists a continuously differentiable element
$u\in\mathscr{U}$ at $x$, then the least concave utility function is also
continuously differentiable at $x$. The proof is very similar to Benveniste-
Scheinkman’s envelope theorem. In this paper, we characterized the existence
of such a continuously differentiable $u\in\mathscr{U}$ in Theorem 3.

##  Appendix A Proofs

###  A.1 Proof of a Lemma

In this subsection, we introduce a property of an Alt’s system $\geq$ on $X$.
This system is said to satisfy second consistency if and only if for every
$x,y,z\in X$,

$x\succsim y\Leftrightarrow[z,y]\geq[z,x].$

We show the following lemma.

Lemma 1. Suppose that $X$ is a Hausdorff and connected topological space and
$\geq$ is an Alt’s system on $X$ that satisfies consistency, the crossover
axiom, and continuity. Then, this system satisfies second consistency.

Proof. We first show the following fact. Suppose that

$[x_{2},y]\geq[z,w]\geq[x_{1},y].$

Then, there exists $x_{3}\in X$ such that

$[x_{3},y]=[z,w].$

To show this, let

$U=\\{x\in X|[x,y]\geq[z,w]\\},\ D=\\{x\in X|[z,w]\geq[x,y]\\}.$

Because $x_{2}\in U$ and $x_{1}\in D$, both $U$ and $D$ are nonempty. By
continuity, both $U$ and $D$ are closed. Since $U\cup D=X$, by the
connectedness of $X$, there exists $x_{3}\in U\cap D$. Then, clearly

$[x_{3},y]=[z,w].$

Note that,

$[x_{2},y]\geq[x_{3},y]\geq[x_{1},y],$

and thus, by consistency, $x_{2}\succsim x_{3}\succsim x_{1}$.

Second, suppose that $x\sim y$. Then, by consistency,

$[x,z]=[y,z].$

Therefore, by the crossover axiom,

$[x,y]=[z,z],$

and again by the crossover axiom,

$[z,y]=[z,x].$

Hence, second consistency holds for every $x,y,z\in X$ such that $x\sim y$.

Now, suppose that second consistency is violated. Then, by the above
arguments, there exists $x,y,z\in X$ such that

$y\succ x,\ [z,y]\geq[z,x].$ (7)

If $y\succ z$, then by consistency, the crossover axiom, and (7),

$[x,x]=[y,y]>[z,y]\geq[z,x].$

Therefore, there exists $w\in X$ such that

$[w,x]=[z,y],\ x\succsim w\succsim z.$

By the crossover axiom,

$[w,z]=[x,y].$

By consistency and the crossover axiom,

$[z,z]=[y,y]>[x,y]=[w,z]\geq[z,z],$

which is a contradiction.

Therefore, $z\succsim y$. By the transitivity of $\succsim$, we have that
$z\succ x$. Thus, by (7), consistency, and the crossover axiom,

$[z,y]\geq[z,x]>[x,x]=[y,y].$

Thus, there exists $w\in X$ such that

$[w,y]=[z,x],\ z\succsim w\succsim y.$

By the crossover axiom,

$[w,z]=[y,x].$

By the crossover axiom and consistency,

$[z,z]\geq[w,z]=[y,x]>[x,x]=[z,z],$

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

###  A.2 Proof of Theorem 1

Suppose that $X$ is a Hausdorff, separable, and path-connected topological
space and $\geq$ is an Alt’s system on $X$. If there exists a continuous
function $u:X\to\mathbb{R}$ that represents $\geq$, then clearly $\geq$
satisfies consistency, the crossover axiom, and continuity. Therefore, it
suffices to show the opposite direction, in other words, that if $\geq$
satisfies consistency, the crossover axiom, and continuity, then there exists
a continuous function $u:X\to\mathbb{R}$ that represents $\geq$.

First, we introduce several lemmas.

Lemma 2. If

$[x_{2},y]\geq[z,w]\geq[x_{1},y],$

then there exists $x_{3}\in X$ such that

$[x_{3},y]=[z,w].$ (8)

Moreover, $x_{2}\succsim x_{3}\succsim x_{1}$, and if $x_{4}$ also satisfies
(8), then $x_{3}\sim x_{4}$.

Proof. We have already proved this result in the proof of Lemma 1, except for
the last claim. Hence, suppose that

$[x_{4},y]=[z,w].$

By the transitivity of $=$,

$[x_{3},y]=[x_{4},y]$

and by consistency,

$x_{3}\sim x_{4},$

as desired. This completes the proof. $\blacksquare$

Lemma 3. Suppose that

$[y,x_{1}]\geq[z,w]\geq[y,x_{2}].$

Then, there exists $x_{3}\in X$ such that

$[y,x_{3}]=[z,w].$ (9)

Moreover, $x_{2}\succsim x_{3}\succsim x_{1}$, and if $x_{4}$ also satisfies
(9), then $x_{3}\sim x_{4}$.

Proof. By using second consistency instead of consistency, we can show this
result using almost the same arguments as in the proof of Lemma 2. Hence, we
omit the proof. $\blacksquare$

Lemma 4. Suppose that $z\succ x$. Then, there exists $y\in X$ such that

$[y,x]=[z,y].$ (10)

Moreover, $z\succ y\succ x$, and if $y^{\prime}$ also satisfies (10), then
$y\sim y^{\prime}$.

Proof. Define

$U=\\{w\in X|[w,x]\geq[z,w]\\},\ D=\\{w\in X|[z,w]\geq[w,x]\\}.$

Then, both $U$ and $D$ are closed. Because $[z,x]>[x,x]=[z,z]$, $z\in U$ and
$x\in D$, and thus both $U$ and $D$ are nonempty. Because $U\cup D=X$, by the
connectedness of $X$, there exists $y\in U\cap D$. Clearly,

$[y,x]=[z,y].$

If $y\succsim z$, then $y\succ x$, and thus

$[y,x]\geq[z,x]>[z,y]=[y,x],$

which is absurd. Therefore, $z\succ y$. Symmetrically, we can show that
$y\succ x$.

Choose any $y^{\prime}\in X$. If $y\succ y^{\prime}$, then by consistency and
second consistency,

$[z,y^{\prime}]>[z,y]=[y,x]>[y^{\prime},x].$

If $y^{\prime}\succ y$, then again by consistency and second consistency,

$[y^{\prime},x]>[y,x]=[z,y]>[z,y^{\prime}].$

Therefore, if $[z,y^{\prime}]=[y^{\prime},x]$, then we must have $y\sim
y^{\prime}$. This completes the proof. $\blacksquare$

Lemma 5.999Note that in the proof of Lemmas 1-4, we only used the
connectedness of $X$; the separability and path-connectedness of $X$ were not
used. However, in the proof of Lemma 5, we must use Debreu’s representation
theorem, and so $X$ must be separable. Suppose that $x\succ y$. If $z\succsim
x$, then there exists a finite sequence $a_{0},a_{1},...,a_{k}\in X$ such
that101010If there is a representation $u:X\to\mathbb{R}$, then this statement
says that $(k-1)(u(x)-u(y))\leq u(z)-u(x)<k(u(x)-u(y))$ for some
$k\in\mathbb{N}$.

$a_{0}=y,\ a_{1}=x,\ [a_{i+1},a_{i}]=[a_{i+2},a_{i+1}]\mbox{ for all }i,$
$z\succsim a_{k},\ [a_{1},a_{0}]>[z,a_{k}].$

Similarly, if $y\succsim w$, then there exists a finite sequence
$a_{1},a_{0},...,a_{-k}$ such that

$a_{0}=y,\ a_{1}=x,\ [a_{i+1},a_{i}]=[a_{i+2},a_{i+1}]\mbox{ for all }i,$
$a_{-k}\succsim w,\ [a_{1},a_{0}]>[a_{-k},w].$

Proof. We treat only the claim on $z$, because the claim on $w$ can be proved
symmetrically. Suppose that such a finite sequence does not exist. Define
$a_{0}=y$ and $a_{1}=x$. By the negation of the claim of this lemma for $k=1$,
we must have

$[z,a_{1}]\geq[a_{1},a_{0}]>[a_{1},a_{1}].$

By Lemma 2, there exists $a_{2}\in X$ such that

$[a_{2},a_{1}]=[a_{1},a_{0}],\ z\succsim a_{2}\succ a_{1}.$

We assume that $a_{0},...,a_{k}$ is already defined, and

$[a_{1},a_{0}]=[a_{i+1},a_{i}]$

for every $i\in\\{1,...,k-1\\}$. By the negation of the claim of this lemma,
we must have

$[z,a_{k}]\geq[a_{1},a_{0}]>[a_{k},a_{k}].$

Therefore, by Lemma 2, there exists $a_{k+1}\in X$ such that

$[a_{k+1},a_{k}]=[a_{1},a_{0}],\ z\succsim a_{k+1}\succ a_{k}.$

Hence, by mathematical induction, we obtain an infinite sequence $(a_{k})$
such that $z\succsim a_{k}$ and

$[a_{1},a_{0}]=[a_{k+1},a_{k}]$

for every $k\in\mathbb{N}$.

We will show that

$[a_{i+k},a_{i}]=[a_{j+k},a_{j}]$ (11)

for every $i,j,k$. If $k=1$, then it has already been proved. By the crossover
axiom,

$[a_{j},a_{i}]=[a_{j+1},a_{i+1}].$

By the transitivity of $=$, for every $k$,

$[a_{j},a_{i}]=[a_{j+k},a_{i+k}].$

Again by the crossover axiom,

$[a_{i+k},a_{i}]=[a_{j+k},a_{j}],$

and thus (11) holds.

Now, by Debreu’s representation theorem,111111This theorem states that if $X$
is a Hausdorff, separable, and connected topological space and $\succsim$ is a
weak order on $X$ that is closed in $X^{2}$, then there exists a continuous
function $v:X\to\mathbb{R}$ such that  $v(x)\geq v(y)\Leftrightarrow x\succsim
y$ for every $x,y\in X$. For the proof of this theorem, see, for example,
Debreu (1954) or Bridges and Mehta (1995). there exists a continuous function
$v:X\to\mathbb{R}$ such that

$v(x^{\prime})\geq v(y^{\prime})\Leftrightarrow x^{\prime}\succsim y^{\prime}$

for every $x^{\prime},y^{\prime}\in X$. Let $c^{*}=\sup_{k}v(a_{k})$. Then,
$v(z)\geq c^{*}>v(a_{0})$, and thus $c^{*}\in\mathbb{R}$. Because $v$ is
continuous and $X$ is connected, $v(X)$ is also a connected set in
$\mathbb{R}$. In $\mathbb{R}$, connectedness is equivalent to convexity.
Therefore, $v(X)$ is convex, and thus there exists $z^{*}\in X$ such that
$v(z^{*})=c^{*}$. By Lemma 4, there exists $w^{*}\in X$ such that
$[w^{*},a_{0}]=[z^{*},w^{*}]$ and $z^{*}\succ w^{*}\succ a_{0}$. Because
$c^{*}=v(z^{*})>v(w^{*})>v(a_{0})$, there exists $k$ such that

$v(a_{k+1})>v(w^{*})\geq v(a_{k}).$

Then, by (11),

$[a_{2k+2},a_{k+1}]=[a_{k+1},a_{0}]>[w^{*},a_{0}]=[z^{*},w^{*}]>[z^{*},a_{k+1}],$

and thus, by consistency,

$a_{2k+2}\succ z^{*}.$

This implies that $v(a_{2k+2})>c^{*}$, which contradicts the definition of
$c^{*}$. This completes the proof. $\blacksquare$

Now, suppose that $x\sim y$ for every $x,y\in X$. Then, $[x,y]=[z,y]=[z,w]$
for every $x,y,z,w\in X$, and thus $u:X\to\mathbb{R}$ represents this system
if and only if $u$ is a constant function. Hence, we hereafter assume that
there exists $x^{*},y^{*}\in X$ such that $x^{*}\succ y^{*}$.

We will recursively construct (possibly finite or infinite) sequences
$(a_{i}^{k})$ for $k\in\mathbb{N}$.121212Later, we construct our utility
function $u$ to satisfy $u(a_{i}^{k})=\frac{i}{2^{k}}$. First, we define
$a_{0}^{0}=y^{*}$ and $a_{1}^{0}=x^{*}$. Suppose that $a_{i}^{0}$ is defined
for some $i\geq 1$. If there exists $x\in X$ such that
$[x,a_{i}^{0}]\geq[a_{1}^{0},a_{0}^{0}]$, then by Lemma 2, there exists
$a_{i+1}^{0}\in X$ such that $[a_{i+1}^{0},a_{i}^{0}]=[a_{1}^{0},a_{0}^{0}]$.
If there is no such $x$, then we do not define $a_{i+1}^{0}$. Similarly,
suppose that $a_{i}^{0}$ is defined for some $i\leq 0$. If there exists $x\in
X$ such that $[a_{i}^{0},x]\geq[a_{1}^{0},a_{0}^{0}]$, then by Lemma 3, there
exists $a_{i-1}^{0}\in X$ such that
$[a_{i}^{0},a_{i-1}^{0}]=[a_{1}^{0},a_{0}^{0}]$. If there is no such $x$, then
we do not define $a_{i-1}^{0}$.

Next, suppose that $(a_{i}^{k})$ is defined for $k=0,...,k^{*}$ and the
following two relationships hold: 1) $a_{i}^{k}=a_{2i}^{k+1}$ and 2)
$[a_{i+1}^{k},a_{i}^{k}]=[a_{1}^{k},a_{0}^{k}]$. We define a sequence
$(a_{i}^{k^{*}+1})$ as follows. First, if $a_{i}^{k^{*}}$ is defined, then we
define $a_{2i}^{k^{*}+1}=a_{i}^{k^{*}}$. Second, if both
$a_{i}^{k^{*}},a_{i+1}^{k^{*}}$ are defined, then by Lemma 4, there exists
$a_{2i+1}^{k^{*}+1}\in X$ such that
$[a_{2i+1}^{k^{*}+1},a_{2i}^{k^{*}+1}]=[a_{2i+2}^{k^{*}+1},a_{2i+1}^{k^{*}+1}]$.
Third, suppose that $a_{i}^{k^{*}}$ is defined and $a_{i+1}^{k^{*}}$ is
undefined. Note that, in this case we have $i\geq 2^{k^{*}}$. If there exists
$x\in X$ such that $[x,a_{i}^{k^{*}}]\geq[a_{1}^{k^{*}+1},a_{0}^{k^{*}+1}]$,
then by Lemma 2, there exists $a_{2i+1}^{k^{*}+1}\in X$ such that
$[a_{2i+1}^{k^{*}+1},a_{2i}^{k^{*}+1}]=[a_{1}^{k^{*}+1},a_{0}^{k^{*}+1}]$. If
such an $x$ does not exist, then we do not define
$a_{2i+1}^{k^{*}+1}$.131313Note that, if $a_{2i+1}^{k^{*}+1}$ is defined and
$a_{i+1}^{k^{*}}$ is undefined, then there is no $x\in X$ such that
$[x,a_{2i+1}^{k^{*}+1}]\geq[a_{1}^{k^{*}+1},a_{0}^{k^{*}+1}]$. Suppose that
such an $x$ exists. Then, we can define $a_{2i+2}^{k^{*}+1}$. By the same
arguments as in the proof of Lemma 5, we can show that (11) holds for
$(a_{i}^{k^{*}+1})$, and thus,
$[a_{1}^{k^{*}},a_{0}^{k^{*}}]=[a_{2i+2}^{k^{*}+1},a_{i}^{k^{*}}],$ which
contradicts the undefinedness of $a_{i+1}^{k^{*}}$. Similarly, suppose that
$a_{i}^{k^{*}}$ is defined and $a_{i-1}^{k^{*}}$ is undefined. Then, we have
that $i\leq 0$. If there exists $x\in X$ such that
$[a_{i}^{k^{*}},x]\geq[a_{1}^{k^{*}+1},a_{0}^{k^{*}+1}]$, then by Lemma 3,
there exists $a_{2i-1}^{k^{*}+1}\in X$ such that
$[a_{2i}^{k^{*}+1},a_{2i-1}^{k^{*}+1}]=[a_{1}^{k^{*}+1},a_{0}^{k^{*}+1}]$. If
such an $x$ does not exist, then we do not define $a_{2i-1}^{k^{*}+1}$.

By the above arguments, we can define $(a_{i}^{k})$ recursively, and these
sequences satisfy $a_{i}^{k}=a_{2i}^{k+1}$ and
$[a_{i+1}^{k},a_{i}^{k}]=[a_{1}^{k},a_{0}^{k}]$. Let $D\subset X$ be the set
of all $a_{i}^{k}$. We need the following lemma.

Lemma 6. If $x\succ y$, then there exists $z\in D$ such that $x\succ z\succ
y$.141414If there is a representation $u:X\to\mathbb{R}$ such that
$u(x^{*})=1$ and $u(y^{*})=0$, then this statement says that there exists
$k\geq 1$ and $i\in\mathbb{Z}$ such that $u(x)>\frac{i}{2^{k}}>u(y)$.

Proof. Suppose that $x\succ y$ and there is no $z\in D$ such that $x\succ
z\succ y$. We treat the case in which $x\succ a_{0}^{0}$ because the remaining
case can be treated symmetrically.

Because of our assumption, we must have $y\succsim a_{0}^{0}$. First, we
assume that there exists $z\in D$ such that $y\succsim z\succ a_{0}^{0}$. By
Lemma 5, there exists a finite sequence $b_{1},b_{0},...,b_{-k}$ such that
$b_{1}=x$, $b_{0}=y$, $[b_{i+1},b_{i}]=[b_{1},b_{0}]$ for all
$i\in\\{-k,...,-1\\}$, $b_{-k}\succsim a_{0}^{0}$ and
$[b_{1},b_{0}]>[b_{-k},a_{0}^{0}]$. On the other hand, again by Lemma 5, for
every $m\geq 0$, there exists $\ell\geq 0$ such that $y\succsim a_{\ell}^{m}$
and $[a_{1}^{m},a_{0}^{m}]>[y,a_{\ell}^{m}]$. If there does not exist
$a_{\ell+1}^{m}$, then by our definition of $(a_{i}^{m})$,
$[a_{1}^{m},a_{0}^{m}]>[x,a_{\ell}^{m}]$. If $a_{\ell+1}^{m}$ exists, then
$a_{\ell+1}^{m}\succ y$. Because $a_{\ell+1}^{m}\in D$ and there is no $z\in
D$ such that $x\succ z\succ y$, we have that $a_{\ell+1}^{m}\succsim x$, and
thus

$[a_{1}^{m},a_{0}^{m}]=[a_{\ell+1}^{m},a_{\ell}^{m}]\geq[x,a_{\ell}^{m}].$

Therefore, in any case, we have that
$[a_{1}^{m},a_{0}^{m}]\geq[x,a_{\ell}^{m}]$.

Because there exists $z\in D$ such that $y\succsim z\succ a_{0}^{0}$,
$\ell\geq 1$ for some $m$. Because $a_{\ell}^{m}=a_{2\ell}^{m+1}$, we can
choose $m$ so large that $\ell>k$. Define $c_{0}=a_{\ell}^{m}$. Then, clearly
$b_{0}\succsim c_{0}\succsim a_{\ell}^{m}$. Note that,

$[a_{1}^{m},a_{0}^{m}]\geq[x,a_{\ell}^{m}]\geq[x,y]=[b_{1},b_{0}].$

On the other hand,

$[b_{0},a_{\ell-1}^{m}]\geq[a_{\ell}^{m},a_{\ell-1}^{m}]=[a_{1}^{m},a_{0}^{m}]\geq[b_{0},b_{0}],$

and thus, by Lemma 3, there exists $c_{1}\in X$ such that

$[b_{0},c_{1}]=[a_{1}^{m},a_{0}^{m}].$

If $k\geq 1$, then

$[b_{0},c_{1}]\geq[b_{1},b_{0}]=[b_{0},b_{-1}],$

and thus, $b_{-1}\succsim c_{1}\succsim a_{\ell-1}^{m}$. Inductively, we can
show that there exists $c_{k}\in X$ such that $b_{-k}\succsim c_{k}\succsim
a_{\ell-k}^{m}$. Then,

$[b_{-k},a_{\ell-k-1}^{m}]\geq[a_{\ell-k}^{m},a_{\ell-k-1}^{m}]\geq[x,y]\geq[b_{-k},b_{-k}],$

and thus, there exists $b_{-k-1}\in X$ such that $b_{-k-1}\succsim
a_{\ell-k-1}^{m}\succsim a_{0}^{0}$ and

$[b_{-k},b_{-k-1}]=[x,y],$

which contradicts our definition of $k$.

Therefore, we can assume that there is no $z\in D$ such that $y\succsim z\succ
a_{0}^{0}$, and thus, there is no $z\in D$ such that $x\succ z\succ
a_{0}^{0}$. By Lemma 5, there exists a finite sequence $b_{0},...,b_{k}$ such
that

$b_{0}=a_{0}^{0},\ b_{1}=x,\ [b_{i+1},b_{i}]=[b_{i+2},b_{i+1}],\
a_{1}^{0}\succsim b_{k},\ [b_{1},b_{0}]>[a_{1}^{0},b_{k}].$

Choose any $m$ such that $k+1<2^{m}$. Then, $a_{1}^{0}=a_{2^{m}}^{m}$. Suppose
that $a_{2^{m}-1}^{m}\succsim b_{k}$. Then,

$[b_{1},b_{0}]>[a_{2^{m}}^{m},b_{k}]\geq[a_{2^{m}}^{m},a_{2^{m}-1}^{m}]=[a_{1}^{m},a_{0}^{m}],$

and thus, $b_{1}\succ a_{1}^{m}\succ b_{0}$, which contradicts our initial
assumption. Therefore, $b_{k}\succ a_{2^{m}-1}^{m}$. Because $k+1<2^{m}$, by
the pigeonhole principle, there exist $i,j$ such that $b_{i+1}\succ
a_{j+1}^{m}\succ a_{j}^{m}\succsim b_{i}$. Then,

$[b_{1},b_{0}]=[b_{i+1},b_{i}]>[a_{j+1}^{m},b_{i}]\geq[a_{j+1}^{m},a_{j}^{m}]=[a_{1}^{m},a_{0}^{m}],$

and thus $b_{1}\succ a_{1}^{m}\succ b_{0}$, which is a contradiction. This
completes the proof. $\blacksquare$

Define

$D^{\prime}=\\{x\in X|\exists x^{\prime}\in D\mbox{ s.t. }x\sim
x^{\prime}\\}.$

Let $x\in D^{\prime}$. Then, there exist $i,k$ such that $x\sim a_{i}^{k}$.
Define

$u(x)=\frac{i}{2^{k}}.$

Clearly, $u(x)$ does not depend on the choice of $k$. Choose any $x,y,z,w\in
D^{\prime}$. Then, there exist $k$ and $i_{x},i_{y},i_{z},i_{w}$ such that

$x\sim a_{i_{x}}^{k},\ y\sim a_{i_{y}}^{k},\ z\sim a_{i_{z}}^{k},\ w\sim
a_{i_{w}}^{k}.$

Suppose that

$u(x)-u(y)\geq u(z)-u(w).$

This means that

$i_{x}-i_{y}\geq i_{z}-i_{w}.$

If $i_{x}-i_{y}\geq 0\geq i_{z}-i_{w}$, then

$[x,y]\geq[y,y]=[z,z]\geq[z,w].$

If $i_{x}-i_{y}\geq i_{z}-i_{w}\geq 0$, define $i_{*}=\min\\{i_{y},i_{w}\\}$.
Then, both $a_{i_{*}+i_{x}-i_{y}}^{k}$ and $a_{i_{*}+i_{z}-i_{w}}^{k}$ are
defined, and by (11),151515Note that, we can show that (11) holds for
$(a_{i}^{m})$ by the same arguments as in the proof of Lemma 5.

$[x,y]=[a_{i_{*}+i_{x}-i_{y}},a_{i_{*}}^{k}]\geq[a_{i_{*}+i_{z}-i_{w}},a_{i_{*}}^{k}]=[z,w].$

By the symmetrical arguments, we can show that if $0\geq i_{x}-i_{y}\geq
i_{z}-i_{w}$, then $[x,y]\geq[z,w]$. In conclusion, $u(x)-u(y)\geq u(z)-u(w)$
implies $[x,y]\geq[z,w]$. Moreover, if

$u(x)-u(y)>u(z)-u(w),$

then $i_{x}-i_{y}>i_{z}-i_{w}$, and by repeating the above arguments, we have
that

$[x,y]>[z,w].$

Therefore, $u$ represents $\geq$ on $D^{\prime}$.

We extend the definition of $u$ on $X$. If $x$ is not a least element of $X$
on $\succsim$, then by Lemma 6, there exists $y\in D$ such that $x\succ y$.
Define

$u(x)=\sup\\{u(y)|y\in D,\ x\succsim y\\}.$

Conversely, if $x$ is not a greatest element of $X$ on $\succsim$, then by
Lemma 6, there exists $y\in D$ such that $y\succ x$. Define

$u(x)=\inf\\{u(y)|y\in D,\ y\succsim x\\}.$

We can easily show that these two definitions of $u$ coincide at each $x$ such
that both can be defined.

If $x\succsim y$, then clearly $u(x)\geq u(y)$. If $x\succ y$, then by Lemma
6, there exists $z,w\in D$ such that $x\succ z\succ w\succ y$. Therefore,

$u(x)\geq u(z)>u(w)\geq u(y),$

and thus $u(x)>u(y)$. Hence,

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

Next, we show the continuity of $u$. For this purpose, we should show that
both $u^{-1}(]-\infty,a])$ and $u^{-1}([a,+\infty[)$ are closed for every
$a\in\mathbb{R}$. If $\inf u<a<\sup u$, then

$u^{-1}([a,+\infty[)=X\setminus\cup_{y\in D:u(y)<a}\\{z\in X|y\succ z\\},$
$u^{-1}(]-\infty,a])=X\setminus\cup_{y\in D:u(y)>a}\\{z\in X|z\succ y\\},$

and both are clearly closed. The case in which either $a\leq\inf u$ or $\sup
u\leq a$ can be treated easily, and thus we omit the proof for such cases.

To prove that $u$ represents $\geq$, it suffices to show that

$[x,y]\geq[z,w]\Leftrightarrow u(x)-u(y)\geq u(z)-u(w).$

For this, it suffices to show the following two claims:

  1. i)

if $u(x)-u(y)\geq u(z)-u(w)$, then $[x,y]\geq[z,w]$, and

  2. ii)

if $u(x)-u(y)>u(z)-u(w)$, then $[x,y]>[z,w]$.

We show that i) implies ii). Suppose that i) is correct. Choose any
$x,y,z,w\in X$ such that $u(x)-u(y)>u(z)-u(w)$. Note that, because $X$ is
connected, $u(X)$ is also connected in $\mathbb{R}$, and thus it is convex. If
$u(x)>\inf u$, then there exists $x^{\prime}\in X$ such that
$u(x^{\prime})-u(y)\geq u(z)-u(w)$ and $x\succ x^{\prime}$. Then,

$[x,y]>[x^{\prime},y]\geq[z,w],$

which implies that $[x,y]>[z,w]$. If $u(x)=\inf u$, then $w>\inf u$ because
$u(x)+u(w)>u(y)+u(z)$. Therefore, there exists $w^{\prime}$ such that
$u(x)-u(y)\geq u(z)-u(w^{\prime})$ and $w\succ w^{\prime}$. Then,

$[x,y]\geq[z,w^{\prime}]>[z,w],$

which implies that $[x,y]>[z,w]$. Therefore, our claim is correct, and thus it
suffices to show i).

We need the following lemma.

Lemma 7. Suppose that $x\in X$. Then, there exist $x^{\prime}\in X$ and a
sequence $(x^{m})$ on $D^{\prime}$ such that $u(x)=u(x^{\prime})$ and
$x^{m}\to x^{\prime}$ as $m\to\infty$. Moreover, if $u(x)<\sup u$ (resp.
$u(x)>\inf u$), then we can assume that $u(x^{m})>u(x)$ (resp.
$u(x^{m})<u(x)$) for every $m$.161616In the proof of Lemma 7, we need the
path-connectedness of $X$. In proving Theorem 1, only Lemma 7 requires the
path-connectedness of $X$, and the remaining part uses only connectedness.
Therefore, if there is a proof of Lemma 7 that does not use the path-
connectedness of $X$, we can extend Theorem 1 to a general Hausdorff,
separable, and connected topological space.

Proof. We treat only the case in which $u(x)<\sup u$, because the remaining
case can be treated symmetrically.

By assumption, there exists $\hat{x}\in X$ such that $u(\hat{x})>u(x)$.
Because $X$ is path-connected, there exists a continuous function $f:[0,1]\to
X$ such that $f(0)=x,f(1)=\hat{x}$. Define
$t^{\prime}=\sup\\{t\in[0,1]|u(f(t))\leq u(x)\\}$ and
$x^{\prime}=f(t^{\prime})$. Clearly, $t^{\prime}<1$ and $u(x^{\prime})=u(x)$.
Let $g(t)=u(f(t))$. Then, we have that if $t^{\prime}<t<1$, then
$g(t)>g(t^{\prime})=v(x)$. Therefore, by the intermediate value theorem, there
exists $t_{1}$ such that $t^{\prime}<t_{1}<1$ and $f(t_{1})\in D^{\prime}$.
Define $x^{1}=f(t_{1})$. Next, suppose that $t_{m}$ and $x_{m}=f(t_{m})$ are
defined, where $t_{m}>t^{\prime}$ and $x_{m}\in D^{\prime}$. Because
$t_{m}>t^{\prime}$, $t^{\prime}<t<t_{m}$ implies $g(t)>u(x)$. Therefore, again
by the intermediate value theorem, there exists $t_{m+1}$ such that
$t^{\prime}<t_{m+1}<\frac{t^{\prime}+t_{m}}{2}$ and $x_{m+1}=f(t_{m+1})\in
D^{\prime}$. By construction,

$0<t_{m}-t^{\prime}<(t_{1}-t^{\prime})/2^{m-1},$

and thus $t_{m}\to t^{\prime}$ as $m\to\infty$. By the continuity of $f$,
$x^{m}\to x^{\prime}$ as $m\to\infty$. This completes the proof.
$\blacksquare$

Assume that

$u(x)-u(y)\geq u(z)-u(w).$

Then,

$u(x)+u(w)\geq u(y)+u(z)$

If $u(y)=u(z)=\inf u$, then clearly

$[x,y]\geq[x,w]\geq[z,w].$

If $u(x)=u(w)=\sup u$, then clearly

$[x,y]\geq[z,y]\geq[z,w].$

Therefore, without loss of generality, we can assume that either $u(x)<\sup u$
or $u(w)<\sup u$, and either $u(y)>\inf u$ or $u(z)>\inf u$. We treat the case
in which $u(x)<\sup u$ and $u(y)>\inf u$, because the remaining cases can be
treated similarly.

Because $u(x)<\sup u$ and $u(y)>\inf u$, by Lemma 7, there exist
$x^{\prime},y^{\prime},z^{\prime},w^{\prime}\in X$ and sequences
$(x^{m}),(y^{m}),(z^{m}),(w^{m})$ on $D^{\prime}$ such that $u(x^{m})>u(x)$
and $u(y^{m})<u(y)$ for all $m$,

$u(x^{\prime})=u(x),\ u(y^{\prime})=u(y),\ u(z^{\prime})=u(z),\
u(w^{\prime})=u(w),$

and

$x^{m}\to x^{\prime},\ y^{m}\to y^{\prime},\ z^{m}\to z^{\prime},\ w^{m}\to
w^{\prime}\mbox{ as }m\to\infty.$

Then, for every $m$,

$u(x^{m})-u(y^{m})>u(z)-u(w),$

and thus, there exists $k(m)\geq m$ such that

$u(x^{m})-u(y^{m})\geq u(z^{k(m)})-u(w^{k(m)}).$

Because $u$ represents $\geq$ on $D^{\prime}$,

$[x^{m},y^{m}]\geq[z^{k(m)},w^{k(m)}].$

Therefore, by continuity,

$[x,y]=[x^{\prime},y^{\prime}]\geq[z^{\prime},w^{\prime}]=[z,w].$

This implies i). Hence, $u$ represents $\geq$ on $X$.

The rest of our claim is the uniqueness of such a $u$ up to a positive affine
transform. Suppose that $v:X\to\mathbb{R}$ is a continuous representation of
$\geq$. If $x\sim y$ for all $x,y\in X$, then $v$ is a constant function, in
which case the uniqueness is obvious. Hence, we assume that there exist
$x^{*},y^{*}\in X$ such that $x^{*}\succ y^{*}$. Define $u:X\to\mathbb{R}$ as
in the above proof. Then, $u$ is a representation of $\geq$ such that
$u(a_{i}^{k})=\frac{i}{2^{k}}$ for all $a_{i}^{k}\in D$. We can easily show
that

$\displaystyle v(a_{i}^{k})=$
$\displaystyle~{}v(a_{i}^{k})-v(a_{0}^{k})+v(y^{*})$ $\displaystyle=$
$\displaystyle~{}i(v(a_{1}^{k})-v(a_{0}^{k}))+v(y^{*})$ $\displaystyle=$
$\displaystyle~{}\frac{i}{2^{k}}(v(x^{*})-v(y^{*}))+v(y^{*})$ $\displaystyle=$
$\displaystyle~{}(v(x^{*})-v(y^{*}))u(a_{i}^{k})+v(y^{*}).$

Because $v$ is continuous, by Lemma 7,

$v(x)=(v(x^{*})-v(y^{*}))u(x)+v(y^{*}),$

as desired. This completes the proof. $\blacksquare$

###  A.3 Proof of Theorem 3

Suppose that $u$ is continuously differentiable on $\mathbb{R}^{n}_{++}$ and
$Du(x)\neq 0$ for every $x\in\mathbb{R}^{n}_{++}$. Define $v(x,y)=u(x)-u(y)$.
Then,

$I=\\{(x,y)\in\mathbb{R}^{2n}_{++}|x\sim y\\}=v^{-1}(0).$

By the preimage theorem, $I$ is a $2n-1$ dimensional $C^{1}$ manifold, and
thus Debreu’s smoothness holds.171717See section 1.4 of Guillemin and Pollack
(1974). Next, recall that $e=(1,1,...,1)$. Define $g(b)=u(be)$. Then, $g$ is a
concave, increasing, and continuously differentiable function defined on
$\mathbb{R}_{++}$, and because $Du(x)\neq 0$, $g^{\prime}(b)=Du(be)e>0$.
Therefore, by the inverse function theorem, $g^{-1}$ is also continuously
differentiable. This implies that

$\lim_{a\downarrow 0}\frac{b-f(a,b)}{a}=0\Leftrightarrow\lim_{a\downarrow
0}\frac{g(b)-g(f(a,b))}{a}=0.$

Because $g$ is concave, $2g(b)\geq g(b+a)+g(b-a)$, and thus $b\geq f(a,b)\geq
b-a$. This implies that

$\liminf_{a\downarrow 0}\frac{g(b)-g(f(a,b))}{a}\geq 0.$

On the other hand,

$\displaystyle 0\leq$ $\displaystyle~{}\limsup_{a\downarrow
0}\frac{g(b)-g(f(a,b))}{a}$ $\displaystyle=$
$\displaystyle~{}\lim_{a\downarrow
0}\frac{g(b)-g(b-a)}{a}-\liminf_{a\downarrow 0}\frac{g(f(a,b))-g(b-a)}{a}$
$\displaystyle=$ $\displaystyle~{}\lim_{a\downarrow
0}\frac{g(b)-g(b-a)}{a}-\liminf_{a\downarrow 0}\frac{g(b+a)-g(f(a,b))}{a}$
$\displaystyle\leq$ $\displaystyle~{}\lim_{a\downarrow
0}\frac{g(b)-g(b-a)}{a}-\liminf_{a\downarrow 0}\frac{g(b+a)-g(b)}{a}$
$\displaystyle=$ $\displaystyle~{}g^{\prime}(b)-g^{\prime}(b)=0,$

which implies that line smoothness is satisfied.

Conversely, suppose that $\geq$ satisfies both Debreu’s smoothness and line
smoothness. First, we show that the function

$g(b)=u(be)$

is continuously differentiable on $\mathbb{R}_{++}$. Choose any $b>0$. Because
$g$ is concave, it is locally Lipschitz, and thus there exist $\varepsilon>0$
and $L>0$ such that $|g(c)-g(b)|\leq L|c-b|$ for all
$c\in[b-\varepsilon,b+\varepsilon]$. Moreover, the left- and right-side
derivatives $D_{-}g(b),D_{+}g(b)$ can be defined and are real numbers, and $g$
is differentiable if and only if $D_{-}g(b)=D_{+}g(b)$. Because $2g(b)\geq
g(b+a)+g(b-a)$, we have that $b-a\leq f(b,a)\leq b$ for all $a>0$, and thus by
monotonicity, $g(b)\geq g(f(a,b))$. Moreover,

$\displaystyle 0\leq$ $\displaystyle~{}\liminf_{a\downarrow
0}\frac{g(b)-g(f(a,b))}{a}$ $\displaystyle\leq$
$\displaystyle~{}\limsup_{a\downarrow 0}\frac{g(b)-g(f(a,b))}{a}$
$\displaystyle\leq$ $\displaystyle~{}L\lim_{a\downarrow
0}\frac{b-f(a,b)}{a}=0,$

and thus

$\lim_{a\downarrow 0}\frac{g(b)-g(f(a,b))}{a}=0.$

Therefore,

$\displaystyle D_{+}g(b)=$ $\displaystyle~{}\lim_{a\downarrow
0}\frac{g(b+a)-g(b)}{a}$ $\displaystyle=$ $\displaystyle~{}\lim_{a\downarrow
0}\frac{g(b+a)-g(f(a,b))}{a}$ $\displaystyle=$
$\displaystyle~{}\lim_{a\downarrow 0}\frac{g(f(a,b))-g(b-a)}{a}$
$\displaystyle=$ $\displaystyle~{}\lim_{a\downarrow 0}\frac{g(b)-g(b-a)}{a}$
$\displaystyle=$ $\displaystyle~{}D_{-}g(b),$

as desired. Because every differentiable and concave function is continuously
differentiable, $g$ is continuously differentiable.

Because $g$ is increasing and concave, we must have $g^{\prime}(b)>0$ for
every $b>0$.

Now, because $u$ is continuous and increasing, for every
$x\in\mathbb{R}^{n}_{++}$, there uniquely exists $a(x)>0$ such that
$u(x)=u(a(x)e)=g(a(x))$. Therefore, it suffices to show that, under
monotonicity and Debreu’s smoothness, $a:x\mapsto a(x)$ is continuously
differentiable and $Da(x)\neq 0$ on $\mathbb{R}^{n}_{++}$. We treat this
result as a lemma.181818If $I$ is a $C^{2}$ manifold, then this result was
obtained by Debreu (1972), although his proof has a gap. If $I$ is a
$C^{\infty}$ manifold, then this result was rigorously proved in ch.7 of
Bridges and Mehta (1995). However, we can only assume that $I$ is $C^{1}$, and
thus these results cannot be applied. Mas-Colell (1977) noted that Moulin
solved this problem in 1973. However, we could not obtain Moulin’s monograph,
and thus this claim is unverifiable. This is why we present the proof of the
smoothness of $a(x)$.

Lemma 8. Suppose that $X$ is either $\mathbb{R}^{n}_{+}$ or
$\mathbb{R}^{n}_{++}$, and $\succsim$ is a closed and monotone weak order on
$X$,191919We say that a weak order $\succsim$ on $X$ is monotone if $x\gg y$
implies that $x\succ y$. and the set

$I=\\{(x,y)\in\mathbb{R}^{2n}_{++}|x\sim y\\}$

is $2n-1$ dimensional $C^{k}$ manifold. Then, for every
$x\in\mathbb{R}^{n}_{++}$, there uniquely exists $a(x)$ such that $x\sim
a(x)e$, where $e=(1,1,...,1)$. Moreover, the function
$a:\mathbb{R}^{n}_{++}\to\mathbb{R}_{++}$ is $C^{k}$, and $Da(x)\neq 0$ for
every $x\in\mathbb{R}^{n}_{++}$.

Proof. For $x\in\mathbb{R}^{n}_{++}$, define $a(x)=\inf\\{c>0|ce\succsim
x\\}$. It is easy to show that $a(x)$ is the unique number $c>0$ such that
$x\sim ce$, and thus we omit its proof.202020See Proposition 3.C.1 of Mas-
Colell et al. (1995).

Choose any $x\in\mathbb{R}^{n}_{++}$, and define

$I_{x}=\\{y\in\mathbb{R}^{n}_{++}|x\sim y\\}.$

We first show that $I_{x}$ is connected. For this, it suffices to show that,
for every $y\in I_{x}$, there exists a continuous function $f:[0,2]\to I_{x}$
such that $f(0)=x$ and $f(2)=y$.

Define $z_{i}=\min\\{x_{i},y_{i}\\}>0$, and

$b(t)=\begin{cases}(1-t)x+tz&\mbox{if }0\leq t\leq 1,\\\
(2-t)z+(1+t)y&\mbox{if }1\leq t\leq 2.\end{cases}$

Let $M=\max_{i}\\{|x_{i}-y_{i}|\\}$. By monotonicity and the intermediate
value theorem, there uniquely exists $c(t)\in[0,M]$ such that $b(t)+c(t)e\in
I_{x}$. Let $(t_{m})$ be the sequence in $[0,2]$ such that $t_{m}\to t$ as
$m\to\infty$. Suppose that $c(t_{m})\not\to c(t)$. Because $c(t_{m})\in[0,M]$,
there exists a subsequence $t_{\ell(m)}$ such that
$c(t_{\ell(m)})\to\alpha\neq c(t)$ as $m\to\infty$. Because $\succsim$ is
closed in $X^{2}$, $b(t)+\alpha e\in I_{x}$, which contradicts the uniqueness
of $c(t)$. Therefore, $c(t)$ is continuous, and $f(t)=b(t)+c(t)e$ satisfies
all of our requirements.

For $x\in\mathbb{R}^{n}_{++}$, define $L_{x}$ as the set of all $y\in I_{x}$
such that the function $a:\mathbb{R}^{n}_{++}\to\mathbb{R}_{++}$ is $C^{k}$
around $y$ and $Da(y)\neq 0$. To prove this lemma, it suffices to show that
$x\in L_{x}$ for all $x\in\mathbb{R}^{n}_{++}$. Fix $x\in\mathbb{R}^{n}_{++}$.
To show that $x\in L_{x}$, it suffices to show that $L_{x}=I_{x}$. Because
$I_{x}$ is connected, it suffices to show that $L_{x}$ is nonempty, open, and
closed with respect to the relative topology of $I_{x}$.

It is clear that $L_{x}$ is open.

Choose any $y\in I_{x}$. Then, $a(y)=a(x)$. Because $I$ is a $2n-1$
dimensional $C^{k}$ manifold, there exist open neighborhoods
$U\subset\mathbb{R}^{2n-1}$ of $0$ and $V\subset I$ of $(y,y)$, and a $C^{k}$
diffeomorphism $\varphi:U\to V$ such that $\varphi(0)=(y,y)$.212121The
statement “$\varphi$ is a $C^{k}$ diffeomorphism” means that $\varphi$ is a
bijection and both $\varphi$ and $\varphi^{-1}$ are $C^{k}$. Note that, $V$ is
open not in $\mathbb{R}^{2n}$ but only in $I$, and thus to understand this
statement rigorously, we must define the differentiability of functions
defined on non-open sets. See section 1.1-1.2 of Guillemin and Pollack (1974).
Let $T_{(y,y)}(I)$ be the image of $\mathbb{R}^{2n-1}$ for the linear operator
$D\varphi(0)$. This linear space is called the tangent space of $I$ at
$(y,y)$. It is known that $v\in T_{(y,y)}(I)$ if and only if there exists a
continuously differentiable function $c:[-\varepsilon,\varepsilon]\to I$ such
that $c(0)=(y,y)$ and $\dot{c}(0)=v$. Because $c(t)=(y+te,y+te)\in I$, we have
that $(e,e)\in T_{(y,y)}(I)$. If $(0,e)\in T_{(y,y)}(I)$, then

$(-e,e)=2(0,e)-(e,e)\in T_{(y,y)}(I),$

and thus, there exists a continuously differentiable function
$c:[-\varepsilon,\varepsilon]\to I$ such that $\dot{c}(0)=(-e,e)$. This means
that there exists $(z,w)\in I$ such that $w\gg y$ and $y\gg z$. However, by
monotonicity,

$w\succ y\succ z\sim w,$

which is a contradiction. Define

$\psi(v,a)=\varphi(v)+(0,ae).$

Then, by the above result, $D\psi(0,0)$ is a regular matrix, and thus, by the
inverse function theorem, there exist open neighborhoods
$U^{\prime}\subset\mathbb{R}^{2n}$ of $(0,0)$ and
$V^{\prime}\subset\mathbb{R}^{2n}_{++}$ of $(y,y)$ such that
$\psi:U^{\prime}\to V^{\prime}$ is a $C^{k}$ diffeomorphism. Note that for
$(z,w)\in V^{\prime}$, $z\sim w$ if and only if $\psi^{-1}(z,w)=(v,0)$ for
some $v$. Let $\chi(z,w)=\psi^{-1}(z,w)$, and $h(z,b)=\chi_{2n}(z,y+be)$.
Then, $h$ is $C^{k}$ and $z\sim(y+be)$ if and only if $h(z,b)=0$.

We show that $D_{y}h(y,0)\neq 0$ and $\frac{\partial h}{\partial b}(y,0)\neq
0$. Because $h(y+be,b)=0$ for all $b$, $D_{y}h(y,0)e=-\frac{\partial
h}{\partial b}(y,0)$, and thus, it suffices to show that $\frac{\partial
h}{\partial b}(y,0)\neq 0$. Suppose that $\frac{\partial h}{\partial
b}(y,0)=0$. This implies that

$D\chi_{2n}(y,y)(0,e)=0.$

Meanwhile,

$(0,e)=D\psi(0,0)D\chi(y,y)(0,e),$

and

$D\psi(0,0)(w,b)=D\varphi(0)w+b(0,e).$

Therefore,

$(0,e)=D\varphi(0)w$

for some $w\in\mathbb{R}^{2n-1}$. This implies that $(0,e)\in T_{(y,y)}(I)$,
which contradicts our previous arguments. Therefore, our claim is correct. By
the implicit function theorem, there exists an open neighborhood $W^{\prime}$
of $y$ and a $C^{k}$ function $b:W^{\prime}\to\mathbb{R}$ such that $z\sim
y+b(z)e$ and $Db(z)\neq 0$ for every $z\in W^{\prime}$, and $b(y)=0$.

If $y=a(x)e$, then $b(z)$ coincides with $a(z)-a(x)$, and thus $a(x)e\in
L_{x}$ and $L_{x}$ is nonempty.

Thus, it suffices to show that $L_{x}$ is closed in $I_{x}$. Choose a sequence
$(y^{m})$ on $L_{x}$ and suppose that $y^{m}\to y\in I_{x}$ as $m\to\infty$.
Consider the function $b:W^{\prime}\to\mathbb{R}$ defined as above. Because of
monotonicity, $b$ is increasing, and thus $Db(z)e>0$ for every $z\in
W^{\prime}$. Because $y^{m}\to y$ as $m\to\infty$, there exists $m$ such that
$y^{m}\in W^{\prime}$. Consider the following equation:

$b(z)-b(y^{m}+te)=0.$

This equation holds for $(z,t)=(y,0)$. By the implicit function theorem, there
exist an open neighborhood $W^{\prime\prime}\subset W^{\prime}$ of $y$ and a
$C^{k}$ function $t:W^{\prime\prime}\to\mathbb{R}$ such that $Dt(z)e\neq 0$
and $b(z)=b(y^{m}+t(z)e)$ for every $z\in W^{\prime\prime}$, and $t(y)=0$.
This implies that

$a(z)=a(y^{m}+t(z)e).$

Because $y^{m}\in L_{x}$, $a$ is $C^{k}$ around $y$ and $Da(y)\neq 0$, which
implies that $y\in L_{x}$ and $L_{x}$ is closed. This completes the proof.
$\blacksquare$

Because of Lemma 8, $u$ satisfies all of our requirements. This completes the
proof. $\blacksquare$

## Acknowledgement

The author would like to express the gratitude to Marcel K. Richter. In 2013,
the author met Marcel at Keio University, and was presented with some
important pieces of advice and suggestions. All of these have been immensely
worthwhile. Unfortunately, Marcel died in 2014. The author prays for the
repose of his soul. The author would also like to thank Mitsunobu Miyake for
many suggestions at the 2015 DC Conference and the 2020 Japan Economic
Association Fall Meeting. The author is also grateful to Chaowen Yu, Shinsuke
Nakamura, and Shinichi Suda for their helpful comments and suggestions. This
work was supported by JSPS KAKENHI Grant Number JP21K01403. Finally, the
author thanks the anonymous reviewers for their helpful comments and
suggestions on this paper.

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