# A Nonlinear Sandwich Theorem

Mario Ghossoub
University of Waterloo
Giulio Principi
New York University
Lorenzo Stanca
Collegio Carlo Alberto and University of Turin
Mario Ghossoub: University of Waterloo – Department of Statistics and
Actuarial Science – 200 University Ave. W. – Waterloo, ON, N2L 3G1 – Canada
<EMAIL_ADDRESS>Giulio Principi: New York University – Department
of Economics – 19 West 4th Street – New York, NY 10012 – USA<EMAIL_ADDRESS>Lorenzo Stanca: Collegio Carlo Alberto and University of Turin – Department of
ESOMAS – Corso Unione Sovietica, 218 Bis – Turin 10134 – Italy
<EMAIL_ADDRESS>
###### Abstract.

We provide a Sandwich Theorem (König, (1972)) for positively homogeneous
functionals that satisfy additivity only on a restricted domain. Our
relaxation of additivity is based on a binary relation called convex-conic
symmetric preorder, whereby additivity is restricted to all couples of
elements that belong to such relation. We then study applications of our
nonlinear Sandwich Theorem, proving extension and envelope representation
results. Finally, we consider some applications to comonotonicity, a key
property in decision theory, risk measurement, and the theory of risk sharing.

###### Key words and phrases:

Sandwich Theorem; Pataraia’s Theorem; C-(super/sub)linearity; convex-conic
symmetric preorder

###### 2020 Mathematics Subject Classification:

Primary: 46A22; Secondary: 06A06, 06F20.

We thank Andrea Aveni, Fabio Maccheroni, and René Pfitscher for useful
discussions, as well as Stephen Simons for providing some relevant references.
Mario Ghossoub acknowledges financial support from the Natural Sciences and
Engineering Research Council of Canada (NSERC) (Grant No. 2018-03961). Giulio
Principi is grateful for the financial support provided by the Henry M.
MacCracken Fellowship at New York University.

##  1\. Introduction

The classic version of the Sandwich Theorem yields the existence of a linear
functional in between a given sublinear functional and a given superlinear
functional. König, (1972) proved this result as a corollary to the Hahn-Banach
extension theorem. In a recent paper, Amarante, (2019) provided an alternative
proof, by combining an argument in Fuchssteiner and Wright, (1977) with
Pataraia’s Fixed Point Theorem (Pataraia, (1997)). In this paper, we show how
the elegant approach developed by Amarante, (2019) can be directly implemented
to retrieve nonlinear versions of the Sandwich Theorem, restricting
(super/sub)linearity only to some portions of the domain of the functionals
involved. Specifically, we define a binary relation (denoted by $\mathbf{C}$)
that we call the convex-conic symmetric preorder, which is a symmetric
preorder closed with respect to positive scalar multiplication and addition.
Using this preorder, we provide a Sandwich Theorem (see Theorem 2) for
$\mathbf{C}$-(super/sub)linear functionals, i.e., functionals that satisfy
positive homogeneity and (super/sub)additivity only with respect to
$(x,y)\in\mathbf{C}$. We then use this result to prove a Hahn-Banach type
extension result (see Corollary 1) and an envelope representation result (see
Corollary 2). We conclude by providing an illustration of the applicability of
our results, with a focus on comonotonic subadditive functionals.

##  2\. Pataraia’s Theorem and Order-Theoretic Terminology

###  2.1. Order-Theoretic Terminology

111We summarize in this subsection all the order-theoretic concepts that we
use in this paper. For a comprehensive treatment of these notions, we refer
the reader to Caspard et al., (2012) and Schröder, (2016).

Fix a nonempty set $S$. By a (binary) relation over $S$ we mean a set
$\mathbf{R}\subseteq S\times S$. For all $x,y,z\in S$, we will often write
$x\mathbf{R}y$ in place of $(x,y)\in\mathbf{R}$, and $x\mathbf{R}y\mathbf{R}z$
instead of $x\mathbf{R}y$ and $y\mathbf{R}z$.

We say that a binary relation $\mathbf{R}$ is reflexive if $x\mathbf{R}x$ for
all $x\in S$, while it is symmetric if $x\mathbf{R}y$ implies $y\mathbf{R}x$,
for all $x,y\in S$. If $x\mathbf{R}y$ and $y\mathbf{R}z$ implies
$x\mathbf{R}z$, for all $x,y,z\in S$, then $\mathbf{R}$ is said to be
transitive. A reflexive and transitive relation is called a preorder. A
preorder $\mathbf{R}$ is a partial order if it is also antisymmetric, that is,
if $x\mathbf{R}y$ and $y\mathbf{R}x$, then $x=y$, for all $x,y\in S$, in which
case we say that $(S,\mathbf{R})$ is a partially ordered set (henceforth,
poset). A partial order $\mathbf{R}$ is said to be a total order if for all
$x,y\in S$, either $x\mathbf{R}y$ or $y\mathbf{R}x$, in which case we say that
$(S,\mathbf{R})$ is a totally ordered set. A totally ordered subset of a poset
will be referred to as a chain.

Given a poset $(S,\mathbf{R})$ we say that an element $x\in S$ is an
$\mathbf{R}$-upper bound of $B\subseteq S$ if $x\mathbf{R}B$, that is,
$x\mathbf{R}y$ for all $y\in B$. We say that $x$ is an $\mathbf{R}$-supremum
for $B\subseteq S$ if it is an $\mathbf{R}$-upper bound of $B$ and
$z\mathbf{R}x$, for all $z\in S$ with $z\mathbf{R}B$. Infima are defined
analogously. Given a poset $(S,\mathbf{R})$, we denote $\mathbf{R}$-suprema
and $\mathbf{R}$-infima of $B\subseteq S$ by $\mathbf{R}$-$\sup B$ and
$\mathbf{R}$-$\inf B$, respectively. When the binary relation is well-
understood within the context we will simply write $\sup$ and $\inf$ to ease
the notation. We say that a poset $(S,\mathbf{R})$ is strictly inductively
ordered if every chain $C\subseteq S$ admits a $\mathbf{R}$-supremum. Given a
poset $(S,\mathbf{R})$, we say that a mapping $F\mathrel{\mathop{\mathchar
58\relax}}S\to S$ is $\mathbf{R}$-inflationary if $F(s)\mathbf{R}s$, for all
$s\in S$. Given a nonempty set $A$, a poset $(S,\mathbf{R})$, and two maps
$f,g\mathrel{\mathop{\mathchar 58\relax}}A\to S$, we say that $f\mathbf{R}g$
if and only if $f(a)\mathbf{R}g(a)$ for all $a\in A$. We will refer to this
order as the $\mathbf{R}$-pointwise order, and, often simply as the pointwise
order, when the underlying relation is clear from the context. It is
immediately noticeable that given a poset $(S,\mathbf{R})$ and a set $A$, the
induced $\mathbf{R}$-pointwise order on $S^{A}$ is also a partial order.

We say that a (real) vector space $V$ is an ordered vector space if it is
endowed with a partial order $\leq$, such that $x\leq y$ if and only if
$x+z\leq y+z$ and $\alpha x\leq\alpha y$, for all $x,y,z\in V$ and all
$\alpha\geq 0$.222We will exclusively focus on real vector spaces, which we
will henceforth refer to simply as vector spaces. In addition, we say that an
ordered vector space $(V,\leq)$ is a Riesz space if it is also a lattice, that
is, $\sup\left\\{x,y\right\\},\inf\left\\{x,y\right\\}\in V$, for all $x,y\in
V$. A subset $B$ of a Riesz space $(V,\leq)$ is said to be bounded from above
if it admits an upper bound in $V$. Finally, we say that a Riesz space
$(V,\geq)$ is Dedekind-complete if each subset of $V$ which is bounded from
above admits a supremum.333We refer to Aliprantis and Burkinshaw, (2006) for a
detailed treatment of Riesz spaces. For any vector space $V$ we will denote by
$\mathbf{0}_{V}$ its null element.

###  2.2. Pataraia’s Fixed Point Theorem

Similarly to Amarante, (2019) we will apply the following version of
Pataraia’s Fixed Point Theorem (see also Escardó, (2003)).

######  Theorem 1 (Pataraia).

Let $(S,\mathbf{R})$ be a poset and $\mathcal{I}$ be the set of all
inflationary mappings on $S$. If $(S,\mathbf{R})$ is strictly inductively
ordered, then $\mathcal{I}$ has a common fixed point, that is, there exists
$x\in S$ such that $f(x)=x$, for every $f\in\mathcal{I}$.

###### Proof.

Since $\mathbf{R}$ is reflexive, $\mathrm{Id}\in\mathcal{I}$. Moreover, given
that all $f\in\mathcal{I}$ are inflationary, it follows that
$f\mathbf{R}\mathrm{Id}$. Suppose that $\mathcal{C}$ is a chain in
$\mathcal{I}$ and define $\overline{f}\mathrel{\mathop{\mathchar
58\relax}}s\mapsto\sup\left\\{f(s)\mathrel{\mathop{\mathchar
58\relax}}f\in\mathcal{C}\right\\}$. Clearly $\overline{f}\in\mathcal{I}$ and
hence it is a $\mathbf{R}$-supremum for $\mathcal{C}$ with respect to the
pointwise order. This yields that $\mathcal{I}$ with the pointwise order is
strictly inductively ordered. Therefore, by Zorn’s Lemma, $\mathcal{I}$ has a
maximal element, say $M\in\mathcal{I}$. Note that for all $s\in S$ and all
$f\in\mathcal{I}$, $f(M(s))\mathbf{R}M(s)$. Thus, since $M$ is maximal and
$\mathbf{R}$ is antisymmetric, we must have $f\circ M=M$. Therefore, for all
$s\in S$, $M(s)$ is a common fixed point of $\mathcal{I}$. ∎

##  3\. Main Results

###  3.1. Convex-Conic Symmetric Preorders

Given a vector space $V$, we say that $X\subseteq V$ is a convex cone if
$\lambda X\subseteq X$ for all $\lambda>0$, and $X+X\subseteq X$. Now let $X$
be a given convex cone.

######  Definition 1.

A binary relation $\mathbf{C}\subseteq X\times X$ is a convex-conic symmetric
preorder if it is reflexive, transitive, symmetric, and it satisfies the
following properties:

  1. (i)

$(\lambda x)\mathbf{C}x$, for all $x\in X$ and $\lambda>0$.

  2. (ii)

$x\mathbf{C}y\mathbf{C}z\ \textnormal{implies}\ (x+y)\mathbf{C}z$, for all
$x,y,z\in X$.

The convex-conic adjective in the definition of a convex-conic symmetric
preorder is due to the fact that $\mathbf{C}(x)=\left\\{y\in
X\mathrel{\mathop{\mathchar 58\relax}}y\mathbf{C}x\right\\}$ is a convex cone.
Note that the symmetry property (ii) in Definition 1 implies that whenever
$x\mathbf{C}y\mathbf{C}z$, we have $(g+u)\mathbf{C}h$, for all
$g,u,h\in\left\\{x,y,z\right\\}$. It is important to observe that whenever
$\left\\{\mathbf{0}_{V}\right\\}\times X\subseteq\mathbf{C}$, for some convex-
conic symmetric preorder $\mathbf{C}$ on $X$, it follows that $X\times
X=\mathbf{C}$. This is a straightforward consequence of symmetry and
transitivity. Because of property (ii), the same would hold if
$(-x)\mathbf{C}x$ for all $x\in X$ with $-x\in X$. For future reference we
recollect these simple observations on the following lemma.

######  Lemma 1.

Let $X$ be a convex cone and $\mathbf{C}\subseteq X\times X$ a convex-conic
symmetric preorder. If either $\left\\{\mathbf{0}_{V}\right\\}\times
X\subseteq\mathbf{C}$ or $(-x)\mathbf{C}x$ for all $x\in X$ with $-x\in X$,
then $X\times X=\mathbf{C}$.

Therefore, at first sight one may worry about how permissive a convex-conic
symmetric preorder is. We show by means of examples (see Section 3.2.2) that
it is neither always trivial nor too restrictive.

###  3.2. A Sandwich Theorem

Let $\left(\mathbb{V},\leq\right)$ be a Dedekind-complete Riesz space.
Following Fuchssteiner and Wright, (1977), we adjoin to $\mathbb{V}$ an
element denoted by $-\infty$, and we extend $\leq$ to
$\mathbb{V}\cup\left\\{-\infty\right\\}$ assuming $-\infty\leq\mathbb{V}$. We
say that a map $F\mathrel{\mathop{\mathchar
58\relax}}X\to\mathbb{V}\cup\left\\{-\infty\right\\}$ is
$\mathbf{C}$-sublinear if

  1. (i)

$F$ is positively homogeneous, i.e., $F(\lambda x)=\lambda F(x)$, for all
$\lambda>0$ and all $x\in X$.

  2. (ii)

$F$ is $\mathbf{C}$-subadditive, i.e., $F(x+y)\leq F(x)+F(y)$, for all $x,y\in
X$ with $x\mathbf{C}y$.

The definitions of $\mathbf{C}$-superlinearity and $\mathbf{C}$-linearity are
analogous. We endow $V$ with a partial order $\preceq$ such that $(V,\preceq)$
is a partially ordered vector space. We say that a function
$F\mathrel{\mathop{\mathchar
58\relax}}X\to\mathbb{V}\cup\left\\{-\infty\right\\}$ is monotone if $x\preceq
y$ implies $F(x)\leq F(y)$, for all $x,y\in X$. We adopt the convention that
$0\cdot(-\infty)=\mathbf{0}_{\mathbb{V}}$. We are now ready to state and prove
our main result.

######  Theorem 2.

Suppose that $P\mathrel{\mathop{\mathchar
58\relax}}X\to\mathbb{V}\cup\left\\{-\infty\right\\}$ is
$\mathbf{C}$-superlinear and $H\mathrel{\mathop{\mathchar
58\relax}}X\to\mathbb{V}\cup\left\\{-\infty\right\\}$ is
$\mathbf{C}$-sublinear and monotone. If $P\leq H$, then there exists a
$\mathbf{C}$-linear map $Q\mathrel{\mathop{\mathchar
58\relax}}X\to\mathbb{V}\cup\left\\{-\infty\right\\}$ such that $P\leq Q\leq
H$.

The proof will be provided in several steps. Before going into its details, we
provide some definitions and simple remarks. Let

$\mathcal{D}_{\mathrm{PH}}=\left\\{Q\mathrel{\mathop{\mathchar
58\relax}}X\to\mathbb{V}\cup\left\\{-\infty\right\\}\mathrel{\mathop{\mathchar
58\relax}}Q\ \textnormal{is}\ \mathbf{C}\textnormal{-superlinear}\ \hbox{and}\
P\leq Q\leq H\right\\}.$

Clearly, $P\in\mathcal{D}_{\mathrm{PH}}\neq\emptyset$. Moreover,
$\mathcal{D}_{\mathrm{PH}}$ is a poset with respect to the pointwise order. If
$\mathcal{C}$ is a chain in $\mathcal{D}_{\mathrm{PH}}$, then
$\overline{Q}\mathrel{\mathop{\mathchar
58\relax}}x\mapsto\sup\left\\{Q(x)\mathrel{\mathop{\mathchar
58\relax}}Q\in\mathcal{C}\right\\}$ is C-superlinear and a supremum of
$\mathcal{C}$. Indeed, for all $(x,y)\in\mathbf{C}$ and all $Q\in\mathcal{C}$,

$\overline{Q}(x+y)\geq Q(x+y)\geq Q(x)+Q(y).$

This implies that $\overline{Q}$ is C-superlinear. Thus
$\mathcal{D}_{\mathrm{PH}}$ is strictly inductively ordered. Let
$A\mathrel{\mathop{\mathchar 58\relax}}Q\mapsto A_{Q}$ be defined as

$A_{Q}(x)=\inf\limits_{y}\left\\{H(x+y)-Q(y)\mathrel{\mathop{\mathchar
58\relax}}y\in\mathbf{C}(x)\ \textnormal{and}\ Q(y)>-\infty\right\\},$

for all $x\in X$ and all $Q\in\mathcal{D}_{\mathrm{PH}}$. Now fix $g,x\in X$
and $Q\in\mathcal{D}_{\mathrm{PH}}$. We also define,

$T_{g}(Q)(x)=\sup\limits_{h,\lambda}\left\\{Q(h)+\lambda
A_{Q}(g)\mathrel{\mathop{\mathchar 58\relax}}h+\lambda g\preceq x,\
h\in\mathbf{C}(x),\ \lambda\geq 0\right\\}.$

We then obtain the following result.

######  Lemma 2.

The following claims hold:

  1. (1)

For all $Q\in\mathcal{D}_{\mathrm{PH}}$, we have $A_{Q}\geq Q$ and $A_{Q}$ is
$\mathbf{C}$-sublinear.

  2. (2)

For all $Q\in\mathcal{D}_{\mathrm{PH}}$ and $g\in X$, we have $T_{g}(Q)$ is
$\mathbf{C}$-superlinear.

  3. (3)

For all $Q\in\mathcal{D}_{\mathrm{PH}}$, we have $Q\leq T_{g}(Q)\leq H$.

  4. (4)

For all $Q\in\mathcal{D}_{\mathrm{PH}}$ and $x\in X$, we have $T_{x}(Q)(x)\geq
A_{Q}(x)$.

###### Proof.

(1). Fix $Q\in\mathcal{D}_{\mathrm{PH}}$ and $x\in X$, then

$H(x+y)-Q(y)\geq Q(x+y)-Q(y)\geq Q(x)+Q(y)-Q(y)=Q(x),$

for all $y\in\mathbf{C}(x)$ with $Q(y)>-\infty$. Therefore, $A_{Q}\geq Q$. Now
suppose that $(x_{1},x_{2})\in\mathbf{C}$ and
$y_{1},y_{2}\in\mathbf{C}(x_{1})$ with $Q(y_{1})>-\infty$, $Q(y_{2})>-\infty$.
Notice that since $y_{1},y_{2}\in\mathbf{C}(x_{1})$ and $\mathbf{C}$ is
symmetric and transitive, we have
$x_{1}\mathbf{C}y_{1}\mathbf{C}y_{2}\mathbf{C}x_{2}$. By symmetry and property
(ii), it follows that

$\left((x_{1}+y_{1}),(x_{2}+y_{2})\right)\in\mathbf{C}.$

These observations imply that $Q(y_{1}+y_{2})\geq Q(y_{1})+Q(y_{2})$, and
hence

$\displaystyle H(x_{1}+x_{2}+y_{1}+y_{2})-Q(y_{1}+y_{2})$ $\displaystyle\leq
H(x_{1}+x_{2}+y_{1}+y_{2})-Q(y_{1})-Q(y_{2})$ $\displaystyle\leq
H(x_{1}+y_{1})-Q(y_{1})+H(x_{2}+y_{2})-Q(y_{2}).$

Therefore, $A_{Q}(x_{1}+x_{2})\leq A_{Q}(x_{1})+A_{Q}(x_{2})$. Consequently,
for all $Q\in\mathcal{D}_{\mathrm{PH}}$, $A_{Q}$ is $\mathbf{C}$-subadditive.
Positive homogeneity follows from the positive homogeneity of $Q$ and $H$, as
well as from the fact that $\mathbf{C}(x)$ is a cone.

(2). Fix $Q\in\mathcal{D}_{\mathrm{PH}}$, $g\in X$, and
$(x_{1},x_{2})\in\mathbf{C}$. Then $T_{g}(Q)(x_{1}+x_{2})\geq
T_{g}(Q)(x_{1})+T_{g}(Q)(x_{2})$. Indeed, notice that if
$h_{1}\in\mathbf{C}(x_{1})$, $h_{2}\in\mathbf{C}(x_{2})$,
$\lambda_{1},\lambda_{2}\geq 0$ satisfy

$h_{1}+\lambda_{1}g\preceq x_{1}\ \mathrm{and}\ h_{2}+\lambda_{2}g\preceq
x_{2},$

then $h_{1}+h_{2}+(\lambda_{1}+\lambda_{2})g\preceq x_{1}+x_{2}$. Since
$x_{1}\mathbf{C}x_{2}$, by symmetry, transitivity, and property (ii) we have
that $x_{1}\mathbf{C}x_{2}\mathbf{C}h_{1}\mathbf{C}h_{2}$ and

$(h_{1}+h_{2},\lambda_{1}+\lambda_{2})\in\left\\{(h,\lambda)\mathrel{\mathop{\mathchar
58\relax}}h+\lambda g\preceq x_{1}+x_{2},\ h\in\mathbf{C}(x_{1}+x_{2}),\
\lambda\geq 0\right\\}.$

This, together with the $\mathbf{C}$-superadditivity of $Q$, yields

$\displaystyle T_{g}(Q)(x_{1}+x_{2})$
$\displaystyle=\sup\limits_{h,\lambda}\left\\{Q(h)+\lambda
A_{Q}(g)\mathrel{\mathop{\mathchar 58\relax}}h+\lambda g\preceq x_{1}+x_{2},\
h\in\mathbf{C}(x_{1}+x_{2}),\ \lambda\geq 0\right\\}$
$\displaystyle\geq\sup\limits_{h_{1},h_{2},\lambda_{1},\lambda_{2}}\left\\{Q(h_{1}+h_{2})+(\lambda_{1}+\lambda_{2})A_{Q}(g)\mathrel{\mathop{\mathchar
58\relax}}\begin{array}[]{l}h_{1}+h_{2}+(\lambda_{1}+\lambda_{2})g\preceq
x_{1}+x_{2},\\\
h_{1}\in\mathbf{C}(x_{1}),h_{2}\in\mathbf{C}(x_{2}),\lambda_{1},\lambda_{2}\geq
0\end{array}\right\\}$ $\displaystyle\geq
Q(h_{1})+Q(h_{2})+\lambda_{1}A_{Q}(g)+\lambda_{2}A_{Q}(g),$

for all $h_{1}\in\mathbf{C}(x_{1})$, $h_{2}\in\mathbf{C}(x_{2})$,
$\lambda_{1},\lambda_{2}\geq 0$ with $h_{1}+\lambda_{1}g\preceq x_{1}\
\mathrm{and}\ h_{2}+\lambda_{2}g\preceq x_{2}.$ Thus,

$T_{g}(Q)(x_{1}+x_{2})\geq T_{g}(Q)(x_{1})+T_{g}(Q)(x_{2}).$

This proves that for all $Q\in\mathcal{D}_{\mathrm{PH}}$ and all $g\in X$, the
mapping $T_{g}(Q)$ is $\mathbf{C}$-superadditive. Positive homogeneity of each
$T_{g}(Q)$ follows from the positive homogeneity of $Q$ and $A_{Q}$ (see the
previous point) and the fact that $\mathbf{C}(x)$ is a convex cone.

(3). Fix $Q\in\mathcal{D}_{\mathrm{PH}}$. It is immediate to see that
$T_{g}(Q)\geq Q$ (take $\lambda=0$ and $h=x$, recalling that C is reflexive).
We show that $T_{g}(Q)\leq H$. First notice that, if $\lambda=0$, then for all
$x\in X$, by the monotonicity of $H$,

$\displaystyle\sup\limits_{h}\left\\{Q(h)+\lambda
A_{Q}(g)\mathrel{\mathop{\mathchar 58\relax}}h+\lambda g\preceq x,\
h\in\mathbf{C}(x)\right\\}=\sup\limits_{h}\left\\{Q(h)\mathrel{\mathop{\mathchar
58\relax}}h\preceq x,\ h\in\mathbf{C}(x)\right\\}$
$\displaystyle\leq\sup\limits_{h}\left\\{H(h)\mathrel{\mathop{\mathchar
58\relax}}h\preceq x,\ h\in\mathbf{C}(x)\right\\}\leq H(x).$

Therefore, we can focus on all $\lambda>0$. In particular, we have that for
all $x\in X$,

$\displaystyle\sup\limits_{h,\lambda}\left\\{Q(h)+\lambda
A_{Q}(g)\mathrel{\mathop{\mathchar 58\relax}}h+\lambda g\preceq x,\
h\in\mathbf{C}(x),\ \lambda>0\right\\}$
$\displaystyle=\sup\limits_{h,\lambda}\inf\limits_{y\in\mathbf{C}(x),Q(y)>-\infty}\left\\{Q(h)+\lambda
H(g+y)-\lambda Q(y)\mathrel{\mathop{\mathchar 58\relax}}h+\lambda g\preceq x,\
h\in\mathbf{C}(x),\ \lambda>0\right\\}$
$\displaystyle=\sup\limits_{h,\lambda}\inf\limits_{y\in\mathbf{C}(x),Q(y)>-\infty}\left\\{Q(h)+H(\lambda
g+\lambda y)-Q(\lambda y)\mathrel{\mathop{\mathchar 58\relax}}h+\lambda
g\preceq x,\ h\in\mathbf{C}(x),\ \lambda>0\right\\}$
$\displaystyle\leq\sup\limits_{h,\lambda}\left\\{H(\lambda
g+h)\mathrel{\mathop{\mathchar 58\relax}}h+\lambda g\preceq x,\
h\in\mathbf{C}(x),\ \lambda>0\right\\}$
$\displaystyle=\sup\limits_{h,\lambda}\left\\{H(\lambda g+\lambda
h)\mathrel{\mathop{\mathchar 58\relax}}\lambda h+\lambda g\preceq x,\
h\in\mathbf{C}(x),\ \lambda>0\right\\}$ $\displaystyle\leq H(x),$

where the last two steps follow from the fact that $\mathbf{C}(x)$ is a cone
and $H$ is monotone. Thus, connecting the two observations for $\lambda=0$ and
all $\lambda>0$, we have that

$T_{g}(Q)(x)=\sup\limits_{h,\lambda}\left\\{Q(h)+\lambda
A_{Q}(g)\mathrel{\mathop{\mathchar 58\relax}}h+\lambda g\preceq x,\
h\in\mathbf{C}(x),\ \lambda\geq 0\right\\}\leq H(x),$

for all $x\in X$, and hence $Q\leq T_{g}(Q)\leq H$.

(4). Let $x\in X$ and $Q\in\mathcal{D}_{\mathrm{PH}}$. Then, since $Q$ is
positively homogeneous, we have

$\displaystyle T_{x}(Q)(x)$
$\displaystyle=\sup\limits_{h,\lambda}\left\\{Q(h)+\lambda
A_{Q}(x)\mathrel{\mathop{\mathchar 58\relax}}h+\lambda x\preceq x,\
h\in\mathbf{C}(x),\ \lambda\geq 0\right\\}$
$\displaystyle\geq\sup\limits_{h,n}\left\\{Q(h)+\frac{n-1}{n}A_{Q}(x)\mathrel{\mathop{\mathchar
58\relax}}h\preceq\frac{1}{n}x,\ h\in\mathbf{C}(x)\right\\}$
$\displaystyle\geq\frac{1}{m}Q\left(x\right)+\frac{m-1}{m}A_{Q}(x)$
$\displaystyle\geq-\frac{1}{m}\left\lvert
Q\left(x\right)\right\rvert+\frac{m-1}{m}A_{Q}(x),$

for all $m\in\mathbb{N}$. Thus, letting $m\to\infty$, since all Dedekind
complete Riesz spaces are Archimedean444We recall that a Riesz space
$(W,\leq)$ is Archimedean if whenever $\mathbf{0}_{W}\leq nx\leq y$ for all
$n=1,2,\ldots$ and some $y\geq\mathbf{0}_{W}$, we have that
$x=\mathbf{0}_{W}$. (see Lemma 8.4 in Aliprantis and Border, (2006)), we have
$T_{x}(Q)(x)\geq A_{Q}(x)$ for all $x\in X$. ∎

This lemma highlights the fact that for all $g\in X$, $T_{g}(\cdot)$ is a
selfmap, as $T_{g}(Q)\in\mathcal{D}_{\mathrm{PH}}$ for all
$Q\in\mathcal{D}_{\mathrm{PH}}$. Now we are ready to provide the proof of
Theorem 2.

###### Proof of Theorem 2.

By Lemma 2, for all $g\in X$, $T_{g}\mathrel{\mathop{\mathchar
58\relax}}\mathcal{D}_{\mathrm{PH}}\to\mathcal{D}_{\mathrm{PH}}$ is
inflationary. Therefore $\left(T_{g}\right)_{g\in X}$ is a family of
inflationary functions. By Theorem 1 the family $\left(T_{g}\right)_{g\in X}$
has a common fixed point $Q^{*}\in\mathcal{D}_{\mathrm{PH}}$. Since for all
$g\in X$ and all $Q\in\mathcal{D}_{\mathrm{PH}}$ we have $T_{g}(Q)\leq H$, it
follows that for all $x\in X$ and $y\in\mathbf{C}(x)$ with $Q^{*}(y)>-\infty$,

$\displaystyle H(x+y)-Q^{*}(y)$ $\displaystyle\geq T_{g}(Q^{*})(x+y)-Q^{*}(y)$
$\displaystyle\geq
T_{g}(Q^{*})(x)+T_{g}(Q^{*})(y)-T_{g}(Q^{*})(y)=T_{g}(Q^{*})(x).$

Thus, $A_{Q^{*}}(x)\geq T_{g}(Q^{*})(x)=Q^{*}(x)$ for all $x,g\in X$.
Moreover, by Lemma 2-(4), we have
$T_{g}(Q^{*})(x)=Q^{*}(x)=T_{x}(Q^{*})(x)\geq A_{Q^{*}}(x)$, for all $x,g\in
X$. Therefore, $A_{Q^{*}}=Q^{*}$. Since $A_{Q^{*}}$ is $\mathbf{C}$-sublinear
and $Q^{*}$ is $\mathbf{C}$-superlinear we have that $Q^{*}$ is
$\mathbf{C}$-linear. To conclude, since $Q^{*}\in\mathcal{D}_{\mathrm{PH}}$,
the claim follows. ∎

####  3.2.1. Extension and Envelope Results

By Theorem 2, we derive an analogous version of the Hahn-Banach Extension
Theorem (see for example Theorem 1.25 in Aliprantis and Burkinshaw, (2006)).

######  Corollary 1.

Let $H\mathrel{\mathop{\mathchar
58\relax}}X\to\mathbb{V}\cup\left\\{-\infty\right\\}$ be
$\mathbf{C}$-sublinear and monotone, and $Y\subseteq X$ be a convex cone. If
$\ell\mathrel{\mathop{\mathchar
58\relax}}Y\to\mathbb{V}\cup\left\\{-\infty\right\\}$ is $\mathbf{C}$-linear
and satisfies $\ell\leq H|_{Y}$, then there exists a $\mathbf{C}$-linear map
$Q\mathrel{\mathop{\mathchar
58\relax}}X\to\mathbb{V}\cup\left\\{-\infty\right\\}$ such that $Q\leq H$ and
$\ell\leq Q|_{Y}$.

###### Proof.

Define $P\mathrel{\mathop{\mathchar
58\relax}}X\to\mathbb{V}\cup\left\\{-\infty\right\\}$ by

$P(x)=\begin{cases}\ell(x)&x\in Y,\\\ -\infty&x\notin Y,\end{cases}$

for all $x\in X$. Thus $P$ is $\mathbf{C}$-superadditive. Indeed, suppose that
$x\mathbf{C}y$. If $x,y\in Y$, then there is nothing to prove. If $x\notin Y$
or $y\notin Y$, then $P(x)+P(y)=-\infty\leq P(x+y)$. It is immediate to see
that $P$ is positively homogeneous and $P\leq H$. Therefore, by Theorem 2,
there exists a $\mathbf{C}$-linear map $Q\mathrel{\mathop{\mathchar
58\relax}}X\to\mathbb{V}\cup\left\\{-\infty\right\\}$ such that $P\leq Q\leq
H$. Therefore, $\ell=P|_{Y}\leq Q|_{Y}$. ∎

Next, as a direct application of Corollary 1, we provide a general envelope
representation result for $\mathbf{C}$-sublinear and monotone maps. For all
$x\in X$, denote by $C_{x}$ the convex cone generated by $x$, that is,
$C_{x}=\left\\{\lambda x\mathrel{\mathop{\mathchar
58\relax}}\lambda>0\right\\}$. Moreover, for all maps
$H\mathrel{\mathop{\mathchar
58\relax}}X\to\mathbb{V}\cup\left\\{-\infty\right\\}$, let

$D(H)=\left\\{Q\mathrel{\mathop{\mathchar
58\relax}}X\to\mathbb{V}\cup\left\\{-\infty\right\\}\mathrel{\mathop{\mathchar
58\relax}}Q\ \textnormal{is}\ \mathbf{C}\textnormal{-linear}\ \textnormal{and\
}Q\leq H\right\\}.$

######  Corollary 2.

If $H\mathrel{\mathop{\mathchar
58\relax}}X\to\mathbb{V}\cup\left\\{-\infty\right\\}$ is
$\mathbf{C}$-sublinear and monotone, then

$H(x)=\sup\limits_{Q\in D(H)}Q(x),\ \hbox{for\ all}\ x\in X.$

###### Proof.

Let $x\in X$ and define $\ell$ over $C_{x}$ as $\ell(\lambda x)=\lambda H(x)$,
for all $\lambda x\in C_{x}$ with $\lambda>0$. Notice that $\ell$ is
$\mathbf{C}$-linear and positively homogeneous. Moreover, by positive
homogeneity we have that $\ell(\lambda x)=\lambda H(x)=H(\lambda x)$ for all
$\lambda x\in C_{x}$. Thus, by Corollary 1, there exists a $\mathbf{C}$-linear
map $Q\mathrel{\mathop{\mathchar
58\relax}}X\to\mathbb{V}\cup\left\\{-\infty\right\\}$ such that $Q\leq H$ and
$\ell\leq Q|_{C_{x}}$. Moreover,

$H(x)=\ell(x)\leq Q(x)\leq H(x),$

and so $Q(x)=H(x)$. Therefore, for all $x\in X$, there exists a
$\mathbf{C}$-linear map $Q_{x}\mathrel{\mathop{\mathchar
58\relax}}X\to\mathbb{V}\cup\left\\{-\infty\right\\}$ such that
$Q_{x}(x)=H(x)$. This implies that

$H(x)=\sup\limits_{Q\in D(H)}Q(x),$

for all $x\in X$. ∎

######  Remark 1.

Suppose that $\mathbb{V}=\mathbb{R}$ and $V$ is a topological vector space.
For any map $G\mathrel{\mathop{\mathchar
58\relax}}V\to\left[-\infty,\infty\right)$, we define the (effective) domain
of $G$ by $\textnormal{dom}G=\left\\{x\in V\mathrel{\mathop{\mathchar
58\relax}}G(x)>-\infty\right\\}$. We say that $G$ is proper if
$\textnormal{dom}G$ is nonempty. It is important to note that Corollary 2
could also be proved by defining each $\ell$ on the vector space generated by
$x\in V$, i.e., $\textnormal{span}\left\\{x\right\\}$, as $\ell(\alpha
x)=\alpha H(x)$ for all $\alpha\in\mathbb{R}$, and then extending each $\ell$
to be equal to $-\infty$ everywhere else. In such a case, we would have that
each extension $Q|_{\textnormal{dom}G}$ is continuous. Therefore, in this
setting, Corollary 2 could be restated, adding a further restriction on
$D(H)$, namely the continuity of its elements when restricted to their
domains. This approach is similar to the one adopted by Roth, (2000) (e.g.,
see Corollary 3.3).

####  3.2.2. Examples

######  Example 1.

Clearly $\mathbf{C}=X\times X$ is a convex-conic symmetric preorder. Under
this convex-conic symmetric preorder $\mathbf{C}$-(super/sub)linearity
corresponds to the classic (super/sub)linearity.

######  Example 2 (Positive homogeneity).

Fix $x\in X$ and define $\mathbf{D}_{x}=\left\\{(\lambda x,\beta
x)\mathrel{\mathop{\mathchar 58\relax}}\lambda,\beta>0\right\\}$. Then
$\mathbf{D}_{x}$ is a convex-conic symmetric preorder. More importantly,
define $\mathbf{D}=\bigcup_{x\in X}\mathbf{D}_{x}$. We now verify that
$\mathbf{D}$ is also a convex-conic symmetric preorder. Since each
$\mathbf{D}_{x}\subseteq\mathbf{D}$, we have that $\mathbf{D}$ is symmetric,
reflexive, and satisfies property (i). Fix arbitrarily $x,y,z\in X$ and
suppose that $x\mathbf{D}y\mathbf{D}z$. Then there exist
$\alpha,\beta,\lambda,\gamma>0$ and $v,w\in X$ such that,

$(x,y)=(\alpha v,\beta v)\ \textnormal{and}\ (y,z)=(\lambda w,\gamma w).$

Therefore, $x=\alpha v=\frac{\alpha}{\beta}y=\lambda\frac{\alpha}{\beta}w$,
and hence $x\mathbf{D}z$. Consequently, $\mathbf{D}$ is transitive. Moreover,

$(x+y,z)=\left(\lambda\left(\frac{\alpha}{\beta}+1\right)w,\gamma w\right),$

and thus $\mathbf{D}$ satisfies property (ii). Note that any positively
homogeneous map $F\mathrel{\mathop{\mathchar 58\relax}}X\to\mathbb{R}$ is
$\mathbf{D}$-linear.

######  Example 3 (Equivalent measures).

Consider a measurable space $(\Omega,\mathcal{F})$ and denote by $X$ the
convex cone of countably additive measures $\mu\mathrel{\mathop{\mathchar
58\relax}}\mathcal{F}\to\left[0,\infty\right]$. For all measures $\mu\in X$ we
define $N_{\mu}=\left\\{A\in\mathcal{F}\mathrel{\mathop{\mathchar
58\relax}}\mu(A)=0\right\\}$, i.e., the collection of $\mu$-null elements of
$\mathcal{F}$. Two measures $\mu,\nu\in X$ are said to be equivalent, denoted
by $\mu\sim\nu$, if $N_{\mu}=N_{\nu}$. It can be verified that $\sim$ is a
convex-conic symmetric preorder. Indeed, letting $\lambda>0$ and $\mu\in X$,
we have $N_{\lambda\mu}=N_{\mu}$. If $\nu,\mu,\eta\in X$, $\nu\sim\mu$, and
$\mu\sim\eta$, then $N_{\mu}=N_{\nu}=N_{\eta}$. Thus $\sim$ is transitive.
Symmetry is also straightforward. To conclude, if $\nu\sim\mu\sim\eta$ for
$\nu,\mu,\eta\in X$, we have that $\nu(A)+\mu(A)=0$ for all $A\in N_{\eta}$,
and the converse holds as well. This yields $N_{\nu+\mu}=N_{\eta}$, and hence
$\sim$ is a convex-conic symmetric preorder.

######  Example 4 (Strict comonotonicity).

Let $(\Omega,\mathcal{F})$ be a measurable space, and denote by
$L^{0}(\Omega,\mathcal{F})$ the space of $\mathcal{F}$-measurable real-valued
functions. For all $x\in L^{0}(\Omega,\mathcal{F})$, denote by $C_{x}$ the
convex cone generated by $x$. We say that $x,y\in L^{0}(\Omega,\mathcal{F})$
are strictly comonotonic if either (1) $x\in C_{y}$; or (2) for all
$\omega_{1}\neq\omega_{2}$ in $\Omega$,

$\left[x(\omega_{1})-x(\omega_{2})\right]\left[y(\omega_{1})-y(\omega_{2})\right]>0.$

The strict comonotonicity relation, denoted as
$\upharpoonleft\upharpoonright$, requires that $x\upharpoonleft\upharpoonright
y$ if and only if $x,y\in L^{0}(\Omega,\mathcal{F})$ are strictly comonotonic.
We now verify that $\upharpoonleft\upharpoonright$ is a convex-conic symmetric
order. Clearly, $(\lambda x)\upharpoonleft\upharpoonright x$, for all
$\lambda>0$, as $\lambda x\in C_{x}$, for all $x\in
L^{0}(\Omega,\mathcal{F})$. Symmetry is immediate, since if $x\in C_{y}$, then
$y\in C_{x}$. To show transitivity, fix $\omega_{1}\neq\omega_{2}$, and
suppose that $x\upharpoonleft\upharpoonright y$ and
$y\upharpoonleft\upharpoonright z$. Then, we have three cases:

  1. (1)

If $x\in C_{y}$ and $y\in C_{z}$, then $x\in C_{z}$. Thus,
$x\upharpoonleft\upharpoonright z$.

  2. (2)

If $x\in C_{y}$ and $y\notin C_{z}$, then

$\left[y(\omega_{1})-y(\omega_{2})\right]\left[z(\omega_{1})-z(\omega_{2})\right]>0\
\textnormal{ and }\ x=\lambda y,$

for some $\lambda>0$. This implies that

$\left[x(\omega_{1})-x(\omega_{2})\right]\left[z(\omega_{1})-z(\omega_{2})\right]>0.$

Thus, $x\upharpoonleft\upharpoonright z$

  3. (3)

If $x\notin C_{y}$ and $y\notin C_{z}$, then

$\left[x(\omega_{1})-x(\omega_{2})\right]\left[y(\omega_{1})-y(\omega_{2})\right]>0\
\textnormal{ and }\
\left[y(\omega_{1})-y(\omega_{2})\right]\left[z(\omega_{1})-z(\omega_{2})\right]>0.$

Therefore, if $x(\omega_{1})-x(\omega_{2})>0$, then
$y(\omega_{1})-y(\omega_{2})>0$, and hence $z(\omega_{1})-z(\omega_{2})>0$.
The same reasoning (with inverted signs) applies to the case where
$x(\omega_{1})-x(\omega_{2})<0$. Thus $x\upharpoonleft\upharpoonright z$.

To conclude, fix $\omega_{1}\neq\omega_{2}$ and suppose that
$x\upharpoonleft\upharpoonright y\upharpoonleft\upharpoonright z$. Then, we
have three cases:

  1. (1)

If $x\in C_{y}$ and $y\in C_{z}$, then $x+y\in C_{z}$. Thus,
$(x+y)\upharpoonleft\upharpoonright z$.

  2. (2)

If $x\in C_{y}$ and $y\notin C_{z}$, then

$\left[y(\omega_{1})-y(\omega_{2})\right]\left[z(\omega_{1})-z(\omega_{2})\right]>0\
\textnormal{ and }\ x=\lambda y,$

for some $\lambda>0$. This implies that

$\left[x(\omega_{1})+y(\omega_{1})-x(\omega_{2})-y(\omega_{2})\right]\left[z(\omega_{1})-z(\omega_{2})\right]>0.$

Thus, $(x+y)\upharpoonleft\upharpoonright z$.

  3. (3)

If $x\notin C_{y}$ and $y\notin C_{z}$, then

$\displaystyle\left[x(\omega_{1})+y(\omega_{1})-x(\omega_{2})-y(\omega_{2})\right]\left[z(\omega_{1})-z(\omega_{2})\right]$
$\displaystyle=\left[x(\omega_{1})-x(\omega_{2})\right]\left[z(\omega_{1})-z(\omega_{2})\right]+\left[y(\omega_{1})-y(\omega_{2})\right]\left[z(\omega_{1})-z(\omega_{2})\right]>0.$

Thus, $(x+y)\upharpoonleft\upharpoonright z$.

Therefore, $\upharpoonleft\upharpoonright$ is a convex-conic symmetric
preorder.

Comonotonicity is an important property in in decision theory, risk
measurement, and the theory of risk sharing. A different example of a convex-
conic preorder concerns vectors that are affinely related.

######  Example 5 (Affinity).

Let $X$ be a vector space, and fix $e\in X$. We say that $x$ and $y$ in $X$
are $e$-affinely related if there exist $\alpha,\beta\neq 0$ such that
$x=\alpha y+\beta e$. In particular, we define the binary relation
$\mathbf{Aff}$ as follows

$x\mathbf{Aff}y\,\Longleftrightarrow\,\exists\,\alpha\neq 0,\exists\
\beta\in\mathbb{R}\mathrel{\mathop{\mathchar 58\relax}}x=\alpha y+\beta e.$

Even if straightforward, we provide the steps proving that $\mathbf{Aff}$ is a
convex-conic symmetric preorder. Reflexivity is immediate, as $x=x$. Passing
to symmetry, if $x\mathbf{Aff}y$, then there exist $\alpha\neq 0$ and
$\beta\in\mathbb{R}$ such that $x=\alpha y+\beta e$, and hence
$y=\frac{1}{\alpha}-\beta e$, implying that $y\mathbf{Aff}x$. Suppose that
$x\mathbf{Aff}y$ and $y\mathbf{Aff}z$. There exist $\alpha_{1},\alpha_{2}\neq
0$ and $\beta_{1},\beta_{2}\in\mathbb{R}$ such that $x=\alpha_{1}y+\beta_{1}e$
and $y=\alpha_{2}z+\beta_{2}e$. Then

$x=\alpha_{1}(\alpha_{2}z+\beta_{2}e)+\beta_{1}e=\alpha_{1}\alpha_{2}z+(\alpha_{1}\beta_{2}+\beta_{1})e,$

proving that $x\mathbf{Aff}z$, and so $\mathbf{Aff}$ is transitive. Now fix
$x\in X$, then clearly $(\lambda x)\mathbf{Aff}x$ for all $\lambda>0$. In
conclusion, if $x\mathbf{Aff}y\mathbf{Aff}z$, then there exist
$\alpha_{1},\alpha_{2}\neq 0$ and $\beta_{1},\beta_{2}\in\mathbb{R}$ such that
$x=\alpha_{1}y+\beta_{1}e$ and $y=\alpha_{2}z+\beta_{2}e$. Then

$\displaystyle x+y$
$\displaystyle=\alpha_{1}y+\beta_{1}e+\alpha_{2}z+\beta_{2}e=\alpha_{1}\alpha_{2}z+\alpha_{1}\beta_{2}e+\beta_{1}e+\alpha_{2}z+\beta_{2}e$
$\displaystyle=\left(\alpha_{1}\alpha_{2}+\alpha_{2}\right)z+\left(\alpha_{1}\beta_{2}+\beta_{1}+\beta_{2}\right)e,$

implying that $(x+y)\mathbf{Aff}z$. Thus $\mathbf{Aff}$ is a convex-conic
symmetric preorder.

To conclude this section we provide some examples of binary relations that are
not convex-conic symmetric preorders. Clearly any irreflexive, asymmetric, or
nontransitive binary relation would work. For instance, probabilistic
independence and orthogonality in the context of inner product spaces are
relevant examples.

######  Example 6 (Inner product spaces).

Consider the Hilbert space
$L^{2}\left(\left[0,1\right],\mathcal{B}\left[0,1\right],\mathrm{Leb}\right)$,
where $\mathcal{B}\left[0,1\right]$ denotes the Borel sigma algebra and
$\mathrm{Leb}$ the Lebesgue measure. We define the relation $\mathbf{Corr}$ by

$f\textnormal{{Corr}}g\Longleftrightarrow\int_{\left[0,1\right]}fg\,\textnormal{d}\mathrm{Leb}\geq
0.$

It is easy to see that $\mathbf{Corr}$ is reflexive and symmetric, and it
satisfies conditions (i) and (ii) of Definition 1, but it fails transitivity.
Indeed, consider the functions

$f=\mathbf{1}_{\left[0,1\right]},\ g=\mathbf{1}_{\left[0,\frac{1}{2}\right]},\
h=-\mathbf{1}_{\left[\frac{1}{2},1\right]}.$

Then,

$f\mathbf{Corr}g\ \textnormal{and}\ g\mathbf{Corr}h,\ \textnormal{but}\
\int_{\left[0,1\right]}fh\,\textnormal{d}\mathrm{Leb}=-\frac{1}{2}<0.$

The next example provides a symmetric preorder that fails property (ii) of
Definition 1.

######  Example 7.

Define the function $\varphi\mathrel{\mathop{\mathchar
58\relax}}\mathbb{R}^{2}\to\mathbb{R}$ by

$\varphi(x,y)=\begin{cases}-1&(x,y)\in\left(\left(\mathbb{R}\setminus\left\\{0\right\\}\right)\times\left\\{0\right\\}\right)\cup\left(\left\\{0\right\\}\times\left(\mathbb{R}\setminus\left\\{0\right\\}\right)\right)\\\
0&\textnormal{otherwise}.\end{cases}$

Consider the binary relation $x{{\bm{\varphi}}}y$ if and only if
$\varphi(x,y)\geq 0$. By definition, $\varphi(x,x)=0$ for all
$x\in\mathbb{R}$, and so ${{\bm{\varphi}}}$ is reflexive. Moreover,
$\varphi(x,y)=\varphi(y,x)$ for all $x,y\in\mathbb{R}$, proving that
${{\bm{\varphi}}}$ is symmetric. Additionally, if $x,y,z\in\mathbb{R}$ with
$\varphi(x,y)\geq 0$ and $\varphi(y,z)\geq 0$, then it must be the case that
either $x,y,z=0$ or $x,y,z\neq 0$, and hence, $\varphi(x,z)=0$. Therefore
${{\bm{\varphi}}}$ is transitive. However, letting $x,z=1,y=-1$, we obtain

$\varphi(x,y)\geq 0,\ \varphi(y,z)\geq 0,\ \textnormal{and}\
\varphi(x+y,z)=\varphi(0,z)<0.$

Thus ${{\bm{\varphi}}}$ is a symmetric preorder that is not a convex-conic
symmetric preorder.

##  4\. Relaxing positive homogeneity

In this section, we weaken the definition of a convex-conic symmetric preorder
by removing property (i) (i.e., positive homogeneity). In particular, we
define a new preorder called summand symmetric preorder. Fix a vector space
$V$ and a subset $X\subseteq V$ such that $X+X\subseteq X$.

######  Definition 2.

A binary relation $\mathbf{S}\subseteq X\times X$ is a summand symmetric
preorder if it is reflexive, transitive, and symmetric, and it satisfies the
following property:

$x\mathbf{S}y\mathbf{S}z\ \textnormal{implies}\ (x+y)\mathbf{S}z,\
\textnormal{for\ all}\ x,y,z\in X.$ (1)

Note that in this case $\mathbf{S}(x)=\left\\{y\in X\mathrel{\mathop{\mathchar
58\relax}}y\mathbf{S}x\right\\}$ is closed with respect to the summation of
its elements. Clearly, if $X$ is a convex cone and $\mathbf{C}$ is a convex-
conic symmetric preorder on $X$, then $\mathbf{C}$ is also a summand symmetric
preorder.

Let $\left(\mathbb{V},\leq\right)$ be a Dedekind-complete Riesz space. We say
that a map $F\mathrel{\mathop{\mathchar
58\relax}}X\to\mathbb{V}\cup\left\\{-\infty\right\\}$ is
$\mathbf{S}$-sublinear if

  1. (i)

$F$ is positively integer homogeneous, i.e., $F(nx)=nF(x)$, for all
$n\in\mathbb{N}$ and all $x\in X$.

  2. (ii)

$F$ is $\mathbf{S}$-subadditive, i.e., $F(x+y)\leq F(x)+F(y)$, for all $x,y\in
X$ with $x\mathbf{S}y$.

The definitions of $\mathbf{S}$-superlinearity and $\mathbf{S}$-linearity are
analogous. We endow $V$ with a partial order $\preceq$ such that $(V,\preceq)$
is a partially ordered vector space. We say that a map
$F\mathrel{\mathop{\mathchar
58\relax}}X\to\mathbb{V}\cup\left\\{-\infty\right\\}$ is monotone if $x\preceq
y$ implies $F(x)\leq F(y)$, for all $x,y\in X$. We adopt the convention that
$0\cdot(-\infty)=\mathbf{0}_{\mathbb{V}}$. Using the exact same steps as in
the proof of Theorem 2, slightly changing the auxiliary maps, we can retrieve
the following version of Theorem 2 for $\mathbf{S}$-(super/sub)linear
functionals.

######  Theorem 3.

Suppose that $X$ is a convex cone, $P\mathrel{\mathop{\mathchar
58\relax}}X\to\mathbb{V}\cup\left\\{-\infty\right\\}$ is
$\mathbf{S}$-superlinear, and $H\mathrel{\mathop{\mathchar
58\relax}}X\to\mathbb{V}\cup\left\\{-\infty\right\\}$ is
$\mathbf{S}$-sublinear and monotone. If $P\leq H$, then there exists an
$\mathbf{S}$-linear map $Q\mathrel{\mathop{\mathchar
58\relax}}X\to\mathbb{V}\cup\left\\{-\infty\right\\}$ such that $P\leq Q\leq
H$.

We omit the proof of this result as it is almost identical to that of Theorem
2. We simply provide the form of one of the auxiliary maps we used in the
previous proofs. In particular, let

$\mathcal{D}_{\mathrm{PH}}=\left\\{Q\mathrel{\mathop{\mathchar
58\relax}}X\to\mathbb{V}\cup\left\\{-\infty\right\\}\mathrel{\mathop{\mathchar
58\relax}}Q\ \textnormal{is}\ \mathbf{S}\textnormal{-superlinear}\ \hbox{and}\
P\leq Q\leq H\right\\}.$

Now, fix $g,x\in X$, and $Q\in\mathcal{D}_{\mathrm{PH}}$. We also define,

$T_{g}(Q)(x)=\sup\limits_{h,n}\left\\{Q(h)+nA_{Q}(g)\mathrel{\mathop{\mathchar
58\relax}}h+ng\preceq x,\ h\in\mathbf{S}(x),\
n\in\mathbb{N}\cup\left\\{0\right\\}\right\\}.$

The auxiliary function $A_{Q}$ is defined as in Section 3.2, using
$\mathbf{S}$ instead of $\mathbf{C}$.

The analogous of Lemma 2 also holds in this setting. Before providing its
proof, we require the following simple result.

######  Lemma 3.

Suppose that $X$ is a convex cone and $\mathbf{S}$ a summand symmetric
preorder on $X$. Then for all $x,y\in X$,

$x\mathbf{S}y\implies\frac{x}{n}\mathbf{S}y\ \hbox{for\ all}\ n\in\mathbb{N}.$

###### Proof.

Fix $x,y\in X$ with $x\mathbf{S}y$, and choose $n\in\mathbb{N}$ arbitrarily.
Since $\mathbf{S}$ is reflexive, it follows that $x/n\mathbf{S}x/n$. Hence, by
property (1), we obtain

$x=\left(\sum_{k=1}^{n}\frac{x}{n}\right)\mathbf{S}\frac{x}{n}.$

Thus, by transitivity and symmetry of $\mathbf{S}$, we have $x/n\mathbf{S}y$.
∎

######  Lemma 4.

The following claims hold:

  1. (1)

For all $Q\in\mathcal{D}_{\mathrm{PH}}$, we have $A_{Q}\geq Q$ and $A_{Q}$ is
$\mathbf{S}$-sublinear.

  2. (2)

For all $Q\in\mathcal{D}_{\mathrm{PH}}$ and $g\in X$, we have $T_{g}(Q)$ is
$\mathbf{S}$-superlinear.

  3. (3)

For all $Q\in\mathcal{D}_{\mathrm{PH}}$, we have $Q\leq T_{g}(Q)\leq H$.

  4. (4)

For all $Q\in\mathcal{D}_{\mathrm{PH}}$ and all $x\in X$, we have
$T_{x}(Q)(x)\geq A_{Q}(x)$.

###### Proof.

We provide a proof only for points (3) and (4). For the remaining points, the
proofs are identical to those of Lemma 2.

(3). It is immediate to see that $T_{g}(Q)\geq Q$, for all
$Q\in\mathcal{D}_{\mathrm{PH}}$ (take $n=0$ and $h=x$). We show that
$T_{g}(Q)\leq H$. In particular, we have that for all $x\in X$,

$\displaystyle T_{g}(Q)(x)$
$\displaystyle=\sup\limits_{h,n}\left\\{Q(h)+nA_{Q}(g)\mathrel{\mathop{\mathchar
58\relax}}h+ng\preceq x,\ h\in\mathbf{S}(x),\
n\in\mathbb{N}\cup\left\\{0\right\\}\right\\}$
$\displaystyle=\sup\limits_{h,n}\inf\limits_{y\in\mathbf{S}(x),Q(y)>-\infty}\left\\{Q(h)+nH(g+y)-nQ(y)\mathrel{\mathop{\mathchar
58\relax}}h+ng\preceq x,\ h\in\mathbf{S}(x),\
n\in\mathbb{N}\cup\left\\{0\right\\}\right\\}$
$\displaystyle=\sup\limits_{h,n}\inf\limits_{y\in\mathbf{S}(x),Q(y)>-\infty}\left\\{Q(h)+H(ng+ny)-Q(ny)\mathrel{\mathop{\mathchar
58\relax}}h+ng\preceq x,\ h\in\mathbf{S}(x),\
n\in\mathbb{N}\cup\left\\{0\right\\}\right\\}$
$\displaystyle\leq\sup\limits_{h,n}\left\\{H(ng+h)\mathrel{\mathop{\mathchar
58\relax}}h+ng\preceq x,\ h\in\mathbf{S}(x),\
n\in\mathbb{N}\cup\left\\{0\right\\}\right\\}$
$\displaystyle=\sup\limits_{h,n}\left\\{H(ng+nh)\mathrel{\mathop{\mathchar
58\relax}}nh+ng\preceq x,\ h\in\mathbf{S}(x),\
n\in\mathbb{N}\cup\left\\{0\right\\}\right\\}$ $\displaystyle\leq H(x),$

where the fourth inequality follows from setting $y=\frac{h}{n}$ (which
belongs to $\mathbf{S}(x)$ by Lemma 3), while the last two steps follow from
the fact that $\mathbf{S}(x)$ is closed with respect to addition and $H$ is
monotone. Thus, $Q\leq T_{g}(Q)\leq H$.

(4). Let $x\in X$, $Q\in\mathcal{D}_{\mathrm{PH}}$, and $m\in\mathbb{N}$.
Then, since $Q$ and $T_{x}(Q)$ are integer positively homogeneous, we have

$\displaystyle T_{x}(Q)(x)$ $\displaystyle=\frac{T_{x}(Q)((m+1)x)}{m+1}$
$\displaystyle=\frac{1}{m+1}\sup\limits_{h,n}\left\\{Q(h)+nA_{Q}(x)\mathrel{\mathop{\mathchar
58\relax}}h+nx\preceq(m+1)x,\ h\in\mathbf{S}(x),\
n\in\mathbb{N}\cup\left\\{0\right\\}\right\\}$
$\displaystyle\geq\frac{1}{m+1}\sup\limits_{h}\left\\{Q(h)+mA_{Q}(x)\mathrel{\mathop{\mathchar
58\relax}}h\preceq x,\ h\in\mathbf{S}(x)\right\\}$
$\displaystyle\geq\frac{1}{m+1}Q\left(x\right)+\frac{m}{m+1}A_{Q}(x)$
$\displaystyle\geq-\frac{1}{m+1}\left\lvert
Q\left(x\right)\right\rvert+\frac{m}{m+1}A_{Q}(x).$

Thus, letting $m\to\infty$, since all Dedekind complete Riesz spaces are
Archimedean (see Lemma 8.4 in Aliprantis and Border, (2006)), we have
$T_{x}(Q)(x)\geq A_{Q}(x)$, for all $x\in X$. ∎

The proof of Theorem 3 is now totally analogous to that of Theorem 2, and it
is omitted.

##  5\. Applications: Comonotonic Subadditivity

In this section, we apply the previous results to the case of comonotonicity
on a specific measurable space. Suppose that $\Omega=\left[0,1\right]$ and
$\mathcal{F}=\mathcal{B}\left[0,1\right]$, the Borel sigma algebra on the unit
interval. Using our results and observations in Example 4, we obtain the
following.

######  Proposition 1.

Suppose that $H\mathrel{\mathop{\mathchar
58\relax}}B(\Omega,\mathcal{F})\to\left[-\infty,\infty\right)$ is
$\lVert\cdot\rVert_{\infty}$-continuous when restricted to its domain,
monotone, positively homogeneous, and strictly comonotonic subadditive. Then

$H(x)=\sup\limits_{Q\in D(H)}Q(x),\ \hbox{for\ all}\ x\in
B(\Omega,\mathcal{F}),$ (2)

where $D(H)$ is a set of maps from $B(\Omega,\mathcal{F})$ to
$\left[-\infty,\infty\right)$ that are comonotonic additive and
$\lVert\cdot\rVert_{\infty}$-continuous when restricted to their domains.
Moreover, $H$ is comonotonic subadditive on its domain.

###### Proof.

By Corollary 2, Remark 1, and Example 4, it follows that (2) holds, where
$D(H)$ is a convex set of functionals from $B(\Omega,\mathcal{F})$ to
$\left[-\infty,\infty\right)$ that are strictly comonotonic additive and
$\lVert\cdot\rVert_{\infty}$-continuous when restricted to their domains. By
Lemma 7 in the Appendix, all elements of $D(H)$ are comonotonic additive on
their domains. Note that Lemma 7 also implies that $H$ is comonotonic
subadditive on its domain. ∎

Comonotonic additive functionals play a central role in the theory of
decision-making under ambiguity. Their study was pioneered by Schmeidler,
(1989), who also provided a representation of such functionals in terms of
Choquet integrals (Schmeidler, (1986)). Such functionals are also relevant for
their connection to the theory of risk measurement (e.g., see Denuit et al.,
(2005) or Föllmer and Schied, (2016)). As for comonotonic subadditivity, less
attention has been devoted to this property. Song and Yan, (2006) provided,
along with further properties, a full characterization of comonotonic
subadditive functionals as envelopes of Choquet integrals, and Song and Yan,
(2009) provided some applications thereof.

######  Remark 2.

Proposition 1 hints towards the possibility of representing continuous,
monotone, positively homogeneous, and strictly comonotonic subadditive
functionals as suprema of signed Choquet integrals. Indeed, as shown by Wang
et al., (2020), comonotonic additive and continuous functionals can be
represented by Choquet integrals with respect to signed capacities.

##  6\. Concluding Remarks and An Open Question

Building upon the work of Amarante, (2019) and Fuchssteiner and Wright,
(1977), we provided a nonlinear version of the classical Sandwich Theorem. We
used this result to retrieve an extension result and an envelope
representation result. Our examples show that the type of nonlinearities that
we introduce include some important cases that have been devoted considerable
attention in decision theory and mathematical finance. Our approach highlights
the possibility of retrieving Hahn-Banach-type extension results that are
widely applied in functional analysis, theoretical economics, and mathematical
finance. We conclude with an (informal) question and two conjectures:

  1. (O1)

To what extent do our results depend on the fact that our maps can take
$-\infty$ as a value? More formally, is it possible to prove the following
reformulations of Corollaries 1 and 2?

######  Conjecture 1.

Let $H\mathrel{\mathop{\mathchar 58\relax}}X\to\mathbb{V}$ be
$\mathbf{C}$-sublinear and monotone, and let $Y\subseteq X$ be a convex cone.
If $\ell\mathrel{\mathop{\mathchar 58\relax}}Y\to\mathbb{V}$ is
$\mathbf{C}$-linear and satisfies $\ell\leq H|_{Y}$, then there exists a
$\mathbf{C}$-linear $Q\mathrel{\mathop{\mathchar 58\relax}}X\to\mathbb{V}$
such that $Q\leq H$ and $\ell\leq Q|_{Y}$.

######  Conjecture 2.

If $H\mathrel{\mathop{\mathchar 58\relax}}X\to\mathbb{V}$ is
$\mathbf{C}$-sublinear and monotone, then

$H(x)=\sup\limits_{Q\in\tilde{D}(H)}Q(x),\ \hbox{for\ all}\ x\in X,$

where

$\tilde{D}(H)=\Big{\\{}Q\mathrel{\mathop{\mathchar 58\relax}}X\to\mathbb{V}\
\big{|}\ Q\ \textnormal{is}\ \mathbf{C}\textnormal{-linear}\ \textnormal{and\
}Q\leq H\Big{\\}}.$

##  Appendix A Additional Results for Section 5

In this section we report some auxiliary results that we applied in Section 5.
In particular, we provide a partial answer to the following

######  Question 1.

If $x,y\in B(\Omega,\mathcal{F})$ are comonotonic, then there exist two
sequences $(x_{n})_{n\in\mathbb{N}},(y_{n})_{n\in\mathbb{N}}$ in
$B(\Omega,\mathcal{F})$ such that:

  1. (1)

$x_{n}\to x$ and $y_{n}\to y$;

  2. (2)

for all $n\in\mathbb{N}$, $x_{n}$ and $y_{n}$ are strictly comonotonic.

It is immediate to see that this does not hold in general measurable spaces.
Indeed, take any $\Omega\neq\emptyset$ and let
$\mathcal{F}=\left\\{\emptyset,\Omega\right\\}$. In this measurable space
$(\Omega,\mathcal{F})$, a function is $\mathcal{F}$-measurable if and only if
it is constant, and therefore there is no injective measurable function. This
observation highlights the fact that Question 1 may admit a positive answer
only if we focus on measurable spaces with a sufficiently sparse sigma-
algebra, where this sparsity depends also on the cardinality of $\Omega$.
Providing a full answer to this question is out of the scope of this paper.
Hence, we focus on a special case. However, we first need some auxiliary
lemmas.

######  Lemma 5.

Let $x,y\in B(\Omega,\mathcal{F})$. The following are equivalent

  1. (i)

$x$ and $y$ are strictly comonotonic with $x\notin C_{y}$.

  2. (ii)

there exist two increasing functions $h,g\mathrel{\mathop{\mathchar
58\relax}}\mathbb{R}\to\mathbb{R}$ and an injective $z\in
B(\Omega,\mathcal{F})$ such that $x=h(z)$, $y=g(z)$, and $h,g$ are injective
over $z(\Omega)$.

###### Proof.

$(i)\Rightarrow(ii)$. Since $x,y$ are comonotonic, there exist increasing
functions $h,g\mathrel{\mathop{\mathchar 58\relax}}\mathbb{R}\to\mathbb{R}$
and $z\in B(\Omega,\mathcal{F})$ such that $x=h(z)$ and $y=g(z)$ (see e.g.,
Denuit et al., (2023), Theorem 2.7). Suppose that $z$ is not injective. Then,
there exist $\omega_{1},\omega_{2}\in\Omega$ such that
$\omega_{1}\neq\omega_{2}$ and $z(\omega_{1})=z(\omega_{2})$. Thus,

$\left[x(\omega_{1})-x(\omega_{2})\right]\left[y(\omega_{1})-y(\omega_{2})\right]=\left[h(z(\omega_{1}))-h(z(\omega_{2}))\right]\left[y(\omega_{1})-y(\omega_{2})\right]=0,$

contradicting the strict comonotonicity of $x,y$. Therefore, $z$ must be
injective. Now suppose that $h$ is not injective over $z(\Omega)$. Then there
exist $\omega_{1},\omega_{2}\in\Omega$ such that $\omega_{1}\neq\omega_{2}$,
$z(\omega_{1})\neq z(\omega_{2})$, and $h(z(\omega_{1}))=h(z(\omega_{2}))$.
Since $z$ is injective, we have that $\omega_{1}\neq\omega_{2}$ and

$\left[x(\omega_{1})-x(\omega_{2})\right]\left[y(\omega_{1})-y(\omega_{2})\right]=\left[h(z(\omega_{1}))-h(z(\omega_{2}))\right]\left[y(\omega_{1})-y(\omega_{2})\right]=0,$

contradicting the strict comonotonicity of $x,y$. Interchanging $h$ with $g$
and $x$ with $y$, the same conclusion holds for $g$ as well.

$(ii)\Rightarrow(i)$. If $\omega_{1}\neq\omega_{2}$ and
$h(z(\omega_{1}))>h(z(\omega_{2}))$, then we must have that
$z(\omega_{1})>z(\omega_{2})$ and hence $g(z(\omega_{1}))>g(z(\omega_{2}))$.
This proves the claim. ∎

######  Lemma 6.

Suppose that $f\mathrel{\mathop{\mathchar
58\relax}}\left[a,b\right]\to\mathbb{R}$ is $1$-Lipschitz and increasing.
Then, there exists a sequence of $1$-Lipschitz and strictly increasing
functions $(f_{n})_{n\in\mathbb{N}}$ from $\left[a,b\right]$ to $\mathbb{R}$
that converges uniformly to $f$.

###### Proof.

Since $f$ is continuous and increasing, there exist at most countably many
disjoint nondegenerate intervals $(I_{n})_{n\in\mathbb{N}}$ over which $f$ is
constant, i.e., $f(I_{n})=\left\\{k_{n}\right\\}$ for all $n\in\mathbb{N}$ and
some $k_{n}\in\mathbb{R}$. Fix $\varepsilon>0$ and take finitely many points,
$a=a_{1}<\ldots<a_{k}=b$ such that $\left\lvert
a_{i}-a_{i+1}\right\rvert<\varepsilon/2$ for all $i=1,\ldots,k-1$. Suppose
that $f(a_{i})=f(a_{j})$ for some $i<j$. Since $f$ is increasing, we have that

$f(a_{i})=f(a_{i+1})=\ldots=f(a_{j}).$

This implies that $a_{i},\ldots,a_{j}\in I_{n}$ for some $n\in\mathbb{N}$.
Thus, to find a strictly increasing approximation we need to modify our vector
of points. In particular, we remove all points $a_{i},\ldots,a_{j-1}$, and we
add one point $\tilde{a}_{i}$ picked from the set

$\left\\{x\in\left[a_{i-1},\inf I_{n}\right)\mathrel{\mathop{\mathchar
58\relax}}f(x)<f(a_{i})\ \textnormal{and}\ \left\lvert
f(x)-f(a_{i})\right\rvert<\frac{\varepsilon}{2}\right\\}.$

Repeating this operation for all points where the function is constant, within
the set $\left\\{a_{1},\ldots,a_{k}\right\\}$, we retrieve (in a finite amount
of operations) a set $\left\\{\tilde{a}_{1},\ldots,\tilde{a}_{m}\right\\}$
with $\tilde{a}_{1}=a$ and $\tilde{a}_{m}=b$ such that

$f\left(\tilde{a}_{1}\right)<\left(\tilde{a}_{2}\right)<\ldots<f(\tilde{a}_{m-1})<f(\tilde{a}_{m}).$

Define the function $f^{m}\mathrel{\mathop{\mathchar
58\relax}}\left\\{\tilde{a}_{1},\ldots,\tilde{a}_{m}\right\\}\to\mathbb{R}$ by
$f^{m}(\tilde{a}_{i})=f(\tilde{a}_{i})$, for all $i=1,\ldots,m$. By linear
interpolation, we can extend $f^{m}$ to the whole interval $\left[a,b\right]$,
and we denote such an extension again by $f^{m}$. Clearly, $f^{m}$ is strictly
increasing, and we now show that it must also be 1-Lipschitz. To this end,
notice that the slope of $f^{m}|_{\left[\tilde{a}_{i},\tilde{a}_{i+1}\right]}$
satisfies

$\frac{f(\tilde{a}_{i+1})-f(\tilde{a}_{i})}{\tilde{a}_{i+1}-\tilde{a}_{i}}\leq
1,$

for all $i=1,\ldots,k-1$, where the inequality follows from the fact that $f$
is $1$-Lipschitz, $\tilde{a}_{i+1}>\tilde{a}_{i}$, and $f$ is increasing. This
implies that $f^{m}|_{\left[\tilde{a}_{i},\tilde{a}_{i+1}\right]}$ is
$1$-Lipschitz for all $i=1,\ldots,k-1$. If $x>y$, then
$x\in\left(\tilde{a}_{i},\tilde{a}_{i+1}\right]$ and
$y\in\left[\tilde{a}_{j},\tilde{a}_{j+1}\right]$ for some $i>j$. This implies
that

$\displaystyle\left\lvert f^{m}(x)-f^{m}(y)\right\rvert$
$\displaystyle=f^{m}(x)-f^{m}(y)$
$\displaystyle=f^{m}(x)-f^{m}(\tilde{a}_{i})+f^{m}(\tilde{a}_{i})-f^{m}(\tilde{a}_{i-1})+\ldots+f^{m}(\tilde{a}_{j+1})-f^{m}(\tilde{a}_{j})+f^{m}(\tilde{a}_{j})-f^{m}(y)$
$\displaystyle\leq
x-\tilde{a}_{i}+\tilde{a}_{i}+\ldots+\tilde{a}_{j+1}-\tilde{a}_{j}+\tilde{a}_{j}-y=\left\lvert
x-y\right\rvert.$

Thus $f^{m}$ is $1$-Lipschitz. Now we prove that $f^{m}$ is
$\varepsilon$-close to $f$. For all
$x\in\left[\tilde{a}_{i-1},\tilde{a}_{i}\right]$ and $i=2,\ldots,m$ we have

$\displaystyle\left\lvert f(x)-f^{m}(x)\right\rvert$
$\displaystyle\leq\left\lvert f(x)-f(\tilde{a}_{i-1})\right\rvert+\left\lvert
f(\tilde{a}_{i-1})-f^{m}(x)\right\rvert$
$\displaystyle\leq\frac{\varepsilon}{2}+\left\lvert
f^{m}(\tilde{a}_{i-1})-f^{m}(\tilde{a}_{i})\right\rvert$
$\displaystyle=\frac{\varepsilon}{2}+\left\lvert
f(\tilde{a}_{i-1})-f(\tilde{a}_{i})\right\rvert$ $\displaystyle<\varepsilon.$

Since $x$ was chosen arbitrarily, it follows that $\lVert
f-f^{m}\rVert_{\infty}\to 0$. Therefore, there exists a sequence of
$1$-Lipschitz and strictly increasing functions $(f_{n})_{n\in\mathbb{N}}$
converging uniformly to $f$. ∎

Now suppose that $\Omega=\left[0,1\right]$ and
$\mathcal{F}=\mathcal{B}\left[0,1\right]$ is the Borel sigma-algebra.555All
the results provided in this section would hold for any closed interval
$I\subseteq\mathbb{R}$.

######  Lemma 7.

If $x,y\in B(\Omega,\mathcal{F})$ are comonotonic, then there exist two
sequences $(x_{n})_{n\in\mathbb{N}},(y_{n})_{n\in\mathbb{N}}$ in
$B(\Omega,\mathcal{F})$ such that:

  1. (1)

$x_{n}\xrightarrow{\lVert\cdot\rVert_{\infty}}x$ and
$y_{n}\xrightarrow{\lVert\cdot\rVert_{\infty}}y$;

  2. (2)

For all $n\in\mathbb{N}$, $x_{n}$ and $y_{n}$ are strictly comonotonic with
$x_{n}\notin C_{y_{n}}$.

###### Proof.

Since $x,y\in B(\Omega,\mathcal{F})$ are comonotonic, there exist two
increasing $1$-Lipschitz functions $h,g\mathrel{\mathop{\mathchar
58\relax}}\mathbb{R}\to\mathbb{R}$ and an $z\in B(\Omega,\mathcal{F})$ such
that $x=h(z)$, $y=g(z)$ (Denuit et al., (2023), Theorem 2.7). We first prove a
claim that will yield the result.

######  Claim.

There exists a sequence of injective and measurable functions converging to
$z$.

###### Proof of the claim..

Let $\left(s_{n}\right)_{n\in\mathbb{N}}\in
B(\Omega,\mathcal{F})^{\mathbb{N}}$ be a sequence of step functions converging
uniformly to $z$. Each $s_{n}$ can be uniquely identified with a partition
$\left(I^{n}_{i}\right)_{i=1}^{k_{n}}$ of nondegenerate subintervals of
$\Omega$, and a vector of values
$\left(a^{n}_{1},\ldots,a^{n}_{k_{n}}\right)$, for some $k_{n}\in\mathbb{N}$.
For all $n\in\mathbb{N}$ and $\varepsilon>0$, we can define the following,

$s^{\varepsilon}_{n}(\omega)=2\varepsilon\left(\frac{\omega-\inf
I^{n}_{i}}{\sup I^{n}_{i}-\inf I^{n}_{i}}\right)+a^{n}_{i}-\varepsilon,$

for all $\omega\in I^{n}_{i}$ and all $i=1,\ldots,k_{n}$. Clearly
$s^{\varepsilon}_{n}$ is an injective Borel measurable function, for all
$\varepsilon>0$ and all $n\in\mathbb{N}$. Intuitively, we are simply rotating
slightly the constant lines of each $s_{n}$ over all their partitions.
Moreover, note that

$\left\lVert s^{\varepsilon}_{n}-s_{n}\right\rVert_{\infty}\leq\varepsilon,$

for all $n\in\mathbb{N}$ and $\varepsilon>0$. This implies that
$\left(s^{1/n}_{n}\right)_{n\in\mathbb{N}}$ converges uniformly to $z$.
Indeed,

$\left\lVert z-s^{1/n}_{n}\right\rVert_{\infty}\leq\left\lVert
z-s_{n}\right\rVert_{\infty}+\left\lVert
s_{n}-s^{1/n}_{n}\right\rVert_{\infty}\leq\left\lVert
z-s_{n}\right\rVert_{\infty}+\frac{1}{n}\to 0.$

Thus, we found a sequence of injective and measurable functions converging
uniformly to $z$. ∎

Given that $z$ is bounded, there exist $m,M\in\mathbb{R}$ such that
$z(\Omega)\subseteq\left(m,M\right)$. Since
$\left(s^{1/n}_{n}\right)_{n\in\mathbb{N}}$ converges uniformly to $z$, there
exists some $N\in\mathbb{N}$ sufficiently large so that

$s_{n}^{1/n}(\Omega)\subseteq\left[m,M\right],$

for all $n\geq N$. Using a slight abuse of notation, we will now identify by
$\left(s^{1/n}_{n}\right)_{n\in\mathbb{N}}$ its subsequence
$\left(s^{1/n_{k}}_{n_{k}}\right)_{k\in\mathbb{N}}$ with $n_{1}=N$,
$n_{2}=N+1$ and so on. By Lemma 6, there exist two sequences of $1$-Lipschitz
and strictly increasing functions $\left(h_{n}\right)_{n\in\mathbb{N}}$ and
$\left(g_{n}\right)_{n\in\mathbb{N}}$ from $\left[m,M\right]$ to $\mathbb{R}$
converging uniformly to $h$ and $g$. For all $n\in\mathbb{N}$, let
$x_{n}=h_{n}\left(s_{n}^{1/n}\right)$ and
$y_{n}=g_{n}\left(s_{n}^{1/n}\right)$. Fix $n\in\mathbb{N}$ arbitrarily. If
$\omega_{1}\neq\omega_{2}$, then $s_{n}^{1/n}(\omega_{1})\neq
s_{n}^{1/n}(\omega_{2})$ since $s_{n}^{1/n}$ is injective, say without loss of
generality that $s_{n}^{1/n}(\omega_{1})>s_{n}^{1/n}(\omega_{2})$. Since
$h_{n}$ and $g_{n}$ are both strictly increasing we have that

$\displaystyle\left[x_{n}(\omega_{1})-x_{n}(\omega_{2})\right]\left[y_{n}(\omega_{1})-y_{n}(\omega_{2})\right]$
$\displaystyle=\left[h_{n}\left(s_{n}^{1/n}\right)(\omega_{1})-h_{n}\left(s_{n}^{1/n}\right)(\omega_{2})\right]\left[g_{n}\left(s_{n}^{1/n}\right)(\omega_{1})-g_{n}\left(s_{n}^{1/n}\right)(\omega_{2})\right]>0.$

Thus, $x_{n},y_{n}$ are strictly comonotonic. Fix $n\in\mathbb{N}$
arbitrarily. Since $h_{n}$ is $1$-Lipschitz, we have

$\displaystyle\left\lVert
h(z)-h_{n}\left(s_{n}^{1/n}\right)\right\rVert_{\infty}$
$\displaystyle\leq\left\lVert
h(z)-h_{n}\left(z\right)\right\rVert_{\infty}+\left\lVert
h_{n}(z)-h_{n}\left(s_{n}^{1/n}\right)\right\rVert_{\infty}$
$\displaystyle=\left\lVert
h(z)-h_{n}\left(z\right)\right\rVert_{\infty}+\sup\limits_{\omega\in\Omega}\left\lvert
h_{n}(z(\omega))-h_{n}\left(s_{n}^{1/n}(\omega)\right)\right\rvert$
$\displaystyle\leq\left\lVert
h(z)-h_{n}\left(z\right)\right\rVert_{\infty}+\sup\limits_{\omega\in\Omega}\left\lvert
z(\omega)-s_{n}^{1/n}(\omega)\right\rvert\to 0.$

Thus $\left(x_{n}\right)_{n\in\mathbb{N}}$ converges uniformly to $x$. The
same holds for $(y_{n})_{n\in\mathbb{N}}$, and the proof is totally analogous.
∎

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