# On convexity properties with respect to a Chebyshev system

Zsolt Páles  Institute of Mathematics, University of Debrecen, Hungary
<EMAIL_ADDRESS>and  Mahmood Kamil Shihab  Doctoral School of
Mathematical and Computational Sciences, University of Debrecen, Hungary;
Department of Mathematics, College of Education for Pure Sciences, University
of Kirkuk, Iraq<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

The main purpose of this paper is to introduce various convexity concepts in
terms of a positive Chebyshev system $\omega$ and give a systematic
investigation of the relations among them. We generalize a celebrated theorem
of Bernstein–Doetsch to the setting of $\omega$-Jensen convexity. We also give
sufficient conditions for the existence of discontinuous $\omega$-Jensen
affine functions. The concept of Wright convexity is extended to the setting
of Chebyshev systems, as well, and it turns out to be an intermediate
convexity property between $\omega$-convexity and $\omega$-Jensen convexity.
For certain Chebyshev systems, we generalize the decomposition theorems of
Wright convex and higher-order Wright convex functions obtained by C. T. Ng in
1987 and by Maksa and Páles in 2009, respectively.

###### Key words and phrases:

Convexity, Jensen convexity and Wright convexity with respect to a Chebyshev
system

###### 2000 Mathematics Subject Classification:

Primary 26A51, 39B62

The research of the first author was supported by the K-134191 NKFIH Grant.

##  1\. Introduction

The simplex of strictly ordered $n$-tuples of a set $H\subset\mathbb{R}$
denoted by $\sigma_{n}(H)$ is defined by

$\displaystyle\sigma_{n}(H):=\\{(x_{1},\dots,x_{n})\in H\quad|\quad
x_{1}<\dots<x_{n}\\}.$

Obviously, the set $\sigma_{n}(H)$ is nonempty set if and only if the
cardinality $|H|$ of $H$ is bigger than or equal to $n$. We adopt that
$|H|\geq n$. Let $\omega=(\omega_{1},\dots,\omega_{n}):H\to\mathbb{R}^{n}$ be
a vector-valued function, and define the functional operator
$\Phi_{\omega}:\sigma_{n}(H)\to\mathbb{R}$ by

$\displaystyle\Phi_{\omega}(x_{1},\dots,x_{n}):=\begin{vmatrix}\omega_{1}(x_{1})\dots\omega_{1}(x_{n})\\\
\vdots\quad\ddots\quad\vdots\\\
\omega_{n}(x_{1})\dots\omega_{n}(x_{n})\end{vmatrix}\qquad((x_{1},\dots,x_{n})\in\sigma_{n}(H)).$

A continuous function $\omega$ is said to be an _$n$ -dimensional positive
(respectively negative) Chebyshev system over_ $H$ if $\Phi_{\omega}$ is
positive (respectively negative) over $\sigma_{n}(H)$. The system $\omega$ is
called an _$n$ -dimensional Chebyshev system over_ $H$ if it is either a
positive or a negative Chebyshev system over $H$. If
$\omega:\mathbb{R}\to\mathbb{R}^{n}$ equals the _$n$ -dimensional standard or
polynomial system_ $\pi_{n}:I\to\mathbb{R}^{n}$, which is defined by

$\displaystyle\pi_{n}(t):=(1,t,\dots,t^{n-1})\qquad(t\in\mathbb{R}),$

then, by computing Vandermonde determinants, one can easily show that it is a
positive Chebyshev system. For standard results and applications related to
Chebyshev systems, we refer to the monographs [4] and [5].

In what follows, we recall some definitions from the paper [12] (see also the
paper [2] for these definitions in the polynomial setting). Let
$I\subset\mathbb{R}$ be a nonvoid interval, $n\in\mathbb{N}$ and let
$\omega:I\to\mathbb{R}$ be an $n$-dimensional positive Chebyshev system over
$I$.

For a given vector $t=(t_{1},\dots,t_{n})\in\mathbb{R}^{n}_{+}$, the function
$f:I\to\mathbb{R}$ is said to be _$(t,\omega)$ -convex_ if

(1)
$\displaystyle\Phi_{(\omega,f)}(x,x+t_{1}h,\dots,x+(t_{1}+\dots+t_{n})h)\geq
0$

holds for all $h>0$, $x\in I$ with $x+(t_{1}+\dots+t_{n})h\in I$. If
$T\subseteq\mathbb{R}_{+}$ and $f$ is $(t,\omega)$-convex for every $t\in
T^{n}$, then $f$ is called _$(T,\omega)$ -convex_.

If $t=(t_{1},\dots,t_{n})\in\mathbb{R}^{n}_{+}$ and (1) is satisfied with
equality, then $f$ is called a _$(t,\omega)$ -affine function_. If
$T\subseteq\mathbb{R}_{+}$ and $f$ is $(t,\omega)$-affine for every $t\in
T^{n}$, then $f$ is called _$(T,\omega)$ -affine_. In particular, we say that
$f$ is _$\omega$ -Jensen convex_ if it is $(\\{1\\},\omega)$-convex, i.e., if

(2) $\displaystyle\Phi_{(\omega,f)}(x,x+h,\dots,x+nh)\geq 0$

holds for all $h>0$, $x\in I$ with $x+nh\in I$. If (2) is valid with equality
instead of inequality, then $f$ is said to be _$\omega$ -Jensen affine_.

A function $f$ is termed _$\omega$ -convex_ if it is
$(\mathbb{R}_{+},\omega)$-convex. It is easy to see that $f$ is
$\omega$-convex on $I$ if and only if

(3) $\displaystyle\Phi_{(\omega,f)}(x_{0},x_{1},\dots,x_{n})\geq
0\qquad((x_{0},x_{1},\dots,x_{n})\in\sigma_{n+1}(I)).$

A function $f$ is called _$\omega$ -affine_ if (3) is satisfied with equality.

We have to mention that in the case when $\omega=\pi_{n}$, then the concepts
of $\omega$-convexity and $\omega$-Jensen convexity, was introduced by Hopf
[3] and Popoviciu [10] (see also the book [6] by Kuczma) and these properties
were called convexity and Jensen convexity of order $(n-1)$, respectively.

The following result shows that, for any nonempty set
$T\subseteq\mathbb{R}_{+}$, $(T,\omega)$-convexity implies $\omega$-Jensen
convexity.

######  Theorem 1.1.

Let $T\subseteq\mathbb{R}_{+}$ be a nonempty set. If a function
$f:I\to\mathbb{R}$ is $(T,\omega)$-convex (resp. $(T,\omega)$-affine), then it
is $(\mathbb{Q}_{+},\omega)$-convex (resp. $(\mathbb{Q}_{+},\omega)$-affine),
in particular, it is $\omega$-Jensen convex (resp. $\omega$-Jensen affine).

###### Proof.

The result immediately follows from [12, Theorem 5]. ∎

In section $2$ we show that to any $\omega$-Jensen convex (resp.
$\omega$-Jensen affine) function there exists a continuous $\omega$-convex
(resp. $\omega$-affine) function so that these two functions coincide on dense
subset of their domain. As a corollary, we obtain that $\omega$-convex
functions are automatically continuous. The Bernstein-Doetsch theorem is
generalized to the $\omega$-convexity with respect to Chebyshev system, that
is, we show that an $\omega$-Jensen convex function which is bounded over some
nonempty open subinterval, is also $\omega$-convex. We also establish
characterizations of $\omega$-Jensen affine functions in several settings. A
sufficient condition on $\omega$ that ensures the existence of discontinuous
$\omega$-Jensen affine (resp. $\omega$-Jensen convex) function is given as
well. Some of the results of this section extend that of the paper [8] by
Matkowski.

In Section 3, we generalize the concept of standard Wright convexity to
$\omega$-Wright convexity with respect to a positive Chebyshev system, and
establish its relationship with $\omega$-convexity and $\omega$-Jensen
convexity. We also generalize Ng’s decomposition theorem to setting of
$\omega$-Wright convexity with respect certain positive Chebyshev systems.
Finally, in the two dimensional case, we show that $\omega$-Wright convexity
could be equivalent to the Wright convexity for certain positive Chebyshev
systems.

##  2\. Results on $\omega$-Jensen functions

In what follows, if $D$ is a subset of $I$ and $f:D\to\mathbb{R}$, then $f$ is
said to be _locally uniformly continuous_ if, for all compact subinterval
$[a,b]$ of $I$ and for all $\varepsilon>0$, there exists $\delta>0$ such that,
for all $x,y\in[a,b]\cap D$ with $|x-y|<\delta$, we have that
$|f(x)-f(y)|<\varepsilon$.

######  Lemma 2.1.

Let $D$ be a dense subset of $I$ and let $f:D\to\mathbb{R}$ be a locally
uniformly continuous function. Then $f$ admits a continuous extension to $I$.

The verification of this lemma is standard, however, for the convenience of
the reader, we provide a proof.

###### Proof.

Let $x\in I$ be a fixed and let $(x_{n})$ be an arbitrary sequence in $D$ such
that $x_{n}\to x$ as $n\to\infty$. We point out first that $(f(x_{n}))$ is a
Cauchy sequence. The set $\\{x_{n}:n\in\mathbb{N}\\}\cup\\{x\\}$ is compact,
therefore there exists a compact subinterval $[a,b]\subset I$ which contains
all the members of the sequence $(x_{n})$.

Let $\varepsilon>0$ be arbitrary. By the local uniform continuity of $f$,
there exists $\delta>0$ such that $|f(u)-f(v)|<\varepsilon$ for all
$u,v\in[a,b]\cap D$ with $|u-v|<\delta$. The sequence $(x_{n})$ is a Cauchy
sequence, there exists $N\in\mathbb{N}$ such that, for all $n,m\geq N$, we
have $|x_{n}-x_{m}|<\delta$. Consequently,

(4) $\displaystyle|f(x_{n})-f(x_{m})|<\varepsilon$

whenever for all $n,m\geq N$. Therefore $(f(x_{n}))$ is a Cauchy sequence
which implies its convergence. Denote its limit by $g(x)$.

We show that $g$ is well defined at $x$. If $(x_{n})$ and $(y_{n})$ are two
sequences in $D$ converging to $x$, then the sequence
$(x_{1},y_{1},x_{2},y_{2},x_{3},y_{3}\dots)$ is also converges to $x$. What we
have proved above implies that
$(f(x_{1}),f(y_{1}),f(x_{2}),f(y_{2}),f(x_{3}),f(y_{3}),\dots)$ is a
convergent sequence and hence the subsequences $(f(x_{n})$ and $(f(y_{n}))$
must have the same limit. Therefore the value of $g$ at $x$ does not depend on
the choice of the sequence converging to $x$.

Clearly, if $x\in D$, then the sequence $x_{n}=x$ converges to $x$, hence
$g(x)=f(x)$. Therefore, $g$ is an extension of $f$.

In order to prove that $g$ is continuous on $I$, we show that it is locally
uniformly continuous on $I$. To see this, let $[a,b]\subseteq I$. Let
$\varepsilon>0$ be arbitrary and, using the local uniform continuity of $f$,
choose $\delta>0$ such that, for all $u,v\in[a,b]\cap D$ with $|u-v|<\delta$,
we have $|f(u)-f(v)|\leq\varepsilon$.

Let $x,y\in[a,b]$ be arbitrary with $|x-y|<\delta$ and we are going to show
that $|g(x)-g(y)|\leq\varepsilon$. Indeed, consider two sequences $(x_{n})$
and $(y_{n})$ in $[a,b]\cap D$ such that $x_{n}\to x$ and $y_{n}\to y$ as
$n\to\infty$. Then $|x_{n}-y_{n}|\to|x-y|<\delta$, hence, there exists
$N\in\mathbb{N}$ such that, for all $n\geq N$, we have $|x_{n}-y_{n}|<\delta$.
Therefore, according to the choice of $\delta$ (with $u:=x_{n}$ and
$v:=y_{n})$, we get that $|f(x_{n})-f(y_{n})|<\varepsilon$ holds for all
$n\geq N$. Upon taking the limit $n\to\infty$, it follows that
$|g(x)-g(y)|\leq\varepsilon$. ∎

######  Theorem 2.2.

Let $n\geq 2$ and let
$\omega=(\omega_{1},\dots,\omega_{n}):I\to\mathbb{R}^{n}$ be an
$n$-dimensional positive Chebyshev system and let $\mathbb{K}$ be a subfield
of $\mathbb{R}$. If a function $f:I\to\mathbb{R}$ is
$(\mathbb{K}_{+},\omega)$-convex (resp. $(\mathbb{K}_{+},\omega)$-affine),
then there exists a continuous $\omega$-convex (resp. $\omega$-affine)
function $g:I\to\mathbb{R}$ such that
$g|_{I\cap\mathbb{K}}=f|_{I\cap\mathbb{K}}$.

###### Proof.

If $f$ is $(\mathbb{K}_{+},\omega)$-convex, then by definition we have the
inequality

(5) $\displaystyle\Phi_{(\omega,f)}(x_{0},x_{1},\dots,x_{n})\geq
0\qquad((x_{0},\dots,x_{n})\in\sigma_{n+1}(I\cap\mathbb{K})).$

Indeed, apply the inequality (1) with $x:=x_{0}$, $h:=1$, and
$t_{i}:=x_{i}-x_{i-1}\in\mathbb{K}_{+}$. If $f$ is assumed to be
$(\mathbb{K}_{+},\omega)$-affine, then (5) is satisfied with equality.

Using the continuity of the function $\omega$, we are going to show that the
restricted function $f|_{I\cap\mathbb{K}}$ is locally uniformly continuous. To
prove this let $[a,b]\subset I$ be arbitrary. Without loss of generality, we
can assume that $a,b\in\mathbb{K}$. We fix some elements
$u_{0}<u_{1}<\dots<u_{n-2}<a$ and $b<v$ of the set $I\cap\mathbb{K}$. Then,
for $x,y\in[a,b]\cap\mathbb{K}$ with $x<y$, we have that
$(u_{0},\dots,u_{n-2},x,y),(u_{0},\dots,u_{n-3},x,y,v)\in\sigma_{n+1}(I\cap\mathbb{K})$,
therefore (5) implies

(6) $\displaystyle\Phi_{(\omega,f)}(u_{0},\dots,u_{n-2},x,y)\geq
0\qquad\mbox{and}\qquad\Phi_{(\omega,f)}(u_{0},\dots,u_{n-3},x,y,v)\geq 0.$

The first of these inequalities implies that

$\displaystyle\begin{vmatrix}\omega_{1}(u_{0})&\dots&\omega_{1}(u_{n-2})&\omega_{1}(x)&\omega_{1}(y)\\\
\vdots&\ddots&\vdots&\vdots&\vdots\\\
\omega_{n}(u_{0})&\dots&\omega_{n}(u_{n-2})&\omega_{n}(x)&\omega_{n}(y)\\\
f(u_{0})&\dots&f(u_{n-2})&f(x)&f(y)\end{vmatrix}\geq 0.$

Developing this determinant by the last row, this inequality is equivalent to

(7) $\displaystyle f(y)P(x)-f(x)P(y)\geq
R(x,y)\qquad((x,y)\in\sigma_{2}(I\cap\mathbb{K})),$

where $P:[a,b]\to\mathbb{R}$ and $R:[a,b]^{2}\to\mathbb{R}$ are defined by

$\displaystyle P(z):=\Phi_{\omega}(u_{0},\dots,u_{n-2},z)$

and

$\displaystyle
R(x,y):=\sum_{i=0}^{n-2}(-1)^{n-1-i}f(u_{i})\Phi_{\omega}(u_{0},\dots,u_{i-1},u_{i-1},\dots,u_{n-2},x,y)$

For $z\in[a,b]$, we have that $(u_{0},\dots,u_{n-2},z)\in\sigma_{n}(I)$.
Therefore, by the positivity and continuity of the Chebyshev system, it
follows that $P$ is positive and continuous over $[a,b]$. We can also see that
$R$ is continuous on $[a,b]^{2}$ and $R(z,z)=0$ for all $z\in[a,b]$.
Substituting $x=a$ into (7), we get that

$\displaystyle
f(y)\geq\frac{R(a,y)+f(a)P(y)}{P(a)}\qquad(y\in[a,b]\cap\mathbb{K}).$

The right hand side of this inequality is a continuous function of $y$ over
the compact interval $[a,b]$, therefore it is bounded from below. Hence
$f|_{[a,b]\cap\mathbb{K}}$ is also bounded from below. Putting $y=b$ into (7),
it follows that

$\displaystyle\frac{f(b)P(x)-R(x,b)}{P(b)}\geq
f(x)\qquad(x\in[a,b]\cap\mathbb{K}).$

Arguing similarly as above, this inequality yields that
$f|_{[a,b]\cap\mathbb{K}}$ is bounded from above hence there exists a positive
number $K$ such that, for $x\in[a,b]\cap\mathbb{K}$, we have $|f(x)|\leq K$.

Now we consider the second inequality in (6). It can be rewritten as

$\displaystyle\begin{vmatrix}\omega_{1}(u_{0})&\dots&\omega_{1}(u_{n-3})&\omega_{1}(x)&\omega_{1}(y)&\omega_{1}(v)\\\
\vdots&\ddots&\vdots&\vdots&\vdots&\vdots\\\
\omega_{n}(u_{0})&\dots&\omega_{n}(u_{n-3})&\omega_{n}(x)&\omega_{n}(y)&\omega_{n}(v)\\\
f(u_{0})&\dots&f(u_{n-3})&f(x)&f(y)&f(v)\end{vmatrix}\geq 0.$

Developing this determinant by its last row, this inequality is equivalent to

(8) $\displaystyle f(x)Q(y)-f(y)Q(x)\geq
S(x,y)\qquad((x,y)\in\sigma_{2}(I\cap\mathbb{K})),$

where $Q:[a,b]\to\mathbb{R}$ and $S:[a,b]^{2}\to\mathbb{R}$ are defined by

$\displaystyle Q(z):=\Phi_{\omega}(u_{0},\dots,u_{n-3},z,v)$

and

$\displaystyle
S(x,y):=\sum_{i=0}^{n-3}(-1)^{n-1-i}f(u_{i})\Phi_{\omega}(u_{0},\dots,u_{i-1},u_{i-1},\dots,u_{n-3},x,y,v)-f(v)\Phi_{\omega}(u_{0},\dots,u_{n-3},x,y).$

The inclusion $(u_{0},\dots,u_{n-3},z,v)\in\sigma_{n}(I\cap\mathbb{K})$, the
positivity and continuity of the Chebyshev system yield that $Q$ is a positive
and continuous function over $[a,b]$. We also have that $S$ is continuous over
$[a,b]^{2}$ and $S(z,z)=0$ for all $z\in[a,b]$.

For $x,y\in[a,b]\cap\mathbb{K}$ the inequalities (7) and (8) imply that

(9) $\displaystyle f(y)-f(x)$
$\displaystyle\geq\frac{P(y)-P(x)}{P(x)}f(x)+\frac{R(x,y)}{P(x)}\geq-K\frac{|P(y)-P(x)|}{P(x)}+\frac{R(x,y)}{P(x)}=:A(x,y),$
$\displaystyle f(y)-f(x)$
$\displaystyle\leq\frac{Q(y)-Q(x)}{Q(x)}f(x)+\frac{S(x,y)}{Q(x)}\leq\frac{|Q(y)-Q(x)|}{Q(x)}K+\frac{S(x,y)}{Q(x)}=:B(x,y).$

The functions $A$ and $B$ defined on the right hand side of these inequalities
are continuous over $[a,b]^{2}$ and hence they are uniformly continuous over
$[a,b]^{2}$. Therefore, for all $\varepsilon>0$, there exists $\delta>0$ such
that, for all $(x,y),(u,v)\in[a,b]^{2}$ with $\|(x,y)-(u,v)\|<\delta$, we have

$\displaystyle\max(|A(x,y)-A(u,v)|,|B(x,y)-B(u,v)|)<\varepsilon.$

In particular, if $|x-y|<\delta$, then substituting $(u,v)=(x,x)$ and using
that $A$ and $B$ vanish at diagonal points of the square $[a,b]^{2}$, the
above inequality implies that

$\displaystyle\max(|A(x,y)|,|B(x,y)|)<\varepsilon.$

Therefore, in view of the inequalities in (9), $x,y\in[a,b]\cap\mathbb{K}$
with $|x-y|<\delta$, we obtain that

$\displaystyle|f(y)-f(x)|<\varepsilon.$

This proves that $f|_{[a,b]\cap\mathbb{K}}$ is uniformly continuous and hence
$f|_{I\cap\mathbb{K}}$ is locally uniformly continuous. If $f$ is
$(\mathbb{K}_{+},\omega)$-affine, then it is also
$(\mathbb{K}_{+},\omega)$-convex, therefore we have the same conclusion.

In view of Lemma 2.1 with the dense set $D=I\cap\mathbb{K}$, there exists a
continuous function $g:I\to\mathbb{R}$ such that
$g(x)|_{I\cap\mathbb{K}}=f(x)|_{I\cap\mathbb{K}}$.

Finally, we show that $g$ is $\omega$-convex (resp. $\omega$-affine). Let
$(y_{0},\dots,y_{n})$ be an arbitrary element of $\sigma_{n+1}(I)$. Then, by
the density of $\mathbb{K}$ in $I$, for each $j\in\\{0,\dots,n\\}$, there
exists a sequence $(x_{k,j})_{k\in\mathbb{N}}$ in
$\sigma_{n+1}(I\cap\mathbb{K})$ converging to $y_{j}$ as $k\to\infty$. Then,
applying the $(\mathbb{K}_{+},\omega)$-convexity (resp. the
$(\mathbb{K}_{+},\omega)$-affinity) of $f$, we have that (5) is valid, we
obtain

$\displaystyle\Phi_{(\omega,g)}(x_{k,0},x_{k,1},\dots,x_{k,n})$
$\displaystyle=\Phi_{(\omega,f)}(x_{k,0},x_{k,1},\dots,x_{k,n})\geq 0$
$\displaystyle(\mbox{resp. }\Phi_{(\omega,g)}(x_{k,0},x_{k,1},\dots,x_{k,n})$
$\displaystyle=\Phi_{(\omega,f)}(x_{k,0},x_{k,1},\dots,x_{k,n})=0).$

By the continuity of $g$, the function $\Phi_{(\omega,g)}$ is continuous. Upon
taking the limit $k\to\infty$, the above inequality (resp. equality) implies
that

$\displaystyle\Phi_{(\omega,g)}(y_{0},\dots,y_{n})\geq 0\qquad(\mbox{resp.
}\Phi_{(\omega,g)}(y_{0},\dots,y_{n})=0).$

Therefore, $g$ is $\omega$-convex (resp. $\omega$-affine). ∎

######  Corollary 2.3.

Let $n\geq 2$ and $\omega=(\omega_{1},\dots,\omega_{n}):I\to\mathbb{R}^{n}$ be
an $n$-dimensional positive Chebyshev system. If $f:I\to\mathbb{R}$ is
$\omega$-convex (resp. $\omega$-affine), then it is continuous on $I$.

###### Proof.

The corollary follows by applying Theorem 2.2 with $\mathbb{K}:=\mathbb{R}$. ∎

######  Corollary 2.4.

Let $n\geq 2$ and $\omega=(\omega_{1},\dots,\omega_{n}):I\to\mathbb{R}^{n}$ be
an $n$-dimensional positive Chebyshev system. If $f:I\to\mathbb{R}$ is
$\omega$-Jensen convex (resp. $\omega$-Jensen affine), then there exists a
continuous $\omega$-convex (resp. $\omega$-affine) function $g:I\to\mathbb{R}$
such that $g|_{I\cap\mathbb{Q}}=f|_{I\cap\mathbb{Q}}$.

###### Proof.

In view of Theorem 1.1, the $\omega$-Jensen convexity (resp. $\omega$-Jensen
affinity) of $f$ implies that it is $(\mathbb{Q}_{+},\omega)$-convex (resp.
$(\mathbb{Q}_{+},\omega)$-affine). Now the statement of the corollary follows
by applying Theorem 2.2 with $\mathbb{K}:=\mathbb{Q}$. ∎

The following statement is the extension of the celebrated Bernstein–Doetsch
theorem [1] to the setting of $\omega$-Jensen convexity.

######  Theorem 2.5.

If $f:I\to\mathbb{R}$ is $\omega$-Jensen convex and bounded on a nonempty open
subset of $I$, then it is continuous on $I$.

###### Proof.

Let $U$ be a nonvoid open subinterval of $I$ such that $f$ is bounded on $U$
by $K\geq 0$. In the first part of the proof, we are going to show that $f$ is
locally bounded on $I$, i.e., for every $v\in I$, there is an open set
$V\subseteq I$ containing $v$ such that $f$ is bounded on $V$.

Let $v\in I$ be arbitrary. If $v\in U$, then the statement holds with $V=U$.
Therefore, we may assume that $v\not\in U$. Choose a closed interval
$[a,b]\subset U$. Then either $v<a$ or $b<v$. We consider now the case when
$v<a$.

We choose some rational numbers $a-v<r_{1}<\dots<r_{n}<b-v$. Then we have that
$v<a<v+r_{1}<\dots<v+r_{n}<b$. One can construct a bounded neighborhood $W$ of
$v$ such that $\overline{W}\subseteq I$ and, for all $x\in W$, we have
$x<a<x+r_{1}<\dots<x+r_{n}<b$. Now the $\omega$-Jensen convexity of $f$ and
Theorem 1.1 yield that $f$ is $(\mathbb{Q}_{+},\omega)$-convex, hence we can
get that

$\displaystyle\Phi_{\omega,f}(x,x+r_{1},\dots,x+r_{n})\geq 0$

for all $x\in W$. This inequality implies that

$\displaystyle\begin{vmatrix}\omega_{1}(x)&\omega_{1}(x+r_{1})&\dots&\omega_{1}(x+r_{n})\\\
\vdots&\vdots&\ddots\\\
\omega_{n}(x)&\omega_{n}(x+r_{1})&\dots&\omega_{n}(x+r_{n})\\\
f(x)&f(x+r_{1})&\dots&f(x+r_{n})\end{vmatrix}\geq 0.$

Developing the determinant by last row, we obtain

(10) $\displaystyle(-1)^{n}f(x)$
$\displaystyle\Phi_{\omega}(x+r_{1},\dots,x+r_{n})$
$\displaystyle+\sum_{i=1}^{n}(-1)^{n+i}f(x+r_{i})\Phi_{\omega}(x,x+r_{1},\dots,x+r_{i-1},x+r_{i+1},\dots,x+r_{n})\geq
0.$

By the boundedness of $f$ on $U$, the inclusion $x+r_{i}\in[a,b]\subseteq U$
we obtain that $(-1)^{n+i}f(x+r_{i})\leq K$. The continuity of $\omega$ and
positivity of Chebyshev system yield that the function

$\displaystyle
x\mapsto\sum_{i=1}^{n}\frac{\Phi_{\omega}(x,x+r_{1},\dots,x+r_{i-1},x+r_{i+1},\dots,x+r_{n})}{\Phi_{\omega}(x+r_{1},\dots,x+r_{n})}$

is bounded from above by a positive number $L$ over the compact set
$\overline{W}$. Therefore, the inequality (10) implies that, for all $x\in W$,

$\displaystyle(-1)^{n-1}f(x)$
$\displaystyle\leq\sum_{i=1}^{n}(-1)^{n+i}f(x+r_{i})\frac{\Phi_{\omega}(x,x+r_{1},\dots,x+r_{i-1},x+r_{i+1},\dots,x+r_{n})}{\Phi_{\omega}(x+r_{1},\dots,x+r_{n})}\leq
KL,$

which implies that $(-1)^{n-1}f$ is bounded from above over $W$.

To prove that $(-1)^{n-1}f$ is bounded from below over a neighborhood
$V\subseteq W$ of $v$, we additionally fix $v_{0}\in I$ such that $v_{0}<v$.
Now choose rational numbers

$\displaystyle
1<\frac{a-v_{0}}{v-v_{0}}<r_{2}<r_{3}<\dots<r_{n}<\frac{b-v_{0}}{v-v_{0}}.$

Then we have

$\displaystyle
a<v_{0}+r_{2}(v-v_{0})<v_{0}+r_{3}(v-v_{0})<\dots<v_{0}+r_{n}(v-v_{0})<b.$

Now one can construct a neighborhood $V$ of $v$ such that $V\subseteq W$ and,
for all $x\in V$, we have

$\displaystyle
v_{0}+r_{0}(x-v_{0})<v_{0}+r_{1}(x-v_{0})<a<v_{0}+r_{2}(x-v_{0})<\dots<v_{0}+r_{n}(x-v_{0})<b,$

where $r_{0}=0$ and $r_{1}=1$. Again applying Theorem 1.1, we conclude that

$\displaystyle\Phi_{(\omega,f)}(v_{0}+r_{0}(x-v_{0}),v_{0}+r_{1}(x-v_{0}),v_{0}+r_{2}(x-v_{0}),\dots,v_{0}+r_{n}(x-v_{0}))\geq
0,$

that is,

$\displaystyle\Phi_{(\omega,f)}(v_{0},x,v_{0}+r_{2}(x-v_{0}),\dots,v_{0}+r_{n}(x-v_{0}))\geq
0,$

for all $x\in V$. This inequality, for $x\in V$, is equivalent to

$\displaystyle\begin{vmatrix}\omega_{1}(v_{0})&\omega_{1}(x)&\omega_{1}(v_{0}+r_{2}(x-v_{0}))&\dots&\omega_{1}(v_{0}+r_{n}(x-v_{0}))\\\
\vdots&\vdots&\vdots&\ddots&\vdots\\\
\omega_{n}(v_{0})&\omega_{n}(x)&\omega_{n}(v_{0}+r_{2}(x-v_{0}))&\dots&\omega_{n}(v_{0}+r_{n}(x-v_{0}))\\\
f(v_{0})&f(x)&f(v_{0}+r_{2}(x-v_{0}))&\dots&f(v_{0}+r_{n}(x-v_{0}))\\\
\end{vmatrix}\geq 0.$

For brevity, we write $s_{i}(x):=v_{0}+r_{i}(x-v_{0})$, $i\in\\{2,\dots,n\\}$.
Developing the determinant by its last row, we obtain

$\displaystyle(-1)^{n}f(v_{0})$
$\displaystyle\Phi_{\omega}(x,s_{2}(x),\dots,s_{n}(x))+(-1)^{n-1}f(x)\Phi_{\omega}(v_{0},s_{2}(x),\dots,s_{n}(x))$
$\displaystyle+\sum_{i=2}^{n}(-1)^{n-i}f(s_{i})\Phi_{\omega}(v_{0},x,s_{2}(x)\dots,s_{i-1}(x),\dots,s_{i+1}(x),\dots,s_{n}(x))\geq
0,$

which yields

(11) $\displaystyle(-1)^{n-1}f(x)$
$\displaystyle\geq(-1)^{n-1}f(v_{0})\frac{\Phi_{\omega}(x,s_{2}(x),\dots,s_{n}(x))}{\Phi_{\omega}(v_{0},s_{2}(x),\dots,s_{n}(x))}$
$\displaystyle\qquad+\sum_{i=2}^{n}(-1)^{n-i-1}f(s_{i})\frac{\Phi_{\omega}(v_{0},x,s_{2}(x)\dots,s_{i-1}(x),\dots,s_{i+1}(x),\dots,s_{n}(x))}{\Phi_{\omega}(v_{0},s_{2}(x),\dots,s_{n}(x))}.$

By the boundedness of $f$ on $U$ and the inclusion $s_{i}(x)\in[a,b]\subseteq
U$, we get $(-1)^{n-i-1}f(s_{i}(x))\geq-K$. The continuity and the positivity
of Chebyshev system $\omega$ yield that the functions

$\displaystyle
x\mapsto\frac{\Phi_{\omega}(x,s_{2}(x),\dots,s_{n}(x))}{\Phi_{\omega}(v_{0},s_{2}(x),\dots,s_{n}(x))}\quad\mbox{and}\quad
x\mapsto\sum_{i=1}^{n}\frac{\Phi_{\omega}(v_{0},x,s_{2}(x)\dots,s_{i-1}(x),\dots,s_{i+1}(x),\dots,s_{n}(x))}{\Phi_{\omega}(v_{0},s_{2}(x),\dots,s_{n}(x))}$

are bounded from above by positive numbers $M$ and $N$, respectively, over the
compact set $\overline{V}\subseteq\overline{W}$. Therefore, the inequality
(11) implies, for all $x\in V$, that

$\displaystyle(-1)^{n-1}f(x)$ $\displaystyle\geq-|f(v_{0})|M-KN,$

which proves that $(-1)^{n-1}f$ is bounded from below over $V$. From the two-
sided boundedness, it follows that $(-1)^{n-1}f$ is bounded over $V$,
consequently, $f$ is also bounded over $V$.

To complete the proof of the theorem, we have to verify the continuity of $f$
at any point of $I$. Let $v\in I$ be arbitrary. Then, according to what we
have proved in the first part, there exists a neighborhood $V\subseteq I$ of
$v$ such that $f$ is bounded on $V$ by $K$. We are going to show that $f$ is
uniformly continuous on every compact subinterval $[a,b]$ of $V$. This, in
particular, implies the continuity of $f$ at $v$.

Let $[a,b]\subseteq V$. We fix additional elements
$a_{1}<b_{1}<\dots<a_{n-1}<b_{n-1}<a<b<a_{n}<b_{n}$ in $V$. Then define the
function
$\Psi_{1}:[a_{1},b_{1}]\times\dots\times[a_{n-1},b_{n-1}]\times[a,b]^{2}\to\mathbb{R}$
and
$\Psi_{2}:[a_{1},b_{1}]\times\dots\times[a_{n-2},b_{n-2}]\times[a,b]^{2}\times[a_{n},b_{n}]\to\mathbb{R}$
by

$\displaystyle\Psi_{1}(v_{1},\dots,v_{n-1},x,y):=\bigg{|}1-\frac{\Phi_{\omega}(v_{1},\dots,v_{n-1},y)}{\Phi_{\omega}(v_{1},\dots,v_{n-1},x)}\bigg{|}+\sum_{i=1}^{n-1}\bigg{|}\frac{\Phi_{\omega}(v_{1},\dots,v_{i-1},v_{i+1}\dots,v_{n-1},x,y)}{\Phi_{\omega}(v_{1},\dots,v_{n-1},x)}\bigg{|}.$

and

$\displaystyle\Psi_{2}(v_{1},\dots,v_{n-2},x,y,v_{n}):=$
$\displaystyle\bigg{|}1-\frac{\Phi_{\omega}(v_{1},\dots,v_{n-2},x,v_{n})}{\Phi_{\omega}(v_{1},\dots,v_{n-2},y,v_{n})}\bigg{|}+\frac{\Phi_{\omega}(v_{1},\dots,v_{n-2},x,y)}{\Phi_{\omega}(v_{1},\dots,v_{n-2},y,v_{n})}$
$\displaystyle+\sum_{i=1}^{n-2}\bigg{|}\frac{\Phi_{\omega}(v_{1},\dots,v_{i-1},v_{i+1}\dots,v_{n-2},x,y,v_{n})}{\Phi_{\omega}(v_{1},\dots,v_{n-2},y,v_{n})}\bigg{|}.$

Then the functions $\Psi_{1}$ and $\Psi_{2}$ are continuous over a compact
rectangle, therefore, they are uniformly continuous. On the other hand,
$\Psi_{1}(v_{1},\dots,v_{n-1},x,x)=\Psi_{2}(v_{1},\dots,v_{n-2},x,x,v_{n})=0$
for all $x\in[a,b]$ and
$(v_{1},\dots,v_{n-1},v_{n})\in[a_{1},b_{1}]\times\dots\times[a_{n-1},b_{n-1}]\times[a_{n},b_{n}]$.

Thus, for every $\varepsilon>0$, there exists $\delta>0$ such that, for all
$x,y\in[a,b]$ with $|x-y|<\delta$ and for all
$(v_{1},\dots,v_{n-1},v_{n})\in[a_{1},b_{1}]\times\dots\times[a_{n-1},b_{n-1}]\times[a_{n},b_{n}]$,

$\displaystyle\Psi_{1}(v_{1},\dots,v_{n-1},x,y)$
$\displaystyle=\Psi_{1}(v_{1},\dots,v_{n-1},x,y)-\Psi_{1}(v_{1},\dots,v_{n-1},x,x)<\frac{\varepsilon}{K},$
$\displaystyle\Psi_{2}(v_{1},\dots,v_{n-2},x,y,v_{n})$
$\displaystyle=\Psi_{2}(v_{1},\dots,v_{n-2},x,y,v_{n})-\Psi_{2}(v_{1},\dots,v_{n-2},x,x,v_{n})<\frac{\varepsilon}{K}.$

Now choose $x,y\in[a,b]$ with $x<y<x+\delta$. Then choose the rational numbers
$\lambda_{1},\dots,\lambda_{n}$ such that

$\displaystyle\frac{a_{i}-x}{y-x}<\lambda_{i}<\frac{b_{i}-x}{y-x},\qquad
i\in\\{1,\dots,n\\}.$

Then, with the notation $v_{i}:=(1-\lambda_{i})x+\lambda_{i}y$, we have that

$\displaystyle a_{i}<v_{i}<b_{i},\quad i\in\\{1,\dots,n\\}.$

First define

$\displaystyle
r_{i}:=\frac{v_{i}-v_{1}}{y-v_{1}}\quad(i\in\\{1,\dots,n-1\\}),\qquad
r_{n}:=\frac{x-v_{1}}{y-v_{1}},\qquad r_{n+1}:=\frac{y-v_{1}}{y-v_{1}}.$

Clearly, due to the chain of inequalities $v_{1}<v_{2}\dots<v_{n-1}<x<y$, we
have that $r_{1}=0<r_{2}<\dots<r_{n}<r_{n+1}=1$. On the other hand, for
$i\in\\{1,\dots,n-1\\}$,

$\displaystyle
r_{i}=\frac{v_{i}-v_{1}}{y-v_{1}}=\frac{(1-\lambda_{i})x+\lambda_{i}y-((1-\lambda_{1})x+\lambda_{1}y)}{y-((1-\lambda_{1})x+\lambda_{1}y))}=\frac{(\lambda_{i}-\lambda_{1})(y-x)}{(1-\lambda_{1})(y-x)}=\frac{\lambda_{i}-\lambda_{1}}{1-\lambda_{1}}\in\mathbb{Q},$

and

$\displaystyle
r_{n}=\frac{x-v_{1}}{y-v_{1}}=\frac{x-((1-\lambda_{1})x+\lambda_{1}y)}{y-((1-\lambda_{1})x+\lambda_{1}y))}=\frac{\lambda_{1}(x-y)}{(1-\lambda_{1})(y-x)}=\frac{-\lambda_{1}}{1-\lambda_{1}}\in\mathbb{Q}.$

Using Theorem 1.1, the $\omega$-Jensen convexity of $f$ implies that

$\displaystyle\Phi_{(\omega,f)}(v_{1}+r_{1}(y-v_{1}),\dots,v_{1}+r_{n-1}(y-v_{1}),v_{1}+r_{n}(y-v_{1}),v_{1}+r_{n+1}(y-v_{1}))\geq
0,$

which, according to the definition of the numbers
$r_{0},r_{1},\dots,r_{n-1},r_{n}$, yields that

$\displaystyle\Phi_{(\omega,f)}(v_{1},\dots,v_{n-1},x,y)\geq 0.$

Now from the above inequality, we get

$\displaystyle\begin{vmatrix}\omega_{1}(v_{1})&\dots&\omega_{1}(v_{n-1})&\omega_{1}(x)&\omega_{1}(y)\\\
\vdots&\ddots&\vdots&\vdots&\vdots\\\
\omega_{n}(v_{1})&\dots&\omega_{n}(v_{n-1})&\omega_{n}(x)&\omega_{n}(y)\\\
f(v_{1})&\dots&f(v_{n-1})&f(x)&f(y)\end{vmatrix}\geq 0.$

Developing this determinant by the last row, we obtain

$\displaystyle
f(y)\Phi_{\omega}(v_{1},\dots,v_{n-1},x)-f(x)\Phi_{\omega}(v_{1},\dots,v_{n-1},y)$
$\displaystyle+\sum_{i=1}^{n-1}(-1)^{n-i-1}f(v_{i})\Phi_{\omega}(v_{1},\dots,v_{i-1},v_{i+1}\dots,v_{n-1},x,y)\geq
0.$

By the positivity of Chebyshev system and the boundedness of $f$ over $V$, the
above inequality yields

(12) $\displaystyle f(x)-f(y)$ $\displaystyle\leq
f(x)\bigg{(}1-\frac{\Phi_{\omega}(v_{1},\dots,v_{n-1},y)}{\Phi_{\omega}(v_{1},\dots,v_{n-1},x)}\bigg{)}$
$\displaystyle\qquad+\sum_{i=1}^{n-1}(-1)^{n-i-1}f(v_{i})\frac{\Phi_{\omega}(v_{1},\dots,v_{i-1},v_{i+1}\dots,v_{n-1},x,y)}{\Phi_{\omega}(v_{1},\dots,v_{n-1},x)}$
$\displaystyle\leq K\Psi_{1}(v_{1},\dots,v_{n-1},x,y)<\varepsilon.$

Secondly define

$\displaystyle
s_{i}:=\frac{v_{i}-v_{1}}{v_{n}-v_{1}}\quad(i\in\\{1,\dots,n-2\\}\cup\\{n\\}),\qquad
s^{\prime}_{n-1}:=\frac{x-v_{1}}{v_{n}-v_{1}},\quad
s^{\prime\prime}_{n-1}:=\frac{y-v_{1}}{v_{n}-v_{1}}.$

Due to the chain inequalities $v_{1}<v_{2}\dots<v_{n-2}<x<y<v_{n}$, we have
that $s_{1}=0<s_{2}<\dots<s^{\prime}_{n-1}<s^{\prime\prime}_{n-1}<s_{n}=1$. On
the other hand, for $i\in\\{1,\dots,n-2\\}\cup\\{n\\}$,

$\displaystyle
s_{i}=\frac{v_{i}-v_{1}}{v_{n}-v_{1}}=\frac{(1-\lambda_{i})x+\lambda_{i}y-((1-\lambda_{1})x+\lambda_{1}y)}{(1-\lambda_{n})x+\lambda_{n}y-((1-\lambda_{1})x+\lambda_{1}y)}=\frac{(\lambda_{i}-\lambda_{1})(y-x)}{(\lambda_{n}-\lambda_{1})(y-x)}=\frac{\lambda_{i}-\lambda_{1}}{\lambda_{n}-\lambda_{1}}\in\mathbb{Q},$
$\displaystyle
s^{\prime}_{n-1}=\frac{x-v_{1}}{v_{n}-v_{1}}=\frac{x-((1-\lambda_{1})x+\lambda_{1}y)}{(1-\lambda_{n})x+\lambda_{n}y-((1-\lambda_{1})x+\lambda_{1}y)}=\frac{-\lambda_{1}(y-x)}{(\lambda_{n}-\lambda_{1})(y-x)}=\frac{-\lambda_{1}}{\lambda_{n}-\lambda_{1}}\in\mathbb{Q}$

and

$\displaystyle
s^{\prime\prime}_{n-1}=\frac{y-v_{1}}{v_{n}-v_{1}}=\frac{y-((1-\lambda_{1})x+\lambda_{1}y)}{(1-\lambda_{n})x+\lambda_{n}y-((1-\lambda_{1})x+\lambda_{1}y)}=\frac{(1-\lambda_{1})(y-x)}{(\lambda_{n}-\lambda_{1})(y-x)}=\frac{1-\lambda_{1}}{\lambda_{n}-\lambda_{1}}\in\mathbb{Q}.$

Now using Theorem 1.1, the $\omega$-Jensen convexity of $f$ implies that

$\displaystyle\Phi_{(\omega,f)}(v_{1}+s_{1}(v_{n}-v_{1}),\dots,v_{1}+s^{\prime}_{n-1}(v_{n}-v_{1}),v_{1}+s^{\prime\prime}_{n-1}(v_{n}-v_{1}),v_{1}+s_{n}(v_{n}-v_{1}))\geq
0,$

which, according to the definition of the numbers
$s_{1},s_{2},\dots,s^{\prime}_{n-1},s^{\prime\prime}_{n-1},s_{n}$, yields that

$\displaystyle\Phi_{(\omega,f)}(v_{1},\dots,v_{n-2},x,y,v_{n})\geq 0.$

This inequality can be written as

$\displaystyle\begin{vmatrix}\omega_{1}(v_{1})&\dots&\omega_{1}(v_{n-2})&\omega_{1}(x)&\omega_{1}(y)&\omega_{1}(v_{n})\\\
\vdots&\ddots&\vdots&\vdots&\vdots&\vdots\\\
\omega_{n}(v_{1})&\dots&\omega_{n}(v_{n-2})&\omega_{n}(x)&\omega_{n}(y)&\omega_{n}(v_{n})\\\
f(v_{1})&\dots&f(v_{n-2})&f(x)&f(y)&f(v_{n})\end{vmatrix}\geq 0.$

Developing this determinant by the last row, we obtain

$\displaystyle
f(v_{n})\Phi_{\omega}(v_{1},\dots,v_{n-2},x,y)-f(y)\Phi_{\omega}(v_{1},\dots,v_{n-2},x,v_{n})+f(x)\Phi_{\omega}(v_{1},\dots,v_{n-2},y,v_{n})$
$\displaystyle+\sum_{i=1}^{n-2}(-1)^{n-i-1}f(v_{i})\Phi_{\omega}(v_{1},\dots,v_{i-1},v_{i+1}\dots,v_{n-2},x,y,v_{n})\geq
0,$

by the positivity of Chebyshev system, the above inequality is equivalent to

(13) $\displaystyle f(y)-f(x)$ $\displaystyle\leq
f(y)\bigg{(}1-\frac{\Phi_{\omega}(v_{1},\dots,v_{n-2},x,v_{n})}{\Phi_{\omega}(v_{1},\dots,v_{n-2},y,v_{n})}\bigg{)}+f(v_{n})\frac{\Phi_{\omega}(v_{1},\dots,v_{n-2},x,y)}{\Phi_{\omega}(v_{1},\dots,v_{n-2},y,v_{n})}$
$\displaystyle\qquad+\sum_{i=1}^{n-2}(-1)^{n-i-1}f(v_{i})\frac{\Phi_{\omega}(v_{1},\dots,v_{i-1},v_{i+1}\dots,v_{n-2},x,y,v_{n})}{\Phi_{\omega}(v_{1},\dots,v_{n-2},y,v_{n})}$
$\displaystyle\leq K\Psi_{2}(v_{1},\dots,v_{n-2},x,y,v_{n})<\epsilon.$

Hence the inequalities (12) and (13) imply that, for all $x,y\in[a,b]$ with
$|x-y|<\delta$,

$\displaystyle|f(x)-f(y)|<\varepsilon$

holds. This proves that $f$ is uniformly continuous on $[a,b]$. The closed
interval $[a,b]\subseteq V$ was arbitrary, therefore $f$ is continuous on $V$.
∎

######  Theorem 2.6.

Let $f:I\to\mathbb{R}$ be a function which is bounded on a nonempty open
subset of $I$. Then it is $\omega$-Jensen affine if and only if
$f=\alpha_{1}\omega_{1}+\dots+\alpha_{n}\omega_{n}$ for some
$\alpha_{1},\dots,\alpha_{n}\in\mathbb{R}$.

###### Proof.

Assume first that $f$ is an $\omega$-Jensen affine function. Then, it is
$\omega$-Jensen convex, and using Theorem 2.5, it follows that $f$ is
continuous on $I$. We show first that, for all $x_{0},x_{1},\dots,x_{n}\in I$,

(14) $\displaystyle\Phi_{\omega,f}(x_{0},x_{1},\dots,x_{n})=0.$

Indeed, if two of points $x_{0},x_{1},\dots,x_{n}$ coincide, then this
equality is obvious. We may assume that these points are pairwise distinct
moreover that $x_{0}<x_{1}<\dots<x_{n}$. Let $0<r_{1,k}<\dots<r_{n,k}$ be
rational sequences converging to $0<x_{1}-x_{0}<\dots<x_{n}-x_{0}$,
respectively. Then, according to Theorem 1.1, the $\omega$-Jensen affinity of
$f$, for all $k\in\mathbb{N}$, yields that

$\displaystyle\Phi_{\omega,f}(x_{0},x_{0}+r_{1,k},\dots,x_{0}+r_{n,k})=0.$

Using the continuity of $f$ and taking the limit $k\to\infty$, it follows that
(14) holds.

Let us fix $x_{1}<\dots<x_{n}$ in $I$ arbitrarily. Then,

$\displaystyle\Phi_{\omega,f}(x,x_{1},\dots,x_{n})=0.$

holds for all $x\in I$, i.e.,

$\displaystyle\begin{vmatrix}\omega_{1}(x)&\omega_{1}(x_{1})&\dots&\omega_{1}(x_{n})\\\
\vdots&\vdots&\ddots&\vdots\\\
\omega_{n}(x)&\omega_{n}(x_{1})&\dots&\omega_{n}(x_{n})\\\
f(x)&f(x_{1})&\dots&f(x_{n})\end{vmatrix}=0.$

Developing this determinant by the first column, we obtain

$\displaystyle\sum_{i=1}^{n}(-1)^{i-1}\omega_{i}(x)\Phi_{\omega_{1},\dots,\omega_{i-1},\omega_{i+1},\dots,\omega_{n},f}(x_{1},\dots,x_{n})+(-1)^{n}f(x)\Phi_{\omega}(x_{1},\dots,x_{n})=0.$

Therefore,

$\displaystyle
f(x)=\sum_{i=1}^{n}(-1)^{n-i}\frac{\Phi_{\omega_{1},\dots,\omega_{i-1},\omega_{i+1},\dots,\omega_{n},f}(x_{1},\dots,x_{n})}{\Phi_{\omega}(x_{1},\dots,x_{n})}\omega_{i}(x).$

Now, with an obvious choice of $\alpha_{1},\dots,\alpha_{n}\in\mathbb{R}$, we
can see that $f=\alpha_{1}\omega_{1}+\dots+\alpha_{n}\omega_{n}$ holds.

The reversed statement is obvious, if $f$ is a linear combination of the
coordinate functions of $\omega$, then $\Phi_{\omega,f}$ is identically zero
by standard properties of determinants. ∎

The following question seems to be important: What is a necessary and
sufficient condition on $\omega$ that ensures the existence of discontinuous
$\omega$-Jensen affine or discontinuous $\omega$-Jensen convex functions? The
following result provides a sufficient condition. To formulate and prove this
sufficient condition we need the following lemma.

######  Lemma 2.7.

Assume that there exist a positive continuous function
$\omega_{0}:I\to\mathbb{R}_{+}$ and a matrix $M=(a_{i,j})_{1\leq i\leq
n,\,0\leq j\leq n-1}\in\mathbb{R}^{n\times n}$ such that

(15)
$\displaystyle\omega_{i}(x)=(a_{i,n-1}x^{n-1}+\dots+a_{i,1}x+a_{i,0})\cdot\omega_{0}(x)\qquad(x\in
I,\,i\in\\{1,\dots,n\\}).$

Then $\det(M)>0$ and, for all $(x_{1},\dots,x_{n})\in\sigma_{n}(I)$,

(16)
$\displaystyle\Phi_{\omega}(x_{1},\dots,x_{n})=\omega_{0}(x_{1})\cdots\omega_{0}(x_{n})\cdot\det(M)\cdot\Phi_{\pi_{n}}(x_{1},\dots,x_{n}).$

Additionally, let $f:I\to\mathbb{R}$. Then, for all
$(x_{0},\dots,x_{n})\in\sigma_{n+1}(I)$,

(17)
$\displaystyle\Phi_{\omega,f}(x_{0},\dots,x_{n})=\omega_{0}(x_{0})\cdots\omega_{0}(x_{n})\cdot\det(M)\cdot\Phi_{\pi_{n},f/\omega_{0}}(x_{0},\dots,x_{n}).$

###### Proof.

The equality in (15) and the product rule for determinants imply, for all
$(x_{1},\dots,x_{n})\in\sigma_{n}(I)$, that

$\displaystyle\Phi_{\omega}$ $\displaystyle(x_{1},\dots,x_{n})$
$\displaystyle=\begin{vmatrix}(a_{1,n-1}x_{1}^{n-1}+\dots+a_{1,1}x_{1}+a_{1,0})\omega_{0}(x_{1})&\dots&(a_{1,n-1}x_{n}^{n-1}+\dots+a_{1,1}x_{n}+a_{1,0})\omega_{0}(x_{n})\\\
\vdots&\ddots&\vdots\\\
(a_{n,n-1}x_{1}^{n-1}+\dots+a_{n,1}x_{1}+a_{n,0})\omega_{0}(x_{1})&\dots&(a_{n,n-1}x_{n}^{n-1}+\dots+a_{n,1}x_{n}+a_{n,0})\omega_{0}(x_{n})\end{vmatrix}$
$\displaystyle=\omega_{0}(x_{1})\cdots\omega_{0}(x_{n})\begin{vmatrix}a_{1,n-1}x_{1}^{n-1}+\dots+a_{1,1}x_{1}+a_{1,0}&\dots&a_{1,n-1}x_{n}^{n-1}+\dots+a_{1,1}x_{n}+a_{1,0}\\\
\vdots&\ddots&\vdots\\\
a_{n,n-1}x_{1}^{n-1}+\dots+a_{n,1}x_{1}+a_{n,0}&\dots&a_{n,n-1}x_{n}^{n-1}+\dots+a_{n,1}x_{n}+a_{n,0}\end{vmatrix}$
$\displaystyle=\omega_{0}(x_{1})\cdots\omega_{0}(x_{n})\begin{vmatrix}a_{1,0}&\dots&a_{1,n-1}\\\
\vdots&\ddots&\vdots\\\
a_{n,0}&\dots&a_{n,n-1}\end{vmatrix}\cdot\begin{vmatrix}x_{1}^{0}&\dots&x_{n}^{0}\\\
\vdots&\ddots&\vdots\\\ x_{1}^{n-1}&\dots&x_{n}^{n-1}\end{vmatrix}$
$\displaystyle=\omega_{0}(x_{1})\cdots\omega_{0}(x_{n})\det(M)\cdot\Phi_{\pi_{n}}(x_{1},\dots,x_{n}),$

which proves (16).

The value of the determinant $\Phi_{\pi_{n}}(x_{1},\dots,x_{n})$ equals
$\prod_{1\leq i<j\leq n}(x_{j}-x_{i})>0$ (because it is of Vandermonde-type).
Therefore, the positivity of the Chebyshev system $\omega$, the positivity of
the function $\omega_{0}$ and the equality (16) yield that $\det(M)>0$.

The equalities in (15) and the product rule for determinants again imply, for
all $(x_{0},\dots,x_{n})\in\sigma_{n+1}(I)$, that

$\displaystyle\Phi_{\omega,f}$ $\displaystyle(x_{0},\dots,x_{n})$
$\displaystyle=\begin{vmatrix}(a_{1,n-1}x_{0}^{n-1}+\dots+a_{1,1}x_{0}+a_{1,0})\omega_{0}(x_{0})&\dots&(a_{1,n-1}x_{n}^{n-1}+\dots+a_{1,1}x_{n}+a_{1,0})\omega_{0}(x_{n})\\\
\vdots&\ddots&\vdots\\\
(a_{n,n-1}x_{0}^{n-1}+\dots+a_{n,0}x_{1}+a_{n,0})\omega_{0}(x_{0})&\dots&(a_{n,n-1}x_{n}^{n-1}+\dots+a_{n,1}x_{n}+a_{n,0})\omega_{0}(x_{n})\\\
f(x_{0})&\dots&f(x_{n})\end{vmatrix}$
$\displaystyle=\omega_{0}(x_{0})\cdots\omega_{0}(x_{n})\begin{vmatrix}a_{1,n-1}x_{0}^{n-1}+\dots+a_{1,1}x_{0}+a_{1,0}&\dots&a_{1,n-1}x_{n}^{n-1}+\dots+a_{1,1}x_{n}+a_{1,0}\\\
\vdots&\ddots&\vdots\\\
a_{n,n-1}x_{0}^{n-1}+\dots+a_{n,1}x_{0}+a_{n,0}&\dots&a_{n,n-1}x_{n}^{n-1}+\dots+a_{n,1}x_{n}+a_{n,0}\\\
\frac{f}{\omega_{0}}(x_{0})&\dots&\frac{f}{\omega_{0}}(x_{n})\end{vmatrix}$
$\displaystyle=\omega_{0}(x_{0})\cdots\omega_{0}(x_{n})\begin{vmatrix}a_{1,0}&\dots&a_{1,n-1}&0\\\
\vdots&\ddots&\vdots\\\ a_{n,0}&\dots&a_{n,n-1}&0\\\
0&\dots&0&1\end{vmatrix}\cdot\begin{vmatrix}x_{0}^{0}&\dots&x_{n}^{0}\\\\[5.69054pt]
\vdots&\ddots&\vdots\\\ x_{0}^{n-1}&\dots&x_{n}^{n-1}\\\\[5.69054pt]
\frac{f}{\omega_{0}}(x_{0})&\dots&\frac{f}{\omega_{0}}(x_{n})\end{vmatrix}$
$\displaystyle=\omega_{0}(x_{0})\cdots\omega_{0}(x_{n})\cdot\det(M)\cdot\Phi_{\pi_{n},f/\omega_{0}}(x_{0},\dots,x_{n}),$

which shows the validity of (17). ∎

######  Theorem 2.8.

Assume that there exist a positive continuous function
$\omega_{0}:I\to\mathbb{R}_{+}$ and a matrix $M=(a_{i,j})_{1\leq i\leq
n,\,0\leq j\leq n-1}\in\mathbb{R}^{n\times n}$ such that (15) holds. Then
$f:I\to\mathbb{R}$ is an $\omega$-Jensen convex (resp. $\omega$-Jensen affine)
function if and only if $\frac{f}{\omega_{0}}$ is a $\pi_{n}$-Jensen convex
(resp. $\pi_{n}$-Jensen affine) function.

###### Proof.

According to formula (17) of Lemma 2.7, for all $h>0$ and $x\in I\cap(I-nh)$,
we have that

$\displaystyle\Phi_{\omega,f}(x,x+h,\dots,x+nh)=\prod_{i=0}^{n}\omega_{0}(x+ih)\cdot\det(M)\cdot\Phi_{\pi_{n},f/\omega_{0}}(x,x+h,\dots,x+nh).$

Due to the positivity of $\det(M)$ and the positivity of the function
$\omega_{0}$, it follows that the inequality
$\Phi_{\omega,f}(x,x+h,\dots,x+nh)\geq 0$ holds if and only if
$\Phi_{\pi_{n},f/\omega_{0}}(x,x+h,\dots,x+nh)\geq 0$ is valid. This shows
that $f$ is $\omega$-Jensen convex if and only if $\frac{f}{\omega_{0}}$ is
$\pi_{n}$-Jensen convex.

Similarly, $\Phi_{\omega,f}(x,x+h,\dots,x+nh)=0$ if and only if
$\Phi_{\pi_{n},f/\omega_{0}}(x,x+h,\dots,x+nh)=0$, which proves that $f$ is
$\omega$-Jensen affine if and only if $\frac{f}{\omega_{0}}$ is
$\pi_{n}$-Jensen affine. ∎

We need to recall the following characterization of $\pi_{n}$-Jensen affine
functions.

######  Theorem 2.9.

A function $f:I\to\mathbb{R}$ is $\pi_{n}$-Jensen affine if and only if there
exist a constant $A_{0}\in\mathbb{R}$, an additive function
$A_{1}:\mathbb{R}\to\mathbb{R}$, a symmetric biadditive function
$A_{2}:\mathbb{R}^{2}\to\mathbb{R}$, …, and a symmetric $(n-1)$-additive
function $A_{n-1}:\mathbb{R}^{n-1}\to\mathbb{R}$ such that

(18) $\displaystyle
f(x)=A_{n-1}(x,\dots,x)+\dots+A_{2}(x,x)+A_{1}(x)+A_{0}\qquad(x\in I).$

######  Corollary 2.10.

Assume that there exist a positive continuous function
$\omega_{0}:I\to\mathbb{R}_{+}$ and a matrix $(a_{i,j})_{1\leq i\leq n,\,0\leq
j\leq n-1}\in\mathbb{R}^{n\times n}$ such that (15) holds. Then
$f:I\to\mathbb{R}$ is an $\omega$-Jensen affine function if and only if there
exist a constant $A_{0}\in\mathbb{R}$, an additive function
$A_{1}:\mathbb{R}\to\mathbb{R}$, a symmetric biadditive function
$A_{2}:\mathbb{R}^{2}\to\mathbb{R}$, …, and a symmetric $(n-1)$-additive
function $A_{n-1}:\mathbb{R}^{n-1}\to\mathbb{R}$ such that

(19) $\displaystyle
f(x)=\big{(}A_{n-1}(x,\dots,x)+\dots+A_{2}(x,x)+A_{1}(x)+A_{0}\big{)}\omega_{0}(x)\qquad(x\in
I).$

###### Proof.

Assume first that $f$ is $\omega$-Jensen affine. Then, by Theorem 2.8,
$\frac{f}{\omega_{0}}$ is $\pi_{n}$-Jensen affine. Hence, according to Theorem
2.9, there exist a constant $A_{0}\in\mathbb{R}$, and additive function
$A_{1}:\mathbb{R}\to\mathbb{R}$, a symmetric biadditive function
$A_{2}:\mathbb{R}^{2}\to\mathbb{R}$, …, a symmetric $(n-1)$-additive function
$A_{n-1}:\mathbb{R}^{n-1}\to\mathbb{R}$ such that

$\displaystyle\frac{f}{\omega_{0}}(x)=A_{n-1}(x,\dots,x)+\dots+A_{2}(x,x)+A_{1}(x)+A_{0}\qquad(x\in
I).$

This proves that $f$ is of the form (19).

To prove the reversed implication, assume that there exist a constant
$A_{0}\in\mathbb{R}$, and additive function $A_{1}:\mathbb{R}\to\mathbb{R}$, a
symmetric biadditive function $A_{2}:\mathbb{R}^{2}\to\mathbb{R}$, …, a
symmetric $(n-1)$-additive function $A_{n-1}:\mathbb{R}^{n-1}\to\mathbb{R}$
such that (19) holds. Then, according to Theorem 2.9, $\frac{f}{\omega_{0}}$
is a $\pi_{n}$-Jensen affine function. In view of Theorem 2.8, this implies
that $f$ is $\omega$-Jensen affine. ∎

##  3\. Wright convexity with respect to extended Chebyshev systems

In 1954, Wright [14] introduced a concept of convexity which is stronger than
Jensen convexity and weaker than convexity. A function $f:I\to\mathbb{R}$ is
called _Wright convex_ if

$\displaystyle f(tx+(1-t)y)+f((1-t)x+ty)\leq f(x)+f(y)\qquad(x,y\in
I,\,t\in[0,1]).$

A characterization and the ultimate understanding of Wright convexity was
established by Ng [9], who proved that $f:I\to\mathbb{R}$ is Wright convex if
and only if it is of the form $f=g+A|_{I}$, where $g:I\to\mathbb{R}$ is convex
and $A:\mathbb{R}\to\mathbb{R}$ is additive.

The higher-order generalizations of Wright convexity were introduced by
Gilányi and Páles [2] as follows. A function $f:I\to\mathbb{R}$ was called
_Wright convex of order $(n-1)$_, if

$\displaystyle\Delta_{h_{1}}\cdots\Delta_{h_{n}}f(x)\geq 0$

holds for all $h_{1},\dots,h_{n}>0$ and $x\in I\cap(I-(h_{1}+\dots+h_{n}))$.
One can easily see that Wright convex of order $1$ is equivalent to Wright
convexity in the standard sense.

In what follows, we extend the notion of higher-order Wright convexity to the
setting of Chebyshev systems. Let
$\omega=(\omega_{1},\dots,\omega_{n}):I\to\mathbb{R}^{n}$ be a positive
$n$-dimensional Chebyshev system. We say that
$\overline{\omega}:I\to\mathbb{R}^{n+1}$ _is an extension of $\omega$_ if
there exists a continuous function $\omega_{n+1}:I\to\mathbb{R}$ such that
$\overline{\omega}:=(\omega_{1},\dots,\omega_{n},\omega_{n+1})$ and
$\overline{\omega}$ is a positive $(n+1)$-dimensional Chebyshev system.

Let $\omega$ be a positive $n$-dimensional Chebyshev system and
$\overline{\omega}$ be an arbitrarily fixed extension of $\omega$. We say that
a function _$f:I\to\mathbb{R}$ is $\overline{\omega}$-Wright convex_ if, for
all $h_{1},\dots,h_{n}>0$ and $x\in I\cap(I-(h_{1}+\dots+h_{n}))$, the
inequality

(20)
$\displaystyle\sum_{(i_{1},\dots,i_{n})}\frac{\Phi_{(\omega,f)}(x,x+h_{i_{1}},\dots,x+h_{i_{1}}+\dots+h_{i_{n}})}{\Phi_{\overline{\omega}}(x,x+h_{i_{1}},\dots,x+h_{i_{1}}+\dots+h_{i_{n}})}\geq
0$

holds, where the summation is taken over all permutation $(i_{1},\dots,i_{n})$
of the elements $\\{1,\dots,n\\}$.

Our first result establishes the connections between $\omega$-convexity,
$\overline{\omega}$-Wright convexity and $\omega$-Jensen convexity.

######  Theorem 3.1.

Let $\omega$ be a positive $n$-dimensional Chebyshev system and
$\overline{\omega}$ be an extension of $\omega$. Then every $\omega$-convex
function is $\overline{\omega}$-Wright convex and every
$\overline{\omega}$-Wright convex function is $\omega$-Jensen convex.

###### Proof.

Assume first that $f:I\to\mathbb{R}$ is $\omega$-convex. Then, for all
$h_{1},\dots,h_{n}>0$ and $x\in I\cap(I-(h_{1}+\dots+h_{n}))$ all permutation
$(i_{1},\dots,i_{n})$ of the elements $\\{1,\dots,n\\}$, we have that

$\displaystyle\Phi_{(\omega,f)}(x,x+h_{i_{1}},\dots,x+h_{i_{1}}+\dots+h_{i_{n}})\geq
0.$

On the other hand, by the positivity of the Chebyshev system
$\overline{\omega}$, we also have that

$\displaystyle\Phi_{\overline{\omega}}(x,x+h_{i_{1}},\dots,x+h_{i_{1}}+\dots+h_{i_{n}})>0.$

These inequalities yield that (20) is valid on the domain indicated and hence
$f$ is $\overline{\omega}$-Wright convex.

To verify the second assertion, assume that $f:I\to\mathbb{R}$ is
$\overline{\omega}$-Wright convex. Taking $h_{1}:=\dots=h_{n}:=h>0$ in
inequality (20), for all $h>0$ and $x\in I\cap(I-nh)$, we obtain that

(21)
$\displaystyle\frac{\Phi_{(\omega,f)}(x,x+h,\dots,x+nh)}{\Phi_{\overline{\omega}}(x,x+h,\dots,x+nh)}\geq
0.$

Due to the positivity of the Chebyshev system $\overline{\omega}$, it follows
that $\Phi_{\overline{\omega}}(x,x+h,\dots,x+nh)$ is positive, therefore, we
can conclude that $\Phi_{(\omega,f)}(x,x+h,\dots,x+nh)\geq 0$, which shows
that $f$ is $\omega$-Jensen convex. ∎

The next theorem describes the connection between $\overline{\omega}$-Wright
convexity and Wright convexity of order $(n-1)$.

######  Theorem 3.2.

Assume that there exist a positive continuous function
$\omega_{0}:I\to\mathbb{R}_{+}$ and a matrix $M:=(a_{i,j})_{1\leq i\leq
n,\,0\leq j\leq n-1}\in\mathbb{R}^{n\times n}$ such that (15) holds. Define
$\omega_{n+1}:I\to\mathbb{R}$ by $\omega_{n+1}(t):=t^{n}\omega_{0}(t)$. Then
$\overline{\omega}:=(\omega,\omega_{n+1})$ is an extension of the Chebyshev
system $\omega$. In addition, we have the following assertions:

  1. (i)

For all $(x_{0},x_{1},\dots,x_{n})\in\sigma_{n+1}(I)$, the equality

(22)
$\displaystyle\frac{\Phi_{(\omega,f)}(x_{0},x_{1},\dots,x_{n})}{\Phi_{\overline{\omega}}(x_{0},x_{1},\dots,x_{n})}=[x_{0},x_{1},\dots,x_{n};f/\omega_{0}]$

holds. Furthermore, a function $f:I\to\mathbb{R}$ is $\omega$-convex if and
only if $f/\omega_{0}$ is convex of order $(n-1)$.

  2. (ii)

For all $h_{1},\dots,h_{n}>0$ and $x\in I\cap(I-(h_{1}+\dots+h_{n}))$, the
equality

(23)
$\displaystyle\sum_{(i_{1},\dots,i_{n})}\frac{\Phi_{(\omega,f)}(x,x+h_{i_{1}},\dots,x+h_{i_{1}}+\dots+h_{i_{n}})}{\Phi_{\overline{\omega}}(x,x+h_{i_{1}},\dots,x+h_{i_{1}}+\dots+h_{i_{n}})}=\frac{\Delta_{h_{1}}\cdots\Delta_{h_{n}}(f/\omega_{0})(x)}{h_{1}\cdots
h_{n}}$

holds. Furthermore, a function $f:I\to\mathbb{R}$ is
$\overline{\omega}$-Wright convex if and only if $f/\omega_{0}$ is Wright
convex of order $(n-1)$.

###### Proof.

We first verify that $\overline{\omega}=(\omega,\omega_{n+1})$ is a positive
Chebyshev system. Let $(x_{0},x_{1},\dots,x_{n})\in\sigma_{n+1}(I)$. Then,
applying the equality (17) of Lemma 2.7 with $f:=\omega_{n+1}$, we get

(24) $\displaystyle\Phi_{\overline{\omega}}(x_{0},x_{1},\dots,x_{n})$
$\displaystyle=\omega_{0}(x_{0})\cdots\omega_{0}(x_{n})\cdot\det(M)\cdot\Phi_{(\pi_{n},\omega_{n+1}/\omega_{0})}(x_{0},x_{1},\dots,x_{n})$
$\displaystyle=\omega_{0}(x_{0})\cdots\omega_{0}(x_{n})\cdot\det(M)\cdot\Phi_{\pi_{n+1}}(x_{0},x_{1},\dots,x_{n})>0.$

The last inequality is due to the fact that
$\Phi_{\pi_{n+1}}(x_{0},x_{1},\dots,x_{n})$ is a Vandermonde determinant and
$x_{0}<x_{1}<\dots<x_{n}$. This proves that $\overline{\omega}$ is a positive
Chebyshev system, indeed.

In the rest of the proof denote $g:=f/\omega_{0}$. To show that assertion (i)
holds, let $(x_{0},x_{1},\dots,x_{n})\in\sigma_{n+1}(I)$ be fixed. In view of
the Lemma 2.7, we have the equality (17). Combining this equality with (24),
we can obtain

$\displaystyle\frac{\Phi_{(\omega,f)}(x_{0},x_{1},\dots,x_{n})}{\Phi_{\overline{\omega}}(x_{0},x_{1},\dots,x_{n})}=\frac{\Phi_{(\pi_{n},g)}(x_{0},x_{1},\dots,x_{n})}{\Phi_{\pi_{n+1}}(x_{0},x_{1},\dots,x_{n})}.$

From the theory of divided differences, we have the identity

$\displaystyle\frac{\Phi_{(\pi_{n},g)}(x_{0},x_{1},\dots,x_{n})}{\Phi_{\pi_{n+1}}(x_{0},x_{1},\dots,x_{n})}=[x_{0},x_{1},\dots,x_{n};g],$

which, together with the previous equality shows that (22) holds.

The function $f$ is $\overline{\omega}$-convex if and only if, for all
$(x_{0},x_{1},\dots,x_{n})\in\sigma_{n+1}(I)$, the left hand side of (22) is
nonnegative. According to this equality, this happens if and only if the right
hand side is nonnegative, i.e., if $f/\omega_{0}$ is convex of order $(n-1)$.

To show assertion (ii), let $h_{1},\dots,h_{n}>0$ and $x\in
I\cap(I-(h_{1}+\dots+h_{n}))$ be fixed. Therefore, with the substitutions
$x_{j}:=x+h_{1}+\dots+h_{j}$, (where $j\in\\{0,\dots,n\\}$), (22) implies that

$\displaystyle\frac{\Phi_{(\omega,f)}(x,x+h_{1},\dots,x+h_{1}+\dots+h_{n})}{\Phi_{\overline{\omega}}(x,x+h_{1},\dots,x+h_{1}+\dots+h_{n})}=[x,x+h_{1},\dots,x+h_{1}+\dots+h_{n};g].$

Applying this equality for $(h_{i_{1}},\dots,h_{i_{n}})$ (instead of
$(h_{1},\dots,h_{n})$), where $(i_{1},\dots,i_{n})$ is an arbitrary
permutation of $(1,\dots,n)$, we can see that

$\displaystyle\sum_{(i_{1},\dots,i_{n})}\frac{\Phi_{(\omega,f)}(x,x+h_{i_{1}},\dots,x+h_{i_{1}}+\dots+h_{i_{n}})}{\Phi_{\overline{\omega}}(x,x+h_{i_{1}},\dots,x+h_{i_{1}}+\dots+h_{i_{n}})}=\sum_{(i_{1},\dots,i_{n})}[x,x+h_{i_{1}},\dots,x+h_{i_{1}}+\dots+h_{i_{n}};g].$

On the other hand, from the paper [2], we have that

$\displaystyle\sum_{(i_{1},\dots,i_{n})}[x,x+h_{i_{1}},\dots,x+h_{i_{1}}+\dots+h_{i_{n}};g]=\frac{\Delta_{h_{1}}\dots\Delta_{h_{n}}g(x)}{h_{1}\cdots
h_{n}}$

holds, which, together with the previous equality implies (23).

The function $f$ is $\overline{\omega}$-Wright convex if and only if, for all
$h_{1},\dots,h_{n}>0$ and $x\in I\cap(I-(h_{1}+\dots+h_{n}))$, the left hand
side of (23) is nonnegative. According to this equality this happens to be
valid if and only if the right hand side is nonnegative, i.e., if
$f/\omega_{0}$ is Wright convex of order $(n-1)$. ∎

In the following result, we establish a characterization theorem for
$\overline{\omega}$-Wright convexity provided that the underlying Chebyshev
system is strongly related to the polynomial one. This result generalizes the
decomposition theorem of Maksa and Páles [7] which is related to the
polynomial system. An alternative and more elementary proof of that theorem
has been recently given by the authors in [11].

######  Theorem 3.3.

Assume that there exist a positive continuous function
$\omega_{0}:I\to\mathbb{R}_{+}$ and a matrix $M:=(a_{i,j})_{1\leq i\leq
n,\,0\leq j\leq n-1}\in\mathbb{R}^{n\times n}$ such that (15) holds. Define
$\omega_{n+1}:I\to\mathbb{R}$ by $\omega_{n+1}(t):=t^{n}\omega_{0}(t)$ and set
$\overline{\omega}:=(\omega,\omega_{n+1})$. Then a function $f:I\to\mathbb{R}$
is $\overline{\omega}$-Wright convex if and only if there exist an
$\omega$-convex function $F:I\to\mathbb{R}$ and, for each
$k\in\\{1,\dots,n-1\\}$, a symmetric $k$-additive mapping
$A_{k}:\mathbb{R}^{k}\to\mathbb{R}$ and a real constant $A_{0}$ such that, for
all $x\in I$,

(25) $\displaystyle
f(x)=F(x)+(A_{0}+A_{1}(x)+\dots+A_{n-1}(x,\dots,x))\omega_{0}(x).$

###### Proof.

To prove the necessity, assume that the function $f$ is
$\overline{\omega}$-Wright convex. Then the assertion (ii) of Theorem 3.2
implies that $f/\omega_{0}$ is Wright convex of order $(n-1)$. The
decomposition theorem of higher order Wright convex functions [7] implies that
there exist a function $G:I\to\mathbb{R}$ which is convex of order $(n-1)$, a
real constant $A_{0}$ and, for each $k\in\\{1,\dots,n-1\\}$, a symmetric
$k$-additive mapping $A_{k}:\mathbb{R}^{k}\to\mathbb{R}$ such that, for all
$x\in I$,

(26)
$\displaystyle\frac{f}{\omega_{0}}(x)=G(x)+A_{0}+A_{1}(x)+\dots+A_{n-1}(x,\dots,x).$

This implies that (25) holds with $F:=G\omega_{0}$ and $F/\omega_{0}$ is
convex of order $(n-1)$. By assertion (i) of Theorem 3.2, it follows that the
function $F$ is $\omega$-convex.

To prove the sufficiency, assume that (25) holds, multiplying (25) by
$1/\omega_{0}(x)$ implies that

$\displaystyle\frac{f}{\omega_{0}}(x)=\frac{F}{\omega_{0}}(x)+(A_{0}+A_{1}(x)+\dots+A_{n-1}(x,\dots,x)).$

Since the function $F$ is $\omega$-convex, therefore the assertion (i) of
Theorem 3.2 implies that $F/\omega_{0}$ is convex of order $(n-1)$. Again, by
the decomposition theorem of higher order Wright convex functions [7], we can
conclude that $f/\omega_{0}$ is Wright convex of order $(n-1)$. Thus, the
assertion (ii) of Theorem 3.2 yields that the function $f$ is
$\overline{\omega}$-Wright convex. ∎

In our subsequent result we will prove that if the extension of the two
dimensional polynomial system is not a polynomial of at most second degree,
then the convexity with respect to the two dimensional polynomial system
(i.e., standard convexity) is equivalent to Wright convexity with respect to
this extension. For the proof of this result, we will need the following
characterization of a polynomial of at most second degree.

######  Lemma 3.4.

Let $\rho:I\to\mathbb{R}$ be a continuous function which satisfies the
functional equation

(27)
$\displaystyle\frac{\rho(z)-\rho(y)}{z-y}=\frac{\rho(z+u)-\rho(y-u)}{(z+u)-(y-u)},\qquad
u\geq 0,\,y,z\in(I+u)\cap(I-u),\,z+u\geq y.$

Then $\rho$ is a polynomial of at most second degree over $I$.

###### Proof.

If $u>0$ and $y\in(I+u)\cap(I-u)$, then the limit of the right hand side
exists as $z\to y$ by the continuity of $\rho$, which shows that $\rho$ is
differentiable at $y$ and we get

(28) $\displaystyle\rho^{\prime}(y)=\frac{\rho(y+u)-\rho(y-u)}{2u},\qquad
u>0,\,y\in(I+u)\cap(I-u).$

Since $u>0$ was arbitrary, it follows that $\rho$ is differentiable everywhere
on $I$. Now the right hand side of the above equality is differentiable with
respect to $y$, which implies that $\rho$ is twice differentiable. Repeating
this argument, it follows that $\rho$ is three times differentiable on $I$. We
are going to show that $\rho^{\prime\prime\prime}$ is identically zero on $I$.

Let $y\in I$ be fixed arbitrarily. Rearranging the equation (28), we can
obtain that

$\displaystyle 2u\rho^{\prime}(y)=\rho(y+u)-\rho(y-u),\qquad
u\in(y-I)\cap(I-y).$

Differentiating this equality three times with respect to $u$, we get that

$\displaystyle
0=\rho^{\prime\prime\prime}(y+u)+\rho^{\prime\prime\prime}(y-u),\qquad
u>0,\,y\in(I+u)\cap(I-u).$

With the substitution $u=0$, we conclude that
$\rho^{\prime\prime\prime}(y)=0$. Therefore, $\rho^{\prime\prime\prime}$ is
identically zero on $I$. This yields that $\rho$ has to be a polynomial of at
most second degree. ∎

######  Theorem 3.5.

Assume that there exist a positive continuous function
$\omega_{0}:I\to\mathbb{R}_{+}$ and a matrix $M:=(a_{i,j})_{1\leq i\leq
2,\,0\leq j\leq 1}\in\mathbb{R}^{2\times 2}$ such that (15) holds for $n=2$.
Assume that $\omega_{3}:I\to\mathbb{R}$ is a continuous function such that
$\overline{\omega}=(\omega_{1},\omega_{2},\omega_{3})$ is an extension of
$\omega=(\omega_{1},\omega_{2})$ and $\omega_{3}/\omega_{0}$ is not a
polynomial of at most second degree. Then every $\overline{\omega}$-Wright
convex function is $\omega$-convex, i.e., $\overline{\omega}$-Wright convexity
is equivalent to $\omega$-convexity.

###### Proof.

Under the conditions of the theorem, assume that $f:I\to\mathbb{R}$ is an
$\overline{\omega}$-Wright convex function. That is, the inequality

$\displaystyle\frac{\Phi_{(\omega,f)}(x,x+h_{1},x+h_{1}+h_{2})}{\Phi_{\overline{\omega}}(x,x+h_{1},x+h_{1}+h_{2})}+\frac{\Phi_{(\omega,f)}(x,x+h_{2},x+h_{1}+h_{2})}{\Phi_{\overline{\omega}}(x,x+h_{2},x+h_{1}+h_{2})}\geq
0$

holds for all $h_{1},h_{2}>0$ and $x\in I\cap(I-(h_{1}+h_{2}))$. Using Lemma
2.7, we can see that this inequality is equivalent to

$\displaystyle\frac{\omega_{0}(x)\omega_{0}(x+h_{1})\omega_{0}(x+h_{1}+h_{2})\cdot\det(M)\cdot\Phi_{\pi_{2},f/\omega_{0}}(x,x+h_{1},x+h_{1}+h_{2})}{\omega_{0}(x)\omega_{0}(x+h_{1})\omega_{0}(x+h_{1}+h_{2})\cdot\det(M)\cdot\Phi_{\pi_{2},\omega_{3}/\omega_{0}}(x,x+h_{1},x+h_{1}+h_{2})}$
$\displaystyle+\frac{\omega_{0}(x)\omega_{0}(x+h_{2})\omega_{0}(x+h_{1}+h_{2})\cdot\det(M)\cdot\Phi_{\pi_{2},f/\omega_{0}}(x,x+h_{2},x+h_{1}+h_{2})}{\omega_{0}(x)\omega_{0}(x+h_{2})\omega_{0}(x+h_{1}+h_{2})\cdot\det(M)\cdot\Phi_{\pi_{2},\omega_{3}/\omega_{0}}(x,x+h_{2},x+h_{1}+h_{2})}\geq
0,$

which simplifies to

(29)
$\displaystyle\frac{\Phi_{\pi_{2},f/\omega_{0}}(x,x+h_{1},x+h_{1}+h_{2})}{\Phi_{\pi_{2},\omega_{3}/\omega_{0}}(x,x+h_{1},x+h_{1}+h_{2})}+\frac{\Phi_{\pi_{2},f/\omega_{0}}(x,x+h_{2},x+h_{1}+h_{2})}{\Phi_{\pi_{2},\omega_{3}/\omega_{0}}(x,x+h_{2},x+h_{1}+h_{2})}\geq
0.$

This means that $g:=f/\omega_{0}$ is $\overline{\pi}_{2}$-Wright convex, where
$\overline{\pi}_{2}(t):=(1,t,\rho(t))$ and $\rho:=\omega_{3}/\omega_{0}$.
Observe that, for $\varphi\in\\{g,\rho\\}$ and $i\in\\{1,2\\}$, we have

$\displaystyle\Phi_{(\pi_{2},\varphi)}(x,x+h_{i},x+h_{1}+h_{2})$
$\displaystyle=\begin{vmatrix}1&1&1\\\ x&x+h_{i}&x+h_{1}+h_{2}\\\
\varphi(x)&\varphi(x+h_{i})&\varphi(x+h_{1}+h_{2})\end{vmatrix}$
$\displaystyle=h_{3-i}\varphi(x)-(h_{1}+h_{2})\varphi(x+h_{i})+h_{i}\varphi(x+h_{1}+h_{1}).$

Using this formula, the inequality (29) now states that

(30)
$\displaystyle\frac{h_{2}g(x)-(h_{1}+h_{2})g(x+h_{1})+h_{1}g(x+h_{1}+h_{2})}{h_{2}\rho(x)-(h_{1}+h_{2})\rho(x+h_{1})+h_{1}\rho(x+h_{1}+h_{2})}$
$\displaystyle\qquad+\frac{h_{1}g(x)-(h_{1}+h_{2})g(x+h_{2})+h_{2}g(x+h_{1}+h_{2})}{h_{1}\rho(x)-(h_{1}+h_{2})\rho(x+h_{2})+h_{2}\rho(x+h_{1}+h_{2})}\geq
0$

holds for all $h_{1},h_{2}>0$ and $x\in I\cap(I-(h_{1}+h_{2}))$.

By our assumptions, $(\pi_{2},\rho)$ is an extension of $\pi_{2}$ and $\rho$
is not a polynomial of at most second degree. Using Lemma 3.4, it follows that
$\rho$ cannot satisfy the functional equation (27), which means that there
exist $u\geq 0$, $y,z\in(I+u)\cap(I-u)$ with $z+u\geq y$ such that

$\displaystyle\frac{\rho(z)-\rho(y)}{z-y}\neq\frac{\rho(z+u)-\rho(y-u)}{(z+u)-(y-u)}.$

Let $x:=y-u$, $h:=z-y+2u$, $t:=z-y+u$. Then $z=x+t$, $y=x+h-t$ and $z+u=x+h$,
therefore the above relations state that

(31) $\displaystyle h(\rho(x+t)-\rho(x+h-t))+(h-2t)(\rho(x+h)-\rho(x))\neq 0$

for some $h>0$, $x\in I\cap(I-h)$ and $t\in(0,h)$.

Let $h>0$ and $x\in I\cap(I-h)$ be fixed such that, for some $t\in(0,h)$, (31)
holds. Define $T\subseteq(0,h)$ to be the set of those values $t$ for which
(31) is valid. Then the set $T$ is nonempty and, by the continuity of $\rho$,
it is also open. Let $T_{+}$ and $T_{-}$ denote the (disjoint) subsets of
those elements $t\in T$, for which the left hand side of (31) is positive and
negative, respectively. Then at least one of these subsets is nonempty (and
also open).

Since $g$ is $\overline{\pi}_{2}$-Wright convex, hence Theorem 3.1 implies
that $g$ is $\pi_{2}$-Jensen convex, i.e., it is Jensen convex in the standard
sense. According to Rodé’s Theorem [13], $g$ is the pointwise maximum of
Jensen affine functions, i.e., for all $p\in I$, there exists an additive
function $A_{p}:\mathbb{R}\to\mathbb{R}$ such that

(32) $\displaystyle g(y)\geq A_{p}(y-p)+g(p)\qquad(p,y\in I).$

Substituting $y:=x+t$ and $y:=x+h-t$, for $t\in T$ and $p\in I$, we get

$\displaystyle g(x+t)\geq A_{p}(x+t-p)+g(p)\qquad g(x+h-t)\geq
A_{p}(x+h-t-p)+g(p).$

Therefore, with $h_{1}:=t$ and $h_{2}:=h-t$, the inequality (LABEL:Efo) yields
that

$\displaystyle\frac{(h-t)g(x)-h\big{(}A_{p}(x+t-p)+g(p)\big{)}+tg(x+h)}{(h-t)\rho(x)-h\rho(x+t)+t\rho(x+h)}$
$\displaystyle\qquad+\frac{tg(x)-h\big{(}A_{p}(x+h-t-p)+g(p)\big{)}+(h-t)g(x+h)}{t\rho(x)-h\rho(x+h-t)+(h-t)\rho(x+h)}\geq
0.$

Using the additivity of $A_{p}$ and moving the terms containing $A_{p}(t)$ to
the right hand side, this inequality is equivalent to

(33)
$\displaystyle\frac{(h-t)g(x)-h\big{(}A_{p}(x-p)+g(p)\big{)}+tg(x+h)}{(h-t)\rho(x)-h\rho(x+t)+t\rho(x+h)}$
$\displaystyle\qquad+\frac{tg(x)-h\big{(}A_{p}(x+h-p)+g(p)\big{)}+(h-t)g(x+h)}{t\rho(x)-h\rho(x+h-t)+(h-t)\rho(x+h)}$
$\displaystyle\geq\frac{h(\rho(x+t)-\rho(x+h-t))+(h-2t)(\rho(x+h)-\rho(x))}{((h-t)\rho(x)-h\rho(x+t)+t\rho(x+h))(t\rho(x)-h\rho(x+h-t)+(h-t)\rho(x+h))}hA_{p}(t).$

This inequality shows that $A_{p}$ is bounded from above on $T_{+}$ and is
bounded from below on $T_{-}$. Therefore, $A_{p}$ is bounded from above or
from below on a nonempty open subset of $T$. In view well-known properties of
additive functions, this implies that $A_{p}$ is continuous, i.e., there
exists a real constant $a_{p}$ such that $A_{p}(x)=a_{p}x$ holds for all
$x\in\mathbb{R}$. Thus, by (32), we can see that $g$ is the pointwise maximum
of continuous affine functions, Therefore, $g$ must be a convex function (in
the standard sense). From this it follows that $f=g\omega_{0}$ is
$\omega$-convex. ∎

It seems to be an open problem whether an analogue of the previous theorem is
valid for the 3- or higher-dimensional setting.

## References

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# LEARNING TO WRITE WITH COHERENCE FROM NEGATIVE EXAMPLES

###### Abstract

Coherence is one of the critical factors that determine the quality of
writing. We propose writing relevance (WR) training method for neural encoder-
decoder natural language generation (NLG) models which improves coherence of
the continuation by leveraging negative examples. WR loss regresses the vector
representation of the context and generated sentence toward positive
continuation by contrasting it with the negatives. We compare our approach
with Unlikelihood (UL) training in a text continuation task on commonsense
natural language inference (NLI) corpora to show which method better models
the coherence by avoiding unlikely continuations. The preference of our
approach in human evaluation shows the efficacy of our method in improving
coherence.

Index Terms—  Text Generation, Contrastive Learning, Coherence, Natural
Language Processing

##  1 Introduction

An open ending leaves the readers lots of possibilities to imagine their own
conclusion to the story. Endings by each writer may vary, but most of the
endings will show coherence with the preceding context, and therefore end up
forming a story.

While the encoder-decoder, or Seq2Seq framework [1] is designed to model the
conditional likelihood of the decoded sentences given a context vector, it is
not enough for modeling coherent semantics of the continuation. To this end,
we introduce Writing Relevance (WR) training which models coherence, “a fit of
the text to its context” [2]. WR training loss regresses the vector
representation of the context and generated sentences toward the positive
continuation example while separating it further from negative examples.

We experiment with text continuation on commonsense NLI corpora, namely the
HellaSWAG [3] and Story Cloze Test [4] datasets. The model is asked to
generate the continuation given the context sentences while the original
datasets require to choose among the provided continuations. Coherence of
continuations is the key due to the large number of possible cohesive (but
incoherent) continuations and arising uncertainty the corpora allows. Both
corpora contain human-written negative examples which are cohesive but diverge
from information relevant to the context and thus not coherent.111Cohesion is
not coherence [5]. Cohesive ties (e.g., use of pronouns, parallelism) may work
as minimal conditions for coherence, but do not warrant delivering the topic
consistently. As each corpus originates from different text sources222the
HellaSWAG corpus is collected from WikiHow, and ActivityNet Captions [6]; the
Story Cloze Test is a collection of human-written 5-sentence long stories., it
helps to examine the extensible use of our training scheme to different
domains.

Fig. 1: WR loss biases encoder-decoder NLG model to generate sentences closer
to the positive than negative example in the representation space.

In summary, our contributions are two-fold:

  * •

We propose a WR training method that models coherence of the continuation by
leveraging negative examples.

  * •

We demonstrate the potential of our WR training scheme to be applied to
various domains by experimenting on two distinct commonsense NLI corpora.

Fig. 2: Loss computation during the writing relevance (WR) training. Cross-
entropy (CELoss) for token prediction is computed as in the pre-training,
triplet loss (TPLoss) for contrasting negatives is added to the total loss.

##  2 Learning Framework

This section covers the two-step procedure for WR training. First, we pre-
train an encoder-decoder model to generate grammatically correct sentences on
a American Literature Short Story (ALSS) corpus that we have
collected.333Details of the collection and pre-processing can be found in
https://github.com/sonsus/american_literature Then WR training takes place to
fine-tune the model to bias its generation toward coherent sentences while
avoiding out-of-place sentences (i.e., negative examples). We describe these
two steps of training for encoder-decoder model in Section 2.1 and 2.2,
including details of the decoding procedure and sentence representation.

###  2.1 Pre-training Encoder-Decoder Model

Here, we train the encoder-decoder NLG model for maximum likelihood estimation
(MLE) of the consecutive tokens given a context as:

$P(Y|X)=\prod_{i}^{T}p(y_{i}|Y_{0:i-1},X)$ (1)

where $Y$ is a consecutive sentence that continues the context $X$. $X$ and
$Y$ both consist of a series of tokens; {$x_{0}$, $x_{1}$, …,
$x_{T^{\prime}}$}, {$y_{0}$, $y_{1}$, …, $y_{T}$}. As we intend for the model
to learn proper grammar for cohesion but not to induce coherence, we split the
ALSS corpus into pairs of successive sentences.

For pre-training to be effective for latter text continuation, we initially
chose Toronto Boot Corpus [7] as it is considered in-domain data for
commonsense NLI corpora we are targeting. However, to circumvent the copyright
issue of the Toronto Book Corpus, we collected the ALSS corpus as a
replacement.

###  2.2 Writing Relevance Training

After pre-training, the model is capable of writing with correct grammar. In
the WR training stage, we adapt the model to each corpus for text
continuation. The model is optimized by sum of the gradient signals from two
losses (Figure 2). One comes from a triplet loss for distinction of coherent
continuations from negative examples, and the other is from an auxiliary token
prediction loss given by the cross-entropy between predictions and the
positive sentence.

The WR loss ($\mathrm{L_{WR}}$) is defined as follows:

$\displaystyle\mathrm{L_{WR}}=~{}\lambda\mathrm{CE}(Y^{*},Y)+\mathrm{TP_{cos}(a,~{}pos,~{}neg)}$
(2)
$\displaystyle\mathrm{TP_{cos}}(a,\text{pos},\text{neg})=\mathrm{max}(0,1+d^{cos}_{a,\text{pos}}-d^{cos}_{a,\text{neg}})$
$\displaystyle\mathrm{a,~{}pos,~{}neg}=g(\mathbf{h_{X;\hat{Y}}}),~{}g(\mathbf{h_{X;Y}}),~{}g(\mathbf{h_{X;N}})$

where cosine distance ($d_{cos}$) is defined as:

$\displaystyle d^{cos}_{x,y}=$ $\displaystyle~{}\frac{\|x\|\|y\|-x\cdot
y}{2\|x\|\|y\|}$ (3)

In Equation 2, $Y^{*},\hat{Y}$ and $Y,N$ denote teacher-forced and greedy-
decoded predictions and positive and negative continuations respectively.
$\lambda$ is a balancing coefficient (hyperparameter) for the cross-entropy
loss ($\mathrm{CE(\cdot)}$) and triplet loss ($\mathrm{TP_{cos}(\cdot)}$).
Inputs of the triplet loss, $g(\mathbf{h_{X;*}})$, are sentence
representations obtained by [CLS] pooling ($\mathbf{h_{X;*}}$) as explained in
the latter part of this section, mapped by $g(\cdot)$. If there is more than
one negative example for a context, we randomly sample one among the
negatives. To sum up, WR loss is composed of an auxiliary token prediction
loss and a triplet loss that regresses sentence representations.

Decoding Strategy We apply a simple heuristic to avoid duplication in
generated sentences as in Equation 4 following GLACNet [8].444With this
heuristic, nucleus or top-k sampling [9] show only marginal difference toward
greedy decoding, so we decided to stick to greedy decoding under the
heuristic.

$\hat{p}(word)=p(word)\times\frac{1}{1+k\cdot count_{word}}$ (4)

The value we use for $k$ in our experiments is 5. Beam search is not applied
since it is known to cause generic sentences in open-ended generation tasks
[10].

Sentence Representation We use a hidden vector of special token [CLS] obtained
by forward pass of $X$ and [CLS];$Y$ as a sentence representation of $X;Y$,
$\mathbf{h_{X;Y}}$ (Figure 2). The [CLS] pooling is mapped by $g(\cdot)$ which
empirically helped triplet loss reduction. We choose the mapping amongst
Hadamard product and linear projection.555We also tested other combinations
such as Euclid distance metric with other $g(\cdot)$’s: identity mapping,
Hadamard product, and linear projection with learnable parameters. Cosine
distance metric with learnable mapping $g(\cdot)$ worked well with WR training
loss.

##  3 Experiments

###  3.1 Task and Datasets

The architecture choice for our experiment is Transformer [11] model. We
perform text continuation to test our approach of learning to write with
coherence. We use three datasets for this: i) the ALSS dataset for pre-
training, which contains 4 295 short stories (that vary from a paragraph to
few pages long) collected from the public archive, ii) the Story Cloze Test,
and iii) the HellaSWAG where the last two have negative examples of quality.
Since the training split of the Story Cloze Test dataset has no negative
examples, we randomly sample positive continuations from other stories to be
used as negative examples after a sanity-check.666Random-sampled negatives
only confused 3.3% of the participants. Each participant scored as follows:
30/30, 29/30, 28/30.

Fig. 3: Examples from adaptaion corpora. Human-written negative examples are
marked as “Wrong”.

HellaSWAG and Story Cloze Test datasets are composed of short pieces of
writings that contain around 2-6 sentences (Figure 3). As a descriptive
example, performing text continuation on the Story Cloze Test is often called
story ending generation. The Story Cloze Test dataset is originally designed
as a binary choice between two human-written candidate endings where only one
is correct. Similarly, the HellaSWAG dataset is comprised of multiple-choice
problems that give a context of 1-5 sentences, letting a machine choose among
four candidate continuations. The human-written negative examples of the
corpora provide appropriate difficulty, thus being effective negative examples
for training to learn coherence of continuations.

###  3.2 Human Evaluation

Following Stephan et al., (1981) [2], a straightforward way of measuring
coherence is to ask readers. We perform an Amazon Mechanical Turk (AMT) survey
to ask people which continuation looks more natural.777We provide a preference
survey rather than using Likert scale based on [12]; Ranking-based assessment
(including binary ranking) is more consistent. Our survey investigates
sentence preference of the respondents. First, we show both context and
generated continuations of two different models to participants and ask: “What
sentence do you prefer as a continuation?”. Participants are also allowed to
check at “BOTH GOOD” or “NEITHER GOOD” (Table 1).

Context Sentences | Choices
---|---
A man breaks into a house and begins to take things. he takes jewelry and games, some cash, and some food. when the family comes home they call the police. the police come and investigate and manage to track him. | the police officer
is arrested.
the police come and
take the man to jail.
BOTH
GOOD
NEITHER
GOOD
Table 1:  Question of the survey for evaluation. Annotators are asked: “What
sentence do you prefer as a continuation?”.

In order to filter inattentive annotators out, we plant attention-check
questions in the middle of each survey form which have a clearly correct
choice. The screening rejected about 14 % (37 out of 156) of the survey
submissions. 107 individuals participated and submitted 119 valid survey forms
(average 1.11 submission per individual), each containing 27-28 questions
accompanied with 3 screening questions.

Fig. 4: Preference survey results (P=.0042, .0001, .0101, .0024). We surveyed
over 107 English speaking individuals on AMT. For both comparisons (WR vs. UL
and WR vs. CE), WR trained results are preferred over the corpora.

##  4 Results

We compare the performance of the NLG model that is trained with WR loss to
the one that is trained using unlikelihood (UL) training loss that penalizes
repetitive n-grams (n$=$4) and cross-entropy (CE) loss baselines. Since the UL
loss with repetitive n-grams as unlikely candidates outperformed taking the
candidates from negative examples888Tokens that occurs only in the accompanied
negative example, but not in the positive were chosen to be negative
candidates., we take the former for the experiments.

Figure 4 summarizes the preference survey conducted. Continuations written by
WR-trained encoder-decoder models are preferred over the other models trained
with UL and CE loss on both adaptation corpora, HellaSWAG, and Story Cloze
Test. Overall, WR $>$ UL or CE always holds. We conjecture that a high portion
of BOTH GOOD answers in the Story Cloze Test evaluation is caused by simple
sentences, and noise from fabricated negative examples.

In Table 2, we show continuations generated by each model. The continuations
from WR-trained model are preferred by participants despite of lower n-gram
scores. While this is not always the case, for near half of the generated
continuations from WR vs. UL survey, n-gram metrics failed to represent the
human judgement of coherence. This reassures that n-gram metrics are
inappropriate measures for open-ended generation tasks as reported in [13].

| Context (StoryClozeTest):
---
a woman sits behind a table dealing cards . she points at
one of the cards .
| WR (proposed)
---
(B1: 50, M: 15.6, 50%)
| UL-rep.4
---
(B1: 50, M: 18.9, 0%)
| she puts the cards in the
---
cards .
| the woman is shown playing a
---
guitar and singing .
| Context (HellaSWAG):
---
eric was helping his dad clear a wooded area . they were
going to put a picnic table there . all of a sudden he was
swarmed by bees . he had accidentally disturbed their nest .
| WR (proposed)
---
(B1: 14.3, M: 8.0, 42.9%)
| UL-rep.4
---
(B1: 18.2, M: 14.5, 0.0%)
| eric ’s dad was so upset he
---
had to go to the hospital.
| eric was able to get a bath
---
for his dad.

Table 2:  Text continuation examples accompanied by BLEU-1 (B1), METEOR (M),
and preference ratio (%). Preference shows the dominance of the proposed
method, while n-gram scores fail to follow human judgements.

##  5 Related Work

There are loss-based approaches that directly incorporate negative examples
into training text continuation like Large Margin LM (LMLM) [14] and
Unlikelihood (UL) training [15]. LMLM uses ranking loss to achieve better
generation quality by enlarging the log-likelihood margin between the
generated sentence from the negative samples. Unlikelihood (UL) training
introduces a novel unlikelihood loss function that penalizes repetitive or
unlikely tokens to remedy degeneracy problems in neural generation. Some works
utilize classifiers to benefit generation. Holtzman et al., (2018) [16]
deploys natural language understanding (NLU) discriminators that leverage what
it learned from negative examples to resolve empirical problems of neural
generation. Gabriel et al. (2019) [17] extends this for modeling narrative
flow in summarization task. Similar to our approach, alignment of sentence
representations for generation can also be found in Lee et al., (2021) [18]
which focuses on perturbing positive and negative examples for an efficient
contrastive learning of text representations for several NLG tasks.

##  6 Conclusion

We proposed the writing relevance (WR) training framework, which effectively
uses negative examples to improve coherence. WR training makes the encoder-
decoder NLG model learn sentence representation in a contrastive manner to
form better coherence in its writing. We demonstrate the potential of our WR
training scheme to various domains.

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