On generalization of Levinson’s inequality involving averages of 3-convex functions

Abstract

By using an integral arithmetic mean, a generalization of Levinson’s inequality given in (Pečarić et al. in Convex Functions, Partial Orderings, and Statistical Applications. Mathematics in Science and Engineering, vol. 187, 1992) and results from (Vukelić in Appl. Anal. Discrete Math. 14:670–684, 2020), we give extension of Wulbert’s result from (Wulbert in Math. Comput. Model. 37:1383–1391, 2003). Also, we obtain inequalities with divided differences for the functions of higher order.

1 Introduction

Suppose f is a continuous function defined on an interval I with a nonempty interior. Then, define

$$ F(x,y)= \textstyle\begin{cases} \frac{1}{y-x}\int _{x}^{y}f(t)\,dt , & x,y\in I,~x\neq y, \\ f(x), & x=y\in I. \end{cases} $$

(1.1)

In a seminal work [14], Wulbert proved that the integral arithmetic mean F, defined in (1.1), exhibits convexity on the interval $I^{2}$ when the underlying function f is convex over the interval I. In a separate study [15], Zhang and Chu independently rediscovered this result without making any reference to Wulbert’s findings. Their work revealed that the convexity of the integral arithmetic mean F hinges on the crucial condition that f must be convex on the interval I.

Since it will hold significant importance for our forthcoming analysis, let us take into consideration a real-valued function f defined on the interval $[a, b]$. The divided difference of order n for the function f at distinct points $x_{0}, x_{1}, \dots , x_{n} \in [a, b]$ is defined recursively (as elucidated in [1, 9]) in the following manner:

$$ f[x_{i}]=f(x_{i})\quad (i=0,\ldots ,n) $$

and

$$ f[x_{0},\ldots ,x_{n}]= \frac{f [x_{1},\ldots ,x_{n}]-f[x_{0},\ldots ,x_{n-1}]}{x_{n}-x_{0}}, \quad n\in\mathbb{N}_{0}. $$

The value $f[x_{0},\ldots ,x_{n}]$ remains invariant regardless of the order in which the points $x_{0},\ldots ,x_{n}$ are arranged.

The definition can be further extended to accommodate scenarios where some (or all) of the points coincide. Provided that $f^{(j-1)}(x)$ exists, we establish the following notation:

$$ f[\underbrace{x,\ldots ,x}_{j\text{-times}}]= \frac{f^{(j-1)}(x)}{(j-1)!}. $$

(1.2)

In the context of divided differences, the following holds:

$$ f[x_{0},\ldots ,x_{n}]=\sum _{i=0}^{n} \frac{f(x_{i})}{\omega '(x_{i})},\quad \text{where } \omega (x)=\prod_{j=0}^{n}(x-x_{j}). $$

In conclusion, it is evident that the following property holds for divided differences:

$$ f[x_{0},\ldots ,x_{n}]=\sum _{i=0}^{n} \frac{f(x_{i})}{\prod_{j=0,j\neq i}^{n}(x_{i}-x_{j})}. $$

Under the condition that the function f has a continuous nth derivative on the interval $[a, b]$, we can represent the divided difference $f[x_{0},\ldots ,x_{n}]$ using integral notation (refer to [9, p. 15]) as

$$ f[x_{0},\ldots ,x_{n}]= \int _{\Delta _{n}}f^{(n)} \Biggl(\sum _{i=0}^{n}u_{i}x_{i} \Biggr)\,du _{0}\cdots \,du _{n-1}, $$

where

$$ \Delta _{n}= \Biggl\{ (u_{0},\ldots ,u_{n-1}) : u_{i}\geq 0,\sum_{i=0}^{n-1}u_{i} \leq 1 \Biggr\} $$

and $u_{n}=1-\sum_{i=0}^{n-1}u_{i}$.

The notion of n-convexity is attributed to Popoviciu [10]. For the present study, we adhere to the definition as presented by Karlin [6].

Definition 1

A function $f: [a,b]\rightarrow \mathbb{R}$ is said to be n-convex on $[a,b]$, $n \geq 0$, if for all choices of $(n+1)$ distinct points in $[a,b]$, the nth order divided difference of f satisfies

$$ f[x_{0},\dots ,x_{n}] \geq 0. $$

It is worth noting that Popoviciu’s work demonstrated the fundamental result that any continuous n-convex function defined on the interval $[a,b]$ can be represented as the uniform limit of a sequence of n-convex polynomials. Moreover, [7] provides an extensive collection of related results and essential inequalities attributed to Favard, Berwald, and Steffensen.

The proof of the Jensen inequality for divided differences can be found in [4]:

Theorem 1

Let f be an $(n+2)$-convex function on $(a,b)$ and ${\mathbf {x}}\in (a,b)^{n+1}$. Then

$$ G(\mathbf{x})=f[x_{0},\ldots ,x_{n}] $$

is a convex function of the vector $\mathbf{x}=(x_{0},\ldots ,x_{n})$. Consequently,

$$ f \Biggl[\sum_{i=0}^{m}a_{i}x^{i}_{0}, \ldots ,\sum_{i=0}^{m}a_{i}x^{i}_{n} \Biggr]\leq \sum_{i=0}^{m}a_{i}f \bigl[x^{i}_{0},\ldots ,x^{i}_{n} \bigr]\quad (i~ \textit{is an upper index}) $$

(1.3)

holds for all $a_{i}\geq 0$ such that $\sum_{i=0}^{m}a_{i}=1$.

In the context of future research, the notion of a generalized divided difference will hold relevance. Provided below is the definition for reference.

Consider a real-valued function $f(x, y)$ defined on $I \times J$ ($I=[a,b]$, $J=[c,d]$). The divided difference of order $(n,m)$ for the function f at distinct points $x_{0}, \ldots , x_{n} \in I$ and $y_{0},\ldots , y_{m}\in J$ is defined as follows (see [9, p. 18]):

$$\begin{aligned} f\begin{bmatrix} x_{0}, \ldots , x_{n} \\ y_{0},\ldots , y_{m}\end{bmatrix} =&f[y_{0},\ldots , y_{m}] [x_{0}, \ldots , x_{n}] \\ =&f[x_{0}, \ldots , x_{n}] [y_{0},\ldots , y_{m}] \\ =&\sum_{i=0}^{n}\sum _{j=0}^{m} \frac{f(x_{i},y_{j})}{\omega '(x_{i})w'(y_{j})}, \end{aligned}$$

(1.4)

where $\omega (x)=\prod_{i=0}^{n}(x-x_{i})$, $w(y)=\prod_{j=0}^{m}(y-y_{j})$.

Following the aforementioned definition, we can establish the concept of $(n,m)$-convexity, which is as follows (see [9, p. 18]):

Definition 2

A function $f:I\times J\rightarrow \mathbb{R}$ is said to be $(n,m)$-convex, or convex of order $(n,m)$, if for all distinct points $x_{0}, \ldots , x_{n} \in I, y_{0},\ldots , y_{m}\in J$,

$$ f\begin{bmatrix} x_{0}, \ldots , x_{n} \\ y_{0},\ldots , y_{m}\end{bmatrix}\geq 0. $$

(1.5)

If this inequality is strict, then f is said to be strictly $(n,m)$-convex.

In [11], Popoviciu presented and proved the following theorem:

Theorem 2

If the partial derivative $f^{(n+m)}_{x^{n}y^{m}}(\partial ^{(n+m)}f/\partial x^{n}\partial y^{m})$ of f exists, then f is $(n,m)$-convex iff

$$ f^{(n+m)}_{x^{n}y^{m}}\geq 0. $$

(1.6)

If the inequality in (1.6) is strict, then f is strictly $(n,m)$-convex.

In this research, we build upon the generalization of Levinson’s inequality, and thus, we begin by stating the fundamental Levinson’s inequality as follows (see [8] and [12]):

Theorem 3

Let f be a real valued 3-convex function on $[0,2a]$. Then for $0\leq x_{k}\leq a$, $p_{k}>0$ ($1\leq k\leq n$), and $P_{k}=\sum_{i=1}^{k}p_{i}$ ($2\leq k\leq n$) we have

$$\begin{aligned}& \frac{1}{P_{n}}\sum_{k=1}^{n}p_{k}f(x_{k})-f \Biggl(\frac{1}{P_{n}} \sum_{k=1}^{n}p_{k}x_{k} \Biggr) \\& \quad \leq \frac{1}{P_{n}}\sum_{k=1}^{n}p_{k}f(2a-x_{k})-f \Biggl( \frac{1}{P_{n}}\sum_{k=1}^{n}p_{k}(2a-x_{k}) \Biggr). \end{aligned}$$

(1.7)

If $f'''>0$, then the equality holds iff $x_{1}=\cdots =x_{n}$.

In [2], Bullen provided a proof for the generalization of Theorem 3:

Theorem 4

• a) Let f be a real-valued 3-convex function on $[a,b]$ and $x_{k}$, $y_{k}$ ($1\leq k\leq n$) be 2n points on $[a,b]$ such that

  $$ \max \{x_{1},\ldots ,x_{n}\}\leq \min \{y_{1},\ldots ,y_{n}\},\qquad x_{1}+y_{1}= \cdots =x_{n}+y_{n}. $$

  (1.8)

  If $p_{k}>0$ ($1\leq k\leq n$), then

  $$ \frac{1}{P_{n}}\sum_{k=1}^{n}p_{k}f(x_{k})-f \Biggl(\frac{1}{P_{n}} \sum_{k=1}^{n}p_{k}x_{k} \Biggr)\leq \frac{1}{P_{n}}\sum_{k=1}^{n}p_{k}f(y_{k})-f \Biggl(\frac{1}{P_{n}}\sum_{k=1}^{n}p_{k}y_{k} \Biggr). $$

  (1.9)

  If f is strictly 3-convex there is equality in (1.9) if and only if $x_{1}=\cdots =x_{n}$.

• b) If (1.9) holds for a continuous function f, (1.8) is satisfied by 2n-distinct points and $p_{k}>0$ for $k\in [1,n]$, then f is 3-convex.

It is shown in [9] that the condition (1.8) can be weakened, i.e., the following result holds:

Theorem 5

Let f be a 3-convex function on $[a,b]$, $p_{i}>0$ $(1\leq i \leq n)$, $x_{k}$, $y_{k}$ $(1\leq k\leq n)$ be points in $[a,b]$ such that

$$ x_{1}+y_{1}=\cdots =x_{n}+y_{n}=2c $$

(1.10)

and

$$\begin{aligned}& x_{i}+x_{n-i+1}\leq 2c, \end{aligned}$$

(1.11)

$$\begin{aligned}& (p_{i}x_{i}+p_{n-i+1}x_{n-i+1})/(p_{i}+p_{n-i+1}) \leq c, \quad \textit{for } 1 \leq i \leq n. \end{aligned}$$

(1.12)

Then (1.9) is valid.

The primary objective of this paper is to provide an extension of Wulbert’s result, as presented in [14], for 3-convex functions. We will also consider relevant findings from [13]. Moreover, we aim to establish an inequality involving divided differences by utilizing the generalization of Levinson’s inequality given in [9]. As a significant outcome, we will demonstrate the convexity of higher order for functions defined by divided differences.

2 Inequalities involving averages

Theorem 6

Let f be a real-valued 3-convex function on $[a,b]$ and let F be defined in (1.1). Then for $p_{i}>0$ ($1\leq i\leq n$), $a\leq x_{k}, \tilde{x}_{k},y_{k}$, $\tilde{y}_{k}\leq b$ ($1\leq k\leq n$) such that

$$\begin{aligned}& x_{1}+y_{1}=\cdots =x_{n}+y_{n}=2c,\qquad \tilde{x}_{1}+\tilde{y}_{1}=\cdots = \tilde{x}_{n}+\tilde{y}_{n}=2c,\\& x_{i}+x_{n-i+1}\leq 2c,\qquad \tilde{x}_{i}+ \tilde{x}_{n-i+1}\leq 2c,\\& \frac{p_{i}x_{i}+p_{n-i+1}x_{n-i+1}}{p_{i}+p_{n-i+1}}\leq c,\qquad \frac{p_{i}\tilde{x}_{i}+p_{n-i+1}\tilde{x}_{n-i+1}}{p_{i}+p_{n-i+1}} \leq c,\quad 1\leq i\leq n, \end{aligned}$$

and $P_{k}=\sum_{i=1}^{k}p_{i}$ ($2\leq k\leq n$) we have

$$ \frac{1}{P_{n}}\sum_{k=1}^{n}p_{k}F(x_{k}, \tilde{x}_{k})-F(\bar{x}, \bar{\tilde{x}})\leq \frac{1}{P_{n}} \sum_{k=1}^{n}p_{k}F(y_{k}, \tilde{y}_{k})-F(\bar{y}, \bar{\tilde{y}}), $$

(2.1)

where $\bar{x}=\frac{1}{P_{n}}\sum_{k=1}^{n}p_{k}x_{k}$, $\bar{\tilde{x}}=\frac{1}{P_{n}}\sum_{k=1}^{n}p_{k}\tilde{x}_{k}$, $\bar{y}=\frac{1}{P_{n}}\sum_{k=1}^{n}p_{k}y_{k}$, and $\bar{\tilde{y}}=\frac{1}{P_{n}}\sum_{k=1}^{n}p_{k}\tilde{y}_{k}$.

Consequently, for $l+m=3$ the integral arithmetic mean (1.1) is $(l,m)$-convex on $[a,b]^{2}$.

Proof

Since the conditions

$$\begin{aligned}& s\tilde{x}_{1}+(1-s)x_{1}+s\tilde{y}_{1}+(1-s)y_{1}= \cdots =s\tilde{x}_{n}+(1-s)x_{n}+s \tilde{y}_{n}+(1-s)y_{n}=2c,\\& s\tilde{x}_{i}+(1-s)x_{i}+s\tilde{x}_{n-i+1}+(1-s)x_{n-i+1} \leq 2c,\\& \frac{p_{i} (s\tilde{x}_{i}+(1-s)x_{i} )+p_{n-i+1} (s\tilde{x}_{n-i+1}+(1-s)x_{n-i+1} )}{p_{i}+p_{n-i+1}} \leq c, \quad 1\leq i \leq n, \end{aligned}$$

from Theorem 5 are satisfied, by using inequality (1.9), we get

$$\begin{aligned}& \frac{1}{P_{n}}\sum_{k=1}^{n}p_{k}F(x_{k}, \tilde{x}_{k})-F(\bar{x}, \bar{\tilde{x}}) \\& \quad = \frac{1}{P_{n}}\sum_{k=1}^{n}p_{k} \int _{0}^{1}f\bigl(s\tilde{x}_{k}+(1-s)x_{k} \bigr)\,ds \\& \quad \quad {} - \int _{0}^{1}f \Biggl(s\frac{1}{P_{n}} \sum_{k=1}^{n}p_{k} \tilde{x}_{k}+(1-s) \frac{1}{P_{n}}\sum _{k=1}^{n}p_{k}x_{k} \Biggr)\,ds \\& \quad = \int _{0}^{1} \Biggl[\frac{1}{P_{n}}\sum _{k=1}^{n}p_{k}f\bigl(s \tilde{x}_{k}+(1-s)x_{k}\bigr)-f \Biggl( \frac{1}{P_{n}}\sum_{k=1}^{n}p_{k} \bigl(s\tilde{x}_{k}+(1-s)x_{k} \bigr) \Biggr) \Biggr]\,ds \\& \quad \leq \int _{0}^{1} \Biggl[\frac{1}{P_{n}}\sum _{k=1}^{n}p_{k}f\bigl(s \tilde{y}_{k}+(1-s)y_{k}\bigr)-f \Biggl( \frac{1}{P_{n}}\sum_{k=1}^{n}p_{k} \bigl(s\tilde{y}_{k}+(1-s)y_{k} \bigr) \Biggr) \Biggr]\,ds \\& \quad = \frac{1}{P_{n}}\sum_{k=1}^{n}p_{k} \int _{0}^{1}f\bigl(s\tilde{y}_{k}+(1-s)y_{k} \bigr)\,ds \\& \quad \quad {} - \int _{0}^{1}f \Biggl(s\frac{1}{P_{n}} \sum_{k=1}^{n}p_{k} \tilde{y}_{k}+(1-s) \frac{1}{P_{n}}\sum _{k=1}^{n}p_{k}y_{k} \Biggr)\,ds \\& \quad = \frac{1}{P_{n}}\sum_{k=1}^{n}p_{k}F(y_{k}, \tilde{y}_{k})-F( \bar{y},\bar{\tilde{y}}). \end{aligned}$$

Now, if we put $n=2$, $x_{1}=x$, $x_{2}=y_{2}=x+\frac{3h}{2}$, $y_{1}=x+3h$, $\tilde{x}_{1}= \tilde{x}_{2}=y$, $\tilde{y}_{1}=\tilde{y}_{2}=y$, $2x+3h=2y=2c$, $p_{1}=1$, $p_{2}=2$, then inequality (2.1) reduces to

$$ \frac{1}{3}F(x,y)-F(x+h,y)\leq \frac{1}{3}F(x+3h,y)-F(x+2h,y). $$

Using the definition in (1.4), we get

$$ 2h^{3}\bigl(F[x, x+h, x+2h, x+3h]\bigr)[y]\geq 0. $$

It is a known fact that if this property holds for all possible $x,y,h > 0$, then F is $(3,0)$-convex, as stated in [11].

If we put $n=2$, $x_{1}=x$, $x_{2}=x+2h_{1}$, $y_{1}=x+2h_{1}$, $y_{2}=x$, $\tilde{x}_{1}= \tilde{x}_{2}=y$, $\tilde{y}_{1}=\tilde{y}_{2}=y+h_{2}$, $p_{1}=p_{2}=1$, $2x+2h_{1}=2y+h_{2}=2c$ then inequality (2.1) reduces to

$$\begin{aligned}& \frac{1}{2} \bigl(F(x,y)+F(x+2h_{1},y) \bigr)-F(x+h_{1},y) \\& \quad \leq \frac{1}{2} \bigl(F(x+2h_{1},y+h_{2})+F(x,y+h_{2}) \bigr)-F(x+h_{1},y+h_{2}). \end{aligned}$$

Using the definition in (1.4), we get

$$ h_{1}^{2}h_{2}\bigl(F[x, x+h_{1}, x+2h_{1}]\bigr)[y, y+h_{2}]\geq 0. $$

Continuing the previous arguments, since this property holds for all possible $x, h_{1}, y, h_{2} > 0$, we can deduce that F is $(2,1)$-convex.

The proofs for $(0,3)$- and $(1,2)$-convexity exhibit similarities, leading us to conclude that F is $(l,m)$-convex on $[a,b]^{2}$ when $l+m=3$. □

Remark 1

Theorem 6 can be regarded as a generalization of Theorem 5 since inequality (2.1) for $x_{k}=\tilde{x}_{k}$ and $y_{k}=\tilde{y}_{k}$, $k=1,\ldots ,n$ reproduces inequality (1.9).

For similar results regarding Jensen’s inequality involving averages of convex functions, refer to [3] and [5].

3 Inequalities for divided differences

Theorem 7

Let f be an $(n+3)$-convex function on $[a,b]$ and ${\mathbf {x}}, {\mathbf {y}}\in [a,b]^{n+1}$. Then for $x^{i}_{k}$, $y^{i}_{k}$, $(0\leq k\leq n)$ (“i” is an upper index), $a_{i}>0$ $(0\leq i\leq m)$, such that $\sum_{i=0}^{m}a_{i}=1$,

$$\begin{aligned}& x^{i}_{0}+y^{i}_{0}=x^{i}_{1}+y^{i}_{1}= \cdots =x^{i}_{n}+y^{i}_{n}=2c,\\& x^{i}_{k}+x^{m-i}_{k}\leq 2c,\\& \bigl(a_{i}x^{i}_{k}+a_{m-i}x^{m-i}_{k} \bigr)/(a_{i}+a_{m-i})\leq c, \end{aligned}$$

we have

$$\begin{aligned} &\sum_{i=0}^{m}a_{i}f \bigl[x^{i}_{0},\ldots ,x^{i}_{n} \bigr]-f \Biggl[\sum_{i=0}^{m}a_{i}x^{i}_{0}, \ldots ,\sum_{i=0}^{m}a_{i}x^{i}_{n} \Biggr] \\ &\quad \leq \sum_{i=0}^{m}a_{i}f \bigl[y^{i}_{0},\ldots ,y^{i}_{n} \bigr]-f \Biggl[ \sum_{i=0}^{m}a_{i}y^{i}_{0}, \ldots ,\sum_{i=0}^{m}a_{i}y^{i}_{n} \Biggr]. \end{aligned}$$

(3.1)

Consequently,

$$ G(\mathbf{x})=f[x_{0},x_{1},x_{2}] $$

is an $(l_{1},l_{2},l_{3})$-convex function of the vector $\mathbf{x}=(x_{0}, x_{1}, x_{2})$, when $l_{1}+l_{2}+l_{3}=3$.

Proof

Since the conditions

$$\begin{aligned}& \sum_{j=0}^{n}u_{j}x_{j}^{0}+ \sum_{j=0}^{n}u_{j}y_{j}^{0}= \cdots = \sum_{j=0}^{n}u_{j}x_{j}^{m}+ \sum_{j=0}^{n}u_{j}y_{j}^{m}=2c,\\& \sum_{j=0}^{n}u_{j}x_{j}^{i}+ \sum_{j=0}^{n}u_{j}x_{j}^{m-i} \leq 2c,\\& \frac{a_{i}\sum_{j=0}^{n}u_{j}x^{i}_{j}+a_{m-i}\sum_{j=0}^{n}u_{j}x^{m-i}_{j}}{a_{i}+a_{m-i}} \leq c, \quad 0\leq i \leq m, \end{aligned}$$

from Theorem 5 are satisfied, by using inequality (1.9), for the 3-convex function $f^{(n)}$, we get

$$\begin{aligned}& \sum_{i=0}^{m}a_{i}f \bigl[x^{i}_{0},\ldots ,x^{i}_{n} \bigr]-f \Biggl[\sum_{i=0}^{m}a_{i}x^{i}_{0}, \ldots ,\sum_{i=0}^{m}a_{i}x^{i}_{n} \Biggr] \\& \quad = \sum_{i=0}^{m}a_{i} \int _{\Delta _{n}}f^{(n)} \Biggl(\sum _{j=0}^{n}u_{j}x_{j}^{i} \Biggr)\,du _{0}\cdots \,du _{n-1} \\& \quad \quad {} - \int _{\Delta _{n}}f^{(n)} \Biggl(\sum _{j=0}^{n}u_{j}\sum _{i=0}^{m}a_{i}x_{j}^{i} \Biggr)\,du _{0}\cdots \,du _{n-1} \\& \quad = \int _{\Delta _{n}} \Biggl[\sum_{i=0}^{m}a_{i}f^{(n)} \Biggl(\sum_{j=0}^{n}u_{j}x_{j}^{i} \Biggr)-f^{(n)} \Biggl(\sum_{i=0}^{m}a_{i} \sum_{j=0}^{n}u_{j}x_{j}^{i} \Biggr) \Biggr]\,du _{0}\cdots \,du _{n-1} \\& \quad \leq \int _{\Delta _{n}} \Biggl[\sum_{i=0}^{m}a_{i}f^{(n)} \Biggl( \sum_{j=0}^{n}u_{j}y_{j}^{i} \Biggr)-f^{(n)} \Biggl(\sum_{i=0}^{m}a_{i} \sum_{j=0}^{n}u_{j}y_{j}^{i} \Biggr) \Biggr]\,du _{0}\cdots \,du _{n-1} \\& \quad = \sum_{i=0}^{m}a_{i} \int _{\Delta _{n}}f^{(n)} \Biggl(\sum _{j=0}^{n}u_{j}y_{j}^{i} \Biggr)\,du _{0}\cdots \,du _{n-1} \\& \quad \quad {} - \int _{\Delta _{n}}f^{(n)} \Biggl(\sum _{j=0}^{n}u_{j}\sum _{i=0}^{m}a_{i}y_{j}^{i} \Biggr)\,du _{0}\cdots \,du _{n-1} \\& \quad = \sum_{i=0}^{m}a_{i}f \bigl[y^{i}_{0},\ldots ,y^{i}_{n} \bigr]-f \Biggl[\sum_{i=0}^{m}a_{i}y^{i}_{0}, \ldots ,\sum_{i=0}^{m}a_{i}y^{i}_{n} \Biggr]. \end{aligned}$$

Now, if we put $n=2$, $m=1$,

$$\begin{aligned}& x^{0}_{0}=y_{0}, \qquad x_{0}^{1}=y_{0}+\frac{3h}{2}, \\& y^{0}_{0}=y_{0}+3h, \qquad y_{0}^{1}=y_{0}+\frac{3h}{2}, \\& x^{0}_{1}=x^{1}_{1}=y^{0}_{1}=y^{1}_{1}=y_{1}, \\& x_{2}^{0}=x_{2}^{1}=y_{2}^{0}=y_{2}^{1}=y_{2}, \\& a_{0}=\frac{1}{3}, \qquad a_{1}= \frac{2}{3}, \end{aligned}$$

then inequality (3.1) reduces to

$$\begin{aligned}& \frac{1}{3}G(y_{0}, y_{1}, y_{2})-G(y_{0}+h,y_{1},y_{2}) \\& \quad \leq \frac{1}{3}G(y_{0}+3h, y_{1}, y_{2})-G(y_{0}+2h, y_{1}, y_{2}). \end{aligned}$$

Using the generalization of definition (1.4), we get

$$ 2h^{3} \bigl( \bigl(G[y_{0},y_{0}+h,y_{0}+2h,y_{0}+3h] \bigr)[y_{1}] \bigr)[y_{2}]\geq 0. $$

As observed in the proof of Theorem 6, since this property holds for all possible $y_{0}, y_{1}, y_{2}, h > 0$, we can conclude that G is $(3,0,0)$-convex.

If we put $n=2$, $m=1$,

$$\begin{aligned}& x^{0}_{0}=y^{1}_{0}=y_{0}, \qquad x_{0}^{1}=y_{0}^{0}=y_{0}+2h_{0}, \\& x^{0}_{1}=x^{1}_{1}=y_{1}, \\& x_{2}^{0}=x_{2}^{1}=y_{2}^{0}=y_{2}^{1}=y_{2}, \\& y^{0}_{1}=y^{1}_{1}=y_{1}+h_{1}, \\& a_{0}=a_{1}=\frac{1}{2} \end{aligned}$$

then inequality (3.1) reduces to

$$\begin{aligned}& \frac{1}{2}G(y_{0},y_{1},y_{2})+ \frac{1}{2}G(y_{0}+2h_{0},y_{1},y_{2})-G(y_{0}+h_{0},y_{1},y_{2}) \\& \quad \leq \frac{1}{2}G(y_{0}+2h_{0},y_{1}+h_{1},y_{2})+ \frac{1}{2}G(y_{0},y_{1}+h_{1},y_{2})-G(y_{0}+h_{0},y_{1}+h_{1},y_{2}). \end{aligned}$$

Using the generalization of definition (1.4), we get

$$ h_{0}^{2}h_{1} \bigl( \bigl(G[y_{0},y_{0}+h_{0},y_{0}+2h_{0}] \bigr)[y_{1},y_{1}+h_{1}] \bigr)[y_{2}]\geq 0. $$

As before, since this holds for all possible $y_{0}, y_{1}, y_{2}, h_{0}, h_{1}>0$, G is $(2,1,0)$-convex.

If we put $n=2$, $m=3$,

$$\begin{aligned}& x^{0}_{0}=x_{0}^{2}=y^{1}_{0}=y_{0}^{3}=y_{0}, \qquad x_{0}^{1}=x_{0}^{3}=y_{0}^{0}=y_{0}^{2}=y_{0}+h_{0}, \\& x^{0}_{1}=x^{3}_{1}=y^{1}_{1}=y^{2}_{1}=y_{1}, \qquad x_{1}^{1}=x_{1}^{2}=y_{1}^{0}=y_{1}^{3}=y_{1}+h_{1} \\& x_{2}^{0}=x_{2}^{1}=y_{2}^{2}=y_{2}^{3}=y_{2}, \qquad x_{2}^{2}=x_{2}^{3}=y_{2}^{0}=y_{2}^{1}=y_{2}+h_{2}, \\& a_{0}=a_{1}=a_{2}=a_{3}= \frac{1}{4}, \end{aligned}$$

then inequality (3.1) reduces to

$$\begin{aligned}& \frac{1}{4}G(y_{0},y_{1},y_{2})+ \frac{1}{4}G(y_{0}+h_{0},y_{1}+h_{1},y_{2}) \\& \quad \quad {} +\frac{1}{4}G(y_{0},y_{1}+h_{1},y_{2}+h_{2})+ \frac{1}{4}G(y_{0}+h_{0},y_{1},y_{2}+h_{2}) \\& \quad \leq \frac{1}{4}G(y_{0}+h_{0},y_{1}+h_{1},y_{2}+h_{2})+ \frac{1}{4}G(y_{0},y_{1},y_{2}+h_{2}) \\& \quad \quad {} +\frac{1}{4}G(y_{0}+h_{0},y_{1},y_{2})+ \frac{1}{4}G(y_{0},y_{1}+h_{1},y_{2}). \end{aligned}$$

Using the generalization of definition (1.4), we get

$$ \frac{1}{4}h_{0}h_{1}h_{2} \bigl( \bigl(G[y_{0},y_{0}+h_{0}] \bigr)[y_{1},y_{1}+h_{1}] \bigr)[y_{2},y_{2}+h_{2}]\geq 0. $$

Continuing the previous arguments, since this property holds for all possible $y_{0}$, $y_{1}$, $y_{2}$, $h_{0}, h_{1}, h_{2} > 0$, we can conclude that G is $(1,1,1)$-convex.

The proofs for $(0,3,0)$-, $(0,0,3)$-, $(1,2,0)$-, $(0,2,1)$-, $(0,1,2)$-, $(2,0,1)$-, and $(1,0,2)$-convexity share similarities. □