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Article

# Weighted Hermite–Hadamard-Type Inequalities by Identities Related to Generalizations of Steffensen’s Inequality

1
Croatian Academy of Sciences and Arts, Trg Nikole Šubića Zrinskog 11, 10000 Zagreb, Croatia
2
Faculty of Civil Engineering, University of Rijeka, Radmile Matejčić 3, 51000 Rijeka, Croatia
3
Faculty of Textile Technology, University of Zagreb, Prilaz Baruna Filipovića 28a, 10000 Zagreb, Croatia
*
Author to whom correspondence should be addressed.
Mathematics 2022, 10(9), 1505; https://doi.org/10.3390/math10091505
Received: 30 March 2022 / Revised: 26 April 2022 / Accepted: 29 April 2022 / Published: 1 May 2022

## Abstract

:
In this paper, we obtain some new weighted Hermite–Hadamard-type inequalities for $( n + 2 ) −$convex functions by utilizing generalizations of Steffensen’s inequality via Taylor’s formula.
MSC:
26D15; 26A51

## 1. Introduction

The Hermite–Hadamard inequality is one of the most important mathematical inequalities. It was discovered independently first by Hermite [1] and later by Hadamard [2]. The classical Hermite–Hadamard inequality provides an estimate from below and above the mean value of convex function f: $[ a , b ] → R$. More precisely, we have the following.
$f a + b 2 ≤ 1 b − a ∫ a b f ( x ) d x ≤ f ( a ) + f ( b ) 2 .$
To illustrate the importance of the Hermite–Hadamard inequality, let us mention that the Hermite–Hadamard inequality can be considered as the necessary and sufficient condition for convexity of a function. Furthermore, the Hermite–Hadamard inequality has an important role in numerical analysis, mathematical analysis and functional analysis. Various generalizations, extensions and applications of the Hermite-Hadamard inequality have appeared in the literature (see [3,4,5,6,7,8]).
In this paper, we consider the weighted Hermite–Hadamard inequality for convex functions given in following theorem (see [8,9,10]).
Theorem 1.
Let p: $[ a , b ] → R$ be a non-negative function. If f: $[ a , b ] → R$ is a convex function, then we have the following:
$f ( m ) ≤ 1 P ( b ) ∫ a b p ( x ) f ( x ) d x ≤ b − m b − a f ( a ) + m − a b − a f ( b )$
or
$P ( b ) f ( m ) ≤ ∫ a b p ( x ) f ( x ) d x ≤ P ( b ) b − m b − a f ( a ) + m − a b − a f ( b ) ,$
where the following is the case.
$P ( t ) = ∫ a t p ( x ) d x a n d m = 1 P ( b ) ∫ a b p ( x ) x d x .$
In 1918, Steffensen proved the following inequality (see [11]).
Theorem 2
([11]). Suppose that f is non-increasing and g is integrable on $[ a , b ]$ with $0 ≤ g ≤ 1$ and $λ = ∫ a b g ( t ) d t .$ Then, we have the following.
$∫ b − λ b f ( t ) d t ≤ ∫ a b f ( t ) g ( t ) d t ≤ ∫ a a + λ f ( t ) d t .$
The inequalities are reversed for f non-decreasing.
Many papers have been devoted to generalizations and refinements of Steffensen’s inequality and its connection to other well-known inequalities such as Gauss–Steffensen’s, Hölder’s, Jenssen-=Steffensen’s and other inequalities. A complete overview of the results related to Steffensen’s inequality can be found in monographs [12,13].
By using the Mitrinović [14] result in which the inequalities in (2) follow from identities:
$∫ a a + λ f ( t ) d t − ∫ a b f ( t ) g ( t ) d t = ∫ a a + λ [ f ( t ) − f ( a + λ ) ] [ 1 − g ( t ) ] d t + ∫ a + λ b [ f ( a + λ ) − f ( t ) ] g ( t ) d t$
and
$∫ a b f ( t ) g ( t ) d t − ∫ b − λ b f ( t ) d t = ∫ a b − λ [ f ( t ) − f ( b − λ ) ] g ( t ) d t + ∫ b − λ b [ f ( b − λ ) − f ( t ) ] [ 1 − g ( t ) ] d t$
and using Taylor’s formulae in points a and b
$f ( x ) = ∑ i = 0 n − 1 f ( i ) ( a ) i ! ( x − a ) i + 1 ( n − 1 ) ! ∫ a x f ( n ) ( t ) ( x − t ) n − 1 d t$
$f ( x ) = ∑ i = 0 n − 1 f ( i ) ( b ) i ! ( x − b ) i − 1 ( n − 1 ) ! ∫ x b f ( n ) ( t ) ( x − t ) n − 1 d t$
in paper [15], the authors proved the following identities related to generalizations of Steffensen’s inequality.
Theorem 3
([15]). Let f: $[ a , b ] → R$ be such that $f ( n − 1 )$ is absolutely continuous for some $n ≥ 2$ and let g: $[ a , b ] → R$ be an integrable function such that $0 ≤ g ≤ 1$. Let $λ = ∫ a b g ( t ) d t$ and let the function $G 1$ be defined by the following.
$G 1 ( x ) = ∫ a x ( 1 − g ( t ) ) d t , x ∈ [ a , a + λ ] , ∫ x b g ( t ) d t , x ∈ [ a + λ , b ] .$
Then, we have the following:
$∫ a a + λ f ( t ) d t − ∫ a b f ( t ) g ( t ) d t + ∑ i = 0 n − 2 f ( i + 1 ) ( a ) i ! ∫ a b G 1 ( x ) ( x − a ) i d x = − 1 ( n − 2 ) ! ∫ a b ∫ t b G 1 ( x ) ( x − t ) n − 2 d x f ( n ) ( t ) d t$
and the following is obtained.
$∫ a a + λ f ( t ) d t − ∫ a b f ( t ) g ( t ) d t + ∑ i = 0 n − 2 f ( i + 1 ) ( b ) i ! ∫ a b G 1 ( x ) ( x − b ) i d x = 1 ( n − 2 ) ! ∫ a b ∫ a t G 1 ( x ) ( x − t ) n − 2 d x f ( n ) ( t ) d t .$
Theorem 4
([15]). Let f: $[ a , b ] → R$ be such that $f ( n − 1 )$ is absolutely continuous for some $n ≥ 2$ and let g: $[ a , b ] → R$ be an integrable function such that $0 ≤ g ≤ 1$. Let $λ = ∫ a b g ( t ) d t$ and let the function $G 2$ be defined by the following.
$G 2 ( x ) = ∫ a x g ( t ) d t , x ∈ [ a , b − λ ] , ∫ x b ( 1 − g ( t ) ) d t , x ∈ [ b − λ , b ] .$
Then, we have the following:
$∫ a b f ( t ) g ( t ) d t − ∫ b − λ b f ( t ) d t + ∑ i = 0 n − 2 f ( i + 1 ) ( a ) i ! ∫ a b G 2 ( x ) ( x − a ) i d x = − 1 ( n − 2 ) ! ∫ a b ∫ t b G 2 ( x ) ( x − t ) n − 2 d x f ( n ) ( t ) d t$
and the following is obtained.
$∫ a b f ( t ) g ( t ) d t − ∫ b − λ b f ( t ) d t + ∑ i = 0 n − 2 f ( i + 1 ) ( b ) i ! ∫ a b G 2 ( x ) ( x − b ) i d x = 1 ( n − 2 ) ! ∫ a b ∫ a t G 2 ( x ) ( x − t ) n − 2 d x f ( n ) ( t ) d t .$
Since, in this paper, we will deal with $n −$convex functions, let us recall the definition of the $n −$convex function. For more details on convex functions, we refer the interested reader to [6,8].
Let f be a real-valued function defined on the segment $[ a , b ]$. The divided difference of order n of the function f at distinct points $x 0 , … , x n ∈ [ a , b ]$ is defined recursively (see [8]) by the following.
$f [ x i ] = f ( x i ) , ( i = 0 , … , n )$
$f [ x 0 , … , x n ] = f [ x 1 , … , x n ] − f [ x 0 , … , x n − 1 ] x n − x 0 .$
The value $f [ x 0 , … , x n ]$ is independent of the order of the points $x 0 , … , x n$.
The definition may be extended to include the case in which some (or all) of the points coincide. Assuming that $f ( j − 1 ) ( x )$ exists, we define the following.
$f [ x , … , x ︸ j − t i m e s ] = f ( j − 1 ) ( x ) ( j − 1 ) ! .$
Definition 1
([8]). A function $f : [ a , b ] → R$ is said to be n-convex on $[ a , b ]$, $n ≥ 0$, if for all choices of $( n + 1 )$ distinct points in $[ a , b ] ,$ the $n − t h$ order divided difference of f satisfies the following.
$f [ x 0 , … , x n ] ≥ 0 .$
Note that $1 −$convex functions are non-decreasing functions and $2 −$convex functions are convex functions. An $n −$convex function need not to be $n −$times differentiable; however, if $f ( n )$ exists, then f is $n −$convex if and only if $f ( n ) ≥ 0$. The following property also holds: if f is an $( n + 2 ) −$convex function, then there exists the $n −$th derivative $f ( n )$, which is a convex function.
The aim of this paper is to use identities related to generalizations of Steffensen’s inequality, obtained by using Taylor’s formula, to prove new weighted Hermite–Hadamard-type inequalities for $( n + 2 ) −$convex functions.

## 2. Main Results

In this section, applying identities given in Theorems 3 and 4 and the properties of $n −$convex functions, we derive new weighted Hermite–Hadamard-type inequalities.
Theorem 5.
Let f: $[ a , b ] → R$ be $( n + 2 ) −$convex on $[ a , b ]$ and $f ( n − 1 )$ absolutely continuous for $n ≥ 2$. Let g: $[ a , b ] → R$ be an integrable function such that $0 ≤ g ≤ 1$ and $λ = ∫ a b g ( t ) d t$. Let function $G 1$ be defined by the following.
$G 1 ( x ) = ∫ a x ( 1 − g ( t ) ) d t , x ∈ [ a , a + λ ] , ∫ x b g ( t ) d t , x ∈ [ a + λ , b ] .$
Then, we have the following:
$P 1 ( b ) · f ( n ) m 1 ≤ ( n − 2 ) ! ∫ a b f ( t ) g ( t ) d t − ∫ a a + λ f ( t ) d t − ∑ i = 0 n − 2 f ( i + 1 ) ( a ) i ! ∫ a b G 1 ( x ) ( x − a ) i d x ≤ P 1 ( b ) · b − m 1 b − a f ( n ) ( a ) + m 1 − a b − a f ( n ) ( b ) ,$
where the following is the case:
$P 1 ( b ) = 1 ( n − 1 ) · n ∫ a b g ( x ) ( x − a ) n d x − λ n + 1 n + 1$
and the following is obtained.
$m 1 = a + 1 ( n − 1 ) · n · ( n + 1 ) · P 1 ( b ) ∫ a b g ( x ) ( x − a ) n + 1 d x − λ n + 2 n + 2 .$
Proof.
Since $f ( n − 1 )$ is absolutely continuous, function f satisfies the conditions of Theorem 3. Therefore, identity (3) holds.
From condition $0 ≤ g ≤ 1$, function $G 1$ defined by (7) is non-negative. Hence, for every $n ≥ 2$, we have the following.
$∫ t b G 1 ( x ) ( x − t ) n − 2 d x ≥ 0 , t ∈ a , b .$
Define
$p ( t ) = ∫ t b G 1 ( x ) ( x − t ) n − 2 d x .$
Since the function f is $( n + 2 ) −$convex, function $f ( n )$ is convex. Furthermore, function p is non-negative, so we can apply Theorem 1 and obtain the following inequality:
$P 1 ( b ) · f ( n ) m 1 ≤ ∫ a b ∫ t b G 1 ( x ) ( x − t ) n − 2 d x f ( n ) ( t ) d t ≤ P 1 ( b ) · b − m 1 b − a f ( n ) ( a ) + m 1 − a b − a f ( n ) ( b ) ,$
where $P 1 ( b )$ and $m 1$ are given by
$P 1 ( b ) = ∫ a b ∫ t b G 1 ( x ) ( x − t ) n − 2 d x d t$
and
$m 1 = 1 P 1 ( b ) ∫ a b ∫ t b G 1 ( x ) ( x − t ) n − 2 d x t d t .$
By calculating $P 1 ( b )$ and $m 1$, we obtain the following:
$P 1 ( b ) = ∫ a b ∫ t b G 1 ( x ) ( x − t ) n − 2 d x d t = ∫ a a + λ ∫ a x ( 1 − g ( s ) ) d s ( x − a ) n − 1 n − 1 d x + ∫ a + λ b ∫ x b g ( s ) d s ( x − a ) n − 1 n − 1 d x = ∫ a a + λ ( x − a ) n n − 1 d x + λ · ∫ a + λ b ( x − a ) n − 1 n − 1 d x − ∫ a b ∫ a x g ( s ) d s ( x − a ) n − 1 n − 1 d x = − λ n + 1 ( n − 1 ) · n · ( n + 1 ) + ∫ a b g ( x ) ( x − a ) n ( n − 1 ) · n d x$
and
$m 1 = 1 P 1 ( b ) ∫ a b ∫ t b G 1 ( x ) ( x − t ) n − 2 d x t d t = 1 P 1 ( b ) ∫ a b G 1 ( x ) ∫ a x ( x − t ) n − 2 · t d t d x = 1 P 1 ( b ) ∫ a b G 1 ( x ) t · − ( x − t ) n − 1 n − 1 | a x + ∫ a x ( x − t ) n − 1 n − 1 d t d x = 1 P 1 ( b ) ∫ a b G 1 ( x ) a · ( x − a ) n − 1 n − 1 + ( x − a ) n ( n − 1 ) · n d x = a + 1 P 1 ( b ) ∫ a b G 1 ( x ) ( x − a ) n ( n − 1 ) · n d x = a + 1 P 1 ( b ) − λ n + 2 ( n − 1 ) · n · ( n + 1 ) · ( n + 2 ) + ∫ a b g ( x ) ( x − a ) n + 1 ( n − 1 ) · n · ( n + 1 ) d x .$
Using identity (3) for the middle part of the inequality (11), inequality (11) becomes inequality (8). Hence, the proof is completed. □
Theorem 6.
Let f: $[ a , b ] → R$ be $( n + 2 ) −$convex on $[ a , b ]$ and $f ( n − 1 )$ absolutely continuous for $n ≥ 2$. Let g: $[ a , b ] → R$ be an integrable function such that $0 ≤ g ≤ 1$ and $λ = ∫ a b g ( t ) d t$. Let function $G 1$ be defined by (7). If the following is the case:
$∫ a t G 1 ( x ) ( x − t ) n − 2 d x ≤ 0 , t ∈ [ a , b ] ,$
then we have the following:
$P 2 ( b ) · f ( n ) m 2 ≤ ( n − 2 ) ! ∫ a b f ( t ) g ( t ) d t − ∫ a a + λ f ( t ) d t − ∑ i = 0 n − 2 f ( i + 1 ) ( b ) i ! ∫ a b G 1 ( x ) ( x − b ) i d x ≤ P 2 ( b ) · b − m 2 b − a f ( n ) ( a ) + m 2 − a b − a f ( n ) ( b ) ,$
where
$P 2 ( b ) = 1 ( n − 1 ) · n ( a − b ) n + 1 − ( a + λ − b ) n + 1 n + 1 + ∫ a b g ( x ) ( x − b ) n d x$
and
$m 2 = b + 1 ( n − 1 ) · n · ( n + 1 ) · P 2 ( b ) × ( a − b ) n + 2 − ( a + λ − b ) n + 2 n + 2 + ∫ a b g ( x ) ( x − b ) n + 1 d x .$
Proof.
If we assume the following:
$∫ a t G 1 ( x ) ( x − t ) n − 2 d x ≤ 0 , t ∈ [ a , b ]$
then we have the following.
$− ∫ a t G 1 ( x ) ( x − t ) n − 2 d x ≥ 0 , t ∈ [ a , b ] .$
Now similarly to the proof of Theorem 5 using the following non-negative function:
$p ( t ) = − ∫ a t G 1 ( x ) ( x − t ) n − 2 d x$
and identity (4), we obtain inequality (12). Similarly, we calculate the expressions for $P 2 ( b )$ and $m 2$ and obtain the following:
$P 2 ( b ) = − ∫ a b ∫ a t G 1 ( x ) ( x − t ) n − 2 d x d t = ∫ a a + λ ∫ a x ( 1 − g ( s ) ) d s ( x − b ) n − 1 n − 1 d x + ∫ a + λ b ∫ x b g ( s ) d s ( x − b ) n − 1 n − 1 d x = ∫ a a + λ ( x − a ) ( x − b ) n − 1 n − 1 d x + λ · ∫ a + λ b ( x − b ) n − 1 n − 1 d x − ∫ a b ∫ a x g ( s ) d s ( x − b ) n − 1 n − 1 d x = ( a − b ) n + 1 ( n − 1 ) · n · ( n + 1 ) − ( a + λ − b ) n + 1 ( n − 1 ) · n · ( n + 1 ) + ∫ a b g ( x ) ( x − b ) n ( n − 1 ) · n d x$
and
$m 2 = − 1 P 2 ( b ) ∫ a b ∫ a t G 1 ( x ) ( x − t ) n − 2 d x t d t = − 1 P 2 ( b ) ∫ a b G 1 ( x ) ∫ x b ( x − t ) n − 2 · t d t d x = − 1 P 2 ( b ) ∫ a b G 1 ( x ) t · − ( x − t ) n − 1 n − 1 | x b + ∫ x b ( x − t ) n − 1 n − 1 d t d x = − 1 P 2 ( b ) ∫ a b G 1 ( x ) − b · ( x − b ) n − 1 n − 1 − ( x − b ) n ( n − 1 ) · n d x = b + 1 P 2 ( b ) ∫ a b G 1 ( x ) ( x − b ) n ( n − 1 ) · n d x = b + 1 ( n − 1 ) · n · ( n + 1 ) · P 1 ( b ) × ( a − b ) n + 2 n + 2 − ( a + λ − b ) n + 2 n + 2 + ∫ a b g ( x ) ( x − b ) n + 1 d x .$
Hence, the proof is completed. □
Theorem 7.
Let f: $[ a , b ] → R$ be $( n + 2 ) −$convex on $[ a , b ]$ and $f ( n − 1 )$ absolutely continuous for $n ≥ 2$. Let g: $[ a , b ] → R$ be an integrable function such that $0 ≤ g ≤ 1$ and $λ = ∫ a b g ( t ) d t$. Let function $G 2$ be defined by the following.
$G 2 ( x ) = ∫ a x g ( t ) d t , x ∈ [ a , b − λ ] , ∫ x b ( 1 − g ( t ) ) d t , x ∈ [ b − λ , b ] .$
Then, the following is obtained:
$P 3 ( b ) · f ( n ) m 3 ≤ ( n − 2 ) ! ∫ b − λ b f ( t ) d t − ∑ i = 0 n − 2 f ( i + 1 ) ( a ) i ! ∫ a b G 2 ( x ) ( x − a ) i d x − ∫ a b f ( t ) g ( t ) d t ≤ P 3 ( b ) · b − m 3 b − a f ( n ) ( a ) + m 3 − a b − a f ( n ) ( b ) ,$
where
$P 3 ( b ) = 1 ( n − 1 ) · n ( b − a ) n + 1 − ( b − λ − a ) n + 1 n + 1 − ∫ a b g ( x ) ( x − a ) n d x$
and
$m 3 = a + 1 ( n − 1 ) · n · ( n + 1 ) · P 3 ( b ) × ( b − a ) n + 2 − ( b − λ − a ) n + 2 n + 2 − ∫ a b g ( x ) ( x − a ) n + 1 d x .$
Proof.
We follow the similar arguments as in the proof of Theorem 5. As function $f ( n − 1 )$ is absolutely continuous, the identity (5) holds. The inequality (14) follows directly from Theorem 1, substituting the non-negative function p by a non-negative function of the following:
$p ( t ) = ∫ t b G 2 ( x ) ( x − t ) n − 2 d x$
and a convex function f by a convex function $f ( n )$, and then using identity (5) for integral $∫ a b ∫ t b G 2 ( x ) ( x − t ) n − 2 d x f ( n ) ( t ) d t$. Furthermore, we calculate $P 3 ( b )$ and $m 3$ as follows.
$P 3 ( b ) = ∫ a b ∫ t b G 2 ( x ) ( x − t ) n − 2 d x d t = ∫ a b − λ ∫ a x g ( s ) d s ( x − a ) n − 1 n − 1 d x + ∫ b − λ b ∫ x b ( 1 − g ( s ) ) d s ( x − a ) n − 1 n − 1 d x = ∫ b − λ b ( b − x ) ( x − a ) n − 1 n − 1 d x − λ · ∫ b − λ b ( x − a ) n − 1 n − 1 d x + ∫ a b ∫ a x g ( s ) d s ( x − a ) n − 1 n − 1 d x = ( b − a ) n + 1 − ( b − λ − a ) n + 1 ( n − 1 ) · n · ( n + 1 ) − ∫ a b g ( x ) ( x − a ) n ( n − 1 ) · n d x ,$
$m 3 = 1 P 3 ( b ) ∫ a b ∫ t b G 2 ( x ) ( x − t ) n − 2 d x t d t = 1 P 3 ( b ) ∫ a b G 2 ( x ) ∫ a x ( x − t ) n − 2 · t d t d x = 1 P 3 ( b ) ∫ a b G 2 ( x ) t · − ( x − t ) n − 1 n − 1 | a x + ∫ a x ( x − t ) n − 1 n − 1 d t d x = 1 P 3 ( b ) ∫ a b G 2 ( x ) a · ( x − a ) n − 1 n − 1 + ( x − a ) n ( n − 1 ) · n d x = a + 1 P 3 ( b ) ∫ a b G 2 ( x ) ( x − a ) n ( n − 1 ) · n d x = a + 1 P 3 ( b ) ( b − a ) n + 2 − ( b − λ − a ) n + 2 ( n − 1 ) · n · ( n + 1 ) · ( n + 2 ) − ∫ a b g ( x ) ( x − a ) n + 1 ( n − 1 ) · n · ( n + 1 ) d x .$
Hence, the proof is completed. □
Theorem 8.
Let f: $[ a , b ] → R$ be $( n + 2 ) −$convex on $[ a , b ]$ and $f ( n − 1 )$ absolutely continuous for $n ≥ 2$. Let g: $[ a , b ] → R$ be an integrable function such that $0 ≤ g ≤ 1$ and $λ = ∫ a b g ( t ) d t$. Let function $G 2$ be defined by (13). If the following is the case:
$∫ a t G 2 ( x ) ( x − t ) n − 2 d x ≤ 0 , t ∈ [ a , b ]$
then we obtain the following:
$P 4 ( b ) · f ( n ) m 4 ≤ ( n − 2 ) ! ∫ b − λ b f ( t ) d t − ∑ i = 0 n − 2 f ( i + 1 ) ( b ) i ! ∫ a b G 2 ( x ) ( x − b ) i d x − ∫ a b f ( t ) g ( t ) d t ≤ P 4 ( b ) · b − m 4 b − a f ( n ) ( a ) + m 4 − a b − a f ( n ) ( b ) ,$
where
$P 4 ( b ) = − 1 ( n − 1 ) · n ( − λ ) n + 1 n + 1 + ∫ a b g ( x ) ( x − b ) n d x$
and
$m 4 = b − 1 ( n − 1 ) · n · ( n + 1 ) · P 4 ( b ) ( − λ ) n + 2 n + 2 + ∫ a b g ( x ) ( x − b ) n + 1 d x .$
Proof.
Under the assumption that $∫ a t G 2 ( x ) ( x − t ) n − 2 d x ≤ 0$, it is obvious that the following is the case:
$p ( t ) = − ∫ a t G 2 ( x ) ( x − t ) n − 2 d x$
where it is a non-negative function. Again, replacing $p ( t )$ in Theorem 1 by (16) and f by $f ( n )$ and then using the identity (6) for
$∫ a b ∫ a t G 2 ( x ) ( x − t ) n − 2 d x f ( n ) ( t ) d t ,$
we obtain the required inequalities (15). Finally, a simple calculation yields the following:
$P 4 ( b ) = − ∫ a b ∫ a t G 2 ( x ) ( x − t ) n − 2 d x d t = ∫ a b − λ ∫ a x g ( s ) d s ( x − b ) n − 1 n − 1 d x + ∫ b − λ b ∫ x b ( 1 − g ( s ) ) d s ( x − b ) n − 1 n − 1 d x = − ∫ b − λ b ( x − b ) n n − 1 d x − λ · ∫ b − λ b ( x − b ) n − 1 n − 1 d x + ∫ a b ∫ a x g ( s ) d s ( x − b ) n − 1 n − 1 d x = − ( − λ ) n + 1 ( n − 1 ) · n · ( n + 1 ) − ∫ a b g ( x ) ( x − b ) n ( n − 1 ) · n d x$
and
$m 4 = − 1 P 4 ( b ) ∫ a b ∫ a t G 2 ( x ) ( x − t ) n − 2 d x t d t = − 1 P 4 ( b ) ∫ a b G 2 ( x ) ∫ x b ( x − t ) n − 2 · t d t d x = − 1 P 4 ( b ) ∫ a b G 2 ( x ) t · − ( x − t ) n − 1 n − 1 | x b + ∫ x b ( x − t ) n − 1 n − 1 d t d x = − 1 P 4 ( b ) ∫ a b G 2 ( x ) − b · ( x − b ) n − 1 n − 1 − ( x − b ) n ( n − 1 ) · n d x = b + 1 P 4 ( b ) ∫ a b G 2 ( x ) ( x − b ) n ( n − 1 ) · n d x = b − 1 P 4 ( b ) ( − λ ) n + 2 ( n − 1 ) · n · ( n + 1 ) · ( n + 2 ) + ∫ a b g ( x ) ( x − b ) n + 1 ( n − 1 ) · n · ( n + 1 ) d x .$
Remark 1.
If function f is $( n + 2 ) −$concave, the inequalities in Theorems 5–8 are reversed. This follows from the fact that for $( n + 2 ) −$concave function, we have $− f ( n + 2 ) ≥ 0$. Hence, $− f ( n )$ is convex and we can apply inequality (1) to function $− f ( n )$.
Remark 2.
The expressions $P i ( b )$ and $m i$ for $i = 1 , … , 4$ can also be achieved by the method introduced in [16]. By this method, we calculate $P 1 ( b )$ and $m 1$. Other expressions can be recaptured in a similar manner.
The value of $P 1 ( b )$ can be obtained from (3) by taking $f ( t ) = ( t − a ) n n !$. Then, $f ( n ) ( t ) = 1$. Thus, we have the following.
$P 1 ( b ) = − ( n − 2 ) ! ∫ a a + λ ( x − a ) n n ! d t − ∫ a b ( x − a ) n n ! g ( t ) d t = − λ n + 1 ( n − 1 ) · n · ( n + 1 ) + ∫ a b ( x − a ) n ( n − 1 ) · n g ( t ) d t .$
Hence, we obtained expression (9).
From Theorem 1, we previously obtained the following.
$m 1 = 1 P 1 ( b ) ∫ a b ∫ t b G 1 ( x ) ( x − t ) n − 2 d x t d t .$
To calculate $m 1$, we take function $f ( t ) = ( t − a ) n + 1 ( n + 1 ) !$. Then, $f ( n ) ( t ) = t − a$. Hence, from the identity (3), we obtain expression (10).

## 3. Conclusions

In this paper, we obtained new weighted Hermite–Hadamard-type inequalities for higher order convex functions. We used previously obtained identities related to the generalizations of Steffensen’s inequality. Results obtained in this paper can be considered as a starting point for some future work.

## Author Contributions

Conceptualization, J.P., A.P.P. and K.S.K.; Writing — original draft, J.P., A.P.P. and K.S.K. All authors have read and agreed to the published version of the manuscript.

## Funding

This research received no external funding.

Not applicable.

Not applicable.

Not applicable.

## Acknowledgments

Croatian Science Foundation (HRZZ 7926) “Separation of parameter influence in engineering modeling and parameter identification”.

## Conflicts of Interest

The authors declare no conflict of interest.

## References

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MDPI and ACS Style

Pečarić, J.; Perušić Pribanić, A.; Smoljak Kalamir, K. Weighted Hermite–Hadamard-Type Inequalities by Identities Related to Generalizations of Steffensen’s Inequality. Mathematics 2022, 10, 1505. https://doi.org/10.3390/math10091505

AMA Style

Pečarić J, Perušić Pribanić A, Smoljak Kalamir K. Weighted Hermite–Hadamard-Type Inequalities by Identities Related to Generalizations of Steffensen’s Inequality. Mathematics. 2022; 10(9):1505. https://doi.org/10.3390/math10091505

Chicago/Turabian Style

Pečarić, Josip, Anamarija Perušić Pribanić, and Ksenija Smoljak Kalamir. 2022. "Weighted Hermite–Hadamard-Type Inequalities by Identities Related to Generalizations of Steffensen’s Inequality" Mathematics 10, no. 9: 1505. https://doi.org/10.3390/math10091505

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