New generalizations of Steffensen's inequality by Lidstone's polynomial.

In this paper, we utilize some known Steffensen-type identities, obtained by using Lidstone's interpolating polynomial, to prove new generalizations of Steffensen's inequality. We obtain these new generalizations by using the weighted Hermite-Hadamard inequality for (2 n + 2) - convex and (2 n + 3) - convex functions. Further, the newly obtained inequalities can be observed as an upper- and lower-bound for utilized Steffensen-type identities.

Keywords: Steffensen's inequality; Generalizations; (2n+2)-\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$(2n+2)-$$\end{document} convex and (2n+3)-\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$(2n+3)-$$\end{document} convex functions; Weighted Hermite-Hadamard inequality; Lidstone interpolating polynomial; Green's function; Primary 26D15; Secondary 26A51

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Introduction

The well-known Taylor series gives an approximation of a function at a single point. In order to obtain an approximation in two points Lidstone [[4]] proved its generalization, called the Lidstone series. The Lidstone series has also intrigued other mathematicians such as Boas [[2]], Poritsky [[12]], Schoenberg [[13]], Whittaker [[15]], Widder [[17]], and many others. In [[1]] Agarwal and Wong gave a comprehensive overview of the Lidstone series and the Lidstone interpolating polynomial representations including its error representations, error estimates and boundary value problems.

Let ${\mathrm{\Lambda }}_n(t)$ be the Lidstone polynomial of degree $2n+1$ , $n\in \mathbb{N}$ , defined by the relations

$$\begin{array}{cc}\hfill {\mathrm{\Lambda }}_0(t)& =t,\hfill \\ \hfill {\mathrm{\Lambda }}_n^{″}(t)& ={\mathrm{\Lambda }}_{n-1}(t)\hfill & \hfill ,\\ \hfill {\mathrm{\Lambda }}_n(0)& ={\mathrm{\Lambda }}_n(1)=0,n\ge 1.\hfill \end{array}$$

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Let us recall the following Lidstone interpolating polynomial representation of the function $f\in C^{(2n)}([0,1])$ proved in [[18]].

Lemma 1.1

If $f\in C^{(2n)}([0,1])$ , then

$$\begin{array}{c}\hfill f(t)=\sum _{k=0}^{n-1}\left[f^{(2k)},(0),{\mathrm{\Lambda }}_k,(1-t),+,f^{(2k)},(1),{\mathrm{\Lambda }}_k,(t)\right]+{\int }_0^1{G}_n(t,s)f^{(2n)}(s)ds,\end{array}$$

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where ${\mathrm{\Lambda }}_n(t)$ is the Lidstone interpolating polynomial and $G_n$ is the Green function defined by

$$\begin{array}{c}\hfill G_1(t,s)=G(t,s)=\left\{\begin{array}{cc}(t-1)s,\hfill & \text{if}s\le t,\hfill \\ (s-1)t,\hfill & \text{if}t\le s,\hfill \end{array}\right)\end{array}$$

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and

1.1 $$\begin{array}{c}\hfill G_n(t,s)={\int }_0^1{G}_1(t,p)G_{n-1}(p,s)dp,n\ge 2.\end{array}$$

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For the application we will need the following expression of the Lidstone interpolating polynomial ${\mathrm{\Lambda }}_n(t)$ in terms of the Green function $G_n(t,s)$

1.2 $$\begin{array}{c}\hfill {\mathrm{\Lambda }}_n(t)={\int }_0^1{G}_n(t,s)sds,n\ge 1.\end{array}$$

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In [[1]] Agarwal and Wong showed that the Green function is given by the following expression:

$$\begin{array}{c}\hfill G_n(t,s)=\left\{\begin{array}{cc}-\sum _{k=0}^{n-1}{\mathrm{\Lambda }}_k(t)\frac{{(1-s)}^{2n-2k-1}}{(2n-2k-1)!},\hfill & t\le s,\hfill \\ -\sum _{k=0}^{n-1}{\mathrm{\Lambda }}_k(1-t)\frac{s^{2n-2k-1}}{(2n-2k-1)!},\hfill & s\le t.\hfill \end{array}\right)\end{array}$$

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Since we will deal with an interval [a, b] instead of the interval [0, 1], for the readers convenience, let us also give the aforementioned results on the Lidstone interpolating polynomial and the Green function on an interval [a, b].

The Lidstone series representation of $f\in C^{(2n)}([a,b])$ is given by:

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill f(x)& =\sum _{k=0}^{n-1}{(b-a)}^{2k}\left[f^{(2k)},(a),{\mathrm{\Lambda }}_k,\left(\frac{b-x}{b-a}\right),+,f^{(2k)},(b),{\mathrm{\Lambda }}_k,\left(\frac{x-a}{b-a}\right)\right]\hfill \\ \hfill & +{(b-a)}^{2n-1}{\int }_a^b{G}_n\left(\frac{x-a}{b-a},,,\frac{s-a}{b-a}\right)f^{(2n)}(s)ds\hfill \end{array}\end{array}$$

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and the expression for Green's function (1.3) on an interval [a, b] is given by:

$$\begin{array}{c}\hfill \begin{array}{cc}& G_n\left(\frac{x-a}{b-a},,,\frac{s-a}{b-a}\right)\hfill \\ \hfill & =\left\{\begin{array}{cc}{\displaystyle -\sum _{k=0}^{n-1}{\mathrm{\Lambda }}_k\left(\frac{x-a}{b-a}\right)\frac{{(b-s)}^{2n-2k-1}}{{\left(b,-,a\right)}^{2n-2k-1}(2n-2k-1)!},}\hfill & x\le s,\hfill \\ {\displaystyle -\sum _{k=0}^{n-1}{\mathrm{\Lambda }}_k\left(\frac{b-x}{b-a}\right)\frac{{(s-a)}^{2n-2k-1}}{{\left(b,-,a\right)}^{2n-2k-1}(2n-2k-1)!},}\hfill & s\le x.\hfill \end{array}\right)\hfill \end{array}\end{array}$$

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Let us recall one of the classical inequalities which was proved by Steffensen [[14]].

Theorem 1.2

([[14]]) Suppose that f is non-increasing and g is integrable on [a, b] with $0\le g\le 1$ and $\lambda ={\int }_a^{b}g(t)dt.$ Then we have

$$\begin{array}{c}\hfill {\int }_{b-\lambda }^{b}f(t)dt\le {\int }_a^{b}f(t)g(t)dt\le {\int }_a^{a+\lambda }f(t)dt.\end{array}$$

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The inequalities are reversed for f non-decreasing.

Over the years, Steffensen's inequality was generalized in many ways. An extensive overview of these generalizations can be found in [[3], [5], [10]]. In papers [[8]] the authors introduced and proved some Steffensen-type identities and generalizations of Steffensen's inequality for $(2n)-$ convex and $(2n+1)-$ convex functions using Lidstone interpolating polynomials. In this paper we will use the following Steffensen-type identities which were obtained in [[9]].

Theorem 1.3

([[9]]) Let $f:I\to \mathbb{R}$ be such that $f^{(2n-1)}$ is absolutely continuous for some $n\ge 1$ , $I\subset \mathbb{R}$ an open interval, $a,b\in I$ , $a<b$ . Let $g,p:[a,b]\to \mathbb{R}$ be integrable functions such that p is positive and $0\le g\le 1$ . Let ${\int }_a^{a+\lambda }p(t)dt={\int }_a^{b}g(t)p(t)dt$ and let the function ${\mathcal{G}}_1$ be defined by

1.5 $$\begin{array}{c}\hfill {\mathcal{G}}_1(x)=\left\{\begin{array}{cc}{\int }_a^x(1-g(t))p(t)dt,\hfill & x\in [a,a+\lambda ],\hfill \\ {\int }_x^{b}g(t)p(t)dt,\hfill & x\in [a+\lambda ,b].\hfill \end{array}\right)\end{array}$$

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Then

1.6 $$\begin{array}{c}\hfill \begin{array}{cc}& {\int }_a^{a+\lambda }f(t)p(t)dt-{\int }_a^{b}f(t)g(t)p(t)dt\hfill \\ \hfill & +\sum _{k=0}^{n-1}{(b-a)}^{2k-1}{\int }_a^b{\mathcal{G}}_1(x)\left[f^{\left(2,k\right)},\left(b\right),{\mathrm{\Lambda }}_k^{\prime },,,\left(\frac{x-a}{b-a}\right),-,f^{\left(2,k\right)},\left(a\right),{\mathrm{\Lambda }}_k^{\prime },,,\left(\frac{b-x}{b-a}\right)\right]dx\hfill \\ \hfill & =-{(b-a)}^{2n-1}{\int }_a^b\left({\int }_a^b,{\mathcal{G}}_1,(x),\frac{d{G}_n}{\mathit{dx}},,\left(\frac{x-a}{b-a},,,\frac{s-a}{b-a}\right),d,x\right)f^{(2n)}(s)ds.\hfill \end{array}\end{array}$$

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Theorem 1.4

([[9]]) Let $f:I\to \mathbb{R}$ be such that $f^{(2n-1)}$ is absolutely continuous for some $n\ge 1$ , $I\subset \mathbb{R}$ an open interval, $a,b\in I$ , $a<b$ . Let $g,p:[a,b]\to \mathbb{R}$ be integrable functions such that p is positive and $0\le g\le 1$ . Let ${\int }_{b-\lambda }^{b}p(t)dt={\int }_a^{b}g(t)p(t)dt$ and let the function ${\mathcal{G}}_2$ be defined by

1.7 $$\begin{array}{c}\hfill {\mathcal{G}}_2(x)=\left\{\begin{array}{cc}{\int }_a^{x}g(t)p(t)dt,\hfill & x\in [a,b-\lambda ],\hfill \\ {\int }_x^b(1-g(t))p(t)dt,\hfill & x\in [b-\lambda ,b].\hfill \end{array}\right)\end{array}$$

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Then

1.8 $$\begin{array}{c}\hfill \begin{array}{cc}& {\int }_a^{b}f(t)g(t)p(t)dt-{\int }_{b-\lambda }^{b}f(t)p(t)dt\hfill \\ \hfill & +\sum _{k=0}^{n-1}{(b-a)}^{2k-1}{\int }_a^b{\mathcal{G}}_2(x)\left[f^{\left(2,k\right)},\left(b\right),{\mathrm{\Lambda }}_k^{\prime },,,\left(\frac{x-a}{b-a}\right),-,f^{\left(2,k\right)},\left(a\right),{\mathrm{\Lambda }}_k^{\prime },,,\left(\frac{b-x}{b-a}\right)\right]dx\hfill \\ \hfill & =-{(b-a)}^{2n-1}{\int }_a^b\left({\int }_a^b,{\mathcal{G}}_2,(x),\frac{d{G}_n}{\mathit{dx}},,\left(\frac{x-a}{b-a},,,\frac{s-a}{b-a}\right),d,x\right)f^{(2n)}(s)ds.\hfill \end{array}\end{array}$$

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Theorem 1.5

([[9]]) Let $f:I\to \mathbb{R}$ be such that $f^{(2n)}$ is absolutely continuous for some $n\ge 1$ , $I\subset \mathbb{R}$ an open interval, $a,b\in I$ , $a<b$ . Let $g,p:[a,b]\to \mathbb{R}$ be integrable functions such that p is positive and $0\le g\le 1$ . Let ${\int }_a^{a+\lambda }p(t)dt={\int }_a^{b}g(t)p(t)dt$ and let the function ${\mathcal{G}}_1$ be defined by (1.5). Then

1.9 $$\begin{array}{c}\hfill \begin{array}{cc}& {\int }_a^{a+\lambda }f(t)p(t)dt-{\int }_a^{b}f(t)g(t)p(t)dt+\sum _{k=0}^{n-1}{(b-a)}^{2k}\hfill \\ \hfill & \times {\int }_a^b{\mathcal{G}}_1(x)\left[f^{\left(2,k,+,1\right)},\left(a\right),{\mathrm{\Lambda }}_k,\left(\frac{b-x}{b-a}\right),+,f^{\left(2,k,+,1\right)},\left(b\right),{\mathrm{\Lambda }}_k,\left(\frac{x-a}{b-a}\right)\right]dx\hfill \\ \hfill & =-{(b-a)}^{2n-1}{\int }_a^b\left({\int }_a^b,{\mathcal{G}}_1,(x),G_n,\left(\frac{x-a}{b-a},,,\frac{s-a}{b-a}\right),d,x\right)f^{\left(2,n,+,1\right)}(s)ds.\hfill \end{array}\end{array}$$

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Theorem 1.6

([[9]]) Let $f:I\to \mathbb{R}$ be such that $f^{(2n)}$ is absolutely continuous for some $n\ge 1$ , $I\subset \mathbb{R}$ an open interval, $a,b\in I$ , $a<b$ . Let $g,p:[a,b]\to \mathbb{R}$ be integrable functions such that p is positive and $0\le g\le 1$ . Let ${\int }_{b-\lambda }^{b}p(t)dt={\int }_a^{b}g(t)p(t)dt$ and let the function ${\mathcal{G}}_2$ be defined by (1.7). Then

1.10 $$\begin{array}{c}\hfill \begin{array}{cc}& {\int }_a^{b}f(t)g(t)p(t)dt-{\int }_{b-\lambda }^{b}f(t)p(t)dt+\sum _{k=0}^{n-1}{(b-a)}^{2k}\hfill \\ \hfill & \times {\int }_a^b{\mathcal{G}}_2(x)\left[f^{\left(2,k,+,1\right)},\left(a\right),{\mathrm{\Lambda }}_k,\left(\frac{b-x}{b-a}\right),+,f^{\left(2,k,+,1\right)},\left(b\right),{\mathrm{\Lambda }}_k,\left(\frac{x-a}{b-a}\right)\right]dx\hfill \\ \hfill & =-{(b-a)}^{2n-1}{\int }_a^b\left({\int }_a^b,{\mathcal{G}}_2,(x),G_n,\left(\frac{x-a}{b-a},,,\frac{s-a}{b-a}\right),d,x\right)f^{\left(2,n,+,1\right)}(s)ds.\hfill \end{array}\end{array}$$

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The aim of this paper is to use the above identities related to Steffensen's inequality to obtain new generalizations of Steffensen's inequality for $(2n+2)-$ convex and $(2n+3)-$ convex functions as a continuation of the results published in [[7]]. We will use the weighted Hermite-Hadamard inequality for convex functions given in the following theorem (see [[6], [10], [19]]):

Theorem 1.7

Let $u:[a,b]\to \mathbb{R}$ be a non-negative function, let $U(b)={\int }_a^{b}u(x)dx$ and let $m=\frac{1}{U(b)}{\int }_a^{b}xu(x)dx.$ If f is a convex function on [a, b], then the following inequalities hold:

$$\begin{array}{c}\hfill U(b)f(m)\le {\int }_a^{b}u(x)f(x)dx\le U(b)\left[\frac{b-m}{b-a},f,(a),+,\frac{m-a}{b-a},f,(b)\right].\end{array}$$

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The inequalities are reversed for f concave.

New generalizations of Steffensen's inequality

In this section we will prove our main results by utilizing the Steffensen-type identities (1.6), (1.8), (1.9) and (1.10) for $(2n+2)-$ convex and $(2n+3)-$ convex functions.

Firstly, let us denote

2.1 $$\begin{array}{c}\hfill \begin{array}{cc}\hfill \mathrm{\Omega }(x)& =\sum _{k=0}^{n-1}\frac{1}{(2n-2k)!}\left[{\mathrm{\Lambda }}_k^{\prime },,,\left(\frac{x-a}{b-a}\right),·,{\left(\frac{b-x}{b-a}\right)}^{2n-2k}\right)\hfill \\ \hfill & \left(-,{\mathrm{\Lambda }}_k^{\prime },,,\left(\frac{b-x}{b-a}\right),·,{\left(\frac{x-a}{b-a}\right)}^{2n-2k}\right].\hfill \end{array}\end{array}$$

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In the following theorem we will prove our first result for $(2n+2)-$ convex functions using the identity from Theorem 1.3.

Theorem 2.1

Let $I\subset \mathbb{R}$ be an open interval and let $a,b\in I$ be such that $a<b$ . Let the function $f:I\to \mathbb{R}$ be such that $f^{(2n-1)}$ is absolutely continuous and f is $(2n+2)-$ convex on I for $n\ge 1$ . Let $g,p:[a,b]\to \mathbb{R}$ be integrable functions such that p is positive and $0\le g\le 1$ . Let ${\int }_a^{a+\lambda }p(t)dt={\int }_a^{b}g(t)p(t)dt$ , let $G_n$ be the Green function given by (1.4), and let ${\mathcal{G}}_1$ and $\mathrm{\Omega }$ be defined by (1.5) and (2.1), respectively.

If

2.2 $$\begin{array}{c}\hfill {\displaystyle -{\int }_a^b{\mathcal{G}}_1(x)\frac{d{G}_n}{\mathit{dx}}\left(\frac{x-a}{b-a},,,\frac{s-a}{b-a}\right)dx\ge 0,s\in [a,b],}\end{array}$$

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then

2.3 $$\begin{array}{c}\hfill \begin{array}{cc}& U_1(b)·f^{(2n)}\left(m_1\right)\le \hfill \\ \hfill & {(b-a)}^{2n-1}\left[{\int }_a^{a+\lambda },f,(t),p,(t),d,t,-,{\int }_a^b,f,(t),g,(t),p,(t),d,t,c,+,\sum _{k=0}^{n-1},{(b-a)}^{2k-1}\right)\hfill \\ \hfill & \left(\times ,{\int }_a^b,{\mathcal{G}}_1,(x),\left[f^{\left(2,k\right)},\left(b\right),{\mathrm{\Lambda }}_k^{{}^{\prime }},\left(\frac{x-a}{b-a}\right),-,f^{\left(2,k\right)},\left(a\right),{\mathrm{\Lambda }}_k^{{}^{\prime }},\left(\frac{b-x}{b-a}\right)\right],d,x\right]\hfill \\ \hfill & \le U_1(b)·\left[\frac{b-m_1}{b-a},f^{(2n)},(a),+,\frac{m_1-a}{b-a},f^{(2n)},(b)\right],\hfill \end{array}\end{array}$$

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where

$$\begin{array}{c}\hfill U_1(b)={\int }_a^b{\mathcal{G}}_1(x)\mathrm{\Omega }(x)dx\end{array}$$

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and

2.4 $$\begin{array}{c}\hfill \begin{array}{cc}\hfill m_1=& a-\frac{b-a}{U_1(b)}{\int }_a^b{\mathcal{G}}_1(x){\mathrm{\Lambda }}_n^{{}^{\prime }}\left(\frac{x-a}{b-a}\right)dx.\hfill \end{array}\end{array}$$

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Proof

Firstly, let us define the function $u_1$ on $\left[a,,,b\right]$ by

$$\begin{array}{c}\hfill {\displaystyle u_1(s)=-{\int }_a^b{\mathcal{G}}_1(x)\frac{d{G}_n}{\mathit{dx}}\left(\frac{x-a}{b-a},,,\frac{s-a}{b-a}\right)dx,s\in [a,b].}\end{array}$$

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Under the assumption (2.2) it is obvious that $u_1$ is a non-negative function. Further, let us note that the function $f^{(2n)}$ is convex. This follows from the fact that the function f is $(2n+2)-$ convex. Hence, by applying the weighted Hermite-Hadamard inequality with non-negative function $u_1$ and convex function $f^{(2n)}$ we obtain

2.5 $$\begin{array}{c}\hfill \begin{array}{cc}\hfill U_1(b)·f^{(2n)}\left(m_1\right)& \le -{\int }_a^b\left({\displaystyle {\int }_a^b{\mathcal{G}}_1(x)\frac{d{G}_n}{\mathit{dx}}\left(\frac{x-a}{b-a},,,\frac{s-a}{b-a}\right)dx}\right)f^{(2n)}(s)ds\hfill \\ \hfill & {\displaystyle \le U_1(b)·\left[\frac{b-m_1}{b-a},f^{(2n)},(a),+,\frac{m_1-a}{b-a},f^{(2n)},(b)\right],}\hfill \end{array}\end{array}$$

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where the expressions $U_1(b)$ and $m_1$ can be calculated as

$$\begin{array}{c}\hfill {\displaystyle U_1(b)=-{\int }_a^b\left({\displaystyle {\int }_a^b{\mathcal{G}}_1(x)\frac{d{G}_n}{\mathit{dx}}\left(\frac{x-a}{b-a},,,\frac{s-a}{b-a}\right)dx}\right)ds}\end{array}$$

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and

$$\begin{array}{c}\hfill {\displaystyle m_1=-\frac{1}{U_1(b)}{\int }_a^b\left({\displaystyle {\int }_a^b{\mathcal{G}}_1(x)\frac{d{G}_n}{\mathit{dx}}\left(\frac{x-a}{b-a},,,\frac{s-a}{b-a}\right)dx}\right)sds.}\end{array}$$

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From (1.4) we have

2.6 $$\begin{array}{c}\hfill \begin{array}{cc}& \frac{d{G}_n}{\mathit{dx}}\left(\frac{x-a}{b-a},,,\frac{s-a}{b-a}\right)\hfill \\ \hfill & =\left\{\begin{array}{cc}{\displaystyle -\sum _{k=0}^{n-1}{\mathrm{\Lambda }}_k^{\prime }\left(\frac{x-a}{b-a}\right)\frac{{(b-s)}^{2n-2k-1}}{{\left(b,-,a\right)}^{2n-2k}(2n-2k-1)!},}\hfill & x\le s,\hfill \\ {\displaystyle \sum _{k=0}^{n-1}{\mathrm{\Lambda }}_k^{\prime }\left(\frac{b-x}{b-a}\right)\frac{{(s-a)}^{2n-2k-1}}{{\left(b,-,a\right)}^{2n-2k}(2n-2k-1)!},}\hfill & s\le x.\hfill \end{array}\right)\hfill \end{array}\end{array}$$

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Now using (2.6) in calculating the above expression for $U_1(b)$ we obtain:

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill U_1(b)& =-{\int }_a^b\left({\displaystyle {\int }_a^b{\mathcal{G}}_1(x)\frac{d{G}_n}{\mathit{dx}}\left(\frac{x-a}{b-a},,,\frac{s-a}{b-a}\right)dx}\right)ds\hfill \\ \hfill & =-{\int }_a^b{\mathcal{G}}_1(x)\left({\int }_a^b,\frac{d{G}_n}{\mathit{dx}},,\left(\frac{x-a}{b-a},,,\frac{s-a}{b-a}\right),d,s\right)dx\hfill \\ \hfill & =-{\int }_a^b{\mathcal{G}}_1(x)\left(\sum _{k=0}^{n-1},{\mathrm{\Lambda }}_k^{\prime },\left(\frac{b-x}{b-a}\right),{\int }_a^x,\frac{{(s-a)}^{2n-2k-1}}{{\left(b,-,a\right)}^{2n-2k}(2n-2k-1)!},d,s\right)\hfill \\ \hfill & \left(-,\sum _{k=0}^{n-1},{\mathrm{\Lambda }}_k^{\prime },\left(\frac{x-a}{b-a}\right),{\int }_x^b,\frac{{(b-s)}^{2n-2k-1}}{{\left(b,-,a\right)}^{2n-2k}(2n-2k-1)!},d,s\right)dx\hfill \\ \hfill & =-{\int }_a^b{\mathcal{G}}_1(x)\left(\sum _{k=0}^{n-1},{\mathrm{\Lambda }}_k^{\prime },\left(\frac{b-x}{b-a}\right),\frac{{(x-a)}^{2n-2k}}{{\left(b,-,a\right)}^{2n-2k}(2n-2k)!}\right)\hfill \\ \hfill & \left(-,\sum _{k=0}^{n-1},{\mathrm{\Lambda }}_k^{\prime },\left(\frac{x-a}{b-a}\right),\frac{{(b-x)}^{2n-2k}}{{\left(b,-,a\right)}^{2n-2k}(2n-2k)!}\right)dx\hfill \\ \hfill & ={\int }_a^b{\mathcal{G}}_1(x)\mathrm{\Omega }(x)dx.\hfill \end{array}\end{array}$$

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Further, we have

2.7 $$\begin{array}{c}\hfill \begin{array}{cc}\hfill m_1& {\displaystyle =\frac{-1}{U_1(b)}{\int }_a^b\left({\int }_a^b,{\mathcal{G}}_1,(x),\frac{d{G}_n}{\mathit{dx}},,\left(\frac{x-a}{b-a},,,\frac{s-a}{b-a}\right),d,x\right)sds}\hfill \\ \hfill & =\frac{-1}{U_1(b)}{\int }_a^b{\mathcal{G}}_1(x)\left({\int }_a^b,\frac{d{G}_n}{\mathit{dx}},,\left(\frac{x-a}{b-a},,,\frac{s-a}{b-a}\right),·,s,,d,s\right)dx.\hfill \end{array}\end{array}$$

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Since (1.2) holds, we have

2.8 $$\begin{array}{c}\hfill {\mathrm{\Lambda }}_n\left(\frac{x-a}{b-a}\right)={\int }_a^b{G}_n\left(\frac{x-a}{b-a},,,\frac{s-a}{b-a}\right)·\frac{s-a}{b-a}·\frac{1}{b-a}ds\end{array}$$

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and the derivative of (2.8) is

$$\begin{array}{c}\hfill {\mathrm{\Lambda }}_n^{\prime }\left(\frac{x-a}{b-a}\right)·\frac{1}{b-a}={\int }_a^b\frac{d{G}_n}{\mathit{dx}}\left(\frac{x-a}{b-a},,,\frac{s-a}{b-a}\right)·\frac{s-a}{b-a}·\frac{1}{b-a}ds,\end{array}$$

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where $\frac{d{G}_n}{\mathit{dx}}\left(\frac{x-a}{b-a},,,\frac{s-a}{b-a}\right)$ is given by (2.6).

Therefore we obtain

2.9 $$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\mathrm{\Lambda }}_n^{\prime }\left(\frac{x-a}{b-a}\right)·(b-a)& +a{\int }_a^b\frac{d{G}_n}{\mathit{dx}}\left(\frac{x-a}{b-a},,,\frac{s-a}{b-a}\right)ds\hfill \\ \hfill & ={\int }_a^b\frac{d{G}_n}{\mathit{dx}}\left(\frac{x-a}{b-a},,,\frac{s-a}{b-a}\right)sds.\hfill \end{array}\end{array}$$

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Using the relation (2.9) in the last term of the right-side in (2.7), we obtain (2.4).

To conclude the proof, let us note that the function f satisfies the conditions of Theorem 1.3, so the identity (1.6) holds. Hence, using the identity (1.6) for the middle part of the inequality (2.5) we obtain the desired inequality (2.3).

Similarly, using Theorem 1.4 we obtain the following result.

Theorem 2.2

Let $I\subset \mathbb{R}$ be an open interval and let $a,b\in I$ be such that $a<b$ . Let the function $f:I\to \mathbb{R}$ be such that $f^{(2n-1)}$ is absolutely continuous and f is $(2n+2)-$ convex on I for $n\ge 1$ . Let $g,p:[a,b]\to \mathbb{R}$ be integrable functions such that p is positive and $0\le g\le 1$ . Let ${\int }_{b-\lambda }^{b}p(t)dt={\int }_a^{b}g(t)p(t)dt$ , let $G_n$ be the Green function given by (1.4), and let ${\mathcal{G}}_2$ and $\mathrm{\Omega }$ be defined by (1.7) and (2.1), respectively. If

$$\begin{array}{c}\hfill {\displaystyle -{\int }_a^b{\mathcal{G}}_2(x)\frac{d{G}_n}{\mathit{dx}}\left(\frac{x-a}{b-a},,,\frac{s-a}{b-a}\right)dx\ge 0,s\in [a,b],}\end{array}$$

Graph

then

2.10 $$\begin{array}{c}\hfill \begin{array}{cc}& U_2(b)·f^{(2n)}\left(m_2\right)\le \hfill \\ \hfill & {(b-a)}^{2n-1}\left[{\int }_a^b,f,(t),g,(t),p,(t),d,t,-,{\int }_{b-\lambda }^b,f,(t),p,(t),d,t,+,\sum _{k=0}^{n-1},{(b-a)}^{2k-1}\right)\hfill \\ \hfill & \left(\times ,{\int }_a^b,{\mathcal{G}}_2,(x),\left[f^{\left(2,k\right)},\left(b\right),{\mathrm{\Lambda }}_k^{{}^{\prime }},\left(\frac{x-a}{b-a}\right),-,f^{\left(2,k\right)},\left(a\right),{\mathrm{\Lambda }}_k^{{}^{\prime }},\left(\frac{b-x}{b-a}\right)\right],d,x\right]\hfill \\ \hfill & \le U_2(b)·\left[\frac{b-m_2}{b-a},f^{(2n)},(a),+,\frac{m_2-a}{b-a},f^{(2n)},(b)\right],\hfill \end{array}\end{array}$$

Graph

where

$$\begin{array}{c}\hfill U_2(b)={\int }_a^b{\mathcal{G}}_2(x)\mathrm{\Omega }(x)dx\end{array}$$

Graph

and

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill m_2=& a-\frac{b-a}{U_2(b)}{\int }_a^b{\mathcal{G}}_2(x){\mathrm{\Lambda }}_n^{{}^{\prime }}\left(\frac{x-a}{b-a}\right)dx.\hfill \end{array}\end{array}$$

Graph

Proof

Similar to the proof of Theorem 2.1 using the identity (1.8).

Remark 2.3

For an $(2n+2)-$ concave function f, the inequalities (2.3) and (2.10) are reversed.

To prove new generalizations of Steffensen's inequality by using the identities (1.9) and (1.10), let us denote

2.11 $$\begin{array}{c}\hfill \begin{array}{cc}\hfill \mathrm{\Phi }(x)& =\sum _{k=0}^{n-1}\frac{1}{(2n-2k)!}\left[{\mathrm{\Lambda }}_k,,,\left(\frac{b-x}{b-a}\right),·,\frac{{(x-a)}^{2n-2k}}{{(b-a)}^{2n-2k-1}}\right)\hfill \\ \hfill & \left(+,{\mathrm{\Lambda }}_k,,,\left(\frac{x-a}{b-a}\right),·,\frac{{(b-x)}^{2n-2k}}{{(b-a)}^{2n-2k-1}}\right].\hfill \end{array}\end{array}$$

Graph

Theorem 2.4

Let $I\subset \mathbb{R}$ be an open interval and let $a,b\in I$ be such that $a<b$ . Let the function $f:I\to \mathbb{R}$ be such that $f^{(2n)}$ is absolutely continuous and f is $(2n+3)-$ convex on I for $n\ge 1$ . Let $g,p:[a,b]\to \mathbb{R}$ be integrable functions such that p is positive and $0\le g\le 1$ . Let ${\int }_a^{a+\lambda }p(t)dt={\int }_a^{b}g(t)p(t)dt$ and let $G_n$ be the Green function given by (1.4), and let ${\mathcal{G}}_1$ and $\mathrm{\Phi }$ be defined by (1.5) and (2.11), respectively.

If

2.12 $$\begin{array}{c}\hfill {\displaystyle -{\int }_a^b{\mathcal{G}}_1(x)G_n\left(\frac{x-a}{b-a},,,\frac{s-a}{b-a}\right)dx\ge 0,s\in [a,b],}\end{array}$$

Graph

then

2.13 $$\begin{array}{c}\hfill \begin{array}{cc}& U_3(b)·f^{(2n+1)}\left(m_3\right)\le \hfill \\ \hfill & {(b-a)}^{2n-1}\left[{\int }_a^{a+\lambda },f,(t),p,(t),d,t,-,{\int }_a^b,f,(t),g,(t),p,(t),d,t,+,\sum _{k=0}^{n-1},{(b-a)}^{2k}\right)\hfill \\ \hfill & \left(\times ,{\int }_a^b,{\mathcal{G}}_1,(x),\left[f^{\left(2,k,+,1\right)},\left(a\right),{\mathrm{\Lambda }}_k,\left(\frac{b-x}{b-a}\right),+,f^{\left(2,k,+,1\right)},\left(b\right),{\mathrm{\Lambda }}_k,\left(\frac{x-a}{b-a}\right)\right],d,x\right]\hfill \\ \hfill & \le U_3(b)·\left[\frac{b-m_3}{b-a},f^{(2n+1)},(a),+,\frac{m_3-a}{b-a},f^{(2n+1)},(b)\right],\hfill \end{array}\end{array}$$

Graph

where

$$\begin{array}{c}\hfill U_3(b)={\int }_a^b{\mathcal{G}}_1(x)\mathrm{\Phi }(x)dx\end{array}$$

Graph

and

2.14 $$\begin{array}{c}\hfill m_3=a-\frac{{(b-a)}^2}{U_3(b)}{\int }_a^b{\mathcal{G}}_1(x){\mathrm{\Lambda }}_n\left(\frac{x-a}{b-a}\right)dx.\end{array}$$

Graph

Proof

Let us begin by defining the function $u_3:[a,b]\to \mathbb{R}$ by

$$\begin{array}{c}\hfill {\displaystyle u_3(s)=-{\int }_a^b{\mathcal{G}}_1(x)G_n\left(\frac{x-a}{b-a},,,\frac{s-a}{b-a}\right)dx,s\in [a,b].}\end{array}$$

Graph

Under the assumption (2.12), the function $u_3$ is non-negative. Further, the function $f^{(2n+1)}$ is convex. Hence, by applying the inequality (1.11) with non-negative function $u_3$ and convex function $f^{(2n+1)}$ we obtain

2.15 $$\begin{array}{c}\hfill \begin{array}{cc}\hfill U_3(b)·f^{(2n+1)}\left(m_3\right)& \le -{\int }_a^b\left({\displaystyle {\int }_a^b{\mathcal{G}}_1(x)G_n\left(\frac{x-a}{b-a},,,\frac{s-a}{b-a}\right)dx}\right)f^{(2n+1)}(s)ds\hfill \\ \hfill & {\displaystyle \le U_3(b)·\left[\frac{b-m_3}{b-a},f^{(2n+1)},(a),+,\frac{m_3-a}{b-a},f^{(2n+1)},(b)\right],}\hfill \end{array}\end{array}$$

Graph

where the expressions $U_3(b)$ and $m_3$ can be calculated as

$$\begin{array}{c}\hfill {\displaystyle U_3(b)=-{\int }_a^b\left({\displaystyle {\int }_a^b{\mathcal{G}}_1(x)G_n\left(\frac{x-a}{b-a},,,\frac{s-a}{b-a}\right)dx}\right)ds}\end{array}$$

Graph

and

2.16 $$\begin{array}{c}\hfill {\displaystyle m_3=-\frac{1}{U_3(b)}{\int }_a^b\left({\displaystyle {\int }_a^b{\mathcal{G}}_1(x)G_n\left(\frac{x-a}{b-a},,,\frac{s-a}{b-a}\right)dx}\right)sds.}\end{array}$$

Graph

Now, using (1.4) and Fubini's theorem we obtain:

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill U_3(b)& =-{\int }_a^b{\mathcal{G}}_1(x)\left({\int }_a^b,G_n,,\left(\frac{x-a}{b-a},,,\frac{s-a}{b-a}\right),d,s\right)dx\hfill \\ \hfill & =-{\int }_a^b{\mathcal{G}}_1(x)\left(-,\sum _{k=0}^{n-1},{\mathrm{\Lambda }}_k,\left(\frac{b-x}{b-a}\right),{\int }_a^x,\frac{{(s-a)}^{2n-2k-1}}{{\left(b,-,a\right)}^{2n-2k-1}(2n-2k-1)!},d,s\right)\hfill \\ \hfill & \left(-,\sum _{k=0}^{n-1},{\mathrm{\Lambda }}_k,\left(\frac{x-a}{b-a}\right),{\int }_x^b,\frac{{(b-s)}^{2n-2k-1}}{{\left(b,-,a\right)}^{2n-2k-1}(2n-2k-1)!},d,s\right)dx\hfill \\ \hfill & ={\int }_a^b{\mathcal{G}}_1(x)\left(\sum _{k=0}^{n-1},{\mathrm{\Lambda }}_k,\left(\frac{b-x}{b-a}\right),\frac{{(x-a)}^{2n-2k}}{{\left(b,-,a\right)}^{2n-2k-1}(2n-2k)!}\right)\hfill \\ \hfill & \left(+,\sum _{k=0}^{n-1},{\mathrm{\Lambda }}_k,\left(\frac{x-a}{b-a}\right),\frac{{(b-x)}^{2n-2k}}{{\left(b,-,a\right)}^{2n-2k-1}(2n-2k)!}\right)dx\hfill \\ \hfill & ={\int }_a^b{\mathcal{G}}_1(x)\mathrm{\Phi }(x)dx.\hfill \end{array}\end{array}$$

Graph

From (2.8) we have

2.17 $$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\mathrm{\Lambda }}_n\left(\frac{x-a}{b-a}\right)·{(b-a)}^2& +a{\int }_a^b{G}_n\left(\frac{x-a}{b-a},,,\frac{s-a}{b-a}\right)ds\hfill \\ \hfill & ={\int }_a^b{G}_n\left(\frac{x-a}{b-a},,,\frac{s-a}{b-a}\right)sds.\hfill \end{array}\end{array}$$

Graph

Using the relation (2.17) in (2.16) (after applying Fubini's theorem), i.e. on

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill m_3& =\frac{-1}{U_1(b)}{\int }_a^b{\mathcal{G}}_1(x)\left({\int }_a^b,G_n,,\left(\frac{x-a}{b-a},,,\frac{s-a}{b-a}\right),·,s,,d,s\right)dx,\hfill \end{array}\end{array}$$

Graph

we arrive at (2.14).

Now, using the identity (1.10) for the middle part of the inequality (2.15) we obtain (2.13) as desired.

Remark 2.5

Let us show that for an odd number $n\ge 2$ , the condition (2.12) is always satisfied.

For $n=1$ , from (1.4) we have:

$$\begin{array}{c}\hfill \begin{array}{c}\hfill G_1\left(\frac{x-a}{b-a},,,\frac{s-a}{b-a}\right)=\left\{\begin{array}{cc}\frac{s-b}{b-a}·\frac{x-a}{b-a},\hfill & x<s,\hfill \\ \frac{x-b}{b-a}·\frac{s-a}{b-a},\hfill & s\le x.\hfill \end{array}\right)\end{array}\end{array}$$

Graph

Clearly, $G_1\left(\frac{x-a}{b-a},,,\frac{s-a}{b-a}\right)\le 0$ for every $x,s\in [a,b]$ . Now, combining this with (1.1) on [a, b], i.e. with

$$\begin{array}{c}\hfill G_n\left(\frac{x-a}{b-a},,,\frac{s-a}{b-a}\right)={\int }_a^b{G}_1\left(\frac{x-a}{b-a},,,\frac{p-a}{b-a}\right)G_{n-1}\left(\frac{p-a}{b-a},,,\frac{s-a}{b-a}\right)dp,\end{array}$$

Graph

for $n\ge 2$ , we conclude that $G_n\left(\frac{x-a}{b-a},,,\frac{s-a}{b-a}\right)\ge 0$ for an even $n\ge 2$ and $G_n\left(\frac{x-a}{b-a},,,\frac{s-a}{b-a}\right)\le 0$ for an odd $n\ge 2$ , for every $x,s\in [a,b]$ .

Since ${\mathcal{G}}_1(x)\ge 0$ , for every $x\in [a,b]$ , we conclude that for an odd number $n\ge 2$ ,

2.18 $$\begin{array}{c}\hfill {\displaystyle -{\int }_a^b{\mathcal{G}}_1(x)G_n\left(\frac{x-a}{b-a},,,\frac{s-a}{b-a}\right)dx\ge 0,s\in [a,b],}\end{array}$$

Graph

i.e. the condition (2.12) is satisfied for every odd number $n\ge 2$ .

Theorem 2.6

Let $I\subset \mathbb{R}$ be an open interval and let $a,b\in I$ be such that $a<b$ . Let the function $f:I\to \mathbb{R}$ be such that $f^{(2n)}$ is absolutely continuous and f is $(2n+3)-$ convex on I for $n\ge 1$ . Let $g,p:[a,b]\to \mathbb{R}$ be integrable functions such that p is positive and $0\le g\le 1$ . Let ${\int }_{b-\lambda }^{b}p(t)dt={\int }_a^{b}g(t)p(t)dt$ , let $G_n$ be the Green function given by (1.4), and let ${\mathcal{G}}_2$ and $\mathrm{\Phi }$ be defined by (1.7) and (2.11), respectively.

If

2.19 $$\begin{array}{c}\hfill {\displaystyle -{\int }_a^b{\mathcal{G}}_2(x)G_n\left(\frac{x-a}{b-a},,,\frac{s-a}{b-a}\right)dx\ge 0,s\in [a,b],}\end{array}$$

Graph

then

2.20 $$\begin{array}{c}\hfill \begin{array}{cc}& U_4(b)·f^{(2n+1)}\left(m_4\right)\le \hfill \\ \hfill & {(b-a)}^{2n-1}\left[{\int }_a^b,f,(t),g,(t),p,(t),d,t,-,{\int }_{b-\lambda }^b,f,(t),p,(t),d,t,+,\sum _{k=0}^{n-1},{(b-a)}^{2k}\right)\hfill \\ \hfill & \left(\times ,{\int }_a^b,{\mathcal{G}}_2,(x),\left[f^{\left(2,k,+,1\right)},\left(a\right),{\mathrm{\Lambda }}_k,\left(\frac{b-x}{b-a}\right),+,f^{\left(2,k,+,1\right)},\left(b\right),{\mathrm{\Lambda }}_k,\left(\frac{x-a}{b-a}\right)\right],d,x\right]\hfill \\ \hfill & \le U_4(b)·\left[\frac{b-m_4}{b-a},f^{(2n+1)},(a),+,\frac{m_4-a}{b-a},f^{(2n+1)},(b)\right],\hfill \end{array}\end{array}$$

Graph

where

$$\begin{array}{c}\hfill U_4(b)={\int }_a^b{\mathcal{G}}_2(x)\mathrm{\Phi }(x)dx\end{array}$$

Graph

and

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill m_4=& a-\frac{{(b-a)}^2}{U_4(b)}{\int }_a^b{\mathcal{G}}_2(x){\mathrm{\Lambda }}_n\left(\frac{x-a}{b-a}\right)dx.\hfill \end{array}\end{array}$$

Graph

Proof

Similar to the proof of Theorem 2.4 using the identity (1.10).

Remark 2.7

Let us show that for an odd number $n\ge 2$ , the condition (2.19) is always satisfied.

As shown in Remark 2.5 we conclude that $G_n\left(\frac{x-a}{b-a},,,\frac{s-a}{b-a}\right)\le 0$ for an odd $n\ge 2$ , for every $x,s\in [a,b]$ . Since ${\mathcal{G}}_2(x)\ge 0$ , for every $x\in [a,b]$ , we conclude that for an odd number $n\ge 2$ ,

$$\begin{array}{c}\hfill {\displaystyle -{\int }_a^b{\mathcal{G}}_2(x)G_n\left(\frac{x-a}{b-a},,,\frac{s-a}{b-a}\right)dx\ge 0,s\in [a,b],}\end{array}$$

Graph

i.e. the condition (2.19) is satisfied for every odd number $n\ge 2$ .

Remark 2.8

For a $(2n+3)-$ concave function f, the inequalities (2.13) and (2.20) are reversed.

Conclusion

Since the Hermite-Hadamard inequality and its weighted version are very important not only in pure mathematics but also in applied mathematics, by utilizing it on some known Steffensen-type identities we contributed to its further applications with the aim of obtaining new generalizations of Steffensen's inequality. Also, by proving new generalizations of Steffensen's inequality we showed that the Steffensen-type identities (1.6), (1.8), (1.9) and (1.10) are bounded from above and from bellow. This can lead to the further investigation of these identities, their properties and new generalizations for other classes of functions and possibly direct research into new applications of Steffensen's inequality.

Author Contributions

All authors contributed equally to the manuscript and accept its final form.

Funding

This research received no external funding.

Data Availability Statement

Not applicable.

Declarations

Conflict of interest

The authors declare that there is no conflict of interest regarding this publication.

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By Josip Pečarić; Anamarija Perušić Pribanić and Ksenija Smoljak Kalamir

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