Open Access

Remarks on some inequalities for s-convex functions and applications

Journal of Inequalities and Applications20152015:333

https://doi.org/10.1186/s13660-015-0851-9

Received: 2 July 2015

Accepted: 30 September 2015

Published: 14 October 2015

Abstract

Some new inequalities for functions whose second derivatives in absolute value at certain powers are s-convex in the second sense are established. Two mistakes in a recently published paper are pointed out and corrected.

Keywords

convex function s-convex functionHölder inequality

MSC

26D15

1 Introduction

It is well known that a function \(f:I\to\mathbf{R}\), \(\emptyset\neq I\subset\mathbf{R}\) is called convex if
$$f\bigl(\lambda x+(1-\lambda)y\bigr)\le\lambda f(x)+(1-\lambda)f(y) $$
holds for all \(x,y\in I\) and \(\lambda\in[0,1]\). If \(-f:I\to\mathbf{R}\) is convex, then we say that \(f:I\to\mathbf{R}\) is concave.
In [1], the class of s-convex function in the second sense is defined in the following way: a function \(f:[0,\infty)\to\mathbf{R}\) is said to be s-convex in the second sense if
$$f\bigl(\lambda x+(1-\lambda)y\bigr)\le\lambda^{s}f(x)+(1- \lambda)^{s}f(y) $$
holds for all \(x,y\in[0,\infty)\), \(\lambda\in[0,1]\) and for some fixed \(s\in(0,1]\). The class of s-convex functions in the second sense is usually denoted by \(K_{s}^{2}\). It can be easily seen that for \(s=1\) s-convexity reduces to ordinary convexity of functions defined on \([0,\infty)\). It is proved in [1] that all functions from \(K_{s}^{2}, s\in(0,1)\) are nonnegative. Similarly, a function \(f:[0,\infty)\to\mathbf{R}\) is said to be s-concave in the second sense for some fixed \(s\in(0,1]\) if \(-f\in K_{s}^{2}\). Thus we can conclude that an s-concave function is always nonpositive for any \(s\in(0,1)\).

Example 1

[1]

Let \(s\in(0,1)\) and \(a,b,c,\in\mathbf{R}\). Define the function \(f:[0,\infty)\to\mathbf{R}\) as
$$f(t)=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} a, &t=0,\\ bt^{s}+c,&t>0. \end{array}\displaystyle \right . $$
It can be easily checked that
  1. (i)

    if \(b\ge 0\) and \(0\le c\le a\), then \(f\in K_{s}^{2}\),

     
  2. (ii)

    if \(b>0\) and \(c<0\), then \(f\notin K_{s}^{2}\).

     

Along this paper, we consider a real interval \(I\subset\mathbf{R}\) and denote that \(I^{\circ}\) is the interior of I.

In a recent paper [2], Özdemir et al. proved the following inequalities for functions whose second derivatives in absolute value at certain powers are s-convex in the second sense.

Theorem 1

([2], Theorem 2)

Let \(I\subset[0,\infty)\), \(f:I\to\mathbf{R}\) be a twice differentiable function on \(I^{\circ}\) such that \(f''\in L^{1}[a,b]\), where \(a,b\in I\) with \(a< b\). If \(|f''|\) is s-convex in the second sense on \([a,b]\) for some fixed \(s\in(0,1]\), then the following inequality holds:
$$\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-f\biggl(\frac{a+b}{2}\biggr)\biggr| \\ &\quad\le\frac{(b-a)^{2}}{8(s+1)(s+2)(s+3)}\biggl[\bigl|f''(a)\bigr|+(s+1) (s+2) \biggl|f''\biggl(\frac{a+b}{2}\biggr)\biggr|+\bigl|f''(b)\bigr| \biggr] \\ &\quad\le\frac{[1+(s+2)2^{1-s}](b-a)^{2}}{8(s+1)(s+2)(s+3)} \bigl[\bigl|f''(a)\bigr|+\bigl|f''(b)\bigr| \bigr]. \end{aligned}$$
(1)

Theorem 2

([2], Theorem 4)

Let \(I\subset[0,\infty)\), \(f:I\to\mathbf{R}\) be a twice differentiable function on \(I^{\circ}\) such that \(f''\in L^{1}[a,b]\), where \(a,b\in I\) with \(a< b\). If \(|f''|^{q}\) is s-convex in the second sense on \([a,b]\) for some fixed \(s\in(0,1]\) and \(q\ge 1\), then the following inequality holds:
$$\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-f\biggl(\frac{a+b}{2}\biggr)\biggr| \\ &\quad\le\frac{(b-a)^{2}}{16}\biggl(\frac{1}{3}\biggr)^{1-\frac{1}{q}}\biggl\{ \biggl(\frac{2}{(s+1)(s+2)(s+3)}\bigl|f''(a)\bigr|^{q}+ \frac{1}{s+3} \biggl|f''\biggl(\frac{a+b}{2} \biggr)\biggr|^{q}\biggr)^{\frac{1}{q}} \\ &\qquad{}+\biggl(\frac{1}{s+3}\biggl|f''\biggl( \frac{a+b}{2}\biggr)\biggr|^{q}+\frac{2}{(s+1)(s+2)(s+3)} \bigl|f''(b)\bigr|^{q} \biggr)^{\frac{1}{q}}\biggr\} . \end{aligned}$$
(2)

However, it is a pity that Theorem 5 in [2] is not valid since nonnegative \(|f''|^{q}\) could not be an s-concave function for any fixed \(s\in(0,1)\) which has been mentioned in [3], and it is the Hölder inequality but not the power mean inequality that has been used in proving Theorem 4 of [2].

In this work, we will first derive a new general inequality for functions whose second derivatives in absolute value at certain powers are s-convex in the second sense, which not only provides generalization of Theorem 1 and Theorem 2 but also gives some other interesting special results. Then we establish another new general inequality for functions whose second derivatives in absolute value at certain powers are s-convex in the second sense, which also gives some interesting special results. Finally, applications to some special means of real numbers are considered.

2 Main results

We first provide a new general inequality for functions whose second derivatives in absolute value at certain powers are s-convex in the second sense, and so we need the following lemma.

Lemma 1

Let \(I\subset\mathbf{R}\), \(f:I\to\mathbf{R}\) be a twice differentiable function on \(I^{\circ}\) such that \(f''\in L^{1}[a,b]\), where \(a,b\in I\) with \(a< b\). Then, for any \(\theta\in[0,1]\), the following equality holds:
$$\begin{aligned} &\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-(1-\theta)f\biggl(\frac{a+b}{2}\biggr)-\theta\frac{f(a)+f(b)}{2} \\ &\quad=\frac{(b-a)^{2}}{16}\biggl[\int_{0}^{1} \bigl(t^{2}-2\theta t\bigr)f''\biggl(t \frac{a+b}{2}+(1-t)a\biggr)\,dt \\ &\qquad{}+\int_{0}^{1} \bigl(t^{2}-2\theta t\bigr)f''\biggl(t \frac{a+b}{2}+(1-t)b\biggr)\,dt\biggr]. \end{aligned}$$
(3)

Proof

Integrating by parts, we have the following identity:
$$\begin{aligned} I_{1} =&\int_{0}^{1} \bigl(t^{2}-2\theta t\bigr)f''\biggl(t \frac{a+b}{2}+(1-t)a\biggr)\,dt \\ =&\bigl(t^{2}-2\theta t\bigr)\frac{2}{b-a}f' \biggl(t\frac{a+b}{2}+(1-t)a\biggr)\bigg|_{0}^{1}- \frac{4}{b-a} \int_{0}^{1}(t- \theta)f'\biggl(t\frac{a+b}{2}+(1-t)a\biggr)\,dt \\ =&\frac{2(1-2\theta)}{b-a}f'\biggl(\frac{a+b}{2}\biggr) -\frac{4}{b-a}\biggl[\frac{2(t-\theta)}{b-a}f\biggl(t\frac{a+b}{2}+(1-t)a \biggr)\bigg|_{0}^{1} \\ &{}-\frac{2}{b-a}\int_{0}^{1}f\biggl(t \frac{a+b}{2}+(1-t)a\biggr)\,dt\biggr] \\ =&\frac{2(1-2\theta)}{b-a}f'\biggl(\frac{a+b}{2}\biggr)- \frac{8(1-\theta)}{ (b-a)^{2}}f\biggl(\frac{a+b}{2}\biggr)-\frac{8\theta}{(b-a)^{2}}f(a) \\ &{}+\frac{8}{(b-a)^{2}}\int_{0}^{1}f\biggl(t \frac{a+b}{2}+(1-t)a\biggr)\,dt. \end{aligned}$$
(4)
Using the change of variable \(x=t\frac{a+b}{2}+(1-t)a\) for \(t\in[0,1]\) and multiplying both sides of (4) by \(\frac{(b-a)^{2}}{16}\), we obtain
$$\begin{aligned} &\frac{(b-a)^{2}}{16}\int_{0}^{1} \bigl(t^{2}-2\theta t\bigr)f''\biggl(t \frac{a+b}{2}+(1-t)a\biggr)\,dt \\ &\quad=\frac{b-a}{8}(1-2\theta)f'\biggl(\frac{a+b}{2} \biggr)-\frac{1-\theta}{2} f\biggl(\frac{a+b}{2}\biggr)-\frac{\theta}{2}f(a)+ \frac{1}{b-a}\int_{a}^{\frac{a+b}{2}}f(x)\,dx. \end{aligned}$$
(5)
Similarly, we observe that
$$\begin{aligned} &\frac{(b-a)^{2}}{16}\int_{0}^{1} \bigl(t^{2}-2\theta t\bigr)f''\biggl(t \frac{a+b}{2}+(1-t)b\biggr)\,dt \\ &\quad=-\frac{b-a}{8}(1-2\theta)f'\biggl(\frac{a+b}{2} \biggr) -\frac{1-\theta}{2}f\biggl(\frac{a+b}{2}\biggr)-\frac{\theta}{2}f(b)+ \frac{1}{b-a}\int_{\frac{a+b}{2}}^{b}f(x)\,dx. \end{aligned}$$
(6)
Thus, adding (5) and (6), we get the required identity (3). □

Theorem 3

Let \(I\subset[0,\infty)\), \(f:I\to\mathbf{R}\) be a twice differentiable function on \(I^{\circ}\) such that \(f''\in L^{1}[a,b]\), where \(a,b\in I\) with \(a< b\). If \(|f''|^{q}\) is s-convex in the second sense on \([a,b]\) for some fixed \(s\in(0,1]\) and \(q\ge 1\), then the following inequalities hold:
$$\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(x)\,dx-(1-\theta)f\biggl(\frac{a+b}{2}\biggr) -\theta\frac{f(a)+f(b)}{2}\biggr| \\ &\quad\le\frac{(b-a)^{2}}{16}\biggl(\frac{8\theta^{3}-3\theta+1}{3}\biggr)^{1-\frac{1}{q}}\biggl\{ \biggl[\frac{2(2\theta)^{s+3}-2(s+3)\theta+s+2}{(s+2)(s+3)}\biggl|f''\biggl( \frac{a+b}{2}\biggr)\biggr|^{q} \\ &\qquad{}+\frac{4(1-2\theta)^{s+2}[(s+1)\theta+1]+2(s+3)\theta-2}{(s+1)(s+2)(s+3)} \bigl|f''(a)\bigr|^{q} \biggr]^{\frac{1}{q}} \\ &\qquad{}+\biggl[\frac{2(2\theta)^{s+3}-2(s+3)\theta+s+2}{(s+2)(s+3)}\biggl|f''\biggl( \frac{a+b}{2}\biggr)\biggr|^{q} \\ &\qquad{}+\frac{4(1-2\theta)^{s+2}[(s+1)\theta+1] +2(s+3)\theta-2}{(s+1)(s+2)(s+3)}\bigl|f''(b)\bigr|^{q} \biggr]^{\frac{1}{q}}\biggr\} \end{aligned}$$
(7)
for \(0\le\theta\le\frac{1}{2}\) and
$$\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(x)\,dx-(1-\theta) f\biggl(\frac{a+b}{2}\biggr)-\theta\frac{f(a)+f(b)}{2}\biggr| \\ &\quad\le\frac{(b-a)^{2}}{16}\biggl(\theta-\frac{1}{3}\biggr)^{1-\frac{1}{q}}\biggl\{ \biggl[\frac{2(s+3)\theta-s-2}{(s+2)(s+3)}\biggl|f''\biggl( \frac{a+b}{2}\biggr)\biggr|^{q} \\ &\qquad{}+\frac{2(s+3)\theta-2}{(s+1)(s+2)(s+3)}\bigl|f''(a)\bigr|^{q} \biggr]^{\frac{1}{q}} +\biggl[\frac{2(s+3)\theta-s-2}{(s+2)(s+3)}\biggl|f''\biggl( \frac{a+b}{2}\biggr)\biggr|^{q} \\ &\qquad{}+\frac{2(s+3)\theta-2}{(s+1)(s+2)(s+3)}\bigl|f''(b)\bigr|^{q} \biggr]^{\frac{1}{q}}\biggr\} \end{aligned}$$
(8)
for \(\frac{1}{2}\le\theta\le 1\).

Proof

In case \(0\le\theta\le\frac{1}{2}\), by Lemma 1 and using the Hölder inequality, we have
$$\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-(1- \theta)f\biggl(\frac{a+b}{2}\biggr)-\theta\frac{f(a)+f(b)}{2}\biggr| \\ &\quad\le\frac{(b-a)^{2}}{16}\biggl[\int_{0}^{1}\bigl|t^{2}-2 \theta t\bigr|\biggl|f''\biggl(t\frac{a+b}{2}+(1-t)a\biggr)\biggr|\,dt \\ &\qquad{}+\int_{0}^{1}\bigl|t^{2}-2\theta t\bigr|\biggl|f''\biggl(t\frac{a+b}{2}+(1-t)b\biggr)\biggr|\,dt \biggr] \\ &\quad\le\frac{(b-a)^{2}}{16}\biggl[\biggl(\int_{0}^{1}\bigl|t^{2}-2 \theta t\bigr|\,dt\biggr)^{1-\frac{1}{q}}\biggl(\int_{0}^{1}\bigl|t^{2}-2 \theta t\bigr|\biggl|f''\biggl(t\frac{a+b}{2}+(1-t)a \biggr)\biggr|^{q} \,dt\biggr)^{\frac{1}{q}} \\ &\qquad{}+\biggl(\int_{0}^{1}\bigl|t^{2}-2 \theta t\bigr|\,dt\biggr)^{1-\frac{1}{q}}\biggl(\int_{0}^{1}\bigl|t^{2}-2 \theta t\bigr|\biggl|f''\biggl(t\frac{a+b}{2}+(1-t)b \biggr)\biggr|^{q} \,dt\biggr)^{\frac{1}{q}}\biggr] \\ &\quad\le\frac{(b-a)^{2}}{16}\biggl[\biggl(\frac{8\theta^{3}-3\theta+1}{3}\biggr) ^{1-\frac{1}{q}} \biggl(\int_{0}^{1}\bigl|t^{2}-2\theta t\bigr|\biggl(t^{s}\biggl|f''\biggl( \frac{a+b}{2}\biggr)\biggr|^{q}\\ &\biggl.\qquad{}+(1-t)^{s}\bigl|f''(a)\bigr|^{q} \biggr)\,dt\biggr)^{\frac{1}{q}} \\ &\qquad{}+\biggl(\frac{8\theta^{3}-3\theta+1}{3}\biggr)^{1-\frac{1}{q}}\biggl(\int _{0}^{1}\bigl|t^{2}-2\theta t\bigr|\biggl(t^{s}\biggl|f''\biggl(\frac{a+b}{2} \biggr)\biggr|^{q}+(1-t)^{s}\bigl|f''(b)\bigr|^{q} \biggr)\,dt\biggr)^{\frac{1}{q}}\biggr] \\ &\quad=\frac{(b-a)^{2}}{16}\biggl[\biggl(\frac{8\theta^{3}-3\theta+1}{3}\biggr)^{1-\frac{1}{q}}\biggl(\int_{0}^{2\theta}t(2\theta-t) \biggl(t^{s}\biggl|f''\biggl(\frac{a+b}{2} \biggr)\biggr|^{q}\\ &\qquad{}+(1-t)^{s}\bigl|f''(a)\bigr|^{q} \biggr)\,dt +\int_{2\theta}^{1}t(t-2\theta) \biggl(t^{s}\biggl|f''\biggl(\frac{a+b}{2} \biggr)\biggr|^{q}+(1-t)^{s}\bigl|f''(a)\bigr|^{q} \biggr)\,dt\biggr)^{\frac{1}{q}} \\ &\qquad{}+\biggl(\frac{8\theta^{3}-3\theta+1}{3}\biggr)^{1-\frac{1}{q}}\biggl(\int _{0}^{2\theta}t(2\theta-t) \biggl(t^{s}\biggl|f'' \biggl(\frac{a+b}{2}\biggr)\biggr|^{q}+(1-t)^{s}\bigl|f''(b)\bigr|^{q} \biggr)\,dt \\ &\qquad{}+\int_{2\theta}^{1}t(t-2\theta) \biggl(t^{s}\biggl|f''\biggl(\frac{a+b}{2} \biggr)\biggr|^{q}+(1-t)^{s}\bigl|f''(b)\bigr|^{q} \biggr)\,dt\biggr)^{\frac{1}{q}}\biggr] \\ &\quad=\frac{(b-a)^{2}}{16}\biggl(\frac{8\theta^{3}-3\theta+1}{3}\biggr)^{1-\frac{1}{q}} \biggl\{ \biggl[\frac{2(2\theta)^{s+3}-2(s+3)\theta+s+2}{(s+2)(s+3)} \biggl|f''\biggl(\frac{a+b}{2}\biggr)\biggr|^{q} \\ &\qquad{}+\frac{4(1-2\theta)^{s+2}[(s+1)\theta+1]+2(s+3) \theta-2}{(s+1)(s+2)(s+3)}\bigl|f''(a)\bigr|^{q} \biggr]^{\frac{1}{q}} \\ &\qquad{}+\biggl[\frac{2(2\theta)^{s+3}-2(s+3)\theta+s+2}{(s+2)(s+3)}\biggl|f''\biggl( \frac{a+b}{2}\biggr)\biggr|^{q} \\ &\qquad{}+\frac{4(1-2\theta)^{s+2}[(s+1)\theta+1]+2(s+3) \theta-2}{(s+1)(s+2)(s+3)}\bigl|f''(b)\bigr|^{q} \biggr]^{\frac{1}{q}}\biggr\} , \end{aligned}$$
where
$$\begin{aligned}& \int_{0}^{1}\bigl|t^{2}-2\theta t\bigr|\,dt=\int _{0}^{2\theta}t(2\theta-t)\,dt+\int_{2\theta}^{1}t(t-2 \theta)\,dt =\frac{8\theta^{3}-3\theta+1}{3}, \\& \int_{0}^{2\theta}t^{s+1}(2\theta-t)\,dt= \frac{(2\theta)^{s+3}}{(s+2)(s+3)}, \\& \int_{2\theta}^{1}t^{s+1}(t-2\theta)\,dt= \frac{(2\theta)^{s+3}-2(s+3)\theta+s+2}{(s+2)(s+3)}, \\& \int_{0}^{2\theta}t(1-t)^{s}(2\theta-t)\,dt= \frac{(1-2\theta)^{s+2}[2(s+1)\theta+2]+2(s+3)\theta-2}{(s+1)(s+2)(s+3)}, \end{aligned}$$
and
$$\int_{2\theta}^{1}t(1-t)^{s}(t-2\theta)\,dt= \frac{(1-2\theta)^{s+2}[2(s+1)\theta+2]}{(s+1)(s+2)(s+3)}. $$
In case \(\frac{1}{2}\le\theta\le 1\), by Lemma 1 and using the Hölder inequality, we have
$$\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-(1- \theta)f\biggl(\frac{a+b}{2}\biggr)-\theta\frac{f(a)+f(b)}{2}\biggr| \\ &\quad\le\frac{(b-a)^{2}}{16}\biggl[\int_{0}^{1}\bigl|t^{2}-2 \theta t\bigr|\biggl|f''\biggl(t\frac{a+b}{2}+(1-t)a\biggr)\biggr|\,dt \\ &\qquad{}+\int_{0}^{1}\bigl|t^{2}-2\theta t\bigr|\biggl|f''\biggl(t\frac{a+b}{2}+(1-t)b\biggr)\biggr|\,dt \biggr] \\ &\quad\le\frac{(b-a)^{2}}{16}\biggl[\biggl(\int_{0}^{1}\bigl|t^{2}-2 \theta t\bigr|\,dt\biggr)^{1-\frac{1}{q}}\biggl(\int_{0}^{1}\bigl|t^{2}-2 \theta t\bigr|\biggl|f''\biggl(t\frac{a+b}{2}+(1-t)a \biggr)\biggr|^{q} \,dt\biggr)^{\frac{1}{q}} \\ &\qquad{}+\biggl(\int_{0}^{1}\bigl|t^{2}-2 \theta t\bigr|\,dt\biggr)^{1-\frac{1}{q}}\biggl(\int_{0}^{1}\bigl|t^{2}-2 \theta t\bigr|\biggl|f''\biggl(t\frac{a+b}{2}+(1-t)b \biggr)\biggr|^{q} \,dt\biggr)^{\frac{1}{q}}\biggr] \\ &\quad\le\frac{(b-a)^{2}}{16}\biggl\{ \biggl(\theta-\frac{1}{3} \biggr)^{1-\frac{1}{q}}\biggl(\int_{0}^{1}\bigl|t^{2}-2 \theta t\bigr|\biggl[t^{s}\biggl|f''\biggl( \frac{a+b}{2}\biggr)\biggr|^{q}+(1-t)^{s}\bigl|f''(a)\bigr|^{q} \biggr]\,dt\biggr)^{\frac{1}{q}} \\ &\qquad{}+\biggl(\theta-\frac{1}{3}\biggr)^{1-\frac{1}{q}}\biggl(\int _{0}^{1}\bigl|t^{2}-2\theta t\bigr|\biggl[t^{s}\biggl|f''\biggl(\frac{a+b}{2} \biggr)\biggr|^{q}+(1-t)^{s}\bigl|f''(b)\bigr|^{q} \biggr]\,dt\biggr)^{\frac{1}{q}}\biggr\} \\ &\quad=\frac{(b-a)^{2}}{16}\biggl(\theta-\frac{1}{3}\biggr)^{1-\frac{1}{q}}\biggl\{ \biggl[\biggl(\int_{0}^{1}t^{s+1}(2 \theta-t)\,dt\biggr)\biggl|f''\biggl(\frac{a+b}{2} \biggr)\biggr|^{q}\\ &\qquad{}+\biggl(\int_{0}^{1}t(1-t)^{s}(2 \theta-t)\,dt\biggr)\bigl|f''(a)\bigr|^{q} \biggr]^{\frac{1}{q}} \\ &\qquad{}+\biggl[\biggl(\int_{0}^{1}t^{s+1}(2 \theta-t)\,dt\biggr)\biggl|f''\biggl(\frac{a+b}{2} \biggr)\biggr|^{q}+\biggl(\int_{0}^{1}t(1-t)^{s}(2 \theta-t)\,dt\biggr)\bigl|f''(b)\bigr|^{q} \biggr]^{\frac{1}{q}}\biggr\} \\ &\quad=\frac{(b-a)^{2}}{16}\biggl(\theta-\frac{1}{3}\biggr)^{1-\frac{1}{q}}\biggl\{ \biggl[\frac{2(s+3)\theta-s-2}{(s+2)(s+3)}\biggl|f''\biggl( \frac{a+b}{2}\biggr)\biggr|^{q} \\ &\qquad{}+\frac{2(s+3)\theta-2}{(s+1)(s+2)(s+3)}\bigl|f''(a)\bigr|^{q} \biggr]^{\frac{1}{q}} +\biggl[\frac{2(s+3)\theta-s-2}{(s+2)(s+3)}\biggl|f''\biggl( \frac{a+b}{2}\biggr)\biggr|^{q} \\ &\qquad{}+\frac{2(s+3)\theta-2}{(s+1)(s+2)(s+3)}\bigl|f''(b)\bigr|^{q} \biggr]^{\frac{1}{q}}\biggr\} , \end{aligned}$$
where
$$\begin{aligned}& \int_{0}^{1}\bigl|t^{2}-2\theta t\bigr|\,dt=\int _{0}^{1}t(2\theta-t)\,dt=\theta-\frac{1}{3}, \\& \int_{0}^{1}t^{s+1}(2\theta-t)\,dt= \frac{2(s+3)\theta-s-2}{(s+2)(s+3)}, \end{aligned}$$
and
$$\int_{0}^{1}t(1-t)^{s}(2\theta-t)\,dt= \frac{2(s+3)\theta-2}{(s+1)(s+2)(s+3)}. $$
The proof is thus completed. □

Remark 1

If we take \(\theta=0\) in (7), then we get a midpoint type inequality
$$\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-f\biggl(\frac{a+b}{2}\biggr)\biggr| \\ &\quad\le\frac{(b-a)^{2}}{16}\biggl(\frac{1}{3}\biggr)^{1-\frac{1}{q}} \biggl\{ \biggl[\frac{1}{s+3}\biggl|f''\biggl(\frac{a+b}{2} \biggr)\biggr|^{q} +\frac{2}{(s+1)(s+2)(s+3)}\bigl|f''(a)\bigr|^{q} \biggr]^{\frac{1}{q}} \\ &\qquad{}+\biggl[\frac{1}{s+3}\biggl|f''\biggl( \frac{a+b}{2}\biggr)\biggr|^{q}+\frac{2}{(s+1)(s+2)(s+3)} \bigl|f''(b)\bigr|^{q} \biggr]^{\frac{1}{q}}\biggr\} . \end{aligned}$$
(9)
If we take \(\theta=1\) in (8), then we get a trapezoid type inequality
$$\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-\frac{f(a)+f(b)}{2}\biggr| \\ &\quad\le\frac{(b-a)^{2}}{16}\biggl(\frac{2}{3}\biggr)^{1-\frac{1}{q}} \biggl\{ \biggl[\frac{s+4}{(s+2)(s+3)}\biggl|f''\biggl(\frac{a+b}{2} \biggr)\biggr|^{q}+\frac{2}{(s+1)(s+3)} \bigl|f''(a)\bigr|^{q} \biggr]^{\frac{1}{q}} \\ &\qquad{}+\biggl[\frac{s+4}{(s+2)(s+3)}\biggl|f''\biggl( \frac{a+b}{2}\biggr)\biggr|^{q}+\frac{2}{(s+1)(s+3)} \bigl|f''(b)\bigr|^{q} \biggr]^{\frac{1}{q}}\biggr\} . \end{aligned}$$
(10)
If we take \(\theta=\frac{1}{3}\) in (7), then we get a Simpson type inequality
$$\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-\frac{1}{6}\biggl[f(a)+4f\biggl(\frac{a+b}{2}\biggr)+f(b)\biggr]\biggr| \\ &\quad\le\frac{(b-a)^{2}}{16}\biggl(\frac{8}{81}\biggr)^{1-\frac{1}{q}} \biggl\{ \biggl[\biggl(\frac{s}{3(s+2)(s+3)}+\frac{2}{(s+2)(s+3)}\biggl(\frac{2}{s+3} \biggr)^{s+3}\biggr)\biggl|f''\biggl( \frac{a+b}{2}\biggr)\biggr|^{q} \\ &\qquad{}+\biggl(\frac{4s+16}{(s+1)(s+2)(s+3)}\biggl(\frac{1}{3}\biggr)^{s+3}+ \frac{2s}{3(s+1)(s+2)(s+3)}\biggr)\bigl|f''(a)\bigr|^{q} \biggr]^{\frac{1}{q}} \\ &\qquad{}\times\biggl[\biggl(\frac{s}{3(s+2)(s+3)}+\frac{2}{(s+2)(s+3)}\biggl( \frac{2}{s+3}\biggr)^{s+3}\biggr)\biggl|f'' \biggl(\frac{a+b}{2}\biggr)\biggr|^{q} \\ &\qquad{}+ \biggl(\frac{4s+16}{(s+1)(s+2)(s+3)}\biggl(\frac{1}{3} \biggr)^{s+3}+\frac{2s}{3(s+1) (s+2)(s+3)}\biggr)\bigl|f''(b)\bigr|^{q} \biggr]^{\frac{1}{q}}\biggr\} . \end{aligned}$$
(11)
If we take \(\theta=\frac{1}{2}\) in (7) or (8), then we get an averaged midpoint-trapezoid type inequality
$$\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-\frac{1}{4}\biggl[f(a)+2f\biggl(\frac{a+b}{2}\biggr)+f(b)\biggr]\biggr| \\ &\quad\le\frac{(b-a)^{2}}{16}\biggl(\frac{1}{6}\biggr)^{1-\frac{1}{q}} \biggl\{ \biggl[\frac{1}{(s+2)(s+3)}\biggl(\biggl|f''\biggl( \frac{a+b}{2}\biggr)\biggr|^{q}+\bigl|f''(a)\bigr|^{q} \biggr)\biggr]^{\frac{1}{q}} \\ &\qquad{}+\biggl[\frac{1}{(s+2)(s+3)}\biggl(\biggl|f''\biggl( \frac{a+b}{2}\biggr)\biggr|^{q}+\bigl|f''(b)\bigr|^{q} \biggr)\biggr]^{\frac{1}{q}}\biggr\} . \end{aligned}$$
(12)

Corollary 1

Let \(I\subset[0,\infty)\), \(f:I\to\mathbf{R}\) be a twice differentiable function on \(I^{\circ}\) such that \(f''\in L^{1}[a,b]\), where \(a,b\in I\) with \(a< b\). If \(|f''|\) is s-convex in the second sense on \([a,b]\) for some fixed \(s\in(0,1]\), then the following inequalities hold:
$$\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(x)\,dx-(1-\theta)f\biggl(\frac{a+b}{2}\biggr)-\theta\frac{f(a)+f(b)}{2}\biggr| \\ &\quad\le \frac{(b-a)^{2}}{8}\biggl[\frac{2(2\theta)^{s+3}-2(s+3)\theta+s+2}{(s+2)(s+3)}\biggl|f'' \biggl(\frac{a+b}{2}\biggr)\biggr| \\ &\qquad{}+\frac{2(1-2\theta)^{s+2}[(s+1)\theta+1] +(s+3)\theta-1}{(s+1)(s+2)(s+3)}\bigl(\bigl|f''(a)\bigr|+\bigl|f''(b)\bigr| \bigr)\biggr] \end{aligned}$$
(13)
for \(0\le\theta\le\frac{1}{2}\) and
$$\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(x)\,dx-(1-\theta)f\biggl(\frac{a+b}{2}\biggr)-\theta\frac{f(a)+f(b)}{2}\biggr| \\ &\quad\le\frac{(b-a)^{2}}{8}\biggl[\frac{2(s+3)\theta-s-2}{(s+2)(s+3)}\biggl|f'' \biggl(\frac{a+b}{2}\biggr)\biggr| \\ &\qquad{}+\frac{(s+3)\theta-1}{(s+1)(s+2)(s+3)}\bigl(\bigl|f''(a)\bigr|+\bigl|f''(b)\bigr| \bigr)\biggr] \end{aligned}$$
(14)
for \(\frac{1}{2}\le\theta\le 1\).

Proof

Inequalities (13) and (14) are immediate by setting \(q=1\) in (7) and (8) of Theorem 3. □

Remark 2

If we take \(\theta=0\) in (13), then we get a midpoint type inequality
$$\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-f\biggl(\frac{a+b}{2}\biggr)\biggr| \\ &\quad\le\frac{(b-a)^{2}}{8}\biggl[ \frac{1}{s+3}\biggl|f''\biggl(\frac{a+b}{2} \biggr)\biggr|+\frac{1}{(s+1)(s+2)(s+3)}\bigl(\bigl|f''(a)\bigr|+\bigl|f''(b)\bigr| \bigr)\biggr]. \end{aligned}$$
(15)
If we take \(\theta=1\) in (14), then we get a trapezoid type inequality
$$\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-\frac{f(a)+f(b)}{2}\biggr| \\ &\quad\le\frac{(b-a)^{2}}{8}\biggl[\frac{s+4}{(s+2)(s+3)}\biggl|f'' \biggl(\frac{a+b}{2}\biggr)\biggr|+\frac{1}{(s+1)(s+3)} \bigl(\bigl|f''(a)\bigr|+\bigl|f''(b)\bigr|\bigr) \biggr]. \end{aligned}$$
(16)
If we take \(\theta=\frac{1}{3}\) in (13), then we get a Simpson type inequality
$$\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-\frac{1}{6}\biggl[f(a)+4f\biggl(\frac{a+b}{2}\biggr)+f(b)\biggr]\biggr| \\ &\quad\le\frac{(b-a)^{2}}{8}\biggl\{ \biggl[\frac{s}{3(s+2)(s+3)}+\frac{2}{(s+2)(s+3)} \biggl(\frac{2}{3}\biggr)^{s+3}\biggr]\biggl|f'' \biggl(\frac{a+b}{2}\biggr)\biggr| \\ &\qquad{}+\biggl[\frac{2s+8}{(s+1)(s+2)(s+3)}\biggl(\frac{1}{3}\biggr)^{s+3} +\frac{s}{3(s+1)(s+2)(s+3)}\biggr]\bigl(\bigl|f''(a)\bigr|+\bigl|f''(b)\bigr| \bigr)\biggr\} . \end{aligned}$$
(17)
If we take \(\theta=\frac{1}{2}\) in (13) or (14), then we get an averaged midpoint-trapezoid type inequality
$$\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-\frac{1}{4}\biggl[f(a)+2f\biggl(\frac{a+b}{2}\biggr)+f(b)\biggr]\biggr| \\ &\quad\le\frac{(b-a)^{2}}{8}\biggl[\frac{1}{(s+2)(s+3)} \biggl|f'' \biggl(\frac{a+b}{2}\biggr)\biggr|+\frac{1}{2(s+2)(s+3)}\biggr]\bigl(\bigl|f''(a)\bigr|+\bigl|f''(b)\bigr| \bigr). \end{aligned}$$
(18)

Remark 3

If we put \(M=\sup_{x\in[a,b]}|f''|\) in (15)-(18), then we have
$$\begin{aligned}& \biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-f\biggl( \frac{a+b}{2}\biggr)\biggr|\le \frac{M(s^{2}+3s+4)(b-a)^{2}}{8(s+1)(s+2)(s+3)}, \end{aligned}$$
(19)
$$\begin{aligned}& \biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt- \frac{f(a)+f(b)}{2}\biggr|\le \frac{M(s^{2}+7s+8)(b-a)^{2}}{8(s+1)(s+2)(s+3)}, \end{aligned}$$
(20)
$$\begin{aligned}& \begin{aligned}[b] &\biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-\frac{1}{6}\biggl[f(a)+4f\biggl(\frac{a+b}{2}\biggr)+f(b)\biggr]\biggr| \\ &\quad\le \frac{M(b-a)^{2}}{24(s+1)(s+2)(s+3)}\biggl[s^{2}+3s+(4s+16) \biggl( \frac{1}{3}\biggr)^{s+2}+6(s+1) \biggl(\frac{2}{3} \biggr)^{s+3}\biggr], \end{aligned} \end{aligned}$$
(21)
and
$$ \biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-\frac{1}{4}\biggl[f(a)+2f\biggl(\frac{a+b}{2}\biggr)+f(b)\biggr]\biggr| \le \frac{M(b-a)^{2}}{4(s+2)(s+3)}. $$
(22)

Remark 4

If we further take \(s=1\) in (19)-(22), i.e., for functions f with convex \(|f''|\), we have
$$\begin{aligned}& \biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-f\biggl( \frac{a+b}{2}\biggr)\biggr|\le\frac{M(b-a)^{2}}{24}, \end{aligned}$$
(23)
$$\begin{aligned}& \biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt- \frac{f(a)+f(b)}{2}\biggr|\le\frac{M(b-a)^{2}}{12}, \end{aligned}$$
(24)
$$\begin{aligned}& \biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt- \frac{1}{6}\biggl[f(a)+4f\biggl(\frac{a+b}{2}\biggr)+f(b)\biggr]\biggr|\le \frac{M(b-a)^{2}}{81}, \end{aligned}$$
(25)
$$\begin{aligned}& \biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt- \frac{1}{4}\biggl[f(a)+2f\biggl(\frac{a+b}{2}\biggr)+f(b)\biggr]\biggr|\le \frac{M(b-a)^{2}}{48}. \end{aligned}$$
(26)

Obviously, (23)-(26) indicate that the Simpson type inequality has the best error estimation for functions f with convex \(|f''|\).

Now we turn to establish another new general inequality for functions whose second derivatives in absolute value at certain powers are s-convex in the second sense, we need the following lemma.

Lemma 2

Let \(I\subset\mathbf{R}\), \(f:I\to\mathbf{R}\) be a twice differentiable function on \(I^{\circ}\) such that \(f''\in L^{1}[a,b]\), where \(a,b\in I\) with \(a< b\). Then, for any \(\theta\in[0,1]\), the following equality holds:
$$\begin{aligned} &\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-(1-\theta) f\biggl(\frac{a+b}{2}\biggr)-\theta\frac{f(a)+f(b)}{2} \\ &\quad=(b-a)^{2}\int_{0}^{1}k(t)f'' \bigl(ta+(1-t)b\bigr)\,dt, \end{aligned}$$
(27)
where
$$k(t)=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} \frac{1}{2}t(t-\theta), & 0\le t\le\frac{1}{2},\\ \frac{1}{2}(1-t)(1-\theta-t), & \frac{1}{2}\le t\le 1. \end{array}\displaystyle \right . $$

Proof

See, e.g., Lemma 2 in [4]. □

Theorem 4

Let \(I\subset[0,\infty)\), \(f:I\to\mathbf{R}\) be a twice differentiable function on \(I^{\circ}\) such that \(f''\in L^{1}[a,b]\), where \(a,b\in I\) with \(a< b\). If \(|f''|^{q}\) is s-convex in the second sense on \([a,b]\) for some fixed \(s\in(0,1]\) and \(q\ge 1\), then the following inequalities hold:
$$\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(x)\,dx-(1-\theta)f\biggl(\frac{a+b}{2}\biggr)-\theta\frac{f(a)+f(b)}{2}\biggr| \\ &\quad\le\frac{(b-a)^{2}}{2}\biggl(\frac{8\theta^{3}-3\theta+1}{24}\biggr)^{1-\frac{1}{q}}\biggl\{ \biggl[\frac{2\theta^{s+3}}{(s+2)(s+3)} \\ &\qquad{}+\frac{2(1-\theta)^{s+2}[(s+1)\theta+2]+(s+3)\theta-2}{(s+1)(s+2)(s+3)}-\frac{1-\theta}{2^{s+1}(s+1)(s+2)}\biggr] \\ &\qquad{}\times\bigl(\bigl|f''(a)\bigr|^{q}+\bigl|f''(b)\bigr|^{q} \bigr)\biggr\} ^{\frac{1}{q}} \end{aligned}$$
(28)
for \(0\le\theta\le\frac{1}{2}\) and
$$\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(x)\,dx-(1-\theta)f\biggl(\frac{a+b}{2}\biggr)-\theta\frac{f(a)+f(b)}{2}\biggr| \\ &\quad\le\frac{(b-a)^{2}}{2}\biggl(\frac{3\theta-1}{24}\biggr)^{1-\frac{1}{q}}\biggl\{ \biggl[\frac{(s+3)\theta-2}{(s+1)(s+2)(s+3)}+\frac{1-\theta}{2^{s+1}(s+1)(s+2)}\biggr] \\ &\qquad{}\times\bigl(\bigl|f''(a)\bigr|^{q}+\bigl|f''(b)\bigr|^{q} \bigr)\biggr\} ^{\frac{1}{q}} \end{aligned}$$
(29)
for \(\frac{1}{2}\le\theta\le 1\).

Proof

In case \(0\le\theta\le\frac{1}{2}\), by Lemma 2 and using the Hölder inequality, we have
$$\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-(1- \theta)f\biggl(\frac{a+b}{2}\biggr)-\theta\frac{f(a)+f(b)}{2}\biggr| \\ &\quad\le(b-a)^{2}\int_{0}^{1}\bigl|k(t)\bigr|\bigl|f'' \bigl(ta+(1-t)b\bigr)\bigr|\,dt \\ &\quad\le(b-a)^{2}\biggl[\int_{0}^{1}\bigl|k(t)\bigr|\,dt\biggr]^{1-\frac{1}{q}} \biggl[\int_{0}^{1}\bigl|k(t)\bigr|\bigl|f'' \bigl(ta+(1-t)b\bigr)\bigr|^{q} \,dt\biggr]^{\frac{1}{q}} \\ &\quad\le(b-a)^{2}\biggl(\frac{8\theta^{3}-3\theta+1}{24}\biggr)^{1-\frac{1}{q}}\biggl(\int_{0}^{1}\bigl|k(t)\bigr|\bigl[t^{s}\bigl|f''(a)\bigr|^{q}+(1-t)^{s}\bigl|f''(b)\bigr|^{q} \bigr]\,dt\biggr)^{\frac{1}{q}} \\ &\quad=\frac{(b-a)}{2}\biggl(\frac{8\theta^{3}-3\theta+1}{24}\biggr)^{1-\frac{1}{q}} \\ &\qquad{}\times\biggl[\biggl(\int_{0}^{\frac{1}{2}}t^{s+1}|t-\theta|\,dt+\int_{\frac{1}{2}}^{1}t^{s}(1-t)|1-\theta-t|\,dt\biggr)\bigl|f''(a)\bigr|^{q} \\ &\qquad{}+\biggl(\int_{0}^{\frac{1}{2}}t(1-t)^{s}|t-\theta|\,dt+\int_{\frac{1}{2}}^{1}(1-t)^{s+1}|1-\theta-t|\,dt\biggr)\bigl|f''(b)\bigr|^{q} \biggr] \\ &\quad=\frac{(b-a)}{2}\biggl(\frac{8\theta^{3}-3\theta+1}{24}\biggr) ^{1-\frac{1}{q}}\biggl[ \biggl(\int_{0}^{\theta}t^{s+1}(\theta-t)\,dt+ \int_{\theta}^{\frac{1}{2}}t^{s+1}(t-\theta)\,dt \\ &\qquad{}+\int_{\frac{1}{2}}^{1-\theta}t^{s}(1-t) (1-\theta-t)\,dt+\int_{1-\theta}^{1}t^{s}(1-t) ( \theta+t-1)\,dt\biggr)\bigl|f''(a)\bigr|^{q} \\ &\qquad{}+\biggl(\int_{0}^{\theta}t(1-t)^{s}(\theta-t)\,dt+\int_{\theta}^{\frac{1}{2}}t(1-t)^{s}(t-\theta)\,dt \\ &\qquad{}+\int_{\frac{1}{2}}^{1-\theta}(1-t)^{s+1}(1- \theta-t)\,dt+\int_{1-\theta}^{1}(1-t)^{s+1}( \theta+t-1)\,dt\biggr)\bigl|f''(b)\bigr|^{q}\biggr] \\ &\quad=\frac{(b-a)^{2}}{2}\biggl(\frac{8\theta^{3}-3\theta+1}{24}\biggr)^{1-\frac{1}{q}}\biggl\{ \biggl[\frac{2\theta^{s+3}}{(s+2)(s+3)} \\ &\qquad{}+\frac{2(1-\theta)^{s+2}[(s+1)\theta+2]+(s+3)\theta-2}{(s+1)(s+2)(s+3)}-\frac{1-\theta}{2^{s+1}(s+1)(s+2)}\biggr]\\ &\qquad{}\times\bigl(\bigl|f''(a)\bigr|^{q}+\bigl|f''(b)\bigr|^{q} \bigr)\biggr\} ^{\frac{1}{q}}, \end{aligned}$$
where
$$\begin{aligned}& \begin{aligned}[b] \int_{0}^{1}\bigl|k(t)\bigr|\,dt={}&\int _{0}^{\frac{1}{2}}\biggl|\frac{1}{2}t(t-\theta)\biggr|\,dt+ \int_{\frac{1}{2}}^{1}\biggl|\frac{1}{2}(1-t) (1- \theta-t)\biggr|\,dt \\ ={}&\frac{1}{2}\biggl[\int_{0}^{\theta}t(\theta-t)\,dt+\int_{\theta}^{\frac{1}{2}} t(t-\theta)\,dt\\ &{}+\int _{\frac{1}{2}}^{1-\theta}(1-t) (1-\theta-t)\,dt+ \int_{1-\theta}^{1}(1-t) (t-1+\theta)\,dt\biggr] \\ ={}&\frac{8\theta^{3}-3\theta+1}{24}, \end{aligned} \\& \int_{0}^{\theta}t^{s+1}(\theta-t)\,dt=\int _{1-\theta}^{1} (1-t)^{s+1}(\theta+t-1)\,dt= \frac{\theta^{s+3}}{(s+2)(s+3)}, \\& \begin{aligned}[b] \int_{0}^{\theta}t(1-t)^{s}(\theta-t)\,dt&=\int_{1-\theta}^{1}(1-t) (\theta+t-1)t^{s} \,dt \\ &=\frac{(1-\theta)^{s+2}[(s+1)\theta+2]+(s+3)\theta-2}{(s+1)(s+2)(s+3)}, \end{aligned} \\& \begin{aligned}[b] \int_{\theta}^{\frac{1}{2}}t^{s+1}(t-\theta)\,dt&=\int_{\frac{1}{2}}^{1-\theta} (1-t)^{s+1}(1- \theta-t)\,dt\\ &=\frac{(2\theta)^{s+3}-2(s+3)\theta+s+2}{2^{s+3}(s+2)(s+3)}, \end{aligned} \end{aligned}$$
and
$$\begin{aligned} \int_{\theta}^{\frac{1}{2}}t(1-t)^{s}(t- \theta)\,dt =&\int_{\frac{1}{2}}^{1-\theta}(1-t) (1- \theta-t)t^{s} \,dt \\ =&\frac{(1-\theta)^{s+2}[(s+1)\theta+2]}{(s+1)(s+2)(s+3)}+\frac{2(s+3)^{2}\theta-s^{2}-7s-14}{2^{s+3}(s+1)(s+2)(s+3)}. \end{aligned}$$
In case \(0\le\theta\le\frac{1}{2}\), by Lemma 2 and using the Hölder inequality, we have
$$\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-(1- \theta)f\biggl(\frac{a+b}{2}\biggr) -\theta\frac{f(a)+f(b)}{2}\biggr| \\ &\quad\le(b-a)^{2}\int_{0}^{1}\bigl|k(t)\bigr|\bigl|f'' \bigl(ta+(1-t)b\bigr)\bigr|\,dt \\ &\quad\le(b-a)^{2}\biggl[\int_{0}^{1}\bigl|k(t)\bigr|\,dt\biggr]^{1-\frac{1}{q}} \biggl[\int_{0}^{1}\bigl|k(t)\bigr| \bigl|f''\bigl(ta+(1-t)b\bigr)\bigr|^{q} \,dt \biggr]^{\frac{1}{q}} \\ &\quad\le(b-a)^{2}\biggl(\frac{3\theta-1}{24}\biggr)^{1-\frac{1}{q}}\biggl(\int_{0}^{1}\bigl|k(t)\bigr|\bigl[t^{s}\bigl|f''(a)\bigr|^{q}++(1-t)^{s}\bigl|f''(b)\bigr|^{q} \bigr]\,dt\biggr)^{\frac{1}{q}} \\ &\quad=\frac{(b-a)}{2}\biggl(\frac{3\theta-1}{24}\biggr)^{1-\frac{1}{q}} \biggl[\biggl(\int_{0}^{\frac{1}{2}}t^{s+1}|t-\theta|\,dt+\int_{\frac{1}{2}}^{1} t^{s}(1-t)|1-\theta-t|\,dt\biggr) \bigl|f''(a)\bigr|^{q} \\ &\qquad{}+\biggl(\int_{0}^{\frac{1}{2}}t(1-t)^{s}|t-\theta|\,dt+\int_{\frac{1}{2}}^{1} (1-t)^{s+1}|1-\theta-t|\,dt\biggr) \bigl|f''(b)\bigr|^{q} \biggr] \\ &\quad=\frac{(b-a)}{2}\biggl(\frac{3\theta-1}{24}\biggr)^{1-\frac{1}{q}} \biggl[\biggl(\int_{0}^{\frac{1}{2}}t^{s+1}(\theta-t)\,dt+\int_{\frac{1}{2}}^{1}t^{s}(1-t) (\theta+t-1)\,dt\biggr) \bigl|f''(a)\bigr|^{q} \\ &\qquad{}+\biggl(\int_{0}^{\frac{1}{2}}t(1-t)^{s}(\theta-t)\,dt+\int_{\frac{1}{2}}^{1}(1-t)^{s+1} (\theta+t-1)\,dt\biggr) \bigl|f''(b)\bigr|^{q} \biggr] \\ &\quad=\frac{(b-a)^{2}}{2}\biggl(\frac{3\theta-1}{24}\biggr)^{1-\frac{1}{q}} \biggl\{ \biggl[\frac{(s+3)\theta-2}{(s+1)(s+2)(s+3)}+\frac{1-\theta}{2^{s+1}(s+1)(s+2)}\biggr]\\ &\qquad{}\times \bigl(\bigl|f''(a)\bigr|^{q}+\bigl|f''(b)\bigr|^{q} \bigr)\biggr\} ^{\frac{1}{q}}, \end{aligned}$$
where
$$\begin{aligned}& \begin{aligned}[b] \int_{0}^{1}\bigl|k(t)\bigr|\,dt&=\int _{0}^{\frac{1}{2}}\biggl|\frac{1}{2}t(t-\theta)\biggr|\,dt+ \int_{\frac{1}{2}}^{1}\biggl|\frac{1}{2}(1-t) (1- \theta-t)\biggr|\,dt \\ &=\frac{1}{2}\biggl[\int_{0}^{\frac{1}{2}}t( \theta-t)\,dt+\int_{\frac{1}{2}}^{1}(1-t) (\theta+t-1)\,dt\biggr] \\ &=\frac{3\theta-1}{24}, \end{aligned} \\& \int_{0}^{\frac{1}{2}}t^{s+1}(\theta-t)\,dt=\int_{\frac{1}{2}}^{1} (1-t)^{s+1}( \theta+t-1)\,dt=\frac{(2s+6)\theta-s-2}{2^{s+3}(s+2)(s+3)}, \end{aligned}$$
and
$$\begin{aligned} \int_{0}^{\frac{1}{2}}t(1-t)^{s}( \theta-t)\,dt =&\int_{\frac{1}{2}}^{1}(1-t) ( \theta+t-1)t^{s} \,dt \\ =&\frac{2^{s+3}[(s+3)\theta-2]-2(s+3)^{2}\theta+s^{2}+7s+14}{2^{s+3}(s+1)(s+2)(s+3)}. \end{aligned}$$

The proof is thus completed. □

Remark 5

If we take \(\theta=0\) in (28), then we get a midpoint type inequality
$$\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-f\biggl(\frac{a+b}{2}\biggr)\biggr| \\ &\quad\le\frac{(b-a)^{2}}{2}\biggl(\frac{1}{24}\biggr)^{1-\frac{1}{q}} \biggl[\frac{2^{s+2}-s-3}{2^{s+1}(s+1)(s+2)(s+3)}\bigl(\bigl|f''(a)\bigr|^{q}+\bigl|f''(b)\bigr|^{q} \bigr)\biggr]^{\frac{1}{q}}. \end{aligned}$$
(30)
If we take \(\theta=1\) in (29), then we get a trapezoid type inequality
$$\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-\frac{f(a)+f(b)}{2}\biggr| \\ &\quad\le\frac{(b-a)^{2}}{2}\biggl(\frac{1}{4}\biggr)^{1-\frac{1}{q}} \biggl[\frac{|f''(a)|^{q}+|f''(b)|^{q}}{(s+2)(s+3)}\biggr]^{\frac{1}{q}}. \end{aligned}$$
(31)
If we take \(\theta=\frac{1}{3}\) in (28), then we get a Simpson type inequality
$$\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-\frac{f(a)+4f(\frac{a+b}{2})+f(b)}{6}\biggr| \\ &\quad\le\frac{(b-a)^{2}}{2}\biggl(\frac{1}{81}\biggr)^{1-\frac{1}{q}} \\ &\qquad{}\times\biggl[\frac{2^{s+4}(s+1)+2^{2s+6}(s+7)-6^{s+2}(6-2s)-8(s+3)3^{s+2}}{6^{s+3}(s+1)(s+2)(s+3)}\bigl(\bigl|f''(a)\bigr|^{q}+\bigl|f''(b)\bigr|^{q} \bigr)\biggr]^{\frac{1}{q}}. \end{aligned}$$
(32)
If we take \(\theta=\frac{1}{2}\) in (28) or (29), then we get an averaged midpoint-trapezoid type inequality
$$\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-\frac{1}{4}\biggl[f(a)+2f\biggl(\frac{a+b}{2}\biggr)+f(b)\biggr]\biggr| \\ &\quad\le\frac{(b-a)^{2}}{2}\biggl(\frac{1}{48}\biggr)^{1-\frac{1}{q}} \biggl[\frac{2^{s+1}(s-1)+s+3}{2^{s+2}(s+1)(s+2)(s+3)}\bigl(\bigl|f''(a)\bigr|^{q}+\bigl|f''(b)\bigr|^{q} \bigr)\biggr]^{\frac{1}{q}}. \end{aligned}$$
(33)

Corollary 2

Let \(I\subset[0,\infty)\), \(f:I\to\mathbf{R}\) be a twice differentiable function on \(I^{\circ}\) such that \(f''\in L^{1}[a,b]\), where \(a,b\in I\) with \(a< b\). If \(|f''|\) is s-convex in the second sense on \([a,b]\) for some fixed \(s\in(0,1]\), then the following inequalities hold:
$$\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(x)\,dx-(1-\theta)f\biggl(\frac{a+b}{2}\biggr)-\theta\frac{f(a)+f(b)}{2}\biggr| \\ &\quad\le\frac{(b-a)^{2}}{2}\biggl[\frac{2\theta^{s+3}}{(s+2)(s+3)} \\ &\qquad{}+\frac{2(1-\theta)^{s+2}((s+1)\theta+2)+(s+3)\theta-2}{(s+1)(s+2)(s+3)}-\frac{1-\theta}{2^{s+1}(s+1)(s+2)}\biggr] \bigl(\bigl|f''(a)\bigr|+\bigl|f''(b)\bigr|\bigr) \end{aligned}$$
(34)
for \(0\le\theta\le\frac{1}{2}\) and
$$\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(x)\,dx-(1-\theta)f\biggl(\frac{a+b}{2}\biggr)-\theta\frac{f(a)+f(b)}{2}\biggr| \\ &\quad\le\frac{(b-a)^{2}}{2}\biggl[\frac{(s+3)\theta-2}{ (s+1)(s+2)(s+3)}+\frac{1-\theta}{2^{s+1}(s+1)(s+2)}\biggr] \bigl(\bigl|f''(a)\bigr|+\bigl|f''(b)\bigr|\bigr) \end{aligned}$$
(35)
for \(\frac{1}{2}\le\theta\le 1\).

Proof

Inequalities (34) and (35) are immediate by setting \(q=1\) in (28) and (29) of Theorem 4. □

Remark 6

If we take \(\theta=0\) in (34), then we get a midpoint type inequality
$$ \biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-f\biggl(\frac{a+b}{2}\biggr)\biggr|\le\frac{2^{s+2}-s-3}{2^{s+2}(s+1)(s+2)(s+3)}(b-a)^{2} \bigl(\bigl|f''(a)\bigr|+\bigl|f''(b)\bigr|\bigr). $$
(36)
If we take \(\theta=1\) in (35), then we get a trapezoid type inequality
$$ \biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-\frac{f(a)+f(b)}{2}\biggr|\le\frac{1}{2(s+2)(s+3)}(b-a)^{2} \bigl(\bigl|f''(a)\bigr|+\bigl|f''(b)\bigr|\bigr). $$
(37)
If we take \(\theta=\frac{1}{3}\) in (34), then we get a Simpson type inequality
$$\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-\frac{1}{6}\biggl[f(a)+4f\biggl(\frac{a+b}{2}\biggr)+f(b)\biggr]\biggr| \\ &\quad\le \frac{(s+1)2^{s+3}+(s+7)2^{2s+5}-(3-s)6^{s+2}-4(s+3) 3^{s+2}}{6^{s+3}(s+1)(s+2)(s+3)} \\ &\qquad{}\times(b-a)^{2}\bigl(\bigl|f''(a)\bigr|+\bigl|f''(b)\bigr| \bigr). \end{aligned}$$
(38)
If we take \(\theta=\frac{1}{2}\) in (34) or (35), then we get an averaged midpoint-trapezoid type inequality
$$\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-\frac{1}{4}\biggl[f(a)+2f\biggl(\frac{a+b}{2}\biggr)+f(b)\biggr]\biggr| \\ &\quad\le\frac{s+3-2^{s+1}(1-s)}{2^{s+3}(s+1)(s+2)(s+3)}(b-a)^{2}\bigl(\bigl|f''(a)\bigr|+\bigl|f''(b)\bigr| \bigr). \end{aligned}$$
(39)

Remark 7

If we put \(M=\sup_{x\in[a,b]}|f''|\) in (36)-(39), then we have
$$\begin{aligned}& \biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-f\biggl( \frac{a+b}{2}\biggr)\biggr|\le \frac{M(2^{s+2}-s-3)(b-a)^{2}}{2^{s+1}(s+1)(s+2)(s+3)}, \end{aligned}$$
(40)
$$\begin{aligned}& \biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt- \frac{f(a)+f(b)}{2}\biggr|\le \frac{M(b-a)^{2}}{(s+2)(s+3)}, \end{aligned}$$
(41)
$$\begin{aligned}& \begin{aligned}[b] &\biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-\frac{1}{6}\biggl[f(a)+4f\biggl(\frac{a+b}{2}\biggr)+f(b)\biggr]\biggr| \\ &\quad\le\frac{(s+1)2^{s+3}+(s+7)2^{2s+5}-(3-s)6^{s+2} -4(s+3)3^{s+2}}{3(s+1)(s+2)(s+3)6^{s+2}}M(b-a)^{2} \end{aligned} \end{aligned}$$
(42)
and
$$\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-\frac{1}{4}\biggl[f(a)+2f\biggl(\frac{a+b}{2}\biggr)+f(b)\biggr]\biggr| \\ &\quad\le\frac{s+3-2^{s+1}(1-s)}{2^{s+2}(s+1)(s+2)(s+3)}M(b-a)^{2}. \end{aligned}$$
(43)

Remark 8

If we further take \(s=1\) in (40)-(43), i.e., for functions f with convex \(|f''|\), then we recapture inequalities (23)-(26).

3 Applications to special means

We consider the means for arbitrary positive numbers α, β (\(\alpha\neq\beta\)) as follows:
  1. (1)
    The arithmetic mean
    $$A(\alpha, \beta)=\frac{\alpha+\beta}{2}. $$
     
  2. (2)
    The geometric mean
    $$G(\alpha, \beta)=\sqrt{\alpha\beta}. $$
     
  3. (3)
    The harmonic mean
    $$H(\alpha, \beta)=\frac{2\alpha\beta}{\alpha+\beta}. $$
     
  4. (4)
    The logarithmic mean
    $$L(\alpha, \beta)=\frac{\beta-\alpha}{\ln\beta-\ln\alpha}. $$
     
  5. (5)
    The generalized log-mean
    $$L_{p}(\alpha,\beta)=\biggl[\frac{\beta^{p+1}-\alpha^{p+1}}{(p+1)(\beta-\alpha)}\biggr] ^{\frac{1}{p}},\quad p\neq -1,0. $$
     
  6. (6)
    The identric mean
    $$I(\alpha,\beta)=\frac{1}{e}\biggl(\frac{\beta^{\beta}}{\alpha^{\alpha}}\biggr) \frac{1}{\beta-\alpha}. $$
     

Proposition 1

Let \(0< a< b\) and \(s\in(0,1)\). Then we have
$$\begin{aligned}& \bigl|A^{s}(a,b)-L_{s}^{s}(a,b)\bigr|\le\frac{s(1-s)(b-a)^{2}}{24a^{2-s}}, \\& \bigl|A\bigl(a^{s},b^{s}\bigr)-L_{s}^{s}(a,b)\bigr| \le\frac{s(1-s)(b-a)^{2}}{12a^{2-s}}, \\& \biggl|\frac{2A^{s}(a,b)+A(a^{s},b^{s})}{3}-L_{s}^{s}(a,b)\biggr|\le\frac{s(1-s)(b-a)^{2}}{81a^{2-s}}, \end{aligned}$$
and
$$\biggl|\frac{A^{s}(a,b)+A(a^{s},b^{s})}{2}-L_{s}^{s}(a,b)\biggr|\le\frac{s(1-s)(b-a)^{2}}{48a^{2-s}}. $$

Proof

The assertion follows from applying inequalities (23)-(26) to the mapping \(f(x)=x^{s}\), \(x\in[a,b]\), which implies that \(|f''(x)|=s(1-s)x^{s-2}\) is convex on \([a,b]\), and we may take \(M=\frac{s(1-s)}{a^{2-s}}\). □

Proposition 2

Let \(0< a< b\). Then we have
$$\begin{aligned}& \bigl|A^{-1}(a,b)-L^{-1}(a,b)\bigr|\le\frac{(b-a)^{2}}{12a^{3}}, \\& \bigl|H^{-1}(a,b)-L^{-1}(a,b)\bigr|\le\frac{(b-a)^{2}}{6a^{3}}, \\& \biggl|\frac{2A^{-1}(a,b)+H^{-1}(a,b)}{3}-L^{-1}(a,b)\biggr|\le\frac{2(b-a)^{2}}{81a^{3}} \end{aligned}$$
and
$$\biggl|\frac{A^{-1}(a,b)+H^{-1}(a,b)}{2}-L^{-1}(a,b)\biggr|\le\frac{(b-a)^{2}}{24a^{3}}. $$

Proof

The assertion follows from applying inequalities (23)-(26) to the mapping \(f(x)=\frac{1}{x}\), \(x\in[a,b]\), which implies that \(|f''(x)|=\frac{2}{x^{3}}\) is convex on \([a,b]\), and we may take \(M=\frac{2}{a^{3}}\). □

Proposition 3

Let \(a,b\in\mathbf{R}\), \(0< a< b\). Then we have
$$\begin{aligned}& \bigl|\ln A(a,b)-\ln I(a,b)\bigr|\le\frac{(b-a)^{2}}{24a^{2}}, \\& \bigl|\ln G(a,b)-\ln I(a,b)\bigr|\le\frac{(b-a)^{2}}{12a^{2}}, \\& \biggl|\frac{2\ln A(a,b)+\ln G(a,b)}{3}-\ln I(a,b)\biggr|\le\frac{(b-a)^{2}}{81a^{2}} \end{aligned}$$
and
$$\biggl|\frac{\ln A(a,b)+\ln G(a,b)}{2}-\ln I(a,b)\biggr|\le\frac{(b-a)^{2}}{48a^{2}}. $$

Proof

The assertion follows from applying inequalities (23)-(26) to the mapping \(f(x)=\ln x\), \(x\in[a,b]\), which implies that \(|f''(x)|=\frac{1}{x^{2}}\) is convex on \([a,b]\), and we may take \(M=\frac{1}{a^{2}}\). □

Declarations

Acknowledgements

The author wishes to thank the referees for their helpful comments and suggestions.

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
Institute of Applied Mathematics, School of Science, University of Science and Technology Liaoning

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Copyright

© Liu 2015