Levinson type inequalities for higher order convex functions via Abel–Gontscharoff interpolation

Abstract

In this paper, Levinson type inequalities are studied for the class of higher order convex functions by using Abel–Gontscharoff interpolation. Cebyšev, Grüss, and Ostrowski-type new bounds are also found for the functionals involving data points of two types.

1 Introduction and preliminaries

The theory of convex functions has encountered a fast advancement. This can be attributed to a few causes: firstly, applications of convex functions are directly involved in modern analysis; secondly, many important inequalities are the results of applications of convex functions, and convex functions are closely related to inequalities (see [1]).

Divided differences are seen to be uncommonly valuable when we are managing limits having assorted diverse of smoothness. In [1, p. 14], the definition of divided difference is given as follows:

• mth-order divided difference:

  Let a function $f:[\zeta _{1},\zeta _{2}] \rightarrow \mathbb{R}$. The mth-order divided difference of a function f at $x_{0},\ldots,x_{m} \in [\zeta _{1},\zeta _{2}]$ is defined recursively by

  $$\begin{aligned}& \begin{gathered}{} [x_{i}; f ] = f (x_{i} ),\quad i = 0,\ldots,m, \\ [ x_{0},\ldots,x_{m};f ] =\frac{ [ x_{1},\ldots,x _{m};f ] - [ x_{0},\ldots,x_{m-1};f ] }{x_{m}-x_{0}}. \end{gathered} \end{aligned}$$

  (1)

  It is easy to see that (1) is equivalent to

  $$ [ x_{0},\ldots,x_{m};f ]=\sum_{i=0}^{m} \frac{f (x_{i} )}{q^{\prime } (x_{i} )},\quad \text{where } q (x )=\prod_{j=0}^{m} (x-x_{j} ). $$

  The following definition of a real-valued convex function is characterized by mth-order divided difference (see [1, p. 15]).

• Higher order convex function:

  A function $f : [\zeta _{1},\zeta _{2}] \rightarrow \mathbb{R}$ is said to be m-convex ($m \geq 0$) if and only if, for all decisions of $(m+1 )$ distinct points $x_{0},\ldots,x _{m} \in [\zeta _{1},\zeta _{2}]$, $[x_{0},\ldots,x_{m};f ] \geq 0$ holds. If this inequality is reversed, then f is said to be m-concave.

• Criteria for m-convex functions:

  In [1, p. 16], the criterion to examine the m-convexity of a function f is given as follows.

Theorem 1

If $f^{(m)}$ exists, then f is m-convex if and only if $f^{(m)} \geq 0$.

In [2] (see also [3, p. 32, Theorem 1]), Ky Fan’s inequality is generalized by Levinson for 3-convex functions as follows.

Theorem 2

Let $f :I=(0, 2\alpha ) \rightarrow \mathbb{R}$ with $f^{(3)}(t) \geq 0$. Let $x_{k} \in (0, \alpha )$ and $p_{k}>0$. Then

$$ \frac{1}{P_{n}}\sum_{i=1}^{n}p_{i}f(x_{i})- f\Biggl(\frac{1}{P_{n}}\sum_{i=1} ^{n}p_{i}x_{i}\Biggr)\leq \frac{1}{P_{n}}\sum _{i=1}^{n}p_{i}f(2\alpha -x _{i})-f\Biggl(\frac{1}{P_{n}}\sum_{i=1}^{n}p_{i}(2 \alpha - x_{i})\Biggr). $$

(2)

Functional form of (2) is defined as follows:

$$\begin{aligned} \mathcal{J}_{1}\bigl(f(\cdot )\bigr) =&\frac{1}{P_{n}}\sum _{i=1}^{n}p_{i}f(2a-x_{i})-f \Biggl(\frac{1}{P _{n}}\sum_{i=1}^{n}p_{i}(2a-x_{i}) \Biggr)-\frac{1}{P_{n}} \sum_{i=1}^{n}p _{i}f(x_{i}) \\ & {} +f\Biggl(\frac{1}{P_{n}}\sum_{i=1}^{n}p_{i}x_{i} \Biggr). \end{aligned}$$

(3)

Working with the divided differences, assumptions of differentiability on f can be weakened. In [4], Popoviciu noted that (2) is valid on $(0, 2a)$ for 3-convex functions, while in [5] (see also [3, p. 32, Theorem 2]) Bullen gave a different proof of Popoviciu’s result and also the converse of (2).

Theorem 3

• (a) Let $f:I=[\zeta _{1}, \zeta _{2}] \rightarrow \mathbb{R}$ be a 3-convex function and $x_{n}, y_{n} \in [\zeta _{1}, \zeta _{2}]$ for $n=1, 2,\ldots, k$ such that

  $$ \max \{x_{1} ,\ldots, x_{k}\} \leq \min \{y_{1} ,\ldots, y_{k}\}, \quad x_{1}+y_{1}= \cdots =x_{k}+y_{k} $$

  (4)

  and $p_{n}>0$, then

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

  (5)

• (b) If f is continuous and $p_{n}>0$, (5) holds for all $x_{k}$, $y_{k}$ satisfying (4), then f is 3-convex.

Functional form of (5) is defined as follows:

$$\begin{aligned} \begin{aligned}[b] \mathcal{J}_{2}\bigl(f(\cdot )\bigr) ={}&\frac{1}{P_{n}}\sum _{i=1}^{n}p_{i}f(y_{i})-f\Biggl( \frac{1}{P _{n}}\sum_{i=1}^{n}p_{i}y_{i} \Biggr)-\frac{1}{P_{n}}\sum_{i=1}^{n}p_{i}f(x _{i}) \\ & {} + f\Biggl(\frac{1}{P_{n}}\sum_{i=1}^{n}p_{i}x_{i} \Biggr). \end{aligned} \end{aligned}$$

(6)

Remark 1

It is essential to take note of the fact that under the suppositions of Theorem 2 and Theorem 3, if the function f is 3-convex, then $\mathcal{J}_{i}(f(\cdot ))\geq 0$ for $i=1, 2$ and $\mathcal{J}_{i}(f(\cdot ))=0$ for $f(x)=x$ or $f(x)=x^{2}$ or f is a constant function.

In [6] (see also [3, p. 32, Theorem 4]), Pečarić weakened assumption (4) and proved that inequality (5) still holds, i.e., the following result holds.

Theorem 4

Let $f:I=[\zeta _{1}, \zeta _{2}] \rightarrow \mathbb{R}$ be a 3-convex function, $p_{i}>0$, and let $x_{i}, y_{i} \in [\zeta _{1}, \zeta _{2}]$ such that $x_{i}+y_{i}=2\breve{c}$ for $i=1,\ldots, n$, $x_{i}+x_{n-i+1} \leq 2\breve{c}$, and $\frac{p_{i}x_{i}+p_{n-i+1}x_{n-i+1}}{p_{i}+p _{n-i+1}} \leq \breve{c}$. Then (5) holds.

In [7], Mercer made a notable work by replacing the condition of symmetric distribution of points $x_{i}$ and $y_{i}$ with symmetric variances of points $x_{i}$ and $y_{i}$, the second condition is a weaker condition.

Theorem 5

Let f be a 3-convex function on $[\zeta _{1}, \zeta _{2}]$, $p_{i}$ be positive such that $\sum_{i=1}^{n}p_{i}=1$. Also let $x_{i}$, $y_{i}$ satisfy $\max \{x_{1} ,\ldots, x_{i}\} \leq \min \{y _{1} ,\ldots, y_{i}\}$ and

$$ \sum_{i=1}^{n}p_{i} \Biggl(x_{i}-\sum_{i=1}^{n}p_{i}x_{i} \Biggr)^{2}= \sum_{i=1}^{n}p_{i} \Biggl(y_{i}- \sum_{i=1}^{n}p_{i}y_{i} \Biggr)^{2}, $$

(7)

then (5) holds.

In [8], Adeel et al. generalized Levinson’s inequality for 3-convex function by using two Green functions. In [9], Pečarić et al. gave a probabilistic version of Levinson’s inequality (2) under Mercer’s assumption of equal variances (but for a different number of data points) for the family of 3-convex functions at a point. They showed that this is the largest family of continuous functions for which inequality (2) holds. An operator version of probabilistic Levinson’s inequality is discussed in [10] (see also [11]).

On the other hand, the error function $e_{F}(t)$ can be represented in terms of the Green functions $G_{F, m}(t, s)$ for the boundary value problem

• $z^{(m)}(t)=0$,

• $z^{(i)}(a_{1}) = 0$, $0 \leq i \leq p$,

• $z^{(i)}(a_{2}) = 0$, $p+1 \leq i \leq m-1$:

  $$ e_{F}(t)= \int ^{\zeta _{2}}_{\zeta _{1}}G_{F, m}(t, s)f^{(m)}(s)\,ds, \quad t \in [\zeta _{1}, \zeta _{2}], $$

  where

  $$\begin{aligned} G_{F, m}(t, s) = \frac{1}{(m-1)!} \textstyle\begin{cases} \sum_{i=0}^{p}\tbinom{ m-1}{ i } (t-\zeta _{1})^{i}(\zeta _{1}-s)^{m-i-1}, & \zeta _{1} \leq s \leq t; \\ -\sum_{i=p+1}^{m-p}\tbinom{ m-1 }{ i }(t-\zeta _{1})^{i}(\zeta _{1}-s)^{m-i-1}, & t \leq s \leq \zeta _{2}. \end{cases}\displaystyle \end{aligned}$$

  (8)

Further $\zeta _{1} \leq t$, $s \leq \zeta _{2}$, the following inequalities hold:

$$\begin{aligned}& (-1)^{m-p-1}\frac{\partial ^{i}G_{F, m}(t, s)}{\partial s^{i}} \geq 0, \quad 0 \leq i \leq p, \end{aligned}$$

(9)

$$\begin{aligned}& (-1)^{m-p}\frac{\partial ^{i}G_{F, m}(t, s)}{\partial s^{i}} \geq 0, \quad p+1 \leq i \leq m-1. \end{aligned}$$

(10)

The following result holds in [12].

Theorem 6

Let $f \in C^{m}[a, b]$, and let $P_{F}$ be its ‘two-point right focal’ interpolating polynomial. Then, for $a \leq \zeta _{1} < \zeta _{2} \leq b$ and $0 \leq p \leq m-2$, the following holds:

$$\begin{aligned} f(t) =&P_{F}(t)+e_{F}(t) \\ =& \sum_{i=0}^{p}\frac{(t-\zeta _{1})^{i}}{i!}f^{(i)}( \zeta _{1}) \\ &{}+ \sum_{j=0}^{n-p-2} \Biggl(\sum _{i=0}^{j}\frac{(t-\zeta _{1})^{p+1+i}( \zeta _{1}-\zeta _{2})^{j-i}}{(p+1+i)!(j-i)!} \Biggr)f^{(p+1+j)}( \zeta _{2} ) \\ &{}+ \int ^{\zeta _{2}}_{\zeta _{1}}G_{F, m}(t, s)f^{(m)}(s)\,ds, \end{aligned}$$

(11)

where $G_{F, m}$ is the Green function defined by (8).

In [13], Butt et al. generalized Popoviciu’s inequality via Abel–Gontscharoff interpolating polynomial for higher order convex functions. In the same year in [14], Tasadduq et al. used Abel–Gontscharoff-type Green’s function for a two-point right focal to a generalized refinement of Jensen’s inequality from convex functions to higher order convex functions. The results in [13] and [14] are only for one type of data points. But Levinson-type inequalities studied for the class of 3-convex functions involve two types of data points. In this paper Levinson-type inequalities are generalized via Abel–Gontscharoff interpolating polynomial involving two types of data points.

2 Main results

Motivated by identity (6), we construct the following identities with the help of (8) and (11).

2.1 Bullen-type inequalities for higher order convex functions

First we define the following functional:

$\mathcal{F}$::

Let $(p_{1},\ldots, p_{n_{1}}) \in \mathbb{R}^{n_{1}}$ and $(q_{1},\ldots, q_{m_{1}}) \in \mathbb{R} ^{m_{1}}$ be such that $\sum_{i=1}^{n_{1}}p_{i}=P_{n_{1}}$, $\sum_{i=1}^{m_{1}}q_{i}=Q_{m_{1}}$, and $x_{i}$, $y_{i}$, $\frac{1}{P _{n_{1}}}\sum_{i=1}^{n_{1}}p_{i}x_{i}$, $\frac{1}{Q_{m_{1}}}\sum_{i=1} ^{m_{1}}p_{i}y_{i} \in I_{1}$. Then

$$ \begin{aligned}[b] \breve{\mathcal{J}}\bigl(f(\cdot )\bigr)={}&\frac{1}{Q_{m_{1}}}\sum _{i=1}^{m_{1}}q_{i}f(y _{i})-f\Biggl(\frac{1}{Q_{m_{1}}}\sum_{i=1}^{m_{1}}q_{i}y_{i} \Biggr)-\frac{1}{P _{n_{1}}} \sum_{i=1}^{n_{1}}p_{i}f(x_{i})\\&{}+ f\Biggl(\frac{1}{P_{n_{1}}}\sum_{i=1}^{n_{1}}p_{i}x_{i} \Biggr).\end{aligned} $$

(12)

Theorem 7

Assume $\mathcal{F}$. Let $f: I_{1}= [\zeta _{1}, \zeta _{2}] \rightarrow \mathbb{R}$ be a function such that $f \in C^{m}[\zeta _{1}, \zeta _{2}]$ ($m \geq 3$) and $G_{F, m}$, $\breve{\mathcal{J}}(f(\cdot ))$ are defined in (8) and (12) respectively. Then

$$\begin{aligned} \breve{\mathcal{J}}\bigl(f(\cdot )\bigr)=\breve{\mathcal{J}} \bigl(P_{F}(\cdot )\bigr)+ \int _{\zeta _{1}}^{\zeta _{2}}\breve{\mathcal{J}} \bigl(G_{F, m}(\cdot , s)\bigr)f ^{(m)}(s)\,ds. \end{aligned}$$

(13)

Proof

Using Abel–Gontscharoff identity (11) in (12), we have

$$\begin{aligned} \breve{\mathcal{J}}\bigl(f(\cdot )\bigr) =& \frac{1}{Q_{m_{1}}}\sum _{k=1}^{m _{1}}q_{k} \Biggl[\sum _{i=0}^{p}\frac{(y_{k}-\zeta _{1})^{i}}{i!}f^{(i)}( \zeta _{1}) \\ &{}+ \sum_{j=0}^{n-p-2} \Biggl(\sum _{i=0}^{j}\frac{(y_{k}-\zeta _{1})^{p+1+i}(\zeta _{1}-\zeta _{2})^{j-i}}{(p+1+i)!(j-i)!} \Biggr)f ^{(p+1+j)}(\zeta _{2}) \\ &{}+ \int _{\zeta _{1}}^{\zeta _{2}}G_{F, m}(y_{k}, s)f^{(m)}(s)\,ds \Biggr] \\ &{}-\sum_{i=0}^{p}\frac{(\frac{1}{Q_{m_{1}}}\sum_{k=1}^{m_{1}}q_{k}y _{k}-\zeta _{1})^{i}}{i!}f^{(i)}( \zeta _{1}) \\ &{}- \sum_{j=0}^{n-p-2} \Biggl(\sum _{i=0}^{j}\frac{(\frac{1}{Q_{m_{1}}} \sum_{k=1}^{m_{1}}q_{k}y_{k}-\zeta _{1})^{p+1+i}(\zeta _{1}-\zeta _{2})^{j-i}}{(p+1+i)!(j-i)!} \Biggr) f^{(p+1+j)}(\zeta _{2}) \\ &{}- \int _{\zeta _{1}}^{\zeta _{2}}G_{F, m}\Biggl( \frac{1}{Q_{m_{1}}}\sum_{k=1} ^{m_{1}}q_{k}y_{k}, s\Biggr)f^{(m)}(s)\,ds-\frac{1}{P_{n_{1}}}\sum _{k=1}^{n _{1}}p_{k} \Biggl[\sum _{i=0}^{p}\frac{(x_{k}-\zeta _{1})^{i}}{i!}f^{(i)}( \zeta _{1}) \\ &{}+ \sum_{j=0}^{n-p-2} \Biggl(\sum _{i=0}^{j}\frac{(x_{k}-\zeta _{1})^{p+1+i}(\zeta _{1}-\zeta _{2})^{j-i}}{(p+1+i)!(j-i)!} \Biggr)f ^{(p+1+j)}(\zeta _{2}) + \int _{\zeta _{1}}^{\zeta _{2}}G_{F, m}(x_{k}, s)f ^{(m)}(s)\,ds \Biggr] \\ &{}+\sum_{i=0}^{p}\frac{(\frac{1}{P_{n_{1}}}\sum_{k=1}^{n_{1}}p_{k}x _{k}-\zeta _{1})^{i}}{i!}f^{(i)}( \zeta _{1}) \\ &{}+ \sum_{j=0}^{n-p-2} \Biggl(\sum _{i=0}^{j}\frac{(\frac{1}{P_{n_{1}}} \sum_{k=1}^{n_{1}}p_{k}x_{k}-\zeta _{1})^{p+1+i}(\zeta _{1}-\zeta _{2})^{j-i}}{(p+1+i)!(j-i)!} \Biggr)f^{(p+1+j)} (\zeta _{2}) \\ &{}+ \int _{\zeta _{1}}^{\zeta _{2}}G_{F, m}\Biggl( \frac{1}{P_{n_{1}}}\sum_{k=1} ^{n_{1}}p_{k}x_{k}, s\Biggr)f^{(m)}(s)\,ds. \end{aligned}$$

Using the definition of $\breve{\mathcal{J}}(\cdot )$, we have

$$\begin{aligned} \breve{\mathcal{J}}\bigl(f(\cdot )\bigr) =& \frac{1}{Q_{m_{1}}}\sum _{k=1}^{m _{1}}q_{k} \Biggl[\sum _{i=3}^{p}\frac{(y_{k}-\zeta _{1})^{i}}{i!}f^{(i)}( \zeta _{1}) \\ &{}+ \sum_{j=0}^{n-p-2} \Biggl(\sum _{i=3}^{j}\frac{(y_{k}-\zeta _{1})^{p+1+i}(\zeta _{1}-\zeta _{2})^{j-i}}{(p+1+i)!(j-i)!} \Biggr)f ^{(p+1+j)}(\zeta _{2}) \Biggr] \\ &{}-\sum_{i=3}^{p}\frac{(\frac{1}{Q_{m_{1}}}\sum_{k=1}^{m_{1}}q_{k}y _{k}-\zeta _{1})^{i}}{i!}f^{(i)}( \zeta _{1}) \\ &{}- \sum_{j=0}^{n-p-2} \Biggl(\sum _{i=3}^{j}\frac{(\frac{1}{Q_{m_{1}}} \sum_{k=1}^{m_{1}}q_{k}y_{k}-\zeta _{1})^{p+1+i}(\zeta _{1}-\zeta _{2})^{j-i}}{(p+1+i)!(j-i)!} \Biggr)f ^{(p+1+j)}(\zeta _{2}) \\ &{}- \frac{1}{P_{n_{1}}}\sum_{k=1}^{n_{1}}p_{k} \Biggl[\sum_{i=3}^{p}\frac{(x _{k}-\zeta _{1})^{i}}{i!}f^{(i)}( \zeta _{1}) + \sum_{j=0}^{n-p-2} \Biggl(\sum_{i=3}^{j} \frac{(x_{k}-\zeta _{1})^{p+1+i}(\zeta _{1}-\zeta _{2})^{j-i}}{(p+1+i)!(j-i)!} \Biggr) \\ &{}\times f^{(p+1+j)}(\zeta _{2}) \Biggr] +\sum _{i=3}^{p}\frac{(\frac{1}{P _{n_{1}}}\sum_{k=1}^{n_{1}}p_{k}x_{k}-\zeta _{1})^{i}}{i!}f^{(i)}(\zeta _{1}) \\ &{}+ \sum_{j=0}^{n-p-2} \Biggl(\sum _{i=3}^{j}\frac{(\frac{1}{P_{n_{1}}} \sum_{k=1}^{n_{1}}p_{k}x_{k}-\zeta _{1})^{p+1+i} (\zeta _{1}-\zeta _{2})^{j-i}}{(p+1+i)!(j-i)!} \Biggr)f ^{(p+1+j)}(\zeta _{2}) \\ &{}+ \int _{\zeta _{1}}^{\zeta _{2}}\breve{\mathcal{J}} \bigl(G_{F, m}(\cdot , s)\bigr)f ^{(m)}(s)\,ds. \end{aligned}$$

After some simple calculations,

$$\begin{aligned} \breve{\mathcal{J}}\bigl(f(\cdot )\bigr) =& \frac{1}{Q_{m_{1}}}\sum _{k=1}^{m _{1}}q_{k} \bigl(P_{F}(y_{k}- \zeta _{1}) \bigr)-P_{F} \Biggl(\frac{1}{Q _{m_{1}}}\sum _{k=1}^{m_{1}}q_{k}y_{k}- \zeta _{1} \Biggr) \\ & {} -\frac{1}{P_{n_{1}}}\sum_{k=1}^{n_{1}}p_{k} \bigl(P_{F}(x_{k}-\zeta _{1}) \bigr)+P_{F} \Biggl(\frac{1}{P_{n_{1}}}\sum _{k=1}^{n_{1}}p_{k}x _{k}-\zeta _{1} \Biggr) \\ & {} + \int _{\zeta _{1}}^{\zeta _{2}}\breve{\mathcal{J}} \bigl(G_{F, m}(\cdot , s)\bigr)f ^{(m)}(s)\,ds. \end{aligned}$$

Again, we use the definition of $\breve{\mathcal{J}}(\cdot )$ to get (13). □

In the next result we have generalizations of Bullen-type inequality for m-convex functions.

Theorem 8

Assume the conditions of Theorem 7 with

$$ \breve{\mathcal{J}}\bigl(G_{F, m}(\cdot , s)\bigr) \geq 0, \quad s \in [\zeta _{1}, \zeta _{2}]. $$

(14)

If f is m-convex such that $f^{(m-1)}$ is absolutely continuous, then we have

$$\begin{aligned} \breve{\mathcal{J}}\bigl(f(\cdot )\bigr) \geq \breve{\mathcal{J}} \bigl(P_{F}(\cdot )\bigr). \end{aligned}$$

(15)

Proof

Since $f^{(m-1)}$ is absolutely continuous on $[\zeta _{1}, \zeta _{2}]$, therefore $f^{(m)}$ exists almost everywhere. By using Theorem 1, we have $f^{(m)}(s) \geq 0$ ($m \geq 3$) a.e. on $[\zeta _{1}, \zeta _{2}]$. Hence we can apply Theorem 7 to get (15). □

If we put $m_{1}=n_{1}=n$, $p_{i}=q_{i}$ and use positive weights in (12), then $\breve{\mathcal{J}}(\cdot )$ is converted to the functional $\mathcal{J}_{2}(\cdot )$ defined in (6), also in this case, (13), (14), and (15) become

$$ \mathcal{J}_{2}\bigl(f(\cdot )\bigr)= \mathcal{J}_{2}\bigl(P_{F}(\cdot )\bigr)+ \int _{\zeta _{1}}^{\zeta _{2}}\mathcal{J}_{2} \bigl(G_{F, m}(\cdot , s)\bigr)f^{(m)}(s)\,ds, $$

(13a)

where

$$\begin{aligned}& \begin{aligned} \mathcal{J}_{2} \bigl(P_{F}(\cdot )\bigr) &= \frac{1}{P_{n}}\sum _{k=1}^{n}p_{k} \bigl(P_{F}(y_{k}- \zeta _{1}) \bigr)-P_{F} \Biggl(\frac{1}{P_{n}} \sum _{k=1}^{n}p_{k}y_{k}- \zeta _{1} \Biggr) \\ &\quad{} -\frac{1}{P_{n}}\sum_{k=1}^{n}p_{k} \bigl(P_{F}(x_{k}-\zeta _{1}) \bigr)+P _{F} \Biggl(\frac{1}{P_{n}}\sum_{k=1}^{n}p_{k}x_{k}- \zeta _{1} \Biggr), \end{aligned} \\& \begin{aligned} \mathcal{J}_{2}\bigl(G_{F, m}(\cdot , s)\bigr) &= \frac{1}{P_{n}}\sum_{k=1}^{n}p _{k}G_{F, m}(y_{k}, s)- G_{F, m} \Biggl( \frac{1}{P_{n}}\sum_{k=1}^{n}p _{k}y_{k}, s \Biggr) \\ &\quad{} -\frac{1}{P_{n}}\sum_{k=1}^{n}p_{k}G_{F, m}(x_{k}, s)+G_{F, m} \Biggl(\frac{1}{P_{n}}\sum_{k=1}^{n}p_{k}x_{k}, s \Biggr), \end{aligned} \\& \mathcal{J}_{2}\bigl(G_{F, m}(\cdot , s)\bigr) \geq 0, \quad s \in [\zeta _{1}, \zeta _{2}], \end{aligned}$$

(14a)

and

$$ \mathcal{J}_{2}\bigl(f(\cdot )\bigr) \geq \mathcal{J}_{2}\bigl(P_{F}(\cdot )\bigr), $$

(15a)

respectively.

In the next result, we give a generalization of Bullen-type inequality for n tuples.

Theorem 9

Let $f \in C^{m}[\zeta _{1}, \zeta _{2}]$ ($m \geq 3$), $\mathbf{p}=(p_{1},\ldots,p _{n})$ be a positive n-tuple such that $\sum_{i = 1}^{n} {{p_{i}}} =P_{n}$. Also let $x_{i},y_{i} \in I_{1}$ such that (4) is valid for $i=1,\ldots, n$. Then for the functional $\mathcal{J}_{2}(f(\cdot ))$ defined in (6), we have the following:

• (i) If n is even and p is odd or p is even and n is odd, then for every m-convex function f, (15a) holds.

• (ii) Let inequality (15a) be satisfied. If $P_{F}(\cdot )$ is 3-convex then (6) is valid.

Proof

(i) By using (9), the following inequality

$$ (-1)^{n-p-1}\frac{\partial ^{3}\mathcal{G}_{F, n}(\cdot , s)}{\partial s^{3}} \geq 0 $$

(16)

holds, therefore it is easy to conclude that if (n = even, p = odd) or (p = even, n = odd), then $\frac{\partial ^{3} \mathcal{G}_{F, n}(\cdot , s)}{\partial s^{3}} \geq 0$, or if (n = odd, p = odd) or (p = even, n = even), then $\frac{\partial ^{3}\mathcal{G}_{F, n}(\cdot , s)}{\partial s^{3}} \leq 0$. So, for the cases (n = even, p = odd) or (p = even, n = odd), $\mathcal{G}_{F, n}(\cdot , s)$ is 3-convex with respect to the first variable, therefore by following Remark 1, inequality (14a) holds for n tuples. Hence, by Theorem 8, inequality (15a) holds.

(ii) Since $P_{F}(\cdot )$ is assumed to be 3-convex, therefore using the given conditions and by following Remark 1, the nonnegativity of the R.H.S. of (15a) is immediate, and we have (6) for n-tuples. □

Next we have a generalized form (for real weights) of Levinson-type inequality for 2n points given in [6](see also [3]).

$\mathcal{I}$::

Let $(p_{1},\ldots, p_{n_{1}}) \in \mathbb{R}^{n_{1}}$, $(q_{1},\ldots, q_{m_{1}}) \in \mathbb{R}^{m _{1}}$ be such that $\sum_{i=1}^{n_{1}}p_{i}=P_{n_{1}}$, $\sum_{i=1} ^{m_{1}}q_{i}=Q_{m_{1}}$, $\frac{1}{Q_{m_{1}}}\sum_{i=1}^{m_{1}}q_{i}y _{i}$, and $\frac{1}{P_{n_{1}}}\sum_{i=1}^{n_{1}}p_{i}x_{i} \in I_{1}$. Also let $x_{1},\ldots, x_{n_{1}}$ and $y_{1},\ldots, y_{m_{1}} \in I_{1}$ such that $x_{i}+y_{i}=2\breve{c}$, $x_{i}+x_{n-i+1} \leq 2\breve{c}$ and $\frac{p_{i}x_{i}+p_{n-i+1}x_{n-i+1}}{p_{i}+p _{n-i+1}} \leq \breve{c}$ for $i=1,\ldots, n$. Then (12) holds.

Theorem 10

Assume $\mathcal{I}$. Let $f: I_{1}= [\zeta _{1}, \zeta _{2}] \rightarrow \mathbb{R}$ be such that $f \in C^{m}[\zeta _{1}, \zeta _{2}]$ ($m \geq 3$), $G_{F, m}$ and $\breve{\mathcal{J}}(f(\cdot ))$ as defined in (8) and (12) respectively. Then identity (13) holds.

Proof

Assume $\mathcal{I}$ in Theorem 7 with the given conditions to get the required result. □

Theorem 11

Assume $\mathcal{I}$. Let $f: I_{1}= [\zeta _{1}, \zeta _{2}] \rightarrow \mathbb{R}$ be such that $f \in C^{m}[\zeta _{1}, \zeta _{2}]$ ($m \geq 3$) and $f^{(m-1)}$ is absolutely continuous. Also let $G_{F, m}$ and $\breve{\mathcal{J}}(f(\cdot ))$ be defined in (8) and (12) respectively. If (14) is valid, then (15) is also valid.

Proof

Proof is similar to Theorem 8. □

Theorem 12

Let $f \in C^{m}[\zeta _{1}, \zeta _{2}]$ ($m \geq 3$), $\mathbf{p}=(p_{1},\ldots,p _{n})$ be a positive n-tuple such that $\sum_{i = 1}^{n} {{p_{i}}} =P_{n}$. Also let $x_{i}, y_{i} \in I_{1}$ such that $x_{i}+y_{i}=2\breve{c}$, $x_{i}+x_{n-i+1} \leq 2\breve{c}$ and $\frac{p_{i}x_{i}+p_{n-i+1}x_{n-i+1}}{p_{i}+p_{n-i+1}} \leq \breve{c}$ for $i=1,\ldots, n$. Then, for the functional $\mathcal{J}_{2}(f(\cdot ))$ defined in (6), we have the following:

• (i) If n is even and p is odd or p is even and n is odd, then for every m-convex function f, (15a) holds.

• (ii) Let inequality (15a) be satisfied. If $P_{F}(\cdot )$ is 3-convex, then (6) is valid.

Proof

In Theorem 9, replace condition (4) for $x_{i}$ and $y_{i}$ with the condition given in the statement to get the required result. □

In [7], Mercer made a significant improvement by replacing condition (4) of symmetric distribution with the weaker one that the variances of the two sequences are equal.

Corollary 1

Let $f: I_{1}= [\zeta _{1}, \zeta _{2}] \rightarrow \mathbb{R}$ be such that $f \in C^{m}[\zeta _{1}, \zeta _{2}]$ ($m \geq 3$), $x_{i}$, $y_{i}$ satisfy (7), and $\max \{x_{1} ,\ldots, x_{n}\} \leq \min \{y_{1} ,\ldots, y_{n}\}$. Also let $(p_{1},\ldots, p_{n}) \in \mathbb{R}^{n}$ such that $\sum_{i=1}^{n}p_{i}=P_{n}$. Then (13a) holds.

2.2 Generalization of Levinson’s inequalities

Motivated by identity (3), we construct the following identities with the help of (8) and (11).

$\mathcal{H}$::

Let $f: I_{2}= [0, 2a] \rightarrow \mathbb{R}$ be a function, $x_{1},\ldots, x_{n_{1}} \in (0, a)$, $(p_{1},\ldots, p_{n_{1}})\in \mathbb{R}^{n_{1}}$, $(q_{1},\ldots, q_{m _{1}})\in \mathbb{R}^{m_{1}}$ be real numbers such that $\sum_{i=1} ^{n_{1}}p_{i}=P_{n_{1}}$ and $\sum_{i=1}^{m}q_{i}=Q_{m_{1}}$. Also let $x_{i}$, $\frac{1}{Q_{m_{1}}} \sum_{i=1}^{m_{1}}q_{i}(2a-x_{i})$ and $\frac{1}{P_{n_{1}}}\sum_{i=1}^{n_{1}}p_{i} \in I_{2}$. Then

$$\begin{aligned} \tilde{\mathcal{J}}\bigl(f(\cdot )\bigr) =&\frac{1}{Q_{m_{1}}}\sum _{i=1}^{m_{1}}q _{i}f(2a-x_{i})-f \Biggl(\frac{1}{Q_{m_{1}}}\sum_{i=1}^{m_{1}}q_{i}(2a-x_{i}) \Biggr)-\frac{1}{P _{n}} \sum_{i=1}^{n_{1}}p_{i}f(x_{i}) \\ &{}+ f\Biggl(\frac{1}{P_{n_{1}}}\sum_{i=1}^{n_{1}}p_{i}x_{i} \Biggr). \end{aligned}$$

(17)

Theorem 13

Assume $\mathcal{H}$ and let $f \in C^{m}[0, 2a]$ ($m \geq 3$). Also let $G_{F, m}$ and $\tilde{\mathcal{J}}(f(\cdot ))$ be defined in (8) and (17) respectively. Then we have

$$\begin{aligned} \tilde{\mathcal{J}}\bigl(f(\cdot )\bigr)=\tilde{\mathcal{J}} \bigl(P_{F}(\cdot )\bigr)+ \int _{\zeta _{1}}^{\zeta _{2}}\tilde{\mathcal{J}} \bigl(G_{F, m}(\cdot , s)\bigr)f ^{(m)}(s)\,ds, \end{aligned}$$

(18)

where $0 \leq \zeta _{1}<\zeta _{2}\leq 2a$.

Proof

Replace $\mathcal{F}$ with $\mathcal{H}$ and $y_{i}$ with $2a-x_{i}$ in Theorem 7, we get the required result. □

Theorem 14

Assume $\mathcal{H}$. Let $f \in C^{m}[0, 2a]$ ($m \geq 3$) with $f^{(m-1)}$ be absolutely continuous. Also let $G_{F, m}$ and $\tilde{\mathcal{J}}(f(\cdot ))$ be defined in (8) and (17) respectively. If

$$ \tilde{\mathcal{J}}\bigl(G_{F, m}(\cdot , s)\bigr) \geq 0, $$

(19)

then

$$\begin{aligned} \tilde{\mathcal{J}}\bigl(f(\cdot )\bigr) \geq \tilde{\mathcal{J}} \bigl(P_{F}(\cdot )\bigr), \end{aligned}$$

(20)

where $0 \leq \zeta _{1}<\zeta _{2}\leq 2a$.

Proof

Replace $\mathcal{F}$, $\breve{\mathcal{J}}(f(\cdot ))$ and $y_{i}$ with $\mathcal{H}$, $\tilde{\mathcal{J}}(f(\cdot ))$, $2a-x_{i}$ respectively in Theorem 8 to get the required result. □

If we put $m_{1}=n_{1}=n$, $p_{i}=q_{i}$ and by using positive weights in (17), then $\tilde{\mathcal{J}}(\cdot )$ is converted to the functional $\mathcal{J}_{1}(\cdot )$ defined in (3). Also in this case, (18), (19), and (20) become

$$ \mathcal{J}_{1}\bigl(f(\cdot )\bigr)= \mathcal{J}_{1}\bigl(P_{F}(\cdot )\bigr)+ \int _{\zeta _{1}}^{\zeta _{2}}\mathcal{J}_{1} \bigl(G_{F, m}(\cdot , s)\bigr)f^{(m)}(s)\,ds, $$

(18a)

where

$$\begin{aligned} \mathcal{J}_{1}\bigl(P_{F}(\cdot )\bigr) =& \frac{1}{P_{n}}\sum_{k=1}^{n}p_{k} \bigl(P_{F}(2a-x_{k}-\zeta _{1}) \bigr)-P_{F} \Biggl(\frac{1}{P_{n}} \sum _{k=1}^{n}p_{k}(2a-x_{k})-\zeta _{1} \Biggr) \\ & {} -\frac{1}{P_{n}}\sum_{k=1}^{n}p_{k} \bigl(P_{F}(x_{k}-\zeta _{1}) \bigr)+P _{F} \Biggl(\frac{1}{P_{n}}\sum_{k=1}^{n}p_{k}x_{k}- \zeta _{1} \Biggr) \end{aligned}$$

and

$$\begin{aligned}& \begin{aligned} \mathcal{J}_{1}\bigl(G_{F, m}(\cdot , s)\bigr) ={}& \frac{1}{P_{n}}\sum_{k=1}^{n}p _{k}G_{F, m}(2a-x_{k}, s)- G_{F, m} \Biggl( \frac{1}{P_{n}}\sum_{k=1} ^{n}p_{k}(2a-x_{k}), s \Biggr) \\ & {} -\frac{1}{P_{n}}\sum_{k=1}^{n}p_{k}G_{F, m}(x_{k}, s)+G_{F, m} \Biggl(\frac{1}{P_{n}}\sum_{k=1}^{n}p_{k}x_{k}, s \Biggr), \end{aligned} \\& \mathcal{J}_{1}\bigl(G_{F, m}(\cdot , s)\bigr) \geq 0, \quad s \in [\zeta _{1}, \zeta _{2}], \end{aligned}$$

(19a)

$$\begin{aligned}& \mathcal{J}_{1}\bigl(f(\cdot )\bigr) \geq \mathcal{J}_{1}\bigl(P_{F}(\cdot )\bigr), \end{aligned}$$

(20a)

respectively.

Theorem 15

Let $f \in C^{m}[0, 2a]$ ($m \geq 3$), $\mathbf{p}=(p_{1},\ldots,p_{n})$ be a positive n-tuple such that $\sum_{i = 1}^{n} {{p_{i}}} =P _{n}$. Then, for the functional $\mathcal{J}_{1}(f(\cdot ))$ defined in (3) and for $0 \leq \zeta _{1}<\zeta _{2}\leq 2a$, we have the following:

• (i) If n is even and p is odd or p is even and n is odd, then for every m-convex function f, (20a) holds.

• (ii) Let inequality (20a) be satisfied. If the $P_{F}(\cdot )$ is 3-convex, the R.H.S of (20a) is nonnegative and (3) is valid.

Proof

Proof is similar to Theorem 9. □

3 New bounds for Levinson-type inequality

For two Lebesgue integrable functions $f_{1}, f_{2}: [\zeta _{1}, \zeta _{2}] \rightarrow \mathbb{R}$, we consider the Čebyšev functional

$$\begin{aligned} \varTheta (f_{1}, f_{2}) =& \frac{1}{\zeta _{2}-\zeta _{2}} \int _{\zeta _{1}} ^{\zeta _{2}}f_{1}(t)f_{2}(t) \,dt \\ &{}- \frac{1}{\zeta _{2}-\zeta _{2}} \int _{\zeta _{1}}^{\zeta _{2}}f_{1}(t)\,dt. \frac{1}{ \zeta _{2}-\zeta _{2}} \int _{\zeta _{1}}^{\zeta _{2}}f_{2}(t)\,dt, \end{aligned}$$

(21)

where the integrals are assumed to exist.

The following two results are given in [15].

Theorem 16

Let $f_{1} : [\zeta _{1}, \zeta _{2}] \rightarrow \mathbb{R}$ be a Lebesgue integrable function and $f_{2} : [\zeta _{1}, \zeta _{2}] \rightarrow \mathbb{R}$ be an absolutely continuous function with $(\cdot , -\zeta _{1})(\cdot , -\zeta _{2})[f'_{2}]^{2} \in L[\zeta _{1}, \zeta _{2}]$. Then we have the inequality

$$\begin{aligned} \bigl\vert \varTheta (f_{1}, f_{2}) \bigr\vert \leq \frac{1}{\sqrt{2}}\bigl[\varTheta (f_{1}, f _{1})\bigr]^{\frac{1}{2}}\frac{1}{\sqrt{\zeta _{2}-\zeta _{2}}} \biggl( \int _{ \zeta _{1}}^{\zeta _{2}} (x-\zeta _{1}) (\zeta _{2}-x)\bigl[f'_{2}(x)\bigr]^{2}\,dx \biggr) ^{\frac{1}{2}}. \end{aligned}$$

(22)

The constant $\frac{1}{\sqrt{2}}$ is the best possible.

Theorem 17

Let $f_{1}: [\zeta _{1}, \zeta _{2}] \rightarrow \mathbb{R}$ be absolutely continuous with $f^{\prime }_{1} \in L_{\infty }[\zeta _{1}, \zeta _{2}]$, and let $f_{2}: [\zeta _{1}, \zeta _{2}] \rightarrow \mathbb{R}$ be monotonic nondecreasing on $[\zeta _{1}, \zeta _{2}]$. Then we have the inequality

$$\begin{aligned} \bigl\vert \varTheta (f_{1}, f_{2}) \bigr\vert \leq \frac{1}{2(\zeta _{2}-\zeta _{1})} \bigl\Vert f^{\prime } \bigr\Vert _{ \infty } \int _{\zeta _{1}}^{\zeta _{2}} (x-\zeta _{1}) (\zeta _{2}-x)\bigl[f'_{2}(x)\bigr]^{2}\,df _{2}(x). \end{aligned}$$

(23)

The constant $\frac{1}{2}$ is the best possible.

To generalize the results given in the previous section for two types of data points, we will consider Theorem 16 and Theorem 17.

Theorem 18

Assume $\mathcal{F}$. Let $f \in C^{m}[\zeta _{1}, \zeta _{2}]$ ($m \geq 3$) and $f^{(m)}$ be absolutely continuous with $(.-\zeta _{1})( \zeta _{2} -\cdot )[f^{(m+1)}]^{2} \in L[\zeta _{1}, \zeta _{2}]$. Also let $G_{F, m}$ and $\breve{\mathcal{J}}(f(\cdot ))$ as defined in (8) and (12) respectively. Then we have

$$\begin{aligned} \breve{\mathcal{J}}\bigl(f(\cdot )\bigr) =& \breve{\mathcal{J}} \bigl(P_{F}(\cdot )\bigr) +\frac{f ^{(m-1)}(\zeta _{2}) - f^{(m-1)}(\zeta _{1})}{(\zeta _{2}-\zeta _{2})} \\ &{}\times \int _{\zeta _{1}}^{\zeta _{2}}\breve{\mathcal{J}} \bigl(G_{F, m}( \cdot , s)\bigr)f^{(m)}(s)\,ds + \mathcal{R}_{m}(\zeta _{1}, \zeta _{2}; f), \end{aligned}$$

(24)

and the remainder $\mathcal{R}_{m}(\zeta _{1}, \zeta _{2}; f)$ satisfies the bound

$$\begin{aligned} \begin{aligned}[b] \bigl\vert \mathcal{R}_{m}(\zeta _{1}, \zeta _{2}; f) \bigr\vert \leq {}&\frac{(\zeta _{2}-\zeta _{2})}{\sqrt{2}} \bigl[\varTheta \bigl(\breve{\mathcal{J}}\bigl(G_{F, m}(\cdot , s)\bigr), \breve{\mathcal{J}}\bigl(G_{F, m}(\cdot , s)\bigr)\bigr) \bigr]^{\frac{1}{2}} \\ &{}\times\frac{1}{\sqrt{\zeta _{2}-\zeta _{1}}} \biggl( \int _{\zeta _{1}}^{\zeta _{2}}(s - \zeta _{1}) (\zeta _{2} - s)\bigl[f^{(m+1)}(s)\bigr]^{2} \,ds \biggr)^{ \frac{1}{2}}. \end{aligned} \end{aligned}$$

(25)

Proof

Setting $f_{1} \mapsto \breve{\mathcal{J}}(G_{F, m}(\cdot , s))$ and $f_{2} \mapsto f^{(m)}$ in Theorem 16, we have

$$\begin{aligned}& \biggl\vert \frac{1}{\zeta _{2}-\zeta _{1}} \int _{\zeta _{1}}^{\zeta _{2}}\breve{\mathcal{J}} \bigl(G_{F, m}(\cdot , s)\bigr)f^{(m)}(s)\,ds - \frac{1}{\zeta _{2}-\zeta _{1}} \int _{\zeta _{1}}^{\zeta _{2}}\breve{\mathcal{J}} \bigl(G_{F, m}(\cdot , s)\bigr)\,ds \\& \qquad {} \times \frac{1}{\zeta _{2}-\zeta _{1}} \int _{\zeta _{1}}^{\zeta _{2}}f^{(m)}(s)\,ds \biggr\vert \\& \quad \leq \frac{1}{\sqrt{2}}\bigl[\varTheta \bigl(\breve{\mathcal{J}} \bigl(G_{F, m}( \cdot , s)\bigr), \breve{\mathcal{J}}\bigl(G_{F, m}( \cdot , s)\bigr)\bigr)\bigr]^{\frac{1}{2}}\frac{1}{\sqrt{ \zeta _{2}-\zeta _{2}}} \biggl( \int _{\zeta _{1}}^{\zeta _{2}}(s - \zeta _{1}) ( \zeta _{2} - s) \\& \qquad {} \times \bigl[f^{(m+1)}(s)\bigr]^{2} \,ds \biggr)^{\frac{1}{2}}, \\& \biggl\vert \frac{1}{\zeta _{2}-\zeta _{1}} \int _{\zeta _{1}}^{\zeta _{2}}\breve{\mathcal{J}} \bigl(G_{F, m}(\cdot , s)\bigr)f^{(m)}(s)\,ds - \frac{f^{(m-1)}(\zeta _{2}) - f^{(m-1)}(\zeta _{1})}{(\zeta _{2}-\zeta _{2})^{2}} \int _{\zeta _{1}}^{\zeta _{2}}\breve{\mathcal{J}} \bigl(G_{F, m}(\cdot , s)\bigr)\,ds \biggr\vert \\& \quad \leq \frac{1}{\sqrt{2}}\bigl[\varTheta \bigl(\breve{\mathcal{J}} \bigl(G_{F, m}( \cdot , s)\bigr), \breve{\mathcal{J}}\bigl(G_{F, m}( \cdot , s)\bigr)\bigr)\bigr]^{\frac{1}{2}}\frac{1}{\sqrt{ \zeta _{2}-\zeta _{2}}} \biggl( \int _{\zeta _{1}}^{\zeta _{2}}(s - \zeta _{1}) ( \zeta _{2} - s) \\& \qquad{} \times \bigl[f^{(m+1)}(s)\bigr]^{2} \,ds \biggr)^{\frac{1}{2}}. \end{aligned}$$

Multiplying $(\zeta _{2}-\zeta _{2})$ on both sides of the above inequality and using the estimation (25), we get

$$\begin{aligned} \int _{\zeta _{1}}^{\zeta _{2}}\breve{\mathcal{J}} \bigl(G_{F, m}(\cdot , s)\bigr)f ^{(m)}\,ds =& \frac{f^{(m-1)}(\zeta _{2}) - f^{(m-1)}(\zeta _{1})}{(\zeta _{2} - \zeta _{1})} \int _{\zeta _{1}}^{\zeta _{2}}\breve{\mathcal{J}}\bigl(G _{F, m}(\cdot , s)\bigr)\,ds \\ & {} + \mathcal{R}_{m}(\zeta _{1}, \zeta _{2}; f). \end{aligned}$$

Using identity (13), we get (24). □

The Grüss-type inequalities can be obtained by using Theorem 17.

Theorem 19

Assume $\mathcal{F}$. Let $f \in C^{m}[\zeta _{1}, \zeta _{2}]$ ($m \geq 3$) with $f^{(m)}$ be absolutely continuous and $f^{(m-1)} \geq 0$ a.e. on $I_{1}$. Then identity (24) holds, where the remainder satisfies the estimation

$$\begin{aligned} \bigl\vert \mathcal{R}_{m}(\zeta _{1}, \zeta _{2}; f) \bigr\vert \leq & (\zeta _{2}-\zeta _{2}) \bigl\Vert \breve{\mathcal{J}}\bigl(G_{F, m}(\cdot , s) \bigr)^{\prime } \bigr\Vert _{ \infty } \biggl[\frac{f^{(m-1)}(\zeta _{2})+f^{(m-1)}(\zeta _{1})}{2} \\ & {} - \frac{f^{(m-1)}(\zeta _{2})- f^{(m-1)}(\zeta _{1})}{\zeta _{2}-\zeta _{2}} \biggr]. \end{aligned}$$

(26)

Proof

Setting $f_{1} \mapsto \breve{\mathcal{J}}(G_{F, m}(\cdot , s))$ and $f_{2} \mapsto f^{(m)}$ in Theorem 17, we get

$$\begin{aligned}& \biggl\vert \frac{1}{\zeta _{2}-\zeta _{2}} \int _{\zeta _{1}}^{\zeta _{2}}\breve{\mathcal{J}} \bigl(G_{F, m}(\cdot , s)\bigr))f^{(m)}(s)\,ds - \frac{1}{\zeta _{2}- \zeta _{1}} \int _{\zeta _{1}}^{\zeta _{2}}\breve{\mathcal{J}} \bigl(G_{F, m}( \cdot , s)\bigr)\,ds \\& \quad \quad {} \times \frac{1}{\zeta _{2}-\zeta _{1}} \int _{\zeta _{1}}^{\zeta _{2}}f^{(m)}(s)\,ds \biggr\vert \\& \quad \leq \frac{1}{2} \bigl\Vert \breve{\mathcal{J}} \bigl(G_{F, m}(\cdot , s)\bigr)^{\prime } \bigr\Vert _{ \infty } \frac{1}{\zeta _{2}-\zeta _{1}} \int _{\zeta _{1}}^{\zeta _{2}}(s - \zeta _{1}) (\zeta _{2} - s)\bigl[f^{(m+1)}(s)\bigr]^{2}\,ds. \end{aligned}$$

(27)

Since

$$\begin{aligned}& \int _{\zeta _{1}}^{\zeta _{2}}(s - \zeta _{1}) (\zeta _{2} - s)\bigl[f^{(m+1)}(s)\bigr]^{2} \,ds = \int _{\zeta _{1}}^{\zeta _{2}}[2s - \zeta _{1} - \zeta _{2}]f^{m}(s)\,ds \\& \quad = (\zeta _{2} - \zeta _{1})\bigl[f^{(m-1)}( \zeta _{2}) + f^{(m-1)}(\zeta _{1})\bigr] - 2 \bigl(f^{(m-1)}(\zeta _{2}) - f^{(m-1)}(\zeta _{1})\bigr), \end{aligned}$$

(28)

using (13), (27), and (28), we have (24) with (26). □

Theorem 20

Assume $\mathcal{F}$. Let $f \in C^{m}[\zeta _{1}, \zeta _{2}]$ ($m \geq 3$) with $f^{(m-1)}$ be absolutely continuous. Also let $G_{F, m}$ and $\breve{\mathcal{J}}(f(\cdot ))$ be as defined in (8) and (12) respectively. Moreover, assume that $(p, q)$ is a pair of conjugate exponents, that is, $1 \leq p, q, \leq \infty$, $\frac{1}{p}+\frac{1}{q}=1$. Let $\vert f^{(m)} \vert ^{p}: [\zeta _{1}, \zeta _{2}] \rightarrow \mathbb{R}$ be a Riemann integrable function. Then

$$\begin{aligned} \bigl\vert \breve{\mathcal{J}}\bigl(f(\cdot )\bigr)- \breve{\mathcal{J}} \bigl(P_{F}( \cdot )\bigr) \bigr\vert \leq \bigl\Vert f^{(m)} \bigr\Vert _{p} \biggl( \int _{\zeta _{1}}^{\zeta _{2}} \bigl\vert \breve{\mathcal{J}} \bigl(G_{F, m}(\cdot , s)\bigr)\,ds \bigr\vert ^{q} \biggr) ^{\frac{1}{q}}. \end{aligned}$$

Proof

For the proof see Theorem 3.5 in [16]. □

Remark 2

Similar work can be done for Levinson’s inequality (2), (one type of data points) for higher order-convex functions.

Remark 3

We can give related mean value theorems by using nonnegative functionals (13) and (18), and we can construct the new families of m-exponentially convex functions ($m \geq 3$) and Cauchy means related to these functionals.