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

  1. (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)
  2. (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:

  1. (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.

  2. (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:

  1. (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.

  2. (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:

  1. (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.

  2. (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.