Complete convergence and complete moment convergence for maximal weighted sums of extended negatively dependent random variables under sub-linear expectations

1.   Introduction

We recall that an interval $\chi\subset\mathfrak{R}$ is convex if for all $x$, $y\in\chi$, we have $tx+\left(1-t\right) y\in\chi$, where $t\in\left[ 0, 1\right]$ and a function $f:\chi\rightarrow \mathfrak{R}$ is convex if for all $x$, $y\in\chi$, the inequality

holds. A function $f:\chi\rightarrow \mathfrak{R}$ is concave if the inequality (1.1) holds in opposite direction.

Convexity is essential to understanding and solving problems pertaining to fractional integral inequalities because of its properties and definition, and it has recently gained in importance. Convex functions have yielded several new integral inequalities, as evidenced by ^{[2,3,8,11,14,15,21,24,43,45,46,47,48]}. Hermite-integral Hadamard's inequalities are most commonly encountered when searching for comprehensive inequalities:

where the function $f:\chi\rightarrow \mathfrak{R}$ is convex on $\chi$ and $f\in L^{1}\left(\left[ u, v\right] \right)$.

There are the following two classical fractional integral inequalities which are defined by Hermite-Hadamard type inequalities:

where the function $f:\chi\rightarrow \mathfrak{R}$ is positive, convex on $\chi$ and $f\in L^{1}\left(\left[ u, v\right] \right)$.

The left-sided and right-sided Riemann-Liouville fractional integrals $J_{u^{+}}^{\nu}f\left(v\right)$ and $J_{v^{-}}^{\nu}f\left(u\right)$ of order $\nu > 0$ in (1.3), are defined respectively as (see ^[5,22]):

and

The extended inequalities for (1.2) and (1.5) are fractional integral inequalities of the Fejér and Hermite-Hadamard-Fejér types, and the results are as follows:

and

respectively, where $f$ is as defined above, the function $\zeta:\left[ u, v\right] \rightarrow \left[ 0, \infty\right)$ is integrable symmetric with respect to $\frac{u+v}{2}$, that is

and $\Gamma\left(\nu\right)$ is the Gamma function defined as $\Gamma\left(\nu\right) = \int_{0}^{\infty}x^{\nu-1}e^{-x}dx$, $\operatorname{Re}\left(\nu\right) > 0$.

In addition to being able to be generalized, convexity and convex functions have several generalizations. One of those generalizations is the concept of harmonic convexity and harmonic convex functions, which can be defined as follows:

Definition 1.1. Define $\chi\subseteq\mathfrak{R}\backslash\left\{ 0\right\}$ as an interval of real numbers. A function $f$ from $\chi$ to the real numbers is considered to be harmonically convex, if

for all $x, y\in\chi$ and $t\in\left[ 0, 1\right]$. Harmonically concave $f$ is defined as the inequality in (1.6) reversed.

Using harmonic-convexity, the Hermite-Hadamard type yields the following result.

Theorem 1.2. Let $f:\chi\subseteq\mathfrak{R}\backslash\left\{ 0\right\} \rightarrow \mathfrak{R}$ be a harmonically convex function and $u, v\in\chi$ with $u < v$.If $f\in L\left(\left[ u, v\right] \right)$ then the followinginequalities hold:

Harmonic symmetricity of a function is given in the definition below.

Definition 1.3. A function $\zeta:\left[ u, v\right] \subseteq\mathfrak{R}\backslash\left\{ 0\right\} \rightarrow \mathfrak{R}$ is harmonically symmetric with respect to $\frac{2uv}{u+v}$ if

holds for all $x\in\left[ u, v\right]$.

Fejér type inequalities using harmonic convexity and the notion of harmonic symmetricity were presented in Chan and Wu ^[7].

Theorem 1.4. Let $f:\chi\subseteq\mathfrak{R}\backslash\left\{ 0\right\} \rightarrow \mathfrak{R}$ be a harmonically convex function and $u, v\in\chi$ with $u < v$.If $f\in L\left(\left[ u, v\right] \right)$ and $\zeta:\left[u, v\right] \subseteq\mathfrak{R}\backslash\left\{ 0\right\} \rightarrow \mathfrak{R}$ is nonnegative, integrable and harmonically symmetric withrespect to $\frac{2uv}{u+v}$, then

Hermite-Hadamard's inequalities for harmonically convex functions in fractional integral form were proved in ^[19].

Theorem 1.5. Let $f:\chi\subseteq\left(0, \infty\right) \rightarrow \mathfrak{R}$ be afunction such that $f\in L\left(\left[ u, v\right] \right)$, where$u, v\in\chi$ with $u < v$. If $f$ is a harmonically convex function on $\left[u, v\right]$, then the following inequalities for fractional integrals hold:

with $\nu > 0$ and $g\left(x\right) = \frac{1}{x}$.

Hermite-Hadamard-Fejér inequality for harmonically convex function in fractional integral form were obtained by İşcan et al. in ^[18].

Theorem 1.6. Let $f:\left[ u, v\right] \rightarrow \mathfrak{R}$ be a harmonically convexfunction with $u < v$ and $f\in L\left(\left[ u, v\right] \right)$. If$\zeta:\left[ u, v\right] \rightarrow \mathfrak{R}$ is nonnegative, integrableand harmonically symmetric with respect to $\frac{2uv}{u+v}$, then thefollowing inequalities for fractional integrals holds:

with $\nu > 0$ and $g\left(x\right) = \frac{1}{x}$, $x\in\left[ \frac{1}{v}, \frac{1}{u}\right]$.

Left-sided and right-sided Riemann-Liouville fractional integrals are generalized in the definition given below:

Definition 1.7. ^[20] Let $\tau\left(x\right)$ be an increasing positive and monotonic function on the interval $\left(u, v\right]$ with a continuous derivative $\tau^{\prime}\left(x\right)$ on the interval $\left(u, v\right)$ with $\tau\left(x\right) = 0$, $0\in\left[ u, v\right]$. Then, the left-side and right-side of the weighted fractional integrals of a function $f$ with respect to another function $\tau\left(x\right)$ on $\left[ u, v\right]$ of order $\nu > 0$ are defined by:

and

where $\zeta^{-1}\left(x\right) = \frac{1}{\zeta\left(x\right) }, \zeta\left(x\right) \neq0$.

The following observations are obvious from the above definition:

● If $\tau\left(x\right) = x$ and $\zeta\left(x\right) = 1$, then the weighted fractional integral operators in the Definition 1.7 reduce to the classical Riemann-Liouville fractional integral operators.

● If $\zeta\left(x\right) = 1$, we get the fractional integral operators of a function $f$ with respect to another function $\tau\left(x\right)$ of order $\nu > 0$, defined in ^[1,38] as follows:

and

The study analyzes several inequalities of the Hermite-Hadamard-Fejér type through weighted fractional operators with positive symmetric weight function in the kernel.

2.   Main results

Throughout this paper, $\mathfrak{R}$ denotes the set of all real numbers, $\chi\subset\mathfrak{R}$ denotes an interval. In addition. for $u, v\in J^{\circ}$ with $u < v,$ the functions $\theta_{u, v}, \theta_{v, u}:\left[ 0, 1\right] \rightarrow \mathfrak{R}$ are defined as:

We start this section with the following Lemma which will be used repeatedly in the sequel.

Lemma 2.1. The following results hold:

(i) If $\zeta:[u, v]\subset\left(0, \infty\right) \rightarrow \lbrack0, \infty)$ be an integrable function and symmetric with respect to$\frac{2uv}{u+v}$, then we have

for each $t\in\lbrack0, 1]$,

(ii) If $\zeta:[u, v]\subset\left(0, \infty\right) \rightarrow \lbrack0, \infty)$ be an integrable and symmetric function with respect to$\frac{2uv}{u+v}$, then we have for $\nu > 0$

where $h\left(x\right) = \frac{1}{x}$, $x\in\left[ \frac{1}{v}, \frac{1}{u}\right]$.

Proof. (ⅰ) Let $x = \theta_{u, v}\left(t\right) = \frac{uv}{tu+\left(1-t\right) v}$. It is clear that $x\in\lbrack u, v]$ for each $t\in\left[ 0, 1\right]$ and then

Hence by the definition of harmonic symmetry, we obtain

(ⅱ) By using the harmonic symmetric property of $\zeta$, we have

for all $t\in\left[ \tau^{-1}\left(\frac{1}{v}\right), \tau^{-1}\left(\frac{1}{u}\right) \right]$.

From this and by setting $\frac {1}{\tau\left(x\right) } = \frac{1}{\frac{1}{u}+\frac{1}{v}-\tau\left(t\right) }$, it follows that

Theorem 2.2. Let $f:I\subseteq\left(0, \infty\right) \rightarrow \mathfrak{R}$ be an $L^{1}$ convex function with $0 < u < v$, $u, v\in I$ and$\zeta:\left[ u, v\right] \rightarrow \mathfrak{R}$ be an integrable, positiveand weighted symmetric function with respect to $\frac{2uv}{u+v}$. If $\tau$is an increasing and positive function on $\left[ u, v\right)$ and$\tau\left(x\right)$ is continuous on $\left(u, v\right)$, then, wehave for $\nu > 0$:

Proof. Since $f$ is a harmonic convex function on $[u, v]$, we have

Choosing $x = \zeta\left(\theta_{u, v}\left(t\right) \right)$ and $y = \zeta\left(\theta_{v, u}\left(t\right) \right)$, we obtain

Multiplying both the sides by $t^{\nu-1}\zeta\left(\theta_{u, v}\left(t\right) \right)$ and then integrating the resultant with respect to "$t$" over $[0, 1]$, we obtain

To prove the first inequality in (2.3), we need to use (2.2)

where $t = \frac{\frac{1}{u}-\tau\left(x\right) }{\frac{1}{u}-\frac{1}{v}}$.

By evaluating the weighted fractional operator, one can observe that

Setting $t_{1} = \frac{\frac{1}{u}-\tau\left(x\right) }{\frac{1}{u}-\frac {1}{v}}$and $t_{2} = \frac{\tau\left(x\right) -\frac{1}{v}}{\frac{1}{u} -\frac{1}{v}}$ and using (2.1)

Using (2.5) and (2.6) in (2.4), we obtain

Thus the first inequality is proved.

To prove the second inequality we use the convexity of $f$

Multiplying both the sides by $t^{\nu-1}\zeta\left(\theta_{u, v}\left(t\right) \right)$ and then integrating the resultant with respect to "$t$" over $[0, 1]$, we obtain

Then by using (2.1) and (2.5) in (2.8), we obtain (2.3).

Remark 1. (more specifically), if we utilize Theorem 1 in Theorem 2.2, then

(1) $\tau\left(x\right) = x$, then (2.3) takes the form

where $_{\zeta}J_{u+}^{\nu}$ and $_{\zeta}J_{v-}^{\nu}$ are the left and right weighted Riemann-Liouville fractional operators of order $\nu > 0$, defined by

and

(2) $\tau\left(x\right) = x$ and $\nu = 1$, then inequality (2.3) transforms into inequality (1.8).

(3) $\tau\left(x\right) = x$ and $\zeta\left(x\right) = 1$, then inequality (2.3) reduces to the inequality (1.9).

(4) $\tau\left(x\right) = x$, $\zeta\left(x\right) = 1$ and $\nu = 1$, then inequality (2.3) becomes the inequality (1.7).

Lemma 2.3. Let $f:I\subseteq\left(0, \infty\right) \rightarrow \mathfrak{R}$ be an $L^{1}$ function with $f^{^{\prime}}\in L^{1}$ for$0 < u < v$, $u, v\in I^{\circ}$ and $\zeta:\left[ u, v\right] \rightarrow \mathfrak{R}$ be an integrable, positive and weighted symmetric function withrespect to $\frac{2uv}{u+v}$. If $\tau$ is an increasing and positive functionon $\left[ u, v\right)$ and $\tau\left(x\right)$ is continuous on$\left(u, v\right)$, then, we have for $\nu > 0$:

where $h\left(x\right) = \frac{1}{x}$, $x\in\left[ \frac{1}{v}, \frac{1}{u}\right]$.

Proof. Setting

By integration by parts, making use of Lemma 2.1, and definitions (1.11) and (1.12), we obtain

and

Now

Since

Thus, we obtain from (2.13) that

Which is the required result.

Remark 2. Particularly, in Lemma 2.3, if we take:

(1) $\tau\left(x\right) = x$, then equality (2.11) becomes

where $_{\zeta}J_{u+}^{\nu}$ and $_{\zeta}J_{v-}^{\nu}$ are defined in Remark 1.

(2) $\tau\left(x\right) = x$ and $\zeta(x) = 1$, then equality (2.11) becomes

(3) $\tau\left(x\right) = x$, $\zeta(x) = 1$ and $\nu = 1$, we obtain

We will use the following notations for the rest of this section:

Theorem 2.4. Let $f:I\subseteq\left(0, \infty\right) \rightarrow \mathfrak{R}$ be an $L^{1}$ function with $f^{^{\prime}}\in L^{1}$ for$0 < u < v$, $u, v\in I^{\circ}$ and $\zeta:\left[ u, v\right] \rightarrow \mathfrak{R}$ be an integrable, positive and weighted symmetric function withrespect to $\frac{2uv}{u+v}$. If $\left\vert \left(f\circ h\right)^{^{\prime}}\right\vert$ is harmonic convex on $\left[ u, v\right]$, $\tau$is an increasing and positive function on $\left[ u, v\right)$, and$\tau\left(x\right)$ is continuous on $\left(u, v\right)$, then, wehave for $\nu > 0$:

where $h\left(x\right) = \frac{1}{x}$, $x\in\left[ \frac{1}{v}, \frac{1}{u}\right]$,

and

Proof. According to (2.11) of Lemma 2.3, we obtain

We know that $\zeta$ is a harmonic symmetric with respect to $\frac{2uv}{u+v}$, we observed that

Hence

From (2.21) we get

for $t\in\left[ \tau^{-1}\left(\frac{1}{v}\right), \frac{\tau^{-1}\left(\frac{1}{u}\right) +\tau^{-1}\left(\frac{1}{v}\right) }{2}\right]$ or

for $t\in\left[ \frac{\tau^{-1}\left(\frac{1}{u}\right) +\tau^{-1}\left(\frac{1}{v}\right) }{2}, \tau^{-1}\left(\frac{1}{u}\right) \right]$.

By applying the harmonic convexity of $\left\vert f^{^{\prime} }\right\vert$ on $\left[ u, v\right]$ for $t\in\left[ \tau^{-1}\left(\frac{1}{v}\right), \tau^{-1}\left(\frac{1}{u}\right) \right]$, we get

Applying (2.22)–(2.24) in (2.20), we obtain

Let us evaluate the first integral in (2.25)

and the value of the second integral in (2.25) is given below

Applying (2.21)–(2.27) in (2.20) to obtain the desired inequality (2.19).

Remark 3. Particularly, in Theorem 2.4, if we take

(1) $\tau\left(x\right) = x$, we have

where

and

(2) $\tau\left(x\right) = x$ and $\zeta(x) = 1$, we get

where $u\left(\nu, u, v\right)$ and $v\left(\nu, u, v\right)$ are defined as above.

(3) $\tau\left(x\right) = x$, $\zeta(x) = 1$ and $\nu = 1$, we obtain

and

Theorem 2.5. Let $f:I\subseteq\left(0, \infty\right) \rightarrow \mathfrak{R}$ be an $L^{1}$ function with $f^{^{\prime}}\in L^{1}$ for$0 < u < v$, $u, v\in I^{\circ}$ and $\zeta:\left[ u, v\right] \rightarrow \mathfrak{R}$ be an integrable, positive and weighted symmetric function withrespect to $\frac{u+v}{2}$. If $\left\vert f^{^{\prime}}\right\vert ^{q}$ isharmonic convex on $\left[ u, v\right]$ for $q\geq1$, $\tau$ is anincreasing and positive function on $\left[ u, v\right)$, and $\tau\left(x\right)$ is continuous on $\left(u, v\right)$, then, we have for $\nu > 0$:

where $h\left(x\right) = \frac{1}{x}$, $x\in\left[ \frac{1}{v}, \frac{1}{u}\right]$,

and $u_{\tau}\left(\nu, u, v\right)$, $v_{\tau}\left(\nu, u, v\right)$ aredefined as in Theorem 2.4.

Proof. Applying power-mean inequality to(2.20) and then using (2.22)–(2.24), we get

Since $\left\vert \left(\zeta\circ h\circ\tau\right) \left(x\right) \right\vert \leq\left\Vert \zeta\circ h\circ\tau\right\Vert _{\infty}$, hence it is easy to observe that

Since for $q\geq1$ and $t\in\left[ \tau^{-1}\left(\frac{1}{u}\right), \tau^{-1}\left(\frac{1}{v}\right) \right]$, $\left\vert f^{^{\prime} }\right\vert ^{q}$ is convex on $\left[ u, v\right]$, we get

Thus, now we are able to evaluate the second integral in (2.32)

Applying (2.33), (2.34) in (2.32), we get (2.31).

Remark 4. In Theorem 2.5, especially when we take

(1) $\tau\left(x\right) = x$, we have

where $h\left(x\right) = \frac{1}{x}$, $x\in\left[ \frac{1}{v}, \frac{1} {u}\right]$,

$u\left(\nu, u, v\right)$ and $v\left(\nu, u, v\right)$ are defined in (1) of Remark 3.

(2) $\tau\left(x\right) = x$ and $\zeta(x) = 1$, we get

where $h\left(x\right) = \frac{1}{x}$, $x\in\left[ \frac{1}{v}, \frac{1} {u}\right]$, $u\left(\nu, u, v\right)$, $v\left(\nu, u, v\right)$ are defined in (1) of Remark 3 and $C\left(\nu, u, v\right)$ is as defined above.

(3) $\tau\left(x\right) = x$, $\zeta(x) = 1$ and $\nu = 1$, we obtain

where $h\left(x\right) = \frac{1}{x}$, $x\in\left[ \frac{1}{v}, \frac{1} {u}\right]$, $u\left(u, v\right)$, $v\left(u, v\right)$ are defined in (3) of Remark 3 and $C\left(u, v\right) = \left(\frac{v-u} {2uv}\right) ^{2}$.

3.   Conclusions

In this study, we proved very important and interesting inequalities of Fejér type for a very fascinating generalized class of functions, namely, harmonic convex functions by using general weighted fractional integral operator which depends upon an increasing function. The results of our study not only generalize a number of findings obtained in ^[17,18,19] but one can obtain a number of new results by choosing a increasing function involved. The results can also be an inspiration for young researchers as well as researcher already working in the field of fractional integral inequalities and can further open up new directions of research in mathematical sciences.

Acknowledgements

Princess Nourah bint Abdulrahman University Researchers Supporting Project number (PNURSP2022R8), Princess Nourah bint Abdulrahman University, Riyadh, Saudi Arabia.

Conflict of interest

The authors declare that they have no conflicts of interest in this paper.