Fejér type inequalities for harmonically convex functions

1.   Introduction

Hermite-Hadamard inequality is a double inequality for convex functions that has a lot of literary value (please see ^[16]).

Let $\mathbb{\zeta}:I\longrightarrow \mathbb{R},$ $\emptyset\neq I\subseteq \mathbb{R}$, $\mathfrak{\varsigma}, \mathbb{\tau}\in I$ with $\mathfrak{\varsigma } < \mathbb{\tau}$, be a convex function. Then

the inequality holds in reversed direction if $\mathbb{\zeta}$ is concave.

Fejér ^[15] established the following double inequality as a weighted generalization of (1.1):

where $\mathbb{\zeta}:I\longrightarrow \mathbb{R},$ $\emptyset\neq I\subseteq \mathbb{R}$, $\mathfrak{\varsigma}, \mathbb{\tau}\in I$ with $\mathfrak{\varsigma } < \mathbb{\tau}$ is any convex function and $\mathbb{\vartheta}:\left[\mathfrak{\varsigma}, \mathbb{\tau}\right] \rightarrow \mathbb{R}$ is non-negative integrable and symmetric with respect to $\mathfrak{w} = \frac{\mathfrak{\varsigma}+\mathbb{\tau}}{2}$.

These inequalities have many extensions and generalizations, see ^{[2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18]} and ^{[19,20,21,22,23,24,25,26,27,28,29,30,31]}.

Consider the following mappings on $\left[ 0, 1\right]$:

and

where $\mathbb{\zeta}:\left[ \mathfrak{\varsigma}, \mathbb{\tau}\right] \rightarrow \mathbb{R}$ is a convex function and $\mathbb{\vartheta}:\left[ \mathfrak{\varsigma }, \mathbb{\tau}\right] \rightarrow \mathbb{R}$ is non-negative integrable and symmetric with respect to $\mathfrak{w} = \frac {\mathfrak{\varsigma}+\mathbb{\tau}}{2}$.

The important results that characterize the properties of the above mappings and inequalities are discussed by a number of mathematicians.

Dragomir ^[2] established the theorem which refines the first inequality of (1.1).

Theorem 1. ^[2] Let $\mathbb{\zeta}:\left[ \mathfrak{\varsigma}, \mathbb{\tau}\right] \rightarrow \mathbb{R}$ be a convex function on $\left[ \mathfrak{\varsigma}, \mathbb{\tau}\right]$. Then $\mathbb{H}$ is monotonically increasing and convex on $\left[ 0, 1\right]$. Moreover, one has thefollowing inequalities

Dragomir et al. ^[7] obtained the refinements of (1.1).

Theorem 2. ^[2] Let $\mathbb{\zeta}$, $\mathbb{H}$ be defined asabove. Then

(i) The following inequality holds

(ii) If $\mathbb{\zeta}$ is differentiable on $\left[\mathfrak{\varsigma}, \mathbb{\tau}\right]$, then for all $\mathfrak{\iota}\in\left[ 0, 1\right]$, one has

and

Theorem 3. ^[7] Let $\mathbb{\zeta}$, $\mathbb{H}$, $\mathcal{\breve{G}}$ be defined as above. We have

(i) $\mathcal{\breve{G}}$ is convex and increasing on $\left[0, 1\right]$.

(ii) The following hold

and

(iii) The inequality

holds for all $\mathfrak{\iota}\in\left[ 0, 1\right]$.

(iv) Then

(v) If $\mathbb{\zeta}$ is differentiable on $\left[\mathfrak{\varsigma}, \mathbb{\tau}\right]$, then for all $\mathfrak{\iota}\in\left[ 0, 1\right]$, one has

Theorem 4. ^[7] Let $\mathbb{\zeta}$, $\mathbb{H}$, $\mathcal{\breve{G}}$, $\mathcal{L}$ be defined as above. Then

(i) $\mathcal{L}$ is convex on $\left[ 0, 1\right]$.

(ii) The inequalities

hold for all $\mathfrak{\iota}\in\left[ 0, 1\right]$ and

(iii) The inequalities

hold for all $\mathfrak{\iota}\in\left[ 0, 1\right]$.

Teseng et al. ^[24] proved the following result.

Lemma 1. ^[24] Let $\mathbb{\zeta}:$ $\left[ \mathfrak{\varsigma}, \mathbb{\tau}\right] \rightarrow \mathbb{R}$ be a convex function and let $\mathfrak{\varsigma}\leq\breve{\varkappa}_{1}\leq\mathfrak{w}_{1}\leq\mathfrak{w}_{2}\leq\breve{\varkappa}_{2}\leq\mathbb{\tau}$ with $\mathfrak{w}_{1}+\mathfrak{w}_{2} = \breve{\varkappa}_{1}+\breve{\varkappa}_{2}$. Then

Yang and Tseng ^[28] proven the theorem by using Lemma 1 which refines the first inequality of (1.2) and generalizes Theorem 1.

Theorem 5. ^[28] Let $\mathbb{\zeta}:\left[ \mathfrak{\varsigma}, \mathbb{\tau}\right] \rightarrow \mathbb{R}$ be a convex function and $\mathbb{\vartheta}:\left[ \mathfrak{\varsigma}, \mathbb{\tau}\right] \rightarrow \mathbb{R}$ is non-negative integrable and symmetric with respect to $\mathfrak{w} = \frac{\mathfrak{\varsigma}+\mathbb{\tau}}{2}$. Then $\mathbb{H}_{\mathbb{\vartheta}}$ is convex, increasing on $\left[ 0, 1\right]$, and for all $\mathfrak{\iota}\in\left[ 0, 1\right]$, we have

One of the generalizations of the convex functions is harmonic functions:

Definition 1. ^[17] Define $I\subseteq \mathbb{R} \backslash\left\{ 0\right\}$ as an interval of real numbers. A function $\mathbb{\zeta}$ from $I$ to the real numbers is considered to be harmonically convex, if

for all $\mathfrak{w}, \breve{\varkappa}\in I$ and $\mathfrak{\iota}\in\left[ 0, 1\right]$. Harmonically concave $\mathbb{\zeta}$ is defined as the inequality in (1.4) reversed.

İşcan ^[17] used harmonic-convexity to develop the inequalities of Hermite-Hadamard type.

Theorem 6. ^[17] Let $\mathbb{\zeta}:I\subseteq \mathbb{R} \backslash\left\{ 0\right\} \rightarrow \mathbb{R}$ be a harmonically convex function and $\mathfrak{\varsigma}, \mathbb{\tau}\in I$ with $\mathfrak{\varsigma} < \mathbb{\tau}$. If $\mathbb{\zeta}\in L\left(\left[ \mathfrak{\varsigma}, \mathbb{\tau}\right] \right)$ then theinequalities

hold.

Let $\mathbb{\zeta}:\left[ \mathfrak{\varsigma}, \mathbb{\tau}\right] \subset\left(0, \infty\right) \rightarrow \mathbb{R}$ be a harmonic convex mapping and let $\mathbb{S}, \mathbb{V}:\left[ 0, 1\right] \rightarrow \mathbb{R}$ be defined by

and

The author obtained the refinement inequalities for (1.5) related to the above mappings:

Theorem 7. ^[21] Let $\mathbb{\zeta}:\left[ \mathfrak{\varsigma}, \mathbb{\tau}\right] \subset\left(0, \infty\right) \rightarrow \mathbb{R}$ be a harmonic convex function on $\left[ \mathfrak{\varsigma}, \mathbb{\tau}\right]$. Then

(i) $\mathbb{S}$ is harmonic convex $\left(0, 1\right]$ andincreases monotonically on $\left[ 0, 1\right]$.

(ii) The following hold:

Theorem 8. ^[21] Let $\mathbb{\zeta}:\left[ \mathfrak{\varsigma}, \mathbb{\tau}\right] \subset\left(0, \infty\right) \rightarrow \mathbb{R}$ be a harmonic convex function on $\left[ \mathfrak{\varsigma}, \mathbb{\tau}\right]$. Then

(i) $\mathbb{V}$ is harmonic convex $\left(0, 1\right]$ andincreases monotonically on $\left[ 0, 1\right]$.

(ii) The following hold:

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

Definition 2. ^[22] A function $\mathbb{\vartheta}:\left[ \mathfrak{\varsigma }, \mathbb{\tau}\right] \subseteq \mathbb{R} \backslash\left\{ 0\right\} \rightarrow \mathbb{R}$ is harmonically symmetric with respect to $\frac{2\mathfrak{\varsigma }\mathbb{\tau}}{\mathfrak{\varsigma}+\mathbb{\tau}}$ if

holds for all $\mathfrak{w}\in\left[ \mathfrak{\varsigma}, \mathbb{\tau }\right]$.

Fejér type inequalities using harmonic convexity and the notion of harmonic symmetricity were presented in Chan and Wu ^[1].

Theorem 9. ^[1] Let $\mathbb{\zeta}:I\subseteq \mathbb{R} \backslash\left\{ 0\right\} \rightarrow \mathbb{R}$ be a harmonically convex function and $\mathfrak{\varsigma}, \mathbb{\tau}\in I$ with $\mathfrak{\varsigma} < \mathbb{\tau}$. If $\mathbb{\zeta}\in L\left(\left[ \mathfrak{\varsigma}, \mathbb{\tau}\right] \right)$ and $\mathbb{\vartheta}:\left[ \mathfrak{\varsigma}, \mathbb{\tau}\right]\subseteq \mathbb{R} \backslash\left\{ 0\right\} \rightarrow \mathbb{R}$ is nonnegative, integrable and harmonically symmetric with respect to $\frac{2\mathfrak{\varsigma}\mathbb{\tau}}{\mathfrak{\varsigma}+\mathbb{\tau}}$, then

Chan and Wu ^[1] also defined some mappings related to (1.8) and discussed important properties of these mappings.

Motivated by the studies conducted in ^[2,21,24,27], we define some new mappings in connection to (1.8) and to prove new Féjer type inequalities which indeed provide refinement inequalities as well.

2.   Main results

To prove the major findings of this work, we employ the given important facts about harmonic convex and convex functions.

Theorem 10. ^[8,9] If $\left[ \mathfrak{\varsigma}, \mathbb{\tau}\right] \subset I\subset\left(0, \infty\right)$ and if we consider thefunction $\breve{h}:\left[ \frac{1}{\mathbb{\tau}}, \frac{1}{\mathfrak{\varsigma}}\right] \rightarrow \mathbb{R}$ defined by $\breve{h}\left(\mathfrak{\iota}\right) = \mathbb{\zeta}\left(\frac{1}{\mathfrak{\iota}}\right)$, then $\mathbb{\zeta}$ is harmonicallyconvex on $\left[ \mathfrak{\varsigma}, \mathbb{\tau}\right]$ if and only if $\breve{h}$ is convex in the usual sense on $\left[ \frac{1}{\mathbb{\tau}}, \frac{1}{\mathfrak{\varsigma}}\right]$.

Theorem 11. ^[8,9] If $I\subset\left(0, \infty\right)$ and $\mathbb{\zeta}$ is convex and nondecreasing function then $\mathbb{\zeta}$ isharmonic convex and if $\mathbb{\zeta}$ is harmonic convex and nonincreasing function then $\mathbb{\zeta}$ is convex.

Let us now define some mappings on $\left[ 0, 1\right]$ related to (1.8) and prove some refinement inequalities.

and

where $\mathbb{\zeta}:\left[ \mathfrak{\varsigma}, \mathbb{\tau}\right] \rightarrow \mathbb{R}$ is a harmonic convex function and $\mathbb{\vartheta}:\left[ \mathfrak{\varsigma}, \mathbb{\tau}\right] \rightarrow \mathbb{R}$ is non-negative integrable and symmetric with respect to $\mathfrak{w} = \frac {2\mathfrak{\varsigma}\mathbb{\tau}}{\mathfrak{\varsigma}+\mathbb{\tau}}$.

Lemma 2. Let $\mathbb{\zeta}:$ $\left[ \mathfrak{\varsigma}, \mathbb{\tau}\right] \subset\left(0, \infty\right) \rightarrow \mathbb{R}$ be a harmonic convex function and let $\mathfrak{\varsigma}\leq\breve{\varkappa}_{1}\leq\mathfrak{w}_{1}\leq\mathfrak{w}_{2}\leq\breve{\varkappa}_{2}\leq\mathbb{\tau}$ with $\frac{\mathfrak{w}_{1}\mathfrak{w}_{2}}{\mathfrak{w}_{1}+\mathfrak{w}_{2}} = \frac{\breve{\varkappa}_{1}\breve{\varkappa}_{2}}{\breve{\varkappa}_{1}+\breve{\varkappa}_{2}}$. Then

Proof. For $\breve{\varkappa}_{1} = \breve{\varkappa}_{2}$, the result is obvious. We observe that

By applying the harmonic convexity, we obtain

We first prove a result similar to (1.3) for harmonically convex functions which provide refinement inequalities for (1.8).

Theorem 12. Let $\mathbb{\zeta}:$ $\left[ \mathfrak{\varsigma}, \mathbb{\tau}\right] \subset\left(0, \infty\right) \rightarrow \mathbb{R}$ be a harmonic convex function, $0 < \mathbb{\rho} < 1$, $0 < \mathbb{\theta} < 1$, $\mathbb{\lambda} = \frac{\mathfrak{\varsigma}\mathbb{\tau}}{\mathbb{\rho}\mathfrak{\varsigma}+\left(1-\mathbb{\rho}\right) \mathbb{\tau}}$, $\mathbb{\tau}_{0} = \left(\frac{\mathfrak{\varsigma}\mathbb{\tau}}{\mathbb{\tau}-\mathfrak{\varsigma}}\right) \min\left\{ \frac{\mathbb{\rho}}{1-\mathbb{\theta}}, \frac{1-\mathbb{\rho}}{\mathbb{\theta}}\right\}$ andlet $\mathbb{\vartheta}:\left[ \mathfrak{\varsigma}, \mathbb{\tau}\right]\rightarrow \mathbb{R}$ be nonnegative and integrable and $\mathbb{\vartheta}\left(\frac{\mathbb{\lambda}}{1-\mathbb{\theta}\mathfrak{\iota}\mathbb{\lambda}}\right) = \mathbb{\vartheta}\left(\frac{\mathbb{\lambda}}{1+\left(1-\mathbb{\theta}\right) \mathfrak{\iota}\mathbb{\lambda}}\right)$, $\mathfrak{\iota}\in\left[ 0, \mathbb{\tau}_{0}\right]$. Then

Proof. For every $\mathbb{\tau}\in\left[ 0, \mathbb{\tau}_{0}\right]$, we have the identity

We now prove that the mapping $\mathcal{W}:\left[ 0, \mathbb{\tau}_{0}\right] \rightarrow \mathbb{R}$ defined by

is harmonic convex $\left(0, \mathbb{\tau}_{0}\right]$ and monotonically increasing on $\left[ 0, \mathbb{\tau}_{0}\right]$.

Since the sum of two harmonic convex functions is a harmonic convex, hence $\mathcal{W}$ is a harmonic convex on $\left(0, \mathbb{\tau}_{0}\right]$. Let $\mathfrak{\iota}\in\left(0, \mathbb{\tau}_{0}\right]$, it follows from the harmonic convexity of $\mathbb{\zeta}$ that

We observed that $0 < \mathbb{\rho}\leq\frac{\mathbb{\rho}\left(\mathbb{\tau }-\mathfrak{\varsigma}\right) +\mathfrak{\varsigma}\mathbb{\tau\theta }\mathfrak{\iota}}{\mathbb{\tau}-\mathfrak{\varsigma}}\leq1$, $0\leq \frac{\left(1-\mathbb{\rho}\right) \left(\mathbb{\tau} -\mathfrak{\varsigma}\right) -\mathbb{\theta}\mathfrak{\iota\varsigma }\mathbb{\tau}}{\mathbb{\tau}-\mathfrak{\varsigma}}\leq1-\mathbb{\rho} < 1$, $0\leq\mathbb{\rho}\leq\frac{\mathbb{\rho}\left(\mathbb{\tau} -\mathfrak{\varsigma}\right) -\left(1-\mathbb{\theta}\right) \mathfrak{\iota\varsigma}\mathbb{\tau}}{\mathbb{\tau}-\mathfrak{\varsigma} }\leq\mathbb{\rho}\leq1$ and $0 < 1-\mathbb{\rho}\leq\frac{\left(1-\mathbb{\rho}\right) \left(\mathbb{\tau}-\mathfrak{\varsigma}\right) +\left(1-\mathbb{\theta}\right) \mathfrak{\iota\varsigma}\mathbb{\tau} }{\mathbb{\tau}-\mathfrak{\varsigma}}\leq1$. Thus, by using the harmonic convexity, we obtain

From (2.3) and (2.4), we obtain

Finally, for $\mathfrak{\iota}_{1}$, $\mathfrak{\iota}_{2}$, such that $0 < \mathfrak{\iota}_{1} < \mathfrak{\iota}_{2}\leq\mathfrak{\varsigma}_{0}$, since $\mathcal{W}\left(\mathfrak{\iota}\right)$ is harmonic convex, it follows from (2.3) that

This shows that $\mathcal{W}$ is increasing on $\left[ 0, \mathbb{\tau} _{0}\right]$.

Since $\mathbb{\vartheta}$ is nonnegative, multiplying (2.5) by $\frac{\mathbb{\vartheta}\left(\frac{\mathbb{\lambda} }{1-\mathbb{\theta}\mathfrak{w}\mathbb{\lambda}}\right) }{\mathfrak{w}^{2}}$, integrating the resulting inequalities over $\left[ 0, \mathfrak{\iota }\right]$ and using $\mathbb{\vartheta}\left(\frac{\mathbb{\lambda} }{1-\mathbb{\theta}\mathfrak{w}\mathbb{\lambda}}\right) = \mathbb{\vartheta }\left(\frac{\mathbb{\lambda}}{1+\left(1-\mathbb{\theta}\right) \mathfrak{w}\mathbb{\lambda}}\right)$, we have

By using the identity (2.2) in (2.6), we obtain (2.1).

Remark 1. If we choose $\mathbb{\rho} = \frac{\mathbb{\vartheta} }{\mathbb{\vartheta}+q}$, $\mathbb{\theta} = \frac{1}{2}$, $\mathfrak{\iota } = 2\breve{\varkappa}$ in Theorem 12, then

Remark 2. If we choose $\mathbb{\rho} =$ $\mathbb{\theta} = \frac{1}{2}$, $\mathfrak{\iota} = \mathbb{\tau}_{0} = \frac{\mathfrak{\varsigma}\mathbb{\tau} }{\mathbb{\tau}-\mathfrak{\varsigma}}$ in Theorem 12, then we get (1.8).

Remark 3. If we choose $\mathbb{\rho} =$ $\mathbb{\theta} = \frac{1}{2}$, $\mathfrak{\iota} = \mathbb{\tau}_{0} = \frac{\mathfrak{\varsigma}\mathbb{\tau} }{\mathbb{\tau}-\mathfrak{\varsigma}}$ and $\mathbb{\vartheta}\left(\mathfrak{w}\right) = 1$, $\mathfrak{w\in}\left[ \mathfrak{\varsigma }, \mathbb{\tau}\right]$ in Theorem 12, then we get (1.5).

Theorem 13. Let $\mathbb{\zeta}$, $\mathbb{\lambda}$ and $\mathbb{\tau}_{0}$ be defined as in Theorem 12, $0 < \mathbb{\rho} < 1$, $0 < \mathbb{\theta} < 1$, $\mathbb{\rho}+\mathbb{\theta}\leq1$ and let $\mathbb{X}$ be defined on $\left[ 0, 1\right]$ as

Then, $\mathbb{X}$ is harmonically convex on $\left(0, 1\right]$ andmonotonically increasing on $\left[ 0, 1\right]$, and

Proof. Since $\mathbb{\zeta}$ is harmonically convex on $\left[ \mathfrak{\varsigma }, \mathbb{\tau}\right]$ this prove the harmonic convexity of $\mathbb{X}$ on $\left(0, 1\right]$. By using the condition $\mathbb{\rho}+\mathbb{\theta }\leq1$ implies that $\mathbb{\tau}_{0} = \frac{\mathbb{\rho}\mathfrak{\varsigma }\mathbb{\tau}}{\left(1-\mathbb{\theta}\right) \left(\mathbb{\tau }-\mathfrak{\varsigma}\right) }$. Since the mapping $\mathcal{W}:\left[ 0, \mathbb{\tau}_{0}\right] \rightarrow \mathbb{R}$ defined by

has been proved to be monotonically increasing on $\left[ 0, \mathbb{\tau} _{0}\right]$, thus the mapping $\mathbb{X}$ is also monotonically increasing on $\left[ 0, 1\right]$.

Because X is monotonically increasing on $\left[ 0, 1\right]$, it follows that the inequalities (2.8) can be deduced from these inequalities (2.5). The proof of the theorem was completed as a result of this.

The next theorem can be proved similarly:

Theorem 14. Let $\mathbb{\zeta}$, $\mathbb{\lambda}$, $\mathbb{\tau}_{0}$, $\mathbb{\rho}$ and $\mathbb{\theta}$ be defined as in Theorem 13. Let $\mathbb{X}_{1}$ be defined on $\left[ 0, 1\right]$ as

Then, $\mathbb{X}_{1}$ is harmonically convex monotonically increasing on $\left[ 0, 1\right]$, and

Remark 4. Taking $\mathbb{\rho} = \mathbb{\theta} = \frac{1}{2}$ in the inequality (2.8) reduces to

Remark 5. Taking $\mathbb{\rho} = \mathbb{\theta} = \frac{1}{2}$ in the inequality (2.10) reduces to

Theorem 15. Let $\mathbb{\zeta}$, $\mathbb{\rho}$, $\mathbb{\theta}$, $\mathbb{\lambda}$, $\mathbb{\tau}_{0}$ be defined as in Theorem 13 and let $\mathbb{\vartheta}$ be defined as in Theorem 12. Let $\mathbb{Y}$ be a function defined on $\left[ 0, 1\right]$ by

for some $s\in\left[ 0, \mathbb{\tau}_{0}\right]$. Then $\mathbb{Y}$ isharmonic convex and monotonically increasing on $\left(0, 1\right]$ and

Proof. Since $\mathbb{\zeta}$ is harmonic convex and $\mathbb{\vartheta}$ is nonnegative, we see that $\mathbb{Y}$ is harmonic convex on $\left(0, 1\right]$. Let $\mathfrak{w}\in\left[ 0, s\right]$, where $s\in\left[ 0, \mathbb{\tau}_{0}\right]$, from Theorem 12 we get $\breve {h}\left(\mathfrak{\iota w}\right) = \left(1-\mathbb{\theta}\right) \mathbb{\zeta}\left(\frac{\mathbb{\lambda}}{1-\mathbb{\theta\lambda }\mathfrak{\iota w}}\right) +\mathbb{\theta\zeta}\left(\frac {\mathbb{\lambda}}{1+\left(1-\mathbb{\theta}\right) \mathbb{\lambda }\mathfrak{\iota w}}\right)$ is increasing for $\mathfrak{\iota}\in\left[ 0, 1\right]$. Therefore the inequalities (2.14) are achieved immediately.

Theorem 16. Let $\mathbb{\zeta}$, $\mathbb{\rho}$, $\mathbb{\theta}$, $\mathbb{\lambda}$, $\mathbb{\tau}_{0}$ be defined as in Theorem 15 and let $\mathbb{\vartheta}$ be defined as in Theorem 12. Let $\mathbb{Y}_{1}$ be a function defined on $\left[ 0, 1\right]$ by

for some $s\in\left[ 0, \mathbb{\tau}_{0}\right]$. Then $\mathbb{Y}_{1}$ isharmonic convex $\left(0, 1\right]$ and monotonically increasing on $\left[ 0, 1\right]$, and

Proof. Since $\mathbb{\zeta}$ is harmonic convex and $\mathbb{\vartheta}$ is nonnegative, we see that $\mathbb{Y}_{1}$ is harmonic convex on $\left(0, 1\right]$. Next, for each $\mathfrak{w}\in\left[ 0, \mathfrak{\iota }\right]$, where $\mathfrak{\iota}\in\left[ 0, \mathbb{\tau}_{0}\right]$, it follows from Theorem 12 that $\breve{h}\left(\mathfrak{\iota }\right) = \left(1-\mathbb{\theta}\right) \mathbb{\zeta}\left(\frac{\mathbb{\lambda}}{1-\mathbb{\theta\lambda}\mathfrak{\iota}}\right) +\mathbb{\theta\zeta}\left(\frac{\mathbb{\lambda}}{1+\left(1-\mathbb{\theta}\right) \mathbb{\lambda}\mathfrak{\iota}}\right)$ and $k\left(\mathfrak{\iota}\right) = s-\left(1-\mathfrak{\iota}\right) \mathfrak{w}$ are increasing on $\left[ 0, \mathbb{\tau}_{0}\right]$ and $\left[ 0, 1\right]$ respectively. Hence

is increasing on $\left[ 0, 1\right]$. Using the identity $\mathbb{\vartheta }\left(\frac{\mathbb{\lambda}}{1-\mathbb{\theta\lambda}\mathfrak{\iota} }\right) = \mathbb{\vartheta}\left(\frac{\mathbb{\lambda}}{1+\left(1-\mathbb{\theta}\right) \mathbb{\lambda}\mathfrak{\iota}}\right)$ we see that $\mathbb{Y}\left(\mathfrak{\iota}\right)$ is increasing on $\left[ 0, 1\right]$. Therefore the inequalities (2.16) follows from

and (2.16).

Remark 6. Choose $\mathbb{\rho} = \mathbb{\theta} = \frac{1}{2}$, $s = \mathbb{\tau}_{0} = \frac{\mathfrak{\varsigma}\mathbb{\tau}}{\mathbb{\tau }-\mathfrak{\varsigma}}$ in Theorems 15 and 16. Then the inequalities (2.14) and (2.16) reduce to

where

and

Remark 7. The inequalities (2.17) provide weighted generalizations of Theorems 9 and 15.

In the coming results we provide weighted generalizations of Theorems 2–4 for harmonic convex functions by using Lemma 2.

Theorem 17. Let $\mathbb{\zeta}$, $\mathbb{\vartheta}$, $\mathbb{S}_{\mathbb{\vartheta}}$ be defined as above. Then

(i) The inequality

holds.

(ii) If $\mathbb{\zeta}$ is differentiable on $\left[\mathfrak{\varsigma}, \mathbb{\tau}\right]$ and $\mathbb{\vartheta}$ isbounded on $\left[ \mathfrak{\varsigma}, \mathbb{\tau}\right]$, then theinequalities

hold for all $\mathfrak{\iota}\in\left[ 0, 1\right]$, where $\left\Vert\mathbb{\vartheta}\right\Vert _{\infty} = \underset{\mathfrak{w}\in\left[\mathfrak{\varsigma}, \mathbb{\tau}\right] }{\sup}\mathbb{\vartheta}\left(\mathfrak{w}\right)$.

(iii) If $\mathbb{\zeta}$ is differentiable on $\left[\mathfrak{\varsigma}, \mathbb{\tau}\right]$, then, for all $\mathfrak{\iota}\in\left[ 0, 1\right]$, then

Proof. (i) Using techniques of integration and the hypothesis of $\mathbb{\vartheta}$, we have the following identities:

and

By using Lemma 2, we observe that the following inequalities hold for all $\mathfrak{\iota}\in\left[ 0, \frac{1}{2}\right]$ and $\mathfrak{w} \in\left[ \mathfrak{\varsigma}, \frac{2\mathfrak{\varsigma}\mathbb{\tau} }{\mathfrak{\varsigma}+\mathbb{\tau}}\right]$:

and

Multiplying the inequalities (2.26)–(2.30) by $\frac {\mathbb{\vartheta}\left(\mathfrak{w}\right) }{\mathfrak{w}^{2}}$ and integrating them over $\mathfrak{\iota}$ on $\left[ 0, \frac{1}{2}\right]$, over $\mathfrak{w}$ on $\left[ \mathfrak{\varsigma}, \frac {2\mathfrak{\varsigma}\mathbb{\tau}}{\mathfrak{\varsigma}+\mathbb{\tau} }\right]$ and using identities (2.22)–(2.25), we derive (2.19).

(ii) Since $\mathbb{\zeta}:\left[ \mathfrak{\varsigma }, \mathbb{\tau}\right] \rightarrow \mathbb{R}$ is harmonic convex on $\left[ \mathfrak{\varsigma}, \mathbb{\tau}\right]$, hence $\breve{h}:\left[ \frac{1}{\mathbb{\tau}}, \frac{1}{\mathfrak{\varsigma }}\right] \rightarrow \mathbb{R}$ defined by $\breve{h}\left(\mathfrak{w}\right) = \mathbb{\zeta}\left(\frac{1}{\mathfrak{w}}\right)$ is convex on $\left[ \frac{1}{\mathbb{\tau} }, \frac{1}{\mathfrak{\varsigma}}\right]$. Thus, by integration by parts, we get that following identity holds:

The equality (2.31) is equivalent to the equality:

Using substitution rules for integration and the hypothesis of $\mathbb{\vartheta}$, we have the following identities:

and

Now, using the convexity of $\breve{h}\left(\mathfrak{w}\right) = \mathbb{\zeta}\left(\frac{1}{\mathfrak{w}}\right)$ on $\left[ \frac {1}{\mathbb{\tau}}, \frac{1}{\mathfrak{\varsigma}}\right]$ and the hypothesis of $\mathbb{\vartheta}$, the following inequality holds for all $\mathfrak{\iota}\in\left[ 0, 1\right]$ and $\mathfrak{w}\in\left[ \frac {1}{\mathbb{\tau}}, \frac{\mathfrak{\varsigma}+\mathbb{\tau}} {2\mathfrak{\varsigma}\mathbb{\tau}}\right]$:

which is equivalent to

Integrating the above inequalities over $\mathfrak{w}$ on $\left[ \frac {1}{\mathbb{\tau}}, \frac{\mathfrak{\varsigma}+\mathbb{\tau}} {2\mathfrak{\varsigma}\mathbb{\tau}}\right]$, we get

After making use of suitable substitution, the inequality (2.37) takes the form:

Inequality (2.20) follows from (2.31)–(2.34) and (2.38).

(iii) We use the fact that $\mathbb{\zeta }:\left[ \mathfrak{\varsigma}, \mathbb{\tau}\right] \rightarrow \mathbb{R}$ is harmonic convex on $\left[ \mathfrak{\varsigma}, \mathbb{\tau}\right]$, hence $\breve{h}:\left[ \frac{1}{\mathbb{\tau}}, \frac{1}{\mathfrak{\varsigma }}\right] \rightarrow \mathbb{R}$ defined by $\breve{h}\left(\mathfrak{w}\right) = \mathbb{\zeta}\left(\frac{1}{\mathfrak{w}}\right)$ is convex on $\left[ \frac{1}{\mathbb{\tau} }, \frac{1}{\mathfrak{\varsigma}}\right]$. Thus

and

Adding the above inequalities

The inequality (2.39) becomes

Multiplying (2.40) both sides by $\frac{\mathbb{\vartheta}\left(\mathfrak{w}\right) }{\mathfrak{w}^{2}}$ and integrating over $\left[ \mathfrak{\varsigma}, \mathbb{\tau}\right]$, we get

From (2.17) and (2.41) we get (2.21).

Corollary 1. Suppose that the assumption of Theorem 17 are satisfiedand $\mathbb{\vartheta}\left(\mathfrak{w}\right) = \frac{\mathfrak{\varsigma}\mathbb{\tau}}{\mathbb{\tau}-\mathfrak{\varsigma}}$, $\mathfrak{w}\in\left[\mathfrak{\varsigma}, \mathbb{\tau}\right]$, then

(i) The inequalities

holds.

(ii) The inequalities

hold for all $\mathfrak{\iota}\in\left[ 0, 1\right]$.

(iii) The inequalities

are valid for all $\mathfrak{\iota}\in\left[ 0, 1\right]$.

In the following theorems, we discuss inequalities for the functions $\mathbb{S}$, $\mathbb{S}_{\mathbb{\vartheta}}$, $\mathcal{\breve{G}}_{1}$, $\mathbb{T}$ and $\mathbb{T}_{\mathbb{\vartheta}}$ as considered above:

Theorem 18. Let $\mathbb{\zeta}$, $\mathbb{\vartheta}$, $\mathcal{\breve{G}}_{1}$, $\mathbb{S}_{\mathbb{\vartheta}}$ be defined as above. Then

(i) The inequality

holds for all $\mathfrak{\iota}\in\left[ 0, 1\right]$.

(ii) The inequalities

hold.

(iii) If $\mathbb{\zeta}$ is differentiable on $\left[\mathfrak{\varsigma}, \mathbb{\tau}\right]$ and $\mathbb{\vartheta}$ isbounded on $\left[ \mathfrak{\varsigma}, \mathbb{\tau}\right]$, then, forall $\mathfrak{\iota}\in\left[ 0, 1\right]$, then

where $\left\Vert \mathbb{\vartheta}\right\Vert _{\infty} = \underset{\mathfrak{w}\in\left[ \mathfrak{\varsigma}, \mathbb{\tau}\right]}{\sup}\mathbb{\vartheta}\left(\mathfrak{w}\right)$.

Proof. (i) Using integration by substitution and the assumptions on $\mathbb{\vartheta}$, we have that the following identity holds on $\left[ 0, 1\right]$:

By Lemma 2, the following inequality holds for all $\mathfrak{w} \in\left[ \mathfrak{\varsigma}, \frac{2\mathfrak{\varsigma}\mathbb{\tau} }{\mathfrak{\varsigma}+\mathbb{\tau}}\right]$ with

Multiplying both sides of (2.49) with $\frac{\mathbb{\vartheta}\left(\mathfrak{w}\right) }{\mathfrak{w}^{2}}$, integrating over $\left[ \mathfrak{\varsigma}, \frac{2\mathfrak{\varsigma}\mathbb{\tau}} {\mathfrak{\varsigma}+\mathbb{\tau}}\right]$ and using (2.34) and (2.49), we obtain (2.45).

(ii) We can observe that

By using harmonic symmetric assumption on $\mathbb{\vartheta}$, we get

We can also see that the following identity holds:

Finally we also have

By Lemma 2, the following inequalities hold for all $\mathfrak{w} \in\left[ \mathfrak{\varsigma}, \frac{2\mathfrak{\varsigma}\mathbb{\tau} }{\mathfrak{\varsigma}+\mathbb{\tau}}\right]$:

The inequality

holds with the choices of $\mathfrak{w}_{1} = \frac{4\mathfrak{\varsigma }\mathbb{\tau}\mathfrak{w}}{2\mathfrak{\varsigma}\mathbb{\tau}+\left(\mathfrak{\varsigma}+\mathbb{\tau}\right) \mathfrak{w}}$, $\mathfrak{w} _{2} = \frac{4\mathfrak{\varsigma}\mathbb{\tau}\mathfrak{w}}{3\left(\mathfrak{\varsigma}+\mathbb{\tau}\right) \mathfrak{w}-2\mathfrak{\varsigma }\mathbb{\tau}}$, $\breve{\varkappa}_{1} = \frac{4\mathfrak{\varsigma }\mathbb{\tau}}{3\mathfrak{\varsigma}+\mathbb{\tau}}$ and $\breve{\varkappa }_{2} = \frac{4\mathfrak{\varsigma}\mathbb{\tau}}{\mathfrak{\varsigma }+3\mathbb{\tau}}$.

The inequality

holds with the choices of $\mathfrak{w}_{1} = \mathfrak{w}_{2} = \frac {4\mathfrak{\varsigma}\mathbb{\tau}}{3\mathfrak{\varsigma}+\mathbb{\tau}}$, $\breve{\varkappa}_{1} = \frac{2\mathfrak{\varsigma w}}{\mathfrak{\varsigma }+\mathfrak{w}}$, $\breve{\varkappa}_{2} = \frac{2\mathfrak{\varsigma }\mathbb{\tau}\mathfrak{w}}{2\mathfrak{\varsigma w}+\mathbb{\tau} \mathfrak{w}-\mathfrak{\varsigma}\mathbb{\tau}}$.

The inequality

holds with the choices of $\mathfrak{w}_{1} = \mathfrak{w}_{2} = \frac {4\mathfrak{\varsigma}\mathbb{\tau}}{\mathfrak{\varsigma}+3\mathbb{\tau}}$, $\breve{\varkappa}_{1} = \frac{2\mathbb{\tau}\mathfrak{w}}{\mathbb{\tau }+\mathfrak{w}}$, $\breve{\varkappa}_{2} = \frac{2\mathfrak{\varsigma }\mathbb{\tau}\mathfrak{w}}{\mathfrak{\varsigma w}+2\mathbb{\tau} \mathfrak{w}-\mathfrak{\varsigma}\mathbb{\tau}}$.

The inequality

holds with the choices of $\mathfrak{w}_{1} = \frac{2\mathfrak{\varsigma w} }{\mathfrak{\varsigma}+\mathfrak{w}}$, $\mathfrak{w}_{2} = \frac {2\mathfrak{\varsigma}\mathbb{\tau}\mathfrak{w}}{2\mathfrak{\varsigma w}+\mathbb{\tau}\mathfrak{w}-\mathfrak{\varsigma}\mathbb{\tau}}$, $\breve{\varkappa}_{1} = \mathfrak{\varsigma}$, $\breve{\varkappa}_{2} = \frac{2\mathbb{\tau}\mathfrak{\varsigma}}{\mathfrak{\varsigma}+\mathbb{\tau} }$.

The inequality

holds with the choices of $\mathfrak{w}_{1} = \frac{2\mathbb{\tau}\mathfrak{w} }{\mathbb{\tau}+\mathfrak{w}}$, $\mathfrak{w}_{2} = \frac{2\mathfrak{\varsigma }\mathbb{\tau}\mathfrak{w}}{\mathfrak{\varsigma w}+2\mathbb{\tau} \mathfrak{w}-\mathfrak{\varsigma}\mathbb{\tau}}$, $\breve{\varkappa}_{1} = \frac{2\mathbb{\tau}\mathfrak{\varsigma}}{\mathfrak{\varsigma}+\mathbb{\tau} }$, $\breve{\varkappa}_{2} = \mathbb{\tau}$.

Multiplying (2.54)–(2.58) by $\mathbb{\vartheta}\left(\mathfrak{w}\right)$, integrating them over $\left[ \mathfrak{\varsigma}, \frac{2\mathfrak{\varsigma }\mathbb{\tau}}{\mathfrak{\varsigma}+\mathbb{\tau}}\right]$ and using (2.50)–(2.53), we get (2.46).

(iii) By integration by parts, we get

Using the convexity of $\breve{h}$ and the hypothesis of $\mathbb{\vartheta}$, the inequality holds for all $\mathfrak{\iota}\in\left[ 0, 1\right]$ and $\mathfrak{w}\in\left[ \frac{1}{\mathbb{\tau}}, \frac{\mathfrak{\varsigma }+\mathbb{\tau}}{2\mathfrak{\varsigma}\mathbb{\tau}}\right]$:

Integrating (2.60), using (2.59) and (2.17), we get (2.47).

Corollary 2. According to the assumptions of Theorem 18 with $\mathbb{\vartheta}\left(\mathfrak{w}\right) = \frac{\mathfrak{\varsigma}\mathbb{\tau}}{\mathbb{\tau}-\mathfrak{\varsigma}}$, $\mathfrak{w}\in\left[\mathfrak{\varsigma}, \mathbb{\tau}\right]$, then

(i) The inequality

holds for all $\mathfrak{\iota}\in\left[ 0, 1\right]$.

(ii) The inequalities

hold.

(iii) The inequality

holds for all $\mathfrak{\iota}\in\left[ 0, 1\right]$.

Theorem 19. Let $\mathbb{\zeta}$, $\mathbb{\vartheta}$, $\mathcal{\breve{G}}_{1}$, $\mathbb{S}_{\mathbb{\vartheta}}$, $\mathbb{T}_{\mathbb{\vartheta}}$ bedefined as above. Then

(i) $\mathbb{T}_{\mathbb{\vartheta}}$ is harmonic convex on $\left(0, 1\right]$.

(ii) The inequalities

and

hold for all $\mathfrak{\iota}\in\left[ 0, 1\right]$.

(iii) The following bound is true:

Proof. (i) Since $\mathbb{\zeta}$ is harmonic convex and $\mathbb{\vartheta}$ is nonnegative, we see that $\mathbb{T}_{\mathbb{\vartheta}}$ is harmonic convex on $\left(0, 1\right]$.

(ii) We observe that the following identity holds on $\left[ 0, 1\right]$:

By Lemma 2, the following inequalities hold for all $\mathfrak{w} \in\left[ \mathfrak{\varsigma}, \frac{2\mathfrak{\varsigma}\mathbb{\tau} }{\mathfrak{\varsigma}+\mathbb{\tau}}\right]$:

with

with

Multiplying the inequalities (2.68) and (2.69) by $\mathbb{\vartheta }\left(\mathfrak{w}\right)$, integrating them over $\mathfrak{w}$ on $\left[ \mathfrak{\varsigma}, \frac{2\mathfrak{\varsigma}\mathbb{\tau} }{\mathfrak{\varsigma}+\mathbb{\tau}}\right]$ and using identities (2.48) and (2.67), we derive the first inequality of (2.63). Using the harmonic convexity of $\mathbb{\zeta}$ and the inequality (2.17), the last part of (2.63) holds. Using again the harmonic convexity of $\mathbb{\zeta}$, we get

From (2.45), (2.63) and 2.70), we get (2.65).

(iii) (2.66) holds due to the inequality (2.63).

3.   Conclusions

The subject of mathematical inequalities using convex functions has been seen to be an emerging topic during the past more than three decades. The researchers are trying to find new generalizations of convex functions and as a result new results are being adding to the theory of inequalities. In the current research we have used harmonic convex functions to generalize a number of results that hold for convex functions. In order to get the novel results in this study, we defined some new mappings over the interval $[0, 1]$. We have discussed some interesting properties of these mappings and obtained new refinements of the Hermite-hadamard and Fejér type inequalities already proven for harmonic convex functions. We believe that the results of this paper could be a source of inspiration for mathematicians working in this field and young researchers thinking to start their career in this fascinating field of mathematics.

Acknowledgments

The author is very thankful to all the anonymous referees for their very useful and constructive comments in order to present the paper in the present form. This work is supported by the Deanship of Scientific Research, King Faisal University under the Nasher Track 2021 (Research Project Number NA000177) which has been converted to Ambitious Researcher Track (Research Project Number GRANT931).

Conflict of interest

The authors declare that there are no conflicts of interest regarding the publication of this article.