In this section, we discuss our main results. We write \(\Vert g \Vert _{\infty}=\sup_{t\in[a,b]} \vert g(t) \vert \) for a continuous function \(g:[a,b]\to\mathbb{R}\).
Lemma 3.1
If
\(g:[a,b]\subseteq\mathbb{R}\setminus \lbrace0 \rbrace\to \mathbb{R}\)
is integrable and harmonically symmetric with respect to
\(\frac{2ab}{a+b}\), then
$$\begin{aligned} \bigl( \varepsilon_{\mu, \nu, l, \omega, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}g\circ h \bigr) \biggl( \frac{1}{a} \biggr) &= \bigl( \varepsilon_{\mu, \nu, l, \omega, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}g\circ h \bigr) \biggl( \frac{1}{b} \biggr) \\ &=\frac{1}{2} \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega, (\frac {a+b}{2ab} )^{+}}^{\gamma, \delta, k}g \circ h \bigr) \biggl( \frac {1}{a} \biggr)+ \bigl( \varepsilon_{\mu, \nu, l, \omega, (\frac {a+b}{2ab} )^{-}}^{\gamma, \delta, k}g \circ h \bigr) \biggl( \frac {1}{b} \biggr) \biggr] , \end{aligned}$$
where
\(h(x)=\frac{1}{x}\), \(x\in [\frac{1}{b}, \frac{1}{a} ] \).
Proof
Since g is harmonically symmetric with respect to \(\frac{2ab}{b-a}\), using Definition 2.2, we have \(g (\frac{1}{t} ) =g ( \frac{1}{\frac{1}{a}+\frac{1}{b}-t} )\) for all \(t\in [\frac {1}{b}, \frac{1}{a} ] \). Hence, in the following integral, setting \(u=\frac{1}{a}+\frac{1}{b}-t\) and \(\mathrm{d}u=-\mathrm{d}t\) gives
$$\begin{aligned} & \bigl( \varepsilon_{\mu, \nu, l, \omega, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}g\circ h \bigr) \biggl( \frac{1}{a} \biggr) \\ &\quad = \int_{\frac{a+b}{2ab}}^{\frac{1}{a}} \biggl(\frac{1}{a}-t \biggr)^{\nu -1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega \biggl( \frac {1}{a}-t \biggr) ^{\mu} \biggr)g \biggl(\frac{1}{t} \biggr)\,\mathrm{d}t \\ &\quad = \int_{\frac{a+b}{2ab}}^{\frac{1}{a}} \biggl(\frac{1}{a}-t \biggr)^{\nu -1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega \biggl( \frac {1}{a}-t \biggr) ^{\mu} \biggr)g \biggl( \frac{1}{\frac{1}{a}+\frac {1}{b}-t} \biggr)\,\mathrm{d}t \\ &\quad = \int_{\frac{1}{b}}^{\frac{a+b}{2ab}} \biggl(u- \frac{1}{b} \biggr)^{\nu -1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega \biggl(u- \frac {1}{b} \biggr) ^{\mu} \biggr)g \biggl( \frac{1}{u} \biggr)\,\mathrm{d}u= \bigl( \varepsilon_{\mu, \nu, l, \omega, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}g\circ h \bigr) \biggl( \frac{1}{b} \biggr) . \end{aligned}$$
This completes the proof. □
Lemma 3.2
Let
\(f:I\subset(0, +\infty)\to\mathbb{R}\)
be a differentiable function on
\(I^{\circ}\), the interior of
I, such that
\(f^{\prime}\in L[a,b]\), where
\(a,b \in I\). If
\(g:[a,b]\to\mathbb{R}\)
is integrable and harmonically symmetric with respect to
\(\frac{2ab}{a+b}\), then the following equality holds:
$$\begin{aligned} &f \biggl(\frac{2ab}{a+b} \biggr) \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}g\circ h \bigr) \biggl( \frac{1}{a} \biggr)+ \bigl( \varepsilon_{\mu, \nu , l, \omega^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}g\circ h \bigr) \biggl( \frac{1}{b} \biggr) \biggr] \\ &\qquad{}- \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}fg\circ h \bigr) \biggl( \frac{1}{a} \biggr)+ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}fg \circ h \bigr) \biggl( \frac{1}{b} \biggr) \biggr] \\ &\quad = \int_{\frac{1}{b}}^{\frac{a+b}{2ab}} \biggl( \int_{\frac{1}{b}}^{t} \biggl(s-\frac{1}{b} \biggr) ^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime} \biggl(s- \frac{1}{b} \biggr) ^{\mu} \biggr) (g\circ h ) (s) \,\mathrm{d}s \biggr) (f\circ h ) ^{\prime }(t)\,\mathrm{d}t \\ &\qquad{}- \int_{\frac{a+b}{2ab}}^{\frac{1}{a}} \biggl( \int_{t}^{\frac {1}{a}} \biggl(\frac{1}{a}-s \biggr) ^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime} \biggl(\frac{1}{a}-s \biggr) ^{\mu} \biggr) (g\circ h ) (s) \,\mathrm{d}s \biggr) (f\circ h ) ^{\prime}(t)\,\mathrm{d}t, \end{aligned}$$
(3.1)
where
\(h(x)=\frac{1}{x}\), \(x\in [\frac{1}{b}, \frac{1}{a} ]\), and
\(\omega^{\prime}= ( \frac{2ab}{b-a} ) ^{\mu}\omega\).
Proof
It suffices to show that
$$\begin{aligned} I&= \int_{\frac{1}{b}}^{\frac{a+b}{2ab}} \biggl( \int_{\frac{1}{b}}^{t} \biggl(s-\frac{1}{b} \biggr) ^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime} \biggl(s- \frac{1}{b} \biggr) ^{\mu} \biggr) (g\circ h ) (s) \,\mathrm{d}s \biggr) (f\circ h ) ^{\prime }(t)\,\mathrm{d}t \\ &\quad{}- \int_{\frac{a+b}{2ab}}^{\frac{1}{a}} \biggl( \int_{t}^{\frac {1}{a}} \biggl(\frac{1}{a}-s \biggr) ^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime} \biggl(\frac{1}{a}-s \biggr) ^{\mu} \biggr) (g\circ h ) (s) \,\mathrm{d}s \biggr) (f\circ h ) ^{\prime}(t)\,\mathrm{d}t \\ &=I_{1}-I_{2}. \end{aligned}$$
(3.2)
By Lemma 3.1, integrating by parts, we have
$$\begin{aligned} I_{1}&= \int_{\frac{1}{b}}^{t} \biggl(s-\frac{1}{b} \biggr) ^{\nu -1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime} \biggl(s- \frac{1}{b} \biggr) ^{\mu} \biggr) (g\circ h ) (s) \,\mathrm{d}s \cdot (f\circ h ) (t) \bigg\vert _{\frac{1}{b}}^{\frac {a+b}{2ab}} \\ &\quad{}- \int_{\frac{1}{b}}^{\frac{a+b}{2ab}} \biggl(t-\frac{1}{b} \biggr) ^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime} \biggl(t- \frac{1}{b} \biggr) ^{\mu} \biggr) (g\circ h ) (t) (f \circ h ) (t)\,\mathrm{d}t \\ &=f \biggl(\frac{2ab}{a+b} \biggr) \int_{\frac{1}{b}}^{\frac {2ab}{a+b}} \biggl(s-\frac{1}{b} \biggr) ^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime} \biggl(s- \frac{1}{b} \biggr) ^{\mu} \biggr) (g\circ h ) (s) \,\mathrm{d}s \\ &\quad{}- \int_{\frac{1}{b}}^{\frac{a+b}{2ab}} \biggl(t-\frac{1}{b} \biggr) ^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime} \biggl(t- \frac{1}{b} \biggr) ^{\mu} \biggr) (fg\circ h ) (t) \,\mathrm{d}t. \end{aligned}$$
(3.3)
Thus
$$\begin{aligned} I_{1}&=f \biggl(\frac{2ab}{a+b} \biggr) \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}g\circ h \bigr) \biggl( \frac{1}{b} \biggr)- \bigl( \varepsilon_{\mu, \nu , l, \omega^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}fg\circ h \bigr) \biggl( \frac{1}{b} \biggr) \\ &=\frac{1}{2}f \biggl(\frac{2ab}{a+b} \biggr) \biggl[ \bigl( \varepsilon_{\mu , \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}g\circ h \bigr) \biggl( \frac{1}{a} \biggr)+ \bigl( \varepsilon _{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}g \circ h \bigr) \biggl( \frac{1}{b} \biggr) \biggr] \\ &\quad{}- \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac {a+b}{2ab} )^{-}}^{\gamma, \delta, k}fg\circ h \bigr) \biggl( \frac {1}{b} \biggr). \end{aligned}$$
(3.4)
Analogously,
$$\begin{aligned} I_{2}&= \int_{t}^{\frac{1}{a}} \biggl(\frac{1}{a}-s \biggr) ^{\nu -1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime} \biggl(\frac {1}{a}-s \biggr) ^{\mu} \biggr) (g\circ h ) (s) \,\mathrm{d}s \cdot (f\circ h ) (t) \bigg\vert _{\frac{a+b}{2ab}}^{\frac {1}{a}} \\ &\quad{}- \int_{\frac{a+b}{2ab}}^{\frac{1}{a}} \biggl(\frac{1}{a}-t \biggr) ^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime} \biggl(\frac{1}{a}-t \biggr) ^{\mu} \biggr) (g\circ h ) (t) (f \circ h ) (t)\,\mathrm{d}t \\ &=-f \biggl(\frac{2ab}{a+b} \biggr) \int_{\frac{a+b}{2ab}}^{\frac{1}{a}} \biggl(\frac{1}{a}-s \biggr) ^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime} \biggl(\frac{1}{a}-s \biggr) ^{\mu} \biggr) (g\circ h ) (s) \,\mathrm{d}s \\ &\quad{}+ \int_{\frac{a+b}{2ab}}^{\frac{1}{a}} \biggl(\frac{1}{a}-t \biggr) ^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime} \biggl(\frac{1}{a}-t \biggr) ^{\mu} \biggr) (fg\circ h ) (t) \,\mathrm{d}t. \end{aligned}$$
(3.5)
Also,
$$\begin{aligned} I_{2}&=-f \biggl(\frac{2ab}{a+b} \biggr) \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}g\circ h \bigr) \biggl( \frac{1}{a} \biggr)+ \bigl( \varepsilon_{\mu, \nu , l, \omega^{\prime}, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}fg\circ h \bigr) \biggl( \frac{1}{a} \biggr) \\ &=-\frac{1}{2}f \biggl(\frac{2ab}{a+b} \biggr) \biggl[ \bigl( \varepsilon _{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}g\circ h \bigr) \biggl( \frac{1}{a} \biggr)+ \bigl( \varepsilon _{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}g \circ h \bigr) \biggl( \frac{1}{b} \biggr) \biggr] \\ &\quad{}+ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac {a+b}{2ab} )^{+}}^{\gamma, \delta, k}fg\circ h \bigr) \biggl( \frac {1}{a} \biggr). \end{aligned}$$
(3.6)
Inserting (3.4) and (3.6) into (3.2), we get (3.1), and the proof is complete. □
Remark
Taking \(\omega=0\) in Lemmas 3.1 and 3.2, we have Lemmas 2 and 3 from [7].
The next result is an Hermite-Hadamard-type inequality for a generalized fractional integral operator containing the generalized Mittag-Leffler function.
Theorem 3.3
Let
\(f:I\subset(0, +\infty)\to\mathbb{R}\)
be a function such that
\(f\in L[a,b]\), where
\(a,b \in I\). If
f
is a harmonically convex function on
\([a,b]\), then the following inequality for fractional integrals holds:
$$ \begin{aligned}[b]&f \biggl(\frac{2ab}{a+b} \biggr) \bigl( \varepsilon_{\mu, \nu, l, \omega ^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}1 \bigr) \biggl(\frac{1}{b} \biggr) \\ &\quad \leq\frac{1}{2} \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}f \circ h \bigr) \biggl(\frac{1}{b} \biggr)+ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}f \circ h \bigr) \biggl(\frac{1}{a} \biggr) \biggr] \\ &\quad \leq\frac{f(a)+f(b)}{2} \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime }, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}1 \bigr) \biggl(\frac {1}{a} \biggr), \end{aligned} $$
(3.7)
where
\(h(x)=\frac{1}{x}\), \(x\in [\frac{1}{b}, \frac{1}{a} ]\), and
\(\omega^{\prime}= ( \frac{2ab}{b-a} ) ^{\mu}\omega\).
Proof
Since f is a harmonically convex function on \([a,b]\), we have
$$ f \biggl(\frac{2xy}{x+y} \biggr) \leq\frac{f(x)+f(y)}{2} $$
(3.8)
for \(x,y\in[a,b]\). Substituting \(x=\frac{ab}{\frac{t}{2}a+\frac{2-t}{2}b}\) and \(y=\frac {ab}{\frac{t}{2}b+\frac{2-t}{2}a} \) into inequality (3.8), we get
$$ f \biggl(\frac{2ab}{a+b} \biggr) \leq\frac{f (\frac{ab}{\frac {t}{2}a+\frac{2-t}{2}b} ) +f (\frac{ab}{\frac{t}{2}b+\frac {2-t}{2}a} ) }{2}. $$
(3.9)
Multiplying both sides of (3.9) by \(t^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} ( \omega t^{\mu} )\) and integrating over \([0,1]\), we get
$$\begin{aligned} &2f \biggl(\frac{2ab}{a+b} \biggr) \int_{0}^{1}t^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \bigl( \omega t^{\mu} \bigr)\,\mathrm{d}t \\ &\quad \leq \int_{0}^{1}t^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \bigl( \omega t^{\mu} \bigr)f \biggl(\frac{ab}{\frac{t}{2}a+\frac {2-t}{2}b} \biggr) \,\mathrm{d}t+ \int_{0}^{1}t^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \bigl( \omega t^{\mu} \bigr)f \biggl(\frac {ab}{\frac{t}{2}b+\frac{2-t}{2}a} \biggr) \,\mathrm{d}t . \end{aligned}$$
Setting
$$ u=\frac{1}{ab} \biggl(\frac{t}{2}a+ \frac{2-t}{2}b \biggr),\qquad v=\frac {1}{ab} \biggl(\frac{t}{2}b+ \frac{2-t}{2}a \biggr), $$
(3.10)
we have
$$\begin{aligned} &2f \biggl(\frac{2ab}{a+b} \biggr) \int_{\frac{1}{b}}^{\frac {a+b}{2ab}} \biggl(v-\frac{1}{b} \biggr) ^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime} \biggl(v-\frac{1}{b} \biggr) ^{\mu} \biggr)\,\mathrm{d}v \\ &\quad \leq \int_{\frac{a+b}{2ab}}^{\frac{1}{a}} \biggl(\frac{1}{a}-u \biggr) ^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime} \biggl(\frac{1}{a}-u \biggr) ^{\mu} \biggr) (f\circ h ) (u)\,\mathrm {d}u \\ &\qquad{}+ \int_{\frac{1}{b}}^{\frac{a+b}{2ab}} \biggl(v-\frac{1}{b} \biggr) ^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime} \biggl(v-\frac{1}{b} \biggr) ^{\mu} \biggr) (f\circ h ) (v)\,\mathrm {d}v. \end{aligned}$$
This implies
$$ \begin{aligned}[b] &f \biggl(\frac{2ab}{a+b} \biggr) \bigl( \varepsilon_{\mu, \nu, l, \omega ^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}1 \bigr) \biggl(\frac{1}{b} \biggr) \\ &\quad \leq\frac{1}{2} \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}f \circ h \bigr) \biggl(\frac{1}{b} \biggr)+ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}f \circ h \bigr) \biggl(\frac{1}{a} \biggr) \biggr]. \end{aligned} $$
(3.11)
On the other hand, the harmonic convexity of f yields
$$ f \biggl(\frac{ab}{\frac{t}{2}a+\frac{2-t}{2}b} \biggr) +f \biggl(\frac {ab}{\frac{t}{2}b+\frac{2-t}{2}a} \biggr)\leq f(a)+f(b) $$
(3.12)
for all \(t\in[0,1]\). Multiplying both sides of (3.12) by \(t^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} ( \omega t^{\mu} )\) and integrating over \([0,1]\), we have
$$\begin{aligned} & \int_{0}^{1}t^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \bigl( \omega t^{\mu} \bigr)f \biggl(\frac{ab}{\frac{t}{2}a+\frac{2-t}{2}b} \biggr) \,\mathrm{d}t+ \int_{0}^{1}t^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \bigl( \omega t^{\mu} \bigr)f \biggl(\frac{ab}{\frac{t}{2}b+\frac {2-t}{2}a} \biggr) \,\mathrm{d}t \\ &\quad \leq \bigl(f(a)+f(b) \bigr) \int_{0}^{1}t^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \bigl( \omega t^{\mu} \bigr)\,\mathrm{d}t . \end{aligned}$$
This implies, using substitutions (3.10),
$$\begin{aligned} & \int_{\frac{a+b}{2ab}}^{\frac{1}{a}} \biggl(\frac{1}{a}-u \biggr) ^{\nu -1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime} \biggl(\frac {1}{a}-u \biggr) ^{\mu} \biggr) (f\circ h ) (u) \,\mathrm{d}u \\ &\qquad{}+ \int_{\frac{1}{b}}^{\frac{a+b}{2ab}} \biggl(v-\frac{1}{b} \biggr) ^{\nu -1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime} \biggl(v-\frac{1}{b} \biggr) ^{\mu} \biggr) (f\circ h ) (v)\,\mathrm {d}v. \\ &\quad \leq \bigl(f(a)+f(b) \bigr) \int_{\frac{a+b}{2ab}}^{\frac{1}{a}} \biggl(\frac{1}{a}-u \biggr) ^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime} \biggl(\frac{1}{a}-u \biggr) ^{\mu} \biggr) (f\circ h ) (u) \,\mathrm{d}u . \end{aligned}$$
So
$$\begin{aligned} &\frac{1}{2} \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}f \circ h \bigr) \biggl(\frac {1}{b} \biggr)+ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}f \circ h \bigr) \biggl(\frac {1}{a} \biggr) \biggr] \\ &\quad \leq\frac{f(a)+f(b)}{2} \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime }, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}1 \bigr) \biggl(\frac {1}{a} \biggr). \end{aligned}$$
(3.13)
Combining (3.11) and (3.13), we get (3.7). □
Remark
In Theorem 3.3, if we take \(\omega=0\), then we get the known inequality of Işan et al. [7]
$$\begin{aligned} &f \biggl(\frac{2ab}{a+b} \biggr)\leq\frac{\Gamma(\nu+1)}{2^{1-\nu }} \biggl[J_{\frac{a+b}{2ab}+}(f \circ h) \biggl(\frac{1}{a} \biggr)+J_{\frac {a+b}{2ab}-}(f\circ h) \biggl( \frac{1}{b} \biggr) \biggr] \leq\frac {f(a)+f(b)}{2}. \end{aligned}$$
Theorem 3.4
Let
\(f:I\subset(0, +\infty)\to\mathbb{R}\)
be a harmonically convex functsuch that
\(f\in L[a,b]\). If
\(g:[a,b]\to\mathbb{R}\)
is nonnegative, integrable, and harmonically symmetric with respect to
\(\frac {2ab}{a+b}\), then the following inequality holds:
$$\begin{aligned} &f \biggl(\frac{2ab}{a+b} \biggr) \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}g\circ h \bigr) \biggl( \frac{1}{a} \biggr)+ \bigl( \varepsilon_{\mu, \nu , l, \omega^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}g\circ h \bigr) \biggl( \frac{1}{b} \biggr) \biggr] \\ &\quad \leq\frac{1}{2} \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime }, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}fg \circ h \bigr) \biggl( \frac{1}{a} \biggr)+ \bigl( \varepsilon_{\mu, \nu, l, \omega ^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}fg \circ h \bigr) \biggl( \frac{1}{b} \biggr) \biggr] \\ &\quad \leq\frac{f(a)+f(b)}{2} \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega ^{\prime}, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}g \circ h \bigr) \biggl( \frac{1}{a} \biggr)+ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}g \circ h \bigr) \biggl( \frac{1}{b} \biggr) \biggr], \end{aligned}$$
(3.14)
where
\(h(x)=\frac{1}{x}\), \(x\in [\frac{1}{b}, \frac{1}{a} ] \), and
\(\omega^{\prime}= ( \frac{2ab}{b-a} ) ^{\mu}\omega\).
Proof
Since f ia a harmonically convex function on \([a,b]\), multiplying both sides of (3.9) by \(2t^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} ( \omega t^{\mu} )g (\frac {ab}{\frac{t}{2}a+\frac{2-t}{2}b} )\) and then integrating the resulting inequality over \([0,1]\), we obtain
$$\begin{aligned} &2f \biggl(\frac{2ab}{a+b} \biggr) \int_{0}^{1}t^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \bigl( \omega t^{\mu} \bigr)g \biggl(\frac {ab}{\frac{t}{2}a+\frac{2-t}{2}b} \biggr) \,\mathrm{d}t \\ &\quad \leq \int_{0}^{1}t^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \bigl( \omega t^{\mu} \bigr)f \biggl(\frac{ab}{\frac{t}{2}a+\frac {2-t}{2}b} \biggr)g \biggl( \frac{ab}{\frac{t}{2}a+\frac{2-t}{2}b} \biggr)\,\mathrm{d}t \\ &\qquad{}+ \int_{0}^{1}t^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \bigl( \omega t^{\mu} \bigr)f \biggl(\frac{ab}{\frac{t}{2}b+\frac {2-t}{2}a} \biggr)g \biggl( \frac{ab}{\frac{t}{2}a+\frac{2-t}{2}b} \biggr)\,\mathrm{d}t . \end{aligned}$$
Since g is harmonically symmetric with respect to \(\frac{2ab}{a+b}\), using Definition 2.2, we have \(g ( \frac{1}{x} )=g (\frac{1}{\frac{1}{a}+\frac{1}{b}-x} ) \) for all \(x\in [\frac{1}{b}, \frac{1}{a} ] \). Setting \(x=\frac{1}{ab} (\frac{t}{2}a+\frac{2-t}{2}b ) \) gives
$$\begin{aligned} &2f \biggl(\frac{2ab}{a+b} \biggr) \int_{\frac{a+b}{2ab}}^{\frac {1}{a}} \biggl(\frac{1}{a}-x \biggr) ^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime} \biggl(\frac{1}{a}-x \biggr)^{\mu} \biggr)g \biggl( \frac{1}{x} \biggr)\,\mathrm{d}x \\ &\quad \leq\frac{1}{2} \biggl[ \int_{\frac{a+b}{2ab}}^{\frac{1}{a}} \biggl(\frac {1}{a}-x \biggr)^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega ^{\prime} \biggl(\frac{1}{a}-x \biggr)^{\mu} \biggr)f \biggl( \frac{1}{x} \biggr)g \biggl(\frac{1}{x} \biggr)\,\mathrm{d}x \\ &\qquad{}+ \int_{\frac{a+b}{2ab}}^{\frac{1}{a}} \biggl(\frac {1}{a}-x \biggr)^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega ^{\prime} \biggl(\frac{1}{a}-x \biggr)^{\mu} \biggr)f \biggl( \frac{1}{\frac {1}{a}+\frac{1}{b}-x} \biggr)g \biggl(\frac{1}{x} \biggr)\,\mathrm {d}x \biggr] . \end{aligned}$$
Using substitution \(u=\frac{1}{a}+\frac{1}{b}-x\), we have
$$\begin{aligned} &2f \biggl(\frac{2ab}{a+b} \biggr) \int_{\frac{a+b}{2ab}}^{\frac {1}{a}} \biggl(\frac{1}{a}-x \biggr) ^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime} \biggl(\frac{1}{a}-x \biggr)^{\mu} \biggr)g \biggl( \frac{1}{x} \biggr)\,\mathrm{d}x \\ &\quad \leq\frac{1}{2} \biggl[ \int_{\frac{a+b}{2ab}}^{\frac{1}{a}} \biggl(\frac {1}{a}-x \biggr)^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega ^{\prime} \biggl(\frac{1}{a}-x \biggr)^{\mu} \biggr)f \biggl( \frac{1}{x} \biggr)g \biggl(\frac{1}{x} \biggr)\,\mathrm{d}x \\ &\qquad{}+ \int_{\frac{1}{b}}^{\frac{a+b}{2ab}} \biggl(u-\frac {1}{b} \biggr)^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega ^{\prime} \biggl(u-\frac{1}{b} \biggr)^{\mu} \biggr)f \biggl( \frac{1}{u} \biggr)g \biggl(\frac{1}{u} \biggr)\,\mathrm{d}x \biggr]. \end{aligned}$$
Hence, using Lemma 3.1, we obtain
$$\begin{aligned} &f \biggl(\frac{2ab}{a+b} \biggr) \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}g\circ h \bigr) \biggl( \frac{1}{a} \biggr)+ \bigl( \varepsilon_{\mu, \nu, l, \omega, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}g\circ h \bigr) \biggl( \frac{1}{b} \biggr) \biggr] \\ &\quad \leq\frac{1}{2} \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime }, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}fg \circ h \bigr) \biggl( \frac{1}{a} \biggr)+ \bigl( \varepsilon_{\mu, \nu, l, \omega ^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}fg \circ h \bigr) \biggl( \frac{1}{b} \biggr) \biggr]. \end{aligned}$$
(3.15)
For the proof of the second inequality in (3.14), we first note that f is a harmonically convex function. Then, multiplying both sides of (3.12) by \(2t^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} ( \omega t^{\mu} )g (\frac{ab}{\frac{t}{2}a+\frac {2-t}{2}b} )\) and integrating the resulting inequality over \([0,1]\), we obtain
$$\begin{aligned} & \int_{0}^{1}t^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \bigl( \omega t^{\mu} \bigr)f \biggl(\frac{ab}{\frac{t}{2}a+\frac{2-t}{2}b} \biggr)g \biggl( \frac{ab}{\frac{t}{2}a+\frac{2-t}{2}b} \biggr)\,\mathrm{d}t \\ &\qquad{}+ \int_{0}^{1}t^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \bigl( \omega t^{\mu} \bigr)f \biggl(\frac{ab}{\frac{t}{2}b+\frac{2-t}{2}a} \biggr)g \biggl( \frac{ab}{\frac{t}{2}a+\frac{2-t}{2}b} \biggr)\,\mathrm{d}t \\ &\quad \leq \bigl(f(a)+f(b) \bigr) \int_{0}^{1}t^{\nu-1}E_{\mu, \nu, l}^{\gamma , \delta, k} \bigl( \omega t^{\mu} \bigr)g \biggl(\frac{ab}{\frac {t}{2}a+\frac{2-t}{2}b} \biggr) \,\mathrm{d}t. \end{aligned} $$
From this, using Lemma 3.1, we get
$$ \begin{aligned} &\frac{1}{2} \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}fg \circ h \bigr) \biggl( \frac{1}{a} \biggr)+ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}fg \circ h \bigr) \biggl( \frac{1}{b} \biggr) \biggr] \\ &\quad \leq\frac{f(a)+f(b)}{2} \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega ^{\prime}, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}g \circ h \bigr) \biggl( \frac{1}{a} \biggr)+ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}g \circ h \bigr) \biggl( \frac{1}{b} \biggr) \biggr]. \end{aligned} $$
(3.16)
From (3.15) and (3.16) we obtain (3.14). The proof is completed. □
Theorem 3.5
Let
\(f:I\subset(0, +\infty)\to\mathbb{R}\)
be a differentiable function such that
\(f^{\prime}\in L[a,b]\), where
\(a,b\in I\), \(a< b\). If
\(\vert f^{\prime} \vert \)
is a harmonically convex function and
\(g:[a,b]\to\mathbb{R}\)
is continuous and harmonically symmetric with respect to
\(\frac{2ab}{a+b}\), then the following inequality holds:
$$\begin{aligned} & \biggl\vert f \biggl(\frac{2ab}{a+b} \biggr) \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}g\circ h \bigr) \biggl( \frac{1}{a} \biggr)+ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta , k}g\circ h \bigr) \biggl( \frac{1}{b} \biggr) \biggr] \\ &\qquad{}- \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime }, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}fg\circ h \bigr) \biggl( \frac{1}{a} \biggr)+ \bigl( \varepsilon_{\mu, \nu, l, \omega ^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}fg \circ h \bigr) \biggl( \frac{1}{b} \biggr) \biggr] \biggr\vert \\ &\quad \leq \biggl\vert f \biggl(\frac{2ab}{a+b} \biggr) \biggr\vert \Vert g \Vert _{\infty} \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega ^{\prime}, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}1 \bigr) \biggl( \frac{1}{a} \biggr)+ \bigl( \varepsilon_{\mu, \nu, l, \omega ^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}1 \bigr) \biggl( \frac{1}{b} \biggr) \biggr] \\ &\qquad{}+S \Vert g \Vert _{\infty} \biggl( \frac{b-a}{2ab} \biggr) ^{\nu}\cdot\frac{\nu+4}{4\nu (\nu+1 ) } \bigl( \bigl\vert f(a) \bigr\vert + \bigl\vert f(b) \bigr\vert \bigr), \end{aligned}$$
(3.17)
where
\(h(x)=\frac{1}{x}\), \(x\in [\frac{1}{b}, \frac{1}{a} ] \), \(\omega^{\prime}= ( \frac{2ab}{b-a} ) ^{\mu}\omega\), and
\(\vert E_{\mu, \nu, l}^{\gamma, \delta, k} ( t ) \vert \leq S\).
Proof
From Lemma 3.2, relationships (3.3) and (3.5), and the property of the modulus we have
$$\begin{aligned} J&= \biggl\vert f \biggl(\frac{2ab}{a+b} \biggr) \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}g\circ h \bigr) \biggl( \frac{1}{a} \biggr)+ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta , k}g\circ h \bigr) \biggl( \frac{1}{b} \biggr) \biggr] \\ &\quad{}- \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime }, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}fg\circ h \bigr) \biggl( \frac{1}{a} \biggr)+ \bigl( \varepsilon_{\mu, \nu, l, \omega ^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}fg \circ h \bigr) \biggl( \frac{1}{b} \biggr) \biggr] \biggr\vert \\ &\leq \biggl\vert f \biggl(\frac{2ab}{a+b} \biggr) \biggr\vert \int_{\frac {1}{b}}^{\frac{2ab}{a+b}} \biggl(s-\frac{1}{b} \biggr) ^{\nu-1} \biggl\vert E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime} \biggl(s- \frac {1}{b} \biggr) ^{\mu} \biggr) \biggr\vert \bigl\vert (g\circ h ) (s) \bigr\vert \,\mathrm{d}s \\ &\quad{}+ \int_{\frac{1}{b}}^{\frac{a+b}{2ab}} \biggl(t-\frac{1}{b} \biggr) ^{\nu -1} \biggl\vert E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime } \biggl(t- \frac{1}{b} \biggr) ^{\mu} \biggr) \biggr\vert \bigl\vert (fg\circ h ) (t) \bigr\vert \,\mathrm{d}t \\ &\begin{aligned} &\quad{}+ \biggl\vert f \biggl(\frac{2ab}{a+b} \biggr) \biggr\vert \int_{\frac {a+b}{2ab}}^{\frac{1}{a}} \biggl(\frac{1}{a}-s \biggr) ^{\nu-1} \biggl\vert E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime} \biggl(\frac{1}{a}-s \biggr) ^{\mu} \biggr) \biggr\vert \bigl\vert (g\circ h ) (s) \bigr\vert \,\mathrm{d}s \\ &\quad{}+ \int_{\frac{a+b}{2ab}}^{\frac{1}{a}} \biggl(\frac{1}{a}-t \biggr) ^{\nu -1} \biggl\vert E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime } \biggl(\frac{1}{a}-t \biggr) ^{\mu} \biggr) \biggr\vert \bigl\vert (fg\circ h ) (t) \bigr\vert \,\mathrm{d}t. \end{aligned} \end{aligned}$$
Since \(\Vert g \Vert _{\infty}=\sup_{t\in[a,b]} \vert g(t) \vert \) and \(\vert E_{\mu, \nu, l}^{\gamma, \delta, k} ( t ) \vert \leq S\), we have
$$\begin{aligned} J&\leq \biggl\vert f \biggl(\frac{2ab}{a+b} \biggr) \biggr\vert \Vert g \Vert _{\infty} \int_{\frac{1}{b}}^{\frac{2ab}{a+b}} \biggl(s-\frac {1}{b} \biggr) ^{\nu-1} E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega ^{\prime} \biggl(s- \frac{1}{b} \biggr) ^{\mu} \biggr)\,\mathrm{d}s \\ &\quad{}+S \Vert g \Vert _{\infty} \int_{\frac{1}{b}}^{\frac {a+b}{2ab}} \biggl(t-\frac{1}{b} \biggr) ^{\nu-1} \bigl\vert (f\circ h ) (t) \bigr\vert \,\mathrm{d}t \\ &\quad{}+ \biggl\vert f \biggl(\frac{2ab}{a+b} \biggr) \biggr\vert \Vert g \Vert _{\infty} \int_{\frac{a+b}{2ab}}^{\frac{1}{a}} \biggl(\frac {1}{a}-s \biggr) ^{\nu-1} E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime} \biggl(\frac{1}{a}-s \biggr) ^{\mu} \biggr)\,\mathrm{d}s \\ &\quad{}+S \Vert g \Vert _{\infty} \int_{\frac{a+b}{2ab}}^{\frac {1}{a}} \biggl(\frac{1}{a}-t \biggr) ^{\nu-1} \bigl\vert (f\circ h ) (t) \bigr\vert \,\mathrm{d}t \\ &= \biggl\vert f \biggl(\frac{2ab}{a+b} \biggr) \biggr\vert \Vert g \Vert _{\infty} \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}g \circ h \bigr) \biggl( \frac{1}{a} \biggr)+ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}g \circ h \bigr) \biggl( \frac{1}{b} \biggr) \biggr] \\ &\quad{}+S \Vert g \Vert _{\infty} (J_{1}+J_{2} ). \end{aligned}$$
(3.18)
Setting \(t=\frac{1}{ab} (\frac{u}{2}a+\frac{2-u}{2}b ) \) and using the harmonic convexity of \(\vert f \vert \), we have
$$\begin{aligned} J_{1}&= \int_{\frac{1}{b}}^{\frac{a+b}{2ab}} \biggl(t-\frac{1}{b} \biggr) ^{\nu-1} \bigl\vert (f\circ h ) (t) \bigr\vert \,\mathrm{d}t= \int _{\frac{1}{b}}^{\frac{a+b}{2ab}} \biggl(t-\frac{1}{b} \biggr) ^{\nu -1} \biggl\vert f \biggl(\frac{1}{t} \biggr) \biggr\vert \,\mathrm{d}t \\ &=\frac{1}{2} \biggl( \frac{b-a}{ab} \biggr) ^{\nu} \int_{1}^{2} \biggl(1-\frac{1}{2}u \biggr) ^{\nu-1} \biggl\vert f \biggl(\frac{ab}{\frac {u}{2}a+\frac{2-u}{2}b} \biggr) \biggr\vert \,\mathrm{d}u \\ &\leq\frac{1}{2} \biggl( \frac{b-a}{ab} \biggr) ^{\nu} \int_{1}^{2} \biggl(1-\frac{1}{2}u \biggr) ^{\nu-1} \biggl( \frac{u}{2} \bigl\vert f(b) \bigr\vert + \frac{2-u}{2} \bigl\vert f(a) \bigr\vert \biggr)\,\mathrm{d}u. \end{aligned}$$
This implies
$$ J_{1}\leq\frac{1}{4} \biggl( \frac{b-a}{ab} \biggr) ^{\nu} \biggl(\frac {2\nu+4}{\nu(\nu+1)} \bigl\vert f(b) \bigr\vert -\frac{1}{\nu+1} \bigl\vert f(a) \bigr\vert \biggr). $$
(3.19)
Similarly, using the substitution \(t=\frac{1}{ab} (\frac {v}{2}b+\frac{2-v}{2}a ) \), we get
$$\begin{aligned} J_{2}&= \int_{\frac{a+b}{2ab}}^{\frac{1}{a}} \biggl(\frac{1}{a}-t \biggr) ^{\nu-1} \bigl\vert (f\circ h ) (t) \bigr\vert \,\mathrm{d}t= \int _{\frac{a+b}{2ab}}^{\frac{1}{a}} \biggl(\frac{1}{a}-t \biggr) ^{\nu -1} \biggl\vert f \biggl(\frac{1}{t} \biggr) \biggr\vert \,\mathrm{d}t \\ &\leq\frac{1}{4} \biggl( \frac{b-a}{ab} \biggr) ^{\nu} \biggl(\frac{2\nu +4}{\nu(\nu+1)} \bigl\vert f(a) \bigr\vert -\frac{1}{\nu+1} \bigl\vert f(b) \bigr\vert \biggr). \end{aligned}$$
(3.20)
Substituting (3.19) and (3.20) into (3.18), we obtain (3.17). The proof is completed. □
Theorem 3.6
Let
\(f:I\subset(0, +\infty)\to\mathbb{R}\)
be a differentiable function on
\(I^{\circ}\), the interior of
I, such that
\(f^{\vert}\in L[a,b]\), where
\(a,b\in I\). If
\(\vert f \vert ^{q}\), \(q\geq1\), is a harmonically convex function on
\([a,b]\), \(g:[a,b]\to\mathbb{R}\)
is continuous and harmonically symmetric with respect to
\(\frac {2ab}{a+b}\), then the following inequality holds:
$$\begin{aligned} & \biggl\vert f \biggl(\frac{2ab}{a+b} \biggr) \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}g\circ h \bigr) \biggl( \frac{1}{a} \biggr)+ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta , k}g\circ h \bigr) \biggl( \frac{1}{b} \biggr) \biggr] \\ &\qquad{}- \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime }, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}fg\circ h \bigr) \biggl( \frac{1}{a} \biggr)+ \bigl( \varepsilon_{\mu, \nu, l, \omega ^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}fg \circ h \bigr) \biggl( \frac{1}{b} \biggr) \biggr] \biggr\vert \\ &\quad \leq \biggl\vert f \biggl(\frac{2ab}{a+b} \biggr) \biggr\vert \Vert g \Vert _{\infty} \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega ^{\prime}, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}1 \bigr) \biggl( \frac{1}{a} \biggr)+ \bigl( \varepsilon_{\mu, \nu, l, \omega ^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}1 \bigr) \biggl( \frac{1}{b} \biggr) \biggr] \\ &\qquad{}+S \Vert g \Vert _{\infty} \biggl( \frac{1}{\nu} \biggr) ^{1-1/q} \biggl( \frac{1}{\nu+1 } \biggr) ^{1/q} \biggl( \frac{1}{2} \biggr) ^{\nu+1/q} \\ &\qquad{}\times \biggl[ \biggl( \frac{1}{\nu} \bigl\vert f(b) \bigr\vert ^{q} + \bigl\vert f(a) \bigr\vert ^{q} \biggr)^{1/q}+ \biggl(\frac{1}{\nu} \bigl\vert f(a) \bigr\vert ^{q} + \bigl\vert f(b) \bigr\vert ^{q} \biggr)^{1/q} \biggr] , \end{aligned}$$
(3.21)
where
\(h(x)=\frac{1}{x}\), \(x\in [\frac{1}{b}, \frac{1}{a} ] \), \(\omega^{\prime}= ( \frac{2ab}{b-a} ) ^{\mu}\omega\), and
\(\vert E_{\mu, \nu, l}^{\gamma, \delta, k} ( t ) \vert \leq S\).
Proof
From (3.18) we have
$$\begin{aligned} J&\leq \biggl\vert f \biggl(\frac{2ab}{a+b} \biggr) \biggr\vert \Vert g \Vert _{\infty} \int_{\frac{1}{b}}^{\frac{2ab}{a+b}} \biggl(s-\frac {1}{b} \biggr) ^{\nu-1} E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega ^{\prime} \biggl(s- \frac{1}{b} \biggr) ^{\mu} \biggr)\,\mathrm{d}s \\ &\quad{}+S \Vert g \Vert _{\infty} \int_{\frac{1}{b}}^{\frac {a+b}{2ab}} \biggl(t-\frac{1}{b} \biggr) ^{\nu-1} \bigl\vert (f\circ h ) (t) \bigr\vert \,\mathrm{d}t \\ &\quad{}+ \biggl\vert f \biggl(\frac{2ab}{a+b} \biggr) \biggr\vert \Vert g \Vert _{\infty} \int_{\frac{a+b}{2ab}}^{\frac{1}{a}} \biggl(\frac {1}{a}-s \biggr) ^{\nu-1} E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime} \biggl(\frac{1}{a}-s \biggr) ^{\mu} \biggr)\,\mathrm{d}s \\ &\quad{}+S \Vert g \Vert _{\infty} \int_{\frac{a+b}{2ab}}^{\frac {1}{a}} \biggl(\frac{1}{a}-t \biggr) ^{\nu-1} \bigl\vert (f\circ h ) (t) \bigr\vert \,\mathrm{d}t \\ &= \biggl\vert f \biggl(\frac{2ab}{a+b} \biggr) \biggr\vert \Vert g \Vert _{\infty} \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}g \circ h \bigr) \biggl( \frac{1}{a} \biggr)+ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}g \circ h \bigr) \biggl( \frac{1}{b} \biggr) \biggr] \\ &\quad{}+S \Vert g \Vert _{\infty} \biggl[ \int_{\frac{1}{b}}^{\frac {a+b}{2ab}} \biggl(t-\frac{1}{b} \biggr) ^{\nu-1} \biggl\vert f \biggl(\frac {1}{t} \biggr) \biggr\vert \,\mathrm{d}t+ \int_{\frac{a+b}{2ab}}^{\frac {1}{a}} \biggl(\frac{1}{a}-t \biggr) ^{\nu-1} \biggl\vert f \biggl(\frac {1}{t} \biggr) \biggr\vert \,\mathrm{d}t \biggr] . \end{aligned}$$
Using the substitutions \(t=\frac{1}{ab} (\frac{u}{2}a+\frac {2-u}{2}b )\) and \(t=\frac{1}{ab} (\frac{v}{2}b+\frac {2-v}{2}a ) \), the power mean inequality, and the harmonicity of \(\vert f \vert ^{q}\), it follows that
$$\begin{aligned} J&\leq \biggl\vert f \biggl(\frac{2ab}{a+b} \biggr) \biggr\vert \Vert g \Vert _{\infty} \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega ^{\prime}, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}g \circ h \bigr) \biggl( \frac{1}{a} \biggr)+ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}g \circ h \bigr) \biggl( \frac{1}{b} \biggr) \biggr] \\ &\quad{}+\frac{S \Vert g \Vert _{\infty}}{2} \biggl(\frac {b-a}{ab} \biggr) ^{\nu} \\ &\quad{}\times \biggl[ \int_{1}^{2} \biggl(1-\frac{1}{2}u \biggr) ^{\nu -1} \biggl\vert f \biggl(\frac{ab}{\frac{u}{2}a+\frac{2-u}{2}b } \biggr) \biggr\vert \,\mathrm{d}u+ \int_{1}^{2} \biggl(1-\frac{1}{2}v \biggr) ^{\nu -1} \biggl\vert f \biggl(\frac{ab}{\frac{v}{2}b+\frac{2-v}{2}a} \biggr) \biggr\vert \,\mathrm{d}v \biggr] \\ &\leq \biggl\vert f \biggl(\frac{2ab}{a+b} \biggr) \biggr\vert \Vert g \Vert _{\infty} \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega ^{\prime}, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}g \circ h \bigr) \biggl( \frac{1}{a} \biggr)+ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}g \circ h \bigr) \biggl( \frac{1}{b} \biggr) \biggr] \\ &\quad{}+\frac{S \Vert g \Vert _{\infty}}{2} \biggl(\frac {b-a}{ab} \biggr) ^{\nu} \biggl[ \biggl( \int_{1}^{2} \biggl(1-\frac{1}{2}u \biggr) ^{\nu -1}\,\mathrm{d}u \biggr) ^{1-1/q} \\ &\quad{}\times \biggl( \int_{1}^{2} \biggl(1-\frac{1}{2}u \biggr) ^{\nu-1} \biggl( \frac{u}{2} \bigl\vert f(b) \bigr\vert ^{q}+\frac {2-u}{2} \bigl\vert f(a) \bigr\vert ^{q} \biggr)\,\mathrm{d}u \biggr) ^{1/q} \\ &\quad{}+ \biggl( \int_{1}^{2} \biggl(1-\frac{1}{2}v \biggr) ^{\nu -1}\,\mathrm{d}v \biggr) ^{1-1/q} \biggl( \int_{1}^{2} \biggl(1-\frac{1}{2}v \biggr) ^{\nu-1} \biggl( \frac{v}{2} \bigl\vert f(a) \bigr\vert ^{q}+\frac {2-v}{2} \bigl\vert f(b) \bigr\vert ^{q} \biggr)\,\mathrm{d}v \biggr) ^{1/q} \biggr]. \end{aligned}$$
Making the substitutions \(x=1-\frac{1}{2}u\) and \(y=1-\frac{1}{2}v \) and simple calculations, we get inequality (3.21). □