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Hadamard k-fractional inequalities of Fejér type for GA-s-convex mappings and applications
Journal of Inequalities and Applications volume 2019, Article number: 264 (2019)
Abstract
The main aim of this paper is to establish some Fejér-type inequalities involving hypergeometric functions in terms of GA-s-convexity. For this purpose, we construct a Hadamard k-fractional identity related to geometrically symmetric mappings. Moreover, we give the upper and lower bounds for the weighted inequalities via products of two different mappings. Some applications of the presented results to special means are also provided.
1 Introduction
Let \(f:[\mu,\nu] \rightarrow\mathbb{R}\) be a convex mapping with \(\mu <\nu\), and let \(g:[\mu,\nu] \rightarrow\mathbb{R}\) be a non-negative, integrable and symmetric mapping corresponding to \(\frac{\mu+\nu}{2}\). Then one has
which is called a Fejér-type inequality.
If we take \(g(x)=1\) in (1.1), then inequality (1.1) reduces to the Hermite–Hadamard inequality,
To see more recent results and the related extensions corresponding to (1.1) and (1.2), we refer the interested reader to [1, 2, 6, 7, 10,11,12, 19, 21, 23,24,25, 28, 29, 34] and the references therein.
Let us recall that Niculescu [26] introduced and considered a class of mappings, called GA-convex mappings, as follows: A mapping \(f:\mathcal{I}\subseteq\mathbb{R}_{+}=(0,\infty)\rightarrow\mathbb{R}\) is said to be GA-convex on \(\mathcal{I}\) if
for all \(\mu,\nu\in\mathcal{I}\) and \(\lambda\in[0,1]\).
Using mappings whose first derivative in absolute value are GA-convex, Latif et al. [22] proved the following estimation-type result for the right-middle part of Fejér-type inequality (1.1).
Theorem 1.1
Let \(f:\mathcal{I}\subseteq\mathbb{R}_{+}=(0,\infty)\rightarrow\mathbb {R}\) be a differentiable mapping on \(\mathcal{I}^{\circ}\) and \(\mu,\nu \in\mathcal{I}^{\circ}\) with \(\mu<\nu\) satisfying that \(f'\in L^{1}([\mu,\nu])\), and let \(g:[\mu,\nu]\rightarrow[0,\infty)\) be a continuous positive mapping geometrically symmetric with respect to \(\sqrt{\mu\nu}\), i.e. \(g (\frac{\mu\nu}{x} )=g (x )\). If \(|f'|^{q}\) is GA-convex on \([\mu,\nu]\) for \(q>1\), then the following inequality holds:
where
and
for \(\rho,\varrho>0\) with \(\rho\neq\varrho\).
More integral inequalities considering GA-convex mappings can be found in [4, 5, 13, 16].
Motivated by the research going on this dynamic field, Shuang et al. [30] presented a new class of GA-convex mappings, which is named the GA-s-convex mapping. For the recent results and details, the interested reader is directed to [14, 17] and the references cited therein.
Definition 1.1
([30])
A mapping \(f: \mathcal{I}\subseteq\mathbb{R}_{+}=(0,\infty)\rightarrow \mathbb{R}\) is named GA-s-convex mapping on \(\mathcal{I}\) if
holds for all \(\mu,\nu\in\mathcal{I}\), \(\lambda\in[0,1]\) and for certain fixed \(s\in(0,1]\).
Fractional calculus, as a very useful tool, shows its significance to implement differentiation and integration of real or complex number orders. This topic has attracted much attention from researchers during the last few decades. Among a lot of the fractional integral operators that appeared, because of applications in many fields of sciences, the Riemann–Liouville fractional integral operator and Hadamard fractional integral operator have been extensively studied.
An important generalization of Hadamard fractional integrals was considered by Iqbal et al. in [15] which is named the Hadamard k-fractional integral operators.
Definition 1.2
Let \(f\in L^{1}[a,b]\), then the left-sided and right-sided Hadamard k-fractional integrals of order \(\alpha\in\mathbb{R}^{+}\) and \(k,a\in \mathbb{R}^{+}\) are defined as
and
respectively, where \(\varGamma_{k}(\alpha)\) is the k-gamma function defined by \(\varGamma_{k}(\alpha)=\int^{\infty}_{0}\tau^{\alpha-1}e^{-\frac {\tau^{k}}{k}}\,\mathrm{d}\tau\). Furthermore, \(\varGamma_{k}(\alpha+k)=\alpha\varGamma_{k}(\alpha)\).
Some important inequalities pertaining Hadamard k-fractional integrals can be found in [3, 27, 32].
The following theorem, involving Hadamard-type k-fractional integral operators, is a direct generalization of Theorem 2.6 established by Kunt et al. in [20].
Theorem 1.2
Let \(f:[\mu,\nu]\subseteq\mathbb{R}_{+}\rightarrow\mathbb{R}\) be a GA-convex mapping, \(\alpha>0\) and \(f\in L^{1}[\mu,\nu]\). If \(g:[\mu ,\nu]\rightarrow\mathbb{R}\) is non-negative, integral and geometrically symmetric corresponding to \(\sqrt{\mu\nu}\), then the following inequality for Hadamard-type k-fractional integral operators holds:
It is easy to observe that, for \(k=1\) in Definition 1.2, we have the definition of the left-sided and right-sided Hadamard fractional integrals, i.e.
and
For more details corresponding to the Hadamard fractional integral inequalities, the interested reader is directed to Refs. [8, 31, 33] and the references cited therein.
Consider the Hadamard fractional inequality of the Fejér type with respect to GA-convexity, Kunt [20] obtained the following theorem related to the right-middle part of inequality (1.1).
Theorem 1.3
Let \(f:\mathcal{I}\subseteq(0,\infty)\rightarrow\mathbb{R}\) be a differentiable mapping on \(\mathcal{I}^{\circ}\) satisfying that \(f\in L^{1}[\mu,\nu]\) where \(\mu,\nu\in\mathcal{I}\) with \(\mu<\nu\) and \(\alpha>0\). If \(|f'|\) is GA-convex on \([\mu,\nu]\) and \(g:[\mu,\nu]\rightarrow\mathbb {R}\) is continuous and geometrically symmetric corresponding to \(\sqrt {\mu\nu}\), then the following inequality via fractional integrals holds:
where
and
Different from [22], our purpose in this paper is to obtain, using the Hadamard k-fractional integrals, certain estimation-type results for the left-middle part of a Fejér-type inequality in terms of GA-s-convexity. We also get the upper and lower bounds for the weighted Hadamard-type inequalities via product of two different mappings.
2 Some preliminary lemmas
In this section, we state the following lemmas, which are useful in the proofs of our main results.
Lemma 2.1
For \(\mathcal{U},\mathcal{V}>0\), we have
-
(i)
$$ \varPhi(\mathcal{U},\mathcal{V}):= \int^{1}_{0}\mathcal{U}^{1-\frac{t}{2}} \mathcal{V}^{\frac{t}{2}}\,\mathrm{d}t= \textstyle\begin{cases} \frac{2 (\mathcal{U}-\sqrt{\mathcal{U}\mathcal{V}} )}{\ln {\mathcal{U}}-\ln{\mathcal{V}}}, & \mathcal{U}\neq\mathcal{V}, \\ \mathcal{U}, &\mathcal{U}=\mathcal{V}, \end{cases} $$(2.1)
-
(ii)
$$ \varPsi(\mathcal{U},\mathcal{V}):=\frac{1}{2} \int^{1}_{0}t\mathcal{U}^{1-\frac{t}{2}} \mathcal{V}^{\frac{t}{2}}\,\mathrm{d}t= \textstyle\begin{cases} \frac{\sqrt{\mathcal{U}} [2\sqrt{\mathcal{U}}+\sqrt {\mathcal{V}}(\ln{\mathcal{V}}-\ln{\mathcal{U}}-2) ]}{(\ln{\mathcal {U}}-\ln{\mathcal{V}})^{2}}, & \mathcal{U} \neq\mathcal{V}, \\ \frac{1}{4}\mathcal{U}, &\mathcal{U}=\mathcal{V}, \end{cases} $$(2.2)
-
(iii)
$$ \begin{aligned} & \int^{1}_{0}t^{\sigma}\mathcal{U}^{1-\frac{t}{2}} \mathcal{V}^{\frac{t}{2}}\,\mathrm{d}t \leq\frac{\sqrt{\mathcal {UV}}(\sigma+1)+\mathcal{U}}{(\sigma+1)(\sigma+2)} :=\varUpsilon( \mathcal{U},\mathcal{V},\sigma), \quad\sigma\neq-1, -2. \end{aligned} $$(2.3)
Proof
The proofs of (i) and (ii) follow from a straightforward computation.
The proof of (iii) is as follows.
Using the inequality of \(u^{s}\leq(u-1)s+1\) for all \(0\leq s \leq1\) with \(u>0\), we have
This ends the proof. □
Lemma 2.2
If \(w:[a,b]\subset(0,\infty)\rightarrow\mathbb{R}\) is integrable and geometrically symmetric corresponding to \(\sqrt{ab}\) with \(a< b\) and \(k, \alpha>0\), then we have
Proof
Using the geometrically symmetry of w with respect to \(\sqrt{ab}\), we have \(w (\frac{ab}{x} )=w(x)\), for all \(x\in[a,b]\). If we set \(x=\frac{ab}{t}\), then we have
□
Lemma 2.3
Let \(f:\mathcal{I}\subseteq\mathbb{R}^{+}=(0,\infty)\rightarrow\mathbb {R}\), be a differentiable mapping on \(\mathcal{I}^{\circ}\) (the interior of \(\mathcal{I}\)), \(a,b\in\mathcal{I}\) with \(a< b\), and let \(g:[a,b]\rightarrow\mathbb{R}\) be a continuous positive mapping geometrically symmetric to \(\sqrt{ab}\). If \(f'\in L^{1}([a,b])\), then the following equality for Hadamard k-fractional integral operators with \(k,\alpha>0\) holds:
where \(L(t)=a^{1-\frac{t}{2}}b^{\frac{t}{2}}\) and \(U(t)=a^{\frac {t}{2}}b^{1-\frac{t}{2}}\).
Proof
Let
and
Integrating by parts, we have
Since g is geometrically symmetric to \(\sqrt{ab}\), one has
By this, we have
Adding (2.9) and (2.10), we get the required identity in (2.6). This ends the proof. □
3 Inequalities involving hypergeometric functions
Our first main result is given by the following theorem. For this purpose, we note that the gamma function, the beta function and the incomplete beta function are defined by
and
The integral form of the hypergeometric function is
For the sake of simplicity, we also denote
unless otherwise specified.
Theorem 3.1
Let \(f:\mathcal{I}\subseteq\mathbb{R}^{+}=(0,\infty)\rightarrow\mathbb {R}\), be a differentiable mapping on \(\mathcal{I}^{\circ}\) (the interior of \(\mathcal{I}\)), \(a,b\in\mathcal{I}^{\circ}\) with \(a< b\), and let \(g:[a,b]\rightarrow \mathbb{R}\) be a continuous positive mapping geometrically symmetric to \(\sqrt{ab}\) such that \(f'\in L^{1}([a,b])\). If \(|f'|^{q}\) for \(q \geq1\) is GA-s-convex on \([a,b]\), then the following Hadamard k-fractional inequality with \(k,\alpha>0\) holds:
where
and \(\varUpsilon(\mathcal{U},\mathcal{V},\sigma)\) is defined by (2.3) in Lemma 2.1.
Proof
If we use Lemma 2.3 and Hölder’s inequality, then we have
Using (iii) of Lemma 2.1, we have
and
Considering GA-s-convexity of \(|f'|^{q}\), we have
Utilizing the inequality of \(u^{\theta}\leq(u-1)\theta+1\) for all \(0\leq \theta\leq1\) with \(u>0\), we have
Similarly,
Using (3.3)–(3.6) in (3.2), we get the required inequality in (3.1). This ends the proof. □
Corollary 3.1
If we take \(q=1\) in Theorem 3.1, then the following inequality holds:
where \(\mathbb{B}_{i}(k,\alpha,s)\), \(i=1,2,3,4\), are defined in Theorem 3.1.
Corollary 3.2
If we take \(s=1\) in Theorem 3.1, then the following inequality holds:
where
and
Corollary 3.3
If we take \(k=1\), \(\alpha=1\) and \(s=1\) in Theorem 3.1, then the following inequality holds:
Theorem 3.2
Let \(f:\mathcal{I}\subseteq\mathbb{R}^{+}=(0,\infty)\rightarrow\mathbb {R}\), be a differentiable mapping on \(\mathcal{I}^{\circ}\), \(a,b\in\mathcal{I}\) with \(a< b\), and let \(g:[a,b]\rightarrow\mathbb{R}\) be a continuous positive mapping geometrically symmetric to \(\sqrt{ab}\) such that \(f'\in L^{1}([a,b])\). If \(|f'|^{q}\) for \(q>1\) is GA-s-convex on \([a,b]\), then the following inequality for Hadamard k-fractional integral operators with \(k,\alpha>0\) holds:
where
and
Proof
Utilizing Lemma 2.3 and Hölder’s inequality, we have
Considering GA-s-convexity of \(|f'|^{q}\), and using the inequality of \(u^{\theta}\leq(u-1)\theta+1\) for all \(0\leq\theta\leq1\) with \(u>0\), we have
and
Also,
The inequality (3.7) is proved by using (3.9), (3.10) and (3.11) in inequality (3.8). This ends the proof. □
Corollary 3.4
If we take \(k=1\) and \(g(x)=1\) in Theorem 3.2, then the following inequality holds:
where \(\mathbb{C}_{i}(q,s)\), \(i=1,2,3,4\), are defined in Theorem 3.2.
Corollary 3.5
If we take \(k=1\), \(\alpha=1\) and \(s=1\) in Theorem 3.2, then the following inequality holds:
Theorem 3.3
Let \(f:\mathcal{I}\subseteq\mathbb{R}^{+}=(0,\infty)\rightarrow\mathbb {R}\), be a differentiable mapping on \(\mathcal{I}^{\circ}\), \(a,b\in\mathcal{I}^{\circ}\) with \(a< b\), and let \(g:[a,b]\rightarrow \mathbb{R}\) be a continuous positive mapping geometrically symmetric to \(\sqrt{ab}\) such that \(f'\in L^{1}([a,b])\). If \(|f'|^{q}\) for \(q>1\) is GA-s-convex on \([a,b]\), then the following inequality for Hadamard k-fractional integral operators with \(k,\alpha>0\) holds:
where \(\mathbb{C}_{i}(q,s)\), \(i=1,2,3,4\) are defined in Theorem 3.2.
Proof
If we use the inequality \(\mu^{r}+\nu^{r}\leq2^{1-r}(\mu+\nu)^{r}\) for \(\mu >0\), \(\nu>0\) and \(r<1\), then we have
Using inequalities (3.9) and (3.10) in (3.13), we have
Applying (3.14) and (3.11) to inequality (3.8) in the proof of Theorem 3.2, we obtain the required inequality in (3.12). This ends the proof. □
Corollary 3.6
If we take \(k=1\) and \(g(x)=1\) in Theorem 3.2, then the following inequality holds:
where \(\mathbb{C}_{i}(q,s)\), \(i=1,2,3,4\) are defined in Theorem 3.2.
Corollary 3.7
If we take \(k=1\), \(\alpha=1\) and \(s=1\) in Theorem 3.2, then the following inequality holds:
Theorem 3.4
Let \(f:\mathcal{I}\subseteq\mathbb{R}^{+}=(0,\infty)\rightarrow\mathbb {R}\), be a differentiable mapping on \(\mathcal{I}^{\circ}\), \(a,b\in\mathcal{I}\) with \(a< b\), and let \(g:[a,b]\rightarrow\mathbb{R}\) be a continuous positive mapping geometrically symmetric to \(\sqrt{ab}\) such that \(f'\in L^{1}([a,b])\). If \(|f'|\) is GA-s-convex on \([a,b]\), then for \(q>1\) the following Hadamard k-fractional inequality with \(k,\alpha>0\) holds:
where
and
Proof
From Lemma 2.3 and the GA-s-convexity of \(|f'|\) on \([a,b]\), we have
Using Hölder’s integral inequality, we have
Similarly, one has
Using (3.17) and (3.18) in (3.16), we obtain the required inequality (3.15). This ends the proof. □
Corollary 3.8
If we take \(k=1\), \(s=1\) and \(g(x)=1\) in Theorem 3.4, then the following inequality holds:
Specially, taking \(\alpha=1\), we have
Theorem 3.5
Let \(f:\mathcal{I}\subseteq\mathbb{R}^{+}=(0,\infty)\rightarrow\mathbb {R}\), be a differentiable mapping on \(\mathcal{I}^{\circ}\), \(a,b\in\mathcal{I}\) with \(a< b\), and let \(g:[a,b]\rightarrow\mathbb{R}\) be a continuous positive mapping geometrically symmetric to \(\sqrt{ab}\) such that \(f'\in L^{1}([a,b])\). If \(|f'|^{q}\) for \(q>1\) is GA-s-convex on \([a,b]\) with \(p^{-1}+q^{-1}=1\), then the following inequality for Hadamard k-fractional integral operators with \(k,\alpha>0\) holds:
where
and \(\varUpsilon(\mathcal{U},\mathcal{V},\sigma)\) is defined by (2.3) in Lemma 2.1.
Proof
Using Lemma 2.3 again, we have
Now, considering the following weighted version of Hölder’s inequality, see [9]:
for \(q>1\), \(p^{-1}+q^{-1}=1\), and h is non-negative on I and provided all the other integrals exist and are finite, we have
Considering GA-s-convexity of \(|f'|^{q}\), we have
Using inequality (2.3) in Lemma 2.1, we have
Applying (3.22) and (3.23) to (3.21), we have
Similarly,
The inequality (3.19) is proved by using (3.24) and (3.25) in (3.20). This ends the proof. □
Corollary 3.9
If we take \(k=1\), \(s=1\) and \(g(x)=1\) in Theorem 3.5, then the following inequality holds:
Specially, taking \(\alpha=1\), we have
where \(\varUpsilon(\mathcal{U},\mathcal{V},\sigma)\) is defined by (2.3) in Lemma 2.1.
4 Inequalities for products of two GA-s-convex functions
Theorem 4.1
Let \(f,g,w:[a,b]\rightarrow\mathbb{R}^{+}\), \(a,b\in(0,\infty)\), \(a< b\), be functions satisfying that w and \(fgw\) are in \(L^{1} ([a,b] )\). If f is GA-\(s_{1}\)-convex on \([a,b]\) for some fixed \(s_{1}\in(0,1]\), g is GA-\(s_{2}\)-convex on \([a,b]\) for some fixed \(s_{2}\in(0,1]\), and if w is geometrically symmetric about \(x=\sqrt{ab}\), then we have
where
and
Proof
Since f is GA-\(s_{1}\)-convex and g is GA-\(s_{2}\)-convex on \([a,b]\), we have
and
for all \(t\in[0,1]\). f and g are non-negative, so
Similarly, we also have
The sum of (4.2) and (4.3) yields
Multiplying both sides of (4.4) by \(t^{\frac{\alpha}{k}-1}w (a^{t}b^{1-t} )\), then integrating the obtained inequality with respect to t from 0 to \(\frac{1}{2}\), we have
By the change of variable \(u=a^{t}b^{1-t}\), we get
and
Using the fact that w is geometrically symmetric and by the change of variable \(u=a^{1-t}b^{t}\), we get
Substituting the four equalities above into (4.5), we have the required inequality in (4.1). □
Corollary 4.1
In Theorem 4.1, if we take \(w(u)=1\), then we have
Corollary 4.2
In Theorem 4.1, if we take \(k=1\), \(\alpha=1\) and \(w(u)=1\) for all \(u\in[a,b]\), then we have
Furthermore, if we choose \(s_{1}=s_{2}=1\), then we have Corollary 3.12 in [18].
Remark 4.1
If we choose \(w(u)=1\) for all \(u\in[a,b]\), \(k=1\) and \(s_{1}=s_{2}=1\) in Theorem 4.1, then we have Theorem 3.9 in [18].
Theorem 4.2
Suppose that conditions of Theorem 4.1 hold, then we have the following inequality:
where
and \(\mathbb{M}(a,b)\), \(\mathbb{N}(a,b)\) are defined in Theorem 4.1.
Proof
Using the GA-\(s_{1}\)-convexity of f and GA-\(s_{2}\)-convexity of g, we have
Considering the GA-\(s_{1}\)-convexity of f and GA-\(s_{2}\)-convexity of g again, we have
Multiplying both sides of (4.7) by \(t^{\frac{\alpha }{k}-1}w(a^{1-t}b^{t})\), and integrating the obtained inequality with respect to t from 0 to \(\frac{1}{2}\), we obtain
Using the change of variable and Lemma 2.2, we have
and
Note that w is geometrically symmetric about \(\sqrt{ab}\), we also have
Substituting the six equalities above into (4.8), we have the required inequality in (4.6). This ends the proof. □
Corollary 4.3
In Theorem 4.2, if we take \(w(v) = 1\), then we have
Corollary 4.4
In Theorem 4.2, if we take \(k = 1\), \(\alpha=1\) and \(w(v)=1\), then we have
Furthermore, if we choose \(s_{1}=s_{2}=1\), then we have Corollary 3.16 in [18].
Remark 4.2
If we choose \(w(v)=1\) for all \(v\in[a,b]\), \(k=1\) and \(s_{1}=s_{2}=1\) in Theorem 4.2, then we have Theorem 3.13 in [18].
5 Applications to special means
For positive numbers \(\mu>0\) and \(\nu>0\) with \(\mu\neq\nu\), let us define
and
with \(p\in\mathbb{R}\), respectively.
Now let \(f(x)=x^{r}\) for \(x>0\), \(r\in\mathbb{R}\) with \(r\neq0\). It is easy to check that \(|f'(x)|^{q}=|r|^{q}x^{q(r-1)}\) is GA-convex on \([a,b]\) for \(q\geq1\) and \(r\neq1\), where \(a,b>0\).
Consider the function
Clearly, \(g(x)\) is geometrically symmetric about \(x=\sqrt{ab}\).
Theorem 5.1
Let \(0< a< b\), \(r\in\mathbb{R}\setminus\{-2,0,1,2\}\) and \(q\geq1\). Then the following inequality holds:
Proof
Applying Corollary 3.3 to the functions
and
we derive the required result. □
Corollary 5.1
Suppose all assumptions of Theorem 5.1 are satisfied and if \(r=-1\), then the following inequality holds:
Corollary 5.2
Under the assumptions of Theorem 5.1 with \(q=1\), the following inequality holds:
Especially for \(r=-1\), we get
Theorem 5.2
Let \(0< a< b\), \(r\in\mathbb{R}\setminus\{-2,0,1,2\}\) and \(q>1\). Then the following inequality holds:
Proof
Using Corollary 3.5 for the functions \(f(x)=x^{r}\), \(x>0\), \(r\in\mathbb{R}\setminus\{-2,0,1,2\}\) and \(g(x)= (\frac{x}{\sqrt{ab}}-\frac{\sqrt{ab}}{x} )^{2}\), \(x\in[a,b]\) with \(0< a< b\), we obtain the required result. □
Corollary 5.3
Suppose the assumptions of Theorem 5.2 are fulfilled and if \(r=-1\), then the following inequality holds:
Theorem 5.3
Let \(0< a< b\), \(r\in\mathbb{R}\setminus\{-2,0,1,2\}\) and \(q>1\). Then the following inequality holds:
Proof
Using Corollary 3.7 for the functions \(f(x)=x^{r}\), \(x>0\), \(r\in\mathbb{R}\setminus\{-2,0,1,2\}\) and \(g(x)= (\frac{x}{\sqrt{ab}}-\frac{\sqrt{ab}}{x} )^{2}\), \(x\in[a,b]\) with \(0< a< b\), we deduce the required result. □
Corollary 5.4
Suppose the assumptions of Theorem 5.3 are satisfied and if \(r=-1\), then the following inequality holds:
6 Conclusion
Utilizing mappings whose first-order derivatives absolute values are GA-s-convex, we establish some new Hadamard k-fractional inequalities of Fejér type associated with geometrically symmetric mappings. For the weighted inequalities via products of two different mappings, we also present their upper and lower bounds, which generalize parts of the results given by İşcan and Kunt [18]. With these techniques and the ideas developed in this paper, we hope to motivate the interested reader to further explore this fascinating field of fractional integral inequalities.
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This work was partially supported by the National Natural Science Foundation of China (No. 11871305 and No. 61374028).
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Lei, H., Hu, G., Cao, ZJ. et al. Hadamard k-fractional inequalities of Fejér type for GA-s-convex mappings and applications. J Inequal Appl 2019, 264 (2019). https://doi.org/10.1186/s13660-019-2216-2
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DOI: https://doi.org/10.1186/s13660-019-2216-2