1 Introduction
The calculus of non-integer order, generally referred to as fractional calculus, deals with the generalization of integrals and derivatives operators, in particular inequalities involving fractional integrals. Many definitions of fractional integral operators exist in the literature, for example: Riemann–Liouville fractional integral, Hadamard integral, Liouville integral, Weyl, Erdélyi–Kober and Katugampola fractional integral [2‐5]. Recently, Khalil et al. [6] and Adeljawad [7] presented a new class of fractional operators, namely the local fractional conformable integral and derivative operators. Using such fractional integral operators, one generalizes the fractional operators by involving new parameters and obtains the related inequalities: Hadamard, Hermite–Hadamard, Opial, Grüss, Ostrowski, among others [8‐14]. For instance, Katugampola [15] suggested a generalized fractional integral operator unifying other well-known existing ones: Riemann–Liouville, Hadamard, Weyl, Liouville, and Erdélyi–Kober. Jarad et al. [1] presented the generalized conformable derivatives and integral operators by the standard fractional calculus iteration procedure on fractional conformable operators. Such generalizations motivate the upcoming research to present more innovative ideas to unify the fractional operators and obtain the inequalities involving such generalized fractional operators.
Applications of integral inequalities are important in numerous fields of science: mathematics, physics, engineering, among others; particularly we mention initial-value problems, stability of linear transformation, integral differential equations, and impulse equations [16, 17]. We refer the readers to [10, 18, 19] for such applications in several branches of mathematics and the references therein. Inequalities regarding fractional integral operators accumulate many functional applications in different areas of science. Moreover, the theory of fractional calculus plays an important role in solving differential equations, integral equations, and integral-differential equations, including many other special function problems.
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Thus, the new results involving integral inequalities have been possible; consequently, some applications have been made [16, 17]. We mention a few of them, i.e., the inequalities of Minkowski, Holder, Hardy, Hermite–Hadamard, Jensen, among others [20‐26]. Such applications of fractional integral operators motivate us to present the generalization of the existing fractional conformable operators and generalize the reverse Minkowski inequality [27‐31] involving generalized k-fractional conformable integrals.
The paper is categorized as follows. In Sect. 2, we exhibit the notations and basic definitions of fractional integrals as well as our newly defined generalized k-fractional conformable integrals. We prove the theorems regarding the reverse Minkowski inequality as well as the appropriate spaces for such operators. In Sect. 3, we propose our main results consisting of the reverse Minkowski inequality via the generalized k-fractional conformable integral. In Sect. 4, we present the related inequalities using this fractional integral. The last section containing concluding remarks closes the article.
2 Notations and preliminaries
This section recalls some notations and useful definitions of classical fractional calculus as well as the reverse Minkowski inequality theorem using the classical Riemann integral addressed by Set et al. [21] and its relevant generalization via Riemann–Liouville and Hadamard fractional integrals which were motivation for the study. Furthermore, the fractional conformable integrals are discussed and a theorem is presented so as to recover particular cases.
Definition 1
A function \(f(z)\) is said to be in \(L_{p}[a,b]\) if
$$ \biggl( \int _{a}^{b} \bigl\vert f(z) \bigr\vert ^{p} \,dz \biggr)^{\frac{1}{p}}< \infty ,\quad 1 \leq p< \infty . $$
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Theorem 1
Let
\(f_{1}, f_{2} \in L_{p}[a,b]\)
be two positive functions, with
\(1\leq p\leq \infty \), \(0<\int _{a}^{b}f_{1}^{p}(t)\,dt<\infty \), and
\(0<\int _{a}^{b}f_{2}^{p}(t)\,dt<\infty \). If
\(0< m\leq \frac{f _{1}(t)}{f_{2}(t)}\leq M\)
for
\(m, M \in \mathbb{R^{+}}\)
and
\(\forall t \in [a,b]\), then
with
\(c_{1}=\frac{M(m+1)+(M+1)}{(m+1)(M+1)}\) [21].
$$ \biggl( \int _{a}^{b}f_{1}^{p}(t)\,dt \biggr)^{\frac{1}{p}}+ \biggl( \int _{a}^{b}f_{2}^{p}(t)\,dt \biggr)^{\frac{1}{p}}\leq c_{1} ( \int _{a}^{b}\bigl(f_{1}^{p}+f_{2}^{p}(t) \,dt \bigr)^{\frac{1}{p}}, $$
(2.1)
Theorem 2
Let
\(f_{1}, f_{2} \in L_{p}[a,b]\)
be two positive functions, with
\(1\leq p\leq \infty \), \(0<\int _{a}^{b}f_{1}^{p}(t)\,dt<\infty \), and
\(0<\int _{a}^{b}f_{2}^{p}(t)\,dt<\infty \). If
\(0< m\leq \frac{f _{1}(t)}{f_{2}(t)}\leq M\)
for
\(m, M \in \mathbb{R^{+}}\)
and
\(\forall t \in [a,b]\), then
with
\(c_{2}=\frac{(m+1)(M+1)}{M}-2\) [21].
$$ \biggl( \int _{a}^{b}f_{1}^{p}(t)\,dt \biggr)^{\frac{2}{p}}+ \biggl( \int _{a}^{b}f_{2}^{p}(t)\,dt \biggr)^{\frac{2}{p}}\geq c_{2} \biggl( \int _{a}^{b}f_{1}^{p}(t)\,dt \biggr)^{\frac{1}{p}} \biggl( \int _{a}^{b}f_{2}^{p}(t)\,dt \biggr)^{\frac{2}{p}}, $$
(2.2)
Definition 2
A function \(f(z)\) is said to be in \(L_{p,s}[a,b]\) if
$$ \biggl( \int _{a}^{b} \bigl\vert f(z) \bigr\vert ^{p} z^{s} \,dz \biggr)^{\frac{1}{p}}< \infty ,\quad 1\leq p< \infty , s\geq 0. $$
Definition 3
The space \(X_{c,p}(a,b)\) for \(c\in \mathbb{R} \), \(a< b\) and \(1\leq p< \infty \) contains those complex valued Lebesgue measurable functions g on \((a,b)\) with \(\|g\|_{X_{c,p}}\), where
and for \(p=\infty \),
$$ \|g\|_{X_{c,p}}= \biggl( \int _{a}^{b} \bigl\vert z^{c}f(z) \bigr\vert ^{p} \frac{dz}{z} \biggr)^{\frac{1}{p}} \quad (1\leq p< \infty ), $$
$$ \|g\|_{X_{c,\infty }}=\mathop{\sup \operatorname{ess}}_{z\in (a,b)} \bigl[z^{c} \bigl\vert g(z) \bigr\vert \bigr]. $$
Particularly, for \(c=\frac{1}{p}\), the function space \(X_{c,p}(a,b)\) concurs with the space \(L_{p}(a,b)\) [2].
Definition 4
For \(\alpha \in \mathbb{C}\), \(\operatorname{Re}(\alpha )>0\), the left Riemann–Liouville fractional integral [2, 4] of order α starting from a has the following form:
while the right Riemann–Liouville fractional integral of [2, 4] order \(\alpha >0\) ending at \(b>a\) is defined by
$$ \bigl(\mathcal{I}_{a^{+}}^{\alpha }f \bigr) ( \mathcal{T})=\frac{1}{ \varGamma {(\alpha )}} \int _{a}^{\mathcal{T}}f(x)\frac{dx}{( \mathcal{T}-x)^{1-\alpha }}, $$
(2.3)
$$ \bigl(\mathcal{I}_{b^{-}}^{\alpha }f \bigr) (\mathcal{T})= \frac{1}{ \varGamma {(\alpha )}} \int _{\mathcal{T}}^{b}f(x)\frac{dx}{(x- \mathcal{T})^{1-\alpha }}. $$
(2.4)
Definition 5
For \(\alpha \in \mathbb{C}\) and \(\operatorname{Re}(\alpha )>0\), the left Riemann–Liouville fractional derivative [1] of order α starting from a is defined below:
Meanwhile the right Riemann–Liouville fractional derivative [1] of order \(\alpha >0\) ending at \(b>a\) takes the form
$$ \bigl(\mathcal{D}_{a^{+}}^{\alpha }f \bigr) (\mathcal{T})= \biggl( \frac{d}{dt} \biggr) ^{n} \bigl(\mathcal{I}_{a^{+}}^{n-\alpha }f \bigr) (\mathcal{T}),\quad n=[\alpha ]+1. $$
(2.5)
$$ \bigl(\mathcal{D}_{b^{-}}^{\alpha }f \bigr) (\mathcal{T})= \biggl(- \frac{d}{dt} \biggr)^{n} \bigl(\mathcal{I}_{b^{-}}^{n-\alpha }f \bigr) ( \mathcal{T}). $$
(2.6)
Definition 6
The left Caputo fractional derivative [1] of order α, \(\operatorname{Re}(\alpha )>0\) starting from a has the form
while the right Caputo fractional derivative [1] of order \(\alpha >0\) ending at \(b>a\) takes the form
$$ \bigl({}^{C}\mathcal{D}_{a^{+}}^{\alpha }f \bigr) ( \mathcal{T})= \bigl(\mathcal{I} _{a^{+}}^{n-\alpha }f^{(n)} \bigr) (\mathcal{T}),\quad n=[\alpha ]+1, $$
(2.7)
$$ \bigl({}^{C}\mathcal{D}_{b^{-}}^{\alpha }f \bigr) ( \mathcal{T})= \bigl(\mathcal{I} _{b^{-}}^{n-\alpha }(-1)^{n}f^{(n)} \bigr) (\mathcal{T}). $$
(2.8)
Definition 7
The left Hadamard fractional integral [1] of order \(\alpha \in \mathbb{C}\), \(\operatorname{Re}(\alpha )>0\) starting from a is given below:
and the right Hadamard fractional integral [1] of order α ending at \(b>a\) takes the form
$$ \bigl(\mathcal{F}_{a^{+}}^{\alpha }f \bigr) (\mathcal{T})= \frac{1}{ \varGamma {(\alpha )}} \int _{a}^{\mathcal{T}} (\ln {\mathcal{T}}- \ln {x} )^{\alpha -1}f(x)\frac{dx}{x}, $$
(2.9)
$$ \bigl(\mathcal{F}_{b^{-}}^{\alpha }f \bigr) (\mathcal{T})= \frac{1}{ \varGamma {(\alpha )}} \int _{\mathcal{T}}^{b} (\ln {x}-\ln {\mathcal{T}} )^{\alpha -1}f(x)\frac{dx}{x}. $$
(2.10)
Definition 8
The left Hadamard fractional derivative [1] of order \(\alpha \in \mathbb{C}\), \(\operatorname{Re}(\alpha )>0\) starting from a can be defined as
and the right Hadamard fractional derivative [1] of order α ending at \(b>a\) takes the form
$$ \bigl(\mathcal{G}_{a^{+}}^{\alpha }f \bigr) ( \mathcal{T})= \biggl(t \frac{d}{dt} \biggr)^{n} \bigl( \mathcal{F}_{a^{+}}^{n-\alpha }f \bigr) ( \mathcal{T}),\quad n=[\alpha ]+1, $$
(2.11)
$$ \bigl(\mathcal{G}_{b^{-}}^{\alpha }f \bigr) (\mathcal{T})= \biggl(-t \frac{d}{dt} \biggr)^{n} \bigl(\mathcal{F}_{b^{-}}^{n-\alpha }f \bigr) ( \mathcal{T}). $$
(2.12)
Definition 9
For a real function \(f\in X_{c,p}(a,b)\), the generalized left and right Katugampola fractional integrals [28] of order \(\alpha \in \mathbb{R}\), \(\rho >0\), \(\operatorname{Re}(\alpha )>0\) take the form
and
respectively.
$$ \bigl(\mathcal{K}_{a^{+}}^{\alpha ,\rho }f \bigr) ( \mathcal{T})=\frac{ \rho ^{1-\alpha }}{\varGamma {(\alpha )}} \int _{a}^{\mathcal{T}}f(x)\frac{x ^{\rho -1}\,dx}{(\mathcal{T}^{\rho }-x^{\rho })^{1-\alpha }} $$
(2.13)
$$ \bigl(\mathcal{K}_{b^{-}}^{\alpha ,\rho }f \bigr) (\mathcal{T})= \frac{ \rho ^{1-\alpha }}{\varGamma {(\alpha )}} \int _{\mathcal{T}}^{b}f(x)\frac{x ^{\rho -1}\,dx}{(x^{\rho }-\mathcal{T}^{\rho })^{1-\alpha }}, $$
(2.14)
Definition 10
The generalized left and right Katugampola fractional derivatives [29] of order \(\alpha \in \mathbb{R}\), \(\rho >0\), \(\operatorname{Re}(\alpha )>0\) are defined below:
and
respectively, where \(\gamma =t^{1-\rho }\frac{d}{dt}\).
$$ \bigl(\mathcal{L}_{a^{+}}^{\alpha ,\rho }f \bigr) (\mathcal{T})=\gamma ^{n} \bigl(\mathcal{K}_{a^{+}}^{n-\alpha ,\rho }f \bigr) ( \mathcal{T})=\frac{ \gamma ^{n}\rho ^{n-\alpha }}{\varGamma {(n-\alpha )}} \int _{a}^{ \mathcal{T}}f(x)\frac{x^{\rho -1}\,dx}{(\mathcal{T}^{\rho }-x^{\rho })^{1+ \alpha -n}} $$
(2.15)
$$ \bigl(\mathcal{L}_{b^{-}}^{\alpha ,\rho }f \bigr) (\mathcal{T})=(- \gamma )^{n} \bigl(\mathcal{K}_{b^{-}}^{n-\alpha ,\rho }f \bigr) ( \mathcal{T})=\frac{(-\gamma )^{n}\rho ^{n-\alpha }}{\varGamma {(n-\alpha )}} \int _{\mathcal{T}}^{b}f(x)\frac{x^{\rho -1}\,dx}{(x^{\rho }- \mathcal{T}^{\rho })^{1+\alpha -n}}, $$
(2.16)
Dahmani [30] verified the reverse Minkowski inequality together with a relevant result to the inequality associated with Riemann–Liouville fractional integral corresponding to the following two theorems.
Theorem 3
For
\(\alpha >0\)
and
\(p\geq 1\). Let
\(f_{1}, f_{2}\in L_{1,s}[a,t]\)
be two positive functions in
\([0,\infty)\)
such that, for all
\(t>a\), \(\mathcal{I}_{a^{+},k}^{\alpha ,s}f_{1}^{p}(t)<\infty \)
and
\(\mathcal{I}_{a^{+},k}^{\alpha ,s}f_{2}^{p}(t)<\infty \). If
\(0< m\leq \frac{f _{1}(x)}{f_{2}(x)}\leq M\)
for
\(m, M\in \mathbb{R^{+}}\)
and for all
\(x\in [a, t]\), then
with
\(c_{1}=\frac{M(m+1)+(M+1)}{(m+1)(M+1)}\) [30].
$$ \bigl(\mathcal{I}_{a^{+},k}^{\alpha ,s}f_{1}^{p}(t) \bigr)^{ \frac{1}{p}}+ \bigl(\mathcal{I}_{a^{+},k}^{\alpha ,s}f_{2}^{p}(t) \bigr) ^{\frac{1}{p}}\leq c_{1} \bigl(\mathcal{I}_{a^{+},k}^{\alpha ,s}(f_{1}+f _{2})^{p}(t) \bigr)^{\frac{1}{p}}, $$
(2.17)
Theorem 4
For
\(\alpha >0\)
and
\(p\geq 1\). Let
\(f_{2}, f_{2}\in L_{1,s}[a,t]\)
be two positive functions in
\([0,\infty)\)
such that, for all
\(t>a\), \(\mathcal{I}_{a^{+},k}^{\alpha ,s}f_{1}^{p}(t)<\infty \)
and
\(\mathcal{I}_{a^{+},k}^{\alpha ,s}f_{2}^{p}(t)<\infty \). If
\(0< m\leq \frac{f _{1}(x)}{f_{2}(x)}\leq M\)
for
\(m, M\in \mathbb{R^{+}}\)
and for all
\(x\in [a, t]\), then
with
\(c_{2}=\frac{(m+1)(M+1)}{M}-2\) [30].
$$ \bigl(\mathcal{I}_{a^{+},k}^{\alpha ,s}f_{1}^{p}(t) \bigr)^{ \frac{2}{p}}+ \bigl(\mathcal{I}_{a^{+},k}^{\alpha ,s}f_{2}^{p}(t) \bigr) ^{\frac{2}{p}}\geq c_{2} \bigl(\mathcal{I}_{a^{+},k}^{\alpha ,s}f_{1} ^{p}(t) \bigr)^{\frac{1}{p}} \bigl(\mathcal{I}_{a^{+},k}^{\alpha ,s}f _{2}^{p}(t) \bigr)^{\frac{1}{p}}, $$
(2.18)
Chinchane et al. [31] and Sabrina et al. [32] established the following two theorems for the reverse Minkowski inequality involving Hadamard fractional integral operator.
Theorem 5
For
\(\alpha >0\)
and
\(p\geq 1\). Let
\(f_{1}, f_{2}\in L_{1,s}[a,t]\)
be two positive functions in
\([0,\infty)\)
such that, for all
\(t>a\), \(\mathcal{F}_{a^{+},k}^{\alpha ,s}f_{1}^{p}(t)<\infty \)
and
\(\mathcal{F}_{a^{+},k}^{\alpha ,s}f_{2}^{p}(t)<\infty \). If
\(0< m\leq \frac{f _{1}(x)}{f_{2}(x)}\leq M\)
for
\(m, M\in \mathbb{R^{+}}\)
and for all
\(x\in [a, t]\), then
with
\(c_{1}=\frac{M(m+1)+(M+1)}{(m+1)(M+1)}\) [31, 32].
$$ \bigl(\mathcal{F}_{a^{+},k}^{\alpha ,s}f_{1}^{p}(t) \bigr)^{ \frac{1}{p}}+ \bigl(\mathcal{F}_{a^{+},k}^{\alpha ,s}f_{2}^{p}(t) \bigr) ^{\frac{1}{p}}\leq c_{1} \bigl(\mathcal{F}_{a^{+},k}^{\alpha ,s}(f_{1}+f _{2})^{p}(t) \bigr)^{\frac{1}{p}}, $$
(2.19)
Theorem 6
For
\(\alpha >0\)
and
\(p\geq 1\). Let
\(f_{2}, f_{2}\in L_{1,s}[a,t]\)
be two positive functions in
\([0,\infty)\)
such that, for all
\(t>a\), \(\mathcal{F}_{a^{+},k}^{\alpha ,s}f_{1}^{p}(t)<\infty \)
and
\(\mathcal{F}_{a^{+},k}^{\alpha ,s}f_{2}^{p}(t)<\infty \). If
\(0< m\leq \frac{f _{1}(x)}{f_{2}(x)}\leq M\)
for
\(m, M\in \mathbb{R^{+}}\)
and for all
\(x\in [a, t]\), then
with
\(c_{2}=\frac{(m+1)(M+1)}{M}-2\) [31, 32].
$$ \bigl(\mathcal{F}_{a^{+},k}^{\alpha ,s}f_{1}^{p}(t) \bigr)^{ \frac{2}{p}}+ \bigl(\mathcal{F}_{a^{+},k}^{\alpha ,s}f_{2}^{p}(t) \bigr) ^{\frac{2}{p}}\geq c_{2} \bigl(\mathcal{F}_{a^{+},k}^{\alpha ,s}f_{1} ^{p}(t) \bigr)^{\frac{1}{p}} \bigl(\mathcal{F}_{a^{+},k}^{\alpha ,s}f _{2}^{p}(t) \bigr)^{\frac{1}{p}}, $$
(2.20)
Chinchane et al. [33] presented the reverse Minkowski inequality by fractional integral of Saigo, and recently Chinchane [34] proved the same inequality via the k-fractional integral. In 2017, Jarad et al. [1] introduced a new fractional integral that generalizes the above mentioned pre-existing fractional integrals. Conclusively, we define the generalization of this integral in the k-analogue form in addition to a theorem for the study of their particular cases.
Definition 11
If \(f\in L_{1}[a,b]\), then the left fractional conformable integral of order \(\alpha \geq 0\) defined by Abdeljawad [7] is given by
$$ I_{a}^{\alpha }f(x)= \int _{a}^{x}(t-a)^{\alpha -1} f(t)\,dt, \quad 0 \leq a< x< b\leq \infty , 0< \alpha < 1. $$
(2.21)
Definition 12
If \(f\in L_{1}[a,b]\), then the right fractional conformable integral of order \(\alpha \geq 0\) defined by Abdeljawad [7] is given by
$$ I_{b}^{\alpha }f(x)= \int _{x}^{b}(b-t)^{\alpha -1} f(t)\,dt, \quad 0 \leq a< x< b\leq \infty , 0< \alpha < 1. $$
(2.22)
Definition 13
If \(f\in L_{1,s}[a,b]\), then the generalized left fractional conformable integral \(\mathfrak{T}_{a}^{\alpha ,s}\) of order \(\alpha \in \mathbb{C}\), \(\operatorname{Re}(\alpha )>0\) and \(s>0\), introduced by Jarad et al. [1], is defined by
where Γ is the Euler gamma function.
$$\begin{aligned}& \mathfrak{T}_{a}^{\alpha ,s}f(t)=\frac{s^{1-\alpha }}{\varGamma (\alpha )} \int _{a}^{t} \bigl((t-a)^{s}-(x-a)^{s} \bigr)^{\alpha -1} (x-a)^{s-1} f(x)\,dx, \\& \quad 0\leq a< t< b\leq \infty , \end{aligned}$$
(2.23)
Definition 14
If \(f\in L_{1,s}[a,b]\), then the generalized right fractional conformable integral \(\mathfrak{T}_{b}^{\alpha ,s}\) of order \(\alpha \in \mathbb{C}\), \(\operatorname{Re}(\alpha )>0\) and \(s>0\), introduced by Jarad et al. [1], is defined by
where Γ is the Euler gamma function.
$$\begin{aligned}& \mathfrak{T}_{b}^{\alpha ,s}f(t)=\frac{s^{1-\alpha }}{\varGamma (\alpha )} \int _{t}^{b} \bigl((b-x)^{s}-(b-t)^{s} \bigr)^{\alpha -1} (b-x)^{s-1} f(x)\,dx, \\& \quad 0\leq a< t< b\leq \infty , \end{aligned}$$
(2.24)
Definition 15
The \((k,s)\)-fractional conformable integrals (left and right) of order \(\alpha \in \mathbb{C}\), \(\operatorname{Re}(\alpha )>0\) of a continuous function \(f(x)\) on \([0,\infty )\), are given as follows:
and
respectively, if integrals exist, where \(k>0\), \(s\in \mathbb{R}\setminus \{0\}\).
$$\begin{aligned}& \mathfrak{F}_{a^{+},k}^{\alpha ,s}f(t)=\frac{(s)^{1-\frac{\alpha }{k}}}{k \varGamma _{k} (\alpha )} \int _{a}^{t} \bigl((t-a)^{s}-(x-a)^{s} \bigr)^{\frac{ \alpha }{k}-1} (x-a)^{s-1} f(x)\,dx, \\& \quad 0\leq a< t< b\leq \infty , \end{aligned}$$
(2.25)
$$\begin{aligned}& \mathfrak{F}_{b^{-},k}^{\alpha ,s}f(t)=\frac{(s)^{1-\frac{\alpha }{k}}}{k \varGamma _{k} (\alpha )} \int _{t}^{b} \bigl((b-x)^{s}-(b-t)^{s} \bigr)^{\frac{ \alpha }{k}-1} (b-x)^{s-1} f(x)\,dx, \\& \quad 0\leq a< t< b\leq \infty , \end{aligned}$$
(2.26)
Theorem 7
For
\(k>0\), \(s\in \mathbb{R}\setminus \{0\}\), \(\alpha >0\)
and
\(p\geq 1\). Then, for
\(f\in L_{1,s}[a,t]\), for all
\(t>a\), we have [15]:
(2)
Theorem 8
Let
\(f\in L_{1} [a,b ]\), \(s\in \mathbb{R}\setminus \{0\}\), and
\(k>0\). Then
\(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f(x) (\mathfrak{F} _{b^{-},k}^{\alpha ,s}f(x) )\)
exists for any
\(x\in [a,b ]\), \(\operatorname{Re} (\alpha )>0 \).
Proof
Let \(\triangle ^{\prime }:= [a,b ]\times [a,b ]\) and \(P^{\prime }:\triangle ^{\prime }\rightarrow \mathbb{R}\) such that
Clearly, it can be seen that
where
and
Since \(P^{\prime }\) is measurable on \(\triangle ^{\prime }\), then it can be written as
By using the double integral, we get
i.e.,
So, the function \(Q^{\prime }:\triangle ^{\prime }\rightarrow \mathbb{R}\) such that \(Q^{\prime }(x,t):=P^{\prime }(x,t)f(x)\) is integrable over \(\triangle ^{\prime }\) by Tonelli’s theorem. Hence, by Fubini’s theorem \(\int _{a}^{b}P ^{\prime }(x,t)f(x)\,dx\) is an integrable function over \([a,b]\) as a function of \(t\in [a,b]\), i.e., \(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f(x)\) exists.
$$ P^{\prime } (x,t )= \bigl((x-a)^{\alpha }-(t-a)^{\alpha } \bigr) ^{\frac{\beta }{k}-1}(t-a)^{\alpha -1}. $$
$$ P^{\prime }=P_{+}^{\prime }+P_{-}^{\prime }, $$
$$ P_{+}^{\prime }(x,t):= \textstyle\begin{cases} ((x-a)^{\alpha }-(t-a)^{\alpha } )^{\frac{\beta }{k}-1}(t-a)^{ \alpha -1}, &a\leq t\leq x\leq b, \\ 0, &a\leq x\leq t\leq b, \end{cases} $$
$$ P_{-}^{\prime }(x,t):= \textstyle\begin{cases} ((t-a)^{\alpha }-(x-a)^{\alpha } )^{\frac{\beta }{k}-1}(x-a)^{ \alpha -1}, &a\leq t\leq x\leq b, \\ 0, &a\leq x\leq t\leq b. \end{cases} $$
$$\begin{aligned} \int _{a}^{b}P^{\prime }(x,t)\,dt =& \int _{a}^{x}P^{\prime }(x,t)\,dt = \int _{a}^{x} \bigl((x-a)^{\alpha }-(t-a)^{\alpha } \bigr)^{\frac{ \beta }{k}-1}(t-a)^{\alpha -1}\,dt \\ =&\frac{\alpha k}{\beta }(x-a)^{\frac{\alpha \beta }{k}}. \end{aligned}$$
$$\begin{aligned} \int _{a}^{b} \biggl( \int _{a}^{b}P^{\prime }(x,t) \bigl\vert f(x) \bigr\vert \,dt \biggr)\,dx =& \int _{a}^{b} \bigl\vert f(x) \bigr\vert \biggl( \int _{a}^{b}P^{\prime }(x,t)\,dt \biggr)\,dx \\ =&\frac{\alpha k}{\beta } \int _{a}^{b}(x-a)^{\frac{\alpha \beta }{k}} \bigl\vert f(x) \bigr\vert \,dx \\ \leq &\frac{\alpha k}{\beta }(b-a)^{\frac{\alpha \beta }{k}} \int _{a}^{b} \bigl\vert f(x) \bigr\vert \,dx, \end{aligned}$$
$$\begin{aligned} \int _{a}^{b} \biggl( \int _{a}^{b}P^{\prime }(x,t) \bigl\vert f(x) \bigr\vert \,dt \biggr)\,dx =& \int _{a}^{b} \bigl\vert f(x) \bigr\vert \biggl( \int _{a}^{b}P^{\prime }(x,t)\,dt \biggr)\,dx \\ \leq &\frac{\alpha k}{\beta }(b-a)^{\frac{\alpha \beta }{k}} \bigl\Vert f(x) \bigr\Vert _{L_{1}[a,b]}< \infty . \end{aligned}$$
The existence of the right k-fractional conformable integral \(\mathfrak{F}_{b^{-},k}^{\alpha ,s}f(x)\) can be proved in a similar manner. □
3 Reverse Minkowski inequality involving generalized k-fractional conformable integrals
This section contains our main contribution of establishing the proof of the reverse Minkowski inequality via our newly defined generalized k-fractional conformable integral (2.25) and a related theorem referred to as the reverse Minkowski inequality.
Theorem 9
For
\(k>0\), \(s\in \mathbb{R}\setminus \{0\}\), \(\alpha >0\)
and
\(p\geq 1\). Let
\(f_{1}, f_{2}\in L_{1,s}[a,t]\)
be two positive functions in
\([0,\infty)\)
such that, for all
\(t>a\), \(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{1}^{p}(t)<\infty \)
and
\(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{2} ^{p}(t)<\infty \). If
\(0< m\leq \frac{f_{1}(x)}{f_{2}(x)}\leq M\)
for
\(m, M\in \mathbb{R^{+}}\)
and for all
\(x\in [a, t]\), then
with
\(c_{1}=\frac{M(m+1)+(M+1)}{(m+1)(M+1)}\).
$$ \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{1}^{p}(t) \bigr)^{ \frac{1}{p}}+ \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{2}^{p}(t) \bigr) ^{\frac{1}{p}}\leq c_{1} \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}(f _{1}+f_{2})^{p}(t) \bigr)^{\frac{1}{p}}, $$
(3.1)
Proof
Under the given conditions \(\frac{f_{1}(x)}{f_{2}(x)}\leq M\), \(a\leq x\leq t\), it can be written
which implies that
By multiplying both sides of (3.2) with \(\frac{s^{1-\frac{\alpha }{k}} ((t-a)^{s}-(x-a)^{s} )^{\frac{\alpha }{k}-1}(x-a)^{s-1}}{k \varGamma _{k}(\alpha )}\) and then integrating with respect to the variable x from a to t, we obtain
Accordingly, it can be written as
In contrast, as \(mf_{2}(x)\leq f_{1}(x)\), it follows
Further, by multiplying both sides of (3.5) with \(\frac{s^{1-\frac{ \alpha }{k}} ((t-a)^{s}-(x-a)^{s} )^{\frac{\alpha }{k}-1}(x-a)^{s-1}}{k \varGamma _{k}(\alpha )}\) and then integrating with respect to the variable x from a to t, we obtain
The required result (3.1) follows from (3.4) and (3.6). □
$$ f_{1}(x)\leq M\bigl(f_{1}(x)+f_{2}(x) \bigr)-Mf_{1}(x), $$
$$ (M+1)^{p}f_{1}^{p}(x)\leq M^{p}\bigl(f_{1}(x)+f_{2}(x)\bigr)^{p}. $$
(3.2)
$$\begin{aligned}& \frac{(M+1)^{p}s^{1-\frac{\alpha }{k}}}{k\varGamma _{k}(\alpha )} \int _{a}^{t} \bigl((t-a)^{s}-(x-a)^{s} \bigr)^{\frac{\alpha }{k}-1}f _{1}^{p}(x) (x-a)^{s-1}\,dx \\& \quad \leq \frac{M^{p}s^{1-\frac{\alpha }{k}}}{k\varGamma _{k}(\alpha )} \int _{a}^{t} \bigl((t-a)^{s}-(x-a)^{s} \bigr)^{\frac{\alpha }{k}-1} (f_{1}+f_{2} )^{p}(x) (x-a)^{s-1}\,dx. \end{aligned}$$
(3.3)
$$ \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{1}^{p}(t) \bigr)^{ \frac{1}{p}}\leq \frac{M}{M+1} \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}(f _{1}+f_{2})^{p}(t) \bigr)^{\frac{1}{p}}. $$
(3.4)
$$ \biggl(1+\frac{1}{m}\biggr)^{p}f_{2}^{p}(x) \leq \biggl(\frac{1}{m} \biggr)^{p}\bigl(f _{1}(x)+f_{2}(x) \bigr)^{p}. $$
(3.5)
$$ \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{2}^{p}(t) \bigr)^{ \frac{1}{p}}\leq \frac{1}{m+1} \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}(f _{1}+f_{2})^{p}(t) \bigr)^{\frac{1}{p}}. $$
(3.6)
Inequality (3.1) is known as the reverse Minkowski inequality involving generalized k-fractional conformable integral.
Theorem 10
For
\(k>0\), \(s\in \mathbb{R}\setminus \{0\}\), \(\alpha >0\)
and
\(p\geq 1\). Let
\(f_{2}, f_{2}\in L_{1,s}[a,t]\)
be two positive functions in
\([0,\infty)\)
such that, for all
\(t>a\), \(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{1}^{p}(t)<\infty \)
and
\(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{2} ^{p}(t)<\infty \). If
\(0< m\leq \frac{f_{1}(x)}{f_{2}(x)}\leq M\)
for
\(m, M\in \mathbb{R^{+}}\)
and for all
\(x\in [a, t]\), then
with
\(c_{2}=\frac{(m+1)(M+1)}{M}-2\).
$$ \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{1}^{p}(t) \bigr)^{ \frac{2}{p}}+ \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{2}^{p}(t) \bigr) ^{\frac{2}{p}}\geq c_{2} \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{1} ^{p}(t) \bigr)^{\frac{1}{p}} \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f _{2}^{p}(t) \bigr)^{\frac{1}{p}}, $$
(3.7)
Proof
Taking the product between (3.4) and (3.6) results in
Involving the Minkowski inequality, on the right side of (3.8), we get
From (3.9), it can be concluded that
□
$$ \frac{(m+1)(M+1)}{M} \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{1}^{p}(t) \bigr) ^{\frac{1}{p}} \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{2}^{p}(t) \bigr) ^{\frac{1}{p}}\leq \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}(f_{1}+f _{2})^{p}(t) \bigr)^{\frac{2}{p}}. $$
(3.8)
$$\begin{aligned}& \frac{(m+1)(M+1)}{M} \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{1}^{p}(t) \bigr) ^{\frac{1}{p}} \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{2}^{p}(t) \bigr) ^{\frac{1}{p}} \\& \quad \leq \bigl( \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{1}^{p}(t) \bigr) ^{\frac{1}{p}}+ \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{2}^{p}(t) \bigr) ^{\frac{1}{p}} \bigr)^{2}. \end{aligned}$$
(3.9)
$$\begin{aligned}& \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{1}^{p}(t) \bigr)^{ \frac{2}{p}}+ \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{2}^{p}(t) \bigr) ^{\frac{2}{p}} \\& \quad \geq \biggl(\frac{(m+1)(M+1)}{M}-2 \biggr) \bigl(\mathfrak{F}_{a^{+},k} ^{\alpha ,s}f_{1}^{p}(t) \bigr)^{\frac{1}{p}} \bigl( \mathfrak{F}_{a ^{+},k}^{\alpha ,s}f_{2}^{p}(t) \bigr)^{\frac{1}{p}}. \end{aligned}$$
4 Related fractional integral inequalities
This section contains the generalization of the results presented by Chinchane, Sulaiman, and Sroysang related to the reverse Minkowski inequality via Riemann integral operator, involving the proposed generalized k-fractional conformable integral (2.25).
Theorem 11
For
\(k>0\), \(s\in \mathbb{R}\setminus \{0\}\), \(\alpha >0\), \(p,q\geq 1\)
and
\(\frac{1}{p}+\frac{1}{q}=1\). Let
\(f_{1}, f_{2}\in L_{1,s}[a,t]\)
be two positive functions in
\([0,\infty)\)
such that, for all
\(t>a\), \(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{1}^{p}(t)<\infty \)
and
\(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{2}^{p}(t)<\infty \). If
\(0< m\leq \frac{f_{1}(x)}{f_{2}(x)}\leq M\)
for
\(m, M\in \mathbb{R^{+}}\)
and for all
\(x\in [a, t]\), then
$$ \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{1}^{p}(t) \bigr)^{ \frac{1}{p}} \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{2}^{p}(t) \bigr) ^{\frac{1}{p}}\leq \biggl(\frac{M}{m} \biggr)^{\frac{1}{pq}} \bigl( \mathfrak{F} _{a^{+},k}^{\alpha ,s}f_{1}^{\frac{1}{p}}(t)f_{2}^{\frac{1}{q}}(t) \bigr). $$
(4.1)
Proof
Under the given conditions \(\frac{f_{1}(x)}{f_{2}(x)}\leq M\), \(a\leq x\leq t\), it can be written
Multiplying both sides of (4.2) by \(f^{\frac{1}{p}}(x)\), we can rewrite it as follows:
Multiplying both sides of (4.3) with \(\frac{s^{1- \frac{\alpha }{k}} ((t-a)^{s}-(x-a)^{s} )^{\frac{\alpha }{k}-1}(x-a)^{s-1}}{k \varGamma _{k}(\alpha )}\) and then integrating with respect to the variable x from a to t, we obtain
Accordingly, it can be written as
In contrast, as \(mf_{2}(x)\leq f_{1}(x)\), it follows
Further, by multiplying both sides of (4.6) by \(f_{2}^{ \frac{1}{q}}(x)\) and invoking the relation \(\frac{1}{p}+\frac{1}{q}=1\), it yields
Multiplying both sides of (4.7) by \(\frac{s^{1-\frac{\alpha }{k}} ((t-a)^{s}-(x-a)^{s} )^{\frac{\alpha }{k}-1}(x-a)^{s-1}}{k \varGamma _{k}(\alpha )}\) and then integrating with respect to the variable x from a to t, we obtain
Conducting the product between (4.5) and (4.8) and using the relation \(\frac{1}{p}+\frac{1}{q}=1\), the required inequality (4.1) can be concluded. □
$$ f_{1}(x)\leq Mf_{2}(x)\quad \Rightarrow \quad f_{2}^{\frac{1}{q}}(x)\geq M^{- \frac{1}{q}}f_{1}^{\frac{1}{q}}(x). $$
(4.2)
$$ f_{1}^{\frac{1}{p}}(x)f_{2}^{\frac{1}{q}}(x) \geq M^{-\frac{1}{q}}f _{1}(x). $$
(4.3)
$$\begin{aligned}& \frac{M^{-\frac{1}{q}}s^{1-\frac{\alpha }{k}}}{k\varGamma _{k}(\alpha )} \int _{a}^{t} \bigl((t-a)^{s}-(x-a)^{s} \bigr)^{ \frac{\alpha }{k}-1}f_{1}(x) (x-a)^{s-1}\,dx \\& \quad \leq \frac{s^{1-\frac{\alpha }{k}}}{k\varGamma _{k}(\alpha )} \int _{a}^{t} \bigl((t-a)^{s}-(x-a)^{s} \bigr)^{\frac{\alpha }{k}-1}f_{1} ^{\frac{1}{p}}(x)f_{2}^{\frac{1}{q}}(x) (x-a)^{s-1}\,dx. \end{aligned}$$
(4.4)
$$ M^{-\frac{1}{pq}} \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{1}(t) \bigr) ^{\frac{1}{p}}\leq \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{1}^{ \frac{1}{p}}(t)f_{2}^{\frac{1}{q}}(t) \bigr)^{\frac{1}{p}}. $$
(4.5)
$$ m^{\frac{1}{p}}f_{2}^{\frac{1}{p}}(x)\leq f_{1}^{\frac{1}{p}}(x). $$
(4.6)
$$ m^{\frac{1}{p}}f_{2}(x)\leq f_{1}^{\frac{1}{p}}(x)f_{2}^{\frac{1}{q}}(x). $$
(4.7)
$$ m^{\frac{1}{pq}} \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{2}(t) \bigr) ^{\frac{1}{q}}\leq \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{1}^{ \frac{1}{p}}(t)f_{2}^{\frac{1}{q}}(t) \bigr)^{\frac{1}{q}}. $$
(4.8)
Theorem 12
For
\(k>0\), \(s\in \mathbb{R}\setminus \{0\}\), \(\alpha >0\), \(p,q\geq 1\)
and
\(\frac{1}{p}+\frac{1}{q}=1\). Let
\(f_{1}, f_{2}\in L_{1,s}[a,t]\)
be two positive functions in
\([0,\infty)\)
such that, for all
\(t>a\), \(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{1}^{p}(t)<\infty \)
and
\(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{2}^{p}(t)<\infty \). If
\(0< m\leq \frac{f_{1}(x)}{f_{2}(x)}\leq M\)
for
\(m, M\in \mathbb{R^{+}}\)
and for all
\(x\in [a, t]\), then
with
\(c_{3}=\frac{2^{p-1}M^{p}}{p(M+1)^{p}}\)
and
\(c_{4}= \frac{2^{q-1}}{q(m+1)^{q}}\).
$$ \mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{1}(t)f_{2}(t) \leq c_{3} \bigl(\mathfrak{F} _{a^{+},k}^{\alpha ,s} \bigl(f_{1}^{p}+f_{2}^{p}\bigr) (t) \bigr)+c_{4} \bigl(\mathfrak{F} _{a^{+},k}^{\alpha ,s} \bigl(f_{1}^{q}+f_{2}^{q}\bigr) (t) \bigr), $$
(4.9)
Proof
Using the hypothesis, we get the following identity:
Multiplying both sides of (4.10) with \(\frac{s^{1- \frac{\alpha }{k}} ((t-a)^{s}-(x-a)^{s} )^{\frac{\alpha }{k}-1}(x-a)^{s-1}}{k \varGamma _{k}(\alpha )}\) and then integrating with respect to the variable x from a to t, we obtain
Accordingly, it can be written as
In contrast, as \(0< m<\frac{f_{1}(x)}{f_{2}(x)}\), \(a< x< t\), it follows
Further, by multiplying both sides of (4.13) by \(\frac{s^{1-\frac{ \alpha }{k}} ((t-a)^{s}-(x-a)^{s} )^{\frac{\alpha }{k}-1}(x-a)^{s-1}}{k \varGamma _{k}(\alpha )}\) and then integrating with respect to the variable x from a to t, we obtain
Taking into account Young’s inequality,
Now multiplying both sides of (4.15) by \(\frac{s^{1-\frac{\alpha }{k}} ((t-a)^{s}-(x-a)^{s} )^{\frac{\alpha }{k}-1}(x-a)^{s-1}}{k \varGamma _{k}(\alpha )}\) and then integrating with respect to the variable x from a to t, we obtain
Invoking (4.12) and (4.14) into (4.16), we obtain
Using the inequality \((x+y)^{s}\leq 2^{s-1}(x^{s}+y^{s})\), \(s>1\), \(x,y>0\), one obtains
and
The proof of (4.9) can be concluded from (4.17), (4.18), and (4.19) collectively. □
$$ (M+1)^{p}f_{1}^{p}(x)\leq M^{p}(f_{1}+f_{2})^{p}(x). $$
(4.10)
$$\begin{aligned}& \frac{(M+1)^{p}s^{1-\frac{\alpha }{k}}}{k\varGamma _{k}(\alpha )} \int _{a}^{t} \bigl((t-a)^{s}-(x-a)^{s} \bigr)^{\frac{\alpha }{k}-1}f _{1}^{p}(x) (x-a)^{s-1}\,dx \\& \quad \leq \frac{M^{p}s^{1-\frac{\alpha }{k}}}{k\varGamma _{k}(\alpha )} \int _{a}^{t} \bigl((t-a)^{s}-(x-a)^{s} \bigr)^{\frac{\alpha }{k}-1}(f _{1}+f_{2})^{p}(x) (x-a)^{s-1}\,dx. \end{aligned}$$
(4.11)
$$ \mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{1}^{p}(t) \leq \frac{M^{p}}{(M+1)^{p}}\mathfrak{F}_{a^{+},k}^{\alpha ,s}(f_{1}+f_{2})^{p}(t). $$
(4.12)
$$ (m+1)^{q}f_{2}^{q}(x)\leq (f_{1}+f_{2})^{q}(x). $$
(4.13)
$$ \mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{2}^{q}(t) \leq \frac{1}{(m+1)^{q}}\mathfrak{F}_{a^{+},k}^{\alpha ,s}(f_{1}+f_{2})^{q}(t). $$
(4.14)
$$ f_{1}(x)f_{2}(x)\leq \frac{f_{1}^{p}(x)}{p}+ \frac{f_{2}^{q}(x)}{q}. $$
(4.15)
$$ \mathfrak{F}_{a^{+},k}^{\alpha ,s}(f_{1}f_{2}) (t)\leq \frac{1}{p} \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{1}^{p}(t) \bigr)+ \frac{1}{q} \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{2}^{q}(t) \bigr). $$
(4.16)
$$\begin{aligned} \mathfrak{F}_{a^{+},k}^{\alpha ,s}(f_{1}f_{2}) (t)&\leq \frac{1}{p} \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{1}^{p}(t) \bigr)+ \frac{1}{q} \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{2}^{q}(t) \bigr) \\ &\leq \frac{M^{p}}{p(M+1)^{p}} \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}(f _{1}+f_{2})^{p}(t) \bigr)+\frac{1}{q(m+1)^{q}} \bigl(\mathfrak{F}_{a ^{+},k}^{\alpha ,s}(f_{1}+f_{2})^{q}(t) \bigr). \end{aligned}$$
(4.17)
$$ \mathfrak{F}_{a^{+},k}^{\alpha ,s}(f_{1}+f_{2})^{p}(t) \leq 2^{p-1} \mathfrak{F}_{a^{+},k}^{\alpha ,s} \bigl(f_{1}^{p}+f_{2}^{p}\bigr) (t) $$
(4.18)
$$ \mathfrak{F}_{a^{+},k}^{\alpha ,s}(f_{1}+f_{2})^{q}(t) \leq 2^{q-1} \mathfrak{F}_{a^{+},k}^{\alpha ,s} \bigl(f_{1}^{q}+f_{2}^{q}\bigr) (t). $$
(4.19)
Theorem 13
For
\(k>0\), \(s\in \mathbb{R}\setminus \{0\}\), \(\alpha >0\), \(p\geq 1\). Let
\(f_{1}, f_{2}\in L_{1,s}[a,t]\)
be two positive functions in
\([0,\infty)\)
such that, for all
\(t>a\), \(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{1}^{p}(t)<\infty \)
and
\(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{2} ^{p}(t)<\infty \). If
\(0< c< m\leq \frac{f_{1}(x)}{f_{2}(x)}\leq M\)
for
\(m, M\in \mathbb{R^{+}}\)
and for all
\(x\in [a, t]\), then
$$\begin{aligned} \frac{M+1}{M-c} \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s} \bigl(f_{1}(t)-cf _{2}(t)\bigr) \bigr)&\leq \bigl( \mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{1}^{p}(t) \bigr) ^{\frac{1}{p}}+ \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{2}^{p}(t) \bigr) ^{\frac{1}{p}} \\ &\leq \frac{m+1}{m-c} \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s} \bigl(f_{1}(t)-cf _{2}(t)\bigr) \bigr)^{\frac{1}{p}}. \end{aligned}$$
(4.20)
Proof
Using the hypothesis \(0< c< m\leq M\), we get
It can be concluded that
Further, we have that
implies
Again, we have that
implies
Multiplying both sides of (4.21) with \(\frac{s^{1- \frac{\alpha }{k}} ((t-a)^{s}-(x-a)^{s} )^{\frac{\alpha }{k}-1}(x-a)^{s-1}}{k \varGamma _{k}(\alpha )}\) and then integrating with respect to the variable x from a to t, we obtain
Accordingly, it can be written as
Repeating the same steps with (4.22), we obtain
The proof of (4.20) can be concluded by adding (4.23) and (4.24). □
$$\begin{aligned} mc\leq Mc&\quad \Rightarrow\quad mc+m\leq mc+M \leq Mc+M \\ &\quad \Rightarrow \quad (M+1) (m-c) \leq (m+1) (M-c). \end{aligned}$$
$$ \frac{(M+1)}{(M-c)}\leq \frac{(m+1)}{(m-c)}. $$
$$ m-c\leq \frac{f_{1}(x)-cf_{2}(x)}{f_{2}(x)}\leq M-c $$
$$ \frac{(f_{1}(x)-cf_{2}(x))^{p}}{(M-c)^{p}}\leq f_{2}^{p}(x)\leq \frac{(f _{1}(x)-cf_{2}(x))^{p}}{(m-c)^{p}}. $$
(4.21)
$$ \frac{1}{M}\leq \frac{f_{2}(x)}{f_{1}(x)}\leq \frac{1}{m}\quad \Rightarrow\quad \frac{m-c}{cm}\leq \frac{f_{1}(x)-cf_{2}(x)}{cf_{1}(x)}\leq \frac{M-c}{cM} $$
$$ \biggl(\frac{M}{M-c} \biggr)^{p} \bigl(f_{1}(x)-cf_{2}(x)\bigr)^{p}\leq f_{1}^{p}(x) \leq \biggl(\frac{m}{m-c} \biggr)^{p}\bigl(f_{1}(x)-cf_{2}(x) \bigr)^{p}. $$
(4.22)
$$\begin{aligned}& \frac{s^{1-\frac{\alpha }{k}}}{(M-c)^{p}k\varGamma _{k}(\alpha )} \int _{a}^{t} \bigl((t-a)^{s}-(x-a)^{s} \bigr)^{\frac{\alpha }{k}-1}\bigl(f _{1}(x)-cf_{2}(x) \bigr)^{p}(x-a)^{s-1}\,dx \\& \quad \leq \frac{s^{1-\frac{\alpha }{k}}}{k\varGamma _{k}(\alpha )} \int _{a}^{t} \bigl((t-a)^{s}-(x-a)^{s} \bigr)^{\frac{\alpha }{k}-1}f_{2} ^{p}(x) (x-a)^{s-1}\,dx. \\& \quad \leq \frac{s^{1-\frac{\alpha }{k}}}{(m-c)^{p}k\varGamma _{k}(\alpha )} \int _{a}^{t} \bigl((t-a)^{s}-(x-a)^{s} \bigr)^{ \frac{\alpha }{k}-1}\bigl(f_{1}(x)-cf_{2}(x) \bigr)^{p}(x-a)^{s-1}\,dx. \end{aligned}$$
$$\begin{aligned} \frac{1}{M-c} \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s} \bigl(f_{1}(t)-cf_{2}(t)\bigr)^{p} \bigr) ^{\frac{1}{p}}&\leq \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{2}(t)^{p} \bigr) ^{\frac{1}{p}} \\ &\leq \frac{1}{m-c} \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s} \bigl(f_{1}(t)-cf_{2}(t)\bigr)^{p} \bigr)^{\frac{1}{p}}. \end{aligned}$$
(4.23)
$$\begin{aligned} \frac{M}{M-c} \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s} \bigl(f_{1}(t)-cf_{2}(t)\bigr)^{p} \bigr) ^{\frac{1}{p}}&\leq \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{1}(t)^{p} \bigr) ^{\frac{1}{p}} \\ &\leq \frac{m}{m-c} \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s} \bigl(f_{1}(t)-cf_{2}(t)\bigr)^{p} \bigr)^{\frac{1}{p}}. \end{aligned}$$
(4.24)
Theorem 14
For
\(k>0\), \(s\in \mathbb{R}\setminus \{0\}\), \(\alpha >0\)
and
\(p\geq 1\). Let
\(f_{1}, f_{2}\in L_{1,s}[a,t]\)
be two positive functions in
\([0,\infty )\)
such that, for all
\(t>a\), \(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{1}^{p}(t)<\infty \)
and
\(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{2} ^{p}(t)<\infty \). If
\(0\leq a\leq f_{1}(x)\leq A\)
and
\(0\leq b\leq f _{2}(x)\leq B\)
for
\(m, M\in \mathbb{R^{+}}\)
and for all
\(x\in [a, t]\), then
with
\(c_{5}=\frac{A(a+B)+B(A+b)}{(A+b)(a+B)}\).
$$ \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{1}^{p}(t) \bigr)^{ \frac{1}{p}}+ \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{2}^{p}(t) \bigr) ^{\frac{1}{p}}\leq c_{5} \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}(f _{1}+f_{2})^{p}(t) \bigr)^{\frac{1}{p}}, $$
(4.25)
Proof
Under the given conditions, it follows that
Conducting the product between (4.26) and \(0< a\leq f_{1}(x) \leq A\), we have
From (4.27), we get
and
By multiplying both sides of (4.28) with \(\frac{s^{1-\frac{ \alpha }{k}} ((t-a)^{s}-(x-a)^{s} )^{\frac{\alpha }{k}-1}(x-a)^{s-1}}{k \varGamma _{k}(\alpha )}\) and then integrating with respect to the variable x from a to t, we obtain
Accordingly, it can be written as
Repeating the same steps with (4.29), we obtain
The proof of (4.25) can be concluded by adding (4.30) and (4.31). □
$$ \frac{1}{B}\leq \frac{1}{f_{2}(t)}\leq \frac{1}{b}. $$
(4.26)
$$ \frac{a}{B}\leq \frac{f_{1}(t)}{f_{2}(t)}\leq \frac{A}{b}. $$
(4.27)
$$ f_{2}^{p}(x)\leq \biggl(\frac{B}{a+B} \biggr)^{p}\bigl(f_{1}(x)+f_{2}(x) \bigr)^{p} $$
(4.28)
$$ f_{1}^{p}(x)\leq \biggl(\frac{A}{b+A} \biggr)^{p}\bigl(f_{1}(x)+f_{2}(x) \bigr)^{p}. $$
(4.29)
$$\begin{aligned}& \frac{s^{1-\frac{\alpha }{k}}}{k\varGamma _{k}(\alpha )} \int _{a} ^{t} \bigl((t-a)^{s}-(x-a)^{s} \bigr)^{\frac{\alpha }{k}-1}f_{2}^{p}(x) (x-a)^{s-1}\,dx \\& \quad \leq \frac{B^{p}s^{1-\frac{\alpha }{k}}}{(a+B)^{p}k\varGamma _{k}(\alpha )} \int _{a}^{t} \bigl((t-a)^{s}-(x-a)^{s} \bigr)^{ \frac{\alpha }{k}-1} \bigl(f_{1}(x)+f_{2}(x) \bigr)^{p}(x-a)^{s-1}\,dx. \end{aligned}$$
$$ \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{2}^{p}(t) \bigr)^{ \frac{1}{p}}\leq \frac{B}{a+B} \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}(f _{1}+f_{2})^{p}(t) \bigr)^{\frac{1}{p}}. $$
(4.30)
$$ \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{1}^{p}(t) \bigr)^{ \frac{1}{p}}\leq \frac{A}{b+A} \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}(f _{1}+f_{2})^{p}(t) \bigr)^{\frac{1}{p}}. $$
(4.31)
Theorem 15
For
\(k>0\), \(s\in \mathbb{R}\setminus \{0\}\), \(\alpha >0\). Let
\(f_{1}, f _{2}\in L_{1,s}[a,t]\)
be two positive functions in
\([0,\infty )\)
such that, for all
\(t>a\), \(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{1}^{p}(t)< \infty \)
and
\(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{2}^{p}(t)<\infty \). If
\(0< m\leq \frac{f_{1}(x)}{f_{2}(x)}\leq M\)
for
\(m, M\in \mathbb{R^{+}}\)
and for all
\(x\in [a, t]\), then
$$ \frac{1}{M} \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{1}(t)f_{2}(t) \bigr) \leq \frac{1}{(m+1)(M+1)} \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}(f _{1}+f_{2})^{2}(t) \bigr) \leq \frac{1}{m} \bigl(\mathfrak{F}_{a^{+},k} ^{\alpha ,s}f_{1}(t)f_{2}(t) \bigr). $$
(4.32)
Proof
Using \(0< m\leq \frac{f_{1}(x)}{f_{2}(x)}\leq M\), it follows that
Also, it follows that \(\frac{1}{M}\leq \frac{f_{2}(x)}{f_{1}(x)} \leq \frac{1}{m}\), which yields
Conducting the product between (4.33) and (4.34), we have
By multiplying both sides of (4.35) with \(\frac{s^{1-\frac{ \alpha }{k}} ((t-a)^{s}-(x-a)^{s} )^{\frac{\alpha }{k}-1}(x-a)^{s-1}}{k \varGamma _{k}(\alpha )}\) and then integrating with respect to the variable x from a to t, we obtain
with \(c_{6}=\frac{1}{(m+1)(M+1)}\). Accordingly, the required result (4.32) can be concluded. □
$$ f_{2}(x) (m+1)\leq f_{2}(x)+f_{1}(x) \leq f_{2}(t) (M+1). $$
(4.33)
$$ f_{1}(x) \biggl(\frac{M+1}{M}\biggr)\leq f_{2}(x)+f_{1}(x)\leq f_{1}(x) \biggl( \frac{m+1}{m}\biggr). $$
(4.34)
$$ \frac{f_{1}(x)f_{2}(x)}{M}\leq \frac{(f_{2}(x)+f_{1}(x))^{2}}{(m+1)(M+1)}\leq \frac{f_{1}(x)f_{2}(x)}{m}. $$
(4.35)
$$\begin{aligned}& \frac{s^{1-\frac{\alpha }{k}}}{Mk\varGamma _{k}(\alpha )} \int _{a} ^{t} \bigl((t-a)^{s}-(x-a)^{s} \bigr)^{\frac{\alpha }{k}-1}f_{1}(x)f _{2}(x) (x-a)^{s-1} \,dx \\& \quad \leq c_{6}\frac{s^{1-\frac{\alpha }{k}}}{k\varGamma _{k}(\alpha )} \int _{a}^{t} \bigl((t-a)^{s}-(x-a)^{s} \bigr)^{\frac{\alpha }{k}-1}\bigl(f _{2}(x)+f_{1}(x) \bigr)^{2}(x-a)^{s-1}\,dx \\& \quad \leq \frac{s^{1-\frac{\alpha }{k}}}{mk\varGamma _{k}(\alpha )} \int _{a}^{t} \bigl((t-a)^{s}-(x-a)^{s} \bigr)^{\frac{\alpha }{k}-1}f_{1}(x)f _{2}(x) (x-a)^{s-1} \,dx \end{aligned}$$
Theorem 16
For
\(k>0\), \(s\in \mathbb{R}\setminus \{0\}\), \(\alpha >0\)
and
\(p\geq 1\). Let
\(f_{1}, f_{2}\in L_{1,s}[a,t]\)
be two positive functions in
\([0,\infty)\)
such that, for all
\(t>a\), \(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{1}^{p}(t)<\infty \)
and
\(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{2} ^{p}(t)<\infty \). If
\(0< m\leq \frac{f_{1}(x)}{f_{2}(x)}\leq M\)
for
\(m, M\in \mathbb{R^{+}}\)
and for all
\(x\in [a, t]\), then
where
\(h(f_{1}(x),f_{2}(x))= \max \{M[ (\frac{M}{m}+1 )f_{1}(t)-Mf _{2}(t)],\frac{(m+M)f_{2}(t)-f_{1}(t)}{m}\}\).
$$ \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{1}^{p}(t) \bigr)^{ \frac{1}{p}}+ \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{2}^{p}(t) \bigr) ^{\frac{1}{p}}\leq 2 \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}h^{p} \bigl(f _{1}(t),f_{2}(t)\bigr) \bigr)^{\frac{1}{p}}, $$
(4.36)
Proof
Under the given conditions \(0< m\leq \frac{f_{1}(x)}{f_{2}(x)}\leq M\), \(a\leq x\leq t\), it can be written
and
From (4.35) and (4.38), we obtain
where \(h(f_{1}(x),f_{2}(x))= \max \{M[ (\frac{M}{m}+1 )f_{1}(t)-Mf _{2}(t)],\frac{(m+M)f_{2}(t)-f_{1}(t)}{m}\}\).
$$ 0< m\leq M+m-\frac{f_{1}(x)}{f_{2}(x)}, $$
(4.37)
$$ M+m-\frac{f_{1}(x)}{f_{2}(x)}\leq M. $$
(4.38)
$$ f_{2}(x)< \frac{(M+m)f_{2}(x)-f_{1}(x)}{m}\leq h\bigl(f_{1}(x),f_{2}(x) \bigr), $$
(4.39)
From hypothesis, it also follows that \(0<\frac{1}{M}\leq \frac{f_{2}(x)}{f _{1}(x)}\leq \frac{1}{m}\) implies that
and
From (4.40) and (4.41), we get
which can be rewritten as
We can write from (4.39) and (4.43)
By multiplying both sides of (4.44) with \(\frac{s^{1-\frac{ \alpha }{k}} ((t-a)^{s}-(x-a)^{s} )^{\frac{\alpha }{k}-1}(x-a)^{s-1}}{k \varGamma _{k}(\alpha )}\) and then integrating with respect to the variable x from a to t, we obtain
Accordingly, it can be written as
Repeating the same procedure as above, for (4.45), we have
The required result (4.36) follows from (4.46) and (4.47). □
$$ \frac{1}{M}\leq \frac{1}{M}+\frac{1}{m}- \frac{f_{2}(x)}{f_{1}(x)} $$
(4.40)
$$ \frac{1}{M}+\frac{1}{m}-\frac{f_{2}(x)}{f_{1}(x)}\leq \frac{1}{m}. $$
(4.41)
$$ \frac{1}{M}\leq \frac{(\frac{1}{M}+\frac{1}{m})f_{1}(x)-f_{2}(x)}{f _{1}(x)}\leq \frac{1}{m}, $$
(4.42)
$$\begin{aligned} f_{1}(x)&\leq M\biggl(\frac{1}{M}+\frac{1}{m} \biggr)f_{1}(x)-Mf_{2}(x) \\ &= \frac{M(M+m)f_{1}(x)-M^{2}mf_{2}(x)}{mM} \\ &= \biggl(\frac{M}{m}+1 \biggr)f_{1}(x)-Mf_{2}(x) \\ &\leq M\biggl[ \biggl(\frac{M}{m}+1 \biggr)f_{1}(x)-Mf_{2}(x) \biggr] \\ &\leq h\bigl(f_{1}(x),f_{2}(x)\bigr). \end{aligned}$$
(4.43)
$$\begin{aligned}& f_{1}^{p}(x)\leq h^{p} \bigl(f_{1}(x),f_{2}(x)\bigr), \end{aligned}$$
(4.44)
$$\begin{aligned}& f_{2}^{p}(x)\leq h^{p} \bigl(f_{1}(x),f_{2}(x)\bigr). \end{aligned}$$
(4.45)
$$\begin{aligned}& \frac{s^{1-\frac{\alpha }{k}}}{k\varGamma _{k}(\alpha )} \int _{a} ^{t} \bigl((t-a)^{s}-(x-a)^{s} \bigr)^{\frac{\alpha }{k}-1}f_{1}^{p}(x) (x-a)^{s-1}\,dx \\& \quad \leq \frac{s^{1-\frac{\alpha }{k}}}{k\varGamma _{k}(\alpha )} \int _{a}^{t} \bigl((t-a)^{s}-(x-a)^{s} \bigr)^{\frac{\alpha }{k}-1}h^{p}\bigl(f _{1}(x),f_{2}(x) \bigr) (x-a)^{s-1}\,dx. \end{aligned}$$
$$ \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{1}^{p}(t) \bigr)^{ \frac{1}{p}}\leq \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}h^{p} \bigl(f_{1}(t),f _{2}(t)\bigr) \bigr)^{\frac{1}{p}}. $$
(4.46)
$$ \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}f_{2}^{p}(t) \bigr)^{ \frac{1}{p}}\leq \bigl(\mathfrak{F}_{a^{+},k}^{\alpha ,s}h^{p} \bigl(f_{1}(t),f _{2}(t)\bigr) \bigr)^{\frac{1}{p}}. $$
(4.47)
By (2.25) and Theorem (7), under the appropriate values of parameters for each individual fractional integral, the preceding introduced and proved theorems (Theorem 9 to Theorem 16) can be deduced as particular cases, each result involving the following fractional integrals: Riemann–Liouville, Hadamard, Liouville, Katugampola, and generalized fractional conformable integrals.
5 Conclusion
This manuscript starts with a concise overview of fractional integrals in the sense of Riemann–Liouville, Hadamard, and Katugampola as well as a new fractional integral operator according to Jarad et al. [1]. We define the formulation of k-fractional conformable integral operators and their existence. We generalize the reverse Minkowski inequality using k-fractional conformable integrals; as a particular case, the inequality involving fractional integrals in the Riemann–Liouville, Hadamard, and Katugampola sense is given [15]. The related important inequalities involving k-fractional conformable integral are also illustrated. Several inequalities can be generalized for the application of these newly introduced fractional integral operators. Amongst them, we cite the Chebyshev inequality, Grüss-type inequality, and Chebyshev–Grüss type inequality recently introduced and proved in [35‐37].
Acknowledgements
The authors would like to express profound gratitude to referees and the editor for deeper review of this paper and their valuable advice.
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