On a generalization of the Opial inequality

: Inequalities are essential in pure and applied mathematics. In particular, Opial ’ s inequality and its generalizations have been playing an important role in the study of the existence and uniqueness of initial and boundary value problems. In this work, some new Opial-type inequalities are given and applied to generalized Riemann-Liouville-type integral operators.


Introduction
Integral inequalities are used in countless mathematical problems such as approximation theory and spectral analysis, statistical analysis, and the theory of distributions.Studies involving integral inequalities play an important role in several areas of science and engineering.
In this work, we obtain new Opial-type inequalities, and we apply them to the generalized Riemann-Liouville-type integral operators defined in [15], which include most of the known Riemann-Liouville-type integral operators.

Preliminaries
One of the first operators that can be called fractional is the Riemann-Liouville fractional derivative of order ∈ α , with ( ) > α Re 0, defined as follows (see [16]). and with ( ) ∈ t a b , .
When ( ) ∈ α 0, 1 , their corresponding Riemann-Liouville fractional derivatives are given by When ( ) ∈ α 0, 1 , Hadamard fractional derivatives are given by the following expressions: : , an integrable function, and ( ) ∈ α 0, 1 a fixed real number.The right and left side fractional integrals in [17] of order α of f with respect to g are defined, respectively, by and with There are other definitions of integral operators in the global case, but they are slight modifications of the previous ones.

General fractional integral of Riemann-Liouville type
Now, we give the definition of a general fractional integral in [15]., with continuous positive derivative on ( ) a b , , and , .
The right and left integral operators, denoted, respectively, by Note that these operators generalize the integral operators in Definitions 1-3: (1) If we choose (3) If we choose a function g with the properties in Definition 4 and , with continuous positive derivative on ( ) a b , , and , its right and left generalized derivatives of order α are defined, respectively, by for each Note that if we choose

RL
. Also, we can obtain Hadamard and other fractional derivatives as particular cases of this generalized derivative.

Muckenhoupt inequality
Let us consider ≤ ≤ < ∞ p q 1 and measures μ μ , where we use the convention ⋅ ∞ = 0 0.Moreover, we can choose Muckenhoupt inequality will play a crucial role to prove the next result, which improves the classical Opial inequality in several ways: .Assume that the constant B defined as follows is finite: Then, for every absolutely continuous function , , , , where the constant C can be chosen as p p q Proof.By the Muckenhoupt inequality, the constant C satisfies , with ( ) = f a 0, we have that there exists , , Hence, the Hölder inequality gives has zero Lebesgue measure, but it is possible to have ( ) > μ S 0 0 and/or ( ) > μ S 0 1 . The argument in the proof of Theorem 1 gives that the inequality holds for any fixed choice of values of ′ f on S.
Theorem 1 has the following direct consequence.
On a generalization of the Opial inequality  5 Corollary 2. Let us consider ≤ ≤ < ∞ p q 1 and a measure μ on [ ] a b , with ({ }) = μ b 0. Assume that the constant B defined as follows is finite: Then, for every absolutely continuous function f on [ ] a b , with ( ) = f a 0, , , , , where the constant C can be chosen as Corollary 2 is a tool to obtain the following result. ,with ({ }) = μ b 0. Assume that the constant B defined as follows is finite: Then, for every absolutely continuous function (2) if = p 1 and μ is a finite measure, Proof.Assume first that < ≤ p 1 2. Let us consider ≥ q 2 such that ∕ + ∕ = p q 1 1 1, and so, ( ) = ∕ − p q q 1 and ( ) = ∕ − q p p 1 .Thus, < ≤ ≤ < ∞ p q 1 2 and Corollary 2 gives the result in part (a), since Assume now that μ is a finite measure and fix an absolutely continuous function . We have proved that Since μ is a finite measure, we have for every < ≤ p p 1 0 , dominated convergence theorem gives Finally, we have and the desired inequality holds if

Let us consider now any absolutely continuous function
, then the inequality is direct.So, we can assume that . Thus, there exists a sequence { } s n of simple functions with Hence, there exists N such that for every ≥ n N .Therefore, for every ≥ n N .
Since μ is a finite measure, if we define ( ) ( ) for every ≥ p 1, and we have proved that Also, for any Applying inequalities ( 12), (13), and ( 14) where appropriate, for every ≥ n N .Hence, On a generalization of the Opial inequality  7 which completes part (b).□ If we choose μ as the Lebesgue measure on [ ] a b , , then we obtain the following results.
, , for every absolutely continuous function f on [ ] a b , with ( ) = a 0.
Proof.Let us compute For each > α 0 and ≥ β 0, consider the function u defined on [ ] a b , as Since ( ) Hence, Corollary 2 gives the result.
2, and for every absolutely continuous function f on [ ] a b , such that ( ) = f a 0.

□
Remark 2. Note that in the second inequality in Corollary 5: the constant 1 multiplying ‖ ‖ does not depend on the length of the interval [ ] a b , .
Corollary 2 and Theorem 3 have, respectively, the following direct consequences for general fractional integrals of Riemann-Liouville type.Proposition 6.Let us consider ≤ ≤ < ∞ p q 1 and assume that the constant B defined as follows is finite:

1 .
Other definitions of fractional operators are the following ones.Definition 2. Let < a b and (The right and left side Hadamard fractional integrals of order α, with Re( ) > α 0, are defined, respectively, by

,
are the right and left Riemann-Liouville fractional integrals + and (2), respectively.Its corresponding right and left Riemann-Liouville fractional derivatives are corresponding right and left Hadamard fractional derivatives are

,, 3 Definition
are the right and left fractional integrals + in (5) and (6), respectively.On a generalization of the Opial inequality 

( 1 )Theorem 1 .
It allows us to integrate with respect to very general measures.(2) The hypotheses ( ) = f b 0 and > f 0 on ( ) a b , are no longer needed.(3) The hypothesis [ ] ∈ f C a b , 1 is replaced by a weaker one: it is sufficient to require f to be absolutely continuous on [ ] a b , .Let us consider ≤ ≤ < ∞ p function u on [ ] a b , .For each absolutely continuous function f on [ ] a b

Theorem 3 .
Let us consider ≤ ≤ p 1 2 and a measure μ on [ ] a b

9 Proposition 7 . 1 2 2
On a generalization of the Opial inequality  Let us consider ≤ ≤ p and assume that the constant B defined as follows is finite:Funding information: Ana Portilla, Jose M. Rodriguez, and Jose M. Sigarreta are supported in part by a grant from Agencia Estatal de Investigación (PID2019-106433GB-I00/AEI/10.13039/501100011033),Spain.Jose M. Rodriguez was also supported by the Madrid Government (Comunidad de Madrid-Spain) under the Multiannual Agreement with UC3M in the line of Excellence of University Professors (EPUC3M23), and in the context of the V PRICIT (Regional Programme of Research and Technological Innovation).