On new Milne-type inequalities and applications

Inequalities play a major role in pure and applied mathematics. In particular, the inequality plays an important role in the study of Rosseland’s integral for the stellar absorption. In this paper we obtain new Milne-type inequalities, and we apply them to the generalized Riemann–Liouville-type integral operators, which include most of the known Riemann–Liouville 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 recent years there has been a growing interest in the study of many classical inequalities applied to integral operators associated with different types of fractional derivatives, since integral inequalities and their applications play a vital role in the theory of differential equations and applied mathematics. Some of the inequalities studied are Gronwall, Chebyshev, Hermite-Hadamard-type, Ostrowski-type, Grüss-type, Hardy-type, Gagliardo-Nirenberg-type, Jensen-type, Opial-type, Milne-type, reverse Minkowski, and reverse Hölder inequalities (see, e.g., [1, 3, 4, 7-9, 11, 12, 14-20]).
In this work we obtain new Milne-type inequalities, and we apply them to the generalized Riemann-Liouville-type integral operators defined in [2], which include most of the known Riemann-Liouville integral operators.

Preliminaries
One of the first operators that can be called fractional is the Riemann-Liouville fractional derivative of order α ∈ C, with Re(α) > 0, defined as follows (see [6]). Definition 1 Let a < b and f ∈ L 1 ((a, b); R). The right-and left-side Riemann-Liouville fractional integrals of order α, with Re(α) > 0, are defined, respectively, by and with t ∈ (a, b).
When α ∈ (0, 1), their corresponding Riemann-Liouville fractional derivatives are given by Other definitions of fractional operators are the following ones. Definition 2 Let a < b and f ∈ L 1 ((a, b); R). The right-and left-side Hadamard fractional integrals of order α, with Re(α) > 0, are defined, respectively, by and with t ∈ (a, b).
When α ∈ (0, 1), the Hadamard fractional derivatives are given by the following expressions: with t ∈ (a, b).  [10] of order α of f with respect to g are defined, respectively, by and with t ∈ (a, b).
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 [2].
The right and left integral operators, denoted, respectively, by J α T,a + and J α T,b -, are defined for each measurable function f on [a, b] as We Note that these operators generalize the integral operators in Definitions 1, 2, and 3: (A) If we choose then J α T,a + and J α T,b -are the right and left Riemann-Liouville fractional integrals RL J α a + and RL J α b -in (1) and (2), respectively. The corresponding right and left Riemann-Liouville fractional derivatives are (B) If we choose then J α T,a + and J α T,b -are the right and left Hadamard fractional integrals H α a + and H α b -in (3) and (4), respectively. The corresponding right and left Hadamard fractional derivatives are (C) If we choose a function g with the properties in Definition 4 and then J α T,a + and J α T,b -are the right and left fractional integrals I α g,a + and I α g,b -in (5) and (6), respectively.
for each t ∈ (a, b).
Note that if we choose . Also, we can obtain Hadamard and others fractional derivatives as particular cases of this generalized derivative.

Milne-type inequalities
Milne proved in 1925 the two following discrete and continuous versions of a useful inequality [13]: The following inequality holds for every
We start with our general version of Proposition 6.
If we define k i,j := c i,j y j /x i , then it suffices to prove that Let us prove (11) by induction on n.
If n = 1, then the inequality (11) holds since, in fact, it is an equality. If n = 2, a i := 1/k i,1 and b i := 1/k i,2 , then the following inequalities are equivalent and this last inequality holds by Proposition 6.
Finally, assume that (11) holds for n -1 ≥ 2. Then, the induction hypothesis and the previous inequality give which completes the proof of (11). Hence, (10) If we take limits as x i → 0 and/or y j → 0 for some indices 1 ≤ i ≤ m, 1 ≤ j ≤ n in (10), we obtain the same conclusion if (10), then we obtain for every n for every n. Then, the result follows if we take limits as n → ∞ in this last inequality.
Remark 10 The argument in the proof of Theorem 9 gives that Proposition 6 is equivalent to the case n = 2 in (10). Hence, this theorem is a generalization of the discrete Milne inequality.
x i > 0 and n j=1 y j > 0, then In order to prove Theorem 20 below, generalizing Proposition 8 (the continuous version of Milne inequality), we need the following technical results.

Proposition 14
Let μ, ν be two measures on the spaces X, Y , respectively, and f n : X × Y → [0, ∞] measurable functions with f n ≤ f n+1 for every n, and let f := lim n→∞ f n . If .
Proof The monotone convergence theorem gives for every x ∈ X.
Since f n ≤ f for every n, if for every n then, by hypothesis. Since f 1 ≤ f n for every n, we have ∈ L 1 (μ), and the dominated convergence theorem gives the result.

Proposition 15
Let μ, ν be two measures on the spaces X, Y , respectively, and f n : X × Y → [0, ∞] measurable functions with f n ≥ f n+1 for every n, and let f := lim n→∞ f n . If for each x ∈ X, then .

Proposition 16
Let μ, ν be two measures on the spaces X, Y , respectively, and f n : measurable functions with f n ≤ f n+1 for every n, and let f := lim n→∞ f n . If .

Proposition 17
Let μ, ν be two measures on the spaces X, Y , respectively, and f n : X × Y → [0, ∞] measurable functions with f n ≥ f n+1 for every n, and let f := lim n→∞ f n . Then, for every n and y ∈ Y , the monotone convergence theorem gives for every n and y ∈ Y , there exists the limit ≥ 0 for every n and y ∈ Y , the Fatou lemma gives .
Let us recall some background in [5]. A measure μ defined on the σ -algebra of all Borel sets in a locally compact Hausdorff space X is called a Borel measure on X. The measure μ is called outer regular on E if A Radon measure on X is a Borel measure that is finite on all compact sets, outer regular on all Borel sets, and inner regular on all open sets. Radon measures are also inner regular on all of their σ -finite sets [5, p. 216].
Recall that a second-countable space is a topological space whose topology has a countable base. In a second-countable, locally compact Hausdorff space, any Borel measure that is finite on compact sets is regular and hence Radon [5, p. 217].

Proposition 18
If μ is a Radon measure on X, then C c (X) is dense in L p (μ) for 1 ≤ p < ∞.

Proposition 19
If X, Y are second-countable, locally compact Hausdorff spaces and μ, ν are Radon measures on X and Y , respectively, then μ × ν is a Radon measure on X × Y .
We can prove now the main result in this work, generalizing Proposition 8, the continuous version of the Milne inequality.
Theorem 20 Let X, Y be second-countable, locally compact metric spaces, and let μ, ν be Borel measures on the metric spaces X, Y , respectively, which are finite on compact sets. If f : X × Y → [0, ∞] is any measurable function, then Proof Since X, Y are σ -compact, there exist two sequences of compact sets {X m }, {Y n }, with By hypothesis, μ(X m ), ν(Y n ) < ∞ for every m, n.
As usual, we denote by B X (x, r) the open ball in X with center x ∈ X and radius r > 0. For each δ > 0 consider the open covering {B X (x, δ/5)} x∈X m of X m . Since X m is a compact set there exists a finite subset {x 1 , . . . , x k } ⊆ X m with X m ⊆ B X (x 1 , δ/5) ∪ · · · ∪ B X (x k , δ/5).

Thus, the measurable sets {X
are a partition of X m and In a similar way we can find a partition of measurable sets {Y j n } j=1 of Y n such that diam(Y j n ) < δ/2 for every 1 ≤ j ≤ . Case A. Let us fix m, n and a continuous function f : X m × Y n → (0, ∞). We are going to prove Since f is a strictly positive, continuous function on the compact set X m × Y n , we have Let us fix ε > 0. Since f is uniformly continuous on the compact set X m × Y n , there exists δ > 0 such that

Consider partitions of measurable sets {X
Thus, for every 1 ≤ i ≤ k and 1 ≤ j ≤ . Let us consider the simple function where χ E denotes the characteristic function of the set E, i.e., the function with χ E = 1 on E and χ E = 0 on X \ E.
We have Thus, the two following inequalities are equivalent: since X m and Y n are compact sets and so, (μ × ν)(X m × Y n ) = μ(X m )ν(Y n ) < ∞. Now, we can choose a sequence of ε converging to 0 and so, the corresponding s ε converge to f . Since μ(X m ), ν(Y n ) < ∞, Propositions 12 and 13 give Hence, there exists a subsequence of {f k }, that we will denote also by {f k } for simplicity, such that lim k→∞ f k = f (μ × ν)-a.e. Let us consider the sequence {F k } defined from {f k } as Thus, {F k } ⊂ C(X m ×Y n ) and we also have lim k→∞ F k = f (μ×ν)-a.e., since |f -F k | ≤ |f -f k | for every k. We have proved that . inequality plays an important role in the study of Rosseland's integral for the stellar absorption. In this paper we obtain the Milne-type inequality , with appropriate hypotheses, and we apply it to the generalized Riemann-Liouville-type integral operators, which include most of the known Riemann-Liouville integral operators.
Although the assumptions of this inequality are not very restrictive, an interesting open problem is to weaken these assumptions for at least one of the two measures.