Equivalent Properties of Two Kinds of Hardy-Type Integral Inequalities

: In this paper, using weight functions as well as employing various techniques from real analysis, we establish a few equivalent conditions of two kinds of Hardy-type integral inequalities with nonhomogeneous kernel. To prove our results, we also deduce a few equivalent conditions of two kinds of Hardy-type integral inequalities with a homogeneous kernel in the form of applications. We additionally consider operator expressions. Analytic inequalities of this nature and especially the techniques involved have far reaching applications in various areas in which symmetry plays a prominent role, including aspects of physics and engineering.


Introduction
In 1925, by introducing one pair of conjugate exponents (p, q), Hardy [1] established a well-known extension of Hilbert's integral inequality as follows.
In this paper, using weight functions as well as employing various techniques from real analysis, we establish a few equivalent conditions of two kinds of Hardy-type integral inequalities with the nonhomogeneous kernel: To prove our results, we also deduce a few equivalent conditions of two kinds of Hardy-type integral inequalities with a homogeneous kernel in the form of applications. We additionally consider operator expressions. Analytic inequalities of this nature and especially the techniques involved have far reaching applications in various areas in which symmetry plays a prominent role, including aspects of physics and engineering.

Two Lemmas
For σ > 0, by the Lebesgue term-by-term integration theorem, we derive that: Setting v = (kλ + σ)(− ln u) in the above integral, we obtain: where: stands for the gamma function and: which is a function very well known for its applications in analytic number theory. (7), we obtain that: In the sequel , we assume that p > 1, 1 Lemma 1. If β > −1, σ, λ > 0, there exists a constant M 1 , such that for any non-negative measurable functions f (x) and g(y) in (0, ∞), the following inequality: holds true. Then, we have σ 1 = σ, and M 1 ≥ k 1 (σ).
Proof. If σ 1 > σ, then for n ≥ 1 σ 1 −σ (n ∈ N), we set the following two functions: and deduce that: Setting u = xy, we obtain: and then by (9), we have: By (10), in view of: , we set the following two functions: and obtain: Setting u = xy, we obtain: and then by Fubini's theorem and (9), we have: Since (σ − σ 1 ) − 1 n ≥ 0, it follows that: By (11), in view of the fact that we obtain that ∞ < ∞, which is a contradiction. Hence, we conclude that σ 1 = σ.
For σ 1 = σ, we reduce (11) as follows: Since: is non-negative and increasing in (0, 1], by Levi's theorem, we derive that: This completes the proof of the lemma. there exists a constant M 2 , such that for any non-negative measurable functions f (x) and g(y) in (0, ∞), the following inequality: holds true. Then, we have σ 1 = σ, and M 2 ≥ k 2 (σ).
Proof. If σ 1 < σ, then for n ≥ 1 σ−σ 1 (n ∈ N), we set two functions f n (x) and g n (y) as in Lemma 1, and derive that: Setting u = xy, we obtain: and then by (13), we deduce that: Since (σ 1 − σ) + 1 n ≤ 0, it follows that: By (14), in view of we have ∞ < ∞, which is a contradiction. If σ 1 > σ, then for n ≥ 1 σ 1 −σ (n ∈ N), we set two sequences of f n (x) and g n (y) as in Lemma 1, and obtain: Setting u = xy, we obtain: and then, by Fubini's theorem and (13), we have: Since (σ − σ 1 ) + 1 n ≤ 0, it follows that By (15), in view of the fact that: we have ∞ < ∞, which is a contradiction. Hence, we conclude the fact that σ 1 = σ. For σ 1 = σ, we reduce (15) as follows: Since: is non-negative and increasing in [1, ∞), still by Levi's theorem, we have: This completes the proof of the Lemma.
(i) There exists a constant M 1 , such that for any f (x) ≥ 0, satisfying: we have the following Hardy-type integral inequality of the first kind with nonhomogeneous kernel: (ii) There exists a constant M 1 , such that for any f (x), g(y) ≥ 0, satisfying: we have the following inequality: (iii) σ 1 = σ. If Condition (iii) holds, then M 1 ≥ k 1 (σ) and the constant factor: in (17) and (18) is the best possible.
When Condition (iii) is satisfied, if there exists a constant factor M 1 ≤ k 1 (σ), such that (18) is valid, then by Lemma 1 we have M 1 ≥ k 1 (σ). Then, the constant factor (18) is the best possible. The constant factor M 1 = k 1 (σ) in (17) is still the best possible. Otherwise, by (19) (for σ 1 = σ), we can conclude that the constant factor (18) is not the best possible.
(i) There exists a constant M 1 , such that for any f (x) ≥ 0, satisfying: we have the following Hardy-type inequality of the first kind with homogeneous kernel: (ii) There exists a constant M 1 , such that for any f (x), g(y) ≥ 0, satisfying: we have the following inequality: If Condition (iii) holds, then we have M 1 ≥ k 1 (σ), and the constant M 1 = k 1 (σ) in (22) and (23) is the best possible.
Similarly, we obtain the following weight function: and then in view of Lemma 2 and in a similar manner, we obtain the following theorem: we have the following Hardy-type inequality of the second kind with the nonhomogeneous kernel: (ii) There exists a constant M 2 , such that for any f (x), g(y) ≥ 0, satisfying: we have the following inequality: If Condition (iii) holds, then we have M 2 ≥ k 2 (σ), and the constant factor: in (24) and (25) is the best possible. Setting: in Theorem 2, then replacing Y (resp. G(Y)) by y (resp. g(y)), we derive the following Corollary.

Corollary 2.
If β > −1, 0 < σ = λ − µ < λ, then the following conditions are equivalent. (i) There exists a constant M 2 , such that for any f (x) ≥ 0, satisfying: we have the following Hardy-type inequality of the second kind with homogeneous kernel: (ii) There exists a constant M 2 , such that for any f (x), g(y) ≥ 0, satisfying we have the following inequality: If Condition (iii) holds, then we have M 2 ≥ k 2 (σ), and the constant M 2 = k 2 (σ) = k 1 (µ) in (26) and (27) is the best possible.
In view of (28), it follows that:
In view of (31), it follows that: