On the Best Constant Problem for a Class of Mixed-Norm Hardy Inequalities

In this paper, we obtain the best constant and the equality condition for a class of mixed-norm Hardy inequalities when the weight is a power function. By building and solving the corresponding Euler equation, we look for the best constant and the optimal function. One of the main ingredients is to introduce two key auxiliary functions so that the corresponding equalities are derived.


Introduction
If the powers of the integrands on both sides of the Hardy inequality are different, we call this kind of Hardy inequality as the mixed-norm Hardy inequality. The resulting method, best constant and condition of equality of the mixed-norm Hardy inequality are rather different from the ones for the usual Hardy inequality.
By our knowledge, the earliest example of the mixed-norm Hardy inequality was given in 1930. In the paper [1], G.H. Hardy and J.E. Littlewood presented a conjecture of the following inequality by adopting the variational method and solving the corresponding Euler equation: Let l > k > 1, r = l k − 1, f is a non-negative and measurable function in (0, ∞) and It is noted that the strict proof of the inequality (1.1) was first given by G.A. Bliss in [2]. In 1958, T.M. FLett gave another mixed-norm Hardy inequality in [3]: Let f (t) ≥ 0 in t ≥ 0, and let F α (t) be the α-th Riemann-Liouville integral of f with origin 0, i.e.
If either q ≥ p ≥ 1 and α > 1 p − 1 q , or q > p > 1 and α = 1 p − 1 q , and if γ > −1, then there exists a positive constant B, such that When q > p, α = 1 − 1 q , γ = − 1 p and p > q, then When q = p > 1, then B = (α) (γ +1) (α+γ +1) . The Young inequality for convolution operator is a suitable tool to deal with the mixednorm Hardy inequality. However, this method can't give the best constant and the condition of equality in general. In the monograph [4] on the integral inequalities, Claude George proved the following inequality by using the Young inequality for convolution operator: Suppose that p ≥ 1, q ≥ 1, r ≥ 1, 1 p + 1 q = 1, 1 p + 1 r − 1 > 0, f is non-negative and measurable in (0, ∞) and satisfies In [5], J.S. Bradley proved the following general conclusions about the weighted mixednorm Hardy inequality, which are the generalizations of the case p = q ≥ 1 in [6]: holds for all admissible f ≥ 0, then holds for all admissible f implies that When p ≤ q, (1.5) is also sufficient for (1.4), to hold for all admissible f .
In [7], P.R. Beesack and H.P. Heinig gave the following two dual conclusions including the negative exponent.
holds for all admissible f , then Moreover, in case q ≤ p < 0, K(r) not only satisfies (1.7) but also is non-decreasing on R + , then (1.6) holds for some C > 0 and all admissible f .
holds for all admissible f , then Moreover, in case q ≤ p < 0, J (r) not only satisfies (1.9) but also is non-increasing on R + , then (1.8) holds for some C > 0 and all admissible f .
holds for all admissible f , then Moreover if 0 < q ≤ p < 1, J (r) not only satisfies (1.11) but also is non-increasing, holds for all admissible f , then Moreover if 0 < q ≤ p < 1, K(r) not only satisfies (1.13) but also is non-decreasing, The above theorems are all programmatic for the weighted mixed-norm Hardy inequalities. Although these theorems do not give the best constants, we intend to solve this for some weighted mixed-norm Hardy inequalities based on these theorems.
In [8], Li and Mao obtained the following conclusion by combining with the measure theory: There are other results on the mixed-norm Hardy inequalities, one can see [9][10][11][12][13][14]. Now, let 1 < p < q < ∞ and λ = 1. According to Theorem 1.1, there exists a constant C(p, q, λ) > 0, such that holds for all admissible f ≥ 0, where Base on the inequality (1.1), when λ = 1 p , the best constant in the inequality (1.14) is and the corresponding condition of equality is According to Theorem 1.1, B is a lower bound, while Bp 1 q (p ) 1 p is an upper bound of the best constant C. However, the best constant C and the corresponding condition of equality are not determined, which is still an open problem [15,16].
We will solve the above problem in this paper. Our main conclusions are as follows:

then we have
(1) There exists the following inequality (2) If and only if there exists two positive constants c, d, such that

the equality holds.
This paper is organized as follows. In Sect. 2, we introduce some necessary notations and preknowledge. In Sect. 3, we give the corresponding Euler equation which the optimal function satisfies, and conjecture the best constant and optimal function. In Sect. 4, we introduce an important auxiliary function φ and derive its some properties. In Sect. 5, an important auxiliary function W is introduced and its some basic properties are given. In Sect. 6, we prove the inequality when f is a continuous function. In Sect. 7, the equality condition of the inequality is determined in the continuous case. In Sect. 8, we prove the inequality when f is a measurable function and determine the equality condition.

Some Necessary Notations and Preknowledge
From now on, we suppose that 1 < p < q < ∞ are two constants, and r = q p − 1. For any constant λ = 1, we denote E(λ) as the set of all non-negative and measurable We further define a functional in E(λ) as: and In fact, C(p, q, λ) is the best constant in the inequality (1.14). For any constant λ = 1, we denote It's easy to know that I (a, b, c) converges if and only if bc > a. When bc > a, we obtain .
Now, let's introduce some necessary lemmas.
then y(x) is finite for every x > 0. Moreover, the following limit formula holds The following lemma reveals the relation between the case λ < 1 and the case λ > 1 in the inequality (1.14). Based on this, the proof of Theorem 1.5 can be simplified.

The Corresponding Euler Equation
With the help of Lemma 2.2, we only need to study the case of λ < 1. Throughout Sects. 3-8, λ < 1 is always assumed. Now we introduce and prove a key lemma.
By direct computation, we can get (3.1) Take derivatives on both side of (3.1) with respect to x 0 , we can get Now, we point out that there is no x 0 > 0 to satisfy the following property (p): Let n → ∞ on both side of the above equality, we can get Since A < 0, x q(λ−1)−1 y q−1 (x) is non-negative and continuous on [x 0 , ∞), then which is a contradiction. Therefore, there is no x 0 > 0 to satisfy the property (p).
Let's assume that there is a x 0 > 0, such that y (x 0 ) = 0. As we know, x 0 can't satisfy the property (p), so there exists a δ > 0, such that y ( Let (a, b) be the largest open interval which contains x 0 and satisfies y (x) = 0 in it, then a > 0 or b < ∞. Therefore, b satisfies the property (p), which is also a contradiction. Therefore, y (x) > 0, x ∈ (0, ∞). Thus, we have proved (1). As Take derivatives on both side of the above equality, we can get Thus, we have proved (2) and completed the proof of the lemma. Now we give a solution of the following Cauchy problem: where A is a negative constant. We further demand that y (x) > 0, x ∈ (0, ∞) and Let y(x) = γ x αβ (1+x α ) β be a solution of (3.3), where α, β, γ are pending and positive numbers. By direct calculation, we can obtain I (a, b, c)(a, b, c > 0, bc > a), we can get

. With the help of the integral
Thus, we can conjecture that k(p, q, λ) is the best constant, while g (x) is an optimal function.

Auxiliary Function φ
Next, we start to prove Theorem 1.5. In this section, we introduce an important auxiliary function φ.
Let φ be the function defined by the following equation Let's discuss some related problems about φ.

Auxiliary Function W
In this section, we construct an auxiliary function W based on φ, and determine some limits and partial derivatives of W .
The following equation  If we treat W as a function of the independent variables x, y, z, u, combining with (4.2), we can get the partial derivative of W with respect to u: Then, combining with the chain rule and (4.5)-(4.7), we can get the partial derivatives of W with respect to x, y, z: (5.6)

Proof for the Equality Condition in Continuous Case
Suppose that f is non-negative and continuous in (0, ∞) and satisfies where y(x), z(x) is determined by (6.1). tions in (0, ∞) and satisfy The values of x 1 , x 2 are still determined by (6.2). Set W (x) = W (x, y(x), z(x)), x ∈ (x 1 , x 2 ), then the equality lim determined by (6.3). Take any x 1 < x 3 < x 4 < x 2 , then y(x), z(x) are all positive and continuous functions on still holds, where y 2 = y(x 2 ). Therefore, we have   ≤ k(p, q, λ). Therefore,  ≤ k(p, q, λ).
Till now, we have proved Theorem 1.5 completely.
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