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On a Hilbert-type inequality with a homogeneous kernel in ℝ2 and its equivalent form
Journal of Inequalities and Applicationsvolume 2012, Article number: 94 (2012)
By using the way of weight functions and the technique of real analysis, a new integral inequality with a homogeneous kernel and the best constant factor in ℝ2 is given. The equivalent form and the reverses are considered.
Mathematics Subject Classification (2000): 26D15.
One hundred years ago, Hilbert proved the following classic inequality 
The inequality (1.1) may be classified into several types (discrete and integral etc.), which is of great importance in analysis and its applications [1, 2]. Ever since the advent of inequality (1.1), all kinds of improvements and extensions can be seen in [3–12]. Note that the kernel of (1.1) is homogeneous of degree -1. In 2009,  reviews the negative degree homogeneous kernel of the parameterized Hilbert-type inequalities.
In recent years, many authors have started on Hilbert-type inequality of 0-degree homo-geneous kernel and non-homogeneous kernel. They even established inequalities in ℝ2. In 2008, Yang  obtained the improved inequality as follows: If p, r > 1, (1/p) + (1/q) = 1, (1/r) + (1/s) = 1, 0 < λ < 1 and the right-hand side integrals are convergent, then
where the constant factor is the best possible.
Motivated by (1.2) and the technique of real analysis, we establish a new inequality in ℝ2 with a homogeneous kernel of 0-degree. Furthermore, the equivalent form and the corresponding reverse inequalities are also considered.
In what follows, α1, α2 will be real numbers such that 0 < α1 < α2 < π.
LEMMA 2.1. If , the weight function
then for all x ∈ (-∞, 0) ∪ (0, ∞)
Proof. If x ∈ (-∞,0), then
Letting u = y/x for the first integrals and u = -y/x for the second integrals gives
Similarly, ϖ(x) = k for x ∈ (0, ∞). Hence (2.2) is valid for x ∈ (-∞, 0) ∪ (0, ∞). □
Note. (i) It is obvious that ϖ(0) = 0. (ii) If α1 = α2 = α, then
and ϖ(x) = 2 ln (2 sin α) + (π - 2α) cot α.
LEMMA 2.2. If and f(x) is a nonnegative measurable function in (-∞,∞), then for all x ∈ (-∞, 0) ∪ (0, ∞)
Proof. By Hölder's inequality with weight  and Lemma 2.1, we obtain
By Fubini theorem, we find
LEMMA 2.3. If and g(x) is a nonnegative measurable function in (-∞,∞), then for all x ∈ (-∞, 0) ∪ (0, ∞)
Proof. It can be completed similarly by following the proof of Lemma 2.2 as long as applying the reverse Hölder's inequality , hence we omit the details. Since q < 0, thus (2.7) takes the positive inequality. □
3. Main results and applications
THEOREM 3.1. If such that and , then we obtain the following equivalent inequalities
where the constant factors and kp are both the best possible.
Proof. If (2.5) takes the form of equality for some y ∈ (-∞,0) ∪ (0,∞), then there exist constants A and B such that they are not all zero and
i.e., A|x|pfp(x) = B|y|qa.e. in (-∞,∞) × (-∞, ∞). We conform that A ≠ 0 (otherwise B = A = 0). Then a.e. in(-∞,∞), which contradicts the fact that . Hence (2.5) takes a strict inequality and the same as (2.4), thus (3.2) is valid.
By Hölder's inequality with weight , we find
By (3.2), we obtain (3.1). On the other hand, suppose that (3.1) is valid. Let
then . In view of (2.4), J < ∞. If J = 0, then (3.2) is naturally valid; if J > 0, by (3.1), then
Hence we obtain (3.2). Thus (3.2) and (3.1) are equivalent.
For any ε > 0, suppose that
Then we get the following inequality
By Fubini theorem , it follows
If the constant factor k in (3.1) is not the best possible, then there exists a constant 0 < M ≤ k, such that
In view of (3.6) and (3.7), we obtain
By (3.8), (2.3) and Fatou lemma , we find
Hence k is the best value of (3.1). We conform that kpis also the best value of (3.2). Otherwise, we can get a contradiction by (3.3) that (3.1) is not the best possible. □
THEOREM 3.2. If such that and , then we have the following equivalent inequalities
where the constant factors , both kpand kqare the best possible.
Proof. By Lemma 2.3, similar to the proof of (3.2), we obtain that (3.10) and (3.11) are valid. In view of the reverse equality of (3.3), (3.9) is valid too. On the other hand, suppose that (3.9) is valid, let g(y) defined as Theorem 3.1, it is obvious J > 0. If J = ∞, then (3.10) is valid naturally; if 0 < J < ∞, then by (3.9), we find
Hence we obtain (3.10). Thus (3.10) and (3.9) are equivalent.
(3.11) and (3.9) are equivalent. In fact, we have proved (3.11) is valid above. On the other hand, suppose that (3.11) is valid, by the reverse Hölder's inequality with weight , we obtain
By (3.11), we obtain (3.9), and it is equivalent between (3.11) and (3.9). Thus (3.9), (3.10), and (3.11) are equivalent.
k is the best value of (3.9). In fact, If there exists a constant M ≥ k, such that (3.9) is still valid as we replace k by M. By the reverse inequality of (3.8), we obtain
Suppose that such that . Letting 0 < ε ≤ ε0 gives and
By Lebesgue control convergent theorem , it follows
Then by Levi theorem , we obtain
By (3.14), if follows that k ≥ M for ε → 0+. Hence k is the best value of (3.9). Furthermore, the constant factors in (3.10) and (3.11) are both the best value too. Otherwise, by (3.3) or (3.13), we may get a contradiction that the constant factor in (3.9) is not the best possible. □
By Note (ii), Theorems 3.1 and 3.2, it follows that
COROLLARY 3.3. If such that and , then we obtain the following equivalent inequalities
where the constant factors k1 = 2 ln (2 sin α) + (π - 2α) cot α and are both the best possible. In particular, for α = π/3 or 2π/3, it reduces to
COROLLARY 3.4. If such that and , then we have the following equivalent inequalities
where the constant factors k1 = 2 ln (2 sin α) + (π - 2α) cot α, and are both the best possible. In particular, for α = π/3 or 2π/3, it reduces to
The study was partially supported by the Emphases Natural Science Foundation of Guangdong Institution of Higher Learning, College and University (No. 05Z026).
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The authors declare that they have no competing interests.