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A half-discrete Hilbert-type inequality with a homogeneous kernel and an extension
Journal of Inequalities and Applications volume 2011, Article number: 124 (2011)
Abstract
Using the way of weight functions and the technique of real analysis, a half-discrete Hilbert-type inequality with a general homogeneous kernel is obtained, and a best extension with two interval variables is given. The equivalent forms, the operator expressions, the reverses and some particular cases are considered.
2000 Mathematics Subject Classification: 26D15; 47A07.
1 Introduction
Assuming that p > 1, , f (≥ 0) ∈ Lp (R+), g(≥ 0) ∈ Lq (R+), , || g || q > 0, we have the following Hardy-Hilbert's integral inequality [1]:
where the constant factor is the best possible. If a m , b n ≥ 0, , , , || b || q > 0, then we still have the following discrete Hardy-Hilbert's inequality with the same best constant factor :
For p = q = 2, the above two inequalities reduce to the famous Hilbert's inequalities. Inequalities (1) and (2) are important in analysis and its applications [2–4].
In 1998, by introducing an independent parameter λ ∈ (0, 1], Yang [5] gave an extension of (1) for p = q = 2. Refinement and generalizing the results from [5], Yang [6] gave some best extensions of (1) and (2) as follows: If λ1, λ2 ∈ R, λ1 + λ2 = λ, kλ(x, y) is a non-negative homogeneous function of degree - λ satisfying for any x, y, t > 0, kλ(tx, ty) = t-λ kλ (x, y), , , , , g(≥ 0) ∈ L q , ψ , || f || p , ϕ , || g || q , ψ > 0, then we have
where the constant factor k(λ1) is the best possible. Moreover, if kλ(x, y) is finite and is decreasing with respect to x > 0(y > 0), then for a m ,b n ≥ 0, , , || a || p , ϕ , || b || q , Ψ > 0, we have
with the best constant factor k(λ1). Clearly, for λ = 1, , , (3) reduces to (1), and (4) reduces to (2). Some other results about Hilbert-type inequalities are provided by [7–15].
On half-discrete Hilbert-type inequalities with the non-homogeneous kernels, Hardy et al. provided a few results in Theorem 351 of [1]. But they did not prove that the constant factors are the best possible. And, Yang [16] gave a result by introducing an interval variable and proved that the constant factor is the best possible. Recently, Yang [17] gave the following half-discrete Hilbert's inequality with the best constant factor B(λ1, λ2)(λ, λ1 > 0, 0 < λ2 ≤ 1, λ1 + λ2 = λ):
In this article, using the way of weight functions and the technique of real analysis, a half-discrete Hilbert-type inequality with a general homogeneous kernel and a best constant factor is given as follows:
which is a generalization of (5). A best extension of (6) with two interval variables, some equivalent forms, the operator expressions, the reverses and some particular cases are considered.
2 Some lemmas
We set the following conditions:
Condition (i) v(y)(y ∈ [n0 - 1, ∞)) is strictly increasing with v(n0 - 1) ≥ 0 and for any fixed x ∈ (b, c), f(x, y) is decreasing for y ∈ (n0 - 1, ∞) and strictly decreasing in an interval of (n0 - 1, ∞).
Condition (ii) is strictly increasing with and for any fixed x ∈ (b, c), f(x, y) is decreasing and strictly convex for .
Condition (iii) There exists a constant β ≥ 0, such that v(y)(y ∈ [n0 - β, ∞)) is strictly increasing with v(n0 - β) ≥ 0, and for any fixed x ∈ (b, c), f(x, y) is piecewise smooth satisfying
where is Bernoulli function of the first order.
Lemma 1 If λ1, λ2 ∈ R, λ1 + λ2 = λ, kλ(x, y) is a non-negative finite homogeneous function of degree - λ in and v(y)(y ∈ [n0, ∞), n0 ∈ N) are strictly increasing differential functions with u(b+) = 0, v(n0) > 0, u(c-) = v(∞) = ∞, setting K(x, y) = kλ(u(x), v(y)), then we define weight functions ω(n) and ϖ(x) as follows:
It follows
Moreover, setting , if k(λ1) ∈ R+ and one of the above three conditions is fulfilled, then we still have
Proof . Setting in (7), by calculation, we have (9).
-
(i)
If Condition (i) is fulfilled, then we have
-
(ii)
If Condition (ii) is fulfilled, then by Hadamard's inequality [18], we have
-
(iii)
If Condition (iii) is fulfilled, then by Euler-Maclaurin summation formula [6], we have
The lemma is proved. ■
Lemma 2 Let the assumptions of Lemma 1 be fulfilled and additionally, p > 0(p ≠ 1), a n ≥ 0, n ≥ n0(n ∈ N), f (x) is a non-negative measurable function in (b, c). Then, (i) for p > 1, we have the following inequalities:
(ii) for 0 < p < 1, we have the reverses of (11) and (12).
Proof(i) By Hölder's inequality with weight [18] and (9), it follows
Then, by Lebesgue term-by-term integration theorem [19], we have
and (11) follows.
Still by Hölder's inequality, we have
Then, by Lebesgue term-by-term integration theorem, we have
and then in view of (9), inequality (12) follows.
-
(ii)
By the reverse Hölder's inequality [18] and in the same way, for q < 0, we have the reverses of (11) and (12). ■
3 Main results
We set , , wherefrom , .
Theorem 1 Suppose that λ1, λ2 ∈ R, λ1 + λ2 = λ, kλ(x, y) is a non-negative finite homogeneous function of degree - λ¸ in and v(y)(y ∈ [n0, ∞), n0 ∈ N are strictly increasing differential functions with u(b+) = 0, v(n0) > 0, u(c-) = v(∞) = ∞, ϖ(x) < k (λ1) ∈ R+(x ∈ (b, c)). If , f ( x ), a n ≥ 0, f ∈ L p Φ ( b , c ), || f || p ,Φ > 0 and || a || q ,Ψ > 0, then we have the following equivalent inequalities:
Moreover, if is decreasing and there exist constants δ < λ1 and M > 0, such that, then the constant factor k(λ1) in the above inequalities is the best possible.
Proof By Lebesgue term-by-term integration theorem, there are two expressions for I in (13). In view of (11), for ϖ(x) < k(λ1) ∈ R+, we have (14). By Hölder's inequality, we have
Then, by (14), we have (13). On the other hand, assuming that (13) is valid, setting
then Jp-1 = || a || q ,Ψ . By (11), we find J < ∞. If J = 0, then (14) is naturally valid; if J > 0, then by (13), we have
and we have (14), which is equivalent to (13).
In view of (12), for [ϖ(x) ]1-q> [k(λ1)]1-q, we have (15). By Hölder's in equality, we find
Then, by (15), we have (13). On the other hand, assuming that (13) is valid, setting
then Lq-1 = || f || p Φ . By (12), we find L < ∞. If L = 0, then (15) is naturally valid; if L > 0, then by (13), we have
and we have (15) which is equivalent to (13).
Hence, inequalities (13), (14) and (15) are equivalent.
There exists an unified constant d ∈ (b, c), satisfying u(d) = 1. For 0 < ε < p(λ1 - δ), setting , x ∈ (b, d); , x ∈ (d, c), and , n ≥ n0, if there exists a positive number k(≤ k(λ1)), such that (13) is valid as we replace k(λ1) by k, then in particular, we find
For , we find
namely A(ε) = O(1)(ε → 0+). Hence, by (18) and (19), it follows
By Fatou Lemma [19], we have , then by (20), it follows k(λ1) ≤ k(ε → 0+). Hence, k = k(λ1) is the best value of (12).
By the equivalence, the constant factor k(λ1) in (14) and (15) is the best possible, otherwise we can imply a contradiction by (16) and (17) that the constant factor in (13) is not the best possible. ■
Remark 1 (i) Define a half-discrete Hilbert's operator T : as: for f ∈ L p Φ (b, c), we define , satisfying
Then, by (14), it follows and then T is a bounded operator with || T || ≤ k(λ1). Since, by Theorem 1, the constant factor in (14) is the best possible, we have || T || = k(λ1).
-
(ii)
Define a half-discrete Hilbert's operator as: for a ∈ l q Ψ , we define , satisfying
Then, by (15), it follows and then is a bounded operator with . Since, by Theorem 1, the constant factor in (15) is the best possible, we have .
In the following theorem, for 0 < p < 1, or p < 0, we still use the formal symbols of and ||a|| q ,Ψ and so on. ■
Theorem 2 Suppose that λ1, λ2 ∈ R, λ1 + λ2 = λ, kλ(x, y) is a non-negative finite homogeneous function of degree -λ in , u(x)(x ∈ (b, c), -∞ ≤ b < c ≤ ∞) and v(y)(y ∈ [n0, ∞), n0 ∈ N) are strictly increasing differential functions with u(b+) = 0, v(n0) > 0, u(c-) = v(∞) = ∞, k(λ1) ∈ R+, θλ(x) ∈ (0, 1), k(λ1)(1 - θλ(x)) < ϖ(x) < k(λ1)(x ∈ (b, c)). If 0 < p < 1, , f(x), a n ≥ 0, , and 0 < ||a|| q ,Ψ < ∞. Then, we have the following equivalent inequalities:
Moreover, if is decreasing and there exist constants δ, δ0 > 0, such that and k(λ1 - δ0) ∈ R+, then the constant factor k(λ1) in the above inequalities is the best possible.
Proof. In view of (9) and the reverse of (11), for ϖ(x) > k(λ1)(1 - θλ(x)), we have (22). By the reverse Hölder's inequality, we have
Then, by (22), we have (21). On the other hand, assuming that (21) is valid, setting a n as Theorem 1, then Jp-1 = ||a|| q Ψ. By the reverse of (11), we find J > 0. If J = ∞, then (24) is naturally valid; if J < ∞, then by (21), we have
and we have (22) which is equivalent to (21).
In view of (9) and the reverse of (12), for [ϖ(x)]1-q> [k(λ1)(1 - θ λ (x))]1-q(q < 0), we have (23). By the reverse Hölder's inequality, we have
Then, by (23), we have (21). On the other hand, assuming that (21) is valid, setting
then . By the reverse of (12), we find . If , then (23) is naturally valid; if , then by (21), we have
and we have (23) which is equivalent to (21). ■
Hence, inequalities (21), (22) and (23) are equivalent.
For 0 < ε < pδ0, setting and as Theorem 1, if there exists a positive number k(≥ k(λ1)), such that (21) is still valid as we replace k(λ1) by k, then in particular, for q < 0, in view of (9) and the conditions, we have
Since we have , t ∈ (0, 1] and
then by Lebesgue control convergence theorem [19], it follows
By (26) and (27), we have
and then k(λ1) ≥ k(ε → 0+). Hence, k = k(λ1) is the best value of (21).
By the equivalence, the constant factor k(λ1) in (22) and (23) is the best possible, otherwise we can imply a contradiction by (24) and (25) that the constant factor in (21) is not the best possible. ■
In the same way, for p < 0, we also have the following theorem.
Theorem 3 Suppose that λ1, λ2 ∈ R, λ1 + λ2 = λ, kλ(x, y) is a non-negative finite homogeneous function of degree -λ¸ in , u(x)(x ∈ (b, c), -∞ ≤ b < c ≤ ∞) and v(y)(y ∈ [n0, ∞), n0 ∈ N) are strictly increasing differential functions with u(b+) = 0, v(n0) > 0, u(c-) = v(∞) = ∞, ϖ(x) < k(λ1) ∈ R + (x ∈ (b, c)). If p < 0, , f(x), a n ≥ 0, 0 < || f || p Φ < ∞ and 0 < ||a|| q ,Ψ < ∞. Then, we have the following equivalent inequalities:
Moreover, if is decreasing and there exists constant δ0 > 0, such that k(λ1 + δ0) ∈ R+, then the constant factor k(λ1) in the above inequalities is the best possible.
Remark 2 (i) For n0 = 1, b = 0, c = ∞, u(x) = v(x) = x, if
then (13) reduces to (6). In particular, for ,(6) reduces to (5).
-
(ii)
For n0 = 1, b = 0, c = ∞, u(x) = v(x) = xα(α > 0), , since
is decreasing for y ∈ (0, ∞) and strictly decreasing in an interval of (0, ∞), then by Condition (i), it follows
Since for , , then by (13), we have the following inequality with the best constant factor :
-
(iii)
For n0 = 1, b = β, c = ∞, , , since for any fixed x ∈ (β, ∞),
is decreasing and strictly convex for , then by Condition (ii), it follows
Since for , , then by (13), we have the following inequality with the best constant factor :
-
(iv)
For n0 = 1, b = 1 - β = γ, c = ∞, u(x) = v(x) = (x - γ), , , , we have
and
Hence, v(y)(y ∈ [γ, ∞)) is strictly increasing with v(1 - β) = v(γ) = 0, and for any fixed x ∈ (γ, ∞), f(x, y) is smooth with
We set