ON BICHENG-DEBNATH ’ S GENERALIZATIONS OF HARDY ’ S INTEGRAL INEQUALITY

We consider Hardy’s integral inequality and we obtain some new generalizations of Bicheng-Debnath’s recent results. We derive two distinguished classes of inequalities covering all admissible choices of parameter k from Hardy’s original relation. Moreover, we prove the constant factors involved in the right-hand sides of some particular inequalities from both classes to be the best possible, that is, none of them can be replaced with a smaller constant. 2000 Mathematics Subject Classification. 26D15.

The relation (1.2), the so-called Hardy's integral inequality, plays a highly important role in mathematical analysis and its applications.Although classical, that result was during the last decade generalized in many different ways by numerous mathematicians.One possibility of generalizing it is to investigate its finite sections, that is, inequalities of the same type, where the outer integrals on both sides of (1.2) are, instead over (0, ∞), taken over some of its subsets.
Very recently two papers, [1,2], appeared dealing with that subject.In [2], which is the earlier one, Y. Bicheng et al. obtained a whole new class of generalizations of Hardy's integral inequality (1.2).Their results can be joined and stated as follows: let 0 ≤ a < b ≤ ∞, 1 < p < ∞, and let f be a nonnegative measurable function such that 0 < b a f p (x)dx < ∞.Then Later on, in [1], Bicheng and Debnath derived a class of dual inequalities to (1.3).More precisely, they proved that if the parameters a, b, and p are as in [2], and f is a nonnegative measurable function such that 0 < b a x p f p (x)dx < ∞, then the inequality Note that Bicheng-Debnath results consider only two particular cases of the parameter k from inequality (1.2): k = p in (1.3), and k = 0 in (1.5).Therefore, the aim of this paper is to generalize these relations to cover all admissible choices of the parameter k from Hardy's integral inequality.
Hence, our main objective will be to obtain two distinguished classes of inequalities: one of them dealing with k > 1 and generalizing the class (1.3), and the other one considering k < 1 and generalizing the class (1.5).Furthermore, we will make a detailed analysis of the weight functions appearing on the right-hand sides of all derived inequalities, what will be used in proving that the obtained constant factor (p/|k − 1|) p is the best possible in cases where a = 0 or b = ∞.Finally, we will show that the best possible constant value in the case where 0 < a < b < ∞ is bounded.The established inequalities will conclude the problem of the Bicheng-Debnath-type finite sections of (1.2).
Techniques that will be used in the proofs are mainly based on classical real analysis, especially on the well-known Hölder's and Bernoulli's inequality, and on Fubini's theorem.

Preliminaries.
First, we present some lemmas which are interesting in their own right and will be used in proofs of our main results.
holds for all x ∈ (a, b).Moreover, if a = 0 or b = ∞, then there exists a number x 0 ∈ (a, b) such that for any x ∈ (x 0 ,b) the inequality (2.1) is strict.
Proof.Let 0 ≤ a < b ≤ ∞ and x ∈ (a, b) be arbitrary.Using Hölder's inequality we have x (k−1)/p n − a (k−1)/p 1/q xn a t p−k+(k−1)/p f p (t)dt 1/p (2.4) holds for all n ∈ N. Therefore, for each n there is the equality in Hölder's inequality, so the functions t t (p+1−k)/q f p (t) and t t −(p+1−k)/p have to be effectively proportional in [a, x n ] (cf.[3, page 140]), that is, there are nonnegative real constants c n and d n , not both zero, such that This contradicts the fact that b a t p−k f p (t)dt < ∞, so the proof is completed.
The following lemma contains a result dual to Lemma 2.1.The proofs of both lemmas are similar. (2.9) As in Lemma 2.1, a necessary and sufficient condition for the equalities in (2.7) and (2.9) is that the equality holds in Hölder's inequality.Now, we consider cases where a = 0 or b = ∞.We need to show that there exists a number x 0 ∈ (a, b) such that for any x ∈ (a, x 0 ) the inequality in (2.7) is strict.Otherwise, there exists a decreasing sequence (x n ) n∈N in (a, b), x n a, such that for any n ∈ N the inequality (2.7), rewritten for x = x n , becomes an equality.Moreover, to every n ∈ N there correspond real constants c n ,d n ≥ 0, not both zero, such that By the same arguments as in the proof of Lemma 2.1 we conclude that c n > 0 for all n ∈ N, and there exists N ∈ N such that the relations for both cases a = 0 and b = ∞.This is contradictory to b a t p−k f p (t) dt < ∞, so the lemma is proved.
However, the case where 0 < c < 1 is a little bit different.Here we have ψ c (0) > 0, but ψ c (c) < 0. Since ψ c is strictly decreasing and continuous on [0,c], there exists a unique point x c ∈ (0,c) such that ψ c (x c ) = 0. Consequently, . It is only left to determine an upper and lower bound for ϕ c (x c ).We have and ϕ c (x c ) > ϕ c (0) = c p /p hold.This proves (ii), so the proof of the lemma is now complete.
Remark 2.5.The case p = 2 will be of special interest, since ϕ c (x c ) can be calculated exactly.Really, in that case x c is the unique root of the quadratic function 3. Some new generalizations of Hardy's integral inequality.This section is dedicated to some new generalizations of the relation (1.3).The parameter k from Hardy's original inequality (1.2) will be chosen greater than one.
We start with the basic result, given by the following theorem.
where the weight function g is defined by (3.11) Observe that the second inequality is a consequence of Remark 2.5.
We continue with analyzing two limit cases of the parameters a and b.First, we consider b = ∞.
where the weight function g is defined by

.13)
The function g is strictly increasing and 1/p < g(x; p, k, a) < 1 holds for all x > a.Moreover, the constant (p/(k − 1)) p is the best possible.
Proof.Using Lemma 2.1 and Fubini's theorem, we have the following sequence of relations: (3.14) so (3.12) is proved.Let u = 1 − (a/t) (k−1)/p .Then g(t; p, k, a) = ϕ 1 (u), where ϕ 1 is defined in Lemma 2.3.According to (i) of Lemma 2.3, the function g is strictly increasing (as composition of such functions ϕ 1 and t u(t)) and the relation 1/p < g(t; p, k, a) < 1 holds for all t > a.
We prove that the constant λ = (p/(k − 1)) p is the best possible for (3.12).Otherwise, there exists a smaller constant C, 0 and, by Bernoulli's inequality, further ( This contradicts (3.15), so λ is the best possible constant for (3.12).The proof of the theorem is now complete.
Remark 3.6.When p = 2, inequality (3.12) reduces to the form Finally, the case when 0 = a < b < ∞ has already been described in our paper [6], but by using an approach via mixed-means inequalities (cf.[5]).Here we give a different proof of this result.
holds.The constant (p/(k − 1)) p is the best possible.
Proof.Directly from Lemma 2.1 (relation (2.3) rewritten with the sign < instead of ≤) and Fubini's theorem, we have b 0 4. Dual inequalities.Our objective in this section is to derive dual inequalities to (3.1), (3.12), and (3.19).In other words, we will consider k < 1 in (1.2) and obtain some new generalizations of (1.5).
The following theorem presents dual result to Theorem 3.1.
Proof.The proof of this theorem is based on Lemmas 2.2 and 2.3, and it is similar to the proof of Theorem 3.1.So, the relation (2.7) and Fubini's theorem imply: where c is defined in the proof of Theorem 4.1.
To conclude this paper, consider limit cases of the parameters a and b.As in the preceding section, we start with b = ∞.It has to be mentioned that an inequality concerning this case was given in [6].It is dual to (3.19) and in [6] it was proved by using the concept of mixed means.Here we restate that result and give its other proof.
holds.The constant (p/(1 − k)) p is the best possible.
Proof.Lemma 2.2, with (2.9) rewritten as strict inequality, and Fubini's theorem yield so (4.9) is proved.The best possible constant for (4.9) is discussed in [6].Now, it is only left to describe the case where a = 0, that is, to obtain a dual result to Theorem 3.7.This is a contradiction, so µ is the best possible constant for (4.11).This completes the proof.

4 )
a,b (x)f p (x)dx, (1.3) where the weight function g p,a,b is given by g p,a,b (x) Of course, g p,0,b = lim a 0 g p,a,b and g p,a,∞ = lim b→∞ g p,a,b .

6 )
a,b (x)x p f p (x)dx(1.5)    holds, where the weight function h p,a,b ish p,a,b (x) Again, h p,0,b = lim a 0 h p,a,b and h p,a,∞ = lim b→∞ h p,a,b .Moreover, in cases where a = 0 or b = ∞ the authors proved that the constant factors involved in the right-hand sides of (1.3) and (1.5) are the best possible, while in the case where 0 < a < b < ∞ they provided some upper and lower bounds for maxima of the functions g p,a,b and h p,a,b .

( 3 . 6 )Remark 3 . 2 . 1 − 1 − 1 − 1 −Remark 3 . 3 .Remark 3 . 4 .
so the first relation in (3.1) is proved.To prove the other inequality in (3.1), denote c = 1−(a/b) (k−1)/p and u = 1−(a/t) (k−1)/p .Then c ∈ (0, 1), u ∈ [0,c], and g(t; p, k, a, b) = ϕ c (u), where ϕ c is the function defined in the statement of Lemma 2.3.Also, t u(t) is a strictly increasing mapping of [a, b] onto [0,c].Therefore, by using (ii) of Lemma 2.3, the weight function g has a unique maximizing point on [a, b].More precisely, there is a unique point t 0 ∈ (a, b) such that g(t 0 ,p,k,a,b) = max t∈[a,b] g(t; p, k, a, b).So, g(t; p, k, a, b) < M(p, k, a, b) a.e. on [a, b].Combining this and the fact that b a x p−k f p (x)dx > 0, we finally havep k − 1 p b a g(t; p, k, a, b)t p−k f p (t)dt < p k − 1 p M(p, k, a, b) b a t p−k f p (t)dt.(3.7)The proof of (3.1) is now completed.The inequality (3.3) holds immediately, if (ii) of Lemma 2.3 is used again.The estimate (3.3) can be easily improved by an application of Remark 2.4, since considering (2(a/b) (k−1)/p p+p/(p−1) (a/b) (k−1)/p 1+p/(p−1) Using Theorem 3.1 it is not hard to convince oneself in the validity of the following inequality: 1)/p p b a x p−k f p (x)dx.(3.9)Unfortunately, from the previous analysis (see(3.1)  and (3.3)) it is also evident that the constant factor (p/(k − 1)) p [1 − (a/b) (k−1)/p ] p is not the best possible value λ(p,k,a,b) for which (3.9) exists.However, Theorem 3.1 shows that λ(p,k,a,b) is bounded and provides an explicit upper bound for this constant.In fact, 0 < λ(p,k,a,b) When p = 2, the inequality (3.1) reduces to the form b a

2 )
; p, k, a, b)x p−k f p (xk f p (x)dx(4.1) holds, where the weight function h is defined by h(x; p, k, a, b) and N(p, k, a, b) = max x∈[a,b] h(x; p, k, a, b).Moreover,

Remark 4 . 3 .
Note that the inequality holds, but the constant appearing on its right-hand side is not the best possible value µ(p, k, a, b) for which this relation exists.Theorem 4.1 implies that µ(p, k, a, b) is bounded and 0