Three weights higher order Hardy type inequalities

We investigate the following three weights higher order Hardy type inequality (0.1) ‖g‖q,u ≤ C ‖D ρg‖p,v, where D ρ denotes the following weighted differential operator: D ρg(t) = dg(t) dti , i = 0, 1, . . . , m − 1, di−m dti−m ρ(t) dg(t) dtm , i = m,m + 1, . . . , k, for a weight function ρ(·) . A complete description of the weights u , v and ρ so that (0.1) holds was given in [4] for the case 1 < p ≤ q < ∞ . Here the corresponding characterization is proved for the case 1 < q < p < ∞ . The crucial step in the proof of the main result is to use a new Hardy type inequality (for a Volterra type operator), which we first state and prove. 164 Three weights higher order Hardy type inequalities

A complete description of the weights u , v and ρ so that (0.1) holds was given in [4] for the case 1 < p ≤ q < ∞ .Here the corresponding characterization is proved for the case 1 < q < p < ∞ .The crucial step in the proof of the main result is to use a new Hardy type inequality (for a Volterra type operator), which we first state and prove.

Introduction
Hardy type inequalities have been studied, developed and applied during the last decade in an almost unbelievable way.See, e.g., the monographs [7,13] and the recent review article [6], completely devoted to this subject.In particular, some present knowledge about the higher order Hardy inequality is described in the book [7].The weighted Hardy inequality for derivatives of order k ≥ 1 has the following form: (1.1) g q,u ≤ C g (k)  p,v .
In this paper we shall consider a fairly new type of generalization of the Hardy type inequality (1.1), namely we investigate the estimate where D i ρ denotes the following weighted differential operator: dt m , i = m, m + 1, . . ., k, for a weight function ρ(•).
The introduced differential operator, where all derivatives are understood in the generalized sense, is called the ρ-weighted derivative of g of the corresponding order i = 0, 1, . . ., k .
This operator (even in more general form) was considered already in the twentieth by G. Polya [14] and, independently, in the thirtieth by G. Mammana [9].See the important paper [8] by A.U. Levin and the references given there.This paper is also a source for us to obtain applications of our results.After this a number of authors have studied this operator; here we just mention works by R. Oinarov and B.L. Baideldinov (see e.g.[2,3]), where it is, in particular, pointed out that this operator is interesting for some applications and even gives the basis for some new function spaces.
We shall use the notations lim t→0+ D i ρ g(t) = D i ρ g(0), i = 0, 1, . . ., k − 1, and consider the problem of finding necessary and sufficient conditions on the weights u , v , and ρ, under which the inequality (1.2) holds for all functions g satisfying the "boundary conditions" D i ρ g(0) = 0, i= 0, 1, . . ., l − 1, (1.3) In the paper [4] this problem was solved for the case 1 < p ≤ q < ∞.In the same paper you can find several references and remarks concerning (1.1).We also refer to the fairly new PhD thesis of M. Nassyrova [10].In particular, in [4] it was noted that in the paper [5] by A. Kufner and H.P. Heinig inequality (1.1) was studied under the setting of "boundary conditions" similar to those of the current consideration.Here we prove some corresponding characterizations for the case 1 < q < p < ∞.In particular, by putting ρ = 1 in each of the characterizations we obtain the previous result of A. Kufner and H.P. Heinig [5].
The crucial step in the proofs of these main results is to use a Hardy type inequality for a Volterra type operator of the following form: (1.6) with the kernel where α > 1 is a real number.
Let μ be a non -negative Borel measure on R + , then • q,μ now stands for Let us remark that in the case dμ(x) = u q (x)dx the notation • q,μ coincides with the notation • q,u introduced above.
The paper is organized as follows: in Section 3 we present and prove our main result concerning the inequality (1.5) (see Theorem 1).In order not to disturb our discussions in the fairly technical proof of this theorem we have formulated and proved three useful lemmas in Section 2, where also some other preliminaries can be found.Finally, in Section 4 we use Theorem 1 to prove our main results connected to the inequality (1.2) (see Theorem 2, 3 and 4).

Preliminaries
Here and in the sequel The symbol X Y means X ≤ cY with some constant c > 0 , and the notation X ≈ Y asserts the existence of the two -sided estimate X Y X .Moreover, Z stands for the set of all integers, and Z + stands for the set of non -negative integers.Above and in the sequel k, m, n, l ∈ Z + .
For 0 It is easy to see that the kernels of the operators (1.6) and (2.1) are continuous, don't decrease in the first argument and don't increase in the second argument.
We now state and prove three useful technical lemmas.Further for the operator (2.1) we shall need the following estimate: where μ is some non -negative Borel measure.The kernel of (2.1) satisfies the condition r(t, s) = r(t, x) + r(x, s), when t ≥ x ≥ s ≥ 0, so in view of results in the paper [11] it follows the two -weighted estimate for (2.1) in the form (1.5).But in our case we need to prove the inequality in the form (2.2) because in the left side of (2.2) we have a Borel measure μ. Proof.
Let f ≥ 0. Then Rf is continuous, non-negative and nondecreasing on R + , in addition, (Rf )(0) = 0 .Hence, there exists a sequence of monotonously increasing points The pointed out relations give: By twice using Hölder's inequality, we get (2.4) Therefore, from (2.4) we have (2.5) By introducing the measure dγ dt, where δ is the delta-function, the expression J 0 2 can be written in the following form: Indeed, By using standard results from the theory of Hardy type inequalities (see e.g.[7]), we find that where . Now, we estimate separately the integral ∞ t dγ 0 (s) in the following way: From this estimate we obtain From (2.3), (2.5) and (2.6) we have that (2.2) holds.
To restrict the writing we suppose Proof.By Fubini's theorem we have Three weights higher order Hardy type inequalities Applying Minkowski's inequality in (2.8) gives Lemma 3. Suppose that the monotone function f : R + → R + does not increase, the monotone function ϕ : R + → R + does not decrease.Moreover, both of the functions are locally absolutely continuous on R + and Then the finiteness of any of the integrals implies the finiteness of the other one.Moreover, Proof.Suppose that (2.9) is finite.Then, for any t ∈ R + , we have and, consequently, lim Integrating by parts gives (2.12) By assumption the integral (2.9) is finite and, hence, the left side of (2.12) has a finite limit when t → +∞.We conclude that each summand on the right side of (2.12) has a finite limit when t → +∞ because they are nonnegative for all t ∈ R + .Thus, the equality holds, i.e., the integral (2.10) is finite.Moreover, from the finiteness of (2.10) it follows that Hence, lim t→+∞ ϕ(t)f (t) = 0 , and this fact together with (2.13) yields (2.11).
Similarly, we can prove the finiteness of (2.9) from the finiteness of (2.10).
Remark 1.In all proofs in the sequel we more than once will use the operation of integration by parts, and for all these cases the conditions of Lemma 3 will hold.This implies that it is not necessary to every time refer to Lemma 3 and the equalities of (2.11) type can be written down automatically.Moreover, the conditions of the fulfillment of known inequalities will not be written down in the classical form but in the transformed form (cf. (2.11)), which is more suitable for our purposes.

The main result for the inequality (1.5)
Let , and Then the estimate (1.5) holds if and only if B < ∞.In addition, B ≈ C , where C is the best constant in (1.5).
Because of the non-negativeness of the kernel of the operator K it is enough to prove the correctness of the inequality (1.5) for non-negative functions (f ≥ 0).
For f ≥ 0 consider F (x) = x 0 K(x, s)f (s)ds.Then the function F (x) is continuous and non-decreasing on R + , and, moreover, F (0) = 0 .Assume that h = max{1, 2 α−2 } .Due to the properties of the function F (x) on R + there exists a sequence of monotonously increasing points As before in Lemma 1 we suppose that if Moreover, since according to (3.1), we have Taking into account (3.2), we can estimate Kf q,u in the following way: x k+1 x k u q (x)dx x k u q (s)ds x k u q (s)ds .
Denote by J 1 and J 2 the first two summands in the brackets of the last expression, respectively, and by J 3 the last summand without the constant h q , and then estimate J 1 , J 2 , and J 3 separately.
We start with J 1 .By using Hölder's inequality, we get x k u q (s)ds) p p−q p−q p .Furthermore, x k+1 x k u q (s)ds and, from (3.3) we get Hence, Next, we note that, by assuming that where δ(•) is the delta-function, J 2 can be written similarly as J 0 2 is written in Lemma 1, namely in the following form: Therefore, where . Now, we estimate separately the integral ∞ t dγ(s) in the following way: x k+1 x k u q (s)ds x k K q (s, t)u q (s)ds From this estimate we obtain Now, we put where δ(•) is the delta-function.Then J 3 has the following form: Indeed, x k+1 x k u q (s)ds) Due to Lemma 1 we have where We estimate separately the following integral in the expression A 2 : x k u q (s)ds x k u q (s)ds This estimate gives We put the obtained estimates for A 2 and A 1 into (3.6) and get (3.7) By combining (3.4), (3.5), and (3.7) we see that (1.5) holds, and, in addition, C B , where C is the best constant in (1.5).Necessity.Suppose that the inequality (1.5) holds with the best constant C > 0 .Then it is well -known that the dual inequality where (see e.g.[7]).
Take the following test function for any a, b : 0 < a < b < ∞, and put it into the right side of the inequality (3.8) and we obtain that (3.9) p−q u q (t)dt.
Next we insert g 1 (•) into the left side of (3.8) and find that Hence, by Fubini's theorem, (3.10) By using this estimate in (3.11) we find that p−q u q (s)ds.
From (3.8), (3.9), and (3.12) it follows b a b s u q (t)dt p−q u q (s)ds If in the last estimated expression we proceed to limits when a → 0 and b → ∞, then, by also taking into account (2.11), we get B 1 C .Next, we consider the test function Put f 1 (•) into the right side of the inequality (1.5): Moreover, we insert f 1 (•) into the left side of (1.5).Arguing as before for the test function g 1 (•), we get (3.14) According to (1.5), (3.13), and (3.14) we have b a b t K q (s, t)u q (s)ds If in the last estimated expression we proceed to limits when a → 0 and b → ∞, then, by taking into account (2.11), we get B 2 C .From (1.5) and (2.7) we get As before, we note that the dual inequality Next, we put the test function into the right and left sides of the inequality (3.15),where a, b : 0 < a < b < ∞.Then, by arguing in the same way as in the previous cases, we obtain From (1.5), (3.16), and (3.17) we get b a b t u q 1 (s)ds If in the last estimated expression we proceed to limits when a → 0 and b → ∞, then, by taking into account (2.11) and In the paper [1] sufficient conditions and in the paper [12] necessary conditions of the validity of the inequality (1.5) were found in the case when the kernel of the integral operator has the form (3.18).In fact, it was proved that the correctness of (1.5) depends on α conditions.However, as we have proved in a direct way in this paper we need only three conditions.Remark 3. Lemma 2 and Theorem 1 can be formulated also for the conjugate integral operators.In fact, the proofs are similar so we omit both the formulations and the proofs.

The main results for the inequality (1.2)
We divide the studying of the problem into three cases: (1) Suppose that the function g satisfies the conditions (1.3) and (1.4).Then the inequality (1.2) holds with a finite constant C > 0 independent of g if and only if the following conditions hold: Three weights higher order Hardy type inequalities and Proof.If we assume F (t) = D k ρ g(t), then, according to conditions (1.3) and (1.4), we have In the paper [4] it was proved that If we use the notations K i F (t), i = 1, 2, 3, 4 , for the each summand of the above expression, then the fulfillment of the inequality (1.2) is equal to the simultaneous fulfillment of four inequalities: We study each of these estimates (4.5) separately.
We start with the inequality (4.5) for the operator dx, then we can rewrite (4.5) in the following form: .
Since the operator K 1 with the kernel K 1 (t, s) = (t − s) m−1 is an operator of Riemann-Liouville type, then, in view of results in the paper [15], the inequality (4.6) holds if and only if Then the considered estimate takes the form: Of course, the inequality (4.9) is a standard Hardy type inequality, which holds if and only if (see e.g.[7]): Consider (4.5) for the operator K 3 .It is obvious that the kernel has the form of the kernel from Theorem 1.Therefore, the inequality (4.5) for i = 3 holds if and only if the condition (4.3) and two conditions It is easy to show that the combination of the conditions (4.7) and (4.12) in fact is equivalent to (4.1), and that the combination of (4.8), (4.10), and (4.11) gives (4.2).Moreover, the inequality (4.5) for i = 4 is a standard Hardy type inequality and for its correctness the condition (4.4) is necessary and sufficient (see e.g.[7]).Theorem 3. Let 1 < q < p < ∞, k ≥ 3 and 1 ≤ l < m < k.Suppose that the function g satisfies the conditions (1.3) and (1.4).Then the inequality (1.2) holds with a finite constant C > 0 independent of g if and only if the following conditions hold: . Then, due to the conditions (1.3) and (1.4), we have In the paper [4] it was proved that Insert this expression instead of g in the inequality (1.2) and we get four (4.5) type inequalities, the simultaneous fulfillment of which is equivalent to the validity of (1.2).
Since the first inequality is a standard Hardy type inequality, the condition (4.16) is necessary and sufficient for it correctness (see e.g.[7]).
The integral operator of the second inequality with the kernel K 2 (τ, t) = (τ − t) n−1 is an operator of Riemann-Liouville type, and, in view of the results in the paper [15], it holds if and only if The third inequality is again a standard Hardy type inequality, which is correct if and only if (see e.g.[7]): of the integral operator of the fourth inequality has the form of the kernel from Theorem 1, but the operator itself is conjugate to the operator from Theorem 1.Therefore, by taking into account Remark 3 we can assert that it is valid if and only if condition (4.15) and the following two conditions It is obvious that the combination of the conditions (4.17), (4.19) and (4.20) is equivalent to (4.13), and that the combination of (4.18) and (4.21) gives (4.14).Theorem 4. Let 1 < q < p < ∞, k ≥ 2 and 1 ≤ m = l < k.Suppose that the function g satisfies the conditions (1.3) and (1.4).Then the inequality (1.2) holds with a finite constant C > 0 independent of g if and only if the following conditions hold: ∞ 0 ∞ t u q (s)(s − t) q(m−1) ds p p−q  From [4] it is known that < ∞.
The second and fourth summands lead to standard Hardy type inequalities, which hold respectively if and only if (see e.g.[7]): p−q (4.28) The combination of the conditions (4.26) and (4.28) is equivalent to the condition (4.23) and the combination of (4.27) and (4.29) gives (4.24).
Using this fact and (3.10) we get (Next we estimate the integral b s g 1 (x)dx in the brackets of (3.11): b