Boundedness of Hardy-type operators with a kernel: integral weighted conditions for the case $$0

Let $$1< p <\infty $$1<p<∞ and $$0<q<p$$0<q<p. We prove necessary and sufficient conditions under which the weighted inequality $$\begin{aligned} \left( \int _0^\infty \left( \int _0^t f(x)U(x,t)\mathrm {\,d}x\right) ^q w(t) \mathrm {\,d}t\right) ^\frac{1}{q}\le C \left( \int _0^\infty f^p(t)v(t)\mathrm {\,d}t\right) ^\frac{1}{p}, \end{aligned}$$∫0∞∫0tf(x)U(x,t)dxqw(t)dt1q≤C∫0∞fp(t)v(t)dt1p,where U is a so-called $$\vartheta $$ϑ-regular kernel, holds for all nonnegative measurable functions f on $$(0,\infty )$$(0,∞). The conditions have an explicit integral form. Analogous results for the case $$p=1$$p=1 and for the dual version of the inequality are also presented. The results are applied to close various gaps in the theory of weighted operator inequalities.


Introduction
Operators of the general form where U is a kernel, play an indispensable role in various areas of analysis. The means of their investigation, naturally, greatly depend on additional properties of the kernel U .
In the present article, we study the so-called Hardy-type operators and where the kernel U : [0, ∞) 2 → [0, ∞) is a measurable function which has the following properties: (i) U (x, y) is nonincreasing in x and nondecreasing in y; (ii) there exists a constant ϑ > 0 such that for all 0 ≤ x < y < z < ∞ it holds that U (x, z) ≤ ϑ (U (x, y) + U (y, z)) ; (iii) U (0, y) > 0 for all y > 0.
If ϑ > 0 and U is a function satisfying the conditions above with the given parameter ϑ in point (ii), then we, for the sake of simplicity, call U a ϑ-regular kernel.
where u is a given nonnegative measurable function. To be formally correct, let us assume that all these kernels are defined by the respective formulas above for 0 ≤ x < y < ∞, and by 0 for 0 ≤ y ≤ x < ∞. Hardy-type operators with these kernels find applications, for instance, in the theory of differentiability of functions, interpolation theory and more topics involving function spaces. The two last-named examples of ϑ-regular kernels prove to be particularly useful in research of the so-called iterated Hardy operators [2,5], for example. The particular aspect we investigate in this paper is boundedness of the operators H and H * with a ϑ-regular kernel U between weighted Lebesgue spaces. In order to define these spaces, we need to introduce several auxiliary terms first.
Throughout the text, by a measurable function we always mean a Lebesgue measurable function (on an appropriate subset of R). The symbol M + denotes the cone of all nonnegative measurable functions on (0, ∞). A weight is a function w ∈ M + on (0, ∞) such that Finally, if v is a weight and p ∈ (0, ∞], then the weighted Lebesgue space L p (v) = L p (v)(0, ∞) is defined as the set of all real-valued measurable functions f on (0, ∞) such that Note that if p ∈ (0, 1), then (L p (v), · L p (v) ) is in general not a normed linear space because of the absence of the Minkowski inequality in this case. However, as we deal only with the case 1 ≤ p < ∞ anyway, this detail is not of our concern here. Throughout the text, if p ∈ (0, 1) ∪ (1, ∞), then p is defined by p = p p−1 . Analogous notation is used for q .
In the following, assume that ϑ ∈ (0, ∞), U is a ϑ-regular kernel, H is the corresponding operator from (1) and v, w are weights. Boundedness of H between L p (v) and L q (w) corresponds, by definition, to validity of the inequality for all functions f ∈ M + , and it was completely characterized for p, q ∈ [1, ∞]. The authors credited for this work are Bloom and Kerman [1], Oinarov [15] and Stepanov [20]. The results of [15], for instance, have the following form.
Moreover, the least constant C such that the inequality holds for all f ∈ M + satisfies C ≈ E 1 + E 2 .
Theorem ([15, Theorem 1.2]) Let 1 < q < p < ∞ and r := pq p−q . Then H : Moreover, the least constant C such that (2) holds for all f ∈ L p (v) satisfies C ≈ The conditions obtained in [1,20] have a slightly different form, a more detailed comparison between them is found in [20].
As for the "limit cases", conditions for the case p = ∞ and q ∈ (0, ∞] are obtained very easily, the same applies to the case q = 1 and p ∈ [1, ∞) in which one simply uses the Fubini theorem. Yet another possible choice of parameters is p = 1 and q ∈ (1, ∞]. It was (at least for q < ∞) included in [15,Theorem 1.2] and the conditions may be recovered from that article by correctly interpreting the expressions involving the symbol p in there. Another option is to follow the more general theorem [9, Chapter XI, Theorem 4].
If 0 < p < 1, then the operator H can never be bounded (provided that U , v, w are nontrivial, which is always assumed here). The problem in here lies in the fact that for each t > 0 there exists f t ∈ L p (v) which is not locally integrable at the point t. For more details, see e.g. [13].
No such difficulty arises if 0 < q < 1 ≤ p < ∞. In this case, H may indeed be bounded between L p (v) and L q (w) and it is perfectly justified to ask for the conditions under which this occurs. As for the known answers to this question, the situation is however much worse than in the other cases.
When assumed U ≡ 1, i.e. for the ordinary Hardy operator, the boundedness characterization was found by Sinnamon [17] and it corresponds to the condition E 3 < ∞ (with U ≡ 1, of course). In the general case, in [20] it was shown that the condition E 3 < ∞ is sufficient but not necessary for H : L p (v) → L q (w) to be bounded, while the condition is necessary but not sufficient. For related counterexamples, see [19]. The fact that the two conditions do not meet is a significant drawback. An equivalent description of the optimal constant C in (2) is usually substantial for the result to be applicable in any way.
Lai [12] found equivalent conditions by proving that, with 0 < q < 1 < p < ∞, the operator H is bounded from L p (v) to L q (w) if and only if as well as The suprema in here are taken over all covering sequences, i.e. partitions of (0, ∞) (see [12] or Sect. 2 for the definitions), and r := pq p−q , as usual. Moreover, these conditions satisfy D 1 + D 2 ≈ C r with the least C such that (2) holds for all f ∈ M + . Corresponding variants for p = 1 are also provided in [12]. The earlier use of similar partitioning techniques in the paper [14] of Martín-Reyes and Sawyer should be also credited.
Unfortunately, even though the D-conditions are both sufficient and necessary, they are only hardly verifiable due to their discrete form involving all possible covering sequences. This fact has hindered their use in various applications (see e.g. [5]). In contrast, in the case 1 < q < p < ∞ it is known (see [12,19]) that D 1 + D 2 ≈ A r 3 + A r 4 . This does not apply when 0 < q < 1 ≤ p < ∞, as shown by the results of [20] mentioned earlier.
Rather recently, Prokhorov [16] found conditions for 0 < q < 1 ≤ p < ∞ which have an integral form but involve a function ζ defined by The conditions presented in [16] even involve this function iterated three times. A similar construction was used in the paper [6], also dealing with the same problem. The presence of such an implicit expression involving the weight w virtually prevents any use of these conditions in applications which require further manipulation w (see Sect. 4 for an example). Finding explicit integral conditions for the case 0 < q < 1 ≤ p < ∞, which would have a form comparable e.g. to E 3 and E 4 , hence remained an open problem.
In this paper, we solve this problem and provide the missing integral conditions. No additional assumptions on the weights v, w and the ϑ-regular kernel U are required here, neither are any implicit expressions. The results are presented in Theorems 8, 9 and Corollaries 10, 11. The proofs are based on the well-known method of dyadic discretization (or blocking technique, see [8] for a basic introduction into this method). The particular variant of the technique employed here is essentially the same as the one used in [11]. It is worth noting that the conditions we present here apply to all parameters p, q satisfying 1 ≤ p < ∞ and 0 < q < p. Therefore, the restriction q < 1 is, in fact, unnecessary.
Concerning the structure of this paper, this introduction is followed by Sect. 2 where additional definitions and various auxiliary results are presented. Section 3 consists of the main results, their proofs and some related remarks. In the final Sect. 4 we present certain examples of applications of the results.

Definitions and preliminaries
Let us first introduce the remaining notation and terminology used in the paper. We say that I ⊆ Z is an index set if there exist k min , k max ∈ Z such that k min ≤ k max and Moreover, we denote Let I be an index set containing at least three indices. Then a sequence of points {t k } k∈I is called a covering sequence if t k min = 0, t k max = ∞ and t k < t (k+1) whenever k ∈ I \ {k max }.
Next, let z ∈ N ∪ {0} and n, k ∈ N are such that 0 ≤ k < n. We write z mod n = k if there exists j ∈ N ∪ {0} such that z = jn + k. In other words, k is the remainder after division of the number z by the number n.
In the next part, we present various auxiliary results which will be needed later. The first of these is a known result concerning the saturation of the Hölder inequality. We present an elementary proof of it as well. (3) Then, by the previous part, for each n ∈ N there exists a measurable function g n such that g n = 0 on (0, ∞) \ E n , E n g p n (s)v(s) ds = 2 −n and E n g n (s)ϕ(s) ds = 1. Define g := n∈N g n . Then it holds that  The next proposition was proved in [7, Proposition 2.1], more comments may be found e.g. in [11]. It is a fundamental part of the discretization method.

Proposition 3
Let 0 < α < ∞ and 1 < D < ∞. Then there exists a constant C α,D ∈ (0, ∞) such that for any index set I and any two nonnegative sequences The following result is an analogy to the previous proposition. We present a simple proof, although the result is also well known (see [4,Lemma 3.3]).

Proposition 4
Let 0 < α < ∞ and 1 < D < ∞. Then there exists a constant C α,D ∈ (0, ∞) such that for any index set I and any two nonnegative sequences {b k } k∈I and {c k } k∈I , satisfying Applying Proposition 3, one obtains the following two results. They are useful to handle inequalities involving ϑ-regular kernels.
Proposition 5 Let 0 < α < ∞ and ϑ ∈ [1, ∞). Let U be a ϑ-regular kernel. Then there exists a constant C α,ϑ ∈ (0, ∞) such that, for any index set I, any increasing sequence {t k } k∈I of points from (0, ∞] and any nonnegative sequence it holds that Proof Naturally, we may assume that I contains at least three indices. Let k ∈ I\{k max }. By iterating the inequality from the definition of the ϑ-regular kernel, we get To get the inequality (8), we used Proposition 3, setting D := 2 and c m := U (t m , t (m+1) ) for the relevant indices m. This proves the statement.
and any nonnegative sequence Assume that I contains at least three indices, and let k ∈ I \ {k max }. By the argument from (7), one gets Hence, Estimate (10) follows from Proposition 5 and the assumption (9). The proof is now complete.
Notice that, by the definitions at the beginning of this section, we consider only finite index sets (and therefore also finite covering sequences later on). However, all the results of this section hold for infinite sequences as well. This may be easily shown by using a limit argument. We will nevertheless continue working with finite index sets and covering sequences only. The notion of supremum is used regularly even where it relates to a finite set and where it therefore could be replaced by a maximum. For further remarks see the last part of Sect. 3.
The final basic result concerns ϑ-regular kernels and reads as follows.
Let U be a ϑ-regular kernel and ψ be a nonincreasing nonnegative function defined on (0, ∞). Then

Main results
This section contains the main theorems and their proofs. Remarks to the results and proof techniques can be found at the end of the section. The notation A B means that A ≤ C B, where the constant C may depend only on the exponents p, q and the parameter ϑ. In particular, this C is always independent on the weights w, v, on certain indices (such as k, n, j, K , N , J , μ, . . . ), on the number of summands involved in sums, etc. We write A ≈ B if both A B and B A.
Let v, w be weights. Let U be a ϑ-regular kernel. Then the following assertions are equivalent: holds for all functions f ∈ M + . (ii) Both the conditions Moreover, if C is the least constant such that (11) holds for all functions f ∈ M + , then The variant of the previous theorem for p = 1 reads as follows.
Theorem 9 Let 0 < q < 1 = p and 0 < ϑ < ∞. Let v, w be weights. Let U be a ϑ-regular kernel. Then the following assertions are equivalent: (i) There exists a constant C ∈ (0, ∞) such that the inequality (11) holds for all Moreover, if C is the least constant such that (11) holds for all functions f ∈ M + , then By performing a simple change of variables t → 1 t , one gets the two corollaries below. They are formulated without the discrete conditions, those corresponding to Corollary 10 were presented in Sect. 1. An interested reader may also derive all the discrete conditions easily from their respective counterparts in Theorems 8 and 9.
Corollary 10 Let 1 < p < ∞, 0 < q < p, r := pq p−q and 0 < ϑ < ∞. Let v, w be weights. Let U be a ϑ-regular kernel. Then the following assertions are equivalent: holds for all functions f ∈ M + . (ii) Both the conditions Moreover, if C is the least constant such that (12) holds for all functions f ∈ M + , then Corollary 11 Let 0 < q < 1 = p and 0 < ϑ < ∞. Let v, w be weights. Let U be a ϑ-regular kernel. Then the following assertions are equivalent: (i) There exists a constant C ∈ (0, ∞) such that the inequality (12) holds for all functions f ∈ M + . (ii) Both the conditions Moreover, if C is the least constant such that (12) holds for all functions f ∈ M + , then The next part contains the proofs. The core components of the discretization method used in this article are summarized in Theorem 12 below. It is presented separately for the purpose of possible future reference since this particular variant of discretization may be used even in other problems (cf. [11]).
Throughout the text, parentheses are used in expressions that involve indices, producing symbols such as t (k+1) , t k (n+1) , etc. The parentheses do not have a special meaning, i.e. t (k+1) simply means t with the index k + 1. They are used to make it easier to distinguish between objects as t k (n+1) and t (k n +1) , which, in general, are different and both of them appear frequently in the formulas.

Theorem 12
Let 0 < q < ∞ and 1 ≤ ϑ < ∞. Define Let U be a ϑ-regular kernel. Let K ∈ Z and μ ∈ Z be such that μ ≤ K − 2. Define the index set Let w be a weight such that for all k ∈ Z such that k ≤ K and t K = ∞. For all k ∈ Z such that k ≤ K − 1, denote Then there exist a number N ∈ N and an index set {k n } N n=0 ⊂ Z μ with the following properties.
If we define then (ii) For every n ∈ N such that n ≤ N − 1 it holds that and (iii) For every n ∈ A it holds that (iv) For every n ∈ N, k ∈ Z μ and t ∈ (0, ∞] such that n ≤ N ,

If the same conditions hold and it is even satisfied that
(v) Define k (−1) := μ − 1. Then for every n ∈ N such that n ≤ N it holds that k n −1 Proof At first, observe that it is indeed possible to choose the sequence {t k } with the required properties because the weight w is locally integrable. Since w may take zero values, the sequence {t k } need not be unique. In that case, we choose one fixed {t k } satisfying the requirements. From (14) we deduce that for all k ∈ Z such that k ≤ K − 1.
We proceed with the construction of the index subset {k n }. Define k 0 := μ and k 1 := μ + 1 and continue inductively as follows.
( * ) Let k 0 , . . . , k n be already defined. Then (a) If k n = K , define N := n − 1 and stop the procedure. (b) If k n < K and there exists an index j such that k n < j ≤ K and then define k (n+1) as the smallest index j for which (25) holds. Then proceed again with step ( * ) with n + 1 in place of n. (c) If k n < K and and (25) holds for no index j such that k n < j ≤ K , then define N := n, k (n+1) := K and stop the procedure.
In this manner, one obtains a finite sequence of indices {k 0 , . . . , k N } ⊆ Z μ and the final index k (n+1) = K . We will call each interval k the k-th segment, and each interval [t k n , t (k n +1) ) the n-th block. If n ∈ N is such that n ≤ N , then the n-th block either consists of the single k n -th segment, in which case it holds that k (n+1) = k n + 1, or the n-th segment contains more than one segment and then k (n+1) > k n + 1, If the n-th block is of the second type, then n ∈ A, according to the definition (16). Hence, (17) is satisfied, even though the set A may be empty. The relation (17) in plain words says that each segment is either the last one (i.e., with the highest index k) in a block, or it belongs to a block consisting of more than one segment and the investigated segment is not the last one of those. We have now proved (i).
The property (18) follows directly from the construction. If n ∈ N is such that n ≤ N , then by iterating (18) one gets

Hence, (19) holds and (ii) is then proved.
Property (iii) is again a direct consequence of the way the blocks were constructed. We proceed with proving (iv). Let n ∈ N, k ∈ Z μ and t ∈ (0, ∞] be such that n ≤ N , k ≤ k (n+1) − 1 and t ∈ (t k , t (k+1) ]. Then the following sequence of inequalities is valid: In here, step (26) follows by (24), and step (27) by Proposition 5. If k ≤ k n , then The second inequality here follows by (19). If k > k n , then n ∈ A, k n + 1 ≤ k ≤ k (n+1) − 1 and it holds that The last inequality is granted by (19) and (20). We have proved that Applying this in the inequality obtained at (27), we get the estimate (21). If we now add the assumption k ≤ k (n+1) − 2, then (21) still holds and, in addition to that, we get In here, the last inequality follows from (19) and (20). Applying this result to (21), we obtain (22) and (iv) is thus proved.
To prove (v), let n ∈ N be such that n ≤ N and observe the following: In the first step, (24) was used. In the last one, we used the inequality t k (n−2) ≤ t (k (n−1) −1) which follows from (15).

Proof of Theorem 8 Without loss of generality, we may assume that ϑ ∈ [1, ∞).
Indeed, if the kernel U is ϑ-regular with ϑ ∈ (0, 1), then U is obviously also 1regular. "(ii) ⇒ (i)". Assume that D 1 < ∞ and D 2 < ∞. Let us prove that (11) holds for all f ∈ M + with the least constant C satisfying C r D 1 + D 2 . At first, let us assume that there exists K ∈ Z such that ∞ 0 w = 2 K . Let μ ∈ Z be such that μ ≤ K − 2 and define Z μ by (13). Let {t k } K k=−∞ ⊂ (0, ∞] be a sequence of points such that t K = ∞ and (24) holds for all k ∈ Z such that k ≤ K . Let {k n } N n=0 ⊂ Z μ be the subsequence of indices granted by Theorem 12. Related notation from Theorem 12 will be used in what follows as well.
Step (32) follows from (14). For formal reasons define k −1 := 0. Then, for B 2 we have For the role of the symbol A, see (16). To get (33), we used (20). Inequality (34) follows from Proposition (3) equipped with (18). In steps (35) and (36) we used the Hölder inequality in its integral and discrete form, respectively. Finally, step (37) follows from (23). We have proved Observe that the constant related to the symbol " " in here does not depend on the choice of μ. The reader may nevertheless notice that the construction of the n-blocks in fact depends on μ. However, the constants in the " "-estimates proved with help of that construction are indeed independent of μ. Hence, we may perform the limit pass μ → −∞. Since t μ → 0 as μ → −∞, the monotone convergence theorem (and taking the q-th root) yields for the fixed function f ∈ M + ∩ L p (v). Since the function f was chosen arbitrarily and the constant represented in " " does not depend on f , the inequality (11) holds with C = (D 1 + D 2 ) 1 r for all functions f ∈ M + . Clearly, if C is the least constant such that (11) holds for all f ∈ M + , then At this point, recall that so far we have assumed that ∞ 0 w(x) dx = K for a K ∈ Z. Let us complete the proof of this part for a general weight w.
At first, if ∞ 0 w(x) dx is finite but not equal to any integer power of , the result is simply obtained by multiplying w by a constant c ∈ (1, 2) such that ∞ 0 cw(x) dx = K for a K ∈ Z, and then using homogeneity of the expressions Obviously, for all m ∈ N it holds that w m ≤ w pointwise, hence D 1,m ≤ D 1 and The constant in " " does not depend on m or f and the latter was arbitrarily chosen. Since w m ↑ w pointwise as m → ∞, the monotone convergence theorem (for m → ∞) yields that (11) holds for all functions f ∈ M + and the best constant C in (11) satisfies (38). The proof of this part is now complete.
"(i) ⇒ (ii)". Suppose that (11) holds for all f ∈ M + and C ∈ (0, ∞) is the least constant such that this is true. We need to show that D 1 + D 2 C r .
Let {t k } k∈I be a covering sequence indexed by a set I = {k min , . . . , k max } ⊂ Z. By Proposition 1, for each k ∈ I 0 there exists a measurable function g k supported in [t k , t (k+1) ] and such that g k L p (v) = 1 as well as By Proposition 2 we can find a nonnegative sequence {c k } k∈I 0 such that k∈I 0 c p k = 1 and Define a function g := k∈I 0 c k g k and recall that each g k is supported in [t k , t (k+1) ]. Hence, Finally, we get the following estimate.
In steps (42), (43), (44) and (45) we used (40), (39), (11) and (41), respectively. Since the covering sequence {t k } k∈I was chosen arbitrarily, by taking supremum over all covering sequences we obtain In what follows, we are going to prove a similar estimate for D 2 . Again, let {t k } k∈I be a covering sequence indexed by a set I = {k min , . . . , k max } ⊂ Z. Proposition 1 yields that for every k ∈ I 0 we can find a function h k supported in [t k , t (k+1) ] and such that By Proposition 2, we may find a nonnegative sequence {d k } k∈I 0 such that k∈I 0 d p k = 1 and Define the function h := k∈I 0 d k h k . Then it is easy to verify that h L p (v) = 1. Moreover, we get the following estimate.
The covering sequence {t k } k∈I was arbitrarily chosen in the beginning, hence we may take the supremum over all covering sequences, obtaining the relation The proof of the implication "(i) ⇒ (ii)" and of the related estimates is then finished.
"(iii) ⇒ (ii)". Assume that A 1 < ∞ and A 2 < ∞. We will prove the inequality Let {t k } k∈I be an arbitrary covering sequence indexed by a set I. Then it holds that Taking the supremum over all covering sequences, we obtain D 1 A 1 . Similarly, for any fixed covering sequence {t k } k∈I we get Once again, taking the supremum over all covering sequences, we get D 2 A 2 + A 1 . Hence, we have shown that D 1 + D 2 A 1 + A 2 and the implication "(iii) ⇒ (ii)" is proved.
Similarly as in the proof of "(ii) ⇒ (i)", let us first assume that ∞ 0 w = 2 K for some K ∈ Z . Let μ ∈ Z be such that μ ≤ K − 2 and define Z μ by (13). Let {t k } K k=−∞ ⊂ (0, ∞] be the sequence of points from Theorem 12 and {k n } N n=0 ⊂ Z μ be the subsequence of indices granted by the same theorem. Then In step (46) we used (24), and inequality (47) follows from Proposition 6. We continue by estimating each of the separate terms.
The term B 4 is estimated as follows.
In (49) we used convexity of the r q -th power. Estimate (50) follows from (20), and inequality (51) from Proposition (3) and (18). Finally, in step (52) one makes use of (23). We have proved In the following part, we are going to perform estimates related to the term A 2 . We N Define also y (−1) := 0 and y (N +2) := ∞.
Inequality (71) is obtained by using (18), and inequality (72) by Proposition 5. The final estimate B 12 D 2 was already proved before. We have obtained Let us return to the term B 14 . It holds that and thus also is true for all t > 0. Naturally, the limit variant of Proposition 1 for p = 1 is used in the proof as well. All the estimates are then analogous to their counterparts in the proof of Theorem 8. Therefore, we do not repeat them in here.
Remark 13 (i) Theorem 8, which relates to the inequality (11), i.e. to the operator H * , is the one proved here, while the result for H (i.e. for (12)) is presented as Corollary 10.
Of course, the opposite order could have been chosen, since the version with H instead of H * can be proved in an exactly analogous way. As mentioned before, the variants for H and H * are equivalent by a change of variables in the integrals. The reason why the proof of the "dual" version is shown here is that the discretization-related notation is then the same as in [11].
(ii) Discretization based on finite covering sequences is used here, although the double-infinite (indexed by Z) variant is far more usual in the literature (cf. [5,12,19]). The advantage of the finite version is that the proof works for L 1 -weights w and then it is easily extrapolated for the non-L 1 weights by the final approximation argument. In order to work with infinite partitions, one needs to assume w / ∈ L 1 . The pass to the L 1 -weights then cannot be done in such an easy way as in the opposite order. The authors usually omit the case w ∈ L 1 (see e.g. [5]). Besides that, there is no essential difference between in the techniques based on finite and infinite partitions.
(iii) In Theorems 8 and 9, the equivalence "(i) ⇔ (ii)" was known before [12] and it is reproved here using another method than in [12]. The main achievement is the equivalence "(i) ⇔ (iii)" which can also be proved directly, by the same technique and without need for the discrete D-conditions (cf. [11]). Doing so would however require constructing more different special functions (such as g and h in the "(i) ⇒ (ii)" part of Theorem 8) and therefore also introducing additional notation.
(iv) The kernel U is not assumed to be continuous. However, for every t > 0 the function U (t, ·) is nondecreasing, hence continuous almost everywhere on (0, ∞). Thus, so is the function U q (t, ·) ∞ · v 1− p (s) ds r p . Therefore, the value of the expression A 2 remains unchanged if "sup z∈[t,∞) " in there is replaced by "ess sup z∈[t,∞) ". Although the latter variant may seem to be the "proper" one, both are correct in this case. Besides that, the range z ∈ [t, ∞) in the supremum or essential supremum may obviously be replaced by z ∈ (t, ∞) without changing the value of A 2 .
(v) There is no use of the assumption q < 1 in the proof of Theorem 8, hence its result is indeed valid for all 1 < p < ∞, 0 < q < p. It implies that A * 1 + A * 2 ≈ E r 3 +E r 4 (notice that A * 1 = E r 4 ) in the range 1 ≤ q < p < ∞. This equivalence is, of course, not true for 0 < q < 1 < p < ∞ (recall that the condition E 3 < ∞ is not necessary in this setting, as shown in [20]).

Applications
The integral conditions for the boundedness H : L p (v) → L q (w) with 0 < q < 1 ≤ p < ∞ may be used to complete [5, Theorem 5.1] with two missing cases. These cases are in fact included in [5] but covered there only by discrete conditions. Another explicit characterization may be obtained using [3], the conditions produced in this way would be more complicated compare to those below (cf. also [11]).
Denote by M ↓ the cone of all nonnegative nonincreasing functions on (0, ∞). The result then reads as follows.
holds for all h ∈ M + . In fact, [5,Theorem 4.1] is stated with the assumption ∞ 0 v(y) dy = ∞ which is, however, not used in the proof in [5]. Validity of (99) for all h ∈ M + is equivalent to the condition A 5 + A 6 < ∞ by Theorem 9, since U (x, y) = y x u(s) ds p is a ϑ-regular kernel (with ϑ = 2 p ).