Boundedness of 𝑛-Multiple Discrete Hardy Operators with Weights for 1<𝑞<𝑝<∞

Abstract

We find necessary and sufficient conditions on weighted sequences 𝜔_𝑖, 𝑖=1,2,…,𝑛−1, 𝑢, and 𝑣, for which the operator (𝑆_𝑛−1𝑓)_𝑖=∑^𝑖_𝑘1=1𝜔_1,𝑘1⋯∑^𝑘𝑛−2_{𝑘𝑛−1=1}𝜔_{𝑛−1,𝑘𝑛−1}∑^𝑘𝑛−1_𝑗=1𝑓_𝑗,𝑖≥1, is bounded from 𝑙_𝑝,𝑣 to 𝑙_𝑞,𝑢 for 1<𝑞<𝑝<∞.

1. Introduction

Let 1<𝑝,𝑞<∞ and 1/𝑝+1/𝑝^′=1. Let 𝑣={𝑣_𝑘}^∞_𝑘=1 be positive, 𝜔_𝑖={𝜔_𝑖,𝑘}^∞_𝑘=1, 𝑖=1,2,…,𝑛−1, and 𝑢={𝑢_𝑘}^∞_𝑘=1 nonnegative number sequences, hereinafter referred to as weights.

Let 𝑙_𝑝,𝑣 be the space of sequences 𝑓={𝑓_𝑘}^∞_𝑘=1, for which the following norm is finite: 𝑓_𝑝,𝑣=^∞_𝑖=1𝑣_𝑖𝑓_𝑖^𝑝1/𝑝. (1)

Consider the operator 𝑆_𝑛−1 that acts on the sequence 𝑓={𝑓_𝑘}^∞_𝑘=1 as follows: 𝑆_𝑛−1𝑓_𝑖=^𝑖_𝑘1=1𝜔_1,𝑘1^𝑘1_𝑘2=1𝜔_2,𝑘2⋯^𝑘𝑛−2_{𝑘𝑛−1=1}𝜔_{𝑛−1,𝑘𝑛−1}^𝑘𝑛−1_𝑗=1𝑓_𝑗,𝑖≥1, (2)

and call it 𝑛-multiple discrete Hardy operator with weights.

By changing the order of summation in (2) we have 𝑆_𝑛−1𝑓_𝑖=^𝑖_𝑗=1𝐴_𝑛−1,1𝑖,𝑗𝑓_𝑗,𝑖≥1, (3) where 𝐴_𝑛−1,1(𝑖,𝑗)≡1 for 𝑛=1 and 𝐴_𝑛−1,1𝑖,𝑗=^𝑖_{𝑘𝑛−1=𝑗}𝜔_{𝑛−1,𝑘𝑛−1}^𝑖_{𝑘𝑛−2=𝑘𝑛−1}𝜔_{𝑛−2,𝑘𝑛−2}⋯^𝑖_𝑘1=𝑘2𝜔_1,𝑘1,𝑖≥𝑗≥1, (4)

for 𝑛>1.

Together with operator (3) we consider its dual operator 𝑆^∗_𝑛−1𝑓_𝑗=^∞_𝑖=𝑗𝐴_𝑛−1,1𝑖,𝑗𝑓_𝑖,𝑗≥1. (5)

When 𝑛=1, the operators 𝑆_𝑛−1 and 𝑆^∗_𝑛−1 are simple discrete Hardy operators (𝑆₀𝑓)_𝑖=∑^𝑖_𝑗=1𝑓_𝑗 and (𝑆^∗₀𝑓)_𝑖=∑^∞_𝑗=𝑖𝑓_𝑗; the problem of boundedness from 𝑙_𝑝,𝑣 to 𝑙_𝑞,𝑢 is investigated in detail in [1–3] for all values of the parameters 𝑝 and 𝑞.

In paper [4] necessary and sufficient conditions of boundedness of operators (3) and (5) from 𝑙_𝑝,𝑣 to 𝑙_𝑞,𝑢 are obtained for the case 1<𝑝≤𝑞<∞. However, the problem of boundedness of these operators has not been studied for the case 1<𝑞<𝑝<∞. In this paper we consider this problem.

Let us notice that the boundedness problem of continuous Hardy-type operators has been studied and developed in an extraordinary depth (see, e.g., [3]). The corresponding results for 𝑛-multiple integral Hardy operators were considered by Baiarystanov in [5]. Namely, he investigated the continuous analogue of operators (3) and (5) and found necessary and sufficient conditions of their boundedness from 𝐿_𝑝 to 𝐿_𝑞 for the case 1<𝑝≤𝑞<∞. Moreover, there he obtained sufficient conditions of the same problem for the case 1<𝑞<𝑝<∞. Later in [6] Sagindykov proved that sufficient conditions found by Baiarystanov for the case 1<𝑞<𝑝<∞ are also necessary. However, the method for the continuous case does not work in the discrete situation as it was mentioned in [4]. Thus, here we present a completely different method.

In the sequel we suppose that the sum ∑^𝑘_𝑖=𝑡 is equal to zero for 𝑡>𝑘; the symbol 𝐹≪𝐸 means 𝐹≤𝑐𝐸, where a positive constant 𝑐 does not depend on arguments of the expressions 𝐹 and 𝐸 but can depend on the parameters 𝑝 and 𝑞. The relationship 𝐹≈𝐸 means 𝐹≪𝐸≪𝐹.

2. Auxiliary Statements

For all 𝑖≥𝑗≥1 we assume that 𝐴_𝑙,𝑚(𝑖,𝑗)≡1 if 𝑙<𝑚 and 𝐴_𝑙,𝑚(𝑖,𝑗)=∑^𝑖_{𝑘𝑙=𝑗}𝜔_{𝑙,𝑘𝑙}∑^𝑖_{𝑘𝑙−1=𝑘𝑙}𝜔_{𝑙−1,𝑘𝑙−1}⋯∑^𝑖_{𝑘𝑚=𝑘𝑚+1}𝜔_{𝑚,𝑘𝑚} if 𝑛−1≥𝑙≥𝑚≥1. Moreover, for all 𝑖<𝑗 we suppose that 𝐴_𝑙,𝑚(𝑖,𝑗)=0 if 𝑙,𝑚≥1.

In paper [4] the following lemma is proved.

Lemma 1 (see [4]). For all 𝑖,𝑗,𝜏∶1≤𝑗≤𝜏≤𝑖 one has max_{𝑚−1≤𝑟≤𝑙}𝐴_𝑟,𝑚(𝑖,𝜏)𝐴_𝑙,𝑟+1𝜏,𝑗≤𝐴_𝑙,𝑚≤𝑖,𝑗_𝑙𝑟=∑_𝑚−1𝐴_𝑟,𝑚(𝑖,𝜏)𝐴_𝑙,𝑟+1𝜏,𝑗, (6) when 𝑛−1≥𝑙≥𝑚≥1.

Lemma 2. Let (𝐵_𝑖,𝑗)^∞_𝑖,𝑗=1 be a nonnegative matrix whose elements do not decrease in the first index and do not increase in the second index. Let 𝛾>0. Then for 1≤𝑡,𝑘≤∞ one has Δ⁺_𝑗^𝑘_𝑖=𝑗𝐵_𝑖,𝑗^𝛾≈^𝑘_𝑖=𝑗𝐵_𝑖,𝑗^𝛾−1Δ⁺_𝑗^𝑘_𝑖=𝑗𝐵_𝑖,𝑗,Δ⁻_𝑗^𝑖_𝑗=𝑡𝐵_𝑖,𝑗^𝛾≈^𝑖_𝑗=𝑡𝐵_𝑖,𝑗^𝛾−1Δ⁻_𝑖^𝑖_𝑗=𝑡𝐵_𝑖,𝑗, (7)(8)
where Δ⁺_𝑗𝐸_𝑖,𝑗=𝐸_𝑖,𝑗−𝐸_𝑖,𝑗+1 and Δ⁻_𝑖𝐸_𝑖,𝑗=𝐸_𝑖,𝑗−𝐸_{𝑖−1,𝑗}.

Generally speaking the statement of Lemma 2 is known, but for more complete presentation let us give its proof. We prove only relation (7). Relation (8) can be proved similarly.

Proof. Let 𝛾≥1. Then Δ⁺_𝑗^𝑘_𝑖=𝑗𝐵_𝑖,𝑗^𝛾=^𝑘_𝑖=𝑗𝐵_𝑖,𝑗^𝛾−^𝑘_𝑖=𝑗+1𝐵_𝑖,𝑗+1^𝛾≤𝛾^𝑘_𝑖=𝑗𝐵_𝑖,𝑗^𝛾−1Δ⁺_𝑗^𝑘_𝑖=𝑗𝐵_𝑖,𝑗,Δ⁺_𝑗^𝑘_𝑖=𝑗𝐵_𝑖,𝑗^𝛾≥^𝑘_𝑖=𝑗𝐵_𝑖,𝑗^𝛾−^𝑘_𝑖=𝑗𝐵_𝑖,𝑗^𝑘𝛾−1_𝑖=𝑗+1𝐵_𝑖,𝑗+1=^𝑘_𝑖=𝑗𝐵_𝑖,𝑗^𝛾−1Δ⁺_𝑗^𝑘_𝑖=𝑗𝐵_𝑖,𝑗. (9)
If 0<𝛾<1, then these inequalities hold in opposite direction. Therefore (7) holds for all 𝛾>0. The proof of Lemma 2 is complete.

Moreover, we need the following obvious relations: ^𝑘_𝑖=𝑡𝐵_𝑖,𝑡^𝛾=^𝑘_𝑗=𝑡Δ⁺_𝑗^𝑘_𝑖=𝑗𝐵_𝑖,𝑗^𝛾,_{𝑘𝑗=𝑡}𝐵_𝑘,𝑗^𝛾=^𝑘_𝑖=𝑡Δ⁻_𝑖^𝑖_𝑗=𝑡𝐵_𝑖,𝑗^𝛾. (10)

We also use the following lemma.

Lemma 3 (see [1]). Let 1<𝑝<𝑞<∞. Then the operator 𝑆^∗₀ is bounded from 𝑙_𝑝,𝑣 to 𝑙_𝑞,𝑢 if and only if 𝐵₀(1)=^∞_𝑖=1^𝑖_𝑗=1𝑢^𝑞_𝑗^{𝑝/(𝑝−𝑞)∞}_𝑘=𝑖𝑣^−𝑝′_𝑘^{𝑝(𝑞−1)/(𝑝−𝑞)}𝑣^−𝑝′_𝑖^{(𝑝−𝑞)/𝑝𝑞}<∞; (11) moreover, 𝐵₀(1)≈‖𝑆^∗₀‖.

3. Main and Associated Results

For 𝑟=0,1,…,𝑛−1, we assume that 𝐴_𝑟(𝑛)=^∞_𝑖=1^∞_𝑗=𝑖𝐴^𝑞_𝑟,1𝑗,𝑖𝑢^𝑞_𝑗^{𝑞/(𝑝−𝑞)}×^𝑖_𝑘=1𝐴^𝑝′_{𝑛−1,𝑟+1}(𝑖,𝑘)𝑣^−𝑝′_𝑘^{𝑞(𝑝−1)/(𝑝−𝑞)}×Δ⁺_𝑖^∞_𝑗=𝑖𝐴^𝑞_𝑟,1𝑗,𝑖𝑢^𝑞_𝑗^{(𝑝−𝑞)/𝑝𝑞},𝐵_𝑟(𝑛)=^∞_𝑖=1^∞_𝑗=𝑖𝐴^𝑝′_𝑟,1𝑗,𝑖𝑣^−𝑝′_𝑗^{𝑝(𝑞−1)/(𝑝−𝑞)}×^𝑖_𝑘=1𝐴^𝑞_{𝑛−1,𝑟+1}(𝑖,𝑘)𝑢^𝑞_𝑘^{𝑝/(𝑝−𝑞)}×Δ⁺_𝑖^∞_𝑗=𝑖𝐴^𝑝′_𝑟,1𝑗,𝑖𝑣^−𝑝′_𝑗^{(𝑝−𝑞)/𝑝𝑞}. (12)

Theorem 4. Let 1<𝑞<𝑝<∞ and 𝑛≥1. Then operator (3) is bounded from 𝑙_𝑝,𝑣 to 𝑙_𝑞,𝑢 if and only if 𝐴(𝑛)=max_{0≤𝑟≤𝑛−1}𝐴_𝑟(𝑛)<∞. Moreover, for the norm of operator (3) from 𝑙_𝑝,𝑣 to 𝑙_𝑞,𝑢 the following relation ‖𝑆_𝑛−1‖≈𝐴(𝑛) holds.

Theorem 5. Let 1<𝑞<p<∞ and 𝑛≥1. Then operator (5) is bounded from 𝑙_𝑝,𝑣 to 𝑙_𝑞,𝑢 if and only if 𝐵(𝑛)=max_{0≤𝑟≤𝑛−1}𝐵_𝑟(𝑛)<∞. Moreover, for the norm of operator (5) from 𝑙_𝑝,𝑣 to 𝑙_𝑞,𝑢 the following relation ‖𝑆^∗_𝑛−1‖≈𝐵(𝑛) holds.

For the proofs of Theorems 4 and 5 we need to establish several statements.

Assume that 𝑢^𝑞_𝑖=Δ⁻_𝑖^𝑖_𝑗=1𝐴^𝑞_{𝑛−1,𝑟+1}𝑖,𝑗𝑢^𝑞_𝑗,𝑆_𝑟𝑓_𝑖=^𝑖_𝑗=1𝐴_𝑟,1𝑖,𝑗𝑓_𝑗,𝑆^∗_𝑟𝑔_𝑗=^∞_𝑖=𝑗𝐴_𝑟,1𝑖,𝑗𝑔_𝑖,𝑖,𝑗≥1,0≤𝑟≤𝑛−1. (13)

Consider the inequality 𝑆^∗_𝑟𝑔_𝑞,𝑢≤𝐶𝑔_𝑝,𝑣. (14)

Lemma 6. Let 1<𝑝,𝑞<∞. Then the inequality 𝑆_𝑟𝑓𝑢^𝑞_{𝑝′,𝑣−1}≤𝐶𝑓_𝑞′,𝑢^𝑞−1,𝑓≥0, (15) is dual to (14) with respect to the linear form ∑^∞_𝑖=1𝑓_𝑖𝑔_𝑖𝑢^𝑞_𝑖.

Proof of Lemma 6. If 𝐶>0 is the best constant in (14), then on the basis of principle of duality in 𝑙_𝑞 we have 𝐶=sup_𝑔≥0𝑆^∗_𝑟𝑔_𝑞,𝑢𝑔_𝑝,𝑣=sup_𝑔≥0sup_𝑓≥0∑^∞_𝑘=1𝑓_𝑘𝑆^∗_𝑟𝑔_𝑘𝑢^𝑞_𝑘𝑓_𝑞′,𝑢^𝑞−1𝑔_𝑝,𝑣=sup_𝑔≥0sup_𝑓≥0∑^∞_𝑖=1𝑔_𝑖𝑆_𝑟𝑓𝑢^𝑞_𝑖𝑔_𝑝,𝑣𝑓_𝑞′,𝑢^𝑞−1=sup_𝑓≥0𝑆_𝑟𝑓𝑢^𝑞_{𝑝′,𝑣−1}𝑓_𝑞′,𝑢^𝑞−1. (16)

Lemma 7. Let 1<𝑞<𝑝<∞ and 0≤𝑟≤𝑛−1. If inequality (14) holds with the least constant 𝐶>0, then 𝐵_𝑟(𝑛)≤𝐶. (17)

Proof of Lemma 7. Suppose that inequality (14) holds. Then due to Lemma 6 inequality (15) also holds with the same best constant as in (14).
Let 𝑡≥𝑥≥1. Introduce the test sequence 𝑓={𝑓_𝑗}^∞_𝑗=1 as follows: 𝑓_𝑗=0 for 𝑗>𝑥 and 𝑓_𝑗=(∑^𝑗_𝑖=1𝑢^𝑞_𝑖)^{(𝑞−1)/(𝑝−𝑞)}(∑^𝑡_𝑘=𝑗𝐴^𝑝′_𝑟,1(𝑘,𝑗)𝑣^−𝑝′_𝑘)^{(𝑝−1)(𝑞−1)/(𝑝−𝑞)} for 1≤𝑗≤𝑥.
If we substitute this sequence in the right side of (15), we have 𝑓_𝑞′,𝑢^𝑞−1=^𝑥_𝑗=1^𝑗_𝑖=1𝑢^𝑞_𝑖^{𝑞/(𝑝−𝑞)}×^𝑡_𝑘=𝑗𝐴^𝑝′_𝑟,1𝑘,𝑗𝑣^−𝑝′_𝑘^{𝑞(𝑝−1)/(𝑝−𝑞)}𝑢^𝑞_𝑗^1/𝑞′. (18) Now we will work with the left side of (15) and the test sequence: 𝑆_𝑟𝑓𝑢^𝑞𝑝′_{𝑝′,𝑣−1}=^∞_𝑖=1𝑣^−𝑝′_𝑖^𝑖_𝑗=1𝐴_𝑟,1𝑖,𝑗𝑓_𝑗𝑢^𝑞_𝑗^𝑖_𝑘=1𝐴_𝑟,1(𝑖,𝑘)𝑓_𝑘𝑢^𝑞_𝑘^𝑝′−1 (19) (reduce the sum in the second brackets) ≥^∞_𝑖=1𝑣^−𝑝′_𝑖^𝑖_𝑗=1𝐴_𝑟,1𝑖,𝑗𝑓_𝑗𝑢^𝑞_𝑗^𝑗_𝑘=1𝐴_𝑟,1(𝑖,𝑘)𝑓_𝑘𝑢^𝑞_𝑘^𝑝′−1 (20) (change order of summation) =^∞_𝑗=1𝑓_𝑗^∞_𝑖=𝑗𝐴_𝑟,1𝑖,𝑗𝑣^−𝑝′_𝑖^𝑗_𝑘=1𝐴_𝑟,1(𝑖,𝑘)𝑓_𝑘𝑢^𝑞_𝑘^𝑝′−1𝑢^𝑞_𝑗 (21) (use nonincreasing of 𝐴_𝑟,1(𝑖,𝑘) in the second argument) ≥^∞_𝑗=1𝑓_𝑗^∞_𝑖=𝑗𝐴^𝑝′_𝑟,1𝑖,𝑗𝑣^−𝑝′_𝑖^𝑗_𝑘=1𝑓_𝑘𝑢^𝑞_𝑘^𝑝′−1𝑢^𝑞_𝑗 (22) (substitute the test sequence) ≥^𝑥_𝑗=1^𝑗_𝑖=1𝑢^𝑞_𝑖^{(𝑞−1)/(𝑝−𝑞)𝑡}_𝑘=𝑗𝐴^𝑝′_𝑟,1𝑘,𝑗𝑣^−𝑝′_𝑘^{(𝑝−1)(𝑞−1)/(𝑝−𝑞)}×^𝑡_𝑖=𝑗𝐴^𝑝′_𝑟,1𝑖,𝑗𝑣^−𝑝′_𝑖×^𝑗_𝑘=1^𝑘_𝑖=1𝑢^𝑞_𝑖^{(𝑞−1)/(𝑝−𝑞)}×^𝑡_𝑖=𝑘𝐴^𝑝′_𝑟,1(𝑖,𝑘)𝑣^−𝑝′_𝑖^{(𝑝−1)(𝑞−1)/(𝑝−𝑞)}𝑢^𝑞_𝑘^𝑝′−1𝑢^𝑞_𝑗 (23) (join together the similar cofactors) ≥^𝑥_𝑗=1^𝑗_𝑖=1𝑢^𝑞_𝑖^{(𝑞−1)/(𝑝−𝑞)𝑡}_𝑘=𝑗𝐴^𝑝′_𝑟,1𝑘,𝑗𝑣^−𝑝′_𝑘^{𝑞(𝑝−1)/(𝑝−𝑞)}×^𝑗_𝑘=1^𝑘_𝑖=1𝑢^𝑞_𝑖^{(𝑞−1)/(𝑝−𝑞)}𝑢^𝑞_𝑘^𝑝′−1𝑢^𝑞_𝑗 (24) (use (8)) ≫^𝑥_𝑗=1^𝑗_𝑖=1𝑢^𝑞_𝑖^{(𝑞−1)/(𝑝−𝑞)𝑡}_𝑘=𝑗𝐴^𝑝′_𝑟,1𝑘,𝑗𝑣^−𝑝′_𝑘^{𝑞(𝑝−1)/(𝑝−𝑞)}×^𝑗_𝑘=1Δ⁻_𝑘^𝑘_𝑖=1𝑢^𝑞_𝑖^{(𝑝−1)/(𝑝−𝑞)𝑝′−1}𝑢^𝑞_𝑗 (25) (simplify) =^𝑥_𝑗=1^𝑡_𝑘=𝑗𝐴^𝑝′_𝑟,1𝑘,𝑗𝑣^−𝑝′_𝑘^{𝑞(𝑝−1)/(𝑝−𝑞)}×^𝑗_𝑖=1𝑢^𝑞_𝑖^{𝑞/(𝑝−𝑞)}𝑢^𝑞_𝑗. (26) Therefore, 𝑆_𝑟𝑓𝑢^𝑞𝑝′_{𝑝′,𝑣−1}≫^𝑥_𝑗=1^𝑡_𝑘=𝑗𝐴^𝑝′_𝑟,1𝑘,𝑗𝑣^−𝑝′_𝑘^{𝑞(𝑝−1)/(𝑝−𝑞)}×^𝑗_𝑖=1𝑢^𝑞_𝑖^{𝑞/(𝑝−𝑞)}𝑢^𝑞_𝑗^1/𝑝′, (27) which, together with (15) and (18), gives 𝐶≫^𝑥_𝑗=1^𝑡_𝑘=𝑗𝐴^𝑝′_𝑟,1𝑘,𝑗𝑣^−𝑝′_𝑘^{𝑞(𝑝−1)/(𝑝−𝑞)}×^𝑗_𝑖=1𝑢^𝑞_𝑖^{𝑞/(𝑝−𝑞)}𝑢^𝑞_𝑗^{(𝑝−𝑞)/𝑝𝑞}, (28)
for all 𝑡≥𝑥≥1.
Now in (28) we approach 𝑡→∞ and use (8) 𝐶≫^𝑥_𝑗=1^∞_𝑘=𝑗𝐴^𝑝′_𝑟,1𝑘,𝑗𝑣^−𝑝′_𝑘^{𝑞(𝑝−1)/(𝑝−𝑞)}×Δ⁻_𝑗^𝑗_𝑖=1𝑢^𝑞_𝑖^{𝑝/(𝑝−𝑞)(𝑝−𝑞)/𝑝𝑞} (29) (then we use the Abelian transformation) =^𝑥_𝑗=1^𝑗_𝑖=1𝑢^𝑞_𝑖^{𝑝/(𝑝−𝑞)}Δ⁺_𝑗^∞_𝑘=𝑗𝐴^𝑝′_𝑟,1𝑘,𝑗𝑣^−𝑝′_𝑘^{𝑞(𝑝−1)/(𝑝−𝑞)}+^𝑥_𝑖=1𝑢^𝑞_𝑖^{𝑝/(𝑝−𝑞)}×^∞_𝑘=𝑥+1𝐴^𝑝′_𝑟,1𝑘,𝑗𝑣^−𝑝′_𝑘^{𝑞(𝑝−1)/(𝑝−𝑞)(𝑝−𝑞)/𝑝𝑞} (30) (and use (7)) ≫^𝑥_𝑗=1^𝑗_𝑖=1𝑢^𝑞_𝑖^{𝑝/(𝑝−𝑞)∞}_𝑘=𝑗𝐴^𝑝′_𝑟,1𝑘,𝑗𝑣^−𝑝′_𝑘^{𝑝(𝑞−1)/(𝑝−𝑞)}×Δ⁺_𝑗^∞_𝑘=𝑗𝐴^𝑝′_𝑟,1𝑘,𝑗𝑣^−𝑝′_𝑘^{(𝑝−𝑞)/𝑝𝑞}. (31) Substituting 𝑢^𝑞_𝑖 and approaching 𝑥→∞, we obtain (17).

Lemma 8. Let 1<𝑞<∞ and 0≤𝑟≤𝑛−1. Then 𝑆^∗_𝑟𝑓_𝑞,𝑢≪𝑆^∗_𝑛−1𝑓_𝑞,𝑢,𝑓≥0. (32)

Proof of Lemma 8. The expression 𝐴_𝑛−1,1(𝑖,𝑗) can be transformed as follows: 𝐴_𝑛−1,1𝑖,𝑗=^𝑖_{𝑘𝑛−1=𝑗}𝜔_{𝑛−1,𝑘𝑛−1}…^𝑖_{𝑘𝑟+1=𝑘𝑟+2}𝜔_{𝑟+1,𝑘𝑟+1}×^𝑖_{𝑘𝑟=𝑘𝑟+1}𝜔_{𝑟,𝑘𝑟}⋯^𝑖_𝑘1=𝑘2𝜔_1,𝑘1=^𝑖_{𝑘𝑛−1=𝑗}𝜔_{𝑛−1,𝑘𝑛−1}⋯^𝑖_{𝑘𝑟+2=𝑘𝑟+3}𝜔_{𝑟+2,𝑘𝑟+2}×^𝑖_{𝑘𝑟+1=𝑘𝑟+2}𝜔_{𝑟+1,𝑘𝑟+1}𝐴_𝑟,1𝑖,𝑘_𝑟+1 (33) (change of order of summation) =^𝑖_{𝑘𝑟+1=𝑗}𝐴_𝑟,1𝑖,𝑘_𝑟+1𝜔_{𝑟+1,𝑘𝑟+1}×^𝑘𝑟+1_{𝑘𝑛−1=𝑗}𝜔_{𝑛−1,𝑘𝑛−1}⋯^𝑘𝑟+1_{𝑘𝑟+2=𝑘𝑟+3}𝜔_{𝑟+2,𝑘𝑟+2}=^𝑖_𝑡=𝑗𝐴_𝑟,1(𝑖,𝑡)𝜔_𝑟+1,𝑡𝐴_{𝑛−1,𝑟+2}𝑡,𝑗. (34) Since Δ⁻_𝑡𝐴_{𝑛−1,𝑟+1}(𝑡,𝑗)=𝜔_𝑟+1,𝑡𝐴_{𝑛−1,𝑟+2}(𝑡,𝑗), then 𝐴_𝑛−1𝑖,𝑗=^𝑖_𝑡=𝑗𝐴_𝑟,1(𝑖,𝑡)Δ⁻_𝑡𝐴_{𝑛−1,𝑟+1}𝑡,𝑗. (35) Let 𝑓≥0; then 𝑆^∗_𝑛−1𝑓^𝑞_𝑞,𝑢=^∞_𝑗=1𝑢^𝑞_𝑗^∞_𝑖=𝑗𝐴_𝑛−1,1𝑖,𝑗𝑓_𝑖^𝑞−1×^∞_𝑘=𝑗𝐴_𝑛−1,1𝑘,𝑗𝑓_𝑘≥^∞_𝑗=1𝑢^𝑞_𝑗^∞_𝑘=𝑗𝐴_𝑛−1,1𝑘,𝑗𝑓_𝑘×^∞_𝑖=𝑘𝐴_𝑛−1,1𝑖,𝑗𝑓_𝑖^𝑞−1 (36) (use (35)) =^∞_𝑗=1𝑢^𝑞_𝑗^∞_𝑘=𝑗𝑓_𝑘^𝑘_𝑡=𝑗𝐴_𝑟,1(𝑘,𝑡)Δ⁻_𝑡𝐴_{𝑛−1,𝑟+1}×𝑡,𝑗_∞𝑖=𝑘𝐴_𝑛−1,1𝑖,𝑗𝑓_𝑖^𝑞−1 (37) (change order of summation) =^∞_𝑗=1𝑢^𝑞_𝑗^∞_𝑡=𝑗Δ⁻_𝑡𝐴_{𝑛−1,𝑟+1}×𝑡,𝑗_∞𝑘=𝑡𝐴_𝑟,1(𝑘,𝑡)𝑓∑_𝑘^∞_𝑖=𝑘𝐴_𝑛−1,1𝑖,𝑗𝑓_𝑖^𝑞−1 (38) (use 𝐴_𝑛−1,1(𝑖,𝑗)≥𝐴_𝑟,1(𝑖,𝑡)𝐴_{𝑛−1,𝑟+1}(𝑡,𝑗), which follows from (6)) ≥^∞_𝑗=1𝑢^𝑞_𝑗^∞_𝑡=𝑗Δ⁻_𝑡𝐴_{𝑛−1,𝑟+1}𝑡,𝑗𝐴^𝑞−1_{𝑛−1,𝑟+1}×𝑡,𝑗_∞𝑘∑_=𝑡𝐴_𝑟,1(𝑘,𝑡)𝑓∑_𝑘^∞_𝑖=𝑘𝐴_𝑟,1(𝑖,𝑡)𝑓_𝑖^𝑞−1 (39) (use (7) and (8)) ≫^∞_𝑗=1𝑢^𝑞_𝑗^∞_𝑡=𝑗Δ⁻_𝑡𝐴^𝑞_{𝑛−1,𝑟+1}𝑡,𝑗^∞_𝑘=𝑡Δ⁺_𝑘^∞_𝑖=𝑘𝐴_𝑟,1(𝑖,𝑡)𝑓_𝑖^𝑞 (40) (use the relation Δ⁺_𝑘(∑^∞_𝑖=𝑘𝐴_𝑟,1(𝑖,𝑡)𝑓_𝑖)=𝐴_𝑟,1(𝑘,𝑡)𝑓_𝑘) =^∞_𝑗=1𝑢^𝑞_𝑗^∞_𝑡=𝑗Δ⁻_𝑡𝐴^𝑞_{𝑛−1,𝑟+1}𝑡,𝑗^∞_𝑖=𝑡𝐴_𝑟,1(𝑖,𝑡)𝑓_𝑖^𝑞 (41) (rearrange the cofactors) =^∞_𝑡=1^∞_𝑖=𝑡𝐴_𝑟,1(𝑖,𝑡)𝑓_𝑖^𝑞𝑡_𝑗=1Δ⁻_𝑡𝐴^𝑞_{𝑛−1,𝑟+1}𝑡,𝑗𝑢^𝑞_𝑗= (42) (take into account that 𝐴_{𝑛−1,𝑟+1}(𝑡−1,𝑡)=0) =^∞_𝑡=1^∞_𝑖=𝑡𝐴_𝑟,1(𝑖,𝑡)𝑓_𝑖^𝑞Δ⁻_𝑡^𝑡_𝑗=1𝐴^𝑞_{𝑛−1,𝑟+1}𝑡,𝑗𝑢^𝑞_𝑗=𝑆^∗_𝑟𝑓_𝑞,𝑢; (43) that is, we have (32).

4. Proofs of Theorems 4 and 5

First we prove Theorem 5.

Proof of Theorem 5. Necessity. Suppose that operator (5) is bounded from 𝑙_𝑝,𝑣 to 𝑙_𝑞,𝑢 that equivalently means the validity of the following inequality: 𝑆^∗_𝑛−1𝑓_𝑞,𝑢≤𝐶𝑓_𝑝,𝑣,𝑓≥0. (44)
Then due to (32) inequality (14) holds for any 𝑟=0,1,…,𝑛−1. Therefore, by Lemma 7 we have 𝐵_𝑟(𝑛)≪𝐶,  0≤𝑟≤𝑛−1, that means 𝐵(𝑛)≪𝐶, where 𝐶 is the best constant in (44); that is, 𝐶=‖𝑆^∗_𝑛−1‖.
Sufficiency. will be proved by the induction method.
For 𝑛=1 the operator 𝑆^∗_𝑛−1=𝑆^∗₀ is the Hardy operator. Thus by Lemma 3 if 𝐵₀(1)<∞, then the operator 𝑆^∗₀ is bounded from 𝑙_𝑝,𝑣 to 𝑙_𝑞,𝑢 with the estimate ‖𝑆^∗₀‖≪𝐵₀(1).
Next we assume that for 𝑛=1,2,…,𝑙, 1≤𝑙<𝑛−1, if 𝐵(𝑛)<∞, then the operator 𝑆^∗_𝑛−1, 𝑛=1,2,…,𝑙, is bounded from 𝑙_𝑝,𝑣 to 𝑙_𝑞,𝑢 with the estimate ‖𝑆^∗_𝑛−1‖≪𝐵(𝑛).
Now we need to prove that for 𝑛=𝑙+1 if 𝐵(𝑙+1)=max_{0≤𝑟≤𝑙}𝐵_𝑟(𝑙+1)<∞, then the operator 𝑆^∗_𝑙 is bounded from 𝑙_𝑝,𝑣 to 𝑙_𝑞,𝑢 with the estimate ‖𝑆^∗_𝑙‖≪𝐵(𝑙+1).
Let 𝑓≥0 and 𝑇_𝑗={𝑘∈ℤ∶2^−𝑘≤(𝑆^∗_𝑙𝑓)_𝑗}, 𝑗∈ℕ. Suppose that 𝑘_𝑗=inf𝑇_𝑗 for 𝑇_𝑗≠∅ and 𝑘_𝑗=∞ for 𝑇_𝑗=∅. When 𝑘_𝑗<∞, then 2^−𝑘𝑗≤(𝑆^∗_𝑙𝑓)_𝑗<2^{−𝑘𝑗+1}. Further, for exception of the trivial case we suppose that (𝑆^∗_𝑙𝑓)₁>0.
Let 𝑚₁=1𝑀₁={𝑗∈ℕ∶𝑘_𝑗=𝑘_𝑚1}. Assume that sup𝑀₁+1=𝑚₂. If sup𝑀₁<∞, then 𝑚₂<∞ and sup𝑀₁=max𝑀₁=𝑚₂−1. It is obvious that 𝑚₁<𝑚₂. Suppose that we have found 𝑚₁<𝑚₂<⋯<𝑚_𝑠<∞, 𝑠>1. Then we determine 𝑚_𝑠+1 as 𝑚_𝑠+1=sup𝑀_𝑠+1, where 𝑀_𝑠={𝑗∈ℕ∶𝑘_𝑗=𝑘_𝑚𝑠}. By the definition 2^{−𝑘𝑚𝑠}≤(𝑆^∗_𝑙𝑓)_𝑗<2^{−𝑘𝑚𝑠+1}, 𝑚_𝑠≤𝑗≤𝑚_𝑠+1−1, and 𝑘_𝑚𝑠<𝑘_𝑚𝑠+1, 𝑠≥1.
Further, for convenience let 𝑘_𝑚𝑠=𝑛_𝑠 and 𝑚_𝑠+1−1=𝑚^′_𝑠+1; then 2^−𝑛𝑠≤𝑆^∗_𝑙𝑓_𝑗<2^{−𝑛𝑠+1},𝑚_𝑠≤𝑗≤𝑚^′_𝑠+1, (45)
and 𝑛_𝑠<𝑛_𝑠+1, 𝑠≥1.
Let ℕ₀={𝑠∈ℕ∶𝑚_𝑠<∞}. Then from (45) it follows ℕ=_𝑠∈ℕ0𝑚_𝑠,𝑚_𝑠+1. (46)
There are two possible cases: (1)ℕ₀=ℕ and (2)ℕ₀={1,2,…,𝑠₀}, 𝑠₀>1.
Case (1)ℕ₀=ℕ. Since 𝑛_𝑠<𝑛_𝑠+1, 𝑠≥1, then −𝑛_𝑠−1≤−𝑛_𝑠+2+1. Therefore, using (45) and (6), we have 2^{−𝑛𝑠−1}=2^−𝑛𝑠−2^{−𝑛𝑠−1}≤2^−𝑛𝑠−2^{−𝑛𝑠+2+1}≤𝑆^∗_𝑙𝑓_{𝑚′𝑠+1}−𝑆^∗_𝑙𝑓_𝑚𝑠+2=^{𝑚′𝑠+2}_{𝑖=𝑚′𝑠+1}𝐴_𝑙,1𝑖,𝑚^′_𝑠+1𝑓_𝑖+^∞_{𝑖=𝑚𝑠+2}𝐴_𝑙,1𝑖,𝑚^′_𝑠+1−𝐴_𝑙,1𝑖,𝑚_𝑠+2𝑓_𝑖≤^{𝑚′𝑠+2}_{𝑖=𝑚′𝑠+1}𝐴_𝑙,1𝑖,𝑚^′_𝑠+1𝑓_𝑖+^𝑙−1_𝑖=0𝐴_𝑙,𝑟+1𝑚_𝑠+2,𝑚^′_𝑠+1×^∞_{𝑖=𝑚𝑠+2}𝐴_𝑟,1𝑖,𝑚_𝑠+2𝑓_𝑖. (47) Using (45), (46), and (47), we get 𝑆^∗_𝑙𝑓^𝑞_𝑞,𝑢=^𝑚_𝑠≥1^′𝑠+1_{𝑖=𝑚𝑠}𝑢^𝑞_𝑖𝑆^∗_𝑙𝑓^𝑞_𝑖<_𝑠2^{−𝑛𝑠+1𝑞𝑚′𝑠+1}_{𝑖=𝑚𝑠}𝑢^𝑞_𝑖=2^2𝑞_𝑠2^{−𝑛𝑠−1𝑞𝑚′𝑠+1}_{𝑖=𝑚𝑠}𝑢_𝑖≪_𝑠^{𝑚′𝑠+2}_{𝑖=𝑚′𝑠+1}𝐴_𝑙,1𝑖,𝑚^′_𝑠+1𝑓_𝑖^{𝑞𝑚′𝑠+1}_{𝑖=𝑚𝑠}𝑢^𝑞_𝑖+^𝑙−1_𝑟=0𝑠𝐴^𝑞_𝑙,𝑟+1𝑚_𝑠+2,𝑚^′_𝑠+1×^∞_{𝑖=𝑚𝑠+2}𝐴_𝑟,1𝑖,𝑚_𝑠+2𝑓_𝑖^{𝑞𝑚′𝑠+1}_{𝑖=𝑚𝑠}𝑢^𝑞_𝑖=𝐼_𝑙+^𝑙−1_𝑖=0𝐼_𝑟. (48) First we estimate 𝐼_𝑙. Using Hölder’s inequality twice, we have 𝐼_𝑙≤_𝑠^{𝑚′𝑠+2}_{𝑖=𝑚′𝑠+1}𝐴^𝑝′_𝑙,1𝑖,𝑚^′_𝑠+1𝑣^−𝑝′_𝑖^𝑞/𝑝′×^{𝑚′𝑠+1}_{𝑖=𝑚𝑠}𝑢^𝑞_𝑖^{𝑚′𝑠+2}_{𝑖=𝑚′𝑠+1}𝑣_𝑖𝑓_𝑖^{𝑝𝑞/𝑝}≤_𝑠^{𝑚′𝑠+2}_{𝑖=𝑚′𝑠+1}𝐴^𝑝′_𝑙,1𝑖,𝑚^′_𝑠+1𝑣^−𝑝′_𝑖^{𝑞(𝑝−1)/(𝑝−𝑞)}×^{𝑚′𝑠+1}_{𝑖=𝑚𝑠}𝑢^𝑞_𝑖^{𝑝/(𝑝−𝑞)(𝑝−𝑞)/𝑝}×_𝑠^{𝑚′𝑠+2}_{𝑖=𝑚′𝑠+1}𝑣_𝑖𝑓_𝑖^{𝑝𝑞/𝑝}≪𝐵_𝑙^{(𝑝−𝑞)/𝑝}𝑓^𝑞_𝑝,𝑣, (49) where 𝐵_𝑙=_𝑠^{𝑚′𝑠+2}_{𝑖=𝑚′𝑠+1}𝐴^𝑝′_𝑙,1𝑖,𝑚^′_𝑠+1𝑣^−𝑝′_𝑖^{𝑞(𝑝−1)/(𝑝−𝑞)}×∑^{𝑚′𝑠+1}_{𝑖=𝑚𝑠}𝑢^𝑞_𝑖^{𝑝/(𝑝−𝑞)}. (50)
Next we need the following obvious inequality: Δ⁺_𝑖^∞_𝑗=𝑖𝐴^𝑝′_𝑙,1𝑗,𝑖𝑣^−𝑝′_𝑗=𝐴^𝑝′_𝑙,1(𝑖,𝑖)+^∞_𝑗=𝑖+1Δ⁺_𝑖𝐴^𝑝′_𝑙,1𝑗,𝑖𝑣^−𝑝′_𝑗≥𝐴^𝑝′_𝑙,1(𝑖,𝑖)+^{𝑚′𝑠+2}_𝑗=𝑖+1Δ⁺_𝑖𝐴^𝑝′_𝑙,1𝑗,𝑖𝑣^−𝑝′_𝑗=Δ⁺_𝑖^{𝑚′𝑠+2}_𝑗=𝑖𝐴^𝑝′_𝑙,1𝑗,𝑖𝑣^−𝑝′_𝑗,𝑚^′_𝑠+1≤𝑖≤𝑚^′_𝑠+2. (51)
Now, taking into account the above inequality (51) and using (10) and (7), we estimate 𝐵_𝑙: 𝐵_𝑙≪_𝑠^{𝑚′𝑠+2}_{𝑖=𝑚′𝑠+1}^{𝑚′𝑠+2}_𝑗=𝑖𝐴^𝑝′_𝑙,1𝑗,𝑖𝑣^−𝑝′_𝑗^{𝑝(𝑞−1)/(𝑝−𝑞)}×^{𝑚′𝑠+1}_{𝑘=𝑚𝑠}𝑢^𝑞_𝑘^{𝑝/(𝑝−𝑞)}Δ⁺_𝑖^{𝑚′𝑠+2}_𝑗=𝑖𝐴^𝑝′_𝑙,1𝑗,𝑖𝑣^−𝑝′_𝑗≤_𝑠^{𝑚′𝑠+2}_{𝑖=𝑚′𝑠+1}^∞_𝑗=𝑖𝐴^𝑝′_𝑙,1𝑗,𝑖𝑣^−𝑝′_𝑗^{𝑝(𝑞−1)/(𝑝−𝑞)}×^𝑖_𝑘=1𝑢^𝑞_𝑘^{𝑝/(𝑝−𝑞)}Δ⁺_𝑖^∞_𝑗=𝑖𝐴^𝑝′_𝑙,1𝑗,𝑖𝑣^−𝑝′_𝑗≪^∞_𝑖=1^∞_𝑗=𝑖𝐴^𝑝′_𝑙,1𝑗,𝑖𝑣^−𝑝′_𝑗^{𝑝(𝑞−1)/(𝑝−𝑞)}×^𝑖_𝑘=1𝑢^𝑞_𝑘^{𝑝/(𝑝−𝑞)}Δ⁺_𝑖^∞_𝑗=𝑖𝐴^𝑝′_𝑙,1𝑗,𝑖𝑣^−𝑝′_𝑗=𝐵_𝑙(𝑙+1)^{𝑝𝑞/(𝑝−𝑞)}. (52)
From (49) and (52) we have 𝐼_𝑙≪𝐵^𝑞_𝑙(𝑙+1)𝑓^𝑞_𝑝,𝑣≤𝐵^𝑞(𝑙+1)𝑓^𝑞_𝑝,𝑣. (53)
Now we estimate 𝐼_𝑟, 𝑟=0,1,…,𝑙−1. Let ℕ₁={𝑘=𝑚_𝑠+2∶𝑠∈ℕ}.
Assume that Δ^𝑞_𝑘=𝐴^𝑞_𝑙,𝑟+1(𝑚_𝑠+2,𝑚^′_𝑠+1)∑^{𝑚′𝑠+1}_{𝑖=𝑚𝑠}𝑢^𝑞_𝑖 if 𝑘=𝑚_𝑠+2∈ℕ₁ and Δ^𝑞_𝑘=0 if 𝑘∉ℕ₁. Then 𝐼_𝑟=^∞_𝑘=1Δ^𝑞_𝑘^∞_𝑖=𝑘𝐴_𝑟,1(𝑖,𝑘)𝑓_𝑖^𝑞=𝑆^∗_𝑟𝑓^𝑞_𝑞,Δ,𝑟=0,1,…,𝑙−1. (54)
The operator 𝑆^∗_𝑟 is the operator 𝑆^∗_𝑛−1 for 𝑛=𝑟+1 and 1≤𝑟+1≤𝑙. Therefore, by our assumption we have ‖𝑆^∗_𝑟‖≪𝐵(𝑟+1)=max_{0≤𝑡≤𝑟}𝐵_𝑡(𝑟+1). Hence, 𝐼_𝑟≪max_{0≤𝑡≤𝑟}𝐵_𝑡(𝑟+1), (55) where 𝐵_𝑡(𝑟+1)^{𝑝𝑞/(𝑝−𝑞)}=^∞_𝑖=1^∞_𝑗=𝑖𝐴^𝑝′_𝑡,1𝑗,𝑖𝑣^−𝑝′_𝑗^{𝑝(𝑞−1)/(𝑝−𝑞)}×^𝑖_𝑘=1𝐴^𝑞_𝑟,𝑡+1(𝑖,𝑘)Δ^𝑞_𝑘^{𝑝/(𝑝−𝑞)}×Δ⁺_𝑖^∞_𝑗=𝑖𝐴^𝑝′_𝑡,1𝑗,𝑖𝑣^−𝑝′_𝑗. (56) We estimate the expression ∑^𝑖_𝑘=1𝐴^𝑞_𝑟,𝑡+1(𝑖,𝑘)Δ^𝑞_𝑘: ^𝑖_𝑘=1𝐴^𝑞_𝑟,𝑡+1(𝑖,𝑘)Δ^𝑞_𝑘=∑_{𝑚𝑠+2≤𝑖}𝐴^𝑞_𝑟,𝑡+1𝑖,𝑚_𝑠+2𝐴^𝑞_𝑙,𝑟+1𝑚_𝑠+2,𝑚∑^′_𝑠+1^{𝑚′𝑠+1}_{𝑗=𝑚𝑠}𝑢^𝑞_𝑗 (57) (use the left side of (6) and nonincreasing of 𝐴^𝑞_𝑙,𝑡+1(𝑖,𝑗) in the second argument) ≤_{𝑚𝑠+2≤𝑖}𝐴^𝑞_𝑙,𝑡+1𝑖,𝑚^′_𝑠+1^{𝑚′𝑠+1}_{𝑗=𝑚𝑠}𝑢^𝑞_𝑗≤_𝑚𝑠+2^𝑚_≤𝑖^′𝑠+1_{𝑗=𝑚𝑠}𝐴^𝑞_𝑙,𝑡+1𝑖,𝑗𝑢^𝑞_𝑗≤^𝑖_𝑗=1𝐴^𝑞_𝑙,𝑡+1𝑖,𝑗𝑢^𝑞_𝑗. (58) Substituting the obtained estimate in (56), we have 𝐵_𝑡(𝑟+1)≤𝐵_𝑡(𝑙+1). Therefore, from (55) we get 𝐼_𝑟≪max_{0≤𝑡≤𝑟}𝐵^𝑞_𝑡(𝑙+1)𝑓^𝑞_𝑝,𝑣≤𝐵^𝑞(𝑙+1)𝑓^𝑞_𝑝,𝑣,𝑟=0,1,…,𝑙−1. (59) Then from (48), (53), and (59) we obtain 𝑆^∗_𝑙𝑓_𝑞,𝑢≪𝐵(𝑙+1)𝑓_𝑝,𝑣,𝑆^∗_𝑙≪𝐵(𝑙+1). (60)(61)
Now we turn to case (2)ℕ₀={1,2,…,𝑠₀}, 1≤𝑠₀<∞. In this case we have 𝑚_𝑠0<∞ and 𝑚_𝑠0+1=∞. Here there are two possible cases: 𝑛_𝑠0<∞ and 𝑛_𝑠0=∞.
Below we suppose that ∑^𝑡_𝑠=𝑘=∑^𝑡_𝑠=1 if 𝑘<0.
Let 𝑛_𝑠0<∞. Then 𝑆^∗_𝑙𝑓^𝑞_𝑞,𝑢=^𝑠0𝑚_𝑠=1^𝑠+1_{𝑗=𝑚𝑠}𝑆^∗_𝑙𝑓^𝑞_𝑗𝑢_𝑗=^{𝑠0−2𝑚}_𝑠=1^′𝑠+1_{𝑗=𝑚𝑠}𝑆^∗_𝑙𝑓^𝑞_𝑗𝑢^𝑞_𝑗+^𝑚′𝑠0_{𝑗=𝑚𝑠0−1}𝑆^∗_𝑙𝑓^𝑞_𝑗𝑢^𝑞_𝑗+∑^∞_{𝑗=𝑚𝑠0}𝑆^∗_𝑙𝑓^𝑞_𝑗𝑢^𝑞_𝑗=𝐽₁+𝐽₂+𝐽₃. (62) If 𝐽₁≠0, then 𝑠₀>2, and as in the case ℕ₀=ℕ using (47), we estimate 𝐽₁ and as a result we get 𝐽₁≪𝐵^𝑞(𝑙+1)𝑓^𝑞_𝑝,𝑣. (63) Using (45) and Hölder’s inequality, we estimate the value 𝐽₂: 𝐽₂<2^𝑞2^{−𝑛𝑠0−1𝑞𝑚′𝑠0}_{𝑗=𝑚𝑠0−1}𝑢^𝑞_𝑗≪^∞_{𝑖=𝑚′𝑠0}𝐴_𝑙,1𝑖,𝑚^′_𝑠0𝑓_𝑖^{𝑞𝑚′𝑠0}_{𝑗=𝑚𝑠0−1}𝑢^𝑞_𝑗≤^∞_{𝑖=𝑚′𝑠0}𝐴^𝑝′_𝑙,1𝑖,𝑚^′_𝑠0𝑣^−𝑝′_𝑖^𝑞/𝑝′×^𝑚′𝑠0_{𝑗=𝑚𝑠0−1}𝑢^𝑞_𝑗^∞_{𝑖=𝑚′𝑠0}𝑣_𝑖𝑓_𝑖^{𝑝𝑞/𝑝}≤^∞_{𝑖=𝑚′𝑠0}𝐴^𝑝′_𝑙,1𝑖,𝑚^′_𝑠0𝑣^−𝑝′_𝑖^{𝑞(𝑝−1)/(𝑝−𝑞)}×^𝑚′𝑠0_{𝑗=𝑚𝑠0−1}𝑢^𝑞_𝑗^{𝑝/(𝑝−𝑞)(𝑝−𝑞)/𝑝}𝑓^𝑞_𝑝,𝑣. (64) Using (7) and (10), we estimate the expression in the square brackets: ^∞_{𝑖=𝑚′𝑠0}𝐴^𝑝′_𝑙,1𝑖,𝑚^′_𝑠0𝑣^−𝑝′_𝑖^{𝑞(𝑝−1)/(𝑝−𝑞)𝑚′𝑠0}_{𝑗=𝑚𝑠0−1}𝑢^𝑞_𝑗^{𝑝/(𝑝−𝑞)}≪^∞_{𝑖=𝑚′𝑠0}^∞_𝑗=𝑖𝐴_𝑙,1𝑗,𝑖𝑣^−𝑝′_𝑗^{𝑝(𝑞−1)/(𝑝−𝑞)}×Δ⁺_𝑖^∞_𝑗=𝑖𝐴_𝑙,1𝑗,𝑖𝑣^−𝑝′_𝑗^𝑚′𝑠0_{𝑗=𝑚𝑠0−1}𝑢^𝑞_𝑗^{𝑝/(𝑝−𝑞)}≤^∞_{𝑖=𝑚′𝑠0}^∞_𝑗=𝑖𝐴_𝑙,1𝑗,𝑖𝑣^−𝑝′_𝑗^{𝑝(𝑞−1)/(𝑝−𝑞)}×^𝑖_𝑗=1𝑢^𝑞_𝑗^{𝑝/(𝑝−𝑞)}Δ⁺_𝑖^∞_𝑗=𝑖𝐴_𝑙,1𝑗,𝑖𝑣^−𝑝′_𝑗≤𝐵_𝑙(𝑙+1)^{𝑝𝑞/(𝑝−𝑞)}. (65) Therefore, 𝐽₂≪𝐵_𝑙(𝑙+1)^𝑞𝑓^𝑞_𝑝,𝑣≤𝐵^𝑞(𝑙+1)𝑓^𝑞_𝑝,𝑣. (66) Now we estimate 𝐽₃. Since 2^−𝑛𝑠0≤(𝑆^∗_𝑙𝑓)_𝑗<2^{−𝑛𝑠0+1} for all 𝑗≥𝑚_𝑠0, then (𝑆^∗_𝑙𝑓)_𝑗≤2(𝑆^∗_𝑙𝑓)_𝑡 for all 𝑗,𝑡≥𝑚_𝑠0. Hence, 𝐽₃≤sup_{𝑡≥𝑚𝑠0}^𝑡_{𝑗=𝑚𝑠0}𝑆^∗_𝑙𝑓^𝑞_𝑗𝑢^𝑞_𝑗≤2^𝑞sup_{𝑡≥𝑚𝑠0}^∞_𝑖=𝑡𝐴_𝑛−1,1(𝑖,𝑡)𝑓_𝑖^𝑞𝑡_{𝑗=𝑚𝑠0}𝑢^𝑞_𝑗. (67) Further, as for the estimate of 𝐽₂ by Hölder’s inequality, we have 𝐽₃≪sup_{𝑡≥𝑚𝑠0}^∞_𝑖=𝑡𝐴^𝑞_𝑙,1(𝑖,𝑡)𝑣^−𝑝′_𝑖^{𝑞(𝑝−1)/(𝑝−𝑞)}×^𝑡_{𝑗=𝑚𝑠0}𝑢^𝑞_𝑗^{𝑝/(𝑝−𝑞)(𝑝−𝑞)/𝑝}𝑓^𝑞_𝑝,𝑣≪𝐵^𝑞_𝑙(𝑙+1)𝑓^𝑞_𝑝,𝑣≤𝐵^𝑞(𝑙+1)𝑓^𝑞_𝑝,𝑣. (68) From (62), (63), (66), and (68) we have (60) and (61). If 𝑛_𝑠0=∞ that means 𝑘_𝑚𝑠=∞, then 𝑇_𝑗=∅ for 𝑗≥𝑚_𝑠0; that is, (𝑆^∗_𝑙𝑓)_𝑗=0 for 𝑗≥𝑚_𝑠0. By the assumption (𝑆^∗_𝑙𝑓)₁>0, therefore, 𝑠₀>1. Then 𝑚₂<∞ and 𝑠₀≥2. Hence, 𝑆^∗_𝑙𝑓^𝑞_𝑞,𝑢=^{𝑠0−2𝑚}_𝑠=1^′𝑠+1_{𝑗=𝑚𝑠}𝑆^∗_𝑙𝑓^𝑞_𝑗𝑢^𝑞_𝑗+^𝑚′𝑠0_{𝑗=𝑚𝑠0−1}𝑆^∗_𝑛𝑓^𝑞_𝑗𝑢^𝑞_𝑗=𝐽₁+𝐽₂. (69)
This, together with estimates (63) and (66), gives (60) and (61). Thus, if 𝐵(𝑙+1)<∞, then the operator 𝑆^∗_𝑙 is bounded from 𝑙_𝑝,𝑣 to 𝑙_𝑞,𝑢 and estimate (61) holds. Consequently, for any 𝑛≥1 from 𝐵(𝑛)<∞ we have that the operator 𝑆^∗_𝑛−1 is bounded from 𝑙_𝑝,𝑣 to 𝑙_𝑞,𝑢 and the estimate ‖𝑆^∗_𝑛−1‖≪𝐵(𝑛) holds. The last estimate and the estimate ‖𝑆^∗_𝑛−1‖≫𝐵(𝑛) obtained in the necessity part yield ‖𝑆^∗_𝑛−1‖≈𝐵(𝑛).

The proof of Theorem 4 follows from Theorem 5, because boundedness of the operator 𝑆_𝑛−1 from 𝑙_𝑝,𝑣 to 𝑙_𝑞,𝑢 is equivalent to boundedness of the operator 𝑆^∗_𝑛−1 from 𝑙_{𝑞′,𝑢−1} to 𝑙_{𝑝′,𝑣−1}, and the respective norms of the operators 𝑆^∗_𝑛−1 and 𝑆_𝑛−1 coincide.

Acknowledgment

The authors were supported by the Scientific Committee of the Ministry of Education and Science of Kazakhstan, Grant no. 1529/GF, on the priority area “intellectual potential of the country.”