#### Abstract

We find necessary and sufficient conditions on weighted sequences , , , and , for which the operator is bounded from to for .

#### 1. Introduction

Let and . Let be positive, , , and nonnegative number sequences, hereinafter referred to as weights.

Let be the space of sequences , for which the following norm is finite:

Consider the operator that acts on the sequence as follows:

and call it -multiple discrete Hardy operator with weights.

By changing the order of summation in (2) we have where for and

for .

Together with operator (3) we consider its dual operator

When , the operators and are simple discrete Hardy operators and ; the problem of boundedness from to is investigated in detail in  for all values of the parameters and .

In paper  necessary and sufficient conditions of boundedness of operators (3) and (5) from to are obtained for the case . However, the problem of boundedness of these operators has not been studied for the case . 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., ). The corresponding results for -multiple integral Hardy operators were considered by Baiarystanov in . Namely, he investigated the continuous analogue of operators (3) and (5) and found necessary and sufficient conditions of their boundedness from to for the case . Moreover, there he obtained sufficient conditions of the same problem for the case . Later in  Sagindykov proved that sufficient conditions found by Baiarystanov for the case are also necessary. However, the method for the continuous case does not work in the discrete situation as it was mentioned in . Thus, here we present a completely different method.

In the sequel we suppose that the sum is equal to zero for ; the symbol means , where a positive constant does not depend on arguments of the expressions and but can depend on the parameters and . The relationship means .

#### 2. Auxiliary Statements

For all we assume that if and if . Moreover, for all we suppose that if .

In paper  the following lemma is proved.

Lemma 1 (see ). For all one has when .

Lemma 2. Let be a nonnegative matrix whose elements do not decrease in the first index and do not increase in the second index. Let . Then for one has
where and .

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 . Then
If , then these inequalities hold in opposite direction. Therefore (7) holds for all . The proof of Lemma 2 is complete.

Moreover, we need the following obvious relations:

We also use the following lemma.

Lemma 3 (see ). Let . Then the operator is bounded from to if and only if moreover, .

#### 3. Main and Associated Results

For , we assume that

Theorem 4. Let and . Then operator (3) is bounded from to if and only if . Moreover, for the norm of operator (3) from to the following relation holds.

Theorem 5. Let and . Then operator (5) is bounded from to if and only if . Moreover, for the norm of operator (5) from to the following relation holds.

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

Assume that

Consider the inequality

Lemma 6. Let . Then the inequality is dual to (14) with respect to the linear form .

Proof of Lemma 6. If is the best constant in (14), then on the basis of principle of duality in we have

Lemma 7. Let and . If inequality (14) holds with the least constant , then

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 . Introduce the test sequence as follows: for and for .
If we substitute this sequence in the right side of (15), we have Now we will work with the left side of (15) and the test sequence: (reduce the sum in the second brackets) (change order of summation) (use nonincreasing of in the second argument) (substitute the test sequence) (join together the similar cofactors) (use (8)) (simplify) Therefore, which, together with (15) and (18), gives
for all .
Now in (28) we approach and use (8) (then we use the Abelian transformation) (and use (7)) Substituting and approaching , we obtain (17).

Lemma 8. Let and . Then

Proof of Lemma 8. The expression can be transformed as follows: (change of order of summation) Since , then Let ; then (use (35)) (change order of summation) (use , which follows from (6)) (use (7) and (8)) (use the relation ) (rearrange the cofactors) (take into account that ) 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 to that equivalently means the validity of the following inequality:
Then due to (32) inequality (14) holds for any . Therefore, by Lemma 7 we have ,  , that means , where is the best constant in (44); that is, .
Sufficiency. will be proved by the induction method.
For the operator is the Hardy operator. Thus by Lemma 3 if , then the operator is bounded from to with the estimate .
Next we assume that for , , if , then the operator , , is bounded from to with the estimate .
Now we need to prove that for if , then the operator is bounded from to with the estimate .
Let and , . Suppose that for and for . When , then . Further, for exception of the trivial case we suppose that .
Let . Assume that . If , then and . It is obvious that . Suppose that we have found , . Then we determine as , where . By the definition , , and , .
Further, for convenience let and ; then
and , .
Let . Then from (45) it follows
There are two possible cases: and , .
Case . Since , , then . Therefore, using (45) and (6), we have Using (45), (46), and (47), we get First we estimate . Using Hölder’s inequality twice, we have where
Next we need the following obvious inequality:
Now, taking into account the above inequality (51) and using (10) and (7), we estimate :
From (49) and (52) we have
Now we estimate , . Let .
Assume that if and if . Then
The operator is the operator for and . Therefore, by our assumption we have . Hence, where We estimate the expression : (use the left side of (6) and nonincreasing of in the second argument) Substituting the obtained estimate in (56), we have . Therefore, from (55) we get Then from (48), (53), and (59) we obtain
Now we turn to case , . In this case we have and . Here there are two possible cases: and .
Below we suppose that if .
Let . Then If , then , and as in the case using (47), we estimate and as a result we get Using (45) and Hölder’s inequality, we estimate the value : Using (7) and (10), we estimate the expression in the square brackets: Therefore, Now we estimate . Since for all , then for all . Hence, Further, as for the estimate of by Hölder’s inequality, we have From (62), (63), (66), and (68) we have (60) and (61). If that means , then for ; that is, for . By the assumption , therefore, . Then and . Hence,
This, together with estimates (63) and (66), gives (60) and (61). Thus, if , then the operator is bounded from to and estimate (61) holds. Consequently, for any from we have that the operator is bounded from to and the estimate holds. The last estimate and the estimate obtained in the necessity part yield .

The proof of Theorem 4 follows from Theorem 5, because boundedness of the operator from to is equivalent to boundedness of the operator from to , and the respective norms of the operators and 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.”