1. Introduction
In [
1], Maddox generalized the spaces
by adding powers
in the definitions of the spaces to the terms of elements of sequences
. In [
2,
3,
4], the corresponding double sequence spaces were defined for the Pringsheim and the bounded Pringsheim convergence. In [
5], we additionally defined the corresponding double sequence space for the regular convergence. In [
6], the authors characterized some classes of matrix transformations
where
E is the space of bounded Pringsheim convergent (to 0) double sequences with powers or the space of uniformly bounded double sequences with powers. But many of their results appeared to be wrong. In [
5], we gave corresponding counterexamples and proved the correct results. In this paper, we characterize matrix transformations
where
E is the space of Pringsheim convergent (to 0) double sequences with powers or the space of ultimately bounded double sequences with powers.
Note that assuming uniform boundedness in double sequence spaces E greatly simplifies their structure, making them more analogous to single sequence spaces and easing the argumentation for four-dimensional matrices. In this paper, however, we drop the assumption of uniform boundedness, which results in more challenging argumentation and more interesting conditions for matrices.
First, let us introduce the notions and notation we need in this paper.
Let e be the double sequence with all elements 1 and with the -th element equal to 1 and others to 0 . Let be the double sequence with elements in the k-th row equal to 1 and to 0 in other rows, and let be the double sequence with elements in the l-th column equal to 1 and to 0 in other columns. By index sequence , we mean an increasing sequence of integers.
The following variable exponent spaces were defined by Maddox [
1] and Nakano [
7]:
where
is the space of all complex (or real) sequences. When all the terms of
are constant, we have
,
and
, where
,
c,
are, respectively, the spaces of bounded, convergent, and null sequences.
These variable exponent spaces allow us to generalize classical convergence notions and accommodate cases with non-uniform growth conditions, making them particularly useful for studying summability transformations.
A double sequence
of real or complex numbers is said to
converge to the limit a in Pringsheim’s sense (shortly,
p-converge to a) if
Pringsheim convergence is one of the fundamental notions of convergence for double sequences, with applications in signal processing, numerical methods, and random matrix theory. If, in addition, , or the limits and exist, then x is said to be boundedly convergent to a in Pringsheim’s sense (shortly, -convergent to a) and regularly convergent to a (shortly, r-convergent to a), respectively.
Let denote the linear space of all double sequences. Linear subspaces of are called double sequence spaces.
For any notion of convergence
, the space of all
-convergent double sequences will be denoted by
and the limit of a
-convergent double sequence
x by
-
. The sum of a double series
is defined by
whenever
. We also use the notations
Given a double sequence
E, we define its
and
dual by
where
. For other notions and notations in the area of double sequences, we refer the reader to [
8].
Let
be a bounded double sequence of strictly positive numbers. Then, we set
In [
5], we verified that
Analogously, we will show that
To see this, we use the same technique as in [
9]. Indeed, if
, then there are some
with
(
); hence
for
. Conversely, if
for some
, then for all
, we have
, which is bounded since
is bounded. Hence,
.
In [
3], the authors verified that
Now, we will find the
-dual of the space
. Since the space
is solid, then
[
10]. Since for every
and
, we have
, so for every
, we have
. Analogously, for every
and
, we have
, where
is the space of all finite sequences.
For , let be the index such that and for . For , we set .
For
, we define the map
in the following way: We set
,
. Then,
,
,
. We put
,
,
. Further, we set
,
,
. Continuing in the same way, we obtain the map
and let
be the corresponding map of indexes. Note that
. So,
iff
Therefore,
where
Since
, we obtain
Now, we will find the
-dual of the space
. The space
is solid so
[
10]. We will verify that
First, in the same way as for , we see that for any , its rows and columns are finite sequences. Next, since , then .
Now, we will prove that if
and
, then
. Let
be such that
. Let
be defined by
for
and
otherwise. Then,
. Hence,
Since
y has finite rows and columns, we have
Hence,
Let
be any four-dimensional scalar matrix and
be some convergence notion of double sequences. We define
The map
is called a matrix map of type
. We use the notation
if and only if
A is a matrix map of type
and
whenever
. If
(for example, if
X is solid), we use the notation
. In addition, if
X is a sequence space,
Y is a double sequence space, and
is a three-dimensional matrix, we use the notation
if and only if
and
whenever
. If
are sequence spaces and
is a two-dimensional matrix, we use the notation
if and only if
exists and
whenever
.
In this paper, we give the conditions for the classes of matrices where and .
2. Main Results
In what follows, we assume that
,
are bounded double sequences of strictly positive numbers. To characterize matrix transformations of double convergent sequences with powers, we will use the following results from [
5] supplementing Corollary 2.3 with characterization of matrices in
where
X is a sequence space.
Lemma 1 [
5]
. Let X be a sequence space,(a) iff .
(b) iff .
Proposition 1 [
5]
. Let . Then,(a) iff index sequences : (b) iff and in : .
(c) iff in : .
(d) iff ∀ index sequences : .
Corollary 1 [
5]
. Let be a three-dimensional matrix and let X be a sequence space with . Then, the following statements hold:(a) iff ∀ index sequences and all these matrices are pairwise consistent on X.
(b) iff and in .
(c) iff ∀ index sequences .
(d) iff and in .
(e) iff in .
(f) iff ∀ index sequences in .
Now, we consider the main results of our paper. We used some conditions in more than one theorem, so we used the same numbering system instead of writing them repeatedly.
Theorem 1. iff the following conditions hold:
(i)
(ii)
(iii) such that for
(iv) such that for
(v)
(vi) .
Proof. Necessity. (i), (ii), and (v) follow since .
Since given , the map , then by Lemma 1 (b), . Let be fixed and consider the matrix . We can identify the double sequence space with where and the three-dimensional matrix C with the two-dimensional matrix . Then, .
By Theorem 6.1 in [
11] (Chap. 4), there exists
such that
for
. Hence,
for
. So (iii) follows. The statement (iv) follows in a similar way.
To prove (vi) in a similar way with the map
, we define the map
and let
be the corresponding map of indexes, where we use
instead of
and
instead of
. We note that
, where
. Let
be the three-dimensional matrix with
. Then,
implies
. We can identify the double sequence space
with
where
and the matrix
B with the two-dimensional matrix
. Then,
. By Theorem 5.1, 13, in [
12], this implies
so (vi) follows.
Sufficiency. By (i), (ii), and (v), it follows that
exists for every
. Suppose, on the contrary, that
for some
. Then,
for some sequences of integers
with
. We consider the maps
and
defined in Necessity proof. Let
be the two-dimensional matrix with
. Let
and
. Then,
. So
. Then, by Theorem 5.1, 13, in [
12], the following condition does not hold:
Hence, the condition (vi) does not hold. So the contradiction follows. □
Theorem 2. () iff (i), (ii), (v), and the following conditions hold:
(vii) :
(viii) such that for
(ix) such that for
(x) :
(xi) : .
() iff (i)-(v), (vii)-(xi) and the following condition hold:
(xii) :
() iff (i)-(v), (vii)-(xi) and the following conditions hold:
(xiii) :
(xiv) :
(xv)
(xvi)
Proof. () Necessity. (vii) follows since . (i), (ii), and (v) follow since .
Suppose, on the contrary, that (viii) does not hold for some
. Then, we can find index sequences
such that
. Then, we define
in the following way: We set
for
,
for
,
, and
Then,
. So
and the contradiction follows. So (viii) follows. (ix) follows in a similar way.
To prove (x), suppose on the contrary that (x) does not hold; then,
Then, we can choose index sequences
such that
Now, by (i), (ii), (viii), and (ix), for every
, there exists
such that
for
, and for every
, there exists
such that
for
.
In a similar way with the map
, we define the map
, where we use
instead of
and
instead of
and let
be the corresponding map of indexes. We note that
, where
. Let
be the two-dimensional matrix with
. Then,
implies
. By Theorem 5.1, 9, in [
12], this implies that
which contradicts (
1).
To prove (xi), suppose on the contrary that (xi) does not hold; then,
Then, we can choose index sequences
such that
Defining the map
,
r and the two-dimensional matrix
as in the proof of (x), we have
. Let
. By Theorem 5.1, 9, in [
12], we have
which contradicts (
2).
Sufficiency. By (i), (ii), and (v), it follows that
exists for every
. Suppose, on the contrary, that
for some
. Then,
for some index sequences
. In the same way as in the necessity proof of (x) and (xi), we define the map
and the matrix map
. Then, by (i), we have
where
. Since
, then
. So by Theorem 5.1, 9, in [
12], at least one of the following conditions does not hold:
(a)
(b)
Hence, at least one of conditions (x) and (xi) does not hold. So the contradiction follows.
() The proof follows from () and Theorem 1.
(
)
Necessity. (i), (ii), (v), (vii), (viii), and (ix)–(xi) follow from (
), and (iii) and (xii) follow from (
), since
Let
be fixed. By (ii) and (viii), for every
, there exists
such that
for
. In a similar way with the map
in the proof of (
), we define the maps
,
and the two-dimensional matrix
with
. Then,
, where
. So
implies
. By Theorem 5.1, 9, in [
12], it follows that
(1)
(2) : .
Let Then, (xiii) and (xv) follow. Analogously, we obtain (xiv) and (xvi).
Sufficiency. From (i), (ii), (v), (vii), (viii), and (ix)-(xi), it follows that for all . Let be fixed. In the same way as in the proof of Necessity, we define the maps , and the 2 dimensional matrix . Then, , where .
Let
Then, by (xiii) and (xv) (cf. the proof of Necessity), the matrix
B satisfies the conditions 1) and 2) in the Necessity proof. Hence, by Theorem 5.1, 9, in [
12],
, implying
for each
. Analogously, we can prove
for each
. Hence, we obtain
for each
. □
Theorem 3. () iff (i), (ii), (v), (viii), (ix), and the following conditions hold:
(xvii)
(xviii)
() iff (i), (ii), (v), (viii), (ix), (xii), and (xvii)–(xviii).
() iff (i), (ii), (v), (viii), (ix), (xiii)–(xvi) and (xvii)–(xviii).
Proof. Since , the proof follows from Theorem 2 by taking . □
Theorem 4. iff (i), (ii), (v), (viii), (ix), and the following condition hold:
(xix) .
Proof. Necessity. (i), (ii), and (v) follow since . Necessity of (viii) and (ix) follows in the same way as in Theorem 2.
To prove (xix) in the same way as in the proof of the necessity of (vi) in Theorem 1, we identify the space
with
and the matrix
A with the three-dimensional matrix
. Then,
implies
. By Corollary 1 (f), this is equivalent to saying that for all index sequences
, the two-dimensional matrix
is in
. By Theorem 5.1, 13, in [
12], this implies
Suppose, on the contrary, that (xix) does not hold. Then, we can find index sequences
such that
Since for
, we obtain
so
which contradicts (
3). So (xix) follows.
Sufficiency. By (i), (ii), and (v), it follows that
exists for every
. Assume that
for some
. Then,
for some index sequences
. If we define the two-dimensional matrix
as in the proof of Necessity, we have
. Then, by Theorem 5.1, 13, in [
12],
, and we obtain
Then, the condition (xix) is not satisfied, so the contradiction follows. □
Theorem 5. iff (i)-(iv) and the following condition hold:
(xx) .
Proof. Necessity. (i) and (ii) follow, since
. Since
, we obtain (iii) and (iv) by Theorem 1. Since
, (xx) follows from Theorem 3.4 in [
5].
Sufficiency. By (i), (ii), and (xx), it follows that
exists for every
. Assume that
for some
. Then,
for some
. If we define a two-dimensional matrix
and
r as in the proof of sufficiency of Theorem 1, we have
. Then, by Theorem 5.1, 15, in [
12],
So the condition (xx) does not hold and the contradiction follows. □
Theorem 6. () iff (i), (ii), (viii), (ix), (xx), and the following conditions hold:
(xxi)
(xxii)
() iff (i)–(iv), (viii), (ix), and (xx)–(xxii).
() iff (i)–(iv), (viii), (ix), (xx), (xxii), (xxii), and the following conditions hold:
(xxiii)
(xxiv)
(xxv)
(xxvi) .
Proof. (
)
Necessity. (i), (ii), and (xx) follow, since
. Since
, we obtain (viii) and (ix) by Theorem 2. Since
, (xxi) and (xxii) follow from Theorem 3.5 in [
5].
Sufficiency. By (i), (ii), and (xx), it follows that
exists for every
. Now, let
be given. Let
be such that
. We define
by
for
and
otherwise. Then,
. Hence,
by Theorem 3.5 in [
5].
To prove that
, we note that
Since
, then
by Theorem 3.5 in [
5],
.
For
from (v), we obtain
so
. Analogously, by (viii),
. Hence,
. So,
.
() The proof follows from () and Theorem 5.
(
)
Necessity. Note that
iff
, for all
, the three-dimensional matrix
maps
to
, and for all
, the three-dimensional matrix
maps
to
. By (
), the first statement is equivalent to (i), (ii), (viii), (ix) (xx), (xxi), and (xxii). Since
, (xxiii)–(xxvi) follow from Theorem 3.5(c) in [
5].
Sufficiency. From (i), (ii), (viii), (ix) (xx), (xxi), and (xxii), it follows that for all . Let be fixed. In the same way as in the proof of Theorem 2, we define the maps , and the two-dimensional matrix . Then, , where .
Let . By (xxiv) and (xxv), the matrix B satisfies the conditions
(1) : .
(2) : .
Then, by Theorem 5.1, 11, in [
12],
, implying
for each
. Analogously, we can prove
for each
. Hence, we obtain
for each
. □
Theorem 7. () iff (i), (ii), (viii), (ix), (xx), and the following condition hold:
(xxvii)
() iff (i), (ii), (viii), (ix), (xx), and (xxvii).
() iff (i), (ii), (viii), (ix), (xx), and (xxiii)–(xxvii).
Proof. (
) The proof follows in the same way as in Theorem 6 by applying Theorem 3.6 in [
5].
() The proof follows from () and Theorem 6 ().
() The proof follows from () and Theorem 6 (). □
Theorem 8. iff (i), (ii), (viii), (ix), (xx), and the following condition hold:
(xxviii) .
Proof. Necessity. The proof follows from Theorem 4 since and .
Sufficiency. By (i), (ii), (viii), (ix), and (xx), it follows that
exists for every
. Assume that
for some
. Then,
for some index sequences
. If we define a two-dimensional matrix
and
r as in the proof of necessity of (x) and (xi) in Theorem 2, we have
. Then, by Theorem 5.1, 15, in [
12], the condition (xxviii) does not hold. So the contradiction follows. □
Applying Theorems 1–3 and the fact that , we obtain the following theorems:
Theorem 9. iff (i)–(vi) and the following condition hold:
(xxix)
Theorem 10. () iff (i), (ii), (v), (vii)–(xi), and the following condition hold:
(xxx) .
() iff (i)–(xii) and (xxix)–(xxx).
() iff (i)–(v), (vii)–(xi), (xiii)–(xvi), and the following conditions hold:
(xxxi)
(xxxii) .
Theorem 11. () iff (i), (ii), (v), (viii), (ix), (xvii), and the following conditions hold:
(xxxiii)
(xxxiv) .
() iff (i), (ii), (v), (vi), (viii), (ix), (xii), (xvii), (xxix), (xxxiii), and (xxxiv).
() iff (i), (ii), (v), (viii), (ix), (xiii)–(xvi), (xvii), (xxxiii), (xxxiv), and the following conditions hold:
(xxxv)
(xxxvi) .
Theorem 12. iff (i), (ii), (v), (vi), (viii), (ix), and (xxix).
In the following examples, we illustrate how the previously discussed theorems can be applied to four-dimensional matrices.
Example 1. Let and . We consider the matrix with for or or or and ; otherwise, (). Then, evidently, (i), (ii), (viii), and (ix) are satisfied. In addition,so (xx) and (xxi) are also satisfied in Theorem 6. For , we havei.e., the condition (xxii) holds. So, by Theorem 6 (), . On the other hand, the conditions (iii) and (iv) are not satisfied for this matrix (for example, ), so for and .
Example 2. Let us change the previous example slightly. Again, let and . We consider the matrix with for or and ; otherwise, (). Now, in addition to (i), (ii), (viii), (ix), (xx), and (xxi), the conditions (iii) and (iv) are satisfied. So by Theorem 6 ().
Obviously, (xxiii) and (xxiv) hold. For and any , we obtainso (xxv) is satisfied. Analogously, (xxvi) holds for So by Theorem 6 (). For the reader’s convenience, we present our results in the diagram below (
Table 1). The diagram indicates the theorem number where the characterization of
can be found.