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On weighted integrability of double sine series
J. Math. Anal. Appl. 472 (2019) 1581–1603
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Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa
On weighted integrability of double sine series Krzysztof Duzinkiewicz, Bogdan Szal ∗ University of Zielona Góra, Faculty of Mathematics, Computer Science and Econometrics, 65-516 Zielona Góra, ul. Szafrana 4a, Poland
a r t i c l e
i n f o
a b s t r a c t
Article history: Received 24 April 2018 Available online 10 December 2018 Submitted by S. Tikhonov
We introduce a new class of double sequences DGM (α, β, γ, r) called Double General Monotone and some new class of weight functions to study the weighted integrability of double sine series. Some results of D. Yu, P. Zhou and S. Zhou are also generalized. © 2018 Elsevier Inc. All rights reserved.
Keywords: Double sine series Weighted integrability Classes of double sequences Generalized monotonicity Weight functions
1. Introduction We consider the double sine series ∞ ∞
aj,k sin jx sin ky,
(1.1)
j=1 k=1
on the positive quadrant T2 = [0, π] × [0, π] of the two-dimensional torus. Denote by f (x, y) the sum of (1.1), at the points (x, y) where it converges. Let for r ∈ N, Δr0 aj,k := aj,k − aj+r,k , Δ0r aj,k := aj,k − aj,k+r and Δrr aj,k := Δr0 (Δ0r aj,k ). In the paper [5] the authors defined the following class of double sequences (see also [6]): Definition 1. ([5]) A double sequence {aj,k }∞ j,k=1 ⊂ C belongs to the class of Double General Monotone Sequences (DGM ) if there exist positive constants C and λ ≥ 2, depending only on {aj,k }∞ j,k=1 , such that: 2m−1
|Δ10 aj,n | ≤
j=m
C m
[λm]
|aj,n |, m ≥ λ, n ≥ 1,
j=[m/λ]
* Corresponding author. E-mail addresses: [email protected] (K. Duzinkiewicz), [email protected] (B. Szal). https://doi.org/10.1016/j.jmaa.2018.12.010 0022-247X/© 2018 Elsevier Inc. All rights reserved.
(1.2)
K. Duzinkiewicz, B. Szal / J. Math. Anal. Appl. 472 (2019) 1581–1603
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2n−1
C n
|Δ01 am,k | ≤
k=n 2m−1 2n−1
[λn]
|Δ11 aj,k | ≤
j=m k=n
|am,k |, n ≥ λ, m ≥ 1,
(1.3)
k=[n/λ]
C mn
[λm]
[λn]
|aj,k |, m, n ≥ λ.
(1.4)
j=[m/λ] k=[n/λ]
It is clear that (see [6]) M DS DGM, where M DS is the class of all monotonically decreasing double sequences, i.e. the class of all sequences of non-negative real numbers {aj,k }∞ j,k=1 such that Δ10 aj,k ≥ 0,
Δ01 aj,k ≥ 0,
Δ11 aj,k ≥ 0,
j, k = 1, 2, . . . .
We say that bm,n ∼ bk,l (k, l, m, n ∈ N), if there are two positive constants C1 and C2 such that C1 bm,n ≤ bk,l ≤ C2 bm,n . We can define bm ∼ bk similarly. We say that a weight function φ ∈ Ψ1 if it is a positive function of two variables and satisfies the following conditions: 1. it holds that φ(x, y) ∼ φm,n := φ
π π , , m n
for all (x, y) ∈
π π π π , × , , m, n ∈ N; m+1 m n+1 n
2. if m ∼ k, n ∼ l, then φm,n ∼ φk,l ;
(1.5)
3. it holds that ∞
j −p−2 φj,n = O(m−p−1 φm,n ),
(1.6)
j p−2 φj,n = O(mp−1 φm,n ),
(1.7)
k−p−2 φm,k = O(n−p−1 φm,n ),
(1.8)
kp−2 φm,k = O(np−1 φm,n ),
(1.9)
j=m m j=1 ∞ k=n n k=1
for all m, n ∈ N.
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It should be noted that the conditions (1.6)–(1.9) can be rewritten as almost monotonicity conditions, as done in the paper [13]. Let for p ∈ [1, ∞) and ζ, η ∈ p1 − 1, p1 + 1 φ(x, y) = x−ζp y −ηp
for x, y ∈ (0, π).
(1.10)
It is clear that φ ∈ Ψ1 . Let Lp T2 , 1 ≤ p ≤ ∞, (L1 T2 = L T2 ) be the space of all 2π-periodic real-valued functions of two variables, integrable in the Lebesgue sense to the p-th power when 1 ≤ p < ∞, and essentially bounded on T2 when p = ∞, with the norm ⎧⎧ ⎫1/p ⎪ ⎬ ⎨ ⎪ ⎪ p ⎪ ⎨ |f (x, y)| dxdy , if 1 ≤ p < ∞, ⎭ f p := ⎩ 2 T ⎪ ⎪ ⎪ ⎪ ess sup | f (x, y) | , if p = ∞. ⎩ (x,y)∈T2
Recently, D. Yu, P. Zhou and S. Zhou [17] proved the following theorem: Theorem 1. ([17]) Let a non-negative sequence {am,n }∞ m,n=1 ∈ DGM and am,n → 0
as
m + n → ∞.
(1.11)
Assume that φ ∈ Ψ1 , (φ)− p−1 ∈ L(T2 ) for 1 < p < ∞, or (φ)−1 ∈ L∞ (T2 ) for p = 1. Let a function f be given by (1.1). Then 1
p
φ |f | ∈ L(T2 ),
1 ≤ p < ∞,
if and only if ∞ ∞
(mn)p−2 φm,n apm,n < ∞.
m=1 n=1
The above theorem generalizes the results of B. Ram and S.S. Bhatia [10], C.P. Chen [1] and M.M.H. Marzuq [9]. Now, we consider the class DGM (α, β, γ, r) defined in [3] as follows. Definition 2. ([3]) A double sequence {aj,k }∞ j,k=1 ⊂ C belongs to the class DGM (α, β, γ, r) (called Double General Monotone), if there exist a positive constant C and an integer λ depending only on {aj,k }∞ j,k=1 , for which: 2m−1
|Δr0 aj,n | ≤ Cαm,n , m ≥ λ, n ≥ 1,
j=m 2n−1
|Δ0r am,k | ≤ Cβm,n , n ≥ λ, m ≥ 1,
k=n 2m−1 2n−1
|Δrr aj,k | ≤ Cγm,n , m, n ≥ λ
j=m k=n ∞ ∞ hold, where α := {αm,n }∞ m,n=1 , β := {βm,n }m,n=1 , γ := {γm,n }m,n=1 are non-negative double sequences and r ∈ N.
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The above class with r = 1 was introduced by L. Leindler in the paper [8], which is a two-dimensional equivalent of the classes defined in the papers: [7], [14], [15], and [16]. A similar class of sequences was also considered at the paper [4]. Using Definition 2 with r = 1 and λ ≥ 2, we have that DGM ≡ DGM (1 α,1 β,1 γ, 1), where 1 α,1 β and 1 γ are the sequences defined by the formulas on the right hand sides of the inequalities (1.2), (1.3), and (1.4), respectively. Next, we define a new class of weight functions connected with the class DGM (α, β, γ, 2), which will be used later in the paper. That is, the square [0, π] × [0, π] is divided into four squares whose side lengths are half (because r = 2) of the length of the side of the initial square. We say that a weight function φ ∈ Ψ2 if it is a positive function of two variables satisfying (1.5)–(1.9) and the following condition: φ(x, y) ∼ φm,n := φ
π π , m+1 n+1
,
(1.12)
for all π π π π , × , , (x, y) ∈ m+2 m+1 n+2 n+1 π π π π (x, y) ∈ π − ,π − × , , m+1 m+2 n+2 n+1 π π π π (x, y) ∈ , × π− ,π − , m+2 m+1 n+1 n+2 π π π π (x, y) ∈ π − ,π − × π− ,π − , m+1 m+2 n+1 n+2
and any m, n ∈ N. Let for p ∈ [1, ∞) and ζ, η ∈ p1 − 1, p1 + 1 ⎧ ⎪ x−ζp y −ηp ⎪ ⎪ ⎨ x−ζp (π − y)−ηp ϕ (x, y) = −ζp −ηp ⎪ (π − x) y ⎪ ⎪ ⎩ −ζp −ηp (π − x) (π − y)
for for for for
x, y ∈ (0, π2 ], x ∈ (0, π2 ] and y ∈ π2 , π , y ∈ (0, π2 ] and x ∈ π2 , π , x, y ∈ π2 , π .
(1.13)
It is clear that ϕ ∈ Ψ2 . Moreover, it is easy to show that the function (1.10) does not belong to Ψ2 and the function (1.13) does not belong to Ψ1 . Thus, the classes Ψ1 and Ψ2 are not comparable. In this paper we generalize Theorem 1. Namely, we weaken the condition {am,n }∞ m,n=1 ∈ DGM (1 α,1 β, ∞ 1 γ, 1) to {am,n }m,n=1 ∈ DGM (1 α,1 β,1 γ, 2), by introducing the newly defined set of sequences called Double General Monotone and assuming φ ∈ Ψ2 instead of φ ∈ Ψ1 . 2. Main results The following results hold true: ∞
Theorem 2. Let 1 ≤ p < ∞, λ ∈ N and φ ∈ Ψ2 . Suppose that a non-negative sequence {am,n }m,n=1 satisfies the conditions: (1.11), Δ20 am,n ≥ 0 or Δ20 am,n ≤ 0 for m, n ∈ N, Δ02 am,n ≥ 0 or Δ02 am,n ≤ 0 for m, n ∈ N,
(2.1)
K. Duzinkiewicz, B. Szal / J. Math. Anal. Appl. 472 (2019) 1581–1603 2m−1
|Δ20 aj,n | ≤
j=m 2n−1 k=n
C m
C |Δ02 am,k | ≤ n
1585
[λm]
|aj,n |, m ≥ λ, n ∈ N,
(2.2)
|am,k |, n ≥ λ, m ∈ N
(2.3)
j=[m/λ] [λn] k=[n/λ]
with some positive constant C and
∞ ∞
p−2
φm,n (mn)
m=1 n=1
∞ ∞
p 1/p |Δ22 ak,l |
< ∞.
(2.4)
k=m l=n p
Then the series (1.1) converges on T2 , φ |f | ∈ L(T2 ), and moreover
π π φ(x, y)|f (x, y)|p dxdy = O 0
∞ ∞
φm,n (mn)p−2
m=1 n=1
0
∞ ∞
p |Δ22 ak,l |
.
(2.5)
k=m l=n
Theorem 3. Let a non-negative sequence {am,n }∞ m,n=1 ∈ DGM (1 α,1 β,1 γ, 2) satisfy the conditions: (1.11) and (2.1). Assume that φ ∈ Ψ2 , (φ)− p−1 ∈ L(T2 ) for 1 < p < ∞ or (φ)−1 ∈ L∞ (T2 ) for p = 1. Then 1
p
φ |f | ∈ L(T2 ),
1 ≤ p < ∞,
if and only if ∞ ∞
(mn)p−2 φm,n apm,n < ∞.
m=1 n=1
Theorem 4. (i) DGM (1 α,1 β,1 γ, 1) ⊂ DGM (1 α,1 β,1 γ, 2). (ii) There exists a double sequence {aj,k }∞ j,k=1 satisfying properties (1.11) and (2.1), which belongs to the class DGM (1 α,1 β,1 γ, 2) and does not belong to the class DGM (1 α,1 β,1 γ, 1). It is easy to show that, (ϕ)− p−1 ∈ L(T2 ) for 1 < p < ∞ and (ϕ)−1 ∈ L∞ (T2 ) for p = 1. Therefore, using Theorem 3 we obtain the following corollary. 1
Corollary 1. Let a non-negative sequence {am,n }∞ m,n=1 ∈ DGM (1 α,1 β,1 γ, 2) satisfy conditions (1.11) and (2.1). If a function f is given by (1.1) and ϕ is defined by (1.13), then p
ϕ |f | ∈ L(T2 ),
1 ≤ p < ∞,
if and only if ∞ ∞ m=1 n=1
mαp+p−2 nβp+p−2 apm,n < ∞.
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3. Auxiliary results Denote, for r ∈ Z {0} and k = 0, 1, 2, . . ., by r k,r (x) = cos(k + 2 )x D r 2 sin( 2 x)
the conjugate Dirichlet type kernel. Lemma 1. ([11], [12]) Let r ∈ N, l ∈ Z, and {ak }∞ k=1 ⊂ C. If x = m
ak sin kx = −
k=n
m
m+r
k,r (x) + Δ r ak D
k=n
2lπ r ,
then for all m ≥ n
k,−r (x) − ak D
k=m+1
n+r−1
k,−r (x), ak D
k=n
where Δr ak = ak − ak+r . Lemma 2. ([3]) Let {aj,k }∞ j,k=1 ⊂ C and m, M, n, N ∈ N such that m ≤ M and n ≤ N . 1. If x ∈ (0, π2 ], then ⎞ ⎛ M M M +2 m+1 π ⎝ aj,k sin jx ≤ |Δ20 aj,k | + |aj,k | + |aj,k |⎠ , 4x j=m j=m j=m j=M +1 and if x ∈ [ π2 , π), then ⎞ ⎛ M M M +2 m+1 π ⎝ aj,k sin jx ≤ |Δ20 aj,k | + |aj,k | + |aj,k |⎠ j=m 4(π − x) j=m j=m j=M +1 for any k ∈ N. 2. If y ∈ (0, π2 ], then N N N +2 n+1 π aj,k sin ky ≤ |Δ02 aj,k | + |aj,k | + |aj,k | , 4y k=n
k=n
k=N +1
k=n
and if y ∈ [ π2 , π), then N N N +2 n+1 π aj,k sin ky ≤ |Δ02 aj,k | + |aj,k | + |aj,k | 4(π − y) k=n
k=n
k=N +1
k=n
for any j ∈ N. Lemma 3. If {aj,k }∞ j,k=1 is a non-negative sequence belonging to the class DGM (1 α,1 β,1 γ, 2) with λ ≥ 2, then for any m, n ≥ λ ⎛ am,n = O ⎝
[λm]
[λn]
j=[m/λ] k=[n/λ]
⎞ aj,k ⎠ . jk
(3.1)
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Proof. Let m, n ≥ λ. For any ν and m < μ ≤ 2m we have μ−1
Δ20 aj,ν = am,ν + am+1,ν − aμ,ν − aμ+1,ν .
j=m
By an analogous argument, we get for any μ and n < ν ≤ 2n ν−1
Δ02 aμ,k = aμ,n + aμ,n+1 − aμ,ν − aμ,ν+1 .
k=n
For μ, ν such that m < μ ≤ 2m and n < ν ≤ 2n, we have μ−1 ν−1
Δ22 aj,k =
j=m k=n
=
⎛
ν−1
⎝
Δ20 aj,k −
j=m
k=n
ν−1
μ−1
μ−1
⎞ Δ20 aj,k+2 ⎠
j=m
(am,k + am+1,k − aμ,k − aμ+1,k − am,k+2 − am+1,k+2 + aμ,k+2 + aμ+1,k+2 )
k=n
= am,n + am,n+1 − am,ν − am,ν+1 + am+1,n + am+1,n+1 − am+1,ν − am+1,ν+1 − aμ,n − aμ,n+1 + aμ,ν + aμ,ν+1 − aμ+1,n − aμ+1,n+1 + aμ+1,ν + aμ+1,ν+1 = −(am,ν + am+1,ν − aμ,ν − aμ+1,ν ) − (aμ,n + aμ,n+1 − aμ,ν − aμ,ν+1 ) − (am,ν+1 + am+1,ν+1 − aμ,ν+1 − aμ+1,ν+1 ) − (aμ+1,n + aμ+1,n+1 − aμ+1,ν − aμ+1,ν+1 ) − (aμ,ν + aμ,ν+1 + aμ+1,ν + aμ+1,ν+1 ) + (am,n + am+1,n + am,n+1 + am+1,n+1 ), whence μ−1 ν−1
μ−1
Δ22 aj,k = −
j=m k=n
ν−1
Δ20 aj,ν −
j=m
Δ02 aμ,k −
μ−1
Δ20 aj,ν+1 −
j=m
k=n
ν−1
Δ02 aμ+1,k
k=n
− (aμ,ν + aμ,ν+1 + aμ+1,ν + aμ+1,ν+1 ) + (am,n + am+1,n + am,n+1 + am+1,n+1 ). Therefore, am,n ≤ am,n + am+1,n + am,n+1 + am+1,n+1 =
μ−1 ν−1 j=m k=n
+
ν−1
Δ22 aj,k +
μ−1
Δ20 aj,ν +
j=m
μ−1
Δ20 aj,ν+1 +
j=m
Δ02 aμ,k
k=n
Δ02 aμ+1,k + aμ,ν + aμ,ν+1 + aμ+1,ν + aμ+1,ν+1 ≤
2m−1 j=m
2m−1 2n−1 j=m k=n
k=n
+
ν−1
|Δ20 aj,ν+1 | +
2n−1 k=n
|Δ02 aμ,k | +
2n−1
|Δ22 aj,k | +
2m−1
|Δ20 aj,ν |
j=m
|Δ02 aμ+1,k | + aμ,ν + aμ,ν+1 + aμ+1,ν + aμ+1,ν+1 .
k=n
(3.2) Adding up all inequalities in (3.2) for μ = m + 1, m + 2 . . . , 2m − 1 and ν = n + 1, n + 2, . . . , 2n − 1 and using the assumption {aj,k }∞ j,k=1 ∈ DGM (1 α,1 β,1 γ, 2), we obtain
K. Duzinkiewicz, B. Szal / J. Math. Anal. Appl. 472 (2019) 1581–1603
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2m−1
2n−1
am,n ≤
μ=m+1 ν=n+1
+
⎛
2n−1
2m−1
2n−1
2m−1
2n−1
⎛
j=[m/λ]
⎛
[λn]
⎝C
μ=m+1 ν=n+1
k=[n/λ]
⎞
[λn]
j=[m/λ] k=[n/λ]
⎞
2m−1
2n−1
⎛
aj,k ⎠ ⎝C + jk μ=m+1 ν=n+1 ⎛
[λn] 2m−1 2n−1 aj,ν+1 ⎠ ⎝C + j μ=m+1 ν=n+1
[λm]
⎝C
[λm]
⎝C
μ=m+1 ν=n+1
μ=m+1 ν=n+1
+
2m−1
k=[n/λ]
⎞
⎞
[λm]
j=[m/λ]
⎞ aj,ν ⎠ j
aμ,k ⎠ k
2m−1 2n−1 aμ+1,k ⎠ (aμ,ν + aμ,ν+1 + aμ+1,ν + aμ+1,ν+1 ). + k μ=m+1 ν=n+1
Hence, we have
[λm]
(m − 1)(n − 1)am,n ≤ (m − 1)(n − 1)C
[λn]
j=[m/λ] k=[n/λ] 2n
+ (m − 1)C
2m−1
2n−1
j=[m/λ]
[λn] 2m−1 aj,ν + (n − 1)C j μ=m+1
[λm]
ν=n+2 j=[m/λ]
+
2n−1 [λm] aj,ν aj,k + (m − 1)C jk j ν=n+1
k=[n/λ]
[λn] 2m aμ,k + (n − 1)C k μ=m+2
k=[n/λ]
aμ,k k
(aμ,ν + aμ,ν+1 + aμ+1,ν + aμ+1,ν+1 )
μ=m+1 ν=n+1
and consequently if λ ≥ 2 then
[λn] [λm] 2n 2m 2m 2n aj,k aj,k aj,k aj,k + 8C + 8C + 64 jk jk jk jk j=m k=[n/λ] j=m k=n j=[m/λ] k=[n/λ] j=[m/λ] k=n ⎛ ⎞ [λm] [λn] aj,k ⎠ ⎝ =O . jk [λm]
am,n ≤ C
[λn]
j=[m/λ] k=[n/λ]
This ends the proof. 2 Denote by Am = {j ∈ N : j ≤ m} and Am = {j ∈ N : j ≥ m} or Am = {j ∈ N : j ≥ m} and Am = {j ∈ N : j ≤ m} for m ∈ N. Analogously, let for n ∈ N Bn = {k ∈ N : k ≤ n} and Bn = {k ∈ N : k ≥ n} or Bn = {k ∈ N : k ≥ n} and Bn = {k ∈ N : k ≤ n}. Lemma 4. ([2]) Let λm,n ≥ 0 for all m, n ∈ N. Assume that there exists a constant γ ∈ [0, ∞) such that
λj,n ≤ γmλm,n ,
m, n ∈ N;
(3.3)
λm,k ≤ γnλm,n ,
m, n ∈ N.
(3.4)
j∈Am
k∈Bn
Then for 1 ≤ p < ∞, there exists a positive constant Cp,γ (= γ 2p p2p ) such that ∞ ∞ m=1 n=1
⎛ λm,n ⎝
j∈Am k∈Bn
for all non-negative sequences {am,n }∞ n,m=1 .
⎞p aj,k ⎠ ≤ Cp,γ
∞ ∞ m=1 n=1
λm,n (mnam,n )p
(3.5)
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Lemma 5. Let φ ∈ Ψ2 , f be a non-negative function of two variables, 1 ≤ p < ∞, and
x y F (x, y) :=
π x, y ∈ 0, . 2
f (u, v) dvdu, 0
(3.6)
0
If φf p ∈ L(T2 ) then π
π
2 2
φ(x, y) 0
F (x, y) xy
p
⎛ π π ⎞
2 2 ⎜ ⎟ dxdy = O ⎝ φ(x, y)(f (x, y))p dxdy ⎠ .
0
0
(3.7)
0
Proof. By (1.12) and (3.6) we get π
π
2 2
φ(x, y) 0
F (x, y) xy
∼
m=1 n=1
π π m+2 n+2
φ(x, y)
F (x, y) xy
p dydx
π π m+1 n+1
φm,n
m=1 n=1
(xy)−p (F (x, y))p dydx
π π m+2 n+2
⎛
⎛ ⎜ = O⎝
n+1 ∞ ∞ m+1
dxdy =
0 ∞ ∞
π
π
p
∞ ∞
⎜ φm,n (mn)p−2 ⎝
m=1 n=1
π π m+1 n+1
0
0
⎞p ⎞ ⎟ ⎟ f (u, v) dvdu⎠ ⎠
⎛
∼
∞ ∞
⎞p π π j+1 k+1 ∞ ∞ ⎜ ⎟ φm,n (mn)p−2 ⎝ f (u, v) dvdu⎠ .
m=1 n=1
j=m k=n
π π j+2 k+2
Set λm,n = φm,n (mn)p−2 , Am = {k ∈ N : k ≤ m}, Bn = {k ∈ N : k ≤ n} and π
π
j+1 k+1 aj,k =
f (u, v) dvdu
where
m ≤ k < ∞, n ≤ j < ∞.
π π j+2 k+2
∞
From (1.7) and (1.9) we obtain that the sequence {λm,n }m,n=1 satisfies the conditions (3.3) and (3.4). Therefore, applying (3.5) we have π
π
2 2
φ(x, y) 0
F (x, y) xy
p
0
⎛ ⎞p ⎞ π π m+1
n+1 ∞ ∞ ⎜ ⎜ ⎟ ⎟ dxdy = O ⎝ φm,n (mn)p−2 ⎝mn f (u, v) dvdu⎠ ⎠ ⎛
m=1 n=1
⎛ π π ⎞p m+1
n+1 ∞ ∞ ⎜ ⎟ f (u, v) dvdu⎠ φm,n (mn)2p−2 . ∼ ⎝ m=1 n=1
π π m+2 n+2
π π m+2 n+2
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If p > 0 then using Hölder’s inequality with q = p/ (p − 1), we get ⎛
π π m+1 n+1
⎜ ⎝
π π m+2 n+2
⎧⎛ ⎞ p1 ⎛ π π ⎞ q1 ⎫p π π ⎪ ⎪ m+1 n+1 m+1 n+1 ⎪ ⎪
⎨ ⎬ ⎟ ⎜ ⎟ ⎜ ⎟ p q f (u, v) dvdu⎠ ≤ ⎝ (f (u, v)) dvdu⎠ · ⎝ 1 dvdu⎠ ⎪ ⎪ ⎪ ⎪ π π ⎩ π π ⎭ ⎞p
m+2 n+2
⎛
π π m+1 n+1
⎜ (f (u, v))p dvdu · ⎝
= π π m+2 n+2
m+2 n+2
π π m+1 n+1
⎞
1 1− p p
⎟ dvdu⎠
⎛ ⎜ = O⎝
π π m+2 n+2
⎞
π π m+1 n+1
⎟ (f (u, v))p dvdu · (mn)−2p+2 ⎠ .
π π m+2 n+2
Hence by (1.12) π
π
2 2
φ(x, y) 0
F (x, y) xy
p
⎞ π π
n+1 ∞ ∞ m+1 ⎟ ⎜ dxdy = O ⎝ (f (u, v))p dvdu · (mn)−2p+2 φm,n (mn)2p−2 ⎠ ⎛
m=1 n=1
0
∼
∞ ∞
π π m+1 n+1
π
π
2 2
(f (u, v))p dvdu ∼
φm,n
m=1 n=1
π π m+2 n+2
π π m+2 n+2
φ(x, y)(f (x, y))p dxdy. 0
0
This ends the proof. 2 4. Proofs of the main results 4.1. Proof of Theorem 2 From (2.4), we see that ∞ ∞
|Δ22 ak,l | < ∞,
k=1 l=1
and using Lemma 2, we derive that the series (1.1) converges on T2 . It is clear that π
π π
π
φ(x, y)|f (x, y)|p dxdy = 0
π
2 2
0
π 2 φ(x, y)|f (x, y)|p dx dy +
0
π 2
0
π
2 π
φ(x, y)|f (x, y)|p dxdy = J1 + J2 + J3 + J4 .
φ(x, y)|f (x, y)| dxdy + 0
π 2
π 2
π 2
We have π
π
2 2
φ(x, y)|f (x, y)|p dxdy
J1 = 0
0
π π p
+
φ(x, y)|f (x, y)|p dxdy
0
⎛
π π p
n+1 n ∞ m ∞ m+1 ⎜ φ(x, y) ak,l sin kx sin ly dydx = O⎝
m=1 n=1
π π m+2 n+2
k=1 l=1
K. Duzinkiewicz, B. Szal / J. Math. Anal. Appl. 472 (2019) 1581–1603 π
+
m=1 n=1
π π m+2 n+2 π
+
π
n+1 ∞ ∞ m+1 m=1 n=1
π π m+2 n+2 π
+
π
n+1 ∞ m+1 ∞
π
n+1 ∞ ∞ m+1 m=1 n=1
π π m+2 n+2
1591
m ∞ p φ(x, y) ak,l sin kx sin ly dydx k=1 l=n+1
∞ p n φ(x, y) ak,l sin kx sin ly dydx k=m+1 l=1
∞ φ(x, y)
∞
k=m+1 l=n+1
⎞ p ⎟ ak,l sin kx sin ly dydx⎠ = J11 + J12 + J13 + J14 .
Using the inequalities | sin kx| ≤ kx, | sin ly| ≤ ly for x, y ∈ 0, π2 , k, l ∈ N, and (1.12), we have π
J11 ≤
m=1 n=1
π
n+1 ∞ ∞ m+1
p
φ(x, y)(xy)
n m
p dydx ∼
klak,l
−p−2
φm,n (mn)
m=1 n=1
k=1 l=1
π π m+2 n+2
∞ ∞
n m
p klak,l
.
k=1 l=1
Set λm,n = φm,n (mn)−p−2 , Am = {k ∈ N : k ≥ m}, Bn = {k ∈ N : k ≥ n}. From (1.6) and (1.8), we see that the sequence {λm,n }∞ m,n=1 satisfies conditions (3.3) and (3.4). Therefore, by (3.5), we get ∞ ∞
−p−2
φm,n (mn)
m=1 n=1
n m
p klak,l
=O
∞ ∞
φm,n (mn)p−2 apm,n
.
(4.1)
m=1 n=1
k=1 l=1
It is clear that ∞ ∞ k=m l=n
∞ ∞ ∞ ∞ |Δ22 ak,l | ≥ Δ22 ak,l = (ak,l − ak+2,l − ak,l+2 + ak+2,l+2 ) k=m l=n
l=n
k=m
∞ ∞ ∞ ∞ ∞ ∞ = Δ02 ak,l − Δ02 ak+2,l = Δ02 ak,l − Δ02 ak,l l=n
k=m
k=m
l=n
k=m
k=m+2
m+1 ∞ ∞ = Δ02 ak,l = (am,l − am,l+2 + am+1,l − am+1,l+2 ) l=n
k=m
l=n
∞ ∞ ∞ ∞ n+1 n+1 = am,l − am,l + am+1,l − am+1,l = am,l + am+1,l l=n
l=n+2
l=n
l=n+2
l=n
l=n
= am,n + am,n+1 + am+1,n + am+1,n+1 .
(4.2)
Hence
J11 ≤
∞ ∞ m=1 n=1
p−2
φm,n (mn)
∞ ∞ k=m l=n
p |Δ22 ak,l |
.
K. Duzinkiewicz, B. Szal / J. Math. Anal. Appl. 472 (2019) 1581–1603
1592
Applying Lemma 2 for y ∈ 0, π2 and the inequality | sin kx| ≤ kx for x ∈ 0, π2 , k ∈ N, we have ∞ m m ∞ π ak,l sin kx sin ly ≤ | sin kx| |Δ02 ak,l | + ak,n+1 + ak,n+2 4y k=1 l=n+1
=O
k=1
l=n+1
m m ∞ x x k|Δ02 ak,l | + k(ak,n+1 + ak,n+2 ) . y y k=1 l=n+1
k=1
It is clear that for k ∈ N ∞ l=n+1
∞ n+2 ∞ |Δ02 ak,l | ≥ ak,l − ak,l = ak,l = ak,n+1 + ak,n+2 , l=n+1
l=n+3
(4.3)
l=n+1
and by (2.3) r+1
∞
|Δ02 ak,l | =
∞ 2
(n+1)−1
⎛ ∞ |Δ02 ak,l | = O ⎝
r=0 l=2r (n+1)
l=n+1
[λ2r (n+1)]
r=0 l=[2r (n+1)/λ]
⎞
⎛
ak,l ⎠ = O⎝ l
⎞
∞
l=[(n+1)/λ]
ak,l ⎠ . l (4.4)
Therefore, using (1.12) we obtain π
J12 ≤
m=1 n=1
∼
π
n+1 ∞ ∞ m+1
∞ ∞
π π m+2 n+2
⎛ p m x ⎝ φ(x, y) y
k=1 l=[(n+1)/λ]
⎛
φm,n m−p−2 np−2 ⎝
m=1 n=1
∼
∞ ∞
=O
m
∞
k=1 l=[(n+1)/λ]
⎛ φm,n m−p−2 np−2 ⎝
m=1 n=1
m
n
k=1 l=[(n+1)/λ]
∞ ∞
⎞p k ak,l ⎠ dydx l
∞
−p−2
φm,n (mn)
m=1 n=1
n m
⎞p k ak,l ⎠ l ⎞p m ∞ k k ak,l + ak,l ⎠ l l k=1 l=n
p
klak,l
+O
∞ ∞
φm,n m
−p−2 p−2
n
m=1 n=1
k=1 l=1
∞ m k
k=1 l=n
l
p ak,l
.
Now, let λm,n = φm,n m−p−2 np−2 , Am = {k ∈ N : k ≥ m}, Bn = {k ∈ N : k ≤ n}. From (1.6) and (1.9), we have that the sequence {λm,n }∞ m,n=1 satisfies the conditions (3.3) and (3.4). Therefore, by (3.5) we have ∞ ∞
φm,n m
−p−2 p−2
n
m=1 n=1
∞ m k k=1 l=n
l
p ak,l
=O
∞ ∞
φm,n (mn)p−2 apm,n
m=1 n=1
Using this, (4.1) and (4.2) we get J12 = O
∞ ∞
m=1 n=1
p−2
φm,n (mn)
∞ ∞
k=m l=n
p |Δ22 ak,l |
.
.
(4.5)
K. Duzinkiewicz, B. Szal / J. Math. Anal. Appl. 472 (2019) 1581–1603
1593
Applying Lemma 2 for x ∈ 0, π2 and the inequality | sin ly| ≤ ly for y ∈ 0, π2 , l ∈ N, we have ∞ ∞ n n π ak,l sin kx sin ly ≤ | sin ly| |Δ20 ak,l | + am+1,l + am+2,l 4x k=m+1 l=1
=O
l=1
k=m+1
∞ n n y y l|Δ20 ak,l | + l(am+1,l + am+2,l ) . x x k=m+1 l=1
l=1
It is clear that for l ∈ N ∞ k=m+1
∞ m+2 ∞ |Δ20 ak,l | ≥ ak,l − ak,l = ak,l = am+1,l + am+2,l , k=m+1
k=m+3
(4.6)
k=m+1
and applying (2.2) r+1
∞
|Δ20 ak,l | =
∞ 2
⎛
(m+1)−1
|Δ20 ak,l | = O ⎝
r=0 k=2r (m+1)
k=m+1
[λ2r (m+1)]
∞
r=0 k=[2r (m+1)/λ]
⎞
⎛
⎞
∞
ak,l ⎠ = O⎝ k
k=[(m+1)/λ]
ak,l ⎠ . k (4.7)
Therefore, by (1.12) π
J13 ≤
m=1 n=1
∼
π
n+1 ∞ ∞ m+1
∞ ∞
φ(x, y)
π π m+2 n+2
y p x
∼
∞ ∞
=O
⎝
⎞p n l ak,l ⎠ dydx k
⎞p n l ak,l ⎠ k
∞
φm,n mp−2 n−p−2 ⎝
k=[(m+1)/λ] l=1
⎛
⎞p n ∞ n l l ak,l + ak,l ⎠ k k
m
φm,n mp−2 n−p−2 ⎝
m=1 n=1
∞
k=[(m+1)/λ] l=1
⎛
m=1 n=1
⎛
k=[(m+1)/λ] l=1
∞ ∞
−p−2
φm,n (mn)
m=1 n=1
n m
k=m l=1
p
lkak,l
+O
∞ ∞
φm,n m
p−2 −p−2
n
m=1 n=1
k=1 l=1
n ∞ l ak,l k
p .
k=m l=1
Set λm,n = φm,n mp−2 n−p−2 , Am = {k ∈ N : k ≤ m}, Bn = {k ∈ N : k ≥ n}. From (1.7) and (1.8), we see that the sequence {λm,n }∞ m,n=1 satisfies the conditions (3.3) and (3.4). Therefore, by (3.5) we have ∞ ∞
φm,n m
p−2 −p−2
n
m=1 n=1
n ∞ l ak,l k
p
=O
∞ ∞
φm,n (mn)p−2 apm,n
m=1 n=1
k=m l=1
Whence by (4.1) and (4.2) we get J13 = O
∞ ∞
m=1 n=1
p−2
φm,n (mn)
∞ ∞
k=m l=n
p |Δ22 ak,l |
.
.
(4.8)
K. Duzinkiewicz, B. Szal / J. Math. Anal. Appl. 472 (2019) 1581–1603
1594
By elementary calculations we have ∞
∞
∞
|Δ22 ak,l | ≥
k=m+1 l=n+1
l=n+1
∞ ∞ ∞ m+2 Δ02 ak,l − Δ02 ak,l = Δ02 ak,l , k=m+1
k=m+3
(4.9)
l=n+1 k=m+1
and ∞
∞
k=m+1 l=n+1
∞ n+2 ∞ ∞ ∞ |Δ22 ak,l | ≥ Δ20 ak,l − Δ20 ak,l = Δ20 ak,l . k=m+1 l=n+1
l=n+3
(4.10)
k=m+1 l=n+1
Using Lemma 1, (1.11), the estimates π , 4x l,±2 (y)| ≤ π |D 4y k,±2 (x)| ≤ |D
(4.11) (4.12)
for x, y ∈ 0, π2 , k, l ∈ N, (2.1), (4.9), (4.10), and (4.2) we get ∞
∞
k=m+1 l=n+1
∞
k,2 (x)| · |D l,2 (y)| |Δ22 ak,l | · |D
k=m+1 l=n+1 ∞
m+2
+
∞
ak,l sin kx sin ly ≤
k,−2 (x)| · |D l,2 (y)| + |Δ02 ak,l | · |D
k=m+1 l=n+1 m+2
+
n+2
∞
n+2
k,2 (x)| · |D l,−2 (y)| |Δ20 ak,l | · |D
k=m+1 l=n+1
k,−2 (x)| · |D l,−2 (y)| |ak,l | · |D
k=m+1 l=n+1
∞ ∞ m+2 ∞ m+2 ∞ n+2 n+2 π2 |Δ22 ak,l | + Δ02 ak,l + Δ20 ak,l + ak,l ≤ 16xy k=m+1 l=n+1 l=n+1 k=m+1 k=m+1 l=n+1 k=m+1 l=n+1 ∞ ∞ π2 ≤ |Δ22 ak,l | . 4xy k=m l=n
Hence by (1.12) we get ⎛
J14
⎞ π π p p
n+1 ∞ ∞ ∞ ∞ m+1 1 ⎜ ⎟ = O⎝ φ(x, y) |Δ22 ak,l | dydx⎠ xy m=1 n=1 π π m+2 n+2
∼
∞ ∞
φm,n (mn)p−2
m=1 n=1
k=m l=n
∞ ∞
p |Δ22 ak,l |
.
k=m l=n
Further we have π
π 2
φ(x, y)|f (x, y)|p dydx
J2 = π 2
0
⎛
∞ ∞ ⎜ = O⎝
π π− m+2
m=1 n=1 π−
π
n+1
π π m+1 n+2
p n m φ(x, y) ak,l sin kx sin ly dydx k=1 l=1
K. Duzinkiewicz, B. Szal / J. Math. Anal. Appl. 472 (2019) 1581–1603
+
∞ ∞
π π− m+2
m=1 n=1 π−
+
∞ ∞
+
∞ ∞
π π m+1 n+2
π π− m+2
m=1 n=1 π−
π
n+1
π π m+1 n+2
π π− m+2
m=1 n=1 π−
π
n+1
π
n+1
π π m+1 n+2
1595
m ∞ p φ(x, y) ak,l sin kx sin ly dydx k=1 l=n+1
p ∞ n φ(x, y) ak,l sin kx sin ly dydx k=m+1 l=1
∞ φ(x, y)
∞
k=m+1 l=n+1
⎞ p ⎟ ak,l sin kx sin ly dydx⎠ = J21 + J22 + J23 + J24 .
Using the inequalities | sin kx| ≤ k (π − x) and | sin ly| ≤ ly for k, l ∈ N, x ∈
π 2,π
, y ∈ 0, π2 we get
m n n m ak,l sin kx sin ly ≤ (π − x)y klak,l . k=1 l=1
k=1 l=1
Whence by (1.12) π π− m+2
∞ ∞
J21 ≤
φ(x, y)((π − x)y)
m=1 n=1 π−
=O
π
n+1
∞ ∞
p
n m
p klak,l
dydx
k=1 l=1
π π m+1 n+2
−p−2
φm,n (mn)
m=1 n=1
n m
p klak,l
.
k=1 l=1
Applying Lemma 2 for y ∈ 0, π2 , the inequality | sin kx| ≤ k (π − x) for x ∈ have
π 2,π
, k ∈ N and (4.3) we
m ∞ ∞ m π ak,l sin kx sin ly ≤ (π − x) k |Δ02 ak,l | + ak,n+1 + ak,n+2 4y k=1 l=n+1
=O
k=1
l=n+1
∞ m π−x k|Δ02 ak,l | . y k=1 l=n+1
Next by (1.12) and (4.4)
J22 ≤
∞ ∞
π π− m+2
m=1 n=1 π−
⎛ = O⎝
π
n+1 φ(x, y)
π π m+1 n+2
∞ ∞
m=1 n=1
π−x y ⎛
φm,n m−p−2 np−2 ⎝
m
p
⎛ ⎝
m
∞
k=1 l=[(n+1)/λ] ∞
k=1 l=[(n+1)λ]
⎞p k ak,l ⎠ dydx l
⎞p ⎞ k ak,l ⎠ ⎠ . l
K. Duzinkiewicz, B. Szal / J. Math. Anal. Appl. 472 (2019) 1581–1603
1596
Using Lemma 2 for x ∈ [ π2 , π), the inequality | sin ly| ≤ ly for y ∈ 0, π2 , l ∈ N, and (4.6) we have ∞ n ∞ n π ak,l sin kx sin ly ≤ y l |Δ20 ak,l | + am+1,l + am+2,l 4(π − x) k=m+1 l=1
=O
l=1
k=m+1
∞ n y l|Δ20 ak,l | π−x k=m+1 l=1
and by (1.12), (4.7) we obtain
J23 ≤
∞ ∞
π π− m+2
∼
φ(x, y)
m=1 n=1 π− ∞ ∞
π
n+1
π π m+1 n+2
y π−x
⎛
∞
⎝
⎞p n l ak,l ⎠ dydx k
k=[(m+1)/λ] l=1
⎞p n l ak,l ⎠ . k
∞
φm,n mp−2 n−p−2 ⎝
m=1 n=1
⎛
p
k=[(m+1)/λ] l=1
Further, we can estimate the quantities J21 , J22 , and J23 similarly as the quantities J11 , J12 , and J13 , respectively. Using Lemma 1, (1.11), the estimate k,±2 (x)| ≤ |D
π π for x ∈ [ , π), k ∈ N, 4 (π − x) 2
(4.13)
(4.12), (2.1), (4.9), (4.10), and (4.2) we get ∞
∞
k=m+1 l=n+1
∞ ∞ 1 ak,l sin kx sin ly = O |Δ22 ak,l | , (π − x)y k=m l=n
and by (1.12) ⎛ J24
∞ ∞ ⎜ = O⎝
π π− m+2
∼
φ(x, y)
m=1 n=1 π− ∞ ∞
π
n+1
π π m+1 n+2
φm,n (mn)p−2
m=1 n=1
∞ ∞
1 (π − x)y
p ∞ ∞
p |Δ22 ak,l |
k=m l=n
p |Δ22 ak,l |
.
k=m l=n
Therefore, J2 = O
∞ ∞
p−2
φm,n (mn)
m=1 n=1
∞ ∞
p |Δ22 ak,l |
.
k=m l=n
The quantity J3 we can estimate in a symmetric way as J2 . Hence, we have J3 = O
∞ ∞
m=1 n=1
p−2
φm,n (mn)
∞ ∞
k=m l=n
p |Δ22 ak,l |
.
⎞ ⎟ dydx⎠
K. Duzinkiewicz, B. Szal / J. Math. Anal. Appl. 472 (2019) 1581–1603
1597
Finally, we have
π π φ(x, y)(f (x, y))p dxdy
J4 = π 2
π 2
⎛
π π π− n+2 π− m+2
∞ ∞ ⎜ = O⎝
m=1 n=1 π−
+
∞ ∞
+
+
π π− n+1
π m+1
π m+1
m=1 n=1 π−
π m+1
π π− n+1
k=1 l=1
m ∞ p φ(x, y) ak,l sin kx sin ly dydx ∞ p n φ(x, y) ak,l sin kx sin ly dydx k=m+1 l=1
π π− n+1
π π π− n+2 π− m+2
p m n φ(x, y) ak,l sin kx sin ly dydx
k=1 l=n+1
π π− n+1
π π π− n+2 π− m+2
m=1 n=1 π− ∞ ∞
π m+1
π π π− n+2 π− m+2
m=1 n=1 π− ∞ ∞
∞ φ(x, y)
∞
k=m+1 l=n+1
⎞ p ⎟ ak,l sin kx sin ly dydx⎠ = J41 + J42 + J43 + J44 .
Using the inequalities | sin kx| ≤ k(π − x) and | sin ly| ≤ l(π − y) for k, l ∈ N, x, y ∈
π 2,π
we have
m n n m ≤ (π − x)(π − y) a sin kx sin ly klak,l . k,l k=1 l=1
k=1 l=1
From (1.12) we get
J41 ≤
∞ ∞
π π π− n+2 π− m+2
φ(x, y)((π − x)(π − y))
p
m=1 n=1π− π π− π m+1 n+1
∼
∞ ∞
n m
p klak,l
dydx
k=1 l=1
−p−2
φm,n (mn)
m=1 n=1
m n
p klak,l
.
k=1 l=1
Applying Lemma 2 for y ∈ [ π2 , π), the inequality | sin kx| ≤ k (π − x) for x ∈ [ π2 , π), k ∈ N and (4.3) we have ∞ m m ∞ π ak,l sin kx sin ly ≤ (π − x) k |Δ02 ak,l | + ak,n+1 + ak,n+2 4(π − y) k=1 l=n+1
=O
k=1
∞ m π−x k|Δ02 ak,l | , π−y k=1 l=n+1
and by (1.12), (4.4) we obtain
l=n+1
K. Duzinkiewicz, B. Szal / J. Math. Anal. Appl. 472 (2019) 1581–1603
1598
π π π− n+2 π− m+2
∞ ∞
J42 ≤
φ(x, y)
m=1 n=1 π− ∞ ∞
∼
π m+1
π π− n+1
⎛
φm,n m−p−2 np−2 ⎝
m=1 n=1
π−x π−y
⎛
p
⎝
k=1 l=[(n+1)/λ]
∞
m
⎞p
∞
m
k=1 l=[(n+1)/λ]
k ak,l ⎠ dydx l
⎞p k ak,l ⎠ . l
Using Lemma 2 for x ∈ [ π2 , π), the inequality | sin ly| ≤ l (π − y) for y ∈ [ π2 , π), l ∈ N and (4.6) we have ∞ ∞ n n π ak,l sin kx sin ly ≤ (π − y) l |Δ20 ak,l | + am+1,l + am+2,l 4(π − x) k=m+1 l=1 l=1 k=m+1 n ∞ π−y l|Δ20 ak,l | . =O π−x k=m+1 l=1
From (1.12) and (4.7) we get ∞ ∞
J43 ≤
π π π− n+2 π− m+2
φ(x, y)
m=1 n=1 π− ∞ ∞
∼
π m+1
π π− n+1
⎛
p
⎛ ⎝
⎞p n l ak,l ⎠ dydx k
∞
k=[(m+1)/λ] l=1
⎞p n l ak,l ⎠ . k
∞
φm,n mp−2 n−p−2 ⎝
m=1 n=1
π−y π−x
k=[(m+1)/λ] l=1
Further, we can estimate the quantities J41 , J42 and J43 analogously as the quantities J11 , J12 and J13 , respectively. Using Lemma 1, (1.11), (4.13), (2.1), (4.9), (4.10), and (4.2) we get ∞
∞
k=m+1 l=n+1
∞ ∞ 1 ak,l sin kx sin ly = O |Δ22 ak,l | . (π − x)(π − y) k=m l=n
Whence by (1.12) ⎛ J44
∞ ∞ ⎜ = O⎝
π π π− n+2 π− m+2
m=1 n=1 π−
∼
∞ ∞
φ(x, y) π m+1
π π− n+1
φm,n (mn)p−2
m=1 n=1
∞ ∞
1 (π − x)(π − y)
p ∞ ∞ k=m l=n
p |Δ22 ak,l |
.
k=m l=n
Collecting the partial estimates we obtain (2.5). This ends the proof. 2 4.2. Proof of Theorem 3 Proof. Necessity. Assume that φ|f |p ∈ L(T2 ),
1 ≤ p < ∞.
p |Δ22 ak,l |
⎞ ⎟ dydx⎠
K. Duzinkiewicz, B. Szal / J. Math. Anal. Appl. 472 (2019) 1581–1603
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Let 1 < p < ∞ and φ−1/(p−1) ∈ L(T2 ). Using Hölder’s inequality with q =
π π
π π |f (x, y)| dx dy =
0
0
≤
0
⎧ π π ⎨ ⎩
0
p p−1 ,
we get
(φ(x, y))− p (φ(x, y)) p |f (x, y)| dxdy 1
1
0
(φ(x, y))− p−1 dxdy 1
0
⎧ ⎫ p−1 ⎬ p ⎨ π π ⎩
⎭
0
φ(x, y)|f (x, y)|p dxdy
0
⎫ p1 ⎬ ⎭
p−1
1
p p p = φ− p−1 L(T 2 ) φ|f | L(T2 ) . 1
If φ−1 ∈ L∞ (T2 ), then
π π
π π |f (x, y)| dxdy =
0
0
0
(φ(x, y))−1 φ(x, y)|f (x, y)| dxdy ≤ φ−1 L∞ (T2 ) φ|f | L(T2 ) .
0
Hence f ∈ L(T2 ) for 1 ≤ p < ∞. Therefore, we have
x y G(x, y) :=
f (u, v) dvdu = 0
∞ ∞
x y aj,k
j=1 k=1
0
sin ju sin kv dvdu = 4 0
∞ ∞ aj,k j=1 k=1
0
jk
sin2
jx ky sin2 . 2 2 (4.14)
From (4.14), we obtain
G
[λm] ∞ ∞ π π aj,k jπ kπ , sin2 sin2 ≥4 =4 2λm 2λn jk 4λm 4λn j=1 k=1
≥ 4 sin
2
π [m λ] 4λ m
j=[m/λ] k=[n/λ]
sin
2
[λn]
π [ nλ ] 4λ n
[λm]
[λn]
j=[m/λ] k=[n/λ]
aj,k jk
and using (3.1), we get π π , . am,n = O G 2λm 2λn Now, let
x y |f (u, v)| dvdu.
H(x, y) := 0
If λ ≥ 2, then
π m+2
≥
π 2λm
and
π n+2
G
≥
π 2λn
0
for m, n ∈ N. Hence
π π π π , , . ≤H 2λm 2λn m+2 n+2
aj,k jπ kπ sin2 sin2 jk 4λm 4λn
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Using this, (1.12) and (3.7), we have ∞ ∞
p−2
(mn)
m=1 n=1
⎛
π
∞ ∞ m+1 ⎜ = O⎝ m=1 n=1
⎛ ⎜ = O⎝
π π p , =O (mn) φm,n G 2λm 2λn m=1 n=1 ⎞ π p
n+1 π π ⎟ , φ(x, y)(xy)−p H dydx⎠ m+2 n+2
φm,n apm,n
π π m+1 n+1
H(x, y) xy
φ(x, y) π π m+2 n+2
p
⎞
⎛ π π ⎞ p
2 2 H(x, y) ⎟ ⎜ ⎟ dydx⎠ = O ⎝ φ(x, y) dxdy ⎠ xy 0
0
⎛ π π ⎞ ⎛ π π ⎞
2 2
⎜ ⎟ = O⎝ φ(x, y)|f (x, y)|p dxdy ⎠ = O ⎝ φ(x, y)|f (x, y)|p dxdy ⎠ . 0
p−2
π π m+2 n+2
∞ ∞
m=1 n=1
∞ ∞
0
0
2
0
Proof. Sufficiency. If {am,n }∞ m,n=1 ∈ DGM (1 α,1 β,1 γ, 2) then ∞ ∞
∞ 2 m−1 n−1 ∞ 2 r+1
|Δ22 aj,k | =
s+1
⎛ ∞ |Δ22 aj,k | = O ⎝
r=0 j=2r m s=0 k=2s n
j=m k=n
⎛
∞
= O⎝
j=[m/λ] k=[n/λ]
[λ2s n]
∞
r=0 j=[2r m/λ] s=0 k=[2s n/λ]
⎞
∞
[λ2r m]
⎞ aj,k ⎠ jk
aj,k ⎠ . jk
Using this and Theorem 2, we have
π π 0
⎛
∞ ∞
φ(x, y)|f (x, y)|p dx dy = O ⎝
⎛ φm,n (mn)p−2 ⎝
m=1 n=1
0
=O
∞ ∞
−p−2
φm,n (mn)
m=1 n=1
+O
n m
p klak,l
φm,n m
n ∞
p−2 −p−2
n
m=1 n=1
∞ ∞
⎞p ⎞ aj,k ⎠ ⎠ jk
φm,n m
−p−2 p−2
n
m=1 n=1
p ak,l
∞
j=[m/λ] k=[n/λ]
+O
k=1 l=1
∞ ∞
∞
+O
∞ ∞
p−2
φm,n (mn)
m=1 n=1
k=m l=1
∞ m
p ak,l
k=1 l=n ∞ ∞ 1 ak,l kl
p .
k=m l=n
Set λm,n = φm,n (mn)p−2 , Am = {k ∈ N : k ≤ m}, Bn = {k ∈ N : k ≤ n}. From (1.7) and (1.9), we obtain that the sequence {λm,n }∞ m,n=1 satisfies conditions (3.3) and (3.4). Therefore, by (3.5), we have ∞ ∞
p−2
φm,n (mn)
m=1 n=1
∞ ∞ 1 ak,l kl
p ≤
∞ ∞
φm,n (mn)p−2 apm,n .
m=1 n=1
k=m l=n
Using this, (4.1), (4.5) and (4.8) we get
π π |f (x, y)| dxdy = O p
0
0
Now, the proof is completed. 2
∞ ∞
p−2
(mn)
m=1 n=1
φm,n apm,n
.
K. Duzinkiewicz, B. Szal / J. Math. Anal. Appl. 472 (2019) 1581–1603
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4.3. Proof of Theorem 4 Proof. We prove that DGM (1 α,1 β,1 γ, 1) ⊂ DGM (1 α,1 β,1 γ, 2). Let {aj,k }∞ j,k=1 ∈ DGM (1 α,1 β,1 γ, 1). It is easy to see that 2m−1
|Δ20 aj,n | ≤
j=m
2m−1 j=m
2C ≤ m
2m+1
C |Δ10 aj,n | + |Δ10 aj,n | ≤ m j=m+1
[mλ]
j=[m/λ]
C |aj,n | + m+1
[(m+1)λ]
|aj,n |
j=[(m+1)/λ]
[mλ ]
|aj,n |
(4.15)
j=[m/λ ]
with λ = 2λ. Similarly as above 2n−1 k=n
2C |Δ02 am,k | ≤ n
[nλ ]
|am,k |.
(4.16)
j=[n/λ ]
Now, we have 2m−1 2n−1
|Δ22 aj,k | ≤
2m−1 2n−1
j=m k=n
+
j=m k=n
2m+1
2n+1
|Δ11 aj,k | ≤
j=m+1 k=n+1
+
C m(n + 1)
4C ≤ mn
|Δ11 aj,k | +
[mλ ]
|Δ11 aj,k | +
j=m+1 k=n
C mn
[mλ]
2m+1 2n−1
[mλ]
[nλ]
|aj,k | +
j=[m/λ] k=[(n+1)/λ]
|Δ11 aj,k |
j=m k=n+1
|aj,k | +
j=[m/λ] k=[n/λ]
[(n+1)λ]
2m−1 2n+1
C (m + 1)(n + 1)
C (m + 1)n
[(m+1)λ]
[nλ]
|aj,k |
j=[(m+1)/λ] k=[n/λ]
[(m+1)λ]
[(n+1)λ]
|aj,k |
j=[(m+1)/λ] k=[(n+1)/λ]
[nλ ]
|aj,k |.
(4.17)
j=[m/λ ] k=[n/λ ]
From (4.15), (4.16), and (4.17) we obtain that {aj,k }∞ j,k=1 ∈ DGM (1 α,1 β,1 γ, 2). Let aj,k =
2 + (−1)j 2 + (−1)k · j2 k2
for j, k ∈ N.
(4.18)
It is clear that the above sequence satisfies (1.11) and (2.1). Now, we show that {aj,k }∞ j,k=1 ∈ DGM (1 α, 1 β, 1 γ, 2). It is easy to see that Δ20 aj,k = aj,k ·
4(j + 1) , (j + 2)2
and for λ ≥ 2 2m−1 j=m
|Δ20 aj,n | ≤
2m−1 4 4 |aj,n | ≤ m j=m m
[mλ]
j=[m/λ]
|aj,n |.
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Similarly as above 4(k + 1) , (k + 2)2
Δ02 aj,k = aj,k · and 2n−1
|Δ02 am,k | ≤
k=n
4 n
[nλ]
|am,k |.
k=[n/λ]
By elementary calculations Δ22 aj,k = aj,k ·
4(j + 1) 4(k + 1) · , (j + 2)2 (k + 2)2
and 2m−1 2n−1
|Δ22 aj,k | ≤ 16
j=m k=n
2m−1 2n−1 j=m k=n
16 1 ≤ |aj,k | jk mn
[mλ]
[nλ]
|aj,k |.
j=[m/λ] k=[n/λ]
∞ Therefore {aj,k }∞ / DGM (1 α, 1 β, 1 γ, 1). j,k=1 ∈ DGM (1 α, 1 β, 1 γ, 2). Further, we show that {aj,k }j,k=1 ∈ We have 2m−1
|Δ10 aj,n | =
j=m
2m−1 (−1)n + 2 (−1)j + 2 2 − (−1)j . − n2 j2 (j + 1)2 j=m
Let Am = {j : m ≤ j ≤ 2m − 1 and j is even} for any m ∈ N such that m > λ. Then 2m−1 j=m
2 1 1 1 m−1 3 ≥ |Δ10 aj,n | ≥ 2 − ≥ 2 2, 2 2 2 2 n j (j + 1) n j 4n m j∈Am
j∈Am
and since C m
[mλ]
|aj,n | ≤
j=[m/λ]
9C mn2
[mλ]
j=[m/λ]
1 9Cλ3 ≤ , j2 (m − λ)2 n2
the inequality 2m−1 j=m
|Δ10 aj,n | ≤
C m
[mλ]
|aj,n |
j=[m/λ]
does not hold for large enough m. This ends the proof. 2 References [1] [2] [3] [4]
C.P. Chen, Weighted integrability and L1 -convergence of multiple trigonometric series, Studia Math. 108 (1994) 177–190. C.P. Chen, D.C. Luor, Two-parameter Hardy–Littlewood inequality and its variants, Studia Math. 139 (2000) 9–27. K. Duzinkiewicz, B. Szal, On the uniform convergence of double sine series, Colloq. Math. 151 (2018) 71–95. M. Dyachenko, S. Tikhonov, A Hardy–Littlewood theorem for multiple series, J. Math. Anal. Appl. 339 (2008) 503–510.
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[5] M. Dyachenko, S. Tikhonov, General monotone sequences and convergence of trigonometric series, in: Topics in Classical Analysis and Applications in Honor of Daniel Waterman, World Sci. Publ., 2008, pp. 88–101. [6] P. Kórus, F. Móricz, On the uniform convergence of double sine series, Studia Math. 193 (2009) 79–97. [7] L. Leindler, A new extension of monotone sequence and its applications, JIPAM. J. Inequal. Pure Appl. Math. 7 (1) (2006) 39. [8] L. Leindler, On the uniform convergence of single and double sine series, Acta Math. Hungar. 140 (2013) 232–242. [9] M.M.H. Marzuq, Integrability theorem of multiple trigonometric series, J. Math. Anal. Appl. 157 (1991) 337–345. [10] B. Ram, S. Bhatia, On weighted integrability of double cosine series, J. Math. Anal. Appl. 208 (1997) 510–519. [11] B. Szal, A new class of numerical sequences and its applications to uniform convergence of sine series, Math. Nachr. 284 (14–15) (2011) 1985–2002. [12] B. Szal, On L-convergence of trigonometric series, J. Math. Anal. Appl. 373 (2011) 449–463. [13] S.Yu. Tikhonov, On the integrability of the trigonometric series, Math. Notes 78 (3–4) (2005) 437–442. [14] S.Yu. Tikhonov, Trigonometric series with general monotone coefficients, J. Math. Anal. Appl. 326 (1) (2007) 721–735. [15] S.Yu. Tikhonov, On the uniform convergence of trigonometric series, Mat. Zametki 81 (2) (2007) 304–310, translation in Math. Notes 81 (1–2) (2007) 268–274. [16] S.Yu. Tikhonov, Best approximation and moduli of smoothness: computation and equivalence theorems, J. Approx. Theory 153 (2008) 19–39. [17] D. Yu, P. Zhou, S. Zhou, Mean bounded variation condition and applications in double trigonometric series, Anal. Math. 38 (2012) 83–104.
Contents lists available at ScienceDirect
Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa
On weighted integrability of double sine series Krzysztof Duzinkiewicz, Bogdan Szal ∗ University of Zielona Góra, Faculty of Mathematics, Computer Science and Econometrics, 65-516 Zielona Góra, ul. Szafrana 4a, Poland
a r t i c l e
i n f o
a b s t r a c t
Article history: Received 24 April 2018 Available online 10 December 2018 Submitted by S. Tikhonov
We introduce a new class of double sequences DGM (α, β, γ, r) called Double General Monotone and some new class of weight functions to study the weighted integrability of double sine series. Some results of D. Yu, P. Zhou and S. Zhou are also generalized. © 2018 Elsevier Inc. All rights reserved.
Keywords: Double sine series Weighted integrability Classes of double sequences Generalized monotonicity Weight functions
1. Introduction We consider the double sine series ∞ ∞
aj,k sin jx sin ky,
(1.1)
j=1 k=1
on the positive quadrant T2 = [0, π] × [0, π] of the two-dimensional torus. Denote by f (x, y) the sum of (1.1), at the points (x, y) where it converges. Let for r ∈ N, Δr0 aj,k := aj,k − aj+r,k , Δ0r aj,k := aj,k − aj,k+r and Δrr aj,k := Δr0 (Δ0r aj,k ). In the paper [5] the authors defined the following class of double sequences (see also [6]): Definition 1. ([5]) A double sequence {aj,k }∞ j,k=1 ⊂ C belongs to the class of Double General Monotone Sequences (DGM ) if there exist positive constants C and λ ≥ 2, depending only on {aj,k }∞ j,k=1 , such that: 2m−1
|Δ10 aj,n | ≤
j=m
C m
[λm]
|aj,n |, m ≥ λ, n ≥ 1,
j=[m/λ]
* Corresponding author. E-mail addresses: [email protected] (K. Duzinkiewicz), [email protected] (B. Szal). https://doi.org/10.1016/j.jmaa.2018.12.010 0022-247X/© 2018 Elsevier Inc. All rights reserved.
(1.2)
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2n−1
C n
|Δ01 am,k | ≤
k=n 2m−1 2n−1
[λn]
|Δ11 aj,k | ≤
j=m k=n
|am,k |, n ≥ λ, m ≥ 1,
(1.3)
k=[n/λ]
C mn
[λm]
[λn]
|aj,k |, m, n ≥ λ.
(1.4)
j=[m/λ] k=[n/λ]
It is clear that (see [6]) M DS DGM, where M DS is the class of all monotonically decreasing double sequences, i.e. the class of all sequences of non-negative real numbers {aj,k }∞ j,k=1 such that Δ10 aj,k ≥ 0,
Δ01 aj,k ≥ 0,
Δ11 aj,k ≥ 0,
j, k = 1, 2, . . . .
We say that bm,n ∼ bk,l (k, l, m, n ∈ N), if there are two positive constants C1 and C2 such that C1 bm,n ≤ bk,l ≤ C2 bm,n . We can define bm ∼ bk similarly. We say that a weight function φ ∈ Ψ1 if it is a positive function of two variables and satisfies the following conditions: 1. it holds that φ(x, y) ∼ φm,n := φ
π π , , m n
for all (x, y) ∈
π π π π , × , , m, n ∈ N; m+1 m n+1 n
2. if m ∼ k, n ∼ l, then φm,n ∼ φk,l ;
(1.5)
3. it holds that ∞
j −p−2 φj,n = O(m−p−1 φm,n ),
(1.6)
j p−2 φj,n = O(mp−1 φm,n ),
(1.7)
k−p−2 φm,k = O(n−p−1 φm,n ),
(1.8)
kp−2 φm,k = O(np−1 φm,n ),
(1.9)
j=m m j=1 ∞ k=n n k=1
for all m, n ∈ N.
K. Duzinkiewicz, B. Szal / J. Math. Anal. Appl. 472 (2019) 1581–1603
1583
It should be noted that the conditions (1.6)–(1.9) can be rewritten as almost monotonicity conditions, as done in the paper [13]. Let for p ∈ [1, ∞) and ζ, η ∈ p1 − 1, p1 + 1 φ(x, y) = x−ζp y −ηp
for x, y ∈ (0, π).
(1.10)
It is clear that φ ∈ Ψ1 . Let Lp T2 , 1 ≤ p ≤ ∞, (L1 T2 = L T2 ) be the space of all 2π-periodic real-valued functions of two variables, integrable in the Lebesgue sense to the p-th power when 1 ≤ p < ∞, and essentially bounded on T2 when p = ∞, with the norm ⎧⎧ ⎫1/p ⎪ ⎬ ⎨ ⎪ ⎪ p ⎪ ⎨ |f (x, y)| dxdy , if 1 ≤ p < ∞, ⎭ f p := ⎩ 2 T ⎪ ⎪ ⎪ ⎪ ess sup | f (x, y) | , if p = ∞. ⎩ (x,y)∈T2
Recently, D. Yu, P. Zhou and S. Zhou [17] proved the following theorem: Theorem 1. ([17]) Let a non-negative sequence {am,n }∞ m,n=1 ∈ DGM and am,n → 0
as
m + n → ∞.
(1.11)
Assume that φ ∈ Ψ1 , (φ)− p−1 ∈ L(T2 ) for 1 < p < ∞, or (φ)−1 ∈ L∞ (T2 ) for p = 1. Let a function f be given by (1.1). Then 1
p
φ |f | ∈ L(T2 ),
1 ≤ p < ∞,
if and only if ∞ ∞
(mn)p−2 φm,n apm,n < ∞.
m=1 n=1
The above theorem generalizes the results of B. Ram and S.S. Bhatia [10], C.P. Chen [1] and M.M.H. Marzuq [9]. Now, we consider the class DGM (α, β, γ, r) defined in [3] as follows. Definition 2. ([3]) A double sequence {aj,k }∞ j,k=1 ⊂ C belongs to the class DGM (α, β, γ, r) (called Double General Monotone), if there exist a positive constant C and an integer λ depending only on {aj,k }∞ j,k=1 , for which: 2m−1
|Δr0 aj,n | ≤ Cαm,n , m ≥ λ, n ≥ 1,
j=m 2n−1
|Δ0r am,k | ≤ Cβm,n , n ≥ λ, m ≥ 1,
k=n 2m−1 2n−1
|Δrr aj,k | ≤ Cγm,n , m, n ≥ λ
j=m k=n ∞ ∞ hold, where α := {αm,n }∞ m,n=1 , β := {βm,n }m,n=1 , γ := {γm,n }m,n=1 are non-negative double sequences and r ∈ N.
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The above class with r = 1 was introduced by L. Leindler in the paper [8], which is a two-dimensional equivalent of the classes defined in the papers: [7], [14], [15], and [16]. A similar class of sequences was also considered at the paper [4]. Using Definition 2 with r = 1 and λ ≥ 2, we have that DGM ≡ DGM (1 α,1 β,1 γ, 1), where 1 α,1 β and 1 γ are the sequences defined by the formulas on the right hand sides of the inequalities (1.2), (1.3), and (1.4), respectively. Next, we define a new class of weight functions connected with the class DGM (α, β, γ, 2), which will be used later in the paper. That is, the square [0, π] × [0, π] is divided into four squares whose side lengths are half (because r = 2) of the length of the side of the initial square. We say that a weight function φ ∈ Ψ2 if it is a positive function of two variables satisfying (1.5)–(1.9) and the following condition: φ(x, y) ∼ φm,n := φ
π π , m+1 n+1
,
(1.12)
for all π π π π , × , , (x, y) ∈ m+2 m+1 n+2 n+1 π π π π (x, y) ∈ π − ,π − × , , m+1 m+2 n+2 n+1 π π π π (x, y) ∈ , × π− ,π − , m+2 m+1 n+1 n+2 π π π π (x, y) ∈ π − ,π − × π− ,π − , m+1 m+2 n+1 n+2
and any m, n ∈ N. Let for p ∈ [1, ∞) and ζ, η ∈ p1 − 1, p1 + 1 ⎧ ⎪ x−ζp y −ηp ⎪ ⎪ ⎨ x−ζp (π − y)−ηp ϕ (x, y) = −ζp −ηp ⎪ (π − x) y ⎪ ⎪ ⎩ −ζp −ηp (π − x) (π − y)
for for for for
x, y ∈ (0, π2 ], x ∈ (0, π2 ] and y ∈ π2 , π , y ∈ (0, π2 ] and x ∈ π2 , π , x, y ∈ π2 , π .
(1.13)
It is clear that ϕ ∈ Ψ2 . Moreover, it is easy to show that the function (1.10) does not belong to Ψ2 and the function (1.13) does not belong to Ψ1 . Thus, the classes Ψ1 and Ψ2 are not comparable. In this paper we generalize Theorem 1. Namely, we weaken the condition {am,n }∞ m,n=1 ∈ DGM (1 α,1 β, ∞ 1 γ, 1) to {am,n }m,n=1 ∈ DGM (1 α,1 β,1 γ, 2), by introducing the newly defined set of sequences called Double General Monotone and assuming φ ∈ Ψ2 instead of φ ∈ Ψ1 . 2. Main results The following results hold true: ∞
Theorem 2. Let 1 ≤ p < ∞, λ ∈ N and φ ∈ Ψ2 . Suppose that a non-negative sequence {am,n }m,n=1 satisfies the conditions: (1.11), Δ20 am,n ≥ 0 or Δ20 am,n ≤ 0 for m, n ∈ N, Δ02 am,n ≥ 0 or Δ02 am,n ≤ 0 for m, n ∈ N,
(2.1)
K. Duzinkiewicz, B. Szal / J. Math. Anal. Appl. 472 (2019) 1581–1603 2m−1
|Δ20 aj,n | ≤
j=m 2n−1 k=n
C m
C |Δ02 am,k | ≤ n
1585
[λm]
|aj,n |, m ≥ λ, n ∈ N,
(2.2)
|am,k |, n ≥ λ, m ∈ N
(2.3)
j=[m/λ] [λn] k=[n/λ]
with some positive constant C and
∞ ∞
p−2
φm,n (mn)
m=1 n=1
∞ ∞
p 1/p |Δ22 ak,l |
< ∞.
(2.4)
k=m l=n p
Then the series (1.1) converges on T2 , φ |f | ∈ L(T2 ), and moreover
π π φ(x, y)|f (x, y)|p dxdy = O 0
∞ ∞
φm,n (mn)p−2
m=1 n=1
0
∞ ∞
p |Δ22 ak,l |
.
(2.5)
k=m l=n
Theorem 3. Let a non-negative sequence {am,n }∞ m,n=1 ∈ DGM (1 α,1 β,1 γ, 2) satisfy the conditions: (1.11) and (2.1). Assume that φ ∈ Ψ2 , (φ)− p−1 ∈ L(T2 ) for 1 < p < ∞ or (φ)−1 ∈ L∞ (T2 ) for p = 1. Then 1
p
φ |f | ∈ L(T2 ),
1 ≤ p < ∞,
if and only if ∞ ∞
(mn)p−2 φm,n apm,n < ∞.
m=1 n=1
Theorem 4. (i) DGM (1 α,1 β,1 γ, 1) ⊂ DGM (1 α,1 β,1 γ, 2). (ii) There exists a double sequence {aj,k }∞ j,k=1 satisfying properties (1.11) and (2.1), which belongs to the class DGM (1 α,1 β,1 γ, 2) and does not belong to the class DGM (1 α,1 β,1 γ, 1). It is easy to show that, (ϕ)− p−1 ∈ L(T2 ) for 1 < p < ∞ and (ϕ)−1 ∈ L∞ (T2 ) for p = 1. Therefore, using Theorem 3 we obtain the following corollary. 1
Corollary 1. Let a non-negative sequence {am,n }∞ m,n=1 ∈ DGM (1 α,1 β,1 γ, 2) satisfy conditions (1.11) and (2.1). If a function f is given by (1.1) and ϕ is defined by (1.13), then p
ϕ |f | ∈ L(T2 ),
1 ≤ p < ∞,
if and only if ∞ ∞ m=1 n=1
mαp+p−2 nβp+p−2 apm,n < ∞.
K. Duzinkiewicz, B. Szal / J. Math. Anal. Appl. 472 (2019) 1581–1603
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3. Auxiliary results Denote, for r ∈ Z {0} and k = 0, 1, 2, . . ., by r k,r (x) = cos(k + 2 )x D r 2 sin( 2 x)
the conjugate Dirichlet type kernel. Lemma 1. ([11], [12]) Let r ∈ N, l ∈ Z, and {ak }∞ k=1 ⊂ C. If x = m
ak sin kx = −
k=n
m
m+r
k,r (x) + Δ r ak D
k=n
2lπ r ,
then for all m ≥ n
k,−r (x) − ak D
k=m+1
n+r−1
k,−r (x), ak D
k=n
where Δr ak = ak − ak+r . Lemma 2. ([3]) Let {aj,k }∞ j,k=1 ⊂ C and m, M, n, N ∈ N such that m ≤ M and n ≤ N . 1. If x ∈ (0, π2 ], then ⎞ ⎛ M M M +2 m+1 π ⎝ aj,k sin jx ≤ |Δ20 aj,k | + |aj,k | + |aj,k |⎠ , 4x j=m j=m j=m j=M +1 and if x ∈ [ π2 , π), then ⎞ ⎛ M M M +2 m+1 π ⎝ aj,k sin jx ≤ |Δ20 aj,k | + |aj,k | + |aj,k |⎠ j=m 4(π − x) j=m j=m j=M +1 for any k ∈ N. 2. If y ∈ (0, π2 ], then N N N +2 n+1 π aj,k sin ky ≤ |Δ02 aj,k | + |aj,k | + |aj,k | , 4y k=n
k=n
k=N +1
k=n
and if y ∈ [ π2 , π), then N N N +2 n+1 π aj,k sin ky ≤ |Δ02 aj,k | + |aj,k | + |aj,k | 4(π − y) k=n
k=n
k=N +1
k=n
for any j ∈ N. Lemma 3. If {aj,k }∞ j,k=1 is a non-negative sequence belonging to the class DGM (1 α,1 β,1 γ, 2) with λ ≥ 2, then for any m, n ≥ λ ⎛ am,n = O ⎝
[λm]
[λn]
j=[m/λ] k=[n/λ]
⎞ aj,k ⎠ . jk
(3.1)
K. Duzinkiewicz, B. Szal / J. Math. Anal. Appl. 472 (2019) 1581–1603
1587
Proof. Let m, n ≥ λ. For any ν and m < μ ≤ 2m we have μ−1
Δ20 aj,ν = am,ν + am+1,ν − aμ,ν − aμ+1,ν .
j=m
By an analogous argument, we get for any μ and n < ν ≤ 2n ν−1
Δ02 aμ,k = aμ,n + aμ,n+1 − aμ,ν − aμ,ν+1 .
k=n
For μ, ν such that m < μ ≤ 2m and n < ν ≤ 2n, we have μ−1 ν−1
Δ22 aj,k =
j=m k=n
=
⎛
ν−1
⎝
Δ20 aj,k −
j=m
k=n
ν−1
μ−1
μ−1
⎞ Δ20 aj,k+2 ⎠
j=m
(am,k + am+1,k − aμ,k − aμ+1,k − am,k+2 − am+1,k+2 + aμ,k+2 + aμ+1,k+2 )
k=n
= am,n + am,n+1 − am,ν − am,ν+1 + am+1,n + am+1,n+1 − am+1,ν − am+1,ν+1 − aμ,n − aμ,n+1 + aμ,ν + aμ,ν+1 − aμ+1,n − aμ+1,n+1 + aμ+1,ν + aμ+1,ν+1 = −(am,ν + am+1,ν − aμ,ν − aμ+1,ν ) − (aμ,n + aμ,n+1 − aμ,ν − aμ,ν+1 ) − (am,ν+1 + am+1,ν+1 − aμ,ν+1 − aμ+1,ν+1 ) − (aμ+1,n + aμ+1,n+1 − aμ+1,ν − aμ+1,ν+1 ) − (aμ,ν + aμ,ν+1 + aμ+1,ν + aμ+1,ν+1 ) + (am,n + am+1,n + am,n+1 + am+1,n+1 ), whence μ−1 ν−1
μ−1
Δ22 aj,k = −
j=m k=n
ν−1
Δ20 aj,ν −
j=m
Δ02 aμ,k −
μ−1
Δ20 aj,ν+1 −
j=m
k=n
ν−1
Δ02 aμ+1,k
k=n
− (aμ,ν + aμ,ν+1 + aμ+1,ν + aμ+1,ν+1 ) + (am,n + am+1,n + am,n+1 + am+1,n+1 ). Therefore, am,n ≤ am,n + am+1,n + am,n+1 + am+1,n+1 =
μ−1 ν−1 j=m k=n
+
ν−1
Δ22 aj,k +
μ−1
Δ20 aj,ν +
j=m
μ−1
Δ20 aj,ν+1 +
j=m
Δ02 aμ,k
k=n
Δ02 aμ+1,k + aμ,ν + aμ,ν+1 + aμ+1,ν + aμ+1,ν+1 ≤
2m−1 j=m
2m−1 2n−1 j=m k=n
k=n
+
ν−1
|Δ20 aj,ν+1 | +
2n−1 k=n
|Δ02 aμ,k | +
2n−1
|Δ22 aj,k | +
2m−1
|Δ20 aj,ν |
j=m
|Δ02 aμ+1,k | + aμ,ν + aμ,ν+1 + aμ+1,ν + aμ+1,ν+1 .
k=n
(3.2) Adding up all inequalities in (3.2) for μ = m + 1, m + 2 . . . , 2m − 1 and ν = n + 1, n + 2, . . . , 2n − 1 and using the assumption {aj,k }∞ j,k=1 ∈ DGM (1 α,1 β,1 γ, 2), we obtain
K. Duzinkiewicz, B. Szal / J. Math. Anal. Appl. 472 (2019) 1581–1603
1588
2m−1
2n−1
am,n ≤
μ=m+1 ν=n+1
+
⎛
2n−1
2m−1
2n−1
2m−1
2n−1
⎛
j=[m/λ]
⎛
[λn]
⎝C
μ=m+1 ν=n+1
k=[n/λ]
⎞
[λn]
j=[m/λ] k=[n/λ]
⎞
2m−1
2n−1
⎛
aj,k ⎠ ⎝C + jk μ=m+1 ν=n+1 ⎛
[λn] 2m−1 2n−1 aj,ν+1 ⎠ ⎝C + j μ=m+1 ν=n+1
[λm]
⎝C
[λm]
⎝C
μ=m+1 ν=n+1
μ=m+1 ν=n+1
+
2m−1
k=[n/λ]
⎞
⎞
[λm]
j=[m/λ]
⎞ aj,ν ⎠ j
aμ,k ⎠ k
2m−1 2n−1 aμ+1,k ⎠ (aμ,ν + aμ,ν+1 + aμ+1,ν + aμ+1,ν+1 ). + k μ=m+1 ν=n+1
Hence, we have
[λm]
(m − 1)(n − 1)am,n ≤ (m − 1)(n − 1)C
[λn]
j=[m/λ] k=[n/λ] 2n
+ (m − 1)C
2m−1
2n−1
j=[m/λ]
[λn] 2m−1 aj,ν + (n − 1)C j μ=m+1
[λm]
ν=n+2 j=[m/λ]
+
2n−1 [λm] aj,ν aj,k + (m − 1)C jk j ν=n+1
k=[n/λ]
[λn] 2m aμ,k + (n − 1)C k μ=m+2
k=[n/λ]
aμ,k k
(aμ,ν + aμ,ν+1 + aμ+1,ν + aμ+1,ν+1 )
μ=m+1 ν=n+1
and consequently if λ ≥ 2 then
[λn] [λm] 2n 2m 2m 2n aj,k aj,k aj,k aj,k + 8C + 8C + 64 jk jk jk jk j=m k=[n/λ] j=m k=n j=[m/λ] k=[n/λ] j=[m/λ] k=n ⎛ ⎞ [λm] [λn] aj,k ⎠ ⎝ =O . jk [λm]
am,n ≤ C
[λn]
j=[m/λ] k=[n/λ]
This ends the proof. 2 Denote by Am = {j ∈ N : j ≤ m} and Am = {j ∈ N : j ≥ m} or Am = {j ∈ N : j ≥ m} and Am = {j ∈ N : j ≤ m} for m ∈ N. Analogously, let for n ∈ N Bn = {k ∈ N : k ≤ n} and Bn = {k ∈ N : k ≥ n} or Bn = {k ∈ N : k ≥ n} and Bn = {k ∈ N : k ≤ n}. Lemma 4. ([2]) Let λm,n ≥ 0 for all m, n ∈ N. Assume that there exists a constant γ ∈ [0, ∞) such that
λj,n ≤ γmλm,n ,
m, n ∈ N;
(3.3)
λm,k ≤ γnλm,n ,
m, n ∈ N.
(3.4)
j∈Am
k∈Bn
Then for 1 ≤ p < ∞, there exists a positive constant Cp,γ (= γ 2p p2p ) such that ∞ ∞ m=1 n=1
⎛ λm,n ⎝
j∈Am k∈Bn
for all non-negative sequences {am,n }∞ n,m=1 .
⎞p aj,k ⎠ ≤ Cp,γ
∞ ∞ m=1 n=1
λm,n (mnam,n )p
(3.5)
K. Duzinkiewicz, B. Szal / J. Math. Anal. Appl. 472 (2019) 1581–1603
1589
Lemma 5. Let φ ∈ Ψ2 , f be a non-negative function of two variables, 1 ≤ p < ∞, and
x y F (x, y) :=
π x, y ∈ 0, . 2
f (u, v) dvdu, 0
(3.6)
0
If φf p ∈ L(T2 ) then π
π
2 2
φ(x, y) 0
F (x, y) xy
p
⎛ π π ⎞
2 2 ⎜ ⎟ dxdy = O ⎝ φ(x, y)(f (x, y))p dxdy ⎠ .
0
0
(3.7)
0
Proof. By (1.12) and (3.6) we get π
π
2 2
φ(x, y) 0
F (x, y) xy
∼
m=1 n=1
π π m+2 n+2
φ(x, y)
F (x, y) xy
p dydx
π π m+1 n+1
φm,n
m=1 n=1
(xy)−p (F (x, y))p dydx
π π m+2 n+2
⎛
⎛ ⎜ = O⎝
n+1 ∞ ∞ m+1
dxdy =
0 ∞ ∞
π
π
p
∞ ∞
⎜ φm,n (mn)p−2 ⎝
m=1 n=1
π π m+1 n+1
0
0
⎞p ⎞ ⎟ ⎟ f (u, v) dvdu⎠ ⎠
⎛
∼
∞ ∞
⎞p π π j+1 k+1 ∞ ∞ ⎜ ⎟ φm,n (mn)p−2 ⎝ f (u, v) dvdu⎠ .
m=1 n=1
j=m k=n
π π j+2 k+2
Set λm,n = φm,n (mn)p−2 , Am = {k ∈ N : k ≤ m}, Bn = {k ∈ N : k ≤ n} and π
π
j+1 k+1 aj,k =
f (u, v) dvdu
where
m ≤ k < ∞, n ≤ j < ∞.
π π j+2 k+2
∞
From (1.7) and (1.9) we obtain that the sequence {λm,n }m,n=1 satisfies the conditions (3.3) and (3.4). Therefore, applying (3.5) we have π
π
2 2
φ(x, y) 0
F (x, y) xy
p
0
⎛ ⎞p ⎞ π π m+1
n+1 ∞ ∞ ⎜ ⎜ ⎟ ⎟ dxdy = O ⎝ φm,n (mn)p−2 ⎝mn f (u, v) dvdu⎠ ⎠ ⎛
m=1 n=1
⎛ π π ⎞p m+1
n+1 ∞ ∞ ⎜ ⎟ f (u, v) dvdu⎠ φm,n (mn)2p−2 . ∼ ⎝ m=1 n=1
π π m+2 n+2
π π m+2 n+2
K. Duzinkiewicz, B. Szal / J. Math. Anal. Appl. 472 (2019) 1581–1603
1590
If p > 0 then using Hölder’s inequality with q = p/ (p − 1), we get ⎛
π π m+1 n+1
⎜ ⎝
π π m+2 n+2
⎧⎛ ⎞ p1 ⎛ π π ⎞ q1 ⎫p π π ⎪ ⎪ m+1 n+1 m+1 n+1 ⎪ ⎪
⎨ ⎬ ⎟ ⎜ ⎟ ⎜ ⎟ p q f (u, v) dvdu⎠ ≤ ⎝ (f (u, v)) dvdu⎠ · ⎝ 1 dvdu⎠ ⎪ ⎪ ⎪ ⎪ π π ⎩ π π ⎭ ⎞p
m+2 n+2
⎛
π π m+1 n+1
⎜ (f (u, v))p dvdu · ⎝
= π π m+2 n+2
m+2 n+2
π π m+1 n+1
⎞
1 1− p p
⎟ dvdu⎠
⎛ ⎜ = O⎝
π π m+2 n+2
⎞
π π m+1 n+1
⎟ (f (u, v))p dvdu · (mn)−2p+2 ⎠ .
π π m+2 n+2
Hence by (1.12) π
π
2 2
φ(x, y) 0
F (x, y) xy
p
⎞ π π
n+1 ∞ ∞ m+1 ⎟ ⎜ dxdy = O ⎝ (f (u, v))p dvdu · (mn)−2p+2 φm,n (mn)2p−2 ⎠ ⎛
m=1 n=1
0
∼
∞ ∞
π π m+1 n+1
π
π
2 2
(f (u, v))p dvdu ∼
φm,n
m=1 n=1
π π m+2 n+2
π π m+2 n+2
φ(x, y)(f (x, y))p dxdy. 0
0
This ends the proof. 2 4. Proofs of the main results 4.1. Proof of Theorem 2 From (2.4), we see that ∞ ∞
|Δ22 ak,l | < ∞,
k=1 l=1
and using Lemma 2, we derive that the series (1.1) converges on T2 . It is clear that π
π π
π
φ(x, y)|f (x, y)|p dxdy = 0
π
2 2
0
π 2 φ(x, y)|f (x, y)|p dx dy +
0
π 2
0
π
2 π
φ(x, y)|f (x, y)|p dxdy = J1 + J2 + J3 + J4 .
φ(x, y)|f (x, y)| dxdy + 0
π 2
π 2
π 2
We have π
π
2 2
φ(x, y)|f (x, y)|p dxdy
J1 = 0
0
π π p
+
φ(x, y)|f (x, y)|p dxdy
0
⎛
π π p
n+1 n ∞ m ∞ m+1 ⎜ φ(x, y) ak,l sin kx sin ly dydx = O⎝
m=1 n=1
π π m+2 n+2
k=1 l=1
K. Duzinkiewicz, B. Szal / J. Math. Anal. Appl. 472 (2019) 1581–1603 π
+
m=1 n=1
π π m+2 n+2 π
+
π
n+1 ∞ ∞ m+1 m=1 n=1
π π m+2 n+2 π
+
π
n+1 ∞ m+1 ∞
π
n+1 ∞ ∞ m+1 m=1 n=1
π π m+2 n+2
1591
m ∞ p φ(x, y) ak,l sin kx sin ly dydx k=1 l=n+1
∞ p n φ(x, y) ak,l sin kx sin ly dydx k=m+1 l=1
∞ φ(x, y)
∞
k=m+1 l=n+1
⎞ p ⎟ ak,l sin kx sin ly dydx⎠ = J11 + J12 + J13 + J14 .
Using the inequalities | sin kx| ≤ kx, | sin ly| ≤ ly for x, y ∈ 0, π2 , k, l ∈ N, and (1.12), we have π
J11 ≤
m=1 n=1
π
n+1 ∞ ∞ m+1
p
φ(x, y)(xy)
n m
p dydx ∼
klak,l
−p−2
φm,n (mn)
m=1 n=1
k=1 l=1
π π m+2 n+2
∞ ∞
n m
p klak,l
.
k=1 l=1
Set λm,n = φm,n (mn)−p−2 , Am = {k ∈ N : k ≥ m}, Bn = {k ∈ N : k ≥ n}. From (1.6) and (1.8), we see that the sequence {λm,n }∞ m,n=1 satisfies conditions (3.3) and (3.4). Therefore, by (3.5), we get ∞ ∞
−p−2
φm,n (mn)
m=1 n=1
n m
p klak,l
=O
∞ ∞
φm,n (mn)p−2 apm,n
.
(4.1)
m=1 n=1
k=1 l=1
It is clear that ∞ ∞ k=m l=n
∞ ∞ ∞ ∞ |Δ22 ak,l | ≥ Δ22 ak,l = (ak,l − ak+2,l − ak,l+2 + ak+2,l+2 ) k=m l=n
l=n
k=m
∞ ∞ ∞ ∞ ∞ ∞ = Δ02 ak,l − Δ02 ak+2,l = Δ02 ak,l − Δ02 ak,l l=n
k=m
k=m
l=n
k=m
k=m+2
m+1 ∞ ∞ = Δ02 ak,l = (am,l − am,l+2 + am+1,l − am+1,l+2 ) l=n
k=m
l=n
∞ ∞ ∞ ∞ n+1 n+1 = am,l − am,l + am+1,l − am+1,l = am,l + am+1,l l=n
l=n+2
l=n
l=n+2
l=n
l=n
= am,n + am,n+1 + am+1,n + am+1,n+1 .
(4.2)
Hence
J11 ≤
∞ ∞ m=1 n=1
p−2
φm,n (mn)
∞ ∞ k=m l=n
p |Δ22 ak,l |
.
K. Duzinkiewicz, B. Szal / J. Math. Anal. Appl. 472 (2019) 1581–1603
1592
Applying Lemma 2 for y ∈ 0, π2 and the inequality | sin kx| ≤ kx for x ∈ 0, π2 , k ∈ N, we have ∞ m m ∞ π ak,l sin kx sin ly ≤ | sin kx| |Δ02 ak,l | + ak,n+1 + ak,n+2 4y k=1 l=n+1
=O
k=1
l=n+1
m m ∞ x x k|Δ02 ak,l | + k(ak,n+1 + ak,n+2 ) . y y k=1 l=n+1
k=1
It is clear that for k ∈ N ∞ l=n+1
∞ n+2 ∞ |Δ02 ak,l | ≥ ak,l − ak,l = ak,l = ak,n+1 + ak,n+2 , l=n+1
l=n+3
(4.3)
l=n+1
and by (2.3) r+1
∞
|Δ02 ak,l | =
∞ 2
(n+1)−1
⎛ ∞ |Δ02 ak,l | = O ⎝
r=0 l=2r (n+1)
l=n+1
[λ2r (n+1)]
r=0 l=[2r (n+1)/λ]
⎞
⎛
ak,l ⎠ = O⎝ l
⎞
∞
l=[(n+1)/λ]
ak,l ⎠ . l (4.4)
Therefore, using (1.12) we obtain π
J12 ≤
m=1 n=1
∼
π
n+1 ∞ ∞ m+1
∞ ∞
π π m+2 n+2
⎛ p m x ⎝ φ(x, y) y
k=1 l=[(n+1)/λ]
⎛
φm,n m−p−2 np−2 ⎝
m=1 n=1
∼
∞ ∞
=O
m
∞
k=1 l=[(n+1)/λ]
⎛ φm,n m−p−2 np−2 ⎝
m=1 n=1
m
n
k=1 l=[(n+1)/λ]
∞ ∞
⎞p k ak,l ⎠ dydx l
∞
−p−2
φm,n (mn)
m=1 n=1
n m
⎞p k ak,l ⎠ l ⎞p m ∞ k k ak,l + ak,l ⎠ l l k=1 l=n
p
klak,l
+O
∞ ∞
φm,n m
−p−2 p−2
n
m=1 n=1
k=1 l=1
∞ m k
k=1 l=n
l
p ak,l
.
Now, let λm,n = φm,n m−p−2 np−2 , Am = {k ∈ N : k ≥ m}, Bn = {k ∈ N : k ≤ n}. From (1.6) and (1.9), we have that the sequence {λm,n }∞ m,n=1 satisfies the conditions (3.3) and (3.4). Therefore, by (3.5) we have ∞ ∞
φm,n m
−p−2 p−2
n
m=1 n=1
∞ m k k=1 l=n
l
p ak,l
=O
∞ ∞
φm,n (mn)p−2 apm,n
m=1 n=1
Using this, (4.1) and (4.2) we get J12 = O
∞ ∞
m=1 n=1
p−2
φm,n (mn)
∞ ∞
k=m l=n
p |Δ22 ak,l |
.
.
(4.5)
K. Duzinkiewicz, B. Szal / J. Math. Anal. Appl. 472 (2019) 1581–1603
1593
Applying Lemma 2 for x ∈ 0, π2 and the inequality | sin ly| ≤ ly for y ∈ 0, π2 , l ∈ N, we have ∞ ∞ n n π ak,l sin kx sin ly ≤ | sin ly| |Δ20 ak,l | + am+1,l + am+2,l 4x k=m+1 l=1
=O
l=1
k=m+1
∞ n n y y l|Δ20 ak,l | + l(am+1,l + am+2,l ) . x x k=m+1 l=1
l=1
It is clear that for l ∈ N ∞ k=m+1
∞ m+2 ∞ |Δ20 ak,l | ≥ ak,l − ak,l = ak,l = am+1,l + am+2,l , k=m+1
k=m+3
(4.6)
k=m+1
and applying (2.2) r+1
∞
|Δ20 ak,l | =
∞ 2
⎛
(m+1)−1
|Δ20 ak,l | = O ⎝
r=0 k=2r (m+1)
k=m+1
[λ2r (m+1)]
∞
r=0 k=[2r (m+1)/λ]
⎞
⎛
⎞
∞
ak,l ⎠ = O⎝ k
k=[(m+1)/λ]
ak,l ⎠ . k (4.7)
Therefore, by (1.12) π
J13 ≤
m=1 n=1
∼
π
n+1 ∞ ∞ m+1
∞ ∞
φ(x, y)
π π m+2 n+2
y p x
∼
∞ ∞
=O
⎝
⎞p n l ak,l ⎠ dydx k
⎞p n l ak,l ⎠ k
∞
φm,n mp−2 n−p−2 ⎝
k=[(m+1)/λ] l=1
⎛
⎞p n ∞ n l l ak,l + ak,l ⎠ k k
m
φm,n mp−2 n−p−2 ⎝
m=1 n=1
∞
k=[(m+1)/λ] l=1
⎛
m=1 n=1
⎛
k=[(m+1)/λ] l=1
∞ ∞
−p−2
φm,n (mn)
m=1 n=1
n m
k=m l=1
p
lkak,l
+O
∞ ∞
φm,n m
p−2 −p−2
n
m=1 n=1
k=1 l=1
n ∞ l ak,l k
p .
k=m l=1
Set λm,n = φm,n mp−2 n−p−2 , Am = {k ∈ N : k ≤ m}, Bn = {k ∈ N : k ≥ n}. From (1.7) and (1.8), we see that the sequence {λm,n }∞ m,n=1 satisfies the conditions (3.3) and (3.4). Therefore, by (3.5) we have ∞ ∞
φm,n m
p−2 −p−2
n
m=1 n=1
n ∞ l ak,l k
p
=O
∞ ∞
φm,n (mn)p−2 apm,n
m=1 n=1
k=m l=1
Whence by (4.1) and (4.2) we get J13 = O
∞ ∞
m=1 n=1
p−2
φm,n (mn)
∞ ∞
k=m l=n
p |Δ22 ak,l |
.
.
(4.8)
K. Duzinkiewicz, B. Szal / J. Math. Anal. Appl. 472 (2019) 1581–1603
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By elementary calculations we have ∞
∞
∞
|Δ22 ak,l | ≥
k=m+1 l=n+1
l=n+1
∞ ∞ ∞ m+2 Δ02 ak,l − Δ02 ak,l = Δ02 ak,l , k=m+1
k=m+3
(4.9)
l=n+1 k=m+1
and ∞
∞
k=m+1 l=n+1
∞ n+2 ∞ ∞ ∞ |Δ22 ak,l | ≥ Δ20 ak,l − Δ20 ak,l = Δ20 ak,l . k=m+1 l=n+1
l=n+3
(4.10)
k=m+1 l=n+1
Using Lemma 1, (1.11), the estimates π , 4x l,±2 (y)| ≤ π |D 4y k,±2 (x)| ≤ |D
(4.11) (4.12)
for x, y ∈ 0, π2 , k, l ∈ N, (2.1), (4.9), (4.10), and (4.2) we get ∞
∞
k=m+1 l=n+1
∞
k,2 (x)| · |D l,2 (y)| |Δ22 ak,l | · |D
k=m+1 l=n+1 ∞
m+2
+
∞
ak,l sin kx sin ly ≤
k,−2 (x)| · |D l,2 (y)| + |Δ02 ak,l | · |D
k=m+1 l=n+1 m+2
+
n+2
∞
n+2
k,2 (x)| · |D l,−2 (y)| |Δ20 ak,l | · |D
k=m+1 l=n+1
k,−2 (x)| · |D l,−2 (y)| |ak,l | · |D
k=m+1 l=n+1
∞ ∞ m+2 ∞ m+2 ∞ n+2 n+2 π2 |Δ22 ak,l | + Δ02 ak,l + Δ20 ak,l + ak,l ≤ 16xy k=m+1 l=n+1 l=n+1 k=m+1 k=m+1 l=n+1 k=m+1 l=n+1 ∞ ∞ π2 ≤ |Δ22 ak,l | . 4xy k=m l=n
Hence by (1.12) we get ⎛
J14
⎞ π π p p
n+1 ∞ ∞ ∞ ∞ m+1 1 ⎜ ⎟ = O⎝ φ(x, y) |Δ22 ak,l | dydx⎠ xy m=1 n=1 π π m+2 n+2
∼
∞ ∞
φm,n (mn)p−2
m=1 n=1
k=m l=n
∞ ∞
p |Δ22 ak,l |
.
k=m l=n
Further we have π
π 2
φ(x, y)|f (x, y)|p dydx
J2 = π 2
0
⎛
∞ ∞ ⎜ = O⎝
π π− m+2
m=1 n=1 π−
π
n+1
π π m+1 n+2
p n m φ(x, y) ak,l sin kx sin ly dydx k=1 l=1
K. Duzinkiewicz, B. Szal / J. Math. Anal. Appl. 472 (2019) 1581–1603
+
∞ ∞
π π− m+2
m=1 n=1 π−
+
∞ ∞
+
∞ ∞
π π m+1 n+2
π π− m+2
m=1 n=1 π−
π
n+1
π π m+1 n+2
π π− m+2
m=1 n=1 π−
π
n+1
π
n+1
π π m+1 n+2
1595
m ∞ p φ(x, y) ak,l sin kx sin ly dydx k=1 l=n+1
p ∞ n φ(x, y) ak,l sin kx sin ly dydx k=m+1 l=1
∞ φ(x, y)
∞
k=m+1 l=n+1
⎞ p ⎟ ak,l sin kx sin ly dydx⎠ = J21 + J22 + J23 + J24 .
Using the inequalities | sin kx| ≤ k (π − x) and | sin ly| ≤ ly for k, l ∈ N, x ∈
π 2,π
, y ∈ 0, π2 we get
m n n m ak,l sin kx sin ly ≤ (π − x)y klak,l . k=1 l=1
k=1 l=1
Whence by (1.12) π π− m+2
∞ ∞
J21 ≤
φ(x, y)((π − x)y)
m=1 n=1 π−
=O
π
n+1
∞ ∞
p
n m
p klak,l
dydx
k=1 l=1
π π m+1 n+2
−p−2
φm,n (mn)
m=1 n=1
n m
p klak,l
.
k=1 l=1
Applying Lemma 2 for y ∈ 0, π2 , the inequality | sin kx| ≤ k (π − x) for x ∈ have
π 2,π
, k ∈ N and (4.3) we
m ∞ ∞ m π ak,l sin kx sin ly ≤ (π − x) k |Δ02 ak,l | + ak,n+1 + ak,n+2 4y k=1 l=n+1
=O
k=1
l=n+1
∞ m π−x k|Δ02 ak,l | . y k=1 l=n+1
Next by (1.12) and (4.4)
J22 ≤
∞ ∞
π π− m+2
m=1 n=1 π−
⎛ = O⎝
π
n+1 φ(x, y)
π π m+1 n+2
∞ ∞
m=1 n=1
π−x y ⎛
φm,n m−p−2 np−2 ⎝
m
p
⎛ ⎝
m
∞
k=1 l=[(n+1)/λ] ∞
k=1 l=[(n+1)λ]
⎞p k ak,l ⎠ dydx l
⎞p ⎞ k ak,l ⎠ ⎠ . l
K. Duzinkiewicz, B. Szal / J. Math. Anal. Appl. 472 (2019) 1581–1603
1596
Using Lemma 2 for x ∈ [ π2 , π), the inequality | sin ly| ≤ ly for y ∈ 0, π2 , l ∈ N, and (4.6) we have ∞ n ∞ n π ak,l sin kx sin ly ≤ y l |Δ20 ak,l | + am+1,l + am+2,l 4(π − x) k=m+1 l=1
=O
l=1
k=m+1
∞ n y l|Δ20 ak,l | π−x k=m+1 l=1
and by (1.12), (4.7) we obtain
J23 ≤
∞ ∞
π π− m+2
∼
φ(x, y)
m=1 n=1 π− ∞ ∞
π
n+1
π π m+1 n+2
y π−x
⎛
∞
⎝
⎞p n l ak,l ⎠ dydx k
k=[(m+1)/λ] l=1
⎞p n l ak,l ⎠ . k
∞
φm,n mp−2 n−p−2 ⎝
m=1 n=1
⎛
p
k=[(m+1)/λ] l=1
Further, we can estimate the quantities J21 , J22 , and J23 similarly as the quantities J11 , J12 , and J13 , respectively. Using Lemma 1, (1.11), the estimate k,±2 (x)| ≤ |D
π π for x ∈ [ , π), k ∈ N, 4 (π − x) 2
(4.13)
(4.12), (2.1), (4.9), (4.10), and (4.2) we get ∞
∞
k=m+1 l=n+1
∞ ∞ 1 ak,l sin kx sin ly = O |Δ22 ak,l | , (π − x)y k=m l=n
and by (1.12) ⎛ J24
∞ ∞ ⎜ = O⎝
π π− m+2
∼
φ(x, y)
m=1 n=1 π− ∞ ∞
π
n+1
π π m+1 n+2
φm,n (mn)p−2
m=1 n=1
∞ ∞
1 (π − x)y
p ∞ ∞
p |Δ22 ak,l |
k=m l=n
p |Δ22 ak,l |
.
k=m l=n
Therefore, J2 = O
∞ ∞
p−2
φm,n (mn)
m=1 n=1
∞ ∞
p |Δ22 ak,l |
.
k=m l=n
The quantity J3 we can estimate in a symmetric way as J2 . Hence, we have J3 = O
∞ ∞
m=1 n=1
p−2
φm,n (mn)
∞ ∞
k=m l=n
p |Δ22 ak,l |
.
⎞ ⎟ dydx⎠
K. Duzinkiewicz, B. Szal / J. Math. Anal. Appl. 472 (2019) 1581–1603
1597
Finally, we have
π π φ(x, y)(f (x, y))p dxdy
J4 = π 2
π 2
⎛
π π π− n+2 π− m+2
∞ ∞ ⎜ = O⎝
m=1 n=1 π−
+
∞ ∞
+
+
π π− n+1
π m+1
π m+1
m=1 n=1 π−
π m+1
π π− n+1
k=1 l=1
m ∞ p φ(x, y) ak,l sin kx sin ly dydx ∞ p n φ(x, y) ak,l sin kx sin ly dydx k=m+1 l=1
π π− n+1
π π π− n+2 π− m+2
p m n φ(x, y) ak,l sin kx sin ly dydx
k=1 l=n+1
π π− n+1
π π π− n+2 π− m+2
m=1 n=1 π− ∞ ∞
π m+1
π π π− n+2 π− m+2
m=1 n=1 π− ∞ ∞
∞ φ(x, y)
∞
k=m+1 l=n+1
⎞ p ⎟ ak,l sin kx sin ly dydx⎠ = J41 + J42 + J43 + J44 .
Using the inequalities | sin kx| ≤ k(π − x) and | sin ly| ≤ l(π − y) for k, l ∈ N, x, y ∈
π 2,π
we have
m n n m ≤ (π − x)(π − y) a sin kx sin ly klak,l . k,l k=1 l=1
k=1 l=1
From (1.12) we get
J41 ≤
∞ ∞
π π π− n+2 π− m+2
φ(x, y)((π − x)(π − y))
p
m=1 n=1π− π π− π m+1 n+1
∼
∞ ∞
n m
p klak,l
dydx
k=1 l=1
−p−2
φm,n (mn)
m=1 n=1
m n
p klak,l
.
k=1 l=1
Applying Lemma 2 for y ∈ [ π2 , π), the inequality | sin kx| ≤ k (π − x) for x ∈ [ π2 , π), k ∈ N and (4.3) we have ∞ m m ∞ π ak,l sin kx sin ly ≤ (π − x) k |Δ02 ak,l | + ak,n+1 + ak,n+2 4(π − y) k=1 l=n+1
=O
k=1
∞ m π−x k|Δ02 ak,l | , π−y k=1 l=n+1
and by (1.12), (4.4) we obtain
l=n+1
K. Duzinkiewicz, B. Szal / J. Math. Anal. Appl. 472 (2019) 1581–1603
1598
π π π− n+2 π− m+2
∞ ∞
J42 ≤
φ(x, y)
m=1 n=1 π− ∞ ∞
∼
π m+1
π π− n+1
⎛
φm,n m−p−2 np−2 ⎝
m=1 n=1
π−x π−y
⎛
p
⎝
k=1 l=[(n+1)/λ]
∞
m
⎞p
∞
m
k=1 l=[(n+1)/λ]
k ak,l ⎠ dydx l
⎞p k ak,l ⎠ . l
Using Lemma 2 for x ∈ [ π2 , π), the inequality | sin ly| ≤ l (π − y) for y ∈ [ π2 , π), l ∈ N and (4.6) we have ∞ ∞ n n π ak,l sin kx sin ly ≤ (π − y) l |Δ20 ak,l | + am+1,l + am+2,l 4(π − x) k=m+1 l=1 l=1 k=m+1 n ∞ π−y l|Δ20 ak,l | . =O π−x k=m+1 l=1
From (1.12) and (4.7) we get ∞ ∞
J43 ≤
π π π− n+2 π− m+2
φ(x, y)
m=1 n=1 π− ∞ ∞
∼
π m+1
π π− n+1
⎛
p
⎛ ⎝
⎞p n l ak,l ⎠ dydx k
∞
k=[(m+1)/λ] l=1
⎞p n l ak,l ⎠ . k
∞
φm,n mp−2 n−p−2 ⎝
m=1 n=1
π−y π−x
k=[(m+1)/λ] l=1
Further, we can estimate the quantities J41 , J42 and J43 analogously as the quantities J11 , J12 and J13 , respectively. Using Lemma 1, (1.11), (4.13), (2.1), (4.9), (4.10), and (4.2) we get ∞
∞
k=m+1 l=n+1
∞ ∞ 1 ak,l sin kx sin ly = O |Δ22 ak,l | . (π − x)(π − y) k=m l=n
Whence by (1.12) ⎛ J44
∞ ∞ ⎜ = O⎝
π π π− n+2 π− m+2
m=1 n=1 π−
∼
∞ ∞
φ(x, y) π m+1
π π− n+1
φm,n (mn)p−2
m=1 n=1
∞ ∞
1 (π − x)(π − y)
p ∞ ∞ k=m l=n
p |Δ22 ak,l |
.
k=m l=n
Collecting the partial estimates we obtain (2.5). This ends the proof. 2 4.2. Proof of Theorem 3 Proof. Necessity. Assume that φ|f |p ∈ L(T2 ),
1 ≤ p < ∞.
p |Δ22 ak,l |
⎞ ⎟ dydx⎠
K. Duzinkiewicz, B. Szal / J. Math. Anal. Appl. 472 (2019) 1581–1603
1599
Let 1 < p < ∞ and φ−1/(p−1) ∈ L(T2 ). Using Hölder’s inequality with q =
π π
π π |f (x, y)| dx dy =
0
0
≤
0
⎧ π π ⎨ ⎩
0
p p−1 ,
we get
(φ(x, y))− p (φ(x, y)) p |f (x, y)| dxdy 1
1
0
(φ(x, y))− p−1 dxdy 1
0
⎧ ⎫ p−1 ⎬ p ⎨ π π ⎩
⎭
0
φ(x, y)|f (x, y)|p dxdy
0
⎫ p1 ⎬ ⎭
p−1
1
p p p = φ− p−1 L(T 2 ) φ|f | L(T2 ) . 1
If φ−1 ∈ L∞ (T2 ), then
π π
π π |f (x, y)| dxdy =
0
0
0
(φ(x, y))−1 φ(x, y)|f (x, y)| dxdy ≤ φ−1 L∞ (T2 ) φ|f | L(T2 ) .
0
Hence f ∈ L(T2 ) for 1 ≤ p < ∞. Therefore, we have
x y G(x, y) :=
f (u, v) dvdu = 0
∞ ∞
x y aj,k
j=1 k=1
0
sin ju sin kv dvdu = 4 0
∞ ∞ aj,k j=1 k=1
0
jk
sin2
jx ky sin2 . 2 2 (4.14)
From (4.14), we obtain
G
[λm] ∞ ∞ π π aj,k jπ kπ , sin2 sin2 ≥4 =4 2λm 2λn jk 4λm 4λn j=1 k=1
≥ 4 sin
2
π [m λ] 4λ m
j=[m/λ] k=[n/λ]
sin
2
[λn]
π [ nλ ] 4λ n
[λm]
[λn]
j=[m/λ] k=[n/λ]
aj,k jk
and using (3.1), we get π π , . am,n = O G 2λm 2λn Now, let
x y |f (u, v)| dvdu.
H(x, y) := 0
If λ ≥ 2, then
π m+2
≥
π 2λm
and
π n+2
G
≥
π 2λn
0
for m, n ∈ N. Hence
π π π π , , . ≤H 2λm 2λn m+2 n+2
aj,k jπ kπ sin2 sin2 jk 4λm 4λn
K. Duzinkiewicz, B. Szal / J. Math. Anal. Appl. 472 (2019) 1581–1603
1600
Using this, (1.12) and (3.7), we have ∞ ∞
p−2
(mn)
m=1 n=1
⎛
π
∞ ∞ m+1 ⎜ = O⎝ m=1 n=1
⎛ ⎜ = O⎝
π π p , =O (mn) φm,n G 2λm 2λn m=1 n=1 ⎞ π p
n+1 π π ⎟ , φ(x, y)(xy)−p H dydx⎠ m+2 n+2
φm,n apm,n
π π m+1 n+1
H(x, y) xy
φ(x, y) π π m+2 n+2
p
⎞
⎛ π π ⎞ p
2 2 H(x, y) ⎟ ⎜ ⎟ dydx⎠ = O ⎝ φ(x, y) dxdy ⎠ xy 0
0
⎛ π π ⎞ ⎛ π π ⎞
2 2
⎜ ⎟ = O⎝ φ(x, y)|f (x, y)|p dxdy ⎠ = O ⎝ φ(x, y)|f (x, y)|p dxdy ⎠ . 0
p−2
π π m+2 n+2
∞ ∞
m=1 n=1
∞ ∞
0
0
2
0
Proof. Sufficiency. If {am,n }∞ m,n=1 ∈ DGM (1 α,1 β,1 γ, 2) then ∞ ∞
∞ 2 m−1 n−1 ∞ 2 r+1
|Δ22 aj,k | =
s+1
⎛ ∞ |Δ22 aj,k | = O ⎝
r=0 j=2r m s=0 k=2s n
j=m k=n
⎛
∞
= O⎝
j=[m/λ] k=[n/λ]
[λ2s n]
∞
r=0 j=[2r m/λ] s=0 k=[2s n/λ]
⎞
∞
[λ2r m]
⎞ aj,k ⎠ jk
aj,k ⎠ . jk
Using this and Theorem 2, we have
π π 0
⎛
∞ ∞
φ(x, y)|f (x, y)|p dx dy = O ⎝
⎛ φm,n (mn)p−2 ⎝
m=1 n=1
0
=O
∞ ∞
−p−2
φm,n (mn)
m=1 n=1
+O
n m
p klak,l
φm,n m
n ∞
p−2 −p−2
n
m=1 n=1
∞ ∞
⎞p ⎞ aj,k ⎠ ⎠ jk
φm,n m
−p−2 p−2
n
m=1 n=1
p ak,l
∞
j=[m/λ] k=[n/λ]
+O
k=1 l=1
∞ ∞
∞
+O
∞ ∞
p−2
φm,n (mn)
m=1 n=1
k=m l=1
∞ m
p ak,l
k=1 l=n ∞ ∞ 1 ak,l kl
p .
k=m l=n
Set λm,n = φm,n (mn)p−2 , Am = {k ∈ N : k ≤ m}, Bn = {k ∈ N : k ≤ n}. From (1.7) and (1.9), we obtain that the sequence {λm,n }∞ m,n=1 satisfies conditions (3.3) and (3.4). Therefore, by (3.5), we have ∞ ∞
p−2
φm,n (mn)
m=1 n=1
∞ ∞ 1 ak,l kl
p ≤
∞ ∞
φm,n (mn)p−2 apm,n .
m=1 n=1
k=m l=n
Using this, (4.1), (4.5) and (4.8) we get
π π |f (x, y)| dxdy = O p
0
0
Now, the proof is completed. 2
∞ ∞
p−2
(mn)
m=1 n=1
φm,n apm,n
.
K. Duzinkiewicz, B. Szal / J. Math. Anal. Appl. 472 (2019) 1581–1603
1601
4.3. Proof of Theorem 4 Proof. We prove that DGM (1 α,1 β,1 γ, 1) ⊂ DGM (1 α,1 β,1 γ, 2). Let {aj,k }∞ j,k=1 ∈ DGM (1 α,1 β,1 γ, 1). It is easy to see that 2m−1
|Δ20 aj,n | ≤
j=m
2m−1 j=m
2C ≤ m
2m+1
C |Δ10 aj,n | + |Δ10 aj,n | ≤ m j=m+1
[mλ]
j=[m/λ]
C |aj,n | + m+1
[(m+1)λ]
|aj,n |
j=[(m+1)/λ]
[mλ ]
|aj,n |
(4.15)
j=[m/λ ]
with λ = 2λ. Similarly as above 2n−1 k=n
2C |Δ02 am,k | ≤ n
[nλ ]
|am,k |.
(4.16)
j=[n/λ ]
Now, we have 2m−1 2n−1
|Δ22 aj,k | ≤
2m−1 2n−1
j=m k=n
+
j=m k=n
2m+1
2n+1
|Δ11 aj,k | ≤
j=m+1 k=n+1
+
C m(n + 1)
4C ≤ mn
|Δ11 aj,k | +
[mλ ]
|Δ11 aj,k | +
j=m+1 k=n
C mn
[mλ]
2m+1 2n−1
[mλ]
[nλ]
|aj,k | +
j=[m/λ] k=[(n+1)/λ]
|Δ11 aj,k |
j=m k=n+1
|aj,k | +
j=[m/λ] k=[n/λ]
[(n+1)λ]
2m−1 2n+1
C (m + 1)(n + 1)
C (m + 1)n
[(m+1)λ]
[nλ]
|aj,k |
j=[(m+1)/λ] k=[n/λ]
[(m+1)λ]
[(n+1)λ]
|aj,k |
j=[(m+1)/λ] k=[(n+1)/λ]
[nλ ]
|aj,k |.
(4.17)
j=[m/λ ] k=[n/λ ]
From (4.15), (4.16), and (4.17) we obtain that {aj,k }∞ j,k=1 ∈ DGM (1 α,1 β,1 γ, 2). Let aj,k =
2 + (−1)j 2 + (−1)k · j2 k2
for j, k ∈ N.
(4.18)
It is clear that the above sequence satisfies (1.11) and (2.1). Now, we show that {aj,k }∞ j,k=1 ∈ DGM (1 α, 1 β, 1 γ, 2). It is easy to see that Δ20 aj,k = aj,k ·
4(j + 1) , (j + 2)2
and for λ ≥ 2 2m−1 j=m
|Δ20 aj,n | ≤
2m−1 4 4 |aj,n | ≤ m j=m m
[mλ]
j=[m/λ]
|aj,n |.
K. Duzinkiewicz, B. Szal / J. Math. Anal. Appl. 472 (2019) 1581–1603
1602
Similarly as above 4(k + 1) , (k + 2)2
Δ02 aj,k = aj,k · and 2n−1
|Δ02 am,k | ≤
k=n
4 n
[nλ]
|am,k |.
k=[n/λ]
By elementary calculations Δ22 aj,k = aj,k ·
4(j + 1) 4(k + 1) · , (j + 2)2 (k + 2)2
and 2m−1 2n−1
|Δ22 aj,k | ≤ 16
j=m k=n
2m−1 2n−1 j=m k=n
16 1 ≤ |aj,k | jk mn
[mλ]
[nλ]
|aj,k |.
j=[m/λ] k=[n/λ]
∞ Therefore {aj,k }∞ / DGM (1 α, 1 β, 1 γ, 1). j,k=1 ∈ DGM (1 α, 1 β, 1 γ, 2). Further, we show that {aj,k }j,k=1 ∈ We have 2m−1
|Δ10 aj,n | =
j=m
2m−1 (−1)n + 2 (−1)j + 2 2 − (−1)j . − n2 j2 (j + 1)2 j=m
Let Am = {j : m ≤ j ≤ 2m − 1 and j is even} for any m ∈ N such that m > λ. Then 2m−1 j=m
2 1 1 1 m−1 3 ≥ |Δ10 aj,n | ≥ 2 − ≥ 2 2, 2 2 2 2 n j (j + 1) n j 4n m j∈Am
j∈Am
and since C m
[mλ]
|aj,n | ≤
j=[m/λ]
9C mn2
[mλ]
j=[m/λ]
1 9Cλ3 ≤ , j2 (m − λ)2 n2
the inequality 2m−1 j=m
|Δ10 aj,n | ≤
C m
[mλ]
|aj,n |
j=[m/λ]
does not hold for large enough m. This ends the proof. 2 References [1] [2] [3] [4]
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1603
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