Journal of Fourier Analysis and Applications (2023) 29:60 https://doi.org/10.1007/s00041-023-10039-x Two-Dimensional Hardy–Littlewood Theorem for Functions with General Monotone Fourier Coefficients Kristina Oganesyan 1,2 Received: 4 November 2022 / Revised: 19 March 2023 / Accepted: 24 July 2023 / Published online: 19 September 2023 © The Author(s) 2023 Abstract We prove the Hardy–Littlewood theorem in two dimensions for functions whose Fourier coefficients obey general monotonicity conditions and, importantly, are not necessarily positive. The sharpness of the result is given by a counterexample, which shows that if one slightly extends the considered class of coefficients, the Hardy– Littlewood relation fails. Keywords Fourier series · General monotone coefficients · Hardy–Littlewood theorem Mathematics Subject Classification 42B05 · 42B35 1 Introduction Establishing interconnections between integrability of functions and summability of their Fourier coefficients is the problem which occupies a special place in harmonic analysis. The celebrated Parseval’s identity enables us to reduce a wide class of prob- lems concerning functions to those concerning their Fourier series, and vice versa. Although we do not have such equalities for the spaces L p , p = 2, we can still obtain equivalences of norms of functions and norms of their Fourier series if we Communicated by Ferenc Weisz. The work was supported by the Moebius Contest Foundation for Young Scientists and the Foundation for the Advancement of Theoretical Physics and Mathematics “BASIS” (Grant no. 19-8-2-28-1). B Kristina Oganesyan oganchris@gmail.com 1 Centre de Recerca Matemàtica, Universitat Autònoma de Barcelona, Barcelona, Spain 2 Moscow Center for Fundamental and Applied Mathematics, Lomonosov Moscow State University, Moscow, Russia 60 Page 2 of 30 Journal of Fourier Analysis and Applications (2023) 29:60 impose some additional requirements. Results of this kind are important, in the first place, due to the fact that once such a relation is found, one becomes free to choose if it is handy to deal with functions or with coefficients in this or that case, as if having Parseval’s identity (see, e.g. [4, Chs. 4–6, 12–13] and [15, Sect. 7] for applications). The following result by Paley [26] can be considered the starting point for the research in this direction. Theorem A (Paley, 1931) Let {φn (x)} be an orthonormal system on [a, b] with |φn (x)| ≤ M for all x ∈ [a, b] and n ∈ N. Then (a) If p ∈ (1, 2], then for any f ∈ L p (a, b) with Fourier coefficients {cn } there holds ∞  p |cn | p n p−2  p,M  f  p . (1) n=1  (b) If p ∈ [2, ∞), then, for any sequence {cn } with ∞ n=1 |cn | n p p−2 < ∞, there exists a function f ∈ L p (a, b) that has {cn } as its Fourier coefficients and ∞  p |cn | p n p−2  p,M  f  p . (2) n=1 Throughout the paper, for two functions f and g, the relation f  g (or g  f ) means that there exists a constant C such that f (x) ≥ Cg(x) for all x, and the relation f  g is equivalent to f  g  f (if we write f a g, this means that the corresponding constant is allowed to depend on a). From now on, we discuss Fourier series only with respect to the trigonometric system. The ranges of p in Theorem A are sharp, therefore to have both (1) and (2) true for all p ∈ (1, ∞), one has to impose some additional requirements. Hardy and Littlewood [17] showed that if we restrict ourselves to sine or cosine series with monotone tending to zero coefficients, then both relations (1) and (2) hold for all p ∈ (1, ∞). In this regard, a natural question to ask was: how much can we release the requirement of monotonicity to have ∞ p |cn | p n p−2  p  f  p (3) n=1 still true? This question in turn motivated creation of various extentions of the class of monotone sequences satisfying (3). One of these classes, the so-called general monotone or just G M class [28, Th. 4.2], consists of all sequences {an } fulfilling the condition 2n |ak − ak+1 |  |an | (4) k=n for all n. Thus, now we dropped not only the monotonicity condition but even the basic requirement of positivity, keeping though some regularity of our sequences. One can see that G M class can yet be generalized (see [29, Th. 6.2(B)] and [32, Th. 1]) by putting a mean value on the right-hand side of (4) instead of |an | as follows: Journal of Fourier Analysis and Applications (2023) 29:60 Page 3 of 30 60  2n λn  |ak | |ak − ak+1 |  (5) n k k=n k= λ for some λ > 1 (see also [14] for some properties of such sequences). Note that these classes and several other ones, defined as (4) but with some other majorants on the right-hand side, in different sources can be also called G M. For a comprehensive survey on the concept of general monotonicity, we refer the reader to [21]. One more direction of extending the obtained results (see [1, 18, 32]) is proving q them for weighted spaces. Define weighted Lebesgue spaces L w( p,q) , p, q ∈ (0, ∞], on [−π, π ], as the set of all measurable functions f with finite norm ⎧ π 1 ⎪ ⎪ q −1 q ⎨ |t| p | f (t)|q dt , if 0 < p, q < ∞,  f L q := −π w( p,q) ⎪ ⎪ 1 ⎩ ess sup |t f (t)|, if 0 < p ≤ ∞, q = ∞. p t∈[−π,π ] q The discrete weighted Lebesgue space lw( p,q) is to be defined in the same way. Now, a weighted version of relation (3) is ∞  q q −1 q {cn }l q := |cn |q n p   f L q , (6) w( p ,q) w( p,q) n=1 where p stands for the conjugate to p, that is, 1/ p +1/ p = 1. Note that if we put q = p, we get the standard Hardy–Littlewood relation (3). From now on, writing Hardy– Littlewood type relations we will omit the dependence on p of the corresponding constants, so this dependence will be taken for granted. The following theorem for weighted Lebesgue spaces was obtained by Sagher [27]. Theorem B (Sagher, 1976) If the sequences {an } and {bn } are monotone and vanishing at infinity and the function f has the Fourier series ∞ a0  + (an cos nx + bn sin nx), 2 n=1 then for p ∈ (1, ∞), q ∈ [1, ∞], there holds  f L q  {an }l q + {bn }l q . w( p,q) w( p ,q) w( p ,q) It turns out that the same holds if we release the monotonicity condition in the theorem above to (5), thus withdrawing the requirement of positivity. This result, along with the similar statement proved for Lorentz spaces, was given by Dyachenko et al. [9]. So, in the one-dimensional case we have quite a complete picture. 60 Page 4 of 30 Journal of Fourier Analysis and Applications (2023) 29:60 The whole scenario becomes more complicated if we step out from the one- dimensional setting to the multidimensional one, and the first question we face is to determine what we should mean by monotonicity if we deal with multiple sequences. The usual one-dimensional monotonicity is characterized by the inequal- ities an ≥ an+1 , or equivalently, an := an − an+1 ≥ 0. These two ways of writing the same property give rise to the following fundamentally different multidimensional monotonicity concepts. Our focus will be on the two-dimensional case. Monotonicity in Each Variable Likewise an ≥ an+1 in one dimension, we can require coordinatewise monotonicity, that is, in two-dimensional case the condition will be amn ≤ am n , for all m ≥ m , n ≥ n . (7) It turns out, however, that for such sequence the Hardy–Littlewood relation (3) does not hold for some values of p > 1, namely, we have the following result proved by Dyachenko [6, 8]. Theorem C (Dyachenko, 1986) (a) [6, Th. 1] If {amn }∞ m,n=1 satisfying (7) and amn → 0, as m + n → ∞, (8) is the sequence of the Fourier coefficients with respect to one of the orthonor- mal systems {einx eimy }∞ ∞ ∞ m,n=1 , {sin nx sin my}m,n=1 , and {cos nx cos ny}m,n=1 , of a function f , then for any p ∈ (1, ∞), ∞  p p amn (mn) p−2   f  p . m,n=1  (b) [8, Cor. 2] Let p > 4/3 and the sequence {amn } satisfy (7) and ∞ p m,n=1 amn (mn) p−2 < ∞ (therefore, (8) as well). Then, for any of the systems above, there exists a function f having {amn } as its Fourier coefficients and satisfying ∞  p p amn (mn) p−2   f  p . (9) m,n=1 8’] For pp ∈ (1, p−2 (c) [6, Ths. 8, 4/3), there exists a sequence {amn } satisfying (7) and (8) with ∞ m,n=1 a mn (mn) < ∞ such that the corresponding trigonometric series diverges by squares almost everywhere on (0, 2π )2 . Note that it was shown by Fefferman [13] that for any p > 1 and any f ∈ L p (0, 2π )2 , the Fourier series of f converges by squares almost everywhere on (0, 2π )2 , thus, the Journal of Fourier Analysis and Applications (2023) 29:60 Page 5 of 30 60 third part of the theorem means that (9) is no longer true for p ∈ (1, 4/3). We also remark that in general d-dimensional case the critical value is 2d/(d + 1) (see [7, Th. 1, Th. 4] and [8, Cor. 2]). Monotonicity by Hardy The next approach to the multiple concept of monotonicity is to consider the mono- tonicity in the so-called sense of Hardy (or Hardy–Krause, see [16] and [19], where this concept initially arises). That is, to introduce the following differences 10 amn := amn − am+1,n , 01 amn := amn − am,n+1 , 11 amn := 01 (10 amn ) = 10 (01 amn ) = amn − am+1,n − am,n+1 + am+1,n+1 , and recalling the one-dimentional condition an ≥ 0, generalize it in the following way 11 amn ≥ 0 for all m, n. (10) Note that under the natural requirement (8), condition (10) implies amn ≥ 0, 10 amn ≥ 0, 01 amn ≥ 0. Here comes the result obtained by Móricz [23, Th. 1,2, Cor. 1]. Theorem D (Móricz, 1990) Let p ≥ 1 and the sequence {amn } satisfy (8) and (10).  (a) If ∞ p m,n=1 amn (mn) p−2 < ∞, then the double sine or cosine series with coeffi- cients {amn } is the Fourier series of its sum f and ∞  p p amn (mn) p−2   f  p . m,n=1 (b) If {amn } is the sequence of double sine or cosine Fourier coefficients of f ∈ L p , then ∞  p p amn (mn) p−2   f  p . m,n=1 The reader can find Theorem D proved for Vilenkin systems (and hence for the Walsh system) in [30, Sec. 6.3] and [31]. Condition (10) is quite restrictive and one of the closest generalizations of it in, say, G M spirit is the following one ∞   ∞ |11 amn |  |amn |. m=k n=l 60 Page 6 of 30 Journal of Fourier Analysis and Applications (2023) 29:60 Note that if the sequence satisfies (10), then the left-hand side above becomes just equal to amn . The next result [10, Th. 6B] (see [11] for the proof) extends the one of Móricz. Theorem E (Dyachenko, Tikhonov, 2007) If a nonnegative sequence {amn } satisfy (8) and the so-called G M 2 condition ∞   ∞ ∞  ∞ ∞ ∞ |aml |  |akn |   |amn | |11 amn |  |akl | + + + , (11) m n mn m=k n=l m=k n=l m=k n=l then the corresponding double sine, cosine, or exponential series converges every- where on (0, 2π )2 and is the Fourier series of its sum. Besides, for any p ∈ (1, ∞), ∞  p p amn (mn) p−2   f  p . m,n=1 It is worth mentioning that  the  part was proved without assuming ∞ amn ≥ 0, moreover, it was shown that if ∞ ∞ p m=k n=l | 11 a |  β , then mn kl m,n=1 βmn (mn) p−2  p  f  p . However, in the proof of the counterpart the requirement of nonnegativity plays a crucial role. It was noted in [12, Th. 4.1] that following the lines of this proof one can adapt it for a  more general ∞class of sequences for which the right-hand side of (11) is replaces by ∞ m= k/λ n= l/λ |amn |/mn, λ > 1. Further, it was shown [33] that some other G M type nonnegative sequences happen to obey the two-sided Hardy–Littlewood relation. We present the result from [33] for weighted spaces. Theorem F (Yu, Zhou, Zhou, 2012) Let {amn } be a nonnegative sequence satisfying (8) and the following G M type conditions  2k λk   2l λl  |aml | |akn | |aml |  , |akn |  , m n m=k m= λ−1 k n=l n= λ−1 l  2k  2l λk  λl  |amn | |amn |  mn m=k n=l m= λ−1 k n= λ−1 l ∞ for some λ ≥ 2, and let f (x, y) := m,n=1 amn sin mx sin ny. Then, for any p ∈ 1 − p−1 [1, ∞), for any function φ ∈  with either φ ∈ L if p > 1, or φ −1 ∈ L ∞ , if p = 1, we have ∞  p φ| f | p ∈ L ⇔ amn φ(1/m, 1/n)(mn) p−2 < ∞. m,n=1 Journal of Fourier Analysis and Applications (2023) 29:60 Page 7 of 30 60 In the above result  stands for some class of power-like positive functions, which we are not going to specify here. A similar result with a more general G M type positive sequences and some other (not comparable) class of power-like functions was obtained in [5]. The main purpose of this work is to show that for some kinds of double G M sequences we can prove the Hardy–Littlewood theorem without restricting ourselves only to positive sequences. We present two G M type classes for which the two-sided Hardy–Littlewood inequality holds true. We write that {amn } ∈ G M1c if it satisfies (8) and  ∞ 2k  ∞   2l |11 amn | + |11 amn | ≤ C|akl |, (12) m=k n=l m=k n=l and {amn } ∈ G M2c , if it satisfies (8) and  ∞ 2k  ∞   2l |11 amn | + |11 amn | ≤ C|a2k,l |, (13) m=k n=l m=k n=l for all k, l ∈ N and some constant C depending only on the sequence {amn }. We remark that the letter c in G M c comes from the word “corner”, since a set of the kind [k, 2k] × [l, ∞) ∪ [k, ∞) × [l, 2l] generates a corner on the plane. Note that G M1c sequences obey the one-dimensional G M conditions (4) in each variable (see (14) in the proof of Lemma 1), while G M2c in one variable satisfy (4), and in another one, the “backward” G M condition. q Note that for [−π, π ]2 the L w( p,q) -norms take the form ⎧ π π 1 ⎪ ⎪ q |ts| p −1 | f (t, s)|q dt ds q , if 0 < p, q < ∞, ⎨  f L q := −π −π w( p,q) ⎪ ⎪ 1 |(ts) p f (t, s)|, if 0 < p ≤ ∞, q = ∞. ⎩ ess sup (t,s)∈[−π,π ]2 From now on, for convenience, we adopt the following notation: using that (sin x)(1) = (sin x) = cos x and (sin x)(0) = sin x, we will write a two-dimensional trigonometric series as  1 ∞  amn sin(i) mx sin( j) ny ij i, j=0 m,n=0 and we will say that {amn }∞ ij m,n=1 , i, j = 0, 1, is the sequence of its coefficients. The main result of the paper is the following. Theorem 1 Let p ∈ (1, ∞), q ∈ [1, ∞], and let each of the sequences {amn }∞ ij m,n=1 , i, j = 0, 1, belong either to G M1 or to G M2 . c c 60 Page 8 of 30 Journal of Fourier Analysis and Applications (2023) 29:60 (a) If {amn }∞ ij m,n=1 , i, j = 0, 1, is the sequence of Fourier coefficients of f ∈ L(−π, π ), then  1 ∞  q q ij −1  f L q  |amn |q (mn) p . w( p,q) i, j=0 m,n=1 1 ∞ ij q q −1 (b) If i, m,n=1 |amn | (mn) < ∞, then the corresponding trigonometric p j=0 series converges everywhere on (0, 2π )2 and is the Fourier series of its sum, moreover,  1 ∞  q q ij −1  f L q  |amn |q (mn) p . w( p,q) i, j=0 m,n=1 Sharpness of Theorem 1 for G M2c sequences is provided by a counterexample in Theorem 2, which shows that if we restrict the sum on the left-hand side of (13) to the rectangle (that is, to the intersection and not the union of the two corresponding strips), which is one of the most natural generalizations of the left-hand side of the G M condition (4), then the  part fails for p > 2 and q ≥ p. 2 Proof of the Hardy–Littlewood Theorem for GMc Sequences For a sequence {amn }∞ m,n=1 , we define Amn := max |akl | := max |akl |. (k,l)∈Q m,n (k,l)∈[2m ,2m+1 ]×[2n ,2n+1 ] Lemma 1 (a) For any sequence {akl }∞ k,l=1 ∈ G M1 , there exist c, v > 0 such that c for any (m, n) with Am−1,n−1 ≤ T Am,n there exist a rectangle Q m−1,n−1 ⊂ Q m−1,n−1 of size 2m−v × 2n−v satisfying  akl > c2m+n Amn , k,l∈ Q m−1,n−1 where c and v depend only on C and T . (b) For any sequence {akl }∞ k,l=1 ∈ G M2 , there exist c, v > 0 such that for any (m, n) c with Am+1,n−1 ≤ T Am,n there exist a rectangle Q m+1,n−1 ⊂ Q m+1,n−1 of size 2m−v × 2n−v satisfying  akl > c2m+n Amn , k,l∈ Q m+1,n−1 where c and v depend only on C and T . Journal of Fourier Analysis and Applications (2023) 29:60 Page 9 of 30 60 Proof Note that (8) and (12) imply that  2k  2l |10 amt | + |01 asn | ≤ C|ak,l | (14) m=k n=l for any k, l ∈ N and (s, t) ∈ [k, 2k] × [l, 2l]. Similarly, (8) along with (13) imply (14) with a2k,l instead of ak,l on the right-hand side. In particular, (14) yields that |as,t | − |ak,l | = |as,t | − |ak,t | + |ak,t | − |a2k,l | ≤ C|ak,l |, so |as,t | ≤ (C + 1)|ak,l | ≤ (C + 1)2 |as t | for any (s , t ) ∈ [0.5k, k] × [0.5l, l]. Considering k = 2m , l = 2n , we get for any (s, t) ∈ Q m−1,n−1 |ast | ≥ (C + 1)−2 Amn =: α Amn . (15) For conditions (8) and (13), the same arguments give |as,t | − |a2k,l | = |as,t | − |a2k,t | + |a2k,t | − |a2k,l | ≤ C|a2k,l |, and |as,t | ≤ (C + 1)|a2k,l | ≤ (C + 1)2 |as t | for any (s , t ) ∈ [2k, 4k] × [0.5l, l]. Once more, considering k = 2m , l = 2n , we get (15) for (s, t) ∈ Q m+1,n−1 instead of Q m−1,n−1 . Thus, any sequence {akl } ∈ G M1c satisfies |akl | ≤ (C + 1)|ak l | for (k , l ) ∈ [0.5k, k] × [0.5l, l] as well as any {akl } ∈ G M2c does for (k , l ) ∈ [k, 2k] × [0.5l, l]. In Lemma 1(a), due to condition (14) and inequality (15), for any (k, l) ∈ Q m−1,n−1 , each one of the sequences a2m−1 ,l , a2m−1 +1,l , . . . , a2m ,l and ak,2n−1 , ak,2n−1 +1 , . . . , ak,2n can have at most C max |akl | (k,l)∈Q m−1,n−1 C Am−1,n−1 CT = ≤ =: b (16) 2α Amn 2α Amn 2α changes of sign. The same holds for Q m+1,n−1 in place of Q m−1,n−1 in Lemma 1(b). Focus now on Lemma 1(a). Consider the rectangle R := Q m−1,n−1 = [2m−1 , 2m ]× [2n−1 , 2n ] on the plane and draw all the segments [(k, l), (k + 1, l)] such that ak,l−1 and ak,l have different signs and all the segments [(k, l), (k, l + 1)] such that ak−1,l and ak,l have different signs (call them marked segments). Then our rectangle R is divided by the marked segments into several connected parts corresponding to the 60 Page 10 of 30 Journal of Fourier Analysis and Applications (2023) 29:60 terms of {akl } of the same sign. The interior part of the union of their boundaries has at most b2n−1 vertical marked segments and at most b2m−1 horizontal ones. Take a positive integer u such that 2u > 8bτ, (17) where τ := 4 T (C + 1)2 + 1. Divide R into 22u equal rectangles of size 2m−1−u × 2n−1+u and consider a half of them in a checkerboard pattern. Suppose that there is no rectangle among them containing at most 2n−1−u /τ vertical marked segments and at most 2m−1−u /τ horizontal ones. Then we must have b2m−1 τ b2n−1 τ 22u−1 ≤ + = 2u+2 bτ ≤ 4bτ 2u , 2m−1−u 2n−1−u which contradicts (17). So, there is a rectangle r = [α1 , α2 ] × [β1 , β2 ] of size 2m−1−u × 2n−1−u with at most 2n−1−u /τ vertical marked segments and at most 2m−1−u /τ horizontal ones inside it. Consider the parts corresponding to the terms of {akl } of the same sign inside r . Call the parts whose boundaries intersect the bound- ary of r by A-parts, the other ones, by B-parts. Note that  there is no marked segment 3β1 +β2 β1 +3β2  of an A-part inside the rectangle r := 3α14+α2 , α1 +3α 4 2 × 4 , 4 . Indeed, otherwise there would exist a broken line of marked segments with either at least 0.25(α2 − α1 ) = 2m−3−u horizontal segments or at least 0.25(β2 − β1 ) = 2n−3−u vertical ones. But this is impossible, since τ > 4. The area of all B-parts does not exceed 2m+n−2−2u /τ 2 . Thus, there are at least 2m+n−4−2u (1 − 4τ −2 ) terms of the same sign in r , so the absolute value of the sum of the terms {akl } in r is at least  4 4 2n+m−2u−4 1 − 2 − 2 T (C + 1)2 α Amn > 2n+m−2u−5 α Amn , τ τ which concludes the proof of Lemma 1(a) with c := 2−2u−5 α and v := u + 1. A similar argument is valid for Q m+1,n−1 in Lemma 1(b), which completes the proof.   Remark 1 In the proof of Lemma 1, for G M1c class we only used its one-dimensional G M properties (14), and for G M2c , the corresponding nonsymmetric relations (namely, (14) with a2k,l in place of ak,l ). Remark 2 The claim of Lemma 1(a) is no longer true if we substitute the G M1c con- dition (12) for  2k  2l |11 amn | ≤ C|akl |. (18) m=k n=l Proof Indeed, consider the sequence (−1)m amn := f m (n), m Journal of Fourier Analysis and Applications (2023) 29:60 Page 11 of 30 60 where f m (n) we define as follows:  2−m+1 , log2 n < m(m+1) , f m (n) = (m+t)2 +m−t 2 (m+t+1)2 +m−t−1 2−m−t , 2 ≤ log 2n < 2 , t ∈ Z+ . For such a sequence, condition (8) obviously holds. Consider a rectangle Smn of the form [m, 2m) × [n, 2n). The only nonzero 11 akl in this rectangle are 11 am −1,n and 11 am n , where n ∈ [n, 2n) : log2 (n ) = log2 (n − 1) + 1, i.e. n is a power of two, and  k(k + 1)  m := min m ∈ N : m = log2 n − , k ∈ Z+ . 2 Note that |akl | ≤ |amn | for k ≥ m, l ≥ n, so |11 am n | ≤ |am n | + |am +1,n | ≤ 2|amn |, which yields condition (18) with C = 2. Assume that the assertion of Lemma 1 holds. Then there must exist a constant c such that for at least cmn squares [k, k + 2) × [l, l + 2) in any Smn there holds |akl + ak,l+1 + ak+1,l + ak+1,l+1 | ≥ c|akl |. (19) Consider a rectangle Smn with t(t + 1) (t + 1)(t + 2) + 2m ≤ log2 n ≤ − 2, 2 2 where t > 4m is a positive integer. For any akl in Smn , we have (−1)k akl = 2−t−1 , k whence for any 2 × 2 square [k, k + 2) × [l, l + 2) ⊂ Smn   1 1 2 2 |akl + ak,l+1 + ak+1,l + ak+1,l+1 | = 2−t−1 · 2 − = |akl | < |akl | k k+1 k+1 m = o(|akl |), as m → ∞, which leads to a contradiction.   Lemma 2 For a function f ∈ L(−π, π ), given the representation  1 f (x, y) = f i j (x, y), f i j (−x, y) = (−1)i f i j (x, y), f i j (x, −y) i, j=0 = (−1) j f i j (x, y), 60 Page 12 of 30 Journal of Fourier Analysis and Applications (2023) 29:60 for any p ∈ (1, ∞), q ∈ [1, ∞], we have  1  f L q   f i j L q . w( p,q) w( p,q) i, j=0 Proof The  part is clear, so we have to prove the reverse. We start with the case q < ∞. Noting that for any pair of functions g1 , g2 there always holds |g1 |q + |g2 |q  |g1 + g2 |q + |g1 − g2 |q and recalling that the weight is an even in each variable function, we obtain q q q  f i0 (x, ·) L q +  f i1 (x, ·) L q  ( f i0 + f i1 )(x, ·) L q w( p,q) w( p,q) w( p,q) q q + ( f i0 − f i1 )(x, ·) L q  ( f i0 + f i1 )(x, ·) L q w( p,q) w( p,q) for i = 0, 1. Similarly,  1 q q q  f ij q   f 00 + f 01 + f 10 + f 11  q +  f 00 + f 01 − f 10 − f 11  q L w( p,q) L w( p,q) L w( p,q) i, j=0   1 q   q  f ij q = f q . L w( p,q) L w( p,q) i, j=0 For q = ∞, the claim follows from the equalities 4 f i j (x, y) ≡ f (x, y) + (−1)i f (−x, y) + (−1) j f (x, −y) + (−1)i+ j f (−x, −y).   Next we prove a two-dimensional analogue of [9, L. 2.2] (see also the one- dimensional result [27, Th. 2.4] for Lorentz spaces). Note that similar multidimen- sional results for Lorentz spaces were obtained in [24] and [25]. Lemma 3 Let {amn }∞ ij m,n=1 , i, j = 0, 1, be the sequence of Fourier coefficients of f ∈ L(−π, π ). Then for any p ∈ (1, ∞), q ∈ [1, ∞], there holds  ∞   1  1 k l q q −1 q sup ast (mn) p   f L q . k≥m, l≥n kl w( p,q) i, j=0 m,n=1 s=1 t=1 Proof of Lemma 3 Note that if we prove the statement of the lemma for odd in each variable functions f ∈ L(−π, π ), then it will be true for any integrable f . Indeed, the relation for such functions implies the same for all functions that are either odd or even in each variable due to the boundedness of the Hilbert transform in weighted Lebesgue spaces. The general case follows then by Lemma 2. Thus, we can assume ij that amn = 0 if (i, j) = (0, 0) and omit the upper indices of amn 00 . Journal of Fourier Analysis and Applications (2023) 29:60 Page 13 of 30 60 According to [9, (2.4), (2.7)], for any 1 < p < ∞, 1 ≤ q ≤ ∞, and m ∈ N, there holds    cos x2 (1 − cos mx) sin mx  −1 Im (x)l p,q :=   x +    m p. m sin 2 m l p,q Therefore, for any 1 < p1 , p2 < ∞, 1 < q ≤ ∞, and m, n ∈ N, by Hölder’s inequality  π π 1  m n akl ≤ | f (x, y)Im (x)In (y)|d xd y mn 0 0 k=1 l=1  π  π q 1  π q 1 −1 q ≤ |In (y)| x p1 | f (x, y)|q d x x p1 |Im (x)|q d x q dy 0 0 0 1  π  π 1 − q −1 q m p1 |In (y)| x p1 | f (x, y)|q d x dy 0 0 − 1  π π q q 1  π q −1 1 −1 −1 q ≤m | f (x, y)|q d xd y |In (y)|d y p1 q x p1 y p2 y p2 0 0 0 − 1 − 1  π  π q q 1 −1 −1 q m p1 n p2 x p1 y p2 | f (x, y)|q d xd y 0 0 1 1 − − =: m p1 n p2  f L q . (20) w(( p1 , p2 ),q) Similarly, if q = 1,  π π 1  m n akl ≤ | f (x, y)Im (x)In (y)|d xd y mn 0 0 k=1 l=1 1 1 ≤ sup x p1 |Im (x)| · sup y p2 |In (y)| x∈[0,π ] y∈[0,π ]  π π 1 1 1 1 − − · x p1 −1 y p2 −1 | f (x, y)| d xd y  m p1 n p2  f L 1 . w(( p1 , p2 ),1) 0 0 Thus, for any 1 < p1 , p2 < ∞, 1 ≤ q ≤ ∞, and m ∈ N, we obtain 1  1 1 k l m p1 sup n p2 sup ast ≤ C f  L q , (21) kl w(( p1 , p2 ),q) n∈N k≥m, l≥n s=1 t=1 where the constant C does not depend on m. Now, in order to prove the desired inequality, we will invoke interpolation theory. Recall that the norm of a sequence c := {ck }∞ k=1 in the discrete Lorentz space l p,q , for 60 Page 14 of 30 Journal of Fourier Analysis and Applications (2023) 29:60 p ∈ (1, ∞) and q ∈ (0, ∞], is defined as follows ⎧ ⎪ ∞ q −1 ∗ q 1 ⎨ k=1 k p |c | q , if q < ∞, k cl p,q := 1 ⎪ ⎩sup k |ck∗ |, if q = ∞, p k≥1 where {ck∗ } stands for the decreasing rearrangement of c. It follows from [3, Th. 5.3.1] that for θ ∈ (0, 1) and q ∈ (0, ∞], for the discrete Lorentz spaces l p1 ,∞ and l p2 ,∞ , 0 < p1 < p2 ≤ ∞, with θ/ p1 + (1 − θ )/ p2 = 1/ p, we have (l p1 ,∞ , l p2 ,∞ )θ,q = l p,q . (22) q q For the Lebesgue spaces L w(( p11 , p21 ),q) and L w(( p21 , p22 ),q) , q ∈ (0, ∞], (see (20)), with θ/ p11 + (1 − θ )/ p12 = 1/ p1 , θ/ p21 + (1 − θ )/ p22 = 1/ p2 , [3, Th. 5.4.1] gives q q q (L w(( p11 , p21 ),q) , L w(( p12 , p22 ),q) )θ,q = L w(( p1 , p2 ),q) . (23)  k  l For any fixed m 0 ∈ N, in light of the monotonicity of sup 1 kl ast in n, k≥m 0 , l≥n s=1 t=1 (21) is equivalent to  1 1  k l ∞   p1  m0  sup ast  ≤ C f  L q . (24) kl n=1 lp w(( p1 , p2 ),q) k≥m 0 , l≥n s=1 t=1 2 ,∞ Fix now p1 , p2 ∈ (1, ∞) and q ∈ [1, ∞]. Take θ ∈ (0, 1) and p11 < p12 , p21 < p22 such that θ/ p11 + (1 − θ )/ p12 = 1/ p1 and θ/ p21 + (1 − θ )/ p22 = 1/ p2 . Note that, for any fixed m 0 , the operator  1  k l ∞ Tm 0 f = sup ast k≥m 0 , l≥n kl n=1 s=1 t=1 is sublinear and that due to (24) q q Tm 0 : L w(( p1 , p21 ),q) → l p21 ,∞ and Tm 0 : L w(( p1 , p22 ),q) → l p22 ,∞ , where the involved constants do not depend on m 0 . Then it follows from [22, Th. 6], (22), and (23) that q q q Tm 0 : L w(( p1 , p2 ),q) = (L w(( p1 , p21 ),q) , L w(( p1 , p22 ),q) )θ,q → (l p21 ,∞ , l p22 ,∞ )θ,q = l p2 ,q , so we arrive at 1  ∞  1  k l   m p1  sup ast    f L q , (25) kl n=1 l p2 ,q w(( p1 , p2 ),q) k≥m, l≥n s=1 t=1 Journal of Fourier Analysis and Applications (2023) 29:60 Page 15 of 30 60 for any m. Now we note that  1  k l ∞   ∞  1  k l   q −1 q 1/q  sup ast  = n p2 sup ast k≥m, l≥n kl n=1 l p2 ,q k≥m, l≥n kl s=1 t=1 n=1 s=1 t=1 is decreasing in m for any p2 ∈ (1, ∞) and that the operator   ∞ q  1  k l q 1/q ∞ −1 Tf = n p2 sup ast k≥m, l≥n kl m=1 n=1 s=1 t=1 is sublinear. Since according to (25) we have q q T : L w(( p11 , p2 ),q) → l p11 ,∞ and T : L w(( p12 , p2 ),q) → l p12 ,q , we can once again apply [22, Th. 6] and obtain q q q T : L w(( p1 , p2 ),q) = (L w(( p11 , p2 ),q) , L w(( p12 , p2 ),q) )θ,q → (l p11 ,∞ , l p12 ,∞ )θ,q = l p1 ,q . The latter means that   ∞  1  k l 1 ∞   q −1 q q   n p2 sup ast    f L q , kl m=1 l p1 ,q w(( p1 , p2 ),q) n=1 k≥m, l≥n s=1 t=1 whence the claim follows by putting p1 = p2 = p.   Proof of Theorem 1 In light of Lemma 2 it suffices to prove the theorem only for either odd or even in each variable functions, omitting therefore the upper indices of amn . We start with the part a). Due to Lemma 3 there holds ∞   1  k l q q q −1  f L q  sup ast (mn) p w( p,q) k≥m, l≥n kl m,n=1 s=1 t=1 ∞   1  k l q ∞  (m+n) pq  2 sup ai j =: Pmn . (26) k≥2m , l≥2n kl m,n=0 i=1 j=1 m,n=0 Denote −1 2 −1 2m+1 n+1 q −1 Wmn := |akl |q (kl) p . k=2m l=2n 60 Page 16 of 30 Journal of Fourier Analysis and Applications (2023) 29:60 Consider first G M1c sequences. Let us fix some T > 1. We call a pair (m, n) good (we write (m, n) ∈ G), if either mn = 0 or Am−1,n−1 ≤ T Amn . We have ∞  q ∞  ∞  −1 |akl |q (kl) p = Wmn ≤ Wm0 k,l=1 m,n=0 m=0 ∞    + W0n + Wmn + Wkl n=0 (m,n)∈G∩N2 (m,n)∈G (k,l)∈Bmn =: J1 + J2 + J3 + J4 , where Bmn , (m, n) ∈ G, stands for the set of all pairs (k, l) ∈ / G such that k = m + t, l = n + t for some t ∈ N. According to the one-dimentional Hardy–Littlewood theorem for G M sequences [9, Th. 1.2], we obtain ∞  ∞  q −1 q q J1 = Wm0 = |ak1 |q k p  g L q   f L q , (27) w( p,q) w( p,q) m=0 k=1 π where g(x) = −π f (x, y) sin y dy. A similar estimate is valid for J2 . Consider a pair (m, n) ∈ G ∩ N2 . Denote the rectangles we constructed in Lemma 1 , s 2 ] × [t 1 , t 2 ], so we have 1a) [smn mn mn mn  1  k l q (m+n−2) pq Pm−1,n−1 = 2 sup ai j k≥2m−1 , l≥2n−1 kl i=1 j=1 ⎛ 1 −1 t 1 −1 1 −1 t 2 smn smn (m+n) pq −(m+n)q   mn q   mn q 2 ⎝ ai j + ai j i=1 j=1 i=1 j=1 2 t 1 −1 2 2 ⎞ smn smn tmn   mn q  q + ai j + ai j ⎠ i=1 j=1 i=1 j=1 2 smn tmn 2 (m+n) pq −(m+n)q   q (m+n) pq q 2 ai j 2 Amn  Wmn . 1 j=t 1 i=smn mn Here we used the inequality |x + y + z + t| + |x + y| + |x + z| + |x| ≥ |z + t| + |z| ≥ |t|, valid for any x, y, z, t ∈ C. Journal of Fourier Analysis and Applications (2023) 29:60 Page 17 of 30 60 Hence, using (26), we obtain   q J3 = Wmn  Pm−1,n−1 ≤  f  L q . (28) w( p,q) (m,n)∈G∩N2 (m,n)∈G∩N2 Finally, combining (27), the similar estimate for J2 , and (28), we derive  ∞  1 T−j ≤ q J4 ≤ Wmn (J1 + J2 + J3 )   f  L q , 1 − T −1 w( p,q) (m,n)∈G j=1 which concludes the proof of the first part for the case of G M1c . If we replace G M1c by G M2c , i.e. (12) by (13), we change the definition of a good pair of numbers to the following one: we call a pair (m, n) good, if either mn = 0 or Am+1,n−1 ≤ T Amn . The rest of the proof is the same in light of Lemma 1b) with the only changes: now Bmn , (m, n) ∈ G, stands for the set of all pairs (k, l) ∈ / G such that k = m − t, l = n + t for some t ∈ N and Pm−1,n−1 in (28) becomes Pm+1,n−1 .  q −1 Turn now to the part b). Note that if {amn }∈G M1c ∪G M2c and ∞ |amn |q (mn) p ∞ ∞ m,n=1 < ∞, then we have k=1 l=1 |11 akl | < ∞, which implies that the corresponding trigonometric series converges in the Pringsheim sense everywhere on (0, 2π )2 and is the Fourier series of its sum (see [6, L. 4]). Indeed, under condition (12) we have by (15) and Hölder’s inequality ∞  ∞  ∞   2 k  2 k ∞  |11 akl |  |a2k ,2k |  |a2k ,2k | (mn)−1  |amn |(mn)−1 k,l=1 k=0 k=0 m=2k−1 n=2k−1 m,n=1 ∞   ∞ q 1   ∞ 1 1 − q1 − p1 − q1 −1 q − qp −1 = |amn |(mn) p (mn)  |amn |q (mn) p (mn) < ∞, q m,n=1 m,n=1 m,n=1 and similarly under (13), ∞   ∞ ∞  ∞  |11 akl |  |a2k+1 ,2k |  |a2k+1 ,2k | k=1 l=1 k=0 k=0 k+2 2  2 k ∞  (mn)−1  |amn |(mn)−1 < ∞. m=2k+1 n=2k−1 m,n=1 We will provide the proof only for the system {sin mx, sin ny}, the other cases will follow then from boundedness   of Hilbert  transform in weighted Lebesgue spaces. π π π For (x, y) ∈ m+1 ,m × n+1 , πn , we have ∞   ∞  m  n | f (x, y)| = akl sin kx sin ly ≤ x y kl|akl | k=1 l=1 k=1 l=1 m ∞  +x k |akl − ak,l+1 || D̃l (y) − D̃n (y)| k=1 l=n 60 Page 18 of 30 Journal of Fourier Analysis and Applications (2023) 29:60  n ∞  +y l |akl − ak+1,l || D̃k (x) − D̃m (x)| l=1 k=m ∞   ∞ 1  m n + |11 akl | · |( D̃k (x) − D̃m (x))( D̃l (y) − D̃n (y))|  kl|akl | mn k=m l=n k=1 l=1  ∞  ∞ ∞  ∞ m   m n n + k |akl − ak,l+1 | + l |akl − ak+1,l | + mn |11 akl |. m n k=1 l=n l=1 k=m k=m l=n Applying condition (12), we derive ∞ ∞ 1  n   m  m n m n | f (x, y)|  kl|akl | + k |ak,2t n | + l |a2t m,l | mn m n k=1 l=1 k=1 t=0 l=1 t=0 ∞ + mn |a2t m,2t n | t=0  m  ∞ ∞ n   |akl | m   n m n 1 |akl |  kl|akl | + k + l mn m l n k k=1 l=1 k=1 l= n/2 l=1 k= m/2 ∞ ∞  |akl | + mn . kl k= m/2 l= n/2 In turn, (13) yields ∞ ∞ 1  n   m  m n m n | f (x, y)|  kl|akl | + k |ak,2t n | + l |a2t+1 m,l | mn m n k=1 l=1 k=1 t=0 l=1 t=0 ∞ + mn |a2t+1 m,2t n | t=0  m  ∞ ∞ n   |akl | m   |akl | n m n 1  kl|akl | + k + l mn m l n k k=1 l=1 k=1 l= n/2 l=1 k=2m ∞ ∞  |akl | + mn . kl k=2m l= n/2 Hence, in both cases we get ∞ ∞ 1  n   |akl | m   m n m n |akl | | f (x, y)|  kl|akl | + k + l mn m l n k k=1 l=1 k=1 l= n/2 l=1 k= m/2 ∞ ∞  |akl | + mn kl k= m/2 l= n/2 =: Im,n 1 + Im,n 2 + Im,n 3 + Im,n 4 . (29) Journal of Fourier Analysis and Applications (2023) 29:60 Page 19 of 30 60 Thus, for q < ∞, denoting α := 1 − q/ p, we obtain π π (x y)−α | f (x, y)|q d xd y q  f L q  p,q 0 0 π π ∞ m  n ∞    (x y)−α (Im,n 1 + Im,n 2 + Im,n 3 + Im,n 4 )q d xd y m=1 n=1 π π m+1 n+1 ∞   ∞  (mn)α−2 ((Im,n 1 )q + (Im,n 2 )q + (Im,n 3 )q + (Im,n 4 )q ). m=1 n=1 Recall the Hardy-type inequalities for power weights (see, for instance, [20, (0.6), (0.10), (1.102)]) for q ≥ 1: ∞   n q ∞  nγ n γ +q an , for γ < −1, q ak q (30) n=1 k=1 n=1 and its dual, ∞   ∞ q ∞  nγ n γ +q an , for γ > −1. q ak q (31) n=1 k=n n=1 Using (30) in each variable we arrive at ∞   ∞ ∞  ∞   n  m q α−2 1 α−2−q α−2−q (mn) (Im,n ) = q m n l k|akl | m=1 n=1 m=1 n=1 l=1 k=1 ∞ ∞  m q ∞   ∞  n α−2+q m α−2−q k|akn |  (mn)α−2+q |amn |q n=1 m=1 k=1 m=1 n=1 and  ∞ ∞  ∞  ∞    ∞ 1 m q (mn)α−2 (Im,n 2 )q = m α−2−q n α−2+q k|akl | l m=1 n=1 m=1 n=1 l= n/2 k=1 ∞ ∞  ∞ 1 m q  m α−2−q n α−2+q k|akl | l m=1 n=1 l=n k=1 ∞ ∞  m q  n α−2+q m α−2−q k|akn | n=1 m=1 k=1 ∞ ∞  (mn)α−2+q |amn |q , m=1 n=1 60 Page 20 of 30 Journal of Fourier Analysis and Applications (2023) 29:60 where we used inequality (15). The similar estimate holds for I 3 . And finally, due to (31)  ∞ ∞  ∞  ∞    ∞ 1 ∞  |akl | q (mn)α−2 (Im,n 4 )q = m α−2+q n α−2+q l k m=1 n=1 m=1 n=1 l= n/2 k= m/2 ∞ ∞   1  |a | q ∞ ∞ m α−2+q α−2+q kl  n l k m=1 n=1 l=n k=m ∞ ∞  ∞ |akn | q  m α−2+q n α−2+q k m=1 n=1 k=m ∞ ∞  (mn)α−2+q |amn |q , m=1 n=1 which completes the proof for the case q ∈ [1, ∞). For q = ∞, using (29) we can write 1 − 1p sup (x y) p | f (x, y)| ≤ (mn) (Im,n 1 + Im,n 2 + Im,n 3 + Im,n 4 ). π π π π (x,y)∈( m+1 ,m ]×( n+1 ,n ] Next, − 1p 1 − 1p −1  m  n (mn) Im,n = (mn) kl|akl | k=1 l=1 m  n  1  1 − 1p −1 1 ≤ (mn) (kl) p sup (kl) p |akl |  sup (kl) p |akl | . k=1 l=1 k,l k,l We also have ∞ n   k m  1 − 1p 2 − 1p (mn) Im,n = (mn) |akl |  sup (kl) p |akl | , m l k,l k=1 l= n/2 and the similar estimate for I 3 . Finally, ∞  ∞  |akl |  1 − 1p 4 − 1p (mn) Im,n = (mn) mn  sup (kl) p |akl | , kl k,l k= m/2 l= n/2 which completes the proof of the theorem.   q q Remark 3 For the spaces L w( p,q) (0, 2π ) in place of L w( p,q) (−π, π ), the assertion of Theorem 1 still holds for q ≤ p but fails for q > p. Journal of Fourier Analysis and Applications (2023) 29:60 Page 21 of 30 60 Indeed, for q > p it suffices to consider the one-dimensional sine series ∞  ∞  − p1 − 1p f (x) := k log (k + 2) sin kx =: ak sin kx. k=1 k=1   −1 −1 We have |ak | p k p−2 = k log (k + 2) = ∞, so by the Hardy–Littlewood / L p , whence  f  L q theorem f ∈ (0,2π )   f  L p (π,2π ) = ∞. On the other hand, w( p,q)  q −1  − qp  f L q  |ak |q k p = k −1 log (k + 2) < ∞. w( p,q) (−π,π ) However, for q ≤ p, there holds x q/ p−1  1, so that  f L q   f L q +  f L q   f L q   f L q . w( p,q) (0,2π) w( p,q) (0,π) w( p,q) (π,2π) w( p,q) (0,π) w( p,q) (−π,π) The reason of the failure of the Hardy–Littlewood relation here is that the function in case is supposed to be periodic, while a power weight is not. Thus, if one deals with weighted Lebesgue spaces on [0, 2π ]2 , it makes more sense to consider a weight of the type | sin x|α in place of |x|α , which was in fact done by many authors. Note that for a power weight, weighted integrability at 2π is equivalent to integrability at zero without weight, so, as in the example above, one has to additionally check integrability at zero. 3 Sharpness of the Result Theorem 2 For p > 2, q ≥ p, the claim of Theorem 1(a) does not hold if we replace the G M2c condition (13) by  2k  2l |11 amn | ≤ C|a2k,l |. (32) m=k n=l Proof Assume that p > 2 and consider the sequence (−1)δm amn := gm (n), mγ where γ > 0, δm ∈ {0, 1} are to be chosen later, and gm (n) = gm (n, p ) we define as follows  −1 (−1)δm m −3 n p , log2 n < m(m + 1) p , gm (n) := 2−(m+t) −3(m+t) , ((m + t)2 + m − t) p ≤ log2 n < ((m + t)2 + 3m + t) p , t ∈ Z+ . 2 60 Page 22 of 30 Journal of Fourier Analysis and Applications (2023) 29:60 Fig. 1 Scheme of changes of absolute values of gm (n) In other words, the functions gm are constructed in the following way. First, we divide [1, ∞) into intervals I j , j = 0, 1, . . . , so that I j := {x : 2 p j ≤ log2 x < 2 p ( j + 1)}. After that consider the lower-triangular infinite down and to the right matrix that is filled by all positive integers in increasing order going down and to the right. 1 2 3 4 5 6 7 8 9 10 .. .. .. .. . . . . Next, for any j we asign it the integer i = i( j) if it is ith column that contains the element j. Fix some m and consider the values gm (1), gm (2), . . .. While i( j) = m, we have gm (n) = (−1)δm m −3 n −1/ p for n ∈ I j . Once i( j) becomes equal to m for the first time, that is, when log2 n ≥ m(m + 1) p for the first time, we get gm (n) = 2−m −3 m 2 and this value does not change till i( j) becomes equal to m again and n ∈ I j . When i( j) becomes equal to m for the (s+1)th time, the value gm (n) changes for 2−(m+s) −3(m+s) 2 (see Fig. 1 for a scheme of changes of absolute values of gm (n)). Fix n ∈ I j for some j and consider g1 (n), g2 (n), . . . Let k be such that gm (n) has type 1 if 1 ≤ m ≤ k and type 0 if m ≥ k + 1. Then Journal of Fourier Analysis and Applications (2023) 29:60 Page 23 of 30 60 |gm (n)|  |gm (n)|, for k + 1 ≤ m < m ≤ 2m. (33) Denote m 0 := i( j + 1). If m 0 = k + 1, then g1 (n) = g2 (n) = · · · = gk (n) = 2−(k+1) −3(k+1) , otherwise, gm (n) = 2−(k+1) −3(k+1) for m ≤ m 0 − 1 and gm (n) = 2 2 2−k −3k for m 0 ≤ m ≤ k. Let us compare gk (n) and gk+1 (n). There are two cases. 2 Case 1. m 0 = i( j + 1) = k + 1. Then − p1 |gk+1 (n)| = (k + 1)−3 n  (k + 1)−3 2−(k+1)(k+2)  2−(k+1) 2 −3(k+1) = gk (n). Case 2. m 0 = i( j + 1) < k + 1. Then − p1 |gk+1 (n)| = (k + 1)−3 n  (k + 1)−3 2−k(k+1)−m 0  2−k 2 −3k = gk (n). Thus, in both cases we obtain 0 < g1 (n) ≤ g2 (n) ≤ · · · ≤ gk (n)  |gk+1 (n)|, whence in light of (33), |gm (n)|  |gm (n)|, for all m < m ≤ 2m. (34) It remains to note that for a fixed m, we have for n m := 2m(m+1) p − 1 that 1 − |gm (n m )| = m −3 n m p  m −3 2−m(m+1)  2−m 3 −3m = gm (n m + 1) and for other n there holds gm (n) ≥ gm (n + 1). So, over all |amn | in rkl := [k, 2k] × [l, 2l], the maximal is up to a constant |a2k,l |. Further we note that the constructed sequence clearly satisfies  (8).2l To prove that our sequence belongs to G M2c , let us estimate 2k m=k n=l | amn |. 11 Consider a quadruple am,n+1 am+1,n+1 amn am+1,n with (m, n) ∈ rkl . Note that it can be only of the following five types 0 0 1 0 1 0 1 1 1 1 0 0 1 0 0 0 1 0 1 1 where 0 stands for the terms with log2 n < m(m + 1) p , while 1, for those with log2 n ≥ m(m + 1) p . We will write (m, n) ∈ Ti , i = 1, . . . , 5, if the corresponding quadruple is of the ith type. Note that if (m, n) ∈ T3 , then (m − 1, n) ∈ T1 and (m + 1, n) ∈ T2 , while if (m, n) ∈ T4 , then (m − 1, n) ∈ T2 and (m + 1, n) ∈ T5 . By the construction, quadruples of the three last types with nonzero 11 amn can appear at most four times in rkl , since any (m, n) ∈ T3 ∪ T4 , as well as (m, n) ∈ T5 with 60 Page 24 of 30 Journal of Fourier Analysis and Applications (2023) 29:60 nonzero 11 amn , satisfies n ∈ I j , n + 1 ∈ I j+1 , for some j, which cannot happen twice in [l, 2l]. If there exists a quadruple of the first type, then   |11 amn | = 11 amn (m,n)∈T1 ∩rkl (m,n)∈T1 ∩rkl  − p1 − p1 < 11 (m −3−γ n ) = k −3−γ l  max |amn |. (m,n)∈rkl m≥k, n≥l As for (m, n) ∈ T2 ∩rkl , they all belong to a strip [k , k +1]×[l, 2 l] for some k . Indeed, otherwise there are m 1 and m 2 ≥ m 1 + 2 belonging to [k, 2k], and n 1 , n 2 ∈ [l, 2 l] such that (m 1 , n 1 ), (m 2 , n 2 ) ∈ T2 . But it follows from (m 1 , n 1 ) ∈ T2 that am 1 +1,k , and hence am 2 ,k , has type 0, while (m 2 , n 2 ) ∈ T2 implies that am 2 ,2k , and hence am 1 +1,2k , has type 1. Thus, there exist two pairs of the form (n, n + 1) inside [l, 2 l] such that n ∈ I j , n +1 ∈ I j+1 , for some j, which cannot be true. Therefore, all (m, n) ∈ T2 ∩rkl do belong to a strip [k , k + 1] × [l, 2 l], whence using |11 amn | ≤ |01 amn | + |01 am+1,n | = 01 |amn | + 01 |am+1,n |, which is true as long as (m, n) ∈ T2 ∩ rkl , we deduce that the sum of |11 amn | over (m, n) ∈ T2 ∩rkl is bounded above by four times the maximal |amn | in rkl . Combining the observations above, we arrive at  2k  2l |11 amn |  max |amn |  |a2k,l |, (m,n)∈rkl m=k n=l which proves (32). Further, for any q > 0, ∞  q −1 |amn |q (mn) p m,n=1 ∞ q ∞  q  −1−γ q −((m+t)2 +3m+3t)q ((m+t) +3m+t) p 2 −1 2((m+t) 2 +3m+t) p  m p 2 2 p m=1 t=0 ∞ q −1−γ q  m p = ∞, m=1 if we set γ = 1/ p . Note that our sequence generates the Fourier sine (or cosine) series of an odd (or even) function f that converges in the Pringsheim sense everywhere on (0, 2π )2 to f according to [6, L. 4]. To prove this, since the sequence fulfils (8), it suffices to show that the following sum is finite Journal of Fourier Analysis and Applications (2023) 29:60 Page 25 of 30 60 ∞    |11 amn | ≤ 11 amn + (|01 amn | + |01 am+1,n |) m,n=1 (m,n)∈T1 (m,n)∈T2 ∪T5  + (|amn | + |am,n+1 | + |am+1,n | + |am+1,n+1 |) (m,n)∈T3 ∪T4  ∞  1+ (01 amn + 01 am+1,n ) + m −3−γ 2−m(m+1) (m,n)∈T2 ∪T5 m=1 ∞  ∞  −(m+t)2 −3(m+t) 1+ 2 + 01 am+1,n m=1 t=0 (m,n)∈T2 ∞ −3−γ −m(m−1) 1+ m 2 < ∞. m=1 Let us stick to the case of an odd f , as for cosine series the argument is exactly the same. Denote for m, n ≥ 1,  amn , if log2 n ≥ m(m + 1) p , cmn := , 0, otherwise, and bmn := amn − cmn . Then   ∞    ∞       f L q ≤ bmn sin mx sin ny  q + cmn sin mx sin ny  q . w( p,q) L w( p,q) L w( p,q) m,n=1 m,n=1 Note that  M  N  M  N −1 bmn sin mx sin ny = sin mx 01 bmn Dn (y) + bm N D N (y) m=1 n=1 m=1 n=1  N M−1 −1  N −1 = 11 bmn Dm (x)Dn (y) + 01 b Mn D M (x)Dn (y) m=1 n=1 n=1  M−1 + 10 bm N Dm (x)D N (y) + b M N D M (x)D N (y) m=1  N M−1 −1 =: 11 bmn Dm (x)Dn (y) + A1 + A2 + A3 . m=1 n=1 q k q −1 q Since Dk  L q  l=1 l p  k p by Theorem B, we have for N0 := max(N − w( p,q) 1, 2 M(M+1) p − 1), 60 Page 26 of 30 Journal of Fourier Analysis and Applications (2023) 29:60  N0 −1− p1 1 − p1 1 A1  L q  M −3−γ n (Mn) p + M −3−γ N0 (M N0 ) p  M −1−γ → 0 w( p,q) n=1 as M → ∞. For M0 := min{m : m(m + 1) p ≥ N },  M−1 − p1 1 −3−γ − p1 1 A2  L q  m −4−γ N (m N ) p + M0 N (M0 N ) p → 0 w( p,q) m=M0 as N → ∞. And finally, − p1 1 A3  L q  M −3−γ N (M N ) p → 0 w( p,q) as M → ∞. Thus,   ∞    ∞       bmn sin mx sin ny  q = 11 bmn Dm (x)Dn (y) q . (35) L w( p,q) L w( p,q) m,n=1 m,n=1 Besides,  M  N  M  N −1 cmn sin mx sin ny = sin mx 01 cmn Dn (y) + cm N D N (y) , m=1 n=1 m=1 n=1 where in light of the inequalities 0 < g1 (n) ≤ · · · ≤ g M0 (n) for M0 defined as above M   M0   1 1  cm N sin mx D N (y) q  |cm N |N p ≤ M0 g M0 (N )N p  M0−2 → 0 L w( p,q) m=1 m=1 as N → ∞. Hence,   ∞    ∞       cmn sin mx sin ny  q = 01 cmn sin mx Dn (y) q . (36) L w( p,q) L w( p,q) m,n=1 m,n=1 Combining (35) and (36) we arrive at   ∞     f L q ≤ 11 bmn Dm (x)Dn (y) q w( p,q) L w( p,q) m,n=1   ∞    + 01 cmn sin mx Dn (y) q =: S1 + S2 . L w( p,q) m,n=1 Journal of Fourier Analysis and Applications (2023) 29:60 Page 27 of 30 60 First, for n m = 2m(m+1) p −1, we see that log n m  m 2 and log n m+1 −log n m  m, so ∞ n −1  1  m − p1 1 S1  m p 11 (m −3−γ n )n p m=1 n=1 ⎞ n m+1 −1  1 1 −1 1 −3−γ − p +  ((m + 1) 01 n )n p + (m −3−γ nm p )n m ⎠ p n=n m ∞  n m −1 n m+1 −1 ∞  1   1 −2−γ  mp m −4−γ n −1 + m −3−γ n −1 + m −3−γ  mp < ∞. m=1 n=1 n=n m m=1 Second, denoting n mt := 2((m+t) +3m+t) p − 1, using cmn = (−1)δm |cmn | and the 2 fact that 01 cmn = 0 only if n = n mt for t ≥ −1, we get for q ≥ p, ∞ ∞  q   (−1)δm sin mx q S2 =  01 |cm,n mt |Dn mt (y) q L w( p,q) m=1 t=−1 π π ∞  ∞  q q q = |y| p −1 |x| p −1 (−1)δm sin mx 01 |cm,n mt |Dn mt (y) d xd y −π −π m=1 t=−1 π π ∞  ∞  q −1 q ≤ |y| p (−1)δm sin mx 01 |cm,n mt |Dn mt (y) d xd y. (37) −π −π m=1 t=−1 By the Khintchine inequality (see e.g. [2, Rem. 1.4]) we have for any real sequence {sk } ∈ l2 and the system of Rademacher functions {rn (t)} that 1  ∞  ∞ q q 2 sk rk (t) q sk2 , 0 k=1 k=1 whence 1 π  ∞ ∞  q rm (t) sin mx 01 |cm,n mt |Dn mt (y) d xdt 0 −π m=1 t=−1 1  ∞ ∞  q  rm (t) 01 |cm,n mt |Dn mt (y) dt 0 m=1 t=−1  ∞   ∞ 2 q 2  01 |cm,n mt |Dn mt (y) , (38) m=1 t=−1 60 Page 28 of 30 Journal of Fourier Analysis and Applications (2023) 29:60 whenever the series on the right-hand side converges. Note that by the Minkowski 1 ((m+t)2 +3 m+t) p inequality and the fact that Dn mt  L q 2 p , we have w( p,q) π q ∞    ∞ q −1 2 01 |cm,n mt |Dn mt (y) 2 |y| p dy −π m=1 t=−1 ∞   ∞ 2 q  2  01 |cm,n mt |Dn mt (y)  q/2 L w( p/2,q/2) m=1 t=−1 ∞   ∞ 2 q   01 |cm,n mt |Dn mt (y)  q 2   L w( p,q) m=1 t=−1 ∞  ∞  q 2 m −2γ 2−m −3m 2m(m+1) + 2−((m+t) +3(m+t)) 2((m+t) +3m+t) 2 2 2 2  m=1 t=0 ∞  2 q − 2  m p < ∞. (39) m=1  ∞ Thus, by (38) and (39), for almost all t, the sum ∞ m=1 rm (t) sin mx t=−1  |cm,n mt 01 |Dn mt (y) converges for almost all y uniformly in x, and moreover, (38) and (39) imply that π π  ∞ ∞ q −1 q |y| p rm (t) sin mx 01 |cm,n mt |Dn mt (y) d xd y < ∞ −π −π m=1 t=−1 for almost all t (denote this set by E ⊂ (0, 1)). Taking any t0 ∈ E \{k2−l }k,l∈N,k<2l , so that rm (t0 ) = ±1 for all m, and setting {δm } according to the equality (−1)δm = rm (t0 ), we obtain in light of (37) that S2 < ∞.   Acknowledgements I am grateful to Mikhail Dyachenko and Sergey Tikhonov for acquainting me with this topic and for the many issues I have learnt from our conversations. I also thank Erlan Nursultanov, Miquel Saucedo, and the anonymous referees for their valuable comments and suggestions. Funding Open Access Funding provided by Universitat Autonoma de Barcelona. Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article’s Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article’s Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/. Journal of Fourier Analysis and Applications (2023) 29:60 Page 29 of 30 60 References 1. 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