Spaces and Its Application
Der-Chen Chang, Dachun Yang and Yuan Zhou
Abstract. Let p ∈ (0, 1]. In this paper, the authors prove that a sublinear operator T (which is originally defined on smooth functions with compact support) can be extended as a bounded sublinear operator from product Hardy spaces Hp(Rn× Rm) to some quasi-Banach
space B if and only if T maps all (p, 2, s1, s2)-atoms into uniformly bounded elements of B.
Here s1≥ bn(1/p − 1)c and s2≥ bm(1/p − 1)c. As usual, bn(1/p − 1)c denotes the maximal
integer no more than n(1/p−1). Applying this result, the authors establish the boundedness of the commutators generated by Calder´on-Zygmund operators and Lipschitz functions from the Lebesgue space Lp(Rn× Rm) with some p > 1 or the Hardy space Hp(Rn× Rm) with
some p ≤ 1 but near 1 to the Lebesgue space Lq(Rn× Rm) with some q > 1.
1. Introduction
The theory of Calder´on-Zygmund operators and Hardy spaces on product spaces has been studied by many mathematicians extensively in the past thirty years, see, for example, [8, 9, 11, 12, 18, 19, 27, 28]. Recently, Ferguson and Lacey [13] characterized the product BMO (R2
+× R2+) by the nested commutator determined by the one-dimensional Hilbert
transform in the jth variable, j = 1, 2. Motivated by this, Chen, Han and Miao in [6] established the boundedness on H1(Rn× Rm) of bi-commutators of fractional integrals
with BMO functions. The boundedness on H1(Rn× Rm) of the Marcinkiewicz integral and its commutator with Lipschitz function was also established in [27].
To establish the boundedness of operators on Hardy spaces on Rn and Rn× Rm, one
usually appeals to the atomic decomposition characterization of Hardy spaces, which means that a function or distribution in Hardy spaces can be represented as a linear combination of atoms; see [7, 21] and [3, 5] respectively. Then, the boundedness of linear operators on Hardy spaces can be deduced from their behavior on atoms in principle.
However, Meyer [22, p. 513] (see also [2, 15]) gave an example of f ∈ H1(Rn), whose norm cannot be achieved by its finite atomic decompositions via (1, ∞)-atoms. Based on
2000 Mathematics Subject Classification: Primary: 42B20; Secondary: 42B30, 42B25, 47B47.
Key words and phrases: product space; Hardy space; Lebesgue space; sublinear operator; commutator;
Calder´on-Zygmund operator; Lipschitz function.
The first author is partially supported by a research grant from United States Army Research Office and a competitive research grant at Georgetown University. The second (corresponding) author is partially supported by National Science Foundation for Distinguished Young Scholars (Grant No. 10425106) and NCET (Grant No. 04-0142) of Ministry of Education of China.
this fact, Bownik [2, Theorem 2] constructed a surprising example of a linear functional defined on a dense subspace of H1(Rn), which maps all (1, ∞)-atoms into bounded scalars, but yet cannot extend to a bounded linear functional on the whole H1(Rn). This implies
that it cannot guarantee the boundedness of linear operator T from Hp(Rn) with p ∈
(0, 1] to some quasi-Banach space B only proving that T maps all (p, ∞)-atoms into uniformly bounded elements of B. This phenomenon has also essentially already been observed by Y. Meyer in [23, p. 19]. Moreover, motivated by this, Yabuta [30] gave some sufficient conditions for the boundedness of T from Hp(Rn) with p ∈ (0, 1] to Lq(Rn)
with q ≥ 1 or Hq(Rn) with q ∈ [p, 1]. However, these conditions are not necessary. In [28], a boundedness criterion was established as follows: a sublinear operator T (which is originally defined on smooth functions with compact support) extends to a bounded sublinear operator from Hp(Rn) with p ∈ (0, 1] to some quasi-Banach spaces B if and
only if T maps all (p, 2)-atoms into uniformly bounded elements of B. This result shows the structure difference between atomic characterization of Hp(Rn) via (p, 2)-atoms and (p, ∞)-atoms. This result is generalized to spaces of homogeneous type in [29].
The purpose of this paper is two folds. We first generalize the boundedness criterion on Rn in [28] to product Hardy spaces on Rn× Rm. Precisely, we prove that a sublinear
operator T (which is originally defined on smooth functions with compact support) extends to a bounded sublinear operator from Hp(Rn× Rm) with p ∈ (0, 1] to some quasi-Banach
spaces B if and only if T maps all (p, 2)-atoms into uniformly bounded elements of B. Invoking this result and motivated by [6, 13, 27], we then establish the boundedness of the commutators generated by Calder´on-Zygmund operators and Lipschitz functions from the Lebesgue space Lp(Rn× Rm) with some p > 1 or the Hardy space Hp(Rn× Rm) with
some p ≤ 1 but near 1 to the Lebesgue space Lq(Rn× Rm) with some q > 1.
To state the main results, we first recall some notation and notions on product Hardy spaces. For n, m ∈ N, denote by S(Rn× Rm) the space of Schwartz functions on Rn× Rm
and by S0(Rn× Rm) its dual space. Let D(Rn× Rm) be the space of all smooth functions
on Rn× Rm with compact support. For s
1, s2 ∈ Z+, let Ds1, s2(Rn× Rm) be the set of all functions f ∈ D(Rn× Rm) with vanishing moments up to order s
1 with respect
to the first variable and order s2 with respect to the second variable. More precisely, if
f ∈ D(Rn× Rm), then for α1∈ Zn+ and α2 ∈ Zm+ with |α1| ≤ s1 and |α2| ≤ s2, one has
Z Rn f (x1, x2)xα11dx1 = 0 for all x2 ∈ Rm, Z Rm f (x1, x2)xα22dx2= 0 for all x1 ∈ Rn.
For s1, s2 ∈ Z+ and σ1, σ2 ∈ [0, ∞), we denote by Ds1, s2; σ1, σ2(Rn× Rm) the space
Ds1, s2(Rn× Rm) endowed with the norm
kf kDs1, s2; σ1, σ2(Rn×Rm)≡ sup
x1∈Rn, x2∈Rm
In articles [3, 4, 5], Chang and Fefferman introduced the following atoms and atomic Hardy spaces on the product space Rn× Rm.
Definition 1.1. Let p ∈ (0, 1], s1 ≥ bn(1/p − 1)c and s2 ≥ bm(1/p − 1)c. A function a
supported in an open set Ω ⊂ Rn× Rm with finite measure is said to be a (p, 2, s 1, s2
)-atom provided that
(AI) a can be written as a =PR∈M(Ω)aR, where M(Ω) denotes all the maximal dyadic
subrectangles of Ω and aRis a function satisfying that
(i) aRis supported on 2R = 2I × 2J, which is a rectangle with the same center as
R and whose side length is 2 times that of R,
(ii) aR satisfies the cancelation conditions that
Z 2I aR(x1, x2)xα11dx1= 0 f or all x2∈ 2J and |α1| ≤ s1, Z 2J aR(x1, x2)xα22dx2 = 0 f or all x1∈ 2I and |α2| ≤ s2;
(AII) a satisfies the size conditions that kakL2(Rn×Rm) ≤ |Ω|1/2−1/p and X R∈M(Ω) kaRk2L2(Rn×Rm) 1/2 ≤ |Ω|1/2−1/p.
Definition 1.2. Let p ∈ (0, 1], s1 ≥ bn(1/p − 1)c and s2 ≥ bm(1/p − 1)c. A distribution f ∈ S0(Rn× Rm) is said to be an element in Hp, 2, s1, s2(Rn× Rm) if there exist a sequence
{λk}k∈N ⊂ C and (p, 2, s1, s2)-atoms {ak}k∈N such that f =
P
k∈Nλkak in S0(Rn× Rm)
with Pk∈N|λk|p < ∞. Moreover, define the quasi-norm of f ∈ Hp, 2, s1, s2(Rn× Rm)
by kf kHp, 2, s1, s2(Rn×Rm) ≡ inf{( P
k∈N|λk|p)1/p}, where the infimum is taken over all the
decompositions as above.
It is well known that Hp, 2, s1, s2(Rn× Rm) = Hp, 2, t1, t2(Rn× Rm) with equivalent norms when s1, t1 ≥ bn(1/p − 1)c and s2, t2 ≥ bm(1/p − 1)c; see [3, 4, 5, 10, 17]. Thus,
we denote Hp, 2, s1, s2(Rn× Rm) simply by Hp(Rn× Rm).
Recall that a quasi-Banach space B is a vector space endowed with a quasi-norm k · kB which is nonnegative, non-degenerate (i. e., kf kB = 0 if and only if f = 0), homogeneous,
and obeys the quasi-triangle inequality, i. e., there exists a constant C0 ≥ 1 such that for
all f, g ∈ B,
Definition 1.3. Let q ∈ (0, 1]. A quasi-Banach spaces Bq with the quasi-norm k · kBq is said to be a q-quasi-Banach space if k·kqBq satisfies the triangle inequality, i. e., kf +gkqBq ≤ kf kqBq+ kgkqBq for all f, g ∈ Bq.
We point out that by the Aoki theorem (see [1] or [16, p. 66]), any quasi-Banach space with the positive constant C0 as in (1.1) is essentially a q-quasi-Banach space with q =
blog2(2C0)c−1. From this, any Banach space is a 1-quasi-Banach space. Moreover, `q,
Lq(Rn× Rm) and Hq(Rn× Rm) with q ∈ (0, 1) are typical q-quasi-Banach spaces.
Let q ∈ (0, 1]. For any given q-quasi-Banach space Bq and linear space Y, an operator
T from Y to Bq is called to be Bq-sublinear if for any f, g ∈ Y and λ, ν ∈ C, we have
kT (λf + νg)kBq ≤ ³
|λ|qkT (f )kqBq+ |ν|qkT (g)kqBq
´1/q
and kT (f ) − T (g)kBq ≤ kT (f − g)kBq; see [28, 29]. Obviously, if T is linear, then T is
Bq-sublinear. Moreover, if Bq is a space of functions and T is sublinear in the classical
sense, then T is also Bq-sublinear.
The following is one of main results in this paper, which generalizes the main result in [28] to product Hardy spaces.
Theorem 1.1. Let p ∈ (0, 1], q ∈ [p, 1] and Bq be a q-quasi-Banach space. Suppose
that s1 ≥ bn(1/p − 1)c and s2 ≥ bm(1/p − 1)c. Let T be a Bq-sublinear operator from Ds1, s2(Rn× Rm) to Bq. Then T can be extended as a bounded Bq-sublinear operator from
Hp(Rn× Rm) to B
q if and only if T maps all (p, 2, s1, s2)-atoms in Ds1, s2(Rn× Rm) into
uniformly bounded elements of Bq.
Theorem 1.1 further complements the proofs of Theorem 1 in [11] and a theorem in [9], whose proof is presented in Section 2 below. The necessity of Theorem 1.1 is ob-vious. To prove the sufficiency, for p ∈ (0, 1], s1 ≥ bn(1/p − 1)c, s2 ≥ bm(1/p − 1)c
and f ∈ Ds1, s2(Rn× Rm), we first prove that f has an atomic decomposition which
converges in Ds1, s2; σ1, σ2(Rn× Rm) for some σ1 ∈ (max{n/p, n + s}, n + s + 1) and
σ2 ∈ (max{n/p, n + s}, n + s + 1) (Lemma 2.3), and then extend T to the whole
Ds1, s2; σ1, σ2(Rn× Rm) boundedly (Lemma 2.4). Finally, we continuously extend T to the whole Hp(Rn× Rm) by using the density of D
s1, s2(Rn× Rm) in Hp(Rn× Rm). Recall that a function a is said to be a rectangular (p, 2, s1, s2)-atom if
(R1) supp a ⊂ R = I × J, where I and J are cubes in Rn and Rm, respectively; (R2) RRna(x1, x2)x1α1dx1 = 0 for all x2 ∈ Rmand |α1| ≤ s1, and
R
Rma(x1, x2)xα22dx2= 0 for all x1 ∈ Rn and |α
2| ≤ s2;
As a consequence of Theorem 1.1, we obtain the following result which includes a fractional version of Theorem 1 in [11] and is known to have many applications in harmonic analysis.
Corollary 1.1. Let q0 ∈ [2, ∞) and T be a bounded sublinear operator from L2(Rn× Rm)
to Lq0(Rn× Rm). Let p ∈ (0, 1] and 1/q − 1/p = 1/q
0− 1/2. If there exist positive
con-stants C and δ such that for all rectangular (p, 2, s1, s2)-atoms a supported in R and all
γ ≥ 8 max{n1/2, m1/2},
Z
(Rn×Rm)\ eRγ
|T a(x1, x2)|qdx1dx2 ≤ Cγ−δ,
where eRγ denotes the γ-fold enlargement of R, then T can be extended as a bounded
sublinear operator from Hp(Rn× Rm) to Lq(Rn× Rm).
The proof of Corollary 1.1 is given in Section 2 below. We point out that if q0 = 2 and
T is linear, then Corollary 1.1 is just Theorem 1 in [11]. Moreover, there exists a gap in
the proof of Theorem 1 in [11] (so is the proof of a theorem in [9]), namely, it was not clear in [11] how to deduce the boundedness of the considered linear operator T on the whole Hardy space Hp(Rn× Rm) from its boundedness uniformly on atoms. Our Theorem 1.1
here seals this gap.
Remark 1.1. Using Corollary 1.1, we now give affirmative answers to the questions in Remark 4.2 and Remark 4.3 of [27]. We use the same notation and notions as in [27]. Particularly, denote by µΩ the Marcinkiewicz integral operator on Rn× Rm with kernel
Ω ∈ Lip (α1, α2; Sn−1, Sm−1), here α1, α2∈ (0, 1]. If max{n/(n+α1), m/(m+α2)} < p ≤ 1,
then in Remark 4.2 of [27], we proved that for all (p, 2, 0, 0) atoms a, kµΩ(a)kLp(Rn×Rm).1. Moreover, let b ∈ Lip (β1, β2; Rn× Rm) with β
1, β2 ∈ (0, 1] satisfying β1/n = β2/m and
Cb(µΩ) be the commutator of b and µΩ. If 1/q = 1/p − β1/n and
max{n/(n + α1), m/(m + α2)} < p ≤ 1,
then in Remark 4.3 of [27], we proved that for all (p, 2, 0, 0) atoms a,
kCb(µΩ)(a)kLq(Rn×Rm).1.
However, in [27], it is not clear how to obtain the boundedness of µΩ from Hp(Rn× Rm)
to Lp(Rn× Rm) and boundedness of C
b(µΩ) from Hp(Rn× Rm) to Lq(Rn× Rm) by these
known facts. Applying Theorem 1.1 here, we now obtain these desired boundedness, and hence answer the questions in Remark 4.2 and Remark 4.3 of [27].
Now we turn to the boundedness of commutators generated by Lipschitz functions and Calder´on-Zygmund operators. We first introduce the notion of Lipschitz functions
on Rn× Rm. Let α ∈ (0, 1]. A function b on Rn is said to belong to Lip (α; Rn) if there
exists a positive constant C such that for all x, x0∈ Rn,
|b(x) − b(x0)| ≤ C|x − x0|α.
Obviously, a function in the space Lip (α; Rn) is not necessary bounded. For example,
|x|α∈ Lip (α; Rn), but |x|α6∈ L∞(Rn).
Definition 1.4. Let α1, α2 ∈ (0, 1]. A function f on Rn× Rm is said to belong to
Lip (α1, α2; Rn× Rm), if there exists a positive constant C such that for all x1, y1 ∈ Rn
and x2, y2 ∈ Rm,
(1.2) |[f (x1, x2) − f (x1, y2)] − [f (y1, x2) − f (y1, y2)]| ≤ C|x1− y1|α1|x
2− y2|α2.
The minimal constant C satisfying (1.2) is defined to be the norm of f in the space Lip (α1, α2; Rn× Rm) and denoted by kf k
Lip (α1, α2; Rn×Rm).
We remark that a function in the space Lip (α1, α2; Rn× Rm) is also not necessary to
be bounded. In fact, if f1 ∈ Lip (α1; Rn) and f2 ∈ Lip (α2; Rm), then it is easy to check
f1(x1)f2(x2) ∈ Lip (α1, α2; Rn× Rm).
In this paper, we consider a class of Calder´on-Zygmund operators T on Rn× Rm, whose kernel K is a continuous function on (Rn× Rn× Rm× Rm) \ {(x
1, y1, x2, y2) : x1=
y1 or x2 = y2} and satisfies that there exist positive constants C and ²1, ²2 ∈ (0, 1] such
that
(K1) for all x1 6= y1 and x26= y2,
|K(x1, y1, x2, y2)| ≤ C 1
|x1− y1|n
1
|x2− y2|m
; (K2) for all x1 6= y1, x26= y2, z1 ∈ Rn and |y1− z1| ≤ |x1− y1|/2,
|K(x1, y1, x2, y2) − K(x1, z1, x2, y2)| ≤ C |y1− z1| ²1
|x1− y1|n+²1
1
|x2− y2|m;
(K3) for all x1 6= y1, x26= y2, z2 ∈ Rm and |y2− z2| ≤ |x2− y2|/2,
|K(x1, y1, x2, y2) − K(x1, y1, x2, z2)| ≤ C|x 1 1− y1|n
|y2− z2|²2
|x2− y2|m+²2;
(K4) for all x1 6= y1, x2 6= y2, z1 ∈ Rn, z2 ∈ Rm, |y1− z1| ≤ |x1− y1|/2 and |y2− z2| ≤
|x2− y2|/2, |[K(x1, y1, x2, y2) − K(x1, z1, x2, y2)] − [K(x1, y1, x2, z2) − K(x1, z1, x2, z2)]| ≤ C |y1− z1|²1 |x1− y1|n+²1 |y2− z2|²2 |x2− y2|m+²2 .
The minimal constant C satisfying (K1) through (K4) is denoted by kKk.
Let α1, α2 ∈ (0, 1], b ∈ Lip (α1, α2; Rn× Rm) and T be any Calder´on-Zygmund
oper-ator with kernel K satisfying the above conditions from (K1) to (K4). For any suitable function f and (x1, x2) ∈ Rn× Rm, define the commutator [b, T ] by
(1.3) [b, T ](f )(x1, x2) =
Z
Rn×Rm
K(x1, y1, x2, y2)
×[b(x1, x2) − b(x1, y2) − b(y1, x2) + b(y1, y2)]f (y1, y2) dy1dy2.
The following result gives the boundedness of the commutator [b, T ] on Lebesgue spaces.
Theorem 1.2. Let ²1, ²2, α1, α2 ∈ (0, 1], α1/n = α2/m, p ∈ (1, n/α1) and 1/q = 1/p −
α1/n. Let b ∈ Lip (α1, α2; Rn× Rm), T be a Calder´on-Zygmund operator whose kernel
K satisfies the conditions from (K1) to (K4), and [b, T ] be the commutator as in (1.3). Then there exists a positive constant C independent of kbkLip (α1, α2; Rn×Rm) and kKk such
that for all f ∈ Lp(Rn× Rm),
k[b, T ](f )kLq(Rn×Rm)≤ CkKkkbkLip (α1, α2; Rn×Rm)kf kLp(Rn×Rm).
Here is another main result of this paper, whose proof depends on Corollary 1.1. Theorem 1.3. Let 0 < α1 ≤ min{n/2, 1}, α1/n = α2/m, ²1, ²2∈ (0, 1],
(1.4) max{n/(n + ²1), n/(n + α1), m/(m + ²2), m/(m + α2)} < p ≤ 1
and 1/q = 1/p − α1/n. Assume that b ∈ Lip (α1, α2; Rn× Rm). Let T be a
Calder´on-Zygmund operator whose kernel K satisfies the conditions (K1) through (K4), and [b, T ] be the commutator defined in (1.3). Then there exists a positive constant C independent of kbkLip (α1, α2; Rn×Rm) and kKk such that for all f ∈ Hp(Rn× Rm),
k[b, T ](f )kLq(Rn×Rm)≤ CkKkkbkLip (α1, α2; Rn×Rm)kf kHp(Rn×Rm). The proofs of Theorem 1.2 and Theorem 1.3 are presented in Section 3.
We finally make some conventions. Throughout this paper, let N = {1, 2, · · · } and Z+= N ∪ {0}. We always use C to denote a positive constant that is independent of main
parameters involved but whose value may differ from line to line. We use f .g to denote
2. Proofs of Theorem 1.1 and Corollary 1.1
As a matter of convenience, in this section, we denote n and m, respectively, by n1
and n2. For i = 1, 2 and si ∈ Z+, denote by Dsi(Rni) the set of all smooth functions with compact support and vanishing moments up to order si. Then there exist functions
ψ(i) ∈ D
si(Rni) and ϕ(i) ∈ S(Rni) such that
(i) supp ψ(i) ⊂ B(i)(0, 1), dψ(i) ≥ 0 and dψ(i)(ξi) ≥ 1
2 if 12 ≤ |ξi| ≤ 2, where and in what
follows B(i)(0, r
i) ≡ {xi∈ Rni : |xi| < ri} and dψ(i) denotes the Fourier transform of ψ(i);
(ii) supp dϕ(i)⊂ {ξ
i∈ Rni : 1/2 ≤ |ξi| ≤ 2} and dϕ(i)≥ 0;
(iii) sup{dϕ(i)(ξi) : 3/5 ≤ |ξi| ≤ 5/3} > C for some positive constant C;
(iv)R0∞ψd(i)(t
iξi)dϕ(i)(tiξi)dttii = 1 for all ξi ∈ Rni\ {0}.
Such ψ(i) and ϕ(i) can be constructed by a slight modification of Lemma (1.2) of [14];
see also Lemma (5.12) in [14] for a discrete variant. Then by an argument similar to the proofs of Theorem (1.3) and Theorem 1 in Appendix of [14], we have that for all
f ∈ S(Rn1 × Rn2) and (x 1, x2) ∈ Rn1 × Rn2, (2.1) f (x1, x2) = Z ∞ 0 Z ∞ 0 (ψt1, t2 ∗ ϕt1, t2 ∗ f )(x1, x2) dt1 t1 dt2 t2
in both L2(Rn1 × Rn2) and pointwise, where and in what follows, for any i = 1, 2,
φ(i) ∈ S(Rni), x
i ∈ Rni and ti ∈ (0, ∞), we always let φ(i)ti (xi) ≡ t
−ni i φ(i)(t−1i xi) and φt1, t2(x1, x2) ≡ φ (1) t1 (x1)φ (2)
t2 (x2). For any set E ⊂ (R
n× Rm), set E{ ≡ (Rn× Rm) \ E.
Lemma 2.1. Let si ∈ Z+, ψ(i) ∈ D
si(Rni) and ϕ(i) ∈ S(Rni) satisfy the above conditions
(i) through (iv), where i = 1, 2. Let 0 < σi < σ0i < ni+ si+ 1 for i = 1, 2. Then for any
f ∈ Ds1, s2(Rn1× Rn2), there exists a positive constant C such that for all ²1, ²2 ∈ (0, 1)
and L1, L2∈ (1, ∞), sup (x1, x2)∈Rn1×Rn2 (1 + |x1|)σ1(1 + |x2|)σ2 × µZ ²1 0 Z ∞ 0 + Z ∞ L1 Z ∞ 0 + Z ∞ 0 Z ²2 0 + Z ∞ 0 Z ∞ L2 ¶ Z Rn1×Rn2|(ϕt1, t2∗ f )(y1, y2)| ×|ψt1, t2(x1− y1, x2− y2)| dy1dy2 dt1 t1 dt2 t2 ≤ C£²1+ ²2+ (L1)σ1−n1−s1−1+ (L2)σ2−n2−s2−1 ¤ , (2.2) sup (x1, x2)∈Rn1×Rn2 (1 + |x1|)σ1(1 + |x 2|)σ2 × Z L1 0 Z ∞ 0 Z [B(1)(0, 2L1)]{×Rn2 |(ϕt1, t2∗ f )(y1, y2)| ×|ψt1, t2(x1− y1, x2− y2)| dy1dy2dtt11dtt22 ≤ C(L1)σ1−σ 0 1
and (2.2) with L1, σ1, n1, s1 and B(1) replaced, respectively, by L2, σ2, n2, s2 and B(2).
In order to prove Lemma 2.1, we need the following technical lemma. For i = 1, 2,
ui≥ 0, let Sui(Rni) ≡ ½ ϕ ∈ S(Rni) : Z Rniϕ(xi)x α i dxi = 0, |α| ≤ ui ¾ .
For any s1, s2 ∈ Z−1 ≡ N ∪ {0, −1}, we denote by Ss1, s2(Rn1 × Rn2) the space of functions in S(Rn1× Rn2) with the vanishing moments up to order s
1 in the first variable and order
s2 in the second variable, where we say that f ∈ S(Rn1× Rn2) has vanishing moments up to order −1 in the first or second variable, if f has no vanishing moment with respect to that variable.
Lemma 2.2. Let si ∈ Z−1, ui ∈ Z−1, σi ∈ [0, ∞) and ϕ(i) ∈ Sui(Rni) for i = 1, 2. For
any f ∈ Ss1, s2(Rn1× Rn2), there exists a positive constant C such that
(i) if u1 > −1, then for all t1∈ (0, 1] and (x1, x2) ∈ Rn1 × Rn2,
|(ϕt(1)1 ∗1f )(x1, x2)| ≤ Ct1u1+1(1 + |x1|)−σ1(1 + |x2|)−σ2,
where and in what follows (ϕ(1)t1 ∗1f )(x1, x2) ≡
R
Rn1ϕ(1)t1 (y1)f (x1− y1, x2) dy1;
(ii) if s1> −1, then for all t1 ∈ [1, ∞) and (x1, x2) ∈ Rn1 × Rn2,
|(ϕ(1)t1 ∗1f )(x1, x2)| ≤ Ct−n1−s1−1 1 µ 1 +|x1| t1 ¶−σ1 (1 + |x2|)−σ2;
(iii) if u1, u2> −1, then for all t1, t2 ∈ (0, 1] and (x1, x2) ∈ Rn1× Rn2,
|(ϕt1, t2∗ f )(x1, x2)| ≤ Ct
u1+1
1 tu22+1(1 + |x1|)−σ1(1 + |x2|)−σ2;
(iv) if u1, s2 > −1, then for all t1∈ (0, 1], t2∈ [1, ∞) and (x1, x2) ∈ Rn1 × Rn2,
|(ϕt1, t2 ∗ f )(x1, x2)| ≤ Ct u1+1 1 t−n2 2−s2−1(1 + |x1|)−σ1 µ 1 +|x2| t2 ¶−σ2 ;
(v) if s1, u2 > −1, then for all t1∈ [1, ∞), t2 ∈ (0, 1] and (x1, x2) ∈ Rn1 × Rn2,
|(ϕt1, t2 ∗ f )(x1, x2)| ≤ Ct −n2−s2−1 1 tu22+1 µ 1 +|x1| t1 ¶−σ1 (1 + |x2|)−σ2;
(vi) if s1, s2 > −1, then for all t1, t2 ∈ [1, ∞) and (x1, x2) ∈ Rn1 × Rn2,
|(ϕt1, t2∗ f )(x1, x2)| ≤ Ct −n1−s1−1 1 t−n2 2−s2−1 µ 1 +|x1| t1 ¶−σ1µ 1 +|x2| t2 ¶−σ2 .
Proof. To prove Lemma 2.2, we use some ideas in the proofs of Lemma 2 and Lemma 4 in Appendix (III) of [14]. To prove (i), byRRn1ϕ(1)(x 1)xα1dx1 = 0 for |α| ≤ u1, we have (ϕt1 ∗1f )(x1, x2) = Z Rn1 ϕ (1) t1 (y1) f(x1− y1, x2) − X |γ|≤u1 1 γ!y γ 1(Dγ1f )(x1, x2) dy1 = Z |y1|<|x1|/2 ϕ(1)t1 (y1) f(x1− y1, x2) − X |γ|≤u1 1 γ!y γ 1(D1γf )(x1, x2) dy1 + Z |y1|≥|x1|/2 · · · ≡ I1+ I2.
For the estimation of I1, noticing that |x1|/2 ≤ |x1 − z1| ≤ 2|x1| for |z1| ≤ |x1|/2, by
|y1| < |x1|/2 and the mean value theorem, we obtain
(2.3) ¯ ¯ ¯ ¯ ¯ ¯f (x1− y1, x2) − X |γ|≤u1 1 γ!y γ 1(D1γf )(x1, x2) ¯ ¯ ¯ ¯ ¯ ¯ = sup |γ|=u1+1 sup |z1|≤|x1−y1| |(D1γf )(x1− z1, x2)||y1|u1+1 .|y1|u1+1 sup |z1|≤|x1|/2 (1 + |x1− z1|)−σ1(1 + |x2|)−σ2 .|y1|u1+1(1 + |x1|)−σ1(1 + |x2|)−σ2, where γ = (γ1, · · · , γn1) ∈ Zn+1, x1 = (x11, · · · , xn11) and D1γ = (∂x∂1 1) γ1· · · ( ∂ ∂xn11 )γn1. This leads to that |I1| .(1 + |x1|)−σ1(1 + |x2|)−σ2 Z |y1|<|x1|/2 |y1|u1+1|ϕ(1)t1 (y1)| dy1 .tu1+1 1 (1 + |x1|)−σ1(1 + |x2|)−σ2 Z Rn1|y1| u1+1|ϕ(1)(y 1)| dy1 .tu1+1 1 (1 + |x1|)−σ1(1 + |x2|)−σ2.
To estimate I2, similarly to (2.3), we have
(2.4) ¯ ¯ ¯ ¯ ¯ ¯f (x1− y1, x2) − X |γ|≤s1 1 γ!y γ 1(Dγ1f )(x1, x2) ¯ ¯ ¯ ¯ ¯ ¯.|y1| u1+1(1 + |x 2|)−σ2. If |x1| ≥ 1 and σ1> 0, by |x1|−1 ≤ 2(1 + |x
1|)−1 and (2.4), for all t1 ∈ (0, 1], we have
|I2| .(1 + |x2|)−σ2
Z
|y1|≥|x1|/2
.(1 + |x2|)−σ2tu11+1 Z |y1|≥|x1|/(2t1) |y1|u1+1|ϕ(1)(y1)| dy1 .tu1+1 1 (1 + |x2|)−σ2 Z ∞ |x1|/(2t1) r−σ1−1 1 dr1 .tu1+1 1 (1 + |x1|)−σ1(1 + |x2|)−σ2. If |x1| ≤ 1 or σ1 = 0, by (2.4), |I2|.tu11+1(1 + |x2|)−σ2 Z Rn1|y1| u1+1|ϕ(1)(y 1)| dy1.tu11+1(1 + |x1|)−σ1(1 + |x2|)−σ2.
Thus combining the estimations for I1 and I2 yields (i).
To prove (ii), since ϕ(1) ∈ S
0(Rn1) and f ∈ Ss1, s2(Rn1 × Rn2), we have (ϕ(1)t1 ∗1f )(x1, x2) = Z Rn1 ϕ(1) t1 (y1) − X |γ|≤s1 1 γ!(y1− x1) γ(Dγ 1ϕ(1)t1 )(x1) f(x1− y1, x2) dy1 = Z |x1−y1|<|x1|/2 ϕ(1) t1 (y1) − X |γ|≤s1 1 γ!(y1− x1) γ(Dγ 1ϕ(1)t1 )(x1) f(x1− y1, x2) dy1 + Z |x1−y1|≥|x1|/2 · · · ≡ J1+ J2.
On the estimation for J1, notice that if |z1| ≤ |x1−y1| < |x1|/2, then |x1|/2 ≤ |x1−z1| ≤
2|x1|. By this and ϕ(1)∈ S0(Rn1), we have
¯ ¯ ¯ ¯ ¯ ¯ϕ (1) t1 (y1) − X |γ|≤s1 1 γ!(y1− x1) γ(Dγ 1ϕ(1)t1 )(x1) ¯ ¯ ¯ ¯ ¯ ¯ . sup |γ|=s1+1 sup |z1|≤|x1−y1| |(D1γϕ(1)t1 )(x1− z1)||x1− y1|s1+1 .t−n1−s1−1 1 sup |z1|≤|x1−y1| µ 1 +|x1− z1| t1 ¶−σ1 |x1− y1|s1+1 .t−n1−s1−1 1 µ 1 +|x1| t1 ¶−σ1 |x1− y1|s1+1. Thus, applying (2.5) |f (x1− y1, x2)|.(1 + |x1− y1|)−n1−s1−2(1 + |x2|)−σ2, we further have |J1| .t−n1−s1−1 1 (1 + |x2|)−σ2 Z Rn1 µ 1 +|x1| t1 ¶−σ1 |x1− y1|s1+1 (1 + |x1− y1|)n1+s1+2 dy1
.t−n1−s1−1 1 µ 1 +|x1| t1 ¶−σ1 (1 + |x2|)−σ2.
To estimate J2, if |x1| > 1 and σ1> 0, using an estimate similar to (2.5) and the estimation
that ¯ ¯ ¯ ¯ ¯ ¯ϕ (1)(y 1) − X |γ|≤s1 1 γ!(y1− x1) γ(Dγ 1ϕ(1)t1 )(x1) ¯ ¯ ¯ ¯ ¯ ¯.t −n1−s1−1 1 |x1− y1|s1+1, we obtain |J2| . Z |y1−x1|≥|x1|/2 (1 + |x2|)−σ2t−n1 1−s1−1 |x1− y1|s1+1 (1 + |x1− y1|)σ1+n1+s1+1dy1 .t−n1−s1−1 1 (1 + |x2|)−σ2 Z ∞ |x1|/2 r−σ1−1 1 dr1 .t−n1−s2−1 1 (1 + |x1|)−σ1(1 + |x2|)−σ2,
where in the last step, we used the fact that |x1|−σ1.(1 + |x1|/t1)−σ1 for t1 ≥ 1. If |x1| ≤ 1
or σ1= 0, by (2.5), we then have |J2| .(1 + |x2|)−σ2t−n1−s1−1 1 Z ∞ 0 rn1+s1 1 (1 + r1)n1+s1+2 dr1 .t−n1−s1−1 1 µ 1 +|x1| t1 ¶−σ1 (1 + |x2|)−σ2.
This gives (ii).
To prove (iii), by an argument similar to (i), we obtain that for all t2 ∈ (0, 1],
(2.6) |(ϕ(2)t2 ∗2f )(x1, x2)|.t2u2+1(1 + |x1|)−σ1(1 + |x2|)−σ2,
where and in what follows (ϕ(2)t2 ∗2f )(x1, x2) ≡ RRn2ϕ(2)t2 (y2)f (x1, x2 − y2) dy2. Thus, if |y1| < |x1|/2, then by the mean value theorem, (2.6) and the fact that |x1− z1| ∼ |x1| for
|z1| ≤ |x1|/2, we have (2.7) ¯ ¯ ¯ ¯ ¯ ¯(ϕ (2) t2 ∗2f )(x1− y1, x2) − X |γ|≤u1 1 γ!(y1− x1) γ∂γ 1(ϕ (2) t2 ∗2f )(x1, x2) ¯ ¯ ¯ ¯ ¯ ¯ ≤ |y1|u1+1 sup |γ|=u1+1 sup |z1|≤|x1|/2 |(ϕ(2)t2 ∗2(Dγ1f ))(x1− z1, x2)| .tu2+1 2 |y1|u1+1(1 + |x1|)−σ1(1 + |x2|)−σ2.
(2.8) ¯ ¯ ¯ ¯ ¯ ¯(ϕ (2) t2 ∗2f )(x1− y1, x2) − X |γ|≤s1 1 γ!(y1− x1) γ∂γ 1(ϕ(2)t2 ∗2f )(x1, x2) ¯ ¯ ¯ ¯ ¯ ¯ .tu2+1 2 |y1|u1+1(1 + |x2|)−σ2. Noticing that (2.9) (ϕt1, t2 ∗ f )(x1, x2) = (ϕ (1) t1 ∗1(ϕ (2) t2 ∗2f ))(x1, x2),
replacing (2.3) and (2.4) respectively by (2.7) and (2.8), and repeating the proof of (i), we obtain (iii).
For (v), by (2.6), we have
|(ϕ(2)t2 ∗2f )(x1− y1, x2)|.t2u2+1(1 + |x2|)−σ2(1 + |x1− y1|)−n1−s1−2
for all t2 ∈ (0, 1]. Replacing (2.5) by this estimate, using (2.9) and repeating the proof of
(ii) lead to (v). A similar argument to (v) yields (iv). To obtain (vi), by an argument similar to (ii), we obtain
|(ϕt(2)2 ∗2f )(x1− y1, x2)|.t2−n1−s1−1(1 + |x1− y1|)−n1−s1−2
µ
1 +|x2|
t2
¶−σ2
for all t2 ∈ [1, ∞). Replacing (2.5) by this, using (2.9) and repeating the proof of (ii) leads
to (vi). This finishes the proof of Lemma 2.2.
Proof of Lemma 2.1. Let ²1 ∈ (0, 1). Notice that for all t1 ∈ (0, ∞), |y1| ≤ t1 and
x ∈ Rn1, we have t
1+ |x1| ≤ 2(t1+ |x1− y1|). By this and Lemma 2.2 (iii) and (iv) , we
have that for any t1 ∈ (0, ²1), t2 ∈ (0, 1), |y1| < t1, |y2| < t2 and (x1, x2) ∈ Rn1× Rn2, (2.10) |(ϕt1, t2∗ f )(x1− y1, x2− y2)|.t1t2(1 + |x1|)
−σ1(1 + |x
2|)−σ2,
and that for any t1 ∈ (0, ²1], t2 ∈ [1, ∞), |y1| < t1, |y2| < t2 and (x1, x2) ∈ Rn1× Rn2,
(2.11) |(ϕt1, t2 ∗ f )(x1− y1, x2− y2)|.t1t
σ2−n2−s2−1
2 (1 + |x1|)−σ1(1 + |x2|)−σ2.
From this and σ2 < n2+ s2+ 1, it follows that
sup (x1, x2)∈Rn1×Rn2 (1 + |x1|)σ1(1 + |x2|)σ2 Z ²1 0 Z ∞ 0 Z Rn1×Rn2|ψt1, t2(y1, y2)| ×|(ϕt1, t2 ∗ f )(x1− y1, x2− y2)| dy1dy2dt1 t1 dt2 t2 . Z ²1 0 Z ∞ 0 Z Rn1×Rn2 1 1 + tn2+s2+2−σ2|ϕt1, t2(y1, y2)| dy1dy2dt1dt2
.²1.
Let L1 > 1. By Lemma 2.1 (v) and (vi), we have that for any t1 ∈ (L1, ∞), t2 ∈ (0, 1),
|y1| < t1, |y2| < t2 and (x1, x2) ∈ Rn1× Rn2,
(2.12) |(ϕt1, t2 ∗ f )(x1− y1, x2− y2)|.tσ1−n1−s1−1
1 t2(1 + |x1|)−σ1(1 + |x2|)−σ2,
and that for any t1 ∈ (L1, ∞), t2∈ [1, ∞), |y1| < t1, |y2| < t2 and (x1, x2) ∈ Rn1× Rn2,
(2.13) |(ϕt1, t2 ∗ f )(x1− y1, x2− y2)|.t
σ1−n1−s1−1
1 tσ22−n2−s2−1(1 + |x1|)−σ1(1 + |x2|)−σ2.
From this, (2.12), σ1< n1+ s1+ 1 and σ2 < n2+ s2+ 1, it follows that
sup (x1, x2)∈Rn1×Rn2 (1 + |x1|)σ1(1 + |x2|)σ2 Z ∞ L1 Z ∞ 0 Z Rn1×Rn2 |ϕt1, t2(y1, y2)| ×|(ϕt1, t2∗ f )(x1− y1, x2− y2)| dy1dy2dt1 t1 dt2 t2 . Z ∞ L1 Z ∞ 0 Z Rn1×Rn2|ϕt1, t2(y1, y2)| dy1dy2 dt1 tn1+s1+2−σ1 1 dt2 1 + tn2+s2+2−σ2 2 .(L1)σ1−n1−s1−1.
Using the symmetry, we then obtain the desired estimates for the cases ²2 ∈ (0, 1), L2 ∈
(1, ∞), (t1, t2) ∈ (0, ∞) × (0, ²2) or (t1, t2) ∈ (0, ∞) × (L2, ∞), which gives the first
inequality of Lemma 2.1.
To prove (2.2), notice that if |y1| > 2L1 > 2 and |x1− y1| < t1 < L1, we have |x1| >
|y1|−|x1−y1| > L1. Then by (2.10) through (2.13) with σireplaced by σi0 ∈ (σi, n1−s1−1),
we have sup (x1, x2)∈Rn1×Rn2 (1 + |x1|)σ1(1 + |x2|)σ2 Z L1 0 Z ∞ 0 Z [B(1)(0, 2L1)]{×Rn2|(ϕt1, t2 ∗ f )(y1, y2)| ×|ψt1, t2(x1− y1, x2− y2)| dy1dy2 dt1 t1 dt2 t2 . sup |x1|>L1, x2∈Rn2 (1 + |x1|)σ1−σ 0 1(1 + |x2|)σ2−σ20 Z ∞ 0 Z ∞ 0 Z Rn1×Rn2 |ψt1, t2(y1, y2)| × 1 1 + tn1−s1+2−σ01 1 1 + tn2−s2+2−σ02 dy1dy2dt1dt2 .(L1)σ1−σ 0 1,
which gives (2.2) and hence completes the proof of Lemma 2.1.
Let p ∈ (0, 1], si ≥ bni(1/p − 1)c and ϕ ∈ Ssi(Rni) such that (2.1) holds for i = 1, 2. For f ∈ S0(Rn1 × Rn2) and (x
1, x2) ∈ Rn1 × Rn2, we define
≡ ÃZ ∞ 0 Z ∞ 0 Z |y1−x1|<t1 Z |y2−x2|<t2 |(ϕt1t2 ∗ f )(y1, y2)|2dy1dy2 dt1 tn1+1 1 dt2 tn2+1 2 !1/2 .
It is well-known that f ∈ Hp(Rn1 × Rn2) if and only if f ∈ S0(Rn1 × Rn2) and S(f ) ∈
Lp(Rn1 × Rn2). Moreover,
kf kHp(Rn1×Rn2)∼ kS(f )kLp(Rn1×Rn2);
see [3, 4, 5, 10]. Using this fact, Lemma 2.1 and some ideas from [3, 4, 5, 10], we obtain the following conclusion.
Lemma 2.3. Let p ∈ (0, 1], si ≥ bni(1/p − 1)c and σi ∈ (max{ni + si, ni/p}, ni+ si+
1) for i = 1, 2. Then for any f ∈ Ds1, s2(Rn1× Rn2), there exist numbers {λk}k∈N ⊂ C and (p, 2, s1, s2)-atoms {ak}k∈N ⊂ Ds1, s2(Rn1 × Rn2) such that f =
P k∈Nλkak in Ds1, s2; σ1, σ2(Rn1× Rn2) and ©P k∈N|λk|p ª1/p ≤ Ckf kHp(Rn1×Rn2), where C is a positive constant independent of f .
Proof. We use R to denote the set of all dyadic rectangles in Rn1× Rn2. For k ∈ Z, let Ωk≡ {(x1, x2) ∈ Rn1× Rn2 : S(f )(x1, x2) > 2k}
and
e
Ωk≡ {(x1, x2) ∈ Rn1 × Rn2 : Ms(χΩk)(x1, x2) > 1/2},
where Ms denotes the strong maximal operator on Rn1 × Rn2. It is easy to see that Ω
k is
bounded set. In fact, observing that 1 + |xi| ≤ ti+ |xi| ∼ ti+ |yi| for |xi− yi| < ti and
ti ≥ 1, by Lemma 2.2 and ni+ si+ 1 − σi> 0, we have
[S(f )(x1, x2)]2 . Z 1 0 Z 1 0 Z |y1−x1|<t1 Z |y2−x2|<t2 (1 + |y1|)−2σ1(1 + |y2|)−2σ2dy1dy2dttn11 1 dt2 tn2 2 + Z ∞ 1 Z 1 0 Z |y1−x1|<t1 Z |y2−x2|<t2 (1 +|y1| t1 ) −2σ1(1 + |y 2|)−2σ2dy1dy2 dt1 t3n1+2s1+3 1 dt2 tn2 2 + Z 1 0 Z ∞ 1 Z |y1−x1|<t1 Z |y2−x2|<t2 (1 + |y1|)−2σ1(1 + |yt2| 2 ) −2σ2dy 1dy2 dttn11 1 dt2 t3n2+2s2+3 2 + Z ∞ 1 Z ∞ 1 Z |y1−x1|<t1 Z |y2−x2|<t2 (1 +|y1| t1 ) −2σ1(1 +|y2| t2 ) −2σ2dy 1dy2 × dt1 t2n1+s1+2 1 dt2 t3n2+2s2+3 2 .(1 + |x1|)−2σ1(1 + |x2|)−2σ2.
For each dyadic rectangle R = I × J, set
A(R) ≡ {(y1, y2, t1, t2) : (y1, y2) ∈ R, √n1|I| < t1≤ 2√n1|I|, √n2|J| < t2 ≤ 2√n2|J|},
and
Rk ≡ {R ∈ R : |R ∩ Ωk| ≥ 1/2, |R ∩ Ωk+1| < 1/2} .
Obviously, for each R ∈ R, there exists a unique k ∈ Z such that R ∈ Rk. From (2.1), for any (x1, x2) ∈ Rn1 × Rn2, it is easy to see that
f (x1, x2) = X k∈Z X R∈Rk Z A(R) ψt1, t2(x1− y1, x2− y2)(ϕt1, t2∗ f )(y1, y2) dy1dy2 dt1 t1 dt2 t2 . Let λk≡ C12k|Ωk|1/p and ak(x1, x2) ≡ λ−1k X R∈Rk Z A(R) ψt1, t2(x1− y1, x2− y2)(ϕt1, t2∗ f )(y1, y2) dy1dy2 dt1 t1 dt2 t2 ,
where C is a positive constant. By the argument used in [3, 4, 5, 10], we see that if we suitably choose the constant C, then {ak}k∈Z are (p, 2, s1, s2)-atoms and
( X k∈Z |λk|p )1/p .kf kHp(Rn1×Rn2).
It remains to prove that f =Pk∈Zλkakconverges in Ds1, s2; σ1, σ2(Rn1× Rn2). Since eΩk is bounded, we may assume that eΩk⊂ B(1)(0, 2L1) × B(2)(0, 2L2). Then for any α ∈ Zn1
+ and β ∈ Zn2 +, by Lemma 2.2, we have X R∈Rk Z A(R) |(∂xα1∂xβ2ψt1, t2)(x1− y1, x2− y2)|(ϕt1, t2 ∗ f )(y1, y2)| dy1dy2 dt1 t1 dt2 t2 . X R∈Rk Z A(R) |(ϕt1, t2 ∗ f )(y1, y2)| dy1dy2 dt1 t1+|α|+n1 1 dt2 t1+|β|+n2 2 . Z B(1)(0, 2L2) Z B(2)(0, 2L2) Z L1 0 Z L2 0 dt1dt2dy1dy2 < ∞,
where (x1, x2) ∈ eΩk. This shows that ak ∈ Ds1, s2, σ1, σ2(Rn1× Rn2). Moreover, assume that supp f ⊂ B(1)(0, r
1) × B(2)(0, r2). For any Ni > 1 + log ri with i = 1, 2, let
EN1, N2 ≡ B(1)(0, 2N1) × B(2)(0, 2N2) × [2−N1, 2N1] × [2−N2, 2N2].
Then there exist finite dyadic rectangles R, whose set is denoted by RN1, N2, such that
Let KN1, N2 be the maximal integer of the absolute values of all such k. Then for K >
KN1, N2, by the facts RN1, N2 ⊂ ∪|k|≤KRkand Lemma 2.1 together with σi< σ0i< ni+si+1
for i = 1, 2, we then have ° ° ° ° ° °f − X |k|≤K λkak ° ° ° ° ° ° Ds1, s2; σ1, σ2(Rn1×Rn2) . sup (x1, x2)∈Rn1×Rn2 (1 + |x1|)σ1(1 + |x2|)σ2 × ÃZ 2−N1 0 Z ∞ 0 + Z ∞ 2N1 Z ∞ 0 + Z ∞ 0 Z 2−N2 0 + Z ∞ 0 Z ∞ 2N2 ! Z Rn1×Rn2 |(ϕt1, t2 ∗ f )(y1, y2)| ×|ϕt1, t2(x1− y1, x2− y2)| dy1dy2 dt1 t1 dt2 t2 +(x1, x2)∈Rsupn1×Rn2 (1 + |x1|)σ1(1 + |x2|)σ2 × Z 2N1 2−N1 Z ∞ 0 Z [B(1)(0, 2N1)]{×Rn2|(ϕt1, t2∗ f )(y1, y2)| ×|ψt1, t2(x1− y1, x2− y2)| dy1dy2dt1 t1 dt2 t2 + sup (x1, x2)∈Rn1×Rn2 (1 + |x1|)σ1(1 + |x 2|)σ2 × Z ∞ 0 Z 2N2 2−N2 Z Rn1×[B(2)(0, 2N2)]{|(ϕt1, t2∗ f )(y1, y2)| ×|ϕt1, t2(x1− y1, x2− y2)| dy1dy2 dt1 t1 dt2 t2 .2−N1+ 2−N2 + 2N1(σ1−σ01)+ 2N2(σ2−σ02).
This implies the desired conclusion and hence, finishes the proof of Lemma 2.3.
The following result plays a key role in the proof of Theorem 1.2. In what follows, for any f ∈ D(Rn1× Rn2), we set
sup
x2∈Rn2
diam ( supp f (·, x2)) ≡ sup x1, y1∈Rn1, x2∈Rn2
{|x1− y1| : f (x1, x2) 6= 0, f (y1, x2) 6= 0} ,
and supx1∈Rn1 diam ( supp f (x1, ·)) is similarly defined by interchanging x1 and x2, and y1
and y2.
Lemma 2.4. Let p ∈ (0, 1], q ∈ [p, 1] and Bq be a q-quasi-Banach space. Let s1, s2 ∈ Z+
and T be a Bq-sublinear operator from Ds1, s2(Rn1 × Rn2) to B
q. If there exists a positive
constant C such that for any f ∈ Ds1, s2(Rn1 × Rn2),
kT f kBq≤ C · sup x2∈Rn2 diam ( supp f (·, x2)) ¸n1/p × · sup x1∈Rn1 diam ( supp f (x1, ·)) ¸n2/p kf kL∞(Rn1×Rn2),
then T can be extended as a bounded Bq-sublinear operator from Ds1, s2; σ1, σ2(Rn1 × Rn2)
to Bq.
Proof. Let ψ ∈ C∞(R) such that 0 ≤ ψ(x) ≤ 1 for all x ∈ R, ψ(x) = 1 if |x| ≤ 1/2 and
ψ(x) = 0 if |x| ≥ 1. Let φ(x) ≡ ψ(x/2) − ψ(x) for all x ∈ R. Then supp φ ⊂ {x ∈ R :
1/2 ≤ |x| ≤ 2} and Pj∈Zφ(2−jx) = 1 for all x ∈ R \ {0}. Let Φ
j(x) ≡ φ(2−jx) for all
x ∈ R and j ∈ N, and Φ0(x) ≡ 1 −
P∞
j=1φ(2−jx) for all x ∈ R. Then
P
j∈Z+Φj(x) = 1 for all x ∈ R.
Let i = 1, 2. For ji ∈ Z+ and xi ∈ Rni, let Φj(i)i(xi) ≡ Φji(|xi|). Then for all xi ∈ R
ni, we havePji∈Z+Φ(i)ji(xi) = 1. Set R(i)0 ≡ B(i)(0, 2) and R(i)ji ≡ {xi ∈ R
ni : 2ji−1 ≤ |x
i| ≤
2ji+1} for j
i ∈ N. Then supp Φ(i)ji ⊂ R
(i)
ji for ji ∈ Z+. For ji ∈ Z+, let { eψ
(i)
ji, αi : |αi| ≤
si} ⊂ C∞(Rn) be the dual basis of {xαii : |αi| ≤ si} with respect to weight Φ(i)ji |R(i)ji |−1,
namely, for all αi, βi∈ Z+ with |αi| ≤ si and |βi| ≤ si, 1 |R(i)ji | Z Rnix βi i ψej(i)i, αi(xi)Φ (i) ji(xi) dxi = δαi,βi. Let ψj(i)i, αi ≡ |R(i)ji |−1ψej(i)i, αiΦ(i)ji . Then for ji∈ N and xi ∈ Rni, we have
ψj(i)i, αi(xi) = 2−(ji−1)(ni+|αi|)ψ1, α(i)i(2
−(ji−1)x
i).
From this, it is easy to see that for all ji∈ Z+ and |αi| ≤ s,
(2.14) kψ(i)ji, αikL∞(Rni).2−ji(ni+|αi|).
For f ∈ Ds1, s2(Rn1× Rn2), assume that supp f ⊂ B(1)(0, 2k1) × B(2)(0, 2k2) for some
k1, k2 ∈ N and kf kDs1, s2; σ1, σ2(Rn1×Rn2)= 1 by the Bq-sublinear property of T . For j1, j2 ∈
Z+, we set fj1, j2 ≡ f Φ(1)j1 Φj(2)2 , and for any (x1, x2) ∈ Rn1× Rn2,
Pj(1)1, j2(x1, x2) ≡ X |α1|≤s1 ψj(1)1, α1(x1) Z Rn1fj1, j2(y1, x2)y α1 1 dy1, Pj(2)1, j2(x1, x2) ≡ X |α2|≤s2 ψ(2)j2, α2(x2) Z Rn2 fj1, j2(x1, y2)y α2 2 dy2 and Pj1, j2(x1, x2) ≡ X |α1|≤s1 X |α2|≤s2 ψ(1)j1, α1(x1)ψ(2)j2, α2(x2) Z Rn1×Rn2fj1, j2(y1, y2)y α1 1 y2α2dy1dy2. Then f = kX1+1 j1=0 kX2+1 j2=0 ³ fj1, j2− Pj(1)1, j2 − Pj(2)1, j2+ Pj1, j2 ´
+ kX1+1 j1=0 kX2+1 j2=0 ³ Pj(1)1, j2− Pj1, j2 ´ + kX1+1 j1=0 kX2+1 j2=0 ³ Pj(2)1, j2 − Pj1, j2 ´ + kX1+1 j1=0 kX2+1 j2=0 Pj1, j2.
By the definition of Ds1, s2; σ1, σ2(Rn1 × Rn2), it is easy to see that (2.15) kfj1, j2kL∞(Rn1×Rn2).2−j1σ12−j2σ2. Using kΦ(i)ji kL∞(Rni)≤ 1, we obtain
(2.16) ° ° ° ° Z Rn1 fj1, j2(y1, ·)y α1 1 dy1 ° ° ° ° L∞(Rn2) .2j1(n1+|α1|−σ1)2−j2σ2, (2.17) ° ° ° ° Z Rn2 fj1, j2(·, y2)yα2 2 dy2 ° ° ° ° L∞(Rn1) .2−j1σ12j2(n2+|α2|−σ2), and (2.18) ¯ ¯ ¯ ¯ Z Rn1×Rn2 fj1, j2(y1, y2)y α1 1 yα22dy1dy2 ¯ ¯ ¯ ¯ .2j1(n1+|α1|−σ1)2j2(n2+|α2|−σ2). By the estimates (2.14) through (2.18), we have
° ° °fj1, j2− P (1) j1, j2 − P (2) j1, j2+ Pj1, j2 ° ° ° L∞(Rn1×Rn2).2 −j1σ12−j2σ2. Since fj1, j2− P (1) j1, j2− P (2)
j1, j2+ Pj1, j2 ∈ Ds1, s2(Rn1 × Rn2), by the assumption of the lemma, we then have ° ° °T ³ fj1, j2− P (1) j1, j2 − P (2) j1, j2+ Pj1, j2 ´° ° ° Bq .2j1(n1/p−σ1)2j2(n2/p−σ2), and hence, by σi > ni/p for i = 1, 2,
(2.19) ° ° ° ° ° °T kX1+1 j1=0 kX2+1 j2=0 ³ fj1, j2− P (1) j1, j2 − P (2) j1, j2+ Pj1, j2 ´ ° ° ° ° ° ° Bq . kX1+1 j1=0 kX2+1 j2=0 2j1q(n1/p−σ1)2j2q(n2/p−σ2) 1/q .1. Moreover, we write kX1+1 j1=0 kX2+1 j2=0 h Pj(1)1, j2(x1, x2) − Pj1, j2(x1, x2) i
= X |α1|≤s1 kX1+1 j1=1 kX2+1 j2=0 kX1+1 `1=j1 h ψ(1)j1, α1(x1) − ψ(1)j1−1, α1(x1) i ·Z Rn1 f`1, j2(y1, x2)yα1 1 dy1 − X |α2|≤s2 ψ(2)j2, α2(x2) Z Rn1 Z Rn2 f`1, j2(y1, y2)yα1 1 yα22dy1dy2 ¸ ≡ X |α1|≤s1 kX1+1 j1=1 kX2+1 j2=0 kX1+1 `1=j1 Aα1, j1, `1, j2(x1, x2). By (2.14), (2.15) and (2.18), we have kAα1, j1, `1, j2kL∞(Rn1×Rn2).2−j1(n1+|α1|)2`1(n1+|α1|−σ1)2−j2σ2.
Noticing that Aα1, j1, `1, j2 ∈ Ds1, s2(Rn1× Rn2), by the assumption of the lemma, we obtain
kT (Aα1, j1, `1, j2)kBq.2j1(n1/p−n1−|α1|)2`1(n1+|α1|−σ1)2j2(n2/p−σ2). Thus by σi∈ (max{ni/p, ni+ si}, ni+ si+ 1) for i = 1, 2, we further have
(2.20) ° ° ° ° ° °T kX1+1 j1=0 kX2+1 j2=0 ³ Pj(1)1, j2 − Pj1, j2 ´ ° ° ° ° ° ° Bq . X |α1|≤s1 kX1+1 j1=1 kX2+1 j2=0 kX1+1 `1=j1 2j1q(n1/p−n1−|α1|)2`1q(n1+|α1|−σ1)2j2q(n2/p−σ2) 1/q .1. Similarly, by symmetry, we have
(2.21) ° ° ° ° ° °T kX1+1 j1=0 kX2+1 j2=0 ³ Pj(2)1, j2 − Pj1, j2 ´ ° ° ° ° ° ° Bq .1. Finally, we write kX1+1 j1=0 kX2+1 j2=0 Pj1, j2 = X |α1|≤s1 X |α2|≤s2 kX1+1 j1=1 kX2+1 j2=1 kX1+1 `1=j1 kX2+1 `2=j2 ³ ψ(1)j1, α1− ψj(1)1−1, α1 ´ × ³ ψ(2)j2, α2 − ψ(2)j2−1, α2 ´ Z Rn1 Z Rn2 f`1, `2(y1, y2)y α1 1 yα22dy1dy2 ≡ X |α1|≤s1 X |α2|≤s2 kX1+1 j1=1 kX2+1 j2=1 kX1+1 `1=j1 kX2+1 `2=j2 Aα1, j1, `1, α2, j2, `2.
From (2.14) and (2.17), it follows that
