# Q

Authors: Jizhen Zhou
Journal: Journal of Function Spaces
 (2014)

---

## Abstract

We introduce a new spaceQK(∂D) of Lebesgue measurable functions on the unit circle connecting closely with the Sobolev space. We obtain a necessary and sufficient condition on K such that QK(∂D)=BMO(∂D), as well as a general criterion on weight functions K1 and K2, K1≤K2, such that QK1(∂D)QK2(∂D). We also prove that a measurable function belongs to QK(∂D) if and only if it is Möbius bounded in the Sobolev space LK2(∂D). Finally, we obtain a dyadic characterization of functions in QK(∂D) spaces in terms of dyadic arcs on the unit circle.


---

## Body

## 1. Introduction

In recent years a new class of Möbius invariant function spaces, calledQ spaces, has attracted a lot of attention. These spaces were originally defined in [1] as spaces of analytic functions in the unit disc D in the complex plane C. Later on, some further generalizations such as F(p,q,s) and QK appeared; see [2, 3], for example. Let ∂D be the boundary of D. For p∈(-∞,∞), Xiao studied the space Qp(∂D) in paper [4], consisting of all Lebesgue measurable functions f:∂D→C with
(1)∥f∥Qp(∂D)=sup⁡I⊂∂D(1|I|p∬I|f(u)-f(v)|2|u-v|2-p|du||dv|)1/2<∞,
where the supremum is taken over all subarcs I⊂∂D and |I| is the arc length of I. A series of results of Qp(∂D) can be found in [4–6]. Note that if p=2, then Qp(∂D) coincides with BMO(∂D), the space of measurable functions of bounded mean oscillation on ∂D introduced by John and Nirenberg in [7]. For any given arc I⊂∂D and L2(∂D) function f, the square mean oscillation of f on I is defined by
(2)Φf(I)=1|I|∫I|f(u)-fI|2|du|,
where
(3)fI=1|I|∫If(u)|du|.
Then a function f∈L2(∂D) is said to belong to the space BMO(∂D) if and only if ∥f∥BMO(∂D)2=sup⁡I⊂∂DΦf(I)<∞.In paper [2], Essén and Wulan studied QK spaces of holomorphic functions on the unit disc D and developed their general theory. Later on, Wulan and Zhou gave a decomposition theorem on QK spaces and built a relationship between QK spaces of analytic functions and the Morrey type space; see [8, 9], for example. Our aim in this paper will be to extend these ideas to the real QK spaces so that we may obtain related results on the “real QK spaces” by using known results on real Hardy spaces. Historically, the “real variable” theory of Hardy spaces has proved to be important in the development of harmonic analysis. We feel that these spaces are intrinsically interesting and that understanding them better will help inform our study of spaces of holomorphic functions.As a continuation of [2], Essén et al. described the boundary values behavior of analytic functions in QK spaces [10] as follows.Theorem EWX.
Let
K
:
[
0
,
∞
)
→
[
0
,
∞
)
be nondecreasing and satisfy the conditions
(4)
∫
0
1
‍
φ
K
(
t
)
d
t
t
<
∞
,
(5)
∫
1
∞
‍
φ
K
(
t
)
d
t
t
2
<
∞
,
where
(6)
φ
K
(
s
)
=
sup
⁡
0
<
t
≤
1
K
(
s
t
)
K
(
t
)
,
0
<
s
<
∞
.
Then
f
∈
H
2
belongs to the space
Q
K
if and only if
(7)
sup
⁡
I
⊂
∂
D
(
∬
I
|
f
(
u
)
-
f
(
v
)
|
2
|
u
-
v
|
2
K
(
|
u
-
v
|
|
I
|
)
|
d
u
|
|
d
v
|
)
1
/
2
<
∞
.The above theorem suggests the following definition ofQK(∂D) spaces on the unit circle. Let K:[0,∞)→[0,∞) be a nondecreasing function. The space QK(∂D) consists of all Lebesgue measurable functions f on ∂D for which (7) holds. If K(t)=tp, 0≤p<∞, QK(∂D) coincides with Qp(∂D). The space QK(∂D) first appeared in [11], where Pau gave that the Szegö projection from QK(∂D) to QK is bounded and surjective. By [10] and [11] we know that QK=H2∩QK(∂D) if the weight function K satisfies conditions (4) and (5).In addition,f≲g (for two functions f and g) means that there is a constant C>0 (independent of f and g) such that f≤Cg. We say that f≈g (i.e., f is comparable with g) whenever f≲g≲f. In the whole paper we assume that K is doubling; that is, K(2t)≈K(t).


## 2. BMO andQK(∂D) Spaces

In this section, we investigate the relationship between spacesQK(∂D) and BMO(∂D) and study how QK(∂D) depends on the weight function K.The following identity is easily verified:(8)1|I|2∬I|f(u)-f(v)|2|du||dv|=2Φf(I).Proposition 1.
Q
K
(
∂
D
) is a subset of BMO(∂D) for all K.Proof.
ForI⊂∂D, it is easy to see that
(9)I×I={(z,w):0<|z-w|<|I|,z,w∈I}⋃‍{(z,z),z∈I}.
Note that the area measure of {(z,z),z∈I} is zero. For z,w∈I, we have
(10){(z,w):0<|z-w|<|I|,z,w∈I}⊂⋃k=1∞‍{(z,w):|I|2k<|z-w|≤|I|2k-1}.
Suppose that f∈QK(∂D). For I⊂∂D and integer k, denote by 2-kI the subarcs of I with arc length 2-k|I|. Then(11)∬I|f(u)-f(v)|2|du||dv|≤∑k=1∞‍∬|I|/2k<|u-v|≤|I|/2k-1‍|f(u)-f(v)|2|du||dv|≲1K(1)∑k=1∞‍(|I|2k)2fdsffdfdff×∬|u-v|≤|I|/2k-1‍|f(u)-f(v)|2|u-v|2K(|u-v|2-k|I|)fgdgfgfdggfgfgfgfggfg×|du||dv|≲1K(1)∑k=1∞‍(|I|2k)2fdsfdfdfdff×2k∬I/2k-1|f(u)-f(v)|2|u-v|2fdfdfdfdfdf×K(|u-v|21-k|I|)|du||dv|≲|I|2∥f∥QK(∂D)2.
We have f∈BMO(∂D) by (8).Corollary 2.
The spaceQK(∂D) is Banach with the norm of ∥f∥=|f(0)|+∥f∥QK(∂D), where ∥f∥QK(∂D) is the supremum of (7).Proof.
Let{fn} be a Cauchy sequence in QK(∂D). By Proposition 1 we know that QK(∂D) is subset of BMO(∂D). Hence {fn} is a Cauchy sequence in BMO(∂D) as well and fn→f in BMO(∂D) for some f. It follows from Fatou’s lemma that, for every integer n≥1,
(12)∥f-fn∥QK(∂D)≤limsup⁡j→∞∥fj-fn∥QK(∂D).
This gives fn→f in QK(∂D).Theorem 3.
Q
K
(
∂
D
)
=
B
M
O
(
∂
D
) if and only if
(13)∫01‍K(s)s2ds<∞.Proof.
Assume thatf∈BMO(∂D) and (13) holds. We use nI for the arc in ∂D which has the same center as I and length n|I| for a nonnegative integer n. For any given I⊂∂D and |t|≤|I|, then
(14)∫I‍|f(ei(θ+t))-f(eiθ)|2dθ≲∫I‍{|f(ei(θ+t))-fI|2+|f(eiθ)-fI|2}dθ≲|I|∥f∥BMO(∂D)2+∫I‍|f(ei(θ+t))-f3I|2dθ+|I||f3I-fI|2≲|I|∥f∥BMO(∂D)2.
By the inequality2x/π<sinx<x for 0<x<π/2 and the above estimate, we have
(15)∬I|f(eiθ)-f(eiφ)|2|eiθ-eiφ|2K(|eiθ-eiφ||I|)dθdφ≲∫|t|≤|I|‍∫I‍|f(ei(θ+t))-f(eiθ)|2|ei(θ+t)-eiθ|2gfdgfgfgff×K(|ei(θ+t)-eiθ||I|)dθdt≲∥f∥BMO(∂D)2|I|∫|t|≤|I|‍K(|sin(t/2)|/|I|)(sin(t/2))2dt≲∥f∥BMO(∂D)2|I|∫0|I|‍K(t/|I|)(t/2)2dt≲∥f∥BMO(∂D)2∫01‍K(s)s2ds.
The above estimate shows that BMO(∂D)⊂QK(∂D). This and Proposition 1 imply BMO(∂D)=QK(∂D).
Conversely, suppose thatQK(∂D)=BMO(∂D). If
(16)∫01‍K(s)s2ds=∞,
we can choose an integer sequence {λj}j=1∞ such that
(17)∫2π2-λj1‍K(s)s2ds≥j,j=1,2,3,….
Define a function f as follows:
(18)f(eiθ)=∑j=1∞‍1jei2λjθ.
Then f∈BMO(∂D) ([12], page 178). By assumption we have f∈QK(∂D). It is easy to see that
(19)∫02π‍|f(ei(θ+t))-f(eiθ)|2dθ=∑j=1∞‍1j2|ei2λjt-1|2.
We give the following estimate which will be proved later:
(20)∫0π‍sin2nt2K(t)t2dt≳∫2π/nπ‍K(t)t2dt,n=1,2,….
By (17), (19), and (20), we have
(21)∬02π|f(eiθ)-f(eiφ)|2|eiθ-eiφ|2K(|eiθ-eiφ|2π)dθdφ=∑j=1∞‍1j2∫02π‍|ei2λjt-1|2|eit-1|2K(|eit-1|2π)dt≳∑j=1∞‍1j2∫0π‍sin2(2λjt/2)sin2(t/2)K(t)dt≳∑j=1∞‍1j2∫0π‍sin22λjt2K(t)t2dt≳∑j=1∞‍1j2∫2π2-λjπ‍K(t)t2dt≳∑j=1∞‍1j=∞.
We now prove (20). Note that j/n≥(j+1)/2n is valid for all j,n=1,2,… and K(2t)≈K(t). Then
(22)∫0π‍sin2nt2K(t)t2dt=∑j=0n-1‍∫jπ/n((j+1)/n)π‍1-cos⁡nt2K(t)t2dt≥∑j=0n-1‍K(jπ/n)((j+1)π/n)2∫jπ/n((j+1)/n)π‍1-cos⁡nt2dt≳∑j=1n-1‍K((j+1)π/n)((j+1)π/n)2π2n≥12∑j=1n-1‍(jj+1)2∫jπ/n((j+1)/n)π‍K(t)t2dt≳∫2π/nπ‍K(t)t2dt.
The proof is complete.It is reasonable to assume thatlim⁡r→0+K(r)=0 for otherwise weight function K basically dose not play any role. Moreover, the function f must be at least locally L2 on the boundary when f belongs to the QK spaces. Therefore the weight function K plays a role only if t is small. Then the following result is obvious.Theorem 4.
Letr0∈(0,1) such that K(r0)>0, and set K1(r)=inf⁡(K(r),K(r0)). Then QK1(∂D)=QK(∂D).Proof.
SinceK1≤K and K1 is nondecreasing, it is easy to see that QK(∂D)⊂QK1(∂D). We now prove QK1(∂D)⊂QK(∂D). Note that there exists an integer m∈N such that m-1≤r0/2. If f∈QK1(∂D), then f∈BMO(∂D) by Proposition 1. For any I⊂∂D, divide I into the m subarcs of length |I|/m. For 1≤j≤m, denote Ij the jth subarcs, arranged in the natural order. Let Ij,k be the smallest subarcs containing Ij and Ik. Then we have
(23)A=∑j,k=1k-1≤j≤k+1m‍∫Ij‍∫Ik‍|f(u)-f(v)|2|u-v|2K(|u-v||I|)|du||dv|=∑j,k=1k-1≤j≤k+1m‍∫Ij‍∫Ik‍|f(u)-f(v)|2|u-v|2K1(|u-v||I|)|du||dv|≤∬I|f(u)-f(v)|2|u-v|2K1(|u-v||I|)|du||dv|≤∥f∥QK1(∂D),B=∑j,k=1j>k+1,j<k-1m‍∫Ij‍∫Ik‍|f(u)-f(v)|2|u-v|2K(|u-v||I|)|du||dv|≤K(1)∑j,k=1j>k+1,j<k-1m‍(mI)2∫Ij‍∫Ik‍|f(u)-f(v)|2|du||dv|≲∑j,k=1j>k+1,j<k-1m‍m|I|(∫Ij‍|f(u)-fIj,k|2|du|fdsfdfdfdfdfdfd+∫Ik|f(v)-fIj,k|2|dv|)≲∥f∥BMO(∂D)2≲∥f∥QK1(∂D)2.
The above estimate gives
(24)∬I|f(u)-f(v)|2|u-v|2K(|u-v||I|)|du||dv|=A+B≲∥f∥QK1(∂D)2.
Hence f∈QK(∂D). So we have QK1(∂D)⊂QK(∂D). The proof is complete.The following result is natural in view of Proposition1 and Theorem 3.Theorem 5.
LetK1≤K2 and assume that K1(r)/K2(r)→0 as r→0. If
(25)∫01‍K2(s)s2ds=∞,
then QK2(∂D)QK1(∂D).Proof.
Obviously, we haveQK2(∂D)⊂QK1(∂D). We assume that QK1(∂D)=QK2(∂D). The open mapping theorem tells us that the identity map from one of those spaces into the other one is continuous. Therefore there exists a constant C such that ∥·∥QK2(∂D)≤C∥·∥QK1(∂D). By the assumption, there exists an integer m such that K1(t)≤(2C)-1K2(t) for t≤m-1. For any I⊂∂D, divide I into the 2m subarcs of length |I|/(2m). For 1≤j≤2m, denote by Ij the jth subarcs, arranged in the natural order. Applying the same manner in handing A and B in the proof of Theorem 4, we can deduce that if f∈QK2(∂D), then
(26)∥f∥QK2(∂D)2≤Csup⁡I⊂∂D∬I|f(eiθ)-f(eiφ)|2|eiθ-eiφ|2sdfdsfsdfdsff×K1(|eiθ-eiφ||I|)dθdφ=Csup⁡I⊂∂D(∑j,k=1k-1≤j≤k+12m‍+∑j,k=1j<k-1,j>k+12m‍)×∫Ij‍∫Ik‍|f(eiθ)-f(eiφ)|2|eiθ-eiφ|2K1(|eiθ-eiφ||I|)dθdφ≤Csup⁡I⊂∂D(12C∑j,k=1k-1≤j≤k+12m‍∫Ij‍∫Ik‍|f(u)-f(v)|2|u-v|2K2fdsfsdfsdfsdffdfdffdffdfd×(|u-v||I|)|du||dv|∑j,k=1k-1≤j≤k+12m‍)+Csup⁡I⊂∂D(∑j,k=1j<k-1,j>k+12m‍∫Ij‍∫Ik‍|f(u)-f(v)|2|u-v|2K1dfsfdsfsdfdsfdsffdfddfdfd×(|u-v||I|)|du||dv|∑j,k=1k-1≤j≤k+12m‍)=12∥f∥QK2(∂D)2+M∥f∥BMO(∂D)2,
where M is a constant which is dependent on C. Consequently, for any f∈QK2(∂D) and I⊂∂D, we have
(27)∬I|f(u)-f(v)|2|u-v|2K2(|u-v||I|)|du||dv|≲∥f∥BMO(∂D)2.
A simple computation shows thatzn∈QK2(∂D) for n=±1,±2,…. So all polynomials belong to QK2(∂D) spaces. For any given g(u)=∑j=-∞∞‍ajuj∈BMO(∂D), denote by gn(u)=∑j=-nn‍ajuj the truncation of the function g. Then gn∈QK2(∂D) and ∥gn∥BMO(∂D)≤∥g∥BMO(∂D). Applying Fatou’s lemma, we deduce that
(28)sup⁡I⊂∂D∬I|g(u)-g(v)|2|u-v|2K2(|u-v||I|)|du||dv|≲∥g∥BMO(∂D)2.
Equation (28) and Proposition 1 show BMO(∂D)=QK2(∂D). It follows from Theorem 3 that the integral (13) with K=K2 must be convergent, which contradicts our assumption. We conclude that we must have QK2(∂D)QK1(∂D).


## 3. Möbius InvariantQK(∂D) Spaces

LetK:[0,∞)→[0,∞) be a nondecreasing function. The Sobolev type space LK2(∂D) consists of those Lebesgue measurable functions f:∂D→C satisfying
(29)∥f∥LK2(∂D)=(∬∂D|f(u)-f(v)|2|u-v|2K(|u-v|)|du||dv|)1/2<∞.
If K(t)=tp,0≤p<∞, then LK2(∂D)=Lp2(∂D) are sobolev spaces and are introduced in [4]. See [13] about the theory of Sobolev spaces. If K(t)=tp,p>1, then L2(∂D) is a subspace of LK2(∂D). From Section 2 it turns out that QK(∂D) is closely related to the Sobolev type space LK2(∂D) on the unit circle. By (7) and (29) it follows that QK(∂D) is a subspace of LK2(∂D). As a matter of fact, we have the following result.Theorem 6.
LetK satisfy condition (4). Then f∈QK(∂D), if and only if
(30)|||f|||QK(∂D)=sup⁡a∈D∥f∘ϕa∥LK2(∂D)<∞,
where ϕa(z)=(a-z)/(1-a¯z) is a Möbius transformation of the unit disk for a∈D.Proof.
We acknowledge that this proof is suggested by the technique of [4]. Firstly, we give the following equality for u=ϕa(z) and v=ϕa(w):
(31)∬∂D|f∘ϕa(z)-f∘ϕa(w)|2|z-w|2K(|z-w|)|dz||dw|=∬∂D|f(u)-f(v)|2|u-v|2gfdgfgf×K(|u-v|(1-|a|2)|1-a¯u||1-a¯v|)|du||dv|.
Sufficiency. Suppose that (30) holds. Choose an arc I⊂∂D. Without loss of generality, we assume |I|<1/4. We choose a point of a∈D(a≠0) such that a/|a| and 2π(1-|a|) are the center and arc length of I, respectively. We have the following estimate:
(32)1|1-a¯u|≈1|I|,u∈I.
Then
(33)∬I|f(u)-f(v)|2|u-v|2K(|u-v||I|)|du||dv|≲∬∂D|f(u)-f(v)|2|u-v|2fdsfdsffd×K(|u-v|(1-|a|2)|1-a¯u||1-a¯v|)|du||dv|.
By (31) we complete the proof of sufficiency.
Necessity. We assume that f⊂QK(∂D). For any a∈D, let Ia be the arc in ∂D with the midpoint of a/|a| and the arc length of 2π(1-|a|). If a=0, we set Ia=∂D. Also, define
(34)In=2nIa,n=0,1,…,N-1,
where N is the smallest integer such that 2N|Ia|≥2π; that is, IN=∂D. Then
(35)∬∂D|f(u)-f(v)|2|u-v|2×K(|u-v|(1-|a|2)|1-a¯u||1-a¯v|)|du||dv|=∬I0+∑n=0N-1‍∑m=0N-1‍∫In+1∖In‍∫Im+1∖Im‍{⋯}≤∥f∥QK(∂D)2+A+B,
where
(36)A=∑n=0N-1∑m≤nN-1∫In+1∖In∫Im+1∖Im|f(u)-f(v)|2|u-v|2fdsfdfsdfdsfdff×K(|u-v|(1-|a|2)|1-a¯u||1-a¯v|)|du||dv|,B=∑n=0N-1∑m>nN-1∫In+1∖In∫Im+1∖Im|f(u)-f(v)|2|u-v|2fdsfsdfsdfsdfdsf×K(|u-v|(1-|a|2)|1-a¯u||1-a¯v|)|du||dv|.
For any given u∈In+1∖In,n=1,2,…, we have
(37)1|1-a¯u|≲12n(1-|a|).
By (32) and (37), we obtain that
(38)∑n=0N-1‍∬In+1∖In|f(u)-f(v)|2|u-v|2dsdsdsdssds×K(|u-v|(1-|a|2)|1-a¯u||1-a¯v|)|du||dv|≲∑n=0N-1‍∬In+1∖In|f(u)-f(v)|2|u-v|2fdsfdfdsfdsfdf×K(|u-v|22n|I0|)|du||dv|≲∑n=0N-1‍φK(12n)∬In+1|f(u)-f(v)|2|u-v|2dasdsadsdsdsdsdsdsdd×K(|u-v|2n+1|I0|)|du||dv|≲∥f∥QK(∂D)2∫01‍φK(s)sds.
On the other hand, by Lemma 2.1 of [10], condition (4) implies that K(t)≲tc for small enough c>0. Then
(39)∑n=1N-1‍∑m<n-1‍∫In+1∖In‍∫Im+1∖Im‍|f(u)-f(v)|2|u-v|2fsdfsdfdsfdsffff×K(|u-v|(1-|a|2)|1-a¯u||1-a¯v|)|du||dv|≲∑n=1N-1‍∑m<n-1‍∫In+1∖In‍∫Im+1∖Im‍|f(u)-f(v)|2(2n-2m)2|I0|2dsadsadsadsadsadsdsds×K((2n-2m)|I0|22m+n|I0|2)|du||dv|≲∑n=1N-1‍∑m<n-1‍K(12m)fdsfsdfdsf×∫In+1∖In‍∫Im+1∖Im‍(|f(u)-fIm|2+|f(v)-fIm|2)fdfsdfsdfsdfdsfdfdffdfdf×(22n|I0|2)-1|du||dv|≲∑n=1N-1‍∑m<n-1‍2m(1-c)2ndsadsdsdsdsd×(∥f∥BMO(∂D)2+12n|I0|fdsfsdfsdfdfsdff×∫In+1∖In‍|f(u)-fIm|2|du|)≲∑n=1N-1‍∑m<n-1‍2m(1-c)2n∥f∥BMO(∂D)2(1+(n-m)2)≲∥f∥BMO(∂D)2∑n=1∞‍n22nc.
Here we apply the following estimate:(40)1|In+1|∫In+1‍|f(u)-fIm|2|du|≤1|In+1|∫In+1‍|f(u)-fIn+1|2|du|+|fIm-fIn+1|2≤1|In+1|∫In+1‍|f(u)-fIn+1|2|du|+(∑j=m+1n+1‍|fIj-1-fIj|)2≲(n-m)2∥f∥BMO(∂D)2.
The above estimate gives that
(41)A=(∑n=0N-1‍∬In+1∖In+∑n=1N-1‍∫In+1∖In‍∫In∖In-1‍){⋯}+∑n=1N-1‍∑m<n-1‍∫In+1∖In‍∫Im+1∖Im‍{⋯}≲∥f∥QK(∂D)2.
Applying the same manner in handingA, we have B≲∥f∥QK(∂D)2. Therefore, we obtain
(42)∬∂D|f(u)-f(v)|2|u-v|2×K(|u-v|(1-|a|2)|1-a¯u||1-a¯v|)|du||dv|≲∥f∥QK2.
The proof is complete.Corollary 7.
Q
K
(
∂
D
) is a Möbius invariant space in the sense that |∥f∥|QK(∂D)=|∥f∘ϕa∥|QK(∂D) for any f∈QK(∂D) and a∈D.Proof.
Corollary7 is obvious by Theorem 6.


## 4. Dyadic Characterization

For given arcI⊂∂D, denote by In the set of the 2n arcs of length 2-n|I| obtained by n successive bipartition of I. The discrete characterization of Qp(∂D) space is given in [5]. We will prove a discrete characterization of QK(∂D) spaces. The following is the principle result of this section.Theorem 8.
LetK satisfy condition (4). Then f∈L2(∂D) belongs to the space QK(∂D), if and only if
(43)sup⁡I⊂∂D∑n=0∞∑J∈In‍K(12n)Φf(J)<∞.We first acknowledge that this proof is suggested by the technique of [5]. To prove Theorem 8, we need the following lemmas.Lemma 9.
LetI⊂∂D be an arc. If f∈L2(∂D), then
(44)Ψf,K(I)≈∑J∈I1‍Ψf,K(J)+∑J∈I1‍|fJ-fI|2,
where
(45)Ψf,K(I)=∑n=0∞‍∑J∈In‍K(12n)Φf(J).Proof.
The following result can be found in [5]:
(46)Φf(I)=12∑J∈I1‍Φf(J)+122∑J∈I1‍|fJ-fI|2.
Note that Ik=∪J∈I1Jk-1 and K(2-k)≈K(2-k-1),k=0,1,2…. By (46), we have
(47)Ψf,K(I)=Φf(I)+∑k=1∞‍∑J∈I1‍∑U∈Ik-1‍K(12k)Φf(U)≈∑J∈I1‍(Ψf,K(J)+Φf(J)+|fJ-fI|2)≈∑J∈I1‍(Ψf,K(J)+|fJ-fI|2).
The proof is complete.Lemma 10.
LetK satisfy condition (4). Let I,I′,I′′ be three arcs of equal length: |I|=|I′|=|I′′|, such that I′ and I′′ are adjacent and I⊂I′∪I′′. Then for any f∈L2(∂D), we have
(48)Ψf,K(I)≤Ψf,K(I′)+Ψf,K(I′′)+|fI′-fI′′|.Proof.
See [5] about the proof of the following inequality:
(49)Φf(I)≤Φf(I′)+Φf(I′′)+|fI′-fI′′|2.
Without loss of generality, we assume that I′=[0,1) and I′′=[1,2). For each integer j≥0, let {Ij,k}k=12j+1 be the set of the 2j+1 dyadic arcs of length 2-j contained in I′∪I′′, arranged in the natural order. If J∈Ij, then J∈Ij,k∪Ij,k+1 for some k; by (48) we have
(50)Φf(Ij)≤Φf(Ij,k)+Φf(Ij,k+1)+|fIj,k-fIj,k+1|2.
The different choices of J∈Ij yield different k. Summing over all j and J, we have
(51)Ψf,K(I)=∑j=0∞‍∑J∈Ij‍K(12j)Φf(J)≤2∑j=0∞‍∑k=12j+1‍K(12j)Φf(Ij,k)+∑j=0∞‍∑k=12j+1-1‍K(12j)|fIj,k-fIj,k+1|2.
It is easy to see that
(52)∑j=0∞‍∑k=12j+1‍K(12j)Φf(Ij,k)≲Ψf,K(I′)+Ψf,K(I′′).
The following estimate about the final double sum first appeared in Lemma 1 of [5]. Consider
(53)∑k=12j+1-1‍|fIj,k-fIj,k+1|2≲∑l=1j∑J∈Ij-l′∪Ij-l′′‍l2Φf(J)+|fI′-fI′′|2.
If K satisfies condition (4), Lemma 2.1 in [10] implies that there exists some small enough c>0 such that t-cK(t) is nondecreasing. Substituting j=m+l and summing over j, we finally obtain
(54)∑j=0∞‍∑k=12j+1-1‍K(12j)|fIj,k-fIj,k+1|2≲∑l=1∞‍∑m=0∞‍∑J∈Im′∪Im′′‍K(12m+l)l2Φf(J)+∑j=0∞‍K(12j)|fI′-fI′′|2≲∑l=1∞‍l22lc∑m=0∞‍∑J∈Im′∪Im′′K(12m)Φf(J)+|fI′-fI′′|2∫01‍φK(t)dtt≲Ψf,K(I′)+Ψf,K(I′′)+|fI′-fI′′|2.
Thus we have proved (49) and hence the proof is complete.Lemma 11.
IfK satisfies condition (4), then there exists a p∈(0,∞) such that K(t)/tp is nonincreasing. Furthermore, K(t)≈K(2t) for any 0<t<∞.Proof.
Lemma11 can be found in [14].Proof of Theorem8.
We now prove the necessity. It is easy to see that(55)1|I|2∬I|f(u)-f(v)|2|du||dv|=2Φf(I).
By (55), we have
(56)Ψf,K(I)=12∑k=0∞‍∑J∈Ik‍K(12k)22k|I|2∬J|f(u)-f(v)|2|du||dv|=∬∂DαI(u,v)|f(u)-f(v)|2|du||dv|,
where
(57)αI(u,v)=12∑k=0∞‍∑J∈Ik‍K(12k)22k|I|2χJ(u)χJ(v)
and χJ(u)=1, for u∈J, and χJ(u)=0, for u∈∂D∖J.
Note that|u-v|≤2-k|I| because of u,v∈J∈Ik. Since K satisfies condition (4), by Lemma 11 we may assume that t-2K(t) is nonincreasing. In fact, if p≥2, we can replace K(t) with t2 by Theorem 3. Then
(58)αI(u,v)=12∑k=0∞‍∑J∈Ik‍K(|I|2k|I|)22k|I|2χJ(u)χJ(v)≲∑2k|u-v|≤|I|‍K(|I|2k|I|)22k|I|2≤K(|u-v|/|I|)|u-v|2.
This gives
(59)Ψf,K(I)≲∬I|f(u)-f(v)|2|u-v|2K(|u-v||I|)|du||dv|.
For sufficiency, we claim that(60)∬I|f(u)-f(v)|2|u-v|2K(|u-v||I|)|du||dv|≲1|I|∫-|I||I|‍Ψf,K({I+t})dt+Ψf,K(I),
where {I+t}={z+eit,z∈I} for I⊂∂D.
In fact, by (56) and Fubini’s theorem, we have
(61)1|I|∫-|I||I|‍Ψf,K({I+t})dt=∬∂D1|I|∫-|I||I|‍α{I+t}(u,v)dtfdf×|f(u)-f(v)|2|du||dv|.
This and (56) show that it suffices to verify
(62)K(|u-v|/|I|)|u-v|2≲1|I|∫-|I||I|‍α{I+t}(u,v)dt+αI(u,v).
First, suppose that u,v∈I with |u-v|≤|I|/2 and let l∈N∪0 be such that 2-l-2|I|<|u-v|≤2-l-1|I|. Noting that u∉{I+t} and thus α{I+t}(u,v)=0 when |t|>|I|,
(63)1|I|∫-|I||I|‍α{I+t}(u,v)dt≥12|I|∫∂D‍∑J∈{I+t}l‍K(12l)1|2-lI|2χJ(u)χJ(v)dt=22l|I|3K(12l)∑J∈Il‍∫∂D‍χ{J+t}(u)χ{J+t}(y)dt≳1|I|K(|u-v|/|I|)|u-v|2∑J∈Il‍∫∂D‍χ{J+t}(u)χ{J+t}(v)dt.
For each J, the final integral equals |J|-|u-v|≥|J|/2. Hence the sum over J is at least |I|/2 and (62) holds for |u-v|≤|I|/2.
Ifu,v∈I with |u-v|>|I|/2, by (57) we have
(64)αI(u,v)≥K(12)1|I|2≳K(|u-v|/|I|)|u-v|2.
Hence (62) holds in this case.
We now assume thatf is defined on R with constant fI outside I. Let I+ and I- be the two arcs of the same length as I that are adjacent to I on the left and right, respectively. Note that Ψf,K(I-)=Ψf,K(I+)=0 and fI+=fI-=fI. Note that {I+t}⊂I∪I+ for 0<t<|I| and {I+t}⊂I∪I- for -|I|<t<0. Lemma 10 and (60) give
(65)∬I|f(u)-f(v)||u-v|2K(|u-v||I|)|du||dv|≲1|I|∫-|I||I|‍Ψf,K({I+t})dt+Ψf,K(I)≲1|I|∫-|I||I|‍Ψf,K(I)dt+Ψf,K(I)≲Ψf,K(I).
The proof is complete.Corollary 12.
Let0<p<∞. Then f∈L2(∂D) belongs to Qp(∂D) if and only if
(66)sup⁡I⊂∂D∑n=0∞‍∑J∈In‍(12n)pΦf(J)<∞.Proof.
The nondecreasing functionK satisfies condition (4) if K(t)=tp,0<p<∞. The desired result follows from Theorem 8.



---

*Source: 234790-2014-08-20.xml*