c 2012 by Institut Mittag-Leffler. All rights reserved
On the Carleson duality
Tuomas Hyt¨onen and Andreas Ros´en
Abstract. As a tool for solving the Neumann problem for divergence-form equations, Kenig and Pipher introduced the spaceX of functions on the half-space, such that the non-tangential maximal function of theirL2 Whitney averages belongs toL2 on the boundary. In this paper, answering questions which arose from recent studies of boundary value problems by Auscher and the second author, we find the pre-dual ofX, and characterize the pointwise multipliers fromXto L2 on the half-space as the well-known Carleson-type space of functions introduced by Dahlberg.
We also extend these results toLpgeneralizations of the spaceX. Our results elaborate on the well-known duality between Carleson measures and non-tangential maximal functions.
1. Introduction
A fundamental estimate in harmonic analysis is Carleson’s inequality for Car- leson measures. See [3, Theorem 2] and [4, Theorem 1] for the original formu- lations and applications in the theory of interpolating analytic functions, or for example Stein [11, Section II.2.2] and Coifman, Meyer and Stein [5] for more re- cent accounts in the framework of real-variable harmonic analysis. This inequality states that for a function f(t, x) and a measure dμ(t, x) in the upper half-space R1+n+ :={(t, x);t>0 andx∈Rn}, one has the estimate
R1+n+
|f(t, x)|dμ(t, x)sup
Q
μ(Q)
|Q| Rn
N∗f(y)dy,
where the supremum is over all cubesQinRnandQ:=(0, (Q)) ×Qis theCarleson box,(Q) and|Q|being the sidelength and measure ofQ. FurthermoreN∗ denotes thenon-tangential maximal function
(N∗f)(y) := sup
{(t,x);|x−y|≤at}|f(t, x)|, y∈Rn,
Andreas Ros´en was earlier named Andreas Axelsson.
where a>0 is a fixed constant determining the aperture of the cone. The exact value ofais less important, since for anya1, a2>0 the corresponding non-tangential maximal functionsN∗f are comparable in theLp(Rn) norm for any 1≤p≤∞. See Fefferman and Stein [7, Lemma 1].
Carleson’s inequality has numerous applications. Motivating for this paper is its applications to boundary value problems for elliptic partial differential equations. A recent application concerns boundary value problems for divergence- form equations
divt,xA(t, x)∇t,xu(t, x) = 0,
with non-smooth coefficients A∈L∞(R1+n+ ;C(1+n)×(1+n)) with uniformly positive real part. To solve the Neumann problem withL2(Rn) boundary data, Kenig and Pipher [9] introduced (a space equivalent to) the function spaceX consisting of func- tionsf(t, x), thought of as gradients of solutionsu(t, x), with N∗(W2f)∈L2(Rn), where
(Wqf)(t, x) :=|W(t, x)|−1/qfLq(W(t,x)), (t, x)∈R1+n+ , is theLq Whitney averaged function, with
W(t, x) :=
(x, y)∈R1+n+ ;|y−x|< c1t andc−01<s t< c0
being the Whitney region around (t, x). (Again, the precise value of the fixed constants c0>1 and c1>0 is less important.) The reason for replacing f by the Whitney average W2f is that, unlike the potential u(t, x), the gradient f(t, x)=
∇t,xu(t, x) does not have classical interior pointwise De Giorgi–Nash–Moser bounds.
In the recent works of the second author with P. Auscher [1] and [2], the func- tion spaceX above is fundamental. In these papers, new methods are developed to solve the Neumann (as well as the Dirichlet) problem for systems of divergence-form equations, which rely on solving certain operator-valued singular integral equations in the function spaceX. Two questions arose, which motivated this paper.
• Which functionsg(t, x) are bounded multipliers X −→L2(R1+n+ ;dt dx), f(t, x)−→g(t, x)f(t, x)?
It was shown [1, Lemma 5.5], using Carleson’s inequality, that g is a multiplier if the modified Carleson norm
(1) sup
Q
1
|Q|
Qb
W∞g(t, x)2dt dx 1/2
is finite. We show in this paper (Theorem3.1) that this modified Carleson norm in fact is equivalent to the multiplier norm
gX→L2(R1+n
+ ;dt dx)= sup
f=0
gfL2(R1+n
+ ;dt dx)
fX
.
The modified Carleson norm (1) has been known for some time to be fundamental in the perturbation theory for divergence-form equations. It was introduced already by Dahlberg [6]. See also Fefferman, Kenig and Pipher [8] and Kenig and Pipher [9]
and [10].
• What is the dual, or pre-dual, space of X? We show in this paper (Theo- rem3.2) thatX is the dual space of the space of functionsg(t, x) such that
Rn
sup
Qz
1
|Q|
Qb
W2g(t, x)dt dx 2
dz <∞. (We here identify a functionf∈X with the functionalg→
R1+n+ f g dt dx.) Theo- rem3.2also shows that the spaceX is not reflexive. The interest in understanding duality for the spaceX comes from the dual relation between the Dirichlet problem with L2(Rn) data and the Dirichlet problem with Sobolev H1(Rn) data. See [2, Theorem 1.4] and [9, Theorem 5.4].
Beyond these two results, we prove more general Lp results for the Carleson duality. On one hand, we consider not only W∞g and W2g, but more general Lq
Whitney averages. On the other hand, we measure the non-tangential maximal function and the Carleson functional in Lp norms. For example, this may have useful applications to boundary value problems withLp data.
In Section2, we first prove the corresponding results for a discrete vector-valued model of the Carleson duality. Then in Section 3, we prove equivalence between dyadic and non-dyadic norms, which yields the non-dyadic results.
The spaces we consider here are closely related to the tent spaces introduced by Coifman, Meyer and Stein [5], and in fact reduce to them for certain choices of the parameters. However, as a whole, the scale of spaces that we consider is new.
Since the precise connection to tent spaces is somewhat technical, we postpone a more detailed commentary until Remark3.3below.
2. A discrete vector-valued model
In this section we study a dyadic model of the problem. We use the following notation. LetD= j∈ZDj denote the dyadic cubes inRn, where
Dj:={2−j(0,1)n+2−jk;k∈Zn}.
LetWQ:=((Q)/2, (Q))×Qdenote the dyadic Whitney region, being in one-to-one correspondence withQ∈D. Note that unlike their non-dyadic counterpartsW(t, x), the regions WQ form a disjoint partition of R1+n+ (modulo zero-sets). Define the dyadic Hardy–Littlewood maximal function
MDh(x) := sup
Q:x∈Q∈D
1
|Q|
Q
h(y)dy, x∈Rn, forh∈Lloc1 (Rn). Recall thatMD is bounded onLp(Rn), 1<p≤∞.
Our discrete vector-valued setup is as follows. We assume that to eachQ∈D, there are two associated Banach spacesXQ andYQ. For a sequence f={fQ}Q∈D, wherefQ∈XQ, we define its non-tangential maximal function
(NXf)(x) := sup
Q:x∈Q∈DfQXQ, x∈Rn.
For fixed 1≤p<∞, let Xp denote the space of all sequences f such that fXp:=
NXfLp(Rn)<∞. For a sequenceg={gQ}Q∈D, wheregQ∈YQ, we define the Car- leson functional
(CYg)(x) := sup
Q:x∈Q∈D
1
|Q|
R⊂Q R∈D
gRYR, x∈Rn.
For a fixed number 1<p≤∞, letYp denote the space of all sequencesg such that gYp:=CYgLp(Rn)<∞. Note that the case p=1 is not interesting, since g=0 necessarily ifCYgL1(Rn)<∞.
We assume that for each Q∈D there is a duality XQ,YQ as below, with constantsC uniformly bounded with respect toQ.
Definition 2.1. LetX and Y be two Banach spaces. By a duality X,Y , we mean a bilinear mapX ×Y (f, g)→f, g ∈Rand a constant 0<C <∞such that
|f, g | ≤CfXgY, f∈X, g∈Y, fX≤C sup
gY=1
f, g , f∈X, gY≤C sup
fX=1
f, g , g∈Y. We prove the following duality result.
Theorem 2.2. Let {XQ}Q∈D and {YQ}Q∈D be pairwise dual Banach spaces as above, and let 1/p+1/p=1, 1≤p<∞. Then there is a constant 0<C <∞ such that
Q∈D
|fQ, gQ | ≤CNXfLp(Rn)CYgLp(Rn), fQ∈XQ, gQ∈YQ, NXfLp(Rn)≤C sup
CYgLp(Rn)=1
Q∈D
fQ, gQ , fQ∈XQ, CYgLp(Rn)≤C sup
NXfLp(Rn)=1
Q∈D
fQ, gQ , gQ∈YQ.
The application we have in mind is the following. For functions f(t, x) in R1+n+ , let fQ:=f|WQ∈Lq(WQ)=:XQ, where the Banach space has normfXQ:=
|WQ|−1/qfQLq(WQ)so that
NLqf= sup
Q:x∈Q∈D|WQ|−1/qfLq(WQ).
For functions g(t, x) in R1+n+ , let gQ:=g|WQ∈Lq˜(WQ)=:YQ, where the Banach space has normgYQ:=|WQ|1−1/˜qgQLq˜(WQ)so that
CLq˜f= sup
Q:x∈Q∈D
1
|Q|
R⊂Q R∈D
|WR|1−1/˜qgLq˜(WR).
We generalize slightly the Carleson functional and define CLr
˜
qf(x) = sup
Q:x∈Q∈D
1
|Q|
R⊂Q R∈D
|WR|(|WR|−1/˜qgLq˜(WR))r 1/r
forx∈Rn and 1≤r<∞.
Corollary 2.3. Let 1/p+1/p=1/q˜ +1/˜q=1/r, with r≤p<∞, r≤q≤∞ and 1≤r<∞. Then there is a constant 0<C <∞ such that
f gLr(R1+n
+ )≤CNLqfLp(Rn)CLr
˜
qgLp˜(Rn), NLqfLp(Rn)≤C sup
CLr
˜
qgLp˜(Rn)=1
f gLr(R1+n+ ), CLr˜qgLp˜(Rn)≤C sup
NLqfLp(Rn)=1
f gLr(R1+n
+ ).
Note that the casep=q=r=2 solves a dyadic version of the multiplier question for the space X from the introduction. In this case ˜p= ˜q=∞. Note also that the casep=q=2,r=1, together with Theorem2.4below, solves a dyadic version of the dual space question for the spaceX from the introduction. In this case ˜p= ˜q=2.
Proof. Replacing |f|r and |g|r by f and g, we see that it suffices to consider the caser=1. In this case, the result follows from Theorem 2.2.
Proof of Theorem2.2. (i) For completeness, we start with the well-known proof of the
Q|fQ, gQ |estimate. It suffices to estimate
QfQ gQ. Note that
R⊂Q
gR ≤ |Q| inf
x∈QCYg(x)≤
Q
CYg dx
for anyQ∈D. Select, for given k∈Z, the maximal dyadic cubes Dk⊂D such that fQ>2k. Then Q∈DkQ={x∈Rn;NXf(x)>2k}, and the cubes inDkare disjoint.
We get
Q:fQ>2k
gQ ≤
Q∈Dk
R⊂Q
gR ≤
Q∈Dk
Q
CYg dx=
x:NXf(x)>2k
CYg dx,
and hence
Q∈D
fQ gQ ≈
Q∈D
k:2k<fQ
2kgQ
=
k∈Z
2k
Q:fQ>2k
gQ
≤
k∈Z
2k
x:NXf(x)>2k
CYg dx
=
Rn
k:2k<NXf(x)
2kCYg dx
≈
Rn
NXf CYg dx
≤ NXfpCYgp.
(ii) Next we prove the estimate ofCYgp. Consider first the casep=∞. Pick Q∈Dsuch that
1
|Q|
R⊂Q
gR ≥12CYg∞.
Then constructf={fR}R∈D, choosingfR∈XRso thatfR=1/|Q|andgR/|Q|≈
fR, gR if R⊂Q, and fR:=0 ifR⊂Q. It follows that CYg∞≈
RfR, gR and NXf1=1 sinceNXf=1/|Q|onQandNXf=0 offQ.
Next consider the case 1<p<∞. Select, for givenk∈Z, the maximal dyadic cubesDk⊂D such that
1
|Q|
R⊂Q
gR>2k.
Then{x∈Rn;CYg(x)>2k}= Q∈DkQ, and the cubes inDkare disjoint. We obtain
|{x;CYg(x)>2k}|=
Q∈Dk
|Q| ≤2−k
Q∈Dk
R⊂Q
gR. Now let
fˆR:= 1
|R|
R
CYg dx.
Note that ˆfR does not depend on k, and that ˆfR>2k for R⊂Q∈Dk. We get
|{x;CYg(x)>2k}|≤2−k
R: ˆfR>2kgR and CYgpp≈
k∈Z
2pk|{x;CYg(x)>2k}|
≤
R∈D
k:2k<fˆR
2(p−1)kgR ≈
R∈D
( ˆfR)p−1gR.
Now constructf={fR}R∈D, choosingfR∈XR such that
fR= ( ˆfR)p−1 and ( ˆfR)p−1gR ≈ fR, gR . We get
NXf(x) = sup
Qx
( ˆfQ)p−1= (MD(CYg)(x))p−1. Sincep(p−1)=p, this gives
NXfpp=MD(CYg)ppCYgpp, and we conclude that
Q
fQ, gQ CYgppCYgpNXfp.
(iii) Next we prove the estimate of NXfp. Consider first the case 1<p<
∞. Select, for given k∈Z, the maximal dyadic cubes Dk⊂D such that fQ>2k.
Then {x∈Rn;NXf(x)>2k}= Q∈DkQ, and the cubes in Dk are disjoint. Write kQ:=maxQ∈Dkk≤log2fQ. We obtain
NXfpp≈
k∈Z
2kp|{x;NXf(x)>2k}|=
Q∈D
|Q|
k:Q∈Dk
2kp
≈
Q∈D
|Q|2kQp=
Q∈D
2kQ|Q|2kQ(p−1)≈
Q∈D
fQ
|Q|
k:Q∈Dk
2k(p−1)
.
Write ˆgQ:=|Q|
k:Q∈Dk2k(p−1)and constructg={gQ}Q∈D, choosinggQ∈YQ such thatgQ= ˆgQ and fQ gQ≈fQ, gQ . Then
1
|Q|
R⊂Q
gR
k∈Z
2k(p−1) 1
|Q|
R⊂Q R∈Dk
|R|
=
k∈Z
2k(p−1) 1
|Q||{x;NXf(x)>2k}∩Q|
≈ 1
|Q|
Q
(NXf)p−1dx
≤inf
Q MD((NXf)p−1),
and therefore CYgpp(NXf)p−1pp=NXfpp, since p(p−1)=p. We conclude
that
Q∈D
fQ, gQ NXfppCYgpNXfp.
(iii) We finally prove the estimate of NXf1, i.e. the case p=1. Let D0 be the 2n dyadic cubes with sidelength 2M and one corner at the origin, whereM is chosen large enough, using the monotone convergence theorem, so thatNXf˜1≥
1
2NXf1, where ˜fQ:=fQ if Q⊂Q0 for some Q0∈D0, and ˜fQ:=0 otherwise. As- suming the estimate proved for ˜f, we have
NXf˜1
Qf˜Q, gQ
CYg∞ ,
where we may assumegQ=0 unless Q⊂Q0 for someQ0∈D0. This yields NXf1≤2NXf˜1
Q
f˜Q, gQ
CYg∞
Q
fQ, gQ
CYg∞.
Thus, replacingf by ˜f, we may assume thatfQ=0 unlessQ⊂Q0 for someQ0∈D0.
Givenf contained byD0 as above, we define recursively sets of disjoint dyadic cubes Dj⊂D, j=1,2,3, ..., as follows. Having constructed Dj, let Q∈Dj. Define DQj+1to be the set of maximal dyadic cubesR∈Dsuch thatR⊂QandfR>2fQ. Then letDj+1:= Q∈DjDQj+1. Furthermore, letDf:= ∞j=1Dj and
E(Q) :=Q\
R∈Dj+1Q
R, Q∈ Dj.
From the above construction, if x∈Qk⊂Qk−1⊂...⊂Q0, where Qj∈Dj, then fQk>2k−1fQ1, k=2,3, ..., wherefQ1>0. Hence, if NXf(x)<∞, then there is a minimalQxwithQ∈Df. For thisQ, we havex∈E(Q) andNXf(x)≤2fQ. Thus
(2) NXf≤2
Q∈Df
fQ1E(Q) a.e.
so that NXf1≤2
Q∈DffQ |Q|. Conversely, if x∈Qk⊂Qk−1⊂...⊂Q0, where Qj∈Dj, are all the selected dyadic cubes containing x, then NXf(x)≥fQk≥ 2fQk−1≥...≥2kfQ0. Thus
Q∈Df
fQ |Q|=
Rn
Q∈Df Qx
fQdx≤
Rn
NXf ∞ j=0
2−jdx≤2NXf1.
Now letc∈(0,1) be a constant, to be chosen below, and define Df1:={Q∈ Df;|E(Q)|> c|Q|} and Df2:=Df\Df1. From (2) we have
NXf1≤2
Q∈Df1
fQ |Q|+2c
Q∈Df2
fQ |Q| ≤2
Q∈Df1
fQ |Q|+4cNXf1.
Choosec=18 to obtainNXf1≤4
Q∈Df1fQ |Q|. Constructg={gQ}Q∈D, choos- inggQ∈YQ such thatgQ=|Q|andfQ, gQ ≈fQ |Q|ifQ∈D1f, andgQ:=0 oth- erwise. ThenNXf1
Q∈Df1fQ, gQ . To estimate 1
|Q|
R⊂Q
gR= 1
|Q|
R⊂Q R∈Df1
|R|,
note that ifR∈Df1∩Dj, then
R∈Dj+1R |R|≤78|R|. Thus 1
|Q|
R⊂Q R∈D1f
|R| ≤ 1
|Q| ∞ j=0
7
8
j
|Q|= 8.
ThusCYg∞≤8. This completes the proof of the theorem.
Consider now a duality X,Y between two Banach spaces X and Y as in Definition 2.1. We define the linear map L: X→Y∗ sending f∈X to the linear functional
Λf:Y −→R, g−→f, g .
The estimate |f, g |≤CfXgY shows that LX→Y∗≤C, whereas it follows from the estimate fX≤Csupg
Y=1f, g that L is injective with closed range L(X)⊂Y∗. Thus the duality gives a topological, but not in general isometric, iden- tification, through L, of X with the closed subspace L(X) of Y∗. The estimate gY≤CsupfX=1f, g furthermore shows that this subspace is “large” in the sense that its pre-annihilator is
⊥L(X) :={g∈ Y; Λg= 0 for all Λ∈L(X)}={0}.
In general we may have that L(X)Y∗, but if Y is reflexive, then necessarily L(X)=Y∗. Below we identifyX andL(X), and thus writeX=Y∗ ifL(X)=Y∗. We also note that the above also holds with the roles ofX andY interchanged, giving an identification ofY with a large closed subspace ofX∗.
The following result describes when the duality in Theorem2.2gives the full dual spaces.
Theorem 2.4. With the above notation, consider the duality Xp,Yp , (f, g)−→
Q∈D
fQ, gQ
from Theorem 2.2. We haveYpXp∗ for any 1≤p<∞, as well as X1Y∞∗. If furthermore the duality XQ,YQ is such thatXQ=YQ∗ for allQ∈D, and if 1<p<∞,thenXp=Yp∗.
Proof. (i) We first prove thatX1Y∞∗ . LetQ1Q2Q3... be dyadic cubes.
Define the functionals Λjg:=fQj, gQj on Y∞, where we have chosen fQj∈XQj
such thatfQj=1/|Qj|. It is clear that ΛjY∞∗ ≈1. Consider the sequence space ∞(Z+) and use Hahn–Banach’s theorem to construct lim∈∞(Z+)∗ such that
lim({xn}∞n=1) = lim
n→∞xn
for all convergent sequences{xn}∞n=1. Set Λg:=lim({Λjg}∞j=1). It is straightforward to verify that Λ∈Y∞∗ \X1.
(ii) We next prove that YpXp∗ for 1≤p<∞. Fix some cube Q0∈D with (Q)=1. Define functionals
Λjf:=
R⊂Q0
(R)=2−j
fR, gR
onXp, where gR∈YR is chosen such thatgR=|R|. Then
|Λjf|
R⊂Q0 (R)=2−j
fR |R| ≤
Q0
NXf dx≤ NXfp.
Define Λf:=lim({Λjf}∞j=1). It is straightforward to verify that Λ∈Xp∗\Yp. (iii) Finally we assume that XQ=YQ∗ and 1<p<∞, and aim to show that Xp=Yp∗. Let Λ∈Yp∗, and let Q∈D. Pick fQ∈XQ=YQ∗ such that fQ, gQ = Λ({gQδQR}R∈D) for all gQ∈YQ, where δQR=1 ifR=QandδQR=0 otherwise. Let f:={fQ}Q∈D. Then
(3) Λg=
Q∈D
fQ, gQ
holds whenevergQ=0 only for finitely many Q. From the monotone convergence theorem it follows thatNXfpΛYp∗ , so that f∈Xp. We now use Lemma 2.5 below to deduce that (3) holds for allg∈Yp by continuity.
Lemma 2.5. Assume that 1<p<∞. Then the subspace consisting of se- quencesg={gQ}Q∈D with gQ=0for finitely many Q∈D, is dense inYp.
Proof. (i) Let g∈Yp and let ε>0. Let Q1, ..., Q2n be the dyadic cubes with one corner at the origin and sidelength 2M. Choose M large enough so that
Rn\Q0|CYg|pdx≤εp, where Q0:=Q1∪...∪Q2n. Set g1Q:=
gQ, Q⊂Q0, 0, Q⊂Q0.
LetQj be a sibling ofQj, 1≤j≤2n. Sinceg1Q=0 forQ⊂Qj, it is clear that we have supQjCYg≤infQ
jCYg. Therefore CYg1pp=
Rn\Q0
|CYg1|pdx+
2n
j=1
Qj
|CYg1|pdx
≤
Rn\Q0
|CYg|pdx+
2n
j=1
Qj
|CYg|pdxεp, asCYg1≤CYg.
(ii) Next we consider small cubes insideQ0. Define Cjh(x) := sup
Qx (Q)≤2−j
1
|Q|
R⊂Q
hR, h∈ Yp.
ThenCjg(x)→0, asj→∞, for almost allx, by Lemma2.6below. SinceCjg≤CYg∈ Lp(Rn), it follows by dominated convergence that we can choose j <∞such that Cjgp≤ε. Next chooseδ >0 such that
R:R⊂Q0,(R)≤δgR≤ε2−nj|Q0|−1/p. Set g2Q:=
gQ, if Q⊂Q0 and(Q)≤δ, 0, otherwise.
We have
CYg2(x) = max
Cjg2(x), sup
Qx (Q)>2−j
1
|Q|
R⊂Q
g2R
≤max(Cjg2(x),min(2nj, d(x, Q0)−n)ε2−nj|Q0|−1/p)
≤max(Cjg2(x), ε|Q0|−1/pmin(1, d(x, Q0)−n)), whered(x, Q0):=infy∈Q0|x−y|. This shows thatCYg2pε, since
min(1, d(x, Q0)−n)p|Q0|1/p.
It follows thatg−g1−g2 is finitely non-zero, withg1+g2Ypε.
Lemma 2.6. Let Q0∈D and assume that
Q⊂Q0aQ<∞, where 0≤aQ<∞ forQ⊂Q0. Then
1
|Q|
R⊂Q
aR→0, asQx, (Q)→0, for almost allx∈Q0.
Proof. We argue by contradiction. Assume there existsδ >0 such that E:=
x∈Q0; lim sup
Qx (Q)→0
1
|Q|
R⊂Q
aR> δ
has positive measure. Let A:=
Q⊂Q0aQ<∞. Choose j <∞ such that we have
Q⊂Q0,(Q)>2−jaQ>A−δ|E|/2. Select the maximal cubes Qk⊂Q0, k=1,2, ..., such that (Qk)≤2−j and
R⊂QkaR>δ|Qk|. We get that E⊂ ∞k=1Qk, where the cubesQk are disjoint. This gives
A=
Q⊂Q0
aQ=
Q⊂Q0 (Q)>2−j
aQ+
Q⊂Q0 (Q)≤2−j
aQ
≥
A−δ|E| 2
+
∞ k=1
δ|Qk| ≥A+δ|E| 2 , which is a contradiction. The conclusion follows.
3. The non-dyadic results
In this section, we derive the corresponding non-dyadic results on the Carleson duality from the dyadic results in Section2. We use the following notation. For fixed constantsc0>1, c1>0 anda>0, we use Whitney regionsW(t, x),Lq Whitney averages Wqf of functions f∈Llocq (R1+n+ ), and non-tangential maximal functions N∗f, as in the introduction. Also define the Carleson functionals
Crg(z) := sup
Qz
1
|Q|
Qb
|g(t, x)|rdt dx 1/r
, z∈Rn,
for 1≤r<∞, and the Hardy–Littlewood maximal function M h(z) := sup
Qz
1
|Q|
Q
h(y)dy, z∈Rn,
forh∈Lloc1 (Rn). Here the suprema are over all (non-dyadic) axis-parallel cubes in Rn containingz. We writeC1g=Cgwhenr=1.
We aim to prove the following non-dyadic version of Corollary2.3.
Theorem 3.1. Let 1/p+1/p=1/q+1/˜ q=1/r,˜ with r≤p<∞, r≤q≤∞ and 1≤r<∞. Then there is a constant 0<C <∞ such that
f gLr(R1+n+ )≤CN∗(Wqf)Lp(Rn)Cr(Wq˜g)Lp˜(Rn),
