On sufficient conditions for the boundedness of the Hardy-Littlewood maximal operator between weighted Lp-spaces with different weights
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Proc. Lond. Math. Society, (3) 71 (1995), 135–157 On sufficient conditions for the boundedness of the Hardy-Littlewood maximal operator between weighted Lp–spaces with different weights Carlos P´erez 1 Introduction and main results Let Mbe the Hardy–Littlewood maximal operator defined for locally integrable functions fby Mf(x) = sup x∈Q 1 |Q|ZQ |f(y)|dy, where the supremum is taken over all the cubes containg x, and let 1 < p < ∞. B. Muckenhoupt [10] characterized the weights wsatisfying the weighted norm inequality ZRn (w(y)Mf(y))pdy ≤cZRn (w(y)f(y))pdy (1) for all nonnegative functions f, as those weights satisfying the Apcondition 1 |Q|ZQ w(y)pdy1/p 1 |Q|ZQ w(y)−p0dy1/p0 ≤c(2) for all cubes Q. It is natural to consider a similar problem for a couple of weights (w, v). However, simple examples show (cf. [6] p. 395) that the analogous necessary condition for (w, v) 1 |Q|ZQ w(y)pdy1/p 1 |Q|ZQ v(y)−p0dy1/p0 ≤c, (3) for all cubes Qis not sufficient for the boundedness of Mfrom Lp(vp) to Lp(wp). E. Sawyer has shown in [13], that the correct necessary and sufficient condition is given by 1
ZQ (w(y)M(v−p0χQ)(y))pdy ≤cZQ v(y)−p0dy, (4) for all cubes Q. E. Sawyer’s condition involves the operator Mitself, and it is interesting to obtain sufficient conditions close in form to the necessary and simpler one (3). The first result in that direction was obtained by C. Neugebauer in [11]. He noticed that if (w, v) is a couple of weights such that for some r > 1 1 |Q|ZQ w(y)pr dy1/pr 1 |Q|ZQ v(y)−p0rdy1/p0r ≤c(5) for all cubes Q, then ZRn (w(y)Mf(y))pdy ≤cZRn (v(y)f(y))pdy (6) for all nonnegative functions f. In this paper we take up this problem and show with a different approach that (6) holds assuming very weak conditions on the weights. We shall see that it is enough to replace the average norm associated to the weight v−1in (3) by a stronger norm defined in terms of any Banach function space whose associated space satisfies certain mapping property. To be precise we let Xbe a Banach function space over Rnwith respect to the Lebesgue measure dx (cf. next section). Given a measurable function fand any cube Qwe define the X–average of fover Qby kfkX,Q = τ`(Q)(f χQ) X ,(7) where τδ,δ > 0, is the dilation operator τδf(x) = f(δx), χEis the characteristic function of Eand `(Q) is the sidelength of the cube Q. We define a natural maximal operator associated to the space X. Definition 1.1 For each locally integrable function fthe maximal operator MXis defined by MXf(x) = sup x∈Q kfkX,Q , where the supremum is taken over all the cubes containing x. 2
Let X=LBbe the Orlicz space defined by the Young function B(cf. section 2 or [7] [8]). Then the maximal operator MX=MBis defined in terms of the average kfkX,Q =kfkB,Q = inf{λ > 0 : 1 |Q|ZQ B|f(y)| λdy ≤1} (cf. [1]). If Xis the Lorentz space X=Ls,q, then the maximal operator is MXf(x) = Ms,qf(x) = sup x∈Q 1 |Q|1/s f χQ s,q (cf. [12], [14] and [3]). Given a Banach function space X,X0will denote its associate space, which is another Banach function space (cf. next section). Theorem 1.2 Let 1< p < ∞, and let Xbe a Banach function space such that MX0:Lp(Rn)→Lp(Rn). Suppose that (w, v)is a couple of weights such that there is a positive constant Kfor which 1 |Q|ZQ w(y)pdy1/p kv−1kX,Q ≤K, (8) for all cubes Q. Then ZRn (w(y)Mf(y))pdy ≤cZRn (v(y)f(y))pdy (9) for all nonnegative functions f. A particular example is when X=Lp0r, with r > 1. In this case the associate space is X0=L(p0r)0whose corresponding maximal operator is given by MX0f(x) = M(p0r)0f(x) = sup x∈Q1 |Q|ZQ |f(y)|(p0r)0dy1/(p0r)0 , which is bounded on Lp(Rn). On the other hand, observe that when r= 1 X0=Lp, whose corresponding maximal function Mpis not bounded on Lp(Rn) since Mitself fails to be bounded on L1(Rn). 3
Corollary 1.3 Let 1< p < ∞, and suppose that (w, v)is a couple of weights such that for some r > 1, there is a positive constant Kfor which 1 |Q|ZQ w(y)pdy1/p 1 |Q|ZQ v(y)−p0rdy1/p0r ≤K, (10) for all cubes Q. Then ZRn (w(y)Mf(y))pdy ≤cZRn (v(y)f(y))pdy (11) for all nonnegative functions f. We can deduce a better result using the scale of Lorentz spaces: if X=Lp0r,∞, then X0=L(p0r)0,1and MX0is bounded on Lp(Rn) (cf. section 6). Hence Corollary 1.4 Let 1< p < ∞, and 1< r < ∞. Suppose that (w, v)is a couple of weights such that there is a positive constant Kfor which 1 |Q|ZQ w(y)pdy1/p 1 |Q|1/rp0 χQv−1 Lrp0,∞≤K, (12) for all cubes Q. Then ZRn (w(y)Mf(y))pdy ≤cZRn (v(y)f(y))pdy (13) for all nonnegative functions f. More interesting examples are provided by the theory of Orlicz spaces. Theorem 1.5 Let 1< p < ∞, and let Bbe a doubling Young function such that Z∞ ctp0 B(t)p−1dt t<∞,(14) for some positive constant c. i)Let (w, v)be a couple of weights such that there is a positive constant Kfor which 1 |Q|ZQ w(y)pdy1/p kv−1kB,Q ≤K, (15) 4
for all cubes Q. Then ZRn (w(y)Mf(y))pdy ≤cZRn (v(y)f(y))pdy (16) for all nonnegative functions f. ii) Condition (14) is also a necessary condition. That is, suppose that Bhas the property that ZRn (w(y)Mf(y))pdy ≤cZRn (v(y)f(y))pdy for all nonnegative functions f, whenever the couple of weights (w, v)satisfies 1 |Q|ZQ w(y)pdy1/p kv−1kB,Q ≤K, for all cubes Q. Then Bsatisfies (14). Particular examples are given by B(t)≈tp0logp0−1+δ(1 + t), or the weaker one B(t)≈tp0logp0−1(1 + t)[log log(1 + t)]p0−1+δ, with δ > 0. The key fact is the boundedness of M¯ Bon Lp(Rn), and the relevant class of Young functions is the following. Definition 1.6 Let 1< p < ∞. We say that a doubling Young function Bsatisfies the Bpcondition if there is a positive constant csuch that Z∞ c B(t) tp dt t≈Z∞ ctp0 ¯ B(t)p−1dt t<∞. Then we have the following characterization. Theorem 1.7 Let 1<p<∞. Suppose that Bis a Young function. Then the following are equivalent. i) B∈Bp; (17) 5
ii) there is a constant csuch that ZRn MBf(y)pdy ≤cZRn f(y)pdy (18) for all nonnegative functions f; iii) there is a constant csuch that ZRn MBf(y)pw(y)dy ≤cZRn f(y)pMw(y)dy (19) for all nonnegative functions fand w; iv) there is a constant csuch that ZRn Mf(y)pw(y) [M¯ B(u1/p)(y)]pdy ≤cZRn f(y)pMw(y) u(y)dy, (20) for all nonnegative functions f,wand u. A consequence is the following inequality: Corollary 1.8 Let 1< p < ∞. Suppose that wis a weight. Then ZRn Mf(y)pM[p0]+1w(y)1−pdy ≤cZRn f(y)pw(y)1−pdy (21) for all nonnegative functions f. As usual [r] denotes the integer part of r. Acnowledgement. We wish to thank the referee for the careful reading of this paper. 2 Preliminaries In this section we provide the necessary background from the theory of function spaces that will be used later. We begin by recalling some basic facts about the theory of Banach function spaces introduced by W.A.J. Luxemburg in [9], and we shall refer the reader to [2] for a complete account. Let (R, µ) be a measure space, and let M+(R) be the cone of µ–measurable functions on Rwhose values lie in [0,∞]. A mapping ρ:M+(R)→[0,∞] is called a Banach function norm if, for all f, g, fn,(n= 1,2,3, . . .) in M+(R), for all constants a≥0, and for all µ–measurable subsets Eof R, the following properties hold: 6
i) ρ(f) = 0 iff f= 0 µ–a.e.; ρ(af) = aρ(f); ρ(f+g)≤ρ(f) + ρ(g) ii) 0 ≤g≤f µ–a.e. implies ρ(g)≤ρ(f) iii) 0 ≤fn↑f µ–a.e. implies ρ(fn)↑ρ(f) iv) µ(E)<∞implies ρ(χE)<∞ v) µ(E)<∞implies REf dµ ≤CEρ(f), for some constant CE, 0 < CE<∞, depending on Eand ρbut independent of f. Let M(R) denote the collection of all µ–measurable functions on R. The collection X=X(ρ) of all functions f∈M(R) for which ρ(|f|) = kfkX<∞is called a Banach function space. The most important property of the Banach function spaces is the generalized H¨older inequality ZR |f(y)g(y)|dµ(y)≤ kfkXkgkX0,(22) where X0is the associate space to X. A Banach function space Xis said to be rearrangement–invariant if whenever f, g ∈Xare equimeasurable, then kfkX=kgkX. Recall that two functions are equimeasurable if µf(t) = µg(t), t > 0, where µf(t) = µ{x∈R:|f(x)|> t},is the distribution of f. Most of the properties of the rearrangement–invariant spaces can be formulated in terms of the fundamental function of X,ϕX, given by ϕX(t) = kχEkX, where µ(E) = t. Observe that the particular choice of the set Ewith µ(E) = tis immaterial by the rearrangement–invariance of X.ϕXis quasiconcave and continuous, except perhaps at the origin. Furthermore, if X0is the associated space of Xthe following identity holds ϕX(t)ϕX0(t) = t, t > 0.(23) Examples of rearrangement–invariant spaces include the Lebesgue Lpspaces, the minimal and maximal Lorentz spaces Λ, M, (cf. [2]). Also, the Orlicz and Ls,q spaces that we are going to describe briefly next. A function B: [0,∞)→[0,∞) is a Young function if it is continuous, convex and increasing satisfying B(0) = 0 and B(t)→ ∞ as t→ ∞. We shall assume that Bis normalized so that B(1) = 1. Also, we shall require that Bsatisfies the doubling condition 7
B(2t)≤C B(t), t > k (24) for some constants C > 0, k≥0. We shall make use of the following property B(t)≈t B0(t), t > 0,(25) and that t→B(t) tis increasing. Each Young function Bhas associated a complementary Young function ¯ Bthat satisfies t≤B−1(t)¯ B−1(t)≤2t, t > 0.(26) Let (X, µ) be a measure space and let Bbe a Young function. The Orlicz space LB(µ) consists of all µ–measurable functions fsuch that ZX B|f(y)| λdµ(y)<∞, for some λ > 0. LB(µ) can be normed by the Luxemburg norm defined by kfkB,µ = inf{λ > 0 : ZX B|f(y)| λdµ(y)≤1}. LB(µ) is a rearrangement–invariant space with fundamental function given by ϕB(t) = ϕLB(µ)(t) = 1 B−1(1 t).(27) In particular if Eis a measurable subset of X, then kχEkB,µ =1 B−1(1 µ(E)).(28) Finally, the associated space to LB(µ) is L¯ B(µ). A function fbelongs to the Lorentz space Ls,q, 0 < s, q ≤ ∞, if kfkLs,q(µ)=qZ∞ 0t µ{x∈Rn:|f(x)|> t}1/sqdt t1/q <∞, whenever q < ∞, and sup 0<t<∞ t µ{x∈Rn:|f(x)|> t}1/s <∞, 8
if q=∞. For each 1 < s, q ≤ ∞ Ls,q is a rearrangement–invariant Banach function space with fundamental function ϕ(t) = t1/s, and associated space Ls0,q0. 3 The general case Let Xbe a Banach function space over Rnwith respect to the Lebesgue measure. Recall that for any measurable function fand arbitrary cube Qwe defined the X–average of fover Qby kfkX,Q = τ`(Q)(f χQ) X ,(29) where τδ,δ > 0, is the dilation operator τδf(x) = f(δx), χEis characteristic function of Eand `(Q) is the sidelength of the cube Q. Note that H¨older’s inequality for Banach function spaces (22) yields after the change of variable y=`(Q)z 1 |Q|ZQ f(y)g(y)dy ≤ kfkX,Q kgkX0,Q .(30) We also introduce the following maximal operator associated to the space X. For each locally integrable function fwe have also defined MXby MXf(x) = sup x∈Q kfkX,Q , where the supremum is taken over all the cubes containing x. Proof of Theorem 1.2: Since the set of bounded functions with compact support is dense in Lp(vp) it is enough to show that there is a constant csuch that ZRn (w(y)Mf(y))pdy ≤cZRn (v(y)f(y))pdy, (31) for each nonnegative bounded function with compact support f. For each integer k, and for any constant a > 2nwe let Ωkand Dkbe the sets Ωk={x∈Rn:ak< Mf(x)}, 9
=CZRn f(y)pdy Z∞ 1/2 B(t) tp dt t=CZRn f(y)pdy, since B∈Bp. This proves that i) implies ii). For the proof that ii) implies iii) we discretize as in Theorem 1.2. We fix a constant a > 2n, and for each integer kwe let Ωk, and Dkbe the sets Ωk={x∈Rn:MBf(x)> ak}, Dk={x∈Rn:Md Bf(x)>ak 4n}. Here Md Bdenotes the dyadic version of MB. Hence, by Lemma 4.1 with t=akthere is a family of maximal nonoverlapping dyadic cubes {Qk,j}for which Ωk⊂ ∪j3Qk,j, Dk=∪jQk,j, and ak 4n<kfkB,Qk,j ≤ak 2n.(47) We shall need the following lemma. Lemma 4.2 Suppose a > 2n. For all integers k, j we let Ek,j =Qk,j −Qk,j ∩Dk+1. Then {Ek,j}is a disjoint family of sets which satisfy |Qk,j ∩Dk+1|<2n a|Qk,j|,(48) and |Qk,j|<1 1−2n a |Ek,j|.(49) We postpone the proof of this also until the end of the proof of the theorem. Now, using (47), and (49) we estimate the left side of (19) as in the proof of Theorem 1.2 by ZRn MBf(y)pw(y)dy =X kZΩk−Ωk+1 MBf(y)pw(y)dy ≤(50) ≤apX k akpw(Ωk)≤CX k,j akpw(3Qk,j)≤ ≤CX k,j kfkp B,Qk,j w(3Qk,j) = CX k,j kfkp B,Qk,j w(3Qk,j) |3Qk,j||Qk,j| ≤ 16
≤CX k,j fw(3Qk,j ) |3Qk,j|1/p p B,Qk,j |Ek,j| ≤ ≤CX k,j ZEk,j MB(f(Mw)1/p)(y)pdy ≤CZRn MB(f(Mw)1/p)(y)pdy ≤ ≤CZRn f(y)pMw(y)dy, since we are assuming ii). This proves iii). Let us assume that iii) holds. Observing that (20) is equivalent with ZRn M(fg)(y)pw(y) [M¯ B(g)(y)]pdy ≤cZRn f(y)pMw(y)dy, for all nonnegative functions f,g, and w, iv) follows immediately from (19) after an application of the inequality M(fg)(y)≤MBf(y)M¯ Bg(y), y ∈Rn, which is a consequence of the generalized H¨older’s inequality (30). To prove that iv) implies i) we let w= 1 in (20) obtaining ZRn Mf(y)p1 [M¯ B(u1/p)(y)]pdy ≤cZRn f(y)p1 u(y)dy, for all nonnegative functions f, and u. Since this is (36) in Theorem 3.1 with X=L¯ B, we can apply that proposition to get a constant c > 0 for which Zc 0 ϕB(t)p t dt t<∞.(51) Here ϕB=ϕLBis the fundamental function of LB. We claim that (51) is equivalent with B∈Bp. Indeed, by (27) and (25) it readily follows that Zc 0 ϕB(t)p t dt t=Zc 0 1 B−1(1 t)p 1 t dt t≈(52) ≈Z∞ c B(t) tp dt t, 17
from which we obtain the claim. This concludes the proof of the Theorem save for the proofs of Lemmas 4.1 and 4.2. Proof of Lemma 4.1: The proof is a simple adaptation of arguments in [6] Ch. 2. Since fis bounded with compact support, say suppf⊂K, kfkB,Q ≤ kfkL∞kχKkB,Q = =kfkL∞ 1 B−1|Q| |Q∩K|, and it follows that kfkB,Q →0 as Q↑Rn. Hence, if there are any dyadic cubes Qwith kfkB,Q > t, they are contained in cubes of this type which are maximal with respect to inclusion. We let Ct={Pj}be the family of the dyadic maximal nonoverlapping cubes satisfying t < kfkB,Pj. Let P0 jbe the only dyadic cube containing Pjwith sidelength twice that of Pj. Then t < kfkB,Pj≤2nkfkB,P 0 j . The last inequality can easily be deduced from the definition of the Luxemburg norm using the fact that t→B(t) tis increasing. Hence by the maximality of the cubes {Pj}we get t < kfkB,Pj≤2nt. (53) Observe that from this discussion it is clear that {y∈Rn:Md Bf(y)> t}=∪jPj.(54) Let x∈Ωt. By definition, there is a cube Rcontaining xsuch that t < kfkB,R .(55) Let kbe the unique integer such that 2−(k+1)n<|R| ≤ 2−kn. There is some dyadic cube with side length 2−k, and at most 2nof them, {Ji:i= 1, . . . , n}, meeting the interior of R. It is easy to see that for one of these cubes, say J1, t 2n< χJ1f B,R .(56) 18
This can be seen as follows. If for each i= 1,...,2nwe had χJif B,R ≤t 2n, we would get since R⊂ ∪2n i=1Jithat kfkB,R = χ∪2n i=1Ji f B,R ≤ ≤ 2n X i=1 χJif B,R ≤2nt 2n=t, contradicting (55). Using that |R| ≤ |J1|<2n|R|one can also show t 4n<kfkB,J1.(57) By letting Ct/(4)n={Qj}, we have by (53) that t 4n<kfkB,Qj≤t 2n,(58) for each j, yielding (44). (46) also follows since {y∈Rn:Md Bf(y)>t 4n}=∪jQj. Also, we see from (57) that J1⊂Qk, for some k, and then R⊂3J1⊂3Qk. This gives Ωt⊂ ∪j3Qj, which is (43). Now, by the left side of the inequality (58), and the definition of kfkB,Q we get |Ωt| ≤ CX j |Qj| ≤ ≤CX jZQj B4nf(y) tdy ≤CZRn Bf(y) tdy. (59) To obtain (45) we just use the standard idea of writing fas f=f1+f2, where f1(x) = f(x) if f(x)>t 2, and f1(x) = 0 otherwise. Then MBf(x)≤MBf1(x) + MBf2(x)≤MBf1(x) + t 2. Finally, since (59) holds for each f≥0, t > 0 we have |Ωt| ≤ {y∈Rn:MBf1(y)>t 2}≤CZRn Bf1(y) tdy = 19
=CZ{y∈Rn:f(y)>t/2} Bf(y) tdy, concluding the proof of Lemma 4.1. 2 We now conclude the proof of the Theorem by proving Lemma 4.2. Proof of Lemma 4.2: The family Ek,j is clearly disjoint. We note that (47) and the definition of the Luxemburg norm implies that 1<1 |Qk,j|ZQk,j B4n akf(y)dy, and 1 |Qk,j|ZQk,j B2n akf(y)dy ≤1. Hence by standard properties of the dyadic cubes we can estimate what portion of Qk,j is covered by Dk+1 as in [4] (cf. [6] p. 398) |Qk,j ∩Dk+1| |Qk,j|=X i |Qk,j ∩Qk+1,i| |Qk,j|= =X i:Qk+1,i⊂Qk,j |Qk+1,i| |Qk,j|< <X i:Qk+1,i⊂Qk,j 1 |Qk,j|ZQk+1,i B4n ak+1 f(y)dy ≤ ≤2n a 1 |Qk,j|ZQk,j ∩∪iQk+1,i B2n akf(y)dy ≤ ≤2n a. Here we have used that B(2n at)≤2n aB(t), t > 0, since 2n a<1, and because t→B(t) t is increasing. This gives (48). Finally |Ek,j| |Qk,j|>1−2n a>0, completing the proof of the Lemma and hence that of Theorem 1.7. 2 20
4.2 Proof of Corollary 1.8 If we let w= 1 and uis replaced by wp−1in (20) we have the weighted inequality ZRn Mf(y)p1 [M¯ B(w(p−1)/p)(y)]pdy ≤cZRn f(y)p1 w(y)p−1dy, when B∈Bp. Let δ= [p0]−p0+ 1 >0, and take B(t)≈tp log1+δ(1+t). Then ¯ B(t)≈ tp0log[p0](1+t) and [M¯ B(w(p−1)/p)(y)]p= [MA(w)(y)]p−1, where A(t)≈tlog[p0](1+t). Then Corollary 1.8 will follow if we prove the pointwise inequality MAw(x)≤C M[p0]+1w(x). It is enough to prove that there is a constant Csuch that for each cube Q kfkA,Q ≤C |Q|ZQ M[p0]f(y)dy. By homogeneity we can assume that the right hand side is equal to C. Then, by the definition of the Luxemburg norm we need to prove 1 |Q|ZQ A(w(y)) dy =1 |Q|ZQ w(y) log[p0](1 + w(y)) dy ≤C. But this is a consequence of iterating the following inequality of E.M. Stein [15] ZQ w(y) logk(1 + w(y)) dy ≤CZQ Mw(y) logk−1(1 + Mw(y)) dy, (60) with k= 1,2,3,· · · . 2 4.3 Some further considerations about the class Bp We observe that 1 < p < q < ∞implies that Bp⊂Bq. A typical Young function that belongs to the class Bpis B(t) = tswith 1 ≤s<p. Another more interesting example is the function Bgiven by B(t)≈tp log1+δ(1 + t), 21
or B(t)≈tp log(1 + t) [log log(1 + t)]1+δ, with δ > 0. Since the function B(t) = tsbelongs to Bpwe have 1 < s < p and so implies that B∈Bp−with 0 < < p −s, one could think that the same property would hold for any Young function in Bp. However, this is false as the following example shows. For δ > 0, consider the example mentioned above B(t)≈tp log1+δ(1 + t). Then, B∈Bp, but it can be easily shown that there is no > 0 for which B∈Bp−. We can remedy this situation if we restrict attention to those Young functions that are submultiplicative. We say that the Young function Bis submultiplicative if B(ts)≤B(t)B(s) for each t, s > 0. Lemma 4.3 Let 1< p < ∞. Assume that Bis a submultiplicative Young function such that B∈Bp. Then there exists > 0for which B∈Bp−. Proof: This is a simple consequence of the fact that B∈Bpif and only ¯α(B)< p. (Cf. for instance [2] Ch. 5.) Here ¯α(B) denotes ¯α(B) = lim t→∞ logB(t) logt =inft>1 logB(t) logt , and it can be shown that the limit exists, is finite, and strictly positive. 2 Let us make the following observation concerning a particular case of (20). Taking the weight w= 1, inequality (20) becomes ZRn Mf(y)p1 [M¯ B(u1/p)(y)]pdy ≤cZRn f(y)p1 u(y)dy, (61) 22
for all nonnegative functions f, and u. Let 1 < r < ∞, and consider B(t) = t(p0r)0. Then B∈Bp, and (61) is ZRn Mf(y)p1 [M(ur(p0−1))(y)](p−1)/r dy ≤cZRn f(y)p1 u(y)dy. However, this estimate follows from well–known results. Indeed, it is enough to show that [M¯ B(u1/p)(y)]−pis an Apweight by the theorem of Muckenhoupt and the Lebesgue differentiation theorem. Now, recall that a weight wbelongs to Apif and only if w=w1w1−p 2where w1and w2are A1weights, and that (Mg)δ∈A1 0< δ < 1 (see [6] p. 436). Then it is clear that [M¯ B(u1/p)(y)]−p= [M(ur(p0−1))(y)1/r](1−p) is an Apweight. This argument may suggest that [M¯ B(u1/p)(y)]−psatisfies the Apcondition for each B∈Bp. However, the following example indicates that this is not true in general, and thus above argument is not sharp enough to get (61). Let 1 < p < ∞,δ > 0, and let Bbe the Young function such that ¯ B(t)≈ tp0logp0−1+δ(1 + t). Then B∈Bpbut w=M¯ B(χQ(0,1) )−p6∈ Ap. Otherwise there would exist > 0 such that w∈Ap−(cf. [6] p. 399.) Hence for each M > 0 we would have Z|y|>M w(y) |y|n(p−)dy < ∞.(62) (cf. [6] p. 412.) However (39), and (27) yield w(y)≈¯ B−1(b|y|n)p,|y|> a, for some positive dimensional constant a, b. Thus using polar coordinates and (25) Z|y|>a w(y) |y|n(p−)dy ≈Z|y|>a ¯ B−1(b|y|n)p |y|n(p−)dy ≈Z∞ a1 ¯ B−1(t)p tp−1− dt t ≈Z∞ a2 tp ¯ B(t)p−1− dt t≈Z∞ a2 tp0 log1+δ(p−1)−(p0−1+δ)(t) dt t=∞, contradicting (62). 23
5 Orlicz spaces and two-weight inequalities Proof of Theorem 1.5 Part i) follows immediately from Theorem 1.2 together with Theorem 1.7. Indeed, let X=LBwith associate space X0=L¯ B. Then the hypothesis MX0=M¯ B:Lp(Rn)→Lp(Rn) in Theorem 1.2 is equivalent with ¯ B∈Bp by Theorem 1.7. Part ii) follows from Proposition 3.2 and the computation in (52). 2 As an easy consequence of this theorem we can obtain sufficient conditions much in the spirit of [5]. Corollary 5.1 Let 1<p<∞. Suppose that ϕ: (0,∞)→(0,∞)is increasing, that ϕ(2t)≤Cϕ(t), t > 0, and that for some positive constant c Z∞ c 1 ϕ(t)p−1 dt t<∞.(63) Assume that (w, v)is a couple of weights which satisfies for some positive constant K 1 |Q|ZQ w(y)pdy1/p "1 |Q|ZQ v(y)−p0ϕ wp(Q) |Q|1/p v(y)−1!dy#1/p0 ≤K, (64) for all cubes Q. Then ZRn (w(y)Mf(y))pdy ≤cZRn (v(y)f(y))pdy, (65) for all nonnegative functions f. Proof: Consider the Young function defined by B(t)≈tp0ϕ(t) Kp0which satisfies (14) (i.e. that ¯ B∈Bp). Now, (64) implies 1 |Q|ZQ B wp(Q) |Q|1/p v(y)−1!dy ≤1, or equivalently wp(Q) |Q|1/p v−1 B,Q ≤1. 24
From this and by homogeneity we get (15). Then Theorem 1.5 applies. 2 6 Lorentz spaces and two-weight inequalities In this section we study two weighted norm inequalities for the Hardy–Littlewood maximal operator whenever the weights satisfy (8) with Xbeing a Lorentz space. We begin by recalling that the maximal operator associated to the Lorentz space Ls,q is given by Ms,qf(x) = sup x∈Q 1 |Q|1/s f χQ s,q. We now state a result similar to Theorem 1.7. Theorem 6.1 Let 1< p, s < ∞, and 1≤q < ∞. Then the following are equivalent. i) s < p ; (66) ii) there is a constant csuch that ZRn Ms,qf(y)pdy ≤cZRn f(y)pdy (67) for all nonnegative functions f; iii) there is a constant csuch that ZRn Ms,qf(y)pw(y)dy ≤cZRn f(y)pMw(y)dy (68) for all nonnegative functions f, and w; iv) there is a constant csuch that ZRn Mf(y)pw(y) [Ms0,q0(u1/p)(y)]pdy ≤cZRn f(y)pMw(y) u(y)dy, (69) for all nonnegative functions f,wand u. 25