Endpoint estimates and weighted norm inequalities for commutators of fractional integrals

Publ. Mat. 47 (2003), 103–131 ENDPOINT ESTIMATES AND WEIGHTED NORM INEQUALITIES FOR COMMUTATORS OF FRACTIONAL INTEGRALS D. Cruz-Uribe, SFO and A. Fiorenza Abstract We prove that the commutator [b, Iα], b∈BMO,Iαthe fractional integral operator, satisfies the sharp, modular weak-type inequality |{x∈Rn:|[b, Iα]f(x)|>t}|≤CΨRn BbBMO |f(x)| tdx, where B(t)=tlog(e+t) and Ψ(t)=[tlog(e+tα/n)]n/(n−α). These commutators were first considered by Chanillo, and our result complements his. The heart of our proof consists of the pointwise inequality, M#([b, Iα]f)(x)≤CbBMO [Iαf(x)+Mα,Bf(x)], where M#is the sharp maximal operator, and Mα,B is a generalization of the fractional maximal operator in the scale of Orlicz spaces. Using this inequality we also prove one-weight inequalities for the commutator; to do so we prove one and two-weight norm inequalities for Mα,B which are of interest in their own right. 1. Introduction Given α,0<α<n, define the fractional integral operator Iαby Iαf(x)=Rn f(y) |x−y|n−αdy. If b∈BMO we define the first order commutator [b, Iα]tobethe operator [b, Iα]f(x)=b(x)Iαf(x)−Iα(bf)(x)=Rn (b(x)−b(y)) f(y) |x−y|n−αdy. 2000 Mathematics Subject Classification. 42B20, 42B25. Key words. Fractional integrals, commutators, BMO, Orlicz spaces, maximal functions, norm inequalities. The authors would like to thank C. P´erez for suggesting this problem, and the referee for a number of insightful comments. 104 D. Cruz-Uribe, SFO, A. Fiorenza Since b∈Lp(K) for any p>1 and Kcompact, this integral converges for all f∈Cc(Rn). The commutators [b, Iα]were introduced by Chanillo [3], who showed that for 1 <p<n/α,1/q =1/p −α/n,[b, Iα]: Lp(Rn)→Lq(Rn). This corresponds to the norm inequalities satisfied by Iα. The fractional integral also satisfies an endpoint inequality: |{x∈Rn:|Iαf(x)|>t}| ≤ C1 tRn|f|dxn/(n−α) . However, a straightforward computation with f(x)=χ[0,1] and b(x)= log(1 + x)χ(1,∞)shows that [b, Iα]isnot weak (1,n/(n−α)). (For a stronger counter-example, see Section 6 below.) Our main result is a sharp endpoint inequality for the commutator. Theorem 1.1. Given α,0<α<n, and a function b∈BMO, let B(t)=tlog(e+t)and Ψ(t)=[tlog(e+tα/n)]n/(n−α). Then there exists aconstant Csuch that |{x∈Rn:|[b, Iα]f(x)|>t}|≤CΨRn BbBMO |f(x)| tdx.(1.1) Furthermore, this result is sharp: if (1.1) holds with Ψreplaced by an increasing function Ψ0, then there exist positive constants γand Ksuch that Ψ(t/γ)≤KΨ0(t),t>0. Remark 1.2.Since Band Ψ are submultiplicative, we could write the righthand side of (1.1) as CΨ(B(bBMO )) Ψ Rn B|f(x)| tdx; this appears more natural, but (1.1) is stronger since it is homogeneous in b:wecan multiply bby a constant without changing the size of the constant C. To prove Theorem 1.1 we first prove a pointwise inequality relating [b, Iα], Iα, and a fractional maximal operator defined using the scale of Orlicz spaces. (For precise definitions, see Section 2 below.) Given a Young function B(for example, B(t)=tlog(e+t)), and α,0≤α<n, define the fractional Orlicz maximal operator Mα,B by Mα,Bf(x)=sup Qx|Q|α/nfB,Q, Commutators of Fractional Integrals 105 where the supremum is taken over all cubes containing x. When B(t)=t this reduces to the classical fractional maximal operator, Mαf(x)=sup Qx |Q|α/n |Q|Q|f|dy. The relationship between Mα,B and [b, Iα]involves the sharp maximal function of Fefferman and Stein [11]. Recall that it is defined by M#f(x)=sup Qx 1 |Q|Q|f(y)−fQ|dy, where fQ=1 |Q|Q f(x)dx. Theorem 1.3. Let B(t)=tlog(e+t). Given α,0<α<n,b∈BMO and a non-negative function f, there exists a constant Csuch that for all x, M#([b, Iα]f)(x)≤CbBMO [Iαf(x)+Mα,Bf(x)].(1.2) Remark 1.4.Theorem 1.3 remains true if B(t)=tlog(e+t)isreplaced by any “larger” Orlicz function. We will make this precise in Section 2. Remark 1.5.Inequalities similar to (1.1) and (1.2) for singular integral operators (which formally correspond to the case α=0)are true. These were first proved by P´erez [22], and our proofs are modeled on his. However, our approach to sharpness is different from his. We can also use Theorem 1.3 to prove one-weight norm inequalities for [b, Iα]. The first is a strong (p, q) inequality due to Segovia and Torrea [26] which generalizes Chanillo’s original result. Theorem 1.6. Given α,0<α<n, and p,1<p<n/α,fixqso that 1/q =1/p −α/n.Letwbeaweight satisfying the Apq condition: for all cubes Q, 1 |Q|Q wqdx1/q 1 |Q|Q w−pdx1/p ≤K<∞.(1.3) Then, given any function b∈BMO, [b, Iα]satisfies the strong (p, q) inequality Rn|[b, Iα]f|qwqdx1/q ≤CbBMO Rn|f|pwpdx1/p .(1.4) Remark 1.7.We can also use Theorem 1.3, together with the ideas in [9] and [6], to prove two-weight norm inequalities for commutators of fractional integrals. These will be treated in a separate paper. 106 D. Cruz-Uribe, SFO, A. Fiorenza The Apq condition governs the strong (p, q) inequalities for Iα; this is due to Muckenhoupt and Wheeden [18]. Given this fact, it seemed natural to conjecture that in the limiting case p=1,q=n/(n−α), the condition which governs the weak (1,n/(n−α)) inequality for Iα, wq∈A1, (also due to Muckenhoupt and Wheeden) would govern a weighted version of (1.1). However, this is not the case. Example 1.8. There exists a function wdefined on Rsuch that wq∈ A1,q=1/(1 −α), but there is no constant Csuch that wq({x∈R:|[b, Iα]f(x)|>t})≤CΨR BbBMO |f(x)| twdx ,(1.5) where Band Ψ are as in Theorem 1.1, holds for all f. This result is very surprising, especially since the analogous weighted inequality with α=0holds for singular integral operators. (See [22].) We are unsure what the correct condition on the weight wshould be for (1.5) to hold. The remainder of this paper is organized as follows. In Section 2 we state some preliminary definitions and results about Orlicz spaces. In Sections 3 and 4 we state and prove an endpoint estimate and weighted norm inequalities for the Orlicz fractional maximal operator Mα,B.We use these, together with Theorem 1.3, to prove Theorems 1.1 and 1.6. We actually prove results which hold for a large class of Young functions B, since we can do so for essentially no more work and they are of independent interest. In Section 5 we prove Theorem 1.3, in Section 6 we prove Theorem 1.1, and in Section 7 we prove Theorem 1.6, and construct Example 1.8. Throughout this paper all notation is standard or will be defined as needed. All cubes are assumed to have their sides parallel to the coordinate axes. Given a cube Qand r>0, rQ will denote the cube with the same center as Qand whose sides are rtimes as long. By weights we will always mean non-negative, locally integrable functions which are positive on a set of positive measure. Given a Lebesgue measurable set Eand a weight w,|E|will denote the Lebesgue measure of Eand w(E)=Ewdx. Given 1 <p<∞,p=p/(p−1) will denote the conjugate exponent of p.Cand cwill denote positive constants whose value may change at each appearance. Finally, we assume that the reader is familiar with the definition and basic properties of the Hardy-Littlewood maximal operator M, its dyadic variant Md, and the Muckenhoupt Apweights, 1 ≤p≤∞.Werefer the reader to [10]or[13] for further information. Commutators of Fractional Integrals 107 2. Background on Orlicz spaces In the following we are going to use some notions of Orlicz space theory. Here we summarize some basic facts; we refer the reader to [16], [17], or [24] for further details. A function B:[0,∞)→[0,∞)isaYoung function if it is continuous, convex and strictly increasing, and if B(0) = 0, B(t)→∞as t→∞. If A,Bare Young functions, we write A(t)≈B(t)ifthere are constants t0,c 1,c 2>0 such that c1A(t)≤B(t)≤c2A(t) for t≥t0. Also, we say that Bdominates A, and denote this by AB,ifthere exists c>0 such that for all t>0, A(t)≤B(ct). If this is true for all t≥t0>0, we say that ABnear infinity. AYoung function Bis said to be doubling if there exists a positive constant Csuch that B(2t)≤CB(t) for all t>0; Bis called submultiplicative if B(st)≤CB(s)B(t) for all s, t > 0. Clearly B(t)=tr, r≥1, is submultiplicative. A straightforward computation shows that B(t)=ta[log(e+t)]b,a≥1, b>0, is also submultiplicative. (For simplicity, hereafter we will omit the brackets and write simply talog(e+t)b.) Given a non-empty open set Ein Rnand a Young function B, the Orlicz space LB(E)isthe Banach space of Lebesgue measurable functions fsuch that B(|f|/λ)is(Lebesgue) integrable on Efor some λ>0. It is equipped with the Luxemburg norm fLB(E)= inf λ>0:E B|f| λdx ≤1. When Ehas finite measure (e.g. it is a cube) we often want to normalize by replacing the measure dx by dx/|E|.Inparticular, given a cube Q, we define the mean Luxemburg norm of fon Qby fB,Q = inf λ>0: 1 |Q|Q B|f| λdx ≤1.(2.1) When B(t)=tr,1≤r<∞, fB,Q =1 |Q|Q|f|rdx1/r , so the Luxemburg norm coincides with the (normalized) Lrnorm. If ABnear infinity then there exists a constant C, depending on A and B, such that for all cubes Qand functions f,fA,Q ≤CfB,Q. This follows from the standard embedding theorem which shows that LB(Q)⊂LA(Q). However, we stress that because this is the mean Luxemburg norm, the constant Cis independent of Q. 108 D. Cruz-Uribe, SFO, A. Fiorenza It follows from this that if ABnear infinity, then Mα,Af(x)≤ CMα,Bf(x). In particular, in Theorem 1.3 we can take Bto be any Young function such that tlog(e+t)Bnear infinity. (This makes precise Remark 1.4.) Given a Young function B, the complementary Young function ¯ Bis defined by ¯ B(t)=sup s>0{st −B(s)},t>0.(2.2) Band ¯ Bsatisfy the following inequality: t≤B−1(t)¯ B−1(t)≤2t. We will need a generalization of H¨older’s inequality to Orlicz spaces due to O’Neil [19]. (Also see [17]or[24].) Lemma 2.1. Given a Young function B, then for all functions fand g and all cubes Q, 1 |Q|Q|fg|dx ≤2fB,Qg¯ B,Q.(2.3) More generally, if A,Band Care Young functions such that for all t>0, B−1(t)C−1(t)≤A−1(t), then fgA,Q ≤2fB,QgC,Q.(2.4) If we set g≡1in(2.3), it immediately follows that for all Young functions B,α,0≤α<n, and x∈Rn, Mαf(x)≤CMα,Bf(x).(2.5) 3. Endpoint inequality for M α,B In this section we prove a modular endpoint inequality for the Orlicz fractional maximal operator which we need for the proof of Theorem 1.1. To state it we need the following definition. Definition 3.1. Given a Young function B, define the function hBby hB(s)=sup t>0 B(st) B(t),0≤s<∞. Remark 3.2.The function hBcould be infinite if s>1, but if Bis doubling then it is finite for all 0 <s<∞. (See [17, Theorem 11.7].) If Bis submultiplicative then hB≈B. More generally, given any B, for all s, t ≥0, B(st)≤hB(s)B(t). Commutators of Fractional Integrals 109 Theorem 3.3. Given α,0≤α<n, let Bbe aYoung function such that B(t)/tn/α is decreasing for all t>0. Then there exists a constant depending only on Bsuch that for all t>0,Mα,B satisfies the modular weak-type inequality Φ(|{x∈Rn:Mα,Bf(x)>t}|)≤CRn Bf(x) tdx,(3.1) for all non-negative f∈LB(Rn), where Φis any function such that Φ(s)≤C1Φ1(s)=   0if s=0 s hB(sα/n)if s>0. Before proving Theorem 3.3 we make a number of observations about its statement. Remark 3.4.When α=0we interpret the growth condition to mean that Bcan be any Young function. Remark 3.5.The function Φ1is well-defined: by Lemma 3.12 below, it follows from the fact that B(t)/tn/α is decreasing that 0 <h B(sα/n)<∞ for all s>0. Also note that if Bis submultiplicative, B≈hB,so Φ1(s)≈s/B(sα/n). Remark 3.6.Suppose Φ is continuous and invertible. If we let Ψ = Φ−1, and if Ψ is doubling, then inequality (3.1) can be written as |{x∈Rn:Mα,Bf(x)>t}| ≤ CΨRn Bf(x) tdx.(3.2) By Lemma 3.12, for any Bsatisfying the assumptions of Theorem 3.3 there exist functions Φ, invertible, satisfying Φ(s)≤CΦ1(s). In the proof of Theorem 1.1 we use (3.2) with B(t)=tlog(e+t). Remark 3.7.We can weaken the growth condition on Bas follows: if I(B)<n/α, where I(B) denotes the upper Boyd index of B, then there exists a Young function B1such that B1≈Band B1(t)/tn/α is decreasing. See [12, Theorem 1.1] for further details. Remark 3.8.When B(t)=t,Φ(t)=t1−α/n (or Ψ(t)=tn/(n−α)in (3.2)), and Theorem 3.3 reduces to the weak (1,n/(n−α)) inequality for the classical fractional maximal operator (cf. [10,p.89]). When α=0, Φ(t)=Ψ(t)=t, and (3.1) becomes the modular endpoint inequality for M0,B due to P´erez [21]. 110 D. Cruz-Uribe, SFO, A. Fiorenza Remark 3.9.In the proof of Theorem 1.3 we need Theorem 3.3 for the case B(t)=tlog(e+t). Since Bis submultiplicative, we can take Φ(t)= t1−α/n log(e+tα/n), or equivalently, Ψ(t)=[tlog(e+tα/n)]n/(n−α). We could replace tα/n by tin the definition of Φ and Ψ; we chose not to since with the given definitions we recapture the case α=0. Remark 3.10.In the limiting case B(t)=tn/α, Theorem 3.3 is trivial: for all x∈Rn, Mα,Bf(x)=sup Qx|Q|α/n 1 |Q|Q|f|n/α dxα/n =Rn|f|n/α dxα/n , and therefore Mα,Bfis constant. Then (3.1) is trivially true with Φ defined by Φ(0) = 0, Φ(s)=1,s>0. The proof of Theorem 3.3 requires four lemmas. Lemma 3.11. If Bis a Young function then hBis nonnegative, submultiplicative, increasing in [0,∞), strictly increasing in [0,1] and hB(1)=1. For the (easy) proof see [6, Lemma 3.1] or [17,p.84]. Lemma 3.12. Given α,0≤α<n, let Bbe aYoung function such that B(t)/tn/α is decreasing for all t>0. Then the function Φ1in Theorem 3.3 is increasing, and Φ1(s)/s is decreasing. Moreover, there exists Φsuch that Φ(s)≤C1Φ1(s)and Φis invertible. Proof of Lemma 3.12: If α=0,the assertion is trivial. If 0 <α<n, for 0 <s<σand t>0, sα/nt<σ α/nt. Since the function B(t)/tn/α is decreasing, B(σα/nt) σtn/α ≤B(sα/nt) stn/α ; therefore, B(σα/nt) σB(t)≤B(sα/nt) sB(t). If we take the supremum over all t, then hB(σα/n) σ≤hB(sα/n) s. It follows immediately that Φ1(s)≤Φ1(σ). Furthermore, by Lemma 3.11, Φ1(s)/s is decreasing. Commutators of Fractional Integrals 111 Finally, if C(s)isany continuous and strictly increasing function such that C(0)=1,C(s)→2ass→+∞, then trivially the function Φ defined by Φ(s)=C(s)Φ1(s)isinvertible and satisfies Φ(s)≤2Φ1(s) since Φ1(s)>0ifs>0. Lemma 3.13. If Φ(t)/t is decreasing, then for any positive sequence {xj}, Φ  j xj ≤ j Φ(xj). For a proof, see [14,p.83, n. 103]. Lemma 3.14. Given a non-negative, locally integrable function fand α,0≤α<n, suppose that for some Young function B, cube Qand t>0, |Q|α/nfB,Q >t. Then there exists a dyadic cube Psuch that Q⊂3Pand a positive constant βα,n, depending only on αand n, such that |P|α/nfB,P >β α,nt. When B(t)=tthis result is proved in [5], and when α=0this is implicit in [21, Lemma 4.1]. Either proof readily adapts and we omit the details. Proof of Theorem 3.3: Fix a non-negative function fin LB(Rn). Fix t>0 and define Et={x∈Rn:Mα,Bf(x)>t}. If tis such that the set Etis empty, we have nothing to prove. Otherwise, for each x∈Etthere exists a cube Qxcontaining xsuch that |Qx|α/nfB,Qx>t. By Lemma 3.14, there is a constant βsuch that for each xthere exists adyadic cube Pxwith Qx⊂3Pxand |Px|α/nfB,Px>βt.(3.3) Since f∈LB(Rn), we can replace the collection {Px}with a maximal disjoint subcollection {Pj}. Each Pjsatisfies (3.3), and by our choice of the Qx’s, Et⊂j3Pj. 118 D. Cruz-Uribe, SFO, A. Fiorenza As we noted above, since Csatisfies the Bpcondition, M0,C is bounded on Lp(see [21]); hence ≤CRn|f|pvpdxq/p . This completes our proof. Proof of Corollary 4.4: Fix α,pand qas in the hypotheses. It will suffice to show that if w∈Apq then the pair (w,w) satisfies the hypotheses of Theorem 4.2. Let r=1+p/q; then we can re-write the Apq condition as 1 |Q|Q w−pdx1 |Q|Q (w−p)1−rdxr−1 ≤K<∞. Hence w−pis in Ar, and so satisfies the reverse H¨older inequality with exponent s>1. Let A(t)=tsp; then A−1(t)=t1/sp. Therefore, if we let C−1(t)= t1/(sp) log(e+t), we have that A−1(t)C−1(t)= t log(t)≈B−1(t). Furthermore, C(t)≈(tlog(e+t))(sp); since sp>p ,(sp)<p, and so Csatisfies the Bpcondition. Inequality (4.1) now follows at once: |Q|α/n+1/q−1/p 1 |Q|Q wqdx1/q v−1A,Q =1 |Q|Q w−pdx1/q 1 |Q|Q w−psdx1/sp ≤C1 |Q|Q w−pdxq1 |Q|Q w−pdx1/p ≤K. Commutators of Fractional Integrals 119 5. Proof of Theorem 1.3 Our proof requires several facts about functions in BMO and about the fractional integral operator. We gather these in two lemmas. Lemma 5.1. The following are true: (1) A function bis in BMO if for each cube Qthere exists a constant cQsuch that sup Q 1 |Q|Q|b(x)−cQ|dx < ∞. Further, this supremum is comparable to bBMO . (2) For each p,1<p<∞, there exists a constant Cpsuch that sup Q1 |Q|Q|b(x)−bQ|pdx1/p ≤CpbBMO . (3) If b∈BMO then there exists a constant Csuch that for every cube Q, 1 |Q|Q exp |b(x)−bQ| CbBMO dx < ∞. (4) If b∈BMO then for any cube Qand k≥0, |b2k+1Q−bQ|≤2(k+1)bBMO . Foraproof of (1)–(3), see [10, Chapter 6]. For a proof of (4), see [27, p. 206]. Lemma 5.2. Given α,0<α<n, and a non-negative function f, the following are true: (1) There exists a constant Csuch that for any cube Q, Q Iαfdx≤C|Q|α/n Rn fdx;(5.1) (2) Iαf∈A1, that is, there exists a constant Csuch that M(Iαf)(x)≤ CIαf(x)for almost every x. Hence, it satisfies the reverse H¨older inequality for some exponent s>1. (3) Iαis weak (1,n/(n−α)): for all λ>0, |{x∈Rn:|Iα(x)|>λ}| ≤ C1 λRn|f(x)|dxn/(n−α) .(5.2) 120 D. Cruz-Uribe, SFO, A. Fiorenza Proof: Inequality (5.1) follows easily from Fubini’s theorem: Q Iαfdx=QRn f(y)dy |x−y|n−αdx =Rn f(y)Q|x−y|α−ndxdy ≤C|Q|α/n Rn f(x)dx. To see that Iαf∈A1,itsuffices to note that |x|α−n∈A1, and the convolution of a non-negative function with an A1weight is again in A1. (Cf. [25].) For the weak (1,n/(n−α)) inequality, see [10,p.89]. Proof of Theorem 1.3: By homogeneity, it will suffice to prove (1.2) for x=0.Further, by Lemma 5.2(2), and Lemma 5.1(1), it will suffice to show that, given a cube Qcentered at the origin, there exists a constant cQsuch that 1 |Q|Q|[b, Iα]f(y)−cQ|dy ≤CbBMO [M(Iαf)(0) + Mα,Bf(0)].(5.3) Decompose fas f1+f2, where f1=fχQ∗, and Q∗is the cube centered at the origin whose sides are 2√ntimes larger. Let cQ=(Iα((b−bQ∗)f2))Q. Then, since [b, Iα]f=[b−bQ∗,I α]f, 1 |Q|Q|[b, Iα]f(y)−cQ|dy ≤1 |Q|Q|(b(y)−bQ∗)Iαf(y)|dy +1 |Q|Q|Iα((b−bQ∗)f1)(y)|dy +1 |Q|Q|Iα((b−bQ∗)f2)(y)−(Iα((b−bQ∗)f2))Q|dy =I1+I2+I3. We estimate each integral in turn. For I1,byLemma 5.2(2), Iαf satisfies the reverse H¨older inequality with exponent s>1. Therefore, Commutators of Fractional Integrals 121 if we apply H¨older’s inequality with exponent s,byLemma 5.1(2), I1≤1 |Q|Q|(b(y)−bQ∗)|sdy1/s 1 |Q|Q Iαf(y)sdy1/s ≤CbBMO 1 |Q|Q Iαf(y)dy ≤CbBMO M(Iαf)(0). To estimate the second integral, note that by Lemma 5.1(3), 1 |Q∗|Q∗ exp |b(y)−bQ∗| CbBMO dy < ∞. Since B(t)=tlog(e+t), ¯ B(t)et−1; therefore, by (2.1), b−bQ∗¯ B,Q∗≤CbBMO . Hence, by Lemma 5.2(1), and by the generalized H¨older inequality (2.3), I2≤C|Q|α/n |Q|Rn|b(y)−bQ∗|f1(y)dy =C|Q∗|α/n |Q∗|Q∗|b(y)−bQ∗|f(y)dy ≤C|Q∗|α/nb−bQ∗¯ B,Q∗fB,Q∗ ≤CbBMOMα,Bf(0). Finally, we estimate the third integral. By the mean-value theorem, if |x|>2|h|then there exists θ,0≤θ≤1, such that  1 |x|n−α−1 |x+h|n−α≤C|h| |x+θh|n−α+1 ≤C|h| |x|n−α+1 . 122 D. Cruz-Uribe, SFO, A. Fiorenza If x, y ∈Qand z∈Rn\2kQ∗,k≥0, then |x−z|>2k+1|y−x|. Therefore, I3≤1 |Q|Q 1 |Q|QRn \Q∗ |(b(z)−bQ∗)f(z)| 1 |y−z|n−α−1 |x−z|n−α dz dx dy ≤1 |Q|2QQ ∞  k=0 2k+1Q∗\2kQ∗|(b(z)−bQ∗)f(z)||y−x| |x−z|n−α+1 dz dx dy ≤C |Q|2QQ ∞  k=0 2−k |2k+1Q∗|1−α/n 2k+1Q∗|(b(z)−bQ∗)f(z)|dz dx dy ≤C ∞  k=0 2−k |2k+1Q∗|1−α/n 2k+1Q∗|(b(z)−b2k+1Q∗)f(z)|dz +C ∞  k=0 2−k |2k+1Q∗|1−α/n 2k+1Q∗|(b2k+1Q∗−bQ∗)f(z)|dz ≤ ∞  k=0 2−k|2k+1Q∗|α/nb−b2k+1Q∗¯ B,2k+1Q∗fB,2k+1 Q∗ +CbBMO ∞  k=0 (k+ 1)2−k |2k+1Q∗|1−α/n 2k+1Q∗|f(z)|dz ≤CbBMO Mα,Bf(0) + CbBMO Mαf(0) ≤CbBMO Mα,Bf(0). The fifth inequality follows from Lemma 5.1(4), and the last inequality follows from (2.5). This completes the proof. 6. Proof of Theorem 1.1 In our proof we need a variant of the good-λinequality of Fefferman and Stein [11] relating the dyadic maximal operator and the sharp maximal operator. (Also see [10,p.121].) This exact result is given by P´erez [22]; the proof is a straightforward modification of the proof of the standard result and so is omitted. Commutators of Fractional Integrals 123 Lemma 6.1. Let ϕbe apositive function on (0,∞)such that ϕ(2t)≤ ϕ(t)for all t>0. Then there exists a positive constant Csuch that sup λ>0 ϕ(λ)|{y∈Rn:Mdf(y)>λ}|≤Csup λ>0 ϕ(λ)|{y∈Rn:M#f(y)>λ}| for all functions such that the lefthand side is finite. Proof of Theorem 1.1: First, fix a function in BMO.IfbBMO =0 then bis constant and the result is trivial. We may therefore assume that bBMO =0. Now fix a function f. Since we can decompose an arbitrary function into the sum of its positive and negative parts, without loss of generality we may assume that fis non-negative. By homogeneity, it will suffice to prove (1.1) when t=1,that is, |{x∈Rn:|[b, Iα]f(x)|>1}| ≤ CΨRn B(bBMO |f(x)|)dx.(6.1) But in this case, |{x∈Rn:|[b, Iα]f(x)|>1}| ≤Ψ(B(1)) sup t>0 1 Ψ(B(1/t))|{x∈Rn:|[b, Iα]f(x)|>t}| ≤Ψ(B(1)) sup t>0 1 Ψ(B(1/t))|{x∈Rn:Md([b, Iα]f)(x)>t}|. Let ϕ(t)=1/Ψ(B(1/t)); then a straightforward but somewhat tedious calculation shows that lim t→0 ϕ(2t) ϕ(t)= lim t→∞ ϕ(2t) ϕ(t)=2 n/(n−α), 124 D. Cruz-Uribe, SFO, A. Fiorenza and so ϕ(2t)≤Cϕ(t) for all t>0. Therefore, by Lemma 6.1 and Theorem 1.3, |{x∈Rn:|[b, Iα]f(x)|>1}| ≤Csup t>0 1 Ψ(B(1/t))|{x∈Rn:M#([b, Iα]f)(x)>t}| ≤Csup t>0 1 Ψ(B(1/t)) x∈Rn:Iαf(x)+Mα,Bf(x)>t CbBMO  ≤Csup t>0 1 Ψ(B(1/t)) x∈Rn:Iαf(x)>t CbBMO  +Csup t>0 1 Ψ(B(1/t)) x∈Rn:Mα,Bf(x)>t CbBMO  . By (5.2) and (3.2), and since Ψ and Bare submultiplicative, ≤Csup t>0 1 Ψ(B(1/t)) bBMO tRn|f(x)|dxn/(n−α) +Csup t>0 1 Ψ(B(1/t))ΨRn BbBMO |f(x)| tdx ≤Csup t>0 1 Ψ(B(1/t)) ·1 tn/(n−α)RnbBMO |f(x)|dxn/(n−α) +Csup t>0 1 Ψ(B(1/t)) ·Ψ(B(1/t))ΨRn B(bBMO |f(x)|)dx =J1+J2. Note that sup t>0 1 Ψ(B(1/t)) ·1 tn/(n−α)≤C,(6.2) since 1 Ψ(B(1/t)) ·1 tn/(n−α)=1 log(e+1/t) log e+ [(1/t) log(e+1/t)]α/n n n−α is continuous and has finite limits as t→0, t→∞(0 and 1 respectively). Commutators of Fractional Integrals 125 Furthermore, since t≤B(t) and tn/(n−α)≤Ψ(t), we have that RnbBMO |f(x)|dxn/(n−α) ≤ΨRn B(bBMO |f(x)|)dx.(6.3) From (6.2) and (6.3) we get that J1+J2≤CΨRn B(bBMO |f(x)|)dx, which proves (6.1). We will now show that (1.1) is sharp, in the sense that if we can replace Ψ by Ψ0, then for all t>0, Ψ(t/γ)≤KΨ0(t). To do so, we will adapt the argument in Remark 3.15, which explored sharpness in Theorem 3.3. Let n=1and 0 <α<1, fix x>0 and let Nbe such that 0<N<x; the exact value of Nwill be chosen below. Let f=χ[0,N]. First, we show that there is a constant K, depending only on αsuch that for all y>N, Mα,Bf(y)≤Kyα B−1(y/N).(6.4) (As we showed in Remark 3.15, the opposite inequality holds with constant 1.) To see this, note that since fis a non-increasing function on [0,∞), Mα,Bf(y)=sup z≥y zαfB,[0,z]= sup z≥y zα B−1(z/N). Therefore, it will suffice to show that there exists K>0, such that if z≥y, zα B−1(z/N)≤Kyα B−1(y/N). Let H(t)=tα/B−1(t/N); since for all t>0, B−1(t)≈t/ log(e+t), we have that H(z) H(y)≤C(z/N)α−1log(e+z/N) (y/N)α−1log(e+y/N). The function log(e+t)/t1−αis either decreasing or has a unique local maximum on [1,∞); it follows, therefore, that the righthand term is dominated by a constant which depends only on α. This establishes (6.4). Now let b(y)=log(e+y/N)χ(N,∞)(y). Since the BMO norm is dilation invariant, the BMO norm of bdoes not depend on N.Forall z, 126 D. Cruz-Uribe, SFO, A. Fiorenza Iα(bf)(z)=0,sofor all ysuch that N<y<x, |[b, Iα]f(y)|=b(y)Iαf(y) = log(e+y/N)N 0 dz |z−y|1−α ≥Nlog(e+y/N) y1−α ≥cyα B−1(y/N); by inequality (6.4), ≥cMα,Bf(y) >cxα B−1(x/N). The constant cdepends only on α. We now consider two cases, depending on whether xis large or small. Suppose first that xis such that cxα>B −1(2), and let N= x/B(cxα)<x/2. Then the above inequality shows that for all y∈ [x/2,x], [b, Iα]f(y)>1. Hence, if inequality (1.1) holds for some increasing function Ψ0then we have that x/2≤|{y∈R:|[b, Iα]f(y)|>1}| ≤CΨ0R B(bBMO f(y)) dy ≤CΨ0(NB(bBMO )). On the other hand, there is a constant γ, depending only on α, such that B(bBMO )N=B(bBMO )x B(cxα)≤γΦ(c1/αx), where Φ = Ψ−1. Therefore, if set the righthand side equal to tand solve for x,weget that c1/αx=Ψ(t/γ). If we combine this with the inequality above we get that for all tsufficiently large, Ψ(t/γ)≤KΨ0(t). We will now show that the same inequality holds for all tsufficiently small. Fix x>0 small and let 0 <N<x/2; the exact value of Nwill be Commutators of Fractional Integrals 127 fixed below. Then, by the above argument we have that if N<y<x, [b, Iα]f(y)≥cxα B−1(x/N). Therefore, if we fix usuch that 2u=cxα B−1(x/N), then arguing as above and as in Remark 3.15, we have that x/2≤CΨ0R BbBMO f(y) udy =CΨ0NBbBMO 2B−1(x/N) cxα. Again arguing as in Remark 3.15 we can choose Nsufficiently small such that ≤CΨ0(CB(bBMO )x1−α) ≤CΨ0(CB(bBMO )Φ(x)). We can now argue as we did above to get that for all tsufficiently small, Ψ(t/γ)≤KΨ0(t). This completes our proof. 7. Proof of Theorem 1.6 and Construction of Example 1.8 Our proof of Theorem 1.6 requires two facts which we give as a lemma. Lemma 7.1. The following are true: (1) If w∈Apfor some p>1, then for all q>0, there exists a constant Cqsuch that Rn (Mdf)qwdx≤CqRn (M#f)qw dx. (2) Given α,0<α<n,p,1<p<n/α, let 1/q =1/p −α/n. Then if w∈Apq, Rn|Iαf|qwqdx1/q ≤CRn|f|pwpdx1/p . The first inequality is due to Journ´e[15]; also see [10,p.144]. The second is due to Muckenhoupt and Wheeden [18].