A TWO-WEIGHT ESTIMATE FOR A CLASS OF FRACTIONAL INTEGRAL OPERATORS WITH ROUGH KERNEL

We prove that the operators in a class of rough fractional integral operators and the related maximal operators are bounded from Lp(vp) to Lq(uq) with weight pair (u,v).


Introduction and results
Suppose that Ω ∈ L s (S n−1 ) (s ≥ 1) is homogeneous of degree zero in R n with zero mean value on S n−1 , then it is well known that the Calder ón-Zygmund singular integral is defined by ( In 1967, Bajšanski and Coifman [1] first discussed the boundedness of operators on a class of singular integral operators T A Ω which are associated with the commutators of the Calder ón-Zygmund singular integral T Ω .The operator T A Ω is defined by where A(x) is defined on R n and R m (A;x, y) denotes the mth remainder of the Taylor series of A at x about y.More precisely, where . Following [1], Cohen and Gosselin studied also the boundedness of T A Ω in [2,3,4].Notice that the fractional integral operator is closely related to the Calder ón-Zygmund singular integral operator, which is defined by for 0 < α < n. (See [5,7,9] for the boundedness of T Ω,α .)Therefore, a natural and interesting problem is whether there are similar mapping properties if we replace the kernel function Ω(x)|x| −n by Ω(x)|x| −(n−α) (0 < α < n) in definition (1.2).In this paper, we will consider the above problem.In a very simple way, which is altogether different from the method in [3,4], we will obtain the weighted (L p ,L q ) boundedness with the weight pair (u,v) for the rough fractional integral operator T A Ω,α , which is defined by where 0 < α < n, Ω ∈ L s (S n−1 ) (s > 1) is homogeneous of degree zero in R n , and R m (A; x, y) is as in (1.3).As a corollary of the above result, in this paper, we show also that the same conclusion holds for the fractional maximal operator M A Ω,α , where Before stating our results, we give some definitions.In the following definitions, p = p/(p − 1) and Q denotes a cube in R n with its sides parallel to the coordinate axes and the supremum is taken over all such cubes. (1.10)

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Now we state the results obtained in this paper as follows. Theorem , and 1/q = 1/r + 1/ p − α/n.Moreover, 1/t = 1/r + 1/ p and 1/l = 1/ p − α/n.If s, l, t, p, q and ω satisfy one of the following conditions: (a) 1 ≤ s < t and (u ls /q ,v ls /q ) ∈ A * (t/s , q/s ), and u ls /q ,v ls /q ∈ A(t/s , q/s ); (b) s > l and (v −t s /p ,u −t s /p ) ∈ A * (l /s , p /s ), and v −t s /p ,u −t s /p ∈ A(l /s , p /s ), then there is a constant C > 0, independent of f , such that The proof of Theorem 1.5 depends on the following weighted (L p ,L q ) boundedness and the weak boundedness of the rough fractional integral operator T Ω,α .

Some elementary conclusions
We begin by giving some properties of the weight classes A p and A * p , which will be applied in the proof of theorem.Some elementary properties of A p and p , then for any r > p, (u,v) ∈ A * r .See [8,6] for the proofs of (i)-(iii), respectively.The relations between A(p, q) and A p , A * (p, q) and A * p are explained as follows (see [6]).Suppose that 0 < α < n, 1 < p < n/α and 1/q = 1/ p − α/n, then 2) The proof of Theorem 1.5 is set up on the conclusions of the following two lemmas.
(a) If 1 ≤ s < t and (u ls /q ,v ls /q ) ∈ A * (t/s , q/s ), then (u s ,v s ) ∈ A * (p/s ,l/s ) and (u s , v s ) ∈ A * (t/s , q/s ).(b) If 1 ≤ s < t and u ls /q ,v ls /q ∈ A(t/s , q/s ), then u s ,v s ∈ A(p/s ,l/s ) and u s ,v s ∈ A(t/s , q/s ).
Lemma 2.2.Suppose that m ≥ 1 and where f * (x) denotes the Hardy-Littlewood maximal function of f (x).

Proof. Obviously, |R
Thus by (2.8), if we can show that then we get (2.6).

Proof of Theorem 1.5
Now we turn to the proof of Theorem 1.5.For m ≥ 1, by (2.6), we get Thus, If r = ∞, then t = p, l = q.From (3.2) and L ∞ boundedness of the Hardy-Littlewood maximal operator, it is easy to see that neither I nor II is larger than By Theorem 1.6, we know that when r = ∞, the conclusion of Theorem 1.5 holds.

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As for the case 1 < r < ∞, we first consider the condition (a).Since 1/l = 1/q − 1/r, we have 1/(r/q) + 1/(l/q) = 1.Using Hölder's inequality, we get (3.4) By 1/l = 1/ p − α/n, s < t < p, and the conclusions (a) and (b) of Lemma 2.1, we have (u s ,v s ) ∈ A * (p/s ,l/s ) and u s ,v s ∈ A(p/s ,l/s ).Applying Theorem 1.6 for (s, p,l) and (u,v) and the L r (r > 1) boundedness of Hardy-Littlewood maximal operator, we have Now we estimate II.Since 1 > 1/t > 1/ p, so 1 < t < p < n/α and 1/q = 1/t − α/n.By s < t, using the conclusions (a) and (b) of Lemma 2.1 again, we have (u s ,v s ) ∈ A * (t/s , q/s ) and u s ,v s ∈ A(t/s , q/s ).Hence by the weighted (L t ,L q ) boundedness of T Ω,α with different weights (Theorem 1.6), the L r (r > 1) boundedness of Hardy-Littlewood maximal operator and using Hölder's inequality for r/t and p/t, we get From (3.2)-(3.6),we prove that T A Ω,α is a bounded operator from L p (v p ) to L q (u q ) when the condition (a) of Theorem 1.5 is satisfied.
Applying the conclusions (c), (d) of Lemma 2.1 and Theorem 1.6 under the condition (b), respectively, and by the same method as above, we may prove that when the condition (b) of Theorem 1.5 is satisfied, the operator T A Ω,α is also bounded from L p (v p ) to L q (u q ).Here we omit the details.Remark 3.1.As a corollary of Theorem 1.5, below we will show that the same conclusions hold also for the multilinear fractional maximal operator M A Ω,α (see (1.6) for its definition).Using the idea of proving [5, Lemma 2], we can obtain the following pointwise relation between M A Ω,α and TA |Ω|,α .Lemma 3.2.Let 0 < α < n, Ω ∈ L 1 (S n−1 ).Then From the process proving Theorem 1.5, it is easy to see that the conclusions of Theorem 1.5 hold still for TA |Ω|,α .Combining this with Lemma 3.2, we can easily obtain the weighted (L p ,L q ) boundedness of M A Ω,α for different wights.
Remark 3.3.When D γ A ∈ BMO(R n ), the weighted boundedness of T A Ω,α with the weight pair (u,v) will be given in the following paper.