IMPROVED MUCKENHOUPT-WHEEDEN INEQUALITY AND WEIGHTED INEQUALITIES FOR POTENTIAL OPERATORS

By a variant of the standard good λ inequality, we prove the Muckenhoupt-Wheeden inequality for measures which are not necessarily in the Muckenhoupt class. Moreover we can deal with a general potential operator, and consequently we obtain a suitable approach to the two weight inequality for such an operator when one of the weight functions satisfies a reverse doubling condition.


Introduction
In this paper dµ, dω are locally finite positive Borel measures of R n , n ≥ 1.For a nonnegative locally-dµ integrable function K(x, y) (a.e.continuous in the first variable) we define the potential operator (T fµ)(x) = y∈R n K(x, y)f (y)dµ(y).
The dual operator T * is the operator defined by the kernel K * (x, y) = K(y, z).The usual fractional integral operator I s , with 0 < s < n, is given by K(x, y) = |x − y| s−n .Other examples of operators T are those introduced by Chanillo-Stromberg-Wheeden [Ch-St-Wh] and given by kernels K(x, y) = a (y,|x−y|)  |x−y| n .Here a is considered as a function defined on balls of R n and which satisfies some growth conditions we precise below.
We are interested in finding a constant C > 0 for which (P T ) T fµ L q ω ≤ C f L p µ for all nonnegative functions f with 1 < p, q < ∞.The constant C depends only on n, p, q, ω, µ and K; and when it is necessary we denote this dependance by writing C = C (n, p, q, K, ω, µ).Here g L r ν = R n |g| r dν 1 r .The inequality (P T ) includes the usual two weight norm inequality since it is sufficient to replace f by fv 1 p−1 , and to take dω = udx, dµ = v − 1 p−1 dx, where dx is the usual Lebesgue measure on R n .Inequality (P T ) with T = I s has been studied extensively by many authors (see for instance [Ke-Sa], [Sa-Wh] and [Pe] and the reference given by them).Kerman and Sawyer [Ke-Sa] solved the problem (P Is ) with dω = dx.This particular case is first interesting since it is the usual form which appears in many mathematic and physic areas.It also appears that the case dω = dx is naturally suitable to be treated.In fact using a good λ-inequality, they proved that the left member of (P T ) is majorized by the L q norm of the fractional maximal function.So (P T ) is reduced to a weighted inequality for maximal operator, whose study was done by the first author [Sa].Problem (P T ) with general measures dµ and dω was solved by Sawyer and Wheeden [Sa-Wh].
Let us consider the operator T = I s with 0 < s < n.We have the pointwise inequality where M s is the fractional maximal operator defined by We generally use the letter Q to denote a cube of R n , and by which we mean a product of n intervals [a i , a i +t] (0 < t < ∞).The Muckenhoupt-Wheeden inequality [Mu-Wh] yields a sort of converse (in norm) of the above inequality, and asserts that for 0 < q < ∞: all functions f whenever the measure dω satisfies the Muckenhoupt condition A ∞ , i.e. there are c = c(ω), δ > 0 such as for all cubes Q and all measurables sets E ⊂ Q.
In his thesis Perez [Pe] gave a weaker condition than the A ∞ 's.He proved the above Muckenhoupt-Wheeden inequality for measures dω satisfying D ∞ and B ρ conditions with ρ > 1 − s n , and which can be noted as dω ∈ D ∞ ∩ B ρ .These conditions respectively mean: (here 2Q is the cube having the same center as Q and the length expanded twice) Contrary to the Muckenhoupt-Wheeden technique [Mu-Wh], the Perez's analysis [Pe] is not based on the standard good-λ inequalities.This last author used some estimates obtained by Frazier and Jawerth [Fr-Ja] for local maximal operator, and moreover he was able to treat the problem with a general convolution operator.
In this paper we also prove the Muckenhoupt-Wheeden inequality for measures which are not necessarily in the Muckenhoupt class (see Corollary 4), and with the general potential operator T described above.We do this, with a sort of a variant of the standard good λ inequality and by introducing a suitable maximal operator M T,ω (see Theorem 1).The additional conditions on the measure dω arise only in order to relate this "exotic" maximal operator to a more standard one like M s (see Theorem 3).Consequently we obtain a suitable approach to the two weight inequality for such an operator when one of the weight functions satisfies a reverse doubling condition (see Theorem 5).

Statements of results
Let us define the dyadic maximal operator where k and a i are integers.Fix q ≥ 1.By using the Holder inequality we can observe that (M d T,ω,µ f )(x) is a.e.finite for all bounded functions with compact supports whenever measures dω and dµ satisfy the condition Our first result is as follow: Theorem 1.Let 0 < q < ∞ and K be a nonnegative kernel satisfying the hypothesis H. Assume the measures dω and dµ satisfy the condition (S T ).Then there is C = C(n, q, K) > 0 so that If M d ω is the dyadic maximal operator defined by and consequently we get Proposition 2. Let K, dω, dµ be as above Then for q > 1 we have Moreover this equivalence also holds for the range of q ∈]0, 1] whenever for all f nonnegative functions g.
The above equivalence means The extra assumption in this result is satisfied for instance for the kernel K(x, y) = |x − y| s−n , with dω = dx the Lebesgue measure, and more generally for measures dω ∈ D ∞ ∩ B ρ with 1 − s n < ρ.Thus in view of Theorem 1, the inequality (P T ) is reduced to the following one, related for In order to get this last one, we impose more hypothesis on the kernel K.So as to simplify, we only deal with kernels where a is a function defined on balls satisfying the following hypotheses H: for all balls B and t ≥ 1.
We also define the function a on cubes by a(Q) = a(B), where B is the smallest ball which contains the cube Q.A suitable dyadic maximal operator related to the potential operator The nondyadic version of M d Φ is merely denoted by M Φ .The measure dω satisfies the condition RD ρ with ρ > 0 (and we write as dω ∈ RD ω ) when there c = c(ω, n) > 0 for which Our second result ensures the link between the two maximal operators we have defined above.
In fact C 2 does not depend on the individual measure dω but only on the RD ρ constant of dω.The claim we announced in the introduction can be stated as This is an immediate consequence of Theorems 1 and 2. Indeed since for all R > 0: < ∞ is reduced to the one written in this corollary.Note also in studying the two weight inequality By Theorems 1 and 3, the problem (P T ) is then reduced to the following maximal inequality By the study of this last case (see [Ra1], or adapt the proof given in [Sa]) then we get Theorem 5. Let 1 < p, q < ∞, and K = K a be a kernel satisfying Hi)-ii) with 0 < λ, σ < 1. Suppose dω ∈ RD ω with (1 − λ) < ρ.Then the inequality where C > 0 is a constant which does not depend of each sequence (Q k ) k of cubes and (ε k ) k of nonnegative reals ε k .Moreover in the case 1 < p ≤ q the condition (2) can be replaced by Sawyer and Wheeden [Sa-Wh] proved that for 1 < p ≤ q and for all general measures dω and dµ, then (P T ) is equivalent to (2 ) and With an additional hypothesis on the measure dµ we can simplify the conditions in Theorem 5. We first consider the case p ≤ q.

Suppose dµ ∈ RD(p). Then the inequality (P T ) holds if and only if
for some m ≥ 4 and C > 0 For dµ ∈ RD ρ ∩ D ε ,p with max(1 − λ, 1 p ε ) < ρ , a necessary and sufficient condition for This equivalence is also true when dµ Remark.Now we show that the use of the sharp maximal M # (see [Ya] for a definition) is not well adapted to weaken the weight condition in the Muckenhoupt-Wheeden inequality (1) Indeed such a purpose is based on the two inequalities: (2) Inequality ( 2) is valid for all functions f with (I s f ) ∈ L 1 loc and was proved in [Ad].Although (3) is well known to be true for w ∈ A ∞ , Yabuta [Ya] had obtained such an inequality with a weak condition he denoted as w ∈ C r (with r > q).Thus we think get (1) with this last condition.But since It was proved in [Ya] that condition like (4) implies necessarily w ∈ A ∞ .

Some Lemmas
We first state two Lemmas we need and then we give their proofs.
Lemma 2. Let T = T a be an operator with the kernel K = K a satisfying Hi)-ii), and let dν be a positive Borel measure. A where and

Proof of Lemma 1:
Let Q be a dyadic cube.Then we have Diving by |Q| ω this inequality and taking the supremum, we obtain the conclusion.

Proof of Lemma 2:
A) Let Q be a cube with center x 0 and length 2R > 0. Then |x − y| ≤ c2R for all x, y ∈ Q with c = c(n).We obtain B) Let Q be a cube as in part A, and let m ≥ 1.We can write where and For S 1 (y) we can observe that for y ∈ (mQ) and x ∈ Q then B(y, |x − y|) ⊂ c(n)Q and consequently we get For S 2 (y) with R = |Q| 1 n , we have The proof of the part C) lies on the same ideas.We leave the detail for the reader.

Proofs of main results
Preliminaries for the proof of Theorem 1.
Let f be a nonnegative function bounded and with support compact.By the hypothesis (S T ) then we can observe that Since T fµ is semicontinuous, so for each k ∈ Z the open set Ω k = {T fµ > 2 k } can be written as j Q jk where the Q jk are the dyadic cubes maximal among those dyadic cubes Q satisfying (RQ) ⊂ Ω k .Choosing R ≥ 3 sufficiently large (depending only on the dimension n), we obtain Let fix m ≥ 2 which we will choose later, and let define Using the hypothesis H on the kernel K, and the Whitney condition we get

Lemma.
1) There is C = C(K, R) > 0 so that for all k, j 2) For a suitable choice of the integer m By the Whitney condition one can find at least one z which belongs to (3RQ jk ) ∩ Ω c k .It first implies: The conclusion 1) appears from these two inequalities, indeed we have The part 2) will be a direct consequence of 1).Since So, by integration with respect to the measure dω, this involves

Proof of the Theorem 1:
Using this Lemma, now we prove the inequality Let β ∈]0, 1[ whose value is to be specified later in the course of the proof.Then we get Since T fµ q L q ω < ∞, then choosing β ∈]0, 1[ and c (q, m)β < 1 2 , we have Therefore the Theorem is proved for each bounded function f with support compact.For a general nonnegative function f , we can also obtain the same conclusion by using the monotone convergence theorem.

Proof of proposition 2:
Since the first inequality is proved in Theorem 1, then we are reduced to get the converse inequality . By the well known arguments (covering lemma using dyadic cubes and interpolation) then this last maximal inequality is valid for all q > 1.The same inequality ( * ) is also true for all q with 0 < q ≤ 1 by the means of the extra-hypothesis Hence, we have ( where c = c(n) ≥ 3.By part B) of Lemma 3, the first member of ($) is essentially dominated by the sum of and So it is clear, that it remains to estimate S 2 (.).If λ = 1, then we immediately get Now for λ ∈]0, 1[ we use the hypothesis dωRD ρ with 1 − λ < ρ.We also note that for x ∈ (3Q) then

Proof of Theorem 5:
It is clear that (2) is a necessary condition for (P T ).To get the condition (1) we first note that for |x| > R (R > 0) and |y| < 1 2 R then |x − y| ≈ |x|, and consequently taking f = 1I B(0,R) in inequality (P T ) we have . Now we suppose the conditions (1) and ( 2) are satisfied.The keys for the converse are the following: Indeed by (i) and (ii) we have T fµ L q ω ≤ c M T,ω,µ f L q ω by Theorem 1 and by using (i) ≤ c M Φ fµ L q ω by Theorem 3 since dω ∈ RD ρ with 1 − λ < ρ ≤ c f L p µ by (ii).
To get the point (i), we note that (T 1I B(0,R) µ)1I |x|<2R L q ω < ∞.On the otherhand, we have By a result in [Ra1], a sufficient (and necessary) condition for the embedding (ii) is By Lemma 2 A) then (M Φ fµ) ≤ c(T fµ) and consequently the condition (2) in Theorem 5 implies the above one.

Proof of Proposition 6:
To prove the first part of Proposition 6, we suppose Since 1 < p ≤ q this condition implies M Φ µ : L p µ → L q ω .And as above to get (i) it is sufficient to prove with a constant m ≥ 4. Using the fact that dµ ∈ RD ρ for some ρ > 0, then by Lemma 2 (part C) we get For the second part of this proposition, the point is to note that M Φ µ : ]).

Proof of Proposition 7:
It is clear that a necessary condition for (P T ) is with m ≥ 4 and for all cubes Q, Q k and all ε k > 0. Conversely we suppose this condition be satisfied and dµ ∈ RD(p).Once we have then (i) and (ii) hold as in proof of Theorem 5, and consequently the inequality (P T ) is satisfied.Now using Part C) of Lemma 2, the above condition ( * ) and the hypothesis dω ∈ RD(p) we have L p µ by the condition ( * ) It is also clear that a necessary condition for (P T ) is Conversely we assume this condition be satified and dµ ∈ D ε ,p ∩ RD ρ with 1 − λ < ρ and ε < pρ .It is sufficient to get the conditions in the first part of the present Proposition.As in the proof of Theorem 3 by using part A) of Lemma 2) and since dµ ∈ D ∞ then and consequently Now using dµ ∈ D ε ,p ∩ RD ρ with ε < pρ we can get the condition dµ ∈ RD(p) as follow: Finally we suppose dµ ∈ D ∞ ∩ RD ρ and dω ∈ D ε,q ∩ RD ρ with 1 − λ < ρ and ε < q(1 − σ).It remains to get the above condition ( * * ).Thus we have