WEIGHTED NORM INEQUALITIES FOR THE GEOMETRIC MAXIMAL OPERATOR

We consider two closely related but distinct operators, M0f(x) = sup I3x exp ( 1 |I| ∫ I log |f | dy ) and M∗ 0 f(x) = lim r→0 sup I3x ( 1 |I| ∫ I |f | dy )1/r . We give sufficient conditions for the two operators to be equal and show that these conditions are sharp. We also prove twoweight, weighted norm inequalities for both operators using our earlier results about weighted norm inequalities for the minimal operator: mf(x) = inf I3x 1 |I| ∫ I |f | dy. This extends the work of X. Shi; H. Wei, S. Xianliang and S. Qiyu; X. Yin and B. Muckenhoupt; and C. Sbordone and I. Wik.

We give sufficient conditions for the two operators to be equal and show that these conditions are sharp.We also prove twoweight, weighted norm inequalities for both operators using our earlier results about weighted norm inequalities for the minimal operator: mf(x) = inf

Introduction
Given a real-valued, measurable function f on R n , the geometric maximal function of f is where the supremum is taken over all cubes I which contain x and whose sides are parallel to the co-ordinate axes.Closely related to the geometric maximal operator is the following sequence of maximal operators: for f as before and for any r > 0 define , where the supremum is again taken over all cubes containing x. Equivalently, M r f = M (f r ) 1/r , where M is the Hardy-Littlewood maximal operator.By Hölder's inequality, for s < r, M s f(x) ≤ M r f(x), so we may define the limiting operator M * 0 by By Jensen's inequality, M 0 f (x) ≤ M * 0 f (x).Since we have the wellknown limit lim (see Rudin [9, p. 74]), it is reasonable to conjecture that for all functions f such that for some r > 0, f r ∈ L 1 loc , M * 0 f (x) = M 0 f(x) a.e.However, as we will show below, this is not true in general.
The purpose of this paper is to study the relation between M 0 and M * 0 , and to prove two-weight, weighted norm inequalities for each operator.These problems have been considered previously, with mixed results.In 1980, X. Shi [11] proved the following one-weight norm inequality.
Theorem 1.1.Given a weight w, the following are equivalent: 1. w ∈ A ∞ : there exists a constant C such that for all cubes I, 2. For 0 < p < ∞ the strong-type norm inequality holds for all f ∈ L p (w).
(The equivalence of the A ∞ condition and the so-called reverse Jensen inequality was not apparently discovered by Shi; it was discovered independently by García-Cuerva and Rubio de Francia [6] and Hrusčev [7].) In 1991, H. Wei, S. Xianliang and S. Qiyu [12] attempted to extend this result to the two-weight case on spaces of homogeneous type.Their proof, however, contained an error.This was pointed out by X. Yin and B. Muckenhoupt [13], who proved the following pair of two-weight norm inequalities on R 1 .Theorem 1.2.Given a pair of weights (u, v), the following are equivalent: 1. (u, v) ∈ W ∞ : there exists a constant C such that for all intervals I 2. For 0 < p < ∞ the weak-type norm inequality holds for all f ∈ L p (v).
Theorem 1.3.Given a pair of weights (u, v), the following are equivalent: 1. (u, v) ∈ W * ∞ : there exists a constant C such that for all intervals Their proofs depend heavily on covering lemmas which are particular to the real line.Therefore it is doubtful that they can be extended to higher dimensions.
Yin and Muckenhoupt also gave a complicated example to show that the class W * ∞ is strictly contained in W ∞ .(Also note that in the twoweight case the class W ∞ is strictly larger than A ∞ = ∪ p A p -a simple example is the pair (e |x| , e |2x| ).)Finally, they assert in passing that M 0 f and M * 0 f are the same "for suitably restricted f ".However, they give no indication of what this means.
Independently of these three papers, in 1994 C. Sbordone and I. Wik [10] published a different proof of Theorem 1.1.Their proof, however, requires that for all locally integrable f , M 0 f (x) = M * 0 f(x), which is false.There is a simple counter-example: let C be a nowhere dense subset of [0, 1] such that |C| = 1/2, and define f = χ C .Then for each interval I ⊂ [0, 1], I log |f | dx = −∞, so M 0 f(x) ≡ 0. But by the Lebesgue differentiation theorem, M r f (x) ≥ f (x) for almost every x ∈ [0, 1] and each r > 0, so M * 0 f (x) = 1 on a set of measure one-half.(The error in their proof is in inequality (2.11), as this example shows.) We prove the following results: in Section 2 we give sufficient conditions on a function f for the equality M 0 f (x) = M * 0 f(x) to hold almost everywhere.Our main result shows that for equality to hold log f must be locally integrable and the size of f at infinity must be controlled.
Neither of these conditions is strictly necessary -counter-examples can be readily constructed using monotonically decreasing functions.However, we give examples to show that if either condition is weakened then equality need not hold in general.
In Section 3 we give new proofs of Theorems 1.1, 1.2 and 1.3.Our proofs depend on the weighted norm inequalities for the minimal operator: given a real-valued, measurable function f on R n , the minimal function of f is where the infimum is taken over all cubes containing x. Intuitively, the minimal operator controls where a function is small, just as the maximal operator controls where it is large.We introduced the minimal operator in [2] in order to study the fine structure of functions which satisfy the reverse Hölder inequality.In that paper we also studied the one-weight norm inequalities which it satisfies.In [3], Cruz-Uribe, Neugebauer and Olesen examined the two-weight norm inequalities for the minimal operator on R 1 .(Additional results about variants of the minimal operator can be found in [4] and [5].) Our approach has two advantages.First, the proofs are considerably simpler, though part of the reason for this is that the work is in the proof of the norm inequalities for the minimal operator.Second, in the two-weight case our proofs extend to higher dimensions, provided that we can characterize the weights governing the norm inequalities for the minimal operator in higher dimensions.We obtained partial results in higher dimensions in [3]: for example, our proof of the two-weight, weak-type norm inequality extends to R n if we assume that u is doubling.More systematic results which yield sufficient conditions (both with and without doubling conditions) for norm inequalities for M 0 and M * 0 will appear in Cruz-Uribe [1].
At the end of Section 3 we give another example (simpler than that of Yin and Muckenhoupt) to show that the class W * ∞ is smaller than W ∞ .In Section 4 we prove results analogous to Theorems 1.1, 1.2 and 1.3 for M * 0 .In examining this operator, a key difficulty was the fact that there exist functions f such that if Q n is the cube of side 2n centered at the origin, then lim for x in a set of positive measure.(In other words, we could not a priori restrict ourselves to functions of compact support and then obtain the final result using the monotone convergence theorem.)For example, let . Then for all n > 0 and all x ∈ (0, 1) it is easy to see that Initially, we avoided this problem by assuming a growth condition on v: we say that v satisfies the I ∞ condition if lim sup where the limit supremum is taken over all cubes I containing the origin and all σ > 0 as |I| tends to infinity and as σ tends to zero.This condition appears unnatural; however, it is the formal limit as p tends to infinity of the condition which Rubio de Francia [8] showed is a necessary and sufficient condition on a weight v for there to exist u such that (u, v) is in the Sawyer class S p .(This class governs the strong-type norm inequalities for the Hardy-Littlewood maximal operator.For details, see García-Cuerva and Rubio de Francia [6].We are grateful to A. de la Torre for pointing this relation out to us.) By assuming the I ∞ condition we were able to reduce first to the case of functions of compact support, and then to the case of functions for which M 0 and M * 0 are equal.In this case we could then apply Theorems 1.1, 1.2 and 1.3.To our surprise, we were able to show that the I ∞ condition is necessary as well.In R 1 we thus proved the following analogues of Theorems 1.2 and 1.3.Theorem 1.5.Given a pair of weights (u, v), then for 0 < p < ∞ the weak-type norm inequality Theorem 1.6.Given a pair of weights (u, v), then for 0 < p < ∞ the strong-type norm inequality In the one-weight case the A ∞ condition implies the I ∞ condition; this gives a result in R n analogous to Theorem 1.1.
Theorem 1.7.Given a weight w, then for 0 < p < ∞ the strong-type norm inequality We conclude Section 4 with an example showing that in the two-weight case the W * ∞ condition does not imply the I ∞ condition.This example has the following interesting consequence: the Sawyer-type condition associated with M * 0 , while necessary, is not sufficient for the strong-type norm inequality for Finally, Section 5 is an appendix which contains a problem about a possible two-weight generalization of the A ∞ condition.
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 I, l(I) will denote the length of its sides.By weights we will always mean non-negative functions which are locally integrable and positive on a set of positive measure.Given a Borel set E and a weight v, |E| will denote the Lebesgue measure of E, v(E) = E v dx, and v/χ E will denote the function equal to v on E and infinity elsewhere.Given 1 < p < ∞, p = p/(p − 1) will denote the conjugate exponent of p. Finally, C will denote a positive constant whose value may change at each appearance.

Conditions for the Equality of
In this section we prove Theorem 1.4.We begin with a simple observation which, since we will use it in later sections, we designate as a lemma.

Lemma 2.1. For all non-negative functions f and all
Proof: For M 0 this follows immediately from the definition.For M * 0 the proof is almost as simple: given x ∈ R n and > 0, for every r > 0 there exists a cube I containing x such that If we take the limit as r tends to zero we get (since is arbitrary) that p .An identical argument gives the reverse inequality, and we are done.

Proof of the Sufficiency of Condition (1):
Without loss of generality we may assume that f is non-negative.Further, by Lemma 2.1 we may also assume that p = 1.Now for each k > 0 define for almost every x.Since kf ∈ L 1 if f is, and since both M 0 and M * 0 are positive homogeneous, we may assume without loss of generality that k = 1.Furthermore, it will suffice to show that M * 0 f 1 (x) ≤ M 0 f 1 (x) a.e.Fix x ∈ R n .There are two cases: If M * 0 f 1 (x) = 1, then by the Lebesgue differentiation theorem (since both f and log f are locally integrable) for almost every such x, , and so Then for any n > 0 and for any cube Hence the cubes used to calculate M 1/n f 1 (x) must be uniformly bounded in volume.In particular, fix δ > 0; then for each n > 0 there exists a cube I n containing x such that |I n | is uniformly bounded and Elementary calculus shows that for all x ≥ 1 and integers n > 0, Since f ∈ L 1 and since Mf 1 (x) ≤ Mf(x) + 1, Mf 1 (x) is finite for almost every x.Further, since the I n 's are uniformly bounded in size and all contain x, by passing to a subsequence we may assume that they converge either to a non-degenerate cube I or to the set {x}.In the first case in the second case, by the Lebesgue differentiation theorem it converges to log f 1 (x) for almost every x.But if a sequence {a n } converges to a and if M ≥ 0, then In either case, therefore, if we take the limit in inequality (2) we have that a.e.Since δ > 0 was arbitrary, this establishes the desired inequality.
To complete the proof, since for each k, e., we only need to show that (3) lim The argument is similar to the one just given.Fix x; since log f is locally integrable, there exists γ such that Therefore, for each δ > 0 there exists a sequence of cubes I k containing x such that ∪ k I k is contained in some cube J, and such that Inequality (3) would follow immediately if we could show that ( 4) To show equation ( 4), first note that ) tends to zero pointwise as k tends to infinity, by the dominated convergence theorem it tends to zero in L 1 norm (on J).By the weak (1, 1) inequality for the Hardy-Littlewood maximal operator, for each t > 0, Therefore the sequence {M (log(f k /f )χ J )} tends to zero in measure, and so has a subsequence which converges to zero pointwise almost everywhere.However, the whole sequence is monotonically decreasing, so in fact (4) holds.This completes the proof of the sufficiency of condition (1).
This proof has the following corollary which we will need below.
Corollary 2.2.Let I 0 be a cube, and suppose supp f = I 0 .If for some p, 0 < p < ∞, f ∈ L p (I 0 ) and log f ∈ L 1 (I 0 ) then M * 0 f (x) = M 0 f(x) for almost every x.
Proof: For x ∈ I 0 the above proof holds with essentially no modification.For x outside the support of f a direct computation shows that Proof of the Sufficiency of Condition ( 2): Fix f ∈ L ∞ ; again we may assume that f is non-negative.If α > 2 then by Hölder's inequality, so without loss of generality we may assume that 1 < α ≤ 2. But then we have the inequality Then for any n > 0, any x and any cube I containing x, Therefore, for each such x we can take the supremum over all I containing x and then the limit as n tends to infinity to get M * 0 g(x) ≤ M 0 g(x) a.e.Then by homogeneity, M * 0 f (x) ≤ M 0 f (x) a.e. and we are done.
Examples.We now give three examples to show that the hypotheses of Theorem 1.4 cannot be weakened in general.For simplicity we construct all the examples on the real line.
First recall the example f = χ C , C nowhere dense and |C| = 1/2, given in Section 1 above.This shows that log f needs to be locally integrable.

Example 2.3.
There exists a non-negative function f such that log f ∈ L 1 loc , f / ∈ L p (R) for any p, 0 < p < ∞, and such that for all x, M * 0 f (x) = ∞ and M 0 f (x) < ∞.
Proof: Define the function f by Then log f is locally integrable but f / ∈ L p (R) for any finite p.Now fix n and let k > n.Then The right-hand side tends to infinity as k tends to infinity.Therefore, for all x ≥ 0, M 1/n f (x) = ∞, and so M * 0 f (x) = ∞.An identical argument holds for x < 0.
To see that M 0 f is everywhere finite, first note that since log f is locally bounded, given x ∈ R, M 0 f (x) will be infinite only if (5) lim sup and it follows from this that the limit supremum in (5) is finite.A similar argument shows that M 0 f (x) < ∞ for all x.
Proof: For each integer n ≥ 0, let a n = 2 −(2 2n−1 −1) , and define f by Since log f is locally bounded, to show that M (log f ) is everywhere finite we only need to show that for any x ∈ R, the limit supremum in ( 5) is finite.Let x = 0. Then Hence M (log f )(0) < ∞.A similar but lengthier argument shows that M (log f )(x) < ∞ for all x.Now for any x < 0, r > 0 and n > 0, The right-hand side tends to 2 as n tends to infinity.Therefore M r f (x) = 2 for all r > 0, so In either case, by our choice of the a n 's, It follows from this that for all x < 0, M 0 f (x) = 1.
Finally, note that an estimate similar to the one in Example 2.3 shows that for all x and all α > 1, M (| log f | α )(x) = ∞.

Norm Inequalities for M 0
In this section we give new proofs of Theorems 1.1, 1.2 and 1.3.For each theorem we restrict ourselves to proving the sufficiency of the given weight classes: the necessity follows at once if we substitute the test function v −1 χ I into the corresponding norm inequality.
Our proofs depend on the weighted norm inequalities for the minimal operator.
Theorem 3.1.Given p > 0 and a pair of weights (u, v) on R, the following are equivalent: 1. (u, v) ∈ W p : there exists a constant C such that given any interval I ⊂ R, 2. the weak-type inequality p : there exists a constant C such that given any interval The constants in (2) and (4) only depend on the constants in (1) and (3) and are independent of p.
In the special case where u = v then W p = W * p = A ∞ and inequalities (2) and (4) hold in R n for all n ≥ 1.
The proof of Theorem 3.1 for equal weights is in Cruz-Uribe and Neugebauer [2].The two-weight case is in Cruz-Uribe, Neugebauer and Olesen [3].
To make the connection between the minimal operator and the geometric maximal operator, we first define the geometric minimal operator: given a function f on R n , the geometric minimal function of f is where the infimum is taken over all cubes I containing x.It is immediate from this definition that (m 0 f ) −1 = M 0 (f −1 ) for all f .Now, as we did for the geometric maximal operator, we define a sequence of minimal operators and a limiting minimal operator (This sequence is decreasing so the limit exists.)In light of the results in Section 2 above, the next result is quite surprising, especially since the proof is so elementary.
Lemma 2. Given a cube I 0 (possibly infinite), let f be a function on R n such that for some r > 0, f r ∈ L 1 loc on I 0 .Then for all x, To see the reverse inequality, fix > 0. If x ∈ I 0 then there exists a cube Since was arbitrary, we are done.
An immediate consequence of Lemma 3.2 and the preceding observation is that if f −1 is locally integrable then for any cube I, the sequence {m r (f −1 /χ I ) −1 } increases to M 0 (fχ I ) for all x.
The weight classes W ∞ and W * ∞ of Theorems 1.2 and 1.3 are the formal limits of the classes W p and W * p as p tends to infinity.Furthermore, by Jensen's inequality, if the pair (u, v) is in W ∞ then it is in W p for all p > 0 with a constant independent of p.Similarly, suppose (u, v) ∈ W * ∞ .Then for all cubes I and all x ∈ I, .
If we substitute this into the W * ∞ condition we see that (u, v) is in W * p for all p > 0, again with a constant independent of p.
The proofs of Theorems 1.1, 1.2 and 1.3 are now straightforward.We will prove Theorem 1.3; the proofs of the other two are identical.By Lemma 2.1 we only need to consider the case p = 1.Fix f ∈ L 1 (v) and for each n > 0, let Since the constant C is independent of r, by Lemma 3.2 and the remark following, if we let r tend to 0, by the monotone convergence theorem we get Since v is locally integrable, the right-hand side is finite, so we can take the limit as tends to 0 to get Since M 0 (fχ In ) increases to M 0 f , the desired inequality follows from the monotone convergence theorem.We conclude this section with an example of a pair of weights (u, v) which is in W ∞ \ W * ∞ .Our example is simpler than the one given by Yin and Muckenhoupt [13].Initially we believed that no such example existed, since for all p > 0 the classes W p and W * p are the same.However, a close examination of the proof that they are the same showed that the constant depended on p. Attempts to eliminate this dependency instead yielded the following example.The underlying idea of the construction is to fix an increasing function v which is not a doubling weight and find the "largest" function u such that (u, v) ∈ W ∞ .

Example 3.3. There exists a pair of weights
Proof: For x > 0 define the the functions and extend them to R by making them identically zero for x ≤ 0. For intervals of the form I = [−s, t], s ≥ 0, t > 0, we have it follows that (u, v) satisfies the W ∞ condition on all such intervals.(Note that when s = 0 equality holds; it is in this sense that u is the largest possible function.) If t ≥ 2s then this inequality is immediate.Now suppose that t ≤ 2s.Since u is an increasing function, the left-hand side of inequality ( 6) is smaller than u(t).Hence it will suffice to show that However, since e x ≥ 1 + x, (The last inequality holds since t ≤ 2s.)Therefore, (u Since the right-hand side tends to infinity as t tends to zero, (u, v) cannot be in W * ∞ .

Norm Inequalities for M * 0
In this section we prove Theorems 1.5, 1.6 and 1.7.Each of these theorems is a consequence of the corresponding norm inequality for M 0 .
Proof of Sufficiency: We begin with three lemmas which together show that the I ∞ condition allows us to reduce the proof to the case of functions of compact support.
where the limit supremum is taken over all cubes I containing x 0 and σ > 0 as |I| tends to infinity and σ tends to zero.
Proof: Suppose to the contrary that there exists an x 0 such that the given limit supremum is infinite.Then there exists a sequence of cubes I k containing x 0 such that |I k | tends to infinity, and a sequence of real numbers σ k tending to zero such that By Hölder's inequality we may assume that the σ k 's tend to zero as slowly as desired.Now let J k be the smallest cube containing both I k and the origin.
Hence the k 's tend to zero, so by the above observation we may assume that σ k ≥ k .But then Lemma 4.2.Suppose f is a non-negative function on R n such that for some r > 0, f r ∈ L 1 loc , and K is any compact set.If f K = fχ R n \K , then for each x 0 ∈ R n , (7) lim sup , where the limit supremum is taken over all cubes I containing x 0 and σ > 0 as |I| tends to infinity and σ tends to zero.
Proof: The left-hand side of equation ( 7) is always greater than or equal to the right-hand side, so we only need to prove the reverse inequality.If the left-hand side equals zero there is nothing to prove, so we may assume that it is equal to some λ > 0. Then there exists a sequence of cubes I k containing x 0 and a sequence of real numbers σ k such that |I k | tends to infinity and σ k tends to zero, such that (8) lim Since λ is the limit supremum over all such I and σ, and since by Hölder's inequality the terms on the left-hand side get larger if we make the σ k 's larger, this limit will still hold if we replace the σ k 's by any larger sequence tending to zero.In particular, we may assume that 1 Since K is compact and f r ∈ L 1 loc , by the dominated convergence theorem, Further, since λ > 0, equation (8) implies that there exists τ , 0<τ <1, such that for all k sufficiently large Therefore, again for all k sufficiently large, The right-hand side of this inequality tends to 1 as k tends to infinity.Therefore equations ( 8) and ( 9) imply that , and this establishes the desired inequality.Proof: Without loss of generality we may assume that f is non-negative.Since v ∈ I ∞ , there exists M > 0 and σ 0 > 0 such that, given a cube I containing the origin with |I| > M, then Therefore, by Hölder's inequality we have that for all such cubes I, Now fix x ∈ R n .Suppose first that there exists N > 0 and a sequence of cubes I k containing x and contained in Q N such that Then for all n ≥ N , M * 0 f (x) = M * 0 (fχ Qn )(x), which establishes equation (10).
If no such sequence of cubes exists, then (11) M * 0 f (x) = lim sup , where the limit supremum is taken over all cubes I containing x and σ > 0 as |I| tends to infinity and σ tends to zero.We will show that this implies that M * 0 f (x) = 0, which in turn implies that equation ( 10) holds.
To see this, fix > 0. Then there exists a compact set K such that where the limit supremum is taken over the same I and σ as in equation (11).We again apply Hölder's inequality: since v ∈ I ∞ , by Lemma 4.1 we have that Since is arbitrary, M * 0 f (x) = 0 and we are done.
We can now prove Theorems 1.5, 1.6 and 1.7.We will only prove Theorem 1.6; the proofs of the other two are identical.(For Theorem 1.7, we note in passing that if w ∈ A ∞ then w ∈ A p for all p sufficiently large, which immediately implies that w ∈ I ∞ .)First, by Lemma 4.3 and the monotone convergence theorem, it will suffice to prove Theorem 1.6 for functions f ∈ L p (v) of compact support.Second, by Lemma 2.1 we may assume that p = 1.Fix such an f and suppose that supp f ⊂ Q N for some N > 0. Define the sequence of functions {f n } by As we showed in the proof of Lemma 4.3, there exists r > 0 such that f r is locally integrable.Therefore, f r n ∈ L 1 (Q N ), and log Since f n ≤ f + 1 n χ Q N and v is locally integrable, by the dominated convergence theorem we can take the limit as n tends to infinity and get the desired inequality.

Proof of Necessity:
The necessity of the W ∞ and W * ∞ conditions in Theorems 1.5, 1.6 and 1.7 follows from their necessity in the corresponding theorems for M 0 .The necessity of the I ∞ condition follows from the next lemma since u is positive on a set of positive measure.

Lemma 4.4. Given a weight
Proof: Since v / ∈ I ∞ , there exists a sequence of cubes I k containing the origin such that |I k | tends to infinity, and a sequence of real numbers σ k tending to zero such that for all k, For each k let a k be such that a k |I k | = 1/k 2 , and define the function f by It is immediate that f ∈ L 1 (v).Now fix x ∈ R n and let J k be the smallest cube containing x and I k .Then, as we showed in Lemma 4.1, The Independence of I ∞ and W * ∞ .We give an example to show that the W * ∞ condition does not imply the I ∞ condition.For simplicity we construct our example on R. Therefore, if J is an interval such that 2 n−1 < |J| ≤ 2 n and which intersects [0, 1], then Hence (u, v) ∈ W * ∞ .However, if we let σ = 1/n, then exp[2 2n−1 (n + 1) log 2] 2 2n 2 +2n+1 .
The right-hand side tends to infinity as n tends to infinity, so v does not satisfy the I ∞ condition.
We conclude with the following observation.In this example both v and v −1 are locally integrable, so by Corollary 2.2, for any interval I, M * 0 (v −1 χ I ) = M 0 (v −1 χ I ) a.e.Hence the pair (u, v) satisfies the Sawyertype condition associated with M * 0 , namely, but the strong-type norm inequality does not hold for M * 0 .Hence this condition is necessary but is not sufficient.

Appendix: A Two-Weight Generalization of A ∞
As we noted in Section 1, the two-weight reverse Jensen inequality, W ∞ , does not characterize the union of the two-weight A p classes.Similarly, the stronger W * ∞ condition does not characterize this union either.The same example shows this: the pair (e |x| , e |2x| ) is in W * ∞ but is not in any A p class.
However, suppose (u, v) ∈ W * ∞ and v ∈ I ∞ .Then by the remarks at the beginning of the proof of Lemma 4.3, for all p > 0 sufficiently large, v satisfies the Rubio de Francia condition (1) mentioned in Section 1.In other words, for all p sufficiently large, there exists a function u p such that (u p , v) ∈ S p ⊂ A p .The function u p need not equal u for any p; however, it is natural to ask the following question.Question 5.1.Is it possible to find functions u p such that (u p , v) ∈ S p and the u p 's converge to u (pointwise or as measures) as p tends to infinity?
If this were true it would establish the two conditions W * ∞ and I ∞ as the "natural" limit of the A p condition and so give a two-weight notion of A ∞ .
This question arose as the final draft of this paper was being written and we have no conjecture as to its veracity.However, a straightforward calculation does show that e |2x| ∈ I ∞ , and that for the pair (e |x| , e |2x| ) we may take u p = e |x| χ [−np,np] for n p sufficiently large.

Lemma 4 . 3 .
Let v ∈ I ∞ and suppose f ∈ L 1 (v).Let Q n be the cube centered at the origin of side-length 2n.Then for all x, Qn )(x).