ON A PAPER OF HUDSON.

GEORGIY ARUTYUNYANTS AND DMITRY RYABOGIN

Abstract. We present the detailed proof of Hudson’s result in [1].

1. Introduction

Let Ω ∈ L^{1}(S^{1}), f ∈ L^{1}(R^{2}) be nonnegative functions and let B_{r}(x) be a ball of radius r with the center at x. Consider the maximal operator

M_{Ω}f(x) = sup_{r>0} \frac{1}{r^{2}} ∫_{B_{r}(x)} f(y) Ω (\frac{x − y}{|x − y|}) dy.

It was proved ([2], [3], [4]) that Ω ∈ L log L(S^{1}) implies a weak-type (1,1) for M_{Ω}. It is still an open question whether the L log L(S^{1}) condition can be weakened to L^{1}(S^{1}). In this note we present the Hudson’s result proved in [1], which shows that M_{Ω} has a weak type (1,1) for some Ω ∉ L^{1} log L(S^{1}). For example, one can take

Ω(x_{1}, x_{2}) = g (\frac{x_{2}}{x_{1}} ) , g(θ) = \frac{χ_{(0,1]}(θ)}{θ log^{2}(θ/2)} .

More precisely, we have Theorem (Hudson). Let g ∈ L^{1}([0, 1]) be monotonically decreasing, such that θg(θ) is monotonically increasing, and let Ω(x_{1}, x_{2}) = g(x_{2}/x_{1}). Then

(1)  λ |{x ∈ R^{2} : M_{Ω}f(x) > λ}| ≤ c (∥g∥_{L^{1}([0,1])} + 1) ∥g∥_{L^{1}([0,1])} ∥f∥_{1}.

The core of the proof is Lemma 2. It represents an independent interest and might aid the study of the general case Ω ∈ L^{1}(S^{1}).

All results can be generalized to higher dimensions.

2. Selection Property.

The following definition was introduced by Hudson. Definition. We say that Ω has the selection property if for any measurable set D ⊂ R^{2} (0 < |D| < ∞) and any positive measurable function r(x) defined on D, there is a measurable subset E ⊆ D such that

(2)  |E| ≥ a|D|,

1991 Mathematics Subject Classification. Primary 42B20. Secondary 42E30.

Key words and phrases. Maximal functions.

(3)  S(E, Ω, r)(y) ≡ ∫_{E ∩B_{r(x)}(y)} Ω (\frac{x − y}{|x − y|}) \frac{dx}{r^{2}(x)} ≤ A

for almost every y ∈ R^{2}.

Constants here do not depend on r(x), D, E.

Lemma 1. If Ω has the selection property, then M_{Ω} is of weak type (1,1).

Proof. Let D = {x ∈ R^{2} : M_{Ω}f(x) > λ}. We may assume that D is bounded. Then

|D| ≤ \frac{1}{a}|E| ≤ \frac{1}{aλ} ∫_{E} M_{Ω}f(x)dx ≤ \frac{c}{aλ} ∫_{E} \frac{dx}{r^{2}(x)} ∫_{B_{r(x)}(x)} f(y) Ω (\frac{x − y}{|x − y|}) dy =

\frac{c}{aλ} ∫_{R^{2}} f(y)dy ∫_{E∩B_{r(x)}(y)} Ω (\frac{x − y}{|x − y|}) \frac{dx}{r^{2}(x)} ≤ c \frac{A}{a} \frac{∥f∥_{1}}{λ} .

□

Thus, to prove the Theorem it is enough to show that Ω satisfies the selection property. We emphersize that the construction of E involves no restriction on Ω ∈ L^{1}(S^{1}) and a ≃ 1/∥g∥_{L^{1}([0,1])}. All geometrical restrictions come from the estimates of (3) with A ≃ ∥g∥_{L^{1}([0,1])}.

3. The Dirac mass estimate

The main idea of the proof is to replace g by Dirac mass δ near the spike. Observe that in this case

(4)  S(♣, δ, r)(y) =  ∫_{\{x∈♣: 0<x_{1}−y_{1}<r(x_{1},y_{2}), x_{2}=y_{2}\}} \frac{dx_{1}}{r(x_{1}, y_{2})}.

Claim 1 . If g is replaced by a Dirac mass supported at 0, then Ω has the selection property. More precisely, for any positive measurable function r(x) defined on a set D ⊂ R^{2} of finite measure there is a set D̃ ⊆ D satisfying the following properties

|D̃| > c |D|,  S(D̃, δ, r)(y) ≤ C  for a. e. y ∈ R^{2}.

This claim follows from the Fubini theorem and the following one-dimensional result.

Lemma 2. Let r(t) be any positive measurable function on a measurable set D ⊂ R^{1}(|D| < ∞). Then there exists a measurable subset D̃ ⊂ D such that

(5)  |D̃| > \frac{1}{20}|D|

(6)  F (y) =  ∫_{D̃∩\{0<t−y<r(t)\}} \frac{dt}{r(t)} ≤ C  for almost all y ∈ R^{1}.

Proof. We will define set D̃ as a union of sets {D̃_{i}}_{i=0}^{∞}. The procedure described below is a modification of the Calderón-Zygmund stopping time argument. We construct D̃_{i} as follows

D̃_{i} = {x ∈ D | x ∈ q ∈ Q_{i−1} and r(x) > |q|}∪
∪{x ∈ D | x ∉ q ∈ Q_{i−1} and r(x) > 2^{−i}},

where Q_{i−1} is a system of dyadic intervals which we define by induction.

Set Q_{−1} = ∅ and assume that Q_{0}, ..., Q_{i−1} have already been constructed. Consider the net of dyadic intervals q with |q| = 2^{−i}. The construction of Q_{i} consists of two steps.

Step 1: We choose from the net all those intervals which do not intersect intervals from Q_{i−1} and for which one of the following conditions holds:

(a) |q|^{−1} ∫_{q} F_{i}(x)dx > 1/2,

where

F_{i}(x) = ∫_{D̃_{i}∩\{0<t−x<r(t)\}} \frac{dt}{r(t)},

(b) \frac{|D̃_{i} ∩ q|}{|q|} > \frac{1}{2}.

Step 2: We add all neighbors from the net to the intervals chosen before (Q_{i−1} ∪ { intervals chosen by step 1 }).

Set Q_{i} = Q_{i−1} ∪ { intervals chosen by step 1 } ∪ { intervals chosen by step 2 }. If interval q satisfies (a) and (b) we say that it is chosen by (a).

We claim that ∪_{i=0}^{∞}D̃_{i} = D̃ is the desired set. First of all

(7)  D̃ ⊂ D ⊂ ⋃_{q∈Q_{i},i=0}^{∞} q.

The first inclusion is obvious. The second one follows from the following argument. Fix any x ∈ D. Assume that r(x) > 2^{−i} for some i and x does not belong to any cube from Q_{i−1} (otherwise we are done). Then x ∈ D̃_{i}. Since almost all points of D̃_{i} are points of density, there is a dyadic interval q^{∗} ∋ x, |q^{∗}| = 2^{−j}, j ≥ i and such that |D̃_{i} ∩ q^{∗}|/|q^{∗}| > 1/2. Since D̃_{i} ⊆ D̃_{j}, |D̃_{j} ∩ q^{∗}|/|q^{∗}| > 1/2. Thus by (b), q^{∗} ⊆ q ∈ Q_{j}.

So (5) will easily follow from (7) and

(8)  ∑_{q∈Q_{i},i=0}^{∞} |q| ≤ 20|D̃|.

To prove (8), let us divide the system {Q_{i}}_{i=0}^{∞} into three disjoint subsystems: K_{1} = { intervals chosen by condition (a)}, K_{2} = { intervals chosen by condition (b)}, K_{3} = { intervals chosen by step 2 }. Then it is obvious that

∑_{q∈K_{3}} |q| ≤ 4 ∑_{q∈K_{1},q∈K_{2}} |q|.

Moreover

∑_{q∈K_{2}} |q| = ∑_{i=0}^{∞} ∑_{q∈(Q_{i}\Q_{i−1})∩K_{2}} |q| ≤ 2 ∑_{i=0}^{∞} ∑_{q∈(Q_{i}\Q_{i−1})∩K_{2}} |D̃_{i} ∩ q| ≤

2 ∑_{i=0}^{∞} ∑_{q∈(Q_{i}\Q_{i−1})∩K_{2}} |D̃ ∩ q| ≤ 2|D̃|.

On the other hand

∑_{q∈K_{1}} |q| ≤ ∑_{i=0}^{∞} ∑_{q∈(Q_{i}\Q_{i−1})∩K_{1}} |q| ≤ 2 ∑_{i=0}^{∞} ∑_{q∈(Q_{i}\Q_{i−1})∩K_{1}} ∫_{q} F_{i}(x)dx ≤

2 ∑_{i=0}^{∞} ∑_{q∈(Q_{i}\Q_{i−1})∩K_{1}} ∫_{q} F (x)dx ≤ 2 ∫_{R} F (x)dx =

= 2 ∫_{R} ∫_{D̃∩\{0<t−x<r(t)\}} \frac{dt}{r(t)}dx = 2 ∫_{D̃} \frac{1}{r(t)}  ∫_{\{0<t−x<r(t)\}} dxdt ≤ 2|D̃|.

So we have

∑_{q∈Q_{i},i=0}^{∞} |q| = ∑_{q∈K_{1}} |q| + ∑_{q∈K_{2}} |q| + ∑_{q∈K_{3}} |q| ≤

≤ 5( ∑_{q∈K_{1}} |q| + ∑_{q∈K_{2}} |q|) ≤ 20|D̃|

and it proves (5).

It remains to show (6). At first we observe that F (y) > 1/2 implies y ∈ q ∈ Q_{i} for some i ≥ 0. Indeed, fix any y such that F (y) > 1/2. Since D̃_{j} ⊂ D̃_{j+1} we have F_{j}(y) > 1/2 for sufficiently large j. By the differentiability of integrals, there is a dyadic interval q^{∗}, such that |q^{∗}|^{−1} ∫_{q^{∗}} F_{j}(x)dx > 1/2 and |q^{∗}| = 2^{−i}, i ≥ j. Now

D̃_{j} ⊂ D̃_{i} implies |q^{∗}|^{−1} ∫_{q^{∗}} F_{i}(x)dx > 1/2 and (a) gives q^{∗} ⊆ q ∈ Q_{i}.

Now we decompose F (y) into two parts. The first part will be estimated pointwise, the second one − by mean. Namely

F (y) =  ∫_{L∩\{0<t−y<r(t)\}} \frac{dt}{r(t)} +  ∫_{H∩\{0<t−y<r(t)\}} \frac{dt}{r(t)} = F_{L}(y) + F_{H}(y),

where, L = D̃ ∩ [y, y + 10^{2}|q|] and H = D̃ ∩ [y + 10^{2}|q|, ∞). First, let us show that

(9)  F_{L}(y) < C.

To do so, let us split L into two subsets L_{1} and L_{2}, where L_{1} = D̃ ∩[y + \frac{|q|}{2}, y +10^{2}|q|] and L_{2} = D̃ ∩ [y, y + \frac{|q|}{2}]. Since r(t) > t − y ≥ \frac{|q|}{2} for t ∈ L_{1} ∩ \{t | 0 < t − y < r(t)\}, it is easy to see that

F_{L_{1}}(y) =  ∫_{L_{1}∩\{0<t−y<r(t)\}} \frac{dt}{r(t)} ≤ \frac{2}{|q|}  ∫_{L_{1}∩\{0<t−y<r(t)\}} dt ≤ \frac{2}{|q|}10^{2}|q| ≤ C.

On the other hand if t ∈ L_{2} then t ∈ q ∈ Q_{i} or t ∈ q̃ ∈ Q_{i+1} because of the step 2 in construction of Q_{i+1}. In any case, r(t) > \frac{|q|}{2} and

F_{L_{2}}(y) = ∫_{L_{2}∩\{0<t−y<r(t)\}} \frac{dt}{r(t)} ≤ \frac{2}{|q|} ∫_{y}^{y+ \frac{|q|}{2}}dt = 1.

Combining these two estimates, we get (9).

To finish the proof of the lemma, we should show that

(10)  F_{H}(y) ≤ C

To prove (10), it is enough to show that F_{H\H_{i−1}}(y) ≤ C and F_{H_{i−1}}(y) ≤ C, where H_{i−1} = D̃_{i−1} ∩ [y + 10^{2}|q|, ∞).

Observe that F_{H\H_{i−1}}(y) = 0. Indeed, for any t ∈ D̃ \ D̃_{i−1}, r(t) ≤ 2^{−i+1}, we have D̃ \ D̃_{i−1} ∩ [y + 10^{2}|q|, ∞) ∩ {t| 0 < t − y < r(t)} = ∅.

What is left to show is

(11)  F_{H_{i−1}} ≤ C.

Let q^{∗} be the closest dyadic interval to q from the right with |q^{∗}| = 2^{−i+1}. Since for any ξ ∈ q^{∗}, H_{i−1} ∩ {t| 0 < t − y < r(t)} ⊂ H_{i−1} ∩ {t| 0 < t − ξ < r(t)}, we have

(12)  F_{H_{i−1}}(y) <  ∫_{H_{i−1}∩\{t| 0<t−ξ<r(t)\}} \frac{dt}{r(t)} ≤ \frac{1}{|q^{∗}|} ∫_{q^{∗}} ∫_{H_{i−1}∩\{0<t−ξ<r(t)\}} \frac{dt}{r(t)}dξ.

Observe that

q^{∗} ∩ p = ∅ for any p ∈ Q_{i−2} ∪ {intervals choisen by step 1 during stage i − 1}.

Otherwise, by step 2 q would be covered by some interval chosen on stage l ≤ i − 1. This contradicts the choice of q. Thus

So by (12) and (a) in the construction we have

F_{H_{i−1}}(y) ≤ \frac{1}{|q^{∗}|} ∫_{q^{∗}} ∫_{H_{i−1}∩\{0<t−ξ<r(t)\}} \frac{dt}{r(t)}dξ ≤ \frac{1}{|q^{∗}|} ∫_{q^{∗}} F_{i−1}(ξ)dξ < \frac{1}{2}.

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4. Auxiliary results

The following result is an analogue of the Calderón-Zygmund stopping time procedure. Below l(q) denotes the sidelength of a dyadic cube q. Claim 2 . Let Ω ∈ L^{1}(S^{1}), and let r(x) be any positive measurable function defined on a set G ⊂ R^{2} of finite measure. Then there exist two systems of sets

{Q_{i}}_{i=0}^{∞}  and  {E_{i}}_{i=0}^{∞},

where Q_{i} is a system of disjoint dyadic cubes with l(q) ≥ 2^{−i}, and

E_{i} = ∪_{q∈Q_{i−1}}{x ∈ G ∩ q : r(x) > l(q)} ∪ {x ∈ G \ Q_{i−1} : r(x) > 2^{−i}}

such that the following holds:

(13)  G ⊆ ∪_{i=0}^{∞}Q_{i},
(14)  | ∪_{i=0}^{∞} E_{i}| ≥ \frac{|G|}{50 ( ∥Ω∥_{L^{1}(S^{1})} + 1 )},

for any ancestor q̃ of q ∈ Q_{i}, and for any neighbor q̃ of any ancestor of q ∈ Q_{i}, we have

(15)  \frac{1}{|q̃|} ∫q̃ S(E_{i}, Ω, r)(y) dy < \frac{1}{2}.

We set

Q ≡ ∪_{i=0}^{∞}Q_{i}, E ≡ ∪_{i=0}^{∞}E_{i} ≡ ∪_{q∈Q}{x ∈ G ∩ q : r(x) > l(q)}.

Proof of Claim 2. The construction of E_{i}, Q_{i} repeats the construction of D̃_{i}, Q_{i} in Lemma 2. One has only to replace the condition (a) in the described stopping time procedure by

\frac{1}{|q|} ∫_{q} S(E_{i}, Ω, r)(y)dy > \frac{1}{2}.

The proof of (13), (14) is similar to the proof of (7), (8). Relation (15) follows by reasons which are similar to those presented at the end of Lemma 2 (see the estimate for F_{H_{i−1}}).

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Claim 3 . Let Ω ∈ L^{1}(S^{1}) and let S(E, Ω, r)(y) > 1/2. Then y ∈ Q. Proof of Claim 3. Since E_{i} ⊂ E_{i+1}, we have S(E_{i}, Ω, r)(y) > 1/2 for sufficiently large i. By the differentiation of integrals 1/|q^{∗}| ∫_{q^{∗}} S(E_{i}, Ω, r)(y)dy > 1/2 for some dyadic cube q^{∗} ∋ y with l(q^{∗}) = 2^{−j}, j ≥ i. Now E_{i} ⫅ E_{j} implies 1/|q^{∗}| ∫_{q^{∗}} S(E_{j}, Ω, r)(y)dy > 1/2 and (a) gives q^{∗} ⊆ q ∈ Q_{j}.

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The proof of the following claim is based on the fact that r(x) ≃ l(q) near cube q containing y. Claim 4 . Let Ω ∈ L^{1}(S^{1}) and let y ∈ q ∈ Q_{i} for some i ≥ 0. Then

(16)  S(E ∩ B_{10 l(q)}(y), Ω, r)(y) ≤ C ∥Ω∥_{L^{1}(S^{1})}.

Proof of Claim 4. Remind that

E ≡ ∪_{q∈Q}{x ∈ G ∩ q : r(x) > l(q)},

and

S(E ∩ B_{10 l(q)}(y), Ω, r)(y) ≡ ∫_{E∩B_{r(x)}(y)∩B_{10 l(q)}(y)} Ω (\frac{x − y}{|x − y|}) \frac{dx}{r^{2}(x)}.

Now x ∈ E ∩B_{r(x)}(y) implies r(x) ≥ l(q)/2. Indeed, this is obvious if |x−y| ≥ l(q)/2, since r(x) > |x−y|. If |x−y| < l(q)/2, then either x ∈ q or x belongs to the neighbor q̃ of q with l(q̃) ≥ l(q)/2 (use Step 2). But x ∈ E, so r(x) ≥ l(q̃).

This gives

S(E ∩ B_{10 l(q)}(y), Ω, r)(y) ≤ (\frac{l(q)}{2})^{−2} ∫_{B_{10 l(q)}(y)} Ω (\frac{x − y}{|x − y|}) dx = C ∥Ω∥_{L^{1}(S^{1})}.

□

5. Estimates on the exceptional set

In this section all proofs strongly depend on the geometric properties of g announced in the Theorem.

The following claim is based on Claim 1 and Claim 2. Let D be any set of finite measure. We apply Claim 1 to get D̃. Now, by Claim 2 we get E from D̃ (take G = D̃). Claim 5 . Let y ∈ q ∈ Q_{i} for some i and let

(17)  P = {x ∈ E : y_{2} ≤ x_{2} ≤ y_{2} + 2 l(q), x_{1} ≥ y_{1} + 2 l(q)}.

Suppose also that g ∈ L^{1}([0, 1]) and θ g(θ) is an increasing function. Then

(18)  S(P, Ω, r)(y) ≤ c∥g∥_{L^{1}([0,1])}.

Proof of Claim 5. Observe that x ∈ P ∩ B_{r(x)}(y) implies

2l(q) ≤ x_{1} − y_{1} ≤ |x − y| < r(x).

This , together with the fact that θ g(θ) is increasing gives

(19)  \frac{1}{r(x)}g (\frac{x_{2} − y_{2}}{x_{1} − y_{1}} ) ≤ \frac{1}{2 l(q)} g (\frac{x_{2} − y_{2}}{2 l(q)} ) .

By (19) we have

S(P, Ω, r)(y) ≡ ∫_{P ∩B_{r(x)}(y)} g (\frac{x_{2} − y_{2}}{x_{1} − y_{1}} ) \frac{dx}{r^{2}(x)} ≤ \frac{1}{2 l(q)} ∫_{P ∩B_{r(x)}(y)} g (\frac{x_{2} − y_{2}}{2 l(q)} ) \frac{dx}{r(x)}.

Now we apply the Fubini Theorem to the right-hand side of the above inequality and observe that it does not exceed

\frac{1}{2 l(q)} ∫_{y_{2}}^{y_{2}+2 l(q)}g (\frac{x_{2} − y_{2}}{2 l(q)} ) dx_{2} ∫_{\{x_{1}>y_{1}+2l(q): x∈P ∩B_{r(x)}(y)\}} \frac{dx_{1}}{r(x_{1}, x_{2})}.

Since P ⊂ E ⊂ D̃, the inner integral is bounded by Claim 1. It remains to make a substitution x_{2} − y_{2} = 2 l(q) θ in the outer integral to get

S(P, Ω, r)(y) ≤ \frac{c}{2 l(q)} ∫_{y_{2}}^{y_{2}+2 l(q)}g (\frac{x_{2} − y_{2}}{2 l(q)} ) dx_{2} = c ∫_{0}^{1} g(θ)dθ.

□

The following statement is an analogue of (10). Claim 6 . Let y ∈ q ∈ Q_{i} for some i and let g ∈ L^{1}([0, 1]) be decreasing. Then

(20)  S(H, Ω, r)(y) ≤ C ∥g∥_{L^{1}([0,1])},

where H = E \ (P ∪ B_{10 l(q)}(y)). Proof of Claim 6. Let H_{i−1} = H ∩ E_{i−1}. We show

ℵ: S(H \ H_{i−1}, Ω, r)(y) ≤ C ∥g∥_{L^{1}([0,1])}.

ℶ: S(H_{i−1}, Ω, r)(y) ≤ C ∥g∥_{L^{1}([0,1])}.

ℵ. By the definition of E_{i−1}, we may assume that r(x) ≤ 2^{−i+1} = 2 l(q). On the other hand, by the reasons which are similar to those in the proof of Claim 4, we have l(q) ≤ r(x). This gives ℵ.

ℶ. We claim that

(21)  S(H_{i−1}, Ω, r)(y) ≤ c \frac{1}{|q^{∗}|} ∫_{q^{∗}} S(H_{i−1}, Ω, r)(ξ) dξ,

where q^{∗} has the same center as q, and l(q^{∗}) = 3l(q).

The above relation reads as follows:

∫_{H_{i−1} ∩B_{r(x)}(y)} g (\frac{x_{2} − y_{2}}{x_{1} − y_{1}} ) \frac{dx}{r^{2}(x)} ≤ c \frac{1}{|q^{∗}|} ∫_{q^{∗}} dξ ∫_{H_{i−1} ∩B_{r(x)}(ξ)} g (\frac{x_{2} − ξ_{2}}{x_{1} − ξ_{1}} ) \frac{dx}{r^{2}(x)}.

By Fubini Theorem,

∫_{H_{i−1} ∩B_{r(x)}(y)} g (\frac{x_{2} − y_{2}}{x_{1} − y_{1}} ) \frac{dx}{r^{2}(x)} ≤ c ∫_{H_{i−1}} \frac{dx}{r^{2}(x)} \frac{1}{|q^{∗}|} ∫_{q^{∗} ∩B_{r(x)}(x)} g (\frac{x_{2} − ξ_{2}}{x_{1} − ξ_{1}} ) dξ.

Thus, to show (21) it is enough to prove that for every x ∈ H_{i−1} ∩ B_{r(x)}(y), we have

(22)  g (\frac{x_{2} − y_{2}}{x_{1} − y_{1}} ) ≤ c \frac{1}{|q^{∗}|} ∫_{q^{∗} ∩B_{r(x)}(x)} g (\frac{x_{2} − ξ_{2}}{x_{1} − ξ_{1}} ) dξ.

In fact, it is enough to prove the last inequality with K instead of q^{∗}, where K ⊂ q^{∗}: |K| ≥ c |q^{∗}|.

Fix any x ∈ H_{i−1} ∩ B_{r(x)}(y). Define K(x) ≡ {ξ ∈ q^{∗}, lying above the line l_{xy}}. Since g(θ) is decreasing observe that

g (\frac{x_{2} − y_{2}}{x_{1} − y_{1}} ) ≤ g (\frac{x_{2} − ξ_{2}}{x_{1} − ξ_{1}} ) ∀ξ ∈ K(x).

Now let K ≡ ∩_{x∈H_{i−1} ∩B_{r(x)}(y)} K(x). By elementary geometry |K| ≥ c |q^{∗}|. This gives (22). Thus we proved (21).

It remains to show that the mean in the right-hand side of (21) is finite. Let N(q) denote the set of dyadic neighbors of q of the sidelength l(q). Then observe that

q^{∗} ⊂ M ≡ N(father of q) ∪ father of q.

Hence there is a cube q̃ ∈ M such that

(23)  \frac{1}{|q^{∗}|} ∫_{q^{∗}} S(H_{i−1}, Ω, r)(ξ) dξ ≤ c \frac{1}{|q̃|} ∫_{q̃} S(H_{i−1}, Ω, r)(ξ) dξ.

We claim that the mean in the right-hand side of (23) is bounded by 1/2. Indeed, observe that

q̃ ∩ p = ∅ ∀ p ∈ Q_{i−2} ∪ {intervals choisen by step 1 during stage i − 1}.

Otherwise, by step 2 q would be covered by some cube chosen on stage l ≤ i − 1, which contradicts the choice of q. This means that

\frac{1}{|q̃|} ∫_{q̃} S(H_{i−1}, Ω, r)(ξ) dξ ≤ \frac{1}{|q̃|} ∫_{q̃} S(E_{i−1}, Ω, r)(ξ) dξ < \frac{1}{2}

by (a). The proof of ℶ is finished.

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6. Proof of the Theorem

By Claim 3 it is enough to consider y ∈ Q_{i} for some i ≥ 0. We have

S(E, Ω, r)(y) = S(E ∩ B_{10 l(q)}(y), Ω, r)(y) + S(P, Ω, r)(y) + S(H, Ω, r)(y)

where P and H are as above. Now claims 4,5, and 6 complete the proof.

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Acknowledgment. We would like to thank Alex Iosevich, Mark Rudelson, and Artem Zvavitch for helpful discussions.

References

[1] Steven Hudson, A covering lemma for maximal operators with unbounded kernels , Michgan Math. J., 34 (1987), 147–151.

[2] Michael Christ, Weak type (1,1) bounds for rough operators, Annals of Math., 128 (1988), 19–42.

[3] Michael Christ, J.-L. Rubio de Francia, Weak type (1,1) bounds for rough operators II, Invent. Math., 93 (1988), 225-237.

[4] Andreas Seeger, Singular integral operators with rough kernels, J. of AMS, 9 (1996), 95-105.

Georgiy Arutyunyants, Dep. of Mathematics, University of Missouri, Columbia, MO 65211, USA

E-mail address: arutyung@math.missouri.edu

Dmitry Ryabogin, Department of Mathematics, University of Missouri, Columbia, MO 65211, USA

E-mail address: ryabs@math.missouri.edu