WEAK-TYPE (p, p) ESTIMATES FOR CERTAIN MAXIMAL OPERATORS

GEORGIY ARUTYUNYANTS AND DMITRY RYABOGIN

Abstract. Let Φ ∈ L^{q}(R^{n}) have a compact support, and let f ∈ L^{p}(R^{n}), p ≥ 1, \frac{1}{p} + \frac{1}{q} = 1. We show that the maximal operator M_{Φ} : f → sup_{t>0} Φ_{t} ∗ f has weak-type (p, p) and lim_{t→0} Φ_{t} ∗ f(x) exists for a.e. x ∈ R^{n}. The result is sharp in the sense that for any 1 ≤ s < q there exists Φ ∈ L^{s}(R^{n}), having a compact support, such that M_{Φ} is not of weak-type (p, p).

1. Introduction

Let Φ, f be nonnegative functions and let

Φ_{t}(x) = \frac{1}{t^{n}} Φ (\frac{x}{t}) , Φ_{t} ∗ f(x) = ∫_{R^{n}} Φ_{t}(x − y)f(y)dy.

Consider the maximal operator

(1)  M_{Φ} : f → M_{Φ}f(x) ≡ sup_{t>0} Φ_{t} ∗ f(x).

This operator controls the pointwise convergence of Φ_{t} ∗ f(x) as t → 0, and comes up in many problems in harmonic analysis and partial differential equations. See [7] and references contained therein.

A basic unsolved problem is to determine the range of boundedness of the maximal operator M_{Φ} on the scale of L^{p} spaces. Our main result is the following. Theorem. Let Φ ∈ L^{q}(R^{n}), f ∈ L^{p}(R^{n}), 1 ≤ p < ∞, \frac{1}{p} + \frac{1}{q} = 1. Suppose also that supp (Φ) ⊆ B_{R}(0). Then

(2)  λ^{p} |{x ∈ R^{n} : M_{Φ}f(x) > λ}| ≤ c_{n} R^{n} ∥Φ∥_{q}^{p} ∥f∥_{p}^{p}.

Moreover, for any 1 ≤ s < q, there exists Φ ∈ L^{s}(R^{n}), having a compact support, such that M_{Φ} is not of weak-type (p, p).

Operators of this type have been studied before by several authors. For example, when Φ decreases at a sufficiently high rate at infinity, then M_{Φ} is majorized by the Hardy-Littlewood maximal operator. If for each x, Φ(rx) is decreasing in r, 0 < r < ∞, then by the method of rotations, M_{Φ} is bounded from L^{p}(R^{n}) to itself, 1 < p ≤ ∞. If one strengthens the integrability assumption on Φ by adding a Dinitype condition, then M_{Φ} is of weak-type (1, 1). See [7], page 72. This conclusion also holds in the case Φ(x) = Ω(x/|x|)χ_{B_{1}(0)}(x), Ω log^{+}Ω ∈ L^{1}(S^{n−1}) (see [4], [5], [6]). Here χ_{B_{1}(0)} is the characteristic function of the ball with center at the origin of radius 1.

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

Key words and phrases. Maximal functions.

The proof of our main result is based on a multidimensional stopping time argument inspired by a result due to S. Hudson ([1], [2]). This approach is of purely geometric nature and is interesting in its own right. For example, this point of view yields yet another proof of weak-type (1, 1) boundedness of the Hardy-Littlewood maximal operator.

The sharpness of the Theorem follows from the following example (cf. [7], page 81). Let 1 ≤ p < ∞ and 1 ≤ s < q, \frac{1}{p} + \frac{1}{q} = 1. Define

Φ(x) = (1 − |x_{1}|)^{ϵ−1}χ_{[−1,1]^{n}}(x), f(x) = x_{1}^{−δ}χ_{[0,1]^{n}}(x).

Then for 1 − 1/s < ϵ < δ < 1/p, we have Φ ∈ L^{s}, f ∈ L^{p} and

M_{Φ}f(x) ≥ Φ_{x_{1}} ∗ f(x) ≡ ∞,

provided x ∈ R^{n} : x_{1} > 1, (x_{2}, x_{3}, ..., x_{n}) ∈ (−x_{1}, x_{1} + 1)^{n−1}.

Standard arguments (see e.g. [8], Ch. I) imply: Corollary. Let Φ ∈ L^{q}(R^{n}) have a compact support, and let f ∈ L^{p}(R^{n}), 1 ≤ p < ∞, \frac{1}{p} + \frac{1}{q} = 1. Then lim_{t→0} Φ_{t} ∗ f(x) exists for almost every x ∈ R^{n}.

This paper is organized as follows. In the first section, entitled ”Selection Property”, we describe a higher dimensional version of the property inspired by ideas of Hudson ([1], [2]). See also ([3]) for a thorough description of related ideas and their applications. In the following section, entitled ”Proof of Lemma 2”, we estimate a key expression resulting from the linearized version of the maximal operator.

Remark: It is possible that Theorem can be proved by an appropriate extension of known results in multi-linear interpolation. More precisely, by viewing M_{Φ} as a sub-bi-linear operator, one can show that this operator maps L^{1} × L^{∞} → L^{1,∞} and L^{∞} × L^{1} → L^{∞}. If one could prove an appropriate bi-linear version of the Marcienkiewicz interpolation theorem, Theorem would follow. However, our goal is to give a direct geometric argument that exposes the nature of the operator.

2. Selection Property.

Definition. We say that Φ has the selection (p, q) property, 1/p + 1/q = 1, if for any positive measurable function t(x) defined on a set D ⊂ R^{n}, (0 < |D| < ∞) there is a measurable subset E ⊆ D such that

(3)  |E| ≥ a|D|,
(4)  ∥S(E, Φ, t)∥_{q} ≤ A ∥Φ∥_{q} |D|^{1/q},

where

(5)  S(E, Φ, t)(y) ≡ ∫_{E} Φ (\frac{x − y}{t(x)} ) \frac{dx}{t^{n}(x)}.

Constants here do not depend on t(x), D, E. Lemma 1. If Φ has the selection (p,q) property, 1 < p < ∞, then M_{Φ} satisfies (2).

Proof. Let D = {x ∈ R^{n} : M_{Φ}f(x) > λ}. Without loss of generality we may assume that D is bounded. Then

|D| ≤ \frac{1}{a}|E| ≤ \frac{1}{aλ} ∫_{E} M_{Φ}f(x)dx ≤ \frac{c}{aλ} ∫_{E} \frac{dx}{t^{n}(x)} ∫_{R^{n}} f(y) Φ (\frac{x − y}{t(x)} ) dy

= \frac{c}{aλ} ∫_{R^{n}} f(y)dy ∫_{E} Φ (\frac{x − y}{t(x)} ) \frac{dx}{t^{n}(x)} ≤ \frac{c}{a λ}∥f∥_{p}∥S(E, Φ, t)∥_{q} ≤ \frac{c A}{a λ}∥f∥_{p}∥Φ∥_{q} |D|^{1/q}.

□

Lemma 1 reduces the proof of the Theorem to the following estimate. Lemma 2. The function Φ = χ_{B_{1}(0)} has the selection (1, ∞) property. More precisely, for any positive measurable function t(x) defined on a set D ⊂ R^{n} of finite measure, there is a subset D̃ ⊆ D satisfying the following properties

(6)  |D̃| > c |D|,
(7)  S(D̃, χ_{B_{1}(0)}, t)(y) ≡ ∫_{D̃∩ B_{t(x)}(y)} \frac{dx}{t^{n}(x)} ≤ C.

Here B_{r}(y) = {x ∈ R^{n} : |x − y| ≤ r}. Lemma 3. Let Φ be as in the Theorem and let q < ∞. Then Φ has the selection (p, q) property with a = c from (6) and A = c R^{n/p}.

Proof. We take E = D̃ from the previous lemma and observe that (7), together with the Jensen inequality and the Fubini Theorem, yield

∥S(E, Φ, t)∥_{q}^{q} ≤ C^{q} R^{nq} ∫_{R^{n}} dy ∫_{E∩ B_{R t(x)}(y)} Φ^{q} (\frac{x − y}{t(x)} ) \frac{dx}{R^{n} t^{n}(x)} =

= C^{q} R^{nq} ∫_{E} \frac{dx}{R^{n} t^{n}(x)} ∫_{B_{R t(x)}(x)} Φ^{q} (\frac{x − y}{t(x)} ) dy = C^{q} R^{nq} |E| ∫_{R^{n}} Φ^{q}(Rξ)dξ.

By making a substitution Rξ = z in the last integral we obtain the desired result.

□

3. Proof of Lemma 2

We will define the 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. By l(q) we denote the side-length of a dyadic cube q. We construct D̃_{i} as follows

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

where Q_{i−1} is a collection of dyadic cubes 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 cubes q with l(q) = 2^{−i}. The construction of Q_{i} consists of two steps.

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

(a) |q|^{−1} ∫_{q} S(D̃_{i}, χ_{B_{1}(0)}, t)(y)dy > 1/2,

where

S(D̃_{i}, χ_{B_{1}(0)}, t)(y) = ∫_{D̃_{i}∩B_{t(x)}(y)} \frac{dx}{t^{n}(x)},

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

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

Set Q_{i} = Q_{i−1} ∪ { cubes chosen by step 1 } ∪ { cubes chosen by step 2 }. If a cube 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

(8)  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 t(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 cube q^{∗} ∋ x, l(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 (6) would easily follow from (8) and

(9)  ∑_{q∈Q_{i},i=0}^{∞} |q| ≤ c_{n} |D̃|.

To prove (9), let us divide the system {Q_{i}}_{i=0}^{∞} into three disjoint subsystems: K_{1} = { cubes chosen by condition (a)}, K_{2} = { cubes chosen by condition (b)}, K_{3} = { cubes chosen by step 2 }. Then it is obvious that

∑_{q∈K_{3}} |q| ≤ 5^{n} ∑_{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} S(D̃_{i}, χ_{B_{1}(0)}, t)(y)dy ≤

2 ∑_{i=0}^{∞} ∑_{q∈(Q_{i}\Q_{i−1})∩K_{1}} ∫_{q} S(D̃, χ_{B_{1}(0)}, t)(y)dy ≤ 2 ∫_{R^{n}} S(D̃, χ_{B_{1}(0)}, t)(y)dy =

= 2 ∫_{R^{n}} ∫_{D̃∩ B_{t(x)}(y)} \frac{dx}{t^{n}(x)}dy = 2 ∫_{D̃} \frac{1}{t^{n}(x)} ∫_{B_{t(x)}(x)} dydx ≤ 2 c_{n} |D̃|.

The third line follows by Fubini theorem, B_{t(x)}(x) = {y ∈ R^{n} : |x − y| ≤ t(x)}. So we have

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

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

and it proves (6).

It remains to show (7). At first we observe that S(D̃, χ_{B_{1}(0)}, t)(y) > 1/2 implies y ∈ q ∈ Q_{i} for some i ≥ 0. Indeed, fix any y such that S(D̃, χ_{B_{1}(0)}, t)(y) > 1/2. Since D̃_{j} ⊂ D̃_{j+1} we have S(D̃_{j}, χ_{B_{1}(0)}, t)(y) > 1/2 for sufficiently large j. By the differentiability of integrals, there is a dyadic cube q^{∗}, such that |q^{∗}|^{−1} ∫_{q^{∗}} S(D̃_{j}, χ_{B_{1}(0)}, t)(x)dx

> 1/2 and l(q^{∗}) = 2^{−m}, m ≥ j. Now D̃_{j} ⊂ D̃_{m} implies |q^{∗}|^{−1} ∫_{q^{∗}} S(D̃_{m}, χ_{B_{1}(0)}, t)(x)dx

> 1/2 and (a) gives q^{∗} ⊆ q ∈ Q_{i}, l(q) = 2^{−i}, i ≤ m.

Now we decompose S(D̃, χ_{B_{1}(0)}, t)(y) into two parts. The first part will be estimated pointwise, the second one − by mean. Namely

S(D̃, χ_{B_{1}(0)}, t)(y) = ∫_{L∩ B_{t(x)}(y)} \frac{dx}{t^{n}(x)} + ∫_{H∩ B_{t(x)}(y)} \frac{dx}{t^{n}(x)}

= S(L, χ_{B_{1}(0)}, t)(y) + S(H, χ_{B_{1}(0)}, t)(y),

where, L = D̃ ∩ B_{100 l(q)}(y) and H = D̃ ∩ (R^{n} \ B_{100 l(q)}(y)). First, we show that

(10)  S(L, χ_{B_{1}(0)}, t)(y) < C.

Observe that x ∈ L∩B_{t(x)}(y) implies t(x) ≥ l(q)/2. This is obvious if |x−y| ≥ l(q)/2, since t(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 ∈ L ⊂ D̃ ≡ ⋃_{q∈Q_{i}\Q_{i−1},i=0}^{∞}{x ∈ D ∩ q : t(x) > l(q)},

so t(x) ≥ l(q̃). This gives

S(L, χ_{B_{1}(0)}, t)(y) ≤ (\frac{l(q)}{2})^{−n} |B_{100 l(q)}(y)| = C,

and we get (10). To finish the proof of the lemma, we should show that

(11)  S(H, χ_{B_{1}(0)}, t)(y) ≤ C.

To prove (11), it is enough to get the following estimates

S(H \ H_{i−1}, χ_{B_{1}(0)}, t)(y) ≤ C, S(H_{i−1}, χ_{B_{1}(0)}, t)(y) ≤ C,

where H_{i−1} = D̃_{i−1} ∩ (R^{n} \ B_{100 l(q)}(y)). Observe that S(H \ H_{i−1}, χ_{B_{1}(0)}, t)(y) = 0. Indeed, for any x ∈ D̃ \ D̃_{i−1}, t(x) ≤ 2^{−i+1} we have (D̃ \ D̃_{i−1}) ∩ (R^{n} \ B_{100 l(q)}(y)) ∩ B_{t(x)}(y) = ∅. It is left to show that

(12)  S(H_{i−1}, χ_{B_{1}(0)}, t)(y) ≤ C.

We claim that

(13)  S(H_{i−1}, χ_{B_{1}(0)}, t)(y) ≤ c \frac{1}{|q^{∗}|} ∫_{q^{∗}} S(H_{i−1}, χ_{B_{1}(0)}, t)(ξ) dξ,

where q^{∗} has the same center as q, and l(q^{∗}) = 3 l(q). Applying Fubini Theorem we see that the left hand side of (13) equals

∫_{H_{i−1} ∩B_{t(x)}(y)} \frac{dx}{t^{n}(x)} ≤ c ∫_{H_{i−1}} \frac{|q^{∗} ∩ B_{t(x)}(x)|}{|q^{∗}|} \frac{dx}{t^{n}(x)}.

But for every x ∈ H_{i−1} ∩ B_{t(x)}(y), we have

\frac{|q^{∗} ∩ B_{t(x)}(x)|}{|q^{∗}|} ≥ c.

This gives the inequality claimed in (13).

It remains to show that the right-hand side of (13) 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

(14)  \frac{1}{|q^{∗}|} ∫_{q^{∗}} S(H_{i−1}, χ_{B_{1}(0)}, t)(ξ) dξ ≤ c \frac{1}{|q̃|} ∫_{q̃} S(H_{i−1}, χ_{B_{1}(0)}, t)(ξ) dξ.

We claim that the mean in the right-hand side of (14) is bounded by 1/2. Indeed, observe that

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

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

\frac{1}{|q̃|} ∫_{q̃} S(H_{i−1}, χ_{B_{1}(0)}, t)(ξ) dξ ≤ \frac{1}{|q̃|} ∫_{q̃} S(D̃_{i−1}, χ_{B_{1}(0)}, t)(ξ) dξ < \frac{1}{2}

by (a) in the construction of Q_{i}. This completes the proof of Lemma 2 and the Theorem.

□

Acknowledgment. We would like to thank Steve Hofmann, Loukas Grafakos, Nigel Kalton, Mark Rudelson, Terence Tao and Artem Zvavitch for helpful discussions. Our special thanks are to Alex Iosevich for his help and encouragement. The first author was partially supported under the auspices of the NSF grant DMS00-87339. The second author was supported under the auspices of a grant 02-253, CQ 759 from the University of Missouri Research Board.

References

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

[2] Georgiy Arutyunyants, Dmitry Ryabogin, On a paper of Hudson, www. math. missouri. edu/ryabs/

[3] Pascal Auscher, Steve Hofmann, Camil Muscalu, Terence Tao, and Christoph Thiele Carleson measures, trees, extrapolation, and T (b) theorems, www.math.ucla.edu/tao/

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

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

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

[7] Elias M. Stein, Harmonic analysis: Real variable methods, orthogonality, and oscillatory integrals, Princeton University Press, Princeton NJ, 1993.

[8] Elias M. Stein, Singular integrals and differentiability properties of functions, Princeton University Press, Princeton NJ, 1970.

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