L^{p} BOUNDS FOR SINGULAR INTEGRALS AND MAXIMAL SINGULAR INTEGRALS WITH ROUGH KERNELS

LOUKAS GRAFAKOS AND ATANAS STEFANOV

Abstract. Convolution type Calderón-Zygmund singular integral operators with rough kernels p.v. Ω(x)/|x|^{n} are studied. A condition on Ω implying that the corresponding singular integrals and maximal singular integrals map L^{p} → L^{p} for 1 < p < ∞ is obtained. This condition is shown to be different from the condition Ω ∈ H^{1}(S^{n−1}).

1. Introduction and statements of results

In this paper, Ω will be a complex-valued integrable function defined on the sphere S^{n−1}, with mean value zero with respect to surface measure. Denote by T_{Ω} the Calderón-Zygmund singular integral operator defined as follows:

(1) (T_{Ω}f)(x) = lim_{ε→0} ∫_{|y|>ε} \frac{Ω(y/|y|)}{|y|^{n}}f(x − y) dy = p.v. ∫_{R^{n}} \frac{Ω(y/|y|)}{|y|^{n}}f(x − y) dy,

for f in the Schwartz class S(R^{n}). The limit in (1) is easily shown to exist for any f a C^{1} function on R^{n} with some decay at infinity.

For ε > 0, denote by

(T_{Ω}^{ε}f)(x) = ∫_{|y|>ε} \frac{Ω(y/|y|)}{|y|^{n}}f(x − y) dy

the truncated singular integral associated with T_{Ω} and by

(T_{Ω}^{∗}f)(x) = sup_{ε>0} |T_{Ω}^{ε}f(x)|

the maximal singular integral operator corresponding to this Ω.

Establishing the a priori bound ∥T_{Ω}^{ε}f∥_{L^{p}} ≤ C∥f∥_{L^{p}} independently of f ∈ S(R^{n}) and of ε > 0, leads to a (unique) extension of T_{Ω}^{ε} on L^{p}(R^{n}). Now, for f ∈ L^{p}(R^{n}), T_{Ω}^{ε}f converges in L^{p} as ε → 0 to some T_{Ω}f (which extends T_{Ω}f defined in (1) for f ∈ S(R^{n})), and by Fatou’s lemma, T_{Ω} is a bounded operator on L^{p}.

A similar a priori bound for T_{Ω}^{∗} implies that for f ∈ L^{p}(R^{n}), T_{ε}f converges (to T_{Ω}f) almost everywhere as ε → 0.

We now discuss L^{p} boundedness properties of these operators. It is well known that if Ω has some smoothness, then both T_{Ω} and T_{Ω}^{∗} extend to bounded operators on L^{p}(R^{n}) for all 1 < p < ∞. See [11] for details. In this paper we shall be concerned with Ω rough. The method of rotations introduced by Calderón and Zygmund [2] implies that that T_{Ω} and T_{Ω}^{∗} map L^{p}(R^{n}) → L^{p}(R^{n}) for any Ω odd in L^{1}(S^{n−1}). The situation for general Ω’s is significantly more involved. Calderón and Zygmund [2] proved that if

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

Key words and phrases. Calderón-Zygmund singular integrals, method of rotations.

Grafakos’s research at MSRI partially supported by the NSF under grant DMS 9623120.

∫_{S^{n−1}} |Ω(θ)| ln(2 + |Ω(θ)|) dθ < ∞, (2)

then T_{Ω} and T_{Ω}^{∗} are bounded operators on L^{p} for 1 < p < ∞.

Some years later, condition (2) above was independently improved by Connett [4] and Ricci and Weiss [9] who showed that if

Ω ∈ H^{1}(S^{n−1}), (3)

then T_{Ω} maps L^{p}(R^{n}) into itself for 1 < p < ∞. H^{1}(S^{n−1}) here denotes the 1-Hardy space on the unit sphere in the sense of Coifman and Weiss [3]; (this paper contains a proof of this result in dimension n = 2). See also [8] for a simple proof of this result on R^{n}.

The H^{1} condition (3) is also sufficient to imply that T_{Ω}^{∗} is bounded on L^{p} for 1 < p < ∞. For a proof of this fact we refer the reader to [8] and also to Fan and Pan [7] who recently obtained this result independently for a more general class of operators.

The main purpose of this paper is to present alternative conditions that imply L^{p} boundedness for T_{Ω} and T_{Ω}^{∗}. If we examine the proof giving the formula of the Fourier transform of p.v. Ω(x)/|x|^{n} we observe that the mild assumption

(4)  sup_{ξ∈S^{n−1}} ∫_{S^{n−1}} |Ω(θ)| ln \frac{1}{|θ · ξ|} dθ < +∞,

suffices to imply that (p.v. Ω(x)/|x|^{n}) is a bounded function, which is equivalent to saying that T_{Ω} maps L^{2}(R^{n}) into itself. It is unknown to us whether condition (4) implies L^{p} boundedness for some p ≠ 2.

Motivated by (4) we consider the family of conditions

(5)  sup_{ξ∈S^{n−1}} ∫_{S^{n−1}} |Ω(θ)| (ln \frac{1}{|θ · ξ|})^{1+α} dθ < +∞.

for α > 0. We can show that if Ω satisfies condition (5) for some α > 0, then T_{Ω} maps L^{p}(R^{n}) into itself for some p ≠ 2. More precisely, we have the following theorem:

Theorem 1. Let α > 0. Let Ω be a function in L^{1}(S^{n−1}) with mean value zero which satisfies condition (5) for some α > 0. Then T_{Ω} extends to a bounded operator from L^{p}(R^{n}) into itself for (2 + α)/(1 + α) < p < 2 + α.

As a corollary we obtain that if Ω satisfies condition (5) for all α > 0, then it maps L^{p}(R^{n}) into itself for all 1 < p < ∞. Regarding T_{Ω}^{∗} we can prove the following:

Theorem 2. Let α > 1. Let Ω be a function in L^{1}(S^{n−1}) with mean value zero which satisfies condition (5) for some α > 0. Then T_{Ω}^{∗} extends to a bounded operator from L^{p}(R^{n}) into itself for 1 + 3/(1 + 2α) < p < 2(2 + α)/3.

L^{p} BOUNDS FOR SINGULAR INTEGRALS AND MAXIMAL SINGULAR INTEGRALS 3

We conclude that if Ω satisfies condition (5) for all α > 0, then T_{Ω}^{∗} maps L^{p} to L^{p} for all 1 < p < ∞. We don’t know whether the ranges of indices in Theorems 1 and 2 are sharp. More fundamentally, we do not know an example of an Ω ∈ L^{1}(S^{n−1}) such that T_{Ω} maps L^{p} → L^{p} for some given p = p_{0} ≥ 2 but not for some other p_{1} > p_{0}.

In section 5, we show that condition (5) for all α > 0 is indeed disjoint from the H^{1} condition (3).

2. Boundedness of singular integrals

The theme of the proof of Theorem 1 is based on ideas developed by J. Duoandikoetxea and J.-L. Rubio de Francia [6] to treat several other operators of this sort. Define

σ_{k}(x) = \frac{Ω(x)}{|x|^{n}}χ_{2^{k}≤|x|≤2^{k+1}}, k ∈ Z.

Observe that σ̂_{k}(ξ) = σ̂_{0}(2^{k}ξ). We calculate σ̂_{0}(ξ). Set ξ′ = ξ/|ξ|. Expressing σ̂_{0} in polar coordinates, we obtain

σ̂_{0}(ξ) = ∫_{S^{n−1}} Ω(θ) [∫_{1}^{2} e^{2πir|ξ|(ξ′·θ)} \frac{dr}{r}] dθ. (6)

Using that Ω has mean value zero, we deduce that

|σ̂_{0}(ξ)| ≤ 2π(ln 2)∥Ω∥_{L^{1}}|ξ| = C|ξ|, (7)

which is a good estimate for |ξ| ≤ 2. For |ξ| ≥ 2 observe the following: The integral inside brackets in (6) is bounded by min (2, 3|ξ′·θ|^{−1}|ξ|^{−1}). (Pick a θ so that ξ′·θ ≠ 0.) Therefore it must satisfy the estimate

|∫_{1}^{2} e^{2πir|ξ|(ξ′·θ)} \frac{dr}{r}| ≤ \frac{2 (ln(\frac{3}{2}|ξ′ · θ|^{−1}))^{1+α}}{(ln |ξ|)^{1+α}} . (8)

It follows from (8) and (5) that

|σ̂_{0}(ξ)| ≤ C(ln |ξ|)^{−1−α} for |ξ| ≥ 2. (9)

Since σ_{k} is obtained from σ_{0} by a suitable dilation, it follows that there exists a constant C > 0, such that for all k ∈ Z the estimates below are valid:

|σ̂_{k}(ξ)| ≤ C(ln |2^{k}ξ|)^{−1−α}, for 2^{k}|ξ| ≥ 2,
|σ̂_{k}(ξ)| ≤ C2^{k}|ξ|,  for 2^{k}|ξ| ≤ 2. (10)

Now let ψ be a C^{∞} function supported in {x ∈ R^{n} : 3/4 ≤ |x| ≤ 9/4} such that ∑_{j∈Z}(ψ(2^{j}ξ))^{2} = 1. Let S_{j} be the operator given on the Fourier transform by multiplication by ψ_{j}(ξ) = ψ(2^{j}ξ). Define

T_{j}f = ∑_{k∈Z} S_{j+k}(σ_{k} ∗ S_{j+k}f).

It is easy to see that the identity

T_{Ω}f = ∑_{j∈Z} T_{j}f

is valid at least for f in the Schwartz class. Using a Fourier transform calculation, (10), and the fact that ψ_{j+k} is supported near the annulus |ξ| ∼ 2^{−j−k}, we obtain that T_{j} are bounded on L^{2}(R^{n}) with bound C2^{−j} for j ≥ 0 and C(|j|)^{−1−α} for j ≤ −1. In short

∥T_{j}f∥_{L^{2}} ≤ C(1 + |j|)^{−1−α}∥f∥_{L^{2}} for all j ∈ Z. (11)

We will also need estimates for the following maximal operator

f → σ^{∗}(f) = sup_{k∈Z}(|σ_{k}| ∗ |f|).

Without loss of generality we can assume that ∥Ω∥_{L^{1}(S^{n−1})} = 1. It follows that

ˆ|σ_{0}|(0) = 1. Introduce a radial function in the Schwartz class Φ, such that Φ̂(ξ) = 1 for |ξ| ≤ 2 and Φ̂(ξ) = 0 for |ξ| > 3. Let us also introduce Φ_{k} defined by Φ̂_{k}(ξ) = Φ̂(2^{k}ξ). Clearly we have

(12) σ^{∗}(f) ≤ sup_{k∈Z} |(|σ_{k}| − Φ_{k}) ∗ |f|| + sup_{k∈Z} |Φ_{k} ∗ |f||.

Denote μ_{k} = |σ_{k}| − Φ_{k}. Since μ̂_{k}(0) = 0, the same proof giving (10) implies that

|μ̂_{k}(ξ)| ≤ C2^{k}|ξ|,  for 2^{k}|ξ| ≤ 2,
|μ̂_{k}(ξ)| ≤ C(log |2^{k}ξ|)^{−1−α}, for 2^{k}|ξ| ≥ 2. (13)

Therefore we obtain from (12) that

σ^{∗}(f) ≤ sup_{k∈Z}(μ_{k} ∗ |f|) + Mf ≤ (∑_{k} |μ_{k} ∗ |f||^{2})^{1/2} + Mf, (14)

where M is the Hardy-Littlewood maximal function. Since for all 1 < r < ∞,

∥( ∑_{k}(μ_{k} ∗ f)^{2})^{1/2}∥_{L^{r}}^{r} = Average ∥ ∑_{k} ε_{k}(μ_{k} ∗ f)∥_{L^{r}}^{r}, (15)

over all choices of signs ε_{k} = ±1, estimates for the square function on the right hand side of (14) can be obtained from estimates on integral operators of the form g → ∑_{k} ε_{k}(μ_{k} ∗ g). Now using (13) and (14) we conclude that σ^{∗} maps L^{2} → L^{2}, whenever α > 0. At this point we recall the following lemma:

Lemma 1. (See [6] p. 544) If ∥σ^{∗}(f)∥_{L^{s}} ≤ C∥f∥_{L^{s}} and \frac{1}{2s} = |\frac{1}{2} − \frac{1}{q}|, then for

arbitrary functions g_{k} we have

∥(∑_{k∈Z} |σ_{k} ∗ g_{k}|^{2})^{1/2}∥_{L^{q}} ≤ C∥(∑_{k∈Z} |g_{k}|^{2})^{1/2}∥_{L^{q}}.

Applying Lemma 1 with s = 2 and q = q_{0} = 4, we obtain that

∥T_{j}f∥_{L^{q_{0}}} ≤ C∥(∑_{k∈Z} |σ_{k} ∗ S_{j+k}f|^{2})^{1/2}∥_{L^{q_{0}}} ≤ C∥(∑_{k∈Z} |S_{j+k}f|^{2})^{1/2}∥_{L^{q_{0}}} ≤ C∥f∥_{L^{q_{0}}}, (16)

where the middle inequality is a consequence of Lemma 1 and the first and last inequalities follow from the Littlewood-Paley theorem.

Interpolating between estimates (11) and (16) we obtain that

∥T_{j}f∥_{L^{p}} ≤ C(1 + |j|)^{−(1+α)θ_{p}}∥f∥_{L^{p}},

where 1/p = θ_{p}/2 + (1 − θ_{p})/q_{0}. Now observe that T_{Ω} = ∑_{j∈Z} T_{j} maps L^{p} → L^{p} for all p’s for which p′_{1} < p < p_{1}, where p_{1} = (4 + 4α)/(2 + α) is the unique solution of the equation (1 + α)θ_{p} = 1. The same argument also gives that T_{ε}f = ∑_{k} ε_{k}(μ_{k} ∗ f) maps L^{p} → L^{p} for p′_{1} < p < p_{1} uniformly on the choice of the signs (ε_{j}), ε_{j} = ±1. It follows that the square function in (15) is also bounded on L^{p} for this range of p’s and hence so is σ^{∗}(f) by the estimate in (14). Thus we are in a position to apply Lemma 1 again with s in the interval (p′_{1}, p_{1}).

Now continue this way. Fix s_{1} ∈ (2, p_{1}) and let q_{1} be the unique number bigger than q_{0} = 4 which satisfies the equation 1/2s′_{1} = |1/2 − 1/q_{1}|. Apply Lemma 1 with s = s′_{1} and q = q_{1}. As before we obtain that T_{Ω} maps L^{p} → L^{p} for p′_{2} < p < p_{2}, where p_{2} is the unique solution of the equation (1 + α)θ_{p} = 1, where θ_{p} is given by 1/p = θ_{p}/2 + (1 − θ_{p})/q_{1} now. This bootstrapping argument leads to an inductive definition of three sequences 2 = p_{0} < p_{1} < …, 2 < s_{1} < s_{2} < …, and 4 = q_{0} < q_{1} < … such that for k = 1, 2, …

p_{k−1} < s_{k} < p_{k},  \frac{1}{p_{k}} − \frac{1}{q_{k−1}} = \frac{1}{1 + α} (\frac{1}{2} − \frac{1}{q_{k−1}} ) ,  \frac{1}{2s′_{k}} = \frac{1}{2} − \frac{1}{q_{k}} .

Let b = sup_{k} p_{k}. The above equations easily imply that b = 2 + α. Therefore T_{Ω} maps L^{p} to L^{p} for 2 ≤ p < 2 + α. The remaining range of p’s follows by duality.

3. Boundedness of maximal singular integrals

We now prove Theorem 2. Below we use the same notation as in the previous section. Let

(T_{k}f)(x) = ∫_{|y|>2^{k}} \frac{Ω(y)}{|y|^{n}}f(x − y) dy = ∑_{j=k}^{∞}(σ_{j} ∗ f)(x),

(T^{∗}f)(x) = sup_{k} |(T_{k}f)(x)|.

If 2^{k−1} ≤ ε < 2^{k}, then

|(T_{Ω}^{ε}f)(x)| ≤ |(T_{k}f)(x)| + | ∫_{ε<|y|<2^{k}} \frac{Ω(y)}{|y|^{n}}f(x − y) dy| ≤ |(T_{k}f)(x)| + (|σ_{k}| ∗ |f|)(x).

From the proof of Theorem 1 we know that σ^{∗} maps L^{p} → L^{p} for (2 + α)/(1 + α) < p < 2 + α. Since

|(T_{Ω}^{∗}f)(x)| ≤ |(T^{∗}f)(x)| + σ^{∗}(|f|)(x),

it suffices to show that T^{∗} : L^{p} → L^{p} for the claimed range of p’s, which is contained in the interval ((2 + α)/(1 + α), 2 + α).

With Φ as in the previous section, estimate

(17)  sup_{k∈Z} |(T_{k}f)(x)| ≤ sup_{k∈Z} |Φ_{k} ∗ ∑_{j=k}^{∞} σ_{j} ∗ f| + sup_{k∈Z} |(δ − Φ_{k}) ∗ ∑_{j=k}^{∞} σ_{j} ∗ f| ,

where δ is Dirac mass at the origin. It is easy to see that

sup_{k∈Z} |Φ_{k} ∗ ∑_{j=k}^{∞} σ_{j} ∗ f| ≤ C (M(T f) + M(f)) , (see [6], p.548)

which implies L^{p} bounds for the first term on the right hand side of (17) for (2 + α)/(1 + α) < p < 2 + α. Control the second term on the right hand side of (17) by

sup_{k∈Z} |(δ − Φ_{k}) ∗ ∑_{j=0}^{∞} σ_{j+k} ∗ f| ≤ ∑_{j=0}^{∞} Q_{j}(f),

where

(Q_{j}f)(x) = sup_{k∈Z} |(δ − Φ_{k}) ∗ σ_{j+k} ∗ f| .

To conclude the proof of Theorem 2, it suffices to show that for j ≥ 0 we have

∥Q_{j}f∥_{L^{p}} ≤ C∥f∥_{L^{p}}, 2 ≤ p < 2 + α, (18)

∥Q_{j}f∥_{L^{2}} ≤ C(1 + j)^{−α}∥f∥_{L^{2}}. (19)

Then, a simple interpolation between (18) and (19) gives that Q_{j} maps L^{p} → L^{p} with bound C_{δ}(1 + j)^{2α(2+α−δ−p)/p(α−δ)}, for any δ > 0 small, and the conclusion of Theorem 2 follows by summing on j.

Now observe that

|Q_{j}f| ≤ sup_{k} |σ_{j+k} ∗ f| + sup_{k} |Φ_{k} ∗ σ_{j+k} ∗ f| ≤ C(σ^{∗}(f) + M(σ^{∗}(f))).

Therefore Q_{j} is bounded on L^{p} whenever σ^{∗} is, that is ∥Q_{j}f∥_{L^{p}} ≤ C∥f∥_{L^{p}} when 2 ≤ p < 2+α and (18) is proved. To prove (19) we need to exploit some orthogonality. We have

∥Q_{j}f∥_{L^{2}}^{2} ≤ ∑_{k} ∥(δ − Φ_{k}) ∗ σ_{j+k} ∗ f∥_{L^{2}}^{2} = Average∥ ∑_{k} ε_{k}((δ − Φ_{k}) ∗ σ_{j+k} ∗ f)∥_{L^{2}}^{2},

where ε = (ε_{k})_{k} is a sequence of ±1’s. For a fixed sequence ε_{k} = ±1, let us denote by

M_{j,k}f = ε_{k}(δ − Φ_{k}) ∗ σ_{j+k} ∗ f.

We will need the following

Lemma 2. Let m ≥ 1, j ≥ 0 and k_{1} ≤ … ≤ k_{2m} be integers. Then

∥M_{j,k_{1}} … M_{j,k_{2m}}∥_{2→2} ≤ C^{2m} ∏_{i=1}^{2m} (\frac{1}{1 + j + k_{i} − k_{1}} )^{1+α} .

Proof. Since Φ̂_{k_{1}}(ξ) vanishes for 2^{k_{1}}|ξ| ≤ 2 we have,

∥M_{j,k_{1}} … M_{j,k_{2m}}f∥_{L^{2}}^{2} = ∫ ∏_{i=1}^{2m} |1 − Φ̂_{k_{i}}(ξ)|^{2} |{ \atop σ_{j+k_{i}}}(ξ)|^{2}|f̂(ξ)|^{2}dξ

≤ (C)^{2m} ∫_{|ξ|≥2^{1−k_{1}}} ∏_{i=1}^{2m} [\frac{1}{log (2^{j+k_{i}}|ξ|)}]^{2+2α} |f̂(ξ)|^{2}dξ

≤ C^{2m} ∏_{i=1}^{2m} [\frac{1}{1 + j + k_{i} − k_{1}} ]^{2+2α} ∥f∥_{L^{2}}^{2},

where we used the first estimate in (10) in the last inequality above.

Now we return to the proof of Theorem 2. We must show that ∥M_{j}^{ε,N} ∥_{2→2} ≤ C(1 + j)^{−α} uniformly on N and ε = (ε_{k}), where

M_{j}^{ε,N}f = ∑_{k=−N}^{N} ε_{k}M_{j,k}.

Since M_{j}^{ε,N} are self adjoint operators, we have

∥M_{j}^{ε,N} ∥_{2→2}^{2m} = ∥(M_{j}^{ε,N})^{2m}∥_{2→2} ≤ ∑_{−N≤k_{1}≤...≤k_{2m}≤N} ∥M_{j,k_{1}} … M_{j,k_{2m}}∥_{2→2}

≤ ∑_{−N≤k_{1}≤...≤k_{2m}≤N} C^{2m} ∏_{i=1}^{2m} (\frac{1}{1 + j + k_{i} − k_{1}} )^{1+α} ≤ \frac{NC^{2m}}{(1 + j)^{1+α}} (\frac{1}{(1 + j)^{α}} )^{2m−1}

≤ N \frac{C^{2m}}{1 + j} (1 + j)^{−2mα}.

Taking (2m)^{th} roots and letting m → ∞ we obtain

∥M_{j}^{ε,N} ∥_{2→2} ≤ C(1 + j)^{−α}.

This concludes the proof of (19) and hence of Theorem 2.

4. Examples

It is easy to see that condition (5) for all α > 0 contains the case Ω ∈ L^{q}(S^{n−1}), q > 1, considered by several authors, including [6]. However, it does not include the condition Ω ∈ LlogL(S^{n−1}) of Calderón and Zygmund. It is therefore natural to ask whether there exist examples of Ω ∉ LlogL(S^{n−1}) which satisfy (5) for all α > 0. In this section we prove something more.

We construct an example to show that there exist integrable functions on S^{n−1} with mean value zero which are not in H^{1}(S^{n−1}) but which satisfy (5) for all α > 0. Then we show that there exist functions that satisfy the converse.

We begin with the converse which is easier. The function

Ω(θ) = ∑_{k=2}^{∞} \frac{e^{ikθ}}{(log k)^{2}}

belongs to H^{1}(S^{1}) but it fails to satisfy condition (5) for any α > 0. Both assertions follow from the fact that Ω(θ) behaves like θ^{−1} log^{−2}(θ^{−1}) as θ → 0+ (See [13] p. 189).

We now construct an Ω ∈ L^{1}(S^{1}) \ H^{1}(S^{1}) with mean value zero which satisfies condition (5) for all α > 0. The example presented below is unavoidably complicated. The problem is that such a function must have an infinite number of spikes which are sufficiently far away from each other and which are (barely) integrable and have mean value zero.

At this point we think of S^{1} as the interval [0, 1] via the identification

(20)  Ω̃(x) = Ω(cos(2πx), sin(2πx)),

where Ω is defined on S^{1} and Ω̃ on [0, 1]. It is not hard to see that under the identification given in (20), the condition Ω ∉ H^{1}(S^{n−1}) is equivalent to the fact that the Hilbert transform of Ω̃χ_{[0,1]} is not in L^{1}(R^{1}), and condition (5) is equivalent to

sup_{0≤z≤1} ∫_{0}^{1} |Ω̃(x)| ln^{1+α} \frac{1}{|x − z|}dx ≤ C_{α} < ∞. (21)

For a detailed justification of these facts see [10]. Now let

a_{n} = (ln n)^{−1}  b_{n} = e^{−γ_{n}}
γ_{n} = e^{(ln n)^{1/2}}  δ_{n} = e^{−γ_{n}^{1/4}}
d_{n} = a_{n} + δ_{n}  c_{n} = a_{n} − δ_{n}
β_{n} = 1 − (ln n + \frac{3}{2} ln γ_{n})γ_{n}^{−1}

Heuristically speaking, a_{n} is a sequence that decays slowly to zero, c_{n} and d_{n} are symmetric points about a_{n} at distance δ_{n}, (c_{n} − b_{n}, c_{n}) and (d_{n} − b_{n}, d_{n}) are small intervals near c_{n} and d_{n} with length b_{n} = e^{−γ_{n}}, where (ln n)^{ε} << γ_{n} << n^{ε} for all ε > 0, and the β_{n}’s are powers that converge to one at a rate ∼ γ_{n}^{−1}. It is easy to see that

\frac{b_{n}^{1−β_{n}}}{1 − β_{n}} = \frac{1}{nγ_{n}^{1/2}(ln n + \frac{3}{2}(ln n)^{1/2})} ∼ \frac{1}{nγ_{n}^{1/2} ln n}, (22)

for n large. Now let

Ω̃(x) = ∑_{n=10^{9}}^{∞} (\frac{1}{|x − c_{n}|^{β_{n}}} χ_{(c_{n}−b_{n},c_{n})}(x) − \frac{1}{|x − d_{n}|^{β_{n}}} χ_{(d_{n}−b_{n},d_{n})}(x)) .

We first verify that condition (21) holds for all α > 0. The worst possible z’s in (21) are the singularities of Ω̃, i.e. the points z = c_{n}, d_{n}, and z = 0. By symmetry we

L^{p} BOUNDS FOR SINGULAR INTEGRALS AND MAXIMAL SINGULAR INTEGRALS 9

consider only z = c_{n} and z = 0. Fix N ≥ 10^{9} and consider z = c_{N}. We have

∫_{0}^{1} |Ω̃(x)| ln^{1+α} \frac{1}{|x − c_{N}|}dx ≤ I_{1}(N) + I_{2}(N) + I_{3}(N) + I_{4}(N),

where

I_{1}(N) = ∑_{n≠N} ∫_{c_{n}−b_{n}}^{c_{n}} \frac{1}{|x − c_{n}|^{β_{n}}} ln^{1+α} \frac{1}{|x − c_{N}|}dx,
I_{2}(N) = ∑_{n≠N} ∫_{d_{n}−b_{n}}^{d_{n}} \frac{1}{|x − d_{n}|^{β_{n}}} ln^{1+α} \frac{1}{|x − c_{N}|}dx,
I_{3}(N) = ∫_{c_{N} −b_{N}}^{c_{N}} \frac{1}{|x − c_{N}|^{β_{N}}} ln^{1+α} \frac{1}{|x − c_{N}|}dx,
I_{4}(N) = ∫_{d_{N} −b_{N}}^{d_{N}} \frac{1}{|x − d_{N}|^{β_{N}}} ln^{1+α} \frac{1}{|x − c_{N}|}dx.

Observe that I_{2}(N) ≤ CI_{1}(N) and that I_{4}(N) ≤ I_{3}(N). Also, it is easy to see that

sup_{N≥10^{9}} I_{3}(N) ≤ C sup_{N≥10^{9}} \frac{b_{N}^{1−β_{N}}}{1 − β_{N}} ln^{1+α} \frac{1}{b_{N}} ≤ C sup_{N≥10^{9}} \frac{γ_{N}^{1+α}}{Nγ_{N}^{1/2} ln N} ≤ C_{α}.

To control sup_{N≥10^{9}} I_{1}(N) we need to show that

sup_{N≥10^{9}} [∑_{n≠N} ∫_{c_{n}−b_{n}}^{c_{n}} \frac{1}{|x − c_{n}|^{β_{n}}} ln^{1+α} \frac{1}{|x − c_{N}|}dx] ≤ C_{α}. (23)

Using that |x − c_{N}| ∼ |c_{n} − c_{N}| ∼ |a_{n} − a_{N}| in the integrand above and (22), we conclude that (23) will be a consequence of

sup_{N≥10^{9}} [∑_{n≠N} \frac{b_{n}^{1−β_{n}}}{1 − β_{n}} ln^{1+α} \frac{1}{|a_{n} − a_{N}|}] ≤ C_{α}. (24)

We have two cases. For n > N, |a_{n}−a_{N}| ≥ |a_{N+1}−a_{N}| ≥ (N ln^{2} N)^{−1} and therefore

sup_{N≥10^{9}} [∑_{n>N} \frac{b_{n}^{1−β_{n}}}{1 − β_{n}} ln^{1+α} \frac{1}{|a_{n} − a_{N}|}] ≤ C sup_{N≥10^{9}} ∑_{n>N} \frac{ln^{1+α}(N ln^{2} N)}{nγ_{n}^{1/2} ln n} ≤ C_{α},

the latter being an easy consequence of the integral test. For 10^{9} ≤ n ≤ N − 1 we have

∑_{n=10^{9}}^{N−1} \frac{b_{n}^{1−β_{n}}}{1 − β_{n}} ln^{1+α} \frac{1}{|a_{n} − a_{N}|}

≤ C ∑_{n=10^{9}}^{N−1} \frac{1}{nγ_{n}^{1/2} ln n} ln^{1+α} \frac{1}{|(ln n)^{−1} − (ln N)^{−1}|} = A(N) + B(N),

where A(N) is the sum above of over the indices 10^{9} ≤ n < γ_{N} and B(N) is the sum over the the indices γ_{N} ≤ n ≤ N − 1. On A(N) we have |(ln n)^{−1} − (ln N)^{−1}|^{−1} ≤ C ln n, and thus A(N) is clearly bounded independently of N. On B(N) we have |(ln n)^{−1} − (ln N)^{−1}|^{−1} ≤ CN(ln N)^{2}. Now estimate sup_{N≥10^{9}} B(N) by

C sup_{N≥10^{9}} ln^{1+α}(N^{2}) ∑_{n≥γ_{N}} \frac{1}{nγ_{n}^{1/2} ln n} ≤ C sup_{N≥10^{9}} \frac{ln^{1+α}(N^{2})}{γ_{γ_{N}}^{1/3}} ≤ C,

where we used the integral test to deduce the first inequality above. This concludes the proof of (21) when z = c_{N}. Condition (21) for z = 0 is is equivalent to the following inequality

∑_{n=10^{9}}^{∞} \frac{ln^{1+α}(ln n)}{nγ_{n}^{1/2} ln n} ≤ C_{α},

which is certainly correct by the choice of our parameters. This proves that Ω̃ satisfies condition (21) for all α > 0.

We now prove that Ω̃ is not in the Hardy space H^{1}. Extend Ω̃ to be equal to zero outside the interval [0, 1]. Let H be the usual Hilbert transform. Fix N ≥ 10^{9} and y ∈ [d_{N}, d_{N} + b_{N}]. Obviously

π|(HΩ̃)(y)| ≥ K_{N}(y) − L_{N}(y), (25)

where

K_{N}(y) = | ∫_{c_{N} −b_{N}}^{c_{N}} \frac{1}{|x − c_{N}|^{β_{N}}} \frac{1}{x − y} dx − ∫_{d_{N} −b_{N}}^{d_{N}} \frac{1}{|x − d_{N}|^{β_{N}}} \frac{1}{x − y} dx| ,

L_{N}(y) = ∑_{n≠N} | ∫_{c_{n}−b_{n}}^{c_{n}} \frac{1}{|x − c_{n}|^{β_{n}}} \frac{1}{x − y} dx − ∫_{d_{n}−b_{n}}^{d_{n}} \frac{1}{|x − d_{n}|^{β_{n}}} \frac{1}{x − y} dx| .

We first prove that

sup_{N≥10^{9}} sup_{y∈[d_{N} ,d_{N} +b_{N}]} L_{N}(y) ≤ C (26)

Indeed,

∫_{c_{n}−b_{n}}^{c_{n}} \frac{1}{|x − c_{n}|^{β_{n}}} \frac{1}{x − y} dx = \frac{b_{n}^{1−β_{n}}}{1 − β_{n}} \frac{1}{(−b_{n} + c_{n} − y)} + smaller term

∫_{d_{n}−b_{n}}^{d_{n}} \frac{1}{|x − d_{n}|^{β_{n}}} \frac{1}{x − y} dx = \frac{b_{n}^{1−β_{n}}}{1 − β_{n}} \frac{1}{(−b_{n} + d_{n} − y)} + smaller term

where the smaller terms are bounded by Cb_{n}^{1−β_{n}}/(1 − β_{n}) and ∑_{n≥10^{9}} b_{n}^{1−β_{n}}/(1 − β_{n}) ≤

C. Therefore

L_{N}(y) ≤ C ∑_{n≠N} \frac{b_{n}^{1−β_{n}}}{1 − β_{n}} \frac{|d_{n} − c_{n}|}{|a_{n} − a_{N}|^{2}} ≤ C ∑_{n≠N} \frac{b_{n}^{1−β_{n}}}{1 − β_{n}} \frac{δ_{n}}{|a_{n} − a_{N}|^{2}}

and thus it remains to prove that

(27)  sup_{N≥10^{9}} ∑_{n≠N} \frac{δ_{n}}{nγ_{n}^{1/2} ln n} \frac{1}{((ln n)^{−1} − (ln N)^{−1})^{2}} ≤ C.

The sum in (27) for n > N is bounded by

∑_{n>N} \frac{δ_{n}}{nγ_{n}^{1/2} ln n} \frac{1}{((ln n)^{−1} − (ln N)^{−1})^{2}} ≤ N^{2} ln^{4} N ∑_{n>N} \frac{δ_{n}}{nγ_{n}^{1/2} ln n} ≤ C,

uniformly in N ≥ 10^{9}. Split the sum in (27) for n < N into the sum A′(N) over the indices 10^{9} ≤ n < γ_{N} and the sum B′(N) over the indices γ_{N} ≤ n ≤ N −1. Using that when 10^{9} ≤ n < γ_{N} we have |(ln n)^{−1}−(ln N)^{−1}|^{−1} ≤ C ln n we conclude that A′(N) is bounded independently of N. When γ_{N} ≤ n ≤ N−1 we have |(ln n)^{−1}−(ln N)^{−1}|^{−1} ≤ CN(ln N)^{2} and hence

sup_{N≥10^{9}} B′(N) ≤ C sup_{N≥10^{9}} N^{5} ∑_{n≥ln N} \frac{1}{nγ_{n}^{1/2}(ln n)e^{γ_{n}^{1/4}}} ≤ C,

which follows from the integral test. This proves (27) and hence L_{N}(y) is bounded uniformly in N.

Now we turn our attention to K_{N}(y). Observe that the following inequality holds

∫_{d_{N} −b_{N}}^{d_{N}} \frac{1}{|x − d_{N}|^{β_{N}}} \frac{1}{y − x}dx ≥ \frac{3}{2} ∫_{c_{N} −b_{N}}^{c_{N}} \frac{1}{|x − c_{N}|^{β_{N}}} \frac{1}{y − x}dx,

because of the proximity of y to the support of the first integral. Therefore

|K_{N}(y)| ≥ c ∫_{d_{N} −b_{N}}^{d_{N}} \frac{1}{|x − d_{N}|^{β_{N}}} \frac{1}{y − x}dx − C

when y ∈ [d_{N}, d_{N} + δ_{N}]. Integrate over this set to obtain

\begin{matrix}∫_{d_{N}}^{d_{N} +δ_{N}}|K_{N}(y)|dy ≥ \\ c | ∫_{d_{N} −b_{N}}^{d_{N}} \frac{1}{|x − d_{N}|^{β_{N}}} ln(d_{N} + δ_{N} − x)dx − ∫_{d_{N} −b_{N}}^{d_{N}} \frac{1}{|x − d_{N}|^{β_{N}}} ln(d_{N} − x)dx−| Cδ_{N}. \end{matrix} (28)

We clearly have that

| ∫_{d_{N} −b_{N}}^{d_{N}} \frac{1}{|x − d_{N}|^{β_{N}}} ln(d_{N} + δ_{N} − x)dx| ≤ C| ln δ_{N}|\frac{b_{N}^{1−β_{N}}}{1 − β_{N}} ≤ \frac{γ_{N}^{1/4}}{Nγ_{N}^{1/2} ln N} , (29)

while the the crucial fact is that

| ∫_{d_{N} −b_{N}}^{d_{N}} \frac{1}{|x − d_{N}|^{β_{N}}} ln(d_{N} − x)dx| ≥ C| ln b_{N}|\frac{b_{N}^{1−β_{N}}}{1 − β_{N}} ≥ \frac{γ_{N}^{1/2}}{N ln N} . (30)

Combining (25), (26), (28), (29), and (30) we obtain

∥HΩ̃∥_{L^{1}} ≥ ∑_{N≥10^{9}} ∫_{d_{N}}^{d_{N} +δ_{N}}|(HΩ̃)(y)|dy

≥ c ∑_{N≥10^{9}} \frac{γ_{N}^{1/2}}{N ln N} − C ∑_{N≥10^{9}} \frac{1}{Nγ_{N}^{1/4} ln N} − C ∑_{N≥10^{9}} δ_{N} = ∞.

This proves that Ω̃ ∉ H^{1}([0, 1]).

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Loukas Grafakos, Department of Mathematics, University of Missouri, Columbia, MO 65211, USA

Current address: MSRI, Berkeley, CA 94720-5070

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

Atanas Stefanov, Department of Mathematics, University of Missouri, Columbia, MO 65211, USA

Current address: MSRI, Berkeley, CA 94720-5070

  E-mail address: astefanov@pascal.math.missouri.edu