L^p estimates for the bilinear Hilbert transform

Abstract

For the bilinear Hilbert transform given by: we announce the inequality ∥H fg∥_p3 ≤ K_p1,p2∥f∥_p1∥g∥_p2, provided 2 < p₁, p₂ < ∞, 1/p₃ = 1/p₁ + 1/p₂ and 1 < p₃ < 2.

We announce a partial resolution to long standing conjectures concerning the operator known as the bilinear Hilbert transform, defined as follows:

This operation is initially defined only for certain functions f and g, for instance those in the Schwartz class on ℝ. The conjectures concern the extension of H to a bounded operator on L^p spaces. We have proved:

Theorem 1. H extends to a bounded operator on L^p₁ × L^p₂ into L^p₃, provided 2 < p₁, p₂ < ∞ and 1 < p₃ < 2, where 1/p₃ = 1/p₁ + 1/p₂.

Some 30 years ago, in connection with the Cauchy integral on Lipschitz curves, Calderón (1) raised the question of H mapping L² × L² into L¹; this inequality is true. Indeed, the bilinear Hilbert transform maps into L^p₃ provided only that p₃ > 2/3.

Study of the bilinear Hilbert transform is intimately related to Carleson’s theorem (2) asserting the pointwise convergence of Fourier series. A seminal result, it has received two proofs, with the alternative proof provided by Fefferman (3). These proofs have provided us with ingenious and complementary methods of time frequency analysis. A similar analysis seems necessary to understand H, and so our proof entails significant aspects of both Carleson’s and Fefferman’s proofs. We give a description of our proof, with details presented in their most concrete form. Complete proofs, which appear in ref. 4, require definitions and constructions somewhat more general than those presented here.

The bilinear Hilbert transform must be broken into scales and the frequency behavior of each scale understood. Hence we replace the kernel 1/y with Σ_j=−∞^∞ 2^jρ(2^jy), where ρ is a Schwartz function with Fourier transform ρ̂(ξ) = ∫ e^−2πixξρ(x) dx supported on [1/2, 2). For each j, consider:

which has bilinear symbol ρ̂(2^−j(ξ − θ)). More specifically,

Therefore, if f is supported in frequency on the interval [n2^j, (n + 1)2^j], then H_j fg(x) acts on the inverse Fourier transform of ĝ(ξ)1[(n + 1/2)2^j, (n + 3)2^j](ξ), and is supported in frequency on the interval [(2n + 1/2)2^j, (2n + 4)2^j]. The differing rates of translation make these three intervals distinct.

It is important to note that the location of the intervals is arbitrary, and therefore, for all j and j′, the inner product of H_j fg and H_j′ fg need not tend to zero as |j − j′| tends to infinity. The analysis of H must be done in terms of both time and frequency.

Instead of proceeding with a decomposition of H, we define a model of it adapted to the combinatorics of the time–frequency plane. Let 𝒟 be a dyadic grid in ℝ. Call I × ω ∈ 𝒟 × 𝒟 a tile if |I|·|ω| = 1. The interval ω is a union of four dyadic subintervals of equal length, ω₁, ω₂, ω₃, and ω₄, which we list in ascending order. Thus, ξ_i < ξ_j for all 1 ≤ i < j ≤ 4 and ξ_j ∈ ω_j. (We will only use ω_j for j = 1, 2, 3.) We adopt the notation t = I_t × ω_t and tj = I_t × ω_tj for j = 1, 2, 3. Fix a Schwartz function φ with φ̂ supported on [−1/8, 1/8], in addition require that ∫ φ(x − 16n)φ(x) dx = 0 for all integers n. Set for all tiles t and j = 1, 2, 3,

where c(J) denotes the center of the interval J.

Then our model of the bilinear Hilbert transform is

which is initially defined only for Schwartz functions f₁ and f₂. We emphasize that the sum extends over all tiles, and hence all scales. The analogue of Theorem 1 is

Theorem 2. ℳ extends to a bounded operator on L^p₁ × L^p₂ into L^p′₃, provided 2 < p₁, p₂ < ∞ and 1 < p′₃ = (1/p₁ + 1/p₂)⁻¹ < 2.

With more liberal notions of “grid,” “tile,” and “φ_tj,” the bilinear Hilbert transform is in the convex hull of terms like our model ℳ.

In the present situation we can give a proof by way of duality. Thus we take f₃ ∈ L^p₃ and show that:

The sum is over positive quantities; namely, the decomposition above already captures all of the cancellation necessary for convergence of the sums. It also shows that the sum defining ℳ is unconditionally convergent in t. And as each f_j ∈ L^p_j, where p_j > 2, it follows that each function is locally square integrable. As it turns out, L² arguments are decisive in proving Theorem 2.

We localize the sum above in the x variable by setting:

Certainly ∫ F_t(x) dx = |I_t|^−1/2 ∏_j=1³ |〈f_j, φ_tj〉|. And so we show that F(x) = Σ_tF_t(x) is integrable. This follows from a weak-type result: for p_j as above, there is a δ > 0 so that for all |r_j − p_j| < δ, 1 ≤ j ≤ 3, the operator F(x) maps L^r₁ × L^r₂ × L^r₃ into L^r,∞, where 1/r = 1/r₁ + 1/r₂ + 1/r₃. Then a variant of the Marcinkiewicz interpolation theorem due to Janson (5) implies the strong-type inequality.

A single instance of the weak-type inequality is:

1

for some constant K. But this inequality implies the weak-type result, because F commutes with dilations by powers of 2, and so it suffices to establish this last inequality. These observations are useful since some of our estimates begin to break down on exceptional sets of small measure. Due the localization of F_t in the time variable and the fact that we only aim for a distributional inequality, we can delete tiles t whose time coordinate falls in a set of bounded measure.

The combinatorics of the time frequency plane enter in by way of the partial order on the tiles given by t < t′ if I_t ⊂ I_t′ and ω ⊃ ω′. Note that t and t′ are not comparable with respect to < if and only if t ∩ t′ = ∅. Being disjoint suggests orthogonality for the functions φ_tj and φ_t′j′, the dominant theme of Lemmas 1–3 we state below.

Call a collection of tiles T a Carleson–Fefferman (CF) set with top q if t < q for all t ∈ T. Thus ω_q ∩ ω_t ≠ ∅ for t ∈ T. Call T a j-CF set if T is a CF-set for which the intervals ω_tj intersect for all t ∈ T. Notice that if T is a 1-CF set, say, then the intervals {ω_tj|t ∈ T} are pairwise disjoint for j = 2, 3. Therefore, by application of Cauchy–Schwartz:

2

Notice that the last two square functions are Littlewood–Paley g functions, albeit conjugated by an exponential to account for the location of the CF set in frequency.

This estimate forms the motivation for Lemma 1 below, which formalizes a decomposition of the set of tiles that is fundamental to our argument.

Lemma 1. Fix p_i > 2. There is a δ > 0 and an ɛ₀ > 0 and a constant K so that for all |r_i − p_i| < δ and 0 < ɛ < ɛ₀, the following holds. The collection of all tiles S is a union:

with these properties. First, S⁰ is trivial in that:

3

Then S_n,i,j is a union of disjoint i-CF sets T_q with tops q ∈ S^*_n,i,j, and:

4

Here, recall that 1/r = Σ_i 1/r_i, which can be taken arbitrarily close to 1. And, most significantly, for t = min_i{p_i/2} − ɛ,

5

With Lemma 1 in place, we estimate:

The last sum is finite as r is arbitrarily close to one, while t + ɛ = min{p_i/2} > 1 is a fixed distance from one. Therefore, with Eq. 3, Eq. 1 holds.

We cannot give the complete construction of the S_n,i,j, but rather the initial steps, in which the nearly orthogonal classes of φ_ti are identified. First, we make an important comparison to a maximal function. If T_q is an i-CF set with top q, we have for j ≠ i,

Here M₂g is the maximal function (M|g|²)^1/2. Thus the set F = ∪_i{M₂f_i > C⁻¹} has bounded measure and we define S⁰ = {s|I_s ⊂ F}, making Eq. 4 trivial. For all i-CF sets T with top q, and T ⊂ S/S⁰, we have Δ(T, j) ≤ 1, for j ≠ i.

The remaining construction is inductive. Assume that the S_m,i,j are defined for all m < n and all i, j, in such a way that for S^r = S/∪_m<n∪_i,jS_m,i,j) we have:

6

and for any i-CF set T_q ⊂ S^r with top q, Δ(T_q, j) ≤ 2^−n/r_j+2, for j ≠ i. As the same inequality applies to each sub-CF set of T, we conclude that:

7

We define S^*_n,1,1 to be the set of maximal tiles q with |〈f₁, φ_q,1〉| ≥ 2^{−n/r₁₋₁}, and take S_n,1,1 to consist of all tiles t so that t1 < q for some q ∈ S^*_n,1,1. These tiles are removed, and then S_n,i,i is defined similarly for i = 2, 3. After the deletion of the tiles D₀ = ∪_i=1³S_n,i,i, we have |〈f_i, φ_ti〉| ≤ 2^−n/r_i−1 for all tiles t ∈ S^r′ = S^r/D₀.

The set S_n,1,2 has a slightly different construction. Consider 1-CF sets T_q ⊂ S^r′ with top q so that Δ(T, 2) ≥ 2^−n/r₂+1. We take T_q to be the maximal 1-CF set with this property. Let q(1) be such a top, which is maximal with respect to <, and in addition sup{ξ|ξ ∈ ω_q} is maximal. Remove the tiles T_q(1), and repeat this procedure to define T_q(2) and so on. S_n,1,2 is then ∪_ℓT_q(ℓ) and S^*_n,1,2 = {q(ℓ)|ℓ ≥ 1}. Observe that for any 1-CF set T ⊂ S^r′/S_n,1,2, we have Δ(T, 2) ≤ 2^−n/r₂+1. These procedures are repeated inductively to define the S_n,i,j for all n, i, j.

With the construction above it is elementary to check that these properties hold.

8

And in the case of i ≠ j, the collection S_n,i,j is a union of disjoint i-CF sets T_q, with q ∈ S^*_n,i,j, for which:

9

These last two bounds differ by a factor of 4, which is relevant below. See the comments concerning the minimal tiles immediately following Lemma 3 below. To achieve Eq. 4, one must delete some tiles t, using Eq. 2, the upper bounds Eqs. 6 and 7, and the control on the number of trees given in Eq. 5.

The essence of the matter lies in the control of the number of CF-sets, which is in the verification of Eq. 5. Eq. 5 relies upon the inequalities in the previous paragraph and Lemma 2 and 3 below, which address the issue of almost orthogonality.

Let us consider S_n,1,1, say. The tiles S^*_n,1,1 are maximal and therefore pairwise disjoint, which suggest weak orthogonality for the collection of functions {φ_q,1|q ∈ S^*_n,1,1}. If they were in fact orthogonal, Bessel’s inequality and Eq. 8 implies:

While f₁ is not in L², this inequality can be strengthened to an analogous form for L^r for r > 2.

However, disjointness of tiles does not imply orthogonality, because the functions φ_q,1 are not compactly supported in the x variable. Indeed, by our choice of φ, for two tiles t and s we have 〈φ_ti, φ_si〉 = 0 if ω_ti ∩ ω_si = ∅ or ω_ti = ω_si. But if ω_ti ⊂_≠ ω_si then for all n ≥ 0:

10

If we assume that a stronger separation of the tiles in the x variable, then we would expect orthogonality. And in this direction we have:

Lemma 2. For n ≥ 1 there are constants K and K_n so that the following holds for all A ≥ 1. Let S be any collection of tiles so that:

11

Here, for an interval I, AI denotes the interval with the same center as I and length A|I|. Set N_S(x) = Σ_t∈S1_It(x). Then:

A further combinatorial lemma asserts that if the tiles {ti|t ∈ S′} are merely disjoint, then after deleting tiles t for which I_t falls in an exceptional set of small measure, S′ is a union of O(A³) collections of tiles S that satisfy the stronger disjointness condition (Eq. 11).

The previous lemma is essential in obtaining Eq. 5 for the classes S_n,i,i. A corresponding lemma is necessary for the S_n,i,j, with i ≠ j, with Eq. 9 replacing the role of Eq. 8. It is:

Lemma 3. For n ≥ 1 there are constants K and K_n so that the following holds for all A ≥ 1. Let S be a union of j-CF sets T_q with tops q ∈ S*. Suppose that:

and for t ∈ T_q, q ∈ S* and i ≠ j fixed,

12

Set N_S(x) = Σ_q∈S* 1_Iq(x). Then:

Notice that in a j-CF set T_q, the tiles {ω_ti|t ∈ T_q} are pairwise disjoint. Thus Eq. 12 is stronger than merely asserting that the tiles {ω_ti|t ∈ T} are pairwise disjoint. With the construction of the S_n,i,j for i ≠ j given above, Eq. 12 is true after deleting the minimal tiles S_n,i,j^min in S_n,i,j. The minimal tiles are controlled with the first half of Eq. 9 and the observation that Σ_{s∈Sn,i,j^min} 1_Is(x) ≤ N_n,i,j(x) for all x.

The method of proof of both Lemmas 2 and 3 is similar. For instance, in Lemma 2, one considers the operator:

If S is finite, this is a compact self-adjoint operator, with maximal eigenvalue B. It suffices to estimate B, as for all f ∈ L², Σ_t∈S |〈f, φ_ti〉|² = 〈f, 𝒮_Sf〉 ≤ B∥f∥₂². Consider a normalized extremal eigenfunction f of 𝒮_S. One then estimates:

which is expanded in diagonal and off-diagonal terms. The diagonal term is Σ_t∈S |〈f, φ_ti〉|² = 〈f, 𝒮_Sf〉 ≤ B, which is an adequate estimate for B². The off-diagonal term is by Cauchy–Schwarz,

The innermost sum is bounded by C_nA⁻ⁿ inf_x∈ItMMf(x). This is seen by invoking the estimate |〈f, φ_si〉| ≤ K inf_x∈IsMf(x), using Eq. 10 and carefully exploiting the geometry of the tiles via the assumption 11.

One then sees that the off-diagonal term is no more than:

This, with the diagonal estimate, proves that B² ≤ B + C_nB^1/2A⁻ⁿ∥N_S∥_∞, whence follows Lemma 2.

The research outlined herein is the product of several years of effort (6, 7).