Bulletin T.CXXIII de l’Académie Serbe des Sciences et des Arts - 2002 Classe des Sciences mathématiques et naturelles Sciences mathématiques, N\underline{o} 27

AN EXTENSION ON TRILINEAR HILBERT TRANSFORM

ANETA BUČKOVSKA

(Presented at the 5th Meeting, held on June 21, 2002)

A b s t r a c t. The Trilinear Hilbert transform H : L^{p} × L^{q} × A → L^{r} is extended to D_{L^{p}} × D_{L^{q}} × D_{A} → D_{L^{r}} as a continuous mapping.

AMS Mathematics Subject Classification (2000): 44A15, 46F12 Key Words: trilinear Hilbert transform, multi-linear operators

1. Introduction

In several papers Lacey and Thiele ([3]-[5]) had studied the continuity of the bilinear Hilbert transform

H_{α}(f, g)(x) = p.v.∫ f(x − y)g(x + αy)\frac{dy}{y} , α ∈ R \ \{0, −1\} ,

where f ∈ L^{2}(R) and g ∈ L^{∞}(R), respectively f ∈ L^{p_{1}}(R) and g ∈ L^{p_{2}}(R),

2/3 < p = \frac{p_{1}p_{2}}{p_{1} + p_{2}} , 1 < p_{1}, p_{2} < ∞, q = p/(p − 1), q_{1} = p_{1}/(p_{1} − 1). (1)

Their main result is the affirmative answer on the Calderon conjecture, first for p_{1} = 2, p_{2} = ∞ ([3]), then for p_{1} and p_{2} with the assumptions given above ([5]). Their main result is

||H_{α}(f, a)||_{L^{p}} ≤ C||f||_{L^{p_{1}}} ||a||_{L^{p_{2}}} , f ∈ L^{p_{1}}, a ∈ L^{p_{2}},  (2)

where C > 0 depends on α, p_{1}, p_{2} (for p_{1} = 2, p_{2} = ∞, p equals 2).

In [1] the bilinear Hilbert transform H_{α} : L^{2} × L^{∞} → L^{2} respectively, H_{α} : L^{p_{1}} × L^{p_{2}} → L^{p}, was extended to D′_{L^{2}} × D_{L^{∞}} → D′_{L^{2}}, respectively, D′_{L^{q}} × D_{L^{p_{2}}} → D′_{q_{1}}, (with suitable parameters) as a hypocontinuous, respectively, continuous mapping and in [2] the Bilinear Hilbert transform of ultradistributions was defined.

In this paper we extend the trilinear Hilbert transform on L^{p} ×L^{q} ×A to L^{r} whenever 1 < p, q ≤ ∞, \frac{1}{p} + \frac{1}{q} = \frac{1}{p}, and \frac{2}{3} < r < ∞ to D_{L^{p}} ×D_{L^{q}} ×D_{A} → D_{L^{r}} as a continuous mapping.

2. Preliminaries

Denote by A the Wiener algebra: the space of functions whose Fourier transform is in L^{1}. Denote by D_{A} a space of smooth functions φ on R such that for every α ∈ N_{0} hold

F(φ^{(α)}) = ξ^{α}φ̂ ∈ L^{1}.

Define ||φ||_{k} = sup||ξ^{α}φ̂||_{L^{1}}, k ∈ N_{0}. In [7] it is proved that the mapping

T : L^{p_{1}} × ⋯ × L^{p_{n−s−1}} × A × ⋯ × A → L^{p′_{n−s}}

is continuous, where \frac{1}{p_{1}} + ⋯ + \frac{1}{p_{n−1}} , 1 < p_{i} ≤ ∞, for i = 1, 2, ..., n − s − 1 and

\frac{1}{p_{i_{1}}} + ⋯ + \frac{1}{p_{i_{r}}} < \frac{(n − s) − 2(k − s) + r}{2}

for all 1 ≤ i_{1} < ⋯ < i_{r} ≤ n − s, 1 ≤ r ≤ n − s.

Thus, for instance Trilinear Hilbert Transform

T (f, g, h) = ∫ f(x − t)g(x + t)h(x + 2t)\frac{dt}{t}

maps L^{p} × L^{q} × A to L^{r} whenever 1 < p, q ≤ ∞, \frac{1}{p} + \frac{1}{q} = \frac{1}{p}, and \frac{2}{3} < r < ∞. That is the case when n = 4, k = 2 and s = 1.

3. Mappings T_{f,g}, T_{f,h} and T_{g,h}

Theorem 3.1Let f ∈ D_{L^{p}}, g ∈ D_{L^{q}} and h ∈ D_{A}. Then for 1 < p, q < ∞ mappings T_{f,g}, T_{f,h} and T_{g,h} from D_{A} to D_{L^{r}}, from D_{L^{q}} to D_{L^{r}} and from D_{L^{p}} to D_{L^{r}} respectively are linear and continuous.

P r o o f. Let f ∈ D_{L^{p}}, g ∈ D_{L^{q}} and h ∈ D_{A}.

T (f, g, h)(x) = lim_{ε→0^{+}} ∫_{|t|>ε} f(x − t)g(x + t)h(x + 2t)\frac{dt}{t}

= ∫ _{|t|>N}f(x − t)g(x + t)h(x + 2t)\frac{dt}{t}

+ lim_{ε→0^{+}}∫ _{N≥|t|>ε}f(x − t)g(x + t)h(x + 2t)\frac{dt}{t}

= ∫ _{|t|>N}f(x − t)g(x + t)h(x + 2t)\frac{dt}{t}

+ lim_{ε→0^{+}} ∫ _{N≥|t|>ε}f(x − t)\frac{g(x + t) − g(x)}{t}h(x + 2t)dt

+ lim_{ε→0^{+}} ∫ _{N≥|t|>ε}f(x − t)g(x)h(x + 2t)\frac{dt}{t}

= ∫ _{|t|>N}f(x − t)g(x + t)h(x + 2t)\frac{dt}{t}

+ lim_{ε→0^{+}} ∫ _{N≥|t|>ε}f(x − t)\frac{g(x + t) − g(x)}{t}

·h(x + 2t)dt + lim_{ε→0^{+}} ∫ _{2N≥|y|>2ε}f(x − \frac{y}{2})g(x)h(x + y)\frac{dy}{y}

= ∫ _{|t|>N}f(x − t)g(x + t)h(x + 2t)\frac{dt}{t}

+ lim_{ε→0^{+}}∫ _{N≥|t|>ε}f(x − t)\frac{g(x + t) − g(x)}{t}h(x + 2t)dt

+ lim_{ε→0^{+}}_{∫ 2N≥|y|>2ε}f(x − \frac{y}{2})g(x)\frac{h(x + y) − h(x)}{y}dy

+ lim_{ε→0^{+}} ∫ _{2N≥|y|>2ε}f(x − \frac{y}{2})g(x)h(x)\frac{dy}{y}

= ∫ _{|t|>N}f(x − t)g(x + t)h(x + 2t)\frac{dt}{t}

+ lim_{ε→0^{+}} ∫ _{N≥|t|>ε}f(x − t)\frac{g(x + t) − g(x)}{t}h(x + 2t)dt

+ lim_{ε→0^{+}}_{∫ 2N≥|y|>2ε}f(x − \frac{y}{2})g(x)\frac{h(x + y) − h(x)}{y}dy

− lim_{ε→0^{+}}_{∫ N≥|z|>ε}f(x + z)g(x)h(x)\frac{dz}{z}

= ∫ _{|t|>N}f(x − t)g(x + t)h(x + 2t)\frac{dt}{t}

+ lim_{ε→0^{+}} ∫ _{N≥|t|>ε}f(x − t)\frac{g(x + t) − g(x)}{t}h(x + 2t)dt

+ lim_{ε→0^{+}}_{∫ 2N≥|y|>2ε}f(x − \frac{y}{2})g(x)\frac{h(x + y) − h(x)}{y}dy

− lim_{ε→0^{+}}_{∫ N≥|z|>ε} \frac{f(x + z) − f(x)}{z} · g(x)h(x)dz

− lim_{ε→0^{+}}_{∫ N≥|z|>ε}f(x)g(x)h(x)\frac{dz}{z}

= ∫ _{|t|>N}f(x − t)g(x + t)h(x + 2t)\frac{dt}{t}

+ lim_{ε→0^{+}} ∫ _{N≥|t|>ε}f(x − t)\frac{g(x + t) − g(x)}{t}h(x + 2t)dt

+ lim_{ε→0^{+}}_{∫ 2N≥|y|>2ε}f(x − \frac{y}{2})g(x)\frac{h(x + y) − h(x)}{y}dy

− lim_{ε→0^{+}}_{∫ N≥|z|>ε} \frac{f(x + z) − f(x)}{z}g(x)h(x)dz

= ∫ _{|t|>N}f(x − t)g(x + t)h(x + 2t)\frac{dt}{t}

+ lim_{ε→0^{+}} ∫ _{N≥|t|>ε}f(x − t)G(x, t)h(x + 2t)dt

+ lim_{ε→0^{+}}_{∫ 2N≥|y|>2ε}f(x − \frac{y}{2})g(x)H(x, y)dy

− lim_{ε→0^{+}}_{∫ N≥|z|>ε}F (x, z)g(x)h(x)dz

where

G(x, t) = \{ \begin{matrix}\frac{g(x+t)−g(x)}{t}, & t ≠ 0 \\ \frac{d}{dx}g(x), & t = 0, x ∈ R, \end{matrix}

H(x, y) = \{ \begin{matrix}\frac{h(x+y)−h(x)}{y}, & y ≠ 0 \\ \frac{d}{dx}h(x), & y = 0, x ∈ R \end{matrix} .

F (x, z) = \{ \begin{matrix}\frac{f(x+z)−f(x)}{z}, & z ≠ 0 \\ \frac{d}{dx}f(x), & z = 0, x ∈ R. \end{matrix}

It is obvious that F (x, z), G(x, t) and H(x, y), along with all of their partial derivatives are continuous functions of x, y, z, t, respectively.

Let κ_{N} be characteristic function of (−∞, −N]∪[N, ∞). Then, according to Muscalu’s proof in [7] we have

||κ_{N}(t)\frac{∂}{∂x}f(x − t)g(x + t)h(x + 2t)\frac{1}{t}||_{r} ≤ C||κ_{N} \frac{∂}{∂x}f(x − ·)||_{p}||g||_{q}||h||_{A},

||κ_{N}(t)f(x − t)\frac{∂}{∂x}g(x + t)h(x + 2t)\frac{1}{t}||_{r} ≤ C||κ_{N}f(x − ·)||_{p}||\frac{∂}{∂x}g||_{q}||h||_{A},

||κ_{N}(t)f(x − t)g(x + t)\frac{∂}{∂x}h(x + 2t)\frac{1}{t}||_{r} ≤ C||κ_{N}f(x − ·)||_{p}||g||_{q}||\frac{∂}{∂x}h||_{A}.

So we get

\frac{d}{dx}T (f, g, h)(x) = ∫ _{|t|>N} \frac{∂}{∂x}f(x − t)g(x + t)h(x + 2t)\frac{dt}{t}

+ ∫ _{|t|>N}f(x − t)\frac{∂}{∂x}g(x + t)h(x + 2t)\frac{dt}{t}

+ ∫ _{|t|>N}f(x − t)g(x + t)\frac{∂}{∂x}h(x + 2t)\frac{dt}{t}

+ lim_{ε→0^{+}} ∫ _{N≥|t|>ε} \frac{∂}{∂x}[f(x − t)G(x, t)h(x + 2t)]dt

+ lim_{ε→0^{+}} ∫ _{2N≥|y|>2ε} \frac{∂}{∂x}[f(x − y/2)g(x)H(x, y)]dy

− lim_{ε→0^{+}} ∫ _{N≥|z|>ε} \frac{∂}{∂x}[F (x, z)g(x)h(x)]dz

= ∫ _{|t|>N} \frac{∂}{∂x}f(x − t)g(x + t)h(x + 2t)\frac{dt}{t}

+∫ _{|t|>N}f(x − t)\frac{∂}{∂x}g(x + t)h(x + 2t)\frac{dt}{t}

+ ∫ _{|t|>N}f(x − t)g(x + t)\frac{∂}{∂x}h(x + 2t)\frac{dt}{t}

+ lim_{ε→0^{+}} ∫ _{N≥|t|>ε} \frac{∂}{∂x}[f(x − t)G(x, t)h(x + 2t)]dt

+ lim_{ε→0^{+}} ∫ _{N≥|t|>ε} \frac{∂}{∂x}f(x − t)G(x, t)h(x + 2t)dt

+ lim_{ε→0^{+}} ∫ _{N≥|t|>ε}f(x − t) \frac{\frac{∂}{∂x}g(x + t) − \frac{∂}{∂x}g(x)}{t}h(x + 2t)dt

+ lim_{ε→0^{+}} ∫ _{N≥|t|>ε}f(x − t)G(x, t)\frac{∂}{∂x}h(x + 2t)dt

+ lim_{ε→0^{+}}_{∫ 2N≥|y|>2ε} \frac{∂}{∂x}f(x − \frac{y}{2})g(x)H(x, y)dy

+ lim_{ε→0^{+}} ∫ _{2N≥|y|>2ε}f(x − \frac{y}{2})\frac{∂}{∂x}g(x)H(x, y)dy

+ lim_{ε→0^{+}}_{∫ 2N≥|y|>2ε}f(x − \frac{y}{2})g(x) \frac{\frac{∂}{∂x}h(x + y) − \frac{∂}{∂x}h(x)}{y}dy

− lim_{ε→0^{+}}_{∫ N≥|z|>ε} \frac{\frac{∂}{∂x}f(x + z) − \frac{∂}{∂x}f(x)}{z}g(x)h(x)dz

− lim_{ε→0^{+}}_{∫ N≥|z|>ε}F (x, z)\frac{∂}{∂x}g(x)h(x)dz

− lim_{ε→0^{+}}_{∫ N≥|z|>ε}F (x, z)g(x)\frac{∂}{∂x}h(x)dz

= ∫ _{|t|>N} \frac{∂}{∂x}f(x − t)g(x + t)h(x + 2t)\frac{dt}{t}

+∫ _{|t|>N}f(x − t)\frac{∂}{∂x}g(x + t)h(x + 2t)\frac{dt}{t}

+∫ _{|t|>N}f(x − t)g(x + t)\frac{∂}{∂x}h(x + 2t)\frac{dt}{t}

+ lim_{ε→0^{+}} ∫ _{N≥|t|>ε} \frac{∂}{∂x}f(x − t)\frac{g(x + t) − g(x)}{t}h(x + 2t)dt

+ lim_{ε→0^{+}}_{∫ N≥|t|>ε}f(x − t)\frac{∂}{∂x}g(x + t)h(x + 2t)\frac{dt}{t}

−\frac{∂}{∂x}g(x) lim_{ε→0^{+}}_{∫ N≥|t|>ε}f(x − t)h(x + 2t)\frac{dt}{t}

+ lim_{ε→0^{+}} ∫ _{N≥|t|>ε}f(x − t)\frac{g(x + t) − g(x)}{t} · \frac{∂}{∂x}h(x + 2t)dt

+ lim_{ε→0^{+}}_{∫ 2N≥|y|>2ε} \frac{∂}{∂x}f(x − \frac{y}{2})g(x)\frac{h(x + y) − h(x)}{y}dy

+\frac{∂}{∂x}g(x) lim_{ε→0^{+}}_{∫ 2N≥|y|>2ε}f(x − \frac{y}{2})\frac{h(x + y) − h(x)}{y}dy

+g(x) lim_{ε→0^{+}} ∫ _{2N≥|y|>2ε}f(x − \frac{y}{2})\frac{∂}{∂x}h(x + y)\frac{dy}{y}

−g(x)\frac{∂}{∂x}h(x) lim_{ε→0^{+}}_{∫ 2N≥|y|>2ε}f(x − \frac{y}{2})\frac{dy}{y}

−g(x)h(x) lim_{ε→0^{+}}_{∫ N≥|z|>ε} \frac{∂}{∂x}f(x + z)\frac{dz}{z}

+\frac{∂}{∂x}f(x)g(x)h(x) lim_{ε→0^{+}}_{∫ N≥|z|>ε} \frac{dz}{z}
−\frac{∂}{∂x}g(x)h(x) lim_{ε→0^{+}}_{∫ N≥|z|>ε} \frac{f(x + z) − f(x)}{z}dz
−g(x)\frac{∂}{∂x}h(x) lim_{ε→0^{+}}_{∫ N≥|z|>ε} \frac{f(x + z) − f(x)}{z}dz

= ∫ _{|t|>N} \frac{∂}{∂x}f(x − t)g(x + t)h(x + 2t)\frac{dt}{t}

+ ∫ _{|t|>N}f(x − t)\frac{∂}{∂x}g(x + t)h(x + 2t)\frac{dt}{t}

+ ∫ _{|t|>N}f(x − t)g(x + t)\frac{∂}{∂x}h(x + 2t)\frac{dt}{t}

+ lim_{ε→0^{+}} ∫ _{N≥|t|>ε} \frac{∂}{∂x}f(x − t)g(x + t)h(x + 2t)\frac{dt}{t}

+ lim_{ε→0^{+}}_{∫ N≥|t|>ε}f(x − t)\frac{∂}{∂x}g(x + t)h(x + 2t)\frac{dt}{t}

+ lim_{ε→0^{+}} ∈t_{N≥|t|>ε}f(x − t)g(x + t)\frac{∂}{∂x}h(x + 2t)\frac{dt}{t}

= lim_{ε→0^{+}}_{∫ |t|>ε} \frac{∂}{∂x}f(x − t)g(x + t)h(x + 2t)\frac{dt}{t}

+ lim_{ε→0^{+}}_{∫ |t|>ε}f(x − t)\frac{∂}{∂x}g(x + t)h(x + 2t)\frac{dt}{t}

+ lim_{ε→0^{+}}_{∫ |t|>ε}f(x − t)g(x + t)\frac{∂}{∂x}h(x + 2t)\frac{dt}{t}

= T (f′, g, h)(x) + T (f, g′, h)(x) + T (f, g, h′)(x)

Using a similar technique, we can show by induction that

[T (f, g, h)]^{(n)}(x) = ∑_{k=0}^{n} ∑_{m=0}^{k} \binom{n}{k}\binom{k}{m}T (f^{(n−k)}, g^{(k−m)}, h^{(m)})(x)

for all n ∈ N. Now we have

\begin{matrix}||[T (f, g, h)]^{(n)}(x)||_{r} ≤ ∑_{k=0}^{n} ∑_{m=0}^{k} \binom{n}{k}\binom{k}{m}||T (f^{(n−k)}, g^{(k−m)}, h^{(m)})(x)||_{r} \\ ≤ ∑_{k=0}^{n} ∑_{m=0}^{k} \binom{n}{k}\binom{k}{m}||f^{(n−k)}||_{p} · ||g^{(k−m)}||_{q} · ||h^{(m)}||_{A} \end{matrix}

Theorem 3.2Bilinear mappings

(f, g) ↦ T_{h}(f, g)fromD_{L^{p}} × D_{L^{q}} → D_{L^{r}},
(f, h) ↦ T_{g}(f, h)fromD_{L^{p}} × D_{A} → D_{L^{r}},
(g, h) ↦ T_{f}(g, h)fromD_{L^{q}} × D_{A} → D_{L^{r}},

are continuous.

P r o o f. According to Theorem 1 we have that these mappings are separately continuous. Then Corollary of Theorem 34.1 from [11] implies that these mappings are continuous. □

Corollary 3.3Mapping

T : D_{L^{p}} × D_{L^{q}} × D_{A} → D_{L^{r}},

is continuous.

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