Unboundedness of the trilinear Hilbert transform under the critical index

https://doi.org/10.4134/JKMS.j200656 pISSN: 0304-9914 / eISSN: 2234-3008

UNBOUNDEDNESS OF THE TRILINEAR HILBERT TRANSFORM UNDER THE CRITICAL INDEX

Yasuo Komori-Furuya

Abstract. Demeter [1] and Kuk and Lee [5] proved the unboundedness of the trilinear Hilbert transforms Ha,b,cunder the critical index 1/2 for some parameters a, b and c. We show the unboundedness of Ha,b,c for any parameters.

1. Introduction

We consider the bilinear and trilinear Hilbert transforms and fractional inte- gral operators. The study of bilinear and multilinear operators is not motivated by a mere quest to generalize the study of linear operators, but rather by their natural appearance in analysis. In a study of the Cauchy integral along a Lipschitz curve, Calder´on introduced the bilinear Hilbert transforms:

H_a,b(f, g)(x) := p.v.

Z

R¹

f (x + ay)g(x + by)

y dy,

and conjectured the following.

(1) kHa,b(f, g)kL^q ≤ Ckf kLp1kgkLp2,

where 1 < p₁, p₂ < ∞ and 1/q = 1/p₁+ 1/p₂. Note that q > 1/2. Lacey and Thiele [6, 7] proved that (1) holds where q > 2/3. When 1/2 < q ≤ 2/3 the problem is still open.

On the other hand, for the bilinear fractional integral operators:

I_a,b^α (f, g)(x) :=

Z

R¹

f (x + ay)g(x + by)

|y|^1−α dy,

Grafakos [2], Grafakos and Kalton [3] and Kenig and Stein [4] proved the following.

(2) kI_a,b^α (f, g)kL^q ≤ Ckf kLp1kgkLp2,

Received December 3, 2020; Accepted February 9, 2021.

2010 Mathematics Subject Classification. Primary 42B20.

Key words and phrases. Bilinear Hilbert transform, trilinear Hilbert transform, trilinear fractional integral operator.

c

2021 Korean Mathematical Society 1299

where 1 < p₁, p₂ < ∞, 0 < α < 1 and 1/q = 1/p₁+ 1/p₂− α > 0. Note that q > 1/(2 − α).

Formally if we take α = 0 in (2), we obtain (1). Therefore it is conjectured that the inequality (1) holds for q > 1/2.

We also consider the trilinear Hilbert transforms:

H_a,b,c(f, g, h)(x) := p.v.

Z

R¹

f (x + ay)g(x + by)h(x + cy)

y dy,

and the trilinear fractional integral operators:

I_a,b,c^α (f, g, h)(x) :=

Z

R¹

f (x + ay)g(x + by)h(x + cy)

|y|^1−α dy.

The following inequalities were conjectured.

(3) kHa,b,c(f, g, h)kL^q≤ Ckf kLp1kgkLp2khkLp3, where 1 < p2, p2, p3< ∞ and 1/q = 1/p1+ 1/p2+ 1/p3, (4) kI_a,b,c^α (f, g, h)kL^q ≤ Ckf kLp1kgkLp2khkLp3,

where 1 < p2, p2, p3< ∞, 0 < α < 1 and 1/q = 1/p1+ 1/p2+ 1/p3− α.

Note that q > 1/3 in (3) and q > 1/(3 − α) in (4) respectively.

Grafakos [2] proved that (4) holds for q ≥ 1. Kuk and Lee [5] proved that (4) holds when q > 1/(2 − α).

Grafakos and Kalton [3] and Kenig and Stein [4] proved the following. An- other fractional integral operator:

I˜_a,b,c^α (f, g, h)(x) :=

Z Z Z

R³

f (x + ay₁)g(x + by₂)h(x + cy₃)

(|y1| + |y2| + |y3 )^3−α dy1dy2dy3

satisfies

k ˜I_a,b,c^α (f, g, h)k_Lq ≤ Ckf k_Lp1kgk_Lp2khk_Lp3, where 1/q = 1/p1+ 1/p2+ 1/p3− α > 0 and q > 1/(3 − α).

However Demeter [1] and Kuk and Lee [5] showed the following counterex- amples.

Theorem A (Demeter [1]). There exists p > 1 such that sup

f,g,h∈L^p f,g,h6=0

kH1,2,3(f, g, h)k_Lp/3

kf kL^pkgkL^pkhkL^p

= ∞.

Theorem B (Kuk and Lee [5]). Let c be an irrational with (c + 1)(c − 1) > 0.

Then

sup

f,g,h∈L^p f,g,h6=0

kH_1,−1,c(f, g, h)k_Lp/3

kf kL^pkgkL^pkhkL^p

= ∞, where 1 < p < 3/2,

and

sup

f,g,h∈L^p f,g,h6=0

kI_1,−1,c^α (f, g, h)kL^q

kf kL^pkgkL^pkhkL^p

= ∞,

where 1/q = 3/p − α and 1/(3 − α) < q < 1/(2 − α).

Their proofs depend heavily on parameters a, b and c. Our result is the following.

Theorem 1. For any nonzero distinct real numbers a, b and c, there exists p = p(a, b, c) > 1 such that

sup

f,g,h∈L^p f,g,h6=0

kHa,b,c(f, g, h)k_Lp/3

kf kL^pkgkL^pkhkL^p

= ∞, (5)

sup

f,g,h∈L^p f,g,h6=0

kI_a,b,c^α (f, g, h)kL^q

kf kL^pkgkL^pkhkL^p

= ∞, (6)

where 1/q = 3/p − α.

Unfortunately we do not know whether (5) and (6) hold for any 1 < p < 3/2.

2. Proofs

We give a proof only for Ha,b,c, since the proof for I_a,b,c^α is the same; see the estimates of (10) and (11) in this section.

By a change of variables and change of functions f, g and h, we can reduce the problem to the following two cases.

Definition 1. We define

He_s,t(f, g, h)(x) := H_1,s,t(f, g, h)(x)

= p.v.

Z

R¹

f (x + y)g(x + sy)h(x + ty)

y dy.

Theorem 2. For any positive distinct real numbers s and t, there exists p = p(s, t) > 1 such that

sup

f,g,h∈L^p f,g,h6=0

k eHs,t(f, g, h)k_Lp/3

kf kL^pkgkL^pkhkL^p

= ∞ when s 6= 1 and t 6= 1,

and

sup

f,g,h∈L^p f,g,h6=0

k eHs,−t(f, g, h)k_Lp/3

kf kL^pkgkL^pkhkL^p

= ∞ when s 6= 1.

In fact we shall prove the following.

Theorem 3. For any positive distinct real numbers s and t, there exists p = p(s, t) > 1 such that

(I) sup

A,B,C

k eHs,t(χA, χB, χC)k_Lp/3

kχ_Ak_Lpkχ_Bk_Lpkχ_Ck_Lp

= ∞ when s 6= 1 and t 6= 1, and

(II) sup

A,B,C

k eHs,−t(χA, χB, χC)k_Lp/3

kχAkL^pkχBkL^pkχCkL^p

= ∞ when s 6= 1, where χE is the characteristic function of a set E.

We use the following notation.

Definition 2. For a set of real numbers X = {x₁, x₂, . . . , x_n}, we define ]X := n and dist(X) := inf

i6=j|x_i− x_j|.

For two sets X, Y and a real number a, we define

X + Y := {x + y; x ∈ X, y ∈ Y } and aX := {ax; x ∈ X}.

For a measurable set E, we denote the Lebesgue measure of E by m(E).

2.1. Proof of Theorem 3(I)

We may assume that 1 < s < t. Let u = (t − 1)/(t − s) > 0. Since su > 1, we consider the cases 1 < su < 3 and su ≥ 3:

(I-1) There exists an integer N ≥ 2 such that N + 2

N ≤ su < N + 1 N − 1. (I-2) There exists an integer N ≥ 3 such that N ≤ su < N + 1.

As we shall see later, the proof for (I-1) is essential.

Proof in the case of (I-1). Following an argument in [1], we define the next four sets.

Definition 3.

A0:= {0, −u, −2u}, B₀:=

 0, 1

N, 2

N, . . . ,N N,N + 1

N

 , C₀:= s − t

s − 1A₀+t − 1

s − 1B₀= t − s

s − 1 − A₀+ uB₀
, D₀:= s

s − 1A₀− 1

s − 1B₀= 1

s − 1 sA₀− B0
.

The next lemma is easy.

Lemma 1.

]A0= 3, ]B0= N + 2, ]C0= 3N + 2, ]D0= 3N + 6, dist(D0) = 1

N (s − 1) and D0⊂



− L

s − 1, 0



where L := 2su + ^{N +1}_N . Proof of Lemma 1. Since

−A0+ uB₀= {0, u/N, 2u/N, . . . , u, u + u/N }

∪ {u, u + u/N , . . . , 2u, 2u + u/N }

∪ {2u, 2u + u/N , . . . , 3u + u/N }, we have ](−A0+ uB0) = 3(N + 2) − 4 = 3N + 2. Since

−N + 1

N − (−su) ≥N + 2

N −N + 1

N = 1

N,

we have dist(D₀) = 1/N (s − 1). 

We construct three sets satisfying (I).

Definition 4. Let M = 3N + 10. For any K ∈ N, we define

AK :=







K

X

j=1

aj

M^j + z; aj ∈ A0, 0 < z < 1 M^K





 ,

BK :=







K

X

j=1

b_j

M^j + z; bj∈ B0, 0 < z < 1 M^K





 ,

C_K :=







K

X

j=1

c_j

M^j + z; c_j∈ C0, 0 < z < 1 M^K





 .

We have

(7) kχA_Kk^p_Lp≤ 3^K

M^K, kχB_Kk^p_Lp≤ (N + 2)^K

M^K and kχC_Kk^p_Lp≤ (3N + 2)^K M^K . We shall prove the following.

(8) lim

K→∞

k eHs,t(χAK, χBK, χCK)k_Lp/3

kχA_KkL^pkχB_KkL^pkχC_KkL^p

= ∞ for some p > 1.

For the proof we define the fourth set D_K as follows. Let eD₀:= sA₀− B0, and

De0,K:=





 1 s − 1

K

X

j=1

dj

M^j; dj∈ eD0, (d1, d2, . . . , dK) 6= (0, 0, . . . , 0)





 .

We have:

Lemma 2.

dist( eD0,K) ≥ L

(s − 1)(M − 1)· 1 M^K, where L is the number defined in Lemma 1.

Proof of Lemma 2. Assume that x, y ∈ eD0,K and x > y. Since L < M , we can write

x = 1 s − 1





j0−1

X

j=1

dj

M^j + dj₀

M^j0 +

K

X

j=j0+1

dj

M^j



,

y = 1 s − 1





j₀−1

X

j=1

dj

M^j + d⁰_j0 M^j0 +

K

X

j=j₀+1

d⁰_j M^j



 for some j0 where dj₀> d⁰_j

0 and dj, d⁰_j∈ eD0. By Lemma 1, we have x − y = 1

s − 1·d_j0− d⁰_j0 M^j0 + 1

s − 1

K

X

j=j0+1

d_j− d⁰_j M^j

≥ 1

N (s − 1)· 1

M^j0 − L s − 1

K

X

j=j₀+1

1 M^j

= 1

s − 1

 1

N − L

M − 1

 1

M^j0 + L

(s − 1)(M − 1)· 1 M^K

≥ L

(s − 1)(M − 1)· 1 M^K,

since M − 1 ≥ LN . 

Let

ε := min

 L

(s − 1)(M − 1), 1

 and

DK :=n

x + z; x ∈ eD0,K, 0 < z < ε 4M^K

o . Then we have:

Lemma 3.

DK ⊂ (−∞, 0) and m(DK) = ε((3N + 6)^K− 1)

4M^K .

Proof of Lemma 3. If x ∈ eD0,K, then x ≤ −1

N · 1

(s − 1)M^K.

Since L/(M − 1) ≤ 1/N , if 0 < z < ε/(4M^K), then we have x + z < 0. By Lemma 2 and the definition of ε, the set DK is a disjoint union of intervals, and ] eD0,K= (3N + 6)^K− 1. Therefore we can calculate m(DK). 

For any x ∈ D_K, there exist a_j ∈ A₀, b_j ∈ B₀ and 0 < z < ε/(4M^K) such that

x = 1 s − 1

K

X

j=1

saj− bj

M^j + z.

Let

x^∗:= 1 s − 1

K

X

j=1

−aj+ bj

M^j , and define the interval

I_x:=



x^∗, x^∗+ 1

4 max(s, t, 1)M^K

 . Then we have:

Lemma 4.

Ix⊂ [0, R], where R := _15(s−1)^2(u+1) +¹₄.

Proof of Lemma 4. Since aj≤ 0 and bj≥ 0, we have x^∗≥ 0. Furthermore x^∗≤ 2u +^{N +1}_N

s − 1

∞

X

j=1

1

M^j ≤ 2u + 2

(s − 1)(3N + 9) ≤ 2(u + 1) 15(s − 1).

 Remark. In fact max(s, t, 1) = t, but this representation is important in the proof of (II). The exact value of R is not important. Note that R is independent of K.

By the definitions we obtain the following.

Lemma 5. If x ∈ DK and y ∈ Ix, then

x + y ∈ A_K, x + sy ∈ B_K and x + ty ∈ C_K. Proof of Lemma 5. We write

x = 1 s − 1

K

X

j=1

saj− bj

M^j + z, where aj∈ A0, bj ∈ B0 and 0 < z < ε/(4M^K), and

y = 1 s − 1

K

X

j=1

−aj+ bj

M^j + z⁰, where 0 ≤ z⁰≤ 1/(4 max(s, t, 1)M^K). Then we have

x + ty = 1 s − 1

K

X

j=1

(s − t)a_j+ (t − 1)b_j

M^j + z + tz⁰, and 0 < z + tz⁰< M^−K. Therefore x + ty ∈ CK.

Proofs for the other cases are similar. 

Lemma 6.

(9) k eH_s,t(χ_AK, χ_BK, χ_CK)k_Lp/3 ≥ Cp,s,t

 (3N + 6)³ M^p+3

^K/p , where Cp,s,t is a constant depending only on p, s and t.

Proof of Lemma 6. If x ∈ DK and y < 0, then χB_K(x + sy) = 0. By Lemmas 4 and 5, we have if x ∈ DK,

He_s,t(χ_AK, χ_BK, χ_CK)(x) = Z ∞

0

χ_AK(x + y)χ_BK(x + sy)χ_CK(x + ty)dy y

≥ Z

I_x

χ_AK(x + y)χ_BK(x + sy)χ_CK(x + ty)dy y

≥ 1 Rm(I_x)

= 1

R· 1

4 max(s, t, 1)M^K. (10)

We obtain

k eHs,t(χA_K, χB_K, χC_K)k_Lp/3

≥ 1

R · 1

4 max(s, t, 1)M^K · m(DK)^3/p

= 1

4R max(s, t, 1)

ε 4

3/p 1 M^K

 (3N + 6)^K− 1 M^K

^3/p

≥ Cp,s,t

 (3N + 6)³ M^p+3

^K/p .

Note that ε depends on N , but C_p,s,t is independent of K.  By using (7) and (9), we can prove (8). It suffices to show

(3N + 6)³

M^p+3 >3(N + 2)(3N + 2)

M³ for some p > 1.

Since M = 3N + 10, it is equivalent to (3N + 6)³

(3N + 10)^p > 3(N + 2)(3N + 2), and we can take p such that

1 < p < 2 log(N + 2) − log(3N + 2) + log 9

log(3N + 10) .

 Remark. In the case of the fractional integral I_1,s,t^α , we need to show

(11)  3N + 6

M^q+1

^1/q

> 3(N + 2)(3N + 2) M³

^1/p . Proof in the case of (I-2). We define four sets.

Definition 5.

A0:= {0, −u, −2u}, B0:= {0, 1, 2, . . . , N − 1}, C₀:= t − s

s − 1 − A₀+ uB₀
, D0:= 1

s − 1 sA0− B0
.

The following argument is the same as the one in the case of (I-1). Therefore we show an outline of the proof. We have

]A₀= 3, ]B₀= N, ]C₀= N + 2, ]D₀= 3N, dist(D₀) = 1

s − 1 and D₀⊂



−2su + N − 1 s − 1 , 0

 .

The definitions of four sets AK, BK, CK and eD0,Kare the same as the ones in Definition 4 with M = 5N . We have

(12) kχAKk^p_Lp≤ 3^K

M^K, kχBKk^p_Lp≤ N^K

M^K and kχCKk^p_Lp≤ (N + 2)^K M^K , and

dist( eD0,K) ≥ 2su + N − 1 (s − 1)(M − 1)· 1

M^K. Let

ε := min

 2su + N − 1 (s − 1)(M − 1), 1



and R := 1 s − 1

 u 7 +1

5

 +1

4, and define

DK :=n

x + z; x ∈ eD0,K, 0 < z < ε 4M^K

o . Then

DK ⊂ (−∞, 0), m(DK) =ε((3N )^K− 1)

4M^K and Ix⊂ [0, R].

We obtain

(13) k eHs,t(χA_K, χB_K, χC_K)k_Lp/3 ≥ Cp,s,t

 (3N )³ M^p+3

^K/p . By using (12) and (13), we obtain the desired result when

1 < p < 2 log(3N ) − log(N + 2) log(5N ) .



2.2. Proof of Theorem 3(II)

We may assume that s > 1, but we cannot assume the large/small relation between s and t. Let u = (t + 1)/(s + t) > 0. Note that su > 1. In the same way as in (I), we consider the following two cases: (II-1) 1 < su < 3 and (II-2) su ≥ 3. Substituting −t for t in the proof of (I), we define four sets, respectively.

(II-1) A₀:= {0, −u, −2u}, B0:=

 0, 1

N, 2

N, . . . ,N

N,N + 1 N

 , C0:= s + t

s − 1A0+−t − 1 s − 1 B0

= −(s + t)

s − 1 − A0+ uB0
, D0:= s

s − 1A0− 1 s − 1B0

= 1

s − 1 sA0− B0
.

(II-2) A0:= {0, −u, −2u}, B₀:= {0, 1, 2, . . . , N − 1}, C0:= −(s + t)

s − 1 − A0+ uB0
, D0:= 1

s − 1 sA0− B0
.

The proof is exactly the same as the one in the case of (I). We use max(s, t, 1) in the proof, see Remark below Lemma 4. In Lemma 5, we have x − ty ∈ CK.

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Yasuo Komori-Furuya Department of Mathematics School of Science

Tokai University

Hiratsuka, Kanagawa 259-1299 Japan Email address: komori@tokai-u.jp