SHARP INEQUALITIES FOR LINEAR COMBINATIONS OF ORTHOGONAL MARTINGALES

Transcription

1 SHARP INEQUALITIES FOR LINEAR COMBINATIONS OF ORTHOGONAL MARTINGALES YONG DING, LOUKAS GRAFAKOS, AND KAI ZHU 1 Abstract. For any two real-valued continuous-path martingales X = {X t } t 0 Y = {Y t } t 0, with X Y being orthogonal Y being differentially subordinate to X, we obtain sharp L p inequalities for martingales of the form ax + by with a, b real numbers. The best L p constant is equal to the norm of the operator ai + bh from L p to L p, where H is the Hilbert transform on the circle or real line. The values of these norms were found by Hollenbeck, Kalton Verbitsky [13]. We also give applications of our martingale inequalities to Riesz transforms some discrete operators. 1. Introduction The research on martingale inequalities was initiated in 1966 by Burkholder [5] was further pursued in [6], [7] [8], where techniques for sharp estimates for them were developed. Martingale inequalities nowdays find applications in probability analysis their impact is quite far-reaching. Based on the Burkholder s method, Bañuelos Wang [4] obtained sharp inequalities for martingales under the assumption of differential subordination orthogonality, used them to provide probabilistic proofs to the results of Pichorides [18] concerning the norm of the Hilbert transform on L p (R of Iwaniec Martin [14] about the norm of the Riesz transforms on L p (R n, 1 < p <. We refer the reader to [1], [3] [17] for more on orthogonal martingales applications. We describe the pertinent framework for this paper. Let (Ω, F, P be a probability space F = (F t t 0 be a nondecreasing family of sub-σ-fields of F. Let X = (X t t 0 Y = (Y t t 0 be two real-valued martingales with respect to F. We say that X is orthogonal to Y if X, Y t = 0 for all t 0, where X, Y t is the quadratic covariation between X Y. We also say that Y is differentially subordinate to X (see [4] if Y 0 X 0 X t Y t is a nondecreasing nonnegative function of t for t 0, where X t is the quadratic variation of X. We assume throughout the paper that Y 0 = 0, this assumption is natural since all applications to Hilbert transform, Riesz transforms discrete Hilbert transform fit it (see [4, 2]. For continuous-path real-valued martingale X = (X t t 0, 1 < p <, define X p = sup X t p, t 0 Mathematics 2010 Subject Classification: Primary 60G44, 42A45. Secondary 60G46, 42A50, 42A99. Keywords phases: Sharp inequalities, Martingales, orthogonal, Hilbert transform, discrete. The first author is supported by NSFC( , , the Fundamental Research Funds for the Central Universities(2014KJJCA10. The second author would like to acknowledge the support of Simons Foundation of the University of Missouri Research Board Research Council. 1 Corresponding author. 1

2 2 YONG DING, LOUKAS GRAFAKOS, AND KAI ZHU 1 where X t p = (E X t p 1/p. For 1 < p <, define the constant n p = cot(π/(2p, where p = max(p, p/(p 1. In [4], Bañuelos Wang obtained the following result: Theorem A. ([4] Let X Y be two real-valued continuous-path martingales such that X Y are orthogonal Y is differentially subordinate to X. Then for 1 < p <, (1.1 Y p n p X p (1.2 (X 2 + Y 2 1/2 p E p X p, where E p = (1 + n 2 p 1/2. The constants are best possible. It is well known that the constant n p is exactly the operator norm of Hilbert transform H on L p (R of the conjugate function H T on L p (T, where T is the unit circle (see Pichorides [18]. The constant n p is also the operator norm of Riesz transform R j on L p (R n (see Iwaniec Martin [14]. So the results of Theorem A can be seen as the martingale analogues of the results of Pichorides [18], Iwaniec Martin [14], Essén [11]. For a, b R, 1 < p <, define the constant (1.3 B p = max x R ax b + (bx + a tan γ p + ax b (bx + a tan γ p x + tan γ p + x tan γ p, where γ = π 2p. By changing variables, B p can be equivalently defined as cos(θ + θ (1.4 B p = (a 2 + b 2 p/2 0 p + cos(θ + θ 0 + π p max p 0 θ 2π cos θ p + cos(θ + π, p p cos(ϑ θ (1.5 B p = (a 2 + b 2 p/2 0 p + cos(ϑ θ 0 + π p max p 0 ϑ 2π cos ϑ p + cos(ϑ + π, p p where tan θ 0 = b/a. These constants appeared in the work of Hollenbeck, Kalton Verbitsky [13] who showed that the norm of ai + bh T from L p (T to L p (T is equal to Bp 1/p, where I is the identity operator H T is the conjugate function operator on the circle. The same assertion is also true for the norm of ai + bh from L p (R to L p (R, where H is the Hilbert transform on real line, through a dilation argument known as blowing up the circle (see [20], Chapter XVI, Theorem 3.8. Recently, the authors [9] provided a direct proof of the sharp L p (R inequality for ai + bh by an argument that uses an explicit formula for a crucial subharmonic majorant. In this work, we prove sharp inequalities for the martingale ax + by, where X Y are as in Theorem A a, b are arbitrary real numbers. Motivated by the usefulness of the explicit formula of the crucial subharmonic majorant G in [9], we derive two alternative explicit expressions for this function centered around two points respectively (Lemma 2.2. These observations of G are important we use them appropriately in the proof of the main estimate (1.6 below.

3 SHARP INEQUALITIES FOR MARTINGALES 3 Theorem 1.1. Let X Y be two real-valued continuous-path martingales such that X Y are orthogonal. Let B p be given by (1.5. If Y is differentially subordinate to X Y 0 = 0, then for a, b R 1 < p < (1.6 ax + by p B 1/p p X p. The constant B 1/p p is the best possible in this inequality. Inequality (1.6 is the martingale analogue of that in Hollenbeck et al. [13] for analytic functions in the unit disc. We now turn to the proof of this theorem. Without loss of generality, we assume that a = cos θ 0, b = sin θ 0, so that a 2 + b 2 = Some Lemmas In this section we discuss some crucial lemmas in the proof of the main theorem. The first lemma is a version of Lemma 4.2 in [13], in which we derive an explicit formula for a subharmonic function G that plays a crucial role in the proof. Lemma 2.1. [9, Lemma 3.2] Let 1 < p <, B p be given by (1.5, T = {re it : r > 0, t 0 < t < t 0 + π}, where t p 0 is the value that makes right part of (1.5 attain its maximum, take ε > 0 such that t 0 ε < t 0 < t 0 + π/p < t 0 + π ε. Let z = re it, z 0 = re it 0, G(z = G(re it be π-periodic in t when t 0 ε < t < t 0 + π ε: B p Rez 0 p 1 sgn(rez 0 Re[( z z 0 p z 0 ] arez 0 + bimz 0 p 1 G(z = sgn(arez 0 + bimz 0 (are[( z z 0 p z 0 ] + bim[( z z 0 p z 0 ], if z T B p Rez p arez + bimz p, if z / T. Then G(z is subharmonic on C satisfies (2.1 a Rez + b Imz p B p Rez p G(z. for all z C. In the next lemma, we provide two other explicit formulas for G centered around the points t 0 u 0 = t 0 + π/p, respectively. Lemma 2.2. Let 1 < p <, B p, T t 0, ε be as in Lemma 2.1. Let z = re it, z 0 = re it 0. Then for z = re it T, G(z in Lemma 2.1 has the following equivalent expressions: (2.2 G(z = r p [ B p p (2.3 G(z = r p [ B p cos u 0 p cos u 0 cos(p(t t 0 +t 0 cos(t 0 θ 0 p cos(t 0 θ 0 cos(p(t u 0 +u 0 cos(u 0 θ 0 p cos(u 0 θ 0 ] cos(p(t t 0 +t 0 θ 0 ] cos(p(t u 0 +u 0 θ 0, where u 0 = t 0 + π/p, tan θ 0 = b/a, G(z is π-periodic in t t 0 ε < t < t 0 + π ε. Proof. Expression (2.2 is just the one given in Lemma 3.2 in [9]. We now prove (2.3. In the proof of Lemma 3.2 in [9], using the notation in that reference, we have (2.4 h(x = f( p t 0 cos(x p t 0 + f +( p t 0 sin(x p t 0,

4 4 YONG DING, LOUKAS GRAFAKOS, AND KAI ZHU 1 where p = p/2, t 0 = 2t 0 (2.5 f(t = Bp cos(t/p p a cos(t/p + b sin(t/p p, if we can prove (2.6 h(x = f( p t 0 + π cos(x p t 0 π + f +( p t 0 + π sin(x p t 0 π, then following the proof of Lemma 3.2 in [9], we deduce (2.3 when z T. To obtain (2.6, in view of (2.4, it is sufficient to show that (2.7 f( p t 0 + f( p t 0 + π = 0 (2.8 f + ( p t 0 + f +( p t 0 + π = 0. In fact, by f( p t 0 + π = f(pt 0 + π = B p cos(t 0 + π/p p a cos(t 0 + π/p + b sin(t 0 + π/p p, (2.9 B p = (a 2 + b 2 p/2 cos(t 0 θ 0 p + cos(t 0 θ 0 + π p p p + cos(t 0 + π p p, where tan θ 0 = b/a, we have B p cos(t 0 + π/p p a cos(t 0 + π/p + b sin(t 0 + π/p p = B p p + a + b sin t 0 p = f(pt 0 = f( p t 0, so we get (2.7. For (2.8, note that for 1 < p <, f(t is pπ-periodic continuously differentiable. By (1.5, g(t = f(t + f(t + π 0 g(t has a minimum at p t 0, so f +( p t 0 + f +( p t 0 + π = g ( p t 0 = 0, Thus the lemma is proved. The preceding lemma indicates that the function G has some symmetry properties in terms of t 0 u Proof of Theorem 1.1 The proof of Theorem 1.1 is based on the techniques of Burkholder; also see [4]. We choose the appropriate function for Theorem 1.1 to be the opposite of function G in Lemma 2.1 use the explicit formulas for G obtained by Lemma 2.2. For x, y R, 1 < p <, set V (x, y = ax + by p B p x p, where x = r cos t, y = r sin t, 0 < t < 2π. Define U(x, y = G(x + iy = G(z, where z = re it G(z is the function in Lemma 2.1. Then by Lemma 2.1, we have (3.1 V U.

5 SHARP INEQUALITIES FOR MARTINGALES 5 Denoting by U xx, U yy the second order partial derivatives of U(x, y, we need only to show that for all h, k R, (3.2 U xx (x, yh 2 + U yy (x, yk 2 c(x, y(h 2 k 2 for (x, y S i, where S i, i 1 is a sequence of open connected sets such that the union of the closure of S i is R 2, c(x, y 0 that is bounded on 1/δ r δ for any δ > 0. In fact, using Proposition 1.2 with Remark 1.1 in [4], by (3.2, we can get EV (X t, Y t EU(X t, Y t EU(X 0, Y 0 0, the last inequality is due to Lemma 4.3 in [13] Y 0 = 0. Thus E ax t + by t p B p E X t p, then we get (1.6. To show (3.2,we split the argument into two cases. First, for z = x + iy / T we have U(x, y = ax + by p B p x p, by a direct calculation we obtain from this that (3.3 U xx (x, y = p(p 1 ( ax + by p 2 a 2 B p x p 2 except on the lines {z : x = 0} {z : ax + by = 0}, (3.4 U yy (x, y = p(p 1 ax + by p 2 b 2 except on the line {z : ax + by = 0}. Then U (3.5 xx (x, yh 2 + U yy (x, yk 2 = p(p 1 ( ax + by p 2 B p x p 2 p(p 1 ax + by p 2 b 2 (h 2 k 2. By the property of G(z, we have (see [13] (3.6 ax + by p 2 B p x p 2 in this region. So by (3.5 (3.6 we get (3.7 U xx (x, yh 2 + U yy (x, yk 2 p(p 1 ax + by p 2 b 2 (h 2 k 2. Then (3.2 holds with obvious choice of c(x, y. We now consider the second case where z T. Recall that t 0 ε < t < t 0 + π ε. We use the expression (2.2 for G(z, then U(x, y = r p [ cos(t 0 θ 0 p cos(t 0 θ 0 Since we get U xx (x, y = p(p 1r p 2 cos(p(t t 0 + t 0 θ 0 B p p r x = cos t, r y = sin t, t x = 1 r sin t, t y = 1 cos t, r h 2 cos(p(t t 0 + t 0 ]. ( cos(t0 θ 0 p cos(2t p(t t 0 (t 0 θ 0 cos(t 0 θ 0 p B p cos(2t p(t t 0 t 0,

6 6 YONG DING, LOUKAS GRAFAKOS, AND KAI ZHU 1 where x = r cos t, y = r sin t, tan θ 0 = b/a, Then U yy (x, y = U xx (x, y. (3.8 U xx (x, yh 2 + U yy (x, yk 2 = U xx (x, y(h 2 k 2. We claim that (3.9 U xx (x, y 0 for z T, where z = x + iy. In fact, U xx (re it 0 = p(p 1r p 2 ( cos(t0 θ 0 p cos(t 0 θ 0 we know from [13, p.249] that cos(t 0 + θ 0 B p p (3.10 a + b sin t 0 p 2 B p p 2, which, since a = cos θ 0, b = sin θ 0, is equivalent to (3.11 cos(t 0 θ 0 p 2 B p p 2. Combining (3.11 with the fact that (3.12 cos(t 0 θ 0 cos(t 0 + θ 0 cos 2 t 0, we have (3.13 U xx (re it 0 0. Now we use the expression (2.3 for G(z to get [ ] U(x, y = r p cos(u0 θ 0 p cos u 0 p cos(p(t u 0 + u 0 θ 0 B p cos(p(t u 0 + u 0, cos(u 0 θ 0 cos u 0 where u 0 = t 0 + π/p, then U xx (x, y = p(p 1r p 2 where x = r cos t, y = r sin t, tan θ 0 = b/a, so U xx (re iu 0 = p(p 1r p 2 ( cos(u0 θ 0 p cos(u 0 θ 0 where u 0 = t 0 + π/p. We know from [13, p.249] that ( cos(u0 θ 0 p cos(2t p(t u 0 (u 0 θ 0 cos(u 0 θ 0 cos u 0 p B p cos(2t p(t u 0 u 0, cos u 0 cos(u 0 + θ 0 B p cos u 0 p (3.14 a cos u 0 + b sin u 0 p 2 B p cos u 0 p 2, which is equivalent to (3.15 cos(u 0 θ 0 p 2 B p cos u 0 p 2. Combining (3.15 with (3.16 cos(u 0 θ 0 cos(u 0 + θ 0 cos 2 u 0, we have (3.17 U xx (re i(t 0+π/p 0. cos u 0 cos u 0,,

7 Write U xx (re it = p(p 1r p 2 u(t, where A = cos(t 0 θ 0 p cos(t 0 θ 0 SHARP INEQUALITIES FOR MARTINGALES 7 u(t = A cos p 2 t + sgn(2 pb sin p 2 t, cos ( (p 1t 0 + θ 0 Bp p cos(p 1t 0, p B = B p sin(p 1t 0 cos(t 0 θ 0 p sin ( (p 1t 0 + θ 0. cos(t 0 θ 0 Then u(t is a p 2 -trigonometric function, thus also p 2 -trigonometrically convex for t 0 < t < t 0 + π/p (see [16, p.54]. We have U xx (re it = p(p 1r p 2 u(t is harmonic thus subharmonic within the angle {z = re it : r > 0, t 0 < t < t 0 + π/p} via a direct computation. Then, by (3.13, (3.17 the Phragmén-Lindelöf theorem for subharmonic functions (see [16, p.49], we have (3.18 U xx (x, y = U xx (re it 0 for z T. We can also use the maximum principle for harmonic functions directly to deduce (3.18. This proves (3.9. Then (3.2 holds with c(x, y = U xx (x, y. This completes the proof of (1.6. A few comments are in order: Remarks. (a The case a = 0, b = 1 of Theorem 1.1 is contained in [4]. (b When a = 0, b = 1, p > 2, the function U = G becomes the function U 2 (x, y in [4]. (c When a = 0, b = 1, 1 < p < 2, the function U = G is used in [18] in [12]. 4. The sharpness of the constant B 1/p p To show that the constant B p is sharp, we apply a similar argument as in [4]. Let f(z = u(z + iv(z be analytic in the unit disc D with f(0 = 0 B t be Brownian motion in D killed upon leaving D. Consider the martingales X t = u(b t Y t = v(b t, we have X, Y t = 0 X t Y t = 0 (see [10]. So X Y are orthogonal with equal quadratic variations. Then the inequality in Theorem 1.1 exactly reduces to the inequality in Theorem 4.1 in [13]. Since Bp 1/p is already the best constant in Theorem 4.1 of [13], we conclude that the constant Bp 1/p cannot be improved in Theorem Examples applications In this section, we give some applications of Theorem 1.1 to Riesz transforms operators related to discrete versions of the Hilbert transform. The Riesz transforms R j, j = 1, 2,..., n are defined by (see [19, p.57] R j f(x = Γ((n + 1/2 p.v. π (n+1/2 R n x j y j f(ydy x y n+1 for f L p (R n. It has been shown that the L p norm for R j is the same as the one for 1-dimensional Hilbert transform, n p, by using the method of rotations applicable to singular integrals with odd kernels (see [14]. But in the linear combination case ai + br j, a, b R, the method of rotations could not be used to deduce its L p norm.

8 8 YONG DING, LOUKAS GRAFAKOS, AND KAI ZHU 1 Our martingale inequalities provide a way to solve this problem. We have the following corollary of Theorem 1.1. Corollary 5.1. For any j = 1, 2,..., n, a, b R 1 < p <, (5.1 (ai + br j f p B 1/p p f p, where B p is given by (1.5. The constant B 1/p p is the best possible in this inequality. We give a sketch proof of Corollary 5.1. Using the same connection between martingale transforms Riesz transforms in [4, p.592] involving linear combinations with the identity operator, Theorem 1.1, we can get (5.1. The constant Bp 1/p is the best possible follows from the fact that ai + br j are extensions, in the Fourier multiplier sense, of ai + bh, where H is the Hilbert transform; see the proof of (47 in [14, p.37] for the full details. Consider the following discrete version of the Hilbert transform (5.2 (Da n := 1 a n k π k, where k runs over all the non-zero integers in Z a = (a n n. Recently, Bañuelos Kwaśnicki [2] proved that the operator norm of D on l p (Z is equal to the operator norm of the continuous Hilbert transform H on L p (R. The proof in [2] is based on Theorem A uses two auxiliary operators J [defined in (5.3] K which satisfies KJ = D. We find out some interesting relationship between K the identity operator I (see (5.13 apply Theorem 1.1 to obtain the following results concerning J K. Proposition 5.1. Let e = (e n n be a sequence in l p (Z, 1 < p <. Let (J e n = J m e n m, where m Z (5.3 J n = 1 ( 2y πn 0 (y 2 + π 2 n 2 sinh 2 y dy for n 0, J 0 = 0. Then for a, b R, (5.4 (ai + bj e p Bp 1/p e p, where B p is given by (1.5 I is the identity operator: the convolution with kernel I 0 = 1, I n = 0 for n 0. The constant Bp 1/p is the best possible in this inequality. Proof. We use the notation in [2]. We only need to redefine the operator in (2.5 in [2], so that let (5.5 (J A e n = E (x0,y 0 ( a A Mζ + ba M ζ Z ζ = (2πn, 0. Since the conditional expectation is a contraction on L p, 1 < p <, it follows from (1.6 in Theorem 1.1 that k 0 (5.6 J A e p Bp 1/p A e p. Let we have H = [ (5.7 J H e p B 1/p p e p. ],

9 SHARP INEQUALITIES FOR MARTINGALES 9 Notice that E (x0,y 0 ( Mζ Z ζ = (2πn, 0 = (Ie n, where I is the identity operator, then following the same proof in [2], we deduce (5.4. The sharpness of the constant is due to the sharpness of Proposition 5.2 the fact that (5.8 (ak + bde p (ai + bj e p for any sequence a l p (Z, 1 < p < a, b R. Proposition 5.2. Let e = (e n n be a sequence in l p (Z, 1 < p <, D be defined in (5.2. Let K be the convolution operator in Section 2.3 in [2] with kernel (K n such that K n 0 for all n the sum of all K n is equal to 1. Then for a, b R, (5.9 (ak + bde p B 1/p p e p, where B p is given by (1.5. The constant B 1/p p Proof. By Section 2.3 in [2], then by Proposition 5.1, we have (De n = (KJ e n, is the best possible in this inequality. (5.10 (ak + bde p = K(aI + bj e p B 1/p p e p. Denote by p,p the operator norm from L p to L p (or l p to l p. To deduce the sharpness, we define the dilation operators T ε for any ε > 0 1 < p < by (T ε f(x = ε 1/p f(εx, then T ε p,p = 1 for all ε > 0. Notice that K is a convolution operator with kernel (K n such that K n 0 for all n n Z K n = 1 (see [2]. Because of Theorem 4.2 in [15], now we can work on the real line replace D K by (5.11 (M D f(x = p.v. 1 f(x m π m m 0 (5.12 (M K f(x = m Z K m f(x m, respectively. It is known by [15] that lim (T 1/εM D T ε f(x = (Hf(x, where H is the Hilbert transform. We claim that (5.13 lim (T 1/ε M K T ε f(x = (If(x, for a.e. x R f L p (R, where I is the identity operator such that (If(x = f(x. In fact, for any f S(R (Schwartz function, we have lim (T 1/εM K T ε f(x = lim m N K mf(x εm + lim m >N K mf(x εm = m N K mf(x + lim m >N K mf(x εm

10 10 YONG DING, LOUKAS GRAFAKOS, AND KAI ZHU 1 for any N > 0. Then f(x lim (T 1/ε M K T ε f(x = m Z K mf(x m N K mf(x lim m >N K mf(x εm = m >N K mf(x lim m >N K mf(x εm lim f(x f(x εm m >N K m C(f m >N K m for any N > 0. Since K n 0 for all n n Z K n = 1, letting N, we get lim (T 1/εM K T ε f(x = f(x for x R, f S(R. For f L p (R, we can get (5.13 for a.e. x R by density. Then, by Fatou s lemma T ε p,p = 1, ai + bh p,p sup T 1/ε (am K + bm D T ε p,p am K + bm D p,p. ε By Theorem 4.2 in [15], am K + bm D p,p = ak + bd p,p, so, using (5.10 the fact that ai + bh p,p = B 1/p p (see [13], we get ak + bd p,p = ai + bh p,p = B 1/p p. For the operator ai + bd, a, b R, applying the method in [15, Lemma 4.3], we immediately obtain ai + bd p,p ai + bh p,p = B 1/p p. We conjecture that ai +bd p,p = ak+bd p,p = Bp 1/p. The solution of this conjecture may require additional ideas as I D are natural projections of nonorthogonal martingales. Another discrete Hilbert transform is defined by (5.14 (D 1/2 a n = p.v. 1 π k Z a n k k + 1/2. More generally, consider the operator defined for α (0, 1 by sin πα a n k (5.15 (D α a n = p.v. π k + α. k Z A natural conjecture is that for 1 < p < α (0, 1, the norms D α p,p coincide with the norms cos παi + sin παh p,p = Bp 1/p ; see Conjecture 5.7 in [15]. It seems that our martingale inequalities have some relationship with this conjecture.

11 SHARP INEQUALITIES FOR MARTINGALES 11 References [1] R. Bañuelos, The foundational inequalities of D. L. Burkholder some of their ramifications, Illinois J. Math. 54 (2010, [2] R. Bañuelos, M. Kwaśnicki, On the l p -norm of the discrete Hilbert transform ( [3] R. Bañuelos, A. Osȩkowski, Sharp martingale inequalities applications to Riesz transforms on manifolds, Lie group Gauss space, J. Funct. Anal. 269 (2015, [4] R. Bañuelos, G. Wang, Sharp inequalities for martingales with applications to the Beurling- Ahlfors Riesz transforms, Duke Math. J. 80 (1995, [5] D. L. Burkholder, Martingale transforms, Ann. Math. Statist. 37 (1966, [6] D. L. Burkholder, Boundary value problems sharp inequalities for martingale transforms, Ann. Probab. 12 (1984, [7] D. L. Burkholder, Sharp inequalities for martingales stochastic integrals, Astérisque 157 (1988, [8] D. L. Burkholder, Explorations in martingale theory its applications, Lecture Notes in Mathematics 1464 (1991, [9] Y. Ding, L. Grafakos, K. Zhu, On the norm of the operator ai + bh on L p (R, Bull. Korean Math. Soc. 55 (2018, [10] R. Durrett, Brownian Motion Martingales in Analysis, Wadsworth, Belmont, California, (1984. [11] M. Essén, A superharmonic proof of the M. Riesz conjugate function theorem, Ark. Mat. 22 (1984, [12] L. Grafakos, Best bounds for the Hilbert transform on L p (R 1, Math. Res. Lett. 4 (1997, [13] B. Hollenbeck, N. J. Kalton, I. E. Verbitsky, Best constants for some operators associated with the Fourier Hilbert transforms, Studia Math. 157 (2003, [14] T. Iwaniec, G. Martin, Riesz transforms related singular integrals, J. Reine Angew. Math. 473 (1996, [15] E. Laeng, Remarks on the Hilbert transform some families of multiplier operators related to it, Collect. Math. 58 (2007, [16] B. Ya. Levin, Lectures on entire functions, Transl. Math. Monogr. 150, Amer. Math. Soc., Providence, RI, [17] A. Osȩkowski, Sharp martingale semimartingale inequalities, Monografie matematyczne 72, Birkhäuser, Springer, Basel, [18] S. K. Pichorides, On the best values of the constants in the theorems of M. Riesz, Zygmund Kolmogorov, Studia Math. 44 (1972, [19] E. M. Stein, Singular integrals differentiability properties of functons, Princeton University Press, Princeton, [20] A. Zygmund, Trigonometric Series, Volumme II, Cambridge UK, Yong Ding, Laboratory of Mathematics Complex Systems, School of Mathematical Sciences, Beijing Normal University, Ministry of Education of China, Beijing , China address: dingy@bnu.edu.cn Loukas Grafakos, Department of Mathematics, University of Missouri, Columbia MO 65211, USA address: grafakosl@missouri.edu Kai Zhu, School of Mathematical Sciences, University of Chinese Academy of Sciences, Beijing , China address: zhukai@ucas.ac.cn