Sharp weak type estimates for Riesz transforms

Let d\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$d$$\end{document} be a given positive integer and let {Rj}j=1d\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\{R_j\}_{j=1}^d$$\end{document} denote the collection of Riesz transforms on Rd\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathbb {R}^d$$\end{document}. For 1<p<∞\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$1<p<\infty $$\end{document}, we determine the best constant Cp\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$C_p$$\end{document} such that the following holds. For any locally integrable function f\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$f$$\end{document} on Rd\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathbb {R}^d$$\end{document} and any j∈{1,2,…,d}\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$j\in \{1,\,2,\,\ldots ,\,d\}$$\end{document}, ||(Rjf)+||Lp,∞(Rd)≤Cp||f||Lp,∞(Rd).\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} ||(R_jf)_+||_{L^{p,\infty }(\mathbb {R}^d)}\le C_p||f||_{L^{p,\infty }(\mathbb {R}^d)}. \end{aligned}$$\end{document}A related statement for Riesz transforms on spheres is also established. The proofs exploit Gundy–Varopoulos representation of Riesz transforms and appropriate inequality for orthogonal martingales.

motivations for this direction of research comes from the fact that good estimates for the L p norm of the Riesz transforms on R d and the Beurling-Ahlfors operator on C have important consequences in the study of quasiconformal mappings and related nonlinear geometric PDEs (cf. [8,12]). The purpose of this paper is to continue this line of research and investigate the action of Riesz transforms on weak spaces L p,∞ .
Let us start with recalling some related results from the literature. The first paper we mention is that of Pichorides [19], who identified the norm of the Hilbert transform as an operator on L p (R), 1 < p < ∞. Recall that the Hilbert transform H on the line is the operator defined by the principal value integral Pichorides' result asserts that where p * = max{ p, p/( p − 1)}. This statement has been extended to the higherdimensional setting by Iwaniec and Martin [13] and, independently, by Bañuelos and Wang [3]. Suppose that d ≥ 2 is a given integer. The counterpart of the Hilbert transform in R d is the collection of Riesz transforms (R j ) d j=1 (see e.g., Stein [21]). This family of operators is given by R d x j − y j |x − y| d+1 f (y)dy, j = 1, 2, . . . , d, (1.2) where the integrals, as in (1.1), are supposed to exist in the sense of Cauchy principal values. The aforementioned result of Iwaniec and Martin [13] and Bañuelos and Wang [3] is the identification of L p norms of Riesz transforms: Thus, the norms do not change when the dimension d increases. This result has been extended in numerous directions; the literature on this subject is very large, we only mention here the sharp weak-type bounds of Davis [6] and Janakiraman [14], and optimal logarithmic estimates due to the author [17].
In the present paper we will be interested in the action of Riesz transforms on the spaces L p,∞ , 1 < p < ∞, equipped with the norm where the supremum is taken over all measurable subsets A of R d satisfying 0 < |A| < ∞. We should mention here that || · || p,∞ is equivalent to the more common norm given by However, we do prefer to work with || · || p,∞ , since it is more convenient for our purposes. To formulate our results, we need some notation. Throughout the paper, for any fixed λ ≥ 0, the function λ : [0, ∞) → [0, ∞) is given by λ (t) = (t −λ) + . For any measurable function f : R d → R, the symbol f * stands for the non-increasing rearrangement of f , defined by Furthermore, f * * : (0, ∞) → R will denote the integral mean of f * , given by In other words, f * * is the Hardy-Littlewood maximal function of f * : It is straightforward to see that the L p,∞ -norm of f is related to f * * by the formula We are ready to state the first result of this paper. and (1.5) The inequalities (1.3) and (1.4) are sharp.
Next, we turn to the case when the underlying measure space is the unit sphere There are many ways of defining Riesz transforms in this setting: see Arcozzi and Li [2] for an overview of the possibilities. We will work with two types R c and R b , the so-called cylindrical Riesz transforms and Riesz transforms of ball type. For the necessary definitions, we refer the reader to Sect. 4 below, and only mention here that for d = 2, both operators reduce to the Hilbert transform on the unit circle S 1 = [0, 2π), given by the singular integral For f : S d−1 → R, its non-increasing rearrangement f * , integral mean f * * and weak norm || f || L p,∞ (S d−1 ) are defined by analogous formulas as in R d . We will establish the following statement for the directional Riesz transforms R c m , R b m with respect to the (x , x m )-plane.
A few words about the proof and the organization of the paper. We will use a probabilistic approach to Riesz transforms, which has its roots at the works of Varopoulos and Gundy. To be more precise, we will first establish an appropriate martingale version of the inequality (1.3): this is done in Sect. 2. Then, in Sect. 3, we deduce (1.3), (1.4) and (1.5), using the stochastic representation of Riesz transforms [10]. In Sect. 4, we use a similar approach to deduce (1.6) and (1.7). The final part of the paper is devoted to the optimality of the constants appearing in Theorems 1.1 and 1.2.

A martingale inequality
The results of this paper depend heavily on an appropriate martingale inequalities. Let us start with introducing the necessary probabilistic background and notation. Assume that ( , F, P) is a complete probability space, equipped with (F t ) t≥0 , a nondecreasing family of sub-σ -fields of F, such that F 0 contains all the events of probability 0. Let X, Y be two adapted real-valued martingales with right-continuous trajectories that have limits from the left. The symbol [X, Y ] will stand for the quadratic covariance process of X and Y , see e.g., Dellacherie and Meyer [7] for details. The martingales X, Y are said to be orthogonal if the process [X, Y ] is constant with probability 1. Following Bañuelos and Wang [3] and Wang [23], we say that Y is differentially subordinate to X , if the process t≥0 is nonnegative and nondecreasing as a function of t.
The differential subordination implies many interesting inequalities comparing the sizes of X and Y . The literature on this subject is quite extensive, we refer the interested reader to the survey [5] by Burkholder, the paper of Wang [23] and the monograph [18]. Here we only mention one result, due to Bañuelos and Wang [23], which will be needed in our further considerations. We use the notation ||X || p = sup t≥0 ||X t || p for 1 ≤ p ≤ ∞. For λ ≥ 0, we define the convex function λ : [0, ∞) → R by (t) = (t − λ) + . The main result of this section is the following.

Theorem 2.2
Suppose that X, Y are orthogonal martingales such that X takes values in [0, 1], Y is differentially subordinate to X and Y 0 ≡ 0. Then for any λ ≥ 0 we have The inequality is sharp: there is a nontrivial pair X, Y for which both sides are equal.
Here Y ∞ stands for the pointwise limits of Y as t → ∞; the existence of this limit follows immediately from the boundedness of X and Y in L 2 (see Theorem 2.1 above).
The proof of Theorem 2.2 will be based on the existence of certain special harmonic functions. Let H = R × (0, ∞) denote the upper half-space and let S = [0, 1] × R stand for the vertical strip in R 2 . For a given λ ≥ 0, define the auxiliary function U λ : H → R by the Poisson integral Obviously, U λ is harmonic on H and satisfies Consider a conformal mapping ϕ(z) = −e −iπ z or, in real coordinates, ϕ(x, y) = −e π y cos(π x), e π y sin(π x) .

This function maps
The function U λ is harmonic on (0, 1)×R and, by (2.2), can be extended to a continuous function on the whole S by setting U λ (0, y) = U λ (1, y) = λ (|y|). Some further properties of U λ are investigated in the lemma below. ds.

(ii) The partial derivative U λx x is nonpositive in the interior of S. (iii) We have the majorization
Proof (i) It suffices to substitute t = se π y in the integral defining U λ .
(ii) By the harmonicity of U λ inside the strip S, it suffices to show that U λyy ≥ 0. But this is clear, in view of (2.4): for any s ∈ R, the function y → λ 1 π log |s| + y is convex.
(iii) By (i), all we need is to show the majorization for x ∈ {0, 1}. However, when x = 0 or x = 1, then both sides of (2.5) are equal.
In the proof of Theorem 2.2 we shall require the following auxiliary fact, which appears (in a slightly different form) as Corollary 1 in Bañuelos and Wang [4].

Lemma 2.4
Suppose that X, Y are real-valued orthogonal martingales such that Y is differentially subordinate to X . Then Y has continuous paths and is orthogonal and differentially subordinate to X c , the continuous part of X .
We are ready to prove the martingale inequality.
Proof of Theorem 2.2 Fix t ∈ (0, ∞) and introduce the process Z = (X, Y ). Since U λ is of class C ∞ in the interior of S, we may apply Itô's formula to obtain where Here X s denotes the jump of X at time s. Note that we have used above the equalities Y s− = Y s and Y = Y c , which are due to the continuity of paths of Y . Let us analyze the above terms separately. First, note that The term I 1 has zero expectation, since both stochastic integrals are martingales. Next, we have I 2 = 0, because of the orthogonality of X c and Y . The differential subordination together with Lemma 2.3 (ii) give Finally, each summand in I 4 is nonpositive, by the concavity of U λ (·, y) for any fixed y ∈ R [again, apply Lemma 2.3 (ii)]. Therefore, using the majorization of that lemma, we obtain the estimate and (

Inequalities for Riesz transforms in R d
There is a well-known representation of Riesz transforms in terms of the so-called background radiation process, introduced by Gundy and Varopoulos in [10]. Let us briefly describe this connection. Throughout this section, d is a fixed positive integer.
Suppose that X is a Brownian motion in R d and let Y be an independent Brownian motion in R (both processes start from the appropriate origins). For any y > 0, introduce the stopping time or, which is the same, V f is the convolution of f with the Poisson kernel where g runs over C ∞ 0 (R d ). Less formally, T y f is given as the following conditional expectation with respect to the measureP = P ⊗ dx (dx denotes Lebesgue's measure on R d ): for any z ∈ R d , See Gundy and Varopoulos [10] for the rigorous statement of this equality. The interplay between the operators T y A and Riesz transforms is explained in the following theorem, consult [10] or Gundy and Silverstein [9].
Then T y A j f → R j f almost everywhere as y → ∞. We shall require the following auxiliary fact, see Lemma 3.2 in [17].
We are ready to establish the inequalities of Theorem 1.1.
Proof of (1.3) Pick j ∈ {1, 2, . . . , d} and suppose that A = A j is the matrix of Theorem 3.1. Fix λ > 0 and an integrable function f : R d → [0, 1]. For any x ∈ R d and y > 0, denote Now, fix x, y and consider the pair ξ = ξ x,y = (ξ t ) t≥0 , ζ = ζ x,y = (ζ t ) t≥0 of martingales given by for t ≥ 0. Then the martingale ζ is differentially subordinate to ξ , since ds is nonnegative and nondecreasing as a function of t. Furthermore, ξ and ζ are orthogonal, which is a direct consequence of the equality Ax, Therefore, by Theorem 2.1, and integrating both sides with respect to x ∈ R d gives by virtue of Fubini's theorem. Similarly, for any h ∈ L q (R d ) (q = p/( p − 1) is the harmonic conjugate to p), Combining these estimates with (3.1) and Hölder's inequality yields This implies T y A f ∈ L p (R d ) and hence g ∈ L p (R d ), by Chebyshev's inequality [recall that g is given by (3.2)]. Thus, by Lemma 3.2, we may write where in the last line we have used the fact that |g| takes values in {0, 1}. Now, by (2.1), we get that The function U λ is nonnegative, so by Fubini's theorem, we may write π ||ξ x,y || 1 (s + 1) 2 s 2 + 2s cos(π ||ξ x,y || 1 ) + 1 · ||ξ x,y || 1 dx ds.
Since f is nonnegative, bounded by 1 and integrable, we have Indeed, with no loss of generality we may assume that f is compactly supported. Then it suffices to use of the following straightforward property of the Poisson kernel: for any positive numbers D and ε, there is a level y 0 such that if y ≥ y 0 , x ∈ R and E is a set of measure D, then E P y (x)dx < ε.
Next, an application of Fubini's theorem gives Thus, by Lebesgue's dominated convergence theorem, Fatou's lemma and Theorem 3.1, we get the bound However, we easily compute that Pick an arbitrary measurable subset E of R d with 0 < |E| < ∞ and decompose it into the union of and, obviously, Adding both above statements, we get an inequality which is equivalent to Now optimize the right-hand side over λ. A direct computation of the derivative with respect to this parameter shows that the minimal value is attained for λ = sinh −1 (2|| f || 1 /|E|) /π . For this choice of λ, we obtain the estimate This is precisely the desired upper bound for R j f * * .
Here in the last passage we have used the Hardy-Littlewood inequality f g ≤ f * g * , which is actually valid on general measure spaces (see [11]). Now, fix 0 < m < M and use integration by parts, then (1.3) and then integration by parts again, to obtain Here in view of (1.3), and which is positive by (1.3). However, by de l'Hospital rule and the assumption || f || L p,∞ (R d ) ≤ 1, we get lim m→∞ I I = 0. Thus, letting m → 0 and M → ∞, we obtain Therefore, integrating by parts and arguing as above, we get However, r 1/ p f * * (r ) ≤ 1, so the substitution r := 2r |E| in the integral above yields since E was arbitrary. The claim is proved.

Inequalities for Riesz transforms on spheres
Now we will analyze the weak-type estimates for Riesz transforms on the unit sphere S d−1 = {x ∈ R d : |x| = 1} equipped with the standard Riemannian metric and normalized SO(d) invariant measure. We will work with two non-equivalent notions of Riesz transforms on the spheres. These two possibilities arise from the fact that there are two natural ways to "fill in" S d−1 so that it is the boundary of an d-dimensional Riemannian manifold (cf. [2]). Let us analyze these two cases separately. Firstly, one can express S d−1 as the boundary of the cylinder S d−1 × [0, ∞), and this leads to the cylindrical Riesz transform R c . For fixed 1 ≤ < m ≤ d, consider the differential operator T m = x ∂ m − x m ∂ . If x + i x m = re iθ , then T m = ∂/∂θ is the derivative with respect to the angular coordinate in the (x , x m ) plane and hence is a well defined vector field on S d−1 . We define the directional Riesz transform (of cylinder type) by where S d−1 denotes the Laplace-Beltrami operator on S d−1 . See Stein [20] for the detailed exposition of the subject. We turn to the second type of Riesz transform on S d−1 (cf. Korányi and Vági [15,16]). Denote by H k the space of spherical harmonics of degree k (see Stein [21]) and let where ν is the outward pointing normal vector to S d−1 . One easily extends (∂/∂ν) −1 to L 2 0 (S d−1 ) by the following formula: if f = k≥1 f k is the decomposition of f into spherical harmonics, then (∂/∂ν) −1 f = k≥1 f k /k. We define the directional Riesz transforms of ball type by the formula An important remark is in order. In general, if M is a complete Riemannian manifold equipped with the corresponding gradient ∇ M and the Laplace-Beltrami operator M , then one defines the associated Riesz transform by Since S d−1 is a Riemannian manifold, this gives rise to the question about the relation of the associated Riesz transform R S d−1 to those introduced above. It turns out that this new operator is strictly related to the cylindrical case. To make this interplay more apparent, observe the identity and thus the analysis of R S d−1 reduces to that of R c .
From Green's formula and elementary properties of the Laplace-Beltrami operator (cf. [20,21], we infer that if R is a directional Riesz transform (of cylinder or ball type) and f, g are sufficiently regular functions on S d−1 , then This will allow us to carry out the appropriate duality argument [see the proof of (1.4) above]. Now we will describe the probabilistic representation of the above Riesz transforms. Let B = (B 1 , B 2 , . . . , B d ) be the standard Brownian motion in R d , starting from 0, and let τ = inf{t ≥ 0 : B t / ∈ B d } be the first exit time of B from the unit ball. Note that B τ has the uniform distribution on S d−1 . Let A be a continuous function on the closed unit ball, with values in the class of d × d matrices. This function gives rise to the following operation on stochastic integrals. For a given f ∈ C ∞ (S d−1 ), let F be its Poisson extension to B d and put We define the A-transform of f by the conditional expectation The connection between the operators T A and directional Riesz transforms is explained in the following statement, see Arcozzi [1].
(ii) Suppose that d ≥ 3 and let ϕ be defined by the formula We are ready to establish the bounds for Riesz transforms. We start with the following analogue of (1.3).
Proof As we have already observed, the random variable B τ is uniformly distributed on S d−1 . Therefore, by conditional version of Jensen's inequality, However, the martingale A * F is orthogonal and differentially subordinate to the martingale (F(B τ ∧t )) t≥0 . Consequently, by (2.1), we may write We have 0) and, of course, Summing these facts, we obtain Optimizing the right-hand side over λ > 0, we get the inequality which is (4.2). We postpone the proof of the sharpness to the next section.
Proof of (1.6) and (1.7) As previously, it suffices to establish the first estimate. The proof is similar to that of (1.4). We fix 1 < p < ∞, a function f ∈ L p,∞ (S d−1 ) of norm one and a set E ⊂ S d−1 of positive measure. We start from observing that Integrating by parts and noting that lim r →0 f * (r )r (Rχ E ) * * (r ) = 0, we get By (4.2), the right-hand side does not exceed where the last three equalities follow again from integration by parts. Combining the above facts yields the inequality (1.6).

Sharpness of (1.3) and (1.4), d = 1
In the one-dimensional case, if we take f = χ E for some measurable E ⊂ R satisfying 0 < |E| < ∞, then both sides of (1.3) are equal. This follows at once from the identity of Stein and Weiss [22]: , t > 0.
We turn to (1.3). Fix 1 < p < ∞, take E = [−1, 1] and consider the function f : R → R given by Let us first compute the L p,∞ -norm of f . Observe that Thus, for a given t > 0, we have The latter equality can be transformed into Thus, we have || f || L p,∞ (R) = 1. The next step in the analysis is to compare f and Hχ E , which is explicitly given by It is clear that these two functions have opposite signs ( f (x)Hχ E (x) < 0 for almost all x ∈ R) and are "equimonotone" in the sense that for any x 1 , The latter property follows from (5.1), the analogous symmetry condition for Hχ E and the fact that on [1, ∞) both functions are decreasing and nonnegative. Combining all the above facts gives Thus, both sides of (1.4) are equal.
Furthermore, rearranging this function appropriately, we may assume that its modulus is "equimonotone" with |H T χ E |; that is, for all s, t ∈ S 1 , we have | f (s)| ≤ | f (t)| if and only if |H T χ E (t)| ≤ |H T χ E (s)|. Finally, we may change the sign of f on the appropriate set so that f (x)H T χ E (x) ≤ 0 for almost all x ∈ S 1 . Then