Sharp inequalities for dyadic A1 weights

We show how the Bellman function method can be used to obtain sharp inequalities for the maximal operator of a dyadic A1 weight on Rn\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathbb{R}^n}$$\end{document} . Using this approach, we determine the optimal constants in the corresponding weak-type estimates. Furthermore, we provide an alternative, simpler proof of the related maximal Lp-inequalities, originally shown by Melas.

for any dyadic cube Q in R n . This is equivalent to saying that where M d is the dyadic maximal operator, given by The smallest C for which ( Fefferman [2] states that any A 1 weight satisfies the reverse Hölder inequality ⎛ (1.3) for certain p > 1 and c ≥ 1 which depend only on the dimension n and the value of [w] 1 . The exact information on the range of possible p's was studied by Melas [3] (see also [1] for related results in the non-dyadic case).
The purpose of this paper is to study the corresponding weak-type estimates. We will prove the following result. Theorem 1.2. Let w be a dyadic weight on R n , and let 1 ≤ p ≤ p 0 (n, [w] 1 ). Then for every dyadic cube Q, we have Both the range of p and the constant 1 are already best possible in the estimate A few words about the proof. Using a standard dilation argument, it is enough to establish (1.5) for Q = [0, 1] n . In fact, we will prove the estimate in a wider context of probability spaces equipped with a tree-like structure similar to the dyadic one. Next, while Melas' proof of Theorem 1.1 is combinatorial and rests on a clever linearization of the dyadic maximal operator, our approach will be entirely different and will exploit the properties of a certain special function. In the literature, this type of argument is called the Bellman function method and has been applied recently in various settings: see e.g. [4][5][6][7][8], and references therein. The paper is organized as follows. Section 2 contains some preliminary definitions. The description of the Bellman method can be found in Sect. 3, and it is applied in two final parts of the paper: in Sect. 4 we present the study of the weak type estimate, while in Sect. 5 we provide an alternative proof of Melas' result.
2. Measure spaces with a tree-like structure. Assume that (X, F, μ) is a given non-atomic probability space. We assume that it is equipped with an additional tree structure. Definition 2.1. Let α ∈ (0, 1] be a fixed number. A sequence T = (T n ) n≥0 of partitions of X is said to be α-splitting if the following conditions hold.
(i) We have T 0 = {X} and T n ⊂ F for all n.
(ii) For any n ≥ 0 and any E ∈ T n , there are pairwise disjoint sets E 1 , E 2 , . . . , E m ∈ T n+1 whose union is E and such that |E i |/|E| ≥ α for all i.
Let us stress that the number m in (ii) may be different for different E.
Example. Assume that X = (0, 1] n is the unit cube of R n with Borel subsets and Lebesgue's measure. Let T k be the collection of all dyadic cubes of volume 2 −kn contained in X (i.e., products of intervals of the form (a2 −k , (a + 1) In what follows, we will restrict ourselves to α ≤ 1/2 since for α > 1/2 there is only one α-splitting tree: T = ({X}, {X}, {X}, . . .). Let us define the maximal operator and A 1 class corresponding to the structure T . Definition 2.2. Given a probability space (X, F, μ) with a sequence T as above, we define the corresponding maximal operator M T as We will also use the notation M n T for the truncated maximal operator, associated with T n = (T 0 , T 1 , . . . , T n−1 , T n , T n , T n , . . .).

Definition 2.3. A nonnegative integrable function w is an
for any E ∈ T . This is equivalent to saying that for almost all x ∈ X. The smallest C for which the above holds is called the A 1 constant of w and will be denoted by [w] 1 .
3. On the method of proof. Now we will describe the technique which will be used to establish the inequalities announced in Section 1. Throughout this section, c > 1, α ∈ (0, 1/2] are fixed constants. Distinguish the following subset of R 3 + : Let Φ, Ψ : R + → R be two given functions, and assume that we want to show that The key idea in the study of this problem is to construct a special function B = B c,α,Φ,Ψ : D → R, which satisfies the following conditions. A few remarks concerning these conditions are in order. The condition 1 • is a technical assumption which enables the proper handling of the maximal operator. The conditions 2 • and 3 • are appropriate majorizations. The most complicated (and most mysterious) condition is the last one. To shed some light on it, observe that it yields the following concavity-type property of B. Lemma 3.1. Let (x, y, z) be a fixed point belonging to D, and let n ≥ 2 be an arbitrary integer. Let α 1 , α 2 , . . . , α n be positive numbers which sum up to 1, α i x i and y = min{y 1 , y 2 , . . . , y n }.
which yields the desired bound x i ≤ c−1+α cα x. We turn to the main result of this section. Proof. Let w be as in the statement. Define two sequences (w n ) n≥0 , (v n ) n≥0 of measurable functions on X as follows. Given an integer n, an element E of T n , and a point x ∈ E, set The following interplay between these objects will be important to us. Let n, E be as above, and let E 1 , E 2 , . . . , E m be the elements of T n+1 whose union is E. Then we easily check that Furthermore, the inequality [w] 1 ≤ c implies that the triple (w n , v n , M n T w) takes values in D. These conditions, combined with Lemma 3.1, yield the inequality and therefore, by induction, However, the left-hand side equals and hence the application of 2 • and 3 • completes the proof of (3.1).

A sharp weak-type estimate.
The principal result of this section is the following.
Theorem 4.1. Suppose that (X, F, μ) is a probability space equipped with an α-splitting tree T . Then for any A 1 weight w with respect to T and any p satisfying The range of p and the constant 1 are already the best possible in Before we proceed, let us establish the following technical fact.

and working a little bit turns this bound into
which is evident: the left-hand side is convex as a function of x, and both sides are equal when x ∈ {0, 1}.

4.1.
Proof of (4.1). We may assume that [w] 1 > 1, since otherwise w is constant and the assertion holds true. If the average X w is at least 1, then the inequality is trivial. So, suppose that X w < 1; then it suffices to prove the weak-type estimate for p = p 0 (α, [w] 1 ). In view of Theorem 3.2, all we need is to construct an appropriate special function corresponding to c = [w] 1 > 1, α ∈ (0, 1/2], Φ(z) = χ {z≥1} and Ψ(x) = x p . Indeed, this will yield (3.1), and letting n go to ∞ will complete the proof. Introduce B = B c,α,Φ,Ψ : D → R + by We will exploit the following auxiliary property of B. For fixed x, z > 0, the function B(x, ·, z) : y → B(x, y, z) is Proof. We may assume that x, z < 1, since otherwise the claim is obvious. Note that for y ≥ x/c we have in light of Lemma 4.2. Furthermore, for any y < 1, ∂ ∂y Since B is continuous, this gives the desired monotonicity. Now we turn to the verification that B satisfies the conditions 1 • − 4 • . The first two of them are obvious, so let us look at 3 • . By Lemma 4.3, it suffices to prove the majorization for y = x/c. But then the estimate is clear: both sides are equal when x < 1, and for x ≥ 1 the inequality takes the form 1 ≤ x p . Finally, we will check 4 • with We may and do assume that x ∨ z < 1 since otherwise the right-hand side of (3.2) is equal to 1 and there is nothing to prove. By the preceding lemma, it suffices to show (3.2) under the assumption that Suppose first that y > c −1 ; then (3.2) becomes If x ≥ 1, then this bound is clear; if x < 1, then y = y (see (4.3)) and thus both sides are equal. Finally, assuming that y ≤ c −1 , we see that (3.2) reads If x ≤ cy, then y = y (see (4.3)) and hence both sides are equal. If x > cy, then (4.3) implies that y = x /c and the inequality becomes or, after the substitution t = x /y, We have t > c by the assumption we have just made above. On the other hand, exploiting the requirements appearing in 4 • , we get It suffices to note that the left-hand side of (4.4) is a convex function of t and both sides are equal for the extremal values of t: t = c and t = (c − 1 + α)/α (the equality for the latter value of t is just the definition of p 0 (α, c)).

Sharpness.
It is obvious that the constant 1 cannot be improved in (4.2): consider a constant weight w ≡ λ > 1, and let λ ↓ 1. To show that the weak-type estimate cannot hold with exponents larger than p 0 (α, c), we will construct an appropriate example; a related object can be found in [3]. Suppose that (X, F, μ) is a probability space equipped with an α-splitting tree T , such that there is a monotone sequence X = E 0 ⊃ E 1 ⊃ E 2 ⊃ . . ., with E n ∈ T n and μ(E n ) = α n . For x ∈ X, put N (x) = sup{n ≥ 0 : x ∈ E n }; this is well-defined since E 0 = X. Moreover, N (x) < ∞ almost everywhere, because the sets E i shrink to a set of measure zero. Define a weight w by In other words, we have w(x) = (c − 1 + α)/(cα) n , where n is the unique number such that x ∈ E n \E n+1 . Then w is in the A 1 class and [w] 1 = c. To see this, pick x ∈ X and let n be the unique integer such that x ∈ E n \E n+1 . The only elements of T which contain x are E 0 , E 1 , . . ., E n , so However, by the definition of w, we easily compute that Putting k = 0 in the above calculation gives X w = c. Now, if we put q = p 0 (α, c), then the expression in the square brackets is equal to 1. Therefore, if q is larger than p 0 (α, c), then the constant on the righthand side explodes as n → ∞. This shows that the threshold p 0 (α, c) in the weak-type estimate cannot be improved.
The estimate (3.2) can be rewritten in the form If x ≤ (x ∨ z), then both sides are equal; if x > (x ∨ z), then the bound becomes or, after the substitution t = x /(x ∨ z), However, we have t > 1 and t ≤ (c − 1 + α)/(cα) (see the assumptions appearing in 4 • ). It suffices to note that the left-hand side of (5.2) is a convex function, and that both sides are equal for t ∈ {1, (c−1+α)/(cα)}. Thus, (3.1) gives the claim for truncated maximal operator, and letting n → ∞ completes the proof, by the use of Lebesgue's monotone convergence theorem.