Weighted Inequalities for the Dyadic Square Function

We study Fefferman–Stein inequalities for the dyadic square function associated with an integrable, Hilbert-space-valued function on the interval [0, 1). The proof rests on a Bellman function method: the estimates are deduced from the existence of certain special functions enjoying appropriate majorization and concavity.


Introduction
Let w be a weight (i.e., a nonnegative, locally integrable function) on R d and let M be the Hardy-Littlewood maximal operator. In 1971, Fefferman and Stein [9] proved the existence of a finite constant c depending only on the dimension such that (throughout the paper, we use the standard notation w(E) = E w(x)dx  [14,18,19].
In the mean-time, many partial or related results in this direction were obtained. In particular, Buckley [2] showed that the conjecture is true for the weights w δ (x) = |x| −d (1−δ) , 0 < δ < 1. See also Pérez' paper [17] as 360 A. Osȩkowski IEOT well as a series of works [10][11][12] by Lerner, Ombrosi and Pérez devoted to a little weaker statements related to (1.1). Chang et al. [5], and Chanillo and Wheeden [6] studied the above problem in the context of square functions. For a given ϕ ∈ C ∞ 0 (R d ) such that R d ϕ = 0, put ϕ t (x) = t −d ϕ(x/t), t > 0, and define the associated area function by the formula Then, as proved in [5], there is a constant C(d, ϕ) depending only on the parameters indicated, such that In [6], Chanillo and Wheeden generalized this result in several directions. First, they showed the corresponding weak-type (1, 1) estimate: there is a finite constant c(d, ϕ) such that Furthermore, the inequality (1.2) extends naturally to L p , 1 < p < 2: we have for some C(p, d, ϕ) independent of f and w. A very interesting fact is that (1.4) does not hold for p > 2. Chanillo and Wheeden offered the following substitution: for some C(p, d, ϕ) depending only on the parameters in the brackets. Our contribution is to study related two-weight inequalities for the dyadic square function associated with an integrable, Hilbert-space-valued function on [0, 1). Let us introduce the necessary background and notation. In what follows, H stands for the separable Hilbert space, with norm | · | and scalar product ·, · (with no loss of generality, we may and do assume that H = 2 ), and D is the collection of all dyadic subintervals of [0, 1). Let (h n ) n≥0 be the standard Haar system, given by and so on. For any I ∈ D and an integrable function f : [0, 1) → H, we will write f I for the average of f over I: that is, f I = 1 |I| I f (throughout, unless stated otherwise, the integration is with respect to the Lebesgue's measure). Furthermore, for any such f and any nonnegative integer n, we use the notation for the projection of f on the subspace generated by the first n + 1 Haar functions (I k is the support of h k ). Then the dyadic square function of f is given by where the summation runs over all nonnegative integers n such that x ∈ I n . Furthermore, the dyadic maximal operator M d acts on f by the formula We will also need the truncated versions of the above operators; for any nonnegative integer m, set where the summation runs over all n ≤ m such that x ∈ I n , and In all the considerations below, the symbol w will denote a weight on [0, 1), i.e., an integrable function w : We are ready to formulate our main results. and As in the context of area functions, we will show that the inequality of the form (1.7) does not hold with any finite constant when 2 < p < ∞. We will also establish the following "mixed-weight" version of (1.8) in the case 1 < p < 2. (1.9) Clearly, this result is qualitatively weaker than (1.7), however, we have decided to include it here since its proof exploits a novel interpolation-type argument which is of independent interest.
We would like to point out that even in the unweighted setting (i.e., for w ≡ 1), the above estimates are quite tight. The best constant in the 362 A. Osȩkowski IEOT unweighted version of (1.6) is equal to C = 1.4623 . . . , as shown by Bollobás [1] and the author [16]. The constant in (1.7) is of order (p − 1) −1 : see Burkholder [3] and Davis [7]. Finally, as p → ∞, the order √ p in (1.8) is optimal: see e.g. Davis [7] for the description of the best constants.
Typically, proofs of weighted inequalities in the analytic context depend heavily on extrapolation and interpolation arguments. Our approach will rest entirely on the so-called Bellman function method. More precisely, we will deduce the validity of the above L p estimates from the existence of certain special functions, enjoying appropriate majorizations and concavity. The technique is described in detail in the next section. Section 3 is devoted to the weak-type inequality (1.6), while Sects. 4 and 5 contain the proof of the L p inequalities in the case 1 < p < 2 and p ≥ 2. The final part of the paper discusses the probabilistic versions of the above results.

On the Method of Proof
Let us describe the technique which will be used to obtain the results announced in the preceding section. Throughout the paper, we will use the notation Suppose that V : D → R is a given function and assume that we want to establish the inequality for any integrable function f : [0, 1) → H and any nonzero weight w (i.e., satisfying w [0,1) > 0). For instance, the choice V (x, y, u, v) = uχ {y≥1} − 2|x|v leads to the weak type inequality (1.6), after a simple limiting argument (see Sect. 3 below); similarly, the function V (x, y, u, v) = y p u − C p p |x| p v corresponds to the strong-type estimates. A key idea in the study of (2.1) is to consider a function U : D → (−∞, ∞], which satisfies the following properties: 1 • For any x ∈ R and any u > 0 we have condition is the third one, and it can be understood as a concavity-type property. The interplay between the existence of such a function and the validity of (2.1) is described in the two statements below. Proof. Fix a nonzero weight w and an integrable function f : [0, 1) → H. First we will show that the sequence ( n≥0 is nonincreasing (there is no problem with the existence of the integral for each n, since f n , S n (f ), w n and M d,n w take only a finite number of values). To achieve this, fix an integer n and let I n+1 be the support of h n+1 . Then Thus, we only need to show an appropriate bound for the integrals over I n+1 . To do this, note that f n , S n (f ), w n and M d,n w are constant on I n+1 . Denote the corresponding values by x, y, u and v; then, clearly, we have (x, y, u, v) ∈ D (the fact that v > 0 follows directly from the assumption w [0,1) > 0). Next, let I − , I + be the left and the right half of I n+1 ; then f n+1 and w n+1 are constant on I ± and the corresponding values can be denoted by x ± d and u ± e, for some d ∈ H and e ∈ [−u, u]. Furthermore, directly from the definition of the truncated square and maximal functions, we see that S n+1 (f ) = y 2 + |d| 2 and M d, is equivalent to (2.4), and therefore the desired monotonicity of the sequence ( To see why the latter bound holds, note that |f 0 | = S 0 (f ) and w 0 = M d,0 w, so in fact we even have the pointwise estimate U (f 0 , S 0 (f ), w 0 , M d,0 w) ≤ 0, due to (2.2). This proves the claim.

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A beautiful fact is that the implication of the above theorem can be reversed; though we will not exploit it, we believe it is worth to be stated and proved. Proof. There is an abstract formula for the special function. Namely, for any (x, y, u, v) ∈ D, put where the supremum is taken over all n, all integrable functions f satisfying f [0,1) = x and all nonzero weights w with w [0,1) = u. Let us verify that this object has all the required properties. The first condition is easy to check: Thus, U (x, |x|, u, u) ≤ 0, since in the light of (2.1), any integral appearing under the supremum defining U (x, |x|, u, u) is nonpositive. The majorization 2 • is also straightforward: considering constant functions f ≡ x and w ≡ u, we see that for all n, U (x + d, y 2 + |d| 2 , u + e, v ∨ (u + e)) − ε (i.e., f + , w + , n are parameters for which the supremum in the definition of (2.7) Replacing f ± by f + n and f − m if necessary, we may assume that Using the same cutting-off procedure, we may also assume that w + n = w + n+1 = w + n+2 = · · · and w − m = w − m+1 = w − m+2 = · · · . Now, let us splice the functions f ± into one function f and the weights w ± into one weight w, with the use of the formula Vol. 85 (2016) Square Function Inequalities 365 That is, we "squeeze" the domain of f + , w + into the interval [0, 1/2), we "squeeze" the domain of f − and w − into the interval [1/2, 1), and then we "glue" f ± into one function, and w ± into one weight on [0, 1). Clearly, we have and hence, for any k, Using the structural properties of the Haar system, it is not difficult to check that if k is sufficiently large, then the last two integrals are equal to righthand sides or (2.6) and (2.7). Since ε was arbitrary, we obtain the desired condition 3 • .

A Weak-Type Inequality
We turn our attention to the weak-type estimate (1.6) which, as we will see, corresponds to the choice V 1 (x, y, u, v) = uχ {y≥1} − 2|x|v. To define the associated special function, consider the splitting of the domain D into the sets D 1 = (x, y, u, v) ∈ D : y < 1 and |x| + 1 + u v (y 2 − 1) ≤ 1 , Let U 1 : D → R be given by We start the analysis with the following majorizations.
(ii) We may assume that (x, y, u, v) ∈ D 1 , since otherwise both sides are equal. Then the inequality is equivalent to y 2 u − u ≤ |x| 2 v − 2|x|v, which can be further transformed into This bound follows directly from the definition of D 1 .
This is equivalent to saying that which holds true due to the above assumptions on x, y, u and v.
Let us check the concavity-type condition. Proof. Let us first consider the case (x, y, u, v) ∈ D 2 , which is much easier. Using (3.1), we write Now we turn to the analysis of the more difficult case (x, y, u, v) ∈ D 1 . Suppose first that y 2 + |d| 2 ≥ 1. Then (2.4) is equivalent to We will show a slightly stronger estimate To see that (3.3) can be deduced from it, note that y < 1 and hence Vol. 85 (2016) Square Function Inequalities 367 as desired. To show (3.4), we note that |x| ≤ 1−y in D 1 . Using this inequality, our assumption y 2 +|d| 2 ≥ 1 and the triangle inequality |x+d|+|x−d| ≥ 2|d|, we get It remains to consider the case (x, y, u, v) ∈ D 1 and y 2 + |d| 2 < 1. First we will prove that It is enough to show the first estimate, the second one follows from the change of signs of d and e. In the light of (3.2), this bound will hold true if we show that But this is simple: (x, y, u, v) ∈ D 1 implies |x| ≤ 1, so Now add the two inequalities in (3.5). We get that the right hand side of (2.4) does not exceed This completes the proof.
Proof of (1.6). Applying Theorem 2.1 to the functions U 1 and V 1 defined above, we obtain Let n go to infinity and apply Lebesgue's monotone convergence theorem to obtain To get the non-strict inequality under the indicator function, fix η ∈ (0, 1) and apply the above bound to f/η; then Letting η → 1 completes the proof. (i) If |d| < |x|, then
Clearly, we have G(0) = 0 and Now, keep t fixed and denote the expression in the square brackets by H(x). Then for any s ∈ (0, t) we have This yields where the latter bound, equivalent to 2 p−1 + 2 ≥ 2 p , follows from This implies that G is nondecreasing on [x, ∞), and hence it is nonnegative.
As we will see, the inequality (1.7) follows from (2.1) with V p (x, y, u, v) = The reason why we choose these particular special constants, will be clarified below, in Remark 4.4. Of course, we have U p (x, y, u, v) = y p u − max C p p |x| p , C p p |x| p + β p |x|y p−1 − y p v. (4.1) Let us now verify that U p , V p enjoy all the required properties. Proof. The first inequality is equivalent to u − (C p p + β p − 1)u ≤ 0, which holds trivially. To show the second bound, observe first that which holds due to So, the majorization is evident for y ≥ β p |x|; for remaining (x, y), the estimate is equivalent to

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A. Osȩkowski IEOT But y < β p |x|, so The proof is complete.
Actually, it is enough to show this bound for u = v. Indeed, having this done, we will get that the expression in the parentheses on the right is nonnegative; thus, taking v larger than u will make the right-hand side even larger than the left-hand side. So, from now on, we focus on the estimate 2((y 2 + |d| 2 ) p/2 − y p ) ≤ C p p |x + d| p + |x − d| p − 2|x| p 2 + β p |x|((y 2 + |d| 2 ) (p−1)/2 − y p−1 ).