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Weighted maximal inequalities for the Haar system
Weighted maximal inequalities for the Haar system
Osękowski, Adam
2018-04-19 00:00:00
Monatsh Math (2018) 186:321–336 https://doi.org/10.1007/s00605-018-1184-9 Adam Osek ˛ owski Received: 29 October 2016 / Accepted: 9 April 2018 / Published online: 19 April 2018 © The Author(s) 2018 Abstract The paper contains the study of weighted maximal L -inequalities for the Haar system, with the optimal dependence on the characteristics of the weights involved. The proofs exploit certain special functions, enjoying appropriate size con- ditions and concavity. Keywords Maximal · Dyadic · Bellman function · Best constants Mathematics Subject Classiﬁcation Primary 42B25; Secondary 46E30, 60G42 1 Introduction The purpose of this paper is to study certain class of weighted inequalities for the Haar system. Let h = (h ) be the collection of Haar functions on [0, 1), given by n n≥0 h = χ , h = χ − χ , 0 [0,1) 1 [0,1/2) [1/2,1) h = χ − χ , h = χ − χ , 2 [0,1/4) [1/4,1/2) 3 [1/2,3/4) [3/4,1) h = χ − χ , h = χ − χ , 4 [0,1/8) [1/8,1/4) 5 [1/4,3/8) [3/8,1/2) h = χ − χ , h = χ − χ 6 [1/2,5/8) [5/8,3/4) 7 [3/4,7/8) [7/8,1) Communicated by A. Constantin. Research supported by the NCN Grant DEC-2014/14/E/ST1/00532. B Adam Ose˛kowski ados@mimuw.edu.pl Department of Mathematics, Informatics and Mechanics, University of Warsaw, Banacha 2, 02-097 Warsaw, Poland 123 322 A. Ose˛kowski and so on. A classical result of Schauder [16] asserts that the Haar system forms p p a basis of L = L (0, 1),1 ≤ p < ∞ (with the underlying Lebesgue measure). Marcinkiewicz showed in [8] that if 1 < p < ∞, then this basis is unconditional: there is a ﬁnite positive constant β depending only on p with the property that if n is a nonnegative integer, a , a ,... , a are real numbers and ε , ε , ..., ε is a sequence 0 1 n 0 1 n of signs, then n n ε a h ≤ β a h . (1.1) k k k p k k k=0 p k=0 p L L One can study a weighted version of this estimate. Here and below, the word “weight” will refer to a positive and integrable function on [0, 1).Given 1 < p < ∞,wesay that a weight w satisﬁes dyadic Muckenhoupt’s condition A (shorter: w is a dyadic A weight) if p−1 1 1 −1/( p−1) [w] := sup w w < ∞. | I | | I | I I I Here the supremum is taken over all dyadic subintervals I of [0, 1) (i.e., all I of the −k −k k form [a2 ,(a + 1)2 ),for some k ≥ 0 and a ∈{0, 1,... , 2 − 1}). There are versions of this deﬁnition for p ∈{1, ∞}: w belongs to the dyadic A class if 1 1 [w] := sup w exp log(1/w) < ∞, | I | | I | I I I and w is a dyadic A weight if [w] = sup w essinf w< ∞. | I | In both conditions, the suprema are taken over all dyadic subintervals I of [0, 1).One easily veriﬁes that [w] ≤[w] if q ≤ p and hence the classes A grow as p A A p p q increases. It follows from the work [10] of Nazarov, Treil and Volberg and the extrapolation theorem of Rubio de Francia [15] that if 1 < p < ∞ and w is an A weight, then there is a constant C depending only on the parameters indicated such that p,w n n ε a h ≤ C a h . k k k p,w k k p p k=0 k=0 L (w) L (w) There is a natural and interesting question concerning the optimal dependence of C on the A -characteristics [w] . More precisely, the problem is to ﬁnd, for each p,w p A 1 < p < ∞, an optimal exponent α = α( p) such that C ≤ C [w] , where p,w p C does not depend on w. This type of question was ﬁrst studied by Buckley [1]in the context of weighted estimates for maximal operators. Wittwer [22] showed that α(2) = 1 which, by the sharp version of the extrapolation theorem of Rubio de Francia, 123 Weighted maximal inequalities for the Haar system 323 established by Dragice ˇ vice ´ tal. [6] (see also Duoandikoetxea [7]), yielded the optimal dependence: n n max{1,1/( p−1)} ε a h ≤ C [w] a h . (1.2) k k k p k k k=0 p k=0 p L (w) L (w) Our ﬁrst result is the following maximal version of the estimate above. Throughout the paper, C denotes the optimal value of the constant appearing in (1.2). Theorem 1.1 Let 1 < p < ∞.If w is a dyadic A weight, N is a nonnegative integer, a , a , ..., a are real numbers and ε , ε , ..., ε is a sequence of signs, then 0 1 N 0 1 N n N max{1,1/( p−1)} 1+1/ p max ε a h ≤ 2 C [w] a h . k k k p k k 0≤n≤ N k=0 p k=0 p L (w) L (w) (1.3) The exponent max{1, 1/( p − 1)} is the best possible. Clearly, only the validity of (1.3) is an issue, the optimality of the exponent follows at once from the fact that the above bound is stronger than (1.2). In the case p = 1, the inequality (1.2) fails even in the unweighted setting, but one can study the substitute in which the maximal function appears on the right. Such an estimate allows much wider class of weights. A simple modiﬁcation of the argument of Burkholder [4] and Coifman [5] shows that if 1 ≤ q < ∞ and w satisﬁes the dyadic condition A , then we have the bound n n max ε a h ≤ c max a h (1.4) k k k q,w k k 0≤n≤ N 0≤n≤ N q q k=0 k=0 L (w) L (w) for N = 0, 1, 2,..., with some c < ∞ depending only on the parameters q,w indicated. Indeed, the aforementioned paper of Burkholder gives an appropriate unweighted good-lambda inequality involving the functions max a h 0≤n≤ N k k k=0 and max ε a h (see (8.13) in [4]), which is then transformed into the 0≤n≤ N k k k k=0 context of A weights by the argument of Coifman. This weighted good-lambda estimate yields in turn the above L -inequality (1.4) by standard integration. Since A ⊂ A for all p, we see that in particular the L -inequality holds true for A p ∞ p weights. Our principal goal is to extract the optimal dependence of c on [w] . q,w A Here is the precise statement. Theorem 1.2 For any parameters 1 ≤ p, q < ∞, there is a constant C depending p,q only on p and q which has the following property. If w is a dyadic A weight, N is a nonnegative integer, a , a , ..., a are real numbers and ε , ε , ..., ε is a 0 1 N 0 1 N sequence of signs, then n n max ε a h ≤ C [w] max a h . (1.5) k k k p,q A k k 0≤n≤ N 0≤n≤ N k=0 q k=0 q L (w) L (w) The linear dependence on the A characteristics is optimal for each p. 123 324 A. Ose˛kowski Since A = A , this gives us an alternative proof of (1.4)for A ∞ p ∞ 1≤ p<∞ weights. There is a natural question whether the dependence of c on [w] is also q,w A linear. We have been unable to answer it, though some information on C indicate p,q that this might not be the case. More precisely, our proof will establish (1.5) with 1/q 1/q −r C = 2 · 6 · inf C + 3 , (1.6) p,q r where C is the best constant in (1.2) and the inﬁmum is taken over all r satisfying r ≥ max{ p, 2} and r > q. Let us provide a more explicit formula for C .Asshown in p,q [2], we have the estimate C ≤ 1109. Now, it follows from the extrapolation theorems of Duoandikoetxea [7] and the sharp weighted bounds for the dyadic maximal operator established in [12], that if r ≥ 2, then √ (r −2)/(r −1) √ C ≤ 8re C ≤ 8reC < 8527r. r 2 2 Modulo the constant factor, this inequality can be reversed. Burkholder [1] proved that the optimal choice for β in (1.1)ismax{ p−1, 1/( p−1)}; this yields C ≥ r −1 ≥ r/2 p r for r ≥ 2. Consequently, we obtain that 1/q 1/q −r C ∼ 2 inf r + 3 p,q (where ‘∼’ means that the ratio of both sides is bounded from below and from above by universal constants). It is easy to see that the expression in the parentheses is an increasing function of r (on the interval [max{ p, q, 2}, ∞)), so the inﬁmum is attained for the choice r = max{ p, q, 2}. Note that if q is ﬁxed and p goes to inﬁnity, then 1+1/q the constant C is of order O( p ); this explosion suggests that the inequality p,q (1.5) in the limit case p =∞ might not hold (i.e., the dependence of the constant on [w] might not be linear). A few words about the proof and the organization of the paper are in order. Our approach will rest on the Bellman function method: we will deduce the validity of (1.3) and (1.5) from the existence of certain special functions, enjoying appropriate size conditions and concavity. The approach originates from the theory of stochastic optimal control, and its fruitful connection with probability and harmonic analysis was ﬁrstly observed by Burkholder in [3], during the study of the sharp version of (1.1). Following the seminal work [3], Burkholder and others applied the method in many semimartingale estimates (see the monograph [11] for details). A decisive step towards wider applications of the technique in harmonic analysis was made by Nazarov, Treil and Volberg [9,10], who put the approach in a more modern and universal form. Since then, the method has been applied in numerous problems arising in various areas of mathematics (cf. e.g. [13,14,17–21] and consult references therein). The Bellman function proof presented in this paper is quite unusual. We start Sect. 2 with a standard statement that a successful treatment of the estimates (1.3) and (1.5) 123 Weighted maximal inequalities for the Haar system 325 requires the construction of a certain function of six variables. However, instead of pro- viding an explicit formula for such an object (which is a typical ingredient of a proof), we propose an abstract two-step reasoning. Namely, ﬁrst we decrease the dimension of the problem, by showing that ﬁnding appropriate functions of four variables is sufﬁcient to deduce the desired estimates. Then, in Sect. 3, we provide these special four-dimensional objects. Again, we do not present explicit formulas (which might be quite complicated, and the analysis of their properties could be delicate). Instead, we manage to get rid of almost all technicalities and extract the existence of these objects from the validity of the inequality (1.2). The ﬁnal part of the paper is devoted to the optimality of the exponents, which is demonstrated by constructing appropriate examples. 2 On the method of proof Throughout this section, 1 < p < ∞,1 ≤ q < ∞ and c ≥ 1 are given and ﬁxed parameters. Introduce the “hyperbolic” set 2 p−1 D = (w, v) ∈ (0, ∞) : 1 ≤ wv ≤ c . p,c This object arises naturally in the analysis of A weights. Next, introduce another domain Dom = R×R×(0, ∞)×[0, ∞)×D , pick a function B : Dom → R p,c p,c p,c and consider the following set of requirements. (i) For any x ∈ R\{0} and any (w, v) ∈ D , p,c B(x, ± x, |x|,(± x) ∨ 0, w, v) ≤ 0. (2.1) (Here and below, a ∨ b = max{a, b}.) (ii) For any (x, y, z, u, w, v) ∈ Dom we have p,c B(x, y, z, u, w, v) = B(x, y, |x|∨ z, y ∨ u, w, v). (2.2) (iii) For any (x, y, z, u, w, v) ∈ Dom we have p,c B(x, y, z, u, w, v) ≥ (y ∨ u) w − ϕ(x, |x|∨ z) w, (2.3) where ϕ : R × (0, ∞) → R is some ﬁxed continuous function depending only on p, q and c. (iv) The function B is midpoint concave in the following sense. Suppose that the points (x, y, w, v), (x , y , w , v ) ∈ R × R × D satisfy the conditions ± ± ± ± p,c |x|≤ z, y ≤ u, (x , y , w , v ) + (x , y , w , v ) + + + + − − − − (x, y, w, v) = 123 326 A. Ose˛kowski and | x − x |=|y − y |. Then + − + − B(x , y , z, u, v , w ) + B(x , y , z, u, v , w ) + + + + − − − − B(x, y, z, u, w, v) ≥ . (2.4) The statement below presents the connection between the list of the above condi- tions with the validity of certain maximal inequalities. Theorem 2.1 If there exists a function B satisfying the conditions (i )–(i v), then we have the estimate n N n 1/q max ε a h ≤ 2 ϕ a h , max a h k k k k k k k 0≤n≤ N 0≤n≤ N q q k=0 k=0 k=0 L (w) L (w) for all N ≥ 0, all sequences a ,a , ...,a ∈ R, ε , ε , ..., ε ∈{−1, 1} and all 0 1 N 0 1 N dyadic A weighs w satisfying [w] ≤ c. p A Remark 2.2 In the considerations below, we will take ϕ(x, z) = K x or ϕ(x, z) = K z, for some constant K : then the assertion corresponds to the estimates (1.3)or (1.5), respectively. Proof It is convenient to split the reasoning into three parts. Step 1 Some reductions and notation. By standard limiting arguments (Fatou’s lemma and Lebesgue’s dominated convergence theorem), we may and do assume that a = 0 (recall that ϕ is assumed to be continuous). Note that it is enough to show the one-sided estimates n N n max ε a h ≤ ϕ a h , max a h , k k k k k k k 0≤n≤ N 0≤n≤ N k=0 k=0 k=0 q + q L (w) L (w) n N n max (−ε )a h ≤ ϕ a h , max a h k k k k k k k 0≤n≤ N 0≤n≤ N k=0 k=0 k=0 + q L (w) L (w) (where a = max{a, 0}), since if we rise their sides to power q and add them, we obtain an estimate which is stronger than the assertion. Furthermore, switching from (ε ) to (−ε ) , we see that it is enough to focus on the ﬁrst bound. For n ≥ 0, k k≥0 k k≥0 introduce the notation n n ∗ ∗ f = a h , g = ε a h , | f | = max | f |, g = max (g ) n k k n k k k k k + n n 0≤k≤n 0≤k≤n k=0 k=0 −1/( p−1) and let w , v denote the projections of w and w on the space generated by the n n ∞ ∞ −1/( p−1) ﬁrst n + 1 Haar functions. That is, if w = b h and w = c h , k k k k k=0 k=0 n n then w = b h and v = c h . Since [w] ≤ c, one easily checks that n k k n k k A k=0 k=0 p for any n the pair (w ,v ) takes values in the set D . n n p,c 123 Weighted maximal inequalities for the Haar system 327 Step 2 Monotonicity property. The main part of the proof is to show that for 0 ≤ n ≤ N − 1wehave ∗ ∗ B f , g , | f | , g ,w ,v ds n n n n n ∗ ∗ ≥ B f , g , | f | , g ,w ,v ds. n+1 n+1 n+1 n+1 n+1 n+1 Note that the integrands are well-deﬁned: as have already observed above, the pairs (w, v) take values in D and the assumption a = 0, imposed at the beginning, p,c 0 guarantees that | f | ≥| f | > 0. n 0 To check the above estimate, let I be the support of h . The functions n+1 ∗ ∗ ∗ ∗ ( f , g , | f | , g ,w ,v ) and ( f , g , | f | , g ,w ,v ) coincide out- n n n n n n+1 n+1 n+1 n+1 n+1 n n+1 side I , so it is enough to show that ∗ ∗ ∗ ∗ B f , g , | f | , g ,w ,v ds ≥ B f , g , | f | , g ,w ,v ds n n n n n n+1 n+1 n+1 n+1 n+1 n n+1 I I ∗ ∗ = B f , g , | f | , g ,w ,v ds, n+1 n+1 n n+1 n+1 where in the latter passage we have exploited (2.2) and the trivial identities | f |= n+1 ∗ ∗ ∗ ∗ ∗ | f |∨| f | , g = g ∨ g . However, f , g , | f | , g , w and v are constant n+1 n n+1 n n n n n n+1 n n on I ; denote the appropriate values by x, y, z, u, w and v, respectively. Then, on I,we have f = x + a h , g = y + ε a h , w = w + b h and n+1 n+1 n+1 n+1 n+1 n+1 n+1 n+1 n+1 n+1 v = v + c h . We see that these four functions, restricted to I , take values n+1 n+1 n+1 in two-point sets: there are two points x with x = (x + x )/2 such that f ∈ ± − + n+1 {x , x }; there are two points y with y = (y + y )/2 such that g ∈{y , y }, − + ± − + n+1 − + and similarly for w and v . Plugging this observation above, we see that the n+1 n+1 monotonicity property reduces to the condition (2.4). Step 3 Completion of the proof. Now, applying (2.3), we get n N n max ε a h − ϕ a h , max a h k k k k k k k 0≤n≤ N 0≤n≤ N k=0 k=0 k=0 q + q L (w) L (w) ∗ q ∗ q = (g ) − ϕ( f , | f | ) wds N N ∗ q ∗ q = (g ) − ϕ( f , | f | ) w ds N N N ∗ ∗ ≤ B( f , g , | f | , g ,w ,v )ds N N N N N ∗ ∗ ≤ B( f , g , | f | , g ,w ,v )ds 0 0 0 0 0 and it remains to note that the latter integrand is nonpositive, due to (2.1). 123 328 A. Ose˛kowski Therefore, we have reduced the problem of showing (1.5) to the construction of an appropriate function of six variables. This seems to be a difﬁcult task; the following theorem allows to decrease the number of variables to four. Theorem 2.3 Let r ≥ q and L > 0 be ﬁxed. Suppose that U : R × R × D → R p,c satisﬁes the majorizations U (x, ± x, w, v) ≤ 0 for all x ∈ R, w, v ∈ D , (2.5) p,c r r r U (x, y, w, v) ≥|y| w − L |x| w for all (x, y, w, v) ∈ R × R × D , (2.6) p,c U (x, y, w, v) ≥ U (x, 0, w, v) for all (x, y, w, v) ∈ R × R × D , (2.7) p,c and the following concavity inequality. If the points (x, y, w, v), (x , y , w , v ) ∈ ± ± ± ± R × R × D satisfy the conditions p,c (x , y , w , v ) + (x , y , w , v ) + + + + − − − − (x, y, w, v) = and | x − x |=| y − y |, then + − + − U (x , y , w , v ) + U (x , y , w , v ) + + + + − − − − U (x, y, w, v) ≥ . (2.8) Then the function B(x, y, z, u, w, v) U (x, y, w, v) + U (x,(u − y) ∨ 0, w, v) q−1 = 2 · r −q r −q (3L) (|x|∨ z) r − q q q − · (6L) (|x|∨ z) w satisﬁes the conditions (i )–(i v) with 2Lx if r = q, ϕ(x, z) = 1/q −r 6L · r/q + 3 z if r > q. Proof We start with (2.1). Observe that a ∨ 0 − a = (− a) ∨ 0, so by (2.5) and (2.7), U (x, ± x, w, v) + U (x,(∓ x) ∨ 0, w, v) q−1 B(x, ± x, |x|,(± x) ∨ 0, w, v) ≤ 2 · r −q r −q (3L) |x| U (x, ± x, w, v) + U (x, ∓ x, w, v) q−1 ≤ 2 · ≤ 0. r −q r −q (3L) |x| 123 Weighted maximal inequalities for the Haar system 329 The condition (2.2) is evident. To show (2.3), suppose ﬁrst that r = q. Then, directly r r −1 r r from (2.6) and the elementary estimate (a + b) ≤ 2 (a + b ), r −1 r r r r B(x, y, z, u, w, v) ≥ 2 |y| w + ((u − y) ∨ 0) w − 2L |x| w r r r ≥ (y ∨ u) w − (2L) |x| w. In the case r > q, note that for any nonnegative numbers A , A we have the estimate 1 2 A A + A ≥ A A .Ifweplug A = (((u − y) ∨ 0)/(3L(|x|∨ z)) and A = 2 2 2 1 2 1 1 3L(|x|∨ z) w, we get ((u − y) ∨ 0) w q q + 3L(|x|∨ z) w ≥ (u − y) ∨ 0 w. r −q r −q (3L) (|x|∨ z) Similarly, one shows that |y| w + 3L(|x|∨ z) w ≥|y| w. r −q r −q (3L) (|x|∨ z) q−1 Multiply these inequalities throughout by 2 , add them and apply the elementary q q−1 q q estimate (a + b) ≤ 2 (a + b ) to obtain r r |y| w + ((u − y) ∨ 0) w q−1 q 2 · ≥ (y ∨ u) w − 6L(|x|∨ z) w. r −q r −q (3L) (|x|∨ z) Consequently, (2.6)gives B(x, y, z, u, w, v) r r r r |y| w + ((u − y) ∨ 0) w − 2L |x| w r − q q−1 q q ≥ 2 · − (6L) (|x|∨ z) w r −q r −q (3L) (|x|∨ z) q q −r q q ≥ (y ∨ u) w − r/q + 3 (6L) (|x|∨ z) w. It remains to verify the concavity inequality (2.4). Fix parameters x, y, z, ... as in the statement of (iv); by symmetry, we may assume that |x |≥|x |. Observe ﬁrst that + − by (2.6), ∂ U (x, y, w, v) + U (x,(u − y) ∨ 0, w, v) r − q q−1 q q 2 · − · (6L) s w r −q r −q ∂s (3L) s q U (x, y, w, v) + U (x,(u − y) ∨ 0, w, v) q−1 =− 2 (r − q) r −q r −q+1 (3L) s q q−1 − (r − q) · (6L) s w |x| q q−1 ≤ (r − q)(6L) s w · − 1 , 3s which is nonpositive for |x|≤ 3s. This calculation will allow us to change appropri- ately the terms |x|∨ z in the formula for B, sometimes with values smaller than |x|. 123 330 A. Ose˛kowski The ﬁrst consequence is that we may assume that |x |≤ z. Indeed, if both |x |, |x | − + − are larger than z, then replacing z by |x | does not change the right-hand side of (2.4) and does not increase the left-hand side, making the inequality stronger. Our next step is to note that since |x |≤ z,wehave |x |≤|x − x|+|x|=|x − x |+|x|≤ 2|x|+|x |≤ 3z + + − − and hence, by the above bound for the partial derivative ∂/∂s, we conclude that B(x , y , z, u, w , v ) ± ± ± ± U (x , y , w , v ) + U (x ,(u − y ) ∨ 0, w , v ) ± ± ± ± ± ± ± ± q−1 ≤ 2 · r −q r −q (3L) z r − q q q − (6L) z w. (2.9) q q q Now, we obviously have z w = (z w + z w )/2 and, by the midpoint concavity + − of U assumed in the statement of the theorem, we know that U (x , y , w , v ) + U (x , y , w , v ) + + + + − − − − U (x, y, w, v) ≥ . Finally, the inequality (2.7) implies U (x ,(u − y ) ∨ 0, w , v ) ≤ U (x , u − y , w , v ) ± ± ± ± ± ± ± ± and hence, applying the concavity of U again, U (x ,(u − y ) ∨ 0, w , v ) + U (x ,(u − y ) ∨ 0, w , v ) + + + + − − − − U (x, u − y, w, v) ≥ . Combining these observations with (2.9) establishes the desired estimate (2.4). 3 An abstract Bellman function of four variables As we have seen in the previous section, having constructed an appropriate special function immediately gives us a desired inequality for the Haar system. A well-known fact in the general Bellman function theory is that this implication can be reversed: the validity of a given estimate implies the existence of the corresponding abstract Bellman function. Our argumentation depends heavily on this phenomenon: the four dimensional U we search for will be extracted from the estimate (1.2). To state things precisely, we ﬁx, throughout this section, the parameters 1 < p < ∞ and 1 ≤ q < ∞.Pick c ≥ 1 and take r ≥ p.Wehave [w] ≤[w] and hence (1.2) A A r p implies n n max{1,1/(r −1)} ε a h ≤ C [w] a h , (3.1) k k k r k k k=0 r k=0 r L (w) L (w) 123 Weighted maximal inequalities for the Haar system 331 for n = 0, 1, 2, .... Deﬁne U = U : R × R × D → R by the formula p,r,c p,c U (x, y, w, v) ⎧ ⎫ r r n n ⎨ ⎬ r max{r,r/(r −1)} = sup y + ε a h − C c x + a h , k k k k k ⎩ ⎭ r r k=1 k=1 L (w) L (w) where the supremum is taken over all n, all sequences a , a , ..., a of real numbers, 1 2 n all sequences ε , ε , ..., ε of signs and all dyadic A weights w satisfying [w] ≤ 1 2 n p A 1 1 −1/( p−1) c, w = w and w = v. To see that the deﬁnition makes sense (the 0 0 supremum is taken over a nonempty set), we need the following. Lemma 3.1 For any (w, v) ∈ D there is a dyadic A weight w on [0, 1) with p,c p 1 1 −1/( p−1) [w] ≤ c, w = w and w = v. p 0 0 Proof It is easy to show, using a Darboux property, that there are two points P = p−1 (x , y ), P = (x , y ) lying at the lower boundary of D (i.e., satisfying x y = 1 1 2 2 2 p,c 1 p−1 x y = 1) such that (w, v) is the middle of the line segment P P . Deﬁne w on 2 1 2 [0, 1) by setting x if s < 1/2, w(s) = x if s ≥ 1/2. 1 1 −1/( p−1) −1/( p−1) −1/( p−1) Then w = (x + x )/2 = w and w = (x + x )/2 = 1 2 0 0 1 2 ( y + y )/2 = v. It remains to verify that [w] ≤ c. By the above calculation, 1 2 A p−1 1 1 −1/( p−1) p−1 w w = wv ≤ c, 0 0 and if I is an arbitrary dyadic, proper subinterval of [0, 1), then w is constant on I,so p−1 1 1 −1/( p−1) w w = 1 ≤ c. | I | | I | I I Hence w has all the desired properties. Now we will show that the abstract function U above possesses all the properties required in Theorem 2.3. max{1,1/(r −1)} Lemma 3.2 The function U satisﬁes (2.5), (2.6) with L = C c , (2.7) and the midpoint concavity (2.8). Proof The ﬁrst condition follows directly from (3.1): all the expressions appearing under the supremum deﬁning U (x, ± x, w, v) are nonpositive. The majorization (2.6) is obtained by considering the sequence a = a = ··· = 0 in the deﬁnition of 1 2 123 332 A. Ose˛kowski U (x, y, w, v). To check (2.7), we will prove that U (x, y, w, v) = U (x, − y, w, v) and that the function y → U (x, y, w, v) is convex. Both these facts are simple. The ﬁrst of them follows by switching from (ε ) to (−ε ) in the deﬁnition of U (x, y, w, v); then it k k is clear that the suprema deﬁning U (x, ± y, w, v) are the same. To prove the convexity of U (x, ·, w, v),pick α ∈ (0, 1), two real numbers y , y and set y = αy + (1 − α)y . 1 2 1 2 If n is a nonnegative integer, a , a , ..., a is a sequence of real numbers and ε , ε , 1 2 n 1 2 ..., ε is an arbitrary sequence of signs, then the deﬁnition of U and the convexity of the function t →|t | implies that r r n n r max{r,r/(r −1)} y + ε a h − C c x + a h k k k k k r r k=1 k=1 L (w) L (w) r r n n ≤ α y + ε a h + (1 − α)y + ε a h 1 k k k 2 k k k k=1 r k=1 r L (w) L (w) r max{r,r/(r −1)} − C c x + a h k k k=1 r L (w) ≤ αU (x, y , w, v) + (1 − α)U (x, y , w, v). 1 2 Therefore, taking the supremum over all n and all sequences (a ), (ε ) gives the k k convexity of U (x, ·, w, v), and hence (2.7) is established. It remains to show (2.8). Fix points (x, y, w, v), (x , y , w , v ) ∈ R × R × D ± ± ± ± p,c ± ± ± + + ± ± as in the statement. Let a , a ,..., a , ε ,ε ,...,ε , w be as in the deﬁnition of 1 2 n 1 2 n + + U (x , y , w , v ) (we may assume that the lengths of the sequences a , a , ... and ± ± ± ± 1 2 − − a , a , ... are the same, adding some zeros if necessary). Let us splice these objects 1 2 using the following procedure: consider the function f :[0, 1) → R given by x + a h (2s) if s < 1/2, + k k=1 k f (s) = n − x + a h (2s − 1) if s ≥ 1/2. − k k=1 Using the self-similarity of the Haar system, we see that N N x + x 1 2 f = + a h = x + a h , k k k k k=1 k=1 for some N and some sequence (a ) :wehave a = x − x = (x − x )/2 and k 1 + + − k=1 ± ± all remaining terms a are either zero or belong to the set {a , a ,..., a }. We can 1 2 n ± ± do the same splicing procedure with the functions y + ε a h , arriving at ± k k=1 k k the function y + ε a h , where N and a are the same as above, and ε , ε , ..., k k k k 1 2 k=1 ε take values in {−1, 1} (here we have used the assumption |x − x |=|y − y |: N + − + − + − it implies that ε ∈{−1, 1}). Similarly, we “glue” the weights w and w into one weight in [0, 1), setting 123 Weighted maximal inequalities for the Haar system 333 w (2s) if s < 1/2, w(s) = w (2s − 1) if s ≥ 1/2. 1 1/2 1 + − This new weight satisﬁes w = w (2s)ds+ w (2s−1)ds = (w +w )/2 + − 0 0 1/2 −1/( p−1) and, analogously, w = v. Furthermore, we have [w] ≤ c. Indeed, ﬁrst note that p−1 1 1 −1/( p−1) p−1 w w = wv ≤ c. 0 0 Next, if I is a dyadic subinterval of [0, 1/2), then p−1 1 1 −1/( p−1) w w | I | | I | I I p−1 1 1 + + −1/( p−1) = w (w ) ≤ c, |2 I | |2 I | 2 I 2 I since [w ] ≤ c; the case when I is a dyadic subinterval of [1/2, 1) is dealt with similarly. Thus, by the very deﬁnition of U (x, y, w, v),wehave U (x, y, w, v) r r N N r max{r,r/(r −1)} ≥ y + ε a h − C c x + a h k k k k k k=1 r k=1 r L (w) L (w) r r n n + + r max{r,r/(r −1)} + = y + ε a h − C c x + a h + k + k k k k r + r + k=1 k=1 L (w ) L (w ) r r n n − − r max{r,r/(r −1)} − + y + ε a h − C c x + a h . − k − k k k k r − r − k=1 k=1 L (w ) L (w ) ± ± ± ± Since a , a , ..., ε , ε , ..., w were arbitrary, the inequality (2.8) follows. 1 2 1 2 We are ready to establish the inequalities announced in the introductory section. Proof of (1.3)Fix 1 < p < ∞, a weight w and let c =[w] . Apply Lemma 3.2 with r = p to obtain an appropriate function U with the majorizing con- max{1,1/( p−1)} stant C c . Plug this function into Theorem 2.3 with q = p to get the max{1,1/( p−1)} Bellman function B with the majorizing function ϕ(x, z) = 2C c x. This function, used in Theorem 2.1, yields the assertion with the desired constant 1/ p max{1,1/( p−1)} 2 · 2C c . Proof of (1.5)Fix 1 ≤ p < ∞ and 1 ≤ q < ∞,takean A weight w and set c =[w] . Suppose further that r ≥ p and r > q. Then Lemma 3.2, applied with this value of r, yields an appropriate function U with the majorizing constant 123 334 A. Ose˛kowski max{1,1/(r −1)} C c . This function can be used in Theorem 2.3 to obtain the Bellman func- 1/q max{1,1/(r −1)} −r tion B with the majorizing function ϕ(x, z) = 6C c · r/q + 3 z. This Bellman function, used in Theorem 2.1, yields the estimate (1.5) with the constant 1/q 1/q −r max{1,1/(r −1)} 2 · 6C · r/q + 3 c . To sharpen the dependence of the constant on the characteristic c =[w] , we impose the additional assumption r ≥ 2. This leads us precisely to the claim, with the constant C given by (1.6). p,q 4 Sharpness of the exponent in (1.5) Since [w] ≤[w] for any p ≥ 1, it is enough to show the optimality of the A A exponent for p = 1. For the sake of clarity, we split the reasoning into three parts. Step 1 Construction. Let N be a positive integer. Deﬁne the coefﬁcients a , a , ..., 0 1 a N by N −1 a h := h − 2h + 2h − 2h + 2h − ··· + 2 · (−1) h k k 0 1 2 4 8 k=0 and the signs ε , ε , ε , ..., ε N by requiring that 0 1 2 ε a h := h + 2h + 2h + 2h + 2h + ··· + 2h N . k k k 0 1 2 4 8 k=0 Observe that | a h |≤ 3 almost everywhere (one easily checks the identity k k k=0 | a h |= χ + 3χ ) and hence also − N −1 − N −1 k k k=0 [0,2 ) [2 ,1) max a h ≤ 3 almost everywhere. (4.1) k k 0≤n≤2 k=0 In addition, we see that ε a h = 1 + 2( N + 1) ≥ 2( N + 1). (4.2) k k k − N −1 [0,2 ) k=0 Next, set a = N /( N + 1) and consider the weight w on [0, 1), given by 2 N w = h + ah + (1 + a)ah + (1 + a) ah + ··· + (1 + a) ah N 0 1 2 4 N +1 k = (1 + a) χ + (1 + a) (1 − a)χ . − N −1 −k−1 −k [0,2 ) [2 ,2 ) k=0 Step 2 Verifying an A condition. We will prove that w is an A weight satisfying 1 1 −1 [w] = (1 − a) = N + 1. To this end, ﬁx a dyadic interval I ⊆[0, 1).If 123 Weighted maximal inequalities for the Haar system 335 − N −1 −1 | I|≤ 2 , then w is constant on I and hence w/ essinf w = 1 ≤ (1 − a) . | I | I − N So, suppose that the length of I is at least 2 ; then there is a nonnegative integer m −m −m m ≤ N and k ∈{0, 1, 2,... , 2 − 1} such that I =[k · 2 ,(k + 1) · 2 ).If 1 −1 k = 1, then w is constant on I and hence w/ essinf w = 1 ≤ (1 − a) ,as | I | previously. If k = 0, then essinf w = (1 + a) (1 − a) and ! " m N +1 − N −1 k −k−1 m w = 2 (1 + a) 2 + (1 + a) (1 − a)2 = (1 + a) , | I | k=m 1 −1 so w/ essinf w = (1 − a) . Finally, if k ≥ 2, then there is a unique such | I | I −−1 − that I ⊂[2 , 2 ). Since w is constant on this larger interval and nondecreasing on [0, 1), we get essinf w = essinf − w and [0,2 ) 1 1 1 w = w ≤ w, −−1 − − −−1 − − | I | [2 , 2 ) [0, 2 ) I [2 ,2 ) [0,2 ) 1 −1 so the estimate w/ essinf w ≤ (1−a) follows from the case k = 0 considered | I | I above (replace k by there). Step 3 Completion of the proof. By the elementary bound e ≥ 1 + x, we get N +1 − N −1 1 + a 1 − a 1 − a −1 = 1 + ≥ exp −( N + 1) ≥ e . 2 1 + a 1 + a Therefore, if we ﬁx q ≥ 1 and apply (4.1), (4.2), we obtain n n max ε a h ≥ max ε a h k k k k k k N N 0≤n≤2 0≤n≤2 k=0 q k=0 1 L (w) L (w) ≥ ε a h k k k k=0 L (w) − N −1 N +1 ≥ 2 · 2( N + 1) · (1 + a) N +1 1 + a =[w] · 2( N + 1)(1 − a) · −1 ≥[w] · 2e −1 ≥ e [w] max a h . A k k 0≤n≤2 k=0 q L (w) The proof is complete. Acknowledgements The author would like to thank an anonymous referee for the very careful reading of the ﬁrst version of the paper and many helpful suggestions which greatly improved the presentation. 123 336 A. 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