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BURKHOLDER’S FUNCTION VIA MONGE–AMPE`RE EQUATION 1 VASILYVASYUNINANDALEXANDERVOLBERG 1 0 2 n Abstract. WewillshowhowtogetBurkholder’sfunctionfrom[Bu1]byusing a J Monge-Ampèreequation. Thismethodisquitedifferentfromthoseintheseries 7 of Burkholder’s papers [Bu1]–[Bu7]. ] P A 1. Introduction . h Bellman function method in Harmonic Analysis was introduced by Burkholder t a for finding the norm in Lp of the Martingale transform. Later it became clear that m [ the scope of the method is quite wide. 2 The technique, originated in Burkholder’s papers [Bu1]–[Bu7], can be credited v for helping to solve several old Harmonic Analysis problems and for unifying ap- 3 3 proach to many others. In the first category one would name the (sharp weighted) 6 2 estimates of such classical operators as the Ahlfors–Beurling transform (Banuelos– . 6 Wang [BaWa1], Banuelos–Janakiraman [BaJa1], Banuelos–Mendez [BaMH], Na- 0 zarov–Volberg [NV1], Petermichl–Volberg [PV], Dragicevic–Volberg [DV2]) and 0 1 the Hilbert and Riesz transforms (Petermichl [P1], [P2]). In the second category : v one can name all kind of dimension free estimates of weighted and unweighted i X Riesz transforms (see a vast literature in [DV1]–[DV3]). Roughly, Bellman func- r a tion method makes apparent the hidden scaling properties of a given Harmonic Analysis problem. Conversely, given a Harmonic Analysis problem with certain scaling properties one can (formally) associate with is a non-linear PDE, the so- called Bellman equation of the problem. Let us recall to the reader that in the series of papers [Bu], [Bu1]–[Bu7] Donald Burkholder investigated Martingale transform and gave the sharp bounds on this operator in various settings–but by similar methods. The methods were so novel and powerfulthat theinfluenceof thesearticles will befelt formany years to come. Research of the first author was partially supported by RFBR grant 08-01-00723. The second author is supported by NSFgrant DMS 0758552. 1 2 VASILYVASYUNINANDALEXANDERVOLBERG Thenovelty was akey. Oneof theleadingmathematician workinginthedomain of Harmonic Analysis told the second author that these papers of Burkholder “spin his head”. In the book of Daniel Strook [Str] many pages are devoted to the technique developed by Burkholder in the abovementioned series of papers, and the reader can sense the same feeling. It is explained in [Str] that the simplest way tounderstandthesharpestimates of Martingale transformobtained byBurkholder is to operate with one of the so-called Burkholder’s function: 1 u (x,y) = p(1 )p−1(y (p∗ 1)x )(x + y )p−1, (1) p − p∗ | |− − | | | | | | here p∗ := max(p, p ), 1 < p < . p−1 ∞ However, the main question is of course how to get this function? Where did it come from? These questions are asked in [Str] as well. Of course, Burkholder explains in many details the way this function (and several of its relatives) are obtained. It is almost (but not quite) the least bi-concave majorant of function y p−1 (p∗ 1)p x p. (2) | | − − | | It is obtained by solving a certain PDE and performingcertain manipulations with the solution after that. The reader will find much more about u after reading this p article, in particular in Section 6. But it seems like the same questions persist even after this explanation. And a new question can appear: how wide is the applicability of the technique that Burkholder elaborated in [Bu1]–[Bu7]? There is a vague feeling that the area of applicability is quite wide. To make this feeling more precise one should look at thefunctionabove closer andseethatitis acreaturefromanother universe, which, initially, does not have too much in common with Harmonic Analysis. Burkholder function is a natural dweller of the area called Stochastic Optimal Control. It is a solution of a corresponding Bellman equation (or a dynamic programming equation) but in the setting, when the differential equations subject to control are not the usual ones. They are stochastic differential equations. The reader can find some notes on this in [VoEcole], [NTV2], [VaVo], [VaVo2], [SlSt]. These notes explainwhyStochasticOptimalControlistherighttooltoworkwithacertainclass of Harmonic Analysis problems. On the other hand, Stochastic Optimal Control problems generically can be reduced to solving a so-called Bellman PDE (and proving the so-called “verification theorems”, but this is a second task). Bellman BURKHOLDER’S FUNCTION 3 PDEsbelongtotheclassoffullynon-linearPDEs. OftentheyarePDEsofMonge– Ampère type. In thepresentarticle wewould like to showthe reader how to obtain Burkhloder functions (the one above and others from [Bu1]–[Bu7]) by reducing the search for them to solving certain Monge–Ampère equations. The scope of the application of the methods of Stochastic Optimal Control to Harmonic Analysis proved to be quite large. After Burkholder the first systematic application of this technique appeared in 1995 in the first preprint version of [NTV1]. It was vastly developed in [NT] and in (now) numerous papers that followed. A small part of this literature can be found in the bibliography below. 2. Notations and definitions We shall say that an interval I and a pair of positive numbers α± such that I α+ +α− = 1 generate a pair of subintervals I+ and I− if I± = α± I (I means | | I | | | | the length of I) and I = I− I+. For a given interval J the symbol = (α) will ∪ J J denote the families of subintervals of J such that J ; • ∈J if I then I± . • ∈ J ∈ J For a special choice if all α± = 1 we get the dyadic family = . Every family I 2 J D has its own set of Haar functions: J α+ I if t I , − − α |I| ∈ I h (t) =  r I ∀ ∈ J I  α− − α+I|I| if t ∈ I+. r I  If the family is such that that the maximal length of the interval of n-th gener- J ation (i.e., after splitting the initial interval J into 2n parts) tends to 0 as n , → ∞ the Haar family forms an orthonormal basis in the space L2(J) const . ⊖{ } Forafunctionf L1(I)thesymbol f meanstheaverageoff overtheinterval ∈ h iI I: 1 f = f(t)dt. h iI I | |ZI Definition. Fix a real p, 1< p < , and let p′ = p , p∗ = max p,p′ . Introduce ∞ p−1 { } the following domain in R3: Ω =Ω(p) = x = (x ,x ,x ): x 0, x p x . 1 2 3 3 1 3 { ≥ | | ≤ } 4 VASILYVASYUNINANDALEXANDERVOLBERG For a fixed partition of an interval J we define two function on this domain J B (x) =B (x;p) = sup g p , max max h| | iJ f,g (cid:8) (cid:9) B (x) = B (x;p) = inf g p , min min f,g h| | iJ (cid:8) (cid:9) wherethesupremumistakenoverallfunctionsf,gfromLp(J)suchthat f = x , h iJ 1 g = x , f p = x , and (f,h ) = (g,h ). We shall refer to any such pair of h iJ 2 h| | iJ 3 | I | | I | functions f, g as to an admissible pair. When (f,h ) = (g,h ) happens for all | I | | I | dyadic intervals inside J we call g a Martingale transform of f. We shall call B (x) (and B (x)) the Bellman functions of the problem of finding the best max min constant for the Martingale transform inequality: g f g p C(p) f p . (3) |h iJ|≤ |h iJ|⇒ h| | iJ ≤ h| | iJ This best constant was found by Burkholder: C(p) = (p∗ 1)p, p∗ := max(p, p ). − p−1 Remark 1. It is amazing that there is no proof that would find this C(p) without findingthefunctionof3variablesB (x)orsomeofitsrelatives(like,forexample, max u from (1)). p Remark 2. Burkholder proved that the functions B do not depend on the initial interval J and on a specific choice of its partition. Below we work only with dyadic partitions. Remark 3. In the case p = 2 the Bellman function are evident: B (x) = B (x) = x2+x x2. max min 2 3− 1 Indeed, since f 2 = J x = J x2+ (f,h )2, k k2 | | 3 | | 1 | I | I∈J X we have 1 1 g 2 = g 2 = x2+ (g,h )2 h| | iJ J k k2 2 J | I | | | | |I∈J X 1 = x2+ (f,h )2 = x2+x x2. 2 J | I | 2 3− 1 | | I∈J X Define the following function on R2 = z = (z ,z ): z > 0 : + { 1 2 i } zp (p∗ 1)pzp , if z (p∗ 1)z , F (z ,z )= 1 − − 2 1 ≤ − 2 (4) p 1 2 p(cid:2)(1 1 )p−1(z +(cid:3)z )p−1 z (p∗ 1)z , if z (p∗ 1)z .  − p∗ 1 2 1− − 2 1 ≥ − 2 (cid:2) (cid:3)  BURKHOLDER’S FUNCTION 5 Note for for p = 2 the expressions above are reduced to F (z ,z )= z2 z2. 2 1 2 1 − 2 3. The main result Now we are ready to state the main result: 1 1 Theorem 1. The equation F (x , x ) = F (xp,Bp) determines implicitly the p | 1| | 2| p 3 1 1 function B = B (x;p) and the equation F (x , x ) = F (Bp,xp) determines min p | 2| | 1| p 3 implicitly the function B= B (x;p). max Remark. Thereadercantakealookatformulae(5.23)–(5.27) onpage660of[Bu1] and recognize that this is how Burkholder describes B . The same is true for max B . min 4. How to find Bellman functions We start from deducing the main inequality for Bellman functions. Introduce new variables y = 1(x +x ), y = 1(x x ), and y = x . In terms of the new 1 2 2 1 2 2 2− 1 3 3 variables we define a function M, M(y ,y ,y )= B(x ,x ,x ) = B(y y ,y +y ,y ), 1 2 3 1 2 3 1 2 1 2 3 − on the domain Ξ= y = (y ,y ,y ): y 0, y y p y . 1 2 3 3 1 2 3 { ≥ | − | ≤ } Since the point of the boundary x = x p (y = y y p) occurs for the only 3 1 3 1 2 | | | − | constant test function f = x (and therefore then g = x is a constant function as 1 2 well), we have B(x ,x , x p) = x p, 1 2 1 2 | | | | or M(y ,y , y y p)= y +y p. (5) 1 2 1 2 1 2 | − | | | Note that the function B is even with respect of x and x , i.e., 1 2 B(x ,x ,x )= B( x ,x ,x )= B(x , x ,x ). 1 2 3 1 2 3 1 2 3 − − It follows from the definition of B if we consider the test functions f˜= f for the − first equality and g˜= g for the second one. For the function M this means that − we have the symmetry with respect to the lines y = y 1 2 ± M(y ,y ,y ) = M(y ,y ,y ) = M( y , y ,y ). (6) 1 2 3 2 1 3 1 2 3 − − 6 VASILYVASYUNINANDALEXANDERVOLBERG Therefore, it is sufficient to find the function B in the domain Ω = Ω (p) = x = (x ,x ,x ) : x 0, x p x , (7) + + 1 2 3 i 1 3 { ≥ | | ≤ } or the function M in the domain Ξ = y = (y ,y ,y ): y 0, y y y , (y y )p y . (8) + 1 2 3 1 1 2 1 1 2 3 { ≥ − ≤ ≤ − ≤ } Then we get the solution in the whole domain by putting B(x ,x ,x ) = B( x , x ,x ). 1 2 3 1 2 3 | | | | Duetothesymmetry (6)wehavethefollowingboundaryconditionsonthe“new part” of the boundary ∂Ξ : + ∂M ∂M = on the hyperplane y = y , 2 1 ∂y ∂y 1 2 (9) ∂M ∂M = on the hyperplane y = y . 2 1 ∂y −∂y − 1 2 If we consider the family of test functions f˜= τf, g˜ = τg together with f and g we come to the following homogeneity condition B(τx ,τx ,τpx ) = τpB(x ,x ,x ), 1 2 3 1 2 3 or M(τy ,τy ,τpy ) = τpM(y ,y ,y ). 1 2 3 1 2 3 We shall use this property in the following form: take derivative with respect to τ and put τ = 1 ∂M ∂M ∂M y +y +py = pM(y ,y ,y ). (10) 1 2 3 1 2 3 ∂y ∂y ∂y 1 2 3 Let us fix two points x± Ω such that x+ x− = x+ x− , for the cor- ∈ | 1 − 1| | 2 − 2| responding points y± Ξ this means that either y+ = y−, or y+ = y−. Then ∈ 1 1 2 2 for an arbitrarily small number ε > 0 by the definition of the Bellman function B = B there exist two couples of test functions f± and g± on the intervals I± max such that f± = x±, g± = x±, f± p = x±, and g± p B(x±) ε. h iI± 1 h iI± 2 h| | iI± 3 h| | iI± ≥ − On the interval I = I+ I− we define a pair of test functions f and g as follows ∪ f I± = f±,g I± = g±. Thisisapairoftestfunctionsthatcorrespondstothepoint | | x = α+x++α−x−,whereα± = I± /I ,becausetheproperty x+ x− = x+ x− | | | | | 1 − 1| | 2 − 2| means (f,h ) = (g,h ). This yields | I | | I | B(x) g p = α+ g+ p ++α− g− p − α+B(x+)+α−B(x−) ε. ≥ h| | iI h| | iI h| | iI ≥ − BURKHOLDER’S FUNCTION 7 Since ε is arbitrary we conclude B(x) α+B(x+)+α−B(x−). (11) ≥ For the function B = B we can get in a similar way min B(x) α+B(x+)+α−B(x−). (12) ≤ Recall that this is not quite concavity (convexity) condition, because we have the restriction x+ x− = x+ x− . But in terms of the function M | 1 − 1| | 2 − 2| M (y) α+M (y+)+α−M (y−), max max max ≥ M (y) α+M (y+)+α−M (y−), min min min ≤ when either y = y+ = y−, or y = y+ = y−, we indeed have the concavity 1 1 1 2 2 2 (convexity) of the function M with respect to y , y under a fixed y , and with 2 3 1 respect to y , y under a fixed y . 1 3 2 Since the domain is convex, under the assumption that the function B are suf- ficiently smooth these conditions of concavity (convexity) are equivalent to the differential inequalities M M M M y1y1 y1y3 0, y2y2 y2y3 0, y Ξ, (13) My3y1 My3y3! ≤ My3y2 My3y3! ≤ ∀ ∈ ∂2M for M= M (here M stand for the partial derivatives ) and max yiyj ∂y ∂y i j M M M M y1y1 y1y3 0, y2y2 y2y3 0, y Ξ, (14) My3y1 My3y3! ≥ My3y2 My3y3! ≥ ∀ ∈ for M= M . min Extremal properties of the Bellman function requires for one of matrices in (13) and (14) to be degenerated. So we arrive at the Monge–Ampère equation: M M = (M )2 (15) yiyi y3y3 yiy3 either for i = 1 or for i = 2. To find a candidate M for the role of the true Bellman function M we shall solve this equation. After finding this solution we shall prove that M = M. The method of solving homogeneous Monge–Ampère equation is described, for example, in [VaVo], [VaVo2], [SlSt]. In particular we know that the solution of the 8 VASILYVASYUNINANDALEXANDERVOLBERG Monge–Ampère equation has to be of the form M = t y +t y +t , (16) i i 3 3 0 where t = M , k = 1,2,3. The solution M is linear along the lines (let us call k yk them extremal trajectories) y dt +y dt +dt = 0. (17) i i 3 3 0 One of the ends of the extremal trajectory has to be a point on the boundary y = y y p, where constant functions are the only test functions corresponding 3 1 2 | − | to these points. Denote this point by U = (y ,u,(y u)p). Note that we write 1 1 − (y u)p instead of y up because the domain Ξ is under consideration. For 1 1 + − | − | the second end of the extremal trajectory we have four possibilities 1) it belongs to the same boundary y = (y y )p; 3 1 2 − 2) it is at infinity (y ,y ,+ ), i.e., the extremal lines goes parallel to the 1 2 ∞ y -axis; 3 3) it belongs to the boundary y = y ; 2 1 4) it belongs to the boundary y = y . 2 1 − The first possibility gives us no solution. Namely, we have the following Proposition. If p = 2, then the function B cannot be equal to B(x) = M(y), max 6 where M is the solution of the Monge–Ampère equation (15) such that one of its extremal trajectory is of type 1) above. The same claim holds for B . min Proof. To check this it is sufficient to verify that the test functions of the type α+βh (t) cannot be an extremal function of our problem with the only exception I of p = 2, when the situation is trivial: B (x) = B (x) = x +x2 x2, and any max min 3 2− 1 pair of test function is extremal. We will show that the Bellman functions being solution of the homogeneous Monge-Ampère equation cannot be linear on a chord [x−,x+] connecting two points x± on the boundary ∂Ω, i.e. such a chord cannot be an extremal trajectory of our Monge-Ampère equation. We assume now that two points x± Ω such that x+ x− = x+ x− ∈ + | 1 − 1| | 2 − 2| are on the boundary x± = (x±)p and x = 1(x+ + x−). We need to show that 3 1 2 1((x+)p +(x−)p) can be the value neither of B (x) nor of B (x). 2 2 2 max min Without lost of generality we may assume that x+ > x−. Let us denote a := 1 1 1(x+ x−) then 2 1 − 1 x± = x a x± = x σa, 1 1± 2 2± BURKHOLDER’S FUNCTION 9 where σ = 1 depending on the direction of our chord: it can be either in the ± plane x x = const (and then σ = 1) or in the plane x +x = const (and then 1 2 1 2 − σ = 1). The pair of the test functions f,g on I = [0,1] that gives the value − 1 1 A(a) =: g p = x+ p+ x− p = x +a p+ x a p h| | iI 2 | 2| | 2| 2 | 2 | | 2− | (cid:0) (cid:1) (cid:0) (cid:1) is f = x +ah g = x +σah . 1 I 2 I Firstofallweassumethatx± Ω ,butx± = 0i.e. x a > 0, x a > 0, because ∈ + i 6 1± 2± if one of x± is zero we are in the cases either 3) or 4) listed before Proposition. i Our aim will be to find another pair of test functions f˜,g˜ corresponding to the same point x, but with g˜p either bigger than A(a) (and then A(a) cannot be h| | iI the value of B (x)) or less than A(a) (and then A(a) cannot be the value of max B (x)). min Let as make here two remarks. First, we see that the expression A(a) does not depend on the direction σ. Therefore, the construction of the desired f˜,g˜ ensures us that the point x cannot be the center of an extremal trajectory with two ends on ∂Ω in any direction σ = 1. Secondly, we note that it is not obligatory to look ± for f˜,g˜ for all a (0,max x ,x ), it sufficient to do this for small values of a/x . 1 2 2 ∈ { } Indeed, suppose that the chord L = [x−,x+] represents an extremal trajectory of the corresponding Monge–Ampère equation. Let us consider the “crescent” betweenthechordLandtheboundary∂Ω. Itshouldbefilledinbychordsonwhich B(x) are linear (this is the property of the solutions of the homogeneous Monge– AmpèreequationexpressedinPogorelov’stheorem,see[Pog]). Amongthesechords we can take one (say L˜ = [x˜−,x˜+]) of arbitrarily small length (arbitrarily small valueofa˜/x˜ ). Therefore, wecan workwithanewpointx˜ andnewchordL˜: would 2 we show that L˜ cannot be an extremal trajectory, the chord L could not be one either. To construct the desired pair f˜,g˜ we use the following family of functions φ s equal to 1 on [0,1/2 s] [1 s,1] and to 1 on (1/2 s,1 s). Let us note that − ∪ − − − − all φ have the same distribution function as h , and s I (φ ,h ) = 1 4s. s I − We are interested in φ := φ . Then 1/8 1 (φ,h ) = , I 2 10 VASILYVASYUNINANDALEXANDERVOLBERG and ψ = φ h has − I 1 (ψ,h )= , (ψ,h )= (φ,h ) I −2 J J for all other dyadic J. So ψ is a martingale transform of φ (it is equal to 0 on [0,3/8] [1/2,7/8] and to 2 on two intervals (7/8,1), (3/8,1/2)) and we can ∪ ± examine the pair f˜= x +aφ, g˜= x +aψ. 1 2 Since fãnd f have the same distribution function, we have f˜p = f p = x , h| | iI h| | iI 3 i.e., f˜,g˜ is a pair of test functions corresponding to the same point x. To investigate the difference g˜p g p we use the function h| | iI −h| | iI 1 3 λ (α) := (1+2α)p +(1 2α)p + (1+α)p +(1 α)p . p 8 − 4 − − Since (cid:0) (cid:1) (cid:0) (cid:1) 1 λ (α) = p(p 1)(p 2)(p 3)α4 +O(α6), p 8 − − − we have λ (α) > 0 for small α if 1 < p < 2 or p > 3 and λ (α) < 0 for small α if p p 2 < p < 3. Recall that x 0< t < 3 2 8 x 2a 3 < t < 1 g˜(t) =  2− 8 2   x2 21 < t < 78 x +2a 7 < t < 1,  2 8   therefore, h|g˜|piI = 18 (x2+2a)p +(x2−2a)p + 43xp2 and a (cid:0) g˜p g p =(cid:1)xpλ . h| | iI −h| | iI 2 x 2 (cid:0) (cid:1) For small α we have a desired example for B if p (1,2) (3, ) (because max ∈ ∪ ∞ λ > 0 and for B if p (2,3) (because λ < 0). p min p ∈ Now we interchange in a sense the roles of fãnd g˜: instead of a we take a new parameter, say a˜, and put f˜= x +a˜ψ, g˜= x +a˜φ. 1 2 Since we have now a˜ f˜p f p = xpλ h| | iI −h| | iI 2 x 1 and the function (cid:0) (cid:1) 1 3 t x+tψ p = (x 2t p+ x+2t p)+ x p 7→ h| | iI 8 | − | | | 4| |