{ESES 1. 200;? ii 5%? (00 W7 7r LERARY Michigan State . University This is to certify that the dissertation entitled Riesz Transforms and the Bellman Function Technique Doctoral presented by Oliver Dragiéevié has been accepted towards fulfillment of the requirements for the degree in Mathematics W ’ Major Professor’s Signature / March 14, 2003 Date MSU is an Affirmative ActiorVEqual Opportunity Institution PLACE IN RETURN BOX to remove this checkout from your record. TO AVOID FINES return on or before date due. MAY BE RECALLED with earlier due date if requested. DATE DUE DATE DUE DATE DUE 6/01 c'JClFlCIDateOuepGSop. 1 5 RIESZ TRANSFORMS AND THE BELLMAN FUNCTION TECHNIQUE By Oliver Dragiéevié A DISSERTATION Submitted to Michigan State University in partial fulfillment of the requirements for the degree of DOCTOR OF PHILOSOPHY Department of Mathematics 2003 ABSTRACT RIESZ TRANSFORMS AND THE BELLMAN FUNCTION TECHNIQUE By Oliver Dragiéevié In the first part we use the technique of averaging to give a sharp weighted esti- mate for the operator of convolution with 2’2 (the Ahlfors—Beurling operator) for an arbitrary planar A2 weight. As an application we touch upon the theory of quasicon- formal mappings on C. The Ahlfors—Beurling operator can be realized as a combination of second-order Riesz transforms. We prove a new Littlewood-Paley-type inequality which is the key to our results in the second part. As one of its consequences, we show that the scalar Riesz transforms and their vector analogues admit dimension free upper estimates of the norms when acting on LP(1R") for arbitrary dimension n and p > 1. The essence of the proof is the utilization of the method of Bellman functions, which, requiring but few assumptions, appears to allow its application to other kinds of Riesz transforms. We present the proof of one such example - dimensionless boundedness of Riesz transforms on Gaussian spaces. Acknowledgements First and foremost, I wish to express my deepest gratitude to my advisor, Professor Alexander Volberg. Not only has he introduced me to my area of research, his attitude towards me has all this time exceeded mere professional courtesy. I feel very honoured to have had him as my teacher. He is one of the few true mathematical and moral authorities I know. My undergraduate advisor, Professor Josip Globevnik of the University of Ljub- ljana, has strongly influenced the choice of my mathematical path and extended his advice and help without hesitation when they were most needed. My family has provided me with unconditional support from the very beginning. I am grateful to have them. I would particularly like to thank Miss Katarina Slapniéar for the most excellent and courageous steering of the boat. I cannot forget numerous occasions on which my friend Bogdan Suceava has helped me, including those when I was not aware I needed help. Nor can I skip the trust- worthy Onur Aglrseven. Prof. Fedor Nazarov showed me what a superb mathematician should be like. His extraordinary classes as well as he himself have had a profoundly strong impact on me. Dr. William Brown, our former Director of Graduate Studies, made the right move at the right time, for which I am thankful. Finally, I am indebted to Sgt. Major (ret.) R.P. Brazee, who was always there. iii Contents 1 Introduction 2 Sharp estimate of the Ahlfors—Beurling transform 2.1 Motivation ................................. 2.2 The main idea ............................... 2.3 The averaging ............................... 2.4 The main calculation ........................... 2.5 Sharpness ................................. 3 Riesz transforms via Bellman functions 3.0.1 New findings ........................... 3.0.2 Further notation and preliminaries ............... 3.1 Bilinear dimensionless Littlewood-Paley type estimate ......... 3.1.1 Bellman function appears .................... 3.2 Estimates for the classical Riesz transforms ............... 3.3 Estimates for the Ornstein-Uhlenbeck semigroup (Gaussian case) . . . 4 Addendum Bibliography iv’ 5 9 11 12 28 33 37 39 40 41 41 49 51 68 72 Chapter 1 Introduction Recall that for a function f E LICK") its Fourier transform is given by flit) = (270“;1 W f (y) 6““> dy. We used the notation (x,y) = mm + + xnyn, where a: = (1:1,...,:z:,,) and y = (y1,,y,,) belong to R". Choose 16 E {1, ,n}. The scalar Riesz transform Rk on LP 2 LP(R") is defined on a test function f (say, belonging to Cf,” or the Schwartz class 8) by (ka)‘(x) = zfiflx). For arbitrary functions f 6 LP we extend this by density. It is our aim to show that these operators are bounded on LP(IR") for p 6 (1, 00) with norms independent of the dimension n. Moreover, the same is true for the vector-valued Operator (R1, ..., R") The constant that appears in all of these esti- mates is 2(p — 1) if p 2 2 and 2/ (p — 1) otherwise. Actually, we will derive this as a consequence of a considerably more general result (Theorem 2), which displays a certain Littlewood-Paley type inequality involving the operator of the harmonic extension onto the upper half-space. Its significance might also lie in that it can apparently serve as a model for similar results in various settings. We examine the case of Gaussian spaces (section 3.3) and prove an analogous result, Theorem 5. The problem of establishing bounds on Riesz transforms on LP has a rather rich history. S.K. Pichorides [24] found the exact norm in one-dimensional case (Hilbert transform). In the same work he gave dimensionless upper estimates on the norms of scalar Riesz transforms in arbitrary dimension. Later on T. Iwaniec and C. Martin [13] proved that his estimates were sharp, thus actually computing the norms. The fact that the norms of vector Riesz transforms can be bounded with estimates independent of the dimension n was first observed by E. Stein [27]. There were quite a few other proofs following, e.g. by G. Pisier [25], T. Iwaniec and G. Martin [13], and R. Bafiuelos and G. Wang [5], some of them improving the constants. Probabilistic methods, applied by RA. Meyer [18], were used to obtain the initial proof of this theorem in the Gaussian setting, whereas in G. Pisier was the first to have found an analytic proof [25]. The question of estimating Riesz transforms in various (and sometimes very diffi- cult) other situations has generated a lot of attention. We refer here to the results of T. Coulhon, D. Miiller and J. Zienkiewicz [8] as well as T. Coulhon and X. T. Duong [7], where Riesz transforms on rather general manifolds are considered. The work of F. Lust-Piquard [16] and [17] regards certain discrete analogues of Riesz transforms and Riesz transforms on Fock spaces, respectively. Our approach is different than any previous, in that it uses the technique of Bellman functions. The Bellman function approach can be viewed as an application of certain ideas 2 from Bellman PDE in stochastic optimal control to harmonic analysis. By its nature it is very well suited to give dimension free estimates. It also usually gives estimates close to the sharp ones. However, in our application to Riesz transforms one uses the Bellman function in a new way, which allows self-improvement of the estimates. The reader will observe that our approach in more general cases reduces the proof to the question of boundedness of certain spectral multipliers on D". However this appears to be the only true obstacle that we are to encounter. For this reason we believe that our technique is transformable within several other settings. The scope of generality will be the subject of further consideration. We hope that the properties of the Bellman function could also be utilized in a way to obtain dimension free estimates of the weak type 1-1. So far this has eluded us. Although the dimensionless boundedness of the Riesz transform has been proven several times by now, the following result gives, besides a new proof, in some cases also the best known estimate on the LP-norm of R. Another essential object of this treatise will be the Ahlfors-Beurlz’ng operator, which will be denoted by T. It is defined as the convolution with the kernel 2‘2 on the complex plane, understood in the Cauchy principal value sense. The relation to the Riesz transforms is hidden in the equality TzRf—Rg—h’RIRg, where R1 and R2 are planar Riesz transforms. Having in mind previously announced results on R.,-, we are immediately able to give the estimate for T on LP(R2). However, in this work we are interested in a different, though related kind of estimates. Namely, we prove that llTllB(L2(w)) S C ° Qw,2 (1.0.1) for some absolute constant C and arbitrary A2 weight w on R2. This result is sharp. Suppose that the function f = u + iv : (I —> C is sufficiently smooth (say, C1) in the region Q C C. Denote by J f its Jacobi determinant um uy v3 1),, It is trivial to verify that Jf = [Bf]2 -— |3f|2. When Jf > 0 in 9, one can show that f preserves the orientation at every 2 E (2. In that case 5f It i: 8—f is a complex function with [a] < 1 everywhere in Q. If Halloo = k < 1, then f is referred to as a K -quasz'foncormal mapping, where The point here is that, as it turns out, the estimate (1.0.1) carries important consequences in the theory of K -quasiconformal mappings on the extended complex plane. It is possible to relate this result to the questions of regularity of solutions of the Beltrami equation. Further definitions, proofs as well as the explanation of the function-theoretical background of this problem await us in the next chapter. Chapter 2 Sharp estimate of the Ahlfors—Beurling transform Our most important object in this chapter will be the so-called two dimensional martingale transform. In order to define it properly we should start with the notion of a Haar basis. We call the family L := {[m 2", (m+1) 2"] ; m, n E Z} the standard dyadic lattice. Observe that O is the only real number that is not contained in the interior of any dyadic interval, that is, any member of L. Each interval I C IR gives rise to its Haar function h], defined by hli=l1l“1/2(XI+ — XL), where I __ and 1+ denote the left and the right half of the interval I respectively, and XE stands for the characteristic function of the set E, as usual. It is a well-known fact that the set {hI ; I E L} forms an orthonormal basis of the space L2(R). At this point we should emphasize that our attention will be concentrated on the planar case. To that purpose we need to introduce a similar basis for the space L2(IR2). This will be described in detail in the continuation of this preface. Now we are able to define the operator T, on L2(R) by Ta’f :: 20(1) {—1, 1} is arbitrary. Such operators are called martingale transforms. Observe that T, is an isometry satisfying T3 = I . The symbol (f) 1 shall stand for III"1 f, f dm, the average of the function f over the interval I. We say that a measurable function 111 : IR -—) R+ satisfies the dyadic A2 condition if Qw,2 1: S111) (w)1(w—1)1 < 00. let. It is well known that for any such weight w the martingale transforms are uniformly bounded on L2(w), i.e. the Hilbert space of measurable complex functions on R, endowed with the scalar product (f,g)w := Af($)g—($—)w(x)dzr. More precisely, in [28] it was shown that there is a constant C' > 0 such that for arbitrary choice of functions w 6 A2, a : L —> {—1, 1} and f E L2(w) we have the condition llTafllL2(w) S CQw,2 llfllL2(w) ' (2-0’1) The estimate (2.0.1) is sharp in the sense that one cannot replace CQw,2 in it by ¢(Qw,2), where d grows slower than a linear function. Our interest will focus on studying the case of R2 instead of IR, and a certain important singular integral Operator on R2 will play the role of To. In the planar 6 case, all the definitions simply proceed from the one dimensional case in a natural way. Thus the term dyadic lattice will now stand for the collection of all squares of the form I x J C R2, where I and J are dyadic intervals of the same length. To each such square Q = I x J we will assign three Haar functions: has») == has) maul-1”, ham) == III‘l/zxds) hJ(t) and hQ(8, t) :2 h1(8)hy(t) . As previously, one can verify that the set {hi ,hé, hQ ; Q E L} builds an orthonormal basis in L2(R2). Now the two-dimensional martingale transform becomes the Operator Taf :2 Z 01(Q) ha + Z UJ(Q) hQa QEL QeL Qer. where, as before, 01,01, 0 : L —> {—1,1} and f E L2(R2) are arbitrary functions. The term (two-dimensional) A2 weight now stands for a positive measurable func- tion it) on 1R2, such that Qw,2 3: SUP Q L2(w) of convolu- tion with the kernel 272, i.e. _ f($—U,y—U) Tum—[m (“+2.”), dudv. Here w is a planar A2 weight, of course. Operator T, sometimes multiplied by 1/7r, is called the Ahlfors-Beurling operator. Our main result of the chapter will be Theorem. T is in the weakly closed linear span of operators of the type To. This gives us an immediate Corollary. HTHLP(w)—+Lp(w) S 009)er It is enough to prove the Corollary for p = 2 (see [23]). This corollary was the main goal of [23], where it has been proved by different methods. It seems to us that the present proof is much more streamlined and maybe more conceptual. The theorem looks slightly unexpected because the same result would not be true for T replaced by the first order Riesz transforms on the plane. In fact, all our Operators T, have symmetric kernels k0, meaning that ka(:r,y) = ka(y,x), but the first order Riesz transforms have antisymmetric kernels. This is why in [22] a completely different set of dyadic singular operators has been used to represent Riesz transforms. The Operators from [22] are slightly complicated. We already explained that in representing first order Riesz transforms one cannot average our 8 simpler operators Ta. But we do not know whether one can average something a) simple as T, b) antisymmetric in order to obtain the first order Riesz transforms. 2.1 Motivation Consider the standard differential operators and 3—12—+i§- —2 6m 0y ° The regularity of solutions of Beltrami equation 3f=a8f (2.1.1) has received a lot Of attention from mathematicians since the 40’s. The function u, called the Beltrami coefiicient, belongs to the space L°°(C). Its norm k is strictly smaller than one. The result of Bojarski, Ahlfors and Bers, Lavren- tiev states that there is a solution to (2.1.1) which belongs to the Sobolev class W12 and is a global homeomorphism of the extended complex plain C. For 2 E C, it maps the infinitely small ellipse, centered at z and with ratio of the axes Egg-)1], into some infinitely small circle, centered at f (.2). For this reason it turns out to be important to consider the constant Every homeomorphism f : C —> C, belonging to W12, for which (2.1.1) is fulfilled, 2 is called a K - quasiconformal mapping. Any local solution of (2.1.1) from W1,loc is called a quasiregular mapping. Denoting g 2 3f, we are able to write f (such that f(z) ~ 2 + C, [z] —+ 00), as 1 9(4) = — dA . f(2) ,/<_z (<)+z+c This is where our operator T begins to play its role. For 8 f = Tg + 1 and so 9 = We + u. or (I-MT)9=M- Since [luTllean g HHHooHTllm—mz g k < 1, the operator (I — ,uT)’l exists and g = (I - MT)“1u. It has been shown that the norm of pT as an operator on LI, is still less than 1 if p is slightly greater than 2. On the other hand, the word “slightly” is rather important. More precisely, it is known [12] that p should not exceed 1+ It“. This fact, combined with the awareness that [|T||L2_,L2 = 1, gives rise to the assumption that llTllLP—iLr = P — 1- For in that case, lluTIILMLv S llulloollTHLMLr < k(1+ k"1-1) =1- 10 This is still an open question. The best known estimate has been obtained by F. Nazarov and A. Volberg in [20]. They proved that [ITHLPELp S 2(p — 1). This improves the previous estimate in [5] (namely, with 4(p — 1)). Recently, the estimate 2(p — 1) was Obtained in [4] by using methods, different from those in [20]. It is relevant that we also study weighted LP spaces. For it was shown in [2] that ||(I - HT)_lllLP—>LP S C(k)llT||Lp(w)_.Lp(w). where w = If, 0 f‘lll"2 for p E (1 + k,1 + k‘l) and f is a quasiconformal home- omorphism, satisfying (2.1.1) together with the normalization f (z) = z + 0(1) as [2] —+ 00. This estimate, together with the Corollary on the page 8, gives the linear growth of [[(I — ,LtT)—1”LP—>Lp, where p ——> 1 + k. One knows (see [2]) that this implies that weakly quasiregular maps on the plane are quasiregular. Thus this geometric fact becomes the corollary of our main theorem about the representation of the Ahlfors- Beurling transform as a closure of the linear span of martingale transforms T, and the correct weighted estimates of To, obtained by Wittwer [28]. 2.2 The main idea As we have announced, we will represent our T as the result of averaging of operators, similar to To. After that, the desired estimate of ||T||L2(,,,)_,L2(w) will follow from the two—dimensional version of (2.0.1) for ||Ta||L2(w)_,L2(w). 11 2.3 The averaging Instead of a dyadic lattice let us for a moment consider a grid 9 of squares. This is a family of squares of the form I x J, where I and J are dyadic intervals of unit length. Furthermore, for t E R2 define S, 2: 9 + t, i.e. the grid of unit squares such that one (in fact, four) of them contain point t as one of their vertices. Now we are ready to introduce our “core” Operators ll”, : L2(w) -—> L2(w) by at := Z l (f. halt. — (f, hint]. QES: Since (f, hhlhh can be written as (1/2) [ = // f(31,32)h£2(31,32)d31d32 = f/f(31,32)h1(31)XJ(32)dSI €132 R R J I and similarly for (f, hilal- Thus for (fixed) f E L2(R2) and a: = ($1,332) 6 R2 we have (lanes) = 2: [0.115% - (f, haht Mar) = QES: : Z [ALf(31v32)hl(31)X1(82)d81d82 ham) Q69: _ [mfgf(31a32)XI(31)hJ(52)d81 €132 -hé($)] = 13 : [Ii/R f(sl,sz) Z [h1(sl)xJ(82)h£?($) — Xr(81)hJ(82)hf2($)l d31d32- QESt The expression under the summation sign in the last row is nonzero for exactly one Q E 9;; namely such that hIQ(CC) ¢ 0 7Q ham). This means that a: = ($1,352) 6 Q, therefore 331 E I and 3:2 6 J. Thus h’Q(2:) = h,(:r1) and ham) 2 hJ(;r2). We get (norm): =[Ht/Rnwa[menxksaman—xi1dsldsz. (2.3.1) Since 9, does not change if we increase or decrease any component of t by 1, we may assume that I = (t1 — 1,t1) and J = (t2 —1,t2). By denoting I0 = [—1/2,1/2], this implies 1 1 I=t1—§+Io and J=t2—§+Io. Now let X0 and ho be as in the formulation Of the PrOposition and let [to :2 —ho. The fact that Io is symmetric with respect to 0 yields the equalities Xr(2) = Xo(t1 -1/2 - Z) h1(z) = ko(t1—1/2 — z) for all z 6 IR. The analogous pair is valid also for J, of course. The point here is that our goal was to modify the expressions on the left to look more like a part of a convolution integral with t1 and t2 as integration variables and 2 as a center of convolving. 14 The last two equalities imply, together with (2.3.1), ()(lPthE //f 81,82 )(lkO tttl—1/2-31)X0(2—1/2—82)k0(1—1/2—331) — X0(t1—1/2 — 81)k0(t2 — 1/2 — 82)k0(t2 — 1/2 — $2)]d81 €182 . Recall that $16 I = (t1 — 1,t1) and $2 6 J = (t2 —1,t2). Hence x, < t,- < :13,- + 1 for i = 1, 2. Averaging in our case means integrating over all admissible ti. Therefore 12+1 n+1 ()(513EPf):/2/(P(t1,)(t2)fx(x)dtldt2- By using the most recent expression for (Pt f )(x) and changing variables (to t,- — 1 / 2) we get x2+1/2 $1+1/2 ()(lEleIE )Z/ / // f(81,82)[A] (1816182 dtl dtg, 112-1/2 11- 1/2 IR IR where A = k0(t1— 5171) k0(t1 - 31) X0(t2 ‘ 32) — ko(t2 - $2) k0(t2 - 32)Xo(t1— 81)- Applying Fubini’s theorem we get 22+1/2 x1+1/2 (lEle)(a:1,:z:2)=/R/l;f(sl,sg)/ / [A]dt1dt2 dsld32. (2.3.2) 2—1/2 1—1/2 This is how we Obtained the candidate for the convolution kernel F of the operator IEIP’. Namely, line (2.3.2) gives us the relation $2+l/2 11+1/2 F(.’131 — 31,232 — 32) =/ / [A] dtl dtg. 2—1/2 1-1/2 15 Taking 31 = 32 = 0 gives F($1,$2) = x2+1/2 11+l/2 2/ / [k0(t1— $1) ko(t1)X0(t2) — (€002 — $2) 19002) Xolt 1)] dli1 dt? 2—1/2 1-1/2 We are able to separate variables t1 and t2, so F(.T1,.’272) = xl+1/2 1:2+1/2 =/ l60(751— $1)k0(t1)dt1 / X0052) dt2 1—1/2 32-1/2 xi+l/2 32+1/2 — / X0(t1)dt1 '/ k0(t2 — $2) k0(t2) dtg . x1—1/2 12—1/2 Observing that, for i = 1, 2, xi+l/2 :L‘i-l-l/Z / X0(ti) dti = / Xo(ti)Xo($i - ti) dti :1: I g-1/2 i‘1/2 and that ho * k0 = ho * ho, we finally get 17(31: $2) = (ho * ho)($1) (X0 * Xo)($2) — (X0 * Xo)($1) (ho * ho)(5132), as desired. [:1 The graphs of the functions a and 5 can be found on figures 2.1 and 2.2. Notice, as a corollary, that (2.0.1) also holds for the operator IEIP in place of To, since it held for all lP’t’s. Instead of the unit grid we may consider a grid of squares with sides of an arbitrary length p > 0. Denote such a grid by 9‘: if t 6 R2 is a vertex of one of its members. 16 {P NIH — — — up N'H Figure 2.1: Graph of oz Figure 2.2: Graph of fl 17 Henceforth we will call p the size of the grid and t its reference point. We obtain another family of Operators, defined by Pff == Z W. have — (f, baht]. Q69? Remark. In order to clarify some basic properties of these Operators, we present the following observations (and omit the easy proofs). For t E R2, p > 0 and any function f on R2 define Sf(:r) = f(p:1: + t). If w is any weight, then S maps L2(w) —> L2(Sw). We also have the identity PfT—S-lllpos. More precisely, Pf [L2(w) is the composition of Operators 2 S 2 IPo 2 S“ 2 L (w) —>L (Sw) —+L (Sw)——>L (w). Since ||Sf||L2(5w) = fillfllpw), it follows that ||S||L2(w)_,L2(5w) = I}? similarly llS—lllL2(Sw)—+L2(w) = \/.5- If w 6 A2, then also Sw 6 A2 and 623,2 2 ng. Hence 1% : L2(Sw) -—> L2(Sw) is a bounded operator, which inherits the estimate for its norm from T,. That is, it satisfies the same inequality as T, does in (2.0.1), according to [28]. These facts combined tell us that every Pf is a bounded operator on L2(w) with ||Pf||L2(,,,)_,L2(w) _<_ C ng, where C is an absolute constant, as usual. Let us use the same procedure of averaging as it has been done earlier for E, but now for Pf. We get lElP”. Applying a similar proof as for Proposition 1, one obtains 18 Proposition 2. Choose p > 0. Averaging the operators Pf over the set (2" := R2 / (pZz) returns a convolution operator with the kernel Fp(.’L‘1,CL'2) I: le-F (Si—1,?) . Here the set (2” is endowed with the normalized planar Lebesgue measure. Thus we have found the kernel Of the operator, resulting from averaging over all grids of a fixed size. Our next step will be to average over all sizes. Let us explain what we mean by that. Take r > O. A lattice of calibre r is said to be a family of intervals (squares), Obtained from the standard dyadic lattice L by dilating it by the factor r and trans- lating by an arbitrary vector t. In other words, such a lattice, call it Lf, is the union of grids of sizes r - 2" , n E Z, having t as their reference point. It is clear that the set of all possible calibres naturally corresponds to the interval [1, 2). For our purpose, the most appropriate measure on this interval turns out to be dr / r. This makes all other possible choices of intervals, e.g. [2",2"+1), have the same measure (log 2). We introduce kernels kr :2 f: Fr'zn. n=—oo This sum is well defined, because as a function it equals, by the last assertion, to °° 1 - 1 °° 1 - "ngFimnl = aged...)- The function F is bounded, thus the series converges absolutely on IR2 and uniformly outside any neighbourhood of the origin. It is easy to see that it converges in the 19 sense of distributions to k’", understood as a distribution in the following sense. Let p be a test function from Schwartz class 8, :1: = (331,152), then (Q5, (Cr) = lim ¢(T)k'(:r)d:r. 5"“ |x|>e We would like to show that k’ defines a bounded convolution Operator, more precisely, that it is a strong limit of its partial sums. The fact that k”* is a sum of operators, obtained by averaging over grids of size r . 2", hints at k’* itself being an average, this time over unions of these grids, i.e. lattices of calibre r. Here we present what exactly we have in mind by that. For M E Z let the M -th partial sum of the series k’ be M k7,, 2: Z Fr'2". n=—oo Lemma 1. The function k3,, defines a bounded convolution operator on L2(w). The limit k'* := limMnoo khan: exists in strong sense and also gives rise to a bounded operator on L2(w). Proof. For the sake of simplicity we will assume that r = 1; the proof does not change at all for general r. We start with a formal definition M M 1 k]M*f———(Z F2") *fz Z 4—n anfdt (2.3.3) nz—oo nz—oo [0’2n12 by Proposition 2. At this point we will need the following observation: For any n, M E Z, n g M, we have the relation 1 .. 1 n — P2fdt=— 1P2fdt. 4n [0,2nl2 t 4M [0‘2M12 t 20 This time we simplify the proof by taking M = 0. Then the square [0,1]2 consists of exactly 4‘” dyadic squares of size 2". The integral f P?" f dt over each of them equals to the integral over [0, 2"]2, because of the invariance of the measure on (22" with respect to the map on £22" E R2/2"Z2, induced by the shift on R2. Since the sum of integrals over these squares equals to the integral over their union, which is [0, 1]2, we have proved this part of the statement. This enables us to rewrite the line (2.3.3) as Z g—M / rim, 0 .5... W which equals to 74-41 :2}... r?" f dt. 0 2MP" Let us explain why we were able exchange the order of integration and summation in the last step. Again we do it for M = 0 only. We need to show that the L2(w) norm of the difference [[01]an rznfdt— gNLPP2"fdt_ f0 sznfdt __ [’21] n<— —N is small if N E N 18 large. Lemma 2. For every function f E L2(w), w 6 A2, and every t E R2 we have Ali—E.” Z Pinfllmw) = 0- lfllZN 21 Proof. A collection {en ; n E Z} of vectors in a Hilbert space 1% is said to give rise to a Riesz basis of 1H if there are M, m > 0 such that for any sequence (An)n€Z of scalars we have a double-sided inequality m: ”Arlen“2 S ll ZAnean S M: llAnenllz- neZ nEZ nEZ It is a known fact that w 6 A2 4:) Hoar functions form a Riesz basis in L2(w). M For every N E N let us define a measurable function gN: R2 ——+ C by sin/(t) = || 2 1P’fnflltz’rwr IRIZN We claim that, for any t E R2, Ilim gN(t) = 0. (2.3.4) In order to prove this statement choose t 6 R2 and consider the translation Operator 5;. Then 2 1P?“ = 3-. Z it"s, IanN IHIZN as we have seen in the Remark on page 18. Thus 9N(t) = N Z 1P’3"(Stflllr.2(s.w) S N Z Pgnlleuusm» llStfllL2(s.w)- IRIZN lfllZN We can estimate the first term on the right by using (2.0.1) to Obtain that it is less or equal to C - Qsmg. Referring to the above-mentioned Remark one more time we 22 get Qstwg = 62.”, so we conclude that g~(t) S C - Qw,2llfllL2(w) (2-3-5) for all t E R2 and all N E N. Taking N = O, the line (*) implies that Z Z (Hangman...) + nominate...) < oo. "61 069?" so - I I J J 2 _ [3530 Z Z (llthlIfl(w)+llOO [0,ll2 [THEN We can do the same for Zn<_N instead of Z|n|>N; and so we justified reversing the order of integration and summation on page 21. So we were indeed right to claim 23 that kM1f=4Mfo Zoor2"fdt. 2Ml2n _ The integrands all satisfy (the two-dimensional version of) the inequality (2.0.1), so we may conclude that k11M=1= does, too. This proves the first part Of Lemma 1. Now take f E L2(w). Let us verify that Us], :1: f ; M E N} form a Cauchy sequence in L2(w). Choose M, N E N, N g M and compute the difference N 1)2" 2" _ (1%,,— k, =4M/02M12mzmlp fdt— —N4/[()2N]2 2 1e, fdt _. "Z—W M 1 .. . —— 1’2 dt—— 1/ 1P2 dt= M [[0 M1,”; or 4—.—. ...,-2 2; f ——1—/ Z Pznfdt. M [0 t M ’2 l2 n=N+l This difference is small if M and N are sufficiently large, which basically follows from Lemma 2, as well. On the other hand, choosing f from the Schwartz class 8 we can use the fact that k], converges to k1 in the sense of distributions, and hence we can conclude that k], =1: f —) k1 =1: f pointwisely. Here k1 * f is understood as the convolution of a distribution and a test function. In particular, for f E S we have 1 __ 2n k .1 f: 1‘].an 4M1/0 2M1? 2201», fdt (2.3.6) and similarly for all other r E [1, 2). Reasoning 1n the same way as before we see that (2.0.1) is fulfilled with k1 in place of T,. This finishes the proof of Lemma 1. Cl 24 Remark. TO understand the distribution It1 better, one can notice that its Fourier transform is a bounded function. In fact, using the previous reasoning with w = 1, we conclude that f +—> k1 =1: f is a bounded operator on L2(R2). Note that the integrand in (2.3.6) is the sum 231 (f. hing - (f, hint], Q where Q runs over a “truncated lattice”, i.e. the union of grids of sizes 2", —00 < n S M, and with reference points in t. Since the square [0,21‘4]2 represents all possible reference points for such unions, the expression under the limit sign in (2.3.6) means exactly averaging over these unions. But when M —> 00, the “truncated” lattice becomes “complete”. Thus we may understand the limit (2.3.6), more generally, the convolution with k’, as the result of averaging over all lattices of calibre r. Since our operators over lattices are bounded in the sense of (2.0.1), the same is true of k"*. Averaging Operators k’*, i.e. integrating k’ with respect to dr/r (again in the strong sense, that is on any fixed test function f), gives us again a convolution operator. Call its kernel k. Then Mac)=/12k"(=v)i;:=/l2 i Fr'2"(x)%:= i [IQFT'2”(2)9T3= nz—oo nz—oo 8 = 23 .. F3i€. (2.3.7) flz—OO Since the integral ff dr/r is finite and the estimate (2.0.1) holds for all k’*’s , it also does for kak. Thus we have been able to represent the operator k* as a result of averaging of our “brick” Operators Pf over lattices of all calibres. 25 Observe that there is a map m : S1 —+ C such that m) = lxlzl (2.3.8) for all a: E R2. By taking unimodular vectors we see that in that case m = My. For the existence of such m it suffices to show that function [2:]2k(a:) depends only on the direction of 2:. So take :1: = (231,232) 6 IR2 and )1 > 0. We use Proposition 2 to compute k(/\x)=/Ooo( F"()\x)— H/oooF( 175;?) = =/.°°i (52—1) (v eveni- Clearly, now we have to introduce a new variable a := p//\. We get 1 °° 2:1 $2 1:1 2:2 du _ 1 W) 72/. l“ (I) f (I) ‘ ‘3 (7:) 0' (Ill a — 23"”), which is what we wanted. Remark. Observe that k is an even function, because so are a and ,6. Therefore m is even on S1 as well. We are able to compute the function m. Namely, the last row above implies that m(e“") = M(cos cp, sin go) - M(sin (p, cos (p) = M(ei‘p) — M(e‘("/2“p)). A computation shows that M (cos (,0, sin 90) = (cot go) , 08290 where 26 a2/6—a3/4 ; 0(a)= a3/12—a2/6-1/(48a)+1/12 ; I/ZSagl, 1/(16a)—1/12 ; a2 1. For a < 0 the function is defined by the requirement that it be even. Finally, let \Il(a) = (1 + a“2)(a). We have acquired the formula m(e“”) = \I'(cotc,o) — \I’(tan (,0) . Clearly, the function m is continuous on the sphere S 1, thus k is continuous on lR2\{0}. Recall that we are aiming at the operator T, which is the convolution with 1/22. Its kernel can be written, in polar coordinates, as e‘2i‘P/r2. Comparing this to line (2.3.8), where k(re“p) = m(e“")/r2, we suspect that it would be useful if we could find a suitable way of transforming m(e“”) into the function 62”. Denoting h(() = (‘2 for C 63 SI, we get (m =1: h)(e““) : j“ m(ei“’)e‘2i(“’_”) dd) 2 = e"2“" f“ m(e”f’)e2‘” dip = Me”) [n m(e“”) [cos 2w + isin 2w] dw. —1|' Since m is an even function on [—7r,7r] (because a and ,8 are even on IR), the integral ff" m(e“”) sin 21,!) dw equals to zero. Therefore our expression simplifies to h(e“") l” m(e‘“’) cos 2w dip. Hence h = , (2.3.9) where 27 c = f” m(e“”) cos 21/2 dd). 2.4 The main calculation We come to the main point of this chapter. It needs to be verified that c 75 0. If this is so (and we will see that it is), then (2.3.9) represents our singular operator T as the “average” of martingale transforms T, (actually their analogues, built with the help of Pf’s). Recall from (2.3.7) and (2.3.8) that Hence 00 Zn i112 c = f f F (8—) 6082” dz/J dp. o o l0 P3 Taking r = 1 / p we get 00 21r c = / / F(rei“’) r cos 2w dw dr. 0 o In cartesian coordinates, this integral reads 172—312 In Proposition 1 we saw that F (x. y) = a($)fl(y) - fi(x)a(y). which leads to expressing our integral as 28 y2 c=2//l;a(:r) ):2+y _y.2d:rdy Since a and 6 are even functions, supported on the interval [—1, 1], we obtain 0: 8/0/a(:1:) y):2+y2 2dxdy. We are going to evaluate c by computing the inner integral first. Figure 2.2 shows that My) 2 1 — y on the interval [0,1]. Combine this with the identity 2:2 _ y2 _ 1 2y“ $2 + 312 — $2 + y2 and the Observation (following from Figure 2.1) that foa( :1:—)d:1: - 0 to get 1 1 =16 2—1/fl1dd. c for/(y )0$2+y2zy For y > 0, a computation returns 0(9) == [01 am ris2+y2 1 1 1 1 = 5 (2 arctan 5!; — arctang) + (5108(3/2 +1) + 3108.7! + 41082 - 2108(492 +1)) - We are only left to evaluate the integral [0 y2(y -1)C(y)dy- We can directly calculate this integral to find that it equals to 1 1 —2 (arctan 2 — 4 arctan 0.5 + 8? 10g5 — 41082) , 29 which is approximately -0.042. This fact enables us to state our main result for this chapter. Theorem 1. For any A2 weight w, the operator T of convolution with kernel 2‘2 satisfies the boundedness condition llTllL2(w)—)L2(w) S CQw,2a where the constant C does not depend on the weight. Proof. For it E [—7r,7r] let U¢ : C -—> C be defined by U¢(§) :2 (e-W’ and denote kw :2 k0 Uy. If we denote by 5.), the mapping 5,), f = f 0 U4, and if K¢ stands for the convolution operator with kernel k¢, then we can easily verify the similarity relation Ky, = SJlKoSI), . Since the Operator K0 of convolution with k is bounded in L2(S¢w), as we saw on page 25, and each 19¢ is an isometry L2(w) ——> L2(S,,,w), it follows that the operators K1), again belong to B(L2(w)), and their norms can be uniformly estimated by the norm of K0. Choose 72 E N and let —7r/2 = 7P0 < 2111 < < 1,0,, = 7r/2 be a subdivision of the interval [-—7r/2,7r/2] so that A1034: wj — $34 < 27r/n forj = 1, ...,n. Put 1 " -2... T1,, 2: 2:6 JijK¢j. 3:1 This definition, applied to all n E N, determines a bounded family of operators in B(L2(w)). Therefore there is a subsequence which converges weakly to some operator, call it T'. 30 Let f, g be two smooth functions whose supports are disjoint compact sets. We would like to show that (T f, g)", = (T’ f, 9),”. First we make the following computation. % I... f(a: — s) f e—Qit'ms) dw dm2(s) = —7|’ -C- sz(a: — S)l—_51./:l2 e'm’m (fie-w) d1/2dm2(s) . From (2.3.9) we see that this equals to So in fact we have l 1 R2 f(rv — 3le ds — (f * —) (x) — (Tf)(rc). After the change of variable we get that (fox) = i [R f (8) f“ 52ka — s) dw dm2(s) . Denoting by 9' and 9 the supports of f and g, respectively, this equality gives us «ring =/§ fan/j; Z‘rkax—awdsmw (x1dx. On the other hand, (T'f,w=/9: /()f(8) _2iw1A1/ijwj(x — 3) ds 3(5)w(:r) da". 31 Thus (Tf,g>w _w = -1— 3 “6‘2"” .r-s — n €2in )- x—s 371122: a: [so/JUN... M W ; Minx )]d9() (1d, hence for every 5 > 0 there is no 6 N such that n 2 no implies ' i s s 2‘ wa: :1: |w-w|geld/gm )ld [919m (>d . This of course means that (T f, 9),” = (T’ f, 9),” for any f,g E C°° with disjoint compact supports. From here it is easy to see that T — T’ = Mw, the multiplication operator by some w E L°°(R2). Note that our w does not depend on w. Finally, HTHL2(w)—+L2(w) S HTIHL2(w)—>L2(w) + lleHL2(w)—+L2(w)- Operator T’ is a weak limit of operators, whose norms are uniformly bounded by C ' ng. AlSO, “MwHL2(w)—+L2(w) S llwlloo S Ilwlloon,2, because 62.03 2 1 for all weights w, by Holder’s inequality. We conclude that there is a constant C" > 0 such that llTllL2(w)—+L2(w) _<_ C, ' Qw,2 for all w 6 A2. 32 We proved our main estimate. To finish the proof of Theorem 1 it is enough to show that w = 0. Fix a compact set K and let R be a large number. Consider 1,03 smooth with compact support, which is 1 on the disc of radius R centered at the origin, 0 outside of the disc of radius R+1, and has a bounded gradient (independently of R). It is easy to see that TZ/JR(LL‘) —-> 0 when R tends to infinity. And it is easy to see that T,’,1/2;;(.r) —> 0 uniformly in n and :r E K. In particular, IngwRHLoo S €(R), IITwRHLoo S 5(R). Fixing a measurable subset E of K and using the fact that (T — T'WJR = cur/JR = w on K, we can write I [wxdel = |<(T — Wm» _<_ 28(R)|El. which tends to zero when R grows to infinity. Hence fE coda: = O for any measurable subset E of K. We conclude that Ma, 2 0. So T = T’, and Theorem 1 is completely proved. C] 2.5 Sharpness We still need to show that this estimate is sharp, in the same sense the estimate (2.0.1) was. For that purpose we shall need the following auxiliary result, which can be proved by direct simple calculations. Lemma 3. Let |a| < 2 and define w : R2 —+ [0,00) by w(:1:) = M". There exist constants M, m > 0, not depending on a, such that Now we are ready to prove the sharpness. Calculations of this kind have already been made for other singular integral operators, but not for T. So we include this calculation for the sake of completeness. 33 Proposition 3. Let (t : R —> (0,00) grow slower than a linear function. There is a weight w 6 A2 and a function f E L2(w) such that llellL2(w) > ¢(Qw,2) “fHL2(w)' (2-5-1) Proof. For |a| < 2 define w(z) = Izl". The restriction on a is needed if w is to satisfy the A2 condition. Furthermore, let E = {(r, cp) ;0 < r < 1, O < g0 < 7r/2}, X = —E, and f(Z) = lzl’0XE. Let us estimate the left and right side (actually, their squares) of the equation (2.5.1). Thus we begin by llellizu.) = / |(f * 2-2)12w(x,y) dandy 2 1R2 2 / X|(f*2‘2)($,y)|2w(x.y)d:vdy= =// ff 1.3:??? hunt» M We use the identity 1 _ (av-Slz-(y-t)2 . 2(:II-SXy-t) [(x — s) + 21y — m2 ‘ [(sr - s)2 + (y — W ’[(a: — .)2 + (y — 2:121? to estimate the square of modulus of the inner integral (i.e. the one over E) from 2 (:1:2 + y2)“'/2 d2: dy. (2.5.2) above by the square of its imaginary part, that is, by (ff [oi—ill 2.1—thy ( 2+’2)_a’2"3d’)2‘ Since (2:,y) E X and (s, t) 6 E, we have ($-8)(y-t) 2 my- 34 The bound for the denominator comes from the triangle inequality: [(33 - S)’ + (y - 10212 = KIM!) - (s,t)l4 S (Way)! + |(8,t)|)4o Therefore 2) -—a/2 2) —a/2 2 W. d... Mfg/E d...) = [($—:)+ (31"t)l2 (15:1)|+|(8t)|)4 _a 2 =7r2:1:2y2 ([1 r 4rdr) . 0 (l($,y)l+7') By taking u = r/|(:I:, y)| we can continue with 2 2 2 2 I/Kmn ul—a 2 7r :1: 11:, “O‘— du > y |( y)| f0 (1+u)4 _ 77$?! [1211—0 duz— ”2 Ivy _ ($2 + y2)a+2 0 (1+1)4 — 256(2 — a)2 (2:2 + y2)a+2 ' Now the integral (2.5.2) can be estimated as I 7r2 > 2 2 2 2a/2-a—2 : .. _256(2_a)2/ny(x +y) drdy (2—a)3’ (253) where 71.2 7r/2 C, 2 E6 0 cosch sin2 2 ¢(Qw,2)2 ' 01' —1— > (wows) 2—a for certain given constant C’. It suffices to show that lim (2 — a) ¢(Qw,2) Z 0 cit—+2- or, equivalently, lim (4 — a2) ¢(Qw,2) = 0. car—>2“ By Lemma 3, Qw,2 X 4—aT Combining this with the assumption on the growth of (b completes the proof. [:1 36 Chapter 3 Riesz transforms via Bellman functions For a test function f on R" we will denote by f its harmonic extension to the upper half-space R" x (0,00). Sometimes the “newly acquired” variable t 6 (0,00) will for the sake of convenience be labeled as sun“. We shall consider functions f = (fl, ..., fM) with values in some CM, M E N. Here f = (fl, ..., fM). By definition, the (Banach) space LP(IR" —-> C”) will consist of all such functions f that satisfy 1/p M ”/2 l/p ||f||p = (A |f|Pdm) = j“; [Z lfi(33)l2] dm(:c) < oo. " " i=1 By J f we shall mean the Jacobi matrix of f, which is defined as .. 3}; J = f [6117]] . Furthermore, H ' “2 will stand for the Hilbert-Schmidt norm on the space of matrices, unless specified otherwise. Throughout the chapter, p will be a number from (1, 00), while q will stand for its conjugate exponent. Let us denote p“ = max{p, q}. We will most often encounter 37 the factor p._1____ 19-1 ; P22 (19-1)'1 ;1 (CM) and g E L"(R" —-> C”). Then 2 f [R IIJf(x,t)||2l|J§(x,t)||2tda:dt s (12* — mummy“.- The proof will be presented in Section 3.1. We believe this theorem displays a useful and rather general example of an in- equality of the Littlewood-Paley type. In particular, its following corollary represents our main inequality for the classical Riesz transforms. Theorem 3. Up 2 2 and f 6 D", then ”(2 lEf|2)1’2|Ip s 2(p*—1)l|f|lp, i=1 and therefore also ||Rk||B(Lp) g 2(p* -~ 1) for k = 1, ...,n. As an example of how our technique can deliver results in non-standard settings, we treat in Section 3.3 spaces endowed with the Gaussian measure. We are able to obtain the analogues to the above formulated results (see Theorems 4 and 5). It is also demonstrated there how closely our original problem of estimating Riesz transforms relates to the boundedness of particular spectral multipliers. We learned recently that the result of Theorem 3 was also obtained by R. Bafiuelos and G. Wang [5]. T. Iwaniec and C. Martin proved [13, Theorem 1.5] that n II(ZIR.-f|2)1’2llp s fiHpm. (3.0.1) i=1 38 where Hp(1) 2 ”Bl + iRgllp and R1, R2 are planar Riesz transforms. As far as we know, the factors Hp(1) have not yet been computed. However, S.K. Pichorides [24] provedthat 7r R < t — f >2, ll kllp_co (2p) orp- where Rk is a scalar Riesz transform on arbitrary R". In particular, it immediately follows that Hp(1) g 2cot (27%) for p _>_ 2. Comparing this with Theorem 3, we see that limsupflflgfl S 2_\/—.Z <1, 10—100 209-1) 77 therefore the estimate (3.0.1) is stronger if p is large enough. 3.0.1 New findings 1. We prove some Littlewood-Paley type estimates which imply dimensionless boundedness of Riesz transforms. These estimates seem to be new both in classical as well as Gaussian setting. 2. We show how the boundedness of Riesz transforms can be reduced to bound- edness of a certain spectral multiplier. 3. The Bellman function can provide a unified approach to many various Riesz transforms (provided that the corresponding spectral multiplier theorem is avail- able). It tends to give quite good estimates in p (depending on how well the Bellman function is found). 4. There is a proof, due to D. Burkholder, of a weak type inequality for the mar- tingale transform which contains a sort of “Bellman function”. This gives rise 39 to hope that our approach could also be useful for dimensionless weak type estimates. 3.0.2 Further notation and preliminaries There is another way to describe Riesz transforms. Let MES-g- i=1 be the classical Laplace Operator on R". For any test function qt we have = / 26 6 26mm:— / IIV¢II2dx Rn i=1 where V denotes the gradient 6 a “(a—at.) Thus A is an unbounded symmetric operator, whose domain can be taken to be the Schwartz class 8. One can show that A is closable and that its closure 3 is a self-adjoint extension of A (it therefore being unique). Since A was a negative operator, so is 3. So there is a unique positive self-adjoint operator A with the property A2 = —-A-. Symbolically, on S we have the formula A = \/ —A . Then one can verify that 8 —1 331,- 0A . Iii: Note that one can define Riesz transforms R.- in many other situations when we have an abstract Laplacian operator. For example, in section 3.3 we also define 40 the Ornstein-Uhlenbeck Riesz transforms with the help of the Ornstein-Uhlenbeck second-order differential operator. We will also use the operator of Poisson extension P to the upper half-plane R" x (0,00). By Pt f (2:) we mean the value of the harmonic extension of f at the point (:6, t) E R" x (0,00). In short, Ptf(:1:) = f(x,t), where f is defined on page 37. It is well known that Pg : C-tA. 3.1 Bilinear dimensionless Littlewood-Paley type estimate This section is devoted to proving the fundamental result of the chapter, Theorem 2. We will need a few auxiliary results, which we will explain en route, as the need for them will arise. 3.1.1 Bellman function appears Take p _>_ 2 and let g be its conjugate exponent. In addition assume that M and N are natural numbers. Define 9 = {(<.n.Z.H) e C“ x C” x R x IR; ICI” s 2,!an s H}. This is a convex domain in CM x C” x R x IR E Rd, at = 2(M + N +1). There is a function B : $2 —> R, such that everywhere on its domain 0 S B(C,n,Z,H)S(P-1)Z1”Hl"’ 41 -d2B(C,n, 2. H) 2 2lalClldnl- This is our Bellman function. By the latter property we mean that for any choice of (C,7),Z,H) EQand (0,0745) ECM XCN xRwae have (-d2B(C,n, Z. H) [a,fl,%5lT, [0, fl, % 5lT> Z 2lallfll- (3-1-1) Naturally, the Hessian d28 can be thought of as a real d x d matrix, whose upper left 2 x 2 submatrix, for instance, equals 828 acf"a<§-" 3 i,j=1,2 where c = (g0), ((111)) and gm = cf” + is”. The existence of a function that satisfies these conditions was proved in [9]. The proof uses a result of D. Burkholder on subordination of martingales. In general we do not know of a formula, i.e. an algebraic expression, which would possess the properties of a Bellman function. The only exception is described in the next Lemma. Lemma 4. In case when p = 2 one possible choice for B is B(Ci7liZ1H) : \/(Z — lCl2)(H _ I’ll2) ' Proof. The proof is given for the case M = N = 1. The function obviously satisfies the first requirement. For the second one, take C = C1 + 2'9, 17 = 771 + i772, Z, H so that (Q17, Z, H) E (2. Write B = B<<,n,z,H> = \/ 0, therefore . b2 (—B . (dZB)v, v) 2 c —— 15' But it turns out that __ i __ 2 H _ 2 2 Z _ 2 c 4a—Ial( ln|)+lfl|( ICI), hence obviously (-B - (d2B)v,v) Z 2lallfilB- Cl Proof of Theorem 2. Choose test functions f : IR" —+ CM and g : R" —+ C”. Take a: E R", t > 0 and define v = WW) = (Ptf(~’v), 119(3), Palfl”(r), Pz|9|q($))- Since the Poisson extension can be expressed as an integral against the Poisson kernel [3, p. 6], applying the Jensen’s inequality shows that v E (2. Introduce the notation R1“ := IR“ x (0,00). Accordingly, for :c’ E IR:+1 write x’ = (x, t) where a: E R" and t > 0. Furthermore, let We have 12: R1“ ——-> QCCMXCNXRXIR (fat) *—> (Fifi-'0),13:9(33),Pt|f|p($),leglq($))- 44 Define b:=Bov: R1+1—>[0,oo). Lemma 5. Under this notation _ / A’b(x,t)tdtdx s (12* — alumna”.- + Proof. Let Q be any region in R1“ which contains point 23’] = {0}" ED {1} 6 IR" x (0, 00) and let G9 be its Green function. The Green formula says that I I I I __ 609071:” I ab LbAGQ( ,fEl) LAbGQ( ,1121) — 39b an aQGQ( ,1171) an . Since Gn(',$'1) E O on 80 and since its Laplacian, multiplied by —1, acts like the Dirac function of :r’l [11, 2.2.], this implies —/A’bGQ(-,x'1) =b(:r'1)+/ hm. (3.1.2) n an an Note that the normal derivative of G9 on the boundary equals to the Poisson kernel [11, 2.2.], again multiplied by —1, so it is negative. Since b is always positive, we get 52 The first property of B leads to the estimate W) S (p‘ - 1)(P1lf|”(0))1/”(P1Igl"(0))‘/" and so —/A'b(a:')Gg(x',x'1)dx' g n 45 1 (1+ly Here on is the normalizing factor for the Poisson kernel, like in [3, p. 6]. l/p 0. Let QR be the rectangle [—R, 12]" x [0,2R] in R1“ and define x’R = {0}" EB {R} E R" x (0,00). One can show that on (12%) = R""1Gm(:c,y). Choosing Q 2 DR 2 R91 and :r’R = Rx’l instead of :c'l, (3.1.3) takes the form —/ A'b(:c') Ga, (1,291) dx' g 93 R 1/p c *_ :0 Rn q 1/q s ..(p 1)(Rn|f(y)| (Bauer-2H”) (...Igon...) . Apply the Lagrange theorem for Gg,(-,:z:1) along the normal vector. Use again that this function is O on the boundary. We get that slow (21 > we MEL—2.: ”(a-"s Here 3:” is some point that lies inside the line segment between a: and :c’ = (x, t). We afforded a slight abuse of notation and wrote :1: instead of (:13, 0) in the line above. —/A’:cb(’) ’aG”‘(’ an’m——1—)(R")tdx’g Rn+1 l/p *_ p q 1/q 0 such that M C QR. Since the integrand is positive, we can estimate the integral on the left from below by the same integral, just over M instead of QR. After sending R to infinity, the term 9&1 will uniformly go to 0, because M was bounded. But 0 lies on the boundary of QR, where W equals the Poisson kernel. So when R -—> 00, it approaches the value of the Poisson kernel for the upper half-space R1“, i.e. p(0,x’1), which is exactly equal to c". On the right side of the inequality, sending R to infinity leaves only cn(p"' -— 1)“ f [[pHqu. Finally we make M large and complete the proof of Lemma 5. Cl In order to proceed we need the following easy calculation. For the sake of sim- plifying the formulae let us temporarily write xn+1 instead of the variable t. Lemma 6. For every (x,t) E R1“, "+1 do do A'b(x, t) = Z2 ;< (vim)- ; However, this inequality can be strengthened, i.e. it actually self-improves, thanks 6 6111' 3 BZP‘flx) - Ptg(:lr) to the next useful result which will be proven in the Addendum. Lemma 7. Let m, n, k E N. Denote d = m + n + k. For arbitrary u E Rm+n+k write 1) = um EB 22,, EB uk, where u, E IRi for i = m, n,k. Let R = Hum”, r = “on”. Suppose matrix A 6 IR“ is such that (Au,u) Z 2Rr (3.1.5) for all u 6 Rd. Then there is r > 0, satisfying 2 1 2 (Av,u) 2 TR + ; r again for all u 6 Rd. In particular, operator AIRm+nx{0}k “becomes” injectiue. We use this lemma for m = 2M, n = 2N and k = 2. By applying it to the operator —d"’B(v) in place of A, where u is as above, we find such 7' z r(:c, t) that we even have 48 _ 2 _ _ > —1 = z< d was...) _ )3 ( ..Ptg +7 {mm ) n+1 2 n+1 2 8 1 (9 —T;:; 333,-th +;; Bic—:Ptf __ n+1 2 1/2 n+1 8 2 1/2 Z 2 (Z; 5:539 ) (:2; a—xiptf ) = 2llJfll2llJ£~lH2- Now use (3.1.4) to finish the proof. D 3.2 Estimates for the classical Riesz transforms This section is devoted to the proof of Theorem 3. First, let us prove the following Lemma. Lemma 8. Let f and g be two test functions and let Rk, 1 S k _<_ n, be any Riesz transform. Then (g. Rf) = 4 / (Aag. am) tdt. o where 8;; = 5%. Proof. Introduce ¢(t) = (Pig. RBI). Integration by parts gives ¢ = [0... Mt) tdt OI‘ °° d2 (Qika)=/ Et-z-(Ptg,Ptka)tdt. o 49 The right side equals to 00 / tdt+2/ tdt+f tdt. 0 0 0 By P,’ we mean 15‘, of course. But then P’= —AP; and so Pt” 2 A2Pt. Thus we can continue the line above by /OOO(A2Ptg,Ptka)tdt + 2/000(APtg,APtka) tdt + fom(Ptg,A2Ptka)tdt. The operator A is symmetric with respect to the pairing (-, -), hence we get 4 f0 ”(Ana/110.12,. f)tdt Since A commutes with P; and 3,, we see that APtRk 2: ABd-A‘l = 6,13,, which yields the desired equality. I] Now (ZIRJI )”2 i=1 = [fi ZIRJI )P”dm[l/p = =[/|(R1f f)( )l”dm($ )[ 1/p= IlfRfllLvmn—m"): where 31 f (R1 f, ,R,l f )By using duality, this IS the same as sap{|(9lf,g>l ;g E L"(R" —> C"). llgllq = 1} - It follows from Lemma 8 that I<92f.g)l = Z( R.f g.) - .. MBPtfdxtdt i=1 n a ”2 < 4/000 / (2:: I—P:g,|:)0/ :(Z: 0 fl?) dxtdt. Finally, we use Theorem 2 for M = 1 and N =—n. Cl 50 3.3 Estimates for the Ornstein—Uhlenbeck semi- group (Gaussian case) In this section we turn R" into a probability space by endowing it with the canonical Gaussian measure In order to get a self-adjoint operator again we deal, instead of A, with the Ornstein- Uhlenbeck differential operator " 6 A I: A— 1' CU git? am.- This time computation gives 7: 62¢ n a¢ .- (Aowfififi) :fnn (Z? ’ina) 4561/1: —/Rn llv¢ll2d#, i=1 ' i=1 which again allows us to define A I: V _AOU and ._ a -l R,- .— 8:13.014 in the same sense as before. The main goal of this paragraph is a new inequality of the Littlewood-Paley type. It is analogous to the one that we have already demonstrated in the standard setting (Theorem 2). Here we state the result. 51 By B we shall mean the AOU—generated extension of function F into the upper half-space 1R1, i.e. R(a:, t) = (PtF)(a:). The letter J stands for the Jacobi matrices, as before, and it is again their Hilbert-Schmidt norms that we consider. Theorem 4. Let f : R" ——> RM and g : R" —> R” be test functions. Under the assumed notation, [W HJf~(-’r, t)l|2ll1§(a:.t)||2 dubs) wt) dt 3 C (p* —1)Hf”pHqu, Here C = 83%(1+ e‘2). As a consequence of this inequality we will deliver a new proof of the well- known problem about dimension free bounds of Riesz transforms associated with the Ornstein-Uhlenbeck operator Aou. Previously, Riesz transforms R,- were defined by R,- := 3% o A“. In order to define A"1 in the Gaussian setting properly, we restrict ourselves to the orthogonal complement of the kernel of on- This is a subspace of L2(1R", u) of co—dimension 1. It consists of functions f which are orthogonal to the constants on R" and are thus of mean zero, i.e. fun f do = 0. We denote by no the orthogonal projection onto this subspace and define 8 R4 := ('93:,- o A‘liro. Our objective will be to estimate the vector-valued Riesz transforms, in other words, we will try to give LID—estimates of the function Rf = (ZIEfI2)1/2- i=1 Theorem 5. [fl < p < 00 and f E LP(R",du), then llRfllp S Cpllfllp, 52 for some constants Cp > 0 that do not depend on the dimension n. One can take 0,, = c,, + C(p“ - 1)][OHLP(u)—)LP(#), where C is the same as in Theorem 4 and _ 1 ;p22 CP‘ 0((p—1r1/2) ; p—>1 ' It was first proved by RA. Meyer [18] that for some 0,, > 0 the inequality above holds. This was done by applying probabilistic methods. As mentioned in the intro- duction, G. Pisier [25] found an analytic proof of the same result, while subsequently several other results and improvements followed. It is again our aim to use the Bellman function as in the previous section. However, there are difficulties which prevent us from reaching exactly the same result. The first of them is the non-commutativity of 8,- and Aou- Namely, one easily verifies that [Aou, 6,] = A0110,- — aiAOU = 81' . (3.31) This obstruction defies our efforts to restore a satisfactory version of Lemma 8. In order to clarify this, let us “repeat” the proof of the Lemma. Take two nice functions f, g. It is not difficult to see that (9.1M) = (Pog.PoRI.f) = 4 / (APtg, 1)./126erl f) t2 dt. 0 By (3.3.1), the integral equals to / (AP,g,P.a,.Af)t2dt— / (AP,g,P,o,.A-1f)t2dt. 0 0 We perform integration by parts on the integral to the right and use (3.3.1) again. We get 1(9) ka) : / (APtg,Pt0kAf) (t2 — 51:4) dt + 'é/ (APtg,Ptka> t4 dt. 0 0 53 Inductively, the equality 1 °° °° was): / N(t)dt—a,./ (APtg,P,ka)t2th. 0 0 holds for every N E N. Here N (PA/(t) = Z ant2" n=1 and (_1)n-—1 . 2271—1 (2n)! an: We could hope to replace N(t) by its limit when N —-) 00. But this limit turns out to be sin2(t), a function which does not have the properties necessary to continue the proof with the Bellman function as we did in the previous section. One way around this is to introduce a certain auxiliary operator 0. In the classical case, Lemma 8 gave us <9. Rm = 4 / = [0000439, BIBOf) «b(t) dt, (3.3.2) where the Operator 0 and the function w are yet to be determined. The eigenvalues of —/_\OU are precisely all natural numbers. The eigenvectors, corresponding to the eigenvalue m E N, are the Hermite polynomials ha, where a = (a1,...,an) E N" is a multiindex of length m, which means |a| :2 a1 + + on = m. If n = 1, then d’" h... =—1m+2 (z) < )e dz, [6+2]; mENU{O}. 54 One can easily see that h;n(a:) = mhm_1(;r) . (3.3.3) For arbitrary n, we define ho, :2 ®?:1ha,. That is, for x = (1'1, ..., :17”) E R" we put ham := aha-1)... han(:z:,,) = I] ha,(z,). We shall denote by Pm the orthogonal projection of L2(R", du) onto the eigenspaces Hm := Lin{ha ; [0| 2 m}. The fact that —-A0U is a self—adjoint operator implies that the polynomials ha are mutually orthogonal, but it actually turns out that they form a complete orthogonal system in L2(lR", du). Since Aouha = —|a| ha, it follows that o Aha = ‘/|a|ha . Pm, = e” Who, and . 6,120 = a, ho, ® «a ha,_1 ® ® ha... thus 81' I Hm —‘)Hm_1. For the last equality we used (3.3.3). This implies that if we test formula (3.3.2) for f 2 ha, we see that 0 must be a multiplier, more precisely, of the form (9220um mEN for a certain collection of scalars om, m E N. 55 Suppose 0m 2 1 for all m E N (i.e. O = I ). One can verify, after the same procedure of testing the formula on ha, that If) in this case must equal w(t) = 2(sint+ sinh t) . This function again does not enable us to continue the proof with the Bellman function, which once more explains why in the Gaussian setting (contrary to the classical one) there is no “suitable” formula of the form <9, Rm -—- [mam am we) dt and similarly with the roles of f and g exchanged in the integral on the right. This hints that the generalization (3.3.2) might indeed be necessary. It turns out that the function w(t) = te"°‘ suits our purpose. In this case com- putation returns _ (fin+\/m—l+a)2 WVm — 1 ' Of course, this is only true if m 2 2. For m = 1 we define 01 := 0, which leads to 0m a slight modification of formula (3.3.2) into <9. Rm = (g. R.P.I> + [000mm, anon at) at. (334) Remark. This relation is valid for all polynomials f in n variables. On the other hand, this domain is sufficient, for one can show that the polynomials form a dense subspace of any LP(u), 1 < p < 00. For reasons which will became apparent further on in this section, this (and not C§° as before) is going to be our family of test functions on the Gaussian space. Proof of Theorem 5. We are going to use (3.3.4). First we estimate the linear part 56 of the polynomial f. Write P1f= Z coho 06H" la|=1 which is the same as P1f(:c) = 2:1 cixi, because h1(3:) : a: in IR. Therefore RkP1f(:c) E ck, which means that [IRPlng = Z]; |c,-|2. But this equals to ||P1f||§, since {ha ; [0| 2 1} are mutually orthogonal in L2(u). Write ) = (f... Itlpdu(t))l/p lllleHi»< 710 *) 7(1)) 7(1) ) Then “Rplfllp— — ||P1f||2— — llfllp (3-3-5) Note that C ._ 709*) _ __ _1/2 as P '- 7(1)) — 0((p 1) ) p _l 1- We remark that the coefficients 7(p) appear in a similar role also in G. Pisier’s article [25]. The more difficult part of the proof is to estimate the integral part of the formula (3.3.4). We would like to show that [0 I| we) at s Cpllfllp for some dimensionless constant Cp. This will immediately follow from our main theorem, that is, our Littlewood-Paley-type inequality for the Gaussian setting. We incorporate its proof here. Proof of Theorem 4. Having prepared the ingredients, we can apply the same Bellman function technique as in the previous section. That is, we deal with the function u : IR" x (0,00) —+ {2 defined by v(x.t) = (Bf($),Pt9($),Ptlf|p($), Pt|9|q($)) 57 and compose it with a Bellman function B for b = B o v . However, in our attempt of repeating the proof we encounter a few problems: 1. We need to explain though, why the function u is well defined, i.e. why it really maps into 0. For that purpose, apply Jensen’s inequality to the Mehler formula for the heat extension Ht f = eAOU ‘ f of f: Hzf(rr) = Rn f(e"x + W1 — 6‘2‘) duty)- We see that [Htflp S Htlflp for any p Z 1. By [18, p. 180], with every t > 0 there is associated a probability measure at on (0,00), such that Pt f = f0°° H3 f duds). Now Holder’s inequality implies that v E Q, as desired. In addition to that, one can show by direct calculation that 1lepr s ((1.. MW) duo)” s ”In... Moreover, by applying Ht to a constant function we actually see that IIHtll B(LP(u)) = 1. Consequently, if p _>_ 1 then Ham. s f... lleflldet(3) s [m llfllpduds) = ”In... (3.3.6) Here we again establish the exact norm of the extension operator by using constant functions. Indeed, A1 = Aho = O by a formula on page 55, so Ptl = e“‘“1 = :00 [~t[”A"1 _ 1 n=0 _ ° 71! 2. In the classical case, it were the expressions of the form dvd_v 2 _ (d 8(2)) d1}, d$i> 58 that linked the Jacobians of P, f and By with [l f I],, and ||g|]q. In the proof of Lemma 6 we used that the components of u were harmonic functions, i.e. that they were in the kernel of the Laplacian on the upper half-space in Rn“. In order to mimic this reasoning in the Gaussian setting, we have to come up with an appropriate differential operator. Since P, = e“’“, it is obvious that 62 523.1% 2 A2Pt in the strong sense, which implies 62 B2 (gt-'2'+AOU)Pt= (by-A2)Pt=0, hence the proper choice is given by 62 IOU:fi+AOU- By “prOper” we of course mean what has just been explained, i.e. that now we have the analogue of Lemma 6 : n+1 do dv gum, t) = Zuzana, 337,-)“ (3.3.7) i=1 The point here is that now we can repeat the reasoning from the classical setting without any modifications and obtain that —A’oub(x.t) 2 2IIJfII2IIJ§II2. (3.3.8) The only property we used for this, apart from the equality (3.3.7), was the charac- teristic estimate involving the Hessian of B. 3. The next step is to prove Lemma 5 in this setting. This means estimating the integral — [W who») undmodt (33-9) 59 from above by C (p)|| f Ilpll gllq, where C (p) is some positive number. Let F : R —+ R be a sufficiently smooth function. Then 2 1'2 I 1' (Noe-2‘) = 0 where BA((, 77, Z, H) = B(/\C,/\‘1n,)\PZ, A‘QH) . This B,\ will be our main tool for the time being. It will temporarily occupy the place of the “true” Bellman function. Accordingly, we denote b), = B,\ 0 ’U . The idea is to work with the concrete b,\ and then minimize the estimates over all /\ > 0. The reason for dealing with these special functions is that they seem to supply us with more information about the behaviour of their gradients than just the abstract functions, provided by the existence theorem. We are going to need this information in order to justify some of our calculations. Fix /\ > 0. Let us begin with the estimate. de/Ax): b,\( (1' ,t) d/J,(.’II )I/It”( )dt _ 0. Consequently, the integral S 8(1)“ 1)2( ApllfIIZ+/\""Hg||;’) / ld)"(t)|dt. on the right converges and equals to / I3—2l6—3d8=26_2+1. 0 In order to estimate E and m, we have to consider the integral / lim bA(:r, t)z[)'(t) du(a:) 1R" —)w for w = 0 and w = 00. We have just proven that 8(1) mm) duct) 3 ——;—9- (Arm; + A-qngng) . Rn The expression on the right is also the bound of our integral when w = 0 (case E). However, in the special case when we actually know a formula for the Bellman function (Lemma 4) we have b(x, 0) = 0. Thus the term E is not the essential part of the estimate of the integral in (3.3.9). On the other hand, limHoo zp’ (t) = 0, therefore / :1 (MIC) = 0- 63 Cases E and I are about estimating integrals 6b /Rn th—{B —67’\(:z:, t)2,/2(t)dp(a:) (3.3.11) where w = 0 and w = 00, respectively. Since 7,!)(w) = 0 we suspect that these integrals might vanish as well. In order to conclude that, it suffices to know that that @55- is not behaving too strangely. The proof of Lemma 6 gave us %($0) = (VBA(UO)7 gholhd = — 2?}. ’U a: -8—B-A- III — ( ac ( o). PtAf( ))W +( an (v0), P.Ag( ))W + +%§Zi(vo>P.AlfIP = llvlglqllz- Where 9 75 0, we can write 9 _ Vlgl" = q—lgl" 1V9- lgl 64 Since q > 1 and g is a polynomial, IIVIquHz < 00. Consequently, lim sup W) [Rn PtAlgl"($) duh) = 0- t—iw Of course, the same is true with |f|P in place of lglq. What about the other terms? The part of (3.3.12), corresponding to 77, gives lim sup 112(1) /n gig-(Mar, t)) PtAg(a:) d,u(:z:). (3.3.13) t—nu 77 We again use (3.3.10) to compute partial derivatives. In this case ~ 6Q _ q_2 3Q where ~ 2 _. 6Q (E-l)qn|n|" 2 when 141lean |C|2(2 - q)n|n|“’ when ICI” S W’- H ICI” S lnl", then 1 _ .. s (2-(1)|77|2P+1 " = (2-q)|77|" ‘- Thus there is a constant M > 0, such that 88 _ < M 0‘1 I 377 . Inl everywhere in 9. Therefore the absolute value of the integral in (3.3.13) is bounded from above by M R IPtg(x)|"'1|PtAg(I)lduh?), which in turn admits estimate from the Holder’s inequality, i.e. —1 M S IIPt9I|g(q_1)HPtA9”2 < 00- 65 Consequently, the term (3.3.13) equals to zero. A similar proof works for the terms with the only remaining variable, C. We conclude that Summary of proofs Combining all our estimates, starting with (3.3.8), we find that 2 [W IIJf(2=.t)H2||J§(:v.t)ll21W) (Mm) dt 5 + _1 2 g _ [RH AIOUW, t) we) at.) dt 3 c' (1;; (Wm: + A-qngnz) for all /\ > 0, where C’ = 16(1 + 6‘2). Taking the minimum in A and using that (p - 1); S 6% completes the proof of Theorem 4. [:1 In order to finish the proof of Theorem 5, apply Theorem 4 with O f in place of f (we can do that, since 0 clearly maps polynomials into polynomials). From here, from (3.3.4) and from (3.3.5) it follows, identically as in the classical setting (page 50), that “Rfllp S 61)“pr + C(P‘ ‘1)”Ofllr The coefficients om of our multiplier O can be written as F(m‘1/2) if m 2 2, where F is a certain function, analytic in the neighborhood of the origin. This implies [18] dimensionless boundedness of O on LP(1R", dp). Hereby the proof is wound up. D Remark. Instead of A00 we could run our proof on other Operators and still ob- tain dimensionless boundedness. From the considerations above it emerged that the properties we require are that these operators should be positive (or negative) and 66 of discrete spectrum {Am ; m E N}. If the growth Am, m ——> 00, is of “proper” pace, then Meyer’s theorem will probably work on multipliers of the form «Am Am- 2 OZZOuma 0m:( + 1+a) - meN V’\m\/ ’\m—l Namely, already the LP—boundedness of such multipliers will be sufficient to run our Bellman function procedure. 67 Chapter 4 Addendum Here we demonstrate Lemma 7. Since the assumption and the conclusion of the Lemma are homogenous inequal- ities with respect to 2), it is equivalent to prove: IfE C H, then there is T > 0 so that E C ET, where E=EA={’UERd;(A’U,’U)S2} H={UER";RT§1} ET={UERd;TR2+%T2_<_2}. Note that E, C H for every 7' > 0. Since A is a positive matrix, E (more precisely, its boundary) is an ellipsoid. The geometrical shape of other two sets is also clear. Suppose that we have the proof in case when k = 0. Now take arbitrary natural numbers m, n,k. If E C H, then E’ C H’, where E’ and H’ are images of E and H, respectively, under the orthogonal projection Rd —> Rm” x {0}". Since E’ is again an ellipsoid, by assumption there is T > 0 such that E’ C E; := {v E 68 Rm” x {0}" ; 7R2 + $13 S 2}. It is clear that this 7' also satisfies E C ET. Hence it is enough to prove the Lemma for k = 0. We may also assume that at some point equality is attained in (3.1.5). This implies that there is v 6 R“ for which RT = 1 and (Av, 21) = 2. In other words, 12 6 8E 03H. For A = W we have AR 2 A‘lr = 1. The operator T = AIRm <8) A‘IIRn leaves H unchanged, whereas it maps E into some other ellipsoid, whose boundary intersects that of H in a point with R = r = 1. Finally, there are rotations Um 6 SO(m) and U" E 80(n), such that Um 8) Un maps this intersecting point into 220 :2 um EB 12,, = (1,0, ...,0 (1,w)€ Rm EB IR" = Rd. Trili-slummarizeritl suffices to solve the case when k = 0 and 6E intersects 6H at 220. Since in this case it is obvious that the only admissible 7' is 1, our task is reduced to proving that E is contained in E1 = J28“ 2: B, i.e. in the closed ball in Rd, centered at O and with radius f2. The intersection 8H0 BB is the “torus” T = {v 6 Rd ; R = r = 1} = Sm"l x 8"“. Let 3’ be the family of all 2-dimensional planes in Rd which pass through 0 and v0. We would like to find a subfamily (P’ of planes that intersect with T in more than the obvious two points. Take P E ’3’. There is u 6 vol such that Hull2 = 2 and P = Lin{vo, u}. Roughly speaking, this establishes a correspondence between (P and a portion of the sphere 6B = fiSd‘l. We can write u = {a} 69126) {—a} EBc for some a 6 IR, b E Rm‘l and c E Rn‘l. pr : Avo + mi 6 P is to intersect T in a point, different than :lzvo, then we must have (A + M)2 + Ilubll2 = (A — W2 + IIALCII2 = 1 for some A, u E R, u ¢ 0 (otherwise p = :tvo). After adding and subtracting 69 equations and using that 2a2 + ||b||2 + ”CH2 = Hull2 = 2, one can see that this set of equations is equivalent to A2 + p2 = 1 4M + M(HbH2 - llClI’) = 0 If a = 0 then we must have Mb” = ”CH. For a yé 0 the system admits solutions ,2 ____ (1— ..2)2 - llbllzllcllg (1+ 02)2 — HbH’HCH2 #2 _ 4a2 (1+ 02V - HbH’HCHZ. Note that W + Ilcll’ z b < H II Men _ 2 1—a2<1+a2, thus the denominators are always positive. Hence the solution does not exist (i.e. P does not belong to 1}”) if and only if a = 0 /\ llbll 75 NC”- This justifies employing identifications {P E {u = u(a, b, c) E 68;u_Lv0} and (P’s{uefP;a790 v ||b||=|lc||}. which in principle imply that the set 1]” is “dense” in (P. 70 We would like to show that E (1 P C B 0 P or, equivalently, BE (1 P C B H P for all P E 1}”. Since BE 0 P is an ellipse, this simply follows from the fact that E C H and that T 0 P contains at least four different points, as has just been shown. 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