Bellman function technique in Harmonic Analysis. Lectures of INRIA Summer School in Antibes, June 2011
aa r X i v : . [ m a t h . P R ] J un BELLMAN FUNCTION TECHNIQUE IN HARMONIC ANALYSIS
ALEXANDER VOLBERG
Abstract.
It is strange but fruitful to think about the functions as random pro-cesses. Any function can be viewed as a martingale (in many different ways)with discrete time. But it can be useful to have continuous time too. Processescan emulate functions, expectation of profit functional on the solution of stochas-tic differential equation can emulate the functional on usual familiar functions.The advantage is that now we have “all” admissible functions “enumerated” assolutions of one stochastic differential equation, and choosing the best functionoptimizing a given functional becomes a problem of choosing the right controlprocess. But such problem has been long since solved in the part of mathematicscalled Stochastic Optimal Control. So-called Bellman equation reduces an infi-nite dimensional problem of choosing the best control to a finite dimensional (butnon-linear as a rule) PDE called Bellman equation. Its solution, called Bellmanfunction of a given optimization problem, gives us a lot of information about op-timum and optimizers. This method gave some interesting results in the classicalHarmonic Analysis, having on the surface nothing to do with probability. Some-times the results obtained by this method did not find “classical” proofs so far. Itis especially well-suited to estimates of singular integrals, probably because of theunderlying probabilistic structure of classical singular integrals. Quasiconformal maps: sharp distortion estimates and sharpregularity
We deal first with Beltrami equation(1) f ¯ z − µf z = 0 , with bounded function µ called Beltrami coefficient , for simplicity µ is compactlysupported on C , f being analytic near ∞ (see (1)) supposed to have the followingLaurent decomposition at infinity(2) f ( z ) = z + c + c − z + . . . . If µ is smooth it is not difficult to see that the solution is smooth on the whole C . But we are interested in just measurable bounded µ :(3) k µ k L ∞ ( C ) = k < . Several natural questions appear:1. What is the smoothness of f depending on µ, k ?2. What are distortion properties of f ? How it distorts the area and othermeasures?3. In what classes (Sobolev, say) we can solve (1) in such a way that it will becontinuous ˆ C → ˆ C , where ˆ C = C ∪ ∞ ? As at infinity it is a perfect holomorphicmap, this question concerns only finite part of C , so it is local. Denote g := f ¯ z . It is a function with compact support. We can restore f :(4) f ( z ) = 1 π Z C ζ − z g ( ζ ) dm ( ζ ) + c + z . We used (2) and we naturally assume integrability of g .Then obviously f z = 1 π Z C ζ − z ) g ( ζ ) dm ( ζ ) + 1 , where the integral is singular and should be understood , e.g. in the sense of principalvalues. This is an important operator called the Ahlfors–Beurling transform (ABtransform): T g := 1 π Z C ζ − z ) g ( ζ ) dm ( ζ ) . Then (1) automatically becomes(5) g − µT g = ( I − µ T ) g = h , where h = µ is bounded with compact support. So in particular h ∈ ∩ p ≥ L p ( C ).It is easy to make Fourier analysis of convolution kernel πz of AB operator, and tosee that it is the Fourier multiplier with symbol ζ / ¯ ζ . Therefore, k T k L ( C ) = 1 andhaving then k µ T k L ( C ) ≤ k <
1, we can conclude that (5) has a solution in L ( C )given by the usual Neumann series:(6) g = h + µT h + µT µT h + . . . , . Notice that g is compactly supported (as µ is). Restore f by (4). The boundednessof T in L implies now that(7) f ∈ W ,loc ( C ) . Let us see now that g given in (6) is actually better than in L . Operator T hasnorm 1 in L and it has norm close to 1 in L p , p > , p ≈
2. In fact, it is an operatorwith Calder´on–Zygmund kernel, and as such it is bounded in all L p . Interpolatingbetween, say L and L , we get(8) k T k L p ( C ) =: n ( p ) → , p → . So we can find such a p = p ( k ) = 2+ ε ( k ) , ε ( k ) > , that the series in (6) convergesin this L ε ( k ) . So g ∈ L ε ( k ) . Again restore f by formula (4) (it is the same f ofcourse), again use that f z = T f ¯ z + 1 = T g + 1, and that T is bounded in all L p being a Calder´on–Zygmund operator. We got that f self-improves from (7) to(9) f ∈ W p ,loc ( C ) , p = 2 + ε ( k ) , ε ( k ) > . We formulate this small fact as a fundamental Ahlfors–Bers–Bojarski’s theorem:
Theorem 1.1.
Any solution of (1) in W ,loc self-improves to being in W ε ( k )1 ,loc , ε ( k ) > . In particular, any such solution is continuous on C and even H¨oldercontinuous. There exists a solution, which is a homeomorphism of ˆ C into itself. ELLMAN FUNCTION TECHNIQUE IN HARMONIC ANALYSIS 3
New questions appear:4. What is the largest 2 + ε ( k )?5. What is n ( p ) in (8)?To this we want to add some more questions. Introduce a constant K = 1 + k − k ∈ [1 , ∞ ); k = K − K + 1 . It has a geometric meaning: it gives the maximal ratio of the axis of infinitesimalellipses obtained as the images of infinitesimal circles by all possible solutions of (1).
Definition.
Any solution of (1) from Theorem 1.1 is called K-quasiregular map.Any homeomorphic solution is called K-quasiconformal map or K-quasiconformalhomeomorphism. It is basically unique because by normalization at infinity it canbe only shifted.Again questions:6. What is the sharp distortion of K-quasiconformal maps? Namely, if f ( D ) = D ,where D denotes the unit disc and f (0) = 0, then what is the sharp (largest)exponent in(10) ∀ E ⊂ D , | f ( E ) | ≤ C K | E | e ( K ) ?Without normalizations, allowing f to be any K-quasiconformal map this becomesthe question what is the best (largest) exponent in(11) ∀ E ⊂ B , | f ( E ) || f ( B ) | ≤ C K (cid:18) | E || B | (cid:19) e ( K ) ?Function f ( z ) := z | z | K − , | z | ≤ , and = z , for | z | > e ( K ) ≤ K . Gehring’s problem: e ( K ) = K . It is equivalent to saying (we will se that) inQuestion 4 the sharp exponent of Sobolev integrability is 2 + ε ( k ) = 1 + k − . Thisis very tough, but it was done by Astala [1].Glance now at (6): it gives that if we want to show that the exponent p of Sobolevintegrability goes up to 1 + k − , it is enough to prove that k T k L /k = 1 /k , in other words that(12) k T k L p = p − , p > . This is very open, we will show how Bellman function gives partial results.
Big Iwaniec’s problem or p − -problem: n ( p ) = max( p − , pp − − W ,loc . How much below this we can start to have the same self-improvement?7. Let f ∈ W q ,loc solves (1), and q ∈ (1 , q = q ( k ) such thatwe still have for each such f self-improvement to W ,loc ? (And then automaticallyto W ε ( k )1 ,loc by Theorem 1.1, and then up to W /k − ,loc by Astala’s [1]?) Iwaniec’s problem: q ( K ) = 1 + k . ALEXANDER VOLBERG
We will prove it here using Bellman function technique and Astala’s sharp distor-tion result [1]. We follow the exposition of [2].1.1.
Invertibility of Beltrami operator.
If Big Iwaniec’s problem were solvedthan we would immediately get(13) If p ∈ [2 , /k ) , then k ( I − µ T ) − k L p ≤ C ( k )1 + k − p . (Actually even with C ( k ) = 1 /k , but this we do not care about as k is fixed and wevary p .)By duality and small talk one would get(14) If p ∈ I k := (1 + k, /k ) , then k ( I − µ T ) − k L p ≤ C ( k ) dist ( p, R \ I k ) . Big Iwaniec’s conjecture is still a conjecture, but this is a Theorem of Petermichl–Volberg [58], which we start to prove now. It will use Bellman function technique.Notice that now there exists an even more precise version of this result, namely, see[3].
Theorem 1.2. If p ∈ I k := (1 + k, /k ) , then k ( I − µ T ) − k L p ≤ C ( k ) dist ( p, R \ I k ) . Let f be a K-quasiconformal homeomorphism, let p ∈ [2 , /k ), where k = K − K +1 .Denote J f = | f z | − | f ¯ z | , the Jacobian of f . We need lemma: Lemma 1.3.
Let f, p be as above, denote w := J − p/ f . Then w ∈ A with [ w ] A ≤ p C ( K )1 + k − p . Remark.
We will give the proof following [2]. There is another very interestingproof in [3].
Proof.
Notice first that(15) (1 − k ) | f z | ≤ J f = | f z | − | f ¯ z | ≤ | f z | ≤ | f z | + | f ¯ z | . That is all this quantities are comparable with C = C ( K ). The next step is toshow that there is C ( K ) such that if B ⊂ C is a disc and if f is a K-quasiconformalhomeomorphism of C , then(16) 1 | B | Z B ( | f z | + | f ¯ z | ) p ≤ pC ( K )1 + k − p (cid:18) | f ( B ) || B | (cid:19) p . Using linear maps to pre-compose and to post-compose with f we reduce it tonormalized case | f ( B ) | = | B | = 1. Apply (11) proved by Astala in [1] to the set E t = { z ∈ B : | f z | + | f ¯ z | ≥ t } , t > | E t | ≤ t Z E t ( | f z | + | f ¯ z | ) dm ≤ K t Z E t ( | f z | − | f ¯ z | ) dm = K t | f ( E t ) | ≤ C ( K ) 1 t | E t | K . ELLMAN FUNCTION TECHNIQUE IN HARMONIC ANALYSIS 5
Therefore, | E t | ≤ min(1 , C ( K ) 1 t KK − ) . This is the same as |{ z ∈ B : | f z | + | f ¯ z | ≥ t } ≤ min(1 , C ( K ) 1 t KK − ) . Distribution function calculation now shows(17) Z B ( | f z | + | f ¯ z | ) p ≤ C ′ + C ′′ p Z ∞ t p − t KK − = C ′ + C ′′ p Z ∞ t k − p dt ≤ C ′ + C ′′ p
11 + k − p , as KK − = 1 + k . This proves (16).Now we are ready to prove Lemma 1.3. Notice that w = J − p/ f = ( J f − ◦ f ) p/ − .Then1 | B | Z w dm = 1 | B | Z B ( J f − ◦ f ) p/ − ( z ) dm ( z ) = 1 | B | Z f ( B ) J p/ f − ( ζ ) J f − ( ζ ) J f − ( ζ ) dm ( ζ ) , where we made the change of variable z = f − ( ζ ). We continue1 | B | Z B w dm = | f ( B ) || B | | f ( B ) | Z f ( B ) J p/ f − ( ζ ) dm ( ζ ) ≤ pC ( K )1 + k − p (cid:18) | B || f ( B ) | (cid:19) p/ | f ( B ) || B | . So(18) 1 | B | Z B w dm ≤ pC ( K )1 + k − p (cid:18) | B || f ( B ) | (cid:19) p/ − . We used here (16) with K- quasidisc f ( B ) instead of a disc B . But this does notmatter as any K − quasidisc (:= the image of a disc by K-quasiconformal map) isan almost disc with constants depending only on K .Now notice that we assumed p ≥
2, so if p n := p − | B | Z B w − dm = 1 | B | Z B ( J f ) p/ − ( z ) dm ( z ) = 1 | B | Z B J p n / f ( z ) dm ( z ) , and we use again (16) with p n replacing p , gives us:(19)1 | B | Z B w − dm ≤ p n C ( K )1 + k − p n (cid:18) | f ( B ) || B | (cid:19) p n / = max( C ′ , p − C ( K )3 + k − p (cid:18) | f ( B ) || B | (cid:19) p/ − . (cid:3) Multiplying (18) and (19) we get Lemma 1.3:(20) w = J − p f , p ∈ [2 , k ) ⇒ [ w ] A ≤ p C ( K )1 + k − p . Now we can reap a first consequence:
Theorem 1.4.
Suppose k µ k ∞ = k < , let p ∈ I k := (1 + k, k ) . Then operators I − µT, I − T µ are boundedly invertible in L p . ALEXANDER VOLBERG
Proof.
We can work with I − µT as I − T µ = T ( I − µT ) T − and T is boundedlyinvertible in each L q , < q < ∞ ( T − is again a Fourier multiplier of Calder´on–Zygmund type).Suppose we know how to prove the estimate from below(21) k ( I − µT ) g k p ≥ c ( p, k ) k g k p , ∀ g ∈ L p ( C ) , p ∈ I k . Then we would now exactly the same for I − µT , and, so, for the adjoint operator( I − µ T ) ∗ . Then I − µT would have dense images in all L p we consider. Joiningthis with the estimate from below (21) we would conclude that I − µT are invertiblein in all L p , p ∈ I k .So it is enough to have (21). And we would like a good estimate for c ( K, p ) in it.It is enough to prove (21) for the dense set of functions g ∈ C ∞ ( C ) , Z C g dm = 0 . Let φ be the Cauchy transform of g : φ = π R g ( ζ ) ζ − z dm ( ζ ).Denoting h := g − µT g we come to equation φ ¯ z − µφ z = h , in which we want to estimate k φ ¯ z k p ≤ C ( K, p ) k h k p if p ∈ I k . By Theorem 1.1 there is a K -qc homeomorphism f satisfying f ¯ z − µf z = 0. Set u = φ ◦ f − , and let us see how equation φ ¯ z − µφ z = h will be transformed by this change ofvariable.We calculate φ ¯ z = ( u z ◦ f ) f ¯ z + ( u ¯ z ◦ f ) ¯ f z ,φ z = ( u z ◦ f ) f z + ( u ¯ z ◦ f ) ¯ f ¯ z ,φ ¯ z − µφ z − h = ( u z ◦ f ) f ¯ z + ( u ¯ z ◦ f ) ¯ f z − µ (( u z ◦ f ) f z + ( u ¯ z ◦ f ) ¯ f ¯ z ) − h =( u ¯ z ◦ f ) ¯ f z − µ ( u ¯ z ◦ f )¯ µ ¯ f z − h = (1 − | µ | )( u ¯ z ◦ f ) ¯ f z − h = 0 . Hence obviously Z | u ¯ z ◦ f ) | p | f z | p ≤ C ( K ) Z | h | p ⇒ Z | u ¯ z ◦ f ) | p J p/ − f J f ≤ C ( K ) Z | h | p And changing variable we get(22) Z | u ¯ z | p | ( J f − ) − p/ = Z | u ¯ z | p | ( J f ◦ f − ) p/ − ≤ C ( K ) Z | h | p On the other hand, Z | u z ◦ f ) | p | f ¯ z | p ≤ k − k Z | u z ◦ f ) | p | J p/ − f J f = k − k Z | u z | p ( J f − ) − p/ Denote by W := ( J f − ) − p/ . It is the one in Lemma 1.3, only f replaced by f − ,which is a K -qc homeomorphism as well.But the last expression above can be written as Z | u z | p ( J f − ) − p/ = Z | T ( u ¯ z ) | p ( J f − ) − p/ = Z | T ( u ¯ z ) | p W =: T U
ELLMAN FUNCTION TECHNIQUE IN HARMONIC ANALYSIS 7 (in fact, u can be restored from u ¯ z by Cauchy integral with no addition because u vanishes at infinity; then u z = T ( u ¯ z ). Notice that (22) says that(23) Z | u ¯ z | p W ≤ C ( K ) | h | p . Combine (23) with (here F is some “unknown” function on [1 , ∞ ), but finite forall finite arguments) T U = Z | T ( u ¯ z ) | p W = F ([ w ] A ) Z | u ¯ z | p W to get(24) k φ ¯ z k p ≤ C ( Z | u ¯ z | p W + Z | T ( u ¯ z ) | p W ) ≤ C ( K )(1 + F ([ w ] A ) k h k pp . Noticing that Lemma 1.3 gives the estimate [ w ] A ≤ p C ( K )1+ k − p , we conclude finallythat k g k p ≤ C ( K ) F ( p C ( K )1 + k − p ) k h k p , if p ∈ [2 , k ) . We need the same estimate now for 1 + k < p ≤
2. We need W ∈ A p (now p ≤
2. But it is the same as to say that W − / ( p − ∈ A p ′ , p ′ = p/ ( p − W := ( J f − ) − p/ , so W − / ( p − will be ( J f − ) p ′ − , which is inverse to theone in Lemma 1.3, so also in A ⊂ A p ′ . We get for the whole interval of p ’s: p ∈ I k = (1 + k, k ) also(25) k g k p ≤ F (max( p C ( K )1 + k − p , p C ( K )1 + k − p ′ )) k h k p = F ( p C ( K ) dist ( p, R \ I k ) ) k h k p . Theorem 1.1 is proved. (cid:3)
In [2] the following conjecture was formulated that claims that function F in(25) is just linear. Notice that this would in fact easily follow from Big Iwaniec’conjecture. Conjecture (26) k g k p ≤ p C ( K ) dist ( p, R \ I k ) ) k h k p , equivalently k ( I − µT ) − k p ≤ p C ( K ) dist ( p, R \ I k ) ) . We will prove now this conjecture using the Bellman function technique. But firstlet us derive the corollary of the conjecture. As always k µ k ∞ = k < Theorem 1.5 (Corollary of the conjecture) . Any solution of F ¯ z − µF z = 0 , which is in W k ,loc is automatically in W ,loc , and so satisfies Theorem 1.1. It auto-matically self-improves then (by Astala’s [1] ) to be in W k − ,loc . First use Conjecture to prove
Lemma 1.6 (Behavior at the end points of interval I k ) . Operators I − µT, I − T µ have dense range in L k and, correspondingly, trivial kernels on L k . ALEXANDER VOLBERG
Proof. By T ( I − µT ) T − = I − T µ and invertibility of T in all spaces L p , < p < ∞ ,it is enough to prove just the dense range of I − µT in L k . Consider ε > φ ε − (1 − ε ) µT φ ε = h for nice h ∈ C ∞ . We want to consider the solution for p = 1 + k . We consider this p in I (1 − ε ) k because k (1 − ε ) µ k ∞ = (1 − ε ) k . Point p is obviously C ( K ) ε close tothe right end point of I (1 − ε ) k .Hence, applying conjecture we conclude that k φ ε k p ≤ C ( K ) ε k h k p . Notice two things: 1) In L the norma of φ ε are uniformly bounded by C ( K ) justby using Neumann series in L in (27); 2) in L p the norms of T ( εφ ε ) are uniformlybounded. It is immediate to conclude from 1) and 2) that εµ T φ ε converges weakly to zero in L p . Rewrite our equation (27) as follows: φ ε − µT φ ε = h − εµT φ ε . The right hand side weakly in L p converges to any function h , whose family isstrongly dense in L p . So the right hand side is weakly dense in L p = L k . But itis in Range ( I − µT ), so this range is weakly dense in L k . Being a linear set thisrange is then strongly dense in L k . Lemma 1.6 is proved (cid:3) The proof of Theorem 1.5.
Consider R ¯ z − µR z = 0 , R ∈ W k ,loc . Choose φ ∈ C ∞ .Set G = φR . Then G ¯ z − G z = ( φ ¯ z − µφ z ) R .
Looking at this formula we can start to think that the support of µ is compact (iscontained in the support of φ ).As G vanishes at infinity it is the Cauchy transform of its ¯ ∂G = G ¯ z , and therefore, G z = T G ¯ z . We can rewrite the equation( I − µT ) ψ = h ; ψ := G ¯ z , h = ( φ ¯ z − µφ z ) R ∈ L k )1 − k ⊂ L ( C ) ∩ L ε ( C ) . The inclusion above for function R is by Sobolev imbedding, in fact, we assumedthat R ∈ W k ,loc , and by the compactness of the support of R . It has been alreadyremarked, that in the last two equation we have the right to think that µ = 0outside of the support of φ . Let us consider the convergent in L ( C ) of the series ofcompactly supported functions: ψ = h + µT h + µT µT h + . . . . It solves our equation, it is in L ( C ) and it is compactly supported, hence it is in L k ( C ). Therefore we got a solution ψ of ( I − µT ) ψ = h , which is in L k ∩ L .But ψ = G ¯ z is also in L k . By Lemma 1.6 we have ψ = ψ ∈ L . It means that R ∈ W ,loc . Theorem 1.5 is proved. (cid:3) ELLMAN FUNCTION TECHNIQUE IN HARMONIC ANALYSIS 9
Example. f = | z | − K z for z ∈ D , f = z outside D . It is a solution of Beltramiequation with µ, k µ k ∞ = k = K − K +1 , and it is in W q ,loc for every q < k . But it isNOT K- quasiregular mapping, it has a singularity at 0.This example shows how sharp is Theorem 1.5. Its proof hinges on Conjecture 26.We prove this Conjecture now using Bellman technique. First we analyze function F from (25). Recall that W := ( J f − ) − p/ , and if p ∈ (1 + k, k ) the estimate in(25) is F p ([ W ] A ), where F p is the best function one can have in the estimate k T k L p ( W ) ≤ F p ([ W ] A p ) . Suppose we can prove
Theorem 1.7. F ( x ) ≤ C x max(1 , / ( p − . Then we recall that for 1 + k < p ≤ , p ′ := pp − we already estimated [ W ] / ( p − A p =[ W − / ( p − ] A p ′ ≤ [ W − / ( p − ] A ≤ p C ( K ) p − − k . And for 1 + k > p ≥
2, [ W ] A ≤ p C ( K )1+ k − p .These estimates were based on sharp distortion theorem of Astala [1]. We madethese estimates in Lemma 1.3. These estimates and Theorem 1.7 then imply triviallyconjecture (26). Hence this Theorem 1.7 is the only ingredient left to be proved tohave Theorem 1.5.2. Linear estimates of weighted Ahlfors–Beurling transform byBellman function technique
Let ω be any weight on R , denote its heat extension into R by ω ( x, t ) = ω ( x , x , t ): ω ( x, t ) = 1 πt Z Z R ω ( y ) exp( − k x − y k t ) dy dy . We define [ ω ] heatA p := sup ( x,t ) ∈ R ω ( x, t ) (cid:16) ω − p − ( x, t ) (cid:17) p − . The weights w with finite [ w ] heatA p are called A p weights. There is an extensivetheory of A p weights, see for example [55],[40]. The usual definition differs from theone above, but it describes the same class of weights. Actually, we will say moreabout the relationship between the classical definition and ours. But first we statetwo more theorems, whose combined use gives Theorem 1.7 at least for p ≥ Remark.
The method called Rubio de Francia extrapolation–one can see its ex-position in [27]–actually shows that to have a full range of p ’s in Theorem 1.7 it isenough to prove it only for p = 2. Theorem 2.1.
For any A p weight w and any p ≥ we have k T k L p ( wdA ) → L p ( wdA ) ≤ C ( p )([ w ] heatA p ) p − . We want to discuss the connection between [ w ] heatA p and [ w ] classA p . Here [ w ] classA p denotes the following supremum over all discs in the plane:[ w ] classA p := sup B ( x,R ) (cid:18) | B ( x, R ) | Z B ( x,R ) ωdA (cid:19) · (cid:18) | B ( x, R ) | Z B ( x,R ) ω − p − dA (cid:19) p − . Obviously, there exists a positive absolute constant a such that for any function w a [ w ] classA p ≤ [ w ] heatA p . Remark . The opposite inequality is easy to prove too. In fact, we have
Theorem 2.2.
There exists a finite absolute constant b such that [ w ] heatA p ≤ b [ w ] classA p . Proof.
Constants will be denoted by the letters c, C ; they may vary from line to lineand even within the same line. We introduce the following notations. B k denotes B (0 , k ) , k = 0 , , , ... , h f i B stands for the average | B | R B f dA , f ( B ) stands for R B f dA . If B = B (0 , r ), then h f i hB stands for π r RR R f ( x ) exp( − k x k r ) dx dx . Lemma 2.3.
Suppose f and g , positive functions on the plane, are such that sup B h f i B h g i B = A , then there exists a finite absolute constant c such that h f i B h g i hB ≤ cA for any disc B .Proof. Scale invariance allows us to prove this only for one disc B = B (0 , h f i B h g i hB ≤ c h f i B Σ k k exp( − k − ) A h f i B k . On the other hand h f i B k > c h f i B k − > . . . c k h f i B (recall that B is the unit disc).Plugging this in the inequality above, we get h f i B h g i hB ≤ c h f i B Σ k C k exp( − k − ) A h f i B . In other words, h f i B h g i hB ≤ cA Σ k C k exp( − k − ) = cA and the lemma is proved. (cid:3) Now we want to prove Theorem 2.2. Fix B . Again by scale invariance it is enoughto consider B = B (0 , h f i B k h g i hB k ≤ cA for any k .Now h f i hB h g i hB ≤ c h g i hB Σ2 k exp( − k − ) h f i B k ≤ c h g i hB Σ2 k exp( − k − ) cA h g i hB k . The last inequality used (28).On the other hand, h g i hB k > c h g i hB k − > . . . c k h g i hB (recall that B is the unit disc).Plugging this in the inequality above, we get h f i hB h g i hB ≤ c h g i hB Σ k C k exp( − k − ) cA h g i hB . In other words,
ELLMAN FUNCTION TECHNIQUE IN HARMONIC ANALYSIS 11 h f i hB h g i hB ≤ c A Σ k C k exp( − k − ) = c A .
Theorem 2.2 is completely proved. (cid:3)
The next result proves Theorem 1.7 for p = 2. We will show later how to extrap-olate just from the result at p = 2 to all possible p ’s. Theorem 2.4.
For any A weight w we have k T k L ( wdA ) → L ( wdA ) ≤ C [ w ] classA . Proof.
There will be many steps. But we are going to prove only Theorem 2.1 andonly for p = 2. By Theorem 2.2 and Rubio de Francia extrapolation this is enough.The operator T is given in the Fourier domain ( ξ , ξ ) by the multiplier ζ ¯ ζ = ζ | ζ | = ( ξ + iξ ) ξ + ξ = ξ ξ + ξ − ξ ξ + ξ + 2 i ξ ξ ξ + ξ . Thus, T can be written as T = R − R + 2 iR R ,where R , R are Riesz transforms on the plane (see [55] for their definition andproperties). Another way of writing T is T = m + im , where m , m are Fourier multiplier operators. Notice that the multipliers them-selves (as functions, not as multiplier operators) are connected by m = m ◦ ρ , where ρ is π/ operators are related by m = U ρ m U − ρ , where U ρ is an operator of ρ -rotation in ( x , x ) plane. But for any operator K wehave k U ρ K U − ρ k L ( wdA ) → L ( wdA ) = k K k L ( w ◦ ρ − dA ) → L ( w ◦ ρ − dA ) . Combining this with the fact that Q heatw, = Q heatw ◦ ρ − , for any rotation, we concludethat we only need the desired estimate of Theorem 2.4 for m = R − R . Actually,we will show that(29) k R i k L ( wdA ) → L ( wdA ) ≤ C Q heatw, , i = 1 , . To prove (2.6) we fix, say, R and two test functions ϕ, ψ ∈ C ∞ . We will be usingheat extensions. For f on the plane, its heat extension is given by the formula f ( y, t ) := 1 π t Z Z R f ( x ) exp( − | x − y | t ) dx dx , ( y, t ) ∈ R . We usually use the same letter to denote a function and its heat extension.
Lemma 2.5.
Let ϕ, ψ ∈ C ∞ . Then the integral RRR ∂ϕ∂x · ∂ψ∂x dx dx dt convergesabsolutely and (30) Z Z R ϕ · ψ dx dx = − Z Z Z ∂ϕ∂x · ∂ψ∂x dx dx dt . Proof.
The proof of this lemma is actually trivial. It is based on the well-known factthat a function is an integral of its derivative, and also involves Parseval’s formula.Consider ϕ, ψ ∈ C ∞ and now Z Z ψR ϕdx dx = Z Z ξ ξ + ξ ˆ ϕ ( ξ , ξ ) ˆ ψ ( − ξ , − ξ ) dξ dξ =2 Z Z Z ∞ e − t ( ξ + ξ ) ξ ˆ ϕ ( ξ , ξ ) ˆ ψ ( ξ , ξ ) dξ dξ dt = − Z ∞ Z Z iξ ˆ ϕ ( ξ , ξ ) e − t ( ξ + ξ ) · iξ ˆ ψ ( − ξ , − ξ ) e − t ( ξ + ξ ) dξ dξ dt = − Z ∞ Z Z ∂ϕ∂x ( x , x , t ) ∂ψ∂x ( x , x , t ) dx dx dt = − Z Z Z R ∂ϕ∂x ( x , x , t ) ∂ψ∂x ( x , x , t ) dx dx dt . Above we used Parseval’s formula twice, and also we used the absolute convergenceof the integrals
Z Z Z R e − t ( ξ + ξ ) ξ ˆ ϕ ( ξ , ξ ) ˆ ψ ( ξ , ξ ) dξ dξ dt , Z Z Z R ∂ϕ∂x ( x , x , t ) ∂ψ∂x ( x , x , t ) dx dx dt . For the first integral this is obvious. The absolute convergence of the second integralcan be easily proved. We leave this as an exercise for the reader . (cid:3)
Our next goal is to estimate the right side of (30) from above.
Theorem 2.6.
For any ϕ, ψ ∈ C ∞ , and any positive function w on the plane wehave Z Z Z R (cid:12)(cid:12)(cid:12)(cid:12) ∂ϕ∂x (cid:12)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12)(cid:12) ∂ψ∂x (cid:12)(cid:12)(cid:12)(cid:12) dx dx dt ≤ A Q heatw, (cid:18)Z Z | ϕ | w dx dx + Z Z | ψ | w dx dx (cid:19) where A is an absolute constant. Bellman function
In the proof we at last use a Bellman function tailored for this problem. It is B from the following theorem. The meaning of Q in the next theorem is Q := [ w ] heatA .We use the notation H f for the Hessian matrix of function f : R k → R (thematrix of second derivatives of f ), and d f for the second differential form, whichis the quadratic form ( H f ( x ) dx, dx ), where ( · , · ) is the usual scalar product in R k , x is a point in the domain of definition of f , dx is an arbitrary vector in R k . ELLMAN FUNCTION TECHNIQUE IN HARMONIC ANALYSIS 13
Theorem 2.7.
For any
Q > define the domain D Q := { < ( X, Y, x, y, r, s ) : x < Xs, y < Y r, < rs < Q } . Let K be any compact subset of D Q . Thenthere exists a function B = B Q,K ( X, Y, x, y, r, s ) infinitely differentiable in a smallneighborhood of K such that
1) 0 ≤ B ≤ Q ( X + Y ) , − d B ≥ | dx || dy | . We prove Theorem 2.7 later. Now we use it to obtain the proof of Theorem 2.6.
Proof.
Given a non-constant smooth w that is constant outside some large ball, weconsider Q = Q heatw, . We treat only the case w ∈ A , that is Q < ∞ , for otherwisethere is nothing to prove. Consider two nonnegative functions ϕ, ψ ∈ C ∞ . Now take B = B Q,K , where a compact K remains to be chosen.We are interested in b ( x, t ) := B (( ϕ w )( x, t ) , ( ψ w − )( x, t ) , ϕ ( x, t ) , ψ ( x, t ) , w ( x, t ) , w − ( x, t )) . This is a well defined function, because the choice of Q ensures that the 6-vector v , consisting of heat extensions of corresponding functions on R , v := (( ϕ w )( x, t ) , ( ψ w − )( x, t ) , ϕ ( x, t ) , ψ ( x, t ) , w ( x, t ) , w − ( x, t ))lies in D Q for any ( x, t ) ∈ R . Also we can fix any compact subset M of the openset R and guarantee that for ( x, t ) ∈ M , the vector v lies in a compact K . In fact,notice that for our w and for compactly supported ϕ, ψ the mapping ( x, t ) → v ( x, t )maps compacts in R to compacts in D Q . Now just take K large enough.The main object we want to study is(31) (cid:18) ∂∂t − ∆ (cid:19) b ( x, t ) . For simplicity we assume that B is already C up to the boundary of D Q . Thetechnical details what to do without this assumption are left to the audience, see[58]. We want to estimate the expression in (31) 1) from above in average and 2)from below in a pointwise way.1) Take a “slab” S ε,H := { ( x, t ) ∈ R : ε ≤ t ≤ H } . Notice that for any fixedpositive t Z R ∆ b ( x, t ) dx = 0 . This is because we assumed B to be smooth and because v ( x, t ) → t when x → ∞ rather fast, and the same is true for ∇ v ( x, t ). Hence, Z S ε,H (cid:18) ∂∂t − ∆ (cid:19) b ( x, t ) dxdt = Z S ε,H ∂∂t b ( x, t ) dxdt = Z R b ( x, H ) dx − Z R b ( x, ε ) dx . Now we recall that b = B ◦ v , that B ≥ B ( X, Y, . . . ) ≤ Q ( X + Y ). Then we get (functions below are heatextensions of the corresponding symbols on R ): Z S ε,H (cid:18) ∂∂t − ∆ (cid:19) b ( x, t ) dxdt ≤ (32) 5 Q Z R ( ϕ w ( x, H ) + ψ w − ( x, H )) dx = 5 Q Z R [( ϕ w )( x ) + ( ψ w − )( x )] dx .
2) Now we make a pointwise estimate of (31) from below. The next calculationis simple but it is key to the proof. In it as everywhere v = (( ϕ w )( x, t ) , ( ψ w − )( x, t ) , ϕ ( x, t ) , ψ ( x, t ) , w ( x, t ) , w − ( x, t )) . Lemma 2.8. (cid:18) ∂∂t − ∆ (cid:19) b ( x, t ) = (cid:18)(cid:0) − d B (cid:1) ∂v∂x , ∂v∂x (cid:19) R + (cid:18)(cid:0) − d B (cid:1) ∂v∂x , ∂v∂x (cid:19) R . Proof. ∂∂t b = ( ∇ B, ∂v∂t ) R , ∆ b = (cid:18) ( d B ) ∂v∂x , ∂v∂x (cid:19) R + (cid:18) ( d B ) ∂v∂x , ∂v∂x (cid:19) R + ( ∇ B, ∆ v ) R . We just used the chain rule. Now (cid:18) ∂∂t − ∆ (cid:19) b = (cid:18) ∇ B, ( ∂v∂t − ∆ v ) (cid:19) R − (cid:18) ( d B ) ∂v∂x , ∂v∂x (cid:19) R − (cid:18) ( d B ) ∂v∂x , ∂v∂x (cid:19) R . However, the first term is zero because all entries of the vector v are solutions of theheat equation. (cid:3) By Theorem 2.7 − d B ≥ | dx || dy | . Therefore, for ( x, t ) Lemma 2.8 gives:(33) (cid:18) ∂∂t − ∆ (cid:19) b ( x, t ) ≥ (cid:12)(cid:12)(cid:12)(cid:12) ∂ϕ∂x (cid:12)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12)(cid:12) ∂ψ∂x (cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12) ∂ϕ∂x (cid:12)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12)(cid:12) ∂ψ∂x (cid:12)(cid:12)(cid:12)(cid:12) . Combining (32) (33) we get(34)
Z Z Z S ε,H (cid:18)(cid:12)(cid:12)(cid:12)(cid:12) ∂ϕ∂x (cid:12)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12)(cid:12) ∂ψ∂x (cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12) ∂ϕ∂x (cid:12)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12)(cid:12) ∂ψ∂x (cid:12)(cid:12)(cid:12)(cid:12)(cid:19) ≤ Q (cid:18)Z Z ϕ w + Z Z ψ w − (cid:19) . Theorem 2.6 is completely proved by using a Bellman function of our problemwhose existence is claimed in Theorem 2.7 (cid:3)
Theorem 2.4 is proved. (cid:3)
Remark.
The proof of Theorem 2.6 is actually “equivalent” to solution of anobstacle problem for a certain fully non-linear PDE. Consider σ = (cid:20) , , (cid:21) . Then we need H B ± σ ≥ D Q . As we a looking for the best possible B satisfying these relationships, it is natural that we should requiredet( H B + σ ) = 0 or det( H B − σ ) = 0 , ELLMAN FUNCTION TECHNIQUE IN HARMONIC ANALYSIS 15 these are Monge–Amp`ere equations.
This approach is used in [59] , [54] . Nowwe use another method to prove the existence of B Q required in Theorem2.7. More Bellman functions to prove the existence of Bellman function B Q from Theorem 2.7. Dyadic shifts. We start with a much simpler “model”operator— T σ . The logic will be the following. We want to get a sharp weightedestimate of k T σ k L ( w ) → L ( w ) via the A characteristic of w . In the paper of Nazarov,Treil, Volberg, see [50], one can find that the norm k T σ k L ( u ) → L ( v ) is attained onsome “simple” test functions—and that this holds for every pair u, v . Thus alsofor u = v = w . However, on the family T of test functions one can compute the N w, ( T σ ) := sup {k T σ t k L ( w ) : t ∈ T , k t k L ( w ) = 1 } . It turns out that Theorem 2.9. N w, ( T σ ) ≈ Q classw, . J. Wittwer does that in [63] basing her approach on [50]; see also [56]. Thus, weget k T σ k L ( w ) → L ( w ) = N w, ( T σ ) ≈ Q classw, .So let us show what the model operator is, what its sharp weighted estimate is,and how one obtains a special function (Bellman function) from this estimate.Consider the family of dyadic singular operators T σ : T σ f = Σ I ∈D σ ( I ) ( f, h I ) h I . Here D is a dyadic lattice on R , h I is a Haar function associated with the dyadicinterval I ( h I is normalized in L ( R , dx )), and σ ( I ) = ±
1. We call the family T σ themartingale transform . It is a dyadic analog of a Calder´on-Zygmund operator. Hereare important questions about T σ , the first one about two-weight estimates and thesecond one about one weight estimates:1) What are necessary and sufficient conditions for sup σ k T σ k L ( u ) → L ( v ) < ∞ ?2) What is the sharp bound on sup σ k T σ k L ( w ) → L ( w ) in terms of w ? How can onecompute sup σ k T σ k L ( w ) → L ( w ) ?These questions are dyadic analogs of notoriously difficult questions about “clas-sical” Calder´on-Zygmund operators like the Hilbert transform, the Riesz transformsand the Ahlfors-Beurling transform. The dyadic model is supposed to be easier thanthe continuous one. This turned out to be true. The answers to the questions aboveappeared in [50], [63]. Moreover these answers are key to answering questions about“classical” Calder´on-Zygmund operators.Strangely enough, the answer to the second question (which seems to be easier,because it is about “one weight”) seems to require the ideas from the “two-weight”case. Here is our explanation of this phenomena. The necessary and sufficientconditions on ( u, v ) to answer the first question were given in [50]. They amountto the fact that sup σ k T σ k L ( u ) → L ( v ) is almost attained on the family of simple testfunctions. This fact has beautiful consequences in the one weight case. For thensup σ k T σ k L ( w ) → L ( w ) is attainable (almost) on the family of simple test functions.One may try to compute sup σ k T σ t k L ( w ) for every element of this test family, thusgetting a good estimate for the norm sup σ k T σ k L ( w ) → L ( w ) . Test functions are rather simple, so this program can be carried out. This has been done in Wittwer’s paper[63]. We will give another proof below. Here is the result. Recall that Q dyadicw, := sup I ∈D h w i I h w − i I . Theorem 2.10. sup σ k T σ k L ( w ) → L ( w ) ≤ A Q dyadicw, . Remark.
We will postpone the proof of Theorem 2.10 (another use of
Bellmanfunction technique ), here we will use it first to finish the proof of the existence of B Q claimed in Theorem 2.6.So we assume now that Theorem 2.10 is already proved. Let us rewrite Theorem2.10 as followssup σ ( I )= ± | Σ I ∈D σ ( I ) ( f, h I ) ( g, h I ) | ≤ AQ dyadicw, k f k L ( w ) k g k L ( w − ) , or Σ I ∈D | ( f, h I ) | | ( g, h I ) | ≤ AQ dyadicw, k f k L ( w ) k g k L ( w − ) . This inequality is scaleless, so we write it as(35) J ∈ D , | J | Σ I ∈D , I ⊂ J |h f i I − − h f i I + | |h g i I − − h g i I + || I | ≤ AQ dyadicw, h f w i / J h g /w i / J . Here I − , I + are the left and the right halves of I , and h·i l means averaging over l as usual. Given a fixed J ∈ D and a number Q >
1, we wish to introduce theBellman function of (35): B ( X, Y, x, y, r, s ) = sup { | J | Σ I ∈D , I ⊂ J |h f i I − − h f i I + | |h g i I − − h g i I + || I | : h f i J = x, h g i J = y, h w i J = r, h w − i J = s, h f w i J = X, h g /w i J = Y, w ∈ A dyadic , Q dyadicw, ≤ Q } . Obviously, the function B does not depend on J , but it does depend on Q . Itsdomain of definition is the following: R Q := { ≤ ( X, Y, x, y, r, s ) , x ≤ Xs, y ≤ Y r, ≤ rs ≤ Q } . By (35) it satisfies(36) 0 ≤ B ≤ AQX / Y / . We are going to prove that it also satisfies the following “differential” inequality.Denote v := ( X, Y, x, y, r, s ), v − = ( X − , Y − , x − , y − , r − , s − ), v + = ( X + , Y + , x + , y + , r + , s + ),let v, v + , v − lie in R Q , and let v = ( v − + v + ). Then(37) B ( v ) −
12 ( B ( v + ) + B ( v − )) ≥ | x + − x − || y + − y − | . ELLMAN FUNCTION TECHNIQUE IN HARMONIC ANALYSIS 17
In fact, let f, g, w almost maximize B ( v ) (on the interval J ), let f + , g + , w + do thisfor B ( v + ), f − , g − , w − do this for B ( v − ). The freedom of scale for B allows us to put f + , g + , w + on J + and f − , g − , w − on J − . Then we have “gargoyle” functions F = ( f + on J + f − on J − G = ( g + on J + g − on J − W = ( w + on J + w − on J − . Obviously, h F i J = ( x + + x − ) = x, h G i J = y, h W i J = r, h W − i J = s, h F W i J = X, h G W − i J = Y . These numbers together form the vector v . In other words F, G, W compete with f, g, w in the definition (35) of Bellman function B ( v ). Bythis definition, B ( v ) ≥ | J | Σ I ∈D , I ⊂ J |h F i I − − h F i I + | |h G i I − − h G i I + || I | . But the almost optimality of f + , g + , w + on J + and f − , g − , w − on J − gives us (recallthat F = f ± on J ± , G = g ± on J ± ): B ( v + ) ≤ ε + 14 | J + | Σ I ∈D , I ⊂ J + |h F i I − − h F i I + | |h G i I − − h G i I + || I | , and B ( v − ) ≤ ε + 14 | J − | Σ I ∈D , I ⊂ J − |h F i I − − h F i I + | |h G i I − − h G i I + || I | . Combining these, we get B ( v ) −
12 ( B ( v + ) + B ( v − )) ≥ − ε + 14 |h F i J − − h F i J + | |h G i J − − h G i J + = − ε + 14 |h f − i J − − h f + i J + | |h g − i J − − h g + i J + = − ε + 14 | x − − x + || y − − y + | . We are done with (37) because ε is an arbitrary positive number. Therefore, our B is a very concave function. We are going to modify B to have its Hessian satisfythe conclusion of Theorem 2.7. To do that we fix a compact K in the interior of R Q , and we choose ε such that 100 ε < dist ( K, ∂R Q ). Consider the convolution of B with ε ϕ ( vε ) , v ∈ R , where ϕ is a bell shape infinitely differentiable function withsupport in the unit ball of R . It is now very easy to see that this convolution (wecall it B K,Q ) satisfies the following inequalities(38) 0 ≤ B K,Q ≤ Q ( X + Y ) , and for any vector ξ = ( ξ , ξ , ξ , ξ , ξ , ξ ) ∈ R ,(39) − ( d B K,Q ξ, ξ ) R ≥ | ξ || ξ | . The factor 2 appears because B ( v ) − ( B ( v + ) + B ( v − )) in (37) corresponds to − d B , and | x − x + | = | x − − x + | (the same being valid with y ’s replacing x ’s and − replacing +).Theorem 2.7 is completely proved modulo the proof of Theorem 2.10. Proof of Theorem 2.10.
To prove Theorem 2.10 we need the following decomposi-tion:
Lemma 2.11. (40) h I = α I h wI + β I χ I √ I , where1) | α I | ≤ p h w i I ,2) | β I | ≤ | ∆ I w |h w i I , where ∆ I w := h w i I + − h w i I + ,3) { h wI } I is an orthonormal basis in L ( w ) ,4) h wI assumes on I two constant values, one on I + and another on I − .Proof. To find α, β we first apply k · k L ( w ) to both parts of (40): h w i I = k h I k L ( w ) = α + β h w i I , and secondly we multiply (40) by χ I / p | I | and integrate with respectto w dx : ( h w i I + − h w i I + ) = β I h w i I . Clearly Lemma is proved. (cid:3) Now let S F := X I c I ( f, h I ) h I , where constants c I are such that | c I | ≤ . Let σ := w − for the rest of the proof. Fix φ ∈ L ( w ) , ψ ∈ L ( σ ). We need toprove(41) | ( S φw, ψσ ) | ≤ C k φ k w k ψ k σ . We estimate ( S φw, ψσ ) as | X I c I ( φw, h I )( ψσ, h I ) | ≤ X I | c I ( φw, h wI ) p h w i I ( ψσ, h σI ) | p h σ i I | + X I | c I h φw i I ∆ I w h w i I ( ψσ, h σI ) p h σ i I √ I | + X I | c I h ψσ i J ∆ I σ h σ i I ( φw, h wI ) p h w i I √ I | + X I | c I h φw i I h ψσ i J ∆ I w h w i I ∆ I σ h σ i I √ I √ I | =: I + II + III + IV .
So we have I ≤ X I ( φw, h wI ) p h w i I · ( ψσ, h σI ) p h σ i I , II ≤ X I ( φw, h wI ) p h w i I ·h ψσ i I | ∆ I σ |h σ i I p | I | ,III ≤ X I h φw i I | ∆ I w |h w i I p | I |· ( ψσ, h σI ) p h σ i I , IV ≤ X I h φw i I | ∆ I w |h w i I p | I |·h ψσ i I | ∆ I σ |h σ i I p | I | . ELLMAN FUNCTION TECHNIQUE IN HARMONIC ANALYSIS 19
The estimate of I is trivial because h wI , h σI are orthonormal systems in L ( w ) , L ( σ )correspondingly:(42) I ≤ sup I p h w i I h σ i I sX I ( φw, h wI ) sX I ( ψσ, h σI ) ≤ [ w ] / A k φ k w k ψ k σ . To estimate the rest let us fix α ∈ (0 , /
2) and introduce(43) µ I := h w i αI h σ i αI (cid:18) | ∆ I w | h w i I + | ∆ I σ | h σ i I (cid:19) | I | . We are going to give a Bellman function proof of the following lemma.
Lemma 2.12.
The sequence { µ I } I ∈ D is a Carleson sequence with Carleson constantat most C [ w ] αA . We take Lemma 2.12 for granted till the end of the proof of Theorem 2.10. Firstintroduce a notation, let µ be a positive measure on R , then M dµ f ( x ) := sup I ∈ D,x ∈ I µ ( I ) Z I | f | dµ . This is called dyadic weighted maximal function. We will use it with µ = wdx or σdx .To estimate IV , II , and symmetric to it III we notice that | ∆ I σ |h σ i I p | I | ≤ h w i − α/ h σ i − α/ √ µ I , so, choosing p ∈ (1 , h ψσ i I | ∆ I σ |h σ i I p | I | ≤ h w i − α/ h σ i − α/ ( h| ψ | p σ i I ) /p h σ i − /p √ µ I ≤h w i − α/ h σ i − α/ I inf x ∈ I ( M dσ | ψ | p ( x )) /p · √ µ I , where M dσ is the dyadic weighted maximal function. Therefore, IV ≤ X I h w i − α h σ i − αI inf I ( M dσ | ψ | p ) /p · inf I ( M dw | φ | p ) /p · µ I .II ≤ X I ( φw, h wI ) h w i − α/ I h σ i − α/ I inf I ( M dσ | ψ | p ) /p h w i / I · √ µ I . The estimate of
III will be totally symmetric, so we omit it. We continue: IV ≤ [ w ] − αA X I inf I ( M dσ | ψ | p ) /p · inf I ( M dw | φ | p ) /p · µ I .II ≤ [ w ] − α/ A sX I ( φw, h wI ) sX I inf I ( M dσ | ψ | p ) /p h w i I · µ I . Choose F = ( M dσ | ψ | p ) /p · ( M dw | φ | p ) /p and G = ( M dσ | ψ | p ) /p and apply the follow-ing simple lemma ( Exercise !) Lemma 2.13.
Let { α L } L ∈ D define Carleson measure with intensity B . Let F be apositive function on the line. Then (44) X L (inf L F ) α L ≤ B Z R F dx . (45) X L inf L G h w i L α L ≤ C B Z R Gw dx .
Then using Lemma 2.12 we get IV ≤ [ w ] − αA [ w ] αA Z R ( M dσ | ψ | p ) /p · ( M dw | φ | p ) /p dx =[ w ] A Z R ( M dσ | ψ | p ) /p · ( M dw | φ | p ) /p w / σ / dx ≤ [ w ] A sZ ( M dw | φ | p ) /p wdx sZ ( M dσ | φ | p ) /p σdx ≤ C [ w ] A k φ k w k ψ k σ . As to II , we have again using Lemma 2.13 (the second part) and Lemma 2.12: II ≤ C [ w ] − α/ A [ w ] α/ A sX I ( φw, h wI ) sZ R ( M dσ | ψ | p ) /p ( x ) w ( x ) dx ≤ C [ w ] A k φ k w sZ R ( M dσ | ψ | p ) /p ( x ) σ ( x ) dx ≤ C [ w ] A k φ k w k ψ k σ . Theorem 2.10 is completely proved, function B Q is constructed. (cid:3) We need only to see the validity of Lemma 2.12. This is done by yet another
Bellman function . Bellman proof of Lemma 2.12.
We prove even a more general statement, namely weprove the version in R d and even in each metric space with geometric doublingcondition and doubling measure µ . So let us have a metric space with geometricdoubling condition, meaning that every ball of radius r can fit only at most K disjoint balls of radius r/ K being independent of the ball and its radius. Suchmetric spaces carry a doubling measure µ by a theorem of Konyagin–Volberg [48],and let D denote the family of “dyadic cubes” on this metric space (constructionsare numerous, the first belongs to M. Christ [26]), and let s i ( I ) are dyadic childrenof I ∈ D . Finally, let I ∈ D and let µ I := ( h w i µ,I h σ i µ,I ) α (cid:18) ( h w i µ,s i ( I ) − h w i µ,I ) h w i µ,I + ( h σ i µ,s i ( I ) − h σ i µ,I ) h σ i µ,I (cid:19) µ ( I ) . Lemma 2.12 becomes the following statement, which we are proving below:(46) ∀ I ∈ D X J ∈ D,J ⊂ I µ J ≤ C α [ w ] αµ,A µ ( I ) . ELLMAN FUNCTION TECHNIQUE IN HARMONIC ANALYSIS 21
Let
Q > , < α < . In domain Ω Q := { ( x, y ) : x > , y > , < xy ≤ Q } function b Q ( x, y ) := x α y α satisfies the following estimate of its Hessian matrix (ofits second differential form, actually) − d b Q ( x, y ) ≥ α (1 − α ) x α y α (cid:18) ( dx ) x + ( dy ) y (cid:19) . The form − d b Q ( x, y ) ≥ x > , y >
0. Also obviously 0 ≤ b Q ( x, y ) ≤ Q α in Ω Q . Proof.
Direct calculation. (cid:3)
Fix now a cube I and let s i ( I ) , i = 1 , ..., M , be all its sons. Let a = ( h w i µ,I , h σ i µ,I ), b i = ( h w i µ,s i ( I ) , h σ i µ,s i ( I ) ), i = 1 , . . . , M , be points–obviously–in Ω Q , where Q tem-porarily means [ w ] A . Consider c i ( t ) = a (1 − t ) + b i t, ≤ t ≤ q i ( t ) := b Q ( c i ( t )).We want to use Taylor’s formula(47) q i (0) − q i (1) = − q ′ i (0) − Z dx Z x q ′′ i ( t ) dt . Notice two things: Sublemma shows that − q ′′ i ( t ) ≥ t ∈ [0 , / − q ′′ i ( t ) ≥ c ( h w i µ,I h σ i µ,I ) α (cid:18) ( h w i µ,s i ( I ) − h w i µ,I ) h w i µ,I + ( h σ i µ,s i ( I ) − h σ i µ,I ) h σ i µ,I (cid:19) This requires a small explanation. If we are on the segment [ a, b i ], then the firstcoordinate of such a point cannot be larger than C h w i µ,I , where C depends only ondoubling of µ (not w ). This is obvious. The same is true for the second coordinatewith the obvious change of w to σ . But there is no such type of estimate from belowon this segment: the first coordinate cannot be smaller than k h w i µ,I , but k may(and will) depend on the doubling of w (so ultimately on its [ w ] A norm). In fact, atthe “right” endpoint of [ a, b i ] the first coordinate is h w i µ,s i ( I ) ≤ R I w dµ/µ ( s i ( I )) ≤ C R I w dµ/µ ( I ) = C h w i µ,I , with C only depending on the doubling of µ . But theestimate from below will involve the doubling of w , which we must avoid. But if t ∈ [0 , / a, b i ] then obviously the firstcoordinate is ≥ h w i µ,I and the second coordinate is ≥ h σ i µ,I .We do not need to integrate − q ′′ i ( t ) for all t ∈ [0 ,
1] in (47). We can only useintegration over [0 , /
2] noticing that − q ′′ i ( t ) ≥ q ′′ i ( t ) = ( b Q ( c i ( t )) ′′ = ( d b Q ( c i ( t ))( b i − a ) , b i − a ) , (where ( · , · ) means the usual scalar product in R ) immediately gives us (48) withconstant c depending on the doubling of µ but independent of the doubling of w .Next step is to add all (47), with convex coefficients µ ( s i ( I )) µ ( I ) , and to notice that P Mi =1 µ ( s i ( I )) µ ( I ) q ′ i (0) = ∇ b Q ( a ) P Mi =1 · ( a − b i ) µ ( s i ( I )) µ ( I ) = 0, because by definition a = M X i =1 µ ( s i ( I )) µ ( I ) b i . Notice that the addition of all (47), with convex coefficients µ ( s i ( I )) µ ( I ) gives us now (wetake into account (48) and positivity of − q ′′ i ( t )) b Q ( a ) − M X i =1 µ ( s i ( I )) µ ( I ) b Q ( b i ) ≥ c c ( h w i µ,I h σ i µ,I ) α M X i =1 (cid:18) ( h w i µ,s i ( I ) − h w i µ,I ) h w i µ,I + ( h σ i µ,s i ( I ) − h σ i µ,I ) h σ i µ,I (cid:19) . We used here the doubling of µ again, by noticing that µ ( s i ( I )) µ ( I ) ≥ c (recall that s i ( I )and I are almost balls of comparable radii). We rewrite the previous inequalityusing our definition of ∆ I w, ∆ I σ listed above as follows µ ( I ) b Q ( a ) − M X i =1 µ ( s i ( I )) b Q ( b i ) ≥ c c ( h w i µ,I h σ i µ,I ) α (cid:18) (∆ I w ) h w i µ,I + (∆ I σ ) h σ i µ,I (cid:19) µ ( I ) . Notice that b Q ( a ) = h w i αµ,I h σ i αµ,I . Now we iterate the above inequality and get forany of dyadic I ’s: X J ⊂ I ,J ∈ D ( h w i µ,J h σ i µ,J ) α (cid:18) (∆ J w ) h w i µ,J + (∆ J σ ) h σ i µ,J (cid:19) µ ( J ) ≤ C Q α µ ( I ) . This is exactly the Carleson property of the measure { µ I } indicated in our Lemma2.12, with Carleson constant C Q α . The proof showed that C depended only on α ∈ (0 , /
2) and on the doubling constant of measure µ . Lemma 2.12 is completelyproved. (cid:3) Estimates for Ahlfors–Beurling operator. Towards the BigIwaniec problem by Bellman footsteps
In the previous section we estimated AB operator T in weighted L ( w ). Theestimate was sharp in [ w ] A :(49) | ( T f, g ) | ≤ C [ w ] A k f k L ( w ) k g k L ( w − ) , it implied a sharp in [ w ] A p estimate in weighted L p ( w ):(50) | ( T f, g ) | ≤ C ( p ) [ w ] max(1 , p − ) A p k f k L p ( w ) k g k L p ′ ( w − / ( p − ) , p ′ := p/ ( p − . But we did not care about
C, C ( p ) at all. Now we consider just w = 1, but wecare about C ( p ) very much. Big Iwaniec’s problem conjectures(51) C ( p ) = max( p, p/ ( p − − p ∗ − . This is open at the moment of writing this phrase. Using various
Bellmanfunctions we will show the row of improvements(52) C ( p ) ≤ p ∗ − . (53) C ( p ) ≤ . p ∗ − . (54) C ( p ) ≤ . p ∗ − . (55) C ( p ) ≤ . p ∗ − . ELLMAN FUNCTION TECHNIQUE IN HARMONIC ANALYSIS 23
Recall things that we already know: 1) T = R − R + 2 iR R , where R i are Riesztransforms = multipliers with symbol ξ i / ( | ξ | + | ξ | ) / , i = 1 , R i f, ¯ g ) = − Z Z R ∂f∂x i ∂g∂x i dxdt , where f, g in the left hand side are from C ∞ ( R ), and f, g in the right hand side areheat extensions of functions in the left.Hence Conjecture 51 is nothing else as the following innocent looking conjecture (57) 2 (cid:12)(cid:12)(cid:12)(cid:12) Z Z R (cid:18) ∂f∂x + i ∂f∂x (cid:19) · (cid:18) ∂g∂x + i ∂g∂x (cid:19) dxdt (cid:12)(cid:12)(cid:12)(cid:12) ≤ ( p − k f k p k g k p ′ , p > . Let complex-valued functions f = u + iv, g = φ + iψ . Consider f := ( u, v ) as amap R → R , do the same with G := ( φ, ψ ). We have Jacobian matrices DF, DG then. These are 2 × too much! ):(58) 2 Z Z R (cid:12)(cid:12)(cid:12)(cid:12) ∂f∂x + i ∂f∂x (cid:12)(cid:12)(cid:12)(cid:12) · (cid:12)(cid:12)(cid:12)(cid:12) ∂g∂x + i ∂g∂x (cid:12)(cid:12)(cid:12)(cid:12) dxdt ≤ ( p − k f k p k g k p ′ , p > . This is exactly(59)2
Z Z R ( | DF | − DF ) / ( | DG | − DG ) / dxdt ≤ ( p − k f k p k g k p ′ , p > , where | · | is the Hilbert-Schmidt norm of the matrix.Nobody can prove (59) and equivalent to it (58). They may be wrong!However, we will start with proving slightly lighter estimates:(60) 2 Z Z R (cid:12)(cid:12)(cid:12)(cid:12) ∂f∂x (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∂g∂x (cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12) ∂f∂x (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∂g∂x (cid:12)(cid:12)(cid:12)(cid:12) dxdt ≤ ( p − k f k p k g k p ′ , p > . Moreover, we will prove a stronger than (60) (but weaker than (58)) estimate(61)2
Z Z R (cid:18)(cid:12)(cid:12)(cid:12)(cid:12) ∂f∂x (cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12) ∂f∂x (cid:12)(cid:12)(cid:12)(cid:12) (cid:19) / (cid:18)(cid:12)(cid:12)(cid:12)(cid:12) ∂g∂x (cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12) ∂g∂x (cid:12)(cid:12)(cid:12)(cid:12) (cid:19) / dxdt ≤ ( p − k f k p k g k p ′ , p > . This will give us (52), (53) correspondingly. To get to (54) and further improve-ments as (55) we will need a bit more (stochastic integrals).Notice that we already know (by (56)) that (60) immediately proves the following
Theorem 3.1. k R − R k p ≤ p − , p ≥ . Because 2 R R = U ◦ ( R − R ) ◦ U − , where U is an isometry in all L p spaces(in fact, U is the rotation of the argument of function by 45 ◦ ), we get (52) fromdoubling the claim of Theorem 3.1. proof of (60) . The first step is by examination of what we already had in Section2.1 after the statement of Theorem 2.10. We do now exactly the same:
Suppose we have the following inequality for functions on interval [0 , provided with dyadic lattice D :(62) Σ I ∈D | ( f, h I ) | | ( g, h I ) | ≤ ( p − k f k L p k g k L p ′ , p ≥ . This inequality is scaleless, so we write it as(63) J ∈ D , | J | Σ I ∈D , I ⊂ J |h f i I − − h f i I + | |h g i I − − h g i I + || I | ≤ ( p − h| f | p i /pJ h| g | p ′ i /p ′ J . Here I − , I + are the left and the right halves of I , and h·i l means averaging over l as usual. Given a fixed J ∈ D , p ≥
2, we wish to introduce the Bellman function of(35): B p ( X, Y, x, y ) = sup { | J | Σ I ∈D , I ⊂ J |h f i I − − h f i I + | |h g i I − − h g i I + || I | : h f i J = x, h g i J = y, h| f | p i J = X, h| g | p ′ i J = Y } . Obviously, the function B does not depend on J , but it does depend on p . Itsdomain of definition is the following: R p := { ( X, Y, x , y ) , | x | p ≤ X, | y | p ′ ≤ Y } . By (62) it satisfies(64) 0 ≤ B ≤ ( p − X /p Y /p ′ . We are going to prove that it also satisfies the following “differential” inequality.Denote v := ( X, Y, x , y ), v − = ( X − , Y − , x − , y − ), v + = ( X + , Y + , x + , y + ), let v, v + , v − lie in R p , and let v = ( v − + v + ). Then(65) B ( v ) −
12 ( B ( v + ) + B ( v − )) ≥ | x + − x − || y + − y − | . The proof is verbatim the same as in Section 2.1. And this inequality in infinites-imal sense becomes(66) d B p ≥ | d x || d y | . Having the function B p satisfying1) 0 ≤ B p ≤ ( p − X /p Y /p ′ ;2) − d B p ≥ | d x || d y | .Assuming that B p is sufficiently smooth (which incidentally it is, one can writethe formula for B p ), we can repeat verbatim we can repeat the proof of Theorem2.6: we start with analyzing ( x = ( x , x ) ∈ R )(67) (cid:18) ∂∂t − ∆ (cid:19) b ( x, t ) ELLMAN FUNCTION TECHNIQUE IN HARMONIC ANALYSIS 25 exactly as in the proof of Theorem 2.6: the only difference that b now is not B Q ◦ v but our B p ◦ v and v also slightly different, it is now v ( x, t ) := ( | f | p ( x, t ) , | g | p ′ ( x, t ) , f ( x, t ) , g ( x, t )) , where these are heat extensions of functions on R with corresponding symbol. Weestimate the expression in (67) in a pointwise way from below using 2), and in theaverage on a slab, using 1) we got exactly (60), Theorem 3.1, and, therefore, (52). Remark.
Notice that variables x , y are complex, they are “bench guards” (“mesto-blyustiteli”) for complex-valued functions f = u + iv, g = φ + iψ . So actually B p isa function of 6 real variables, and, hence, (66) should be understood as(68) − d B p ( X ; Y ; u, v ; φ, ψ ) = ( H B p h, h ) ≥ √ du + dv p dφ + dψ , where u, v, φ, ψ are just real variables (they are “bench guards” for functions withthe same symbols and their heat extensions), and h = ( dX, dY, du, dv, dφ, dψ ) is anotation (strange may be) for an arbitrary vector in R . (cid:3) To obtain (53) we notice first that in Theorem 3.1 we can use R cos θ − R sin θ inplace of R , and R sin θ + R cos θ in place of R . In fact this is just application ofrotation on θ in arguments. Then we notice that ( R cos θ − R sin θ ) − ( R sin θ + R cos θ ) = ( R − R ) cos 2 θ − R R sin 2 θ . Therefore, we got Theorem 3.2.
For any φ ∈ (0 , π ] , k ( R − R ) cos φ − R R sin φ k p ≤ p − if p ≥ . We notice that a certain estimate of T = ( R − R ) + 2 iR R can be obtained ifwe answer the following question. Suppose A, B are two operators in L p ( µ ), and forany angle k A cos φ − B sin φ k p ≤
1, then what is the estimate of k A − iB k p ?This is easy on real functions, let f ∈ L preal ( µ ), and let A, B map real functionsto real functions ( A = R − R , B = 2 R R are such). In fact, Z | f | p dµ ≥ Z | ( Af )( x ) cos φ + ( Bf )( x ) sin φ | p dµ = Z ( | Af | + | Bf | | p/ | cos( a ( x ) − φ ) | p dµ ( x ) . Integrate this over π R π . . . , by Fubini’ theorem we will get(69) Z | f | p dµ ≥ Z ( | Af | + | Bf | ) p/ dµ · π Z π | cos φ | p dφ . Put τ ( p ) := (cid:18) π Z π | cos φ | p dφ (cid:19) /p , then on real functions (70) k A + iB k p ≤ sup φ k A cos φ + B sin φ k p /τ ( p ) . Unfortunately this was in real category. We do not know how obtain (70)–orsomething like that–for general operators
A, B on complex function. May be this isalso an exercise?
However, we will obtain now (53). First we need
The proof of (61) . We use the following elementary lemma from Linear Algebra:
Lemma 3.3 (Linear Algebra lemma) . Let
A, B, C be nonnegative matrices of size d × d . Let (71) ( Ah, h ) ≥ Bh, h ) / ( Ch, h ) / , ∀ h ∈ C d . Then there exists τ ∈ (0 , ∞ ) independent of h such that ( Ah, h ) ≥ τ ( Bh, h ) + 1 τ ( Ch, h ) , ∀ h ∈ C d . Proof.
Exercise. (cid:3)
We apply this lemma separately to h , h , where ( f = u + iv, g = φ + iψ , ) h = ( ∂ x | f | p ( x, t ) , ∂ x | g | p ′ ( x, t ) , ∂ x u ( x, t ) , ∂ x v ( x, t ) , ∂ x φ ( x, t ) , ∂ x ψ ( x, t )) ,h = ( ∂ x | f | p ( x, t ) , ∂ x | g | p ′ ( x, t ) , ∂ x u ( x, t ) , ∂ x v ( x, t ) , ∂ x φ ( x, t ) , ∂ x ψ ( x, t )) , and A = H B p ( | f | p ( x, t ) , | g | p ′ ( x, t ) , u ( x, t ) , v ( x, t ) , φ ( x, t ) , ψ ( x, t )), and B consisting ofall zeros except 3 , , C consisting of all zerosexcept 5 , , (cid:3) The proof of (53) . We use the previous notations. We want a better estimate of
T f = ( A + iB )( u + iv ) = Au − Bv + i ( Av + Bu ). Using the trick above (69) we canaverage the following equality over (0 , π ) Z | ( Au − Bv )( x ) cos φ + ( Av + Bu )( x ) sin φ | p dµ = Z ( | Au − Bv | + | Av + Bu | | p | cos( a ( x ) − φ ) | p dµ ( x ) . Then we get τ ( p ) · ( Z | T f | p ) /p ≤ sup φ ( Z | ( Au − Bv )( x ) cos φ + ( Av + Bu )( x ) sin φ | p ) /p =sup φ sup real ψ , k ψ k p ′ ≤ Z [( Au − Bv )( x ) cos φ + ( Av + Bu )( x ) sin φ ] ψ ( x ) dx =: E However the last expression can be rewritten using (56) and integration by parts asfollows: E = 2 ℜ Z Z R ( ∂ x + i∂ x ) f ( x, t )( ∂ x + i∂ x ) e − iφ ψ ( x, t ) dxdt ≤ √ Z Z (cid:0) | ∂ x f | + | ∂ x f | (cid:1) / (cid:0) ( ∂ x ψ ) + ( ∂ x ψ ) (cid:1) / ≤ √ p − k f k p , p > . We used (61). Here √ | ( ∂ x + i∂ x ) f ( x, t ) | ≤ √ (cid:0) | ∂ x f | + | ∂ x f | (cid:1) / . Finally we get(72) k T k p ≤ √ p − (cid:16) π R π | cos φ | p dφ (cid:17) /p , p > . ELLMAN FUNCTION TECHNIQUE IN HARMONIC ANALYSIS 27
Asymptotically this is 1 . ... ( p − p , interpolating between L ,where the norm of T is 1 and the estimate (72) for this large p , then optimizing bythe choice of p one can get (53) ( exercise! ). (cid:3) Notice that (61) immediately proves the following
Theorem 3.4. k T : L preal → L p k ≤ √ p − , p ≥ ;2) | ( T f, g ) | ≤ ( p − k f k p k g k p ′ , p ≥ , if f, g are real valued . Proof.
Just look at (59), compare it with (61) and the fact that ( | DF | − DF ) / ≤√ | DF | , F = ( u, v ). We also need to notice that in this inequality for real valued f = u + i DF = 0 and the constant √ exercise . (cid:3) So everything above hinges on inequality (62). This inequality was proved byBurkholder in mid 80’s and it is one of the remarkable inventions . It is done byuse of
Bellman function technique .3.1.
The proof of inequality (62) . Burkholder’s Bellman function.
We fol-low [19], [23], [25]–but loosely. See also the exposition in the review paper [4].Let f be real valued on [0 ,
1] =: I . Let { h I } I ∈D be the usual Haar functions on I normalized in L . Consider an operator T ε f = X I ∈D ε I ( f, h I ) h I , ε := { ε I } I , ε I = ± . This family is called martingale transforms.Burkholder proved the following remarkable
Theorem 3.5. sup ε k T ε k p = p ∗ − p, p/ ( p − − . He gave several proofs, all difficult, to be found in [19]–[25]. Another proof byVasyunin–Volberg see arxiv: 1006.2633, [59].In all these proofs the following object is indispensable. It is Burkholder’s
Bell-man function .Let Ω := { ( x, y, z ) : | x | p ≤ z } and let B ( x, y, z ) := sup {k g k pp : h f i I = x, h g i I = y, h| f | p i I = z, ∀ I ∈ D | ( g, h I ) | = | ( f, h I ) |} . Symmetries: (73) B ( tx, ty, t p z ) = t p B ( x, y, z ) , B ( − x, y ) = B ( x, y ) , B ( x, − y ) = B ( x, y ) . Burkholder found the formula for B :Consider for positive x, yF p ( x, y ) = y p − ( p ∗ − p x p , if y ≤ ( p ∗ − x ; p (cid:16) − p ∗ (cid:17) p − ( y + x ) p − ( y − ( p ∗ − x ) , if y ≥ ( p ∗ − x . Consider the solution of an implicit equation: F p ( | x | , | y | ) = F p ( z /p , B /p ( x, y, z )) . If p ≥ p ∈ (1 , F p ( | y | , | x | ) = F p ( B /p ( x, y, z ) , z /p ).Obviously one gets a Theorem 3.6. B (0 , ,
1) = ( p ∗ − p , which gives Theorem 3.5, from which we get that (62) is proved right away. In fact, The proof of (62) . We write sup ε | ( T ε f, g ) | ≤ ( p ∗ − k f k p k g k p ′ , which follows fromTheorem 3.5. But this supremum obviously is equal toΣ I ∈D | ( f, h I ) | | ( g, h I ) | . Therefore (62) is proved. (cid:3)
Remarks.
1) As soon as (62) is proved we have our
Bellman function B p .2) It gives all our inequalities like (61) and its consequences like (53).3) It is not Burkholder’s function.4) The existence of our Bellman function B p follows from the existence of Burkholder’sBellman function. These are demographic creatures, they create one another–we saw this in previous sections too.
We are left to prove Theorem 3.5. Instead of finding exact formula for B ( x, y, z )listed above we will use a certain shortcut (invented already by Burkholder himself).Suppose Burkholder’s B is finite. The shortcut proof of Theorem 3.5.
Along with symmetries (73) it has very goodconcavity properties:(74) B ( x, y, z ) −
12 ( B ( x + α, y + α, z + β ) + B ( x − α, y − α, z − β )) ≥ , if all points lie in Ω. Also(75) B ( x, y, z ) −
12 ( B ( x + α, y − α, z + β ) + B ( x − α, y + α, z − β )) ≥ , if all points lie in Ω.Inequalities (74), (75) are left as exercise .Notice that this means that M ( a, b, c ) := B ( a + b, a − b, c )is concave in ( a, c ), and in ( b, c ). Definition . Such M is called bi-concave. Definition.
Function ϕ on R is called zigzag concave if ϕ ( x, y ) −
12 ( ϕ ( x + α, y + α ) + ϕ ( x − α, y − α ) ≥ ,ϕ ( x, y ) −
12 ( ϕ ( x + α, y − α ) + ϕ ( x − α, y + α ) ≥ , or, which is the same as, ϕ ( x, y ) −
12 ( ϕ ( x + , y + ) + ϕ ( x − , y − )) ≥ , if | x + − x − | = | y + − y − | , x = 12 ( x + + x − ) , y = 12 ( y + + y − ) . ELLMAN FUNCTION TECHNIQUE IN HARMONIC ANALYSIS 29
Theorem 3.7.
Put ϕ ( x, y ) := sup ( x,y,z ) ∈ Ω [ B ( x, y, z ) − ( p ∗ − p z ] . It is zigzag con-cave. It is the least zigzag concave majorant of h ( x, y ) := | y | p − ( p ∗ − p | x | p .There is no zigzag concave majorant ψ such that ψ ( tx, ty ) = t p ψ ( x, y ) of function h c := | y | p − c | x | p if c < ( p ∗ − p .Proof. Put c p = ( p ∗ − p . Fix ( x − , y − ) and ( x + , y + ). Find z − which almost givessupremum in ϕ ( x − , y − ) = sup[ B ( x − , y − , z ) − c p z ]. Do the same for ϕ ( x + , y + ) to find z + . Then B ( x − , y − , z − ) − c p z − ≤ ϕ ( x − , y − ) ≤ B ( x − , y − , z − ) − c p z − + ε ,B ( x + , y + , z + ) − c p z + ≤ ϕ ( x + , y + ) ≤ B ( x + , y + , z + ) − c p z + + ε . Let x = ( x + + x − ) , y = ( y + + y − ) and put z = ( z + + z − ). Then ϕ ( x, y ) = sup · · · ≥ B ( x, y, z ) − c p z = B ( x, y, z ) − c p
12 ( z + + z − ) ≥
12 ( B ( x − , y − , z − ) − c p z − ) + 12 ( B ( x + , y + , z + ) − c p z + ) ≥
12 ( ϕ ( x − , y − ) + ϕ ( x + , y + )) − ε . So ϕ is zigzag concave. Also ϕ ( x, y ) = sup · · · ≥ lim z →| x | p + [ B ( x, y, z ) − c p z ] ≥ | y | p − c p | x | p = h ( x, y ) . So ϕ is a zigzag concave majorant of h . Why the least? Let ψ be any zigzag concavefunction such that h ≤ ψ . Put Ψ := ψ ( x, y ) + c p z . Then it is easy to see that Ψ satisfies (74), (75). Also on ∂ Ω = { z = | x | p } we haveΨ( x, y, z ) ≥ h ( x, y ) + c p z = h ( x, y ) + c p | x | p = | y | p . Then combination of the last inequality and the fact that Ψ satisfies (74), (75) gives(attention exercise! ) Ψ( x, y, z ) ≥ B ( x, y, z ) . This a non-trivial exercise. But then trivially for every ( x, y ) ψ ( x, y ) = sup z :( x,y,z ) ∈ Ω [Ψ( x, y, z ) − c p z ] ≥ sup z :( x,y,z ) ∈ Ω [ B ( x, y, z ) − c p z ] = ϕ ( x, y ) . We need now to prove that h c , c < c p does not have zigzag concave homogeneousmajorant.This and more is done in Lemma 3.8.
Function h c , c < c p does not have zigzag concave homogeneous majo-rant. If c = c p , then the function h c p =: h has such majorant given by Φ ( x, y ) := | y | p − ( p ∗ − p = h ( x, y ) , if h ≤ p (cid:16) − p ∗ (cid:17) p − ( | y | + | x | ) p − ( | y | − ( p ∗ − | x | ) , if h > . Another zigzag concave majorant of h = h c p (but not the least) is given by Φ( x, y ) := p (cid:18) − p ∗ (cid:19) p − ( | y | + | x | ) p − ( | y | − ( p ∗ − | x | ) . Remark.
The fact that Φ ( x, y ) ≤ | x | ≥ | y | will be crucial for the proof ofTheorem 3.5. The proof of Lemma 3.8.
We work in the firs quadrant. Homogeneous ϕ can bewritten as ϕ ( x, y ) = ( x + y ) ϕ ( xx + y , yx + y ) , s := y − xy + x , then 1 + s yx + y , − s xx + y .s ′ x = − sx + y , s ′ y = 1 − sx + y . We put g ( s ) := ϕ ( − s , s ). Next we list some results of computations: ϕ x = p ( x + y ) p − g ( s )+( x + y ) p − g ′ ( s )(1+ s ) , ϕ y = p ( x + y ) p − g ( s )+( x + y ) p − g ′ ( s )(1 − s ) .ϕ xx = p ( p − x + y ) p − g ( s ) − p − x + y ) p − g ′ ( s )(1 + s ) + ( x + y ) p − g ′′ ( s )(1 + s ) ,ϕ yy = p ( p − x + y ) p − g ( s ) + 2( p − x + y ) p − g ′ ( s )(1 − s ) + ( x + y ) p − g ′′ ( s )(1 − s ) ,ϕ xy = p ( p − x + y ) p − g ( s ) − p − x + y ) p − g ′ ( s ) s − ( x + y ) p − g ′′ ( s )(1 − s ) . So on x + y = 1 ϕ xy = − [(1 − s ) g ′′ ( s ) + 2( p − sg ′ ( s ) − p ( p − g ( s ) ,ϕ xx + ϕ yy = 2(1 + s ) g ′′ ( s ) − p − sg ′ ( s ) + 2 p ( p − g ( s ) . So combining the two:( ∂ x − y ) ϕ = ϕ xx − ϕ xy + ϕ yy = 4 g ′′ ( s ) . ( ∂ x + y ) ϕ = ϕ xx +2 ϕ xy + ϕ yy = 4 g ′′ ( s )+4 ϕ xy = 4( s g ′′ ( s )+( p − − sg ′ ( s )+ pg ( s ))) . Zigzag concave means the last two lines have sign ≤
0. To find ϕ satisfying thesetwo ≤ g ! Then put(76) g ( s ) = a (cid:18) s − ρ − s (cid:19) . Then the second inequality s g ′′ ( s ) + ( p − − sg ′ ( s ) + pg ( s )) ≤ sg ′ ( s ) − pg ( s ) ≥ , on [ − , . It is satisfied (as g is linear) if and only if it is satisfied in − a (1 + ρ − p ) ≥ , − a (1 + ρ + pρ ) ≥ . As g is greater than (cid:0) s (cid:1) p − c (cid:0) − s (cid:1) p , it is positive at s = 1, so a >
0. then we getfrom previous inequalities that ρ ≥ max (cid:18) p − , p − (cid:19) = p ∗ − . Let us try linear g with ρ = p ∗ −
1. So g has zero at s p such that p ∗ − s p − s p . But if h ( x, y ) = ( x + y ) p H ( s ) , H ( s ) := (cid:18) s (cid:19) p − c p (cid:18) − s (cid:19) p , then H has zero at the same point s p . Now let us find a from the condition H ′ ( s p ) = g ′ ( s p ) ⇒ a = p (cid:18) − p ∗ (cid:19) p − . ELLMAN FUNCTION TECHNIQUE IN HARMONIC ANALYSIS 31
Where H is concave on [ − , i p is such that (cid:18) s (cid:19) p − − c p (cid:18) − s (cid:19) p − = 0 . So it is clear that it is always > s p . As H is concave on [ − , i p ] it is concave on[ − , s p ] (and a little bit on the right of s p too).It is also easy to see that on [ − , s p ](78) s H ′′ ( s ) + ( p − − sH ′ ( s ) + pH ( s )) ≤ . This is an exercise .Put now ˜ g ( s ) = ( our linear g ( s ) s ∈ [ s p ,
1] ; H ( s ) , s ∈ [ − , s p ] . Then Φ ( x, y ) = ( x + y ) p ˜ g ( xx + y , yx + y ) is exactly the same Φ as in Lemma 3.8’sstatement. We just checked that it is a zigzag concave majorant of h ( x, y ). Wealso checked that Φ( x, y ) = ( x + y ) p g ( xx + y , yx + y ), where g is our linear function builtabove, is zigzag concave majorant of h ( x, y ) as well. It is exactly function Φ as inLemma 3.8’s statement.Now let c < c p . Linear function cannot be higher than corresponding H c on [ − , αs + β is higher, then α + β >
0. Also (77) gives2 sα − p ( α + β ) ≥ ⇒ α (2 − p ) s − pβ ≥ , α ( p − − pβ ≥ . Then β < , α >
0. So linear function is positive in 1 and negative at zero. So itmust vanish on [ − , H c can have only c ≥ c p . In fact, we remember that ρ ≥ p ∗ − ≥ c p . But if our linear function is amajorant of H c with c < c p it is also a majorant of H c p . Therefore, its zero must be < c p . This is a contradiction, and a linear solution of two differential inequalitieswill not have minorant with c < c p . Concave solution will not have such minorantseither. Exercise .Lemma 3.8 is finished. (cid:3)
Theorem 3.7 is completely proved. (cid:3)
Finishing the proof of Theorem 3.5. The real case.
Now that we have function Φ(Φ will work too) that is1) zigzag concave on the plane,2) is such that Φ( x, y ) ≥ h ( x, y ) := | y | p − ( p ∗ − p | x | p ,we can do the following. Fix f, g step functions on I := [0 , P = ( x, y ) = ( h f i I , h g i I ), P + = ( x + , y + ) = ( h f i I + , h g i I + ), P − = ( x − , y − ) =( h f i I − , h g i I − ). Notice that of course P = ( P + + P − ). Also | x + − x − | = | y + − y − | because this differences are √ | I | | ( f, h I ) and √ | I | | ( g, h I ) correspondingly, and we as-sumed in Theorem 3.5 that for every dyadic interval | ( f, h I ) | = | ( g, h I ) | . Let also | x | ≥ | y | (for example both are zeros) Then we can use properties of Φ:0 ≥ Φ( x, y ) ≥ Φ( x + , y + ) | I + | + Φ( x − , y − ) | I − | . As intervals I + , I − are as good as I we can repeat this for them. Just iterating thisprocedure and denoting by I σ dyadic intervals of size 2 − n with σ being any stringof ± of length n we get Σ σ Φ( x σ , y σ ) | I σ | ≤ . Combine this with property 2) above. ThenΣ σ | y σ | p | I σ | ≤ ( p ∗ − p Σ σ | x σ | p | I σ | . But by our construction y σ = h g i I σ , x σ = h f i I σ . So we getΣ σ |h g i I σ | p | I σ | ≤ ( p ∗ − p Σ σ |h f i I σ | p | I σ | . Going to the limit when n → ∞ we get h| g | p i I ≤ ( p ∗ − p h| f | p i I , which gives theclaim of Theorem 3.5 in the case of real-valued f, g . (cid:3) Finishing the proof of Theorem 3.5. The complex-valued and Hilbert-valued cases.
Acertain “miracle” happens: Φ , Φ have extra properties of symmetry, not apparentat this moment. Extra symmetry.
Consider ϕ ( x, y ) = Φ( p x + x , p y + y ). We use standardnotations, now x, y are vectors, k · k is the norm of a vector, dx := ( dx , dx ) , dy :=( dy , dy ) are also arbitrary vectors.We want to see that1) − d ϕ := − ( H ϕ (cid:20) dxdy (cid:21) , (cid:20) dxdy (cid:21) ) ≥
0, if k dx k = k dy k . This is “zigzag concavity”direct analog.2) ϕ ( x, y ) ≥ h ( k x k , k y k ).The second is obvious, but the first happens by a “miracle”. Let us prove it andsee, where the “miracle” happens.Calculations (really abusing the language we understand that Φ x , Φ y are partialderivatives of Φ with respect to the first and the second variables): ϕ x = Φ x · x p x + x , ϕ x = Φ x · x p x + x .ϕ x x = Φ xx · x x + x + Φ x x ( x + x ) / , ϕ x x = Φ xx · x x + x + Φ x x ( x + x ) / .ϕ x x = Φ xx · x x x + x − Φ x x x ( x + x ) / . Symmetrically for y derivatives. Also ϕ x i y j = Φ xy · x i y j p x + x p y + y , i, j = 1 , . Therefore, − ( H ϕ (cid:20) dxdy (cid:21) , (cid:20) dxdy (cid:21) ) = Φ x k x k (cid:18) x dx − x dx k x k (cid:19) + Φ y k y k (cid:18) y dy − y dy k y k (cid:19) +Φ xx (cid:18) x dx + x dx k x k (cid:19) +2 ϕ xy (cid:18) x dx + x dx k x k (cid:19) (cid:18) y dy + y dy k y k (cid:19) + ϕ yy (cid:18) y dy + y dy k y k (cid:19) ELLMAN FUNCTION TECHNIQUE IN HARMONIC ANALYSIS 33 = Φ x k x k k ˆ dx k + Φ y k y k k ˆ dy k + Φ xx ( dx, x k x k ) + 2Φ xy ( dx, x k x k )( dy, y k y k ) + Φ yy ( dy, y k y k ) , where ˆ dx, ˆ dy are projections of vectors dx, dy on direction orthogonal to x, y corre-spondingly.Recall that up to a positive constant (which we drop now abusing the language)Φ( x, y ) = ( y − ( p − x )( x + y ) p − , if p ≥ , and ( p − x, y ) = − ( x − ( p − y )( x + y ) p − , if p ≤ . Let us consider p ≥
2, the other case being similar. Looking at the formulae abovewe get by direct calculation with formula for Φ that for any numbers h ′ , k ′ Φ xx h ′ +2Φ xy h ′ k ′ +Φ yy k ′ = − p ( p − x + y ) p − ( h ′ − k ′ ) − p ( p − p − x ( x + y ) p − ( h ′ + k ′ ) . (By the way we immediately see that this form is ≤ | k ′ | = | h ′ | , which is infini-tesimal version of zigzag concavity.)Now let us combine our formulae, putting h ′ = ( dx, x k x k ) , k ′ = ( dy, y k y k ). Then( H ϕ (cid:20) dxdy (cid:21) , (cid:20) dxdy (cid:21) ) = Φ x k x k k ˆ dx k + Φ y k y kk ˆ dy k − p ( p − k x k + k y k ) p − ( h ′ − k ′ ) − p ( p − p − k x k ( k x k + k y k ) p − ( h ′ + k ′ ) . Let us look at the first line of the last formula. CalculateΦ x k x k k ˆ dx k + Φ y k y kk ˆ dy k = (cid:18) Φ x k x k + Φ y k y k (cid:19) k ˆ dy k + Φ x k x k ( k ′ − h ′ ) + T erm , where
T erm := Φ x k x k ( k h k − k k k ). This is just because k ˆ dx k + h ′ = k h k , and thesame is true for k . In particular,(79) T erm = 0 if k h k = k k k , and T erm ≤ , if k h k ≥ k k k , . In fact,(80) Φ x k x k = − p ( p − k x k + k y k ) p − < . Combine three last formulae. Then we have( H ϕ (cid:20) dxdy (cid:21) , (cid:20) dxdy (cid:21) ) = (cid:18) Φ x k x k + Φ y k y k (cid:19) k ˆ dy k + T erm − p ( p − p − k x k ( k x k + k y k ) p − ( h ′ + k ′ ) . The second line is obviously negative (see (79)). To have the first line negative it isnecessary and sufficient to have(81) (cid:18) Φ x k x k + Φ y k y k (cid:19) ≤ . Calculate: Φ y k y k = p ( k y k − ( p − k x k )( k x k + k y k ) p − . Combine this with (80) to get (cid:18) Φ x k x k + Φ y k y k (cid:19) = − ( k x k + k y k ) p − (cid:18) p ( p − − p + p ( p − k x kk y k (cid:19) = (82) − p ( p −
2) ( k x k + k y k ) p − k y k ≤ , if p ≥ . The case p < exercise ! (cid:3) Theorem is finally completely proved. (cid:3)
We want to remember a formula that has been just obtained ( p ≥ H ϕ (cid:20) dxdy (cid:21) , (cid:20) dxdy (cid:21) ) = − p ( p −
2) ( k x k + k y k ) p − k y k k ˆ dy k − p ( p − k x k + k y k ) p − ( k dx k −k dy k )(83) − p ( p − p − k x k ( k x k + k y k ) p − (cid:18) ( dx, x k x k ) + ( dy, y k y k ) (cid:19) , where k ˆ dy k = k dy k − ( dy, y k y k ) . Also(84) p (cid:18) − p ∗ (cid:19) p − ϕ ≥ k y k p − ( p ∗ − p k x k p . On the other hand, if 1 < p <
2, we know that ϕ ( x, y ) = ( k y k p − ( p ∗ − k x k )( k x k + k y k ) p − will satisfy( H ϕ (cid:20) dxdy (cid:21) , (cid:20) dxdy (cid:21) ) = − p (2 − p ) ( k x k + k y k ) p − k x k k ˆ dx k − p ( p − k x k + k y k ) p − ( k dy k −k dx k )(85) − p ( p − − p ) k x k ( k x k + k y k ) p − (cid:18) ( dx, x k x k ) + ( dy, y k y k ) (cid:19) , where k ˆ dx k = k dx k − ( dx, x k x k ) . The same majorization (84) happens for 1 < p < stochastic integrals .4. Stochastic Integrals. Itˆo’s formula
Let w ( s ) := w s denote Brownian motion started at 0, that is w = 0, and for all t < t < t , random variables w t − w t , w t − w t are Gaussian independent withzero average and variances √ t − t , √ t − t correspondingly.We want to understand what does it mean Z ba ξ ( t ) dw t . It is not the Riemann sum defintion as the following example shows.
ELLMAN FUNCTION TECHNIQUE IN HARMONIC ANALYSIS 35
Example.
Consider two simplest Riemann sums built on a partition of the interval[ a, b ]: Σ := m X i =1 w ( t i − )( w ( t i ) − w ( t i − )) , Σ := m X i =1 w ( t i )( w ( t i ) − w ( t i − )) . If refinement is small, we should have had (if stochastic integral were a Riemannsum thing) that these two random variables Σ and Σ are close. Let us see, whetherthis is the case.Notice that uniformly (when the partition changes) they are in L (Ω , F , P ), where(Ω , F , P ) is the probability space on which Brownian motion is given. E (Σ ) = X i 0, then k f n k k L ( P ) → E Σ = 0 , E Σ = E Σ + E ( X i ( w ( t i ) − w ( t i − )) = X i ( t i − t i − ) = b − a = 0 . We understand now that stochastic integral R ba ξ ( t ) dw ( t ) is a much more subtlething than Riemann sum integral. Stochastic integrals were understood by KioshiItˆo. A bit on Itˆo’s definition. .Let B F be a sigma algebra of sets A ⊂ R × Ω such that for every t ∈ [ a, b ] we have A ∩ (( −∞ , t ] × Ω) is in B t × F t , where B t is Borel sigma algebra on ( −∞ , t ], F t is asigma algebra generated by { w s } s ≤ t . Let M [ a, b ] is the set of functions measurablewith respect to B F such that(a) f ( t ) is measurable with respect to F t for each t ,(b) with probability 1, R ba | f ( t ) | dt < ∞ .For all such random functions (random processes) Itˆo defines(86) Z ba f ( t ) dw ( t ) . Definition. f ∈ M [ a, b ] is called a step function if there exists a partition suchthat f ( t ) = f ( t i )( ω ) , for t ∈ [ t i , t i +1 ). We introduce the stochastic integral for themin a natural way Z ba f dt := X i f ( t i )( ω ) · ( w ( t i +1 ) − w ( t i )) . Lemma 4.1. For every f ∈ M [ a, b ] there exits a sequence of step functions as abovesuch that with probability n →∞ Z ba | f ( t ) − f n ( t ) | dt = 0 . Moreover if in addition E Z ba | f ( t ) | dt < ∞ , then step functions can be chosen to have lim n →∞ E Z ba | f ( t ) − f n ( t ) | dt = 0 . We need the following Lemma. Lemma 4.2. Let ϕ be a step function as above. Let δ, ǫ > . Then P {| Z ba ϕ ( t ) dw ( t ) | > ε } ≤ δǫ + P { Z ba | ϕ ( t ) | dt > δ } . This lemma immediately gives the following reasoning. If–as above– f ∈ M [ a, b ]and f n are step functions from Lemma 4.1, then P − lim n →∞ Z ba | f ( t ) − f n ( t ) | dt = 0 . Then P − lim n,m →∞ Z ba | f m ( t ) − f n ( t ) | dt = 0 . By definition ∀ ε > , P { Z ba | f m ( t ) − f n ( t ) | dt > ε } → , m, n → ∞ . ELLMAN FUNCTION TECHNIQUE IN HARMONIC ANALYSIS 37 Now we use Lemma 4.2 to havelim sup m,n →∞ P {| Z ba f n ( t ) dw ( t ) − Z ba f m ( t ) dw ( t ) | > ε } ≤ δε for any δ > 0. So the sequence of random variables ξ n := R ba f n ( t ) dw ( t ) is Cauchyconvergent in measure (in probability). So in probability it converges to a certainrandom variable ξ . This ξ is by definition R ba f dw ( t ). Itˆo’s stochastic integralis constructed. This integral has many nice properties:If E R ba | f ( t ) | dt < ∞ , then E R ba f ( t ) dw ( t ) = 0 and E ( Z ba f ( t ) dw ( t )) = E Z ba | f ( t ) | dt . If in addition E R ba | g ( t ) | dt < ∞ then(87) E ( Z ba f ( t ) dw ( t ) · Z ba g ( t ) dw ( t )) = E Z ba f ( t ) · g ( t ) dt . ( Integral of the product is the product of integrals. )4.2. Stochastic differential. Let b ( t ) ∈ M [ a, b ], and a ( t ) be measurable withrespect to F t for every t , and Z ba | a ( t ) | dt < ∞ . Suppose ζ ( t ) is a random process such that for all t , t such that a ≤ t ≤ t ≤ bζ ( t ) − ζ ( t ) = Z t t a ( t ) dt + Z t t b ( t ) dw ( t ) . Then we write the above line as stochastic differential : dζ ( t ) = a ( t ) dt + b ( t ) dw ( t ) . Remark. If a = 0 this integral is a martingale (obviously) on the filtration {F t } t> of sigma algebras generated by Brownian motions.4.3. Itˆo’ formula. Let ζ have the stochastic differential in the sense above and let u ( t, x ) be a (several times) smooth function. Consider new process η ( t ) := u ( t, ζ ( t )) . Theorem 4.3. Then η also has stochastic differential and dη ( t ) = [ u ′ t ( t, ζ ( t ) + u ′ x ( t, ζ ( t )) a ( t ) + 12 u ′′ xx ( t, ζ ( t )) · b ( t )] dt + u ′ x ( t, ζ ( t )) · b ( t ) · dw ( t ) . Proof. The proof is quite subtle. See [45], [62]. (cid:3) Matrix Itˆo’s formula also exists and will be used. Let a be m × σ is a m × k matrix of processes (with entries in M [ a, b ]). Let W ( t ) bea column of k independent Brownian motion. Let ζ ( t ) be a m × dζ ( t ) = a ( t ) dt + σ dW ( t ) . Let u ( t, x ) be a smooth function, where x ∈ R m . Let η ( t ) = u ( t, ζ ( t )). Then η alsohas stochastic differential, and matrix Itˆo’s formula gives(88) dη ( t ) = [ ∂u/∂t + ∇ x u ( t, ζ ) · a ( t ) + 12 trace( σ H u ( t, ζ ) σ ∗ )] dt + ∇ x u · σdW ( t ) . Here · is the scalar product in R m .4.4. Space-time Brownian motion. Let us discuss Theorem 4.3. If a = 0, thenthe process ζ is a martingale (see Remark before the theorem). However, it is quiteunrealistic to expect that if we consider the composition of a non-linear function u and a martingale, then we would get another martingale. And in fact, if a = 0 theformula in Theorem 4.3 becomes (if a = 0)(89) dη ( t ) = [ u ′ t ( t, ζ ( t ) + 12 u ′′ xx ( t, ζ ( t )) · b ( t )] dt + u ′ x ( t, ζ ( t )) · b ( t ) · dw ( t ) , and the “non-martingale” part (called drift ) in square brackets is very much present.But there is one very important exception.Suppose f ∈ C ∞ and u f ( t, x ) is the heat extension of f , in other words, thesolution of the heat equation:(90) (cid:18) ∂∂t − ∂ ∂x (cid:19) u f = 0 , u f (0 , x ) = f ( x ) . Fix large positive T and consider function of ( t, x ) given by u = u f ( T − t, x ). Wewant to compose it with stochastic process as in Theorem 4.3, with a = 0 , b = 1.Then we get the process η := u f ( T − t, w t ) . It will be a martingale on [0 , T ]. In fact, we can use (89) to get dη ( t ) = [ − ∂u f ∂t ( T − t, w t ) + 12 ∂ u f ∂x ( T − t, w t )] dt + ∂u f ∂x ( T − t, w t ) dw t , and by (90) the drift term in the brackets disappears.If we work with heat extension for functions on R k the same will be true. NowBrownian motion W t is k -dimensional (just k independent Brownian motions) and u f is the solution of heat equation(91) (cid:18) ∂∂t − 12 ∆ (cid:19) u f = 0 , u f (0 , x ) = f ( x ) . Then we get the following stochastic differential(92) d u f ( T − t, W t ) = ∇ x u f ( T − t, W t ) · dW t , where · is the scalar product in R k . ELLMAN FUNCTION TECHNIQUE IN HARMONIC ANALYSIS 39 We are interested now in the case of complex valued function f on R , so k = 2.Thinking that gradient is always column vector and W t is 2-dimensional row vectorit is convenient to rewrite (92) as(93) f ( W T ) − u f ( T, 0) = Z T dW t · (cid:20) ∂ x ∂ y (cid:21) u f ( T − t, W t ) . Definition. The expressions R T dW t · (cid:20) ∂ x ∂ y (cid:21) u f ( T − t, W t ) will be called heat mar-tingales .But we will need a bigger class, where heat martingales are supplemented bytheir martingale transforms . The simplest martingale transforms are given byexpressions Z T dW t · A (cid:20) ∂ x ∂ y (cid:21) u f ( T − t, W t ) , where A is a fixed matrix not depending neither on ω (elementary event) nor ontime t .Consider a special matrix(94) A := (cid:20) , ii, − (cid:21) Then we get Z T dW t · (cid:20) ∂ x + i∂ y i ( ∂ x + i∂ y ) (cid:21) u f ( T − t, W t ) , which is(95) 2 Z T dW t · (cid:20) ¯ ∂i ¯ ∂ (cid:21) u f ( T − t, W t ) . This is quite suggestive. In fact, denoting temporarily the Ahlfors–Beurling trans-form R − R + 2 iR R by symbol AB , we recall that AB ¯ ∂ = ∂ . The followingtheorem holds. Theorem 4.4. 1) lim T →∞ E ( Z T dW t · (cid:20) ∂ x ∂ y (cid:21) u f ( T − t, W t ) | W T = z ) = f ( z ) , 2) lim T →∞ E ( Z T dW t · (cid:20) ∂ x + i∂ y i ( ∂ x + i∂ y ) (cid:21) u f ( T − t, W t ) | W T = z ) = AB ( f )( z ) . Proof. Let us consider a test function g and build a heat martingale X ( t ) , ≤ t ≤ T ,by formula 1), but with f replaced by g : X ( t ) := g ( T, 0) + Z t dW s · (cid:20) ∂ x ∂ y (cid:21) u g ( T − s, W s ) . Let Y ( t ) , ≤ t ≤ T , denote the martingale in formula 2): Y ( t ) := Z t dW s · (cid:20) ∂ x + i∂ y i ( ∂ x + i∂ y ) (cid:21) u f ( T − s, W s ) . Then by “rule” that the product of stochastic integrals is “the integral of the prod-uct”, we get (below k ( t ; x, y ) := π t e − x y t )2 π T E ( Y ( T ) · X ( T )) = 2 π T Z T Z Z R ¯ ∂u f ( T − t ; x, y ) ¯ ∂u g ( T − t ; x, y ) k ( t ; x, y ) dxdy dt = − π T Z T Z Z R ¯ ∂u f ( t ; x, y ) ¯ ∂u g ( t ; x, y ) k ( T − t ; x, y ) dxdy dt . Notice that 2 π T k ( T − t ; x, y ) → T goes to infinity. It is not then difficult tosee that the last expression becomes very close to Z T Z Z R ¯ ∂u f ( t ; x, y ) ¯ ∂u g ( t ; x, y ) dxdy dt , when T goes to infinity.Recall formula (30) and formula AB = R − R + 2 iR R .( Number 2 in (30)should be dropped now as we are working with extensions with respect to ∂∂t − ∆unlike before formula (30), where we worked with ∂∂t − ∆.) Combined they give usthat the last expression would be equal to ( AB ( f ) , g ) if the integration would be R ∞ ...dt and not R T ...dt . But as T is large and f, g are nice the “error” goes to zerowhen T goes to infinity. So(96) 2 πT E ( Y ( T ) · X ( T )) = ( AB ( f ) , g ) + o (1) . On the other hand, X ( T ) = g ( W T ) by (93) with f replaced by g . Therefore, for anytest function g πT E ( Y ( T ) · g ( W T )) = 2 πT Z C dµ T ( z ) E ( Y ( T ) | W T = z ) g ( z ) , where dµ T = π T e − | z | T dm ( z ) is given by the density distribution of W T . Now usingthe facts that g is a nice test function and that 2 πT dµ T ( z ) dm ( z ) → T → ∞ we obtain Z C E ( Y ( T ) | W T = z ) g ( z ) dm ( z ) = 2 πT E ( Y ( T ) · g ( W T )) + o (1) . Comparing this with (96) we get the formula(97) AB ( f )( z ) = lim T →∞ E ( Z T dW t · A ∇ x,y u f ( T − t ; W t ) | W T = z ) . Theorem is proved. (cid:3) Remark . It is very easy to see now that for martingale { Y ( t ) } ≤ t ≤ T constructedabove k AB ( f ) k pL p ( C ,dm ) ≤ lim T →∞ π T E | Y ( T ) | p for any p . It is a sort of averaging operator. Moreover, for martingale { X ( t ) } ≤ t ≤ T we obviously have limiting equality k g k pL p ( C ,dm ) = lim T →∞ π T E | X ( T ) | p . ELLMAN FUNCTION TECHNIQUE IN HARMONIC ANALYSIS 41 This is trivial from (93): just raise both part to the power p and take the expectation(first conditioning over W T = z , then integrating with respect to dµ T ( z )) and useagain the fact that 2 πT dµ T ( z ) dm ( z ) → g = f . We see that k AB ( f ) k p ≤ M p k f k p follows from E | Y | p ≤ M p E | X | p for martingales Y , X . Notice that Y is just themartingale transform of X with the help of matrix A , whose norm is 2. Thisexplains the constant 2 in the above display inequality. This is why we study below X, Y and their relationship. Remark. The reader can find many interesting examples, references and explana-tions in recent review of Banuelos devoted to Burkholder’s estimate: [4].4.5. Orthogonal (conformal) martingales. Introducing two martingales on thefiltration of Brownian motion X ( t ) := Z t dW s · ∇ x,y u f ( T − s ; W s ) , ≤ t ≤ T ; Y ( t ) := Z t dW s · A ∇ x,y u f ( T − s ; W s ) =: A ⋆ X ( t ) , ≤ t ≤ T , and using the previous remark, we get that it might be a fruitful idea to look for asharp martingale transform inequality (98) E k A ⋆ X ( t ) k p ≤ M p E k X k p . Let f = φ + iψ . Introduce the notations H s = (cid:2) u φx ( T − s ; W s ) , u ψx ( T − s ; W s ) (cid:3) ,H s = (cid:2) v φy ( T − s ; W s ) , v ψy ( T − s ; W s ) (cid:3) .K s = (cid:2) u φx ( T − s ; W s ) − u ψy ( T − s ; W s ) , u φy ( T − s ; W s ) + u ψx ( T − s ; W s ) (cid:3) ,K s = (cid:2) − u φy ( T − s ; W s ) − u ψx ( T − s ; W s ) , u φx ( T − s ; W s ) − u ψy ( T − s ; W s ) (cid:3) , we can write complex martingale X = X + iX , Y = A ⋆ X = Y + iY in the form(below dW s is a 2-row-vector)(99) (cid:2) X ( t ) , X ( t ) (cid:3) = Z t dW s (cid:20) H s H s (cid:21) = Z t ( H s dw s + H s dw s ) . (100) (cid:2) Y ( t ) , Y ( t ) (cid:3) = Z t dW s (cid:20) K s K s (cid:21) = Z t ( K s dw s + K s dw s ) . Properties of H, K . Vector processes H and K are related by(101) k K s k + k K s k ≤ k H s k + k H s k )for all elementary events ω and all times s .Relationship (101) is called differential subordination of martingale Y to mar-tingale 2 X . Theorem 4.5 (Burkholder’s theorem) . If martingale M is differentially subordi-nated to martingale N , then E k M ( t ) k p ≤ ( p ∗ − E k N ( t ) k p , p ∗ := max( p, p/p − . The constant is sharp. In particular,(102) E k A ⋆ X k p ≤ p ∗ − E k X ( t ) k p . But the constant is not sharp! We followed the probabilistic proof in [7], which“randomize” the idea of [58]. There is an analytic proof following [58] more directly,see [49]. More properties of H, K . Vector processes K have extra properties:(103) K s · K s = 0 , k K s k = k K s k for all elementary events ω and all times s . Such martingales are called orthogonalor conformal . Theorem 4.6 (Banuelos–Janakiraman’s theorem) . We make the exposition of [5] using the notations above. If martingale M is differentially subordinated to martin-gale N , and martingale M is conformal and k M (0) k ≤ k N (0) k then E k M ( t ) k p ≤ r p ( p − E k N ( t ) k p , p > . In particular,(104) E k A ⋆ X k p ≤ p p ( p − E k X ( t ) k p , p > . But the constant is not sharp!However, this inequality gives k T k p ≤ p p ( p − , p > . Interpolation between p = 2 and large p with this estimate, optimization in thislarge p , will give (54). The proof of Theorem 4.6. Our main tool will be formula (83). Trivial renormaliza-tion shows that to prove Theorem 4.6 it is enough to prove that if M, N are twomartingales on the filtration of 2-dimensional Brownian motion and M is differen-tially subordinated to q p − p · N , p > 2, and M is conformal then(105) E k M ( t ) k p ≤ ( p − E k N ( t ) k p , p > . Consider such M, N , and their H , H , K , K . We know that(106) k K k ≤ p − p k H k , where k K k := k K k + k K k , k H k := k H k + k H k , and k · k + k · k = 0 . This and equality k K k = k K k easily implies(107) k · k + k · k = 0 . Let V ( M, N ) := k M k p − ( p − k N k p , p > ϕ ( M, N ) := p (1 − /p ) p − ( k M k + k N k ) p − ( k M k − ( p − k N k ).We would like to prove that E ( V ( M ( t ) , N ( t )) ≤ 0. But it has been proved that V ≤ ϕ . So it is enough to prove(108) E ( ϕ ( M ( t ) , N ( t ))) ≤ . ELLMAN FUNCTION TECHNIQUE IN HARMONIC ANALYSIS 43 To prove (108) we use:(109) ϕ ( M ( t ) , N ( t )) = ϕ ( M (0) , N (0)) + Z t dϕ ( M ( s ) , N ( s )) . To compute E dϕ we use Itˆo’s formula, which of course involves Hessian H ϕ . Moreprecisely, dϕ ( s ) will involve ( H ϕ (cid:20) H s K s (cid:21) , (cid:20) H s K s (cid:21) )+( H ϕ (cid:20) H s K s (cid:21) , (cid:20) H s K s (cid:21) )Now we look at formula (83), which gives dϕ = p (1 − /p ) p − ( A + B + C + D ) , A := − p ( p − k M ( s ) k + k N ( s ) k ) p − ( k H k −k K k ) ,B := − p ( p − k M ( s ) k + k N ( s ) k ) p − k M ( s ) k − "(cid:18) M k − M k k M k (cid:19) + (cid:18) M k − M k k M k (cid:19) , where we need to recall that M = M + iM , M , M being its real and imaginaryparts. Part C comes from the last part of formula (83), and, obviously, C ≤ ,D = ...dw s + ...dw s , where ... involve functions of k sij , h sij and ∇ ϕ ( M ( s ) , N ( s )). This shows that R t D ( s )is a martingale starting at 0 and so(110) E Z t D ( s ) = 0 . We open the brackets in B , use (107), and the fact that k + k = k + k , toget B ≤ − p ( p − k K k )( k M k + k N k ) p − . Now A + B = − p ( k M k + k N k ) p − [( p − k H k − p k K k ] ≤ , if (106) is valid. Term C is non-positive. Term D disappears after integration E R t ,As a result we come to (see (109)): E Z t dϕ ( M ( s ) , N ( s )) = E ϕ ( M (0) , N (0)) = ϕ ( M (0) , N (0)) ≤ , because if x := k M (0) k ≤ y := k N (0) k then ϕ ( x, y ) ≤ 0, which is obvious from theformula for ϕ . (cid:3) As we already mentioned, this proves (54). To prove (55) one needs even morecareful stochastic analysis, and we leave this for the next round of lectures somewherein the future. Bellman function of Stochastic Optimal Control problems Let W s be d dimensional Brownian motion. Let x ( t ) is a d -dimensional processgiven by(111) x ( t ) = x + Z t b ( α ( s ) , x ( s )) ds + Z t σ ( α ( s ) , x ( s )) dW s , in other words the process starts at x ∈ R d and satisfies a stochastic differentialequation dx ( t ) = b ( α ( t ) , x ( t )) dt + σ ( α ( t ) , x ( t )) dW t , where α is a d -dimensional control process, we can choose it ourselves, but it mustbe adapted, that is α ( s ) has to be measurable with respect to sigma algebra F s generated by W t , ≤ t ≤ s . Also values of the process α are often restricted: α ( s, ω ) ∈ A ⊂ R d .Matrix function σ is smooth and d × d -dimensional, and b is a smooth columnfunction of size d . Everything happens in Ω ⊂ R d (often = R d ).The choice of adapted process α ( s ) gives us different motions, all started at thesame initial x ∈ R d .This is a “broom” of motions, hidden elementary even ω gives “one stem of abroom”.Suppose we are given the profit function f ( α, x ), meaning that on a trajectory of x ( t ), for the time interval [ t, t + ∆ t ], the profit is f ( α ( t ) , x ( t )) + o (∆ t ). So on thewhole trajectory we earn Z ∞ f ( α ( t ) , x ( t )) dt . We are also given the pension–we call it bonus function F –how much one is givenat the end of the life. We want to choose a control process α = α ( s ) to maximizethe average profit :(112) v α ( x ) := E Z ∞ f ( α ( t ) , x ( t )) dt + lim sup t →∞ E F ( x ( t )) . If b = 0 and F is convex then one ca write lim instead of lim sup. The optimal average gain , or(113) v := sup α v α ( x )is called the Bellman function of stochastic optimal control problem (111), (112).Usually the analysis consists ofa) writing Bellman PDE on v ;b) solving it;c) using “verification theorem”, which says that under certain conditions on data σ, b, F, f, Ω , A the classical solution of Bellman PDE is exactly v from (113). ELLMAN FUNCTION TECHNIQUE IN HARMONIC ANALYSIS 45 Writing Bellman PDE.. This consists of a) Itˆo’s formula, b) Bellman’s prin-ciple of dynamic programming.Using Itˆo’s formula (88) we get dv ( x ( s )) = d X k =1 ∂v∂x k ( x ( s )) d X j =1 σ kj ( α ( s ) , x ( s )) dw js + d X k =1 ∂v∂x k ( x ( s )) b k ( α ( s ) , x ( s )) ds +12 d X i,j =1 ∂ v∂x i ∂x j ( x ( s )) a ij ( α ( s ) , x ( s )) , where a ij ( α, x ) := d X k =1 σ ik ( α, x ) σ kj ( α, x )is i, j matrix element of d × d matrix σσ ∗ .Introduce two linear differential operators with non-constant coefficients: L ( α, x ) := d X k =1 b k ( α, x ) ∂∂x k , L ( α, x ) := d X i,j =1 a ij ( α, x ) ∂ ∂x i ∂x j , and L ( α, x ) := L ( α, x ) + L ( α, x ) . Let us hit our formula for dv ( x ( t )) above by the expectation E , then the first linebecomes 0, and we get E (cid:20) ddt v ( x ( t )) (cid:21) = E [ L ( α ( t ) , x ( t )) v ]( x ( t )) . Or(114) E v ( x ( t )) = v ( x ) + E Z t [ L ( α ( s ) , x ( s )) + L ( α ( s ) , x ( s ))] v ( x ( s )) ds . Now we need the second ingredient to write the Bellman equation: the Bellmanprinciple= dynamic programming principle . It is in this next equality: v ( x ) = sup α E [ Z ∞ f ( α ( t ) , x ( t )) dt + lim sup t →∞ . . . ](115) = sup α E [ Z t f ( α ( t ) , x ( t )) dt + v ( x ( t ))] , ∀ t > . A minute though shows that this reflects the stationarity of Brownian motion andthe fact that to be perfect one has to be perfect every second.Now plug E v ( x ( t )) from (114) into (115). We get α E [ Z ∞ f ( α ( t ) , x ( t )) + L ( α ( t ) , x ( t )) v ( x ( t ))] dt , ∀ t > . Divide by t and tend t to zero. We “obtain” Bellman equation :(116) sup α ∈ A [( L ( α, x ) v )( x ) + f ( α, x )] = 0 . Positivity (usually present) of f and convexity (usually present) of F imply (ifthere is no drift, that is if b ( α, x ) = 0) obstacle condition :(117) v ( x ) ≥ F ( x ) , ∀ x ∈ Ω . Often it becomes boundary condition :(118) v ( x ) = F ( x ) , ∀ x ∈ ∂ Ω . The definition of v in domain (not in the whole R d ) should be slightly changed. Theintegration of profit function now is not from zero to infinity, but from zero to thestopping time of the first hit of ∂ Ω by the trajectory x ( t ).See details of obtaining (116) in the beautiful book of N. Krylov [47].In applications one is also interested in supersolutions of the Bellman equation(116) :(119) ( sup α ∈ A [ L ( α, x ) V ( x ) + f ( α, x )] ≤ , x ∈ Ω ,V ( x ) ≥ F ( x ) , x ∈ Ω . Lemma : Let V solves (4) and let v be the Bellman function,then V ≥ v in Ω. Proof : Equation (116) states that −L ( α, x ) V ( x ) ≥ f ( α, x ). Using (114) for V andthen (119), one gets V ( x ) = E V ( x ( t )) − E Z t ( L ( α ( s ) , x ( s )) V )( x ( s )) ds ≥ E F ( x ( t )) + E Z t f ( α ( s ) , x ( s )) ds. Writing lim t →∞ of both parts,we get V ( x ) ≥ v α ( x ). It rests to take the supremum overthe control process α .5.2. Special matrices σ bring us to Harmonic Analysis. Let us consider avery simple matrix σ not depending on x :(120) d = 1 , σ ( α, x ) = α ... α d =: α . If on the top of that b = 0 then operator L just involves Hessian matrix H v offunction v : ( L ( α ) v )( x ) = 12 d X i,j =1 ∂ ∂x i ∂x j v ( x ) = 12 ( H v ( x ) α, α ) . ELLMAN FUNCTION TECHNIQUE IN HARMONIC ANALYSIS 47 We claim that this is the generic case of Harmonic Analysis problems in R . Equation(116) becomes(121) ( sup α ∈ A [ h H v ( x ) α, α i + f ( α, x )] = 0 , x ∈ Ω ,v ( x ) ≥ F ( x ) , x ∈ Ω . If b = 0 then we just add the first order differential operator (called drift ):(122) ( sup α ∈ A [ h H v ( x ) α, α i + P dk =1 b k ( α, x ) ∂∂x k v ( x ) + f ( α, x )] = 0 , x ∈ Ω ,v ( x ) ≥ F ( x ) , x ∈ Ω . Harmonic analysis on R “becomes” the analysis of the following Bellman equation(and this is exactly what we did in the sections above devoted to the analysis of theAhlfors–Beurling operator):(123) d = 2 , σ ( α, x ) = α α ... ... α d α d =: α . Conformal restrictions : Matrix α can have restrictions α ∈ A of the type thatthe first row is orthogonal to the second row and that the norms of the rows areequal. The reader can notice that these are Cauchy–Riemann conditions, andthe corresponding solution of (111) will be a conformal martingale (again if b = 0).Bellman equation becomes(124) ( sup α ∈ A [ trace( α ∗ H v ( x ) α ) + P dk =1 b k ( α, x ) ∂∂x k v ( x ) + f ( α, x )] = 0 , x ∈ Ω ,v ( x ) ≥ F ( x ) , x ∈ Ω . Remarks. 1) This is (exactly as (122)) very non-linear (actually an example ofso-called fully non-linear ) equation of the second order.2) This equation is much more difficult to analyze than (122). On the other hand,we can easily notice that conformal restrictions on α makes clear that Hessian of v should be replaced by Laplacian of v (or some kind of semi-Laplacian-semi-Hessian).6. Examples showing almost perfect analogy between StochasticOptimal Control and Harmonic Analysis A ∞ weights and associated Carleson measures. Buckley’s inequality. We call a nonnegative function on R an A ∞ weight (dyadic A ∞ weight actually) if(125) h w i J ≤ C e h log w i J , ∀ J ∈ D . Here D is a dyadic lattice on R , h · i J is the averaging over J .We are going toillustrate our use of Bellman function technique by a collection of examples, thefirst of which is the result of Buckley that can be found (along with “continuousanalogs”) in the paper of Fefferman-Kenig-Pipher [39]. Theorem 6.1. Let w ∈ A ∞ . Then (126) ∀ I ∈ D , I | I | X ℓ ⊆ I , ℓ ∈D (cid:18) h w i ℓ + − h w i ℓ − h w i ℓ (cid:19) | ℓ | ≤ C , Where C depends only on C in (125) . Here ℓ ± are right and left sons of ℓ ∈ D . Who moves ? x , x = h w i J , h log w i J α = h w i son of J − h w i J ⇒ | α | = 12 |h w i J − − h w i J + | . Function of profit can be read off (126) if one notices that | I | P ℓ ⊆ I , ℓ ∈D · · · is theaverage over the lines of life. Each line of life initiates at I and then proceeds to I ε ( ε = +1 or ε = − I ε ε ( ε = +1 or ε = − | I | P ℓ ⊆ I , ℓ ∈D · · · plays the role of E R ∞ · · · . This allows us to choose the correctprofit function f ( α, x ) = 4 α x . Bonus function F ≡ α =( α ,α ) (cid:20) h H v α, α i + 8 α x (cid:21) = 0to be solved in(128) Ω = (cid:8) ( x , x ) : 1 ≤ x e − x ≤ c (cid:9) with the obstacle condition(129) v ( x ) ≥ ∀ x ∈ Ω . Compare with (121)!6.2. A two-weight inequality. ∀ J ∈ D h u i J h v i J ≤ ⇒ ∀ I ∈ D | I | X ℓ ⊆ I , ℓ ∈D |h u i ℓ + − h u i ℓ ||h v i ℓ + − h v i ℓ − | || ℓ |≤ C h u i / I h v i / I . Who moves ? x , x = h u i J , h v i J . As in the previous problem f α ( x ) is easy to find : f ( α, x )) = 4 | α | | α | . ELLMAN FUNCTION TECHNIQUE IN HARMONIC ANALYSIS 49 Bonus function F ≡ α =( α ,α ) ∈ R [ h H v α, α i + 8 | α | | α | ] = 0 , v ≥ { x = ( x , x ) : 0 ≤ x , x ; x x ≤ } . Compare with (121)!6.3. John-Nirenberg inequality : Bellman equation with a drift but with f ( α, x ) ≡ . ∀ J ∈ Dh| ϕ − h ϕ i J | i J ≤ δ ⇒ ∀ I ∈ Dh e ϕ i I ≤ C δ e h ϕ i I . Who moves ? x = h ϕ i J , x = h| ϕ − h ϕ i J | i J == 1 | J | X I ⊆ J , I ∈D (cid:26) h ϕ i I + − h ϕ i I − (cid:27) | I | . Notice that ( t = n ): x t − E ( x t +12 | x t ) = x − x +2 + x − (cid:18) x +1 − x − (cid:19) = ( α t ) . But x t − E ( x t +11 | x t ) = x − x +1 + x − . On the other hand, x t +1 = x t + Z t +1 t σdw s + Z t +1 t bds . Thus drift b stands for E ( x t +1 | x t ) − x t (in the case of discrete time). Therefore, b ( α, x ) = (cid:18) − α (cid:19) in our case. Notice that f ( α, x ) ≡ | I | P ℓ ⊆ I ... in thefunctional. Bellman equation in this case has a formsup α =( α ,α ) (cid:20) h d vα, α i − ∂v∂x α (cid:21) = 0 . Compare with (122)!In other words :(130) ∂ v∂x − ∂v∂x ∂ v∂x ∂x ∂ v∂x ∂x ∂ v∂x ≤ , det ∂ v∂x − ∂v∂x ∂ v∂x ∂x ∂ v∂x ∂x ∂ v∂x = 0 . in Ω δ = { x = ( x , x ) , x ∈ R , ≤ x ≤ δ } . The obstacle condition is(131) v ( x ) ≥ F ( x ) ≡ e x in Ω δ Denote B dδ the dyadic Bellman function of a corresponding problem. There aremany solutions of the above equation in Ω δ which satisfy the obstacle condition ≥ e x in Ω δ and even satisfying the boundary condition = e x on x = 0. These are ϕ ε,q ( x , x ) = q (1 − √ ε − x )1 − √ ε e x + √ ε − x −√ ε , δ ≤ ε < , q ≥ . One can compute v δ -the smallest solution of the above equation satisfying the ob-stacle condition. v δ = 1 − √ δ − x − √ δ e x + √ δ − x −√ δ . This is not B dδ ! In fact, B dδ > v δ . However, v δ is the Bellman function for non-dyadic John–Nirenberg inequality !!! The rest is in Vasyunin’s lectures and in[53].6.4. Burkholder-Bellman function. ∀ I ∈ D |h g i I + − h g i I − | ≤ |h f i I + − h f i I − |⇒ ∀ I ∈ D such that |h g i I | ≤ |h f i I | one has h| g | p i I ≤ ( p − p h| f | p i I p ≥ . The constant ( p − p is sharp. This is a famous theorem of Burkholder which heproved by constructing the corresponding Bellman function. He found it by solvinga corresponding Bellman PDE - a complicated one. We would like to show a simple“heuristic” method of solution. Who moves ? x = h g i J , x = h f i J x = h| f | p i J . Our rules say that f ( α, x ) = 0, E F ( x t , x t , x t ) ≈ E | g | p . Denoting by F t the σ -algebragenerated by dyadic subintervals of I of length 2 − n | I | , t = 2 n , we can write E | g | p ≈E | ( E x |F t ) | p = E | x t | p which gives us the correct bonus function F ( x , x , x ) = | x | p .Notice that A = { α = ( α , α , α ) : | α | ≤ | α |} now.This is because | α | = |h g i J + − h g i J − | , | α | = |h f i J + − h f i J − | , and we are giventhat the first quantity is always majorized by the second one.So we have the Bellman equationsup | α |≤| α | , α h H v α, α i = 0in Ω = { x : ( x , x , x ) : | x | p ≤ x } (convex), with obstacle condition v ( x , x , x ) ≥ | x | p . Compare with (121)!This example is interesting because we have a non-trivial set of restrictions A for“control” α .Solutions were given by Burkholder [19] (see also [20]–[25]) and also (a differentapproach using Monge–Amp`ere equation) can be found in [59]. See also a veryinteresting review [4]. ELLMAN FUNCTION TECHNIQUE IN HARMONIC ANALYSIS 51 An interesting Bellman function built by the use of Monge–Amp`ere equation canbe also found in [60], [61].7. The technique of laminates, Bellman function, and estimates ofsingular integrals from below Definition. Laminate on M s × is a positive finite measure on symmetric real ma-trices M s × such that(132) f ( A ) ≥ Z f ( A + M ) dν ( M )for all rank 1 concave functions f . Theorem 7.1. Any laminate on M s × can be approximated weakly by the push for-ward of Lebesgue measure on the plane by the Hessian of smooth compactly supportedfunctions, in other words, for any good F Z F ( M ) dν ( M ) ≈ Z R F ( Du ) dxdy , where Du ( x, y ) := (cid:20) u xx u xy u yx u yy (cid:21) . Observation. Laminates supported by diagonal matrices (cid:20) X Y (cid:21) are just exactlyexactly the measures on R such that ( z = ( X, Y ))(133) f ( a ) ≥ Z C f ( a + z ) dν ( z )for all bi-concave (meaning separately concave in X and Y ) function f . Definition. ( R X dν, R Y dν ) is called baricenter of a laminate ν supported ondiagonal matrices.Fix p > p η = p + η, η > 0. Put s := 1 − p η , K := p η p η − , p − η − K + 1 K − . We are going to construct very interesting laminates supported on Y = KX , Y + 1 K X . Fix p ≥ 2, fix small η > 0, put p η := p + η , (134) s := 1 − p η , K := 1 s = p η p η − , p η = 2 KK − , p η − K + 1 K − , 1) supportedby lines L K : Y = KX , L /K : Y = 1 K X . Let f be a bi-concave function and(135) f ( z ) = O ( | z | p ) , z → ∞ . Then concavity in horizontal variable gives(136) f ( t, t + h ) ≥ t − K ( t + h ) t + h − K ( t + h ) f ( t + h, t + h ) + ht + h − K ( t + h ) f ( 1 K ( t + h ) , t + h ) . Rewrite it as(137) f ( t + h, t + h ) ≤ t + h − K ( t + h ) t − K ( t + h ) f ( t, t + h ) − ht − K ( t + h ) f ( 1 K ( t + h ) , t + h ) . The concavity in vertical variable gives(138) f ( t, t ) ≥ t − K tt − K t + h f ( t, t + h ) + ht − K t + h f ( t, K t ) . From (137), (138) we obtain (of course we divide by h , and next, we will make h tend to 0) f ( t + h, t + h ) − f ( t, t ) h ≤ h (cid:20) t + h − K ( t + h ) t − K ( t + h ) − − t − K tt − K t + h (cid:21) f ( t, t + h ) − t − K ( t + h ) f ( 1 K ( t + h ) , t + h ) − t − K t + h f ( t, K t ) . Make h → 0. Then(139) f ′ ( t, t ) − KK − f ( t, t ) t ≤ − KK − f ( K t, t ) t − KK − f ( t, K t ) t . We recall (134) and multiply by 1 /t p η . Notice that after that LHS = (cid:18) f ( t,t ) t pη (cid:19) ′ .We integrate from 1 to ∞ and use (135) to forget the term at infinity. Then weobtain for any bi-concave function on the plane(140) − f (1 , ≤ − KK − Z ∞ f ( 1 K t, t ) dtt p η +1 − KK − Z ∞ f ( t, K t ) dtt p η +1 Introduce ν K,η : Z R φ dν K,η = KK − Z ∞ φ ( 1 K t, t ) dtt p + η +1 . It is a laminate supported by L K : Y = KX . And introduce ν /K,η : Z R φ dν /K,η = KK − Z ∞ φ ( t, K t ) dtt p + η +1 . It is a laminate supported by L /K : Y = K X . Now (140) can be rewritten as(141) f (1 , ≥ Z f ( dν K,η + dν /K,η ) ELLMAN FUNCTION TECHNIQUE IN HARMONIC ANALYSIS 53 If all our concavity in getting (141) become linearities then we have equality in(141).So dν K,η + dν /K,η is a laminate with baricenter (1 , , µ K,η = 14 ( dν K,η + dν /K,η ) + 14 δ ( − , + 12 δ (0 , . Test it on φ ( X, Y ) = | X + Y | p , φ ( X, Y ) = | X − Y | p . Then(142) R φ dµ K,η R φ dµ K,η = K (( K + 1) p + ( K + 1) p /K p ) η − + ( K − K (( K − p + ( K − p /K p ) η − + ( K − 1) + p ( K − . Choosing η > R φ dµ K,η R φ dµ K,η ≥ (cid:18) K + 1 K − (cid:19) p − Cη . Notice that we can consider a bit different laminate than µ K,η , Namely let us pushforward µ K,η by the map X → X, Y → − Y . The new measure is called σ K,η . Then(143) transforms to(144) R φ dσ K,η R φ dσ K,η ≥ (cid:18) K + 1 K − (cid:19) p − Cη . Now we use Theorem 7.1. It implies that there exist smooth functions withcompact support on the plane such that(145) R | u xx − u yy | p dm R | u xx + u yy | p dm ≥ (cid:18) K + 1 K − (cid:19) p − Cη . Notice that K depends on η (see (134)) but K + 1 K − → p − , η → . Thus from (145) we get the estimate(146) k R − R k p ≥ p − , if p ≥ p ∗ − smallest constant c = c p such that the function h c ( X, Y ) = | Y + X | p − c p | Y − X | p has a bi-concave majorant. Definition. Let us call ϕ p ( X, Y ) the smallest bi-concave majorant of h c ( X, Y ) = | Y + X | p − c p | Y − X | p for the smallest (as we know) possible c = c p = p ∗ − ϕ p in the next section.Now let us consider a different family (it is a perturbation of h c ): h c,τ := | (( Y + X ) + τ ( X − Y ) ) / | p − c p | Y − X | p . Here is a result proved in [18]. Theorem 7.2. For sufficiently small universal τ > , any p ∈ (1 , ∞ ) , and any τ ∈ [ − τ , τ ] , the smallest c for which there exists a bi-concave majorant of h c,τ is c p ( τ ) = (( p ∗ − + τ ) / . Using the same considerations with laminates as above (especially Theorem 7.1)we can prove the following estimate from below for “quantum” linear combinationof secon order Riesz transforms: Theorem 7.3. For sufficiently small τ and any small positive ǫ one can find g ∈ L p ( m ) such that k ( | ( R − R ) g | + τ | ( R + R ) g | ) / k p ≥ (( p ∗ − + τ ) / k g k p − ǫ . This gives rise to the following problem: Problem. For sufficiently small τ k (cid:20) R − R τ I (cid:21) : L p ( m ) → L p ( R , l ) k = (( p ∗ − + τ ) / ?The answer is affirmative, see [18]. Notice that for p ∈ (1 , 2) and large τ this isno longer true. Somewhere we have a “phase transition” of the sharp constant. Itis not clear what is the critical τ ( p ).7.1. “Explanation” of laminates above via Burkholder’s function ϕ p ( X, Y ) and its properties. We introduce coordinates ( x, y ): Y = y + x, X = y − x . Let γ p = p (1 − p ∗ ) p − . In the first and second quadrants of xy , Burkholder’s function in these coordinatesis equal to (here the reader should glance at (134) and make η = 0 in it, s and k below are as in (134), but with η = 0)(147) ϕ p ( x, y ) := ( γ p ( y − ( p ∗ − | x | )( | x | + y ) p − , if y −| x | y + | x | ≥ s := 1 − p = k y p − ( p ∗ − p | x | p , if − ≤ y −| x | y + | x | ≤ s . Now extend ϕ p ( x, y ) to the whole plane by ϕ p ( x, y ) = ϕ p ( − x, − y ) . Burkholder proved [19] Theorem 7.4. Such a function coincides with the smallest majorant of h c ( x, y ) = | y | p − c p | x | p , c = p ∗ − bi-convex in X, Y coordinates. For c ∈ [0 , p ∗ − there isno such bi-concave majorant of h c . Observation 2 . We use here both coordinates ( X, Y ) and ( x, y ). In the cone X ≤ Y ≤ KX function ϕ p is linear along Y = const segments. Similarly, In thecone K X ≤ Y ≤ X function ϕ p is linear along X = const segments.This linearity allows to calculate (we are in ( X, Y ) now) ϕ p ( t + h, t + h ) − ϕ p ( t, t ) ELLMAN FUNCTION TECHNIQUE IN HARMONIC ANALYSIS 55 virtually without any loss if we use the T -shape 4-tuple of points in R : (( K ( t + h ) , t + h ) , ( t, t + h ) , ( t + h, t + h ) , ( t, K t )) as in Section 7.If we move one of the lines L K , L /K then two things may happen: 1) we go outsideof these linearity cones, and subsequently we get strict inequality for ϕ p (1 , ϕ p = h p ∗ − equality because by the definition of of ϕ p (see (147)) this equality holds only onthe boundary of and outside of the union of linearity cones.Notice also that on these lines L K , L /K (recall that K = p + ηp + η − if p ≥ 2) we havethat ϕ p ≈ h p ∗ − ≈ , ϕ p ≥ h p ∗ − . For c larger than p ∗ − h c coincides with itsbi-convex majorant, but then they will be quite negative there and integration of h c along a laminate supported on such lines cannot be almost positive as it was thecase above. 8. Stochastic Calculus and / quasiconvexity We have a bijection of matrices M = (cid:20) a, bc, d (cid:21) onto ( z, w ): z = a + d + i ( b − c ) , w = a − d + i ( b + c ). Notice that 2 det M = | z | − | w | .Recall that Sverak’s function is the following “simple” object S ( z, w ) := ( | z | − | w | , | z | + | w | ≤ | z | − , otherwiseFunction ψ p ( z, w ) := (( p − | z | − | w | )( | z | + | w | ) p − , p ≥ , can be easily obtained from S using the idea of Iwaniec, see e.g. [9]. The process isa certain averaging. Therefore, the fact that S is rank-1 convex implies that ψ p isalso rank-1 convex.To solve the Big Iwaniec problem of the previous sections it would be enough thatany of these functions is quasiconvex at zero matrix. 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Department of Mathematics, Michigan State University, East Lansing, MI 48824,USA E-mail address ::