A characterization of the two weight norm inequality for the Hilbert transform

Transcription

1 A characterization of the two weight norm inequality for the Hilbert transform Ignacio Uriarte-Tuero (joint works with Michael T. Lacey, Eric T. Sawyer, and Chun-Yen Shen) September 10, 2011

2 A p condition Definition A weight is w 0 L 1 loc (Rn ). Notation: w(e) = E w = E w. 1 < p < : w A p means there exists C > 0 such that for all Q ( ) ( ) 1 1 p 1 w w 1 p 1 C (0.1) Q Q Q Q Theorem (Coifman, Fefferman, Hunt, Muckenhoupt, Wheeden) 1 < p <, M is the Hardy-Littlewood maximal operator M : L p (w) L p (w) w A p i.e. R n Mf p w C p R n f p w ( =) is also true for T Calderón-Zygmund operator (= ) Non-degen T : L p (µ) L p (µ) = dµ = w dx, w A p. But (= ) assumes µ(b) > 0 for any ball B (0.2)

3 Two weights Th p w C h p v Standard trick: h = f σ σ = v 1 p 1 ) (= σ p v = v 1 p 1 Get: T (f σ) p w C f p σ (In general necessary for testing conditions to be sufficient) Maximal Function 1 Mν(x) = sup ν, x Q Q Q x R n Supremum over cubes Q with sides parallel to coordinate axes containing x

4 Theorem on Maximal Function Inequalities σ, w positive locally finite Borel measures on R n 1 < p < Muckenhoupt 72 - Two Weight A p - (Sawyer s reformulation) The maximal operator M satisfies the weak type two weight norm inequality M(f σ) L p, (ω) sup λ>0 1 p λ {M (f σ) > λ} ω C f L p (σ), f Lp (σ), (0.4) if and only if the two weight A p condition holds for all cubes Q: ( ) 1 ( ) 1 1 p 1 p dω dσ C 2. (0.5) Q Q Q Q

5 Theorem on Maximal Function Inequalities Sawyer 82 - Two Weight Norm Inequality for M The maximal operator M satisfies the two weight norm inequality M(f σ) L p (ω) C f L p (σ), f Lp (σ), (0.7) if and only if for all cubes Q, M (χ Q σ) (x) p dω(x) C 1 Q Q dσ(x). (0.8) Only need to test the strong type inequality for functions of the form χ Q σ. The full L p (ω) norm of M(χ Q σ) need not be evaluated. There is a corresponding weak-type interpretation of the two-weight A p condition (0.5).

6 Conjecture I Conjecture σ and ω positive Borel measures on R n, 1 < p <, and T a standard singular integral operator on R n. The following two statements are equivalent: [ 1 Q T (f σ) p ω(dx) C ] 1 Q [ ] 1 dω p 1 Q Q dσ p C, Q T χ Qσ p ω(dx) C Q σ(dx), Q T χ Q ω p σ(dx) C Q ω(dx), f p σ(dx), f C 0, cubes Q. In our case: H(σf ) L 2 (ω) C f L 2 (σ). (0.10)

7 Conjecture II Conjecture is most important when T is Hilbert transform, Beurling Transform, or Riesz Transforms. This question occurs in different instances: Sarason Conjecture on composition of Hankel operators, semi-commutator of Toeplitz operators, perturbation theory of some self-adjoint operators. Conjecture only verified for positive operators: Poisson integrals, fractional integral operators (Sawyer), and well-localized (or band) operators (i.e. operators which in the Haar basis have a matrix supported in a bounded number of diagonals ) (Nazarov, Treil and Volberg) Cotlar and Sadosky proved the two-weight Helson-Szego Theorem: completely solves the L 2 case of the Hilbert transform, but not real-variable.

8 Milestone: Nazarov, Treil, Volberg I Let p = 2 (most relevant case). For an interval I and measure ω, consider fattened (Poisson) A 2 condition: where P(I, ω) R sup P(I, ω) P(I, σ) = A 2 2 <, (0.11) I I ( I + dist(x, I )) 2 ω(dx) ( I dω ) I (0.12) Assume P < s.t. for all intervals I 0, and decompositions {I r : r 1} of I 0 into disjoint intervals I r I 0 pivotal condition: ω(i r )P(I r, χ I0 σ) 2 P 2 σ(i 0 ) (0.13) r=1 And dual pivotal condition (reverse roles of σ and ω): σ(i r )P(I r, χ I0 ω) 2 (P ) 2 ω(i 0 ) (0.14) r=1

9 Milestone: Nazarov, Treil, Volberg II There are two obviously necessary conditions for the two weight Hilbert inequality (0.10) to hold (Sawyer-type testing conditions). Uniformly over intervals I : H(1 I σ) 2 ω(dx) H 2 σ(i ), (0.15) I H(1 I ω) 2 σ(dx) (H ) 2 ω(i ). (0.16) I

10 Milestone: Nazarov, Treil, Volberg III Theorem (Nazarov, Treil, Volberg) Suppose the pair of weights σ, ω satisfy the pivotal condition and the dual pivotal condition, that is, both P and P are finite. Then, the two weight inequality for the Hilbert transform H(σf ) L 2 (ω) C f L 2 (σ). (0.17) holds if and only if the three quantities A 2, H and H are all finite. Nazarov, Treil, and Volberg further stated that the pivotal conditions might be necessary for the Hilbert transform two weight bound (0.17).

11 Further conditions I Energy condition. This is key improvement with respect to NTV. Condition that accounts for the distribution / concentration of the weights inside intervals. Given interval I, define ( E(I, ω) E ω(dx) I E ω(dx ) I ( x x ) ) 2 1/2. (0.18) Note E(I, ω) 1, but can be quite small if ω is highly concentrated inside I. E.g. E(I, ω) = 0 if ω = δ a for a I. The energy functional E(I, ω) improves smoothness of the kernel estimates. Here, E ω I φ ω(i ) 1 φ ω (dx), (0.19) thus, E ω I f is the average value of f with respect to the weight ω on the interval I. I I

12 Further conditions II For a parameter 0 ε 2, the ε-energy condition is that there exists E ε < s.t. ω(i r )[E(I r, ω)] ε [P(I r, χ I0 σ)] 2 Eε 2 σ(i 0 ). (0.20) r=1 We require that (0.20) hold for all intervals I 0, and strict partitions {I r : r 1} of I 0. Let E ε be the smallest constant for which these inequalities are uniformly true. E ε is the best constant in the same inequality with the roles of ω and σ reversed.

13 Further conditions III The ε-energy condition (0.20) on a pair of measures increases in strength as ε decreases. The (ε = 2)-energy condition is necessary: E 2 A 2 + H, and similarly for E 2. Proof of this fact uses that ν(i ) = 0 implies P (I ; ν) I inf x,y I Hν (x) Hν (y). (0.21) x y That the constant E 0 < (i.e. E(I r, ω) 1 in (0.20)) is the pivotal condition (0.13) introduced by Nazarov, Treil and Volberg.

14 Main Theorem (Lacey, Sawyer, UT 10) Let ω and σ be locally finite positive Borel measures on R with no point masses in common, ω ({x}) σ ({x}) = 0 for all x R. Suppose that for some 0 ε < 2, E ε + E ε <. Then H(σf ) L 2 (ω) C f L 2 (σ) holds if and only if sup P(I, ω) P(I, σ) = A 2 2 <, (0.11) I H(1 I σ) 2 ω(dx) H 2 σ(i ), (0.15) I H(1 I ω) 2 σ(dx) (H ) 2 ω(i ) (0.16). I Moreover, H( σ) L 2 (σ) L 2 (ω) max {A 2, H, H, E ε, E ε }. The case of general measures requires a modification of the A 2 condition.

15 Counterexamples I Recall middle-third Cantor set E and Cantor measure ω. ω assigns equal weight to these intervals

16 Counterexamples II I 1,1 I 2,1 I 2,2 I 3,1 I 3,2 I 3,3 I 3,4 σ = k,j s k δ zk,j, z k,j I k,j. s k satisfy a precursor to the A 2 condition: s k ω(3i k,j ) 3I k,j 2 = 1, s k = 2k 3 2k.

17 Counterexamples III Place the z k,j at the center of the I k,j : z 1,1 z 2,1 z 2,2 z 3,1 z 3,2 z 3,3 z 3,4 For this σ, the pair (ω, σ) satisfies the A 2 condition (0.11), the forward testing condition (0.15), but fails the backwards testing condition (0.16). (Alternative example to Nazarov-Volberg and Nikolski-Treil.) Hω1 I 2 σ(dx) (H ) 2 ω(i )???? (0.16) I

18 Counterexamples IV Consider Hω on the intervals I k,j : z 1,1 z 3,4 Place the z k, at the unique zeros of Hω on I k,j! This pair of weights DOES satisfy the two-weight inequality for the Hilbert transform (0.17), but DOES NOT satisfy the dual pivotal condition (0.14).

19 SKIP: Some comments on proofs I How does the energy condition appear in the sufficiency proof? For a toy model case, containing idea, recall the familiar argument (with Lebesgue measure): c j center of I j, b j dx = 0, supp (b j ) I j (bad function) Matters reduced to Hb j (y) dy = bj (x) R\2I j R\2I j x y dx dy ( 1 = b R\2I j j (x) I j x y 1 ) x dx y dy (x I j, say x = c j ) I j b j (x) dy dx I j R\2I j y x 2 b j (x) dx P(I j, χ R\Ij dy) I j

20 SKIP: Some comments on proofs II Now the weighted version: b j dω(x) = 0, supp (ωb j ) I j (bad function) H(ωb j )(y) dσ(y) = bj (x) R\2I j R\2I j x y dω(x) dσ(y) ( = b j (x) E ω(dx ) 1 I j I j x y 1 ) x dω(x) y dσ(y) R\2I j ( if ω = δ a, a I j, then = 0!! Choice of x I j now matters!!) ( write 1 x y 1 x y = x x ) I j I j (x y)(x y) ( b j (x) E ω(dx ) x x ) I j ω(dx) P(I j, χ I j I j R\Ij σ(dy))

21 SKIP: Some comments on proofs III Last line of previous slide was: ( b j (x) E ω(dx ) x x ) I j ω(dx) P(I j, χ I j I j R\Ij σ(dy)) { [ ( b j L 2 (ω) E ω(dx ) x x 1 )] 2 2 I j ω(dx)} P(I j, χ I j R\Ij I j = b j L 2 (ω) ω(i j ) 1 2 E(Ij, ω) P(I j, χ R\Ij σ(dy)) σ(dy)) And now insert this idea into Nazarov-Treil-Volberg s huge machine...

22 H(σf),φ ω duality A 1 2 A 1 1 W A 2 A 2 3 A 2 1 A 2 2 E E A 3 2 Corona E A 3 1 A 4 1 A 4 2 A 4 3 H + E Paraproducts A 5 1 A 5 2 E projection A 6 1 A 6 2 A 6 3 H + E E A 7 1 A 7 2

23 Further refinements I Nazarov-Treil-Volberg approach: consider the bilinear form B(f, φ) = H σ f, φ w = H(σf ), φ w. Appropriately (randomly) choose a dyadic grid D, define Haar bases adapted to the two weights. E.g. {hj w : J D}, for the weight w: Given a dyadic interval I (I D) and a weight w, if w(i )w(i + ) > 0, (otherwise hi w = 0), supp(h w I ) = I, hi w is constant on I and I +, hi w dw = 0, I (hi w ) 2 dw = 1. I h w I = 1 w(i ) 1/2 [ (w(i+ ) ) 1/2χI w(i ) ( w(i ) ) ] 1/2χI+ w(i + )

24 Further refinements II The set {h w I } I D forms a basis of L 2 (w). Notation ( w I f = fhi w I ) dw hi w = f, hi w w hi w Note that (martingale difference) w I f = χ I+ E w I + f + χ I E w I f χ I E w I f Then f = I w I f and f 2 L 2 (w) = I D f, h w I w 2

25 Further refinements III Then B(f, φ) = I D f, hi σ σ H σ hi σ, hw J w φ, hj w w. J D Motivated by T 1 theorem, split the sum, according to the whether or not I is bigger than J. This creates two forms, called B abv (f, φ) and B blw (f, φ), plus diagonal term. B blw (f, φ) : I D J D : J < I f, h σ I σ H σ h σ I, hw J w φ, h w J w. Note that B(f, φ) {B blw (f, φ) + B abv (f, φ)} = Diagonal term (A 2 + H + H ) f σ φ w

26 Further refinements IV Previous proof assumes conditions that guarantee that there is C < such that B blw (f, φ) C f σ φ w, which can be verified in any standard T 1 theorem. But it is not clear that in the two weight setting the boundedness of B(f, φ) implies that of B blw (f, φ). If this implication does not hold for all pairs of weights, then the current approach can not resolve the conjecture!

27 Further refinements V Write the norm of the bilinear form B blw as B blw. Main Theorem (Lacey, Sawyer, Shen, UT 11) The following inequalities and their duals hold, for any pair of weights w, σ which do not share a common point mass. B blw + H + A 1/2 2 H + F + BF + A 1/2 2. (0.22) That is, modulo the Nazarov-Treil-Volberg conjecture, the above and below splitting is completely understood in terms of the functional energy F and bounded fluctuation BF conditions. Theorem (Lacey, Sawyer, Shen, UT 11) Let (w, σ) be any pair of weights with σ doubling. Then the two weight inequality (0.17) holds if and only if the quantities A 2, H, H, F and BF are all finite.

28 Further refinements VI Definition Given interval I 0, set F(I 0 ) to be the maximal dyadic subintervals F such that E σ F f > 4Eσ I 0 f. Set F 0 = {I 0 }, and inductively set F j+1 := F F j F(F ). The collection of f -stopping intervals is F := j=0 F j. Take f non-negative and supported on an interval I 0, and f -stopping intervals as above. Let {g F : F F} be a collection of functions in L 2 (w) which is F-adapted to F., i.e., essentially, for each F, 1 g F is supported on F and constant on F Child F (F ); 2 E w J g F = 0 for each J J (F ), where J (F ) are the maximal dyadic J F such that 3J F.

29 Further refinements VII Let F be the smallest constant in the inequality below, holding for all non-negative f L 2 (σ), and collections {g F } as just described. F F J J (F ) P(f (R F )σ, J ) x J, g F J w F f σ We refer to this as the functional energy condition. Here sets are identified with their characteristic functions. [ ] 1/2 g F 2 w. F F (0.23)

30 Further refinements VIII Write f BF F (F ), and say that f is of bounded fluctuation if (i) f is supported on F, (ii) f is constant on each F Child F (F ), and (ii) for each dyadic interval I F, which is not contained in some F Child F (F ), we have E σ I f 1. We then denote as BF the best constant in the inequality I : π F I =F J : J I π F J=F E σ I J σ f H σ I J, σ J g w BF{σ(F ) 1/2 + f σ } g w (0.24) where f BF F (F ), and g is F-adapted to F. Note: the two terms σ(f ) 1/2 and f σ on the right above are in general incomparable. This is the bounded fluctuation condition. The role of the constant one in the inequalities E σ I f 1 is immaterial. It can be replaced by any fixed constant.