Rectifiability and Singular Integrals: solving DavidSemmes problem - PowerPoint PPT Presentation

frame 13. We get � � � � � � | R H ( ν a ) | 2 | ψ | dm d ≤ 6 C 1 λ + R H ( | ψ | dm d ) · R H ( ν a ) d ν a � � 2 H ( a ) 1 / 2 � � √ � 1 / 2 | R H ( | ψ | dm d ) | 2 d ν ≤ 6 C 1 λ + The last integral can be taken “layer” by “layer” as d ν = d dm d × δ − 1 dx d +1 . On each layer we use that R H is bounded in L 2 ( m d ). Hence, we continue � | R H ( ν a ) | 2 | ψ | dm d ≤ 6 C 1 λ + 2 λ 1 / 2 mass 1 / 2 � � � 1 / 2 . | ψ | 2 dm d H Temporarily normalize by | E | ≤ 1 ⇒ mass ≤ 1. Then Cauchy � R H ∗ ψ d ν a | = inequality gives mass = mass ( ν a ) ≤ | � � | R H ( ν a ) || ψ | dm d ≤ C ( λ + λ 1 / 2 ) 1 / 2 ≤ C λ 1 / 4 . R H ( ν a ) ψ dm d | ≤ | Alexander Volberg Solving a problem of David and Semmes

frame 14. Therefore, using the assumption | E | ≤ 1 we finally get the � estimate on λ from below λ ≥ c ( mass ) 4 = c ( d dm d ) 4 . This gives us immediately the following estimate on the Riesz energy from below (see (1) with minimal λ ): � � � 5 . E ( f , E ) ≥ c f dm d To get rid of the assumption | E | ≤ 1 we just use the scaling invariance to get �� � 4 � f dm d = mass 5 f dm d E ( f , E ) ≥ c | E | 4 . (3) | E | Theorem 3 is proved. Alexander Volberg Solving a problem of David and Semmes

frame 15. David-Semmes lattice Let µ be a d -dimensional AD regular measure in R d +1 . Let E = supp µ . then there exists a family D of sets Q ⊂ R d +1 with the following properties: The family D is the union of families D k (families of level k cells), k ∈ Z . If Q ′ , Q ′′ ∈ D k , then either Q ′ = Q ′′ or Q ′ ∩ Q ′′ = ∅ . Each Q ′ ∈ D k +1 is contained in some Q ∈ D k (necessarily unique due to the previous property). The cells of each level cover E , i.e., ∪ Q ∈D k Q ⊃ E for every k . For each Q ∈ D k , there exists z Q ∈ Q ∩ E (the “center” of Q ) such that B ( z Q , 2 − 4 k − 3 ) ⊂ Q ⊂ B ( z Q , 2 − 4 k +2 ) . For each Q ∈ D k and every ε > 0, we have µ { x ∈ Q : dist( x , R d +1 \ Q ) < ε 2 − 4 k } ≤ C ε γ µ ( Q ); C ,γ = C , γ ( d , reg ) Alexander Volberg Solving a problem of David and Semmes

frame 16. Sketch of construction Since all cells in D k have approximately the same size 2 − 4 k , it will be convenient to introduce the notation ℓ ( Q ) = 2 − 4 k where k is the unique index for which Q ∈ D k . Let Z k be a maximal 2 − 4 k -separated set in E = supp µ . Then { B ( z , 2 − 4 k ) } z ∈ Z k , cover E . For each z ∈ Z k consider Voronoi cell V z := { x ∈ R d +1 : | x − z | = min | x − z ′ |} . z ′ ∈ Z k Then 1) V z ⊂ B ( z , 2 − 4 k ), 2) { V z } z ∈ Z k cover E , 3) dist ( z , � z ′ ∈ Z k , z ′ � = z V z ′ ) ≥ 2 − 4 k − 1 . The last one because Z k is 2 − 4 k -separated, the first one because Z k is maximal such. Also 4) There are only finitely many w ∈ Z k − 1 such that V z ∩ V w � = ∅ . We say that w ∈ Z k is a descendant of z ∈ Z ℓ , ℓ ≥ k , if there is a chain z k = z , z ℓ = w , z j ∈ Z j such that V z j ∩ V z j +1 � = ∅ . D ( z ) is the set f all descendants of z and � � V z := V w . w ∈ D ( z ) Alexander Volberg Solving a problem of David and Semmes

frame 17. Note that � V z contains V z and is contained in the 2 � ℓ> k 2 − 4 ℓ = 2 15 2 − 4 k -neighborhood of V z . Thus, V z ′ ) ≥ 2 − 4 k − 1 − 2 152 − 4 k > 2 − 4 k − 2 . dist( z , ∪ z ′ ∈ Z k \{ z } � (4) Nobility order. There exists a partial order ≺ on ∪ k Z k such that each Z k is linearly ordered under ≺ and the ordering of Z k +1 is consistent with that of Z k in the sense that if z ′ , z ′′ ∈ Z k +1 and z ′ ≺ z ′′ , then for every w ′ ∈ Z k such that V w ′ ∩ V z ′ � = ∅ , there exists w ′′ ∈ Z k such that V w ′′ ∩ V z ′′ � = ∅ and w ′ � w ′′ . In other words, the ordering we are after is analogous to the classical “nobility order” in the society: comparing maximally “noble” ancestors one generation up defines “nobility”. Put now for each z ∈ Z k � E z := � � V z \ V z ′ . z ′ ∈ Z k , z ≺ z ′ Alexander Volberg Solving a problem of David and Semmes

frame 18. By (4) we have the left inclusion (the right one is clear too) B ( z , 2 − 4 k − 2 ) ∩ E ⊂ E z ⊂ B ( z , 2 − 4 k +1 ) for all z ∈ Z k . Next goal is to show the tiling: that for every z ∈ Z k +1 there exists w ∈ Z k such that E z ⊂ E w . For a given z ∈ Z k +1 choose w to be the largest in ≺ element of Z k . Let w ′ ∈ Z k be such that w ≺ w ′ . Let z ′ ∈ Z k +1 , z ′ ∈ D ( w ′ ). Automatically z ≺ z ′ . And so { z ′ ∈ Z k +1 : z ′ ∈ D ( w ′ ) } ⊂ { z ′ ∈ Z k +1 : z ≺ z ′ } . On the other hand, by definition V w ′ ⊂ � z ′ ∈ Z k +1 : z ′ ∈ D ( w ′ ) V z ′ , and so � � � � � V z ′ ⊂ V w ′ = V z ′ z ′ ∈ Z k +1 : z ′ ∈ D ( w ′ ) z ′ ∈ Z k +1 , z ≺ z ′ � � � V z ′ ⊂ � � � V z \ V w \ V w ′ z ′ ∈ Z k +1 , z ≺ z ′ w ′ ∈ Z k , w ≺ w ′ This is exactly E z ⊂ E w . Alexander Volberg Solving a problem of David and Semmes

frame 19. Carleson families. For us this will be the right notion of sparse, rare family of cells. From now on, we will fix a good AD regular in the entire space R d +1 measure µ and a David-Semmes lattice D associated with it. Definition A family F ⊂ D is called Carleson with Carleson constant C > 0 if for every P ∈ D , we have � µ ( Q ) ≤ C µ ( P ) , Q ∈F P where F P = { Q ∈ F : Q ⊂ P } . Alexander Volberg Solving a problem of David and Semmes

frame 20. Non-BAUP cells. Actually non-OUWGL We will start with the definition of a δ -non-BAUP cell. Definition Let δ > 0 . We say that a cell P ∈ D is δ -non-BAUP if there exists a point x ∈ P ∩ supp µ such that for every hyperplane L passing through x, there exists a point y ∈ B ( x , ℓ ( P )) ∩ L for which B ( y , δℓ ( P )) ∩ supp µ = ∅ . Note that in this definition the plane L can go in any direction. In what follows, we will need only planes parallel to certain H but, since H is determined by the flatness direction of some unknown cell P , we cannot fix the direction of the plane L in the definition of non-BAUPness from the very beginning. Theorem (David–Semmes) Let µ be AD-regular. If for all δ > 0 the family of δ -non-BAUP cells is a Carleson family, then µ is rectifiable. Alexander Volberg Solving a problem of David and Semmes

frame 21. Main Theorem Theorem Let µ be an AD regular measure of dimension d in R d +1 . If the associated d-dimensional Riesz transform operator x f �→ R ∗ ( f µ ) , where R ( x ) = | x | d +1 , is bounded in L 2 ( µ ) , then the non-BAUP cells in the David-Semmes lattice associated with µ form a Carleson family. Proposition 3.18 of David–Semmes 1993 (page 141) asserts that this condition “implies the WHIP and the WTP” and hence, by Theorem 3.9 (pages 137), the uniform rectifiability of µ . Alexander Volberg Solving a problem of David and Semmes

frame 22. Idea Using the boundedness of R µ in L 2 ( µ ) we will establish the abundance of flat cells . On the other hand, if non-BAUP cells are not rare (not Carleson) they will be also abundant. Then we will be able to build intermitting layers of flat and non-BAUP cells. This will allow us to construct an analog of vector field ψ on non-BAUP scales. This is because non-BAUP cell has holes in supp µ in it! Flat cells will play the role of the set E (which was totally flat). Then Riesz energy concentrated on each flat layer will be sufficiently large (the non-BAUP layer encompassing a flat layer and ψ of this non-BAUP layer ensures that). Then we will need that flat layers are almost orthogonal . Adding huge amount of not-so-small Riesz energies we get estimate from below on � | R µ 1 | 2 d µ as large as we wish. Contradiction. Alexander Volberg Solving a problem of David and Semmes

frame 23. The flatness condition and its consequences We shall fix a linear hyperplane H ⊂ R d +1 . Let z ∈ R d +1 , A , α, ℓ > 0 (we view A as a large number, α as a small number, and ℓ as a scale parameter). We want the measure µ to be close inside the ball B ( z , A ℓ ) to a multiple of the d -dimensional Lebesgue measure m L on the hyperplane L containing z and parallel to H . We say that a measure µ is geometrically ( H , A , α )-flat at the point z on the scale ℓ if every point of supp µ ∩ B ( z , A ℓ ) lies within distance αℓ from the affine hyperplane L containing z and parallel to H and every point of L ∩ B ( z , A ℓ ) lies within distance αℓ from supp µ . We say that a measure µ is ( H , A , α )-flat at the point z on the scale ℓ if it is geometrically ( H , A , α )-flat at the point z on the scale ℓ and, in addition, for every Lipschitz function f supported on � B ( z , A ℓ ) such that � f � Lip ≤ ℓ − 1 and f dm L = 0, we have � � � � � � ≤ αℓ d . � � f d µ � Alexander Volberg Solving a problem of David and Semmes

frame 24. Note that the geometric ( H , A , α )-flatness is a condition on supp µ only. It doesn’t tell one anything about the distribution of the measure µ on its support. The latter is primarily controlled by the second, analytic, condition in the full ( H , A , α )-flatness. These two conditions are not completely independent: if, say, µ is AD regular, then the analytic condition implies the geometric one with slightly worse parameters. However, it will be convenient for us just to demand them separately. The flatness means the possibility of mass transporting µ | B ( z , A ℓ ) to c · m L | B ( z , A ℓ ) with small cost α . Flatness allows to switch integration over µ to that over c · m L . Below are technical but very useful lemmas estimating the error of such switching. Alexander Volberg Solving a problem of David and Semmes

frame 25. Lemma Let µ be a nice measure (estimate from above). Assume that µ is ( H , A , α ) -flat at z on scale ℓ with some A > 5 , α ∈ (0 , 1) . Let ϕ be any non-negative Lipschitz function supported on B ( z , 5 ℓ ) with � ϕ dm L > 0 . Put �� � − 1 � a = ϕ dm L ϕ d µ, ν = a ϕ m L . Let Ψ be any function with � Ψ � Lip ( supp ϕ ) < + ∞ . Then � � � � � � � � ≤ C αℓ d +2 � Ψ � Lip ( supp ϕ ) � ϕ � Lip . Ψ d ( ϕµ − ν ) � � � � | Ψ | p d ( ϕµ − ν ) � � ≤ As a corollary, for every p ≥ 1 , we have C ( p ) αℓ d +2 � Ψ � p − 1 L ∞ ( supp ϕ ) � Ψ � Lip ( supp ϕ ) � ϕ � Lip . Alexander Volberg Solving a problem of David and Semmes

frame 26. Lemma Assume in addition to the conditions of Lemma 9 that ϕ ∈ C 2 , and that the ratio of integrals a is bounded from above by some known constant. Then � � � � � � � Ψ ϕ [ R H ( ϕµ − ν )] d µ � � � � 1 d +2 ℓ d +2 � ϕ � 2 ≤ C α � Ψ � L ∞ ( supp ϕ ) + ℓ � Ψ � Lip ( supp ϕ ) Lip . where C > 0 may, in addition to the dependence on d, which goes without mentioning, depend also on the growth constant of µ and the upper bound for a. Alexander Volberg Solving a problem of David and Semmes

frame 27. Disclaimer: Integral should be understood first. Split it as � � Ψ ϕ [ R H ( ϕµ )] d µ − Ψ ϕ [ R H ν ] d µ . Then R H ν = a R H ( ϕ dm L ) and so is a smooth function as ϕ is smooth. The first term should be understood as a form by using anti-symmetry of R H . The first lemma is just by definition. In the second Lemma choose d +2 and split R H = R H 1 δℓ + R H ,δℓ . Then δ = α � � Ψ ϕ [ R H R H δℓ ( ϕµ − ν )] d µ = − δℓ (Ψ ϕ d µ ) d ( ϕµ − ν ) is estimated by the first Lemma using � R H δℓ � Lip ≤ δ − ( d +1) ℓ − ( d +1) and � R H δℓ (Ψ ϕ d µ ) � Lip ≤ � R H δℓ � Lip � Ψ ϕ � L 1 ( µ ) . Alexander Volberg Solving a problem of David and Semmes

frame 28. � Ψ ϕ [ R H ,δℓ ( ϕµ − ν )] d µ essentially reduces The short range term to estimate: �� � � 1 � � R H ( x − y )(Ψ( x ) − Ψ( y )) ϕ ( x ) ϕ ( y ) d µ ( x ) d µ ( y ) � ≤ � 2 | x − y |≤ δℓ �� ≤ 1 d µ ( x ) d µ ( y ) 2 � Ψ � Lip ( supp ϕ ) � ϕ � 2 ≤ | x − y | d − 1 L ∞ x , y ∈ supp ϕ, | x − y | <δℓ ≤ C δℓ d +3 � Ψ � Lip ( supp ϕ ) � ϕ � 2 Lip . Alexander Volberg Solving a problem of David and Semmes

frame 29. Geometric Flattening Lemma We are heading to the proof that the boundedness of R µ in L 2 ( µ ) implies flatness of abundant family of cells. The first step is the following analysis-to-geometry Lemma. Fix some continuous function ψ 0 : [0 , + ∞ ) → [0 , 1] such that ψ 0 = 1 on [0 , 1] and ψ 0 = 0 on [2 , + ∞ ). For z ∈ R d +1 , 0 < r < R , define � | x − z | � � | x − z | � ψ z , r , R ( x ) = ψ 0 − ψ 0 . R r Alexander Volberg Solving a problem of David and Semmes

frame 30. Lemma (Geometric Flattening Lemma) c , � Fix five positive parameters A , α, β, � C > 0 . There exists ρ > 0 depending only on these parameters and the dimension d such that the following implication holds. Suppose that µ is a � C-good measure on a ball B ( x , R ) centered at a point x ∈ supp µ that is AD regular in B ( x , R ) with lower regularity constant � c. Suppose also that | [ R ( ψ z ,δ R , ∆ R µ )]( z ) | ≤ β for all ρ < δ < ∆ < 1 2 and all z ∈ B ( x , (1 − 2∆) R ) such that dist ( z , supp µ ) < δ 4 R. Then there exist a scale ℓ > ρ R, a point z ∈ B ( x , R − ( A + α ) ℓ ) , and a linear hyperplane H such that µ is geometrically ( H , A , α ) -flat at z on the scale ℓ . Alexander Volberg Solving a problem of David and Semmes

Replacing µ by R − d µ ( x + R · ) if necessary, we may assume without loss of generality that x = 0 , R = 1. The absence of geometric flatness and also the boundedness of [ R ( ψ z ,δ, ∆ µ )]( z ) are inherited by weak limits. More precisely, let ν k be a sequence of � C -good measures on B (0 , 1) and AD -regular there with lower regularity constant � c . Assume that ν is another measure on B (0 , 1) and ν k → ν weakly in B (0 , 1). Lemma If for some A ′ > A and 0 < α ′ < α , the measure ν is geometrically ( H , A ′ , α ′ ) -flat on the scale ℓ > 0 at some point z ∈ B (0 , 1 − ( A ′ + α ) ℓ ) , then for all sufficiently large k, the measure ν k is geometrically ( H , A , α ) -flat at z on the scale ℓ . If for some 0 < δ < ∆ < 1 2 and some z ∈ B (0 , 1 − 2∆) with dist ( z , supp ν ) < δ 4 , we have | [ R ( ψ z ,δ, ∆ ν )]( z ) | > β , then for all sufficiently large k, we also have dist ( z , supp ν k ) < δ 4 and | [ R ( ψ z ,δ, ∆ ν k )]( z ) | > β . Alexander Volberg Solving a problem of David and Semmes

frame 32. So suppose that with fixed 5 constants as above and with smaller and smaller ρ k we still have µ k ’s with the absence of geometric flatness and at the same time with the boundedness of R [( ψ z ,δ, ∆ ν )]( z ), 0 < ρ k < δ < ∆ < 1 / 2, for all z ∈ B (0 , 1 − 2∆), dist ( z , supp µ k ) < δ 4 by the same β . Then we can come to a weak limit, and get that this limit µ negates the following Alternative . Alternative If ν is any good measure on B (0 , 1) that is AD regular there, then either for every A , α > 0 there exist a scale ℓ > 0 , a point z ∈ B (0 , 1 − ( A + α ) ℓ ) and a linear hyperplane H such that ν is geometrically ( H , A , α ) -flat at z on the scale ℓ , or | [ R ( ψ z ,δ, ∆ ν )]( z ) | = + ∞ . sup 0 <δ< ∆ < 1 2 z ∈ B (0 , 1 − 2∆) , dist ( z , supp ν ) < δ 4 We are left to prove the Alternative. Alexander Volberg Solving a problem of David and Semmes

frame 33. Sketching the proof of the Alternative The negation of every of the two condition of the Alternative is inherited by all tangent measures of ν . Since ν is finite and AD regular in B (0 , 1), its support is nowhere dense in B (0 , 1). Take any point z ′ ∈ B (0 , 1 2 ) \ supp ν . Let z be a closest point to z ′ in supp ν . Note that since 0 ∈ supp ν , we have | z − z ′ | ≤ | z ′ | , so | z | ≤ 2 | z ′ | < 1. Also, the ball B = B ( z ′ , | z − z ′ | ) doesn’t contain any point of supp ν . Let n be the outer unit normal to ∂ B at z . Consider the blow-ups ν z ,λ of ν at z . As λ → 0, the supports of ν z ,λ lie in a smaller and smaller neighborhood of the half-space S = { x ∈ R d +1 : � x , n � ≥ 0 } bounded by the linear hyperplane H = { x ∈ R d +1 : � x , n � = 0 } . So, every tangent measure of ν at z must have its support in half-space S . Thus, starting with any measure ν that gives a counterexample to the alternative we are trying to prove, we can modify it so that it is supported on a half-space. But this is impossible: either support is then on the boundary of S (then geometric flatness “almost” follows) or if � � x , n � otherwise, then the integral | x | d +1 d ν ( x ) blows up. B (0 , ∆) Alexander Volberg Solving a problem of David and Semmes

frame 34. The flattening lemma Major step in the argument: from geometric flatness and the absence of large oscillation of R H µ on supp µ near some fixed point z on scales ≍ ℓ to the flatness of µ at z on scale ℓ . Lemma C. ∃ A ′ , α ′ > 0 depending on c , � Fix four positive parameters A , α, � c , � C and d such that: if H is a linear hyperplane in R d +1 , A , α, � z ∈ R d +1 , L is the affine hyperplane containing z and parallel to C-good finite measure in R d +1 that is AD H, ℓ > 0 , and µ is a � regular in B ( z , 5 A ′ ℓ ) with the lower regularity constant � c. Assume that µ is geometrically ( H , 5 A ′ , α ′ ) -flat at z on the scale ℓ and, in addition, for every (vector-valued) Lipschitz function g with � supp g ⊂ B ( z , 5 A ′ ℓ ) , � g � Lip ≤ ℓ − 1 , and g d µ = 0 , one has µ 1 , g � µ | ≤ α ′ ℓ d . |� R H Then µ is ( H , A , α ) -flat at z on the scale ℓ . Alexander Volberg Solving a problem of David and Semmes

frame 35. Discussion The first step in proving the rectifiability of a measure is showing that its support is almost planar on many scales in the sense of the geometric ( H , 5 A ′ , α ′ )-flatness in the assumptions of the Flattening Lemma. This step is not that hard and we will carry it next. The second assumption involving the Riesz transform means, roughly speaking, that R H µ 1 is almost constant on supp µ ∩ B ( z , A ′ ℓ ) in the sense that its “wavelet coefficients” near z on the scale ℓ are small. There is no canonical smooth wavelet system in L 2 ( µ ) when µ is an arbitrary measure but mean zero Lipschitz functions serve as a reasonable substitute. The boundedness of R H µ in L 2 ( µ ) implies that R H µ 1 ∈ L 2 ( µ ) (because for finite measures µ , we have 1 ∈ L 2 ( µ )), so an appropriate version of the Bessel inequality can be used to show that large wavelet coefficients have to be rare and the balls satisfying the second assumption should also be viewed as typical. Alexander Volberg Solving a problem of David and Semmes

frame 36. Fix A ′ > 1 , α ′ ∈ (0 , 1) , β > 0 to be chosen later. We want to show first that if N > N 0 ( A ′ , α ′ , β ), then there exists a Carleson family F 1 ⊂ D and a finite set H of linear hyperplanes such that every cell P ∈ D \ F 1 contains a geometrically ( H , 5 A ′ , α ′ ) -flat cell Q ⊂ P at most N levels down from P for some linear hyperplane H ∈ H that may depend on P . Let R = 1 16 ℓ ( P ). According to Geometric Flattening Lemma, we can choose ρ > 0 so that either 1) there is a scale ℓ > ρ R and a point z ∈ B ( z P , R − 16[(5 A ′ + 5) + α ′ 3 ] ℓ ) ⊂ P such that µ is geometrically ( H ′ , 16(5 A ′ + 5) , α ′ 3 )-flat at z on the scale ℓ for some linear hyperplane H ′ , 2) or there exist ∆ ∈ (0 , 1 2 ), δ ∈ ( ρ, ∆) and a point z ∈ B ( z P , (1 − 2∆) R ) with dist ( z , supp µ ) < δ 4 R such that | [ R ( ψ z ,δ R , ∆ R µ )]( z ) | > β where ψ z ,δ R , ∆ R is the function introduced on frame 29. Alexander Volberg Solving a problem of David and Semmes

frame 37. In the first case, take any point z ′ ∈ supp µ such that | z − z ′ | < α ′ 3 ℓ and choose the cell Q with ℓ ( Q ) ∈ [ ℓ, 16 ℓ ) that contains z ′ . Since z ′ ⊂ B ( z P , R ) ⊂ P and ℓ ( Q ) < ℓ ( P ), we must have Q ⊂ P . Also, since | z Q − z ′ | ≤ 4 ℓ ( Q ), we have | z − z Q | < 4 ℓ ( Q ) + α ′ 3 ℓ < 5 ℓ ( Q ). Note now that, if µ is geometrically ( H , 16 A , α )-flat at z on the scale ℓ , then it is geometrically ( H , A , α )-flat at z on every scale ℓ ′ ∈ [ ℓ, 16 ℓ ). Note also that the geometric flatness is a reasonably stable condition with respect to shifts of the point and rotations of the plane. Applying these observations with ℓ ′ = ℓ ( Q ), z ′ = z Q , ε = α ′ 3 A , and choosing any finite ε -net Y on the unit sphere, we see that µ is geometrically ( H , 5 A ′ , α ′ )-flat at z Q on the scale ℓ ( Q ) with some H whose unit normal belongs to Y . Note also that the number of ℓ ( Q ) ≤ log 16 ρ − 1 + C . ℓ ( P ) levels between P and Q in this case is log 16 Alexander Volberg Solving a problem of David and Semmes

Explanation of shifting and rotating More precisely, if µ is geometrically ( H ′ , A + 5 , α )-flat at z on the scale ℓ , then it is geometrically ( H , A , 2 α + A ε )-flat at z ′ on the scale ℓ for every z ′ ∈ B ( z , 5 ℓ ) ∩ supp µ and every linear hyperplane H with unit normal vector n such that the angle between n and the unit normal vector n ′ to H ′ is less than ε . To see it, it is important to observe first that, despite the distance from z to z ′ may be quite large, the distance from z ′ to the affine hyperplane L ′ containing z and parallel to H ′ can be only αℓ , so we do not need to shift L ′ by more than this amount to make it pass through z ′ . Combined with the inclusion B ( z ′ , A ℓ ) ⊂ B ( z , ( A + 5) ℓ ), this allows us to conclude that µ is ( H ′ , A , 2 α )-flat at z ′ on the scale ℓ . After this shift, we can rotate the plane L ′ around the ( d − 1)-dimensional affine plane containing z ′ and orthogonal to both n and n ′ by an angle less than ε to make it parallel to H . Again, no point of L ′ ∩ B ( z , A ℓ ) will move by more than A εℓ and the desired conclusion follows. Alexander Volberg Solving a problem of David and Semmes

frame 38. In the second case of frame 36, there is a point z ∈ B ( z P , (1 − 2∆) R ) and a point z ′ ∈ supp µ , such that | [ R ( ψ z ,δ R , ∆ R µ )]( z ) | > β , | z − z ′ | < δ 4 R , where � | x − z | � � | x − z | � ψ z , r , R ( x ) = ψ 0 − ψ 0 . R r Let now Q and Q ′ be the largest cells containing z ′ under the restrictions that ℓ ( Q ) < ∆ 32 R and ℓ ( Q ′ ) < δ 32 R . Since both bounds are less than ℓ ( P ) and the first one is greater than the second one, we have Q ′ ⊂ Q ⊂ P . Now we want to show that the difference of averages of R µ χ E over Q and Q ′ is at least β − C in absolute value for every set E ⊃ B ( z , 2 R ) and, thereby, for every set E ⊃ B ( z P , 5 ℓ ( P )). Here C depends only on the norm of operator R µ in L 2 ( µ ). Alexander Volberg Solving a problem of David and Semmes

frame 39. Estimate of � R µ χ E � Q − � R µ χ E � Q ′ We can write χ E = ψ z ,δ R , ∆ R + f 1 + f 2 where | f 1 | , | f 2 | ≤ 1 and supp f 1 ⊂ ¯ B ( z , 2 δ R ), supp f 2 ∩ B ( z , ∆ R ) = ∅ . � � | R µ f 1 | 2 d µ ≤ C | f 1 | 2 d µ ≤ C ( δ R ) d ≤ C ℓ ( Q ′ ) d ≤ C µ ( Q ′ ) , Hence |� R µ f 1 � Q | , |� R µ f 1 � Q ′ | are bounded by some C . Note also that Q ⊂ B ( z ′ , 8 ℓ ( Q )) ⊂ B ( z ′ , ∆ 4 R ) ⊂ B ( z , ∆ 2 R ), so the distance from Q to supp f 2 is at least ∆ 2 R > ℓ ( Q ). Thus, � R µ f 2 � Lip ( Q ) ≤ C ℓ ( Q ) − 1 the difference of the averages of R µ f 2 over Q and Q ′ is bounded by some C . Alexander Volberg Solving a problem of David and Semmes

frame 40. We are left to to estimate the difference of averages � R µ ψ z ,δ R , ∆ R � Q − � R µ ψ z ,δ R , ∆ R � Q ′ . For this L 2 ( µ ) ≤ C (∆ R ) d ≤ C ℓ ( Q ) d ≤ C µ ( Q ) , � R µ ψ z ,δ R , ∆ R � 2 L 2 ( µ ) ≤ C � ψ z ,δ R , ∆ R � 2 so the average over Q is bounded by a constant. On the other hand, Q ′ ⊂ B ( z ′ , 8 ℓ ( Q ′ )) ⊂ B ( z ′ , δ 4 R ) ⊂ B ( z , δ 2 R ) . Again dist ( supp ψ z ,δ R , ∆ R , Q ′ ) ≥ δ 2 R . Therefore, 2 R )) ≤ C ( δ R ) − 1 . � R ( ψ z ,δ R , ∆ R µ ) � Lip ( B ( z , δ But this means that |� R µ ψ z ,δ R , ∆ R � Q ′ − R µ ψ z ,δ R , ∆ R ( z ) | ≤ C ( δ ). The quantity | R µ ψ z ,δ R , ∆ R ( z ) | is large! It is bigger than β . Alexander Volberg Solving a problem of David and Semmes

frame 41. We finally get |� R µ χ E � Q − � R µ χ E � Q ′ | ≥ β − 3 C . This conclusion can be rewritten as µ ( P ) − 1 d 2 |� R µ χ E , ψ P � µ | ≥ c ρ 2 ( β − C ) where � � 1 1 d µ ( Q ) χ Q − ψ P = [ ρℓ ( P )] µ ( Q ′ ) χ Q ′ 2 This is for any E , B ( z P , 5 ℓ ( P )) ⊂ E . Let us show that the set of cells satisfying the latter property can be only rare , that is Carleson family . Alexander Volberg Solving a problem of David and Semmes

frame 42. Carlesonness of cells P satisfying the second case of frame 36. Fix any cell P 0 and consider the cells P satisfying the second case of frame 36 (there exists a point z such that | R µ ψ z ,δ R , ∆ R ( z ) | > β with certain position of z inside P , R ≈ ℓ ( P ), and large β ). Then B ( z P , 5 ℓ ( P )) ⊂ B ( z P 0 , 50 ℓ ( P 0 )). Also we saw on the previous frame 41 that µ ( P ) ≤ C ( ρ, β ) |� R µ χ B ( z P 0 , 50 ℓ ( P 0 )) , ψ P � µ | 2 . Here ψ P form Haar system of depth N ≈ log ℓ ( P ) ℓ ( Q ′ ) , ℓ ( Q ′ ) ≈ δℓ ( P ) , δ ∈ ( ρ, 1 / 2), so N ≤ c log 1 ρ . Any Haar system of depth N is a Riesz system . By the property of Riesz system (see frames 44–46 below) we get that � � |� R µ χ B ( z P 0 , 50 ℓ ( P 0 )) , ψ P � µ | 2 ≤ µ ( P ) ≤ C ( ρ, β ) P ⊂ P 0 P ⊂ P 0 C � R µ χ B ( z P 0 , 50 ℓ ( P 0 )) � 2 µ . The latter is smaller than C µ ( B ( z P 0 , 50 ℓ ( P 0 ))) ≤ C ′ µ ( P 0 ), and we established Carleson property of P ’s as above. Alexander Volberg Solving a problem of David and Semmes

frame 43. Abundance of geometrically flat cells is already obtained. Fix A , α > 0. We shall say that a cell Q ∈ D is (geometrically) ( H , A , α )-flat if the measure µ is (geometrically) ( H , A , α )-flat at z Q on the scale ℓ ( Q ). We have just shown (modulo estimates of Riesz system that follows) that there exists an integer N , a finite set H of linear hyperplanes in R d +1 , and a Carleson family F ⊂ D (depending on A , α ) such that for every cell P ∈ D \ F , there exist H ∈ H and a geometrically ( H , A , α ) -flat cell Q ⊂ P that is at most N levels down from P . Alexander Volberg Solving a problem of David and Semmes

frame 44. Riesz systems: Haar system, Lipschitz wavelet system. Let ψ Q ( Q ∈ D ) be a system of Borel L 2 ( µ ) functions Definition The functions ψ Q form a Riesz family with Riesz constant C > 0 if � � 2 � � � � � � � � a 2 a Q ψ Q ≤ C � � Q � � Q ∈D L 2 ( µ ) for any real coefficients a Q . Note that if the functions ψ Q form a Riesz family with Riesz constant C , then for every f ∈ L 2 ( µ ), we have � |� f , ψ Q � µ | 2 ≤ C � f � 2 L 2 ( µ ) . Q ∈D Alexander Volberg Solving a problem of David and Semmes

frame 45. Assume next that for each cell Q ∈ D we have a set Ψ Q of L 2 ( µ ) functions associated with Q . Definition The family Ψ Q (Q ∈ D ) of sets of functions is a Riesz system with Riesz constant C > 0 if for every choice of functions ψ Q ∈ Ψ Q , the functions ψ Q form a Riesz family with Riesz constant C. Riesz systems are useful because of the following Lemma. Lemma Suppose that Ψ Q is any Riesz system. Fix A > 1 . For each Q ∈ D , define µ ( Q ) − 1 / 2 |� R µ χ E , ψ � µ | . ξ ( Q ) = E : B ( zQ , A ℓ ( Q )) ⊂ E ,µ ( E ) < + ∞ sup inf ψ ∈ Ψ Q Then ∀ δ > 0 , F δ := { Q ∈ D : ξ ( Q ) ≥ δ } is Carleson. Alexander Volberg Solving a problem of David and Semmes

frame 46. The one line proof of Lemma is on frame 42. But it is important that there are two natural classes of Riesz systems: Haar systems of fixed depth Ψ H ( N ), and Lipschitz wavelet systems Ψ L ( A ). Let now N be any positive integer. For each Q ∈ D , define the set of Haar functions Ψ h Q ( N ) of depth N as the set of all functions ψ that are supported on Q , are constant on every cell Q ′ ∈ D that is � � ψ 2 d µ ≤ C . The N levels down from Q , and satisfy ψ d µ = 0, Riesz property follows immediately from the fact that D can be represented as a finite union of the sets D ( j ) = ∪ k : k ≡ j mod N D k ( j = 0 , . . . , N − 1) and that for every choice of ψ Q ∈ Ψ h Q ( N ), the functions ψ Q corresponding to the cells Q from a fixed D ( j ) form a bounded orthogonal family. � � d 1 1 µ ( Q ) χ Q − Our ψ P = [ ρℓ ( P )] µ ( Q ′ ) χ Q ′ from frame 41 are 2 obviously from Ψ H ( N ) with N ≤ c log 1 ρ , so the main inequality of frame 42 is done, and YES the abundance of geometrically flat cells is completely established. Alexander Volberg Solving a problem of David and Semmes

frame 47. Lipschitz wavelet systems. We will need them to establish the abundance of ( H , A , α ) flat (not just geometrically flat) cells. In the Lipschitz wavelet system, the set Ψ ℓ Q ( A ) consists of all Lipschitz functions ψ supported on B ( z Q , A ℓ ( Q )) such that � ψ d µ = 0 and � ψ � Lip ≤ C ℓ ( Q ) − d 2 − 1 . Since µ is nice, we � | ψ | 2 d µ ≤ C ( A ) ℓ ( Q ) − d µ ( Q ) ≤ C ( A ) in this automatically have case. If Q , Q ′ ∈ D and ℓ ( Q ′ ) ≤ ℓ ( Q ), then, for any two functions ψ Q ∈ Ψ ℓ Q ( A ) and ψ Q ′ ∈ Ψ ℓ Q ′ ( A ), we can have � ψ Q , ψ Q ′ � µ � = 0 only if B ( z Q , A ℓ ( Q )) ∩ B ( z Q ′ , A ℓ ( Q ′ )) � = ∅ , in which case, � ℓ ( Q ′ ) � d 2 +1 |� ψ Q , ψ Q ′ � µ | ≤ � ψ Q � Lipdiam ( Q ′ ) � ψ Q ′ � L 1 ( µ ) ≤ C ( A ) . ℓ ( Q ) THEN: Alexander Volberg Solving a problem of David and Semmes

frame 48. � � 2 � � � � � � � � a Q ψ Q ≤ 2 | a Q |·| a Q ′ |·|� ψ Q , ψ Q ′ � µ | � � � � Q ∈D Q , Q ′ ∈D , ℓ ( Q ′ ) ≤ ℓ ( Q ) L 2 ( µ ) � ℓ ( Q ′ ) � d 2 +1 � ≤ C ( A ) | a Q | · | a Q ′ | ℓ ( Q ) Q , Q ′ ∈D , ℓ ( Q ′ ) ≤ ℓ ( Q ) B ( zQ , A ℓ ( Q )) ∩ B ( zQ ′ , A ℓ ( Q ′ )) � = ∅ �� ℓ ( Q ′ ) � � d +1 � | a Q | 2 + ℓ ( Q ′ ) ℓ ( Q ) | a Q ′ | 2 ≤ C ( A ) . ℓ ( Q ) Q , Q ′ ∈D , ℓ ( Q ′ ) ≤ ℓ ( Q ) B ( zQ , A ℓ ( Q )) ∩ B ( zQ ′ , A ℓ ( Q ′ )) � = ∅ � ℓ ( Q ′ ) � d +1 � � ≤ C , ≤ C . ℓ ( Q ) Q ′ ∈D : ℓ ( Q ′ ) ≤ ℓ ( Q ) Q ∈D : ℓ ( Q ′ ) ≤ ℓ ( Q ) B ( zQ , A ℓ ( Q )) ∩ B ( zQ ′ , A ℓ ( Q ′ )) � = ∅ B ( zQ , A ℓ ( Q )) ∩ B ( zQ ′ , A ℓ ( Q ′ )) � = ∅ are bounded by some constants independent of Q and Q ′ Alexander Volberg Solving a problem of David and Semmes respectively.

frame 49. We recall Flattening Lemma from frame 34, that says how to get flat cell if it is already geometrically flat. Lemma C. ∃ A ′ , α ′ > 0 depending on c , � Fix four positive parameters A , α, � c , � C and d such that: if H is a linear hyperplane in R d +1 , A , α, � z ∈ R d +1 , L is the affine hyperplane containing z and parallel to C-good finite measure in R d +1 that is AD H, ℓ > 0 , and µ is a � regular in B ( z , 5 A ′ ℓ ) with the lower regularity constant � c. Assume that µ is geometrically ( H , 5 A ′ , α ′ ) -flat at z on the scale ℓ and, in addition, for every (vector-valued) Lipschitz function g with � supp g ⊂ B ( z , 5 A ′ ℓ ) , � g � Lip ≤ ℓ − 1 , and g d µ = 0 , one has µ 1 , g � µ | ≤ α ′ ℓ d . |� R H Then µ is ( H , A , α ) -flat at z on the scale ℓ . Alexander Volberg Solving a problem of David and Semmes

frame 50. We already saw that all cells P (except for a rare (Carleson) family F 1 ) are such that not more than N generation down inside P a cell Q lies, which is ( H , A ′ , α ′ )-geometrically flat, where A ′ , α ′ depend on A , α as Flattening Lemma requires, and H = H Q belongs to H , a finite family (cardinality of it depends on A , α too). Given P , we find such Q , and Flattening Lemma applied to any µ := µ · 1 E , E ⊃ B ( z Q , 100 A ′ ℓ ( Q )), shows that either Q is ( H , A , α )-flat, or for each such E there exists a function g = g E � such that it is supported on B ( z Q , 5 A ′ ℓ ( Q )), g d µ = 0, Lipschitz with norm at most 1 /ℓ ( Q ) ≤ C ( N ) /ℓ ( P ) and µ 1 E , g � ≥ α ′ ℓ ( Q ) d = c ( N ) α ′ ℓ ( P ) d . � R H d 2 . They form a Lipschitz wavelet Consider ψ P = ψ P , E = g /ℓ ( P ) system Ψ L ( C ), as on frame 47. Therefore, ξ ( P ) = µ ( P ) − 1 |� R µ 1 E , ψ �| ≥ C ( N ) α ′ . inf sup 2 E : E ⊃ B ( z p , C ℓ ( P ) ψ ∈ Ψ L ( C ) Alexander Volberg Solving a problem of David and Semmes

frame 51. We know that such P can form only a rare (Carleson) family if R µ is a bounded operator in L 2 ( µ ). Call it F 2 . So by the exception of two rare families F 1 , F 2 , any other cell P ∈ D will have inside it and not more than N (fixed number depending on A , α ) generations down a sub-cell Q , which is ( H , A , α )-flat. Here H will be chosen from a finite family H of hyperplanes (having fixed cardinality depending on A , α ). The abundance of flat cells is completely proved. Alexander Volberg Solving a problem of David and Semmes

frame 52. A bit of combinatorics. Recall that our goal is to prove that the family of all non-BAUP cells P ∈ D is Carleson. In view of the just proved abundance of flat cells in several fixed directions , it will suffice to show that we can choose A , α > 0 such that for every fixed linear hyperplane H and for every integer N , the corresponding family F = F ( A , α, H , N ) of all non-BAUP cells P ∈ D containing an ( H , A , α )-flat cell Q at most N levels down from P is Carleson. The idea . Suppose this is not the case. Then there will be P from F =family of non-BAUP cells containing a flat cell in a fixed direction at most N generations down such that it can be tiled (up to tiny measure) by arbitrarily large number of layers of non-BAUP cells. Use now the abundance of flat cells. We can also tile the same cell P by layers of ( H , A , α )-flat cells Q (up to tiny measure) also with as many layers as we wish. Moreover we can alternate layers. Namely: Alexander Volberg Solving a problem of David and Semmes

frame 53. Alternating non-BAUP and flat layers. Lemma If F is not Carleson, then for every positive integer K and every η > 0 , there exist a cell P ∈ F and K + 1 alternating pairs of finite layers P k , Q k ⊂ D (k = 0 , . . . , K) such that P 0 = { P } . P k ⊂ F P for all k = 0 , . . . , K. All layers Q k consist of ( H , A , α ) -flat cells only. Each individual layer (either P k , or Q k ) consists of pairwise disjoint cells. If Q ∈ Q k , then there exists P ′ ∈ P k such that Q ⊂ P ′ (k = 0 , . . . , K). If P ′ ∈ P k +1 , then there exists Q ∈ Q k such that P ′ ⊂ Q (k = 0 , . . . , K − 1 ). � Q ∈ Q K µ ( Q ) ≥ (1 − η ) µ ( P ) . Alexander Volberg Solving a problem of David and Semmes

frame 54. Sketch of the proof. Suppose F is not Carleson. For every η ′ > 0 and every positive integer M , we can find a cell P ∈ F and M + 1 layers L 0 , . . . , L M ⊂ F P that have the desired Cantor-type hierarchy and satisfy � P ′ ∈L M µ ( P ′ ) ≥ (1 − η ′ ) µ ( P ). We will go now from the layer L m to L m + SN , where S = S ( N ) will be large and M ≈ KSN , where K is from the previous frame. We take P ′ ∈ L m and choose Q ( P ′ ) less than N generations down, which is flat. Those P ′′ ∈ L M + N that are inside such Q ( P ′ ) we color white, the collection of Q ( P ′ ) we color blue. Notice that at this moment the mass of all non-colored P ′′ ∈ L m + N is ≤ (1 − c 4 − 4 dN ) µ ( P ). In those P ′′ ∈ L m + N that are not colored again we will have Q ( P ′′ ) less than N generations down that are flat, color them blue, color white those P ′′′ ∈ L m +2 N that are in some of Q ( P ′′ ). Notice that at this moment the mass of all non-colored P ′′′ ∈ L m +2 N is ≤ (1 − c 4 − 4 dN ) 2 µ ( P ). Alexander Volberg Solving a problem of David and Semmes

frame 55. Non-colored follow non-colored, and in S steps (if S = S ( N ) is sufficiently large) the portion of µ ( P ) of non-colored cells become very small. Then we stop and put m new := m + SN , we consider only the part of layer L m + SN , namely those cells of it that lie in some white colored cells. Call it L ′ m new . So if L m was P k , then L ′ m new will be our P k +1 . Consider all blue cells we have on the road. Take the family of maximal blue cells out of those which we just constructed. This will be layer of disjoint ( H , A , α )-flat cells and this will be our layer Q k . Given K , we choose S very large, η ′ very small. Then we make the error in tiling µ ( P ) only of the order ( K + 1)[ η ′ + (1 − c 4 − 4 dN ) S ], which is as small as we wish. Lemma of frame 53 is proved. Alexander Volberg Solving a problem of David and Semmes

frame 56. Almost orthogonality. Flat layers. Fix K . Choose ε > 0, A , α > 0, η > 0 in this order. Construct layers as on frame 53. Consider flat layers Q k ignoring the non-BAUP layers P k almost entirely. For a cell Q ∈ D and t > 0, define Q t = { x ∈ Q : dist ( x , R d +1 \ Q ) ≥ t ℓ ( Q ) } . Note that µ ( Q \ Q t ) ≤ Ct γ µ ( Q ) for some fixed γ > 0. This is stated on frame 15. Let ϕ 0 be any C ∞ function supported on � B (0 , 1) and such that ϕ 0 dm = 1 where m is the Lebesgue measure in R d +1 . Put � � · 1 ϕ Q = χ Q 2 ε ∗ ( εℓ ( Q )) d ϕ 0 . εℓ ( Q ) Then ϕ Q = 1 on Q 3 ε and supp ϕ Q ⊂ Q ε . In particular, the diameter of supp ϕ Q is at most 8 ℓ ( Q ). Alexander Volberg Solving a problem of David and Semmes

frame 57. In addition, C C �∇ 2 ϕ Q � L ∞ ≤ � ϕ Q � L ∞ ≤ 1 , �∇ ϕ Q � L ∞ ≤ εℓ ( Q ) , ε 2 ℓ ( Q ) 2 . From now on, we will be interested only in the cells Q from the flat layers Q k . With each such cell Q we will associate the corresponding approximating plane L ( Q ) containing z Q and parallel to H and the approximating measure ν Q = a Q ϕ Q m L ( Q ) where a Q is chosen so that � ν Q ( R d +1 ) = ϕ Q d µ . � � Both integrals ϕ Q dm L ( Q ) and ϕ Q d µ are comparable to ℓ ( Q ) d , provided that ε < 1 48 , say. In particular, in this case, the normalizing factors a Q are bounded by some constant. Alexander Volberg Solving a problem of David and Semmes

frame 58. Define � ϕ Q R H [ ϕ Q µ − ν Q ] , G k = k = 0 , . . . , K . Q ∈ Q k Now put F k = G k − G k +1 when k = 0 , . . . , K − 1 , F K = G K . Note that K � F m = G k . m = k Alexander Volberg Solving a problem of David and Semmes

frame 59. “Orthogonality” of telescopic layers. This is almost orthogonality of “errors” (errors between genuine and flat situations). Lemma Assuming that ε < 1 48 , A > 5 , and α < ε 8 , we have |� F k , G k +1 �| ≤ σ ( ε, α ) µ ( P ) for all k = 0 , . . . , K − 1 , where σ ( ε, α ) is some positive function such that ε → 0+ [ lim lim α → 0+ σ ( ε, α )] = 0 . Long and difficult. Frames 25, 26 are constantly used. And the boundedness of R µ in L 2 ( µ ) is used. Alexander Volberg Solving a problem of David and Semmes

frame 59A. A flavor of the reasoning of almost orthogonality. This is a typical block in the proof of almost orthogonality. It is clear that it uses frame 26 and smallness of α in flatness. Lemma Suppose that Q ∈ Q k . Then � |� R H ( χ Q \ Q ′ µ ) , ϕ Q ′ R H ( ϕ Q ′ µ − ν Q ′ ) � µ | Q ′ ∈ Q k +1 , Q ′ ⊂ Q 1 d +2 ε − 3 µ ( Q ) . ≤ C α Alexander Volberg Solving a problem of David and Semmes

frame 60. Now notice that G 0 = � Q ∈ Q 0 ϕ Q R H [ ϕ Q µ − ν Q ] . We want to see first that � G 0 � 2 µ ≤ C µ ( P ) . As the summands have pairwise disjoint supports, it will suffice to prove the inequality � ϕ Q R H ( ϕ Q µ − ν Q ) � 2 L 2 ( µ ) ≤ C µ ( Q ) for each individual Q ∈ Q 0 and then observe that � Q ∈ Q k µ ( Q ) ≤ µ ( P ). Of course � ϕ Q R H ( ϕ Q µ � 2 L 2 ( µ ) ≤ C µ ( Q ) by the boundedness of R µ . But the estimate � ϕ Q R H ( ν Q ) � 2 L 2 ( µ ) ≤ C µ ( Q ) is not so trivial because we start with flat measure ν Q := a Q ϕ Q m L but we send it by R H into L 2 ( µ ). Such an estimate can be obtained however by using an error estimate of frame 26. Alexander Volberg Solving a problem of David and Semmes

frame 61. At this point, we need to know that the non-BAUPness condition depends on a positive parameter δ . We will fix that δ from now on in addition to fixing the measure µ . Note that despite the fact that we need to prove that the family of non-BAUP cells is Carleson for every δ > 0, the David-Semmes uniform rectifiability criterion does not require any particular rate of growth of the corresponding Carleson constant as a function of δ . Alexander Volberg Solving a problem of David and Semmes

frame 62. We have the identity � � 2 � � K − 1 K K � � � � � � G 0 � 2 � F k � 2 L 2 ( µ ) = F k = L 2 ( µ ) + 2 � F k , G k +1 � µ . � � � � k =0 k =0 k =0 L 2 ( µ ) As we have seen on a couple of previous frames, � G 0 � 2 L 2 ( µ ) ≤ C µ ( P ), and the scalar products can be made arbitrarily small by first choosing ε > 0 small enough and then taking a sufficiently small α > 0 depending on ε . So we will get a contradiction if we are able to bound � F k � 2 L 2 ( µ ) for k = 0 , . . . , K − 1 from below by τ 2 µ ( P ), with some τ = τ ( δ ) > 0 (as usual, the dependence on the dimension d and the regularity constants of µ is suppressed). We choose very large K , then we choose A > A 0 ( δ ) , ε < ε 0 ( δ ) , η < η 0 ( ε ) , α < α 0 ( ε, δ ).. Alexander Volberg Solving a problem of David and Semmes

frame 63. Densely packed cells. Fix k ∈ { 0 , 1 , . . . , K − 1 } . We can write the function F k as � F Q F k = Q ∈ Q k where � F Q = ϕ Q R H ( ϕ Q µ − ν Q ) − ϕ Q ′ R H ( ϕ Q ′ µ − ν Q ′ ) . Q ′ ∈ Q k +1 , Q ′ ⊂ Q We shall call a cell Q ∈ Q k densely packed if � Q ′ ∈ Q k +1 , Q ′ ⊂ Q µ ( Q ′ ) ≥ (1 − ε ) µ ( Q ). Otherwise we shall call the cell Q loosely packed. The main claim of this section is that the loosely packed cells constitute a tiny minority of all cells in Q k if η ≤ ε 2 . Alexander Volberg Solving a problem of David and Semmes

frame 64. Indeed, we have     � µ ( Q ) ≤ ε − 1 � �  Q \  Q ′   µ Q ∈ Q k Q ∈ Q k Q ′ ∈ Q k +1 , Q ′ ⊂ Q Q is packed loosely    � � = ε − 1 µ ( Q ′ )  µ ( Q ) − Q ∈ Q k Q ′ ∈ Q k +1   �  ≤ η ≤ ε − 1  µ ( P ) − µ ( Q ′ ) εµ ( P ) ≤ εµ ( P ) . Q ′ ∈ Q k +1 Alexander Volberg Solving a problem of David and Semmes

frame 65. We can immediately conclude from here that � � � µ ( Q ) = µ ( Q ) − µ ( Q ) Q ∈ Q k Q ∈ Q k Q ∈ Q k Q is densely packed Q is loosely packed ≥ (1 − η ) µ ( P ) − εµ ( P ) ≥ (1 − 2 ε ) µ ( P ) . From now on, we will fix the choice η = ε 2 . Alexander Volberg Solving a problem of David and Semmes

frame 66. We claim now that to estimate � F k � 2 L 2 ( µ ) from below by τ 2 µ ( P ), it suffices to show that for every densely packed cell Q ∈ Q k , we have � F Q � 2 L 2 ( µ ) ≥ 2 τ 2 µ ( Q ). To see it, just write � � � F k � 2 � F Q � 2 � F Q � 2 L 2 ( µ ) = L 2 ( µ ) ≥ L 2 ( µ ) Q ∈ Q k Q ∈ Q k Q is densely packed � 2 τ 2 µ ( Q ) ≥ 2(1 − 2 ε ) τ 2 µ ( P ) ≥ τ 2 µ ( P ) , ≥ Q ∈ Q k Q is densely packed provided that ε < 1 4 . Alexander Volberg Solving a problem of David and Semmes

frame 67. Modification of measure process. From now on, we will fix k ∈ { 0 , . . . , K − 1 } and a densely packed cell Q ∈ Q k . We denote by Q the set of all cells Q ′ ∈ Q k +1 that are contained in the cell Q . Lemma The goal of this section is to show that there exists a subset Q ′ of Q such that � Q ′ ∈ Q ′ µ ( Q ′ ) ≥ (1 − C ε ) µ ( Q ) and � � F Q � L 2 ( µ ) ≥ 1 2 � R H ( ν − ν Q ) � L 2 ( ν ) − σ ( ε, α ) µ ( Q ) , where ν = � Q ′ ∈ Q ′ ν Q ′ and σ ( ε, α ) is some positive function such that lim ε → 0+ [lim α → 0+ σ ( ε, α )] = 0 . Alexander Volberg Solving a problem of David and Semmes

frame 68. The proof is long and technical, but looking at � F Q = ϕ Q R H ( ϕ Q µ − ν Q ) − ϕ Q ′ R H ( ϕ Q ′ µ − ν Q ′ ) Q ′ ∈ Q k +1 , Q ′ ⊂ Q we see that the claim is at least natural, as it says that µ “cancels” out inside F Q , and we also replace µ outside in � F Q � L 2 ( µ ) by the measure ν consisting of flat pieces parallel and close to flat ν Q . The statement of the lemma holds with γ 1 2 ε − 2 d +3 + αε − d − 3 ] . 4 + α σ ( ε, α ) = C [ ε 2 Alexander Volberg Solving a problem of David and Semmes

frame 69. Next measure modification. Reflection trick. Fix a hyperplane L parallel to H at the distance 2∆ ℓ ( Q ) from supp µ ∩ Q . Number ∆ is small compared to ε and large compared to α . Let S be the (closed) half-space bounded by L that contains supp µ ∩ Q . For x ∈ S , denote by x ∗ the reflection of x about L . Define the kernels R H ( x , y ) = R H ( x − y ) − R H ( x ∗ − y ) , � x , y ∈ S R H the corresponding operator. We will assume and denote by � that α << ∆, so the approximating hyperplanes L ( Q ′ ) ( Q ′ ∈ Q ′ ) and L ( Q ), which lie within the distance αℓ ( Q ) from supp µ ∩ Q are contained in S and lie at the distance ∆ ℓ ( Q ) or greater from the boundary hyperplane L . Alexander Volberg Solving a problem of David and Semmes

frame 70. Lemma The goal of this section is to show that, for some appropriately chosen ∆ = ∆( α, ε ) > 0 , and under our usual assumptions about ε , A, and α , we have � � R H ( ν − ν Q ) � L 2 ( ν ) ≥ � � R H ν � L 2 ( ν ) − σ ( ε, α ) µ ( Q ) where, again, σ ( ε, α ) is some positive function such that ε → 0+ [ lim lim α → 0+ σ ( ε, α )] = 0 . Let T be operator with kernel R H ( x ∗ − y ), � R H with R H ( x − y ) − R H ( x ∗ − y ). To compare � R H ( ν − ν Q ) � with � � R H ν � one needs to estimate � R H ν Q − T ν � , so one needs two smallnesses: 1) � R H ν Q − T ν Q � ; 2) � T ( ν − ν Q ) � (all norms in L 2 ( ν )). Alexander Volberg Solving a problem of David and Semmes

frame 71. Explanation of � R H ν Q − T ν Q � smallness. Elementary estimates: a) � R H ν Q � Lip ≤ C ε 2 ℓ ( Q ) (smoothness of ϕ Q is needed), therefore b ) | R H ν Q ( x ) − T ν Q ( x ) | ≤ � R H ν Q � Lip | ∆ ℓ ( Q ) | ≤ C ∆ ε 2 . � Thus, � R H ν Q − T ν Q � L 2 ( ν ) ≤ C ∆ µ ( Q ). ε 2 Alexander Volberg Solving a problem of David and Semmes

frame 72. Explanation of � T ( ν − ν Q ) � smallness. Obviously 1 C � R H ( · ∗ − y ) � L ∞ ( S ) ≤ ∆ d ℓ ( Q ) d and � R H ( · ∗ − y ) � Lip ( S ) ≤ ∆ d +1 ℓ ( Q ) d +1 . Hence, � T ( ν Q − ν ) � Lip ( S ) � R H ( · ∗ − y ) � Lip ( S ) ( � ν � + � ν Q � )) ≤ sup y ∈ ( supp ν ∪ supp ν Q ) C C ≤ ∆ d +1 ℓ ( Q ) d +1 µ ( Q ) ≤ ∆ d +1 ℓ ( Q ) . C Similarly, � T ( ν Q − ν ) � L ∞ ( S ) ≤ ∆ d . Thus, by frame 25 � � � � � � ≤ C αℓ ( Q ′ ) d +2 1 1 1 | T ( ν Q − ν ) | 2 d ( ϕ Q ′ µ − ν Q ′ ) � � � ∆ d +1 ℓ ( Q ) εℓ ( Q ′ ) ∆ d ≤ C α ∆ − 2 d − 1 ε − 1 ℓ ( Q ′ ) d ≤ C α ∆ − 2 d − 1 ε − 1 µ ( Q ′ ) . Alexander Volberg Solving a problem of David and Semmes

frame 73. Explanation of � T ( ν − ν Q ) � L 2 ( ν ) smallness. Now we want to sum up over Q ′ ∈ Q ′ , where the last set was mentioned on frame 67: for our goals now it can be considered as the whole collection of Q ′ ∈ Q k +1 lying inside our Q ∈ Q k . Let Φ := � Q ′ ∈ Q ′ ϕ Q ′ . Summing over Q ′ ∈ Q ′ , we get � � | T ( ν Q − ν ) | 2 d ν ≤ | T ( ν Q − ν ) | 2 d (Φ µ )+ C α ∆ − 2 d − 1 ε − 1 µ ( Q ) , The last integrand we write as T ( ν Q − ν ) = ( T ν Q − T ϕ Q µ ) + ( T ϕ Q µ − T Φ µ ) + ( T Φ µ − T ν ) . We estimate the smallness of each term in L 2 (Φ µ ) (or, which is practically the same, L 2 ( µ Q )) on the next 3 slides. Alexander Volberg Solving a problem of David and Semmes

frame 74. Explanation of � T ( ϕ Q µ − ν Q ) � L 2 (Φ µ ) smallness. On the other hand, applying frame 25 again, we see that for every x ∈ supp µ Q , � � � � � � � � � R H ( x ∗ − · ) d ( ϕ Q µ − ν Q ) � � � [ T ( ϕ Q µ − ν Q )]( x ) � = � � ≤ C αℓ ( Q ) d +2 � R H ( x ∗ − · ) � Lip ( S ) � ϕ Q � Lip 1 1 ≤ C αℓ ( Q ) d +2 εℓ ( Q ) ≤ C α ∆ − d − 1 ε − 1 ∆ d +1 ℓ ( Q ) d +1 as an obvious estimate (from frame 72) is used with exchanged x , y : | R H ( x ∗ − · ) � Lip ( S ) ≤ C ∆ d +1 ℓ ( Q ) d +1 Hence, � T ( ϕ Q µ − ν Q ) � L 2 (Φ µ ) ≤ C α ∆ − d − 1 ε − 1 � µ ( Q ). Alexander Volberg Solving a problem of David and Semmes

frame 75. Explanation of � T (Φ µ − ν ) � L 2 (Φ µ ) smallness. Similarly to the previous slide, for every Q ′ ∈ Q ′ , we have � � � � � � � � � R H ( x ∗ − · ) d ( ϕ Q ′ µ − ν Q ′ ) � � � [ T ( ϕ Q ′ µ − ν Q ′ )]( x ) � = � � ≤ C αℓ ( Q ′ ) d +2 � R H ( x ∗ − · ) � Lip ( S ) � ϕ Q ′ � Lip 1 1 ≤ C αℓ ( Q ′ ) d +2 ∆ d +1 ℓ ( Q ) d +1 εℓ ( Q ′ ) ≤ C α ∆ − d − 1 ε − 1 ℓ ( Q ′ ) d ℓ ( Q ) d ≤ C α ∆ − d − 1 ε − 1 µ ( Q ′ ) µ ( Q ) . Summing these inequalities over Q ′ ∈ Q ′ , we get | [ T (Φ µ − ν )]( x ) | ≤ C α ∆ − d − 1 ε − 1 ∀ x ∈ supp µ Q , therefore � T Φ µ − T ν � L 2 (Φ µ ) ≤ C α ∆ − d − 1 ε − 1 � µ ( Q ). Alexander Volberg Solving a problem of David and Semmes

frame 76. Explanation of � T (Φ µ − ϕ Q µ ) � L 2 (Φ µ ) smallness. Since the operator norm of T in L 2 ( µ Q ) is bounded by a constant, (this is because R H ( x ∗ − y ) = R H ∆ ℓ ( Q ) ( x − y ) + [ R H ( x ∗ − y ) − R H ∆ ℓ ( Q ) ( x − y )] and the last kernel has Poisson estimate in absolute value) we have 2 � γ � T (( ϕ Q − Φ) µ ) � L 2 ( µ Q ) ≤ � ϕ Q − Φ � L 2 ( µ ) ≤ C ε µ ( Q ) by the fact that the union of Q ′ ∈ Q ′ take (1 − C ε )-portion of measure µ of cell Q (see Lemma on frame 67). Thus, we finally get � R H ( ν − ν Q ) � L 2 ( ν ) � 2 + α ∆ − d − 1 ε − 1 � � γ 2 + ∆ ε − 2 + α 1 2 ∆ − 2 d +1 2 ε − 1 ≥ � � R H ν � L 2 ( ν ) − C ε µ ( Q ) . Alexander Volberg Solving a problem of David and Semmes

frame 77. Now we choose ∆ = ε 3 and α = ε C with large C . We come to the point that we need to estimate from below the Riesz Energy � � R H ν � L 2 ( ν ) , where ν := � Q ′ ∈ Q ′ , Q ′ ⊂ Q ν Q ′ . It is truly desirable to have � � R H ν � L 2 ( ν ) ≥ ... using another than ε constant. To give a δ -breath. The subset Q ′ of the set { Q ′ : Q ′ ⊂ Q , Q ′ ∈ Q k +1 } is chosen in Lemma on frame 67. In fact, it is almost the whole { Q ′ : Q ′ ⊂ Q , Q ′ ∈ Q k +1 } , the difference being the use of a certain Marcinkiewicz function to choose Q ′ . To estimate Riesz Energy we need function ψ, � R H ψ = 1 on ν and such that: see frame 88. For that we need first non-BAUP layer P k +1 tiling Q over Q k +1 and special family of cells in it. Alexander Volberg Solving a problem of David and Semmes

frame 78. A collection of P ’s (inside Q ) of non-BAUP layer P k +1 . One can construct (under our usual assumptions of ε is sufficiently small in terms of δ , A is sufficiently large in terms of δ , α is sufficiently small in terms of ε and δ ), a family P ′ ⊂ P k +1 such that Every cell P ′ ⊂ P ′ is contained in Q ε and satisfies ℓ ( P ′ ) ≤ 2 αδ − 1 ℓ ( Q ). � P ′ ∈ P ′ µ ( P ′ ) ≥ c µ ( Q ). The balls B ( z P ′ , 10 ℓ ( P ′ )), P ′ ∈ P ′ are pairwise disjoint. The function � � d +1 � ℓ ( P ′ ) h ( x ) = ℓ ( P ′ ) + dist ( x , P ′ ) P ′ ∈ P ′ satisfies � h � L ∞ ≤ C . Alexander Volberg Solving a problem of David and Semmes

79. Figure. We start with showing that every δ -non-BAUP cell P ′ contained in Q has much smaller size than Q . Indeed, we know that supp µ ∩ B ( z Q , A ℓ ( Q )) is contained in the αℓ ( Q )-neighborhood of L ( Q ) and that B ( y , αℓ ( Q )) ∩ supp µ � = ∅ for every y ∈ B ( z Q , A ℓ ( Q )) ∩ L ( Q ). Suppose that P ′ ⊂ Q is δ -non-BAUP. If A > 5, then B ( x P ′ , ℓ ( P ′ )) ⊂ B ( z Q , 5 ℓ ( Q )) ⊂ B ( z Q , A ℓ ( Q )) . Moreover, since y P ′ − x P ′ ∈ H , we have dist ( y P ′ , L ( Q )) = dist ( x P ′ , L ( Q )) ≤ αℓ ( Q ) . Let y ∗ P ′ be the projection of y P ′ to L ( Q ). Then | y ∗ P ′ − y P ′ | ≤ αℓ ( Q ) and | y ∗ P ′ − z Q | ≤ | y P ′ − z Q | < A ℓ ( Q ). Thus, the ball B ( y P ′ , 2 αℓ ( Q )) ⊃ B ( y ∗ P ′ , αℓ ( Q )) intersects supp µ , so δℓ ( P ′ ) < 2 αℓ ( Q ), i.e., ℓ ( P ′ ) ≤ 2 αδ − 1 ℓ ( Q ). Alexander Volberg Solving a problem of David and Semmes

frame 80. Let now P = { P ′ ∈ P k +1 : P ′ ⊂ Q } . Consider the Marcinkiewicz function � � d +1 � ℓ ( P ′′ ) g ( P ′ ) = D ( P ′ , P ′′ ) P ′′ ∈ P The standard argument with integration it over Q shows that � g ( P ′ ) µ ( P ′ ) ≤ C 1 µ ( Q ) P ′ ∈ P for some C 1 > 0 depending on the dimension d and the goodness parameters of µ only. Define P ∗ = { P ′ ∈ P : P ′ ⊂ Q ε , g ( P ′ ) ≤ 3 C 1 } . Alexander Volberg Solving a problem of David and Semmes

frame 81. Note that � � � � µ ( P ′ ) ≥ µ ( P ′ ) − µ ( P ′ ) − µ ( P ′ ) . P ′ ∈ P ∗ P ′ ∈ P P ′ ∈ P : P ′ �⊂ Q ε P ′ ∈ P : g ( P ′ ) > 3 C 1 However, � � µ ( P ′ ) ≥ µ ( Q ′ ) ≥ (1 − ε ) µ ( Q ) . P ′ ∈ P Q ′ ∈ Q Further, since the diameter of each P ′ ∈ P is at most 8 ℓ ( P ′ ) ≤ 8 αδ − 1 ℓ ( Q ), every cell P ′ ∈ P that is not contained in Q ε is contained in Q \ Q 2 ε , provided that α < 1 8 εδ . Thus, under this restriction, � µ ( P ′ ) ≤ µ ( Q \ Q 2 ε ) ≤ C ε γ µ ( Q ) . P ′ ∈ P : P ′ �⊂ Q ε Alexander Volberg Solving a problem of David and Semmes

frame 82. Further, since the diameter of each P ′ ∈ P is at most 8 ℓ ( P ′ ) ≤ 8 αδ − 1 ℓ ( Q ), every cell P ′ ∈ P that is not contained in Q ε is contained in Q \ Q 2 ε , provided that α < 1 8 εδ . Thus, under this restriction, � µ ( P ′ ) ≤ µ ( Q \ Q 2 ε ) ≤ C ε γ µ ( Q ) . P ′ ∈ P : P ′ �⊂ Q ε Finally, by Chebyshev’s inequality, � µ ( P ′ ) ≤ µ ( Q ) . 3 P ′ ∈ P : g ( P ′ ) > 3 C 1 Bringing these three estimates together, we get the inequality � P ′ ∈ P ∗ µ ( P ′ ) ≥ 1 2 µ ( Q ), provided that A , ε, α satisfy some restrictions of the admissible type. Alexander Volberg Solving a problem of David and Semmes

frame 83. Vitali’s lemma sparceness. Now we will rarefy the family P ∗ a little bit more. Consider the balls B ( z P ′ , 10 ℓ ( P ′ )), P ′ ∈ P ∗ . By the classical Vitali covering theorem, we can choose some subfamily P ′ ⊂ P ∗ such that the balls B ( z P ′ , 10 ℓ ( P ′ )), P ′ ∈ P ′ are pairwise disjoint but � � � P ′ . B ( z P ′ , 30 ℓ ( P ′ )) ⊃ B ( z P ′ , 10 ℓ ( P ′ )) ⊃ P ′ ∈ P ′ P ′ ∈ P ∗ P ′ ∈ P ∗ Then we will still have � � µ ( P ′ ) ≥ c ℓ ( P ′ ) d P ′ ∈ P ′ P ′ ∈ P ′ � � µ ( B ( z P ′ , 30 ℓ ( P ′ ))) ≥ c µ ( P ′ ) ≥ c µ ( Q ) . ≥ c P ′ ∈ P ′ P ′ ∈ P ∗ � � d +1 The estimate on h = � ℓ ( P ′ ) follows from P ′ ∈ P ′ ℓ ( P ′ )+ dist ( x , P ′ ) the Marcinkiewicz choice of P ∗ . Alexander Volberg Solving a problem of David and Semmes

� x η . frame 84. Die Zubereitung f¨ ur ψ , η : ψ = ∆ Fix the non-BAUPness parameter δ ∈ (0 , 1). Fix any C ∞ radial function η 0 supported in B (0 , 1) such that 0 ≤ η 0 ≤ 1 and η 0 = 1 2 ). For every P ′ ∈ P ′ , define on B (0 , 1 � � � � 1 1 η P ′ ( x ) = η 0 δℓ ( P ′ )( x − x P ′ ) − η 0 δℓ ( P ′ )( x − y P ′ ) . Note that η P ′ is supported on the ball B ( z P ′ , 6 ℓ ( P ′ )). This ball is contained in Q , provided that 12 αδ − 1 < ε (recall that ℓ ( P ′ ) ≤ 2 αδ − 1 ℓ ( Q ) and P ′ ⊂ Q ε ). Also η P ′ ≥ 1 on B ( x P ′ , δ 2 ℓ ( P ′ )) and the support of the negative part of η P ′ is disjoint with supp µ . Put � η = η P ′ . P ′ ∈ P ′ Since even the balls B ( z P ′ , 10 ℓ ( P ′ )) corresponding to different P ′ ∈ P ′ are disjoint, we have − 1 ≤ η ≤ 1. � We want to show that η d ν ≥ c ( δ ) µ ( Q ) with some c ( δ ) > 0. Alexander Volberg Solving a problem of David and Semmes

frame 85. � η d µ ≥ c ( δ ) µ ( Q ) with some c ( δ ) > 0. This is Obviously, because of the choice of P ′ and because, where η is negative does not carry any mass µ . Moreover, � � � � � � ≥ c ( δ ) µ ( Q ) − ε γ µ ( Q ) ≥ c � � η Φ d µ ≥ η + d µ − ( χ Q − Φ) d µ 2 µ ( Q ) . � � So we need to estimate as a small thing η ( d Φ µ − ν ), which is � the sum over Q ′ ∈ Q ′ of η ( d ϕ Q ′ µ − ν Q ′ ). By frame 25 we have � η ( d ϕ Q ′ µ − ν Q ′ ) | ≤ C αℓ ( Q ′ ) d +2 � ϕ Q ′ � Lip � η � Lip ( supp ϕ Q ′ ) ≤ | ℓ ( Q ′ ) C αε − 1 ℓ ( Q ′ ) d +1 � η � Lip ( supp ϕ Q ′ ) ≤ C αε − 1 µ ( Q ′ ) sup δℓ ( P ′ ) . P ′ : B ( z P ′ , 6 ℓ ( P ′ )) ∩ Q ′ ε � = ∅ For Q ′ ⊂ P ′ , fine. Otherwise Q ′ ∩ P ′ = ∅ , B ( z P ′ , 6 ℓ ( P ′ )) ∩ Q ′ ε � = ∅ give C ℓ ( P ′ ) ≥ εℓ ( Q ′ ). And again smallness of α kills all ε − 2 δ − 1 . Alexander Volberg Solving a problem of David and Semmes

frame 86. Vector field ψ . Fix P ′ ∈ P ′ . Let e P ′ be the unit vector in the direction y P ′ − x P ′ . Put � 0 u P ′ ( x ) = η P ′ ( x + te P ′ ) dt . −∞ Let us think that H is parallel to x d +1 = 0 and that e 1 = e P ′ (this is without loss of generality). Then ∂ 1 u = η . But R H = ( ∂ 1 , . . . , ∂ d ) 1 | x | d − 1 . Therefore, � � � 1 R H ∆ u = R H ∆ η = ( ∂ 1 , . . . , ∂ d ) | x | d − 1 ⋆ ∆ η = ( ∂ 1 , . . . , ∂ d ) η . We showed that � R H , 1 ∆ u = ∂ 1 η = η . Alexander Volberg Solving a problem of David and Semmes

frame 87. Since the restriction of η P ′ to any line parallel to e P ′ consists of two opposite bumps, the support of u P ′ is contained in the convex hull of B ( x P ′ , δℓ ( P ′ )) and B ( y P ′ , δℓ ( P ′ )). Also, since �∇ j η P ′ � L ∞ ≤ C ( j )[ δℓ ( P ′ )] − j and since supp η P ′ intersects any line parallel to e P ′ over two intervals of total length 4 δℓ ( P ′ ) or less, we have � 0 C ( j ) |∇ j u P ′ ( x ) | ≤ | ( ∇ j η P ′ )( x + te P ′ ) | dt ≤ [ δℓ ( P ′ )] j − 1 −∞ for all j ≥ 0. Define the vector fields � ψ P ′ = (∆ u P ′ ) e P ′ , ψ = ψ P ′ . P ′ ∈ P ′ Alexander Volberg Solving a problem of David and Semmes

frame 88. Then, clearly, ( R H ) ∗ ( ψ m ) = η and all below are satisfied ( m := m d +1 ): ψ = � P ′ ∈ P ′ ψ P ′ , supp ψ ⊂ S , dist ( supp ψ, L ) ≥ ∆ ℓ ( Q ) = ε 3 ℓ ( Q ). ψ P ′ is supported in the 2 ℓ ( P ′ )-neighborhood of P ′ and satisfies � C C ψ P ′ = 0 , � ψ P ′ � L ∞ ≤ δℓ ( P ′ ) , � ψ P ′ � Lip ≤ δ 2 ℓ ( P ′ ) 2 . � | ψ | dm ≤ C δ − 1 µ ( Q ). ( R H ) ∗ ( ψ m ) = η . � T ∗ ( ψ m ) � L ∞ ( supp ν ) ≤ C αδ − 2 ε − 3 d − 3 . R H ( | ψ | m ) � L 2 ( ν ) ≤ C δ − 1 � � � µ ( Q ) . Alexander Volberg Solving a problem of David and Semmes

frame 89. In fact, � � � � [ δℓ ( P ′ )] − 1 m ( B ( z P ′ , 6 ℓ ( P ′ ))) | ψ | dm = | ψ P ′ | dm ≤ C P ′ ∈ P ′ P ′ ∈ P ′ ≤ C δ − 1 � ℓ ( P ′ ) d ≤ C δ − 1 � µ ( P ′ ) ≤ C δ − 1 µ ( Q ) . P ′ ∈ P ′ P ′ ∈ P ′ To get the uniform estimate for T ∗ ( ψ m ), note that � � � � � � R H ( x ∗ − · ) , ψ P ′ � dm | [ T ∗ ( ψ P ′ m )]( x ) | = � � � ≤ C δ − 1 � R H ( x ∗ −· ) � Lip ( S � · ℓ ( P ′ ) d +1 ≤ C δ − 1 ∆ − d − 1 ℓ ( P ′ ) d +1 ℓ ( Q ) d +1 ≤ C αδ − 2 ∆ − d − 1 µ ( P ′ ) µ ( Q ) for every x ∈ supp ν (we remind the reader that ℓ ( P ′ ) ≤ 2 αδ − 1 ℓ ( Q )). Adding up and recalling our choice ∆ = ε 3 : � T ∗ ψ � L ∞ ( supp ν ) ≤ C αδ − 2 ε − 3 d − 3 � µ ( P ′ ) µ ( Q ) ≤ C αδ − 2 ε − 3 d − 3 . P ′ ∈ P ′ Alexander Volberg Solving a problem of David and Semmes

frame 90. The bound of � � R H ( | ψ | dm ) � L 2 ( ν ) . First we estimate � � R H ( | ψ | dm ) � L 2 ( µ Q ) . And then use our transfer estimates midificating the measure as it has been already done many time before. � Recall that for every P ′ ∈ P ′ , we have | ψ P ′ | dm ≤ C δ − 1 ℓ ( P ′ ) d . Hence, we can choose constants b P ′ ∈ (0 , C δ − 1 ) so that | ψ P ′ | m − b P ′ χ P ′ µ is a balanced signed measure, i.e., � � | ψ P ′ | dm = b P ′ χ P ′ d µ . Let � f = b P ′ χ P ′ . P ′ ∈ P ′ Our goal is to prove first � R H ( | ψ | m ) | ≤ C δ − 1 + | � | � R H ( f µ ) | + χ V ( P ′ ) | � R H ( b P ′ χ P ′ µ ) | , P ′ ∈ P ′ where for each P ′ ∈ P ′ , denote by V ( P ′ ) the set of all points x ∈ R d +1 such that dist ( x , P ′ ) ≤ dist ( x , P ′′ ) for all P ′′ ∈ P ′ . Alexander Volberg Solving a problem of David and Semmes

frame 91. This estimate of | � R H ( | ψ | m ) | from the previous slide converts to converts into � � R H ( | ψ | m ) � 2 L 2 ( µ Q )   �  δ − 2 µ ( Q ) + � f � 2 � b P ′ χ P ′ � 2  ≤ C δ − 2 µ ( Q ) , ≤ C L 2 ( µ ) + L 2 ( µ ) P ′ ∈ P ′ which we wanted. To get the pointwise estimate of frame 90 we write for x ∈ V ( P ′ ). [ � R H ( | ψ | m − f µ )]( x ) = [ � R H ( | ψ P ′ | m )]( x ) − [ � R H ( b P ′ χ P ′ µ )]( x ) � [ � R H ( | ψ P ′′ | m − b P ′′ χ P ′′ µ )]( x ) . + P ′′ ∈ P ′ , P ′′ � = P ′ Alexander Volberg Solving a problem of David and Semmes

frame 92. If x ∈ V ( P ′ ) and cells are Vitali disjoint, then dist ( x , P ′′ ) ≥ c ℓ ( P ′′ ) and so � � � � � � � � R H ( | ψ P ′′ | m − b P ′′ χ P ′′ µ )]( x ) � K H ( x − · ) d ( | ψ P ′′ | m − b P ′′ χ P ′′ µ ) � = � � � � � � � [ K H ( x − · ) − K H ( x − z P ′′ )] d ( | ψ P ′′ | m − b P ′′ χ P ′′ µ ) � = � � � ≤ 2 � K H ( x − · ) − K H ( x − z P ′′ ) � L ∞ ( P ′′ ) | ψ P ′′ | dm � � d +1 C ℓ ( P ′′ ) ℓ ( P ′′ ) dist ( x , P ′′ ) d +1 δ − 1 ℓ ( P ′′ ) d ≤ C δ − 1 ≤ , ℓ ( P ′′ ) + dist ( x , P ′′ ) and the same for R H ( x ∗ − y ). Hence all this huge sum on the previous slide is ≤ δ − 1 h ( x ) ≤ C /δ by the Marcinkiewicz choice of P ′ , see slide 78. Alexander Volberg Solving a problem of David and Semmes

frame 93. Note also that � � R H ( | ψ P ′ | m ) � L ∞ ≤ C δ − 1 (this is just the trivial bound C ℓ ( P ′ ) for the integral of the absolute value of the kernel over a set of diameter 12 ℓ ( P ′ ) C δℓ ( P ′ ) for the maximum of | ψ P ′ | ). multiplied by the bound Therefore, � � R H ( | ψ | m ) � 2 L 2 ( µ Q ) ≤ C δ − 2 µ ( Q ) is proved, and then we (non-trivially, but habitually) transfer this into � � R H ( | ψ | m ) � 2 L 2 ( ν ) ≤ C δ − 2 µ ( Q ) . Alexander Volberg Solving a problem of David and Semmes

frame 94. Smearing of the measure ν Exactly as in the beginning of the lectures we replace the measure ν by a compactly supported measure � ν that has a bounded density with respect to the ( d + 1)-dimensional Lebesgue measure m in R d +1 . More precisely, for every κ > 0, we will construct a measure ν with the following properties: � � ν is absolutely continuous and has bounded density with respect to m . supp � ν ⊂ S and dist ( supp � ν, L ) ≥ ∆ ℓ ( Q ). ν ( S ) = ν ( S ) ≤ µ ( Q ). � � � η d � ν ≥ η d ν − κ . � � R H ( | ψ | m ) | 2 d � R H ( | ψ | m ) | 2 d ν + κ . | � | � ν ≤ � � ν | 2 d � R H ν | 2 d ν + κ . | � | � R H � ν ≤ Alexander Volberg Solving a problem of David and Semmes

frame 95. Suppose � � R H ν � L 2 ( ν ) < λµ ( Q ) with tiny λ . Then, choosing sufficiently small smearing parameter we get very small κ > 0 and we can ensure that the measure � ν constructed in the previous section, satisfies � � � ν | 2 d � R H ( | ψ | m ) | 2 d � | � R H � | � ν < λµ ( Q ) , η d � ν ≥ θµ ( Q ) , ν ≤ Θ µ ( Q ) where θ, Θ > 0 are two quantities depending only on δ (plus, of course, the dimension d and the goodness and AD-regularity constants of µ ). Our aim is to show that if λ = λ ( δ ) > 0 is chosen small enough, then these three conditions are incompatible. Then of course � � R H ν � L 2 ( ν ) ≥ λµ ( Q ) with not-so-tiny λ , and almost orthogonality finishes the contradiction. Alexander Volberg Solving a problem of David and Semmes

frame 96. Extremal problem. For non-negative a ∈ L ∞ ( m ), define � ν a = a � ν and consider the extremal problem � | � R H � ν a | 2 d � Ξ( a ) = λµ ( Q ) � a � L ∞ ( m ) + ν a → min � ν a ≥ θµ ( Q ). Note that since � under the restriction η d � ν is absolutely continuous and has bounded density with respect to m , the measure � ν a is well defined and has the same properties. The first goal is to show that the minimum is attained and for every minimizer a , we have � a � L ∞ ( m ) ≤ 2 and ν a | 2 + 2( � | � R H � R H ) ∗ [( � R H � ν a ] ≤ 6 λθ − 1 ν a ) � everywhere in S . This is done precisely as in the beginning of the lectures. Review. Alexander Volberg Solving a problem of David and Semmes