Fitting a Sobolev function to data II

Transcription

1 Fitting a Sobolev function to data II Charles Fefferman 1 Arie Israel 2 Garving Luli 3 DEPARTMENT OF MATHEMATICS, PRINCETON UNIVERSITY, FINE HALL, WASHING- TON ROAD, PRINCETON, NEW JERSEY address: cf@math.princeton.edu DEPARTMENT OF MATHEMATICS, UNIVERSITY OF TEXAS AT AUSTIN, ROBERT LEE MOORE HALL, 2515 SPEEDWAY, AUSTIN, TEXAS address: arie@math.utexas.edu DEPARTMENT OF MATHEMATICS, UNIVERSITY OF CALIFORNIA AT DAVIS, ONE SHIELDS AVENUE, DAVIS, CALIFORNIA address: kluli@math.ucdavis.edu 1 Research supported by NSF grant DMS and AFOSR grant FA Research supported by an NSF postdoctoral fellowship, DMS Research supported by NSF grant DMS

2 ABSTRACT. In this paper and two companion papers, we produce efficient algorithms to solve the following interpolation problem: Let m 1 and p > n 1. Given a finite set E R n and a function f : E R, compute an extension F of f belonging to the Sobolev space W m,p (R n ) with norm having the smallest possible order of magnitude; secondly, compute the order of magnitude of the norm of F. The combined running time of our algorithms is at most CN log N, where N denotes the cardinality of E, and C depends only on m,n, and p. Keywords: Whitney extension problem, interpolation, efficient algorithms AMS2010 Classifications: 65D05, 65D17

3 Contents Introduction 5 Chapter 5. Proof of the Main Technical Results Starting the Induction The Induction Step An Approximation to the Sigma Tools to Fill the Gap Between Geometrically Interesting Cubes Computing Lengthscales Passing from Lengthscales to CZ Decompositions Completing the Induction 95 Chapter 6. Proofs of the Main Theorems Extension in Homogeneous Sobolev Spaces Extension in Inhomogeneous Sobolev Spaces 117 Bibliography 127 3

4

5 Introduction Continuing from [21], we interpolate data by a function F : R n R whose Sobolev norm has the least possible order of magnitude. More precisely, let m 1 and p > n 1. Given a function f : E R with E R n finite, we compute a function F W m,p (R n ) such that F = f on E, and F W m,p C F W m,p for any competing function F W m,p (R n ) such that F = f on E. Here, C depends only on m, n, and p. Our computations consist of efficient algorithms to be implemented on an (idealized) von Neumann computer. We study two distinct models of computation. In the first model ( infinite precision ), we assume that our computer deals with exact real numbers, without roundoff error. Our second, more realistic model of computation assumes that our machine handles only S-bit machine numbers, for some fixed, large S. In our previous paper [21], we stated our main results for infinite precision, and developed technical tools to be used in the proof of those results. Here, we complete the proof of our results for infinite precision. Issues arising from the finite-precision model of computation will be addressed in [22]. Chapters 1 through 4 of our text form the content of [21]. In this paper, we present Chapters 5 and 6. Chapter 7 will appear in [22] 5

6

7 CHAPTER 5 Proof of the Main Technical Results We will prove the Main Technical Results by induction on A (see Chapter 3). Recall the order relation < on multiindex sets A M defined in Section 2.6. In particular, recall that A = M is minimal under <. Fix a finite subset E 1 32, where denotes the unit cube [0, 1) n. We assume that N = #(E) Starting the Induction We first establish the base case of the induction. This corresponds to proving the Main Technical Results for A = M. (See Chapter 3.) Let CZ(M) be the collection of maximal dyadic cubes such that #(E 3) 1. Using one time-work at most CN log N in space CN, we produce a CZ(M)-ORACLE that answers queries as follows. A query consists of a point x. The response to the query x is a list of all the cubes CZ(M) such that x 65. The work and storage required to answer a query are at most C log N. We simply apply the PLAIN VANILLA CZ-ORACLE from Section 4.6.3; see Remark Since #(E) 2 and E, the collection CZ(M) does not contain the cube. Therefore, each CZ(M) is a strict subcube of, hence has a dyadic parent + such that #(3 + E) 2 (because is maximal), and so in particular (5.1.1) #(9 E) 2 for all CZ(M). LEMMA If, CZ(M) and then 1 2 δ δ 2δ. PROOF. We proceed by contradiction. Suppose that and δ 1δ 4 for some, CZ(M). Then 3 + 3, and hence #(E 3 + ) #(E 3 ) 1. However, this contradicts that #(3 + E) 2, completing the proof of the lemma. 7

8 LEMMA There exists ɛ 1 > 0, depending only on m, n, and p, such that 9 is not tagged with (M, ɛ 1 ) for any CZ(M). PROOF. Assume that ɛ 1 (0, 1) is less than a small enough universal constant. Let CZ(M). It suffices to show that σ(9) does not have an (M, x, ɛ 1, δ 9 )-basis, thanks to (5.1.1). We argue by contradiction. Suppose that (P α ) α M is an (M, x, ɛ 1, δ 9 )-basis for σ(9). Therefore, P 0 (x ) = 1, and α P 0 (x ) = 0 for α M, α 0. In other words, P 0 1. Moreover, there exists ϕ 0 X such that ϕ 0 = 0 on E 9 and δ n/p m ϕ 0 X(9) + δ m 9 ϕ 0 P 0 L p (9) ɛ 1 δ n/p m 9. We know that #(E 9) 2. Fix x E 9. By Lemma we have ϕ 0 (x) P 0 (x) C { ϕ 0 X(9) + δ m ϕ } 0 P 0 L p (9) C ɛ 1 δ n/p m 9. But ϕ 0 (x) = 0, and thus P 0 (x) C ɛ 1. However, if we take ɛ 1 < 1/C, then this inequality contradicts the fact that P 0 1. Recall that #(3 E) 1 for each CZ(M). This implies the next result. LEMMA If CZ(M) then 3 is tagged with (M, 1/2). We have thus established properties (CZ1-CZ5) for the decomposition CZ(M). Indeed, (CZ1), (CZ2), and (CZ4) are consequences of Lemmas 5.1.1, 5.1.2, and 5.1.3, respectively. Note that (CZ3) and (CZ5) are vacuously true because we are treating the base case A = M. We next associate an extension operator and a linear functional to each of the nontrivial cubes in CZ(M). More precisely, we define CZ main (M) := { CZ(M) : (65/) E }. For each CZ main (M) there is a unique point x() E 65. (Recall that #(E 3) 1 for each CZ(M).) For each CZ main (M), we define the following objects: A linear map T (,M) : X( 65 E) P X given by (5.1.2) T (,M) (f, P) = P + f(x()) P(x()). 8

9 A list Ξ(, M) = {ξ }, where (5.1.3) ξ (f, P) = ( f(x()) P(x()) ) δ n/p m. A list of assist functionals Ω(, M), which we take to be empty. Clearly, the functional ξ and map T (,M) both have Ω(, M)-assisted bounded depth (bounded depth). ALGORITHM: FIND MAIN-CUBES AND COMPUTE EXTENSION OPERATORS (BASE CASE). We compute a list of the cubes in CZ main (M). For each CZ main (M), we compute a short form description of the bounded depth functional ( ) 65 ξ : X E P R. We give a query algorithm, which requires work at most C log N to answer queries. A query consists of a cube CZ main (M) and point x. The response to the query (, x) is a short form description of the linear map (f, P) J x T (,M) (f, P). These computations require one-time work at most CN log N in space CN. EXPLANATION. We compute a list of cubes CZ main (M) and associated points x() E 65. This computation requires work at most CN log N in space CN; see the algorithm FIND MAIN-CUBES in Section For each in the list CZ main (M), we compute the linear functional ξ (f, P) = { f(x()) P(x()) } δ n/p m. There are at most CN such functionals, and we compute each one using work and storage at most C. Given (, x) CZ main (M), we use a binary search to determine the position of in the list CZ main. We then compute the linear map (f, P) J x T (,M) (f, P) = P + f(x()) P(x()). This requires work at most C log N per query. 9

10 LEMMA There exists C 1, depending only on m, n, and p, such that for each CZ main (M), the following properties hold. T (,M) (f, P) = f on 65 E. T (,M) (f, P) X( 65 ) + δ m T (,M)(f, P) P L p ( 65 ) C ξ (f, P). C 1 (f, P) 65 ξ (f, P) C (f, P) 65. PROOF. Note that E 65 = {x()} and T (,M)(f, P)(x()) = f(x()) for each CZ main (M). This implies the first bullet point. Recall that T (,M) (f, P) P, hence T (,M) (f, P) X( 65 ) = 0. Moreover, δ m T (,M)(f, P) P L p ( 65 This implies the second bullet point. ) = δ m f(x()) P(x()) L p ( 65 ) From the first and second bullet points we have Cδ m+n/p f(x()) P(x()) = C ξ (f, P). (f, P) 65 T (,M)(f, P) X( 65 ) + (δ 65 ) m T (,M) (f, P) P L p ( 65 ) C ξ (f, P). Let F X satisfy F = f on 65 δ n/p m (f P)(x()) = δ n/p m (F P)(x()) C E. Then the Sobolev inequality implies that ( ) F X( 65 ) + δ m F P L p ( 65 ) Taking the infimum over such F, we obtain the estimate ξ (f, P) C (f, P) 65. Thus we obtain the third bullet point, and this completes the proof of the lemma. This completes the proof of the base case of the induction. In the next section we start to prove the induction step The Induction Step Fix a set of multiindices A M with (5.2.1) A M. We assume by induction that we have already carried out the Main Technical Results for each A < A. Our goal is to find suitable constants a(a), ɛ 1 (A), ɛ 2 (A), c (A), S(A) and to carry out the Main Technical Results for A. 10

11 Let A < A be the maximal mutiindex set with respect to the order relation < on 2 M. (See Section 2.6 for the definition of the order relation.) By induction hypothesis, we have already carried out the Main Technical Results for A. (See Chapter 3.) We have thus produced the following: A decomposition CZ(A ) of into dyadic cubes, with the following properties. If with, CZ(A ), then and 1 2 δ δ 2δ. The collection of cubes { 65 : CZ(A )} has bounded overlap, meaning that there exists a constant C = C(n) such that, for each CZ(A ) there are at most C cubes CZ(A ) with (65/) (65/). From (5.1.1) and since CZ(M) refines CZ(A ) (see the induction hypothesis) we know that (5.2.2) ( E is nearby ) #(E 9) 2 for each CZ(A ). An oracle that accepts queries x and returns the list of all cubes CZ(A ) such that x 65. A list CZ main (A ) consisting of all the CZ(A ) such that 65 E. For each CZ main (A ), a list of assists Ω(, A ) [X(E)]. For each CZ main (A ), a list of Ω(, A )-assisted bounded depth linear functionals Ξ(, A ) [ X( 65 E) P] written in short form, as well as a linear extension operator ( ) 65 T (,A ) : X E P X, which we compute in the sense that (after one-time work) we can answer queries: In response to a query x we return a short form description of the Ω(, A )-assisted bounded depth linear map (f, P) J x T (,A )(f, P). These objects and algorithms have good properties as part of the induction assumption on A. We listed some of these properties just above. The remaining properties are mentioned later, as required. We denote (5.2.3) a = a(a ), the geometric constant used in the Main Technical Results for A. 11

12 ALGORITHM: APPROXIMATE OLD TRACE NORM. For each CZ main (A ), we compute linear functionals ξ 1,..., ξ D on P, such that D (5.2.4) ξ(0, P) p and ξ i (P) p (P P) ξ Ξ(,A ) i=1 differ by at most a factor of C. We carry this out using work and storage CN. EXPLANATION. For each ξ in the list Ξ(, A ), we compute the map P ξ(0, P) using work and storage at most C, by examining the short form description of (f, P) ξ(f, P) that has been computed. Applying COMPRESS NORMS (see Section 2.8), we compute linear functionals ξ 1,, ξ D such that (5.2.4) holds, using work and storage at most C # [Ξ(, A )]. By the inductive hypothesis, we know that the sum of # [Ξ(, A )] over all CZ main (A ) is bounded by CN, hence the work and storage guarantees are met The Non-monotonic Case. Here, we assume that A M is not monotonic and prove the Main Technical Results for A. See Section 2.6 for the definition of monotonic sets. We define CZ(A) = CZ(A ) and ɛ 2 (A) = ɛ 2 (A ), c (A) = c (A ), a(a) = a(a ), and S(A) = S(A ). The constant ɛ 1 (A) is chosen later in this section. We define Ω(, A) := Ω(, A ), Ξ(, A) := Ξ(, A ) and T (,A) := T (,A ) for each CZ main (A) = CZ main (A ). The properties of Ω(, A), Ξ(, A) and T (,A) asserted in the Main Technical Results for A are immediate from the corresponding properties of Ω(, A ), Ξ(, A ) and T (,A ) asserted in the Main Technical Results for A. Next, we prove properties (CZ1-CZ5) for the label A. Note that (CZ1) for A follows from (CZ1) for A. Also, note that (CZ5) for A holds because CZ(A) = CZ(A ). 12

13 Note that (CZ3) for A holds vacuously: There do not exist cubes CZ(A), CZ(A ) which satisfy the hypotheses of (CZ3). This follows because CZ(A) = CZ(A ). We need not check (CZ4), since A M; see (5.2.1). It remains to prove (CZ2) for A, which we accomplish in the next lemma. We determine ɛ 1 (A) = ɛ 1 in the lemma below. LEMMA There exists a universal constant ɛ 1 > 0 such that the following holds. Suppose that CZ(A) and δ c (A). Then S(A) is not tagged with (A, ɛ 1 ). PROOF. We assume that ɛ 1 > 0 is less than a small enough universal constant. Let CZ(A) satisfy δ c (A). Assume for the sake of contradiction that S(A) is tagged with (A, ɛ 1 ). If #(S(A) E) 1 then S(A ) = S(A) is tagged with (A, ɛ 1 (A )). However, this contradicts the induction hypothesis. Hence, we may assume from now on that #(S(A) E) 2. Thus, σ(s(a)) has an (Ã, x, ɛ 1, δ S(A) )-basis for some à A. Hence, Lemma implies that there exists κ [κ 1, κ 2 ] such that σ(s(a)) has an (A, x, ɛ κ 1, δ S(A), Λ)-basis, with A A and ɛ κ 1Λ 100D ɛ κ/2 1. Here, κ 1, κ 2 > 0 are universal constants. Suppose for the moment that A < A. Then S(A) is tagged with (A, ɛ κ 1 ). Note that ɛ κ 1 ɛκ 1 1 ɛ 1 (A ), for small enough ɛ 1. Thus, S(A ) = S(A) is tagged with (A, ɛ 1 (A )). However, this contradicts the induction hypothesis. Hence, Thus, there exists (P α ) α A with σ(s(a)) has an (A, x, ɛ κ 1, δ S(A), Λ)-basis. (5.2.5) P α ɛ κ 1 (δ S(A) ) n p + α m σ(s(a)) (α A) β P α (x ) = δ βα (β, α A) β P α (x ) ɛ κ 1 (δ S(A)) α β (α A, β M, β > α) β P α (x ) Λ (δ S(A) ) α β (α A, β M). 13

14 We are assuming that A is not monotonic. Thus we can pick α 0 A and γ M such that α 0 + γ M \ A. We define α = α 0 + γ and A = A {α}. Note that A A = {α} with α A. Consequently, A < A. We define P α = P α0 x q, where q(y) = α 0! (y x α! ) γ. That is, P α (y) = α 0! 1 α! ω! ω P α0 (x )(y x ) ω+γ. Note that P α = q Pα main 0, where and that P main α 0 = ω m 1 γ ω m 1 γ R α0 := P α0 P main α 0 = 1 ω! ω P α0 (x )(y x ) ω, ω >m 1 γ 1 ω! ω P α0 (x )(y x ) ω. In the above sum for R α0, since ω > m 1 γ α 0 we have ω > α 0, and so ω P α0 (x ) Cɛ κ 1 δ α 0 ω. Consequently, R α0 L p (S(A)) C ɛ κ 1 δn/p+ α 0. The bullet point properties of P α0 now yield the following properties of P α. α P α (x ) = 1 β P α (x ) Cɛ κ 1 δ α β (β M, β > α) β P α (x ) CΛδ α β (β M). We now show that P α Cɛ κ 1 (δ ) n p + α m σ(s(a)). To start, (5.2.5) implies that there exists ϕ X with ϕ = 0 on S(A) E and ϕ X(S(A)) + δ m ϕ P α 0 L p (S(A)) Cɛ κ 1 (δ ) n p + α 0 m. Applying R α0 L p (S(A)) Cɛ κ 1 δ n p + α 0, we see that ϕ P main α 0 X(S(A)) + δ m ϕ Pmain α 0 L p (S(A)) Cɛ κ 1 (δ ) n p + α0 m. 14

15 Moreover, the Leibniz Rule shows that q (ϕ Pα main 0 ) X(S(A)) + δ m q (ϕ Pmain α 0 m C (δ ) k+ γ m k (ϕ Pα main 0 Note that q P main α 0 k=0 C (δ ) γ ( ϕ P main α 0 ) L p (S(A)) ) L p (S(A)) X(S(A)) + δ m ϕ Pmain (by Lemma 2.3.2) Cɛ κ 1 (δ ) γ (δ ) n p + α 0 m = Cɛ κ 1 (δ ) n p + α m. P, hence q ϕ X(S(A)) + δ m q ϕ P α L p (S(A)) Cɛ κ 1(δ ) n p + α m. α 0 ) L p (S(A)) Since q ϕ = 0 on S(A) E, we have shown that P α Cɛ κ 1 (δ ) n p + α m σ(s(a)). This proves all the bullet point properties of P α. The bullet point properties of the P α (α A) imply that ( β P α (x )) α,β A is (Cɛ κ 1, CΛ, δ )- near triangular. Inverting the matrix ( β P α (x )) α,β A, we obtain a matrix (M αω ) α,ω A such that and β P α (x )M αω = δ βω (β, ω A) α A M αω δ αω { Cɛ κ 1 ΛD δ ω α : if α, ω A, α ω CΛ D δ ω α : if α, ω A. Set P # ω = α A P αm αω. The bullet point properties of (P α ) α A imply that P # ω Cɛ κ 1 Λ2D δ n/p+ ω m σ(s(a)) (ω A) β P # ω(x ) = δ βω (β, ω A) For ω A and β M with β > ω, we write (5.2.6) β P # ω(x ) = α<β β P α (x )M αω + α β β P α (x )M αω. An arbitrary term in the first sum in (5.2.6) is bounded by Hence, this sum is at most C ɛ κ 1 ΛD δ ω β. [ Cɛ κ 1 δ α β ] [ CΛ D δ ω α If α β, then [ α > ω, since ] [ β > ω. Thus, ] an arbitrary term in the second sum in (5.2.6) is bounded by CΛδ α β Cɛ κ 1 ΛD δ ω α. Hence, this sum is at most C ɛ κ 1 ΛD+1 δ ω β. 15 ].

16 Thus, β P # ω(x ) Cɛ κ 1 Λ2D δ ω β (β M, ω A, β > ω). According to the bullet point properties of (P # ω) ω A, we see that σ(s(a)) has an (A, x, Cɛ κ 1 Λ2D, δ )-basis, hence σ(s(a)) has an (A, x, C ɛ κ 1 Λ2D, δ S(A) )-basis. (See Remark ) For small enough ɛ 1 we have C ɛ κ 1 Λ2D C ɛ κ/2 1 ɛ κ 1/4 1 ɛ 1 (A ), hence σ(s(a)) has an (A, x, ɛ 1 (A ), δ S(A) )-basis. Hence, S(A) is tagged with (A, ɛ 1 (A )). However, since δ c (A ) and S(A) = S(A ), this contradicts the induction hypothesis. This completes the contradiction, and with it, the proof of the lemma. We have thus proven the Main Technical Results for A in the non-monotonic case The Monotonic Case. From this point onward, we assume that A is monotonic. (See Section 2.6 for the definition of monotonic multiindex sets.) We drop this assumption when we prove our main theorem in Chapter 6. We will now begin the task of carrying out the induction step by proving the Main Technical Results for A. (See Chapter 3.) We begin by treating a preliminary case. LEMMA Suppose that δ 1 for all 4 CZ(A ). Then the Main Technical Results for A imply the Main Technical Results for A. PROOF. We take CZ(A) to equal CZ(A ). The other objects and algorithms in the Main Technical Results for A are copies of the corresponding objects and algorithms in the Main Technical Results for A. By making at most C calls to the CZ(A )-ORACLE, we can check whether the hypothesis of Lemma holds. This takes one-time work at most C log N. In the sequel, we assume that we are in the case that (5.2.7) δ 1/8 for some CZ(A ). Recall that the decomposition CZ(A ) has the following properties: 16

17 CZ(A ) is a finite partition of = [0, 1) n into pairwise disjoint dyadic subcubes. If, CZ(A ) and then δ /δ {1/2, 1, 2}. If CZ(A ) then #(9 E) 2. (See (5.2.2).) LEMMA If CZ(A ) and dist(, R n \ ) = 0 then δ { 1, 1, 1} PROOF. Let CZ(A ) with dist(, R n \ ) = 0. Recall that δ 1, because CZ(A ) { } (see (5.2.7)). We need to show that δ 1. For the sake of contradiction assume that δ 8 1. Then 16 since dist(, R n \ ) = 0, we have 9 R n \ 1 10, hence 9 R n \E. But #(E 9) 2, according to the above bullet points. This contradiction completes the proof of Lemma We now pass from the decomposition CZ(A ) of to a decomposition CZ(A ) of R n. PROPOSITION There exists a decomposition CZ(A ) of R n into pairwise disjoint dyadic cubes, with the following properties: (a) CZ(A ) CZ(A ). (b) If, CZ(A ) and then 1 8 δ δ 8δ ( good geometry ). Moreover, the collection of cubes { 65 : CZ(A )} has bounded overlap (each cube intersects a bounded number of other cubes). (c) If CZ(A ) \ CZ(A ), then 65 E =. (d) If CZ(A ) \ CZ(A ), then 100 intersects cubes in CZ(A ) with sidelength less than δ. (e) If CZ(A ) \ CZ(A ), then δ 1. (f) If CZ(A ) then #(9 E) 2. We produce a CZ(A )-ORACLE. The CZ(A )-ORACLE accepts a query consisting of a point x R n. The response to a query x is the list of cubes CZ(A ) such that x 65. The work and storage required to answer a query are at most C log N. PROOF. Let consist of the maximal dyadic cubes R n satisfying the condition [δ 1 or 0 / 2]. A dyadic cube R n belongs to if and only if (5.2.8) δ = 1 or 0 / 2, 17

18 and (5.2.9) δ + 2 and Here, as usual, + denotes the parent of a dyadic cube. For any x R n, there exists a dyadic cube containing x such that δ 2 and 0 2. Hence, each x R n is contained in some cube. Hence, partitions R n into pairwise disjoint dyadic cubes. Note that the cube = [0, 1) n belongs to. We now establish good geometry of (with constant 1/4). We prove that if, and then 1δ 4 δ 4δ. Assume for the sake of contradiction that there exist cubes, with δ 1δ 8 and. By (5.2.9), we have δ + 2 and Moreover, note that (since and δ 1δ 8, hence + and δ + 1δ 4 ). Hence, 0 2. Moreover, δ 4δ + 8. However, since, the analogue of (5.2.8) with replaced by must hold. This yields a contradiction. This completes the proof that the cubes in have good geometry. We define the collection CZ(A ) to consist of all the cubes except for =, together with all the cubes CZ(A ). Since partitions R n and CZ(A ) partitions, we see that CZ(A ) partitions R n into pairwise disjoint dyadic cubes. Moreover, property (a) clearly holds. If, CZ(A ), and, then both and touch the boundary of. We prove the claim that contains all 4 n of the dyadic cubes [ 2, 2) n with δ = 1. Indeed, we have + [ 2, 2) n, δ + = 2 and for any such. Hence, each satisfies (5.2.8) and (5.2.9), which implies that belongs to. This proves our claim. Hence, in particular, any that intersects the boundary of = [0, 1) n must satisfy δ = 1. Moreover, by Lemma 5.2.3, any CZ(A ) that intersects the boundary of must satisfy δ {1/2, 1/4, 1/8}. Hence, the previous two statements imply that for any and CZ(A ) with we have 1δ 8 δ δ. Finally, for, CZ(A ) with, we have 1δ 2 δ 2δ, by good geometry of the cubes in CZ(A ). 18

19 Recall that the cubes in satisfy good geometry (with constant 1/4). Thus, combining the previous three statements, for any, CZ(A ) with, we have 1 8 δ δ 8δ. The above property shows that CZ(A ) satisfies the hypothesis of Lemma with γ = 1/8. Hence, for, CZ(A ) with 65 65, we have. It follows that the collection { 65 : CZ(A )} has bounded overlap. This completes the proof of property (b). From (5.2.9), each CZ(A ) \ CZ(A ) satisfies δ 1. This proves property (e). We now prove property (c). Let CZ(A ) \ CZ(A ). Then and R n \. According to property (e), there are only two cases to consider If δ = 1, then 65 cannot intersect 1 32 (because = and δ = δ = 1). Since E 1 32, we conclude that 65 E =. If δ 2, then 0 / 2 thanks to (5.2.8). Assume for the sake of contradiction that Since and are disjoint, we conclude that 1 δ 1 = δ Hence, 0 2 (since and δ 16, and = [0, 1) n ). Hence, we derive a contradiction. Thus, 65 cannot intersect Since E 1 32, we conclude that 65 E =. This completes the proof of property (c). Property (d) is easy to prove. Let. Then thanks to (5.2.9). Hence, 0 6 (since 2 + 6). Since δ 1, this implies 9 (recall that = [0, 1) n ). Together with (5.2.7), this implies property (d). We now prove property (f). Let CZ(A ) be given. If CZ(A ) then #(9 E) 2, thanks to (5.2.2). If, then 9, hence #(9 E) = #(E) 2. This concludes the proof of property (f). We prepare to describe the construction of the CZ(A )-ORACLE. We can determine whether a dyadic cube R n belongs to CZ(A ) using work and storage at most C log N. We explain the procedure below. Let R n be given. 19

20 First, suppose that. Then CZ(A ) if and only if CZ(A ). We can determine whether CZ(A ) by using the CZ(A )-ORACLE to produce a list of all the cubes CZ(A ) satisfying x 65. (Recall, x denotes the center of.) Then CZ(A ) if and only if belongs to the aforementioned list. Thus, in this case, we can determine whether CZ(A ) using work at most C log N. Next, suppose that. Then can never belong to CZ(A ). Lastly, suppose that R n \. Then CZ(A ) if and only if. Recall from (5.2.8) and (5.2.9) that if and only if [δ = 1 or 0 / 2] and [δ + 2 and ]. We can check each of these conditions using at most C computer operations. Thus, in this case we can determine whether CZ(A ) using work at most C. Hence, we can determine whether a given cube belongs to CZ(A ) using work at most C log N. We next explain how to compute the unique cube x CZ(A ) containing x. It will then not be difficult to produce a list of the cubes CZ(A ) satisfying x 65. We describe this step at the very end. We check whether or not x. We split into cases depending on the result. First, suppose that x. We then compute the cube CZ(A ) containing x using the CZ(A )-ORACLE. We set x =. Now suppose that x R n \. Let be the unique cube in \ { } containing x. We will explain how to compute. We compute the dyadic cube R n such that δ = 1 and x. We test to see whether 0 2. We can do that using at most C computer operations. If 0 2 then is a maximal dyadic cube satisfying the condition [δ 1 or 0 / 2 ]. Hence, in that case, is the unique cube in containing x. We set x =. Now suppose that 0 / 2. Thus, satisfies (5.2.8). Since and are intersecting dyadic cubes (they both contain x), and since is maximal with respect to the property (5.2.8), we conclude that. Assume that =. Then 0 / 2, by assumption. On the other hand, suppose that. Then δ > 1 (since δ = 1). Since satisfies (5.2.8), we conclude that 0 / 2. 20

21 Thus, in the case where 0 / 2, we know that 0 / 2. Since x this shows that x 1 4 δ. Moreover, since satisfies (5.2.9) we know that 0 9. Hence, (5.2.10) for the unique cube containing x. 1 4 δ x 9δ There are no more than C dyadic cubes R n satisfying (5.2.10) with x ; moreover, it takes work at most C to list all these cubes. We examine each cube and test to see whether it belongs to CZ(A ). We set aside the unique cube that passes the test. We set x =. We have just explained how to compute the cube x CZ(A ) containing a given point x R n. The work requires is at most C log N. We now explain how to construct the CZ(A )-ORACLE. Suppose that CZ(A ) satisfies x 65. Then (5.2.11) x and 1 8 δ x δ 8δ x. This is a consequence of condition (b) in Proposition and an application of Lemma (with γ = 1/8). We produce a list of all the dyadic cubes that satisfy both (5.2.11) and x 65. There are at most C such cubes and it takes work at most C to list them all. We examine each cube to see whether it belongs to CZ(A ). We return the list of all those cubes that belong to CZ(A ). This completes the description of the CZ(A )-ORACLE. This completes the proof of the proposition Keystone Cubes. We define integer constants S 0 := S(A ), (5.2.12) S 1 := the smallest integer greater than 100, 10 5 S 0, and 2 [c (A )] 1, S 2 := the smallest odd integer greater than 10 5 S 1. We let ɛ > 0 be a small parameter. We assume in what follows that (5.2.13) ɛ > 0 is less than a small enough universal constant. 21

22 We eventually fix ɛ to be a universal constant, but only much later in the proof. We will take ɛ 2 (A) = ɛ κ and ɛ 1 (A) = ɛ 1/κ for a small universal constant κ. The discussion of the final choice of the numerical constants relevant to the Main Results for A occurs in Section See also (5.6.4). We next define the keystone cubes associated to the decomposition CZ(A ). We will prove a few basic properties of the keystone cubes and introduce the relevant algorithms. DEFINITION A cube # CZ(A ) is keystone if and only if δ δ # for every CZ(A ) that meets S 2 #. LEMMA The collection {S 1 # : # CZ(A ) keystone} has bounded overlap. Moreover, each keystone cube # CZ(A ) belongs to CZ(A ). δ # 1 PROOF. Suppose that # 1, # 2 are keystone cubes such that S 1 # 1 S 1 # 2 and δ #. Then # 1 S 2 # 2, since S S 1. Therefore, δ 2 # 1 the keystone cubes. δ #, by definition of 2 Consequently, δ # = δ 1 # whenever S 1 # 1 S 1 # 2. Thus, no more than C cubes 2 S 1 # 2 can intersect any given cube S 1 # 1. This implies the first conclusion of Lemma Finally, observe that no cube in CZ(A ) \ CZ(A ) can be keystone, thanks to condition (d) in Proposition and the fact that S This completes the proof of the lemma. The definition of keystone cubes written above agrees with the definition in Section 4.5, where we set K = S 2 and let A be a large universal constant in Section 4.5. The MAIN KEYSTONE CUBE ALGORITHM in Section 4.5 says the following. Given CZ(A ), we can compute a keystone cube K() CZ(A ) such that the following condition holds. There exists a sequence S = ( 1, 2, L ) of CZ(A ) cubes such that and such that = 1 2 L = K(), δ l C (1 c) l l δ l for 1 l l L. We do not compute the sequence S, we just claim its existence. We now modify the sequence S to consist only of cubes from CZ(A ) while maintaining the important properties of S. 22

23 We first discuss the case in dimension n = 1. We let S denote the sequence formed by omitting from S all the cubes that belong to CZ(A ) \ CZ(A ). Recall that all the cubes in CZ(A ) are contained in = [0, 1) and all the cubes in CZ(A ) are contained in R\[0, 1). Consider a maximal subsequence k1,, k2 of cubes in S that belong to CZ(A ). Then, by connectedness, each k (k 1 k k 2 ) is contained in either [1, ) or (, 0). Assume for sake of definiteness that each k is contained in [1, ). Then k1 1 and k2 +1 are the same cube in CZ(A ), namely the unique cube in CZ(A ) that meets the endpoint x = 1. (This is because the sequence must exit and reenter [0, 1) using the same cube that borders the endpoint x = 1.) Thus we can remove the aforementioned subsequence from S and obtain a connected path of cubes. The resulting sequence is exponentially decreasing with the same constants C and c above. The same argument shows that we can remove every maximal subsequence of S consisting of cubes in CZ(A ) \ CZ(A ). We now handle the case when the dimension n is at least 2. Suppose that some of the cubes in S belong to CZ(A ). Let k1 and k2 denote the first and last cubes in the sequence S belonging to CZ(A )\CZ(A ). Let S sub = ( k1,, k2 ) denote the corresponding subsequence of S. We know that 1 = and L = K() both belong to CZ(A ). Hence, 1 < k 1 k 2 < L. Note that both k1 1 and k2 1 intersect the boundary of and belong to CZ(A ). We join k1 1 and k2 +1 with a sequence S sub properties. = ( k1,, k3 ) with the following The cubes k CZ(A ) intersect the boundary of. k1 k1 1, k3 k2 +1, and k k+1 for k 1 k k 3 1, k 3 k 1 is bounded by a universal constant. Each k has sidelength between 1/2 and 1/8. These properties can be arranged due to Lemma We replace the subsequence S sub with the sequence S sub in S. We obtain a sequence S = ( 1, 2,, L ) of cubes in CZ(A ) such that 1 = and L = K(); moreover, l l+1 (1 l L 1) and δ l C (1 c ) l l δ l (1 l l L). 23

24 Indeed, the fact that S satisfies the exponentially decreasing property follows directly from the construction: We removed a subsequence of connected cubes in S and replaced it with a subsequence of bounded length consisting of cubes of size {1/2, 1/4, 1/8}. This has no effect on the fact that the sidelengths are exponentially decreasing in the sense of the above estimate. Hence, the sequence S joining and K() is exponentially decreasing. We never actually compute the sequences S or S, we just claim their existence. Using the above analysis, the MAIN KEYSTONE CUBE ALGORITHM and the algorithm LIST ALL KEYSTONE CUBES in Section 4.5, we obtain the following result. ALGORITHM: KEYSTONE-ORACLE. After one-time work at most CN log N in space CN we produce the following outcomes: We list all the keystone cubes # in CZ(A ). We can answer queries: A query consists of a cube CZ(A ), and the response to a query is a keystone cube K() to which is connected by an exponentially decreasing path = 1 2 L = K() with δ l C (1 c) l l δ l for 1 l l L. We guarantee that l CZ(A ) and 65 l C for each l. We guarantee that S 1 K() C; also that K() = if is keystone. The work required to answer a query is at most C log N. We list all (, ) CZ(A ) CZ(A ) such that and K( ) K( ). Let BD(A ) (the border disputes ) denote the set of all such pairs (, ). We guarantee that the cardinality of BD(A ) is at most CN. REMARK Let 1 L be as above. For fixed, we can have = l for at most C distinct l. This is because the path 1 2 L is exponentially decreasing. REMARK We do not attempt to compute the sequence of cubes 1,, L 1 - we only guarantee that this sequence exists, and we guarantee that we can compute the keystone cube K() = L located at the end of the sequence. 24

25 5.3. An Approximation to the Sigma We begin the proof of the Main Technical Results for A. We recall that A M is a monotonic set. In Sections and 5.2.3, we have defined a dyadic decomposition CZ(A ) of R n and a notion of keystone cubes in CZ(A ). We cannot compute all the cubes in CZ(A ) since there are infinitely many. Instead, we have access to a CZ(A )-ORACLE and the KEYSTONE-ORACLE. The integer constants S 0, S 1.S 2 relating to the keystone cubes are defined in (5.2.12). According to the Main Technical Results for A (see Chapter 3), for each CZ main (A ) the functional (5.3.1) M (,A )(f, R) := satisfies ξ Ξ(,A ) ξ(f, R) p (5.3.2) c (f, R) (1+a) M (,A )(f, R) C (f, R) 65. Recall that a is the constant a(a ) from the Main Technical Results; see (5.2.3). For each CZ(A )\CZ main (A ), we define Ξ(, A ) := and M (,A )(f, R) := 0. By definition of the collection CZ main (A ) and by property (c) in Proposition 5.2.1, we have 65 E =. Thus, (f, R) (1+a) = 0 for any CZ(A ) \ CZ main (A ). Thus, we see that (5.3.2) holds for all CZ(A ). 1/p Assigning Jets to Keystone Cubes. Let # CZ(A ) be a keystone cube. We define its associated CZ cubes to be the collection (5.3.3) I( # ) := { CZ(A ) : S 0 # }. We note that the cubes in I( # ) belong to CZ(A ) rather than CZ(A ). Hence, the cubes in I( # ) are contained in R n, and may not be contained in. LEMMA Let A 1 be given. Assume that, CZ(A ) and A. Then (5.3.4) (5.3.5) δ 10 3 Aδ, and A. 25

26 PROOF. We first prove (5.3.4). Assume for the sake of contradiction that δ > 10 3 Aδ for some, CZ(A ) with A. Then 65. However, this contradicts the good geometry of the cubes in CZ(A ) (see Proposition 5.2.1). This completes the proof of (5.3.4) by contradiction. Lastly, (5.3.5) follows from (5.3.4) and our assumption that A. By Lemma 5.3.1, the CZ cubes associated to a given # satisfy the following property: for each I( # ) we have (5.3.6) (5.3.7) δ 10 3 S 0 δ #, and 65 S 1 #. (Recall (5.2.12) which states that S S 0.) REMARK The definition of keystone cubes shows that δ δ # whenever I( # ). Hence, (5.3.6) implies that the cardinality of I( # ) is bounded by C for each keystone cube #. If I( # 1 ) and I(# 2 ) then (5.3.7) implies that S 1 # 1 S 1 # Recall that the cubes S 1 # ( # keystone) have bounded overlap. (See Lemma ) Thus, each CZ(A ) belongs to I( # ) for at most C distinct keystone cubes #. ALGORITHM: MAKE NEW ASSISTS AND ASSIGN KEYSTONE JETS. For each keystone cube #, we compute a list of new assists Ω new ( # ) [ X(S 1 # E) ], written in short form, and we produce an Ω new ( # )-assisted bounded depth linear map R # # : X(S 1 # E) P P, written in short form. Furthermore, we guarantee that the following conditions are met. The sum of depth(ω) over all ω in Ω new ( # ), and over all keystone cubes #, is bounded by CN. Given (f, P) X(S 1 # E) P, denote R # = R # # (f, P). Then α ( R # P ) 0 for all α A (recall, A is monotonic; see Remark 2.6.1). Let R P, with β (R P) 0 for all β A. Then (5.3.8) ξ(f, R # ) p C I( # ) ξ Ξ(,A ) I( # ) ξ Ξ(,A ) ξ(f, R) p. To compute the assists Ω new ( # ) and the short form of the maps R # # (for all the keystone cubes # ) requires work at most CN log N, and storage at most CN. 26

27 EXPLANATION. Given P P, we define V P to be the affine subspace consisting of all polynomials R P satisfying α (R P) 0 for all α A. We note that R V P α (R P)(0) = 0 for all α A, since A is monotonic. We introduce coordinates on V P, defined by w = (w 1,, w k ) R k = R w (x) = α A α P(0) x α + α! k j=1 w j xα j α j!, where we write M \ A = {α 1,, α k }. We consider the sum of the p-th powers of the functionals ξ(f, R w ) over all ξ Ξ(, A ) and I( # ). We want to minimize this expression with respect to w R k. We can approximately solve this minimization problem using the algorithm OPTIMIZE VIA MATRIX from Section 2.8. We describe the process below. For each CZ main (A ) with S 0 #, we have δ # δ by definition of keystone cubes. Hence, from (5.3.6) we have (5.3.9) S 0 # and δ # δ C δ # for a universal constant C. We list all the dyadic cubes that satisfy (5.3.9). There are at most C cubes in this list. We test each to see whether it belongs to CZ main (A ). Thus, we can compute the list L = { CZ main (A ) : S 0 # }. The list L contains all the cubes that participate in (5.3.8) for which Ξ(, A ). (Recall that Ξ(, A ) = for CZ(A ) \ CZ main (A ).) We list all the functionals ξ appearing in Ξ(, A ) for some L. From the Main Technical Results for A (see Chapter 3), each such ξ is given in the form I ξ(f, R w ) = λ(f) + µ a ω a (f) + ˇλ(( α P(0)) α A ) + a=1 k ˇµ j w j, j=1 where λ and ˇλ are linear functionals; ω a Ω(, A ) for some L; µ a and ˇµ j are real coefficients; and depth(λ) = O(1), I = O(1). In this discussion, we write X = O(Y) to indicate that X CY for a universal constant C. 27

28 Processing each functional ξ this way takes work O(1) per functional. Thus, with work O(L) (see below), we obtain a list of all the above ξ s, written as (5.3.10) ξ l (f, R w ) = λ l (f) + I l a=1 µ la ω la (f) + ˇλ l (( α P(0)) α A ) + for l = 1,, L; here, L = k ˇµ lj w j j=1 I( # ) # [ Ξ(, A ) ]. Here, each I l = O(1), each λ l has bounded depth, and each ω la belongs to Ω( la, A ) for some la L. Of course the la need not be distinct, and k dim(p) = D. Processing the functionals w ξ l (f, R w ) in (5.3.10) with the algorithm OPTIMIZE VIA MATRIX (see Section 2.8), we compute a matrix (b jl ) with the following property. The sum of the absolute values of the p-th powers of the functionals ξ l (f, R w ) (1 l L) is essentially minimized for fixed f, ( α P(0)) α A by setting w = w = (w 1,, w k ), where ] L I l (5.3.11) w j = b jl [λ l (f) + µ la ω la (f) + ˇλ l (( α P(0)) α A ) l=1 a=1 { L = b jl [λ l (f) + l=1 We have thus defined new assists ω new j I l a=1 µ la ω la (f) ]} + L b jl ˇλ l (( α P(0)) α A ) l=1 ω new j (f) + λ new j (( α P(0)) α A ). and new functionals λ new j. We may therefore take R # (f, P) := R # w with w j = ω new j (f) + λ new j (( α P(0)) α A ) (1 j k) and we obtain the estimate (5.3.8). Note that R # has assisted bounded depth, with # assists ω new j (j = 1,, k). Indeed, { ] (5.3.12) [R α # (f, P) (0) = # We can compute the new functionals λ new j requires work ω new j (f) + λ new j (( α P(0)) α A ) if α = α j, j {1,, k} α P(0) if α A. ( O(L) = O I( # ) We will now express the new assists ω new j We write ω new j = ω new,1 j (5.3.13) and (5.3.16)). (1 j k) using the obvious method. This # [ Ξ(, A ) ]). in short form. + ωj new,2, where ω new,1 j 28 and ω new,2 j are defined below (see

29 Each λ l (f) has bounded depth, so the functional (5.3.13) ω new,1 j : f can be computed in short ( form using ) work O(L log L) = O log N # [Ξ(, A )] I( # ) L b jl λ l (f) l=1 ( and storage O(L) = O I( # ) This computation follows by a simple sorting procedure. We provide details below. ) # [Ξ(, A )]. Recall that λ l has bounded depth and is given in short form (without assists): (5.3.14) λ l (f) = x S l c l (x) f(x), where #(S l ) C. (5.3.15) Thus, we can express the functional (5.3.13) as ω new,1 j : f x S d j (x) f(x), where S = L S l and l=1 d j (x) = L b jl c l (x) for x S. We compute the weights d j (x) by sorting. More precisely, we sort the points of S. We make an array indexed by S. We initialize the array to have all zero entries. We loop over l = 1,, L, and we loop over all the points y S l. We determine the position of each y in the list S, and we add the number b jl c l (y) at the relevant position in the array. This requires work at most C log(s) C log L for a fixed pair (l, y). Hence, the total work required is at most CL log L, since the number of relevant pairs (l, y) is L l=1 #(S l) L l=1 C CL. Similarly, the total storage is at most C L l=1 #(S l) CL. l=1 Thus, we can compute the functional (5.3.13) using work O(L log L) and storage O(L). It remains to compute the functional (5.3.16) ω new,2 j : f L l=1 b jl I l a=1 We recall that each ω la belongs to µ la ω la (f) in short form. (Recall, each I l = O(1).) I( # ) Ω(, A ). 29

30 We can express the functional (5.3.16) in the form (5.3.17) ω new,2 j : f q jω ω(f), with ( work O(L log L) = O log N I( # ) ω I( # ) Ω(,A ) ) ( # [Ξ(, A )] and storage O(L) = O We can compute the relevant numbers q jω by sorting, since (5.3.18) q jω = b jl µ la. (l,a):ω la =ω I( # ) ) # [Ξ(, A )]. Finally, once our functional is in the form (5.3.17), we can easily write it in short form (5.3.19) ω new,2 j with work O log N I( # ) ω Ω(,A ) Again, we perform a sort to carry this out. We compute ω new j (5.3.15) and (5.3.19). : f x S k j (x) f(x) depth(ω) and storage O I( # ) ω Ω(,A )) depth(ω). = ω new,1 j +ωj new,2 in short form by adding the short form expressions Altogether, we obtain the new assists ω new j # using work at most C (log N) and storage at most C I( # ) I( # ) 1 + #[ Ξ(, A ) ] #[ Ξ(, A ) ] + and the new functionals λ new j ω Ω(,A ) ω Ω(,A ) depth(ω) depth(ω). (Again, recall that Ξ(, A ) = Ω(, A ) = for any CZ(A ) \ CZ main (A ).) for a given Each CZ main (A ) belongs to I( # ) for at most C distinct # (see Remark 5.3.1). Therefore, we can compute the new assists and the new functionals for all the keystone 30

31 cubes # using work at most [ C (log N) #{ Keystone Cubes # } + + CZ main (A ) ω Ω(,A ) CZ main (A ) ] depth(ω), # [ Ξ(, A ) ] which is at most CN log N by the induction hypothesis and the statement of the KEYSTONE- ORACLE (which guarantees that the number of keystone cubes is bounded by CN). Similarly, we see that all the new assists can be computed using storage at most CN. Finally, note that there are at most D new assists for each given keystone cube # CZ(A ), and each such assist has depth at most #(S 1 # E). By the bounded overlap of the cubes S 1 # (see Lemma 5.2.4), we see that the sum of the depths of all the new assists is at most C #(E) = CN. This completes the explanation of the algorithm. Let # CZ(A ) be a keystone cube. For each (f, R) X(S 1 # E) P, we define [ ] p M # (f, R) := ξ(f, R) p = [ # M(,A )(f, R) ] p (5.3.20). I( # ) ξ Ξ(,A ) I( # ) The terms ξ(f, R) appearing above are well-defined, since (5.3.7) states that 65 S 1 # for each I( # ). (Recall that the domain of each functional ξ in Ξ(, A ) is X((65/) E) P.) We now show that the keystone functional defined in (5.3.20) is comparable to the trace semi-norm near the given keystone cube. LEMMA Let # be a keystone cube. Then for all (f, R) X(S 1 # E) P. c (f, R) S0 # M# # (f, R) C (f, R) S1 # PROOF. From (5.3.7) we learn that (1 + a) (65/) S 1 # for any I( # ). (Recall that a = a(a ) 1/; see (5.2.3).) Let (f, R) X(S 1 # E) P be given. 31

32 For each I( # ), by definition of the seminorm (, ) (1+a), we can choose F X with F = f on E (1 + a) and F X((1+a)) + δ m (1+a) F R L p ((1+a)) 2 (f, R) (1+a). From the left-hand estimate in (5.3.2) we have (f, R) (1+a) C M (,A )(f, R). Thus, by definition (5.3.20) we have (5.3.21) F X((1+a)) + δ m F R L p ((1+a)) C M # # (f, R) for I( # ). The assumptions in Sections and are valid, where CZ = CZ(A ) and = I( # ). = S 0 # ; r = a, and A = C for a large enough universal constant C. We exhibited a CZ(A )-ORACLE and proved good geometry for CZ(A ), which is a decomposition of R n, in Proposition We proved the properties of the collection stated in Section 4.6.5: by definition of I( # ) in (5.3.3) and since CZ(A ) is a partition of R n, we obtain (4.6.4); also, from (5.3.6) we obtain (4.6.5). We may thus apply the results stated in Section By Lemma 4.6.2, there exists a partition of unity θ # Cm (R n ) ( I( # )) such that θ # = 1 on S 0 #, I( # ) where supp θ # (1 + a) and α θ # (x) C δ α for x (1 + a), α m. Define F := I( # ) F θ #. Since the cardinality of I( # ) is at most C (see Remark 5.3.1), Lemma and (5.3.21) show that F X(S0 # ) C M # # (f, R). Moreover, since δ # δ Cδ # for all I( # ), we have δ m F R # L p (S 0 # ) C δ m F R L p ((1+a)) θ # I( # ) L ((1+a)) C M # # (f, R) (see (5.3.21)). 32

33 Because supp(θ # ) (1 + a) and F = f on E (1 + a) we see that F = f on E S 0 #. Hence, the above estimates imply that (f, R) S0 # F X(S 0 # ) + δ m S 0 # F R L p (S 0 # ) M # # (f, R). This proves one inequality in the statement of the lemma. Next, using the right-hand estimate in (5.3.2), we see that [ ] p [ M # (f, R) = # M(,A )(f, R) ] p C I( # ) CZ main (A ) I( # ) (f, R) p 65. Recall that I( # ) contains at most C cubes. Thus, by Lemma 2.4.1, where we use the estimate δ # δ Cδ # and that 65 S 1 # for each I( # ), the right-hand side in the above estimate is bounded by C (f, R) p S 1 #. This completes the proof of the lemma. The parameter ɛ > 0 now makes its first appearance. Recall that ɛ is assumed to be less than a small enough universal constant. See (5.2.13). Then PROPOSITION Let be a dyadic subcube of, such that 3 is tagged with (A, ɛ). Assume also that # CZ(A ) is a keystone cube, and that S 1 # 65. Suppose that H X satisfies H = f on E S 1 # and α H(x #) = α P(x #) for all α A. (5.3.22) δ m # H R # # L p (S 1 # ) C H X(S1 # ). Here, C 1 is a universal constant; and R # # = R # # (f, P). (See the algorithm MAKE NEW ASSISTS AND ASSIGN KEYSTONE JETS.) PROOF. Recall that S 0 # 65 and 3 is tagged with (A, ɛ). Thus, Lemma implies that S 0 # is tagged with (A, ɛ κ ) for some universal constant κ > 0. Recall that S 1 2[c (A )] 1 ; see (5.2.12). Thus, since S 1 # 65 we have (5.3.23) δ # S 1 1 δ 65 c (A ). Hence, the induction hypothesis implies that (5.3.24) S 0 # is not tagged with (A, ɛ 1 (A )) for any A < A. In particular, S 0 # is not tagged with (A, ɛ κ ) for any A < A. 33

34 Since S 0 # is tagged with (A, ɛ κ ) but not with (A, ɛ κ ) for any A < A, we see that σ(s 0 # ) has an (A, x #, ɛ κ, δ S0 #)-basis. Thus there exist polynomials (P α) α A such that (5.3.25) (5.3.26) (5.3.27) P α ɛ κ [ δ S0 # ] α +n/p m σ(s0 # ) for α A, β P α (x #) = δ αβ for β, α A, β P α (x #) ɛ κ [ δ S0 # ] α β for β M, α A, β > α. To start, we prove the following statement. Suppose that (5.3.28) (5.3.29) R σ(s 0 # ), and α R(x #) = 0 for all α A. Then, for some W = W(m, n, p) 0 we have (5.3.30) β R(x #) W δ m n/p β # for β M. For the sake of contradiction, suppose that (5.3.28) and (5.3.29) do not imply (5.3.30). Then, for some large constant Ŵ 0, which will be determined later, there exists R σ(s 0 # ) satisfying (5.3.29) and (5.3.31) max β M β R(x #) (δ #) n p + β m = Ŵ. For each integer l 0, define the multi-index set { } l = β M : β R(x #) (δ #) n p + β m ) Ŵ(2 l. Note that 0 thanks to (5.3.31), and also l l+1 for l 0. Since #M = D and l M is an increasing sequence, there is an index l {0,, D} such that l = l +1. Pick the maximal element α l (under the standard order on multi-indices defined in Section 2.6). Since α l we have (5.3.32) α R(x #) (δ #) n p + α m Ŵ(2 l ). Now, if β M and β > α, then β / l = l +1 by the maximality of α. Hence, (5.3.33) β R(x #) (δ #) n p + β m Ŵ(2 l 1 ) for each β M with β > α. 34

35 We define Z and A by setting Ŵ = ZA, with A = 2 l and 0 l D. Then (5.3.31) - (5.3.33) state that α R(x #) Z [δ #] m n/p α, β R(x #) Z 1/2 [δ #] m n/p β for β M, β > α, and β R(x #) Z A [δ #] m n/p β for β M. Define P α := ( α R(x #)) 1 R. Then (5.3.29) implies that (5.3.34) β P α (x #) = δ αβ for β A {α}. Since R σ(s 0 # ), (5.3.35) P α Z [ 1 α +n/p m δ #] σ(s0 # ), and also (5.3.36) (5.3.37) β P α (x #) Z 1/2 [δ #] α β for β M, β > α, and β P α (x #) Z A [δ #] α β for β M. Set A := {α A : α < α} {α}. Since α R(x #) = 0 for all α A, we see that α / A. Thus, (5.3.38) A < A. For α A with α < α, we set P α := P α α P α (x #)P α. Note that (5.3.39) (5.3.40) (5.3.41) P α Cɛ κ Z A [ δ #] α +n/p m σ(s0 # ) (by (5.3.25), (5.3.27), (5.3.35)), β P α (x #) = δ αβ for β A (by (5.3.26), (5.3.34)), and β P α (x #) Cɛ κ Z A [ δ #] α β for β M, β > α (by (5.3.27), (5.3.37)). Examining (5.3.34)-(5.3.41), we see that (5.3.42) (P α ) α A forms an (A, x #, C (S 0 ) m max{ɛ κ Z A, Z 1/2 }, δ S0 #)-basis for σ(s 0 # ). Recall that Z = Ŵ1/A with A = 2 l and 0 l D. We pick Ŵ to be a large enough universal constant, and assume that ɛ is less than a small enough universal constant. Then (5.3.38) and (5.3.42) imply that S 0 # is tagged with (A, ɛ 1 (A )). But this contradicts 35

36 (5.3.24). This completes our proof that (5.3.30) holds whenever the polynomial R satisfies (5.3.28) and (5.3.29). We now prove the main assertion in Proposition Suppose that H X satisfies H = f on E S 1 # and α H(x #) = α P(x #) for all α A. Then α (J x H P) 0 for all # α A. (Recall, A is monotonic; see Remark ) We apply the estimate (5.3.8) followed by Lemma 5.3.2, and hence, we see that M # # (f, R # # ) C M # # (f, J x # H) C (f, J x # H) S 1 # C H X(S 1 # ), which implies that M # # (0, R # # J x # H) C H X(S 1 # ). Thus, Lemma implies that (0, R # # J x # H) S 0 # C H X(S 1 # ), hence By the defining properties of R # # R # # J x # H C H X(S 1 # ) σ(s 0 # ). (see the algorithm MAKE NEW ASSISTS AND ASSIGN KEYSTONE JETS), and by our assumption on α H(x #), we have Thus, (5.3.30) shows that α (R # # J x # H)(x #) = α (P P)(x #) = 0 for all α A. β (J x # H R# # )(x #) C (δ #) m n/p β H X(S1 # ) for all β M. Hence, by the Sobolev inequality we have δ m # H R # # L p (S 1 # ) C H X(S1 # ). That proves (5.3.22) and completes the proof of Proposition Marked Cubes. We summarize various objects that we have computed in previous sections of the paper. This is meant to serve as a reference for the reader. The main cubes: We compute the collection of cubes CZ main (A ), each marked with pointers to the following objects. The list Ω(, A ) of assist functionals on X( 65 E), expressed in short form. The list Ξ(, A ) of functionals on X( 65 E) P, which have Ω(, A )- assisted bounded depth, expressed in short form in terms of assists Ω(, A ). 36

37 The list of functionals ξ 1,, ξ D on P. (See the Main Technical Results for A and the algorithm APPROXIMATE OLD TRACE NORM in Section 5.2.) The keystone cubes: We list all the keystone cubes # for CZ(A ), each marked with pointers to the following objects. The list Ω new ( # ) of new assist functionals on X(S 1 # E), expressed in short form. The linear map R # # : X(S 1 # E) P P, which has Ω new ( # )-assisted bounded depth, and is expressed in short form in terms of assists Ω new ( # ). (See MAKE NEW ASSISTS AND ASSIGN KEYSTONE JETS in Section ) The border-dispute pairs: We list all the border-dispute pairs (, ) BD(A ). (See the KEYSTONE-ORACLE in Section ) We store these cubes in memory along with their markings Testing Cubes. Let be a dyadic subcube of. Since CZ(A ) is a dyadic decomposition of, one and only one of the following alternatives holds. (A) is a disjoint union of cubes from CZ(A ). (B) is strictly contained in one of the cubes of CZ(A ). DEFINITION Let be a dyadic cube. If alternative (A) holds, we call a testing cube. Let 0 < λ < 1. We say that a testing cube is λ-simple if δ λ δ for any CZ(A ) with (65/). We introduce a geometric parameter (5.3.43) t G R, which is an integer power of two. We assume that 0 < t G < c, where c is a small enough constant determined by m, n, p. We will later determine a(a) to be an appropriate constant depending on t G. For the main conditions satisfied by a(a), see the fourth and fifth bullet points in Chapter 3. Near the end of Section 5.3 we determine t G to be a constant depending only on m, n, and p - but not yet. 37