(E Finite) The Heart of the Matter! Charles Fefferman. April 5, 2013

Transcription

1 C m (R n ) E (E Finite) The Heart of the Matter! Charles Fefferman April 5, 2013

2 Recall Notation X = C m (R n ) F X = sup x R n max α m α F(x) J x (F) = (m 1)-rst order Taylor poly of F at x J x (F) P P = vector space of (m 1)-rst degree (real) polynomials on R n

3 E R n X(E) = {F E : F X} f X(E) = inf{ F X : F X,F = f on E}. (X = C m (R n )) Constants denotes c, C, C etc always depend only on m and n

4 Today we deal only with (arbitrarily large) finite sets E.

5 In Talk 1, we posed the problem of deciding whether there exists F X such that and F X M F(x) f(x) Mσ(x) for all x E (f,σ,e,m given)

6 We defined convex sets Γ l (x,m) P (for x E, M > 0, l 0) by induction on l.

7 Base case: l = 0 Γ 0 (x,m) consists of all P P such that α P(x) M for α m and P(x) f(x) Mσ(x)

8 Inductive step Fix l 0, M > 0. Suppose we have computed Γ l (x,m), for all x E. We will compute Γ l+1 (x,m) for all x E.

9 We take Γ l+1 (x,m) to consist of all P Γ l (x,m) such that for each y E,there exists P Γ l (y,m) such that for α m 1. α (P P )(x) M x y m α

10 The heart of the matter is the following result.

11 Main Theorem For constants l, C depending only on m, n, the following holds: Let E R n (finite), f : E R, σ: E [0, ), M > 0 be given.

12 Fix x o E. If Γ l (x 0,M), then there exists F X such that F X CM and F(x) f(x) CMσ(x) for all x E.

13 More precisely, given P 0 Γ l (x 0,M), there exists F X such that F X CM and F(x) f(x) CMσ(x) for all x E J x0 (F) = P 0.

14 The proof of the Main Theorem is constructive. With extra care, we can make F depend linearly on f (or on (f,p 0 )).

15 In particular, the proof of the Main Theorem yields a Linear Extension Operator (It has Bounded Depth)

16 So, as promised in the last talk, Theorem on Γ l (x,m) Existence of Linear Extension Operator of Bounded Depth.

17 Sketch of the proof of Main Theorem. (Contains Lies!)

18 Local Problem. In addition to E,f,σ,M, suppose we are given a cube Q a point x 0 E 3Q a jet P 0 P. Problem: Find F X(3Q) such that F X(3Q) CM F(x) f(x) CMσ(x) for all x E 3Q J x0 (F) = P 0

19 Call that the Local Interpolation Problem LIP(Q,x 0,P 0 )

20 Our Main Theorem asserts that we can solve LIP(Q,x 0,P 0 ) for Q 0 = unit cube whenever P 0 Γ l (x 0,M).

21 We will prove that any LIP(Q,x 0,P 0 ) can be solved, provided P 0 Γ l (x 0,M).

22 We will measure the difficulty of a Local Interpolation Problem by assigning to it a Label A

23 Here, a Label A is any subset of M = {All multi-indices α = (α 1,...,α n ) of order α = α 1 + +α n m 1}

24 Labels A come with a (total) order relation <

25 Roughly speaking, if A < B, then we think that an interpolation problem carrying the lable A is easier than one carrying the label B.

26 If then A B, B < A. Accordingly, labels the hardest problems, and M labels the easiest ones.

27 To assign label A to a LIP(Q,x 0,P 0 ), we look at the convex set Γ l(a) (x 0,M) l(a) = integer constant depending on A.

28 We suppose P 0 Γ l(a) (x 0,M), else we wouldn t be trying to solve LIP(Q,x 0,P 0 ).

29 Very roughly speaking, we attach Label A to LIP(Q,x 0,P 0 ) provided P 0 +MP α Γ l(a) (x 0,M) for each α A, where P α (y) = δ m α Q (y x 0 ) α. Γ l(a) (x 0,M) contains room to maneuver

30 Main Properties of Labels Any LIP(Q,x 0,P 0 ) that carries Label A also carries Label B for any B A. (That is why we should say A < B when B A.)

31 Every LIP(Q,x 0,P 0 ) carries the label. Only the most trivial LIP(Q,x 0,P 0 ) carry the lable M. (e.g. E 3Q has only 1 point) We may solve the problem by taking F = P 0.

32 Another Crucial Property of Labels The attaching of a label A to a Local Interpolation Problem LIP(Q,x 0,P 0 ) depends on Q and x 0 but not on P 0. (That will rescue us later on.)

33 Using Labels to Prove the Main Theorem

34 By induction on the Label A (with respect to the order relation <), we will prove a Main Lemma for A.

35 Main Lemma for A Let LIP(Q,x 0,P 0 ) be a Local Interpolation Problem. Suppose P 0 Γ l(a) (x 0,M) and LIP(Q,x 0,P 0 ) carries the label A. Then the problem LIP(Q,x 0,P 0 ) has a solution.

36 Since every Local Interpolation Problem carries the Label, the Main Lemma for tells us that any LIP(Q,x 0,P 0 ) for which P 0 Γ l( ) (x 0,M) can be solved. Thus, Main Lemma for A = Main Theorem.

37 Proof of Main Lemma (A) by Induction on A

38 Base Case: A = M. (M is minimal under <) Recall: If LIP(Q,x 0,P 0 ) carries label M, then may just use F = P 0. If then it works. (So set l(m) = 1) P 0 Γ 1 (x 0,M),

39 Induction Step Fix a Label A = M. Assume the Main Lemma for B holds for all B < A. We will then prove the Main Lemma for A.

40 That will complete the induction prove the Main Lemma for A (any label A) establish the Main Theorem prove all 6 Theorems from Talk 3.

41 There are 2 cases for the proof of the Induction Step: A is monotonic (the important case) or A is non-monotonic (the trivial case).

42 Recall A is a set of multi-indices of order m 1. A is called monotonic if the following holds: Let α, β be multi-indices. If α A and α+β m 1, then α+β A.

43 Why the non-monotonic case is essentially trivial

44 Suppose G X, G X 1, G(x) σ(x) for all x E J x0 (G) = the polynomial y (y x 0 ) α. Let θ(x) be a harmless cutoff function supported in a ball of radius 1 about x 0. Define G(x) = θ(x) (x x 0 ) β G(x) for all x R n. (Assume α+β m 1.)

45 Then G X, G X C, G(x) Cσ(x) for all x E J x0 ( G) = the polynomial y (y x 0 ) α+β. So G is just like G, with α replaced by α+β.

46 Using that remark, we can prove: Let A be a non-monotonic label. Let α A, β M, with α+β M\A. Any LIP(Q,x 0,P 0 ) that carries the label A also carries the label B = A {α+β}.

47 Consequently, the Main Lemma for A follows from the Main Lemma for B, which we are assuming as part of our induction hypothesis. (Note B < A.)

48 So, as promise, the Induction Step is essentially trivial in the non-monotonic case.

49 Induction Step in the monotonic case

50 Fix A = M monotonic. Assume Main Lemma for B (all B < A). Fix LIP 0 = LIP(Q 0,x 0,P 0 ). Assume LIP 0 carries the label A, but does not carry any label B < A. P 0 Γ l(a) (x 0,M). Must solve LIP 0.

51 The Plan

52 We will solve LIP(Q 0,x 0,P 0 ) by Cutting 3Q 0 into subcubes {Q ν } ν=1,...,νmax. Using Inductive Assumption (i.e. Labels < A) to solve a Local Interpolation Problem on each 3Q ν, then Patching the local Interpolants together.

53 Here goes...

54 Step 1: Make a Calderon-Zygmund decomposition of 3Q 0 into finitely many subcubes Q ν (ν = 1,...,ν max )

55 Show picture on page 55.

56 Step 2: For each CZ cube Q ν such that E 3Q ν, we pick a point x ν E 3Q ν, and find a jet We then consider P ν Γ l(a) 1 (x ν,m). LIP ν = LIP(Q ν,x ν,p ν ).

57 Show picture on page 57.

58 Step 3: The CZ decomposition in Step 1 is defined to guarantee that each LIP ν carries a label B ν < A. Hence by induction assumption, we can solve each LIP ν from Step 2.

59 Step 4: For each CZ cube Q ν such that 3Q ν E, let F ν be the solution of LIP ν from Step 2. For each CZ cube Q ν such that 3Q ν E =, let F ν = P 0. (Recall: LIP 0 = LIP(Q 0,x 0,P 0 ).)

60 Now, for each Q ν, we have found F ν X(3Q ν ) such that F ν X(3Qν ) CM F ν (x) f(x) CMσ(x) for all x E 3Q ν J xν (F ν ) = P ν. These are our local interpolants. We will patch them together.

61 Step 5: Let {θ ν } ν=1,...,νmax be the Whitney Partition of Unity arising from the cubes Q ν. ν θ ν = 1 on 3Q 0 suppθ ν (1.01)Q ν α θ ν Cδ α Q ν for α m.

62 We then define our interpolant: F = ν θ ν F ν on 3Q 0. We hope it works.

63 Can it work? By setting F = ν θ ν F ν with θ ν living on the tiny lengthscale δ Qν, we run the risk that α F may be huge, because α θ ν is huge.

64 As in the proof of the Whitney s Theorem, we write F = Fˆν + ν θ ν (F ν Fˆν ) on each Qˆν, and hope that the derivatives of F ν Fˆν are small whenever Q ν and Qˆν touch.

65 Since and J xν (F ν ) = P ν J xˆν (Fˆν ) = Pˆν, that amounts to saying that α (P ν Pˆν )(xˆν ) CMδ m α Qˆν whenever Q ν and Qˆν touch. The P ν (ν = 1,...,ν max ) are mutually consistent.

66 So we need to pick the jets P ν from Step 2 in such a way that we can gurantee How? mutually consistency of the P ν.

67 Carrying out the Plan

68 Step 1: (The CZ decomposition) Given a cube Q, we (like it and keep it) or (dislike it and bisect it) according to the following rule:

69 We like and keep Q Either or #(E 3Q) 1, for some x E 3Q, the problem LIP(Q,x,P) carries some label < A. (Which P? We don t care!)

70 The CZ decomposition in Step 1 behaves as promised.

71 Step 2: Let Q ν be a CZ cube. We must pick x ν E 3Q ν (unless E 3Q ν = ) and we must pick P ν Γ l(a) 1 (x ν,m).

72 Now, Q ν is a CZ cube, so we like Q ν, but we don t like its parent Q ν +.

73 Either Case (a): #(E 3Q ν ) 1 or Case (b): LIP(Q ν,x ν,p) carries a label A for some x ν E 3Q ν. Case (a): Local Interpolation on 3Q ν is trivial. Case (b): We ve found our x ν!

74 So far, in the non-trivial case, we ve found our x ν E 3Q ν. Now we have to find P ν Γ l(a) 1 (x ν,m).

75 Recall the definition of the Γ l. If P Γ l+1 (x,m) and y E, then there exists P Γ l (y,m) such that α (P P )(x) M x y m α for α m 1.

76 We have P 0 Γ l(a) (x 0,M) and x ν E. Hence, there exists P ν Γ l(a) 1 (x 0,M) such that α (P ν P 0 )(x 0 ) M x ν x 0 m α for α m.

77 Is that the P ν we are looking for? Not yet. It s too arbitrary, because Γ l(a) 1 (x ν,m) has room to maneuver, which means that we may not achieve mutual consistency of the P ν s.

78 Key idea: Can pick P ν Γ l(a) 1 (x ν,m) such that α (P ν P 0 )(x 0 ) M x ν x 0 m α for α m 1, as before, and also α (P ν P 0 ) 0 for all α A.

79 This is possible, because there s room to maneuver and because A is monotonic. With this choice of P ν, we will achieve mutual consistency of the P ν.

80 Mutual consistency of the P ν s holds because there isn t too much room to maneuver, since we didn t like the parents of the CZ cubes Q ν.

81 Understand all that is the hardest part of the proof. Let s declare victory: We have accomplished Step 2 of the Plan, and achieved mutual consistency.

82 Note that we have looked separately at each Q ν to pick P ν. Yet P ν and Pˆν are consistent when Q ν touches Qˆν.

83 Step 3: As promised, except in trivial cases, we have carried out Steps 1 and 2 to guarantee that the local problems LIP ν = LIP(Q ν,x ν,p ν ) carries Labels < A.

84 Therefore, we can solve each LIP ν and obtain Local Interpolants F ν. That takes care of Step 3.

85 Step 4 was to pass from the cubes Q ν to the partition of unity 1 = ν θ ν. No problem! (Same in Whitney s proof)

86 Step 5 was to patch the local solutions together: F = ν θ ν F ν on 3Q 0. No problem! We are done with Step 5.

87 We have now carried out all the steps (1... 5) of the plan, and guaranteed the required mutual consistency.

88 It works!

89 This concludes the Sketch of Proof of the Main Theorem.

90 We got through it!

91 The next (and last) talk will be much easier!