The Main results for C m (R n ) E (E Finite, Large) Charles Fefferman. April 5, 2013

Transcription

1 The Main results for C m (R n ) E (E Finite, Large) Charles Fefferman April 5, 2013

2 Great Credit goes to Whitney Y. Brudnyi and P. Shvarstman Bo az Klartag

3 Notation m,n 1 fixed X := C m (R n ) F X = sup x R n max α m α F(x) E R n X(E) = {F E : F X} f X(E) = inf{ F X : F X,F = f on E}

4 J x (F) = (m 1)-rst order Taylor polynomial of F at x J x (F)(y) = 1 α! α F(x) (y x) α J x (F) P α m 1 P = vector space of (m 1)-rst degree (real) polynomials on R n

5 X = C m (R n ) Constants denotes c, C, C etc always depend only on m and n A B means A CB A B means ca B CA (A and B have the same order of magnitude )

6 Theorems from Talk 1 Let E R n, #(E) = N <. Theorem 1 Given f : E R, we can compute the order of magnitude of f X(E) in CN logn computer operations. Computation of the norm

7 Theorem 2 Given f : E R, we can efficiently* compute a function F X such that F = f on E and F X C f X(E). * Efficiently means One-time work CN logn Query work C logn Storage CN Computation of extension

8 Given f : E R, given M > 0, we defined convex subsets by induction on l. Γ l (x,m) P, (x E,l 0) P Γ 0 (x,m) { α P(x) M (x E, α m 1) P(x) = f(x)

9 P Γ l+1 (x,m) y E, P Γ l (y,m) such that for α m 1. α (P P )(x) M x y m α

10 Theorem 3 For large enough l depending only on m, n, the following holds: Let f : E R (E finite), M > 0. Suppose Γ l (x,m) for x E. Then The Gammas f X(E) CM.

11 Theorems Not in Talk 1

12 Theorem 4 Let E R n, N = #(E) <. Then there exists a linear map T: X(E) X such that for any f X(E), we have Tf = f on E, and Tf X C f X(E).

13 We can take T to have a simple form. Given x R n, we can write Tf(x) = y E λ(x,y)f(y) for coefficients λ(x,y) independent of f. For each x, we will have at most C non-zero λ(x,y) ( Bounded depth )

14 T can be computed efficiently. After one-time work CN logn using storage CN we can answer queries. A query consists of a point x R n. The response to a query x is the family of non-zero coefficients λ(x, y). The work to answer a query is C logn.

15 Theorem 5 Let E R n, N = #(E) <. Then there exist subsets S 1,S 2,...,S K E such that K CN #(S k ) C for each k = 1,2,...,K For any f : E R, we have f X(E) max k=1,...,k f X(S k ) Sharp Finiteness Theorem

16 Moreover, the sets S 1,...,S K can be computed from E by an efficient algorithm { At most CN log N computer operations, At most CN storage

17 Recall, if #(S) C, then we can compute the order of magnitude of by Linear Algebra, f X(S) thanks to Whitney Extension Theorem.

18 Removing Outliers

19 Given f : E R with #(E) = N <, an integer Z, 0 Z N, want to find S E with #(S) Z and f X(E\S) as small as possible.

20 Easier Problem Compute SZ E such that #(SZ) C Z and for any S E with #(S) Z, we have f X(E\S Z ) C f X(E\S) The Unfair Competition

21 Theorem 6 Such an S Z can be computed using C ZNlogN operations and storage CN

22 Moreover, we can list the elements of E, E = {x 1,x 2,...,x N } in a way such that for any Z, we can take SZ = {x 1,x 2,...,x C Z } Removing Outliers

23 Logical Relationships among Theorems

24 Theorem on convex sets Γ l (x,m) Thm on Linear Ext operator of Bounded Depth Compute Extension (obvious) Computation of the norm (obvious) Sharp Finiteness Principle Remove Outliers

25 Sharp Finiteness Theorem Remove Outliners

26 Recall: f X(E) max k=1,...,k f X(S k ) S 1,...,S K E K CN #(S k ) C Can compute S 1,...,S K For each k, can compute Y k f X(Sk )

27 We present pseudocode for an algorithm OUTLIERS(E,f,Z)

28 The input for OUTLIERS(E,f,Z) consists of A finite set E R n A function f : E R An integer Z 0

29 OUTLIERS(E,f,Z) prints out the elements of a subset S Z E that wins the unfair competition

30 OUTLIERS(E,f,Z) { While (E and Z 0) {Compute S 1,...,S K as in Sharp Finiteness Thm. For k = 1,...K, compute Y k f X(Sk ). Find ˆk such that Yˆk = max{y 1,...,Y k }. Print out the elements of Sˆk. Replace (E,f,Z) by (E \Sˆk,f,Z 1). E\Sˆk } } ( loop body in purple color above)

31 Why it works Let SZ be the set of all the elements of E printed out by OUTLIERS(E,f,Z). Must show #(SZ ) C Z If S E and #(S) Z, then f X(E\S Z ) C f X(E\S)

32 First Bullet Point Each time we execute the loop body, we print out at most C elements of E, and then decrease Z by 1. Once we reach Z = 0 (or earlier), we terminate the procedure. So, obviously, #(S Z ) C Z.

33 The proof of the second bullet point is an induction on Z. Z = 0 S Z = and the second bullet point holds trivially. Assume the second bullet point for Z 1, will prove it for Z.

34 Second Bullet Point Let S E with #(S) Z. Must show Cases: f X(E\S Z ) C f X(E\S). (a) f X(E\S) f X(E) (S is a weak competitor ) (b) f X(E\S) f X(E) (S is a strong competitor )

35 Suppose S is a weak competitor. Then f X(E\S Z ) f X(E) f X(E\S) proving the second bullet point in this easy case.

36 Suppose S is a strong competitor. Let s see what happens the first time we execute the loop body. We compute S 1,...,S K, Y 1,...,Y K, ˆk. Recall f X(E) max k=1,...,k f X(S k ) f X(Sk ) Y k for each k Yˆk = max{y 1,...,Y k }.

37 Crucial observation: S Sˆk. Otherwise, Sˆk E \S, hence contradiction! Yˆk f X(Sˆk) f X(E\S) f X(E) max k=1,...,k f X(S k ) Yˆk,

38 It is now easy to decide from the second bullet point the induction assumption (Second bullet point for Z 1)

39 Want to show f X(E\S Z ) C f X(E\S). Let Ẽ = E \Sˆk, f = f Ẽ, S = S Ẽ. Our observation #( S) Z 1. Let S Z be the set of points printed out by OUTLIERS(Ẽ, f,z 1).

40 Then one checks that E \SZ = Ẽ \ S Z, and E \S = Ẽ \ S. Since #( S) Z 1, inductive assumption (2nd bullet point for Z 1) f X( Ẽ\ S Z ) C f X(Ẽ\ S).

41 Therefore f X(E\S Z ) C f X(E\S). (The constant C did not grow.) This completes the proof of the 2nd bullet point, and shows that the algorithm OUTLIERS(E,f,Z) performs as promised.

42 So, as claimed, Sharp Finiteness Theorem Can Remove Outliners

43 Linear Extension opreator of Bounded Depth Sharp Finiteness Theorem

44 The Plan Use well-separated pairs decomposition Use Linear Extension operator of Bounded Depth Put it all together

45 Warm-up Exercise Recall the well-separated pairs decomposition: Given E R n, #(E) = N <. Then E E \Diagonal is partitioned into Cartesian products and E ν E ν, ν = 1,...,ν max, ν max CN, diam(e ν)+diam(e ν) 10 3 dist(e ν,e ν).

46 For each ν = 1,...,ν max, we pick a representative (x ν,x ν ) E ν E ν.

47 IOU from Talk 1: Let f : E R. Then { } f(x) f(y) max : x,y E,x y x y f(x ν 100 max ) f(x ν ) ν=1,...,ν max x ν x ν.

48 Proof: Suppose f(x ν ) f(x ν ) x ν x ν for each ν. Must show f(x) f(y) 100 x y for all x,y E.

49 Suppose not. Pick x,x E such that f(x ) f(x ) > 100 x x with x x as small as possible. Note { x y < x x x,y E f(x) f(y) 100 x y Inductive hypothesis

50 Note (x,x ) E E \Diagonal, since f(x ) f(x ) > 100 x x. So, for some ν, (x,x ) E ν E ν. Also, and (x ν,x ν ) E ν E ν, diam(e ν )+diam(e ν ) 10 3 dist(e ν,e ν ).

51 Consequently, { x x ν + x x ν 10 3 x x x x ν + x x ν 10 3 x ν x ν. In particular, x x ν and x x ν are strictly less than x x. Inductive hypothesis { f(x ) f(x ν) 100 x x ν f(x ) f(x ν) 100 x x ν.

52 We now know that Therefore, f(x ) f(x ν) 100 x x ν f(x ) f(x ν ) 100 x x ν f(x ν) f(x ν) x ν x ν f(x ) f(x ) 100 x x ν +100 x x ν + x ν x ν 101 x x ν +101 x x ν + x x.

53 Almost done! We know that Also, f(x ) f(x ) 101 x x ν +101 x x ν + x x. Therefore, x x ν + x x ν 10 3 x x. f(x ) f(x ) 2 x x, contradicting the assumption that f(x ) f(x ) > 100 x x.

54 That completes the proof of our assertion on the Lipschitz constant. End of Warm-up Exercise

55 Return to the deduction of the Sharp Finiteness Theorem from Linear Extension operators of Bounded Depth.

56 In the same spirit as the Warm-up Exercise, we can prove the following lemma on Whitney Fields.

57 The Glorified Warm-up Exercise.

58 Lemma: Let E R n, #(E) = N <. Let {E ν E ν : ν = 1,...,ν max } be a well-separated pairs decomposition for E E \Diagonal. Let (x ν,x ν) E ν E ν be representatives (ν = 1,...,ν max ). Let P = (P x ) x E be a Whitney field. Let M > 0 be a real number.

59 Suppose that α (P x ν P x ν )(x ν ) M x ν x ν m α for ν = 1,...,ν max and α m 1. Then α (P x P y )(x) CM x y m α for x,y E and α m 1.

60 Bringing in Extension operators of Bounded Depth.

61 Suppose T: X(E) X is a linear extension operator of bounded depth. Given x R n, the jet J x (Tf) is computed from f S(x), where S(x) E has at most C points.

62 We know that Tf = f on E and that Tf X C f X(E).

63 Lemma: Let f : E R. Let x,x E. Suppose f S(x ) S(x ) 1. Then α [J x (Tf) J x (Tf)](x ) C x x m α for α m 1.

64 Show picture on page 64.

65 Proof of the Lemma: Since f S(x ) S(x ) 1, there exists F X such that F X 2 and F = f on S(x ) S(x ). Let f = F E. Then f = f on S(x ) S(x ), and F X 2, F = f on E, hence f X(E) 2.

66 Let F = T f. Note that J x ( F) = J x (Tf) and J x ( F) = J x (Tf), since J x (T f) is determined by f = f S(x S(x ) ) and similarly for J x (T f).

67 Since f X(E) 2 and T: X(E) X has norm C, we have F X = T f X C f X(E) 2C. Taylor s theorem α[ ] J x ( F) J x ( F) (x ) C x x m α for α m 1.

68 Since J x ( F) = J x (Tf) and J x ( F) = J x (Tf), we conclude that α [J x (Tf) J x (Tf)](x ) C x x m α for α m 1. That s the conclusion of the Lemma.

69 Putting it all together.

70 E R n, #(E) = N < Well-separated pairs decomposition with representatives (x ν,x ν ) E ν E ν, ν = 1,...,ν max. ν max CN.

71 T: X(E) X Linear extension operator of bounded depth. J x (Tf) determined by f S(x), where S(x) E has at most C points. Tf = f on E.

72 Goal: Sharp Finiteness Theorem. Will define subsets S 1,S 2,...,S νmax E with ν max CN, #(S ν ) C for each ν. Will then show that f X(E) C max ν=1,...,ν max f X(Sν ).

73 For ν = 1,...,ν max, we define We already know that S ν = S(x ν) S x ν. ν max CN. Also, S(x ν) and S(x ν) each contain C points, so S ν contains at most C points.

74 It remains to show that for any f : E R. f X(E) C max ν=1,...,ν max f X(Sν )

75 Let f : E R. Suppose We will show that f X(Sν ) 1 for each ν. f X(E) C. That will complete the proof of the sharp finiteness theorem (assume there exists linear operator of bounded depth...)

76 For x E, defined P x = J x (Tf). Since Tf = f on E, we have P x (x) = f(x) for all x E.

77 Recall S ν = S(x ν) S x ν by definition, f X(Sν ) 1 by assumption. Applying the preceding Lemma, we find that [ α J x ν (Tf) J x (Tf)] (x ν ν) C x ν x ν m α for α m 1.

78 That is, [ ] α P x ν P x ν (x ν ) C x ν x ν m α for α m 1. This holds for each (x ν,x ν ) from the well-separated pairs decomposition.

79 The Glorified Warm-up Exercise now shows that [P α x P x ] (x ) C x x m α for all x,x E, α m 1.

80 One can also show that α P x (x) C for all x E, α m 1. That is similar in spirit to the proof of [P α x P x ] (x ) etc, but MUCH EASIER.

81 Now one knows that α P x (x) C for all x E, α m 1, α[ P x P x ] (x ) C x x m α for all x,x E, α m 1, and for all x E. P x (x) = f(x)

82 Whitney s extension theorem now shows that there exists F X with norm such that for all x E. In particular, F X C, J x (F) = P x F(x) = f(x) for all x E, since P x (x) = f(x).

83 Thus F X, F X C, F = f on E. Conclusion: f X(E) C. That s what we wanted to prove.

84 So we have reduced the sharp finiteness theorem to the existence of linear extension operators of bounded depth.

85 Next talk sketches: Proof of the theorem on Γ l (x,m) (the Heart of the Matter) Why theorem on Γ l (x,m) gives us linear extension operators of bounded depth.