TAILIEUCHUNG - Đề tài "A sharp form of Whitney’s extension theorem "

In this paper, we solve the following extension problem. Problem 1. Suppose we are given a function f : E → R, where E is a given subset of Rn. How can we decide whether f extends to a Cm−1,1 function F on Rn ? Here, m ≥ 1 is given. As usual, Cm−1,1 denotes the space of functions whose (m − 1)rst derivatives are Lipschitz 1. We make no assumption on the set E or the function f. This problem, with Cm in place of Cm−1,1, goes back to Whitney [15], [16], [17]. To answer it, we prove the following sharp form of the Whitney extension theorem | Annals of Mathematics A sharp form of Whitney s extension theorem By Charles L. Fefferman Annals of Mathematics 161 2005 509 577 A sharp form of Whitney s extension theorem By Charles L. Fefferman Contents 0. Introduction 1. Notation 2. Statement of results 3. Order relations involving multi-indices 4. Statement of two main lemmas 5. Plan of the proof 6. Starting the main induction 7. Nonmonotonic sets 8. A consequence of the main inductive assumption 9. Setup for the main induction 10. Applying Helly s theorem on convex sets 11. A Calderon-Zygmund decomposition 12. Controlling auxiliary polynomials I 13. Controlling auxiliary polynomials II 14. Controlling the main polynomials 15. Proof of Lemmas and 16. A rescaling lemma 17. Proof of Lemma 18. Proofs of the theorems 19. A bound for k References I am grateful to the Courant Institute of Mathematical Sciences where this work was carried out. Partially supported by NSF grant DMS-0070692. 510 CHARLES L. FEFFERMAN 0. Introduction In this paper we solve the following extension problem. Problem 1. Suppose we are given a function f E R where E is a given subset of R . How can we decide whether f extends to a Cm-1 1 function F on Rn Here m 1 is given. As usual Cm-1 1 denotes the space of functions whose m 1 rst derivatives are Lipschitz 1. We make no assumption on the set E or the function f . This problem with Cm in place of Cm-1 1 goes back to Whitney 15 16 17 . To answer it we prove the following sharp form of the Whitney extension theorem. Theorem a. Given m n 1 there exists k depending only on m and n for which the following holds. Let f E R be given with E an arbitrary subset of Rn. Suppose that for any k distinct points x1 . xk 6 E there exist m 1 rst degree polynomials P1 . Pk on Rn satisfying a Pi xi f Xi for i 1 . k b ỡ Pi xi M for i 1 . k and p m 1 and c ỡ Pi Pj xi M xi Xj m-l l for i j 1 . k and p m 1 with M independent of x1 . xk. Then f extends to a Cm-1 1 function on Rn. The converse of Theorem

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