Extensions of functions - lecture notes Krzysztof J. Ciosmak University of Oxford, Mathematical Institute, Oxford, United Kingdom Email address: [email protected] Abstract. The aim of the course is to present several results on extensions of functions. Among the most important are Kirszbraun's and Whitney's the- orems. They provide powerful technical tools in many problems of analysis. One way to view these theorems is that they show that there exists an in- terpolation of data with certain properties. In this context they are useful in computer science, e.g. in clustering of data and in dimension reduction. Contents Chapter 1. Extensions of Lipschitz maps 5 1. McShane's theorem 5 2. Kirszbraun's theorem 6 3. Kneser{Poulsen conjecture 8 4. Continuity of extensions 9 Chapter 2. Whitney's extension theorem 13 1. Covering theorems 13 2. Partitions of unity 15 3. Whitney's extension theorem 16 Chapter 3. Minimal Lipschtiz extensions to differentiable functions 19 1. Affine jets 19 2. Extension problem 19 3. Proofs 20 Chapter 4. Ball's extension theorem 27 1. Markov type and cotype 27 2. Statement of the theorem 30 3. Examples 36 Chapter 5. Assesment and references 37 1. Reading list 37 2. Assesment 37 Additional references 37 Bibliography 39 Additionnal Topics and References 41 3 CHAPTER 1 Extensions of Lipschitz maps The first lecture is devoted to study of extensions of Lipschitz maps. Suppose that we are given two metric spaces (X; dX ) and (Y; dY ), then we say that a map f : X ! Y Lipschitz provided that there exists a finite constant L such that dY (f(x1); f(x2)) ≤ LdX (x1; x2) for all x1; x2 2 X: An optimal constant is the Lipschitz constant of f. If L is the optimal constant then we say that f is L-Lipschitz. 1. McShane's theorem In this section we shall consider functions defined on an arbitrary metric space (X; d) with values in the real line. Before we prove the main theorem we shall need simple lemmata. Lemma 1.1. Let L ≥ 0. Suppose that F is a family of L-Lipschitz functions f : X ! R. Then supff j f 2 Fg and infff j f 2 Fg are L-Lipschitz. Proof. Take any f 2 F. Then for any x; y 2 X there is f(x) − f(y) ≤ Ld(x; y): Therefore f(x) ≤ Ld(x; y) + supff(y) j f 2 Fg: Taking the supremum on the left-hand side proves that supff j f 2 Fg is L- Lipschitz. The proof for infff j f 2 Fg follows analogous lines. Lemma 1.2. Let y 2 X. Then the function d(·; y): X ! R is 1-Lipschitz. Proof. This follows immediately from the triangle inequality: jd(x1; y) − d(x2; y)j ≤ d(x1; x2): The following is a theorem of McShane, see [13]. Theorem 1.3 (McShane). Suppose that A ⊂ X and that f : A ! R is a Lipschitz function. Then there exists an extension of f, i.e. a function f~: X ! R ~ such that fjA = f, with with the Lipschitz constant as that of f. Moreover, there exists the smallest and the greatest functions f1 and f2 respec- tively that satisfy these properties. They are given by the formulae f1(x) = supff(a) − Ld(x; a) j a 2 Ag for x 2 X and f2(x) = infff(a) + Ld(a; x) j a 2 Ag for x 2 X: 5 6 1. EXTENSIONS OF LIPSCHITZ MAPS The proof resembles the proof of the Hahn{Banach theorem. Proof. Let L be the Lipschitz constant of f. We want to find an L-Lipschitz f~ such that for x 2 X; a 2 A (1.1) jf~(x) − f(a)j ≤ Ld(x; a): Observe that this already implies that f~ is an extension of f. Now, condition (1.1) is equivalent to that for all a1; a2 2 A and x 2 X there is ~ (1.2) f(a1) − Ld(x; a1) ≤ f(x) ≤ f(a2) + Ld(x; a2): From Lemma 1.2 we infer that functions x 7! f(a1) − Ld(x; a1) and x 7! f(a2) + Ld(a2; x) are both L-Lipschitz. Lemma 1.1 tells us that both f1(x) = supff(a) − Ld(x; a) j a 2 Ag and f2(x) = infff(a) + Ld(a; x) j a 2 Ag are L-Lipschitz. Function f1 satisfies the left-hand side inequality of (1.2) while f2 satisfies the right-hand side inequality of (1.2). Also, as f is L-Lipschtiz f1 ≤ f2. Moreover, if x 2 A, putting a = x in the formulae for f1; f2 yields f(a) ≤ f1(a) ≤ f2(a) ≤ f(a): Thus we have here equalities. The proof is complete. 2. Kirszbraun's theorem Here we consider more delicate question of extending Lipschitz maps with values in a vector space. Suppose that we have map f : A ! `1(N), on a subset A ⊂ X of a metric space (X; d) with values in the space `1(N) of bounded sequences. Then applying McShane's theorem to each coordinate of f we may extend it to the entire space X, preserving its Lipschitz constant. However, this is not true for maps with values in arbitrary metric spaces. Another important positive example is provided by the Kirszbraun's theorem, which addresses the problem in the case of Euclidean spaces. We refer the reader to original paper of Kirszbraun [10], proofs of Valentine [18], of Schoenberg [17] and a proof in the discrete setting by Brehm [6]. Theorem 2.1 (Kirszbraun). Suppose that A ⊂ Rn and that f : A ! Rm is a Lipschitz map with respect to Euclidean metrics on A and on Rm. Then there exists an extension f~: Rn ! Rm of f with the same Lipschitz constant. Proof. Observe first that it is enough to consider a situation when f is 1- Lipschitz. Suppose that A = fx1; : : : ; xkg is a finite set. Suppose that x2 = A and ~ ~ n ~ we want to find a point f(x) such that f : A[fxg ! R is 1-Lipschitz and fjA = f. That is we want f~(x) to belong to the set k \ B(f(xi); kx − xik): i=1 The assumption is that kf(xi) − f(xj)k ≤ kxi − xjk for all i; j = 1; : : : ; k. Define a function on Rm by the formula nky − f(x )k o g(y) = max i j i = 1; : : : ; k : kx − xik 2. KIRSZBRAUN'S THEOREM 7 Clearly, it converges to infinity, as y converges to infinity. Therefore there exists n m y0 2 R such that g(y0) = minfg(y) j y 2 R g. Pick indices i1; : : : ; il for which ky0 − f(xij )k (2.1) = g(y0) for j = 1; : : : ; l: kx − xij k We may assume that y0 2 Convff(xij ) j j = 1; : : : ; lg: Indeed, if not, then there exists a separating hyperplane, i.e. a unit vector v 2 Rm such that for all j = 1; : : : ; l there is hv; y0 − f(xij )i > 0 for some > 0. Set y0 = y0 − v. Let P be the orthogonal projection onto the space perpendicular to v. Then 0 2 2 2 2 ky0 − f(xij )k = kP (y0 − f(xij )k + (hv; y0 − f(xij )i − ) < ky0 − f(xij )k : This contradicts the fact that at y0 the minimum of g is attained. Therefore there exist non-negative real numbers λi1 ; : : : ; λil that sum up to one such that l X y0 = λij f(xij ): j=1 Define a random vector X, with values in Rn so that P(X = xij ) = λij for j = 1; : : : ; l: Then y0 = Ef(X) and 2 2 2 (2.2) kX − xk = g(y0) f(X) − Ef(X) : Let X0 be an independent copy of X. Then 2 1 0 2 1 0 2 2 X − X = kX − X k and kf(X) − f(X )k = f(X) − f(X) : E E 2E 2E E E Taking the expectation of (2.2) yields 2 2 2 2 g(y0) E f(X) − Ef(X) = EkX − xk ≥ E X − EX : Therefore, as f is 1-Lipschitz, 1 1 1 kf(X) − f(X0)k2 ≤ kX − X0k2 ≤ g(y )2 kf(X) − f(X0)k2: 2E 2E 0 2E Hence g(y0) ≤ 1, unless f(X) is constant, so that y0 = f(X). In this case g(y0) = 0. We have proven that for any x 2 Rn and any finite set A the interesection of closed balls \ B(f(y); kx − yk) 6= ;: y2A By compactness, such intersection is also non-empty for any infinite set A. We partially order subsets of Rn containing A and admitting a 1-Lipschitz extension of f by inclusion. By the Kuratowski{Zorn lemma, there exists a maximal element Z of this ordering. If Z 6= Rn, then we could have extended the 1-Lipschitz map to an element x2 = Z, contradicting the choice of Z. This completes the proof. 8 1. EXTENSIONS OF LIPSCHITZ MAPS The Kirszbraun theorem holds true also for Hilbert spaces, with the same proof. 3. Kneser{Poulsen conjecture The Kirszbraun theorem admits also another formulation. n Theorem 3.1. Suppose that x1; : : : ; xk 2 R and let r1; : : : ; rk ≥ 0. Suppose that k \ B(xi; ri) 6= ;: i=1 m Then for any points y1; : : : ; yk 2 R such that kyi − yjk ≤ kxi − xjk for all i; j = 1; : : : ; k there is k \ B(yi; ri) 6= ;: i=1 The above theorem says that if we shrink the centres of the balls, with non- empty intersection, then also the intersection of balls with new centres will be non-empty.