Extensions of Functions - Lecture Notes

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 theorems. 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

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 ∈ 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 sup{f | f ∈ F} and inf{f | f ∈ F} are L-Lipschitz. Proof. Take any f ∈ F. Then for any x, y ∈ X there is f(x) − f(y) ≤ Ld(x, y). Therefore f(x) ≤ Ld(x, y) + sup{f(y) | f ∈ F}. Taking the supremum on the left-hand side proves that sup{f | f ∈ F} is L- Lipschitz. The proof for inf{f | f ∈ F} follows analogous lines. Lemma 1.2. Let y ∈ X. Then the function d(·, y): X → R is 1-Lipschitz. Proof. This follows immediately from the triangle inequality:

|d(x1, y) − d(x2, y)| ≤ 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 f|A = f, with with the Lipschitz constant as that of f. Moreover, there exists the smallest and the greatest functions f1 and f2 respectively that satisfy these properties. They are given by the formulae

f1(x) = sup{f(a) − Ld(x, a) | a ∈ A} for x ∈ X and f2(x) = inf{f(a) + Ld(a, x) | a ∈ A} for x ∈ 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 ∈ X, a ∈ A (1.1) |f˜(x) − f(a)| ≤ 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 ∈ A and x ∈ 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) = sup{f(a) − Ld(x, a) | a ∈ A} and f2(x) = inf{f(a) + Ld(a, x) | a ∈ A} 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 ∈ 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 → `∞(N), on a subset A ⊂ X of a metric space (X, d) with values in the space `∞(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 = {x1, . . . , xk} is a finite set. Suppose that x∈ / A and ˜ ˜ n ˜ we want to find a point f(x) such that f : A∪{x} → R is 1-Lipschitz and f|A = 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 | 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 ∈ R such that g(y0) = min{g(y) | y ∈ R }. 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 ∈ Conv{f(xij ) | j = 1, . . . , l}. Indeed, if not, then there exists a separating hyperplane, i.e. a unit vector v ∈ 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 ∈ Rn and any ﬁnite set A the interesection of closed balls \ B(f(y), kx − yk) 6= ∅. y∈A 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 x∈ / 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 ∈ R and let r1, . . . , rk ≥ 0. Suppose that k \ B(xi, ri) 6= ∅. i=1 m Then for any points y1, . . . , yk ∈ 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.

Proof. Let k \ x ∈ B(xi, ri). i=1

Then kx − xik ≤ ri for each i = 1, . . . , k. By the Kirszbraun theorem there exists k k \ \ y ∈ B(yi, kx − xik) ⊂ B(yi, ri). i=1 i=1 . The Kneser–Poulsen conjecture, see e.g. [4], is a form of quantiﬁcation of the above theorem. That is, it says that non only will the shrinked balls have non- empty intersection, but also that the volume of the intersection will be estimated from below by the volume of the intersection of original balls.

n Conjecture 3.2 (Kneser–Poulsen). Suppose that x1, . . . , xk ∈ R and let m r1, . . . , rk ≥ 0. Then for any points y1, . . . , yk ∈ R such that kyi − yjk ≤ kxi − xjk for all i, j = 1, . . . , k there is k k \ \ λ B(xi, ri) ≤ λ B(yi, ri) . i=1 i=1 Moreover k k [ [ λ B(xi, ri) ≥ λ B(yi, ri) . i=1 i=1 Here λ stands for the Lebesgue measure. In [4], the conjecture has been answered in the aﬃrmative for n = m = 2. If n = m and k ≤ n + 1, then in [9] the conjecture has been proven for the volumes of intesections. We include the references [11], [16] for the original statement of the conjecture. 4. CONTINUITY OF EXTENSIONS 9

4. Continuity of extensions

In this section we shall be concerned with the following question. Let (X, dX ) and (Y, dY ) be metric spaces and let A ⊂ X. Suppose that we are given 1-Lipschitz maps v : X → Y and u: A → Y . Does there exist a 1-Lipschitz extensionu ˜ of u such that the uniform distance ofu ˜ to v is equal to the uniform distance from u to v on A? For a thorough discussion of this and related questions we refer the reader to [7]. Let A ⊂ B ⊂ Rn, n ∈ N. We shall prove that given any 1-Lipschitz maps v : Rn → Rm, for m ∈ N, and u: A → Rm, such that sup{ku(x) − v(x)k | x ∈ A} ≤ δ, there exists a 1-Lipschitz extensionu ˜: B → Rm of u, that isu ˜(x) = u(x) for x ∈ A, such that p 2 (4.1) sup{kv(x) − u˜(x)k | x ∈ B} ≤ δ + 2δdv(A, B).

Here by dv(A, B) we denote the number (4.2) sup{kv(x) − v(y)k | x ∈ A, y ∈ B}.

Note that for 1-Lipschitz functions v we have dv(A, B) ≤ diam(B). We shall also give an example of functions u, v such that the bound is attained. This is to say, u, v are such that for any 1-Lipschitz extensionu ˜ of u we have equality in (4.1). Moreover, as we shall show, we cannot hope, in general, for any bound, if dv(A, B) is inﬁnite.

Proposition 4.1 (C.). Let A ⊂ B ⊂ Rn and let m m u: A → R , v : B → R be 1-Lipschitz maps. Assume that ku(x) − v(x)k ≤ δ for x ∈ A. Then there exists a 1-Lipschitz map u˜: B → Rm such that u˜(x) = u(x) for x ∈ A and p 2 kv(x) − u˜(x)k ≤ δ + 2δdv(A, B) for all x ∈ B. 2 2 Proof. Let = δ + 2δdv(A, B). Let us deﬁne a map m h: B × {0} ∪ A × {} → R by the formulae h(x, 0) = v(x) for x ∈ B and h(x, ) = u(x) for x ∈ A. Then h is a 1-Lipschitz map on a subset of Rn+1. Indeed, if x ∈ A and y ∈ B, then kh(y, 0) − h(x, )k2 = kv(y) − u(x)k2 = = kv(y) − v(x)k2 + kv(x) − u(x)k2 + 2hv(y) − v(x), v(x) − u(x)i ≤ 2 2 2 2 ≤ kx − yk + δ + 2δdv(A, B) = kx − yk + . For other points of the domain of h the 1-Lipschitz condition follows from 1- Lipschitzness of u and v. Using Theorem 2.1 we may extend h to a 1-Lipschitz map h˜ : Rn+1 → Rm. Set for x ∈ B,u ˜(x) = h˜(x, ). Thenu ˜ is a 1-Lipschitz extension of u and moreover, for x ∈ B, kv(x) − u˜(x)k = kh˜(x, 0) − h˜(x, )k ≤ . 10 1. EXTENSIONS OF LIPSCHITZ MAPS

The following example [7] illustrates that the answer to the question posed at the begining of the current section is in general negative.

n x+y Example 4.2 (C.). Let m > 1 and let x, y ∈ R , x 6= y, z = 2 . Let a = kx − zk = ky − zk, let δ > 0. Define u: {x, y} → Rm by setting u(x) and u(y) so that ku(x) − u(y)k = kx − yk. Map u defined in this way is 1-Lipschitz. For the definition of v consider the triangle whose vertices are u(x), u(y) and a point, called v(z), such that kv(z) − u(x)k = kv(z) − u(y)k = a + δ. Set v(x), v(y) to be the points on the triangle’s edges containing u(x) and u(y) respectively such that kv(x) − u(x)k = δ and kv(y) − u(y)k = δ. If we define v : {x, y, z} → Rm in this manner, then it is 1-Lipschitz. By Kirszbraun theorem we may extend it to Rn in such a way that the extension is still 1-Lipschitz. We shall call this extension v : Rn → Rm. Moreover, sup{ku(t) − v(t)k | t ∈ A} = δ. Here A = {x, y}. Observe that any 1-Lipschitz extensionu ˜ of u to the point z u(x)+u(y) must satisfyu ˜(z) = 2 . Thus, if we set B = {x, y, z}, then any 1-Lipschitz extensionu ˜ of u to B satisfies p kv(z) − u˜(z)k = δ2 + 2δa. The situation is illustrated below. v(z)

a a

v(x) v(y) δ δ u(x) a u(x)+u(y) a u(y) 2 Note now that 1 r 1 1 a = d (A, B) if δ ≥ a and a = d (A, B) + d (A, B)2 + d (A, B)δ if δ ≤ a. v 4 v 16 v 2 v

This exhibits that the bound (4.1) is indeed sharp, if δ ≥ dv(A, B). Note that δ ≤ a if and only if δ ≤ dv(A, B). Hence we have shown that for all extensionsu ˜ of u we have p 2 sup{kv(z) − u˜(z)k | z ∈ B} = δ + 2δdv(A, B) if δ ≥ dv(A, B) and s 1 r1 sup{kv(z) − u˜(z)k | z ∈ B} = δ2 + δd (A, B) + δ d (A, B)2 + 2d (A, B)δ 2 v 4 v v if δ ≤ dv(A, B). Let us consider the case when the target space is one-dimensional. 4. CONTINUITY OF EXTENSIONS 11

Proposition 4.3 (C.). Let X be a metric space. Let v : X → R be a 1-Lipschitz function. Then for any set A ⊂ X and for any 1-Lipschitz function u: A → R such that for all x ∈ A (4.3) |u(x) − v(x)| ≤ δ, there exists 1-Lipschitz extension u˜: X → R of v such that for all x ∈ X (4.4) |v(x) − u˜(x)| ≤ δ.

Proof. Take any 1-Lipschitz extensionu ˜0 : X → R of u. Existence of such function follows from Theorem 1.3. Deﬁne now 1

u˜(x) =u ˜0(x) ∧ (v(x) + δ) ∨ (v(x) − δ). Employing Lemma 1.1 it is readily verifiable thatu ˜ satisfies the desired properties. The following theorem characterises the situation for Euclidean spaces. Note that the situation in the multi-dimensional case differs strikingly from the one- dimensional case (see [7]). Theorem 4.4 (C.). Let X,Y be real Hilbert spaces such that Y is of dimension at least two. Let v : X → Y be a map. The following conditions are equivalent: i) for any A ⊂ X and for any 1-Lipschitz map u: A → Y there exists a 1-Lipschitz extension u˜: X → Y of u such that for all x ∈ X v(x) − u˜(x) ∈ Convv(z) − u(z) | z ∈ A . ii) for any δ > 0, any A ⊂ X and for any 1-Lipschitz map u: A → Y such that for all x ∈ A kv(x) − u(x)k ≤ δ, there exists 1-Lipschitz extension u˜: X → Y of u such that for all x ∈ X kv(x) − u˜(x)k ≤ δ, iii) v is affine and 1-Lipschitz.

1Here, symbols a ∧ b and a ∨ b stand for minimum and maximum of two real numbers a, b respectively.

CHAPTER 2

Whitney’s extension theorem

In this lecture we shall consider another problem concerning extensions of functions. The exposition is based on [8]. Suppose that f : Rn → R is continuously differentiable. We shall denote by Df the derivative of f. Then for any compact set K ⊂ Rn there is n f(y) − f(x) − Df(x)(y − x) o lim sup | 0 < kx − yk ≤ δ, x, y ∈ K = 0. δ→0+ ky − xk It turns out that this condition for any compact set K ⊂ C ⊂ Rn implies that f, defined on a set C ⊂ Rn, is a restriction of a continuously differentiable function defined on the entire Rn. This is the statement of the Whitney’s extension theorem. Theorem 0.1 (Whitney). Suppose that C ⊂ Rn is a closed set. Suppose that f : C → R and v : C → (Rn)∗ are continuous and that for any compact set K ⊂ C there is n f(y) − f(x) − v(x)(y − x) o lim sup | 0 < kx − yk ≤ δ, x, y ∈ K = 0. δ→0+ ky − xk

Then there exists a continuosly diﬀerentiable extension f˜: Rn → R of f such that Df˜ = v on C.

1. Covering theorems Before we proceed to a proof of the Whintey’s extension theorem we shall construct a suitable covering which shall allow us to construct an extension with the desired properties. 1.1. Vitali’s covering theorem. We shall ﬁrst provide a construction of the Vitali’s covering theorem. If B is a closed ball in Rn we shall deonte by Bˆ a concentric ball of radius ﬁve times the radius of B. If F is a collection of balls we shall denote by Fˆ the family {Bˆ | B ∈ F}.

Definition 1.1. Let A ⊂ Rn. Then a family F of balls in Rn is a covering of A if A ⊂ S F. We shall say that F is a ﬁne covering of A if additionally there is inf{diamB | x ∈ B,B ∈ F} = 0 for each x ∈ A. Theorem 1.2 (Vitali). Let F be any family of non-degenerate closed balls in Rn with bounded radii. Then there exists a countable subfamily G of F of pairwise disjoint balls in F such that [ [ (1.1) F ⊂ Gˆ.

13 14 2. WHITNEY’S EXTENSION THEOREM

Proof. Let D be the upper bound for the radii of balls in F. Define for j = 0, 1, 2,... n 1 1 o F = B ∈ F | D < diamB ≤ D . j 2j 2j−1 Define Gj for j = 0, 1, 2,... in the following manner. Let G0 be a maximal pairwise disjoint subfamily of balls in F0. Suppose that we have already chosen G0,..., Gk−1. Let now Gk be a maximal Sk−1 S pairwise disjoint subfamily of balls in Fk that are disjoint from j=0 Gj. S∞ Define G = j=0 Gj. Then G is a subfamily of balls in F that are pairwise disjoint. We need to prove (1.1). Take any ball B ∈ F. Then B ∈ Fj for some j = 0, 1, 2,... . Then, by the maximality of the considered families, there exists a ball 0 Sj 0 0 1 B ∈ l=0 Gl such that B ∩ B 6= ∅. Then diamB > 2j D. Since B ∈ Fj, diamB ≤ 1 0 ˆ0 2j−1 D, so that diamB < 2diamB . Therefore B ⊃ B. The family G is at most countable by separability of Rn. The proof is complete. 1.2. Whitney’s covering theorem. We refer the reader to [20] for the original formulation of the following theorem. The original formulation deals with cubes in Euclidean spaces. Below we shall provide a formulation that suffices for a proof of the Whitney’s extension theorem and that can be extended to the setting of metric-measure spaces that satisfy the doubling condition.

Theorem 1.3 (Whitney). Let U ⊂ Rn be an open set. Let c ∈ (0, 1) and λ > 0. Then there exists a family F of pairwise disjoint balls in U such that U = Fˆ. For y ∈ U let r(y) = c(1 ∧ dist(y, U c)) and

Sy = {B(x, r) ∈ F | B(x, λr) ∩ B(y, λr(y)) 6= ∅}. Then 1 + λcn1 + λcn #S ≤ λ + (1 + λ) . y 1 − λc 1 − λc

Moreover, for each ball B(x, r) ∈ Sy there is 1 − λc r 1 + λc ≤ ≤ . 1 + λc r(y) 1 − λc Proof. Consider a family of balls F0 = B(x, r(x)) | x ∈ U .

By an application of Vitali’s covering theorem, there exists a countable subfamily F of F0 such that [ U = Fˆ. and the balls in F are pairwise disjoint. Suppose that B(x, λr(x)) ∩ B(y, λr(y)) 6= ∅ for some x, y ∈ U. Then |r(x) − r(y)| ≤ ckx − yk ≤ cλ(r(x) + r(y)) Therefore 1 − λc 1 + λc r(y) ≤ r(x) ≤ r(y). 1 + λc 1 − λc 2. PARTITIONS OF UNITY 15

Moreover 1 + λc B(x, r(x)) ⊂ B y, λ + (1 + λ) r(y) , 1 − λc as 1 + λc kx − yk + r(x) ≤ λ(r(x) + r(y)) + r(x) ≤ λ + (1 + λ) r(y) 1 − λc The fact that the balls in F are pairwise disjoint and the fact that radii of balls in 1−λc F that intersect B(y, λr(y)) are bounded from below by 1+λc r(y) imply that 1 + λcn1 + λcn #S ≤ λ + (1 + λ) . y 1 − λc 1 − λc This follows by a calculation of the Lebesgue measure. 1.3. Besicovitch’s covering theorem. A much more complicated and in- tricate is Besicovitch’s covering theorem. Here we state it without a proof. For a proof we refer the reader to [8].

Theorem 1.4 (Besicovitch). For each n ∈ N there exists a number Nn such that if F is a family of non-degenerate closed balls in Rn with bounded radii and if

A is the set of centres of balls in F, then there exist subfamilies G1,..., GNn ⊂ F of pairwise disjoint balls in F such that

N [n [ A ⊂ Gi. i=1

2. Partitions of unity Using the Whitney’s covering theorem we shall construct a smooth partition of unity, subordinate to the covering.

Lemma 2.1. Let ρ > 1. There exists a smooth function µ: R → R such that 0 ≤ µ ≤ 1, µ(t) = 1 for t ≤ 1, µ(t) = 0 for t ≥ ρ.

Lemma 2.2. Suppose that U is an open set in Rn. Then there exist smooth, ∞ n non-negative functions (vj)j=1 on R such that ∞ ∞ X X C v = 1, Dv = 0 on U and kDv (x)k ≤ for x ∈ U. j j j c(1 ∧ dist(x, U c)) j=1 j=1 Moreover, for any x ∈ U, the number of indices j = 1, 2, 3,... such that the support of vj intersects B(x, ρr(x)) is at most C. Proof. Let c ∈ (0, 1) and ρ > 1, λ = 5ρ. Pick a covering F of U constructed as in Theorem 1.3. Let µ be a function of Lemma 2.1 for ρ. For a ball B ∈ F define kx − xBk n uB(x) = µ , x ∈ R , 5rB where xB is the centre of B and rB is its radius. Then each uB is a smooth function with values in [0, 1], uB = 1 on B(xB, 5rB) and uB = 0 on B(xB, 5ρrB). Moreover C C0 (2.1) kDuB(x)k ≤ ≤ rB r(x) 16 2. WHITNEY’S EXTENSION THEOREM whenever B(x, λr(x))∩B(xB, λrB) 6= ∅, and uB = 0 on B(x, λr(x)) if B(x, λr(x))∩ B(xB, λrB) = ∅. Set X σ = uB. B∈F Then σ is a smooth function on U, as the sum is locally finite, σ ≥ 1 on U and for x ∈ U C kDσ(x)k ≤ . r(x) uB Set vB = σ . Observe that Du u Dσ Dv = B − B , B σ σ2 so the bound on the derivative of vB is satisfied. 3. Whitney’s extension theorem In this section we shall provide a proof of the Whitney’s extension theorem. Proof of Theorem 0.1. Pick a covering F of the complement U of the set C, constructed as in Theorem 1.3 and an associated partition of unity (vB)B∈F of Lemma 2.2. For each centre xB of a ball B ∈ F pick sB ∈ C such that

kxB − sBk = dist(xB,C). Deﬁne f˜: Rn → R by the formula f˜(x) = f(x) for x ∈ C and ˜ X f(x) = f(sB) + v(xB)(x − sB) vB(x) for x ∈ U. B∈F Clearly, f˜ is a smooth function on U with ˜ X Df(x) = f(sB) + v(xB)(x − sB) DvB(x) + vB(x)v(xB) for x ∈ U. B∈F We claim that Df˜ = v on C. Fix a point a ∈ C and let K = C ∩ B(a, 1). Then K is a compact set. Deﬁne for δ > 0 n f(y) − f(x) − v(x)(y − x) o φ(δ) = sup +kv(x)−v(y)k | x, y ∈ K, 0 < kx−yk ≤ δ . ky − xk

X ≤ vB(x) f(sB) − f(a) + v(sB)(x − sB) − v(a)(x − a) ≤ B∈F X X vB(x)|f(sB) − f(a) − v(sB)(a − sB)| + vB(x)|(v(sB) − v(a))(x − a)|. B∈F B∈F 3. WHITNEY’S EXTENSION THEOREM 17

If B(xB, λrB) ∩ B(x, λr(x)) 6= ∅, then

ka − sBk ≤ ka − xBk + kxB − sBk ≤ 2ka − xBk ≤ 2(ka − xk + kx − xBk) ≤

≤ 2(ka − xk + λ(r(x) + rB)) ≤ Ckx − ak, for some constant C, depending on c and ρ. Hence |f˜(x) − f˜(a) − v(a)(x − a)| ≤ Cφ(Ckx − ak)kx − ak. Therefore f˜ is diﬀerentiable at a and Df˜(a) = v(a). We need now show that f˜ has continuous derivative. Fix a ∈ C and x ∈ Rn, kx − ak < 1. If x ∈ C, then kDf˜(x) − Df˜(a)k = kv(x) − v(a)k ≤ φ(kx − ak). Suppose that x ∈ U. Take b ∈ C such that kx − bk = dist(x, C). Then kDf˜(x) − Df˜(a)k = kDf˜(x) − v(a)k ≤ kDf˜(x) − v(b)k + kv(b) − v(a)k. As kb − ak ≤ kb − xk + kx − ak ≤ 2kx − ak, we see that kv(b) − v(a)k ≤ φ(2kx − ak). We need to estimate the ﬁrst summand. We calculate

˜ X kDf(x) − v(b)k = f(sB) + v(sB)(x − sB) DvB(x) + vB(x)(v(sB) − v(b)) ≤ B∈F

X ≤ − f(b) + f(sB) + v(sB)(b − sB) DvB(x) + B∈F

X X + (v(sB) − v(b))(x − b) DvB(x) + vB(x)(v(sB) − v(b)) ≤ B∈F B∈F C X X ≤ φ(kb − s k)kb − s k + kx − bk + φ(kb − s k). r(x) B B B B∈Sx B∈Sx Now, kx − bk ≤ kx − ak < 1, so that r(x) = c(1 ∧ kx − bk) < c. Moreover, if 0 B(xB, λrB) ∩ B(x, λr(x)) 6= ∅, then rB ≤ Cr(x) ≤ C . Therefore,

kb − sBk ≤ kb − xk + kx − xBk + kxB − sBk ≤ 1 1 ≤ r(x) + λ(r(x) + r ) + r ≤ Cr(x) = Ckx − bk ≤ Ckx − ak. c B c B We infer that kDf˜(x) − d(b)k ≤ Cφ(ckx − ak) and in turn kDf˜(x) − Df˜(a)k ≤ Cφ(ckx − ak). This proves the asserted continuity and completes the proof of the theorem. Remark 3.1. The constants in the proof of the above theorem, in Theorem 1.3 and in Lemma 2.2 depend on ρ > 1 and c ∈ (0, 1), which might be picked arbitrarily.

CHAPTER 3

Minimal Lipschtiz extensions to diﬀerentiable functions

In this chapter we shall provide another proof of the Whitney’s extension theorem. The proof will be in the spirit of the proof of Kirszbraun’s theorem. Our exposition is based on [12].

1. Affine jets The formulation that we shall deal with considers fields of affine jets. Such a field is an association to any point of its domain of an affine, real-valued function. Let A ⊂ B ⊂ Rn. Suppose that we are given a field of affine jets T on A. That is, n T is a map A 3 a 7→ Ta, where Ta is an affine function on R . We say that a field U, defined on B, extends T whenever for any a ∈ A, Ta = Ua. A differentiable function u on Rn extends a field T provided that the field of Taylor expansions of u extends T .

2. Extension problem

Suppose that T is a field of affine jets defined on some set A ⊂ Rn. The Lipschitz extension problem is to find necessary and sufficient condition on the field T that will ensure that T admits an extension to a differentiable function on Rn which has Lipschitz derivative. For a field T defined on A ⊂ Rn define T (y) − T (y) Γ1(T ) = 2 sup a b | a 6= b, a, b ∈ A, y ∈ n . ka − yk2 + kb − yk2 R We shall show that if T is the field of Taylor expansions of a differentiable function u, then Γ1(T ) is equal to the Lipschitz constant of the derivative Du of u. Let us now turn to the minimal Lipschitz extension problem. A differentiable function u on Rn is said to be a minimal Lipschitz extension of a field T if the field of its Taylor expansions extends T and for any other extension v of T , the Lipschtiz constant of Dv is at least as large as that of Du. We will show that Γ1(T ) is equal to the infimum of all Lipschitz constants of derivatives of functions extending T . The aim of this lecture is to prove the following theorem. Theorem 2.1. Γ1 is a unique functional such that: i) if U extends T , then Γ1(U) ≥ Γ1(T ), ii) if U is defined on Rn and Γ1(U) < ∞, then the function defined by the formula n u(x) = Ux(x) for x ∈ R is differentiable and its derivative is Lipschitz,

19 20 3. MINIMAL LIPSCHTIZ EXTENSIONS TO DIFFERENTIABLE FUNCTIONS iii) if u is differentiable function on Rn with Lipschitz derivative, then Γ1(U) is equal to the Lipschitz constant of Du, where U is the field of jets of Taylor expansions of u, iv) for any field T such that Γ1(T ) < ∞, there exists an extension of T to a field of affine jets defined on Rn with Γ1(U) = Γ1(T ). Corollary 2.2. If T is a filed of affine jets satisfying Γ1(T ) < ∞, then there exists a differentiable function u defined on Rn, which is a minimal extension of T and such that its derivatve has Lipschitz constant equal to Γ1(T ) The above theorem holds true also in the case of Hilbert spaces, separable or not. The previous proof of the Whitney’s extension theorem does not convey to this setting. The proof of i) is immediate, as well as the proof of uniqueness. The proof of the main part of the theorem, iv), relies on an approach similar to the one employed in the proof of Kirszbraun’s theorem.

3. Proofs In what follows we shall denote by H a Hilbert space equipped with scalar product h·, ·i. For any ﬁeld T of aﬃne jets we write

Ta(x) = ua + hDau, x − ai for all a in the domain of T and x ∈ H, for some ua ∈ R and some Dau ∈ H. Definition 3.1. For a field of affine jets T defined on A ⊂ H we define its Lipschitz constant Γ1(T ) by the formula T (y) − T (y) Γ1(T ) = 2 sup a b | y ∈ H, a 6= b, a, b, ∈ A , ka − yk2 + kb − yk2 whenever A has at least two elements and Γ1(T ) = 0 if has not. Proposition 3.2. If A has at least two elements, then

1 nq 2 2 o Γ (T ) = sup Aa,b + Ba,b + |Aa,b| | a 6= b, a, b ∈ A , where 2(u − u ) + hD u + D u, b − ai D u − D u A = a b a b ,B = a b . a,b ka − bk2 a,b ka − bk Proof. For a 6= b and y ∈ H denote T (y) − T (y) g (y) = a b . a,b ka − yk2 + kb − yk2 We need to prove that for any a 6= b there is

q 2 2 sup{ga,b(y) | y ∈ H} = Aa,b + Ba,b + |Aa,b|. For any y ∈ H set 2 a + b t = y − . ka − bk 2 Then 1 1 T (y) − T (y) = ka − bk2A + ka − bkhD u − D u, ti a b 2 a,b 2 a b 3. PROOFS 21 and 1 ka − yk2 + kb − yk2 = ka − bk21 + ktk2. 2 Thus hDau−Dbu,ti |Aa,b + | sup |g (y)| | y ∈ H = sup ka−bk | t ∈ H . a,b 1 + ktk2

Suppose that Dau = Dbu. Then the maximum is attained at t = 0 and is equal to |Aa,b|. Let now Dau 6= Dbu. Let D u − D u e = a b . kDau − Dbuk Then we may write t = αe + f for some α ∈ R and f perpendicular to e. Then |A + αB | sup |g (y)| | y ∈ H = sup a,b a,b | α ∈ . a,b 1 + α2 R An elementary calculation completes the proof. Remark 3.3. In the above proof for computation of the supremum it suffices to take |α| ≤ 1, so that ktk ≤ 1 also suffices. Thus T (y) − T (y) a + b ka − bk Γ1(T ) = 2 sup a b | y ∈ B , . ka − yk2 + kb − yk2 2 2 Proposition 3.4. Let u be differentiable function on H with Lipschitz derivative Du. Then Γ1(U) is equal to the Lipschitz constant of Du. Here U is the field of Taylor expansions of u. Proof. Let L denote the Lipschitz constant of Du. For x, y ∈ H we have

Ux(y) = ux + hDxu, y − xi. We may write Z 1 ux − uy = hDy+t(x−y)u, x − yidt. 0 For any x, y, z ∈ H there is Z 1 1 2 |Ux(z)−uz| = |ux−uz+hDxu, z−xi| = hDz+t(x−z)u−Dxu, x−zidt ≤ Lkz−xk . 0 2 Therefore 1 |U (z) − U (z)| ≤ |U (z) − u | + |u − U (z)| ≤ L(kx − zk2 + kx − yk2). x y x z z y 2 Hence Γ1(U) ≤ L. Conversely, by Proposition 3.2, there is kD u − D uk Γ1(U) ≥ B = x y x,y kx − yk 1 for all x, y ∈ H, x 6= y. Therefore Γ (U) ≥ L. Proof of iv) of Theorem 2.1. Let T be a field of affine jets such that 1 1 1 Γ (T ) < ∞. Let K = 2 Γ (T ) and let A be the domain of T , i.e., the set where T is defined. As in the proof of the Kirszbraun theorem, it is enough to extend T to 22 3. MINIMAL LIPSCHTIZ EXTENSIONS TO DIFFERENTIABLE FUNCTIONS a point x∈ / A. Thus, we need to show that there exist ux ∈ R,Dxu ∈ H such that for all a ∈ A and all y ∈ H there is T (y) − T (y) (3.1) −K ≤ x a ≤ K, kx − yk2 + ka − yk2 where Tx(y) = ux + hDxu, y − xi. If K = 0, we see that T is constant on A, so that the extension to x is trivial. In what follows we assume that K > 0. Then (3.1) is, by Proposition 3.2, equivalent to q 2 2 |Aa,x| + Aa,x + Ba,x ≤ 2K for any a ∈ A. In other words B2 (3.2) |A | ≤ K − a,x . a,x 4K Note that ux appears only in Aa,x, so that it will be possible to eliminate it in the following way. The condition (3.2) is equivalent to that for any a ∈ A there is 1 K 1 u ≤ u + hD u + D u, x − ai + ka − xk2 − kD u − D uk2 x a 2 a x 2 8K a x and that for any b ∈ A there is 1 K 1 u ≥ u + hD u + D u, x − bi − kb − xk2 + kD u − D uk2. x b 2 b x 2 8K b x There exists such ux if and only if for all a, b ∈ A there is 1 K 1 u + hD u + D u, x − bi − kb − xk2 + kD u − D uk2 ≤ b 2 b x 2 8K b x 1 K 1 ≤ u + hD u + D u, x − ai + ka − xk2 − kD u − D uk2. a 2 a x 2 8K a x We need to prove existence of Dx that satisfies the above inequality. The above inequality, for any a, b ∈ A, may be restated that some quadratic form in Dxu is non-positive. More precisely, the above is equivalent to 2 (3.3) kDxu − Va,bk ≤ αa,b + βa,b for all a, b ∈ A, where 1 V = (D u + D u) + K(b − a), a,b 2 a b 1 α = 4K(u − u ) + 2KhD u + D u, b − ai − kD u − D uk2 + 2K2ka − bk2, a,b a b a b 2 a b 1 2 β = (D u − D u) + K(2x − a − b) . a,b 2 a b Observe that 1 α = 4KA − B2 + 4K2ka − bk2 ≥ 0, a,b 2 a,b a,b p by the definition of K, see (3.2). Set ra,b = αa,b + βa,b. Then (3.3) becomes

kDxu − Va,bk ≤ ra,b for all a, b ∈ A, T or, in other words, Dxu ∈ a,b∈A B(Va,b, ra,b). By compactness it is enough to veriify this condition for any ﬁnite subset of A. Thus, without of generality, we assume that A is ﬁnite. In what follows we shall need two lemmata below. 3. PROOFS 23

For a, b, c, d ∈ A and y ∈ H set 2 2 2 2 Φ((a, b), (c, d)) = ra,b + rc,d − kVa,bk − kVc,dk and 1 1 X = D u + K(y − a),Y = D u + K(b − y). a,y 2 a b,y 2 b Lemma 3.5. For any a, b, c, d ∈ A and y ∈ H there is

Φ((a, b), (c, d)) ≥ −4hXa,y,Yd,yi − 4hXc,y,Yb,yi.

Proof. We write αa,b + αc,d = δ11 + δ12 + δ2 + δ3 + δ4, where 1 δ = 4K(u − u ) + 2KhD u + D u, d − ai − kD u − D uk2, 11 a d a d 2 a d 1 δ = 4K(u − u ) + 2KhD u + D u, b − ci − kD u − D uk2, 12 c b c d 2 c b δ2 = 2K hDau, b − di + hDbu, c − ai + hDcu, d − bi + hDdu, a − ci , 1 δ = − kD u − D uk2 − kD u − D uk2 + kD u − D uk2 + kD u − D uk2, 3 2 a b c d a d c b 2 2 2 δ4 = 2K (ka − bk + kc − dk ). Now, using (3.2) for pairs (a, d) and (b, c), we infer that 2 2 2 δ11 + δ12 ≥ −2K kd − ak + kb − ck . Therefore αa,b + αc,d ≥ δ2 + δ3 + ρ1, where 2 2 2 2 2 ρ1 = 2K ka − bk + kc − dk − ka − dk − kb − ck . Observe that ρ1 + δ2 + δ3 = 4 − hXa,y,Yd,yi − hXc,y,Yb,yi + hXa,y,Yb,yi + hXc,y,Yd,yi . Since 2 2 βa,b + βc,d = kXa,y − Yb,yk + kXc,y − Yd,yk , and 2 2 2 2 −kVa,bk − kVc,dk = −kXa,y + Yb,yk − kXc,y + Yd,xk , we have 2 2 βa,b + βc,d − kVa,bk − kVc,dk = −4 hXa,y,Yb,yi + hXc,y,Yd,yi . This completes the proof.

Lemma 3.6. Suppose that there is a, b ∈ A such that ra,b = 0. Then [ B(Vc,d, rc,d) 6= ∅. c,d∈A Proof. Let c, d ∈ A. Then 2 2 2 (3.4) ra,b + rc,d − kVa,b − Vc,dk = Φ((a, b), (c, d)) + 2hVa,b,Vc,di.

The condition ra,b = 0 implies that

Xa,y = Yb,y and Va,b = Xa,y + Yb,y. Lemma 3.5 implies that Φ((a, b), (c, d)) ≥ −4 hXa,y,Yd,yi + hXc,y,Yb,yi = −2hVc,d,Va,bi. 24 3. MINIMAL LIPSCHTIZ EXTENSIONS TO DIFFERENTIABLE FUNCTIONS

By (3.4) and the above we see that

kVa,b − Vc,dk ≤ rc,d for all c, d ∈ A.

Hence Va,b belongs to the intersection in the statement of the lemma.

By the lemma above, it is now enought to handle situation when ra,b > 0 for all a, b ∈ A. Proof of iv) of Theorem 2.1. From the proof of Kirszbraun’s theorem it follows that if λ ≥ 0 is a minimal number such that \ B(Va,b, λra,b) 6= ∅, a,b∈A then the intersection contains a single element Vm which belongs to the convex hull of Va,b, with (a, b) in the set

E = {(a, b) ∈ A × A | kVm − Va,bk = λra,b}.

Therefore there exist non-negative ξa,b that sum up to one such that X (3.5) Vm = ξa,bVa,b. a,b∈E We want to prove that λ ≤ 1. By (3.5) we see that X (3.6) 0 = ξa,bξc,dhVm − Va,b,Vm − Vc,di. (a,b),(c,d)∈E Note that 2 2 2 kVc,d − Va,bk = kVm − Va,bk + kVm − Vc,dk − 2hVm − Va,b,Vm − Vc,di. Therefore for (a, b), (c, d) ∈ E we have 2 2 2 2 2 kVc,d − Va,bk = λ ra,b + λ rc,d − 2hVm − Va,b,Vm − Vc,di.

Multipying these equations by respective coeﬃcients ξa,bξc,d and employing (3.6) we get X 2 2 2 2 2 0 = ξa,bξc,d − kVc,d − Va,bk + λ ra,b + λ rc,d . (a,b),(c,d)∈E Let us denote X 2 2 2 ∆ = ξa,bξc,d − kVc,d − Va,bk + ra,b + rc,d . (a,b),(c,d)∈E Then 2 X 2 2 ∆ = (1 − λ ) ξa,bξc,d(ra,b + rc,d). (a,b),(c,d)∈E The assumption on the radii implies that X 2 2 ξa,bξc,d(ra,b + rc,d) > 0. (a,b),(c,d)∈E To complete the proof we thus need to show that ∆ ≥ 0. Observe that 2 X ∆ = 2kVmk + ξa,bξc,dΦ((a, b), (c, d)). (a,b),(c,d)∈E 3. PROOFS 25

Set X 1 X X = ξ D u + K(y − a) = ξ X a,b 2 a a,b a,x (a,b)∈E (a,b)∈E and X 1 X Y = ξ D u + K(b − y) = ξ Y . a,b 2 b a,b b,y (a,b)∈E (a,b)∈E Then X X + Y = ξa,bVa,b, (a,b)∈E so that 2 2 kX + Y k = kVmk . By Lemma 3.5 it follows that X ξa,bξc,dΦ((a, b), (c, d)) ≥ −8hX,Y i. (a,b),(c,d)∈E Therefore ∆ ≥ 2kX + Y k2 − 8hX,Y i = 2kX − Y k2 ≥ 0. This completes the proof.

CHAPTER 4

Ball’s extension theorem

In this lecture we shall provide an extension theorem in the situations where it is not possible to preserve the Lipshitz constant.

Example 0.1. An example of such situation is the case of the space `1. Suppose 1 we consider a map f : {x1, x2, x3} → ` , deﬁned on the set {x1, x2, x3} of vertices of an equilateral triangle of sidelength 2 by the formulae

f(xi) = ei for i = 1, 2, 3, 3 1 where (ei)i=1 are basis elements of ` . Then f is an isometry. However, it does not extend to a 1-Lipschitz map defined on {x , x , x , c} where c is the barycentre of 1 2 3 √ √ 3 3 {x1, x2, x3}. For if such f(c) existed, then kf(c)−eik1 ≤ . Then |f(c)i −1| ≤ , √ √ 3 √3 3 3 2 3 so f(c)i ≥ 1− 3 for i = 1, 2, 3. We get a contradiction 3 ≥ kf(c)−e1k1 ≥ 2− 3 . Therefore in such situations we cannot hope to preserve the Lipschitz constant, and we need to allow for more flexibility. Note that the techniques of extension of a function point by point do not apply readily. We need to come up with another approach. Our exposition will be based on [1] and [15]. The approach in this lecture will be analgous to the one used in the proof of the classical theorem of Maurey concerning extension of linear operators between Banach spaces enjoying finite type or cotype. The type and cotype describe the behaviour of sums of independent random variables in Banach spaces. The analogues of type and cotype introduced in [1] describe the behaviour of Markov chains in metric spaces. Let us recall that the theorem of Maurey says that whenever X,Y are Banach spaces, X having type 2 with constant T2(X) and Y having cotype 2 with constant C2(Y ), then any bounded operator T : Z → Y on a subspace Z ⊂ X admits an extension to a bounded operator on X with norm at most T2(X)C2(Y )kT k. Let us note a connection of these topics with the Ribe programme, proposed by Bourgain [5]; see also [2] and [14].

1. Markov type and cotype Let us recall the deﬁnitions of type and cotype.

Definition 1.1. Let X be a Banach space. Let 1, 2,... be independent Bernoulli random variables. Let p ∈ (1, 2]. We say that X has type p whenever there is K < ∞ such that for all n ∈ N and all x1, x2,... ∈ X there is n p n X p X p E ixi ≤ K kxik . i=1 i=1

27 28 4. BALL’S EXTENSION THEOREM

Similarly, X has cotype q ∈ [2, ∞), whenever for some constant L, all n ∈ N and all sequences x1, x2,... ∈ X there is n q n X q X q E ixi ≥ L kxik . i=1 i=1 In what follows we shall denote n n n X o ∆ = (x1, . . . , xn) ∈ [0, 1] | xi = 1 . i=1 We think of ∆ as the space of probabilites on the set {1, . . . , n}. We say that an n × n matrix A is stochastic if it has non-negative entries and all of its rows add up to one. Given π ∈ ∆ a stochastic matrix A is reversible relative to π if for all i, j = 1, . . . , n there is πiaij = πjajii. ∞ These notions stem from the Markov chains theory. A sequence (Xi)i=0 of random variables with values in {1, . . . , n} is a Markov chain provided that

P(Xk+1 = tk+1 | Xk = tk,...,X0 = t0) = P(Xk+1 = tk+1 | Xk = tk) for all t0, . . . , tk+1 ∈ {1, . . . , n}. A Markov chain is called stationary if for all k, l ∈ N and all t0, . . . , tk+1 ∈ {1, . . . , n} there is

P(Xk+1 = tk+1,Xk = tk,...,X0 = t0) = P(Xk+l+1 = tk+1,Xk+l = tk,...,Xl = t0) Then a stochastic matrix A is its transition matrix if for all i, j ∈ {1, . . . , n} there is aij = P(Xk+1 = i | Xk = j) for all k ∈ N. A measure π on the state space {1, . . . , n} is a stationary measure of the chain if any Xj is distributed according to π, for j ∈ N. A stationary Markov chain with stationary measure π and transition matrix A is reversible if the detailed balance condition holds true. That is if

πiaij = πjaji for i, j = 1, . . . , n. Definition 1.2. Let (X, d) be a metric space. We say that X has a Markov type p ∈ (0, ∞) with constant M if for every stationary and reversible Markov chain ∞ (Xi)i=0 on {1, . . . , n} and any function f : {1, . . . , n} → X, any k ∈ N, there is p p p Ed(f(Xk), f(X0)) ≤ M kEd(f(X1), f(X0)) . An optimal constant M is the Markov type-p constant of X. We denote it by Mp(X). Clearly, any metric space has Markov type 1 with constant 1, by the triangle inequality. Remark 1.3. The above deﬁnition may be reformulated as follows. Metric space X has Markov type p ∈ (0, ∞) with constant M if for any π ∈ ∆, any n × n stochastic matrix A reversible relative to π and any x1, . . . , xn there is n n X k p p X p πiAijd(xi, xj) ≤ M k πaijd(xi, xj) i,j=1 i,j=1 for all k ∈ N. 1. MARKOV TYPE AND COTYPE 29

Lemma 1.4. Let H be a Hilbert space. Then H has Markov type 2 with constant M2(H) = 1. Proof. Let π ∈ ∆ and let A be an n × n stochastic matrix reversible relative to π. We need to prove that n n X k 2 X 2 (1.1) πAijkxi − xik ≤ t πaijkxi − xik . i,j=1 i,j=1 Clearly, we may assume that H is one-dimensional, as the general case follows from this one. Consider the space L2(π); i.e. the space Rn with scalar product deﬁned by the formula n X n hv, wi = viwiπi for v, w ∈ R . i=1 Then the right-hand side of (1.1) may be written as, by reversibility and stochasticity, n n X 2 2 X 2 k πiaij xi + xj − 2xixj = 2k πiaijxi − 2hAx, xi = 2kh(Id − A)x, xi. i,j=1 i,j=1 Similarly we express the left-hand side, with A replaced by Ak. Then the inequality we are to prove is hk(Id − A) − (Id − Ak)x, xi ≥ 0. The reversibility of A is equivalent to A being self-adjoint on L2(π). Therefore it is diagonalisable and has real eigenvalues. It is therefore enough to prove that for any eigenvalue λ of A there is k(1 − λ) ≥ 1 − λk.

k−1 1−λk It is enough to prove that λ ≤ 1, as then k ≥ 1 + λ + ... + λ = 1−λ . For this we employ stochasticity of A: n n n n X X X X kAykL1(π) = πi aijyj ≤ πiaij|yj| = πjaji|yj| = kykL1(π). i=1 j=1 i,j=1 i,j=1

Thus |λ| ≤ 1 and the lemma is proven. For the extension theorem we shall also need a dual notion of Markov cotype, analogously to the situation of extension of operators acting on Banach spaces.

Definition 1.5. A metric space (X, d) has metric Markov cotype q ∈ (0, ∞) with constant N if for every n, k ∈ N, every π ∈ ∆, any stochastic n × n matrix A reversible relative to π, any x1, . . . , xn ∈ X there exist y1, . . . , yn ∈ X such that

n n n k X X X 1 X π d(x , y )q + k π a d(y , y )q ≤ N q π Al d(x , x )q. i i i i ij i j i k ij i j i=1 i,j=1 i,j=1 l=1 An optimal constant is the metric Markov cotype-q constant of X and is denoted by Np(X). 30 4. BALL’S EXTENSION THEOREM

Remark 1.6. In the language of Markov chains, the above condition may be rewritten in the following form: for any f : {1, . . . , n} → X, any stationary and ∞ reversible Markov chain (Xi)i=0 there exists a function g : {1, . . . , n} → X such that for all k ∈ N there is k 1 X d(f(X ), g(X ))q + k d(g(X ), g(X ))q ≤ N q d(f(X ), f(X ))q. E 0 0 E 1 0 k E l 0 l=1

Remark 1.7. Note the apperance of function g, or points y1, . . . , yn ∈ X in the above definition. If g = f, or yi = xi for i = 1, . . . , n, then we would simply get a reverse inequality to the one defining the Markov type, but with Cesàroaverage, for technical reasons. However, this introduction is necessary, as otherwise the inequality would be false, by consideration of constant chain distributed on two points of X. Definition 1.8. Let (X, d) be a metric space and let P(X) denote the space of finitely supported probability measures on X. We say that (X, d) is W p-barycentric with constant Γ if there exists a map B : P(X) → X, called the barycentre map, such that for all x ∈ X there is B(δx) = x and n n p n X X p X p d B λiδxi ,B λiδyi ≤ Γ λid(xi, yi) i=1 i=1 i=1 for all xi, yi ∈ X, i = 1, . . . , n and all non-negative λi, i = 1, . . . , n that sum up to one. Note that Banach spaces are 1-barycentric with constant 1. The barycentre map is simply the expectation of the measure.

2. Statement of the theorem In this section we shall state and prove a generalisation by Naor and Mendel,see e.g. [15], of the Ball’s extension theorem of [1].

Theorem 2.1. Let (X, dx) and (Y, dY ) be two metric spaces and let p ∈ (0, ∞). Suppose that X has metric Markov type p with constant Mp(X) and Y has metric p Markov cotype p with constant Np(Y ) and that Y is W -barycentric with constant Γ. Suppose that Z ⊂ X and that f : Z → Y is a Lipschitz map with Lipschitz constant L. Then for any finite set S ⊂ X there is a map fS : S → Y such that fS|S∩Z = fS∩Z and the Lipschitz constant of fS is at most CΓMp(X)Np(Y )L, where C is a universal constant. Remark 2.2. If Y is a reflexive Banach space, then the finite extension property ensures that there also exist an extension to entire space X.

Lemma 2.3. Suppose that (X, dX ) and (Y, dY ) are two metric spaces such that Y is W p-barycentric with constant Γ, Z ⊂ X, f : Z → Y and > 0. Suppose also that there exists a constant K > 0 such that for any n ∈ N, any x1, . . . , xn ∈ X and every symmetric n × n matrix H with non-negative entries H there exists Φ : {x1, . . . , xn} → Y such that: i) H Φ |{x1,...,xn}∩Z = f{x1,...,xn}∩Z , 2. STATEMENT OF THE THEOREM 31

ii) n n X H H p p X p hijdY (Φ (xi), Φ (xj)) ≤ K L hijdX (xi, xj) , i,j=1 i,j=1 where L is the Lipschitz constant of f.

Then there exists F : {x1, . . . , xn} → Y , which agrees with f on {x1, . . . , xn} ∩ Z and whose Lipschitz constant is at most (1 + )ΓKL.

Proof. Let us ﬁx points x1, . . . , xn ∈ X. Consider sets C,D of n×n symmetric matrices given by the formulae n n o C = dY (Φ(xi), Φ(xj)) i,j=1 | Φ: {x1, . . . , xn} → Y agrees with f on {x1, . . . , xn} ∩ Z , D = M | M has non-negative entries . Let E be the closed convex hull of the sum C + D. Consider matrix T with entries p p p tij = K L dX (xi, xj) for i, j = 1, . . . , n. We need to prove that T ∈ E. Indeed, if T ∈ E, then there are non-negative λ1, . . . , λm that add up to one and maps Φ1,..., Φm : {x1, . . . , xm} → Y , which all agree with f on {x1, . . . , xn} ∩ Z and such that for each i, j = 1, . . . , n there is m X p p p p λkdY (Φk(xi), Φk(xj)) ≤ (1 + ) K L dX (xi, xj) . k=1 Pm Deﬁne now µi = k=1 λkδΦk(xi) for each i = 1, . . . , n and F : {x1, . . . , xn} → Y by the formula F (xi) = B(µi), where B is the barycentre map of Y . Then F agrees with f on {x1, . . . , xn} ∩ Z, as all Φ1,..., Φm did. Moreover, m p p X p p p p p p dY (F (xi),F (yi)) ≤ Γ λkdY (Φk(xi), Φk(xj)) ≤ (1 + ) Γ K L dX (xi, xj) . k=1 This is the desired inequality. We shall now prove that T ∈ E. If not, then by the Hahn–Banach theorem n there would exist a symmetric matrix H = (hij)i,j=1) such that n n X n X o hijtij < inf hijmij | M ∈ E . i,j=1 i,j=1 ∗ ∗ Testing the above inequality by a matrix ρ(eiej + ejei ) for large ρ, we see that necessarily there is hij ≥ 0 for all i, j = 1, . . . , n. Moreover, as C ⊂ E, for any Φ: {x1, . . . , xn} → Y that agrees with f on {x1, . . . , xn} ∩ Z there is n n p p X p X p K L hijdX (xi, xj) < hijdY (Φ(xi), Φ(xj)) i,j=1 i,j=1 which stands in contradiction with the assumption. Before we pass to the proof of the Ball’s extension theorem, let us prove the following lemma.

Lemma 2.4. Let m, n ∈ N, p ≥ 1. Let C be an n × n stochastic matrix which is reversible relative to some π ∈ ∆. Let also B be an m × n stochastic matrix. Then 32 4. BALL’S EXTENSION THEOREM for any metric space (X, dX ), any z1, . . . , zm ∈ X there are w1, . . . , wn ∈ X such that m n m n p X ∗ p n X X p X po 3 (B DπCB)ijdX (zi, zj) ≥ max πibirdX (wi, zr) , πicijdX (wi, wj) . i,j=1 i=1 r=1 i,j=1

Here Dπ denotes the diagonal matrix with elements π1, . . . , πn on the diagonal. ∞ Proof. Let f : {z1, . . . , zm} → ` be an isometric embedding, given e.g. by m the formula f(z) = (d(z, zi))i=1. Deﬁne m X yi = birf(zr) for i = 1, . . . , n. r=1

Let wi ∈ {z1, . . . , zm} be such that kyi − wik`∞ = min kyi − f(zj)k`∞ | j = 1, . . . , m . By the triangle inequality and by Jensen’s inequality we have p p p−1 p p p dX (wi, wj) = kf(wi)−f(wj)k`∞ ≤ 3 kf(wi)−yi)k`∞ +kf(wj)−yjk`∞ +kyi−yjk`∞ . By the reversibility and stochasticity of C we get n X p πicijidX (wi, wj) ≤ i,j=1 (2.1) n n p−1 X p p−1 X p ≤ 3 πicijkyi − yjk`∞ + 2 · 3 πikyi − f(wi)k`∞ . i,j=1 i=1 Therefore n n m p X p X X πicijkyi − yjk`∞ = cij birbjs f(zr) − f(zs) ≤ `∞ i,j=1 i,j=1 r,s=1 n m m X X p X ∗ p ≤ πicijbirbjskf(zr) − f(zs)k`∞ = (B DπCB)rsdX (zr, zs) . i,j=1 r,s=1 r,s=1

To bound the second sum in (2.1) we take into account that kyi − f(wi)k`∞ ≤ kyi − f(zr)k`∞ for any i = 1, . . . , n and r = 1, . . . , m. Moreover, as matrices C,B are stochastic, m X (CB)ir = 1 for each i = 1, . . . , n. r=1 Thus, by Jensen’s inequality, n n m X p X X p πikyi − f(wi)k`∞ = πi(CB)irkyi − f(wi)k`∞ ≤ i=1 i=1 r=1 n m n m m p X X p X X X ≤ πi(CB)irkyi − f(zr)k`∞ = πi(CB)ir bis f(zs) − f(zr) ≤ `∞ i=1 r=1 i=1 r=1 s=1 n m m X X p X ∗ p ≤ πi(CB)irbiskf(zs) − f(zr)k`∞ = (B DπCB)rsdX (zr, zs) . i=1 r,s=1 r,s=1 This proves the bound for the second term in the inequality in the statement of the lemma. We now pass to a proof of the bound for the ﬁrst term. 2. STATEMENT OF THE THEOREM 33

By the triangle inequality and Jensen’s inequality we have for each i, j = 1, . . . , n p p−1 p p p dX (wi, zr) ≤ 3 kf(wi) − yik`∞ + kyj − f(zr)k`∞ + kyi − yjk`∞ . Since C is stochastic, we get n m n m X X p X X p πibirdX (wi, zr) = πibircijdX (wi, zr) . i=1 r=1 i,j=1 r=1 Observe that the quantities n m n X X p X p πibircijkf(wi) − yik`∞ = πikf(wi) − yik`∞ i,j=1 r=1 i=1 and n m n X X p X p πibircijkyi − yjk`∞ = πikyi − yjk`∞ i,j=1 r=1 i=1 have already been appropriately bounded. It is enough thus to bound n m n m m p X X p X X X πibircijkf(zr) − yjk`∞ = πibircij bjs f(zs) − f(zr) ≤ `∞ i,j=1 r=1 i,j=1 r=1 s=1 n m m X X p X ∗ p ≤ πibircijbjskf(zs) − f(zr)k`∞ = (B DπCB)rsdX (zr, zs) . i,j=1 r,s=1 r,s=1 The proof is complete. We shall now provide a proof of the generalised Ball’s extension theorem.

Proof of Theorem 2.1. Let M > Mp(X), N > Np(Y ). Fix some m, n ∈ N, some x1, . . . , xn ∈ X \ Z and z1, . . . , zm ∈ Z. We shall prove that the assumptions of Lemma 2.3 are satisﬁed. Consider a (n + m) × (n + m) symmetric matrix H with non-negative entries. Let V (H),U(H) be the n × n and m × m symmetric block-diagonal submatrices of H and let W (H) be the oﬀ-diagonal n × m matrix. Set m n m n X p X X p X p RH = ursdX (zr, zs) + 2 wirdX (xi, zr) + vijdX (xi, xj) r,s=1 i=1 r=1 i,j=1 and for y1, . . . , yn ∈ Y let m n m n X p X X p X p LH (y1, . . . , yn) = ursdY (f(zr), f(zs)) +2 wirdY (yi, f(zr)) + vijdY (yi, yj) . r,s=1 i=1 r=1 i,j=1

By Lemma 2.3 it suffices to prove that for some y1, . . . , yn ∈ Y there is 18p (2.2) L (y , . . . , y ) ≤ ΛR , where Λ = (N p + 1)M pLp. H 1 n H 3 Let δ > 0. We first observe that m m X p p X p ursdY (f(zr), f(zs)) ≤ L ursdX (zr, zs) , r,s=1 r,s=1 34 4. BALL’S EXTENSION THEOREM so it suffices to prove existence of y1, . . . , yn ∈ Y that satisfy (2.3) n m n n m n X X p X p X X p X p 2 wirdY (yi, f(zr)) + vijdY (yi, yj) ≤ Λ 2 wirdX (xi, zr) + vijdX (xi, xj) +δ . i=1 r=1 i,j=1 i=1 r=1 i,j=1 Since metrics vanish on the diagonal, we may assume that V (H),U(H) vanish on the diagonal as well. We claim that for t > 0 large enough there exists θ > 0 and π ∈ ∆ such that

wir = θπibir for every i = 1, . . . , n and r = 1, . . . , m and

vij = θtπiaij for every i, j = 1, . . . , n, i 6= j. Here A, B are two stochastic matrices of dimensions n × n and m × n. Moreover, A is reversible relative to π. Indeed, we take

n r m X X 1 X wir θ = w , π = w , b = for i = 1, . . . , n, r = 1, . . . , m ir i θ is ir Pm w i=1 j=1 s=1 s=1 is and Pn 1 j=1 vij 1 vij aii = 1 − Pm for all i = 1, . . . , n and aij = Pm for i 6= j. t r=1 wir t r=1 wir Under this change of parameters the left-hand side of the inequality (2.3) may be rewritten as n m n X X p X p (2.4) θ 2 πibirdY (yi, f(zr)) + t πiaijdY (yi, yj) . i=1 r=1 i,j=1

t 1 Pτ s Denote by τ = d 2p e and by Cτ (A) = τ s=1 A the Ces`aroaverages. Now Cτ (A) is reversible relative to π, so Lemma 2.4 guarantees existence of w1, . . . , wn ∈ Y such that

m p X ∗ p 3 (B DπCτ (A)B)rsdY (f(zr), f(zs)) ≥ r,s=1 n m n n X X p X po ≥ max πibirdY (wi, f(zr)) , πiCτ (A)ijdY (wi, wj) . i=1 r=1 i,j=1

From the deﬁnition of the Markov cotype there exist y1, . . . , yn ∈ Y such that n n n X p X p p X q (2.5) πidY (wi, yi) + τ πiaijdY (yi, yj) ≤ N πiCτ (A)ijdY (wi, wj) . i=1 i,j=1 i,j=1

We will show that if t is large enough, then the appropriate y1, . . . , yn will suﬃce. Observe that

p p−1 p p dY (yi, f(zr)) ≤ 2 (dY (yi, wi) + dY (wi, f(zr)) ), 2. STATEMENT OF THE THEOREM 35 so we may bound the ﬁrst sum in (2.4) as follows

n m n m X X p p X X p p 2 πibirdY (yi, f(zr)) ≤ 2 πibir dY (yi, wi) + dY (wi, f(zr)) = i=1 r=1 i=1 r=1 n n m p X p p X X p = 2 πidY (yi, wi) + 2 πibirdY (wi, f(zr)) . i=1 i=1 r=1

Now, using corollary of Lemma 2.4 and the Markov cotype property, we get that the left-hand side of (2.4) may be bounded by

n n n m p X p X p p X X p θ 2 πidY (yi, wi) + τ πiaijdY (yi, yj) + 2 πibirdY (wi, f(zr)) ≤ i=1 i,j=1 i=1 r=1 n n m p X p p X X p ≤ θ (2N) πiCτ (A)ijdY (wi, wj) + 2 πibirdY (wi, f(zr)) ≤ i,j=1 i=1 r=1 m p p X ∗ ≤ θ 6 (N + 1) (B DπCτ (A)B)rsdY (f(zr), f(zs)) ≤ r,s=1 m p p p X ∗ ≤ θ 6 (N + 1)L (B DπCτ (A)B)rsdX (zr, zs) . r,s=1

Now, we use the triangle inequality

p p−1 p p p dX (zr, zs) ≤ 3 dX (zr, xi) + dX (xi, xj) + dX (xj, zs) and the identitiy

n ∗ X (B DπCτ (A)B)rs = πibirbjsCτ (A)ij. i,j=1

They give that the left-hand side of (2.4) may be continued to be bounded by

n m 18p X X (N p + 1)Lpθ π b b C (A) d (z , x )p + d (x , x )p + d (x , z )p. 3 i ir js τ ij X r i X i j X j s i,j=1 r,s=1

We have therefore three sums to deal with. We deal with the ﬁrst and the third one analogously:

n m n m X X p X X p πibirbjsCτ (A)ijdX (zr, xi) = πibirCτ (A)ijdX (zr, xi) = i,j=1 r,s=1 i,j=1 r=1 n m n m X X 1 X X = π b d (z , x )p = w d (z , x )p. i ir X r i θ ir X r i i=1 r=1 i=1 r=1 36 4. BALL’S EXTENSION THEOREM

The third sum gives rise to the same quantity, by reversibility of Cτ (A) with respect to π. The second sum we bound using the Markov type property of X as follows n m n X X p X p πibirbjsCτ (A)ijdX (xi, xj) = πiCτ (A)ijdX (xi, xj) = i,j=1 r,s=1 i,j=1 τ n τ n 1 X X 1 X X = π Aσ d (x , x )p ≤ M pσ π a d (x , x )p = τ i ij X i j τ i ij X i j σ=1 i,j=1 σ=1 i,j=1 n n τ + 1 X vij 1 τ + 1 X = M p d (x , x )p = M p v d (x , x )p. 2 θt X i j θ 2t ij X i j i,j=1 i,j=1 Thus (2.4) is bounded from above by n m n 18p X X X (N p + 1)Lp 2 w d (z , x )p + M p v d (x , x )p . 3 ir X r i ij X i j i=1 r=1 i,j=1 Thus the assumptions of Lemma 2.3 are satisﬁed and the theorem is proven. 3. Examples In this section we shall provide some examples where the assumptions of Theo- rem 2.1 are satisﬁed, so that it is possible to extend Lipschitz maps without losing too much on their Lipschitz constant. Examples of spaces satisfying the Markov type p property include p-smooth Banach spaces. Recall that a Banach space X is called p-smooth provided that there is a constant C > 0 such that 1 kx + τyk + kx − τyk − 1 ≤ Cτ p 2 for all τ > 0 and all unit vectors x, y ∈ X Similarly, a Banach space is called q-convex whenever there is C > 0 such that x + y Cq ≤ 1 − 2 for all unit vectors x, y such that kx − yk = . Examples of spaces satisfying the Markov cotype q property are q-convex Ba- nach spaces. CHAPTER 5

Assesment and references

1. Reading list The reading list consists of all the papers cited above, lecture notes [15], and parts of books [19, 3].

2. Assesment Students are encouraged to give a short presentation on a topic related to the content of the course. The timing of this will be arranged by the lecturer with the student group. Suggested topics include: i) Brehm’s theorem [6], ii) continuity of Kirszbraun’s extension theorem [38], iii) Kirszbraun’s theorem for Alexandrov spaces [39, 21], iv) two-dimensional Kneser-Poulsen conjecture [4], v) origami [30], vi) absolutely minimising Lipschitz extensions and inﬁnity Laplacian [35, 47, 48, 22, 23, 25, 24], vii) Fenchel duality and Fitzpatrick functions [46, 27], viii) sharp form of Whitney’s extension theorem [31], ix) Whitney’s extension theorem for Cm [32], x) Markov type and cotype calculation [15, 1, 44], xi) extending Lipschitz functions via random metric partitions [41, 15].

Additional references Additional references for lectures: i) Kirszbraun’s theorem [10, 18], ii) Kneser-Poulsen conjecture [11, 16, 9], iii) Whitney’s extension theorem [20]. References regarding applications: i) clustering of data [14, 40], ii) dimension reduction [33].

Bibliography

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