Extension of Lipschitz Functions

University of Innsbruck Faculty of Mathematics, Computer Science and Physics Department of Mathematics

Lorenz Oberhammer

Master’s Thesis

Submitted to the Director of Studies of the University of Innsbruck in partial fulﬁlment of the requirements for the degree of Diplom-Ingenieur

Supervisor:

Univ.-Prof. Dr. Eva Kopecká

Innsbruck, 2016

Abstract

In this master’s thesis two recent results related to the problem of the extension of Lipschitz functions are presented. The ﬁrst one deals with functions deﬁned on subsets of metric spaces with values in the real numbers that are both Lipschitz and continuous with respect to some given topology on the metric space (not necessarily the topology induced by the metric). A condition is given when such functions admit extensions to the whole space that preserve both the Lipschitz condition and continuity with respect to the topology. The second one examines the possibility of “continuous” selections of extensions for Lipschitz functions between Hilbert spaces.

iii

Acknowledgments

First and foremost, I would like to thank my supervisor, Univ.-Prof. Dr. Eva Kopecká, for giving me the opportunity to write about such beautiful mathematics, for her continuous support, and for her patience during the sometimes lengthy process of finishing this master’s thesis. Many thanks also to my colleagues and friends for some interesting discussions we had, in particular Noema Nicolussi, who is so incredibly talented at finding counterexamples. Parts of the initial work on this master’s thesis have been carried out during my stay at Aristotle University of Thessaloniki from September 2014 to June 2015. I would like to thank all people who contributed to making this stay such a rewarding experience. Furthermore, I am grateful for the financial support I recieved during that period through the Erasmus+ Programme of the European Commission. Last but not least, I would also like to thank my family, for being always there for me.

Contents

1 Introduction 1

2 Extension Theorems for Lipschitz Functions 5 2.1 Basic deﬁnitions ...... 5 2.2 A word of caution ...... 5 2.3 The Kirszbraun intersection property ...... 7 2.4 An extension theorem of McShane ...... 8 2.5 Coordinate-wise extension ...... 10 2.6 Injective metric spaces ...... 11 2.7 Kirszbraun’s Theorem ...... 12 2.8 When the nearest point mapping is a retraction ...... 16 2.9 Preserving the closed convex hull of the image ...... 17 2.10 What about Banach spaces? ...... 17

3 Urysohn’s Lemma and Tietze’s Extension Theorem 19 3.1 Normal topological spaces ...... 19 3.2 A generalized variant of recursion ...... 21 3.3 Urysohn’s Lemma ...... 21 3.4 Tietze’s Extension Theorem ...... 24

4 Double Extensions 27 4.1 A counterexample ...... 27 4.2 Matoušková’s positive result ...... 28 4.3 Compact Hausdorﬀ spaces ...... 41 4.4 Duals of Banach spaces ...... 42 4.5 Reﬂexive Banach spaces ...... 43 4.6 Closed points ...... 44 4.7 Unbounded functions ...... 45 4.8 A second proof for Theorem 4.2 ...... 47 4.9 But is it really weaker? ...... 54

5 Michael’s Selection Theorem 57 5.1 Paracompact topological spaces ...... 57 5.2 The relation of paracompactness to other properties ...... 57

vii Contents

5.3 Partitions of unity ...... 62 5.4 Michael’s Selection Theorem ...... 62

6 Continuous Selections 67 6.1 Closed and convex sets ...... 67 6.2 Compact sets in Euclidean space ...... 68 6.3 Continuous selections ...... 76

List of Symbols 81

Bibliography 83

Index 85

viii List of Figures

2.1 Illustration for Example 2.1...... 6 2.2 The Kirszbraun intersection property...... 7 2.3 Injective metric spaces...... 11

3.1 Illustration for the proof of Urysohn’s Lemma...... 23

4.1 Illustration for Lemma 4.3...... 29 4.2 Illustration for Lemma 4.4...... 30 4.3 Illustration for Lemma 4.5...... 31

5.1 The relation of paracompactness to other properties...... 61 5.2 Lower semi-continuity and upper semi-continuity...... 64

6.1 Extending gA when Lip(gA) is small...... 77 6.2 Extending gA when Lip(gA) is large...... 78

Chapter 1

Introduction

The aim of this master’s thesis is to present some recent results of research on two aspects of the following general question.

Problem. Let (M, dM ) and (N, dN ) be metric spaces. Assume that S ⊆ M and that f : S → N is L-Lipschitz, for some L > 0. Does there exist an extension of f to ˜ ˜ M, i.e. a function f : M → N such that f|S = f, that is again L-Lipschitz? Not surprisingly, this is not always possible. However, there are two classic theorems which answer the above question positively for all subsets S and all functions f if the spaces (M, dM ) and (N, dN ) are members of certain classes of metric spaces. The following result, which is due to E. J. McShane [13], implies that the above question can always be answered positively when (N, dN ) is the real line with the usual metric, regardless of what the space (M, dM ) is. Theorem (McShane 1934). Let (X, d) be a metric space, S ⊆ X and f : S → R a function that satisﬁes the Lipschitz condition |f(x) − f(y)| ≤ Ld(x, y) on S, for some L > 0. Then the function Φ(x) := sup(f(¯x) − Ld(¯x, x)) (x ∈ X) x¯∈S extends f and |Φ(x) − Φ(y)| ≤ Ld(x, y) for all x, y ∈ X. Likewise, by Kirszbraun’s Theorem [6] it is possible to ﬁnd an extension as above when both M and N are Hilbert spaces and dM and dN are the metrics induced by the inner products of M and N, respectively.

Theorem (Kirszbraun 1934). Let H1 and H2 be Hilbert spaces, S ⊆ H1 and g :

S → H2 a function with the property that kg(x) − g(y)kH2 ≤ Lkx − ykH1 for all x, y ∈ S, for some L > 0. Then there exists a function f : H1 → H2 such that kf(x) − f(y)kH2 ≤ Lkx − ykH1 for all x, y ∈ H1, and such that f|S = g. We begin our discussion of the subject with those two theorems, which will be presented with full proof in Chapter 2. Occasionally we will also make some remarks on possible generalizations. Other extension theorems, for the special case where the range space is R, are Urysohn’s Lemma [20] and Tietze’s Extension Theorem [19].

1 1. Introduction

Theorem (Urysohn 1925). Let X be a normal topological space. Then for each pair of closed and disjoint subsets A and B of X, there exists a continuous function f : X → [0, 1] such that A ⊆ f −1({0}) and B ⊆ f −1({1}).

Theorem (Tietze 1915). Let X be a normal topological space. If F is a closed subset of X and f : F → R is continuous, then there exists a continuous extension ˜ ˜ f : X → R of f. Furthermore, the extension can be chosen such that infF f ≤ f ≤ supF f on X. Note that the assumptions in the second theorem are more general than the assumptions in the result of McShane mentioned above, as the domain space is allowed to be a member of the larger class of normal topological spaces (instead of a metric space) and just continuity is required for the function f (instead of Lipschitz continuity). Also the additional requirement that f is deﬁned on a closed subset of the space is not really a restriction, as Lipschitz continuous functions into complete metric spaces allow a unique Lipschitz continuous extension to the closure. However, the conclusion gets weaker, too, as the extension provided by the theorem is only continuous (not Lipschitz continuous). It is the purpose of Chapter 3 to establish those results. The ﬁrst major problem dealt with in this master’s thesis is the question whether it is possible to combine the theorems of McShane and Tietze.

Problem. Let (X, τ) be a normal topological space and d a metric on X. Let F be a subset of X which is closed with respect to τ, and let g be a real-valued function deﬁned on F that is both τ|F -continuous and L-Lipschitz, for some L > 0. Then by the result of McShane there exists an extension f1 of g to the whole space that is L-Lipschitz, and by Tietze’s Extension Theorem there exists an extension f2 of g to the whole space that is τ-continuous and such that infF g ≤ f2 ≤ supF g on X. Is it possible to have f1 = f2? Again not really surprisingly, it turns out that in general this is not possible. However, E. Matoušková [11] could show that it is possible if we restrict ourselves to bounded functions and require that the topology τ and the metric d are “compatible” in a certain sense.

Theorem (Matoušková 1997). Let (X, τ) be a normal topological space and d a metric on X such that the set B(A, α) = {x ∈ X; d-dist(A, x) ≤ α} is τ-closed for every τ-closed A ⊆ X and every α > 0. Suppose that F ⊆ X is τ-closed and g : F → R is a bounded τ|F -continuous function that is L-Lipschitz in d for some L > 0. Then there exists a τ-continuous function f : X → R such that f|F = g, infF g ≤ f ≤ supF g on X, and that is L-Lipschitz in d. The idea here is to adapt the proof of Urysohn’s Lemma. As corollaries to this theorem one gets that such “double extensions” in particular exist in compact Hausdorﬀ spaces with lower semi-continuous metrics and in duals

2 of Banach spaces with the weak* topology, which is also a canonical example of spaces that naturally carry a topology (the weak* topology) and a metric (the norm-distance).

Corollary (Matoušková 1997). Let (X, τ) be a compact Hausdorﬀ space, d a τ- lower semi-continuous metric on X, F ⊆ X τ-closed, g ∈ C(F ) and, for some L > 0, L-Lipschitz in d. Then there exists f ∈ C(X) such that f = g on F , minF g ≤ f ≤ maxF g, and f is L-Lipschitz in d. Corollary (Matoušková 1997). Let X be a Banach space. Let F be a w∗-closed subset of its dual X∗, g a bounded, w∗-continuous function on F that is L-Lipschitz in the norm-metric on X∗ for some L > 0. Then there exists a w∗-continuous ∗ function f on X such that f = g on F , infF g ≤ f ≤ supF g, and f is L-Lipschitz in the norm-metric on X∗.

In Chapter 4 we will prove the above theorem and its two corollaries. Also a problem with an alternative proof of it which was discovered during the work on this master’s thesis is discussed. In some sense related to the problem of extension of functions is the so-called “selection problem”.

Problem. Assume that X and Y are topological spaces. Let Φ: X → P(Y ) be a carrier, i.e. a function that assignes to each x ∈ X a nonempty subset Φ(x) ⊆ Y . Under what conditions on X, Y and Φ does there exist a continuous selection for Φ, i.e. a continuous function f : X → Y such that f(x) ∈ Φ(x) for all x ∈ X?

Michael’s Selection Theorem [14] answers this question positively for lower semi- continuous carriers from paracompact Hausdorﬀ spaces into Banach spaces that have closed and convex images.

Theorem (Michael 1956). Let X be a paracompact Hausdorﬀ space, Y a Banach space, and Φ: X → P(Y ) a lower semi-continuous carrier such that Φ(x) is a closed and convex subset of Y for each x ∈ X. Then Φ admits a continuous selection.

As a corollary one also gets that continuous selections for carriers can be extended from closed subsets to the whole space.

Corollary (Michael 1956). Let X be a paracompact Hausdorﬀ space, Y a Banach space, Φ: X → P(Y ) a lower semi-continuous carrier such that Φ(x) is closed and convex for every x ∈ X, and A ⊆ X closed. Then every continuous selection for Φ|A can be extended to a continuous selection for Φ. The above results as well as the necessary topological background are presented in Chapter 5. Finally, the second major problem dealt with in this master’s thesis is the following.

3 1. Introduction

Problem. Let H1 and H2 be Hilbert spaces and S ⊆ H1. By Kirszbraun’s Theorem every L-Lipschitz function f : S → H2 can be extended to a L-Lipschitz function deﬁned on all of H1. However, the extensions are not unique. Is it possible to assign extensions in a continuous way?

If X is a Hilbert space and A ⊆ X, we consider the following function spaces. By N (A) we denote the set of all bounded nonexpansive mappings from A into X. By L(A) we denote the set of all bounded Lipschitz continuous mappings from A into X. Both spaces are subsets of C(A, X), the Banach space of all bounded continuous functions from A to X with the supremum norm. For the case that X is ﬁnite-dimensional and A ⊆ X is compact, E. Kopecká [8] explicitly constructed uniformly continuous extension operators from N (A) to N (X) that preserve Lipschitz constants.

Theorem (Kopecká 2012). Let X be a Euclidean space and A ⊆ X compact. Then there exists a uniformly continuous function F : N (A) → N (X) such that, if f ∈ N (A), then F (f)|A = f, and if f is L-Lipschitz, for some L > 0, then F (f) is also L-Lipschitz.

In order to get a similar result in the general case, the “non-constructive” way via Michael’s Selection Theorem is used. Let Ψ: L(A) → P(C(X,X)) be the map that assigns to each f ∈ L(A) the set of all its extensions in L(X) that have the same Lipschitz constant. E. Kopecká [7] could show that the carrier Ψ is lower semi-continuous. Thus by Michael’s Selection Theorem one gets the following result.

Theorem (Kopecká 2012). Let X be a Hilbert space and ∅ 6= A ⊆ X. Then the carrier Ψ: L(A) → P(C(X,X)) admits a continuous selection F : L(A) → C(X,X). Furthermore, for every closed subset D of L(A) and every continuous selection F : D → C(X,X) of the extension mapping Ψ, there is a continuous selection F˜ : L(A) → C(X,X) which extends F .

All of the above can be found in more detail in Chapter 6.

4 Chapter 2

Extension Theorems for Lipschitz Functions

We examine under what conditions Lipschitz mappings between metric spaces can be extended from a subset of some metric space to the whole space, preserving the Lipschitz condition. In general this is not possible (Example 2.1). However, there are two classic theorems which answer the above question positively for all subsets and all Lipschitz functions deﬁnded on them if the domain and range space are members of certain classes of metric spaces. A result of McShane (Theorem 2.3) says that this is always possible when the range space is the real line. By Kirszbraun’s Theorem (Theorem 2.9) Lipschitz function between Hilbert spaces can be extended from arbitrary subsets to the whole space. We also discuss some generalizations of those theorems.

2.1 Basic deﬁnitions

Let (M, dM ) and (N, dN ) be metric spaces. A function f : M → N is said to be Lipschitz continuous if and only if there exists a constant L ≥ 0 such that

dN (f(x), f(y)) ≤ LdM (x, y) for all x, y ∈ M. The Lipschitz constant of f, which will be denoted by Lip(f), is the smallest constant L ≥ 0 such that the above relation holds, that is,

Lip(f) = min{L ≥ 0; ∀x, y ∈ M : dN (f(x), f(y)) ≤ LdM (x, y)}.

If L ≥ 0 is the Lipschitz constant of f, we sometimes say that f is L-Lipschitz. When Lip(f) ≤ 1, f is called nonexpansive. When Lip(f) < 1, f is called a contraction.

2.2 A word of caution

Let us begin with a word of caution. It is not always possible to extend Lipschitz mappings with the same Lipschitz constant [23, Example 10.2].

5 2. Extension Theorems for Lipschitz Functions

C x1 − x2 H B [0, 1 ] k·kC 2 1 Bk·kC [x1, 2 ]

x1 0 x

2x x1 + x2

1 Bk·kC [x2, 2 ]

Figure 2.1: Illustration for Example 2.1.

Example 2.1. In R2 let H be a regular hexagon with center at the origin and C 2 a circle inscribed in H. Let k · kH and k · kC be the two norms on R whose unit circles are H and C respectively. Suppose x1 and x2 are two consecutive points of contact between H and C and let S = {0, x1, x2}. Deﬁne T on S by T (0) = 0, 2 T (x1) = x1 and T (x2) = x2. Then T is a nonexpansive mapping from (R , k · kH ) 2 1 into (R , k · kC ). However, T cannot be extended to 3 (x1 + x2) in a way such that it remains nonexpansive.

Note that the norms k·kH and k·kC are the corresponding Minkowski functionals for H and C respectively. That is, kxkH = inf{t > 0; x ∈ tH} and kxkC = inf{t > 0; x ∈ tC}.

Proof. As x1 and x2 are points of contact between H and C,

kx1kH = kx2kH = kx1kC = kx2kC = 1.

By symmetry, x1 − x2 is a point of contact as well, so we also have kx1 − x2kH = 2 kx1 − x2kC = 1. This proves that T is a nonexpansive map from (R , k · kH ) to 2 1 (R , k · kC ). Now put x = 3 (x1 + x2). A geometrical argument (see Figure 2.1) 1 shows that kxkH = kx − x1kH = kx − x2kH = 2 . So any choice y = T (x) such that 1 1 T remains nonexpansive on S ∪ {x} would have to satisfy kykC ≤ 2 , ky − x1kC ≤ 2 , 1 ky − x2kC ≤ 2 . But this clearly is impossible.

6 2.3. The Kirszbraun intersection property

x2 r2 y2 r2

r1 r1

x x1 y y1

r3 r3 y3 x3

Figure 2.2: The Kirszbraun intersection property.

2.3 The Kirszbraun intersection property

Nevertheless, in many important cases L-Lipschitz functions on subspaces can be extended to L-Lipschitz functions on the whole space. We try to characterize the spaces where this works. For simplicity, we will only consider nonexpansive mappings. If the range space is a normed linear space, this is no restriction at all, as in that case one can always divide functions by their Lipschitz constants. A pair (X,Y ) of metric spaces is said to have the nonexpansive extension property if corresponding to every nonexpansive map T from an arbitrary subset ˜ ˜ S of X into Y there exists a nonexpansive map T of X into Y such that T |S = T , that is, the restriction of T˜ to S is T . There is a geometric characterization of the pairs of spaces that have this property. Let us agree that a pair (X,Y ) of metric spaces has the Kirszbraun intersection property (or property (K) for short) provided that, whenever (B[xi, ri])i∈I and (B[yi, ri])i∈I are two families of closed balls in X and Y respectively, each in- dexed over the same set I, such that

dY (yi, yj) ≤ dX (xi, xj)(i, j ∈ I), (2.1) then \ B[xi, ri] 6= ∅ (2.2) i∈I implies \ B[yi, ri] 6= ∅. (2.3) i∈I Then the following holds [23, Theorem 11.1]. Theorem 2.2. A pair (X,Y ) of metric spaces has the nonexpansive extension property if and only if it has property (K).

7 2. Extension Theorems for Lipschitz Functions

Proof. First suppose that (X,Y ) has the nonexpansive extension property and that (B[xi, ri])i∈I and (B[yi, ri])i∈I are two families of closed balls in X and Y respectively T satisfying (2.1) and (2.2). Let x ∈ i∈I B[xi, ri]. The map T deﬁned by T (xi) = yi (i ∈ I) is, by (2.1), a nonexpansive map from S = {xi; i ∈ I} to Y . By assumption, T can be extended to all of X and, in particular, to x. Thus there exists y ∈ Y such that

dY (y, yi) ≤ dX (x, xi) ≤ ri (i ∈ I) T and, therefore, y ∈ i∈I B[yi, ri]. So (2.3) is satisﬁed. For the converse, let T : S → Y be a nonexpansive map from the proper subset S of X. We extend T point by point. Choose x ∈ X \ S. The collection of closed balls (B[w, dX (x, w)])w∈S and (B[T (w), dX (x, w)])w∈S satisfy (2.1) because T T is nonexpansive. Since x ∈ w∈S B[w, dX (x, w)] and (X,Y ) has property (K), there T exists y ∈ w∈S B[T (w), dX (x, w)]. Setting T (x) = y then clearly extends T as a nonexpansive map to S ∪ {x}. To ﬁnish the argument, note that

0 0 0 0 0 P := {T : S → Y ; S ⊆ S ⊆ X,T |S = T,T nonexpansive} is nonempty, becomes a partially orderd set if it is ordered in the usual way (T 0 ≤ T 00 if and only if T 00 extends T 0), and that every chain in P has an upper bound in P (the union of the members of the chain). So by Zorn’s Lemma there exists some maximal element T˜ of P, and the domain of T˜ clearly is X, as otherwise one could extend the domain to one more point, using the above procedure, contradicting the maximality of T˜.

The Kirszbraun intersection property is named after M. D. Kirszbraun, who used it in [6] to show that for every positve integer n the pair (Rn, Rn) has the nonexpansive extension property.

2.4 An extension theorem of McShane

The following theorem for the special case that the range space is R is due to E. J. McShane [13]. It does not rely on the Kirszbraun intersection property but rather gives an explicit formula of the extension.

Theorem 2.3 (McShane 1934). Let (X, d) be a metric space, S ⊆ X and f : S → R a function that satisﬁes the Lipschitz condition |f(x) − f(y)| ≤ Ld(x, y) on S, for some L > 0. Then the function

Φ(x) := sup(f(¯x) − Ld(¯x, x)) (x ∈ X) (2.4) x¯∈S extends f and |Φ(x) − Φ(y)| ≤ Ld(x, y) for all x, y ∈ X.

We will use the following little lemma in the proof of Theorem 2.3.

8 2.4. An extension theorem of McShane

Lemma 2.4. Let S be a set and g, h : S → R functions such that supx∈S g(x) and supx∈S h(x) are ﬁnite. Then sup g(x) − sup h(x) ≤ sup(g(x) − h(x)). (2.5) x∈S x∈S x∈S

Proof. If supx∈S(g(x) − h(x)) = +∞, this obviously holds. So let us assume that supx∈S(g(x) − h(x)) < +∞. For all y ∈ S we have g(y) − sup h(x) ≤ g(y) − h(y) ≤ sup(g(x) − h(x)), x∈S x∈S hence g(y) ≤ sup(g(x) − h(x)) + sup h(x). x∈S x∈S It follows that sup g(x) ≤ sup(g(x) − h(x)) + sup h(x), x∈S x∈S x∈S and this directly implies (2.5). Proof of Theorem 2.3. Let x ∈ S. Then, as f is L-Lipschitz on S,

f(¯x) − Ld(¯x, x) ≤ f(x) for all x¯ ∈ S. Hence Φ(x) ≤ f(x), and, as the inequality becomes an equality for x¯ = x, Φ(x) = f(x). A priori, we have Φ: X → R ∪ {+∞}, as the supremum in the definition of Φ might not exist for some x ∈ X. However, in fact the function is everywhere finite, as for fixed x ∈ X we can choose some x0 ∈ S and then have

f(¯x) − Ld(¯x, x) = = [(f(¯x) − Ld(¯x, x)) − (f(¯x) − Ld(¯x, x0))] + (f(¯x) − Ld(¯x, x0)) = = L(d(¯x, x0) − d(¯x, x)) + (f(¯x) − Ld(¯x, x0)) ≤ Ld(x, x0) + Φ(x0) for all x¯ ∈ S, and thus Φ(x) ≤ Ld(x, x0) + Φ(x0) < +∞. It remains to show that Φ is L-Lipschitz on X. According to Lemma 2.4, if x and y are members of X such that (without loss of generality) Φ(x) ≥ Φ(y), then

|Φ(x) − Φ(y)| = Φ(x) − Φ(y) = = sup(f(¯x) − Ld(¯x, x)) − sup(f(¯x) − Ld(¯x, y)) ≤ x¯∈S x¯∈S ≤ sup[(f(¯x) − Ld(¯x, x)) − (f(¯x) − Ld(¯x, y))] = sup L(d(¯x, y) − d(¯x, x)) ≤ x¯∈S x¯∈S ≤ sup Ld(x, y) = Ld(x, y), x¯∈S and Φ is L-Lipschitz. The following example shows how an extension produced by (2.4) may look like.

9 2. Extension Theorems for Lipschitz Functions

Example 2.5. Let X = R, S = [0, 1], f(x) = x and L = 1. Then  x x ≤ 1, Φ(x) = 2 − x x > 1.

Proof. It is suﬃcient to consider x ∈ R \ [0, 1]. If x < 0, then Φ(x) = sup0≤x¯≤1(¯x − (¯x − x)) = x, and if x > 1, then Φ(x) = sup0≤x¯≤1(¯x − (x − x¯)) = 2 − x.

2.5 Coordinate-wise extension

If the range space is `∞, a generalization of Theorem 2.3 is available by applying it coordinate-wise.

Theorem 2.6. Let (X, d) be a metric space, S ⊆ X and f : S → `∞ a function that satisﬁes the Lipschitz condition |f(x) − f(y)| ≤ Ld(x, y) on S, for some L > 0. Then f can be extended to all of X such as to satisfy the same Lipschitz condition.

Proof. Let `∞ = `∞(I), for some set I. As f is L-Lipschitz, all the functions fi : S → R, x 7→ fi(x) = f(x)i, for i ∈ I, also have this property. Thus, by Theorem ˜ 2.3, they can be extended to all of X, preserving the Lipschitz condition. Let fi be the extension of fi that is obtained in this way, for every i ∈ I. Then ˜ ˜ ˜ f : X → `∞, x 7→ f(x) = (fi(x))i∈I extends f and is L-Lipschitz. In the above context it may be interesting to mention the following fact about embedding of metric spaces into `∞, which is due to Kuratowski [10]. Theorem 2.7. Let (X, d) be a metric space. Then X can be isometrically embedded into `∞(X).

Proof. Pick some x0 ∈ X and consider the map

i : X → `∞(X), x 7→ (d(x, z) − d(x0, z))z∈X .

As supz∈X |d(x, z) − d(x0, z)| ≤ supz∈X d(x, x0) = d(x, x0), it is well-deﬁned. To see that i is an isometry, note that

ki(x) − i(y)k = sup |(d(x, z) − d(x0, z)) − (d(y, z) − d(x0, z))| = z∈X = sup |d(x, z) − d(y, z)| ≤ sup d(x, y) = d(x, y) z∈X z∈X for any two elements x, y ∈ X, where equality is attained in the case z = x or z = y.

Clearly, unless i(X) lies “nicely” in `∞(X) (e.g. i(X) is a 1-Lipschitz retract of `∞(X)), the extensions of Lipschitz mappings into some metric space X obtained by combining Theorem 2.7 and Theorem 2.6 may no longer be mappings into X.

10 2.6. Injective metric spaces

T T˜

S X i

Figure 2.3: Injective metric spaces.

2.6 Injective metric spaces

Theorem 2.3 and Theorem 2.6 in particular imply that (X, R) and (X, `∞) have the nonexpansive extension property for every metric space X. Spaces with that property appear frequently enough that they have aquired their own name. A metric space Y with the property that the pair (X,Y ) has the nonexpansive extension property for every metric space X is called injective. That is, a metric space Y is injective if and only if for every metric space X, every subspace S of X and every nonexpansive map T : S → Y there exists a nonexpansive map T˜ : X → Y such that the diagram which is shown in Figure 2.3 commutes. Those spaces have an interesting characterization in terms of geometric properties, too. A metric space (Y, dY ) is called metrically convex if y1, y2 ∈ Y and 0 < λ < 1 imply the existence of z ∈ Y with dY (y1, z) = λdY (y1, y2) and dY (y2, z) = (1 − λ)dY (y1, y2). Y has the binary intersection property if every collection of mutually intersecting closed balls in Y has nonempty intersection. Then the following holds [23, Theorem 11.2]. Theorem 2.8. Let Y be a metric space. Then Y is injective if and only if Y is metrically convex and has the binary intersection property. Proof. Suppose ﬁrst that Y is injective. Then (X,Y ) has the nonexpansive extension property for every metric space X. Let y1, y2 ∈ Y and 0 < λ < 1. If r1, r2 ∈ R are chosen so that |r1 − r2| = dY (y1, y2), then the map T (ri) = yi (i = 1, 2) is nonexpansive as a function from the subset {r1, r2} of R to (Y, dY ). As the pair (R,Y ) has the nonexpansive extension property, T can be extended as a nonexpansive map to s = (1 − λ)r1 + λr2. Let z = T (s). Then

dY (y1, z) ≤ |r1 − s| = λ|r1 − r2| = λdY (y1, y2) and dY (y2, z) ≤ |r2 − s| = (1 − λ)|r1 − r2| = (1 − λ)dY (y1, y2). If any of the above inequalities was sharp, we would get

dY (y1, y2) ≤ dY (y1, z) + dY (z, y2) < dY (y1, y2),

11 2. Extension Theorems for Lipschitz Functions

which clearly is a contradiction. Thus dY (y1, z) = λdY (y1, y2) and dY (y2, z) = (1 − λ)dY (y1, y2). It follows that Y is metrically convex. In order to show that Y has the binary intersection property, let (B[yi, ri])i∈I be a collection of mutually intersecting closed balls in Y . Then dY (yi, yj) ≤ ri + rj for all i, j ∈ I as B[yi, ri] ∩ B[yj, rj] 6= ∅. Consider on X = I ∪ {I} the metric dX which is deﬁned by

• dX (x, x) = 0,

• dX (i, j) = ri + rj (i, j ∈ I, i 6= j),

• dX (i, I) = dX (I, i) = ri (i ∈ I).

Then the collections of closed balls (B[i, ri])i∈I and (B[yi, ri])i∈I satisfy (2.1) and (2.2). But since by assumption the pair (X,Y ) has the nonexpansive extension T property, that is, by Theorem 2.2, property (K), this implies i∈I B[yi, ri] 6= ∅. Conversely, suppose that Y is metrically convex and has the binary intersection property. By Theorem 2.2, it suﬃces to show that (X,Y ) has property (K) for every metric space X. Let X be a metric space and (B[xi, ri])i∈I and (B[yi, ri])i∈I two families of closed balls in X and Y respectively satisfying (2.1) and (2.2). Since T i∈I B[xi, ri] 6= ∅, there exists some element z ∈ X such that z ∈ B[xi, ri]∩B[xj, rj] for all i, j ∈ I, and it follows that

dX (xi, xj) ≤ dX (xi, z) + dX (z, xj) ≤ ri + rj for all i, j ∈ I. Consequently dY (yi, yj) ≤ ri + rj for all i, j ∈ I, and since Y is metrically convex, this implies that the collection (B[yi, ri])i∈I is mutually intersecting. In fact, for given i, j ∈ I one can put λ = ri ∈ (0, 1) and then there exists z ∈ Y ri+rj such that dY (yi, z) = λdY (yi, yj) ≤ ri and dY (yj, z) = (1 − λ)dY (yi, yj) ≤ rj. T From the binary intersection property it now follows that i∈I B[yi, ri] 6= ∅ and (X,Y ) has property (K).

2.7 Kirszbraun’s Theorem

As was already said, the nonexpansive extension property was ﬁrst established for (Rn, Rn) by M. D. Kirszbraun [6]. A few years later, his results were independently rediscoverd by F. A. Valentine [21, 22], who showed the nonexpansive extension property ﬁrst only for the pair (R2, R2), but then also for (Rn, Rn) and, even slightly more generally than Kirszbraun’s original result, for any pair of Hilbert spaces. Nevertheless, all those theorems are usually referred to as “Kirszbraun’s Theorem”. We show the general version for Hilbert spaces.

12 2.7. Kirszbraun’s Theorem

Theorem 2.9 (Kirszbraun 1934). Let H1 and H2 be Hilbert spaces, S ⊆ H1 and g : S → H2 a function with the property that kg(x) − g(y)kH2 ≤ Lkx − ykH1 for all x, y ∈ S, for some L > 0. Then there exists a function f : H1 → H2 such that kf(x) − f(y)kH2 ≤ Lkx − ykH1 for all x, y ∈ H1, and such that f|S = g. According to Theorem 2.2, in order to establish Theorem 2.9 it is sufficient to show that the pairs (H1,H2) have property (K). This will follow at once from the sequence of lemmata below. Apart from some minor modifications, the exposition follows [4]. Lemma 2.10. Let H be a Hilbert space and F ⊆ H nonempty and finite. Then conv F is compact. Proof. Observe that conv F is the continuous image of the set

F X S = {(λx)x∈F ∈ [0, 1] ; λx = 1} x∈F

P F under the map (λx)x∈F 7→ x∈F λxx, which is compact, as [0, 1] is compact and S is closed in [0, 1]F . Lemma 2.11. Let H be a Hilbert space, C ⊆ H nonempty, closed and convex, b ∈ H and b0 ∈ C the nearest point projection of b onto C. Then

hz − b, b0 − bi ≥ kb0 − bk2 (2.6) for all z ∈ C. Proof. Let z ∈ C be arbitrary. Then, by convexity of C, we have δz + (1 − δ)b0 ∈ C for all δ ∈ (0, 1], hence

kb0 − bk2 ≤ kδz + (1 − δ)b0 − bk2 = kδ(z − b) + (1 − δ)(b0 − b)k2 = = δ2kz − bk2 + 2δ(1 − δ)hz − b, b0 − bi + (1 − δ)2kb0 − bk2, after subtraction of kb0 − bk2 on both sides

0 ≤ δ2kz − bk2 + 2δ(1 − δ)hz − b, b0 − bi − 2δkb0 − bk2 + δ2kb0 − bk2, after division by 2δ 1 1 0 ≤ δkz − bk2 + (1 − δ)hz − b, b0 − bi − kb0 − bk2 + δkb0 − bk2, 2 2 ﬁnally after some rearranging of terms 1 kb0 − bk2 − hz − b, b0 − bi ≤ δ(kz − bk2 − 2hz − b, b0 − bi + kb0 − bk2). 2 Thus for δ → 0 we get kb0 − bk2 − hz − b, b0 − bi ≤ 0, or, equivalently, (2.6).

13 2. Extension Theorems for Lipschitz Functions

Lemma 2.12. Let H1 and H2 be Hilbert spaces, F ⊆ H1 nonempty and ﬁnite, and g : F → H2 a function such that 0 ∈ conv g(F ). Then either kg(x)−g(y)k > kx−yk for some x, y ∈ F or kg(x)k ≤ kxk for some x ∈ F .

Proof. Suppose that both kg(x)−g(y)k ≤ kx−yk for all x, y ∈ F and kg(x)k > kxk for all x ∈ F . Then 1 hg(x), g(y)i = (kg(x)k2 + kg(y)k2 − kg(x) − g(y)k2) > 2 1 > (kxk2 + kyk2 − kx − yk2) = hx, yi 2 for any x, y ∈ F . As by assumption 0 ∈ conv g(F ), there is a family (λx)x∈F of P P nonnegative real numbers such that x∈F λx = 1 and 0 = x∈F λxg(x). But then

2 X X X X 0 = k0k = h0, 0i = h λxg(x), λyg(y)i = λxλyhg(x), g(y)i > x∈F y∈F x∈F y∈F X X X X X 2 > λxλyhx, yi = h λxx, λyyi = k λxxk ≥ 0, x∈F y∈F x∈F y∈F x∈F which clearly is a contradiction.

Lemma 2.13. Let H1 and H2 be Hilbert spaces, S ⊆ H1 nonempty and ﬁnite, g : S → H2 a function such that kg(x) − g(y)k ≤ kx − yk for all x, y ∈ S, and (rx)x∈S a T T family of positive real numbers such that x∈S B[x, rx] 6= ∅. Then x∈S B[g(x), rx] 6= ∅. T Proof. Let a ∈ x∈S B[x, rx]. Without loss of generality we can assume a∈ / S. Otherwise the assertion is obviously true. Let C = conv g(S). Then C is, by Lemma 2.10, a compact subset of H2. Consider the function

kz − g(x)k Φ: C → [0, +∞), z 7→ max . x∈S ka − xk

kz−g(x)k As for every x ∈ S the function z 7→ ka−xk from C to [0, +∞) is continuous, Φ is continuous. In particular, it attains its minimum on the compact set C, so there is some member b of C such that Φ(b) ≤ Φ(z) for all other members z of C. Put γ := Φ(b) and kb − g(x)k T := {x ∈ S; = γ}. ka − xk Then T is a nonempty subset of S. Denote D = conv g(T ). Again, by Lemma 2.10, D is compact. In particular it is closed. Let b0 be the nearest point projection of b onto D. We will show that b = b0. Assume the converse holds. Consider the points

0 0 bδ = (1 − δ)b + δb = b + δ(b − b)

14 2.7. Kirszbraun’s Theorem

0 for δ ∈ (0, 1]. Obviously, as both b and b are members of C and C is convex, bδ ∈ C for all δ ∈ (0, 1]. If x ∈ T , then, using the inequality from Lemma 2.11, we get

2 2 kg(x) − bδk = k(g(x) − b) − (bδ − b)k = 2 2 = kg(x) − bk − 2hg(x) − b, bδ − bi + kbδ − bk = = kg(x) − bk2 − 2δhg(x) − b, b0 − bi + δ2kb0 − bk2 ≤ ≤ kg(x) − bk2 − 2δkb0 − bk2 + δ2kb0 − bk2 = = kg(x) − bk2 + δ(δ − 2)kb0 − bk2 < kg(x) − bk2 for all δ ∈ (0, 1], in particular

kb − g(x)k kb − g(x)k δ < = γ. ka − xk ka − xk

If x ∈ S \ T , then

kb − g(x)k kb − g(x)k lim δ = < γ, δ→0 ka − xk ka − xk so there exists some δx ∈ (0, 1] such that kb − g(x)k δ < γ ka − xk

∗ ∗ for all δ ∈ (0, δx]. Finally, if we choose δ ∈ (0, 1] such that δ ≤ δx for all x ∈ S \ T , then

kbδ∗ − g(x)k Φ(bδ∗ ) = max < γ, x∈S ka − xk a contradiction. Hence our assumption that b 6= b0 was false and b = b0. In particular, b ∈ D. Put now T 0 := {x − a; x ∈ T } and

0 0 g : T → H2, x 7→ g(x + a) − b.

As b ∈ D, there exists a family of real numbers (λx)x∈T with λx ≥ 0 for all x ∈ T P P and x∈T λx = 1, such that b = x∈T λxg(x). It follows

X 0 X X λx+ag (x) = λx+a(g(x + a) − b) = λx(g(x) − b) = x∈T 0 x∈T 0 x∈T X X = λxg(x) − λxb = b − b = 0, x∈T x∈T that is, 0 is contained in the convex hull of g0(T 0). Furthermore, for all x, y ∈ T 0,

kg0(x) − g0(y)k = k(g(x + a) − b) − (g(y + a) − b)k =

15 2. Extension Theorems for Lipschitz Functions

= kg(x + a) − g(y + a)k ≤ k(x + a) − (y + a)k = kx − yk.

Hence by Lemma 2.12 there exists some x0 ∈ T 0 with kg0(x0)k ≤ kx0k. As

kg0(x0)k = kg(x0 + a) − bk = kb − g(x0 + a)k = γka − (x0 + a)k = γkx0k and x0 6= 0, because a∈ / T , this implies γ ≤ 1. But then Φ(b) ≤ 1, i.e. kb − g(x)k ≤ T ka − xk ≤ rx for all x ∈ S, that is, b ∈ x∈S B[g(x), rx].

Lemma 2.14. Let H1 and H2 be Hilbert spaces, S ⊆ H1 nonempty, g : S → H2 a function such that kg(x) − g(y)k ≤ kx − yk for all x, y ∈ S, and (rx)x∈S a family of T T positive real numbers such that x∈S B[x, rx] 6= ∅. Then x∈S B[g(x), rx] 6= ∅.

∗ ∗ Proof. By Alaoglu’s Theorem, the closed unit ball in H2 is w -compact. The map ∗ H2 → H2 , x 7→ h·, xi is linear, bijective, an isometry, and also a homeomorphism ∗ ∗ of the topological spaces (H2, w) and (H2 , w ), so the closed unit ball in H2 is w- compact. Finally, as translation and multiplication by a scalar are continuous in (H2, w), for each x ∈ S, the closed ball B[g(x), rx] is w-compact. Fix some x0 ∈ S. Then, by Lemma 2.13, the sets

B[g(x), rx] ∩ B[g(x0), rx0 ]

for x ∈ S are closed subsets of the compact space B[g(x0), rx0 ] such that any ﬁnite subfamily of the family of all those sets has nonempty intersection. Hence

\ \ (B[g(x), rx] ∩ B[g(x0), rx0 ]) = B[g(x), rx] 6= ∅ x∈S x∈S by the ﬁnite intersection property.

2.8 When the nearest point mapping is a retraction

Let X be a topological space and A a subspace of X. A continuous map r : X → A such that r|A is the identity on A is called a retraction. In this situation the subspace A is also called a retract of X. A particular example of such a function is the nearest point projection onto closed and convex subsets of a Hilbert space. It has the additional property that it is 1-Lipschitz.

Proposition 2.15. Let H be a Hilbert space and C ⊆ H closed and convex. Then the nearest point projection π : H → C is a 1-Lipschitz retraction.

Proof. That π(x) = x for x ∈ C is clear. So let x, y ∈ H. By Lemma 2.11 we have

hπ(y) − x, π(x) − xi ≥ kπ(x) − xk2,

16 2.9. Preserving the closed convex hull of the image as π(y) ∈ C, and hπ(x) − y, π(y) − yi ≥ kπ(y) − yk2, as π(x) ∈ C. This reformulates as

hπ(y) − π(x), π(x) − xi ≥ 0 hπ(x) − π(y), π(y) − yi ≥ 0 without any eﬀort. Addition of the above inequalities yields

hπ(x) − π(y), (x − y) − (π(x) − π(y))i ≥ 0 or, equivalently, kπ(x) − π(y)k2 ≤ hπ(x) − π(y), x − yi. From this, by Cauchy- Schwarz, we get kπ(x) − π(y)k2 ≤ kπ(x) − π(y)kkx − yk, and ﬁnally kπ(x) − π(y)k ≤ kx − yk, as required.

2.9 Preserving the closed convex hull of the image

Note that one can also show that the extensions from Kirszbraun’s Theorem can be chosen such that their image is contained in the closed convex hull of the image of the original mapping.

Theorem 2.16. Let H1 and H2 be Hilbert spaces, S ⊆ H1 and g : S → H2 a function with the property that kg(x)−g(y)kH2 ≤ Lkx−ykH1 for all x, y ∈ S, for some L > 0.

Then there exists a function f : H1 → H2 such that kf(x) − f(y)kH2 ≤ Lkx − ykH1 for all x, y ∈ H1, f|S = g, and such that f(H1) ⊆ conv g(S).

Proof. Let f0 : H1 → H2 be a function such that kf0(x)−f0(y)kH2 ≤ Lkx−ykH1 for all x, y ∈ H1, and such that f0|S = g, which exists by Theorem 2.9. By Proposition 2.15, the nearest point projection π : H2 → conv g(S) is a 1-Lipschitz retraction. Put f := π ◦ f0. Then

kf(x) − f(y)kH2 = kπ(f0(x)) − π(f0(y))kH2 ≤ kf0(x) − f0(y)kH2 ≤ Lkx − ykH1 for all x, y ∈ H1, f|S = (π ◦ f0)|S = π ◦ f0|S = π ◦ g = g and by deﬁnition f(H1) ⊆ conv g(S).

2.10 What about Banach spaces?

We have just seen that it is always possible to extend nonexpansive maps on Hilbert spaces. In Example 2.1, however, we have encountered a pair of Banach spaces and a nonexpansive map that could not be extended. So does it only work in Hilbert spaces? With an additional assumption on the range space, this is in fact the case.

17 2. Extension Theorems for Lipschitz Functions

A Banach space (X, k · k) is called strictly convex if and only if, for any x ∈ X kxk and any y ∈ X \{0}, kx + yk = kxk + kyk only for x = kyk y. Then the following holds [23, Theorem 12.3], which we just cite without proof.

Theorem 2.17. Suppose X and Y are Banach spaces, Y is strictly convex and dim(Y ) ≥ 2. Then (X,Y ) has the nonexpansive extension property if and only if X and Y are Hilbert spaces.

18 Chapter 3

Urysohn’s Lemma and Tietze’s Extension Theorem

We prove Urysohn’s Lemma (Theorem 3.5) and Tietze’s Extension Theorem (The- orem 3.8) about continuous functions on normal topological spaces.

3.1 Normal topological spaces

Recall that a topological space (X, τ) is said to be normal if and only if for each disjoint pair of closed sets A, B ⊆ X there exist open sets U, V such that A ⊆ U, B ⊆ V , and U ∩ V = ∅. A normal space that has the Hausdorff property is called a T4 space. Let us first give some examples of normal topological spaces. As usual, a metrizable topological space is a space (X, τ) such that there exists a metric d on X such that τ and the topology induced by the metric d coincide. Theorem 3.1. Every metrizable space is normal. The proof of this seems to be standard. We follow [9]. Proof. Let (X, d) be a metric space. If A, B are disjoint closed subsets of X, put 1 U := [ B(a, dist(B, a)), a∈A 3 1 V := [ B(b, dist(A, b)). b∈B 3 Then U and V are open, U ⊇ A, V ⊇ B, and U ∩ V = ∅ by the triangle inequality. 1 In fact, if x ∈ U ∩ V , there exist a ∈ A such that d(a, x) < 3 dist(B, a) and b ∈ B 1 such that d(b, x) < 3 dist(A, b), and then 1 1 2 d(a, b) ≤ d(a, x) + d(x, b) < dist(B, a) + dist(A, b) ≤ d(a, b), 3 3 3 a contradiction. A topological space (X, τ) is called compact if and only if every open cover of X has a finite subcover.

19 3. Urysohn’s Lemma and Tietze’s Extension Theorem

Theorem 3.2. Every compact Hausdorﬀ space is normal.

The proof seems to be standard as well. Again we follow [9].

Proof. Let X be a compact Hausdorff space and A, B ⊆ X closed with A ∩ B = ∅. Then, as X is Hausdorff, for every a ∈ A and every b ∈ B, there exist open neighborhoods Ua,b of a and Va,b of b, such that a ∈ Ua,b, b ∈ Va,b and Ua,b ∩Va,b = ∅. S Now, for fixed a ∈ A, B ⊆ b∈B Va,b and, as B is compact as a closed subset of a (a) (a) Sna (a) compact space, there exist elements b1 , . . . , bna ∈ B such that B ⊆ i=1 V . Let a,bi

n [a Va := Va,b(a) i=1 i and n \a Ua := Ua,b(a) i=1 i for all a ∈ A. Then a ∈ Ua, B ⊆ Va, Ua and Va are open and Ua ∩ Va = ∅. S As A ⊆ a∈A Ua and A is compact, there exist elements a1, . . . , an ∈ A such that Sn Sn Tn A ⊆ i=1 Uai . Let U := i=1 Uai and V := i=1 Vai . Then A ⊆ U, B ⊆ V , U and V are open and U ∩ V = ∅. Also recall that a topological space (X, τ) is said to be regular if and only if for each point x ∈ X and each neighborhood U of x there exists a closed neighborhood V of x such that V ⊆ U, or, equivalently, for each point x ∈ X and each closed set A ⊆ X such that x∈ / A, there are disjoint open sets U and V such that x ∈ U and A ⊆ V . A regular Hausdorﬀ space is called a T3 space. A Lindelöf space is a topological space (X, τ) such that every open cover of X has a countable subcover.

Theorem 3.3. Each regular Lindelöf space is normal.

This is taken from [5, p. 113], where it is – without further reference – attributed to Tychonoﬀ. Proof. Let (X, τ) be a regular Lindelöf space and A, B closed subsets of X such that A ∩ B = ∅. For every point a of A, as X \ B is a neighborhood of a and X is regular, there exists a closed neighborhood of a that is contained in X \ B. Similarly, for every point b of B, as X \ A is a neighborhood of b and X is regular, there exists a closed neighborhood of b that is contained in X \ A. Put

U = {U ∈ τ; U ∩ B = ∅}, V = {V ∈ τ; V ∩ A = ∅}. Then, by the above considerations, U is an open cover of A, V is an open cover of B, and hence U ∪ V ∪ {X \ (A ∪ B)} is an open cover of X. So as X is Lindelöf,

20 3.2. A generalized variant of recursion

there will be members Un ∈ U, n = 1, 2,... , and members Vn ∈ V, n = 1, 2,... , S∞ S∞ such that A ⊆ n=1 Un and B ⊆ n=1 Vn. Let 0 [ Un := Un \ Vk, 1≤k≤n 0 [ Vn := Vn \ Uk. 1≤k≤n 0 0 S∞ 0 S∞ 0 Then the sets Un and Vn are again open, n=1 Un ∩ n=1 Vn = ∅ by construction, S∞ 0 S∞ 0 and we still have A ⊆ n=1 Un and B ⊆ n=1 Vn by the deﬁning properties of U and V. But this means that we have found disjoint open neigborhoods of A and B.

3.2 A generalized variant of recursion

In the proofs of the following two theorems – and also on various occasions later in the text – a slightly more general variant of the usual method of deﬁning functions by recursion will be used. More precisely, in contrast to the “traditional” method, which is for example used in set theory to introduce the arithmetic operations on the set of the natural numbers, or to deﬁne certain sequences of integers, the generalized method can be described as follows.

Proposition 3.4. Let X be a set, let Pn be an n-ary condition for n = 1, 2,... and suppose that

• there exists x1 ∈ X such that P1(x1) is true,

• for all n = 1, 2,... , if x1, x2, . . . , xn ∈ X are such that P1(x1), P2(x1, x2), ..., Pn(x1, x2, . . . , xn) are all true, then there exists xn+1 ∈ X such that Pn+1(x1, x2, . . . , xn+1) is true.

Then there exists a function Φ: {1, 2,... } → X such that Pn(Φ(1), Φ(2),..., Φ(n)) is true for all n = 1, 2,... . The “traditional” method relies only on the Peano Axioms. For the variant above, the Axiom of Choice (Zorn’s Lemma) seems to be necessary.

3.3 Urysohn’s Lemma

The following result on a continuous extension of a function taking just two values is usually known as “Urysohn’s Lemma”. The proof given here – while apparently not much diﬀerent from the original one [20] – is taken from [9]. See also [5, pp. 113–115]. Theorem 3.5 (Urysohn 1925). Let X be a normal topological space. Then for each pair of closed and disjoint subsets A and B of X, there exists a continuous function f : X → [0, 1] such that A ⊆ f −1({0}) and B ⊆ f −1({1}).

21 3. Urysohn’s Lemma and Tietze’s Extension Theorem

We ﬁrst show two lemmata. Lemma 3.6. Let X be a set, D ⊆ (0, +∞) dense and suppose that for each t ∈ D S there is a set Ut ⊆ X such that Us ⊆ Ut whenever s < t and t∈D Ut = X. Deﬁne f : X → [0, +∞) by f(x) := inf{t ∈ D; x ∈ Ut}. Then [ {x ∈ X; f(x) < s} = Ut D3t~~s for all s ∈ R. Proof. This is nothing but a simple computation. In fact,~~

~~{x ∈ X; f(x) < s} = {x ∈ X; inf{t ∈ D; x ∈ Ut} < s} = [ = {x ∈ X; ∃D 3 t < s : x ∈ Ut} = Ut D3t~~

{x ∈ X; f(x) ≤ s} = {x ∈ X; inf{t ∈ D; x ∈ Ut} ≤ s} = \ = {x ∈ X; ∀D 3 t > s : x ∈ Ut} = Ut. D3t>s Lemma 3.7. Let X be a topological space, D ⊆ (0, +∞) dense and suppose that for each t ∈ D there is an open set Ut ⊆ X such that Us ⊆ Ut whenever s < t and S t∈D Ut = X. Deﬁne f : X → [0, +∞) by

f(x) := inf{t ∈ D; x ∈ Ut}. Then f is continuous. Proof. As the intervals of the form (−∞, s) and (s, +∞) for s ∈ R are a subbase for the topology of the reals, in order to prove this claim, it is suﬃcient to check that their preimages {x ∈ X; f(x) < s} and {x ∈ X; f(x) > s} are open. But in view of Lemma 3.6 we have [ {x ∈ X; f(x) < s} = Ut, D3t ~~s} = X \{x ∈ X; f(x) ≤ s} = X \ Ut = X \ Ut, D3t>s D3t>s which is also open.~~

~~22 3.3. Urysohn’s Lemma~~

~~U 1 2~~

~~A U 3 U 1 4 4 X~~

~~Figure 3.1: Illustration for the proof of Urysohn’s Lemma.~~

Proof of Theorem 3.5. Let X be a normal topological space and A, B closed and disjoint subsets. We will construct a continuous function f : X → [0, 1] such that A ⊆ f −1({0}) and B ⊆ f −1({1}) from sets that correspond (approximately) to the “level sets” {x ∈ X; f(x) < t}, using the procedure outlined in Lemma 3.6 and Lemma 3.7. In fact, if we have a family (Ut)t∈D of open subsets of X for some dense subset D of (0, +∞) such that

~~• Us ⊆ Ut whenever s < t,~~

~~S • t∈D Ut = X,~~

~~• inf{t ∈ D; x ∈ Ut} = 0 for all x ∈ A,~~

~~• inf{t ∈ D; x ∈ Ut} = 1 for all x ∈ B,~~

~~• 0 ≤ inf{t ∈ D; x ∈ Ut} ≤ 1 for all x ∈ X, then, by Lemma 3.7, the function f : X → R deﬁned by~~

f(x) := inf{t ∈ D; x ∈ Ut} satisﬁes all the conditions. So let Q = {q1, q2,... } be the set of all rational numbers in (0, 1) and put D := Q ∪ [1, +∞), such that D is a dense subset of (0, +∞). Open sets Uq1 ,Uq2 ,... will be constructed recursively such that, for each n = 1, 2,... ,

~~(1) A ⊆ Uqn ,~~

~~(2) Uqn ⊆ X \ B,~~

~~(3) Us ⊆ Ut whenever s, t ∈ {q1, q2, . . . , qn} and s < t.~~

~~23 3. Urysohn’s Lemma and Tietze’s Extension Theorem~~

In order to construct the set Uq1 , put A1 := A and B1 := B and choose, using normality of the space X, Uq1 open such that A1 ⊆ Uq1 ⊆ Uq1 ⊆ X \ B1. Clearly, this open set Uq1 then satisﬁes the above conditions (1) - (3). If open sets Uq1 ,Uq2 ,...,Uqn−1 already have been constructed for some n ≥ 2, satisfying the above conditions (1) - (3), put [ An := A ∪ Uqk k∈{1,2,...,n−1}:qkqn

Then An and Bn are closed and disjoint. So, because of normality of the space X, one can choose Uqn open such that An ⊆ Uqn ⊆ Uqn ⊆ X \ Bn. This open set Uqn then satisﬁes again the conditions (1) - (3) above. Finally, let U1 := X \ B and Ut := X for t > 1. Then the family (Ut)t∈D has the required properties. Note that the conclusion of Urysohn’s Lemma actually characterizes normal topological spaces. In fact, suppose a topological space X has the property that for each pair of closed and disjoint subsets A and B, there exists a continuous function f : X → [0, 1] which is zero on A and one on B. Then X must be normal, for if −1 1 −1 1 A and B are closed and disjoint, we have A ⊆ f ([0, 2 )), B ⊆ f (( 2 , 1]), and the −1 1 −1 1 sets U := f ([0, 2 )), V := f (( 2 , 1]) are open and disjoint.

~~3.4 Tietze’s Extension Theorem~~

Another useful property of normal topological spaces is that continuous functions deﬁned on closed subsets can be extended continuously to the entire space. This question was ﬁrst investigated by H. Tietze in [19] for the special case of metric spaces. Later P. Urysohn used Theorem 3.5 to show the general version for normal topological spaces [20]. Nevertheless, the result is usually known as “Tietze’s Extension Theorem”. The exposition here follows again [9].

Theorem 3.8 (Tietze 1915). Let X be a normal topological space. If F is a closed subset of X and f : F → R is continuous, then there exists a continuous extension ˜ ˜ f : X → R of f. Furthermore, the extension can be chosen such that infF f ≤ f ≤ supF f on X. We begin with the following lemma.

Lemma 3.9. Let X be a normal topological space, F ⊆ X closed and f : F → [0, α] 1 continuous for some α > 0. Then there exists a continuous function g : X → [0, 3 α] 2 such that 0 ≤ f − g ≤ 3 α on F . −1 1 −1 2 Proof. Let A := f ([0, 3 α]) and B := f ([ 3 α, α]). Then A and B are closed and disjoint subsets of F . By the closedness of F , they are also closed in X. So by

~~24 3.4. Tietze’s Extension Theorem~~

Urysohn’s Lemma (Theorem 3.5) there exists a continuous function g0 : X → [0, 1] 1 which is zero on A and one on B. Let g := 3 αg0. Then obviously g is a continuous 1 function from X to [0, 3 α]. Furthermore, if x ∈ A, then 1 2 f(x) − g(x) = f(x) ∈ [0, α] ⊆ [0, α], 3 3 if x ∈ B, then 1 1 2 2 f(x) − g(x) = f(x) − α ∈ [ α, α] ⊆ [0, α], 3 3 3 3

−1 1 2 and if x ∈ F \ (A ∪ B) = f (( 3 α, 3 α)), then 2 2 f(x) − g(x) ∈ (0, α) ⊆ [0, α]. 3 3 Proof of Theorem 3.8. Let us first consider the case that f is bounded. Then, without loss of generality, we can assume that f : F → [0, 1]. We recursively construct continous functions 2 f : F → [0, ( )n−1], n = 1, 2,..., n 3 and 1 2 g : X → [0, ( )n−1], n = 1, 2,..., n 3 3 2 n such that 0 ≤ fn − gn ≤ ( 3 ) on F as follows. Let f1 := f. Then f1 : F → [0, 1] 1 is continuous and by Lemma 3.9 there exists g1 : X → [0, 3 ] continuous with 0 ≤ 2 2 n−2 1 2 n−2 f1 − g1 ≤ 3 on F . If fn−1 : F → [0, ( 3 ) ] and gn−1 : X → [0, 3 ( 3 ) ] such that 2 n−1 0 ≤ fn−1 − gn−1 ≤ ( 3 ) on F are already constructed for some n ≥ 2, we define 2 n−1 fn := fn−1 − gn−1|F : F → [0, ( 3 ) ] and can find, according to Lemma 3.9, a 1 2 n−1 continuous function gn : X → [0, 3 · ( 3 ) ] with the property that 0 ≤ fn − gn ≤ 2 2 n−1 P∞ 3 · ( 3 ) on F . Let g := n=1 gn. As

∞ ∞ n−1 X X 1 2 1 1 kgnk∞ ≤ = 2 = 1, n=1 n=1 3 3 3 1 − 3 g is the uniform limit of continuous functions, hence itself continuous, and in particular g : X → [0, 1]. If x ∈ F , for any positive integer N,

∞ X |f(x) − g(x)| = |f(x) − gn(x)| ≤ n=1 ∞ X ≤ |f(x) − g1(x) − g2(x) − · · · − gN−1(x)| + |gn(x)| = n=N ∞ N−1 ∞ n−1 X 2 X 1 2 = |fN (x)| + |gn(x)| ≤ + = n=N 3 n=N 3 3

~~25 3. Urysohn’s Lemma and Tietze’s Extension Theorem~~

N−1 2 N−1 N−1 2 1 ( 3 ) 2 = + 2 = 2 , 3 3 1 − 3 3 so f(x) = g(x). Hence g is the extension we have been looking for. π π If f is unbounded, then arctan f : F → (− 2 , 2 ) is continuous and bounded, so, π π according to the above considerations, there exists g0 : X → [− 2 , 2 ] continuous −1 π π with g0 = arctan f on F . Let A := g0 ({− 2 , 2 }). Then A is a closed subset of X, disjoint from the closed subset F , so by Urysohn’s Lemma (Theorem 3.5) there exists h : X → [0, 1] continuous such that h = 0 on A and h = 1 on F . Deﬁne π π g := g0 · h. Then g : X → (− 2 , 2 ) is continuous with g = arctan f on F . Hence we can take tan g as our extension. Again, the conclusion of the theorem actually characterizes normal topological spaces. In fact, assume that some topological space X has the property that for every closed subset F and for every continuous function f : F → R there exists a ˜ ˜ continuous extension f of f to X such that infF f ≤ f ≤ supF f on X. Then if A and B are closed and disjoint subsets of X, consider F := A ∪ B and f := χB, that is,  0 x ∈ A, f(x) = 1 x ∈ B. Obviously, F is closed and f is continuous. If f˜ is an extension as above, let ˜−1 1 1 ˜−1 3 5 U = f ((− 4 , 4 )) and V = f (( 4 , 4 )). Then U and V are open, A ⊆ U, B ⊆ V , and U ∩ V = ∅.

~~26 Chapter 4~~

~~Double Extensions~~

Let us consider the following situation which canonically appears in the dual of a Banach space with the norm-metric and the weak* topology. On some normal topological space (X, τ) also a metric d is given, and a real-valued function g deﬁned on a τ-closed subset F of X is both τ|F -continuous and Lipschitz. According to what we have just seen, there are two (possibly diﬀerent) ways to extend g.

~~• By Theorem 2.3, there exists an extension f1 : X → R that is again Lipschitz, with the same Lipschitz constant.~~

• By Tietze’s Extension Theorem (Theorem 3.8), there exists an extension f2 : X → R that is again τ-continuous and such that infF g ≤ f2 ≤ supF g on X. But does there exist an extension that satisfies both of the above conditions? That is, is it possible to have f1 = f2? We will see shortly that in general it is not possible to meet this requirement (Example 4.1). However, if we restrict ourselves to bounded functions and require that the topology τ and the metric d are “compatible” in a sense to be specified, then a result of Matoušková [11, 12] guarantees the existence of such “double extensions” (Theorem 4.2). A large part of the present chapter will be devoted to the presentation of this result. Finally, also two interesting applications to compact Hausdorff spaces (Corollary 4.6) and duals of Banach spaces (Corollary 4.8) will be discussed.

~~4.1 A counterexample~~

As was already mentioned above, without further restrictions it may happen that the collections of extensions provided by Theorem 2.3 and by Tietze’s Extension Theorem, respectively, are mutually disjoint. The following example is the result of joint work with Noema Nicolussi. Example 4.1. Let X be the space of the reals and d the metric on X given by the formula  0 x = y  d(x, y) = 1 x 6= y & x, y ∈ (−1, 1)  2 1 else

27 4. Double Extensions for x, y ∈ X. Then the set F = {0, 1} ⊆ X is closed with respect to the usual topology for the reals and the function g : F → R, 0 7→ 0, 1 7→ 1, is continuous with respect to the usual topology for the reals relativized and 1-Lipschitz with respect to the metric d. However, there is no extension f : X → R of g that is at the same time continuous, and such that 0 ≤ f ≤ 1, and 1-Lipschitz.

Proof. Assume that such a function f exists. Then, for all x ∈ (−1, 1), we have 1 1 f(x) = |f(x) − f(0)| ≤ d(x, 0) ≤ 2 , so, by continuity, f ≤ 2 on [−1, 1]. However, as f is an extension of g, f(1) = g(1) = 1. Contradiction.

~~Note that the topology τd induced by the metric d is the discrete topology, that is, τd = P(X), the set of all subsets, and hence τ = τ|·| ⊆ τd.~~

~~4.2 Matoušková’s positive result~~

Nevertheless, E. Matoušková has shown in [11] that under certain conditions “double extensions” as above do exist. Namely, we have to restrict our attention to bounded functions, and we have to ensure that the topology and the metric are “compatible”, in the following sense. If (X, d) is a metric space, S a subset of X and r ≥ 0, we write B(S, r) = {x ∈ X; d-dist(S, x) ≤ r}. The “compatibility condition” between the topology and the metric then is that the sets B(A, α) must be closed with respect to the topology for every subset A of X that is closed with respect to the topology, and for every α > 0. Thus we have the following theorem [11, Theorem 2.3]. See also [2, pp. 55–57].

Theorem 4.2 (Matoušková 1997). Let (X, τ) be a normal topological space and d a metric on X such that the set B(A, α) = {x ∈ X; d-dist(A, x) ≤ α} is τ-closed for every τ-closed A ⊆ X and every α > 0. Suppose that F ⊆ X is τ-closed and g : F → R is a bounded τ|F -continuous function that is L-Lipschitz in d for some L > 0. Then there exists a τ-continuous function f : X → R such that f|F = g, infF g ≤ f ≤ supF g on X, and that is L-Lipschitz in d. As we will see shortly, the main idea is to start with the construction from the proof of Urysohn’s Lemma (Theorem 3.5), that is, to construct an extension from sets that correspond (approximately) to its “level sets”, and to modify this construction a bit. In fact, for the resulting function to be not only continuous, but also an extension, and to satisfy the given Lipschitz condition, some additional conditions have to be imposed on those “level sets”. The following two lemmata will outline the strategy.

~~Lemma 4.3. Let X be a set, Y ⊆ X a subset, g : Y → [0, +∞) a function, D ⊆ (0, +∞) dense and (Ut)t∈D a family of subsets of X such that Us ⊆ Ut whenever~~

~~28 4.2. Matoušková’s positive result~~

~~X \ Ua~~

~~{y ∈ Y ; g(y) > b}~~

~~{y ∈ Y ; g(y) < a} Y~~

~~U b X~~

~~Figure 4.1: Illustration for Lemma 4.3.~~

~~S s < t and t∈D Ut = X. Deﬁne f(x) := inf{t ∈ D; x ∈ Ut} for x ∈ X. Then if ε > 0 and {y ∈ Y ; g(y) < a} ⊆ Ub ε for all a < b in some 5 -net of D, f ≤ g + ε on Y , and if ε > 0 and~~

~~{y ∈ Y ; g(y) > b} ⊆ X \ Ua~~

~~ε for all a < b in some 5 -net of D, g − ε ≤ f on Y . In particular, if both of the above conditions hold for all a < b in D, then f = g on Y .~~

~~Proof. Assume that the ﬁrst condition holds for some ε > 0. Fix y0 ∈ Y . Choose a ε and b from the 5 -net of D such that g(y0) < a < b < g(y0) + ε. Then it follows that~~

y0 ∈ {y ∈ Y ; g(y) < a} ⊆ Ub, so f(y0) ≤ b < g(y0)+ε. Similarly, assume that the second condition holds for some ε ε > 0. Fix y0 ∈ Y . Choose a and b from the 5 -net of D such that g(y0) − ε < a < b < g(y0). Then it follows that

~~y0 ∈ {y ∈ Y ; g(y) > b} ⊆ X \ Ua, so g(y0) − ε < a ≤ f(y0).~~

Lemma 4.4. Let (X, d) be a metric space, D ⊆ (0, +∞) dense and (Ut)t∈D a family S of subsets of X such that Us ⊆ Ut whenever s < t and t∈D Ut = X. Assume further that d-dist(Us,X \ Ut) ≥ t − s whenever s < t, Us 6= ∅ and X \ Ut 6= ∅. Then the function f : X → [0, +∞) deﬁned as f(x) := inf{t ∈ D; x ∈ Ut} is 1-Lipschitz.

~~29 4. Double Extensions~~

~~X \ Ut~~

~~> t − s~~

~~U s X~~

~~Figure 4.2: Illustration for Lemma 4.4.~~

Proof. Let x, y ∈ X. Without loss of generality we can assume that f(x) < f(y). Then, for all s ∈ D with s > f(x), x ∈ Us, in particular Us 6= ∅, and, for all t ∈ D with t < f(y), y ∈ X \ Ut, in particular X \ Ut 6= ∅, hence

~~d(x, y) ≥ sup d-dist(Us,X \ Ut) ≥ f(x)~~

~~Lemma 4.5. Let (X, τ) be a normal topological space and d a metric on X such that the set B(A, α) is τ-closed for every τ-closed A ⊆ X and every α > 0.~~

~~(1) If 0 < s < r < M are some numbers, E is a τ-closed subset of X and (Ft)0 r − t for each 0 < t ≤ s, then~~

~~[ E ∩ B(Ft, s − t) = ∅. 0~~

~~(2) If 0 < r < s < M are some numbers, E is a τ-closed subset of X and (Ft)s≤t t − r for each s ≤ t < M, then~~

~~[ E ∩ B(Ft, t − s) = ∅. s≤t~~

~~30 4.2. Matoušková’s positive result~~

~~B(Ft, s − t) B(Ft, t − s)~~

~~Ft E E Ft~~

~~0 t s r M 0 r s t M~~

~~> r − t > t − r (a) Part (1). (b) Part (2).~~

~~Figure 4.3: Illustration for Lemma 4.5.~~

Proof. For the first part, define si := i(r − s) for i = 0, 1,... and choose j ≥ 0 so that 2s − r ≤ sj < s. Note that such a j always exists as s = (2s − r) + (r − s). Then consider the set [ C = B(Fs, r − s) ∪ B(Fsi , r − si). 1≤i≤j C is τ-closed as the finite union of τ-closed sets. Furthermore, by construction of the set, C ∩ E = ∅. Finally, we will show that B(Ft, s − t) ⊆ C for all t ∈ (0, s], such that also the closure of the union of all those sets is contained in C and hence disjoint from E. We consider two cases. If 2s − r ≤ t ≤ s, then s − t ≤ r − s and since Ft ⊆ Fs we have B(Ft, s − t) ⊆ B(Fs, r − s) ⊆ C. If si−1 < t ≤ si for some 1 ≤ i ≤ j, then

~~s − t < s − si−1 = s − (i − 1)(r − s) = r − i(r − s) = r − si~~

and since Ft ⊆ Fsi we have B(Ft, s − t) ⊆ B(Fsi , r − si) ⊆ C. For the second part, deﬁne si := M − i(s − r) for i = 0, 1,... and choose j ≥ 0 so that s < sj ≤ 2s − r. Note that such a j always exists as s = (2s − r) − (s − r). Then consider the set [ C = B(Fs, s − r) ∪ B(Fsi , si − r). 1≤i≤j C is τ-closed as the ﬁnite union of τ-closed sets. Furthermore, by construction of the set, C ∩ E = ∅. Finally, we will show that B(Ft, t − s) ⊆ C for all t ∈ [s, M), such that also the closure of the union of all those sets is contained in C and hence disjoint from E. We consider two cases. If s ≤ t ≤ 2s − r, then t − s ≤ s − r and since Ft ⊆ Fs we have B(Ft, t − s) ⊆ B(Fs, s − r) ⊆ C. If si ≤ t < si−1 for some 1 ≤ i ≤ j, then

~~t − s < si−1 − s = M − (i − 1)(s − r) − s = M − i(s − r) − r = si − r~~

~~and since Ft ⊆ Fsi we have B(Ft, t − s) ⊆ B(Fsi , si − r) ⊆ C.~~

~~31 4. Double Extensions~~

Proof of Theorem 4.2. Let (X, τ) be a normal topological space, d a metric on X such that the sets B(A, α) are τ-closed for every τ-closed A ⊆ X and every α > 0, F ⊆ X a τ-closed subset and g : F → R a bounded τ|F -continuous function that is L-Lipschitz in d on F for some L > 0. Without loss of generality we can assume that infF g = 0 and that L = 1. Let M = supF g. If M = 0 we can take f = 0. Otherwise, let Q = {q1, q2,... } be the set of all rational numbers in (0,M). Suppose we can construct a family (Ur)r∈Q of τ-open subsets of X such that

~~•{ x ∈ F ; g(x) < r} ⊆ Ur for all r ∈ Q,~~

~~•{ x ∈ F ; g(x) > r} ⊆ X \ Ur for all r ∈ Q,~~

• d-dist(Us,X \ Ut) ≥ t − s whenever s, t ∈ Q and s < t, where it should be noted that the last condition makes sense as the ﬁrst two imply ∅ $ Ur $ X for all r ∈ Q, and put Ut = X for t ≥ M. Then by Lemma 3.7, Lemma 4.3 and Lemma 4.4, the function f : X → R deﬁned by

f(x) := inf{t ∈ Q ∪ [M, +∞); x ∈ Ut}, is τ-continuous, coincides with g on F and is 1-Lipschitz in d. Clearly we also have 0 ≤ f ≤ M. So all we have to do is to construct such a family of sets.

~~The sets Uq1 ,Uq2 ,... will be constructed recursively such that for each n = 1, 2,...~~

~~(1) {x ∈ F ; g(x) < qn} ⊆ Uqn ,~~

~~(2) {x ∈ F ; g(x) > qn} ⊆ X \ Uqn ,~~

~~(3) d-dist(Uqk ,X \ Uql ) ≥ ql − qk if 1 ≤ k, l ≤ n and qk < ql,~~

~~(4) d-dist(Vt,X \ Uql ) ≥ ql − t if 0 < t < ql, for all 1 ≤ l ≤ n,~~

~~(5) d-dist(Uqk ,F \ Vt) ≥ t − qk if qk < t < M, for all 1 ≤ k ≤ n, where Vλ = {x ∈ F ; g(x) < λ} for λ ∈ R.~~

~~Basis. We begin with the case n = 1. Let~~

~~[ A = B(Vt, q1 − t) 0~~

~~32 4.2. Matoušková’s positive result~~

~~The functions ψj,m. For each j = 1, 2,... with qj > q1 and m = 1, 2,... choose a τ-continuous function ψj,m on X so that 0 ≤ ψj,m ≤ 1 and~~

~~ 1 1, x ∈ B(F \ Vqj , (1 − m )(qj − q1)), ψj,m(x) = (4.1) 0, x ∈ A.~~

~~This is possible by Urysohn’s Lemma (Theorem 3.5) since both of the above sets are τ-closed subsets of the normal space X and disjoint, as an application of Lemma 4.5 (1) with~~

~~• s = q1,~~

~~1 • r ∈ (q1, qj − (1 − m )(qj − q1)), • M = M,~~

~~1 • E = B(F \ Vqj , (1 − m )(qj − q1)),~~

~~• Ft = Vt shows. In fact, Vt = {x ∈ F ; g(x) < t} ⊆ {x ∈ F ; g(x) ≤ t} by continuity of g,~~

F \ Vqj = {x ∈ F ; g(x) ≥ qj}, so, exploiting the fact that g is 1-Lipschitz on F , d-dist(Vt,F \ Vqj ) ≥ qj − t and hence 1 d-dist(F ,E) = d-dist(V ,B(F \ V , (1 − )(q − q ))) ≥ t t qj m j 1 1 ≥ q − t − (1 − )(q − q ) > r − t. j m j 1

~~The functions ψi,m. Similarly, for each i = 1, 2,... with qi < q1 and m = 1, 2,... , choose a τ-continuous function ψi,m on X so that −1 ≤ ψi,m ≤ 0 and~~

~~ 1 −1, x ∈ B(Vqi , (1 − m )(q1 − qi)), ψi,m(x) = (4.2) 0, x ∈ B.~~

~~Again this is possible by Urysohn’s Lemma (Theorem 3.5) since both of the above sets are τ-closed subsets of the normal space X and disjoint, as an application of Lemma 4.5 (2) with~~

~~1 • r ∈ (qi + (1 − m )(q1 − qi), q1),~~

~~• s = q1, • M = M,~~

~~1 • E = B(Vqi , (1 − m )(q1 − qi)),~~

~~• Ft = F \ Vt~~

~~33 4. Double Extensions~~

shows. In fact, F \ Vt = {x ∈ F ; g(x) ≥ t}, Vqi = {x ∈ F ; g(x) < qi} ⊆ {x ∈ F ; g(x) ≤ qi} by continuity of g, so, exploiting the fact that g is 1-Lipschitz on F , d-dist(F \ Vt, Vqi ) ≥ t − qi and hence 1 d-dist(F ,E) = d-dist(F \ V ,B(V , (1 − )(q − q ))) ≥ t t qi m 1 i 1 ≥ t − q − (1 − )(q − q ) > t − r. i m 1 i ∞ Let (ai)i=1 be some enumeration of {1, 2,... } × {1, 2,... }. Put ∞ X −i h(x) := 2 ψai (x). i=1 Then h is clearly τ-continuous. Deﬁne

~~Uq1 := {x ∈ X; h(x) < 0}. Then this set is clearly τ-open. We check that the conditions (1) - (5) from above are satisﬁed.~~

~~(1) Let x ∈ F such that g(x) < q1. Choose i ∈ {1, 2,... } such that g(x) < qi < q1.~~

~~Then x ∈ Vqi , so ψi,m(x) = −1 for all m, ψj,m(x) = 0 for all j with qj > q1~~

~~and all m and hence h(x) < 0. But this implies x ∈ Uq1 .~~

~~(2) Let x ∈ F such that g(x) > q1. Choose j ∈ {1, 2,... } such that g(x) > qj > q1.~~

~~Then x ∈ F \ Vqj , so ψj,m(x) = 1 for all m, ψi,m(x) = 0 for all i with qi < q1~~

~~and all m and hence h(x) > 0. But this implies x ∈ X \ Uq1 . (3) This is trivially satisﬁed. (4) It is suﬃcient to show that~~

~~d-dist(Vqi ,X \ Uq1 ) ≥ q1 − qi~~

for all i such that 0 < qi < q1, as the general case follows then from this using a density argument. So ﬁx such an i. Assume that 1 X \ U ∩ B(V , (1 − )(q − q )) 6= q1 qi m 1 i ∅ for some m. Let x be a point in the intersection. Then clearly h(x) ≥ 0 and 1 ψi,m(x) = −1, but as x ∈ B(Vqi , (1 − m )(q1 − qi)) ⊆ B(Vqi , q1 − qi) ⊆ A, ψj,m(x) = 0 for all j with qj > q1 and all m, a contradiction. So 1 X \ U ∩ B(V , (1 − )(q − q )) = q1 qi m 1 i ∅ for all m and we have proved our claim.

~~34 4.2. Matoušková’s positive result~~

~~(5) Again we only show that~~

~~d-dist(Uq1 ,F \ Vqj ) ≥ qj − q1~~

for all j such that q1 < qj < M, as the general case can then be easily obtained from this using a density argument. So ﬁx such a j. Assume that 1 U ∩ B(F \ V , (1 − )(q − q )) 6= q1 qj m j 1 ∅ for some m. Let x be a point in the intersection. Then clearly h(x) ≤ 0 and 1 ψj,m(x) = 1, but as x ∈ B(F \ Vqj , (1 − m )(qj − q1)) ⊆ B(F \ Vqj , qj − q1) ⊆ B, ψi,m(x) = 0 for all i with qi < q1 and all m, a contradiction. So 1 U ∩ B(F \ V , (1 − )(q − q )) = q1 qj m j 1 ∅ for all m and we have proved our claim.

~~Step. Suppose now that Uq1 ,Uq2 ,...,Uqn−1 have already been constructed for some n ≥ 2, satisfying the conditions (1) - (5) from above. Let [ A = B(Vt, qn − t), 0qn~~

~~The functions ϕl,m. For each l < n with ql > qn and m = 1, 2,... choose a τ-continuous function ϕl,m on X so that 0 ≤ ϕl,m ≤ 1 and~~

~~ 1 1, x ∈ B(X \ Uql , (1 − m )(ql − qn)), ϕl,m(x) = (4.3) 0, x ∈ P ∪ A. This is possible since by condition (3) for n − 1~~

d-dist(Uqk ,X \ Uql ) ≥ ql − qk for all k < n such that qk < qn and hence 1 B(X \ U , (1 − )(q − q )) ∩ P = ql m l n ∅ and also 1 B(X \ U , (1 − )(q − q )) ∩ A = , ql m l n ∅ as an application of Lemma 4.5 (1) with

~~35 4. Double Extensions~~

~~• s = qn,~~

~~1 • r ∈ (qn, ql − (1 − m )(ql − qn)), • M = M,~~

~~1 • E = B(X \ Uql , (1 − m )(ql − qn)),~~

~~• Ft = Vt shows. In fact, by condition (4) for n − 1,~~

d-dist(Vt,X \ Uql ) ≥ ql − t for all 0 < t < ql and hence 1 d-dist(F ,E) = d-dist(V ,B(X \ U , (1 − )(q − q ))) ≥ t t ql m l n 1 ≥ q − t − (1 − )(q − q ) > r − t l m l n for all 0 < t ≤ qn. The functions ϕk,m. Similarly, for each k < n with qk < qn and m = 1, 2,... , choose a τ-continuous function ϕk,m on X so that −1 ≤ ϕk,m ≤ 0 and

~~ 1 −1, x ∈ B(Uqk , (1 − m )(qn − qk)), ϕk,m(x) = (4.4) 0, x ∈ R ∪ B.~~

~~Again this is possible since by condition (3) for n − 1~~

d-dist(Uqk ,X \ Uql ) ≥ ql − qk for all l < n such that ql > qn and hence 1 B(U , (1 − )(q − q )) ∩ R = qk m n k ∅ and also 1 B(U , (1 − )(q − q )) ∩ B = , qk m n k ∅ as an application of Lemma 4.5 (2) with

~~1 • r ∈ (qk + (1 − m )(qn − qk), qn),~~

~~• s = qn, • M = M,~~

~~1 • E = B(Uqk , (1 − m )(qn − qk)),~~

~~36 4.2. Matoušková’s positive result~~

~~• Ft = F \ Vt shows. In fact, by condition (5) for n − 1,~~

d-dist(Uqk ,F \ Vt) ≥ t − qk for all qk < t < M and hence 1 d-dist(F ,E) = d-dist(F \ V ,B(U , (1 − )(q − q ))) ≥ t t qk m n k 1 ≥ t − q − (1 − )(q − q ) > t − r k m n k for all qn ≤ t < M. The functions ψj,m. For each j = 1, 2,... with qj > qn and m = 1, 2,... also choose a τ-continuous function ψj,m on X so that 0 ≤ ψj,m ≤ 1 and

~~ 1 1, x ∈ B(F \ Vqj , (1 − m )(qj − qn)), ψj,m(x) = (4.5) 0, x ∈ P ∪ A.~~

~~This is possible since by condition (5) for n − 1~~

d-dist(Uqk ,F \ Vqj ) ≥ qj − qk for every k < n such that qk < qn and hence 1 B(F \ V , (1 − )(q − q )) ∩ P = qj m j n ∅ and also 1 B(F \ V , (1 − )(q − q )) ∩ A = , qj m j n ∅ as an application of Lemma 4.5 (1) with

~~• s = qn,~~

~~1 • r ∈ (qn, qj − (1 − m )(qj − qn)), • M = M,~~

~~1 • E = B(F \ Vqj , (1 − m )(qj − qn)),~~

~~• Ft = Vt shows. In fact, Vt = {x ∈ F ; g(x) < t} ⊆ {x ∈ F ; g(x) ≤ t} by continuity of g,~~

F \ Vqj = {x ∈ F ; g(x) ≥ qj}, so, exploiting the fact that g is 1-Lipschitz on F , d-dist(Vt,F \ Vqj ) ≥ qj − t and hence 1 d-dist(F ,E) = d-dist(V ,B(F \ V , (1 − )(q − q ))) ≥ t t qj m j n

~~37 4. Double Extensions~~

~~1 ≥ q − t − (1 − )(q − q ) > r − t. j m j n~~

~~The functions ψi,m. Similarly, for each i = 1, 2,... with qi < qn and m = 1, 2,... , also choose a τ-continuous function ψi,m on X so that −1 ≤ ψi,m ≤ 0 and~~

~~ 1 −1, x ∈ B(Vqi , (1 − m )(qn − qi)), ψi,m(x) = (4.6) 0, x ∈ R ∪ B. Again this is possible since by condition (4) for n − 1~~

d-dist(Vqi ,X \ Uql ) ≥ ql − qi for every l < n with ql > qn and hence 1 B(V , (1 − )(q − q )) ∩ R = qi m n i ∅ and also 1 B(V , (1 − )(q − q )) ∩ B = , qi m n i ∅ as an application of Lemma 4.5 (2) with

~~1 • r ∈ (qi + (1 − m )(qn − qi), qn),~~

~~• s = qn, • M = M,~~

~~1 • E = B(Vqi , (1 − m )(qn − qi)),~~

• Ft = F \ Vt shows. In fact, F \ Vt = {x ∈ F ; g(x) ≥ t}, Vqi = {x ∈ F ; g(x) < qi} ⊆ {x ∈ F ; g(x) ≤ qi} by continuity of g, so, exploiting the fact that g is 1-Lipschitz on F , d-dist(F \ Vt, Vqi ) ≥ t − qi and hence 1 d-dist(F ,E) = d-dist(F \ V ,B(V , (1 − )(q − q ))) ≥ t t qi m n i 1 ≥ t − q − (1 − )(q − q ) > t − r. i m n i

If α ∈ {1, 2,... } × {1, 2,... } is such that ϕα was not yet deﬁned, put ϕα = 0. ∞ Let (ai)i=1 be some enumeration of {1, 2,... } × {1, 2,... }. Deﬁne ∞ X −i h(x) := 2 (ϕai (x) + ψai (x)). i=1 Then h is clearly τ-continuous. Let

~~Uqn := {x ∈ X; h(x) < 0}. Then this set is obviously τ-open. We check that the conditions (1) - (5) from above are satisﬁed.~~

~~38 4.2. Matoušková’s positive result~~

~~(1) Let x ∈ F such that g(x) < qn. Choose i ∈ {1, 2,... } such that g(x) < qi < qn.~~

~~Then x ∈ Vqi , so ψi,m(x) = −1 for all m, ψj,m(x) = 0 for all j with qj > qn and all m and also ϕl,m(x) = 0 for all l < n with ql > qn and all m, so h(x) < 0.~~

~~But this implies x ∈ Uqn .~~

~~(2) Let x ∈ F such that g(x) > qn. Choose j ∈ {1, 2,... } such that g(x) > qj >~~

~~qn. Then x ∈ F \ Vqj , so ψj,m(x) = 1 for all m, ψi,m(x) = 0 for all i with qi < qn and all m and also ϕk,m(x) = 0 for all k < n with qk < qn and all m,~~

~~so h(x) > 0. But this implies x ∈ X \ Uqn .~~

~~(3) If k, l < n then nothing has to be shown. If k = n, then we have to show that~~

~~d-dist(Uqn ,X \ Uql ) ≥ ql − qn~~

~~for all l < n with ql > qn. Fix such an l. Assume that~~

~~1 U ∩ B(X \ U , (1 − )(q − q )) 6= qn ql m l n ∅~~

for some m. Let x be a point in the intersection. Then clearly h(x) ≤ 0 and 1 ϕl,m(x) = 1, but as x ∈ B(X \ Uql , (1 − m )(ql − qn)) ⊆ B(X \ Uql , ql − qn) ⊆ R, ϕk,m(x) = 0 for all k < n with qk < qn and all m and also ψi,m(x) = 0 for all i with qi < qn and all m, a contradiction. So

~~1 U ∩ B(X \ U , (1 − )(q − q )) = qn ql m l n ∅~~

~~for all m and condition (3) is satisﬁed. If l = n, then we have to show that~~

~~d-dist(Uqk ,X \ Uqn ) ≥ qn − qk~~

~~for all k < n with qk < qn. Fix such a k. Assume that~~

~~1 X \ U ∩ B(U , (1 − )(q − q )) 6= qn qk m n k ∅~~

for some m. Let x be a point in the intersection. Then clearly h(x) ≥ 0 and 1 ϕk,m(x) = −1, but as x ∈ B(Uqk , (1 − m )(qn − qk)) ⊆ B(Uqk , qn − qk) ⊆ P , ϕl,m(x) = 0 for all l < n with ql > qn and all m and also ψj,m(x) = 0 for all j with qj > qn and all m, a contradiction. So

~~1 X \ U ∩ B(U , (1 − )(q − q )) = qn qk m n k ∅~~

~~for all m and condition (3) is satisﬁed.~~

~~39 4. Double Extensions~~

~~(4) If l < n, obviously nothing has to be shown. So we only have to check that~~

~~d-dist(Vt,X \ Uqn ) ≥ qn − t~~

~~for 0 < t < qn. By using a density argument, it is suﬃcient to show this inequality for rational t only, i.e. that~~

~~d-dist(Vqi ,X \ Uqn ) ≥ qn − qi~~

for all i such that 0 < qi < q1. So ﬁx such an i. Assume that 1 X \ U ∩ B(V , (1 − )(q − q )) 6= qn qi m n i ∅ for some m. Let x be a point in the intersection. Then clearly h(x) ≥ 0 and 1 ψi,m(x) = −1, but as x ∈ B(Vqi , (1 − m )(qn − qi)) ⊆ B(Vqi , qn − qi) ⊆ A, ϕl,m(x) = 0 for all l < n with ql > qn and all m and also ψj,m(x) = 0 for all j with qj > qn and all m, a contradiction. So 1 X \ U ∩ B(V , (1 − )(q − q )) = qn qi m n i ∅ for all m and we have proved our claim. (5) If k < n, obviously nothing has to be shown. So we only have to check that

~~d-dist(Uqn ,F \ Vt) ≥ t − qn~~

~~for qn < t < M. Like before, by using a density argument, it is suﬃcient to show this inequality only for rational t, i.e. that~~

~~d-dist(Uqn ,F \ Vqj ) ≥ qj − qn~~

for all j such that qn < qj < M. So ﬁx such a j. Assume that 1 U ∩ B(F \ V , (1 − )(q − q )) 6= qn qj m j n ∅ for some m. Let x be a point in the intersection. Then clearly h(x) ≤ 0 and 1 ψj,m(x) = 1, but as x ∈ B(F \ Vqj , (1 − m )(qj − qn)) ⊆ B(F \ Vqj , qj − qn) ⊆ B, ϕk,m(x) = 0 for all k < n with qk < qn and all m and also ψi,m(x) = 0 for all i with qi < q1 and all m, a contradiction. So 1 U ∩ B(F \ V , (1 − )(q − q )) = qn qj m j n ∅ for all m and we have proved our claim.

~~40 4.3. Compact Hausdorﬀ spaces~~

~~4.3 Compact Hausdorﬀ spaces~~

We will now see some applications of Theorem 4.2. Compact Hausdorﬀ spaces are in particular normal spaces, as we have seen above (Theorem 3.2) and it turns out that the topologies of such spaces and lower semi-continuous metrics deﬁned on them are always “compatible” in the sense of Theorem 4.2 [11, Corollary 2.4].

Corollary 4.6 (Matoušková 1997). Let (X, τ) be a compact Hausdorﬀ space, d a τ-lower semi-continuous metric on X, F ⊆ X τ-closed, g ∈ C(F ) and, for some L > 0, L-Lipschitz in d. Then there exists f ∈ C(X) such that f = g on F , minF g ≤ f ≤ maxF g, and f is L-Lipschitz in d.

~~This is the immediate consequence of the following lemma.~~

Lemma 4.7. Let (X, τ) be a compact Hausdorﬀ space and d a τ-lower semi- continuous metric on X. Then for every τ-closed A ⊆ X and every α > 0 the set B(A, α) = {x ∈ X; d-dist(A, x) ≤ α} is τ-closed.

~~Proof. Let A ⊆ X be τ-closed and α > 0. If x ∈ X, then d-dist(A, x) = infA×{x} d and the inﬁmum is attained at some point, as~~

~~\ {d|A×{x} = inf d} = {d|A×{x} ≤ δ} = A×{x} δ>infA×{x} d = \ (A × {x}) ∩ {d ≤ δ},~~

δ>infA×{x} d and the intersection is not empty because of the ﬁnite intersection property, i.e. if a family of closed subsets of a compact space has the property that every ﬁnite intersection of members of the family has non-void intersection, then also the intersection of all members of the family contains some point. But now we have

~~B(A, α) = {x ∈ X; d-dist(A, x) ≤ α} = {x ∈ X; ∃a ∈ A : d(a, x) ≤ α} =~~

~~= pr2({(a, x) ∈ A × X; d(a, x) ≤ α}) =~~

= pr2((A × X) ∩ {(x1, x2) ∈ X × X; d(x1, x2) ≤ α}), where pr2 denotes the projection on the second coordinate, and this set is τ-closed. In fact, (A × X) ∩ {(x1, x2) ∈ X × X; d(x1, x2) ≤ α} is closed in X × X, hence compact, and as pr2 is continuous, also its image

~~pr2((A × X) ∩ {(x1, x2) ∈ X × X; d(x1, x2) ≤ α}) is τ-compact, in particular τ-closed.~~

~~41 4. Double Extensions~~

~~4.4 Duals of Banach spaces~~

Another application of Theorem 4.2 is in the context of Banach spaces. Banach spaces are a canonical example of a space with both a metric (the norm-distance) and a topology (the weak or weak* topology). Again one can show that these are “compatible” in the sense of Theorem 4.2 [11, Corollary 2.5].

Corollary 4.8 (Matoušková 1997). Let X be a Banach space. Let F be a w∗-closed subset of its dual X∗, g a bounded, w∗-continuous function on F that is L-Lipschitz in the norm-metric on X∗ for some L > 0. Then there exists a w∗-continuous ∗ function f on X such that f = g on F , infF g ≤ f ≤ supF g, and f is L-Lipschitz in the norm-metric on X∗.

In order to apply Theorem 4.2, we ﬁrst have to see why the dual of a Banach space with the weak* topology is a normal topological space. We will show this using Theorem 3.3. So we have to check that the dual of a Banach space with the weak* topology is a regular Lindelöf space.

~~Proposition 4.9. Let X be a Banach space. Then (X∗, w∗) is regular.~~

Proof. This follows from the regularity of R and from the fact that products and subspaces of regular spaces are again regular, as the weak* topology on X∗ is the relativization to X∗ of the product topology on RX .

~~Proposition 4.10. Let X be a Banach space. Then (X∗, w∗) is a Lindelöf space.~~

∗ ∗ Proof. Consider a cover (Ui)i∈I of X by w -open sets. We have to show that there exists a countable subcover. For each positive integer n, by Alaoglu’s Theorem and as multiplication by a scalar is w∗-continuous, B[0, n] is a w∗-compact subspace of ∗ X , so there is a ﬁnite subset In of I such that (Ui)i∈In covers B[0, n]. But then

~~∞ ∗ [ [ X = Ui, n=1 i∈In such that (Ui)i∈S∞ I is the countable subcover we have been looking for. n=1 n Thus, to sum up, we have shown the following.~~

~~Theorem 4.11. Let X be a Banach space. Then (X∗, w∗) is a normal topological space.~~

What remains to check for Corollary 4.8 is that the sets B(A, α) with respect to the norm-metric on X∗ are w∗-closed for every w∗-closed A ⊆ X∗ and every α > 0. This is contained in the following lemma which will thus make the proof of Corollary 4.8 complete.

~~42 4.5. Reﬂexive Banach spaces~~

Lemma 4.12. Let X be a Banach space. Suppose that A is a w∗-closed subset of its dual X∗ and α > 0. Then the set B(A, α) = {x∗ ∈ X∗; k · k-dist(A, x∗) ≤ α} is w∗-closed. Proof. We will show that B(A, α) = A+B[0, α]. The latter is the sum of a w∗-closed and a w∗-compact set, by Alaoglu’s Theorem and as multiplication by a scalar is ∗ ∗ ∗ w -continuous, and hence w -closed. Indeed, let (xi )i∈I be a net in A + B[0, α] that converges to some point x∗ of X∗ with respect to w∗. Then it can be written as

~~∗ ∗ ∗ (xi )i∈I = (yi )i∈I + (zi )i∈I ,~~

∗ ∗ ∗ where (yi )i∈I is a net in A and (zi )i∈I is a net in B[0, α]. As B[0, α] is w -compact, ∗ ∗ ∗ (zi )i∈I has some subnet (zj )j∈J that converges to a point z of B[0, α] with respect ∗ ∗ ∗ ∗ ∗ to w . Hence also (yj )j∈J = (xj )j∈J − (zj )j∈J converges with respect to w , and since A is w∗-closed, its limit, say y∗, lies again in A. But then we have x∗ = y∗ + z∗ ∈ A + B[0, α], and A + B[0, α] is w∗-closed. For the above equality let now ﬁrst x∗ ∈ A + B[0, α]. Then one can write x∗ = y∗ + z∗ for some y∗ ∈ A and z∗ ∈ B[0, α], such that ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ky − x k = ky − (y + z )k = kz k ≤ α, k · k-dist(A, x ) = infy∗∈A ky − x k ≤ α, and x∗ ∈ B(A, α). Conversely, assume that x∗ ∈ X∗ and k · k-dist(A, x∗) ≤ α. Put C = A ∩ B[x∗, 2α]. Then C is nonempty, w∗-compact, by Alaoglu’s Theorem and as translation and multiplication by a scalar is w∗-continuous, and we have k · k-dist(C, x∗) ≤ α. Con- sider the function h : X∗ → R, y∗ 7→ ky∗ − x∗k. By Alaoglu’s Theorem and as translation and multiplication by a scalar is w∗- continuous, h is w∗-lower semi-continuous. In particular, h attains its minimum on C. Let y∗ ∈ C be a minimizer. Then ky∗ − x∗k = k · k-dist(C, x∗) ≤ α. But this implies that x∗ = y∗ + (x∗ − y∗) ∈ A + B[0, α].

~~4.5 Reﬂexive Banach spaces~~

Note that if X is reflexive Corollary 4.8 applies to (X, w) itself. Actually this is a characterization of reflexivity [11, Theorem 3.1]. See also [2, pp. 57–58]. Theorem 4.13. Let X be a Banach space. Then X is not reflexive if and only if there exists a bounded, weakly closed subset F of X and a weakly continuous function g on F which is 1-Lipschitz in norm such that no continuous extension of g on X is c-Lipschitz for any c > 0.

~~43 4. Double Extensions~~

~~4.6 Closed points~~

According to Theorem 4.2, closedness of the sets B(A, α) is a suﬃcient condition for the existence of “double” extensions of functions on normal topological spaces where in addition some metric is given. But is this also necessary? In general, the answer is negative, as the following example shows.

~~Example 4.14. Let X = {0, 1, 2} and τ = {∅,X, {0, 1}}. Then (X, τ) is a normal topological space. Furthermore, let d be the metric on X given by the formula~~

 0 x = y  d(x, y) = 1 x 6= y & x, y ∈ {1, 2}  2 1 else for x, y ∈ X. Then there exist a τ-closed subset A ⊆ X and α > 0 such that B(A, α) is not τ-closed. However, every bounded τ-continuous real-valued function deﬁned on a τ-closed subset of X that is Lipschitz in d can be extended to all of X preserving τ-continuity, bounds, and Lipschitz continuity.

Proof. Normality can be checked by direct computation, as the only τ-closed subsets 1 in X are ∅, X and {2}. If A = {2} and α = 2 , then B(A, α) = {1, 2}, which is not τ-closed. Finally, the only nontrivial case for an extension is that of a function given on {2}. But here one can take the constant extension.

Of course, the above example is a bit awkward, as the topological space is not Hausdorﬀ. If we consider only Hausdorﬀ spaces, we get that closedness of the sets B(A, α) is in fact also a necessary condition [11, p. 7].

Theorem 4.15. Let (X, τ) be a T4 space and d a metric on X such that there exist a τ-closed set A and α > 0 such that B(A, α) is not τ-closed. Then there exist a τ-closed subset F ⊂ X and a bounded, τ|F -continous and 1-Lipschitz function g : F → R such that no extension f of g to X is both τ-continuous, with infF g ≤ f ≤ supF g, and 1-Lipschitz. Proof. Choose some z ∈ B(A, α) \ B(A, α) and put R = d-dist(A, z). Then R > α and the function  0 x ∈ A g(x) = R x = z is a bounded, τ|F -continuous and 1-Lipschitz function on the τ-closed set F = A ∪ {z}. Suppose that g admits a τ-continuous extension f to X with infF g ≤ f ≤ supF g that is 1-Lipschitz. If u ∈ B(A, α) and δ > 0 there exists v ∈ A such that d(u, v) < α + δ, hence

~~f(u) = f(u) − f(v) = |f(u) − f(v)| ≤ d(u, v) < α + δ.~~

~~44 4.7. Unbounded functions~~

~~Since δ > 0 can be chosen arbitrarily small, f ≤ α on B(A, α), and by τ-continuity of f, it follows that f ≤ α on B(A, α). But this is clearly impossible as f(z) = g(z) = R > α.~~

~~Thus, to sum up, we have got the following “improved” version of Theorem 4.2.~~

Theorem 4.16. Let (X, τ) be a T4 space and d a metric on X. Then for every τ-closed F ⊆ X and every bounded τ|F -continuous function g : F → R that is L-Lipschitz in d for some L > 0 exists a τ-continuous function f : X → R such that f|F = g, infF g ≤ f ≤ supF g, and that is L-Lipschitz in d if and only if the set B(A, α) is τ-closed for every τ-closed A ⊆ X and every α > 0.

It may be of interest at this point to note that there exist pairs of T4 spaces and metrics that do not have this property. For example, if we take X = R and d is the metric deﬁned by  0 x = y  d(x, y) = 1 x 6= y & x, y ∈ (−1, 1)  2 1 else

~~1 for x, y ∈ R, then B({0}, 2 ) = (−1, 1) is not closed with respect to the usual topology for the reals. In fact, this is exactly the way how Example 4.1 was constructed.~~

~~4.7 Unbounded functions~~

In the hypothesis of Theorem 4.2 the function g has to be bounded. Tietze’s Ex- tension Theorem (Theorem 3.8) and Theorem 2.3, however, also “work” with unbounded functions. So can Theorem 4.2 be further improved, possibly dropping the requirement of boundedness? We will see by an example that without additional assumptions this is not possible. The example will be in the Hilbert space `2 with the weak topology and the norm-distance, so we ﬁrst have to see why Theorem 4.2 applies in this situation.

~~Proposition 4.17. Let H be a Hilbert space. Then (H, w) is a T4 space and if A ⊆ H is weakly closed and α > 0, then the set~~

~~B(A, α) = {x ∈ H; k · k-dist(A, x) ≤ α} is weakly closed.~~

Proof. It is well-known that if H is a Hilbert space and H∗ is its dual, then the map ϕ : H → H∗, x 7→ h·, xi is linear, bijective, an isometry, and a homeomorphism ∗ ∗ ∗ ∗ of the topological spaces (H, w) and (H , w ). So (H, w) is a T4 space as (H , w ) is such a space by Theorem 4.11, and if A ⊆ H is w-closed and α > 0, then B(A, α) = ϕ−1(B(ϕ(A), α)) is w-closed by Lemma 4.12.

~~45 4. Double Extensions~~

In particular, if it was possible to have “double” extensions such as in Theorem 4.2 regardless of the boundedness of the original function, then this would have to be possible also in the Hilbert space `2 with with the weak topology and the norm- distance. However, the following example [11, Example 3.2] shows that this is not the case.

Example 4.18. There exists a weakly closed subset F of the Hilbert space `2 and an unbounded weakly continuous function g : F → R which is 1-Lipschitz in norm, such that no extension of g on `2 is both weakly continuous and 1-Lipschitz.

~~n ∞ Proof. Let (e )n=0 be the canonical basis of `2. Put~~

~~n 1 0 1 n x = n 4 e + n 2 e (n ≥ 1) and F = {xn; n ≥ 1}. Then F is w-closed, for if y ∈ F w, then one can ﬁnd ε > 0 such that the w-open neighborhood~~

~~0 {x ∈ `2; |hx − y, e i| = |x0 − y0| < ε}~~

1 0 1 n of y and F have exactly one point in common, say n 4 e + n 2 e , and then y = 1 0 1 n n 1 n 4 e + n 2 e ∈ F . Also, if we deﬁne g : F → R by g(x ) = n 2 , this function is weakly continuous, and since for n > m

n m 1 1 1 |g(x ) − g(x )| = n 2 − m 2 ≤ (n + m) 2 ≤ 1 1 2 1 n m ≤ ((n 4 − m 4 ) + n + m) 2 = kx − x k, it is 1-Lipschitz in norm. Suppose now that f is a weakly continuous and 1-Lipschitz 1 1 1 1 extension of g on `2. Choose N ≥ 1 such that n 2 − n 4 = n 4 (n 4 − 1) ≥ f(0) + 1 whenever n ≥ N. Deﬁne n 1 n y = n 2 e (n ≥ N). Then f(xn) − f(yn) ≤ |f(xn) − f(yn)| ≤ kxn − ynk and hence

n n n n 1 1 f(y ) ≥ f(x ) − kx − y k = n 2 − n 4 ≥ f(0) + 1 for all n ≥ N. However, 0 is a weak accumulation point of {yn; n ≥ N} and hence w contained in {yn; n ≥ N} , which leads to a contradiction, as then f(0) ≥ f(0) + 1. k Indeed, let U be an arbitrary weak neighborhood of 0. Then there exist K, α ∈ `2, 0 ≤ k ≤ K, and ε > 0 such that

~~k 0 ∈ {x ∈ `2; |hx, α i| < ε, 0 ≤ k ≤ K} ⊆ U.~~

~~P 1 Let us agree that a subset S of {n; n ≥ N} is called “good” if n∈S n < ∞ and “bad” otherwise. Then, for each k ∈ {0,...,K}, the set~~

~~1 n k 2 k {n ≥ N; |hy , α i| = |n αn| ≥ ε}~~

~~46 4.8. A second proof for Theorem 4.2~~

~~k − 1 is good. Otherwise, for each n in a bad subset of {n; n ≥ N}, |αn| ≥ εn 2 and k α ∈/ `2. But now, as~~

K {n ≥ N; |hyn, αki| < ε, 0 ≤ k ≤ K} = \ {n ≥ N; |hyn, αki| < ε} = k=0 K = {n; n ≥ N}\ [ {n ≥ N; |hyn, αki| ≥ ε}, k=0 as ﬁnite unions of good sets are again good, as the complement of a good set is necessarily bad, and bad sets are in particular always nonempty, there exists some n ≥ N such that yn ∈ U.

~~4.8 A second proof for Theorem 4.2~~

In [12] a different attempt to prove Theorem 4.2 was made. It uses a slightly different method which is also interesting. However, during the work on this master’s thesis a problem with an argument in that proof was found. At the time of this writing it is not clear whether this argument can be “fixed”. So we just add the assertion in question as an additional assumption to the theorem and prove this – possibly weaker – version of Theorem 4.2. A function d : X × X → R will be called a not necessarily symmetric pseudometric on the set X, provided that d ≥ 0, d(x, x) = 0 and d(x, y) ≤ d(x, z)+d(z, y) for all x, y, z ∈ X, without possibly d(x, y) = d(y, x) for all x, y ∈ X. If d is a not necessarily symmetric pseudometric on some set X and Y , Z are subsets of X, we write d-dist(Y,Z) = inf{d(y, z); y ∈ Y, z ∈ Z} and d-dist(Z,Y ) = inf{d(z, y); z ∈ Z, y ∈ Y }, as we usually also do in (pseudo-)metric spaces, with the difference, however, that here d-dist might be “order-sensitive”. Finally, a topological space (X, τ) is said to have property (P) if, whenever d is a not necessarily symmetric pseudometric on X such that, for every τ-closed A ⊆ X and every α > 0, both the sets

{x ∈ X; d-dist(A, x) ≤ α} and {x ∈ X; d-dist(x, A) ≤ α} are τ-closed, and whenever f : X → R is a bounded τ-continuous function that satisﬁes the Lipschitz condition f(x) − f(y) ≤ d(x, y) for all x, y ∈ X, then the not necessarily symmetric pseudometric d0 on X that is given by

~~d0(x, y) = d(x, y) − (f(x) − f(y))~~

47 4. Double Extensions for x, y ∈ X also has the property that, for every τ-closed A ⊆ X and every α > 0, both the sets {x ∈ X; d0-dist(A, x) ≤ α} and {x ∈ X; d0-dist(x, A) ≤ α} are τ-closed. The statement which is actually shown in [12] then is the following.

Theorem 4.19. Let (X, τ) be a normal topological space that has property (P) and d a metric on X such that the set B(A, α) = {x ∈ X; d-dist(A, x) ≤ α} is τ-closed for every τ-closed A ⊆ X and every α > 0. Suppose that F ⊆ X is τ-closed and f : F → R is a bounded τ|F -continuous function that is L-Lipschitz in d on F for some L > 0. Then there exists a τ-continuous function f˜ : X → R such that ˜ ˜ f|F = f, infF f ≤ f ≤ supF f on X, and that is L-Lipschitz in d on X. Note the additional assumption that (X, τ) has property (P), which makes Theo- rem 4.19 possibly a weaker variant of Theorem 4.2. This was not originally included in [12], where an argument is given that every topological space has property (P). However, it is in that argument [12, p. 209] where a problem was discovered. We begin with two lemmata.

Lemma 4.20. Let (X, τ) be a topological space and d a not necessarily symmetric pseudometric on X with the property that if A ⊆ X is τ-closed and α > 0 then both the sets {x ∈ X; d-dist(A, x) ≤ α} and {x ∈ X; d-dist(x, A) ≤ α} are τ-closed. Then for every τ-closed subset A of X and every α > 0 the sets {x ∈ X; d-dist(A, x) < α} and {x ∈ X; d-dist(x, A) < α} are Fσ sets.

~~Proof. We readily compute 1 {x ∈ X; d-dist(A, x) < α} = [ {x ∈ X; d-dist(A, x) ≤ α − } 1 n n: n <α and 1 {x ∈ X; d-dist(x, A) < α} = [ {x ∈ X; d-dist(x, A) ≤ α − }. 1 n n: n <α~~

Lemma 4.21. Let X be a normal topological space. Suppose that A and B are Fσ sets in X such that A ∩ B = A ∩ B = ∅. Then there exist open sets U and V such that A ⊆ U, B ⊆ V , and U ∩ V = ∅.

S∞ ∞ Proof. Write A = n=1 An for a countable family (An)n=1 of closed sets in X and S∞ ∞ B = n=1 Bn for a countable family (Bn)n=1 of closed sets in X. For each n = 0 1, 2,... , choose, using the normality of X, disjoint open sets Un and Un such that

~~48 4.8. A second proof for Theorem 4.2~~

0 0 An ⊆ Un and B ⊆ Un, and disjoint open sets Vn and Vn such that Bn ⊆ Vn and 0 A ⊆ Vn. Then, let ˆ [ Un := Un \ Vk 1≤k≤n and ˆ [ Vn := Vn \ Uk. 1≤k≤n ˆ ˆ Obviously, Un and Vn are again open sets. Furthermore, by construction, all sets Vk 0 0 are contained in X \ Vk, and hence in X \ A, and all sets Uk are contained in X \ Uk, ˆ ˆ and hence in X \ B, such that Un still contains An and Vn still contains Bn. Also S∞ ˆ S∞ ˆ we clearly have n=1 Un ∩ n=1 Vn = ∅. So if we put ∞ [ ˆ U := Un n=1 and ∞ [ ˆ V := Vn, n=1 those sets do what we want. The next theorem is a generalization of Urysohn’s Lemma. Theorem 4.22. Let (X, τ) be a normal topological space and d a not necessarily symmetric pseudometric on X with the property that if A ⊆ X is τ-closed and α > 0 then both the sets {x ∈ X; d-dist(A, x) ≤ α} and {x ∈ X; d-dist(x, A) ≤ α} are τ-closed. Suppose F0 and F1 are τ-closed disjoint subsets of X with the property that d-dist(F1,F0) ≥ 1. Then there exists a τ-continuous function f : X → [0, 1], taking the value 0 on F0 and taking the value 1 on F1, such that f(x)−f(y) ≤ d(x, y) for all x, y ∈ X. Proof. Like in the proof of Urysohn’s Lemma (Theorem 3.5), the function f will be constructed from sets that correspond (approximately) to the “level sets” {x ∈ X; f(x) < t}. For the resulting function to satisfy the condition that f(x) − f(y) ≤ d(x, y) for all x, y ∈ X, an additional condition has to be imposed on those sets, though. Suppose that a family (Ut)t∈D of τ-open subsets of X, for some dense subset D of (0, +∞), is given, such that

~~• Us ⊆ Ut whenever s < t, S • t∈D Ut = X,~~

~~49 4. Double Extensions~~

~~• inf{t ∈ D; x ∈ Ut} = 0 for all x ∈ F0,~~

~~• inf{t ∈ D; x ∈ Ut} = 1 for all x ∈ F1,~~

~~• inf{t ∈ D; x ∈ Ut} ≤ 1 for all x ∈ X,~~

~~• d-dist(X \ Ut, Us) ≥ t − s whenever s, t ∈ D, s < t, Us 6= ∅ and X \ Ut 6= ∅. Then, by Lemma 3.7, the function f : X → R deﬁned by~~

~~f(x) := inf{t ∈ D; x ∈ Ut}~~

−1 −1 will be τ-continuous. Also, F0 ⊆ f ({0}), F1 ⊆ f ({1}), and 0 ≤ f ≤ 1. Finally, the newly added last condition guarantees that f(x)−f(y) ≤ d(x, y) for all x, y ∈ X. Indeed, let x, y ∈ X. If f(x) ≤ f(y) nothing has to be shown. So we can assume that f(x) > f(y). In this case, for all s ∈ D with s > f(y), y ∈ Us, in particular Us 6= ∅, and, for all t ∈ D with t < f(x), x ∈ X \ Ut, in particular X \ Ut 6= ∅, hence

~~d(x, y) ≥ sup d-dist(X \ Ut, Us) ≥ f(y)~~

It remains to show that such a family of sets exists. Let Q = {q1, q2,... } be the set of all rational numbers in (0, 1). Put D := Q ∪ [1, +∞). Then D is a dense subset of (0, +∞). A sequence of τ-open subsets Uq1 ,Uq2 ,... of X will be constructed recursively such that, for each n = 1, 2,... ,

~~(1) d-dist(X \ Uqn ,F0) ≥ qn,~~

~~(2) d-dist(F1, Uqn ) ≥ 1 − qn,~~

~~(3) d-dist(X \ Ut, Us) ≥ t − s whenever s, t ∈ {q1, q2, . . . , qn} and s < t.~~

~~Finally, let U1 := X \ F1 and Ut := X for t > 1. Then the family (Ut)t∈D has all the required properties.~~

~~Basis. In order to construct the set Uq1 , put~~

~~A := {x ∈ X; d-dist(x, F0) < q1}, 0 A := {x ∈ X; d-dist(x, F0) ≤ q1},~~

~~B := {x ∈ X; d-dist(F1, x) < 1 − q1}, 0 B := {x ∈ X; d-dist(F1, x) ≤ 1 − q1}.~~

~~0 0 By Lemma 4.20, A and B are Fσ sets, and we have A ⊆ A for the τ-closed set A and B ⊆ B0 for the τ-closed set B0.~~

~~50 4.8. A second proof for Theorem 4.2~~

0 Suppose that A ∩ B 6= ∅. Then there exists x ∈ X with d-dist(x, F0) ≤ q1 and d-dist(F1, x) < 1 − q1. Let ε > 0 such that d-dist(F1, x) < 1 − q1 − ε. Then there exist y ∈ F0 with d(x, y) < q1 + ε and z ∈ F1 with d(z, x) < 1 − q1 − ε, such that

~~d(z, y) ≤ d(z, x) + d(x, y) < 1 − q1 − ε + q1 + ε = 1.~~

0 However, d-dist(F1,F0) ≥ 1. So A ∩ B = ∅. 0 Similarly, suppose that A∩B 6= ∅. Then there exists x ∈ X with d-dist(x, F0) < q1 and d-dist(F1, x) ≤ 1 − q1. Let ε > 0 such that d-dist(x, F0) < q1 − ε. Then there exist y ∈ F0 with d(x, y) < q1 − ε and z ∈ F1 with d(z, x) < 1 − q1 + ε, such that

~~d(z, y) ≤ d(z, x) + d(x, y) < 1 − q1 + ε + q1 − ε = 1.~~

~~0 However, d-dist(F1,F0) ≥ 1. So A ∩ B = ∅. Therefore, by Lemma 4.21, there exist disjoint τ-open sets U and V such that A ⊆ U and B ⊆ V . Let~~

Uq1 := U. We have to check that this set satisﬁes the conditions (1) - (3) from above. Suppose that d-dist(X \ Uq1 ,F0) < q1. Then there exist y ∈ X \ Uq1 and z ∈ F0 such that d(y, z) < q1, in particular, y ∈ A. However, A ⊆ Uq1 , a contradiction. So Uq1 satisﬁes condition (1). Suppose that d-dist(F1, Uq1 ) < 1 − q1. Then there exist y ∈ F1 and z ∈ Uq1 such that d(y, z) < 1 − q1, in particular, z ∈ B. However,

~~Uq1 ⊆ X \ B, a contradiction. So Uq1 satisﬁes condition (2). Condition (3) is trivial for n = 1.~~

~~Step. If open sets Uq1 ,Uq2 ,...Uqn−1 , satisfying the above conditions (1) - (3), already have been constructed for some n ≥ 2, put~~

~~A := {x ∈ X; d-dist(x, F0) < qn}∪ [ ∪ {x ∈ X; d-dist(x, Uqk ) < qn − qk}, k∈{1,2,...,n−1}:qk~~

B := {x ∈ X; d-dist(F1, x) < 1 − qn}∪ [ ∪ {x ∈ X; d-dist(X \ Uql , x) < ql − qn}, l∈{1,2,...,n−1}:ql>qn 0 B := {x ∈ X; d-dist(F1, x) ≤ 1 − qn}∪ [ ∪ {x ∈ X; d-dist(X \ Uql , x) ≤ ql − qn}. l∈{1,2,...,n−1}:ql>qn

~~0 Then, again, by Lemma 4.20, both A and B are Fσ sets, and A ⊆ A as well as B ⊆ B0, where A0, B0 are τ-closed. A similar analysis as in the case n = 1, using~~

~~51 4. Double Extensions~~

~~(1) - (3), shows that A0 ∩ B = A ∩ B0 = ∅. Therefore, by Lemma 4.21, there exist disjoint τ-open sets U and V such that A ⊆ U and B ⊆ V . Let~~

Uqn := U. Again, conditions (1) - (3) are satisﬁed, as can be seen by a direct calculation, similar to the case n = 1. The next theorem is a generalization of Tietze’s Extension Theorem. Theorem 4.23. Let (X, τ) be a normal topological space which has property (P). Let d be a metric on X such that the set B(A, α) is τ-closed for each τ-closed A ⊆ X and every α > 0. Assume further that g, h : X → R, g ≤ h, are bounded, satisfy, for some L > 0, the inequality g(x) − h(y) ≤ Ld(x, y) for all x, y ∈ X, g is τ- upper semi-continuous, h is τ-lower semi-continuous. Then there exists a function f : X → R which is τ-continuous, L-Lipschitz in d, and such that g ≤ f ≤ h. Proof. By adding a constant and multiplying by a constant of g and h we can suppose that −1 ≤ g ≤ h ≤ 1. By multiplying the metric d by a constant, we can suppose that L = 1. We will construct recursively not necessarily symmetric pseudometrics dn, n = 1, 2,... , and functions gn, hn : X → R, n = 1, 2,... , such that, for n = 1, 2 ... , (1) whenever A ⊆ X is τ-closed and α > 0, both the sets

~~{x ∈ X; dn-dist(A, x) ≤ α} and {x ∈ X; dn-dist(x, A) ≤ α} are τ-closed,~~

~~(2) gn ≤ hn,~~

~~2n−1 (3) gn ≤ 3n−1 ,~~

~~2n−1 (4) hn ≥ − 3n−1 ,~~

~~(5) gn is τ-upper semi-continuous,~~

~~(6) hn is τ-lower semi-continuous,~~

~~(7) gn(x) − hn(y) ≤ dn(x, y) for all x, y ∈ X.~~

Put d1 := d, g1 := g, h1 := h. Then d1, g1, h1 obviously satisfy the above conditions (1) - (7). So suppose now that dn−1, gn−1, hn−1 have already been constructed, for some n ≥ 2, satisfying the above conditions (1) - (7). Put ( 2n−2 ) G := x ∈ X; g (x) ≥ n−1 3n−1

52 4.8. A second proof for Theorem 4.2 and ( 2n−2 ) H := x ∈ X; h (x) ≤ − . n−1 3n−1 2n−1 Then G and H are τ-closed, disjoint, and dn−1(x, y) ≥ 3n−1 for any x ∈ G and y ∈ H. Hence by Theorem 4.22 there exists a τ-continuous function ψn−1 : X → R 2n−2 2n−2 2n−2 2n−2 such that − 3n−1 ≤ ψn−1 ≤ 3n−1 , ψn−1 = 3n−1 on G, ψn−1 = − 3n−1 on H, and such that ψn−1(x) − ψn−1(y) ≤ dn−1(x, y) for all x, y ∈ X. Put

dn(x, y) := dn−1(x, y) − (ψn−1(x) − ψn−1(y)) for x, y ∈ X, gn := gn−1 − ψn−1, hn := hn−1 − ψn−1. Let us check that dn, gn, hn satisfy the conditions (1) - (7). Condition (1) is a consequence of the topological space (X, τ) having property (P). Condition (2) is obvious. Conditions (3) and (4) can be checked by direct computation. Conditions (5) and (6) are clear as well. Finally, condition (7) follows from

~~gn(x) − hn(y) = (gn−1(x) − ψn−1(x)) − (hn−1(y) − ψn−1(y)) =~~

~~= (gn−1(x) − hn−1(y)) − (ψn−1(x) − ψn−1(y)) ≤~~

≤ dn−1(x, y) − (ψn−1(x) − ψn−1(y)) = dn(x, y) P∞ for x, y ∈ X. Put now f = n=1 ψn. Then, as ∞ ∞ n−1 X X 2 1 1 kψnk∞ ≤ n = 2 = 1, n=1 n=1 3 3 1 − 3 f is well-deﬁned and τ-continuous, as the uniform limit of τ-continuous functions. Also note that we have −1 ≤ f ≤ 1. From the construction it follows that, for all n = 1, 2,... , n−1 n−1 X 2 g − ψk = gn ≤ n−1 k=1 3 and n−1 n−1 X 2 h − ψk = hn ≥ − n−1 , k=1 3 such that g ≤ f ≤ h. Finally, since n X ψn+1(x) − ψn+1(y) ≤ dn+1(x, y) = d(x, y) − (ψk(x) − ψk(y)) k=1 and hence n+1 X (ψk(x) − ψk(y)) ≤ d(x, y) k=1 for all n = 1, 2,... and all x, y ∈ X, we get that f is 1-Lipschitz in d. Proof of Theorem 4.19. Deﬁne functions g and h on X such that g = h = f on F , g = infF f on X \ F , h = supF f on X \ F . Then g and h satisfy the conditions in Theorem 4.23 and hence there exists a τ-continuous function f˜ : X → R such that ˜ ˜ f = f on F , infF f ≤ f ≤ supF f on X, and that is L-Lipschitz in d.

~~53 4. Double Extensions~~

~~4.9 But is it really weaker?~~

As was already said, the actual problem in the proof for Theorem 4.2 given in [12] is in an argument which says that every topological space has property (P) [12, p. 209]. In its published form it doesn’t seem to be valid. And while currently no counterexample is known, it seems like this assertion could be wrong in general. In the following example, which is due to Eva Kopecká, a topological space is constructed that fails to have a slightly stronger property than property (P). A topological space (X, τ) is said to have property (P’) if, whenever d is a not necessarily symmetric pseudometric on X such that, for every τ-closed A ⊆ X and every α > 0, both the sets

{x ∈ X; d-dist(A, x) ≤ α} and {x ∈ X; d-dist(x, A) ≤ α} are τ-closed, and whenever f : X → R is a τ-continuous function that satisﬁes the Lipschitz condition f(x) − f(y) ≤ d(x, y) for all x, y ∈ X, then the not necessarily symmetric pseudometric d0 on X that is given by

d0(x, y) = d(x, y) − (f(x) − f(y)) for x, y ∈ X also has the property that, for every τ-closed A ⊆ X and every α > 0, both the sets {x ∈ X; d0-dist(A, x) ≤ α} and {x ∈ X; d0-dist(x, A) ≤ α} are τ-closed. That is, property (P’) is almost the same as property (P), but without the requirement that the function f must be bounded.

Example 4.24. For every w-closed subset A of the Hilbert space `2 and every α > 0, the set B(A, α) = {x ∈ `2; k · k-dist(A, x) ≤ α} is w-closed, and the function 0 f : `2 → R, x 7→ hx, e i = x0 is w-continuous and 1-Lipschitz in the norm metric on `2. However, if we put

~~ρ(x, y) = kx − yk − (f(x) − f(y)), then there exist a w-closed subset A0 of `2 and α0 > 0 such that the set~~

~~{x ∈ `2; ρ-dist(A0, x) ≤ α0} is not w-closed. In particular, the topological space (`2, w) does not have property (P’).~~

~~54 4.9. But is it really weaker?~~

Proof. That the sets B(A, α) are w-closed is a consequence of Proposition 4.17. It 0 is obvious that f : `2 → R, x 7→ hx, e i = x0 is w-continuous and 1-Lipschitz in the norm metric. So it remains to check the last claim, i.e. to ﬁnd a w-closed subset A0 of `2 and α0 > 0 such that

{x ∈ `2; ρ-dist(A0, x) ≤ α0} is not w-closed. Put 0 √ n ∞ A0 := {ne + ne }n=1. w Then A0 is w-closed, for if y ∈ A0 , then one can ﬁnd ε > 0 such that the w-open 0 neighborhood {x ∈ `2; |hx − y, e i| = |x0 − y0| < ε} of y and A0 have exactly one 0 √ n 0 √ n point in common, say ne + ne , and then y = ne + ne ∈ A0. Also consider the set √ n ∞ B := { ne }n=1. It has the property that 0 ∈ Bw. To see this, let

~~i W = {x ∈ `2; |hx, w i| < εi, i = 1, 2, . . . , k},~~

i where k ≥ 1 and εi > 0, w ∈ `2, i = 1, 2, . . . , k, be an arbitrary w-open neighborhood of 0 and assume that B ∩ W = ∅. Then, for all n = 1, 2,... , there exists √ n i some i ∈ {1, 2, . . . , k} such that |h ne , w i| ≥ εi. Put ε = mini εi and denote by √ n i Ii√, for i = 1, 2,√ . . . , k, the set of all n = 1, 2,... such that |h ne , w i| ≥ ε. As n i i i √ε |h ne , w i| = n|wn| we have that n ∈ Ii if and only if |wn| ≥ n , so, because of i P 1 w ∈ `2, i = 1, 2, . . . , k, we have that n∈Ii n < ∞, i = 1, 2, . . . , k. However, this contradicts the divergence of the harmonic series. Because of √ √ √ ρ(nen + nen, 0) = kne0 + nenk − (n − 0) = n2 + n − n, √ 2 1 for all n = 1, 2,... and limn→∞ n + n − n = 2 , there exists α0 > 0 such that ρ-dist(A0, 0) > α0, that is, 0 ∈/ {x ∈ `2; ρ-dist(A0, x) ≤ α0}. However, √ √ ρ(ne0 + nen, nen) = kne0k − (n − 0) = 0,

~~√ n and hence ρ-dist(A0, ne ) = 0 for all n = 1, 2,... , in particular B ⊆ {x ∈ `2; ρ-dist(A0, x) ≤ α0}. It follows that~~

~~w w 0 ∈ B ⊆ {x ∈ `2; ρ-dist(A0, x) ≤ α0} and {x ∈ `2; ρ-dist(A0, x) ≤ α0} is not w-closed.~~

~~Chapter 5~~

~~Michael’s Selection Theorem~~

~~We prove Michael’s Selection Theorem (Theorem 5.10) about continuous selections for carriers from paracompact Hausdorﬀ spaces into Banach spaces.~~

~~5.1 Paracompact topological spaces~~

A topological space X is called paracompact if and only if every open cover of X has a locally finite refinement, i.e., an open cover of X consisting of subsets of the members of the “original” cover, and such that every point has a neighborhood that intersects only finitely many members of the cover. If every countable open cover of X has a locally finite refinement, X is called countably paracompact.

~~5.2 The relation of paracompactness to other properties~~

~~Let us ﬁrst see how the notion of paracompactness is related to some other properties of topological spaces we have already encountered.~~

~~Theorem 5.1. Every metrizable space is paracompact.~~

This is a “hard” theorem. The ﬁrst proof given seems to be in [18], while a considerably shorter one was found later [17]. We follow [1, pp. 81–83]. If (X, d) is a metric space, A ⊆ X nonempty and ε > 0, we consider

~~Oε(A) = {x ∈ X; dist(A, x) < ε}, the “outer ε-set” of A, and~~

Iε(A) = {x ∈ X; B(x, ε) ⊆ A} = {x ∈ X; dist(X \ A, x) ≥ ε}, the “inner ε-set” of A. We also deﬁne Oε(∅) := ∅, and Iε(∅) := ∅. In the following lemma we collect a few properties of those sets for later use.

~~Lemma 5.2. Let (X, d) is a metric space, A ⊆ X, and ε > 0. Then the following holds.~~

~~(1) Oε(A) is open and Iε(A) is closed.~~

~~57 5. Michael’s Selection Theorem~~

~~(2) Iε(A) ⊆ A ⊆ Oε(A) and Oε(Iε(A)) ⊆ A.~~

~~(3) If x ∈ Iε(A) and y ∈ X \ A, then d(x, y) ≥ ε.~~

(4) If x ∈ X satisfies B(x, ε) ∩ Iε(A) 6= ∅, then x ∈ A. Proof. (1) follows from the continuity of x 7→ dist(A, x). The first chain of inclusions in (2) is obvious. If x ∈ Oε(Iε(A)), then by definition dist(Iε(A), x) < ε. So there is some y ∈ Iε(A) such that d(y, x) < ε. But then x ∈ B(y, ε) ⊆ A, as y ∈ Iε(A). (3) is obvious from the definition. For (4), assume that x ∈ X \A. Then d(y, x) ≥ ε for all y ∈ Iε(A). In particular, this holds for any y ∈ B(x, ε) ∩ Iε(A). But that is clearly not possible. S Proof of Theorem 5.1. Let (X, d) be a metric space and let X = i∈I Vi be an open cover of X. We have to show that (Vi)i∈I has a locally finite refinement, that is, an open cover of X consisting of subsets of the the sets Vi, and such that every point has a neighborhood that intersects only finitely many members of the cover. This will be done in a few steps. But first note that, by the “Well-ordering Theorem”, there exists a well-ordering of I. In the following, let be such a well-ordering. n The sets Si . For each n = 1, 2,... , we recursively define a function I → n P(X), i 7→ Si by the formula   n [ n Si = I2−n Vi \ Sj  . j≺i

~~We claim that ∞ [ [ n X = Si . n=1 i∈I~~

Indeed, if x ∈ X, put i0 := min{i ∈ I; x ∈ Vi} and choose n ≥ 1 such that −n n n −n B(x, 2 ) ⊆ Vi0 . Then x ∈ Si0 . Otherwise, from the deﬁnition of Si0 , B(x, 2 ) 6⊆ S n −n −n S n Vi0 \ j≺i0 Sj , that is, using the fact that B(x, 2 ) ⊆ Vi0 , B(x, 2 ) ∩ j≺i0 Sj 6= ∅. −n n If follows that B(x, 2 ) ∩ Sj 6= ∅ for some j ≺ i0. But then x ∈ Vj by Lemma 5.2 (4), a contradiction. Also note that, if i ≺ j,

~~n [ n n Sj ⊆ Vj \ Sk ⊆ X \ Si . k≺j~~

n n S n −n So if x ∈ Sj and y ∈ Si , we have y∈ / Vj \ k≺j Sk , and consequently d(x, y) ≥ 2 by Lemma 5.2 (3). The same holds also for arbitrary i 6= j, as one can simply interchange the roles of i and j if necessary. n n The sets Ci and Ui . For each n = 1, 2,... and each i ∈ I we deﬁne

~~n n Ci := O2−(n+3) (Si )~~

58 5.2. The relation of paracompactness to other properties and n n Ui := O2−(n+2) (Si ). n n n n n n Clearly, Ci is closed, Ui is open, and Ci ⊆ Ui . If i ≺ j, x ∈ Uj and y ∈ Ui , n −(n+2) n pick some u ∈ Sj such that d(x, u) < 2 and some v ∈ Si such that d(y, v) < 2−(n+2). Then we get

2−n ≤ d(u, v) ≤ d(u, x) + d(x, y) + d(y, v) < d(x, y) + 2−(n+1), so d(x, y) > 2−(n+1). Again the latter inequality holds also for arbitrary i 6= j, as one can simply interchange the roles of i and j if necessary. The sets Cn. For n = 1, 2,... let

~~[ n Cn := Ci . i∈I~~

These unions of closed sets are closed, as for every x ∈ X the open ball B = −(n+2) n n n B(x, 2 ) intersects at most one of the sets Ci , i ∈ I. Indeed, if y ∈ B∩Ci ⊆ Ui n n and z ∈ B ∩ Cj ⊆ Uj for some i 6= j, we get

2−(n+1) < d(y, z) ≤ d(y, x) + d(x, z) < 2−(n+2) + 2−(n+2) = 2−(n+1), a contradiction. n 1 1 The sets Wi . Finally, for n = 1 and each i ∈ I we deﬁne Wi := Ui , and for n = 2, 3,... and each i ∈ I we deﬁne

~~n−1 n n [ Wi := Ui \ Ck. k=1~~

n We claim that the family (Wi )(n,i)∈{1,2,... }×I is a locally finite refinement of (Vi)i∈I . n It is clear that the sets Wi are open. The family also is a cover of X. To see this, fix some x ∈ X. As the family n n n n (Si )(n,i)∈{1,2,... }×I covers X and Si ⊆ Ci , also the sets Ci cover X. Put

~~n k := min{n ≥ 1; ∃i ∈ I : x ∈ Ci }.~~

k k Let i = i(k) be the unique element of I such that x ∈ Ci . Then x ∈ Wi . This is 1 1 1 clear for k = 1, as then x ∈ Ci ⊆ Ui = Wi , and for k > 1 this follows again from k k x ∈ Ci ⊆ Ui and from the fact that x∈ / Cl for all l < k by the construction of k. S∞ S n So X = n=1 i∈I Wi , as desired. n To see that (Wi )(n,i)∈{1,2,... }×I is a reﬁnement of (Vi)i∈I , note that

~~n n n n Wi ⊆ Ui = O2−(n+2) (Si ) ⊆ O2−n (Si ) ⊆ O2−n (I2−n (Vi)) ⊆ Vi by Lemma 5.2 (2).~~

~~59 5. Michael’s Selection Theorem~~

n It remains to show locally ﬁniteness. Fix some x ∈ X. As the sets Si cover X, n0 −(n0+3) there exist n0 and i0 such that x ∈ Si0 . Let B = B(x, 2 ). We will show that n B intersects only ﬁnitely many of the sets Wi . As

~~n0 n0 n0 −(n +3) −(n +3) B ⊆ O2 0 (Si0 ) ⊆ O2 0 (Si0 ) = Ci0 ⊆ Cn0~~

n n we get that B∩Wi = ∅ for all n > n0 and all i ∈ I. If 1 ≤ n ≤ n0, then B∩Ui 6= ∅ n (and thus B ∩ Wi 6= ∅) for at most one i ∈ I. In fact, assume that for i 6= j there n n n n exist y ∈ B ∩ Ui ⊆ Ui and z ∈ B ∩ Uj ⊆ Uj . Then

~~2−(n0+1) ≤ 2−(n+1) < d(y, z) ≤ d(y, x) + d(x, z) < 2−(n0+3) + 2−(n0+3) = 2−(n0+2), a contradiction. So B intersects only ﬁnitely many such sets.~~

~~Theorem 5.3. Every compact Hausdorﬀ space is paracompact.~~

~~This directly follows from the deﬁnition.~~

~~Proof. A finite subcover is in particular a locally finite refinement.~~

Thus, to sum up, we have seen so far that the notion of paracompactness gen- eralizes both metrizable spaces and compact Hausdorff spaces. In Chapter 3 we have seen that metrizable spaces and compact Hausdorff spaces are also examples of normal spaces. So what about the relation of paracompactness and normality? The answer to this question is given next. The presentation follows [3, p. 300]. We begin with the following result on the closure of the union of a locally finite familiy of sets.

~~Lemma 5.4. Let X be a topological space and let (As)s∈S be a locally ﬁnite family S S of subsets of X. Then s∈S As = s∈S As.~~

S 0 S 0 Proof. As As0 ⊆ s∈S As for all s ∈ S, we get that As0 ⊆ s∈S As for all s ∈ S and S S thus s∈S As ⊆ s∈S As. S Conversely, let x ∈ s∈S As. Then there exists a neighborhood U of x such that the set S0 = {s ∈ S; As ∩ U 6= ∅} is ﬁnite. As

~~[ [ [ x ∈ As = As ∪ As, s∈S s∈S0 s∈S\S0~~

~~S S S S but x∈ / s∈S\S0 As, we conclude x ∈ s∈S0 As = s∈S0 As ⊆ s∈S As.~~

~~The next lemma is now only stated explicitly to avoid “repetition” in the following. It contains the method of proof for Theorem 5.6 and Theorem 5.7, which is almost identical.~~

~~60 5.2. The relation of paracompactness to other properties~~

~~metrizable compact Hausdorﬀ~~

~~paracompact Hausdorﬀ~~

~~normal Hausdorﬀ (T4)~~

~~regular Hausdorﬀ (T3)~~

~~Figure 5.1: The relation of paracompactness to other properties.~~

Lemma 5.5. Let X be a paracompact space and A, B a pair of closed subsets of X. If for every x ∈ B there exist open sets Ux, Vx such that A ⊆ Ux, x ∈ Vx and Ux ∩ Vx = ∅, then there also exist open sets U, V such that A ⊆ U, B ⊆ V and U ∩ V = ∅.

Proof. The sets Vx, x ∈ B, together with X \ B, form an open cover of X. As X is paracompact, it has a locally ﬁnite reﬁnement (Ws)s∈S. Let S0 = {s ∈ S; Ws ∩ S B 6= ∅}. If one puts V := s∈S0 Ws, then clearly V is open and we have that B ⊆ V . Obviously, the set U := X \ V is also open. Furthermore, by Lemma 5.4, S S V = s∈S0 Ws = s∈S0 Ws, which implies that A ⊆ U, because Ws ∩ A = ∅ for all s ∈ S0 (every set Ws, s ∈ S0, is a subset of some Vx, x ∈ B, and then Ws is a subset of X \ Ux ⊆ X \ A). It is clear that U ∩ V = ∅. The following two theorems are now immediately clear.

~~Theorem 5.6. Every paracompact Hausdorﬀ space is regular.~~

~~Proof. If one takes A to be a singleton and B a closed set disjoint from A, the prerequisites of Lemma 5.5 are satisﬁed and its application shows that points and closed sets can be separated.~~

~~Theorem 5.7. Every paracompact Hausdorﬀ space is normal.~~

~~Proof. If A and B are disjoint closed subsets, using the fact that the space is regular, the prerequisites of Lemma 5.5 are satisﬁed and its application shows that closed sets can be separated.~~

~~For convenience, an overview of the relations we just established is shown in Figure 5.1.~~

~~61 5. Michael’s Selection Theorem~~

~~5.3 Partitions of unity~~

Let us now review one of the most important features of paracompact Hausdorﬀ spaces. A family (fs)s∈S of continuous functions from a topological space X to the unit interval [0, 1] is called a locally ﬁnite partition of unity on X if and only if, for every x ∈ X,

~~• there exists a neighborhood U of x such that all but ﬁnitely many of the functions fs vanish on U,~~

~~P • s∈S fs(x) = 1.~~

A locally ﬁnite partition of unity (fs)s∈S on a topological space X is said to be −1 subordinated to an open cover V of X if the cover (fs ((0, 1]))s∈S is a reﬁnement of V. Then the following holds.

~~Theorem 5.8. Let X be a paracompact Hausdorﬀ space. Then for every open cover of X there exists a locally ﬁnite partition of unity subordinated to it.~~

~~Again we follow [3, pp. 301–302].~~

Proof. Let V be an open cover of X. As X is paracompact, there exists a locally finite refinement W of V. The regularity of X (Theorem 5.6) implies that each x ∈ X has an open neighborhood Ux such that Ux ⊆ W for some W ∈ W with x ∈ W . This gives a refinement {Ux; x ∈ X} of W. Let U be a locally finite refinement of {Ux; x ∈ X}. Then for every U ∈ U there exists WU ∈ W such that S U ⊆ WU . For each W ∈ W, put FW := {U; U ∈ U,WU = W }. Then FW is closed S (Lemma 5.4), FW ⊆ W , and W ∈W FW = X. By Urysohn’s Lemma, for every W ∈ W, we can find a continuous function gW : X → [0, 1] such that gW (x) = 1 for P x ∈ FW and gW (x) = 0 for x ∈ X \ W . Then g := W ∈W gW is well-defined (W is S locally finite), strictly positive ( W ∈W FW = X) and continuous. Put fW := gW /g for W ∈ W. Then (fW )W ∈W is a locally finite partition of unity subordinated to the cover V.

~~5.4 Michael’s Selection Theorem~~

Let X and Y be topological spaces. A function Φ: X → P(Y ), that assignes to each x ∈ X a nonempty subset Φ(x) ⊆ Y shall be called a carrier.A continuous selection for Φ is a continuous function f : X → Y such that f(x) ∈ Φ(x) for all x ∈ X. The so-called “selection problem” is now concerned with the question under what conditions on X, Y and Φ continuous selections exist, or, slightly more general, considering subsets A ⊆ X, under what conditions on X, Y , Φ and A, continuous selections for Φ|A can be extended to continuous selections for Φ.

~~62 5.4. Michael’s Selection Theorem~~

In the series of papers [14, 15, 16] E. Michael investigated this question and gave – among with other results – a condition when carriers from paracompact Hausdorﬀ spaces into Banach spaces allow continuous selections. A ﬁrst observation in this context is the following [14, Proposition 2.2].

Proposition 5.9. Let X and Y be topological spaces. Assume that Φ: X → P(Y ) is a carrier such that, for every x0 ∈ X and y0 ∈ Φ(x0), there exists a continuous selection f for Φ|U , where U is some neighborhood of x0, such that f(x0) = y0. Then, for every open set V ⊆ Y , the set G = {x ∈ X; Φ(x) ∩ V 6= ∅} is open in X.

Proof. For each x0 ∈ G pick a y0 ∈ Φ(x0) ∩ V . Then, by assumption, there exists a neighborhood U of x0 and a continuous selection f for Φ|U such that f(x0) = y0. Put 0 0 0 U := {x ∈ U; f(x) ∈ V }. Then U is a neighborhood of x0 in X and U ⊆ G.

Thus it seems reasonable to restrict our attention to carriers that have the property from the conclusion of Proposition 5.9. If X and Y are topological spaces, a carrier Φ: X → P(Y ) is called lower semi-continuous if for all open sets V ⊆ Y the set {x ∈ X; Φ(x) ∩ V 6= ∅}, the so-called “large pre-image” of V , is open in X. Similarly, a carrier Φ: X → P(Y ) is called upper semi-continuous if for all open sets V ⊆ Y the set

{x ∈ X; Φ(x) ⊆ V } is open in X. However, the latter notion will not be used in the following. There are also equivalent point-wise characterizations of the above notions. Namely, a carrier Φ: X → P(Y ) is lower semi-continuous if and only if, for any ﬁxed x0 ∈ X, if V ⊆ Y is open such that Φ(x0) ∩ V 6= ∅, then there exists a neighborhood U of x0 such that Φ(x) ∩ V 6= ∅ for all x ∈ U. Analogously, a carrier Φ: X → P(Y ) is upper semi-continuous if and only if, for any ﬁxed x0 ∈ X, if V ⊆ Y is open such that Φ(x0) ⊆ V , then there exists a neighborhood U of x0 such that Φ(x) ⊆ V for all x ∈ U. Figure 5.2 illustrates the concept of lower semi-continuous and upper semi- continuous carriers. We are now ready to state and prove the main theorem of this section, which is also known as “Michael’s Selection Theorem” [14, Proposition 3.2”].

Theorem 5.10 (Michael 1956). Let X be a paracompact Hausdorﬀ space, Y a Banach space, and Φ: X → P(Y ) a lower semi-continuous carrier such that Φ(x) is a closed and convex subset of Y for each x ∈ X. Then Φ admits a continuous selection.

~~We roughly follow the lines of the original proof given in [14]. See also [1].~~

~~63 5. Michael’s Selection Theorem~~

~~Y Y~~

~~X X U x0 U x0 (a) A lower semi-continuous carrier that (b) An upper semi-continuous carrier is not upper semi-continuous. that is not lower semi-continuous.~~

~~Figure 5.2: Lower semi-continuity and upper semi-continuity.~~

Lemma 5.11. Let X be a paracompact Hausdorﬀ space, Y a normed linear space, Φ: X → P(Y ) a lower semi-continuous carrier such that Φ(x) is a convex subset of Y for all x ∈ X, and ε > 0. Then there exists a continuous function f : X → Y such that dist(Φ(x), f(x)) < ε for all x ∈ X.

~~Proof. For every y ∈ Y let~~

~~Uy = {x ∈ X; Φ(x) ∩ B(y, ε) 6= ∅}.~~

~~Then every Uy is open and the family of all such sets is an open cover of X. Let (pi)i∈I be a locally ﬁnite partition of unity subordinated to this cover. (It exists by~~

~~Theorem 5.8.) For each i ∈ I choose some yi ∈ Y such that pi vanishes outside Uyi . Let X f(x) := pi(x)yi (x ∈ X). i∈I~~

As (pi)i∈I is locally ﬁnite, f is well-deﬁned and continuous. Fix some x ∈ X. If pi(x) > 0, then surely x ∈ Uyi . The condition that Φ(x) ∩ B(yi, ε) 6= ∅ can be equivalently rewritten as yi ∈ Φ(x) + B(0, ε). Both Φ(x) and B(0, ε) are convex, so this is the sum of two convex sets, which is thus also convex. But f(x) is a convex combination of elements of this set, so we have that f(x) ∈ Φ(x) + B(0, ε). But this is the same as dist(Φ(x), f(x)) < ε.

Lemma 5.12. Let X be a paracompact Hausdorﬀ space, Y a normed linear space, Φ: X → P(Y ) a lower semi-continuous carrier such that Φ(x) is a convex subset of Y for all x ∈ X, ε > 0 and f : X → Y continuous such that dist(Φ(x), f(x)) < ε for all x ∈ X. Then

~~Ψ(x) := Φ(x) ∩ B(f(x), ε)(x ∈ X) is also a lower semi-continuous carrier and Ψ(x) is convex for all x ∈ X.~~

~~64 5.4. Michael’s Selection Theorem~~

Proof. First of all note that Ψ(x) 6= ∅ for all x ∈ X as dist(Φ(x), f(x)) < ε. It is also clear that the sets Ψ(x) for x ∈ X are convex. Let us see why this carrier is lower semi-continuous. Fix some x0 ∈ X and let V ⊆ Y be open such that

Ψ(x0) ∩ V = Φ(x0) ∩ B(f(x0), ε) ∩ V 6= ∅. Let y be a member of this set. Then there exists some 0 < r < ε such that y ∈ Φ(x0) ∩ B(f(x0), r) ∩ V , which is therefore also non-empty. Thus by the lower semi-continuity of Φ, there exists a neighborhood U1 of x0 such that

Φ(x) ∩ (B(f(x0), r) ∩ V ) 6= ∅ 0 0 for all x ∈ U1. Let r > 0 be such that r + r < ε. By continuity of f, there exists 0 a neighborhood U2 of x0 such that f(x) ∈ B(f(x0), r ) for all x ∈ U2. Let now x ∈ U1 ∩ U2. Then

0 ∅ 6= Φ(x) ∩ B(f(x0), r) ∩ V ⊆ Φ(x) ∩ B(f(x), r + r ) ∩ V ⊆ Ψ(x) ∩ V. Proof of Theorem 5.10. We will construct recursively lower semi-continuous carriers Φ0, Φ1,... from X into Y with convex values, and continuous functions f0, f1,... from X into Y that “approximate” the carriers as follows. Let Φ0 := Φ and let f0 be an arbitrary continuous function from X to Y such that dist(Φ0(x), f0(x)) < 1 for all x ∈ X (Lemma 5.11). If Φn and fn already have been constructed for some n ≥ 0, such that Φn is a lower semi-continuous carrier −n with convex values, fn is continuous, and such that dist(Φn(x), fn(x)) < 2 for all x ∈ X, let −n Φn+1(x) := Φn(x) ∩ B(fn(x), 2 )(x ∈ X), such that Φn+1 is again a lower semi-continuous carrier from X into Y with convex values (Lemma 5.12), and let fn+1 : X → Y be a continuous function such that −(n+1) dist(Φn+1(x), fn+1(x)) < 2 for all x ∈ X (Lemma 5.11). The so-constructed functions f0, f1,... have the following pleasing properties. −n Firstly, because of Φ(x) = Φ0(x) ⊇ Φ1(x) ⊇ ... , dist(Φn(x), fn(x)) < 2 implies that −n dist(Φ(x), fn(x)) < 2 ∞ for all n ≥ 0 and all x ∈ X. Secondly, (fn)n=0 is uniformly Cauchy. In fact, if n ≥ 0 and x ∈ X, choose some y ∈ Φn+2(x). Then, as

~~−(n+1) Φn+2(x) = Φn+1(x) ∩ B(fn+1(x), 2 ) = −n −(n+1) = Φn(x) ∩ B(fn(x), 2 ) ∩ B(fn+1(x), 2 ),~~

~~n−1 n−1 X X |fn(x) − fm(x)| = fk+1(x) − fk(x) ≤ |fk+1(x) − fk(x)| ≤ k=m k=m~~

~~65 5. Michael’s Selection Theorem~~

n−1 ∞ ≤ X 2−k + 2−(k+1) ≤ X 2−k + 2−(k+1), k=m k=N which tends to zero as N tends to inﬁnity. If f is the limit, f is continuous, and for all x ∈ X we have dist(Φ(x), f(x)) ≤ 2−n for all n ≥ 0, that is, f(x) ∈ Φ(x), as Φ(x) is closed. Sometimes the following easy consequence [14, Proposition 1.4] is also useful.

Corollary 5.13 (Michael 1956). Let X be a paracompact Hausdorﬀ space, Y a Banach space, Φ: X → P(Y ) a lower semi-continuous carrier such that Φ(x) is closed and convex for every x ∈ X, and A ⊆ X closed. Then every continuous selection for Φ|A can be extended to a continuous selection for Φ.

~~Proof. Let g be a continuous selection for Φ|A. We must extend g to a continuous selection for Φ. Let Ψ: X → P(Y ) be deﬁned as follows.  {g(x)} x ∈ A, Ψ(x) := Φ(x) x ∈ X \ A.~~

Then Ψ is a lower semi-continuous carrier with closed and convex values. In fact, if x0 ∈ A and V ⊆ Y is open such that Ψ(x0)∩V 6= ∅, that is, g(x0) ∈ V , 0 then, by continuity of g, there is a neighborhood U1 of x0 in A such that g(x) ∈ V 0 0 for all x ∈ U1. We can write U1 = U1 ∩ A for a neighborhood U1 of x0 in X. Also, as Ψ(x0) ∩ V 6= ∅ implies in particular that Φ(x0) ∩ V 6= ∅, and Φ is lower semi- continuous at x0, there exists a neighborhood U2 of x0 in X such that Φ(x)∩V 6= ∅ for all x ∈ U2. Let now U := U1 ∩ U2. Then U is a neighborhood of x0 in X and Ψ(x) ∩ V 6= ∅ for all x ∈ U. Similarly, if x0 ∈ X \ A and V ⊆ Y is open such that 0 Ψ(x0) ∩ V 6= ∅, that is, Φ(x0) ∩ V 6= ∅, one cand ﬁnd a neighborhood U of x0 in X such that Φ(x) ∩ V 6= ∅ for all x ∈ U 0. Put U := U 0 ∩ X \ A. Then U is again a neighborhood of x0 in X and we have that Ψ(x) ∩ V 6= ∅ for all x ∈ U. But now by Theorem 5.10 the carrier Ψ admits a continuous selection. It is obvious that this is then a continuous selection of Φ that extends g.

~~66 Chapter 6~~

~~Continuous Selections~~

By Kirszbraun’s Theorem (Theorem 2.9), the extension problem for Lipschitz functions in particular has a solution when both the domain space and the range space are Hilbert spaces with the metrics induced by their inner products. However, the extensions obtained by Kirszbraun’s Theorem are not unique. We discuss some results of E. Kopecká [8, 7], who studied how extensions can be assigned in a continuous way. In Euclidean space we explicitly construct Lipschitz constant preserving extension operators for nonexpansive mappings on compact subsets (Theorem 6.2). For the general case, the “non-constructive” approach via Michael’s Selection The- orem is used. That is, we study the map that assigns to a function deﬁned on some subset of the Hilbert space the set of all its Lipschitz constant preserving extensions and check that it satisﬁes the requirements for Michael’s Selection Theorem. It follows that there exists a continuous selection of this map (Theorem 6.7).

~~6.1 Closed and convex sets~~

As a motivating example, we consider the following fact. It is known that in a Hilbert space there exists a unique nearest point projection onto closed convex subsets. The projection is also 1-Lipschitz (Proposition 2.15). So the composition yields an extension operator for Lipschitz functions that preserves Lipschitz constants.

Example 6.1. Let X be a Hilbert space and let C ⊆ X be closed and convex. Denote by π : X → C the nearest point projection. Then if, for some L > 0, f : C → X is L-Lipschitz, the function f ◦ π : X → X extends f and is again L-Lipschitz. Furthermore, if g : C → X and h : C → X are L-Lipschitz, then supX kg ◦ π − h ◦ πk = supC kg − hk.

~~Proof. If f : C → X is as stated, then f ◦ π(x) = f(x) for all x ∈ C and~~

~~kf ◦ π(x1) − f ◦ π(x2)k ≤ Lkπ(x1) − π(x2)k ≤ Lkx1 − x2k~~

~~for all x1, x2 ∈ X, as required. The second claim is obvious.~~

~~67 6. Continuous Selections~~

~~6.2 Compact sets in Euclidean space~~

Explicit extension operators for Kirszbraun’s Theorem are also available in another, less obvious situation. Namely, this is also possible in Euclidean space, and when extensions are made from compact subsets. Before we start, we set some notation. If X is a Hilbert space and A ⊆ X, we consider the following function spaces. By N (A) we denote the set of all bounded nonexpansive mappings from A to X, that is, functions f : A → X such that kf(x) − f(y)k ≤ kx − yk for all x, y ∈ A. In the special case where 0 ∈ A, we also consider the set of all functions f ∈ N (A) such that f(0) = 0. This set will by denoted by N0(A). The space N (A) is clearly a subset of C(A, X), the Banach space of all bounded continuous functions from A to X with the supremum norm. So N (A) naturally is a metric space. The same is true also for N0(A). Then the following holds [8, Theorem 1.3]. Theorem 6.2 (Kopecká 2012). Let X be a Euclidean space and A ⊆ X compact. Then there exists a uniformly continuous function F : N (A) → N (X) such that, if f ∈ N (A), then F (f)|A = f, and if f is L-Lipschitz, for some L > 0, then F (f) is also L-Lipschitz. We will not show this directly, but the following slightly weaker statement, from which we will then deduce Theorem 6.2. Theorem 6.3. Let X be a Euclidean space and A a closed subset of B[0, 1] such that 0 ∈ A. Then there exists a uniformly continuous function F0 : N0(A) → N0(B[0, 1]) such that, if f ∈ N0(A), then F0(f)|A = f, and if f is L-Lipschitz, for some L > 0, then F0(f) is also L-Lipschitz. A ﬁrst step towards this is the uniform convexity of Hilbert spaces. Recall that a Banach space (X, k · k) is called uniformly convex if and only if for all ε > 0 there exists some δ > 0 such that, for all x, y ∈ B[0, 1], if kx − yk ≥ ε, then 1 k 2 (x + y)k ≤ 1 − δ. Proposition 6.4. Let X be a Hilbert space. Then X is uniformly convex. Proof. Note that it is suﬃcient to consider only the case where ε ≤ 2. By the parallelogram identity,

kx + yk2 + kx − yk2 = 2(kxk2 + kyk2), we have 1 1 1 k (x + y)k2 = (kxk2 + kyk2) − kx − yk2 2 2 4 for all x, y ∈ X. So if x, y ∈ B[0, 1] and kx − yk ≥ ε, then 1 1 k (x + y)k2 ≤ 1 − ε2, 2 4

~~68 6.2. Compact sets in Euclidean space~~

q 1 2 and one can take δ := 1 − 1 − 4 ε . Next we show two rather “technical” lemmata on the following special function. If X is a Hilbert space, A is a closed and bounded subset of X, f : A → X is Lipschitz continuous, and z ∈ X \ A, we consider the function ΦA,f,z : X → R, which is deﬁned by kx − f(a)k ΦA,f,z(x) = sup . a∈A kz − ak The study of this function is motivated by the following idea. If f is L-Lipschitz on A, for some L > 0, and x ∈ X is such that ΦA,f,z(x) ≤ L, then we can extend f to z, keeping it L-Lipschitz, by letting f(z) be that point x. By Kirszbraun’s Theorem (Theorem 2.9), such a point always exists, and, in fact, a function very similar to the above appeared in its proof. In order to construct an explicit extension operator as required by Theorem 6.3, we would need now a way to choose one particular point with that property. A “natural” approach here seems to minimize the functions ΦA,f,z. Lemma 6.5. Let X be a Hilbert space, A a closed subset of B[0, 1] such that 0 ∈ A, f ∈ N0(A), and z ∈ B[0, 1] \ A. Then the function Φ = ΦA,f,z has the following properties. (1) Φ is convex. (2) Φ is dist(A, z)−1-Lipschitz. (3) Φ attains its minimum at exactly one point m = m(A, f, z). (4) If f is L-Lipschitz, for some L > 0, then Φ(m) ≤ L. (5) The minimizer m belongs to B[0, 1]. (6) If m 6= w ∈ X, then kw − mk2 ≤ 4(Φ(w)2 − Φ(m)2). Proof. (1) For every a ∈ A, x, y ∈ X, and λ ∈ [0, 1], kλx + (1 − λ)y − f(a)k kx − f(a)k ky − f(a)k ≤ λ + (1 − λ) , kz − ak kz − ak kz − ak kx−f(a)k such that the function x 7→ kz−ak is convex. The supremum of a family of convex functions is again convex, as can be easily checked. Hence Φ is convex. (2) For every a ∈ A, and x, y ∈ X,

~~kx − f(a)k ky − f(a)k −1 − ≤ dist(A, z) kx − yk, kz − ak kz − ak~~

kx−f(a)k −1 such that the function x 7→ kz−ak is dist(A, z) -Lipschitz. The supremum of a family of dist(A, z)−1-Lipschitz functions is again dist(A, z)−1-Lipschitz, as can be easily checked. Hence Φ is dist(A, z)−1-Lipschitz.

~~69 6. Continuous Selections~~

~~−1 (3) For t ≥ 0 let Kt = Φ ((−∞, t]) be the sublevel sets of Φ. Note that, for every t ≥ 0, \ Kt = B[f(a), tkz − ak], a∈A~~

which implies that Kt is weakly compact. Moreover Ks ⊆ Kt if s < t. Let λ = inf Φ. Then \ Kλ = Kt t>λ is nonempty, by the ﬁnite intersection property. So Φ attains its minimum. As for uniqueness of the minimizer, assume m ∈ Kλ and w ∈ X, w 6= m. Clearly we have Φ(w) ≥ λ. For the sake of contradiction, assume that Φ(w) = λ. Put kw − mk ε := . λ supa∈A kz − ak Then, by Proposition 6.4, there exists some δ > 0 such that, for all x, y ∈ 1 B[0, 1], if kx − yk ≥ ε, then k 2 (x + y)k ≤ 1 − δ. In particular, for every a ∈ A, since for w − f(a) x = λkz − ak and m − f(a) y = λkz − ak we have that x ∈ B[0, 1], y ∈ B[0, 1], and kx − yk ≥ ε, 1 1 k (x + y)k = (λkz − ak)−1k (w + m) − f(a)k ≤ 1 − δ, 2 2 that is k 1 (w + m) − f(a)k 2 ≤ (1 − δ)λ. kz − ak 1 But then Φ( 2 (w + m)) ≤ (1 − δ)λ < λ, a contradiction. (4) If f is L-Lipschitz, for some L > 0, by Kirszbraun’s Theorem (Theorem 2.9), f can be extended to z such that it is again L-Lipschitz. That is, there exists kx−f(a)k some x ∈ X such that kx − f(a)k ≤ Lkz − ak for all a ∈ A, or kz−ak ≤ L, and hence kx − f(a)k Φ(x) = sup ≤ L. a∈A kz − ak But then clearly also Φ(m) ≤ L. (5) As f is 1-Lipschitz, λ ≤ 1, and hence \ Kλ = B[f(a), λkz − ak] ⊆ B[0, λkzk] ⊆ B[0, 1], a∈A so the minimizer m belongs to B[0, 1].

~~70 6.2. Compact sets in Euclidean space~~

~~(6) Let m 6= w ∈ X. For the sake of contradiction assume that kw − mk2 > 4(Φ(w)2 − Φ(m)2). Then kw − mk2 = 4(Φ(w)2 − Φ(m)2 + α) for some α > 0. If a ∈ A, then kz − ak ≤ 2, and~~

(Φ(w)2 − Φ(m)2 + α)kz − ak2 ≤ kw − mk2 = = k(w − f(a)) − (m − f(a))k2 = = kw − f(a)k2 + km − f(a)k2 − 2hw − f(a), m − f(a)i ≤ ≤ Φ(w)2kz − ak2 + Φ(m)2kz − ak2 − 2hw − f(a), m − f(a)i.

~~Hence 2hw − f(a), m − f(a)i ≤ (2Φ(m)2 − α)kz − ak2. Let now x = (1 − s)m + sw for some s ∈ (0, 1). Then, for all a ∈ A,~~

kx − f(a)k2 = k(1 − s)(m − f(a)) + s(w − f(a))k2 = = (1 − s)2km − f(a)k2 + s2kw − f(a)k2 + 2(1 − s)shm − f(a), w − f(a)i ≤ ≤ kz − ak2((1 − s)2Φ(m)2 + s2Φ(w)2 + (1 − s)s(2Φ(m)2 − α)) = = kz − ak2(Φ(m)2 − s(α − s(Φ(w)2 − Φ(m)2 + α))).

~~If, for example, α s = , 2(Φ(w)2 − Φ(m)2 + α) then Φ(x) < Φ(m), a contradiction.~~

~~Lemma 6.6. Let X be a Hilbert space, A a closed subset of B[0, 1] such that 0 ∈ A, and z ∈ B[0, 1] \ A.~~

~~−1 (1) The mapping ΦA,·,z : N0(A) → C(B[0, 1]), f 7→ ΦA,f,z is dist(A, z) -Lipschitz.~~

~~(2) The mapping m(A, ·, z): N0(A) → B[0, 1], f 7→ m(A, f, z) is uniformly continuous.~~

~~Proof. (1) Let f, g ∈ N0(A) and x ∈ B[0, 1]. Then, for every a ∈ A,~~

~~kx − f(a)k kx − g(a)k −1 − ≤ dist(A, z) kf − gk∞, kz − ak kz − ak so, by Lemma 2.4,~~

~~kx − f(a)k kx − g(a)k ΦA,f,z(x) − ΦA,g,z(x) = sup − sup ≤ a∈A kz − ak a∈A kz − ak ! kx − f(a)k kx − g(a)k −1 ≤ sup − ≤ dist(A, z) kf − gk∞ a∈A kz − ak kz − ak~~

~~71 6. Continuous Selections~~

and kx − g(a)k kx − f(a)k ΦA,g,z(x) − ΦA,f,z(x) = sup − sup ≤ a∈A kz − ak a∈A kz − ak ! kx − g(a)k kx − f(a)k −1 ≤ sup − ≤ dist(A, z) kf − gk∞, a∈A kz − ak kz − ak and thus −1 |ΦA,f,z(x) − ΦA,g,z(x)| ≤ dist(A, z) kf − gk∞. −1 In particular, kΦA,f,z − ΦA,g,zk∞ ≤ dist(A, z) kf − gk∞.

~~ε2 (2) For given ε > 0, let δ0 = min{1, 32 } and put δ = dist(A, z)δ0. Then, if f, g ∈ N0(A) are such that kf − gk∞ ≤ δ,~~

~~|ΦA,f,z(x) − ΦA,g,z(x)| ≤ δ0~~

~~for all x ∈ B[0, 1]. Denote ϕ = ΦA,f,z, ψ = ΦA,g,z, and u = m(A, f, z), v = m(A, g, z). Then~~

~~ϕ(v) ≤ ψ(v) + δ0 ≤ ψ(u) + δ0 ≤ ϕ(u) + 2δ0,~~

~~by using twice the above inequality and by exploiting the fact that v is the minimizer of ψ, and~~

2 2 2 2 2 kv − uk ≤ 4(ϕ(v) − ϕ(u) ) ≤ 4((ϕ(u) + 2δ0) − ϕ(u) ) = 2 2 = 16ϕ(u)δ0 + 16δ0 ≤ 32δ0 ≤ ε , by Lemma 6.5 (6), by the above inequality, and using the fact that ϕ(u) ≤ 1 and δ0 ≤ 1. So kv − uk ≤ ε.

Proof of Theorem 6.3. If A = B[0, 1], nothing has to be shown. So we will assume A $ B[0, 1] in the sequel. As was already said above, the main idea for the construction of the operator F0 : N0(A) → N0(B[0, 1]) is to minimize functions of the type ΦA,f,z. More precisely, we will pick some countable dense subset {z1, z2,... } of B[0, 1] \ A, extend each f ∈ N0(A) ﬁrst recursively to a function on A ∪ {z1, z2,... }, which we will for simplicity again call f, by the condition

f(zn) = m(A ∪ {z1, z2, . . . , zn−1}, f, zn), and then let F0(f) be the unique continuous extension of f from A ∪ {z1, z2,... } to B[0, 1]. As for the recursion, this works because in every step the minimizer

~~m(A ∪ {z1, z2, . . . , zn−1}, f, zn)~~

~~72 6.2. Compact sets in Euclidean space~~

of ΦA∪{z1,z2,...,zn−1},f,zn exists and is unique by Lemma 6.5 (3). Furthermore, note that the resulting extensions will remain nonexpansive, by Lemma 6.5 (4), and that they will satisfy every Lipschitz condition that the initial functions possibly satisﬁed, again by Lemma 6.5 (4). As for the continuous extension from the dense subset of B[0, 1] \ A to B[0, 1], ∞ note that, if x ∈ B[0, 1] and (xn)n=1 is a sequence in A ∪ {z1, z2,... } such that ∞ limn→∞ xn = x, then, as f is nonexpansive on A∪{z1, z2,... }, (f(xn))n=1 is Cauchy, ∞ so in particular has a limit limn→∞ f(xn). Also, if (yn)n=1 is another sequence in A∪{z1, z2,... } such that limn→∞ yn = x, and ε > 0, then, if N is chosen suﬃciently large such that

~~ε •k limn→∞ f(xn) − f(xN )k ≤ 4 ,~~

~~ε •k f(yN ) − limn→∞ f(yn)k ≤ 4 ,~~

~~ε •k xN − xk ≤ 4 ,~~

~~ε •k x − yN k ≤ 4 , we have~~

~~k lim f(xn) − lim f(yn)k ≤ n→∞ n→∞~~

≤ k lim f(xn) − f(xN )k + kf(xN ) − f(yN )k + kf(yN ) − lim f(yn)k ≤ n→∞ n→∞ ε ε ε ε ≤ + kx − y k + ≤ + kx − xk + kx − y k + ≤ ε, 4 N N 4 4 N N 4 hence limn→∞ f(xn) = limn→∞ f(yn). So we can put

~~F0(f)(x) = lim f(xn) n→∞~~

∞ for x ∈ B[0, 1] and any sequence (xn)n=1 in A ∪ {z1, z2,... } with the property that limn→∞ xn = x. The resulting function F0(f) clearly is an extension of f. Also, if f ∈ N0(A) ∞ ∞ is L-Lipschitz, for some L > 0, x, y ∈ B[0, 1], (xn)n=1, (yn)n=1 are sequences in A ∪ {z1, z2,... } such that limn→∞ xn = x, limn→∞ yn = y, and ε > 0, then, if N is chosen suﬃciently large such that

~~ε •k limn→∞ f(xn) − f(xN )k ≤ 4 ,~~

~~ε •k f(yN ) − limn→∞ f(yn)k ≤ 4 ,~~

~~ε •k xN − xk ≤ 4L ,~~

~~ε •k y − yN k ≤ 4L ,~~

~~73 6. Continuous Selections then~~

~~kF0(f)(x) − F0(f)(y)k = k lim f(xn) − lim f(yn)k ≤ n→∞ n→∞~~

≤ k lim f(xn) − f(xN )k + kf(xN ) − f(yN )k + kf(yN ) − lim f(yn)k ≤ n→∞ n→∞ ε ε ≤ + Lkx − y k + ≤ 4 N N 4 ε ε ≤ + Lkx − xk + Lkx − yk + Lky − y k + ≤ Lkx − yk + ε, 4 N N 4 such that F0(f) is also L-Lipschitz. So it remains to show that the mapping f 7→ F0(f) is uniformly continuous. For this we have to show that for every ε > 0 there exists some δ > 0 such that, for all f, g ∈ N0(A), if kf − gk∞ ≤ δ, then kF0(f) − F0(g)k∞ ≤ ε. So ﬁx ε > 0. As X is Euclidean, for the chosen dense subset {z1, z2,... } of B[0, 1] \ A, there exists an ε index N such that {z1, z2, . . . , zN } is an 3 -net of B[0, 1] \ A. For n = 1, 2,...,N, denote An = A∪{z1, z2, . . . , zn}. Then, by Lemma 6.6 (2), there exists some δN > 0 such that ε |m(A , f, z ) − m(A , g, z )| ≤ N−1 N N−1 N 3 if f, g ∈ N0(AN−1) and kf − gk∞ ≤ δN . Without loss of generality we can assume ε that δN ≤ 3 . Similarly, again by Lemma 6.6 (2), there exists some δN−1 > 0 such that |m(AN−2, f, zN−1) − m(AN−2, g, zN−1)| ≤ δN if f, g ∈ N0(AN−2) and kf − gk∞ ≤ δN−1. Again we can assume that δN−1 ≤ δN . Now we continue like this, i.e. we choose δn > 0 such that

~~|m(An−1, f, zn) − m(An−1, g, zn)| ≤ δn+1 for all f, g ∈ N0(An−1) with kf −gk∞ ≤ δn, and δn ≤ δn+1, for n = N −2,N −3,... , until ﬁnally we ﬁnd some δ1 > 0 such that~~

~~74 6.2. Compact sets in Euclidean space~~

~~But now if x ∈ B[0, 1], either x ∈ A, and ε kF (f)(x) − F (g)(x)k = kf(x) − g(x)k ≤ δ = δ ≤ · · · ≤ δ ≤ ≤ ε, 0 0 1 N 3 or x ∈ B[0, 1] \ A, and~~

~~kF0(f)(x) − F0(g)(x)k ≤~~

~~≤ kF0(f)(x) − F0(f)(zi)k + kF0(f)(zi) − F0(g)(zi)k + kF0(g)(zi) − F0(g)(x)k ≤~~

~~≤ kx − zik + kF0(f)(zi) − F0(g)(zi)k + kzi − xk ≤ ε,~~

~~ε if zi ∈ {z1, z2, . . . , zN } is chosen such that kzi − xk ≤ 3 .~~

Proof of Theorem 6.2. If A ⊆ X is compact, there exist a0 ∈ A and r > 0 such 1 0 1 that r (A − a0) ⊆ B[0, 1]. Denote A = r (A − a0). By Theorem 6.3, we have a map 0 0 F0 : N0(A ) → N0(B[0, 1]) that is uniformly continuous, such that, if f ∈ N0(A ), then F0(f)|A0 = f, and if f is L-Lipschitz, for some L > 0, then F0(f) is also L-Lipschitz. Put

~~0 0 F : N (A ) → N (B[0, 1]), f 7→ F0(f − f(0)) + f(0).~~

Then F 0 is again uniformly continuous, as can be easily checked, F 0(f) extends f, for every f ∈ N (A0), and F 0 preserves Lipschitz constants. According to Proposition 2.15, the nearest point projection π : X → B[0, 1] is a 1-Lipschitz retraction. Hence we naturally get an extension operator

~~π : N (B[0, 1]) → N (X), f 7→ f ◦ π as in Example 6.1. Finally, let~~

1 ϕ : N (A) → N (A0), f 7→ [x 7→ f(rx + a )] 1 r 0 and 1 ϕ : N (X) → N (X), f 7→ [x 7→ rf( (x − a ))]. 2 r 0 1 As kϕ1(f)−ϕ1(g)k∞ = r kf−gk∞ for f, g ∈ N (A) and kϕ2(f)−ϕ2(g)k∞ = rkf−gk∞ for f, g ∈ N (X), both ϕ1 and ϕ2 are Lipschitz continuous, in particular uniformly continuous. Also note that both ϕ1 and ϕ2 preserve Lipschitz constants. Now if we deﬁne F : N (A) → N (X) to be the composition of mappings

~~0 N (A) →ϕ1 N (A0) →F N (B(0, 1)) →π N (X) →ϕ2 N (X),~~

~~F has the desired properties.~~

~~75 6. Continuous Selections~~

~~6.3 Continuous selections~~

Finally we will see yet another approach to this problem. In [7] Michael’s Selection Theorem is used to show the existence of continuous extension operators that preserve Lipschitz constants also for the general case of arbitrary subsets of a Hilbert space and without the restriction to some subfamily of mappings with an upper bound for the Lipschitz constant. Let X be a Hilbert space. By L(A) we denote the space of all bounded Lipschitz continuous functions from a subset A ⊆ X into X, which is naturally a metric space considered as a subset of C(A, X), the space of all bounded continuous functions from A into X with the supremum norm. Furthermore, we consider the map

Ψ: L(A) → P(C(X,X)), g 7→ {f ∈ C(X,X); f|A = g, f ∈ L(X), Lip(f) = Lip(g)}, which assignes to each g ∈ L(A) the set of all its bounded Lipschitz continuous extensions to X that have the same Lipschitz constant. Note that by Theorem 2.16 such a function always exists. The main result of this section then is the following.

Theorem 6.7 (Kopecká 2012). Let X be a Hilbert space and ∅ 6= A ⊆ X. Then the carrier Ψ: L(A) → P(C(X,X)) admits a continuous selection F : L(A) → C(X,X). Furthermore, for every closed subset D of L(A) and every continuous selection F : D → C(X,X) of the extension mapping Ψ, there is a continuous selection F˜ : L(A) → C(X,X) which extends F . As was already said, the idea here is to use Michael’s Selection Theorem (Theo- rem 5.10). So we have to show that Ψ is lower semi-continuous and has closed and convex values. This will be done in the following sequence of lemmata.

Lemma 6.8. Let X be a Hilbert space, ∅ 6= A ⊆ X. Let f ∈ L(X) such that Lip(f) = Lip(f|A). Then for every ε > 0 there exists δ > 0 with the following property. If gA ∈ L(A) is such that kf − gAkA < δ, then gA admits an extension gX ∈ L(X) so that kf − gX kX < ε and Lip(gX ) = Lip(gA).

Proof. If f is constant, i.e. Lip(f) = 0, then one can put δ = ε and take for gX the composition of any Lipschitz constant preserving extension of gA with the retraction on the ε-ball around the unique value of f. So assume f is not constant. Then in particular Lip(f) > 0. Let ε > 0 and put 0 ε ε := 3 . Then, choose s < 1 close enough to 1 such that

~~2 ε02 • 4(1 − s)kfkX ≤ 8 ,~~

~~ε0 • (1 − s)kfkX ≤ 4 , 2 1 • s ≥ 2 ,~~

~~76 6.3. Continuous selections~~

~~A × {η} X × {η} h = gA X × {0} h = sf~~

~~Figure 6.1: Extending gA when Lip(gA) is small. and, depending on s, δ > 0 such that~~

~~• Lip(gA) ≥ s Lip(f|A) for all gA ∈ L(A) with kf − gAkA < δ,~~

~~ε02 • 4δkfkX ≤ 8 ,~~

~~ε0 • δkfkX ≤ 4 , ε • δ < 4 .~~

~~ε0 Furthermore, put η := Lip(f) . Now let gA ∈ L(A) with kf−gAkA < δ. We distinguish two cases.~~

~~(1) s Lip(f) ≤ Lip(gA) < 2 Lip(f)~~

~~(2) 2 Lip(f) ≤ Lip(gA)~~

~~First case. Here we consider the function~~

~~h :(X × {0}) ∪ (A × {η}) → X on a subset of the `2-sum X × R, which is given by~~

~~h(x, 0) = sf(x)(x ∈ X) and h(y, η) = gA(y)(y ∈ A).~~

~~See Figure 6.1. We show that it is Lipschitz continuous with Lip(h) = Lip(gA). As Lip(sf) = s Lip(f) ≤ Lip(gA), it is suﬃcient to estimate for x ∈ X and y ∈ A, where we get~~

~~2 2 kh(x, 0) − h(y, η)k = ksf(x) − gA(y)k ≤ 2 ≤ (skf(x) − f(y)k + kf(y) − gA(y)k + (1 − s)kf(y)k) ≤ 2 ≤ (skf(x) − f(y)k + δ + (1 − s)kfkX ) ≤~~

~~77 6. Continuous Selections~~

~~˜ 2δ A A Lip(gA) g˜A = gA g˜A = f~~

~~Figure 6.2: Extending gA when Lip(gA) is large.~~

2 2 2 ≤ s kf(x) − f(y)k + 4skfkX (δ + (1 − s)kfkX ) + (δ + (1 − s)kfkX ) ≤  ε02  02 ! 2 2 2 2 2 2 ε ≤ s Lip(f) kx − yk +  ≤ Lip(gA) kx − yk + = s2 Lip(f)2 Lip(f)2 2 2 2 2 2 = Lip(gA) (kx − yk + η ) = Lip(gA) k(x, 0) − (y, η)kX×R, as required. So by Kirszbraun’s Theorem h can be extended to a bounded Lipschitz continuous function on X × R with Lipschitz constant Lip(gA). For simplicity, denote this extension again by h. Then we deﬁne the extension gX of gA by

~~gX (x) := h(x, η)(x ∈ X).~~

~~It is clear that gX ∈ L(X) and that Lip(gX ) = Lip(gA). So it remains to check that kf − gX kX < ε. Let x ∈ X. Then~~

~~kgX (x) − f(x)k ≤ kgX (x) − sf(x)k + (1 − s)kfkX ≤ 0 0 0 0 ≤ kh(x, η) − h(x, 0)k + ε ≤ Lip(gA)η + ε < 2 Lip(f)η + ε = 3ε = ε.~~

Second case. Here we extend the function gA in two steps. Consider the set 2δ A˜ = {x ∈ X; dist(A, x) ≥ }. Lip(gA) ˜ ˜ First we extend gA to A ∪ A by the function g˜A : A ∪ A → X which is given by  gA(x) x ∈ A, g˜A(x) = f(x) x ∈ A.˜

~~See Figure 6.2. In order to see that Lip(˜gA) = Lip(gA), it is suﬃcient to estimate for x ∈ A˜ and y ∈ A. Here~~

~~kg˜A(x) − g˜A(y)k = kf(x) − gA(y)k ≤ kf(x) − f(y)k + kf(y) − gA(y)k ≤~~

~~78 6.3. Continuous selections~~

Lip(g )! ≤ Lip(f)kx − yk + δ ≤ Lip(f) + A kx − yk ≤ Lip(g )kx − yk, 2 A as required. Let now gX be any bounded Lipschitz continuous extension of g˜A to X so that Lip(gX ) = Lip(˜gA) = Lip(gA). It remains to check that kf − gX kX < ε. If ˜ ε ˜ x ∈ A ∪ A, kf(x) − gX (x)k is either 0 or bounded by δ < 4 . If x ∈ X \ (A ∪ A), then there exists y ∈ A such that kx − yk < 2δ and Lip(gA)

kgX (x) − f(x)k ≤ kgX (x) − gX (y)k + kgX (y) − f(y)k + kf(y) − f(x)k ≤ 2δ 2δ ≤ Lip(gA) + δ + Lip(f) ≤ 4δ < ε. Lip(gA) Lip(gA) Lemma 6.9. Let X be a Hilbert space, ∅ 6= A ⊆ X. Then the carrier Ψ: L(A) → P(C(X,X)) is lower semi-continuous. Proof. We use the point-wise characterization of lower semi-continuity. That is, we have to show that for every x0 ∈ L(A) and every open subset V of C(X,X) such that Ψ(x0) ∩ V 6= ∅ there exists a neighborhood U of x0 in L(A) such that Ψ(x) ∩ V 6= ∅ for all x ∈ U. But that is exactly what Lemma 6.8 says. Lemma 6.10. Let X be a Hilbert space, ∅ 6= A ⊆ X. Then values of the carrier Ψ: L(A) → P(C(X,X)) are closed. Proof. The uniform limit of extensions of some g ∈ L(A) is again an extension of g and again Lipschitz continuous with the same Lipschitz constant as g.

~~Lemma 6.11. Let X be a Hilbert space, ∅ 6= A ⊆ X. Then values of the carrier Ψ: L(A) → P(C(X,X)) are convex.~~

~~Proof. Let f1, f2 ∈ Ψ(g), for some g ∈ L(A), and λ ∈ (0, 1). Then~~

~~λf1(y) + (1 − λ)f2(y) = λg(y) + (1 − λ)g(y) = g(y) for all y ∈ A, such that λf1 + (1 − λ)f2 is an extension of g, and~~

~~k(λf1 + (1 − λ)f2)(x) − (λf1 + (1 − λ)f2)(y)k ≤~~

≤ λkf1(x) − f1(y)k + (1 − λ)kf2(x) − f2(y)k ≤ ≤ λ Lip(g)kx − yk + (1 − λ) Lip(g)kx − yk = Lip(g)kx − yk for all x, y ∈ X, such that λf1 + (1 − λ)f2 is Lipschitz continuous with Lipschitz constant Lip(g). Proof of Theorem 6.7. The metric space L(A) is paracompact by Theorem 5.1. C(X,X) is a Banach space. By Lemma 6.9, Lemma 6.10 and Lemma 6.11, the carrier Ψ is lower semi-continuous and has closed and convex values. So Michael’s Selection Theorem (Theorem 5.10) yields continuous selections. For the second claim, use Corollary 5.13.

~~List of Symbols~~

P(X) the set of all subsets of X N the set of the natural numbers Z the set of the integers Q the set of the rational numbers R the set of the real numbers C the set of the complex numbers B(x, r) the open ball with radius r around x B[x, r] the closed ball with radius r around x Lip(f) the Lipschitz constant of f conv S the convex hull of S conv S the closed convex hull of S C(X,Y ) the set of all bounded continous functions from X into Y C(X) the set of all bounded continous functions from X into R N (A, X) the set of all bounded nonexpansive mappings from A into X N0(A, X) the set of all bounded nonexpansive mappings from A into X that ﬁx the origin L(A, X) the set of all bounded Lipschitz continuous mappings from A into X

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~~84 Index~~

Banach space T3, 20 strictly convex, 18 T4, 19 uniformly convex, 68 compact, 19 countably paracompact, 57 Carrier, 62 Lindelöf, 20 lower semi-continuous, 63 metrizable, 19 upper semi-continuous, 63 normal, 19 Continuous selection, 62 paracompact, 57 Contraction, 5 regular, 20 Kirszbraun intersection property, 7

~~Lipschitz constant, 5 Lipschitz continuity, 5 Locally ﬁnite partition of unity, 62 subordinated to a cover, 62~~

~~Metric space injective, 11 metrically convex, 11 with the binary intersection property, 11~~

~~Nonexpansive extension property, 7 Nonexpansive map, 5~~

~~Property (K), 7 Property (P’), 54 Property (P), 47 Pseudometric not necessarily symmetric, 47~~

~~Retract, 16 Retraction, 16~~

~~Topological space~~