The Chebyshev Set Problem

James FletcherDepartment of MathematicsThe University of Auckland

Supervisor: Warren B. Moors

A thesis submitted in fulfillment of the requirements for the degree of Master of Sciencein Mathematics, The University of Auckland, 2013.

Abstract

A Chebyshev set is a subset of a normed linear space that admits unique best approxima-tions. In the first part of this thesis we focus on some positive results concerning Cheby-shev sets, deriving properties of the metric projection, sufficient conditions for a subset ofa normed linear space to be a Chebyshev set, and sufficient conditions for a Chebyshevset to be convex. In the second half of the thesis we look at the so called ‘Chebyshev setproblem’, constructing a highly non-trivial example of a non-convex Chebyshev set in aninner product space.

i

Acknowledgements

I would like to thank my supervisor Warren Moors for all of his friendly assistance andmotivation throughtout the year, his advice on doing and writing mathematics, and forproposing such an interesting research topic. Lastly, thank you to my family for all ofyour support and encouragement throughout the year.

ii

Contents

Abstract i

Acknowledgements ii

Glossary iv

1 Introduction 11.1 Motivation and historical background . . . . . . . . . . . . . . . . . . . 11.2 Outline of thesis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

2 Positive results 32.1 Definitions and examples . . . . . . . . . . . . . . . . . . . . . . . . . . 32.2 Necessary conditions for a proximinal/Chebyshev set . . . . . . . . . . . 52.3 Properties of the metric projection . . . . . . . . . . . . . . . . . . . . . 72.4 Sufficient conditions for a proximinal/Chebyshev set . . . . . . . . . . . 102.5 The convexity of Chebyshev sets . . . . . . . . . . . . . . . . . . . . . . 232.6 Continuity of the metric projection . . . . . . . . . . . . . . . . . . . . . 33

3 A non-convex Chebyshev set in an inner product space 413.1 History and motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . 413.2 The construction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

3.2.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . 423.2.2 Smoothness and unique nearest points . . . . . . . . . . . . . . . 503.2.3 The inductive step . . . . . . . . . . . . . . . . . . . . . . . . . 623.2.4 The non-convex Chebyshev set . . . . . . . . . . . . . . . . . . . 76

4 Future research 80

Bibliography 80

iii

Glossary

• For any set X , P(X) is the set of all subsets of X .

• For any subset A of a topological space (X, τ), we define

– int(A), called the interior of A, is the union of all open sets contained in A,

– A, called the closure of A, is the intersection of all closed sets containing A,

– ∂A, called the boundary of A, is A \ int(A).

• Throughout this thesis all normed linear spaces will be assumed to be over R.

• For any normed linear space (X, ‖·‖), we define

– B[x, r] := {y ∈ X : ‖x− y‖ ≤ r}, for any x ∈ X and r ≥ 0. Note: thisimplies that B[x, 0] = {x}, for any x ∈ X ,

– B(x, r) := {y ∈ X : ‖x− y‖ < r} , for any x ∈ X and r > 0,

– BX := B[0, 1],

– SX := {x ∈ X : ‖x‖ = 1} .

• For any inner product space (X, 〈·, ·〉), ∠xyz will denote the angle between thevectors (y − x) and (y − z), where x, y, z are distinct points in X . That is,

∠xyz := cos−1(〈y − x, y − z〉‖y − x‖ ‖y − z‖

),

where ‖·‖ is the norm induced by the inner product.

iv

Chapter 1

Introduction

1.1 Motivation and historical backgroundA famous result in approximation theory, due to Chebyshev [27], states that if (C[0, 1], ‖·‖∞)is the normed linear space consisting of all continuous functions on [0, 1] equipped withthe uniform norm and Pn is the subspace of C[0, 1] consisiting of all poynomials of de-gree no larger than n ∈ N, then for every member of C[0, 1] there exists a unique bestapproximation in Pn .

Motivated by this, a subset K of a metric space (X, d) will be called a Chebyshev setif, for every point in X , there exists a unique nearest point in K. Since the geometry ofsuch sets is our primary interest we will limit our investigations to normed linear spaces.

Whilst working in approximation theory, Stechkin asked whether the set of Chebyshevrational functions forms a Chebyshev set in Lp with p > 1. Since it was known that anonempty closed convex set in a Hilbert space is a Chebyshev set, and the set of Chebyshevrational functions is not convex, the answer appeared to be no (at least in L2). Thus arosethe famous Chebyshev set problem - must a Chebyshev set in a Hilbert space be convex?

The first attempt at a solution came from Motzkin [23], who proved that any Chebyshevset in R2 (equipped with the Euclidean norm) must be convex. Unfortunately, his methoddid not generalise to higher dimensions, and so it was left to Bunt [6] and Kritikos [22],working independently, to supply the first proof that a Chebyshev set in Rn (equippedwith the Euclidean norm) is convex for any n ∈ N. Alternative proofs were also given byJessen [15] and Busemann [7]. Shortly afterwards, these results were extended to arbitrarysmooth finite-dimensional spaces. For the interested reader, the master’s thesis of Kelly[19] provides an excellent survey of the finite-dimensional case.

It was quickly realised that continuity properties of the metric projection play a keyrole in determining the convexity of a Chebyshev set. Amongst others, Klee [20] [21], As-

1

plund [2] and Vlasov [28], assuming continuity of the metric projection, produced variousimportant results, including Vlasov’s theorem, which provides some of the most generalconditions that ensure a Chebyshev set is convex. Following on from this, Balaganskii [3]proved that, working in a Hilbert space, if the set of discontinuities of the metric projectionis countable then the Chebyshev set is convex. Frerking and Westphal [13], making use ofthe theory of monotone operators, generalised this result to a connectedness property ofthe set of discontinuities of the metric projection.

Ficken [20] and Efimov and Stechkin [12], using inversion in spheres to view the prob-lem in terms of farthest points, proved that every compact Chebyshev set in a Hilbert spaceis necessarily convex. In [17], Johnson provided an example of a non-convex Chebyshevset in an incomplete inner product space. So far, attempts to generalise this constructionto a Hilbert space have been unsuccessful. Despite all of these positive results, the prob-lem of classifying which smooth Banach spaces admit only convex Chebyshev sets stillremains open to this day.

1.2 Outline of thesisThe first half of this thesis deals with positive results regarding Chebyshev sets. Firstly,we show that closure is a necessary condition for a subset of a normed linear space to beproximinal/Chebyshev. Following this we derive some properties of the metric projection,culminating in an important theorem - the metric projection is continuous in finite dimen-sions. After this we find sufficient conditions for a subset to be a proximinal/Chebyshevset, introducing concepts such as strict convexity, uniform convexity, and smoothness.Next we find conditions that guarantee a Chebyshev set is convex. The main theoremof this section, due to Vlasov, states that a Chebyshev set in a normed linear space withstrictly convex dual and continuous metric projection is convex. Inspired by this result, wethen derive conditions that guarantee a Chebyshev set has continuous metric projection.Throughout the chapter examples are given to help familiarise the reader with the conceptsdiscussed.

In the second half of the thesis we construct an example of a non-convex Chebyshev setin an incomplete inner product space. This example was first presented by Johnson [17],although his very algebraic proof contains errors which were later corrected by Jiang [16].Our construction is based on that given in the survey article of Balaganskii and Vlasov [4].Whilst the proof presented in this thesis is longer, we hope that the extra detail elucidatesthe geometry behind the construction and provides the necessary background informationfor the reader to understand the numerous technical steps in the proof.

2

Chapter 2

Positive results

2.1 Definitions and examplesWe begin this section by formally defining how we measure the distance of a point to asubset of a normed linear space. From this we define the metric projection, which in turnis used to formulate the concepts of proximinal sets and Chebyshev sets. A few simpleexamples of such sets are then given to familiarise the reader with these objects.

Definition 1. Let (X, ‖·‖) be a normed linear space and K be a nonempty subset of X .For any point x ∈ X we define d(x,K) := infy∈K ‖x− y‖, and call this the distance fromx to K. We will also refer to the map

x 7→ d(x,K)

as the distance function for K.

The following simple result regarding the distance function will be used repeatedlythroughout the remainder of the thesis.

Proposition 2.1.1. Let K be a nonempty subset of a normed linear space (X, ‖·‖). Thenthe distance function for K is nonexpansive (and hence continuous).

Proof. Let x, y ∈ X and k ∈ K. By the triangle inequality

d(x,K) ≤ ‖x− k‖ ≤ ‖x− y‖+ ‖y − k‖ .

Rearranging this we getd(x,K)− ‖x− y‖ ≤ ‖y − k‖ .

3

Since k ∈ K was arbitrary and the left hand side of the previous expression is a fixedquantity, we have

d(x,K)− ‖x− y‖ ≤ d(y,K).

By symmetry d(y,K)− ‖x− y‖ ≤ d(x,K). Thus,

|d(x,K)− d(y,K)| ≤ ‖x− y‖

and we have the nonexpansive property.

The crucial concept behind proximinal and Chebyshev sets is the existence of so callednearest points. The following definition makes this idea precise.

Definition 2. Let (X, ‖·‖) be a normed linear space and K be a nonempty subset of X .We define the set valued mapping PK : X → P(X) by

PK(x) := {y ∈ K : ‖x− y‖ = d(x,K)},

and refer to the elements of PK(x) as the best approximations (or nearest points) to x inK. We say K is a proximinal set if PK(x) is nonempty for each x ∈ X and that K is aChebyshev set if PK(x) is a singleton for each x ∈ X . For a Chebyshev set we define themap pK : X → X as the map that sends x ∈ X to the unique element of PK(x). We willrefer to both PK and pK as the metric projection (for K).

To familiarise the reader with these concepts we now present a few examples.

Example 1. Consider K := R2 \B(0, 1) ⊆ R2 equipped with the Euclidean norm.

K

x

0

PK(x)

PK(0)

4

It is easy to check that for any x ∈ B(0, 1) \ {0},

PK(x) =

{x

‖x‖

},

whilstPK(0) = {y ∈ R2 : ‖y‖ = 1}.

Hence K is proximinal, but not a Chebyshev set.

Example 2. Consider K := B[0, 1] ⊆ R2 equipped with the 3-norm.

K

x

PK(x)

It is straightforward to check that for any x ∈ R2 \K ,

PK(x) =

{x

‖x‖

}.

Therefore, K is a Chebyshev set.

Example 3. Let n ∈ N. Consider K := B(0, 1) ⊆ Rn equipped with the Euclidean norm.Choose x ∈ Rn such that ‖x‖ = 1. Clearly, d(x,K) = 0, but since x 6∈ K, PK(x) = ∅,and so K is not proximinal.

2.2 Necessary conditions for a proximinal/Chebyshev setThe bulk of this thesis is concerned with finding necessary and sufficient conditions for aset to be a proximinal or Chebyshev set. Our first necessary condition, as suggested byExample 3, is that a proximinal set must be closed.

5

Proposition 2.2.1. Let K be a proximinal set in a normed linear space (X, ‖·‖). Then Kis closed.

Proof. Suppose, looking for a contradiction, that K is not closed. Then we can find asequence (xn)

∞n=1 in K such that limn→∞ xn = x for some x ∈ X \K. For each n ∈ N,

d(x,K) ≤ ‖x− xn‖ and so d(x,K) = 0. However, 0 < ‖x− y‖ for any y ∈ K, sincex 6∈ K. Hence, PK(x) is empty, which contradicts the proximinality of K.

Of critical importance in the study of Chebyshev sets is the notion of convexity.

Definition 3. Let X be a vector space and C a subset of X . We say that C is convex if,for any a, b ∈ C and any 0 ≤ λ ≤ 1, we have λa + (1 − λ)b ∈ C. We say C is midpointconvex if, for for any a, b ∈ C, we have a+b

2∈ C.

Looking at Example 1 we may be tempted to conclude that convexity is a necessarycondition for a set to be Chebyshev. The following example demonstrates that this is notthe case. Later we will find some sufficient conditions for a Chebyshev set to be convex.

Example 4. Define the function f : R→ R by

f(x) :=1

2d(x, 4Z)

for all x ∈ R.

40 8−4−8

1

y

pK(y)K

z

B[z, ‖z − pK(z)‖1]

pK(z)

Then, K := Graph(f), viewed as a subset of (R2, ‖·‖1), is a (nonconvex) Chebyshevset. Indeed, for any z := (a, b) ∈ R2, pK(z) = (a, f(a)).

The following result, in combination with Proposition 2.2.1, says that when consider-ing the convexity of a proximinal set we need only check if is it is midpoint convex.

6

Proposition 2.2.2. A closed subset of a normed linear space (X, ‖·‖) is convex if, andonly if, it is midpoint convex.

Proof. Clearly a convex set is midpoint convex, so suppose C ⊆ X is a closed midpointconvex set. Let x, y ∈ C. By induction, we can show that

x+t

2k(y − x) ∈ C

for any k ∈ N ∪ {0} and t ∈ {0, 1, . . . , 2k}. It is straightforward to show that⋃k∈N∪{0}

{t

2k: t ∈ {0, 1, . . . , 2k}

},

the set of dyadic rationals in [0, 1], is dense in [0, 1]. Since C is closed, it follows thatx+ λ(y − x) ∈ C for all λ ∈ [0, 1]. Hence, C is convex.

Remark: having closure in the hypotheses of Proposition 2.2.2 is essential. To see this,observe that Q ⊆ R is midpoint convex but not convex.

2.3 Properties of the metric projectionThe following result is very useful when working with nearest points and will be usedthroughout the remainder of the thesis.

Proposition 2.3.1. Let K be a nonempty subset of the normed linear space (X, ‖·‖). Sup-pose x ∈ X and z ∈ PK(x). For any y ∈ [x, z], z ∈ PK(y).

Proof. Let y ∈ [x, z] and w ∈ K. Then

‖y − z‖ = ‖x− z‖ − ‖x− y‖ , since y ∈ [x, z]

≤ ‖x− w‖ − ‖x− y‖ , since z ∈ PK(x)≤ ‖y − w‖ , by the reverse triangle inequality.

As this inequality holds for all w ∈ K, d(y,K) = ‖y − z‖, and so z ∈ PK(y).

We now proceed to establish a few more properties of the metric projection. Afterdefining what it means for a multivalued map to be continuous we prove a fundamentalresult: in finite dimensional spaces the metric projection for a proximinal set is continuouswherever it is single valued.

7

Lemma 2.3.2. LetK be a proximinal set in a normed linear space (X, ‖·‖). If (xn, yn)∞n=1

is a sequence in X × X , with yn ∈ PK(xn) for all n ∈ N and limn→∞(xn, yn) = (x, y),then y ∈ PK(x).

Proof. Let (xn, yn)∞n=1 be a sequence in X × X converging to some (x, y), with yn ∈PK(xn) for all n ∈ N. Firstly, y ∈ K, since K is closed by Proposition 2.2.1. Since thedistance function for K is non-expansive, we have that∣∣∣ ‖xn − yn‖ − d(x,K)

∣∣∣ = |d(xn, K)− d(x,K)| ≤ ‖xn − x‖

for all n ∈ N. Taking the limit of both sides of this inequality we see that∣∣∣ ‖x− y‖ − d(x,K)∣∣∣ ≤ 0,

and so‖x− y‖ = d(x,K).

Thus, y ∈ PK(x).

Corollary 2.3.3. The metric projection for a Chebyshev set has closed graph.

Proposition 2.3.4. Let K be a nonempty subset of a normed linear space (X, ‖·‖). Sup-pose x ∈ X \K and z ∈ PK(x). Define y(λ) := z + λ(x− z) for each λ ∈ R. Then

I := {λ ≥ 0 : z ∈ PK(y(λ))}

is a nonempty closed interval.

Proof. By Proposition 2.3.1, if α ∈ I and 0 ≤ β ≤ α, then β ∈ I , which establishesthat I is an interval. Clearly, 1 ∈ I , so I is nonempty. Finally, suppose that (λn)∞n=1 isa sequence in I converging to some λ ∈ R. Clearly, λ ≥ 0 and limn→∞ y(λn) = y(λ).Since z ∈ PK (y(λn)) for all n ∈ N, Lemma 2.3.2 says that z ∈ PK (y(λ)). Therefore,λ ∈ I , and so I is closed.

Definition 4. Let (X, ‖·‖) and (Y, ‖·‖′) be normed linear spaces and f : X → P(Y ) afunction. We say f is locally bounded if, for every x0 ∈ X , there exists r,M > 0 such that,for any x ∈ B(x0, r), if y ∈ f(x), then ‖y‖′ ≤M , i.e. f (B(x0, r)) is a bounded subset ofY .

Lemma 2.3.5. Let K be a proximinal set in a normed linear space (X, ‖·‖). Then themetric projection x 7→ PK(x) is locally bounded.

8

Proof. Fix x0 ∈ X and r > 0. Suppose x ∈ B(x0, r) and y ∈ PK(x). By Proposi-tion 2.1.1 and the triangle inequality, we have

‖y‖ ≤ ‖y − x‖+ ‖x− x0‖+ ‖x0‖= d(x,K) + ‖x− x0‖+ ‖x0‖≤ d(x0, K) + 2 ‖x− x0‖+ ‖x0‖< d(x0, K) + 2r + ‖x0‖ .

Since this upper bound does not depend on x or y, PK(B(x0, r)) is bounded, and so themetric projection is locally bounded.

Definition 5. Let (X, ‖·‖) and (Y, ‖·‖′) be normed linear spaces and f : X → P(Y ). Wesay f is continuous at x ∈ X if f(x) is a singleton and, for any sequences (xn)∞n=1, (yn)

∞n=1

in X and Y respectively, with limn→∞ xn = x and yn ∈ f(xn) for all n ∈ N, we havelimn→∞ yn = z, where f(x) = {z}.

Remark: It is clear that when f(x) is a singleton for each x ∈ X this defintion agreeswith our standard defintion of continuity where we view f as a function into Y .

Theorem 2.3.6. Let K be a proximinal set in a finite dimensional normed linear space(X, ‖·‖). If PK(x) is a singleton for some x ∈ X , then the metric projection y 7→ PK(y)is continuous at x.

Proof. Let x ∈ X and suppose PK(x) = {z} for some z ∈ K. Suppose, looking for acontradiction, that the metric projection is not continuous at x. Thus, there exists ε > 0and sequences (xn)∞n=1, (yn)

∞n=1 in X and K respectively such that limn→∞ xn = x, yn ∈

PK(xn) for all n ∈ N, and ‖z − yn‖ > ε for all n ∈ N. We now show that the sequence(yn)

∞n=1 is bounded. By Lemma 2.3.5, the metric projection for K is locally bounded.

Thus, there exists r,M > 0 such that for all n ∈ N, such that xn ∈ B(x, r), we have‖yn‖ ≤ M . Since limn→∞ xn = x, the set of terms in the sequence (xn)

∞n=1 that are not

in B(x, r) must be finite, say {x1, . . . , xk}. Therefore, ‖yn‖ ≤ max{‖y1‖ , . . . , ‖yk‖ ,M}for all n ∈ N. Since K is closed and X is finite dimensional, (yn)∞n=1 has a convergentsubsequence, converging to a point in K. By passing to a subsequence, we may assumethat limn→∞ yn = y for some y ∈ K. Clearly, y 6= z. However, by Proposition 2.3.2,

y ∈ PK(x) = {z},

which is impossible. Hence, the metric projection is continuous at x.

Corollary 2.3.7. The metric projection for a Chebyshev set in a finite dimensional normedlinear space is continuous.

9

2.4 Sufficient conditions for a proximinal/Chebyshev setIn the following section we find sufficient conditions for a subset of a normed linear spaceto be proximinal/Chebyshev. To begin we introduce a restriction on the norm that goessomeway towards ensuring that no points in the space have more than one nearest point.

Definition 6. A normed linear space (X, ‖·‖) is said to be strictly convex if, for any x, y ∈X , if

‖x‖ = ‖y‖ =∥∥∥∥x+ y

2

∥∥∥∥ = 1,

then x = y.

Example 5. Let K be a compact, completely Hausdorff space containing at least twopoints. Then (C(K), ‖·‖∞), where C(K) is the set of continuous functions from K to R,is not strictly convex.

Proof. Let x and y be distinct elements of K. Since K is completely Hausdorff thereexists a continuous function f : K → [0, 1] such that f(x) = 1 and f(y) = 0. Also, defineg : K → R by g(z) := 1 for all z ∈ K. Clearly, f, g ∈ SC(K) and f 6= g. However,(f + g)(x) = 2, and so

∥∥f+g2

∥∥∞ = 1.

Later in the section we will see that Lp and `p spaces are strictly convex for any1 < p <∞.

Sometimes in practice it is easier to make use of the following equivalent characterisa-tions of strict convexity.

Lemma 2.4.1. A normed linear space (X, ‖·‖) is strictly convex if, and only if, for alldistinct x, y ∈ X and r, R > 0 such that ‖x− y‖ = r +R, we have

B[x, r] ∩B[y,R] = {z},

where

z :=

(R

r +R

)x+

(r

r +R

)y.

Proof. Suppose (X, ‖·‖) is strictly convex, x, y ∈ X , r, R > 0 and

‖x− y‖ = r +R.

10

Since

‖z − x‖ =∥∥∥∥( R

r +R

)x+

(r

r +R

)y − x

∥∥∥∥ =

(r

r +R

)‖x− y‖ = r

and

‖z − y‖ =∥∥∥∥( R

r +R

)x+

(r

r +R

)y − y

∥∥∥∥ =

(R

r +R

)‖x− y‖ = R

we havez ∈ B[x, r] ∩B[y,R].

Looking for a contradiction, suppose there exist distinct u, v ∈ B[x, r] ∩ B[y,R]. Setw := u+v

2. We now show that ‖x− w‖ < r. If ‖x− u‖ < r or ‖x− v‖ < r then

‖x− w‖ =∥∥∥∥x− u+ v

2

∥∥∥∥ ≤ ∥∥∥∥x− u2

∥∥∥∥+ ∥∥∥∥x− v2

∥∥∥∥ < r,

so suppose ‖x− u‖ = ‖x− v‖ = r. Since∥∥x−u

r

∥∥ =∥∥x−v

r

∥∥ = 1, x−ur6= x−v

r, and

(X, ‖·‖) is strictly convex, we have∥∥∥∥x− wr∥∥∥∥ =

∥∥∥∥x− u+v2

r

∥∥∥∥ =

∥∥∥∥∥(x−ur

)+(x−vr

)2

∥∥∥∥∥ < 1,

implying ‖x− w‖ < r as required. An almost identical proof shows that ‖y − w‖ < R.Therefore,

‖x− y‖ ≤ ‖x− w‖+ ‖w − y‖ < r +R = ‖x− y‖ ,

which is clearly impossible. Hence, B[x, r] ∩B[y,R] = {z}.

For the converse, suppose (X, ‖·‖) is not strictly convex. Hence, there exist distinctx, y ∈ X such that

‖x‖ = ‖y‖ =∥∥∥∥x+ y

2

∥∥∥∥ = 1.

Clearly x 6= −y, whilst it is straightforward to check that{0,x− y2

}⊆ B[x, 1] ∩B[−y, 1],

which completes the proof.

11

Proposition 2.4.2. A normed linear space (X, ‖·‖) is strictly convex if, and only if, for anyx, y ∈ X \ {0}, if ‖x+ y‖ = ‖x‖+ ‖y‖ then x = αy for some α > 0.

Proof. Suppose that (X, ‖·‖) has the property given above. Let x, y ∈ X with

‖x‖ = ‖y‖ =∥∥∥∥x+ y

2

∥∥∥∥ = 1.

Then ‖x+ y‖ = 2 = ‖x‖ + ‖y‖, so x = αy for some α > 0. Since ‖x‖ = ‖y‖ = 1,α = 1, and so x = y. Thus, (X, ‖·‖) is strictly convex.

Conversely, suppose (X, ‖·‖) is strictly convex and let x, y ∈ X \ {0} be such that‖x+ y‖ = ‖x‖ + ‖y‖. Since x 6= −y and ‖x− (−y)‖ = ‖x‖ + ‖y‖, it follows byLemma 2.4.1 that

B [x, ‖x‖] ∩B [−y, ‖y‖] ={(

‖y‖‖x‖+ ‖y‖

)x−

(‖x‖

‖x‖+ ‖y‖

)y

}.

However, it is also clear that

0 ∈ B [x, ‖x‖] ∩B [−y, ‖y‖] .

Therefore, (‖y‖

‖x‖+ ‖y‖

)x−

(‖x‖

‖x‖+ ‖y‖

)y = 0,

which rearranges to give

x =‖x‖‖y‖

y.

We will need the following definition later in the thesis.

Definition 7. Let (X, ‖·‖) and (Y, ‖·‖′) be normed linear spaces and U be an open subsetof X . We say a function f : U → Y is Gateaux differentiable at x ∈ U if there exists abounded linear operator Tx : X → Y such that

limt→0

f(x+ th)− f(x)t

= Tx(h)

for all h ∈ X . The operator Tx is called the Gateaux derivative of f at x. We say f isGateaux differentiable if it is Gateaux differentiable at all points in U .

12

A related concept is smoothness. We introduce it here because of its relationship tostrict convexity.

Definition 8. We say a normed linear space (X, ‖·‖) is smooth if, for each x ∈ SX , thereexists a unique f ∈ SX∗ such that f(x) = 1.

Remark: the existence of one such functional f is guaranteed by the Hahn-Banachextension theorem. It is also clear that smoothness is equivalent to the norm being Gateauxdifferentiable on X \ {0} ([14], p.124).

Example 6. Let K be a compact, completely Hausdorff space containing at least twopoints. Then (C(K), ‖·‖∞) is not smooth.

Proof. It is straightforward to check that for any z ∈ K the functional δz : C(K) → R,defined by

δz(h) := h(z)

for all h ∈ C(K), is in SC(K)∗ . Let x and y be distinct elements ofK and define f, g ∈ SCK

as in Example 5. Sinceδx(f) = 1 6= 0 = δy(f),

we see that δx 6= δy. However,

δx(g) = 1 = δy(g),

so (C(K), ‖·‖∞) cannot be smooth.

The following result makes clear the link between strict convexity and smoothness.

Proposition 2.4.3. Let (X, ‖·‖) be a normed linear space. If (X∗, ‖·‖′) (where ‖·‖′ isthe dual norm) is strictly convex, then (X, ‖·‖) is smooth. If (X∗, ‖·‖′) is smooth, then(X, ‖·‖) is strictly convex.

Proof. Suppose that (X, ‖·‖) is not smooth. Hence, for some x ∈ SX there exist distinctf, g ∈ SX∗ such that

f(x) = g(x) = 1.

Then

1 =

(f + g

2

)(x) ≤

∥∥∥∥f + g

2

∥∥∥∥′ ≤ ‖f‖′2+‖g‖′

2= 1.

As equality must hold throughout the above expression, we have∥∥∥∥f + g

2

∥∥∥∥′ = ‖f‖′ = ‖g‖′ = 1,

13

and so (X∗, ‖·‖′) is not strictly convex.

Conversely, suppose that (X, ‖·‖) is not strictly convex. Hence, there exist distinctx, y ∈ X such that ‖x‖ = ‖y‖ =

∥∥x+y2

∥∥ = 1. By the Hahn-Banach extension theoremthere exists f ∈ SX∗ such that

f

(x+ y

2

)=

∥∥∥∥x+ y

2

∥∥∥∥′ = 1.

Therefore,

1 = f

(x+ y

2

)=f(x)

2+f(y)

2

Since f(x) ≤ 1 and f(y) ≤ 1 it must be the case that f(x) = f(y) = 1. Thus

x(f) = y(f) = ‖f‖′ ,

whilst x, y are distinct elements of SX∗∗ . Hence, (X∗, ‖·‖′) is not smooth.

The following result shows that when a space is reflexive, strict convexity and smooth-ness are dual properties.

Corollary 2.4.4. Let (X, ‖·‖) be a reflexive normed linear space. Then (X, ‖·‖) is strictlyconvex (smooth) if, and only if, (X∗, ‖·‖′) is smooth (strictly convex).

Proof. We already have one direction of each claim from Proposition 2.4.3. If (X, ‖·‖) isstrictly convex then so is X = X∗∗. Using Proposition 2.4.3 again, (X∗, ‖·‖′) is smooth.If X is smooth then so is X = X∗∗. It then follows that X∗ is strictly convex.

As mentioned before, strict convexity plays an important role in determining when asubset of a normed linear space admits no more than one best approximation to each pointin the space. The following result makes this idea precise.

Proposition 2.4.5. Let K be a nonempty, convex subset of a strictly convex normed linearspace (X, ‖·‖). Then for each x ∈ X , PK(x) contains at most one element.

Proof. Let x ∈ X. Looking for a contradiction, suppose that there exists distinct y1, y2 ∈PK(x). Since K is convex, y1+y2

2∈ K. As∥∥∥∥x− y12

∥∥∥∥ =

∥∥∥∥x− y22

∥∥∥∥ =d(x,K)

2> 0,

14

Proposition 2.4.2 tells us that

d(x,K) ≤∥∥∥∥x− (y1 + y2

2

)∥∥∥∥ =

∥∥∥∥(x− y12

)+

(x− y2

2

)∥∥∥∥<‖x− y1‖

2+‖x− y2‖

2= d(x,K),

which is impossible. Hence, y1 = y2, and so |PK(x)| ≤ 1.

We now introduce a more stringent restriction on the norm than strict convexity.

Definition 9. A normed linear space (X, ‖·‖) is said to be uniformly convex if, for anyε > 0, there exists a δ > 0 such that ‖x− y‖ < ε whenever x, y ∈ BX and∥∥∥∥x+ y

2

∥∥∥∥ > 1− δ.

The following result gives us our first example of a uniformly convex space.

Proposition 2.4.6. Any inner product space is uniformly convex.

Proof. Let (X, 〈·, ·〉) be an inner product space and ε > 0. Suppose x, y ∈ BX and‖x− y‖ ≥ ε. Since ‖x− y‖ ≤ ‖x‖ + ‖y‖ ≤ 2 we must have that ε ≤ 2. Defineδ := 1− 1

2

√4− ε2 > 0. By the parallelogram law,

‖x+ y‖2 = 2 ‖x‖2 + 2 ‖y‖2 − ‖x− y‖2

≤ 4− ‖x− y‖2

≤ 4− ε2

= 4(1− δ)2.

Hence,∥∥x+y

2

∥∥ ≤ 1− δ, and so (X, 〈·, ·〉) is uniformly convex.

The following two results will be useful later on.

Proposition 2.4.7. A normed linear space (X, ‖·‖) is uniformly convex if, and only if, forall pairs of sequences (xn)∞n=1, (yn)

∞n=1 in BX , if limn→∞

∥∥xn+yn2

∥∥ = 1, then

limn→∞

‖xn − yn‖ = 0.

15

Proof. Clearly the above statement holds if (X, ‖·‖) is uniformly convex. For the con-verse, suppose (X, ‖·‖) is not uniformly convex. Thus there exists some ε > 0 and se-quences (xn)∞n=1, (yn)

∞n=1 in BX such that ‖xn − yn‖ ≥ ε for all n ∈ N and

limn→∞

∥∥∥∥xn + yn2

∥∥∥∥ = 1.

Thus limn→∞ ‖xn − yn‖ 6= 0.

Corollary 2.4.8. Let (X, ‖·‖) be a uniformly convex space. If (xn)∞n=1 is a sequence inBX and

limn→∞m→∞

∥∥∥∥xn + xm2

∥∥∥∥ = 1,

then (xn)∞n=1 is a Cauchy sequence.

Proposition 2.4.9. Any uniformly convex space is strictly convex. Any finite dimensionalstrictly convex space is uniformly convex.

Proof. Firstly, suppose (X, ‖·‖) is a uniformly convex space. Let x, y ∈ X and suppose

‖x‖ = ‖y‖ =∥∥∥∥x+ y

2

∥∥∥∥ = 1.

Let ε > 0 and choose δ > 0, as in Definition 9 . Since∥∥x+y

2

∥∥ = 1 > 1 − δ, we have‖x− y‖ < ε. But ε > 0 was arbitrary so x = y and we have strict convexity.

Secondly, suppose (X, ‖·‖) is a finite dimensional non-uniformly convex space. Thusthere exists ε > 0 and sequences (xn)∞n=1, (yn)

∞n=1 in BX such that∥∥∥∥xn + yn

2

∥∥∥∥ > 1− 1

n

and‖xn − yn‖ > ε

for all n ∈ N. Since (xn)∞n=1, (yn)

∞n=1 are bounded sequences in a finite dimensional

space, we may assume (by passing to a subsequence if necessary) that they converge tosome x, y ∈ BX respectively. By taking the limit of the previous inequalities we see that∥∥x+y

2

∥∥ = 1 and ‖x− y‖ ≥ ε. Hence, x 6= y. As ‖x‖ = ‖y‖ = 1, (X, ‖·‖) is not strictlyconvex.

16

Remark: To see that finite dimensionality is essential consider (C[0, 1], ‖·‖∞). Sincethis space is separable, it can be equivalently renomed to be strictly convex ([14], p.107).However, since this renormed space will still be complete and non-reflexive it cannot beuniformly convex, by the following theorem.

We will later make use of the following theorem.

Theorem 2.4.10 (Milman-Pettis). Every uniformly convex Banach space is reflexive.

Proof. For a proof see, [8]

Remark: the converse to the Milman-Pettis theorem is false. Indeed there exist sepa-rable, reflexive, strictly convex Banach spaces which are not isomorphic to any uniformlyconvex space [10]. Completeness is essential to the theorem: incomplete inner productspaces are uniformly convex, but cannot be reflexive [5].

We would like to find conditions on a normed linear space (X, ‖·‖) and a subset K ⊆X that guarantee K is a Chebyshev set. First we need some preliminary results.

Lemma 2.4.11. Let C be a closed convex subset of a normed linear space (X, ‖·‖). ThenC is weakly closed.

Proof. If C is empty or the whole space it is weakly closed so let us suppose otherwise.Let x0 ∈ X \C. Since C is closed and convex, we have by the Hahn-Banach theorem, theexistence of an fx0 ∈ X∗ such that

fx0(x0) > supx∈C

fx0(x).

Thus,

x0 ∈ f−1x0

((supx∈C

fx0(x),∞)

),

which, being the inverse image of an open set, is weakly open. It is then straightforwardto check that

X \ C =⋃

x0∈X\C

f−1x0

((supx∈C

fx0(x),∞)).

Hence, X \ C, being the union of weakly open sets, is weakly open. We conclude that Cis weakly closed.

We will need the following result later.

17

Theorem 2.4.12. The closed unit ball BX in a reflexive normed linear space (X, ‖ · ‖) iscompact with respect to the weak topology on X . Thus, every bounded sequence in X hasa weak cluster point.

Proof. The mapping x 7→ x from (BX ,weak) to(BX ,weak

∗) is a homeomorphism. Since(X, ‖·‖) is reflexive, BX∗∗ = BX . By the Banach-Alaoglu theorem, BX∗∗ is compact withrespect to the weak∗ topology. Hence, BX is weakly compact.

Remark: whilst Theorem 2.4.12 is sufficient for our purposes, a slightly stronger resultexists. The Eberlein-Smulian theorem states that in a Banach space, for any K ⊆ X , thefollowing are equivalent

(i) the weak closure of K is weakly compact,

(ii) Every sequence in K has a weakly convergent subsequence,

(iii) Every sequence in K has a weak accumulation point,

irrespective of whether the weak topology is metrisable (which occurs if, and only if, X isfinite dimensional) [1].

Proposition 2.4.13. Let K be a nonempty, closed convex subset of a reflexive Banachspace (X, ‖·‖). Then K is proximinal.

Proof. Let x ∈ X . Since K is weakly closed by Lemma 2.4.11, and, for all n ∈ N,B[x, d(x,K) + 1

n

]is weakly compact by Theorem 2.4.12, it follows that

B

[x, d(x,K) +

1

n

]∩K

is weakly compact (and nonempty) for all n ∈ N. Furthermore, since

B

[x, d(x,K) +

1

n+ 1

]⊆ B

[x, d(x,K) +

1

n

]for all n ∈ N, we have that

k⋂n=1

(B

[x, d(x,K) +

1

n

]∩K

)6= ∅

for all k ∈ N. By the finite intersection property of compact sets,∞⋂n=1

(B

[x, d(x,K) +

1

n

]∩K

)6= ∅.

18

Finally, note that if y ∈⋂∞n=1

(B[x, d(x,K) + 1

n

]∩K

), then ‖x− y‖ = d(x,K) and

y ∈ K, so y ∈ PK(x).

Theorem 2.4.14. Every nonempty, closed, convex subset of a reflexive, strictly convexBanach space (X, ‖·‖) is a Chebyshev set.

Proof. By Proposition 2.4.5 and Proposition 2.4.13, any nonempty closed convex subsetof X is a Chebyshev set.

Corollary 2.4.15. Every nonempty, closed, convex subset of a uniformly convex Banachspace is a Chebyshev set.

The following example, due to Deutsch [11], demonstrates that having completenessin the hypotheses of Corollary 2.4.15 is essential.

Example 7. Consider

M := {f ∈ C[−1, 1] :∫ 1

0

f(t) dt = 0}.

Then M is a nonempty, closed subspace of the (uniformly convex) incomplete inner prod-uct space (C[−1, 1], ‖·‖2), where

‖f‖2 :=

√∫ 1

−1|f(x)|2 dx

for any f ∈ C[−1, 1], which is not proximinal.

Proof. It is clear that M is nonempty (0 ∈ M) and closed under addition and scalarmultiplication. Suppose (fn)

∞n=1 is a sequence in M converging to some f ∈ C[−1, 1].

19

Then, for any n ∈ N, by the Cauchy-Schwarz inequality,∣∣∣∣∫ 1

0

f(t) dt

∣∣∣∣ = ∣∣∣∣∫ 1

0

f(t) dt−∫ 1

0

fn(t) dt

∣∣∣∣=

∣∣∣∣∫ 1

0

f(t)− fn(t) dt∣∣∣∣

≤∫ 1

0

|f(t)− fn(t)| dt

≤∫ 1

−1|f(t)− fn(t)| dt

≤(∫ 1

−11 dt

) 12(∫ 1

−1|f(t)− fn(t)|2 dt

) 12

=√2 ‖f − fn‖2 .

Taking the limit we see that∣∣∣∫ 1

0f(t) dt

∣∣∣ = 0, and so f ∈ M , which shows that M isclosed.

Now define f ∈ C[−1, 1] by f(t) = 1 for all t ∈ [−1, 1]. Since∫ 1

0f(t) dt = 1,

f 6∈M . Then, for any g ∈M,

‖f − g‖2 =∫ 1

−1|f(t)− g(t)|2 dt

=

∫ 1

0

|1− g(t)|2 dt+∫ 0

−1|1− g(t)|2 dt

=

∫ 1

0

1− 2g(t) + g2(t) dt+

∫ 0

−1|1− g(t)|2 dt

= 1 +

∫ 1

0

g2(t) dt+

∫ 0

−1|1− g(t)|2 dt

≥ 1.

Thus, d(f,M) ≥ 1. Equality holds in the previous calculation if, and only if,∫ 0

−1|1− g(t)|2 =

∫ 1

0

g2(t) dt = 0,

which occurs if, and only if, g|(−1,0) = 1 and g|(0,1) = 0. In such a case, g is not continuous,and so g 6∈M . Hence, ‖f − g‖2 > 1 for all g ∈M .

20

For any 0 < ε < 1 define gε : [−1, 1]→ R by

gε(t) :=

1 if − 1 ≤ t ≤ −ε−tε

if − ε < t < 00 if 0 ≤ t ≤ 1.

It is straightforward to check that gε ∈M . Then

‖f − gε‖2 =∫ 1

−1|f(t)− gε(t)|2 dt

=

∫ 0

−ε

∣∣∣∣1 + t

ε

∣∣∣∣2 dt+ ∫ 1

0

1 dt

= 1 +

[t+

t2

ε+

t3

3ε2

]0−ε

= 1 +ε

3

Since this holds for all 0 < ε < 1, we have that d(f,M) ≤ 1. Thus,

d(f,M) = 1 < ‖f − g‖2

for all g ∈M . Hence, f does not have a nearest point in M , so M is not proximinal.

The following result shows how the parallelogram law generalises to Lp spaces. Wewill use it to identify some more examples of uniformly convex spaces.

Lemma 2.4.16 (Clarkson’s inequalities). [9] Let 1 < p < ∞. Then for any f, g ∈(Lp(µ), ‖·‖p

), where µ is a σ-finite measure, we have

‖f + g‖pp + ‖f − g‖pp ≤ 2p−1

(‖f‖pp + ‖g‖

pp

), if 2 ≤ p <∞,

or‖f + g‖qp + ‖f − g‖

qp ≤ 2

(‖f‖pp + ‖g‖

pp

)q−1, if 1 < p ≤ 2,

where q := pp−1 .

Proof. For a proof see, [8]

Remark: setting p = 2 in the above inequalities we recover the parallelogram law.

Proposition 2.4.17 (Clarkson). [9] Let 1 < p < ∞. Then (Lp(µ), ‖·‖p) is uniformlyconvex.

21

Proof. Let ε > 0 and f, g ∈ BLp be such that ‖f − g‖p ≥ ε. As before, we must haveε ≤ 2. If p ≥ 2 let

δ := 1− p

√1−

(ε2

)p> 0.

By the first Clarkson inequality, we have that

‖f + g‖pp ≤ 2p−1(‖f‖pp + ‖g‖

pp

)− ‖f − g‖pp

≤ 2p − ‖f − g‖pp≤ 2p − εp

= 2p(1− δ)p.

If 1 < p ≤ 2 let

δ := 1− q

√1−

(ε2

)q> 0.

By the second Clarkson inequality, we have that

‖f + g‖qp ≤ 2(‖f‖pp + ‖g‖

pp

)q−1− ‖f − g‖qp

≤ 2q − ‖f − g‖qp≤ 2q − εq

= 2q(1− δ)q.

In each case we have∥∥f+g

2

∥∥p≤ 1− δ, and so the space is uniformly convex.

Corollary 2.4.18. Any nonempty, closed, convex subset of a Hilbert space or Lp(µ) space,equipped with the p-norm, with 1 < p <∞, is a Chebyshev set.

The following result will be used later and also gives us some examples of smoothspaces.

Theorem 2.4.19. For any 1 < p <∞, the dual of Lp(µ) is Lq(µ), where q := pp−1 .

Proof. See, [8]

Corollary 2.4.20. For any 1 < p <∞, Lp(µ) , equipped with the p-norm, is smooth.

22

2.5 The convexity of Chebyshev setsAs shown by Example 4, a Chebyshev set need not be convex. In the following section wefind sufficient conditions on a Chebyshev set K, and the underlying normed linear space(X, ‖·‖), that ensures K is convex. First we require a number of preliminary results.

Lemma 2.5.1. LetK be a proximinal set in a normed linear space (X, ‖·‖). Let x ∈ X\Kand z ∈ PK(x). Let λ > 0 and define xλ := x+ λ(x− z). Suppose zλ ∈ PK(xλ). Then

d(xλ, K) ≥ d(x,K) + ‖xλ − x‖(1− ‖z − zλ‖‖x− z‖

)Proof. Setting α := λ

1+λwe see that x = αz + (1 − α)xλ. Rearranging this expression

and taking norms gives

‖x− z‖ = (1− α) ‖xλ − z‖‖x− xλ‖ = α ‖z − xλ‖ .

Later we will want to eliminate α from our calculations. Rearranging the preceding twoexpressions gives

α =‖x− xλ‖‖z − xλ‖

(2.1)

1

1− α=‖xλ − z‖‖x− z‖

. (2.2)

We introduce an auxillary point w between xλ and zλ, with the distances from w to xλ andzλ having the same ratio as the distances from x to xλ and z, i.e.

w := αzλ + (1− α)xλ.

23

xλ

x

zλ ∈ PK (xλ)

z ∈ PK (x)

w

Next we will remove w from our calculations. For this we need the following, whichare straightforward to verify,

‖w − zλ‖ = (1− α) ‖xλ − zλ‖‖x− w‖ = α ‖z − zλ‖ .

Hence,

‖x− z‖ ≤ ‖x− zλ‖ , since z ∈ PK(x)≤ ‖x− w‖+ ‖w − zλ‖ , by the triangle inequality

= α ‖z − zλ‖+ (1− α) ‖xλ − zλ‖ .

24

Rearranging this gives

d(xλ, K) = ‖xλ − zλ‖

≥ 1

1− α‖x− z‖ − α

1− α‖z − zλ‖

= ‖xλ − z‖ −‖x− xλ‖‖x− z‖

‖z − zλ‖ , by (2.1) , (2.2)

= (‖xλ − x‖+ ‖x− z‖)−‖x− xλ‖‖x− z‖

‖z − zλ‖ , since x ∈ [xλ, z]

= d(x,K) + ‖xλ − x‖(1− ‖z − zλ‖‖x− z‖

), since z ∈ PK(x)

as required.

Corollary 2.5.2. Let K be a proximinal set in a normed linear space (X, ‖·‖). If themetric projection is continuous at x ∈ X \K then

limλ→0+

‖xλ − zλ‖ − ‖x− z‖‖xλ − x‖

= limλ→0+

d(xλ, K)− d(x,K)

‖xλ − x‖= 1,

where {z} = PK(x) and xλ := x+ λ(x− z), zλ ∈ PK(xλ) for each λ > 0.

Proof. We see that for any λ > 0,

1 =‖xλ − x‖‖xλ − x‖

≥ d(xλ, K)− d(x,K)

‖xλ − x‖, since the distance function forK is nonexpansive

≥ 1− ‖z − zλ‖‖x− z‖

, by Lemma 2.5.1.

Sincelimλ→0+

xλ = x

and the metric projection is, by assumption, continuous at x, the right hand side of theabove inequality converges to 1 as λ→ 0+. Hence, by the squeeze theorem,

limλ→0+

d(xλ,M)− d(x,M)

‖xλ − x‖= 1.

25

Definition 10. Let (X, τ) be a topological space. We say a function

f : X → R ∪ {−∞,∞}

is lower semi-continuous if, for every α ∈ R,

{x ∈ X : f(x) ≤ α}

is a closed set. Similarly, we say f is upper semi-continuous if, for every α ∈ R,

{x ∈ X : f(x) ≥ α}

is a closed set.

Remark: It is clear that a function is continuous if, and only if, it is both upper andlower semi-continuous.

The following result will be used later.

Lemma 2.5.3. Let (X, ‖ · ‖) be a normed linear space. Then the norm is lower semi-continuous with respect to the weak topology.

Proof. Let α ∈ R. We need to show that the set A := {x ∈ X : ‖x‖ ≤ α} is closed withrespect to the weak topology. It is straightforward to check that A is convex and, since thenorm is continuous,

A = ‖·‖−1 ([0, α])

is closed with respect to the norm topology. By Lemma 2.4.11, A is weakly closed.

We require the following variational tool from optimisation theory.

Theorem 2.5.4 (Primitive Ekeland Theorem). Let (X, d) be a complete metric space andf : X → R ∪ {±∞} be a lower semi-continuous function that is bounded below and notidentically equal to∞. Given any ε > 0 and x1 ∈ X there exists an x0 ∈ X such that

(i) f(x0) + εd(x0, x1) ≤ f(x1),

(ii) f(x) > f(x0)− εd(x0, x), for all x ∈ X \ {x0}

Proof. See [14]

Lemma 2.5.5. Let K be a Chebyshev set in a Banach space (X, ‖·‖) with continuousmetric projection. Then for any x ∈ X \K, r > 0, and σ > 1 there exists an x0 ∈ X suchthat

26

(i) d(x,K) + (1/σ) ‖x− x0‖ ≤ d(x0, K),

(ii) d(y,K) < d(x0, K) + (1/σ) ‖y − x0‖ , for all y ∈ B[x, r] \ {x0},

(iii) ‖x− x0‖ = r.

Proof. Let x ∈ X \ K, r > 0, σ > 1 be given. Consider the function f : B[x, r] → Rdefined by

f(y) := −d(y,K)

for all y ∈ B[x, r]. Since X is complete, B[x, r] is a complete metric space. Furthermore,since the map y 7→ d(y,K) is nonexpansive, f is lower semi-continuous and boundedbelow. By the Primitive Ekeland Theorem, there exists an x0 ∈ B[x, r] such that

f(x0) +1

σ‖x− x0‖ ≤ f(x)

andf(y) > f(x0)−

1

σ‖y − x0‖ , for all y ∈ B[x, r] \ {x0}.

Thus,

d(x0, K) ≥ d(x,K) +1

σ‖x− x0‖

andd(y,K) < d(x0, K) +

1

σ‖y − x0‖ , for all y ∈ B[x, r] \ {x0},

which proves parts (i) and (ii) of the Lemma. Using (i) we see that

d(x0, K) ≥ d(x,K) > 0,

implying that x0 6∈ K. Now suppose, looking for a contradiction, that ‖x− x0‖ < r.Consider the points

x0,λ := x0 + λ (x0 − pK(x0))

where

0 < λ ≤ r − ‖x0 − x‖‖x0 − pK(x0)‖

.

It is straightforward to check that x0,λ 6= x0 and x0,λ ∈ B[x, r]. Rearranging (ii) we have

1 >1

σ>d(x0,λ, K)− d(x0, K)

‖x0,λ − x0‖,

27

which is impossible since

limλ→0+

d(x0,λ, K)− d(x0, K)

‖x0,λ − x0‖= 1

by Corollary 2.5.2. Therefore, ‖x− x0‖ = r as required.

Rather than looking directly at convexity, it is often easier to look for a weaker prop-erty, which we call almost convexity.

Definition 11. We will say a closed subset A of a normed linear space (X, ‖·‖) is almostconvex if, for any closed ball B[x, r] ⊆ X \A and N > 0, there exists x′ ∈ X and r′ > Nsuch that

B[x, r] ⊆ B[x′, r′] ⊆ X \ A.

Theorem 2.5.6. Let K be a Chebyshev set in a Banach space (X, ‖·‖). If the metricprojection for K is continuous then K is almost convex.

Proof. Suppose that B[x, β] ⊆ X \K for some x ∈ X and β > 0, and N > 0. Clearly,x ∈ X \K and β < d(x,K). Choose α > max{d(x,K), N}. Then select σ > 1, r > 0so that

σ(α− d(x,K)) < r < α− β.

Note: this selection is possible since α− d(x,K) < α− β. By Lemma 2.5.5, there existsx0 ∈ X such that

r = ‖x− x0‖ ≤ σ(d(x0, K)− d(x,K)).

Since ‖x− x0‖ = r < α − β (i.e. ‖x− x0‖ + β < α), we have B[x, β] ⊆ B[x0, α].Furthermore, since

σ(α− d(x,K)) < r ≤ σ(d(x0, K)− d(x,K)),

we have α < d(x0, K), and therefore B[x0, α] ⊆ X \K. Hence, K is almost convex.

We now establish a condition that guarantees when an almost convex set is convex.First, some more preliminary results.

Lemma 2.5.7. Let (X, ‖·‖) be a normed linear space with a strictly convex dual spaceand f ∈ SX∗ . If (B[zn, rn])

∞n=1 is a sequence of closed balls in X such that

(i) B[zn, rn] ⊆ B[zn+1, rn+1] for all n ∈ N,

(ii) B[zn, rn] ⊆ {x ∈ X : f(x) ≤ 1} for all n ∈ N,

28

(iii) limn→∞ rn =∞,

then there exists some r ≤ 1 such that⋃n∈N

B[zn, rn] = {z ∈ X : f(z) ≤ r},

i.e.⋃n∈NB[zn, rn] is a closed half space.

Proof. Letr := sup

x∈⋃

n∈NB[zn,rn]

f(x) ≤ 1.

Suppose, looking for a contradiction, that⋃n∈N

B[zn, rn] 6= {x ∈ X : f(x) ≤ r} .

Thus, there exists some z ∈ {x ∈ X : f(x) ≤ r} such that z 6∈⋃n∈NB[zn, rn]. Since⋃

n∈NBn is nonempty, closed and convex there exists some g ∈ SX∗ such that

supx∈

⋃n∈NB[zn,rn]

g(x) < g(z).

Clearly, f 6= g. Observe that

limn→∞

(sup

x∈B[zn,rn]

f(x)

)= sup

x∈⋃

n∈NB[zn,rn]

f(x)

and

limn→∞

(sup

x∈B[zn,rn]

g(x)

)= sup

x∈⋃

n∈NB[zn,rn]

g(x).

Also, since ‖f‖ = 1, we have that

supx∈B[zn,rn]

f(x) = f(zn) + rn supx∈BX

f(x)

= f(zn) + rn

and similarlysup

x∈B[zn,rn]

g(x) = g(zn) + rn,

29

for any n ∈ N. Sincelimn→∞

rn =∞,

we may assume, without loss of generality, that rn 6= 0 for all n ∈ N. Now consider thesequence (xn)

∞n=1 in X defined by

xn :=z1 − znrn

.

Since z1 ∈ B[z1, r1] ⊆ B[zn, rn] for all n ∈ N, it follows that

‖xn‖ =‖z1 − zn‖

rn≤ 1

for all n ∈ N. Thus, (xn)∞n=1 is a sequence in BX . For any n ∈ N we have

f(xn) =f(z1)− f(zn)

rn

=supx∈B[z1,r1] f(x)− supx∈B[zn,rn] f(x) + rn − r1

rn

= 1 +supx∈B[z1,r1] f(x)− supx∈B[zn,rn] f(x)− r1

rn.

Sincelimn→∞

supx∈B[zn,rn]

f(x) = supx∈

⋃n∈NB[zn,rn]

f(x) <∞

andlimn→∞

rn =∞,

we have thatlimn→∞

f(xn) = 1.

Similarly,limn→∞

g(xn) = 1.

Thus,

limn→∞

(f + g

2

)(xn) = 1,

and so ∥∥∥∥f + g

2

∥∥∥∥ = 1,

which contradicts the strict convexity of X∗.

30

Lemma 2.5.8. Let C be a closed convex subset of a normed linear space (X, ‖·‖). ThenC is contained in a half space or is the whole space.

Proof. Suppose C 6= X . Let x ∈ X \ C. By the Hanh-Banach theorem there exists ahyperplane separating x and C. Thus C is contained in one of the half spaces determinedby this hyperplane.

Lemma 2.5.9. Let J be a closed halfspace of a normed linear space (X, ‖·‖). If for somex, y ∈ X , z := x+y

2∈ int(J) then either x or y is in int(J).

Proof. Since J is a closed halfspace there exists f ∈ X∗ and α ∈ R such that

J = f−1 ((−∞, α]) .

Thus, int(J) = f−1 ((−∞, α)). Suppose x, y ∈ X \ int(J). Therefore,

f(z) = f

(x+ y

2

)=f(x)

2+f(y)

2≥ α

2+α

2= α,

and so z 6∈ int(J).

Lemma 2.5.10. Let (Cn)∞n=1 be an expanding sequence of convex sets in a normed linear

space (X, ‖·‖). Suppose int (Ck) is nonempty for some k ∈ N. Then

int

(∞⋃n=1

Cn

)=∞⋃n=1

int (Cn) .

Proof. Clearly,∞⋃n=1

int (Cn) ⊆ int

(∞⋃n=1

Cn

),

so it suffices to show set inclusion in the reverse direction. Let x ∈ int(⋃∞

n=1Cn

). Then

there exists r > 0 such that B(x, r) ⊆ int(⋃∞

n=1Cn

). Moreover, there exists a k ∈ N,

s > 0, and y ∈ Ck such that B(y, s) ⊆ Ck. Now choose λ > 0 sufficiently small so that

B (x+ λ(x− y), λs) ⊆ B(x, r) ⊆

(∞⋃n=1

Cn

).

31

Hence, there exists z ∈ B (x+ λ(x− y), λs) and m ≥ k such that z ∈ Cm. It is straight-forward to verify that

x ∈ B(z + λy

1 + λ,λs

1 + λ

)=

(1

1 + λ

)z +

(λ

1 + λ

)B(y, s).

Since z ∈ Cm, B(y, s) ⊆ Ck ⊆ Cm and Cm is convex, we conclude that x ∈ int (Cm) ⊆⋃∞n=1 int (Cn) .

Theorem 2.5.11. Let (X, ‖·‖) be a normed linear space with strictly convex dual space.Then any closed almost convex subset of X is convex.

Proof. Looking for a contradiction, suppose C ⊆ X is a closed almost convex set thatis not convex. By Proposition 2.2.2, C is not midpoint convex. Therefore there existx, y ∈ C such that c := x+y

26∈ C. As C is closed d(c,M) > 0. Since

B

[c,d(c,M)

2

]⊆ X \ C

and C is almost convex, there exist sequences (zn)∞n=1 and (rn)

∞n=1, in X and [0,∞) re-

spectively, such that

B

[c,d(c,M)

2

]⊆ B[zn, rn] ⊆ B[zn+1, rn+1]

andB[zn, rn] ⊆ X \ C

for all n ∈ N, and limn→∞ rn =∞. If⋃n∈NB[zn, rn] = X , Lemma 2.5.10 tells us that

⋃n∈N

B(zn, rn) = int

(⋃n∈N

B[zn, rn]

)= int(X) = X.

Thus, x ∈ B (zk, rk) for some k ∈ N, which is impossible since x ∈ C. Therefore,⋃n∈NB[zn, rn] is not the whole space, and so by Lemma 2.5.8, is contained in a half

space. Hence, by Lemma 2.5.7,⋃n∈NB[zn, rn] is a closed half space. Since

c ∈ int

(B

[c,d(c,M)

2

])⊆ int

(∪n∈NB[zn, rn]

),

it follows by Lemma 2.5.9 and Lemma 2.5.10 that either x or y is in

int

(⋃n∈N

B[zn, rn]

)=⋃n∈N

B(zn, rn),

32

which is impossible since x, y ∈ C. Hence, C is convex.

Theorem 2.5.12 (Vlasov). Let K be a Chebyshev set with continuous metric projection ina Banach space (X, ‖·‖) with stricly convex dual space. Then K is convex.

Proof. Since K has continuous metric projection it is almost convex by Theorem 2.5.6.Since any Chebyshev set is necessarily closed and X∗ is strictly convex, K is convex byTheorem 2.5.11.

The strict convexity of the dual is essential here. Consider Example 4. By Corol-lary 2.3.7, the metric projection for K is continuous, whilst it is straightforward to checkthat K is almost convex but clearly not convex.

Corollary 2.5.13. In a Hilbert space every Chebyshev set with continuous metric projec-tion is convex.

Proof. By the previous theorem all we need check is that a Hilbert space has strictly con-vex dual. This follows since the dual of a Hilbert space is again a Hilbert space and so isstrictly convex by Propositions 2.4.9 and 2.4.6.

Corollary 2.5.14. For any 1 < p < ∞, any Chebyshev set in Lp(µ) with continuousmetric projection is convex.

Proof. Again we need only check that Lp has a strictly convex dual for 1 < p < ∞ . ByTheorem 2.4.19, the dual of Lp is Lq (where q := p

p−1 ), which is uniformly convex (andhence stictly convex) by Proposition 2.4.17.

Corollary 2.5.15. Let (X, ‖·‖) be a smooth, finite-dimensional normed linear space. IfK ⊆ X is a Chebyshev set, then K is convex.

Proof. By Theorem 2.3.6, the metric projection for K is continuous. Furthermore, sinceX is finite-dimensional it is complete and reflexive. As X is smooth, X∗ is strictly convexby Corollary 2.4.4. Vlasov’s theorem then ensures the convexity of K.

2.6 Continuity of the metric projectionSection 2.5 demonstrated that continuity of the metric projection plays a key role in estab-lishing the convexity of a Chebyshev set, so it is natural to ask whether every Chebyshevset has a continuous metric projection. Unfortunately, finding examples of Chebyshev setswith discontinuous metric projections is not particuarly straightforward. To begin this sec-tion we present an example of such a set in a reflexive (and hence complete) normed linearspace.

33

Example 8. There exists a reflexive space containing a convex Chebyshev set with discon-tinuous metric projection.

Proof. Consider the Hilbert space `2(Z). Define the linear map T : `2(Z) → `2(Z)componentwise by

T (x)n :=

{λn if n ≤ 0λ−n +

1nλn if 0 < n

where x := (. . . , λ−1, λ0, λ1, . . .) ∈ `2(Z) and n ∈ Z.It is straightforward to check that T is injective. Furthermore, T is continuous. To see

this suppose x ∈ B`2(Z), i.e. ‖x‖2 =∑

n∈Z λ2n ≤ 1. Making use of the Cauchy-Schwarz

inequality we have

‖T (x)‖22 =∑n≤0

λ2n +∑n>0

(λ−n +

λnn

)2

=∑n≤0

λ2n +∑n>0

λ2−n +∑n>0

(λnn

)2

+ 2∑n>0

λ−nλnn

≤∑n≤0

λ2n +∑n>0

λ2−n +∑n>0

(λnn

)2

+ 2

√√√√(∑n>0

λ2−n

)(∑n>0

(λnn

)2)

≤ 5

and so ‖T‖2 ≤√5.

Define a new norm ‖·‖′ on `2(Z) by

‖x‖′ = max{‖x‖2 , 2|λ0|}+ ‖T (x)‖2 .

It is straightforward to check that this is an equivalent norm, since for any x ∈ `2(Z)

‖x‖2 ≤ ‖x‖′ ≤ (2 + ‖T‖2) ‖x‖2 .

Therefore,(`2(Z), ‖·‖′

)is reflexive and complete. Since ‖·‖2 is strictly convex (it is in-

duced by an inner product) and T is injective it follows that ‖·‖′ is strictly convex.Define K := span{en : 1 ≤ n}, where en is the element of `2(Z) with 1 as its nth

coordinate and zeros everywhere else. By Propositions 2.4.5 and 2.4.13, K, viewed as asubset of

(`2(Z), ‖·‖′

), is a Chebyshev set. We now show that K has discontinuous metric

projection.

34

Let y :=∑k

n=1 αnen ∈ span{en : 1 ≤ n} for some 1 ≤ k and α1, . . . , αk ∈ R. Then

‖e0 − y‖′ =

∥∥∥∥∥e0 −k∑

n=1

αnen

∥∥∥∥∥′

= max

{∥∥∥∥∥e0 −k∑

n=1

αnen

∥∥∥∥∥2

, 2

}+

∥∥∥∥∥T(e0 −

k∑n=1

αnen

)∥∥∥∥∥2

= max

2,

√√√√1 +k∑

n=1

α2n

+

√√√√1 +k∑

n=1

(αnn

)2.

This quantity can be minimised by setting αn = 0 for all n ∈ {1, . . . , k}. Thus

d (e0, span{en : 1 ≤ n}) = ‖e0 − 0‖′ = 3.

Since d (e0, span{en : 1 ≤ n}) = d (e0, span{en : 1 ≤ n}) and K is a Chebyshev set, itfollows that pK (e0) = 0.

Fix j > 0. We may assume, without loss of generality, that j ≤ k. Then∥∥∥∥(e0+ e−jj)− y∥∥∥∥′=

∥∥∥∥∥e0+ e−jj −k∑

n=1

αnen

∥∥∥∥∥′

=max

{∥∥∥∥∥e0+ e−jj −k∑

n=1

αnen

∥∥∥∥∥2

, 2

}+

∥∥∥∥∥T(e0 +

e−jj−

k∑n=1

αnen

)∥∥∥∥∥2

= max

2,√√√√1+

1

j2+

k∑n=1

α2n

+

√√√√√1+1

j2+

(1− αjj

)2

+k∑

n=1n6=j

(αnn

)2.

This quantity can be minimised by setting αj = 1 and αn = 0 for all n 6= j. Thus,

d

(e0 +

e−jj, span{en : 1 ≤ n}

)=

∥∥∥∥e0 + e−jj− ej

∥∥∥∥′ = 2 +

√1 +

1

j2.

Since d(e0 +

e−j

j, span{en : 1 ≤ n}

)= d

(e0 +

e−j

j, span{en : 1 ≤ n}

)andK is a Cheby-

shev set, it follows that pK(e0 +

e−j

j

)= ej .

Finally , observe that

limj→∞

(e0 +

e−jj

)= e0,

35

whilst

limj→∞

(pK

(e0 +

e−jj

))= lim

j→∞ej 6= 0 = pK(e0),

showing that the metric projection onto K is not continuous.

We now find conditions on the Chebyshev set as well as the underlying space thatguarantee the metric projection is continuous. To do this we begin by introducing thefollowing restriction on the norm.

Definition 12. Let (X, ‖·‖) be a normed linear space. We say ‖·‖ is a Kadec norm if forevery sequence (xn)

∞n=1 in X , if limn→∞ xn = x with respect to the weak topology and

limn→∞ ‖xn‖ = ‖x‖ then limn→∞ xn = x with respect to the norm topology.

Theorem 2.6.1. Let K be a weakly closed Chebyshev set in a reflexive normed linearspace (X, ‖·‖) with Kadec norm. Then K has continuous metric projection.

Proof. Let x ∈ X and (xn)∞n=1 be a sequence in X converging to x in norm. Observe that

d(x,K) = ‖x− pK(x)‖ ≤ ‖x− pK(xn)‖≤ ‖x− xn‖+ ‖xn − pK(xn)‖= ‖x− xn‖+ d(xn, K)

for all n ∈ N. By Lemma 2.1.1, this last quantity converges to d(x,K) and so, by thesqueeze theorem,

limn→∞

‖x− pK(xn)‖ = d(x,K) = ‖x− pK(x)‖ .

Since the sequence (‖x− pK(xn)‖)∞n=1 is convergent, we have that (pK(xn))∞n=1 is bounded.As (X, ‖·‖) is reflexive, Theorem 2.4.12 gives the existence of a subsequence (pK(xnk

))∞k=1,converging weakly to some v ∈ X . As K is weakly closed, v ∈ K. We now show that‖x− v‖ ≤ d(x,K). Let ε > 0. Then there exists some M ∈ N such that for all k ≥ M ,we have ‖x− pK(xk)‖ ≤ d(x,K) + ε. In other words,

{x− pK(xk) : k ≥M} ⊆ {y ∈ X : ‖y‖ ≤ d(x,K) + ε} .

This latter set is weakly closed, since the norm is weakly lower s-emicontinuous, so mustcontain its weak limit points. Hence, ‖x− v‖ ≤ d(x,K) + ε. Since ε > 0 was arbitrary,we conclude that ‖x− v‖ ≤ d(x,K). Therefore, v = pK(x). Furthermore, since the normis Kadec and

limk→∞‖x− pK(xnk

)‖ = ‖x− pK(x)‖ ,

36

whilst x−pK(xnk) converges weakly to x−pK(x), we have that x−pK(xnk

) converges innorm to x− pK(x). Hence (pK(xnk

))∞k=1 converges to pK(x) in norm and so (pK(xn))∞n=1

converges to pK(x) in norm. Thus, the metric projection is continuous.

Proposition 2.6.2. Let (X, ‖·‖) be a uniformly convex Banach space. Then ‖·‖ is a Kadecnorm.

Proof. Suppose (xn)∞n=1 is a sequence in X converging weakly to some x ∈ X such that

limn→∞ ‖xn‖ = ‖x‖. We need to show that (xn)∞n=1 converges in norm to x. If x = 0then limn→∞ ‖xn‖ = ‖x‖ = 0 and so (xn)

∞n=1 converges in norm to x = 0. So suppose

x 6= 0. Then, since limn→∞ ‖xn‖ = ‖x‖ 6= 0, we may assume that xn 6= 0 for all n ∈ N.Define y := x

‖x‖ and yn := xn‖xn‖ for each n ∈ N. Since (xn)

∞n=1 converges weakly to

x, the sequence (yn)∞n=1 will converge weakly to y. It is also clear that to show (xn)

∞n=1

converges to x in norm it is sufficient to show (yn)∞n=1 converges to y in norm. Choose

f ∈ SX∗ such that f(y) = ‖y‖ = 1. Then

f

(yn + ym

2

)≤∥∥∥∥yn + ym

2

∥∥∥∥ ≤ 1

for all n,m ∈ N. Since (yn)∞n=1 converges weakly to y we have that

limn→∞m→∞

f

(yn + ym

2

)= lim

n→∞

f(yn)

2+ lim

m→∞

f(ym)

2= f(y) = 1,

and so

limn→∞m→∞

∥∥∥∥yn + ym2

∥∥∥∥ = 1.

By Corollary 2.4.8, (yn)∞n=1 is Cauchy and so must converge in norm to some point in X .Since norm convergence implies weak convergence, weak limits are unique, and (yn)

∞n=1

converges weakly to y we conclude that (yn)∞n=1 converges in norm to y. Hence, ‖·‖ is aKadec norm.

Theorem 2.6.3. Let K be a weakly closed Chebyshev set in a smooth uniformly convexBanach space (X, ‖·‖). Then K is convex.

Proof. By Proposition 2.6.2, (X, ‖·‖) has Kadec norm and is reflexive by the Milman-Pettis theorem. Hence, by Corollary 2.4.4, the space has strictly convex dual. Since K isweakly closed it has continuous metric projection by Theorem 2.6.1. By Vlasov’s theoremK is convex.

37

Corollary 2.6.4. Any weakly closed Chebyshev set in a Hilbert space or Lp space (1 <p <∞) is convex.

A classical result, due to Phelps [24], shows that the convexity of a Chebyshev set inan inner product space is equivalent to the the metric projection being nonexpansive. Toprove this fact we first need a result that characterises best approximations for convex setsin inner product spaces.

Proposition 2.6.5. Let K be a convex subset of an inner product space (X, 〈·, ·〉) andx ∈ X, y ∈ K. Then y ∈ PK(x) if, and only if,

〈x− y, z − y〉 ≤ 0,

for all z ∈ K.

Proof. Let x ∈ X, y ∈ K and suppose the above inequality holds. If x = y then there isnothing to prove so assume otherwise. For any z ∈ K

‖x− y‖2 = 〈x− y, x− y〉= 〈x− y, x− z〉+ 〈x− y, z − y〉≤ 〈x− y, x− z〉≤ ‖x− y‖ ‖x− z‖ , by the Cauchy − Schwarz inequality.

Thus, ‖x− y‖ ≤ ‖x− z‖ for all z ∈ K and so y ∈ PK(x).

Conversely, suppose that for some z ∈ K, 〈x− y, z − y〉 > 0. Clearly z 6= y. Choose

0 < λ < min

{1,

2〈x− y, z − y〉‖z − y‖2

},

which guarantees2〈x− y, z − y〉 − λ ‖z − y‖2 > 0.

Since K is convex yλ := λz + (1− λ)y ∈ K. Then

‖x− yλ‖2 = 〈x− yλ, x− yλ〉= 〈x− y − λ(z − y), x− y − λ(z − y)〉= ‖x− y‖2 − λ

(2〈x− y, z − y〉 − λ ‖z − y‖2

)< ‖x− y‖2 .

Thus, ‖x− yλ‖ < ‖x− y‖ , and so y 6∈ PK(x).

38

Theorem 2.6.6 (Phelps). Let K be a Chebyshev set in an inner product space (X, 〈·, ·〉).Then K is convex if, and only if, the metric projection is nonexpansive, that is

‖pK(x)− pK(y)‖ ≤ ‖x− y‖ ,

for all x, y ∈ X.

Proof. Suppose K is convex and x, y ∈ X . Clearly the required inequality holds ifpK(x) = pK(y) so assume otherwise. By Proposition 2.6.5

〈x− pK(x), pK(y)− pK(x)〉 ≤ 0

and〈y − pK(y), pK(x)− pK(y)〉 ≤ 0.

Combining these two gives

〈x− y + pK(y)− pK(x), pK(y)− pK(x)〉 ≤ 0

and so〈pK(y)− pK(x), pK(y)− pK(x)〉 ≤ 〈y − x, pK(y)− pK(x)〉.

Therefore,

‖pK(y)− pK(x)‖2 = 〈pK(y)− pK(x), pK(y)− pK(x)〉≤ 〈y − x, pK(y)− pK(x)〉≤ ‖y − x‖ ‖pK(y)− pK(x)‖ .

Thus, ‖pK(y)− pK(x)‖ ≤ ‖x− y‖ as required.

For the converse suppose the metric projection is nonexpansive but K is not convex.Since K is closed it is not midpoint convex so we can find x, y ∈ K such that z := x+y

26∈

K. Clealry x 6= y. Set r := ‖x−y‖2

> 0. We claim

pK(z) ∈ B[x, r] ∩B[y, r].

This follows since

‖pK(z)− x‖ = ‖pK(z)− pK(x)‖ ≤ ‖z − x‖ = r

and‖pK(z)− y‖ = ‖pK(z)− pK(y)‖ ≤ ‖z − y‖ = r.

39

Since (X, 〈·, ·〉) is strictly convex (as it is an inner product space) and ‖x− y‖ = 2r,Lemma 2.4.1 tells us that

B[x, r] ∩B[y, r] =

{x+ y

2

}= {z}.

This forces pK(z) = z ∈ K, which is impossible. Hence K is convex.

40

Chapter 3

A non-convex Chebyshev set in an innerproduct space

3.1 History and motivationIn this chapter we will construct a non-convex Chebyshev set in the inner product spaceconsisting of all real sequences with only finitely many non-zero terms equipped with theEuclidean inner product. This construction was first proposed by Johnson [17]. Whilstthe motivation behind this construction is certainly geometric, Johnson’s proof is predom-inantly algebraic and analytic, and contained two errors that were later corrected by Jiang[16]. The proof presented in this thesis is more geometric in nature, based on a proof ofBalaganskii and Vlasov [4]. We hope that the more detailed proof presented here assiststhe reader in understanding the numerous technical details of the construction as well asappreciating the general results that contribute to the construction.

Throughout the remainder of the chapter we will frequently be working with Euclideanspace, which we denote by E (or En if we wish to make the dimension explicit). That isRn, for some n ∈ N ∪ {0}, equipped with the standard Euclidean inner product. We willalso make use of the following notation: for any x ∈ E we will write x′ for (x, 0) ∈ E×R.

The idea behind the construction is as follows. We start with M1 := E1 \ (−2, 1),which is clearly not a Chebyshev set in E1 since PM1

(−12

)= {−2, 1}. On the other hand,

PM1(0) = {0} is a singleton. That is every point in E0×{0} has a unique nearest point inM1. We now construct a non-convex setM2 ⊆ E2 such thatM1×{0} =M2∩(E1 × {0}),every point in E1 × {0} has a unique nearest point in M2, and PM1(0) × {0} = PM2(0).Since M2 is non-convex, by Corollary 2.5.15, it is not a Chebyshev set in E2.

This gives us an idea of how to proceed. Construct a non-convex set M3 ⊆ E3 so thatM2 × {0} =M3 ∩ (E2 × {0}), every point in E2 × {0} has a unique nearest point in M3,

41

and PM2(x) × {0} = PM3(x′) for all x ∈ E1 × {0}. Repeating this process indefinitely

(formally we will use induction) produces a sequence of non-convex sets (Mn)∞n=1 such

that Mn ⊆ En, Mn × {0} = Mn+1 ∩ (En × {0}), each point in En × {0} has a uniquenearest point in Mn+1, and PMn(x) × {0} = PMn+1(x

′) for all x ∈ En−1 × {0}, for alln ∈ N.

But how do we actually construct Mn+1 given Mn, for some n ∈ N, such that it hasthe aforementioned properties? Whilst much of the arguments are quite technical, the keyidea is given in Theorem 3.2.20, which states that if K is a nonempty subset of a strictlyconvex normed linear space (X, ‖·‖) and x ∈ X is a point of Gateaux differentiabilityof the distance function for K, then PK(x) has at most one element. Of course this begsthe question, how do we prove that a given set has a differentiable distance function? Toget around this problem, rather than starting with a set and trying to show that its distancefunction is differentiable, we will start with a differentiable function, and using a clevertrick, construct a set that has this function as its distance function.

3.2 The construction

3.2.1 PreliminariesWe now define various objects that will be used throughout the rest of the chapter.

Definition 13. Let M ⊆ E and b be some positive real number. We say M has the b-boundary property if, for any y ∈ ∂M , there exists x ∈ E such that

‖x− y‖ = d(x,M) = b.

Definition 14. We define the following subset of a Euclidean space E,

E(0) := {x ∈ E : xn = 0},

where n := dim(E).

Definition 15. Let M be a nonempty, closed subset of E. Suppose that M has the b-boundary property for some b > 0, E \M is convex and bounded, 0 6∈M , and every pointin E(0) has a unique nearest point in M . We define the following objects:

C := {(x, r) ∈ E × R : x ∈ E \M, |r| ≤ d(x,M)},

A := {(x, r) ∈ C : r ≥ 0},

42

R : C → R, by R(w) :=d(x,M)− r√

2, for all w := (x, r) ∈ C,

ξ :=d√8

b, where d := diam(E \M) = sup{‖x− y‖ : x, y ∈ E \M},

Q := {w ∈ C : d(w,E(0) × R) ≤ ξR(w)},

D := (E × [0,∞))⋂(⋃

w∈Q

B(w,R(w))

).

Remark: It is clear that b ≤ d2.

Before establishing some further properties of these objects, we briefly describe themotivation behind their definitions. The set C is simply the volume contained between thegraph of the distance function for M , restricted to E \M , and its reflection in E × 0. Thefunction R measures the distance from a point w ∈ C to the graph of the distance functionfor M , restricted to E \M . Rewriting the definition of Q gives

Q =

{(x, r) ∈ C : r ≤ d(x,M)−

√2

ξd(x,E(0)

)},

which shows that Q is just C with points removed in a linear manner as we move awayfrom E0. The inclusion of the d(·, E(0)) term in the definition of Q will later be usedto show that our metric projections coincide on E(0). Now that we have a proper subsetof C, we are able to construct the ‘smooth’ set D inside A (it looks like A without asharp point at the top) by taking the union of balls of an appropraite radius (given by thefunction R). This ‘smooth’ set is then used to construct a Gateaux differentiable function,ρ : E \M → R, defined by

ρ(x) := max{r : (x, r) ∈ D}

(we will prove later that this is well defined). From this we construct a set M ′′ in E × Rthat has ρ as its distance function (for at least for some of its points). It is essential thatour smoothed set D lies inside A, since we require M × {0} = M ′′ ∩ (E × {0}). Thedifferentiability of this distance function then enables us to prove the uniqueness of nearestpoints.

The first results shows that M is proximinal.

Proposition 3.2.1. Let K be a nonempty closed subset of a finite dimensional normedlinear space (X, ‖·‖). Then K is proximinal.

43

Proof. Let x ∈ X and (yn)∞n=1 be a sequence inK such that limn→∞ ‖x− yn‖ = d(x,K).

Since K is closed, K ∩B[x, d(x,M) + 1] is compact and furthermore contains an infinitenumber of terms of (yn)

∞n=1. Hence, there exists a subsequence (ynk

)∞k=1 converging tosome y ∈ K. Thus,

‖x− y‖ =∥∥∥x− lim

k→∞ynk

∥∥∥ = limk→∞‖x− ynk

‖ = d(x,K),

and so y ∈ PK(x).

Definition 16. Let K be a convex subset of a vector space X . We say a function f : K →R is convex if,

f (λx+ (1− λ)y) ≤ λf(x) + (1− λ)f(y)

for all x, y ∈ K and 0 ≤ λ ≤ 1, and concave if,

f (λx+ (1− λ)y) ≥ λf(x) + (1− λ)f(y)

for all x, y ∈ K and 0 ≤ λ ≤ 1.

The following general result looks at the distance function for a set with convex com-plement.

Lemma 3.2.2. Let K be a nonempty proper subset of a normed linear space (X, ‖·‖) suchthat X \K is convex. Then the distance function for K, restricted to X \K, is concave.

Proof. Let x, y ∈ X \K and 0 ≤ λ ≤ 1. Clearly,

B[x, d(x,K)], B[y, d(y,K)] ⊆ X \K.

Since X \K is convex, it follows that

B[λx+ (1− λ)y, λd(x,K) + (1− λ)d(y,K)] = λB[x, d(x,K)] + (1− λ)B[y, d(y,K)]

⊆ X \K.

Therefore,λd(x,K) + (1− λ)d(y,K) ≤ d(λx+ (1− λ)y,K),

which establishes the result.

Proposition 3.2.3. C and A are closed, convex, and bounded (and hence compact).

44

Proof. Since the distance function for M is continuous, it is straightforward to check thatC and A are closed. To see the convexity of C suppose (x, r), (y, s) ∈ C and 0 ≤ λ ≤ 1.Since E \M is convex by assumption, we have λx+ (1− λ)y ∈ E \M . Furthermore,

|λr + (1− λ)s| ≤ λ |r|+ (1− λ) |s|≤ λd(x,M) + (1− λ)d(y,M)

≤ d(λx+ (1− λ)y,M),

since the distance function for M restricted to E \M is concave by Lemma 3.2.2. Thusλ(x, r) + (1− λ)(y, s) ∈ C, and so C is convex. An almost identical proof shows that Ais convex. Finally, to see that C (and hence A ⊆ C) is bounded, observe that since E \Mis bounded and the distance function for M is continuous we have

r := max{d(x,M) : x ∈ E \M} <∞

and soE \M ⊆ B(y, s)

for some y ∈ E, s > 0. Therefore,

C ⊆ B(y′,√r2 + s2),

and so C is bounded.

Proposition 3.2.4. The map R : C → R is continuous and concave.

Proof. Suppose w := (x, r), v := (y, s) ∈ C and 0 ≤ λ ≤ 1. Since x, y ∈ E \M ,Lemma 3.2.2 gives

λR(w) + (1− λ)R(v) = λ

(d(x,M)− r√

2

)+ (1− λ)

(d(y,M)− s√

2

)=λd(x,M) + (1− λ)d(y,M)− (λr + (1− λ)s)√

2

≤ d(λx+ (1− λ)y,M)− (λr + (1− λ)s)√2

= R(λw + (1− λ)v).

Thus, R is concave. Continuity of R follows by the continuity of the distance function forM .

Proposition 3.2.5. Q is closed and bounded (and hence compact).

45

Proof. Let (wn)∞n=1 be a sequence in Q converging to some w ∈ E×R. Since C is closedand Q ⊆ C, w ∈ C. By the continuity of both R and the distance function for E(0) × Rwe have

d(w,E(0) × R) = limn→∞

d(wn, E(0) × R) ≤ lim

n→∞ξR(wn) = ξR(w).

Thus, w ∈ Q, and so Q is closed. Moreover, since C is bounded, so is Q.

Proposition 3.2.6. The closure of D is given by

D = (E × [0,∞))⋂(⋃

w∈Q

B[w,R(w)]

).

Proof. It is clear that set inclusion holds in one direction so let x ∈ D and (xn)∞n=1 be a

sequence in D converging to x. Clearly x ∈ E × [0,∞). By the definition of D, thereexists a sequence (wn)

∞n=1 in Q such that

‖xn − wn‖ < R(wn)

for all n ∈ N. Since Q is compact, we may assume, without loss of generality, thatlimn→∞wn = w, for some w ∈ Q. As R is continuous, it follows that

‖x− w‖ =∥∥∥ limn→∞

xn − limn→∞

wn

∥∥∥ = limn→∞

‖xn − wn‖ ≤ limn→∞

R(wn) = R(w).

Hence x ∈ B[w,R(w)] and we’re done.

Lemma 3.2.7. For any x ∈ E \M we have

d(x′, E(0) × R) ≤ d.

Proof. Let x ∈ E \M . Since 0 ∈ E(0) × R and 0 ∈ E \M , we have

d(x′, E(0) × R) ≤ ‖x′ − 0‖ = ‖x− 0‖ ≤ diam(E \M) =: d

Lemma 3.2.8. Let (X, 〈·, ·〉) be an inner product space and K ⊆ X be such that X \Kis convex. Let x ∈ K and suppose x ∈ PK(y) and x ∈ PK(z) for some y, z ∈ X . Thenx, y, z are collinear.

46

Proof. If x = y or x = z the result holds trivially, so suppose otherwise. We now showthat X \ K has nonempty interior. Clearly, y 6∈ K. Moreover, if y ∈ ∂ (X \K) = ∂Kthen d(y,K) = 0, which is impossible. Hence,

y ∈ (X \K) \ ∂ (X \K) = int (X \K) .

Let H1 be the hyperplane in X passing through x that is normal to [x, y] and H2 thehyperplane through x normal to [x, z]. Clearly H1 supports B(y, ‖x− y‖) at x and H2

supports B(z, ‖x− z‖) at x. Now let H3 support X \ K at x (this is possible since x ∈∂K = ∂ (X \K) and X \K is convex with nonempty interior). Since

B(y, ‖x− y‖), B(z, ‖x− z‖) ⊆ X \K,

H3 also supports the two balls at x. As (X, 〈·, ·〉) is smooth, we must haveH1 = H3 = H2.Therefore, [x, y] and [x, z] are linearly dependent and so x, y, z are collinear.

Proposition 3.2.9. (E \M)× {0} ⊆ D.

Proof. Let x′ ∈ (E \M) × {0}. Clearly x′ ∈ E × [0,∞) and x′ ∈ C. Firstly, supposethat d(x,M) ≥ b

2, so that R(x′) = d(x,M)√

2≥ b√

8> 0. Using Lemma 3.2.7, we have

d(x′, E(0) × R) ≤ d =d√8

b· b√

8= ξ

b√8≤ ξR(x′).

Thus, x′ ∈ Q. Since x′ ∈ B(x′, R(x′)), it follows that x′ ∈ D. Alternatively, d(x,M) < b2.

Let y ∈ PM(x). Since M has the b-boundary property, there exists z ∈ E such thatd(z,M) = ‖z − y‖ = b. Thus, z ∈ PM(y), and so by Lemma 3.2.8, x, y, z are collinear.Since

0 < ‖x− y‖ < b

2< b = ‖y − z‖ ,

we have x ∈ (y, z), and so‖x− z‖ < ‖y − z‖ = b.

Let w := (z,−b). We will now show that w ∈ Q. Firstly, it is clear that w ∈ C. Then

R(w) =d(z,M)− (−b)√

2= b√2,

and so

d(w,E(0) × R) ≤ d ≤ d√8

b· b√2 = ξR(w).

47

as required. Finally, we have

‖x′ − w‖2 = ‖(x, 0)− (z,−b)‖2 = ‖x− z‖2 + b2 < 2b2 = R(w)2.

Therefore, x′ ∈ B(w,R(w)) and so x′ ∈ D.

Proposition 3.2.10. D ⊆ A.

Proof. Let

z := (y, t) ∈ D = (E × [0,∞))⋂(⋃

w∈Q

B[w,R(w)]

).

Therefore, t ≥ 0 and there exists some w := (x, r) ∈ Q ⊆ C such that ‖z − w‖ ≤ R(w).Suppose, looking for a contradiction, that d(x,M) < t. As w ∈ C, it follows that

r ≤ |r| ≤ d(x,M) < t.

Then, since‖z − w‖2 = ‖y − x‖2 + (t− r)2 ≤ R(w)2,

we have

t ≤ r +R(w) = r +d(x,M)− r√

2≤ r + d(x,M)− r = d(x,M),

which is impossible. Thus, d(x,M) ≥ t. Observe that

(d(x,M)− t)2 −

((d(x,M)− r√

2

)2

− (t− r)2)

= (d(x,M)− r − (t− r))2 −

((d(x,M)− r√

2

)2

− (t− r)2)

=

(d(x,M)− r√

2

)2

+ 2(t− r)2 − 2 (d(x,M)− r) (t− r)

=

(d(x,M)− r√

2−√2(t− r)

)2

≥ 0.

Rearranging and taking the square root of both sides gives

d(x,M)− t ≥

√(d(x,M)− r√

2

)2

− (t− r)2.

48

Using this and the non-expansivity of the distance function for M gives

d(y,M) ≥ d(x,M)− ‖y − x‖≥ d(x,M)−

√R(w)2 − (r − t)2

= d(x,M)−

√(d(x,M)− r√

2

)2

− (r − t)2

≥ t ≥ 0.

To complete the proof we need to show that y ∈ E \M . By the above working, if t > 0then d(y,M) > 0, so we may as well suppose t = 0. Since

d(x,M)2 −R(w)2 + r2 =d(x,M)2

2+ d(x,M)r +

r2

2=

(d(x,M) + r√

2

)2

≥ 0,

we have that‖x− y‖2 ≤ R(w)2 − r2 ≤ d(x,M)2.

As w ∈ C, x ∈ E \M, and so we conclude that y ∈ E \M . Hence z := (y, t) ∈ A andso D ⊆ A.

The following result will be used to show that Q and D are convex.

Lemma 3.2.11. Let C be a convex subset of a normed linear space (X, ‖·‖). Then thedistance function for C is convex.

Proof. Let x, y ∈ X , 0 ≤ λ ≤ 1, and ε > 0. Hence, there exists w, z ∈ K such that‖x− w‖ ≤ d(x,K)+ε and ‖y − z‖ ≤ d(y,K)+ε. SinceK is convex, λw+(1−λ)z ∈ K.Therefore,

d(λx+ (1− λ)y,K) ≤ ‖λx+ (1− λ)y − (λw + (1− λ)z)‖≤ λ ‖x− w‖+ (1− λ) ‖y − z‖≤ λd(x,K) + (1− λ)d(y,K) + ε.

Since ε > 0 was arbitrary, it follows that

d(λx+ (1− λ)y,K) ≤ λd(x,K) + (1− λ)d(y,K),

and so the distance function for C is convex.

Proposition 3.2.12. Q and D are convex.

49

Proof. Let 0 ≤ λ ≤ 1 and suppose w := (x, r), v := (y, s) ∈ Q. SinceE(0)×R is convex,by Proposition 3.2.4 and Lemma 3.2.11 we have

d(λw + (1− λ)v, E(0) × R) ≤ λd(w,E(0) × R) + (1− λ)d(v, E(0) × R)≤ λξR(w) + (1− λ)ξR(v)≤ ξR(λw + (1− λ)v).

Since C is convex, λw+ (1− λ)v ∈ Q, and so Q is convex. Now let x, y ∈ D. Therefore,x, y ∈ E × [0,∞) and there exist w, v ∈ Q such that ‖x− w‖ < R(w) and ‖y − v‖ <R(v). Making use of the concavity of R again, we see that

‖λx+ (1− λ)y − (λw + (1− λ)v)‖ = ‖λ(x− w) + (1− λ)(y − v)‖≤ λ ‖x− w‖+ (1− λ) ‖y − v‖< λR(w) + (1− λ)R(v)≤ R(λw + (1− λ)v).

Since Q and E × [0,∞) are convex, we conclude that D is convex.

3.2.2 Smoothness and unique nearest pointsIn Chapter 2 we introduced the related concepts of Gateaux differentiability and smooth-ness of normed linear spaces. We now define what it means for a subset of a normed linearspace to be smooth.

First we introduce the concept of a supporting functional/hyerplane.

Definition 17. Let K be a subset of a normed linear space (X, ‖·‖) and x ∈ K. We saythat f ∈ X∗ supports K at x if

f(y) ≤ f(x)

for all y ∈ K. We also say that the hyperplane defined by {y ∈ X : f(y − x) = 0}supports K at x.

Definition 18. Let K be a subset of a normed linear space (X, ‖·‖) and x ∈ K. We saythat K is smooth at x if there exists a unique f ∈ SX∗ such that f supports K at x.

Remark: it is straightforward to show that a normed linear space is smooth, in the sensethat the norm is Gateaux differentiable everywhere except at 0, if, and only if, the closedunit ball is smooth at every point of SX .

50

We now prove a number of results that will later be used to prove a smoothness condi-tion for D. First we introduce the concept of a cone in Euclidean space and derive someits properties.

Definition 19. Let x ∈ E and r ≥ 0. Define the cone of height and base radius r as

K[x, r] := {(y, t) ∈ E × [0,∞) : ‖y − x‖+ t ≤ r} ⊆ E × R.

That is, K[x, r] is the right-circular cone with vertex (x, r) and base B[x, r]× {0}.

The following result gives a different way of describing a cone, and will be used toprove a smoothness condition for cones.

Proposition 3.2.13. For any x ∈ E and r > 0

K[x, r] = E × [0,∞) ∩⋃

λ∈[0,1]

B[(x, (1− 2λ)r), λ√2r].

Proof. Let (y, t) ∈ K[x, r]. Hence, t ≥ 0 and ‖y − x‖+ t ≤ r. Define

λ :=r + ‖y − x‖ − t

2r.

Since‖y − x‖ − t ≤ ‖y − x‖+ t ≤ r

and0 ≤ 2 ‖y − x‖ = ‖y − x‖+ t+ ‖y − x‖ − t ≤ r + ‖y − x‖ − t,

we see that λ ∈ [0, 1]. It is straightforward to show that

‖(y, t)− (x, (1− 2λ)r)‖2 = ‖y − x‖2 + (t− (1− 2λ)r)2

= 2 ‖y − x‖2 ,

whilst (λ√2r)2

=1

2(r + ‖y − x‖ − t)2

≥ 2 ‖y − x‖2 .

Therefore,(y, t) ∈ B[(x, (1− 2λ)r), λ

√2r]

51

and we have set inclusion in one direction.

For the opposite direction, suppose

(y, t) ∈ Rn × [0,∞) ∩⋃

λ∈[0,1]

B[(x, (1− 2λ)r), λ√2r].

Therefore, t ≥ 0 and

‖(y, t)− (x, (1− 2λ)r)‖2 = ‖y − x‖2 + (t− (1− 2λ)r)2 ≤ (√2λr)2

for some λ ∈ [0, 1]. Since

(r − t)2 −((√2λr)2 − (t− (1− 2λ)r)2

)=(√

2(λ− 1)r +√2t)2≥ 0,

it follows that‖y − x‖2 ≤ (r − t)2.

Finally, for λ ∈ [0, 1],

t ≤ (1− 2λ)r + λ√2r = r + λ(

√2− 2)r ≤ r,

and so we conclude that‖y − x‖+ t ≤ r.

Therefore, (y, t) ∈ K[x, r] and we have set inclusion in the opposite direction.

Corollary 3.2.14. Let x ∈ E and r > 0. Then K[x, r] is smooth at any (y, t) ∈ E× (0, r),such that ‖y − x‖+ t = r.

Proof. Let (y, t) ∈ E × (0, r) and suppose ‖y − x‖ + t = r. Clearly, (y, t) ∈ ∂K[x, r].By Proposition 3.2.13, there exists λ ∈ [0, 1] such that

(y, t) ∈ E × [0,∞) ∩B[(x, (1− 2λ)r), λ√2r] ⊆ K[x, r].

Since t 6= r, we have λ > 0, and so

B[(x, (1− 2λ)r), λ√2r]

is smooth. Furthermore, since t 6= 0, (y, t) is a smooth point of

E × [0,∞) ∩B[(x, (1− 2λ)r), λ√2r].

Therefore, K[x, r] is smooth at (y, t).

52

Proposition 3.2.15. Let x ∈ E \M and suppose (x, d(x,M)) ∈ D. Then

K[x, d(x,M)] ⊆ D.

Proof. Since D is convex and K[x, d(x,M)] is simply the convex hull of the set

{(x, d(x,M))} ∪B[x, d(x,M)]× {0}

we need only show that B[x, d(x,M)] × {0} ⊆ D. This follows from Proposition 3.2.9and the fact that B[x, d(x,M)] ⊆ E \M.

We now prove some results regarding the Gateaux differentiability of the distance func-tion.

Theorem 3.2.16. Let K be a nonempty, closed subset of a smooth normed linear space(X, ‖·‖). Suppose that x ∈ X \K is such that PK(x) is nonempty and

limt→0+

d(x+ tv,K)− d(x,K)

t= 1

for some v ∈ SX . Then the distance function for K is Gateaux differentiable at x.

Proof. Let y ∈ PK(x). Clearly, x 6= y, and so ‖·‖ is Gateaux differentiable at (y − x).Hence, there exists a unique f ∈ SX∗ such that

f(y − x) = ‖y − x‖ = d(x,K).

To prove the theorem we will show that −f is the Gateaux derivative of the distancefunction for K at x. For any t > 0,

d(x+ tv,K)− d(x,K)

t− 1 ≤ d(x+ tv,K)− d(x,K)

t+ f(v)

≤ ‖(y − x)− tv‖ − ‖y − x‖ − f(−tv)t

.

As t→ 0+ the left hand side of this inequality converges to zero by the hypotheses, whilstthe right hand side goes to zero since f is the derivative of ‖·‖ at (y − x). Hence, by thesqueeze theorem,

f(v) = − limt→0+

d(x+ tv,K)− d(x,K)

t= −1.

53

Thus f(−v) = ‖v‖ = 1, and so the Gateaux derivative of ‖·‖ at −v is f . Next, let h ∈ Xand t > 0. Then

d(x+ th,K)− d(x,K)− (−f)(th)t

≤ ‖(x− y)− th‖ − ‖x− y‖ − (−f)(th)t

,

and as t→ 0+ the right hand side converges to zero. Thus,

lim supt→0+

d(x+ th,K)− d(x,K)− (−f)(th)t

≤ 0.

Let ε > 0 be given. Since f is the derivative of ‖·‖ at −v, there exists δ > 0 such that∣∣∣∣‖−v + δh‖ − ‖−v‖ − f(δh)δ

∣∣∣∣ < ε.

Furthermore, there exists δ′ > 0 such that when 0 < t ≤ δ′,∣∣∣∣d(x+ tv,K)− d(x,K)

t− 1

∣∣∣∣ < εδ.

Thus, if 0 < t ≤ δδ′ and s := tδ, we have that∣∣∣∣s− ‖th− sv‖+ f(th)

t

∣∣∣∣ =∣∣∣∣∣1−

∥∥ tsh− v

∥∥+ f(tsh)

ts

∣∣∣∣∣=

∣∣∣∣∣∥∥−v + t

sh∥∥− ‖−v‖ − f( t

sh)

ts

∣∣∣∣∣=

∣∣∣∣‖−v + δh‖ − ‖−v‖ − f(δh)δ

∣∣∣∣< ε

and ∣∣∣∣d(x+ sv,K)− d(x,K)

s− 1

∣∣∣∣ st < εδ · 1δ= ε.

Also, since the distance function for K is nonexpansive, we have that for all 0 < t ≤ δδ′,

d(x+ th,K)− d(x,K)− (−f)(th)t

≥ d(x+ sv,K)− ‖th− sv‖ − d(x,K)− (−f)(th)t

=

(d(x+ sv,K)− d(x,K)

s− 1

)s

t+s− ‖th− sv‖+ f(th)

t

> −2ε.

54

Thus,

lim inft→0+

d(x+ th,K)− d(x,K)− (−f)(th)t

≥ 0.

Therefore,

limt→0+

d(x+ th,K)− d(x,K)

t= (−f)(h).

Since this limit holds for any h ∈ X , we have

limt→0−

d(x+ th,K)− d(x,K)

t= lim

t→0+

d(x+ (−t)h,K)− d(x,K)

−t

= − limt→0+

d(x+ t(−h), K)− d(x,K)

t

= −(−f)(−h)= (−f)(h).

Thus, the map y 7→ d(y,K) is Gateaux differentiable at x with derivative −f .

Corollary 3.2.17. Let K be a nonempty, proximinal subset of a smooth normed linearspace (X, ‖·‖) and x ∈ X \K be such that the metric projection for K is continuous atx. Then the distance function for K is Gateaux differentiable at x.

Proof. Let v := x−z‖x−z‖ ∈ SX , where {z} = PK(x). By Corollary 2.5.2, setting λ := t

‖x−z‖gives

limt→0+

d(x+ tv,K)− d(x,M)

t= lim

λ→0+

d(xλ, K)− d(x,K)

‖xλ − x‖= 1,

where xλ = x+ λ (x− z). Theorem 3.2.16 then shows that the distance function for K isGateaux differentiable at x.

Finally, we are ready to prove the smoothness condition on D. This result will berequired later to prove the uniqueness of nearest points in our construction.

Proposition 3.2.18. D is smooth at any (x, r) ∈ ∂D such that r > 0.

Proof. Let w := (x, r) ∈ ∂D with r > 0. By Proposition 3.2.6, there exists some w0 :=(x0, r0) ∈ Q such that ‖w − w0‖ = R(w0). If R(w0) > 0, then any hyperplane supportingD at w will also support the closed ball B[w0, R(w0)] ⊆ D at w. Since (E, 〈·, ·〉) is aninner product space, it is smooth, and so this hyperplane must be unique. Alternatively,R(w0) = 0, which implies w = w0 ∈ Q and d(x,M) = r > 0. Hence, x 6∈ M andw ∈ ∂A. By the definition of Q, d(w,E(0) × R) ≤ ξR(w) = 0, and since E(0) × R is

55

closed, we must have w ∈ E(0) × R. Hence, x ∈ E(0), and so PM(x) = {x} for somex ∈M . Theorem 2.3.6 tells us that the metric projection for M is continuous at x, and so,by Corollary 3.2.17, the distance function for M is Gateaux at x. Since

A ={(y, s) : y ∈ E \M, 0 ≤ s ≤ d(y,M)

},

it follows that w ∈ ∂A is a smooth point forA. LetH be the unique supporting hyperplanefor A at w. Since D ⊆ A, H also supports D at w. Suppose, looking for a contradiction,that D is not smooth at w. Hence, there exists a hyperplane H1 6= H , that also supportsD at w. Next we implement a neat trick, which enables us to view the problem in justtwo dimensions. Consider the two dimensional plane Z in E × R containing the pointsx′, x′, and w (it is clear that these points are not collinear so we do indeed have a twodimensional plane). Suppose, looking for a contradiction, that H ∩ Z = H1 ∩ Z. UsingCorollary 3.2.17 again, we see that H has slope 1 in the direction x− x, and so w+x′

2∈ H .

Since w+x′

2∈ Z also, w+x′

2∈ H1. As H1 supports D at w and, by Proposition 3.2.15,

K[x, d(x,M)] ⊆ D, it follows that H1 supports K[x, d(x,M)] at w+x′

2. Similarly, since

H supports A at w and K[x, d(x,M)] ⊆ D ⊆ A, H supports K[x, d(x,M)] at w+x′

2as

well. However,w + x′

2=

(x+ x

2,d(x,M)

2

),

and since 0 < d(x,M)2

< r, Corollary 3.2.14 implies that the cone K[x, d(x,M)] is smoothat w+x

′

2. Thus, H = H1, which is impossible. Therefore, H ∩ Z 6= H1 ∩ Z.

Next consider the sequences (xn)∞n=1 and (rn)∞n=1 defined by

xn := x+1

n(x− x)

and

rn :=

(1− 1

n

)d(x,M),

for all n ∈ N, respectively. Define the sequence (wn)∞n=1 by

wn := (xn, rn)

for all n ∈ N. Clearly, we have limn→∞wn = w. Let 0 < α ≤ π2

be the angle betweenH ∩ Z and H1 ∩ Z. For each n ∈ N, let un be the nearest point on H1 ∩ Z to wn. Thus,〈w − un, wn − un〉 = 0 for all n ∈ N, and taking the limit of this expression showslimn→∞ un = w. For each n ∈ N define

w†n := (x, rn).

56

Clearly, limn→∞w†n = w.

H1 ∩ Z

H ∩ Z

A ∩ Z

w = (x, r)

x′ = w†1

x′

x′1 = w1 x′2 x′3

w2

w3

w†2

w†3

u1u2 u3

α

Figure 3.1: Diagram showing the first few terms of each sequence.

Therefore, for all n ∈ N,∠ww†nwn =

π

2,∥∥w†n − wn∥∥ = ‖x− xn‖ = d(x,M)− rn = ‖w − x′‖ − rn =

∥∥w − w†n∥∥ ,∠w†nwwn =

π

4,

∠x′wx′ =π

4,

∠x′wwn =π

2.

Thus, since 0 < α ≤ π2,

1 > cos(α) =‖un − wn‖‖w − wn‖

for all n ∈ N. Furthermore,√2 ‖x− xn‖ = ‖w − wn‖

57

andrn = d(x,M)− ‖x− xn‖ ,

for all n ∈ N. Hence, for all n ∈ N,

R(wn) =d(xn,M)− rn√

2=d(xn,M)− d(x,M) + ‖x− xn‖√

2.

By Theorem 2.3.6 and Corollary 2.5.2,

limn→∞

d(xn,M)− d(x,M)

‖xn − x‖= 1.

Therefore,

limn→∞

R(wn)

‖w − wn‖= lim

n→∞

d(xn,M)− d(x,M) + ‖x− xn‖√2 ‖w − wn‖

= limn→∞

d(xn,M)− d(x,M) + ‖x− xn‖2 ‖x− xn‖

= limn→∞

d(xn,M)− d(x,M)

2 ‖x− xn‖+

1

2

= 1.

As x ∈ E(0), w†n ∈ E(0) × R for all n ∈ N, and so

d(wn, E(0) × R)

R(wn)≤∥∥wn − w†n∥∥R(wn)

≤ ‖w − wn‖√2R(wn)

for all n ∈ N. Since b ≤ d2, our earlier working gives

limn→∞

‖w − wn‖√2R(wn)

=1√2<d√8

b= ξ.

Hence, there exists some N ∈ N, such that for all n > N ,

d(wn, E(0) × R) ≤ ξR(wn).

Since

d(xn,M) ≥ d(x,M)− ‖x− xn‖ =(1− 1

n

)d(x,M) = rn = |rn| ,

58

wn ∈ C for all n ∈ N, and so wn ∈ Q for all n > N . Hence, by Proposition 3.2.6,

B[wn, R(wn)] ⊆ D

for all n > N . Since un is the nearest point on H1 ∩ Z to wn for any n ∈ N, and H1

supports D we have

R(wn)

‖w − wn‖≤ ‖un − wn‖‖w − wn‖

= cos(α) < 1

for all n > N , which contradicts limn→∞R(wn)‖w−wn‖ = 1. Hence, D is smooth at w.

We now work toward the key theorem for this chapter.

Lemma 3.2.19. Let K be a nonempty subset of a normed linear space (X, ‖·‖). Supposex ∈ X \ K is a point of Gateaux differentiability of the distance function for K, withGateaux derivative f ∈ X∗, and y ∈ PK(x). Then ‖f‖ = 1 and f(x − y) = ‖x− y‖ =d(x,K).

Proof. Since x 6∈ K, d(x,K) = ‖x− y‖ > 0. Let 0 ≤ t ≤ 1. Since x+ t(y− x) ∈ [x, y],Proposition 2.3.1 tells us that y ∈ PK (x+ t(y − x)) , and so

d(x+ t(y − x), K) = ‖x+ t(y − x)− y‖ = (1− t) ‖x− y‖ .

Therefore,

f(y − x) = limt→0+

d(x+ t(y − x), K)− d(x,K)

t

= limt→0+

(1− t) ‖x− y‖ − ‖x− y‖t

= limt→0+

−t ‖x− y‖t

= −‖x− y‖ .

Thus, f(x− y) = ‖x− y‖ = d(x,K). Proposition 2.1.1 then implies that, for any z ∈ X ,

|f(z)| =∣∣∣∣limt→0

d(x+ tz,K)− d(x,M)

t

∣∣∣∣= lim

t→0

∣∣∣∣d(x+ tz,K)− d(x,K)

t

∣∣∣∣≤ lim

t→0

‖x+ tz − x‖|t|

= ‖z‖ .

Thus, ‖f‖ = 1.

59

Theorem 3.2.20. Let K be a nonempty subset of a strictly convex normed linear space(X, ‖·‖). Suppose x ∈ X is a point of Gateaux differentiability of the distance functionfor K. Then PM(x) contains at most one element.

Proof. If x ∈ K, then PK(x) is either empty or {x}, so we may assume x 6∈ K. Suppposethat y, z ∈ PK(x) and let f ∈ X∗ be the Gateaux derivative of the distance function for Kat x. Applying Lemma 3.2.19 gives

f

(x− y‖x− y‖

)= ‖f‖ = f

(x− z‖x− z‖

)As (X, ‖·‖) is strictly convex, f can attain its norm at only one point of SX . Therefore,

x− y‖x− y‖

=x− z‖x− z‖

.

Since ‖x− y‖ = ‖x− z‖ = d(x,K), we have y = z, which completes the proof.

Lemma 3.2.21. Suppose u := (u1, . . . un+1), v := (v1, . . . vn+1) ∈ En+1, ‖u‖ = ‖v‖ , and(u1, . . . , un) = (v1, . . . , vn). If un+1, vn+1 ≥ 0 or un+1, vn+1 ≤ 0 then u = v.

Proof. We have

u2n+1 = ‖u‖2 −

n∑k=1

u2k

= ‖v‖2 −n∑k=1

v2k

= v2n+1.

Since un+1 and vn+1 have the same sign, it follows that un+1 = vn+1, and so u = v.

The following result is a slight modification of Theorem 3.2.20. We we will make useof it later to prove the uniqueness of nearest points.

Proposition 3.2.22. Let K be a nonempty closed subset of E and K ′ a nonempty subsetE × [0,∞). Define φ : E \ K → R by φ(x) := d(x′, K ′). If x ∈ E \ K is a point ofGateaux differentiability of φ then PK′(x′) contains at most one element.

Proof. Clearly, if PK′(x′) is empty or {x′} we’re done, so we may as well assume thaty ∈ PK′(x′) and y 6= x′. For any z ∈ E and λ > 0 we have

‖x′ + λz′ − y‖ ≥ d(x′ + λz′, K ′).

60

Since E \K is open, for sufficiently small λ > 0, x + λz ∈ E \K. Thus for sufficientlysmall λ > 0,

‖x′ + λz′ − y‖ − ‖x′ − y‖λ

≥ d(x′ + λz′, K ′)− d(x′, K ′)λ

=φ(x+ λz)− φ(x)

λ.

Taking the limit of both sides as λ→ 0+ gives

g(z′) ≥ f(z),

where g ∈ (E × R)∗ is the Gateaux derivative of the norm at (x′ − y) 6= 0 (which existssinceE×R is smooth) and f ∈ E∗ is the Gateaux derivative of φ at x. It is straightforwardto show that

g(v) =x′ − y‖x′ − y‖

· v

for any v ∈ E ×R. Since all of the above inequalities still hold if we replace z by −z andz ∈ E was arbitrary we have

g(z′) = f(z)

for all z ∈ E. Thus, if w ∈ PK′(x′) is another best approximation to x′ in K ′, we have

g(z′) = f(z) = h(z′)

for all z ∈ E, where h ∈ (E × R)∗ is the Gateaux derivative of the norm at x′ − w 6= 0.As before

h(v) =x′ − w‖x′ − w‖

· v

for all v ∈ E × R. Therefore,

x′ − y‖x′ − y‖

∣∣∣∣E×{0}

=x′ − w‖x′ − w‖

∣∣∣∣E×{0}

,

i.e. these points agree apart from at possibly there last coordinate. Since∥∥∥∥ x′ − y‖x′ − y‖

∥∥∥∥ = 1 =

∥∥∥∥ x′ − w‖x′ − w‖

∥∥∥∥and x′ − y, x′ − w ∈ E × [0,−∞), it follows by Lemma 3.2.21 that

x′ − y‖x′ − y‖

=x′ − w‖x′ − w‖

.

As‖x′ − y‖ = ‖x′ − w‖ = d(x′, K ′),

we have y = w, which completes the proof.

61

3.2.3 The inductive stepFirstly, we define a function on E \M using D and derive some of its properties.

Definition 20. Define ρ : E \M → R by

ρ(x) := max{r : (x, r) ∈ D

}for all x ∈ E \M .

Remark: this function is well defined since D is closed and bounded, and (E \M) ×{0} ⊆ D.

Remark: It is straightforward to show that for any non-negative, concave functionf : U → R, where U is a convex subset of E, with non-empty interior, the hypograph

hyp(f) := {(x, r) : x ∈ U, 0 ≤ r ≤ f(x)}

is smooth at (x, f(x)), for some x ∈ int(U), if, and only if, f |int(U) is Gateaux differen-tiable at x.

Proposition 3.2.23. For all x ∈ E \M ,

0 < ρ(x) ≤ d(x,M).

Furthermore, ρ is concave, ρ|∂M = 0, and ρ|E\M is Gateaux differentiable.

Proof. Let x ∈ E \M . By the definition of ρ and Proposition 3.2.10,

(x, ρ(x)) ∈ D ⊆ A.

Hence by the definition of A, 0 ≤ ρ(x) ≤ d(x,M). Thus for any x ∈ ∂M ,

0 ≤ ρ(x) ≤ d(x,M) = 0,

and so ρ|∂M = 0.To see the concavity of ρ, suppose x, y ∈ E \M and 0 ≤ λ ≤ 1. Since (x, ρ(x)) , (y, ρ(y)) ∈

D and D is convex we have

(λx+ (1− λ)y, λρ(x) + (1− λ)ρ(y)) = λ (x, ρ(x)) + (1− λ) (y, ρ(y)) ∈ D.

Thus,λρ(x) + (1− λ)ρ(y) ≤ ρ (λx+ (1− λ)y) .

62

Suppose, looking for a contradiction, that ρ(x) = 0 for some x ∈ E \ M . Sinceρ is concave and ρ|∂M = 0, this implies that ρ = 0. We now show that ρ(0) > 0,which is of course a contradiction. As 0 ∈ E \ M , we have (0, d(0,M)) ∈ C andd(0,M) > 0. Furthermore, since 0 ∈ E(0), (0, d(0,M)) ∈ Q. By Proposition 3.2.6, wehave (0, d(0,M)) ∈ D and so ρ(0) ≥ d(0,M) > 0. Thus, ρ(x) > 0 for all x ∈ E \M.

Finally, we check that ρ|E\M is Gateaux differentiable. Let x ∈ E\M . Since ρ(x) > 0,(x, ρ(x)) ∈ ∂D, and D = hyp(ρ), Proposition 3.2.18 and the previous remark tell us thatρ is Gateaux differentiable at x.

Proposition 3.2.24. Let x ∈ E \M .

(i) If d(x,M) ≤ b2

then ρ(x) = d(x,M).

(ii) If d(x,M) ≥ b2

then ρ(x) ≥ b2.

Proof. (i) Let x ∈ E \M be such that d(x,M) ≤ b2

and y ∈ PM(x). SinceM has the b-boundary property, there exists some z ∈ E \M such that ‖y − z‖ = d(z,M) = b.Since y ∈ PM(z), Lemma 3.2.8 tells us that x, y, z are collinear. Let w := y+z

2.

Since w ∈ [y, z], y ∈ PM(w) and d(w,M) = ‖y − w‖ = ‖y−z‖2

= b2. We now show

that ρ(w) = b2. Since ρ(w) ≤ d(w,M), it is sufficicient to show that

(w, b

2

)∈ D.

Clearly, z′ ∈ C and R(z′) = d(z,M)√2

= b√2. As E(0) ∩ (E \M) 6= ∅ we have that

d(z′, E(0) × R) = d(z, E(0))

≤ d(z, E(0) ∩ (E \M))

≤ d

≤ d√8

b· b√

2

= ξR(z′),

and so z′ ∈ Q. Also,∥∥∥∥(w, b2)− z′

∥∥∥∥2 = ‖w − z‖2 + ( b2)2

=

(b

2

)2

+

(b

2

)2

=b2

2.

Hence,∥∥(w, b

2

)− z′

∥∥ = b√2= R(z′), and so

(w, b

2

)∈ B[z,R(z′)] ⊆ D as required.

Finally, since d(x,M) = ‖x− y‖ ≤ b2= ‖y − w‖ and x, y, z are collinear, there

63

exists some 0 ≤ λ ≤ 1 such that x = λy + (1− λ)w. As ρ is concave and ρ(y) = 0(as y ∈ ∂M ), we have

ρ(x) ≥ λρ(y) + (1− λ)ρ(w) = (1− λ)ρ(w) = (1− λ) b2= ‖y − x‖ = d(x,M).

Since ρ(x) ≤ d(x,M) by Proposition 3.2.23, ρ(x) = d(x,M) as required.

(ii) Let x ∈ E \M be such that d(x,M) ≥ b2. Take any straight line in E passing

through x. Since E \M is bounded and convex, this line must intercept ∂M at twodistinct points y1 and y2. As d(y1,M) = d(y2,M) = 0 and the distance function forM is continuous, the intermediate value theorem tells us there exists z1 ∈ [x, y1) andz2 ∈ [x, y2) such that d(z1,M) = d(z2,M) = b

2. By (i), ρ(z1) = ρ(z2) =

b2. Since ρ

is concave and x ∈ [z1, z2], it follows that ρ(x) ≥ b2.

We now use a clever trick to construct a set in E × [0,∞) such that the distancefunction for this set, restricted to E \M , is given by the smooth function ρ. Making useof Proposition 3.2.22, we will be able to prove the uniqueness of nearest points.

Definition 21. Define the set

M ′ := (E × [0,∞)) \⋃

x∈E\M

B (x′, ρ(x)) .

Proposition 3.2.25. M ′ has the following properties.

(i) M × {0} = M ′ ∩ (E × {0}), M ′ is closed in E × R, and (E × [0,∞)) \ M ′ isnonempty, convex and bounded.

(ii) For every x ∈ E there exists a unique y ∈M ′ such that

‖x′ − y‖ = d(x′,M ′).

Furthermore, for x ∈ E \M ,

d(x′,M ′) = ρ(x).

(iii) 0 6∈M ′.

(iv) For every y ∈ ∂M ′\(int(M)× {0}) there exists x ∈ E\M such that PM ′(x′) = {y}.

64

Proof. (i) Firstly, we show thatM × {0} ⊆M ′.

Suppose, looking for a contradiction, that there exists x ∈ M such that x′ 6∈ M ′.Since x′ ∈ E × [0,∞), it follows that

x′ ∈⋃

y∈E\M

B (y′, ρ(y)) .

Hence, for some y ∈ E \M we have

‖x− y‖ = ‖x′ − y′‖ < ρ(y) ≤ d(y,M),

which is impossible since x ∈ M . Therefore, M × {0} ⊆ M ′. Secondly we showthat

(E \M)× {0} ⊆ (E × [0,∞)) \M ′.

It is straightforward to show that

(E × [0,∞)) \M ′ = (E × [0,∞)) ∩⋃

x∈E\M

B (x′, ρ(x)) .

Let x′ ∈ (E \M)×{0}. Clearly, x′ ∈ E× [0,∞). By Proposition 3.2.23, ρ(x) > 0,and so

x′ ∈ B(x′, ρ(x)) ⊆⋃

z∈E\M

B (z′, ρ(z)) .

Thus, x′ ∈ E × [0,∞) \M ′. Combining these two set inclusions gives

M × {0} =M ′ ∩ (E × {0})

as required. Furthermore, since 0 ∈ E \M , (E × [0,∞)) \M ′ is nonempty.

To see that M ′ is closed in E × R observe that

(E × R) \M ′ = (E × (−∞, 0)) ∪⋃

x∈E\M

B (x′, ρ(x))

is the union of open sets.

The convexity of

(E × [0,∞)) \M ′ = (E × [0,∞)) ∩⋃

x∈E\M

B (x′, ρ(x))

65

follows from the concavity of ρ. Let x, y ∈ (E × [0,∞)) \ M ′ and 0 ≤ λ ≤ 1.Clearly λx+ (1− λ)y ∈ E× [0,∞). Moreover, there exists w, z ∈ E \M such that‖x− w′‖ < ρ(w) and ‖y − z′‖ < ρ(z). Then∥∥λx+ (1− λ)y − (λw + (1− λ)z)′

∥∥ = ‖λ(x− w′) + (1− λ)(y − z′)‖≤ λ ‖x− w′‖+ (1− λ) ‖y − z′‖< λρ(w) + (1− λ)ρ(z)≤ ρ(λw + (1− λ)z).

Therefore,

λx+ (1− λ)y ∈ B((λw + (1− λ)z)′, ρ(λw + (1− λ)z)).

SinceE \M is convex, it follows that λw+(1−λ)z ∈ E \M and so λx+(1−λ)y ∈(E × [0,∞)) \M ′ as required.

Finally, since E \M is bounded and ρ(x) ≤ d(x,M) for all x ∈ E \M , it followsthat ρ, is bounded and so (E × [0,∞)) \M ′ is bounded.

(ii) Let x ∈ E. If x ∈M then x′ ∈M ′, and so x′ has itself as its unique nearest point inM ′. Alternatively, x ∈ E \M . We will now find a point y ∈M ′ such that

‖x′ − y‖ = ρ(x) = d(x′,M ′).

Observe that d(x′,M ′) ≥ ρ(x) by the definition of M ′, so it is sufficient to findy ∈M ′ such that ‖x′ − y‖ = ρ(x).

Let H be a hyperplane in E × R, supporting the convex set D at (x, ρ(x)) ∈ ∂D.

Firstly, suppose H is parallel to E×{0}. Let y := (x, ρ(x)). Since ρ is concave, andsmooth on E \M , ρ must have a global maximum at x. Clearly, ‖x′ − y‖ = ρ(x) soall we need show is that y ∈ M ′. Suppose, looking for a contradiction, that y 6∈ M ′.Since ρ(x) > 0, y ∈ E × [0,∞), and so y ∈ B(w′, ρ(w)) for some w ∈ E \M .Then

‖y − w′‖2 = ‖(x, ρ(x))− (w, 0)‖2

= ‖w − x‖2 + ρ(x)2

≥ ρ(x)2,

and soρ(x) ≤ ‖y − w′‖ < ρ(w),

66

which contradicts the maximality of ρ(x). Hence, y ∈M ′, and so y ∈ PM ′(x′).Alternatively, H and E × {0} are not parallel. The intersection of these two hyper-planes is a translate of a codimension 2 subspace of E × R, which we call N . Sup-pose, looking for a contradiction, that there exists some y′ ∈ N ∩ ((E \M)× {0}).Since y′ ∈ (E \M)× {0} ⊆ D and H supports D, we have that ρ(y) = 0, which isimpossible by Proposition 3.2.23. Thus,

N ∩ ((E \M)× {0}) = ∅.

As x ∈ E \M , this implies that H is not perpendicular to E × {0}. Let H1 be ahyperplane inE×R passing throughN and tangent toB[x′, ρ(x)] such that the pointof contact y is in E × [0,∞). This is possible since

B[x′, ρ(x)] ∩ (E × {0}) ⊆ E \M × {0}.

Clearly ‖x′ − y‖ = ρ(x) > 0, so again it is sufficient to show that y ∈ M ′. SupposeH and E ×{0} form an angle 0 < α ≤ π

4, whilst H1 and E ×{0} form an angle α1.

Since H1 and the vector y−x′ must be orthogona,l it follows that tan(α) = sin(α1).Therefore, the distance from any point (w, r) ∈ E × R on H to E × {0}, which isjust r, is equal to the distance from w′ to H1. Looking for a contradiction, supposey 6∈ M ′. By construction, y ∈ E × [0,∞) so there exists z ∈ E \ M such that‖y − z′‖ < ρ(z). Let v be the point directly above z′ lying on H . Since H supportsD, (z, ρ(z)) ∈ D, and y ∈ H1, it follows by the previous observation that

ρ(z) ≤ d (v, E × {0}) = d(z′, H1) ≤ ‖y − z′‖ < ρ(z),

which is clearly impossible. Therefore, y ∈M ′ as required.

We have now established that d(z′,M ′) = ρ(z) for all z ∈ E \ M . As ρ|E\M issmooth, Proposition 3.2.22 tells us that PM ′(x) is a singleton.

(iii) By assumption, 0 ∈ E \M . By (ii) and Proposition 3.2.23, we have

d(0,M ′) = ρ(0) > 0.

Hence, 0 6∈M ′.

(iv) We begin by showing that⋃x∈E\M

B (x′, ρ(x)) =⋃

x∈E\M

B [x′, ρ(x)] .

67

Let y ∈⋃x∈E\M B (x′, ρ(x)). Hence, there exists sequences (yn)∞n=1 and (xn)

∞n=1 in

E × R and E \M respectively, such that limn→∞ yn = y and

‖yn − x′n‖ < ρ(xn)

for all n ∈ N. Since E \M is a compact we may assume, without loss of generality,that (xn)∞n=1 converges to some x ∈ E \M . Firstly, suppose x ∈ E \M . Then

‖y − x′‖ =∥∥∥ limn→∞

yn − limn→∞

x′n

∥∥∥ = limn→∞

‖yn − x′n‖ ≤ limn→∞

ρ(xn) = ρ(x),

since ρ is smooth (and hence continuous) on E \M . Thus, y ∈ B [x′, ρ(x)]. Alter-natively, x ∈ ∂M . Therefore,

‖y − x′‖ ≤ limn→∞

ρ(xn) ≤ limn→∞

d(xn,M) = d(x,M) = 0

and so y = x′ ∈ ∂M × {0} . Since M has the b-boundary property, there existsz ∈ E \M such that

‖y − z‖ = d(z,M) = b.

Consider the point w := y+z2

. Since y, z ∈ E \M , which is convex, we have w ∈E \M . As w ∈ [y, z],

d(w,M) = ‖y − w‖ =∥∥∥∥y − z2

∥∥∥∥ =b

2> 0

by Proposition 2.3.1, and so w ∈ E \M. By Proposition 3.2.24,

ρ(w) = d(w,M) =b

2> 0.

Therefore,y ∈ B [w, d(w,M)]× {0} ⊆ B [w′, ρ(w)] .

Thus, ⋃x∈E\M

B (x′, ρ(x)) ⊆⋃

x∈E\M

B [x′, ρ(x)] .

Set inclusion obviously holds in the reverse direction, which establishes the requiredequality.

Now suppose w ∈ ∂M ′ \ (int(M)× {0}). By what we’ve just shown there existssome x ∈ E\M such that w ∈ B [x′, ρ(x)]. Thus, ‖x′ − w‖ = ρ(x), since otherwisew 6∈ M ′. By (ii), there exists a unique point in M ′ with distance ρ(x) = d(x′,M ′)from x′. Hence, {w} = PM ′(x

′) as required.

68

Proposition 3.2.26. For any x ∈ ∂M ′ \ (int(M)× {0}) there exists y ∈ E × [0,∞) suchthat

‖x− y‖ = d(y,M ′) =b

2.

Proof. Let x := (v, r) ∈ ∂M ′ \ (int(M)× {0}). By Proposition 3.2.25, part (iv) thereexists w ∈ E \M such that

PM ′(w′) = {x},

and by part b) of the same theorem,

‖w′ − x‖ = d(w′,M ′) = ρ(w).

Firstly, suppose d(w′,M ′) ≥ b2. Then there exists y ∈ [w′, x] such that

‖y − x‖ = d(y,M ′) =b

2.

Clearly, y ∈ E× [0,∞) so we’re done. Alternatively, suppose d(w′,M ′) = ρ(w) < b2. By

Proposition 3.2.24 part (ii),

d(w,M) <b

2,

and hence, by part (i) of the same result,

d(w,M) = ρ(w) = d(w′,M ′).

Since M × {0} ⊆ M ′, M is proximinal, and PM ′(w′) = {x} we must have that x :=(v, 0) ∈M × {0}. Clearly, v ∈ ∂M and so there exists z ∈ E \M such that

‖v − z‖ = d(z,M) = b,

since M has the b-boundary property. Let

y :=

(v + z

2

)′∈ (E \M)× {0} ⊆ E × [0,∞).

Since

d

(v + z

2,M

)=

∥∥∥∥v + z

2− v∥∥∥∥ =

b

2,

Proposition 3.2.24 and Proposition 3.2.25 part (ii) tell us that

b

2= d

(v + z

2,M

)= ρ

(v + z

2

)= d (y′,M ′) .

69

Whilst we have shown that M ′ has convex complement in E× [0,∞), its complementin E × R is not convex. We use the following flip-stretch operator to extend M ′ to a setwith convex complement in E × R.

Definition 22. For θ > 1, define ψθ : E × R→ E × R by

ψθ(x, r) = (x,−θr)

for all (x, r) ∈ E × R.

Lemma 3.2.27. The map ψθ is a bijective linear map, with inverse given by

ψ−1θ (x, r) =

(x,−rθ

), for all (x, r) ∈ E × R.

We also have the following inequalites

‖(x, r)‖ ≤ ‖ψθ(x, r)‖ ≤ θ ‖(x, r)‖

for any (x, r) ∈ E × R, with the first inequality being strict unless r = 0 and the secondstrict unless x = 0. Thus, both ψθ and its inverse are bounded and hence continuous.

Proof. Linearity and bijectiveness are trivial. For the inequalilty, observe that for any(x, r) ∈ E × R,

‖(x, r)‖2 = ‖x‖2 + r2 ≤ ‖x‖2 + (θr)2 = ‖ψθ(x, r)‖2 ≤ (θ ‖x‖)2 + (θr)2 = θ2 ‖(x, r)‖2 .

Replacing (x, r) by ψ−1θ (x, r) in these inequalities shows that

1

θ‖(x, r)‖ ≤

∥∥ψ−1θ (x, r)∥∥ ≤ ‖(x, r)‖

for all (x, r) ∈ E × R. The first inequality is strict unless x = 0 and the second strictunless r = 0. Hence, ψθ and ψ−1θ are bounded, and so continuous.

Definition 23. Let 1 < θ ≤ 2 and define M ′′ ⊆ E × R by

M ′′ :=M ′ ∪ ψθ(M ′).

We would like to show that the properties of M ′ established in Proposition 3.2.25 areinherited by this extension. The next result is used to show that M ′′ has the b

4-boundary

property.

70

Lemma 3.2.28. Let S be the ellipsoidal surface in Rn defined by the equation

x21α2

+x22β2

+ · · ·+ x2nβ2

= 1,

where α > β > 0. For any y ∈ S there exists w ∈ Conv(S) such that

‖w − y‖ = d(y, S) =β2

α.

Proof. Let y := (y1, . . . , yn) ∈ S, and consider y :=(y1

(1− β2

α2

), 0, . . . , 0

)∈ Conv(S).

We will show that y is a nearest point to y in S. To do this we will minimise the functionf : Rn → R defined by

f(x1, x2, . . . , xn) :=

∥∥∥∥(x1, x2, . . . , xn)− (y1(1− β2

α2

), 0, . . . , 0

)∥∥∥∥2 ,for all (x1, x2, . . . , xn) ∈ Rn, subject to the constraint that

x21α2

+x22β2

+ · · ·+ x2nβ2

= 1.

Since

f(x1, x2, . . . , xn) =

(x1 − y1

(1− β2

α2

))2

+ x22 + · · ·+ x2n,

substituting in the constraint shows we need only minimise (without constraint) the func-tion g : R→ R, defined by

g(x1) :=

(x1 − y1

(1− β2

α2

))2

+ β2 − x21β2

α2.

As 1− β2

α2 > 0,

g′(x1) = 2

(x1 − y1

(1− β2

α2

))− 2x1β

2

α2

is zero only when x1 = y1. Furthermore, since

g′′(x1) = 2

(1− β2

α2

)> 0,

71

we see that g is minimised at y1. Therefore, f , subject to the given constraint, is minimisedat (y1, x2, . . . , xn), where y21

α2 +x22β2 + · · ·+ x2n

β2 = 1. Hence, f is minimised (amongst otherpoints) at (y1, y2, . . . yn), and so y is a nearest point to y in S. Finally, since∥∥∥∥(y1, y2, . . . , yn)− (y1(1− β2

α2

), 0, . . . , 0

)∥∥∥∥2 = (y1β2

α2

)2

+ y22 + · · ·+ y2n

≥(y1β

2

α2

)2

+y22β

2

α2+ · · ·+ y2nβ

2

α2

=

(β2

α

)2(y21α2

+y22β2

+ · · ·+ y2nβ2

)=

(β2

α

)2

,

there exists w ∈ [y, y] ⊆ Conv(S) such that ‖w − y‖ = d(y, S) = β2

α.

Proposition 3.2.29. M ′′ has the following properties

(i) M × {0} = M ′′ ∩ (E × {0}), M ′′ is closed in E × R, 0 6∈ M ′′, and (E × R) \M ′′

is nonempty, convex and bounded.

(ii) For every x ∈ E there exists a unique y ∈M ′′ such that

‖x′ − y‖ = d(x′,M ′′).

Furthermore, when x ∈ E \M ,

d(x′,M ′′) = ρ(x).

(iii) M ′′ has the b4-boundary property.

Proof. (i) These statements all follow from Proposition 3.2.25. Since

ψθ (E × {0}) = (E × {0}) ,

we haveM × {0} =M ′ ∩ (E × {0}) =M ′′ ∩ (E × {0}) .

Also,∅ 6= (E × [0,∞)) \M ′ ⊆ (E × R) \M ′′.

72

It is straightforward to verify that

(E × R) \M ′′ = (E × [0,∞)) \M ′ ∪ ψθ ((E × [0,∞)) \M ′) .

Since ψ−1θ is continuous and M ′ is closed in E × R, the above set is the finite unionof closed sets in E × R, and so is closed in E × R.

As 0 6∈M ′ and ψθ(0) = 0, it follows that 0 6∈M ′′.

Since (E × [0,∞))\M ′ is convex and ψθ is linear, ψθ (E × [0,∞) \M ′) is also con-vex. Thus, to prove that (E × R) \M ′′ is convex we need only show that it contains[x, y] where x := (u, r) ∈ (E × [0,∞))\M ′ and y := (v, s) ∈ ψθ ((E × [0,∞)) \M ′).Since x 6∈M ′,

‖u′ − z‖ ≤ ‖x− z‖ < ρ(z) ≤ d(z,M)

for some z ∈ E \M . Therefore, u′ ∈ E \M . Similarly, v′ ∈ E \M . Since r ≥ 0and s ≤ 0, whilst E \M is convex, there exists 0 ≤ λ ≤ 1 such that

λx+ (1− λ)y ∈ E \M × {0}.

As(E \M)× {0} ⊆ ((E × [0,∞)) \M ′) ∩ ψθ ((E × [0,∞)) \M ′)

and the two sets in this intersection are convex, it follows that [x, y] ⊆ (E × R)\M ′′

as required.

Finally, since ψθ is bounded, it follows that (E × R) \ M ′′ is the finite union ofbounded sets, so is bounded.

(ii) Let x ∈ E. If x ∈ M then x′ ∈ M ′′ and the claim holds trivially, so we may as wellassume x ∈ E \M . By part (ii) of Proposition 3.2.25, there exists a unique y ∈ M ′

such that‖x′ − y‖ = d(x′,M ′) = ρ(x).

Looking for a contradiction, suppose there exists z ∈M ′′ \M ′, such that ‖x′ − z‖ ≤d(x′,M ′). Since x′ − z 6∈ E × {0}, Lemma 3.2.27 gives∥∥x′ − ψ−1θ (z)

∥∥ =∥∥ψ−1θ (x′)− ψ−1θ (z)

∥∥ , since x′ ∈ E × {0}=∥∥ψ−1θ (x′ − z)

∥∥< ‖x′ − z‖≤ d(x,M ′),

which is impossible since ψ−1θ (z) ∈M ′. Hence, y is the unique element of M ′′ suchthat

‖x′ − y‖ = d(x′,M ′′) = ρ(x).

73

(iii) Let x ∈ ∂M ′′. It is straightforward to verify that

∂M ′′ = ∂M ′ \ (int(M)× {0}) ∪ ψθ (∂M ′ \ (int(M)× {0})

If x ∈ ∂M ′ \ (int(M)× {0}), by Proposition 3.2.26, there exists y ∈ E × [0,∞)such that

‖x− y‖ = d(y,M ′) =b

2.

Using the same argument as in part (ii), we conclude that

‖x− y‖ = d(y,M ′′) =b

2.

Therefore, ∥∥∥∥x− x+ y

2

∥∥∥∥ = d

(x+ y

2,M ′′

)=b

4.

Alternatively, x ∈ ψθ (∂M′ \ (int(M)× {0})). As before, we can find y ∈ E ×

[0,∞) such that ∥∥ψ−1θ (x)− y∥∥ = d(y,M ′′) =

b

2.

It follows that

ψθ

(B[y,

b

2]

)⊆ (E × R) \M ′′

is an ellipsoid centred at ψθ(y) with minor axes of length b2, and a major axis of

length bθ2

, such that

x ∈ ∂(ψθ

(B[y,

b

2]

)).

By Lemma 3.2.28, there exists w ∈ ψθ(B[y, b

2])

such that

B

[w,

b

2θ

]⊆ ψθ

(B[y,

b

2]

)and x ∈ B

[w, b

2θ

]. Hence,

‖w − x‖ = d(w,M ′′) =b

2θ.

Since θ ≤ 2, b4≤ b

2θ, and so we can find z ∈ [w, x] ⊆ E × R such that

‖z − x‖ = d(z,M ′′) =b

4.

74

The next result explains why we chose to define Q as in Defintion 15.

Proposition 3.2.30. For any x ∈ E(0),

PM(x)× {0} = PM ′′(x′).

Proof. Let x ∈ E(0). If x ∈ M then there is nothing to prove, so assume x 6∈ M . Byassumption, PM(x) is a singleton. Proposition 3.2.29 tells us that PM ′′(x′) is a singletonand d(x′,M ′′) = ρ(x).

We now show that d(x,M) = ρ(x). Since ρ(x) ≤ d(x,M), it is sufficient to checkthat d(x,M) ≤ ρ(x), which is true provided w := (x, d(x,M)) ∈ D. Clearly, w ∈ C,with R(w) = 0. Since x ∈ E(0), we have w ∈ E(0) × R and so

d(w,E(0) × R) = 0 = ξR(w).

Thus, w ∈ Q, and so w ∈ D by Proposition 3.2.6. Finally, since M × {0} ⊆ M ′′, itfollows that PM(x)× {0} = PM ′′(x

′).

The following two results will be used to prove that our non-convex Chebyshev set hasbounded complement.

Proposition 3.2.31.

sup{‖x‖ : x ∈ (E × R) \M ′′} ≤ 3θ sup{‖x‖ : x ∈ E \M}

Proof. To begin with let

x ∈ (E × [0,∞)) \M ′ = (E × [0,∞)) ∩⋃

z∈E\M

B(z′, ρ(z)).

Hence, there exists y ∈ E \M such that ‖y′ − x‖ < ρ(y). Since the distance function forM is non-expansive,

‖x‖ < ‖y′‖+ ρ(y)

≤ ‖y‖+ d(y,M)

≤ 2 ‖y‖+ d(0,M)

≤ 3 sup{‖z‖ : z ∈ E \M}.

Thus,sup{‖x‖ : x ∈ (E × [0,∞)) \M ′} ≤ 3 sup{‖x‖ : x ∈ E \M}.

Since(E × R) \M ′′ = ((E × [0,∞)) \M ′) ∪ ψθ ((E × [0,∞)) \M ′)

and θ > 1, the result follows.

75

Theorem 3.2.32. Let (an)∞n=1 be a sequence with non-negative terms. Then the infinite

product∏∞

n=1 (1 + an) coverges if, and only if, the infinte series∑∞

n=1 an converges.

Proof. For a proof see, [25].

3.2.4 The non-convex Chebyshev setDefinition 24. For any K ⊆ En, define

K ⊗ 0 := {(x1, . . . , xn, 0, 0, . . .) : (x1, . . . , xn) ∈ K}.

Similarly, for any x := (x1, . . . , xn) ∈ En, define

x⊗ 0 := (x1, . . . , xn, 0, 0, . . .) ∈ En ⊗ 0.

Definition 25. Let E be the set of all real sequences with only finitely many non-zeroterms, that is

E :=∞⋃n=1

(En ⊗ 0) .

Equip E with the inner product 〈·, ·〉 defined by

〈x, y〉 :=∞∑k=1

xkyk

for all x := (x1, x2, . . .), y := (y1, y2, . . .) ∈ E. The induced norm ‖·‖ on E is then givenby

‖x‖ :=√〈x, x〉

for all x := (x1, x2, . . .) ∈ E.

Theorem 3.2.33. In the (incomplete) inner product space (E, 〈·, ·〉) there exists a non-convex Chebyshev set with bounded convex complement.

Proof. Firstly, we inductively define a sequence of sets (Mn)∞n=1 such that, for all n ∈ N,

Mn is a nonempty closed subset of En, En \Mn is convex, 0 6∈ Mn, every point in E(0)n

has a unique nearest point in Mn, and Mn has the(14

)n−1-boundary property.To begin let

M1 =: E1 \ (−2, 1).It is clear that M1 is a closed, nonempty subset of E1. Furthermore, it is straightforwardto check that 0 6∈ M1, E1 \M1 is bounded and convex, M1 has the 1-boundary property,and every point in E(0)

1 = {0} has a unique nearest point in M1.

76

Suppose then thatMk is defined for some k ∈ N, such thatMk has the properties listedabove. Construct M ′

k as in Definition 21. Then let

Mk+1 :=M ′′k

whereM ′′

k :=M ′k ∪ ψθk (M ′

k)

andθk := 1 +

1

k2.

By Proposition 3.2.29, it follows that Mk+1 has the required properties, which completesthe construction. Furthermore, by Proposition 3.2.29 , Proposition 3.2.30, and Proposi-tion 3.2.31, we have

Mn × {0} =Mn+1 ∩ (En × {0}) ,

PMn(x)× {0} = PMn+1(x′)

for all x ∈ E(0)n , and

sup{‖x‖ : x ∈ En+1 \Mn+1} ≤ 3θ sup{‖x‖ : x ∈ En \Mn}

for all n ∈ N.We now show that

M :=∞⋃n=1

(Mn ⊗ 0)

is a Chebyshev set in (E, 〈·, ·〉). Firstly, we show that

M ∩ (Ek ⊗ 0) =Mk ⊗ 0

for all k ∈ N. Let k ∈ N. It is clear that Mk ⊗ 0 ⊆ M ∩ (Ek ⊗ 0), so suppose x ∈M ∩ (Ek ⊗ 0). Thus, for some n ∈ N, x ∈ (Mn ⊗ 0) ∩ (Ek ⊗ 0) . Repeatedly using thefact that Mm × {0} =Mm+1 ∩ (Em × {0}) for all m ∈ N, we conclude that

(Mn ⊗ 0) ∩ (Ek × 0) ⊆Mk ⊗ 0.

Hence, x ∈Mk ⊗ 0, and we’re done.Now let x ∈ E. Since x can have only finitely many non-zero components, there exists

n ∈ N such that x ∈ E(0)n ⊗ 0. By construction,

PMn⊗0(x) = {y}

77

for some y ∈Mn ⊗ 0 ⊆M . Furthemore, since

PMn⊗0(x) = PMn+1⊗0(x)

and E(0)n ⊗ 0 ⊆ E

(0)n+1 ⊗ 0, it follows by induction that

PMm⊗0(x) = PMn⊗0(x) = {y}

for all m ≥ n. To show that y is the unique nearest point to x in M , suppose there existsz ∈ M such that ‖x− z‖ ≤ ‖x− y‖. Let k ≥ n be such that z ∈ Ek ⊗ 0. Thus,PMk⊗0(x) = {y}. Since

z ∈ (Ek ⊗ 0)) ∩M =Mk ⊗ 0.

and‖x− z‖ ≤ ‖x− y‖ = d(x,Mk ⊗ 0),

it follows that z = y. Therefore, PM(x) = {y}, and so M is a Chebyshev set.We now show that M is non-convex. Clearly PM1⊗0(0) = {(1)⊗ 0}. By the previous

working, PM(0) = {(1)⊗0}, and so 0 6∈M . However, (−2)⊗0, (1)⊗0 ∈M1⊗0 ⊆M ,so M is not convex.

Making use of the fact that (Mn ⊗ 0) ∩ (Ek × 0) ⊆ Mk ⊗ 0 for all k, n ∈ N, we havethat

E \M =

(∞⋃n=1

(En ⊗ 0)

)\

(∞⋃n=1

(Mn ⊗ 0)

)

=∞⋃n=1

(En ⊗ 0) \ (Mn ⊗ 0)

=∞⋃n=1

((En \Mn)⊗ 0) .

Hence X \M is the union of expanding convex sets, so is convex. Furthermore, since

sup{‖z‖ : z ∈ (En+1 \Mn+1)⊗ 0} ≤ 3θk sup{‖z‖ : z ∈ (En \Mn)⊗ 0}

for all k ∈ N, we have by induction that

sup{‖z‖ : z ∈ E \M} ≤ 3

(∞∏n=1

θn

)sup{‖z‖ : z ∈ (E1 \M1)⊗ 0}

= 6∞∏n=1

(1 +

1

n2

).

78

By Theorem 3.2.32, the infinite product∏∞

n=1

(1 + 1

n2

)converges, since the series

∑∞n=1

1n2

converges. Thus, the right hand side of the previous expression is finite, and so E \M isbounded.

79

Chapter 4

Future research

Unfortunately, since any infinite-dimensional Banach space must have uncountable dimen-sion [8], trying to construct a non-convex Chebyshev set in a Hilbert space, by methodssimilar to those used in the previous chapter, would appear to be doomed to failure.

However, looking at Theorem 3.2.33, it is natural to ask whether the closure of Min the completion of E (which is just `2(N)) is a Chebyshev set in the completion of E.Johnson, in [18], derives a sufficient condition for a point in the completion of E to havea unique nearest point in the closure of M .

Theorem 4.0.34. If (xn)∞n=1 is a sequence in E such that

limn→∞

xn = x ∈ `2(N)

andlimn→∞

pM(xn) = y,

then pN(x) = y, where N is the closure of M in `2(N).

Johnson also gives an example of such a sequence of nearest points, which fails toconverge. It would be interesting to see if there are any further restrictions that can beplaced on the set M that guarantees the convergence of such sequences.

We also note another attempt at a solution to the Chebyshev set problem. In [26], Ric-ceri proposes the study of a conjecture whose positive solution would imply the existenceof a non-convex Chebyshev set in a Hilbert space.

80

Bibliography

[1] Fernando Albiac and Nigel J. Kalton. Topics in Banach space theory, volume 233 ofGraduate Texts in Mathematics. Springer, New York, 2006.

[2] Edgar Asplund. Cebysev sets in Hilbert space. Trans. Amer. Math. Soc., 144:235–240, 1969.

[3] V. S. Balaganskiı. Approximation properties of sets in Hilbert space. Mat. Zametki,31(5):785–800, 815, 1982.

[4] V. S. Balaganskiı and L. P. Vlasov. The problem of the convexity of Chebyshev sets.Uspekhi Mat. Nauk, 51(6(312)):125–188, 1996.

[5] Bela Bollobas. Linear analysis. Cambridge University Press, Cambridge, secondedition, 1999. An introductory course.

[6] L.N.H. Bunt. Bitrage tot de theorie der kovekse puntverzamelingen. PhD thesis,University of Groningen, 1934.

[7] Herbert Busemann. Note on a theorem on convex sets. Mat. Tidsskr. B., 1947:32–34,1947.

[8] N. L. Carothers. A short course on Banach space theory, volume 64 of LondonMathematical Society Student Texts. Cambridge University Press, Cambridge, 2005.

[9] James A. Clarkson. Uniformly convex spaces. Trans. Amer. Math. Soc., 40(3):396–414, 1936.

[10] Mahlon M. Day. Reflexive Banach spaces not isomorphic to uniformly convexspaces. Bull. Amer. Math. Soc., 47:313–317, 1941.

[11] Frank Deutsch. Best approximation in inner product spaces. CMS Books in Mathe-matics/Ouvrages de Mathematiques de la SMC, 7. Springer-Verlag, New York, 2001.

81

[12] N. V. Efimov and S. B. Steckin. Cebysev sets in Banach spaces. Dokl. Akad. NaukSSSR, 121:582–585, 1958.

[13] J. Frerking and U. Westphal. On a property of metric projections onto closed subsetsof Hilbert spaces. Proc. Amer. Math. Soc., 105(3):644–651, 1989.

[14] John R. Giles. Convex analysis with application in the differentiation of convex func-tions, volume 58 of Research Notes in Mathematics. Pitman (Advanced PublishingProgram), Boston, Mass., 1982.

[15] B. Jessen. To saetninger om konvekse punktmaengder. Mat. Tidsskrift B, pages66–70, 1940.

[16] Miaohua Jiang. On Johnson’s example of a nonconvex Chebyshev set. J. Approx.Theory, 74(2):152–158, 1993.

[17] Gordon G. Johnson. A nonconvex set which has the unique nearest point property.J. Approx. Theory, 51(4):289–332, 1987.

[18] Gordon G. Johnson. Closure in a Hilbert space of a prehilbert space Chebyshev set.Topology Appl., 153(2-3):239–244, 2005.

[19] B.F. Kelly. The convexity of chebushev sets in finite dimensional normed linearspaces. Master’s thesis, The Pennsylvania State University, 1978.

[20] Victor Klee. Convexity of Chevyshev sets. Math. Ann., 142:292–304, 1960/1961.

[21] Victor L. Klee, Jr. Convex bodies and periodic homeomorphisms in Hilbert space.Trans. Amer. Math. Soc., 74:10–43, 1953.

[22] M. Kritikos. Sur quelques proprietes des ensembles convexes. Bull. Math. de la Soc.Roumaine des Sciences, 40:87–92, 1938.

[23] T.S. Motzkin. Sur quelques proprietes caracteristiques des ensembles bornes nonconvexes. Rend. Acad. dei Lincei, 21:562–567, 1935.

[24] R. R. Phelps. Convex sets and nearest points. Proc. Amer. Math. Soc., 8:790–797,1957.

[25] Murali Rao and Henrik Stetkær. Complex analysis. An invitation. World ScientificPublishing Co. Inc., Teaneck, NJ, 1991. A concise introduction to complex functiontheory.

82

[26] Biagio Ricceri. A conjecture implying the existence of non-convex Chebyshev sets ininfinite-dimensional Hilbert spaces. Matematiche (Catania), 65(2):193–199, 2010.

[27] Theodore J. Rivlin. An introduction to the approximation of functions. Dover Publi-cations Inc., New York, 1981. Corrected reprint of the 1969 original, Dover Bookson Advanced Mathematics.

[28] L. P. Vlasov. On Cebysev sets. Dokl. Akad. Nauk SSSR, 173:491–494, 1967.

83