Non-Linear Maps Between Subsets of Banach Spaces A

NON-LINEAR MAPS BETWEEN SUBSETS OF BANACH SPACES

A dissertation submitted to Kent State University in partial fulﬁllment of the requirements for the degree of Doctor of Philosophy

Reema Sbeih

December, 2009 Dissertation written by

Reema Sbeih

B.S., Birzeit University, West Bank, Palestine 2000

M.S., Youngstown State University, 2002

M.A., Kent State University, 2008

Ph.D., Kent State University, 2009

Approved by

Dr. Per Enﬂo , Chair, Doctoral Dissertation Committee

Dr. Andrew Tonge , Member, Doctoral Dissertation Committee

Dr. Morley Davidson , Member, Doctoral Dissertation Committee

Dr. Kenneth Batcher , Member, Outside Discipline

Dr. Peter Tandy , Member, Graduate Representative

Accepted by

Dr. Andrew Tonge , Chair, Department of Mathematical

Sciences

Dr. John R. D. Stalvey , Dean, College of Arts and Sciences

ii TABLE OF CONTENTS

ACKNOWLEDGEMENTS ...... iv

1 INTRODUCTION ...... 1

2 BASIC DEFINITIONS AND NOTATION ...... 3

3 PROJECTIONS ONTO THE UNIT BALL IN LP SPACES...... 7

4 SCALING DOWN MAPS BETWEEN S(l∞) AND S(l1) ...... 14

5 SCALING DOWN MAPS BETWEEN S(l∞) AND S(l2) ...... 24

6 SCALING DOWN MAPS BETWEEN S(l∞) AND S(lq) as q → ∞ . . . . 37

7 CONCLUSION AND FUTURE PROJECTS ...... 48

BIBLIOGRAPHY ...... 50

iii ACKNOWLEDGEMENTS

I owe this work to the two people from whom I received my life’s blood, my parents.

My ultimate inspiration comes from my advisor, Dr. Per Enﬂo, the wonderful mathematician who guided through this incredible and long journey.

iv CHAPTER 1

INTRODUCTION

In this dissertation we study non-linear maps between subsets of Banach spaces.

As a background for this study we should mention two areas of mathematical research:

Geometry of Banach Spaces and Approximation Theory.

One way to study and compare the geometry of diﬀerent Banach spaces is to study linear maps between the spaces and to study how much these maps distort distances between points. There is an extensive literature on this [2], [3], [4], [5].

Another less studied way to compare the geometry of diﬀerent Banach spaces is to study maps between diﬀerent subsets of the spaces. In this case the maps are usually non-linear and the subsets are often unit balls or unit spheres of Banach spaces [6],

[7], [10], [11].

In Approximation Theory, the following problem is important: how well can objects (points) from a set A be approximated by objects (points) from a set B? One important way to study this approximation is to ﬁnd for every point in A the nearest point in B, i.e., to ﬁnd the nearest point map, also called the metric projection, and investigate its properties [10], [12], [13]. The maps that we study in this dissertation, the scaling down projection and the scaling down maps, are, as we shall see, nearest point maps.

The dissertation is organized as follows: In Chapter 2 we give the basic deﬁnitions and notation that we use in the rest of the paper. We also show that the scaling down projection and the scaling down maps are nearest point maps.

In Chapter 3, we show that the scaling down projection onto the unit sphere in any

1 2

Banach space has Lipschitz norm at most 2 and that 2 is not attained. We also show that in L1(0, 1), the scaling down projection has norm 2, and that the scaling down projection in L1(0, 1) is the best projection in the sense that all other projections have norm ≥ 2. We also give estimates for the norm of the scaling down projection in Lp for 1 < p < 2.

In chapters 4, 5, and 6 we study scaling down maps between the unit sphere of

∞ p ln , being in some sense the largest unit sphere, and the unit spheres of ln-spaces. In particular we study the Banach-Mazur norms of these maps. Our results show that the Banach-Mazur norms of these maps are much larger than the Banach-Mazur norms of the simplest linear maps between the spaces. CHAPTER 2

BASIC DEFINITIONS AND NOTATION

In this chapter we give the basic deﬁnitions and notation that we use in this paper.

We also show that the scaling down projection and the scaling down map are nearest point maps.

Deﬁnition. [17] A vector space X is said to be a normed space if for every x ∈ X there is associated a nonnegative real number ||x||, called the norm of x, in such a way that

(a) ||x + y|| ≤ ||x|| + ||y|| for all x and y in X,

(b) ||αx|| = |α|||x|| if x ∈ X and α is a scalar,

Every normed space maybe regarded as a metric space, in which the distance d(x, y) between x and y is ||x − y||.

Deﬁnition. [17] A Banach space is a normed space which is complete in the metric deﬁned by its norm; this means that every Cauchy sequence is required to converge.

Let (X, d) be a normed space.

Deﬁnition. The unit ball of X, denoted by B(X) is

B(X) = {x ∈ X : d(x, 0) ≤ 1} .

Deﬁnition. The unit sphere of X, denoted by S(X) is

S(X) = {x ∈ X : d(x, 0) = 1} .

3 4

Deﬁnition. A projection is a map P : X −→ X such that P ◦ P = P.

Deﬁnition. We say that M ⊆ X is contractive if there is a projection P (in general nonlinear) from X onto M such that

d(P x, P y) ≤ d(x, y) for all x, y ∈ X.

Deﬁnition. Let M,N ⊆ X. A map T : M −→ N is a nearest point map if

d(x, T x) ≤ d(x, z) for all z ∈ N.

Deﬁnition. A map P on X is called a Lipschitz map if there is a constant C such that

(2.1) ||P x − P y|| ≤ C||x − y|| for all x, y ∈ X.

The smallest constant C that satisfies (2.1) is called the Lipschitz norm of P , and ||P x − P y|| is denoted by ||P ||. It is easy to see that ||P || = sup . x,y∈X ||x − y|| Definition. The map P on X defined by  x  if ||x|| > 1 P x = ||x||   x if ||x|| ≤ 1 is called the scaling down projection.

Consider the norms || · ||1 and || · ||2 on a vector space. Assume ||x||1 ≤ ||x||2 for every x. Let X, Y be the normed spaces associated with ||x||1 and ||x||2 respectively. We have the following deﬁnition. 5

Deﬁnition. The scaling down map T : S(X) → S(Y ) is the map T that takes x 7→ γxx where ||x||1 = 1 and ||γxx||2 = 1.

Lemma 2.1 The scaling down projection is a nearest point map from X onto B(X).

Proof. Let x ∈ X and let y = P x.

If ||x|| ≤ 1, then y = x and ||x − y|| = 0. x So suppose that ||x|| > 1; then y = . ||x|| We need to show that if z is any point in B(X) then ||x − y|| ≤ ||x − z||. We have ¯¯ ¯¯ ¯¯ µ ¶¯¯ µ ¶ ¯¯ x ¯¯ ¯¯ 1 ¯¯ 1 ||x − y|| = ¯¯x − ¯¯ = ¯¯x 1 − ¯¯ = 1 − ||x|| = ||x|| − 1, and ¯¯ ||x||¯¯ ¯¯ ||x|| ¯¯ ||x|| ||x − z|| ≥ ||x|| − ||z|| ≥ ||x|| − 1.

Therefore ||x − y|| ≤ ||x − z|| for all z ∈ B(X). ¥

Corollary 2.2 The scaling down map and its inverse are nearest point maps.

Deﬁnition. A normed space X is strictly convex if ||tx1 + (1 − t)x2|| < 1 whenever x1 and x2 are diﬀerent points of S(X) and 0 < t < 1.

Deﬁnition. A normed space X is reﬂexive if every bounded sequence has a weakly convergent subsequence.

Deﬁnition. A normed space X is smooth if every point on the unit sphere of X has a unique supporting hyperplane.

Next we will give some of the notation used in this paper. R p p L (E) = {f : E |f| < ∞}, 0 < p < ∞. ¡R ¢ 1 p p ||f||p = E |f| , 0 < p < ∞.

||f||∞ = inf{α : m({x ∈ E : |f(x)| > α}) = 0}.

∞ L (E) = {f : ||f||∞ < ∞}. 6

Let a = {ak} be a sequence of real or complex numbers. Then we have the following deﬁnitions:

P 1 p p ||a||p = ( k |ak| ) , 0 < p < ∞.

||a||∞ = supk|ak|. p l = {a : ||a||p < ∞}, 0 < p < ∞.

∞ l = {a : ||a||∞ < ∞}. CHAPTER 3

PROJECTIONS ONTO THE UNIT BALL IN LP SPACES.

It is known that unit balls in Hilbert spaces and spaces of continuous functions are contractive. In this chapter, we show that in any Banach space, the scaling down projection has Lipschitz norm ≤ 2, and has norm equal to 2 in L1(0, 1). We also show that in L1(0, 1), any projection other than the scaling down projection has Lipschitz norm at least 2. We also estimate the norm of the scaling down projection in Lp(0, 1),

1 < p < 2.

In a Hilbert space, a closed, convex set is contractive since the “nearest point map” is a contractive projection.

Theorem 3.1 [1] (Beauzamy-Maurey result) Let X be reﬂexive, strictly convex, smooth

Banach space of dimension 3 or greater. If the unit ball of X is contractive then X is a Hilbert space.

Remark. Since l∞ and C(0, 1) are not strictly convex, we cannot apply the Beauzamy-

Maurey result. However, in C(0, 1) the unit ball is contractive. To see this, let f, g ∈ C(0, 1).

7 8

Let    f if −1 ≤ f ≤ 1  P f = 1 if f ≥ 1    −1 if f ≤ −1

And    g if −1 ≤ g ≤ 1  P g = 1 if g ≥ 1    −1 if g ≤ −1

It is easy to see that ||P f − P g|| ≤ ||f − g||.

Let X be a normed space.

Lemma 3.2 Let x, y ∈ X such that ||x|| = 1 + ² and ||y|| = 1. Let P be the scaling

down projection on X. Then

||P x − P y|| < 2||x − y||

Proof. Let z = P x. Then

||x − (1 + ²)y|| = ||(1 + ²)z − (1 + ²)y||

= (1 + ²)||z − y||.

Also

||x − (1 + ²)y|| = ||x − y − ²y||

≤ ||x − y|| + ².

(1 + ²)||z − y|| ≤ ||x − y|| + ². 9

Divide both sides of this inequality by (1 + ²)||x − y||. We get

||z − y|| 1 ² ≤ + ||x − y|| 1 + ² (1 + ²)||x − y|| 1 1 ≤ + ( since ||x − y|| ≥ ²) 1 + ² 1 + ² 2 = . 1 + ² This gives

||P x − P y|| 2 (3.1) ≤ . ||x − y|| 1 + ²

Theorem 3.3 Let x, y ∈ X, x 6= y, and let P be the scaling down projection on X.

Then

(3.2) ||P x − P y|| < 2||x − y||.

Proof. Let x, y ∈ X with ||x|| > 1 and ||y|| > 1. Suppose w.l.o.g. that ||x|| > ||y||.

Then ° ° ° x y ° (3.3) ° − ° < ||x − y||. °||y|| ||y||°

So it is enough to prove the theorem for the case where ||y|| = 1 which is done in

Lemma 3.2. ¥

In the next theorem, we show that in L1(0, 1), the Lipschitz norm of the scaling

down projection is 2.

Theorem 3.4 In L1(0, 1), the Lipschitz norm of the scaling down projection is 2.

Proof. From Theorem 3.3 we have that the Lipschitz norm ≤ 2. To show that the

Lipschitz norm is 2, it is enough to ﬁnd one function f(x), outside the unit ball of 10

L1(0, 1) that cannot be projected to the unit ball without getting the Lipschitz norm arbitrarily close to 2. Take f(x) = 1 + ² on [0, 1]. Let h = P f, ||h|| = 1, then h ≡ 1 on [0, 1]. Deﬁne the function g as follows:   1  1 + ² on A, for some A ⊂ [0, 1], m(A) = g(x) = 1 + ²   0 on Ac then ||g|| = 1. And

1 ||f − g|| = (1 + ²)(1 − ) = ² 1 + ² µ ¶ 1 1 2² ||h − g|| = ² · + 1 1 − = . 1 + ² 1 + ² 1 + ² This gives ||h − g|| 2 = → 2 as ² → 0. ||f − g|| 1 + ² ¥

Theorem 3.5 Let P be a projection onto the unit ball in L1(0, 1). If P is not the scaling down projection, then ||P || ≥ 2.

Proof. We show that for every projection P that is not the scaling down projection, there are g and f, ||g|| = 1, ||f|| = 1 + ² such that

||P f − P g|| 2 (3.4) ≥ . ||f − g|| 1 + 2²

Consider f ≡ 1 + ² on [0, 1]. Let h = P f, ||h|| = 1. Let B be a set of measure ² such Z Z that |h(t)| dt ≥ ². Then |h(t)| dt ≤ 1 − ². Put B [0,1]−B    0 if t ∈ B g(t) = 1  if t ∈ [0, 1] − B 1 − ² 11

We can see by the deﬁnition of g that ||g|| = 1. Then Z Z ||g − f|| = |g(t) − f(t)| dt + |g(t) − f(t)| dt B [0,1]−B µ ¶ 1 = ²(1 + ²) + (1 − ²) − (1 + ²) 1 − ² = ²(1 + ²) + ²2 = ²(1 + 2²).

We have ||P g − P f|| = ||g − h|| since P g = g and P f = h. Now Z Z ||g − h|| = |g(t) − h(t)| dt + |g(t) − h(t)| dt. B [0,1]−B Z Z Z We need to ﬁnd |g(t)−h(t)| dt and |g(t)−h(t)| dt. We start with |g(t)− B [0,1]−B Z Z B h(t)| dt. We know that g(t) = 0 on B and that |h(t)| dt ≥ ². So |g(t)−h(t)| dt ≥ Z B B ². Now we need to ﬁnd |g(t) − h(t)| dt. [0,1]−B Z Z Z |g(t) − h(t)| dt ≥ |g(t)| dt − |h(t)| dt [0,1]−B [0,1]−B [0,1]−B

≥ 1 − (1 − ²) = ², since Z 1 g(t) = on [0, 1] − B and |h(t)| dt ≤ 1 − ². 1 − ² [0,1]−B So Z Z ||P g − P f|| = |g(t) − h(t)| dt + |g(t) − h(t)| dt ≥ 2². B [0,1]−B This gives ||P f − P g|| 2² 2 ≥ = . ||f − g|| ²(1 + 2²) 1 + 2²

If f is not mapped onto the unit sphere but ||P f|| < 1 then let Qf be the point on the line segment tf + (1 − t)P f, 0 < t < 1 such that ||Qf|| = 1. Then, as above 12

||Qf − g|| 2 we ﬁnd a g, ||g|| = 1 such that ≥ . Obviously P g = g. Since Qf is ||f − g|| 1 + 2² on the line segment between f and P f we obviously have ||Qf − g|| ≤

||P f − P g|| 2 max(||f − g||, ||P f − g||) = ||P f − g||. For this case ≥ , and the ||f − g|| 1 + 2² proof of the theorem is complete. ¥

We do not know if the scaling down projection in Lp(0, 1), 1 < p < 2 has minimal

Lipschitz norm. The proof we gave for the case where p = 1 does not generalize. In the next theorem, we give an estimate from below of the norm of the scaling down projection in Lp(0, 1) for 1 < p < 2. We observe that this estimate goes to 2 as p → 1 and goes to 1 as p → 2.

Theorem 3.6 The Lipschitz norm of the scaling down projection in Lp(0, 1), 1 < p < 2 is greater than or equal to

1 1 p−1 1 p (2 − p n )(p n − 1) + (2 − p n )

1 where n is arbitrary such that 1 < p n < 2.

Proof. Take f(x) ≡ 1 + ² on [0, 1]. Let h = P f, then h ≡ 1 on [0, 1].

Deﬁne the function g as follows,   1  1 + ² on A ⊂ [0, 1], m(A) = 2 − p n g(x) =   1 − δ on Ac where δ is such that ||g||p = 1.

p 1 p 1 We have ||g||p = 1 which implies (1 + ²) (2 − p n ) + (1 − δ) (p n − 1) = 1. 1 ²(2 − p n ) So we get δ ≈ 1 . p n − 1 Where we used the approximation (1 + ²)p ≈ 1 + ²p and (1 − δ)p ≈ 1 − δp, the error 13

being of order of magnitude ²2 and δ2, respectively. We get Ã ! 1 p ²(2 − p n ) 1 p n ||g − f||p = ² + 1 (p − 1), p n − 1 and Ã ! 1 p 1 ²(2 − p n ) 1 p p n n ||g − h||p = ² (2 − p ) + 1 (p − 1). p n − 1

Therefore Ã ! 1 p n p 1 ²(2 − p ) 1 ² (2 − p n ) + (p n − 1) p 1 ||g − h|| p n − 1 p = Ã ! ||g − f||p 1 p p ²(2 − p n ) 1 n ² + 1 (p − 1) p n − 1 Ã ! 1 p 1 2 − p n 1 n n 2 − p + 1 (p − 1) p n − 1 = µ ¶p 1 1 n 1 (p − 1) p n − 1

1 p 1 (2 − p n ) n 2 − p + 1 (p n − 1)p−1 = 1 1 (p n − 1)p−1

1 1 p−1 1 p = (2 − p n )(p n − 1) + (2 − p n ) .

By choosing n large, we get that as p → 2, the expression above goes to 1. And by letting n → 1, the expression above goes to 2 as p → 1. ¥ CHAPTER 4

SCALING DOWN MAPS BETWEEN S(l∞) AND S(l1)

∞ In this chapter, we study scaling down maps between the unit spheres of ln and

1 ln. These are, in a sense, the simplest possible maps between these unit spheres and have the property that both the map and its’ inverse are nearest point maps. We show that the Lipschitz norms of these maps are of the order of magnitude n, so the

Banach-Mazur norm, as deﬁned below, is of the order of magnitude n2. This should be compared to the simplest possible linear maps where the Banach-Mazur norm is n.

We start with some deﬁnitions. Let (X, d1), (Y, d2) be metric spaces and let M ⊆ X,N ⊆ Y .

Deﬁnition. A map T : M → N is called a Lipschitz map if there is a constant C such that

(4.1) d2(T x, T y) ≤ Cd1(x, y) for all x, y ∈ M.

The smallest constant C that satisﬁes (4.1) is called the Lipschitz norm of T , and is denoted by ||T ||. d (T x, T y) It’s easy to see that ||T || = sup 2 . x,y∈M d1(x, y) Deﬁnition. If T : M → N is a one-to-one and onto Lipschitz map, and T −1 is also

Lipschitz, we deﬁne the Banach-Mazur norm of T as

kT kkT −1k.

14 15

∞ 1 Theorem 4.1 If T is the scaling down map from S(l2 ) to S(l2) then ||T || = 2.

d (T x, T y) Proof. We need to show that ||T || = sup 2 = 2, where x and y are points x,y d1(x, y) ∞ ∞ on the unit sphere of l2 . Let d1 denote the distance in l2 and d2 denote the distance

1 in l2. Take x = (1, 0) and y = (0, 1). Then d1(x, y) = 1 and d2(T x, T y) = d2(x, y) = 2. Which implies ||T || ≥ 2. d (T x, T y) To show that ||T || = 2, we show that 2 ≤ 2 whenever x 6= (1, 0) and y 6= d1(x, y) (0, 1). There are four cases to consider:

1. x = (1, a), y = (b, 1), 0 ≤ a, b ≤ 1, a ≤ b.

2. x = (1, −a), y = (b, 1), 0 ≤ a, b ≤ 1.

3. x = (1, −a), y = (−b, 1), 0 ≤ a, b ≤ 1, a ≤ b.

4. x = (1, a), y = (1, b), 0 ≤ a ≤ 1, −1 ≤ b ≤ 1, a ≤ |b|.

d (T x, T y) We show that if x and y satisfy any of the four cases above, then 2 ≤ 2. d1(x, y)

Case 1. x = (1, a), y = (b, 1), 0 ≤ a, b ≤ 1, a ≤ b.

This gives

d1(x, y) = 1 − a. µ ¶ 1 1 a T x = (1, a) = , . 1 + a 1 + a 1 + a µ ¶ 1 b 1 T y = (b, 1) = , . 1 + b 1 + b 1 + b Thus µ ¶ µ ¶ 1 b 1 a d (T x, T y) = − + − 2 1 + a 1 + b 1 + b 1 + a 1 − a 1 − b = + 1 + a 1 + b 1 − ab = 2 . (1 + a)(1 + b) 16

d2(T x, T y) 1 − ab = 2 2 ≤ 2. d1(x, y) (1 − a )(1 + b)

Case 2. x = (1, −a), y = (b, 1), 0 ≤ a, b ≤ 1.

This gives

d1(x, y) = 1 + a. µ ¶ 1 1 −a T x = (1, −a) = , . 1 + a 1 + a 1 + a µ ¶ 1 b 1 T y = (b, 1) = , . 1 + b 1 + b 1 + b Thus 1 b 1 a d (T x, T y) = − + + 2 1 + a 1 + b 1 + b 1 + a 1 + a 1 − b = + 1 + a 1 + b 2 = . (1 + b) So d (T x, T y) 2 2 = ≤ 2. d1(x, y) (1 + a)(1 + b)

Case 3. x = (1, −a), y = (−b, 1), 0 ≤ a, b ≤ 1, a ≤ b.

This gives

d1(x, y) = 1 + b. µ ¶ 1 1 −a T x = (1, −a) = , . 1 + a 1 + a 1 + a µ ¶ 1 −b 1 T y = (−b, 1) = , . 1 + b 1 + b 1 + b Thus 1 b 1 a d (T x, T y) = + + + 2 1 + a 1 + b 1 + b 1 + a = 2. 17

So d (T x, T y) 2 2 = ≤ 2. d1(x, y) (1 + b)

Case 4. x = (1, a), y = (1, b), 0 ≤ a ≤ 1, − 1 ≤ b ≤ 1, a ≤ |b|.

This gives

d1(x, y) = |a − b|. µ ¶ 1 1 a T x = (1, a) = , . 1 + a 1 + a 1 + a µ ¶ 1 1 b T y = (1, b) = , . 1 + |b| 1 + |b| 1 + |b| Thus ¯ ¯ ¯ ¯ 1 1 ¯ b a ¯ d (T x, T y) = − + ¯ − ¯ 2 1 + a 1 + |b| ¯1 + |b| 1 + a¯

≤ |a − b| + |a − b| = 2|a − b|. So d (T x, T y) 2 ≤ 2. d1(x, y) Cases 1 through 4 show that ||T || = 2. This proves Theorem (4.1). ¥

∞ 1 Theorem 4.2 If T is the scaling down map from S(ln ) to S(ln), n ≥ 3, then

n − 1 ≤ ||T || ≤ 2n.

Proof. We startµ by showing that ||T || ≥¶n − 1. 1 1 Take x = 1, ,..., , a(n − 1) a(n − 1) µ ¶ −1 −1 and y = 1, ,..., , a(n − 1) a(n − 1) where a is a large positive number. This gives

2 d (x, y) = . 1 a(n − 1) 18

µ ¶ a 1 1 T x = 1, ,..., . a + 1 a(n − 1) a(n − 1) µ ¶ a −1 −1 T y = 1, ,..., . a + 1 a(n − 1) a(n − 1)

So µ ¶ a 1 a 1 2 d (T x, T y) = · + · (n − 1) = . 2 a + 1 a(n − 1) a + 1 a(n − 1) a + 1 Thus µ ¶ d (T x, T y) a 2 = (n − 1) → n − 1 as a → ∞. d1(x, y) a + 1

d (T x, T y) So ||T || = sup 2 ≥ n − 1. x,y d1(x, y) Now we need to show that ||T || ≤ 2n. To do that, take two points x and y on the

∞ unit sphere of ln .

Let x = (a1, a2, . . . , an), − 1 ≤ ai ≤ 1 for i = 1, . . . , n and ai = 1 for some 1 ≤ i ≤ n, and y = (b1, b2, . . . , bn), − 1 ≤ bi ≤ 1 for i = 1, . . . , n and bi = 1 for some 1 ≤ i ≤ n.

Here d1(x, y) = max |ai − bi|. 1≤i≤n Xn Xn Let |ai| = A and |bi| = B, then i=1 i=1 ³a a a ´ T x = 1 , 2 ,... n . A A A µ ¶ b b b T y = 1 , 2 ,... n . B B B Thus ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯a1 b1 ¯ ¯a2 b2 ¯ ¯an bn ¯ d2(T x, T y) = ¯ − ¯ + ¯ − ¯ + ··· + ¯ − ¯ . A ¯ B ¯ A B A B Xn ¯a b ¯ = ¯ i − i ¯ ¯ A B ¯ i=1 19

Notice that ¯ ¯ ¯ ¯ ¯a b ¯ ¯a B − b A¯ ¯ i − i ¯ = ¯ i i ¯ ¯ A B ¯ ¯ AB ¯ ¯ ¯ ¯a B − b B + b B − b A¯ = ¯ i i i i ¯ ¯ AB ¯ ¯ ¯ ¯a B − b B + b (B − A)¯ = ¯ i i i ¯ ¯ AB ¯ |a − b | |b (B − A)| ≤ i i + i . A AB So ¯ ¯ Xn ¯a b ¯ Xn |a − b | Xn |b (B − A)| ¯ i − i ¯ ≤ i i + i ¯ A B ¯ A AB i=1 i=1 i=1 Xn |a − b | Xn |b ||B − A| = i i + i A AB i=1 i=1 Xn |a − b | |B − A| = i i + . A A i=1 Now ¯ ¯ ¯ n ¯ ¯X ¯ |B − A| = ¯ (|b | − |a |)¯ ¯ i i ¯ i=1 Xn ¯ ¯ ≤ ¯|bi| − |ai|¯. i=1 ¯ ¯ ¯ ¯ Claim: ¯|bi| − |ai|¯ ≤ |ai − bi|. ¯ ¯ ¯ ¯ Proof. |ai −bi| ≥ |ai|−|bi| and |bi −ai| ≥ |bi|−|ai|. Therefore |ai −bi| ≥ ¯|ai|−|bi|¯. ¥ Xn So |B − A| ≤ |ai − bi|. i=1 This gives ¯ ¯ Xn ¯a b ¯ d (T x, T y) = ¯ i − i ¯ 2 ¯ A B ¯ i=1 Xn |a − b | ≤ 2 i i A i=1

≤ 2n max |ai − bi|. 1≤i≤n 20

d (T x, T y) So 2 ≤ 2n. d1(x, y) This gives ||T || ≤ 2n. ¥

∞ 1 Theorem 4.3 If T is the scaling down map from S(ln ) to S(ln), n ≥ 2, then

n ≤ ||T −1|| ≤ 2n. 2 n Proof. We start by showing that ||T −1|| ≥ . 2 d (T x, T y) 2 To do this, we show that inf 2 ≤ where x and y are point on the unit x,y d1(x, y) n ∞ d2(T x, T y) 2 sphere of ln . To show that inf ≤ , we need to ﬁnd points x and y on x,y d1(x, y) n ∞ d2(T x, T y) 2 the unit sphere of ln satisfying ≤ . d1(x, y) n Take x = (1,..., 1), and y = (1, 1 − δ1, 1 − δ2,..., 1 − δn−1), where δi → 0, i = 1, . . . , n − 1.

Assume δi < δ1 < 2δi ∀i, 2 ≤ i ≤ n − 1. This gives

d1(x, y) = δ1. µ ¶ 1 1 1 T x = (1,..., 1) = ,..., . n n n 1 T y = (1, 1 − δ ,..., 1 − δ ). Xn−1 1 n−1 n − δi i=1 Xn−1 Let k = n − δi. i=1 21

Then µ ¶ µ ¶ 1 1 1 1 − δ1 d2(T x, T y) = − + − + µk n ¶n kµ ¶ 1 1 − δ 1 1 − δ − 2 + ··· + − n−1 n k  n k  Xn−1 µ ¶  n − 1 − δi    1 1 n − 1 i=1  = − +  −  k n  n k 

Xn−1 n − δi 1 1 n − 1 1 = − + + − i=1 . k n n k k Xn−1 n − δi 2 n − 2 Notice that i=1 = 1. This gives d (T x, T y) = + − 1. k 2 k n Now   Xn−1  δi    2 2 1  i=1  = ≈ 2 · 1 +  k Xn−1 n  n  n − δi i=1 Xn−1 2 δi 2 = + i=1 . n n2 So Xn−1 2 δi 2 n − 2 d (T x, T y) ≈ + i=1 + − 1. 2 n n2 n 2 Xn−1 = δ n2 i i=1 We get µ ¶ d2(T x, T y) 2 δ2 δn−1 ≈ 2 1 + + ··· d1(x, y) n δ1 δ1 2 ≤ (1 + 1 + ··· 1) n2 2(n − 1) 2 = ≈ for large n. n2 n 22

d (T x, T y) 2 So inf 2 ≤ . x,y d1(x, y) n n This gives ||T −1|| ≥ . 2 −1 ∞ To show that ||T || ≤ 2n, take two points x1, x2 on the unit sphere of ln such that

||x1 − x2||∞ = a, where 0 < a < 1.

∞ Let T x1 = y1 and T x2 = y2. We know that x1 is on the unit sphere of ln , so

||x1||∞ = 1, which implies that ||x1||1 ≤ n. Now y1 = c1x1 for some constant c1, 1 which implies ||y || = c ||x || , we get c ||x | = 1, which implies c ≥ since 1 1 1 1 1 1 | 1 1 n

||x1||1 ≤ n. 1 Similarly y = c x with c ≥ . 2 2 2 2 n 1 1 Suppose c1 ≥ c2 and consider the points z1 = y1 and z2 = y2. c1 c1 By Theorem 3.3:

||x − x || a 1 2 ∞ ≤ 2, which implies ||z − z || ≥ since ||x − x || = a. ||z − z || 1 2 ∞ 2 1 2 ∞ 1 2 ∞ ¯¯ ¯¯ ¯¯ 1 1 ¯¯ 1 But ||z1 − z2||∞ = ¯¯ y1 − y2¯¯ = ||y1 − y2||∞. c c ∞ c 1 1 a 1 Thus ||y − y || = c ||z − z || ≥ c . 1 2 ∞ 1 1 2 ∞ 2 1 a So ||y − y || ≥ c . 1 2 1 2 1

||y1 − y2||1 c1 1 1 So ≥ ≥ since c1 ≥ . ||x1 − x2||∞ 2 2n n This gives ||T −1|| ≤ 2n. ¥

∞ 1 Theorem 4.4 The Banach-Mazur norm of the scaling down map from S(l2 ) to S(l2) is greater than or equal to 4 and less than or equal to 8.

∞ 1 Proof. Let T be the scaling down map from S(l2 ) to S(l2). In Theorem4.1, we showed that ||T || = 2.

In Theorem4.3, we showed that 2 ≤ ||T −1|| ≤ 4.

So 4 ≤ kT kkT −1k ≤ 8. ¥ 23

∞ 1 Theorem 4.5 The Banach-Mazur norm of the scaling down map from S(ln ) to S(ln) n is greater than or equal to (n − 1) and less than or equal to 4n2. 2

∞ 1 Proof. Let T be the scaling down projection from S(ln ) to S(ln). In Theorem4.2, we showed that n − 1 ≤ ||T || ≤ 2n. n In Theorem4.3, we showed that ≤ ||T −1|| ≤ 2n. 2 n So (n − 1) ≤ kT kkT −1k ≤ 4n2. ¥ 2 CHAPTER 5

SCALING DOWN MAPS BETWEEN S(l∞) AND S(l2)

In this chapter, we do a similar study as in chapter four, but comparing instead

∞ 2 ln and ln.

√ ∞ 2 Theorem 5.1 If T is the scaling down map from S(l2 ) to S(l2), then ||T || = 2.

d (T x, T y) √ Proof. We need to show that ||T || = sup 2 = 2, where x and y are x,y d1(x, y) ∞ ∞ points on the unit sphere of l2 . Here, d1 denotes the distance in l2 and d2 denote

2 the distance in l2. Take x = (1, 0) and y = (0, 1). Then d1(x, y) = 1 and d2(T x, T y) = √ d2(T x, T y) √ √ d2(x, y) = 2. So = 2. This gives ||T || ≥ 2. To show that ||T || = d1(x, y) √ d (T x, T y) √ 2, we show that 2 ≤ 2 whenever x 6= (1, 0) and y 6= (0, 1). There are d1(x, y) ﬁve cases to consider:

1. x = (1, a), y = (b, 1), 0 ≤ a, b ≤ 1, a ≤ b.

2. x = (1, −a), y = (b, 1), 0 ≤ a, b ≤ 1.

3. x = (1, −a), y = (−b, 1), 0 ≤ a, b ≤ 1, a ≤ b.

4. x = (1, a), y = (1, −b), 0 ≤ a ≤ 1, 0 ≤ b ≤ 1, a ≤ b.

5. x = (1, a), y = (1, b), 0 ≤ a ≤ 1, 0 ≤ b ≤ 1, a ≤ b.

Case 1. x = (1, a), y = (b, 1), 0 ≤ a, b ≤ 1, a ≤ b.

This gives

d1(x, y) = 1 − a. 1 T x = √ (1, a). 1 + a2

24 25

1 T y = √ (b, 1). 1 + b2 Thus µ ¶ µ ¶ 1 b 2 1 a 2 d2(T x, T y) = √ − √ + √ − √ 2 1 + a2 1 + b2 1 + b2 1 + a2 1 2b b2 1 = − √ √ + + 1 + a2 1 + a2 1 + b2 1 + b2 1 + b2 2a a2 −√ √ + 1 + a2 1 + b2 1 + a2 2(a + b) = 2 − p (1 + a2)(1 + b2) Ã ! a + b = 2 1 − p . (1 + a2)(1 + b2)

So s √ a + b d2(T x, T y) = 2 1 − p . (1 + a2)(1 + b2)

d (T x, T y) √ To show that 2 ≤ 2, we need to show that d1(x, y) s a + b 1 − p ≤ 1 − a for 0 ≤ a, b ≤ 1, a ≤ b. Notice that (1 + a2)(1 + b2) s a + b 1 − p is largest when a = b. So (1 + a2)(1 + b2)

s r a + b 2a 1 − p ≤ 1 − (1 + a2)(1 + b2) 1 + a2 1 − a = p ≤ 1 − a. (1 + a2) So √ d (T x, T y) 2(1 − a) √ 2 ≤ = 2. d1(x, y) 1 − a 26

Case 2. x = (1, −a), y = (b, 1), 0 ≤ a, b ≤ 1.

This gives

d1(x, y) = 1 + a. 1 T x = √ (1, −a). 1 + a2 1 T y = √ (b, 1). 1 + b2 Thus µ ¶ µ ¶ 1 b 2 1 a 2 d2(T x, T y) = √ − √ + √ + √ 2 1 + a2 1 + b2 1 + b2 1 + a2 1 2b b2 1 = − √ √ + + (1 + a2) 1 + a2 1 + b2 1 + b2 1 + b2 2a a2 +√ √ + 1 + a2 1 + b2 1 + a2 2(a − b) = 2 + p (1 + a2)(1 + b2) Ã ! a − b = 2 1 + p . (1 + a2)(1 + b2)

So s √ a − b d2(T x, T y) = 2 1 + p . (1 + a2)(1 + b2)

d (T x, T y) √ To show that 2 ≤ 2, we need to show that d1(x, y) s a − b 1 + p ≤ 1 + a for 0 ≤ a, b ≤ 1. (1 + a2)(1 + b2) s a − b Notice that 1 + p is largest when b = 0. (1 + a2)(1 + b2) s r a − b a So 1 + p ≤ 1 + √ ≤ 1 + a. (1 + a2)(1 + b2) 1 + a2 27

Thus √ d (T x, T y) 2(1 + a) √ 2 ≤ = 2. d1(x, y) 1 + a

Case 3. x = (1, −a), y = (−b, 1), 0 ≤ a, b ≤ 1, a ≤ b.

This gives

d1(x, y) = 1 + b. 1 T x = √ (1, −a). 1 + a2 1 T y = √ (−b, 1). 1 + b2

So µ ¶ µ ¶ 1 b 2 1 a 2 d2(T x, T y) = √ + √ + √ + √ 2 1 + a2 1 + b2 1 + b2 1 + a2 1 2b b2 1 = + √ √ + + 1 + a2 1 + a2 1 + b2 1 + b2 1 + b2 2a a2 +√ √ + 1 + a2 1 + b2 1 + a2 2(a + b) = 2 + p (1 + a2)(1 + b2) Ã ! a + b = 2 1 + p . (1 + a2)(1 + b2)

So s √ a + b d2(T x, T y) = 2 1 + p . (1 + a2)(1 + b2) d (T x, T y) √ To show that 2 ≤ 2, we need to show that d1(x, y) s a + b 1 + p ≤ 1 + b. (1 + a2)(1 + b2) s a + b Notice that 1 + p is largest when a = b. (1 + a2)(1 + b2) 28

So s r a + b 2b 1 + p ≤ 1 + (1 + a2)(1 + b2) 1 + b2 1 + b = √ ≤ 1 + b. 1 + b2 Thus d (T x, T y) √ 2 ≤ 2. d1(x, y)

Case 4. x = (1, a), y = (1, −b), 0 ≤ a ≤ 1, 0 ≤ b ≤ 1, a ≤ b.

This gives

d1(x, y) = a + b. 1 T x = √ (1, a). 1 + a2 1 T y = √ (1, −b). 1 + b2

So µ ¶ µ ¶ 1 1 2 a b 2 d2(T x, T y) = √ − √ + √ + √ 2 1 + a2 1 + b2 1 + a2 1 + b2 2 − 2ab = 2 − p (1 + a2)(1 + b2) Ã ! 1 − ab = 2 1 − p . (1 + a2)(1 + b2)

d (T x, T y) √ To show that 2 ≤ 2, we need to show that d1(x, y) 1 − ab 1 − p ≤ (a + b)2. (1 + a2)(1 + b2) Notice that 1 − ab 1 − ab 1 − p ≤ 1 − since a ≤ b (1 + a2)(1 + b2) 1 + b2 b2 + ab = 1 + b2 b(a + b) = ≤ (a + b)2. 1 + b2 29

Thus d (T x, T y) √ 2 ≤ 2. d1(x, y)

Case 5. x = (1, a), y = (1, b), 0 ≤ a ≤ 1, 0 ≤ b ≤ 1, a ≤ b.

This gives

d1(x, y) = b − a. 1 T x = √ (1, a). 1 + a2 1 T y = √ (1, b). 1 + b2

So µ ¶ µ ¶ 1 1 2 a b 2 d2(T x, T y) = √ − √ + √ − √ 2 1 + a2 1 + b2 1 + a2 1 + b2 2(1 + ab) = 2 − p (1 + a2)(1 + b2) Ã ! 1 + ab = 2 1 − p . (1 + a2)(1 + b2)

d (T x, T y) √ 1 + ab To show that 2 ≤ 2, we need to show that 1−p ≤ (b−a)2. d1(x, y) (1 + a2)(1 + b2) 1 + ab 1 − p ≤ (b − a)2. (1 + a2)(1 + b2) 1 + ab ⇐⇒ 1 − (b − a)2 ≤ p . (1 + a2)(1 + b2) 1 + 2ab + a2b2 ⇐⇒ 1 − 2(b − a)2 + (b − a)4 ≤ . (1 + a2)(1 + b2) ⇐⇒ (1 − 2(b − a)2 + (b − a)4)(1 + a2)(1 + b2) ≤ 1 + 2ab + a2b2.

⇐⇒ (1 + a2 + b2 + a2b2)(1 − 2(b − a)2 + (b − a)4) ≤ 1 + 2ab + a2b2.

⇐⇒ 1 + a2 + b2 + a2b2 − 2(b − a)2(1 + a2 + b2 + a2b2) + (b − a)4(1 + a2 + b2 + a2b2)

≤ 1 + 2ab + a2b2.

⇐⇒ a2 − 2ab + b2 ≤ 2(b − a)2(1 + a2 + b2 + a2b2) − (b − a)4(1 + a2 + b2 + a2b2). Here, a2 − 2ab + b2 = (b − a)2, divide both sides of the last inequality by (b − a)2, we 30

get 1 ≤ 2(1 + a2 + b2 + a2b2) − (b − a)2(1 + a2 + b2 + a2b2).

⇐⇒ 1 ≤ (2 − (b − a)2)(1 + a2 + b2 + a2b2). This inequality holds which implies that the inequality 1 + ab 1 − p ≤ (b − a)2 also holds. (1 + a2)(1 + b2)

So d (T x, T y) √ 2 ≤ 2. d1(x, y) √ Cases 1 through 5 show that ||T || = 2. This proves Theorem (5.1). ¥

∞ 2 Theorem 5.2 If T is the scaling down map from S(ln ) to S(ln), n ≥ 3, then √ √ n − 1 ≤ ||T || ≤ 2n.

√ Proof. We start by showing that ||T || ≥ n − 1. Ã ! 1 1 Take x = 1, p ,..., p , a(n − 1) a(n − 1) Ã ! −1 −1 and y = 1, p ,..., p , a(n − 1) a(n − 1) where a is a large positive number. This gives 2 d1(x, y) = p . a(n − 1) Ã ! 1 1 1 T x = r 1, p ,..., p , a + 1 a(n − 1) a(n − 1)) a Ã ! 1 −1 −1 T y = r 1, p ,..., p , a + 1 a(n − 1) a(n − 1) a r r 1 a + 1 since ||x|| = ||y|| = 1 + (n − 1)( ) = . 2 2 a(n − 1) a 31

So v u Ã ! u 2 t 1 1 d2(T x, T y) = (n − 1) · p + p (a + 1)(n − 1) (a + 1)(n − 1) r 1 2 = 2 (n − 1) · = √ . (a + 1)(n − 1) a + 1

Thus p d (T x, T y) 2 a(n − 1) 2 = √ · d1(x, y) a + 1 2 r a √ = (n − 1) → n − 1 as a → ∞. a + 1 d (T x, T y) √ So ||T || = sup 2 ≥ n − 1. x,y d1(x, y) √ ∞ Now we need to show that ||T || ≤ 2n. To do this, take two points x and y on S(ln ).

Let x = (a1, a2, . . . , an), − 1 ≤ ai ≤ 1 for i = 1, . . . , n and ai = 1 for some 1 ≤ i ≤ n,

and y = (b1, b2, . . . , bn), − 1 ≤ bi ≤ 1 for i = 1, . . . , n and bi = 1 for some 1 ≤ i ≤ n. Ã ! 1 Ã ! 1 Xn 2 Xn 2 2 2 Here d1(x, y) = max |ai − bi|. Let |ai| = A and |bi| = B, then 1≤i≤n µ i=1 ¶ i=1 ³a a a ´ b b b T x = 1 , 2 ,... n and T y = 1 , 2 ,... n . A A A B B B " µ ¶ # 1 Xn a b 2 2 Thus d (T x, T y) = i − i 2 A B i=1 " µ ¶ # 1 Xn a B − b A 2 2 = i i AB i=1 " µ ¶ # 1 Xn (a − b )B + b (B − A) 2 2 = i i i AB i=1 " µ ¶ # 1 Xn a − b b (B − A) 2 2 = i i + i A AB i=1 " µ ¶# 1 Xn 2(a − b )2 2b2(B − A)2 2 ≤ i i + i , here we used the A2 A2B2 i=1 inequality (a + b)2 ≤ 2a2 + 2b2 for any two real numbers a and b. 32

" Ã ! # 1 Xn 2(a − b )2 Xn (B − A)2 2 = i i + 2 b2 A2 i A2B2 i=1 i=1 " # 1 Xn 2(a − b )2 2(B − A)2 2 = i i + A2 A2 i=1 " # 1 Xn 2(a − b )2 Xn (a − b )2 2 ≤ i i + 2 i i , since A2 A2 i=1 i=1

Ã ! 1 Ã ! 1 2 Xn 2 Xn 2 2 2 2 (B − A) =  (bi) − (ai)  . i=1 i=1 Take the square root of both sides of the equality above, we get

Ã ! 1 Ã ! 1 Xn 2 Xn 2 2 2 B − A = (bi) − (ai) i=1 i=1 = ||y|| − ||x||. Ã ! 1 Xn 2 2 But ||y|| − ||x|| ≤ ||x − y|| = (ai − bi) . i=1 Ã ! 1 Xn 2 2 Thus (B − A) ≤ (ai − bi) . i=1 Xn 2 2 This gives (B − A) ≤ (ai − bi) . i=1 So Ã ! 1 √ Xn (a − b )2 2 d (T x, T y) ≤ 2 i i 2 A2 i=1 Ã ! 1 √ Xn 2 2 ≤ 2 (ai − bi) i=1 µ ¶ 1 √ 2 2 ≤ 2 n. max (ai − bi) 1≤i≤n √ ≤ 2n max |ai − bi|. 1≤i≤n So d (T x, T y) √ 2 ≤ 2n. d1(x, y) 33

√ This gives ||T || ≤ 2n. ¥

∞ 2 Theorem 5.3 If T is the scaling down map from S(ln ) to S(ln), n ≥ 2, then n √ √ ≤ ||T −1|| ≤ 2 n. n − 1 n Proof. We ﬁrst show that ||T −1|| ≥ √ . n − 1 √ d (T x, T y) n − 1 To do this, we show that inf 2 ≤ where x and y are points on the x,y d1(x, y) n √ ∞ d2(T x, T y) n − 1 unit sphere of ln . To show that inf ≤ , we need to ﬁnd points x,y d1(x, y) n √ ∞ d2(T x, T y) n − 1 x and y on the unit sphere of ln satisfying ≤ . d1(x, y) n Take x = (1,..., 1), and y = (1, 1 − δ, . . . , 1 − δ), where δ → 0. Then d1(x, y) = δ. 1 T x = √ (1,..., 1). n 1 T y = p (1, 1 − δ, . . . , 1 − δ). 1 + (n − 1)(1 − δ)2 Ã !2 1 1 Which gives d2(T x, T y) = p − √ 2 1 + (n − 1)(1 − δ)2 n Ã !2 1 1 − δ +(n − 1) √ − p . n 1 + (n − 1)(1 − δ)2 d To see what 2 approaches, we will use the following approximations: d1 p p · 1 + (n − 1)(1 − δ)2 = 1 + (n − 1)(1 − 2δ + δ2) p ≈ 1 + (n − 1)(1 − 2δ) p = n − (2n − 2)δ r √ 2n − 2 = n 1 − δ µ n ¶ √ n − 1 ≈ n 1 − δ . n  2 Ã !2 1 1  1 1  · p − √ ≈  µ ¶ − √  , 1 + (n − 1)(1 − δ)2 n √ n − 1 n n 1 − δ n 34

p this follows from the above approximation for 1 + (n − 1)(1 − δ)2.

Now   2  2  1 1  1  1   µ ¶ − √  =  − 1 √ n − 1 n n n − 1 n 1 − δ 1 − δ n n µ ¶ 1 n − 1 2 ≈ 1 + δ − 1 n n µ ¶ 1 n − 1 2 = δ2. n n  2 Ã !2 1 1 − δ  1 1 − δ  · √ − p ≈ √ − µ ¶ , n 1 + (n − 1)(1 − δ)2  n √ n − 1  n 1 − δ n p again this follows from the above approximation for 1 + (n − 1)(1 − δ)2.

Now   2  2  1 1 − δ  1  1 − δ  √ − µ ¶ = 1 −   n √ n − 1  n n − 1 n 1 − δ 1 − δ n n µ µ ¶¶ 1 n − 1 2 ≈ 1 − (1 − δ) 1 + δ n n µ ¶ 1 n − 1 2 ≈ 1 − 1 − δ + δ n n µ ¶ 1 δ 2 = . n n

This gives µ ¶ 1 (n − 1)2 1 δ 2 d2(T x, T y) ≈ δ2 + (n − 1) · 2 n n2 n n (n − 1)2 n − 1 = δ2 + δ2 n3 n3 µ ¶ (n − 1)2 n − 1 = δ2 + n3 n3 n − 1 = δ2 · . n2 35

Recall that d1(x, y) = δ, which implies that

2 d2(T x, T y) n − 1 2 ≈ 2 . d1(x, y) n √ d (T x, T y) n − 1 So 2 ≈ . d1(x, y) n √ d (T x, T y) n − 1 So inf 2 ≤ . x,y d1(x, y) n n This gives ||T −1|| ≥ √ . n − 1

−1 √ ∞ To show that ||T || ≥ 2 n, take two points x1, x2 on the unit sphere of ln such that ||x1 − x2||∞ = a, where 0 < a < 1. Let T x1 = y1, and T x2 = y2. We know that

∞ √ x1 is on the unit sphere of ln , so ||x1||∞ = 1, which implies that ||x1||2 ≤ n. Now y1 = c1x1 for some constant c1, which implies ||y1||2 = c1||x1||2, we get c1||x1||2 = 1 1 √ since y is on the unit sphere of l2 . Which implies c ≥ √ since ||x || ≤ n. 1 n 1 n 1 2 1 Similarly y = c x with c ≥ √ . Suppose c ≥ c and consider the points z = 2 2 2 2 n 1 2 1 1 1 ||x1 − x2||∞ a y1 and z2 = y2. By Theorem 3.3, ≤ 2. So ||z1 −z2||∞ ≥ since ||x1 − c1 c1 ° ||z1 −°z2||∞ 2 ° ° ° 1 1 ° 1 x2||∞ = a. But ||z1 − z2||∞ = ° y1 − y2° = ||y1 − y2||∞. Thus ||y1 − y2||∞ = c1 c1 ∞ c1 a a ||y1 − y2||2 c1 1 c1||z1 − z2||∞ ≥ c1. So ||y1 − y2||2 ≥ c1. So ≥ ≥ √ since c1 ≥ 2 2 ||x1 − x2||∞ 2 2 n 1 √ √ . This gives ||T −1|| ≤ 2 n. ¥ n

∞ 2 Theorem 5.4 The Banach-Mazur norm of the scaling down map from S(l2 ) to S(l2) √ is greater than or equal to 2 2 and less than or equal to 4.

∞ 2 Proof. Let T be the scaling down map from S(l2 ) to S(l2). √ In Theorem5.1, we showed that ||T || = 2. √ In Theorem5.3, we showed that 2 ≤ ||T −1|| ≤ 2 2. √ So 2 2 ≤ kT kkT −1k ≤ 4. ¥ 36

∞ Theorem 5.5 The Banach-Mazur norm of the scaling down map from S(ln ) to √ 2 S(ln), n ≥ 3 is greater than or equal to n and less than or equal to 2 2n.

∞ 2 Proof. Let T be the scaling down map from S(ln ) to S(ln), n ≥ 3. √ In Theorem5.2, we showed that n − 1 ≤ ||T || ≤ 2n. n √ In Theorem5.3, we showed that √ ≤ ||T −1|| ≤ 2 n. √ n − 1 So n ≤ kT kkT −1k ≤ 2 2n. ¥ CHAPTER 6

SCALING DOWN MAPS BETWEEN S(l∞) AND S(lq) as q → ∞

In this chapter, we do a similar study as in chapters four and ﬁve, but comparing

∞ q instead ln and ln as q → ∞. We start with some deﬁnitions. Let (X, d1), (Y, d2) be metric spaces and let M ⊆ X,N ⊆ Y .

Deﬁnition. Given x0 ∈ X. Deﬁne

B²(x0) = {x ∈ X : d1(x0, x) < ²} .

Deﬁnition. A map T : M −→ N is called locally Lipschitz at x0 if T is Lipschitz on

B²(x0) for some ² > 0.

Deﬁnition. Given a locally Lipschitz map T : M −→ N at x0. Deﬁne,

d2(T x, T y) ||T ||²,x0 = sup . x,y∈B²(x0) d1(x, y)

Deﬁnition. The local Lipschitz norm at xo of a locally Lipschitz map T : M −→ N

at x0, denoted by ||T ||loc x0 , is

||T ||loc x = lim ||T ||²,x . 0 ²→0 0

Deﬁnition. If T : M −→ N is one to one and onto and locally Lipschitz at x0, and

−1 if T is also locally Lipschitz at T x0, we deﬁne the local Banach-Mazur norm of T

37 38

at x0 as

−1 ||T ||loc x0 ||T ||loc T x0 .

∞ q Theorem 6.1 If T is the scaling down map from S(ln ) −→ S(ln), n ≥ 2, then

lim kT kkT −1k ≥ 2. q→∞

Proof. We ﬁrst prove the theorem in dimension 2 in order to introduce the technique that will be used and then move to higher dimensions.

∞ q Let T : S(l2 ) −→ S(l2) be the scaling down map. We show that:

1. lim ||T || ≥ 1. q→∞

2. lim ||T −1|| ≥ 2. q→∞

We will start with 1.

Take x = (1, 0) and y = (0, 1). Then d1(x, y) = 1, and d2(T x, T y) = d2(x, y) =

1 1 d2(T x, T y) 1 1 ((1 − 0)q + (1 − 0)q) q = 2 q . So = 2 q . So ||T || ≥ 2 q . This gives lim ||T || ≥ d1(x, y) q→∞ 1.

a 2. Take x = (1, 1) and y = (1, 1 − δ), where δ = , 0 < a < 1. q

Then d1(x, y) = δ. T x = (1 − t)(1, 1), 0 < t < 1 such that (1 − t)q + (1 − t)q = 1. µ ¶ 1 1 q This gives 1 − t = . 2 Ãµ ¶ 1 µ ¶ 1 ! 1 q 1 q So T x = , . 2 2

Similarly T y = (1 − s)(1, 1 − δ), 0 < s < 1 such that

(1 − s)q + (1 − s)q(1, 1 − δ)q = 1. 39

µ ¶ 1 1 q This gives 1 − s = . 1 + (1 − δ)q

Ãµ ¶ 1 µ ¶ 1 ! 1 q (1 − δ)q q So T y = , . 1 + (1 − δ)q 1 + (1 − δ)q

"µ ¶ 1 µ ¶ 1 #q "µ ¶ 1 µ ¶ 1 #q 1 q 1 q 1 q (1 − δ)q q Thus dq(T x, T y) = − + − . 2 1 + (1 − δ)q 2 2 1 + (1 − δ)q

d To see what 2 approaches as q → ∞, we will use few approximations. d 1 a As noted earlier, δ = , 0 < a < 1. q

So q   1 q   µ ¶q  1  q a q  µ ¶ 1  µ ¶ 1 1 −   1  1 q   1 q  q   dq(T x, T y) =  µ ¶  −  +  −  µ ¶   . 2  a q  2   2  a q   1 + 1 − 1 + 1 − q q

d We are interested in lim 2 . q→∞ d1 Note that: 1 1 ea · lim µ ¶q = = . q→∞ a 1 ea + 1 1 + 1 − 1 + q ea µ ¶ a q 1 − q 1 ea 1 · lim µ ¶q = · = . q→∞ a ea ea + 1 ea + 1 1 + 1 − q "µ ¶ 1 µ ¶ 1 #q "µ ¶ 1 µ ¶ 1 #q ea q 1 q 1 q 1 q So dq ≈ − + − 2 ea + 1 2 2 ea + 1 · ¸ · ¸ 1 ea 1 1 q 1 1 1 1 q ≈ ln − ln + ln − ln . q ea + 1 q 2 q 2 q ea + 1

Where we used the fact that ln(1 + b) ≈ b, in other words ln(b) ≈ b − 1 for small b. 40

So µ ¶ Ã · ¸ · ¸ ! 1 q ea 1 q 1 1 q dq ≈ ln − ln + ln − ln 2 q ea + 1 2 2 ea + 1 µ ¶ Ã · ¸ · ¸ ! 1 q 2ea q ea + 1 q = ln + ln . q ea + 1 2 Ã · ¸ · ¸ ! 1 1 2ea q ea + 1 q q This gives d ≈ ln + ln . 2 q ea + 1 2

Now we use the fact that

1 q q q lim (a1 + a2) = max{a1, a2}. q→∞

We need to ﬁnd a such that:

2ea ea + 1 ln ≈ ln . ea + 1 2 2ea ea + 1 To do that we need to solve = for a. ea + 1 2 2ea ea + 1 = . ea + 1 2 ⇐⇒ 4ea = (ea + 1)2.

⇐⇒ 4ea = e2a + 2ea + 1.

⇐⇒ e2a − 2ea + 1 = 0.

⇐⇒ (ea − 1)2 = 0.

⇐⇒ a = 0. 2ea ea + 1 So ln ≈ ln when a is small. ea + 1 2 1 ea + 1 So for small a, d ≈ ln . 2 q 2 ea + 1 a + 2 ea + 1 Note that ln ≈ ln , here we used the fact that ea ≈ a + 1. So, ln ≈ 2 2 2 a + 2 ³a ´ a ln = ln + 1 ≈ , using the fact that ln(b + 1) ≈ b for small b. 2 2 2 1 a So d ≈ · . 2 q 2 41

d2 2q 1 Thus ≈ a = . d1 2 q This gives lim ||T −1|| ≥ 2. q→∞

Thus lim kT kkT −1k ≥ 2. q→∞ This proves the theorem for dimension 2.

Next, we will prove the theorem in dimension n.

∞ q Let T : S(ln ) −→ S(ln) be the scaling down map. We will show that:

1. lim ||T || ≥ 1. q→∞

2. lim ||T −1|| ≥ 2. q→∞

We will start with 1.

Take x = (1, 0,..., 0) and y = (0,..., 0, 1). Then d1(x, y) = 1 and d2(T x, T y) =

1 1 d2(x, y) = 2 q . Thus ||T || ≥ 2 q . This gives lim ||T || ≥ 1. q→∞

2. Take x = (1,..., 1) and y = (1, 1 − δ, 1 − 2δ, . . . , 1 − (n − 1)δ) where a δ = , 0 < a < 1. q

Then d1(x, y) = (n − 1)δ. T x = (1 − t)(1,..., 1), 0 < t < 1 such that (1 − t)q + ··· + (1 − t)q = 1. µ ¶ 1 1 q This gives 1 − t = . n

Ãµ ¶ 1 µ ¶ 1 ! 1 q 1 q So T x = ,..., . n n

Similarly T y = (1 − s)(1, 1 − δ, . . . , 1 − (n − 1)δ), 0 < s < 1 such that

(1 − s)q + (1 − s)q(1 − δ)q + ··· + (1 − s)q(1 − (n − 1)δ)q = 1. · ¸ 1 1 q This gives 1 − s = . 1 + (1 − δ)q + ··· + (1 − (n − 1)δ)q 42

Let A = 1 + (1 − δ)q + ··· + (1 − (n − 1)δ)q.

Ãµ ¶ 1 µ ¶ 1 µ ¶ 1 ! 1 q (1 − δ)q q (1 − (n − 1)δ q Then T y = , ,..., . A A A

We need to ﬁnd d2(T x, T y).

"µ ¶ 1 µ ¶ 1 #q "µ ¶ 1 µ ¶ 1 #q 1 q 1 q 1 q (1 − δ)q q dq(T x, T y) = − + − + ··· 2 A n n A

"µ ¶ 1 µ ¶ 1 #q 1 q (1 − (n − 1)δ q + − . n A

d To see what 2 approaches as q → ∞, we use few approximations. As noted earlier, d a 1 δ = , 0 < a < 1. We get: q µ ¶ a q 1 · lim (1 − δ)q = lim 1 − = . q→∞ q→∞ q ea µ ¶ a q 1 · lim (1 − (n − 1)δ)q = lim 1 − (n − 1) = . q→∞ q→∞ q e(n−1)a

1 1 e(n−1)a · lim = = . q→∞ A 1 1 1 + ea + ··· + e(n−1)a 1 + + ··· + ea e(n−1)a "µ ¶ 1 µ ¶ 1 #q e(n−1)a q 1 q So dq ≈ − 2 1 + ea + ··· + e(n−1)a n

"µ ¶ 1 µ ¶ 1 #q 1 q e(n−2)a q + − + ··· n 1 + ea + ··· + e(n−1)a

"µ ¶ 1 µ ¶ 1 #q 1 q 1 q + − . n 1 + ea + ··· + e(n−1)a

Let B = 1 + ea + ··· + e(n−1)a.

"µ ¶ 1 µ ¶ 1 #q "µ ¶ 1 µ ¶ 1 #q e(n−1)a q 1 q 1 q e(n−2)a q So dq ≈ − + − + ··· 2 B n n B

"µ ¶ 1 µ ¶ 1 #q 1 q 1 q + − . n B 43

· ¸ · ¸ 1 e(n−1)a 1 1 q 1 1 1 e(n−1)a q So dq ≈ ln − ln + ln − ln + ··· 2 q B q n q n q B · ¸ 1 1 1 1 q + ln − ln . q n q B

Where we used the fact that ln(b) ≈ b − 1 for small b.

µ ¶ Ã · ¸ · ¸ · ¸ ! 1 1 q ne(n−1)a q B q B q q So d ≈ ln + ln + ··· + ln . 2 q B ne(n−2)a n

d To ﬁnd lim 2 , we use the fact that q→∞ d1

1 q q q q lim (a1 + a2 + ··· + an) = max {ai}. q→∞ 1≤i≤n

All the terms inside the parenthesis of the expression for d2 are approximately equal when a is small.

So 1 B d ≈ ln 2 q n 1 1 + ea + ··· + e(n−1)a = ln q n 1 1 + 1 + a + ··· + 1 + (n − 1)a ≈ ln . q n

Where we used the fact that eb ≈ 1 + b for small b.

So (n − 1)na 1 n + d ≈ ln 2 2 q n µ ¶ 1 n − 1 = ln 1 + a q 2 1 n − 1 ≈ a, using the fact that ln(1 + b) ≈ b for small b. q 2 44

d 1 (n − 1)a q 1 Finally 2 ≈ · = . d1 q 2 (n − 1)a 2 This gives lim ||T −1|| ≥ 2. q→∞ Therefore lim kT kkT −1k ≥ 2. q→∞ ¥

∞ Deﬁnition. A point x ∈ S(ln ) is a smooth point if one coordinate of x is 1 and the other coordinates have absolute value < 1.

In Theorem 6.2, we will prove that if we can consider points that are smooth on

∞ the unit sphere of ln , then locally around those points we will have

−1 lim kT klockT kloc = 1. q→∞

∞ q Theorem 6.2 If T : S(ln ) −→ S(ln) is the scaling down map, then for every smooth

∞ point x in S(ln ), we have

−1 lim kT kloc xkT kloc T x = 1. q→∞

Proof. We ﬁrst prove the theorem in dimension 2.

∞ q Let T : S(l2 ) −→ S(l2) be the scaling down map. Take x = (1, 1 − δ) and y = (1, 1 − δ − γ) where 0 < δ < 1 is ﬁxed and γ → 0 as

−1 q → ∞. We show that lim kT kloc xkT kloc T x = 1. q→∞

For the points x, y, we have d1(x, y) = γ. And T x = (1 − t)(1, 1 − δ), 0 < t < 1 such that (1 − t)q + (1 − t)q(1 − δ)q = 1. µ ¶ 1 1 q This gives 1 − t = . 1 + (1 − δ)q Ãµ ¶ 1 µ ¶ 1 ! 1 q (1 − δ)q q Thus T x = , . 1 + (1 − δ)q 1 + (1 − δ)q 45

Similarly T y = (1 − s)(1, 1 − δ − γ), 0 < s < 1 such that

(1 − s)q + (1 − s)q(1 − δ − γ)q = 1. µ ¶ 1 1 q This gives 1 − s = . 1 + (1 − δ − γ)q Ãµ ¶ 1 µ ¶ 1 ! 1 q (1 − δ − γ)q q Thus T y = , . 1 + (1 − δ − γ)q 1 + (1 − δ − γ)q

d We need to ﬁnd an estimate for 2 as q → ∞. d1 We have d1 = γ, we need to ﬁnd d2.

"µ ¶ 1 µ ¶ 1 #q 1 q 1 q dq(T x, T y) = − 2 1 + (1 − δ)q 1 + (1 − δ − γ)q

"µ ¶ 1 µ ¶ 1 #q (1 − δ)q q (1 − δ − γ)q q + − . 1 + (1 − δ)q 1 + (1 − δ − γ)q

d To see what 2 approaches as q → ∞, we make some observations. d1 Note that (1 − δ)q → 0 as q → ∞ since 0 < δ < 1. Similarly (1 − δ − γ)q → 0 as µ ¶ 1 µ ¶ 1 1 q 1 q q → ∞. So −→ 1 as q → ∞. And −→ 1 as 1 + (1 − δ)q 1 + (1 − δ − γ)q "µ ¶ 1 µ ¶ 1 #q 1 q 1 q q → ∞. Hence − −→ 0 as q → ∞. 1 + (1 − δ)q 1 + (1 − δ − γ)q

q This gives an estimate for the ﬁrst term in d2. We need to ﬁnd an estimate for the

"µ ¶ 1 µ ¶ 1 #q (1 − δ)q q (1 − δ − γ)q q term − . 1 + (1 − δ)q 1 + (1 − δ − γ)q

µ ¶ 1 µ ¶ 1 (1 − δ)q q 1 q Note that = (1 − δ) −→ 1 − δ as q −→ ∞. 1 + (1 − δ)q 1 + (1 − δ)q

µ ¶ 1 µ ¶ 1 (1 − δ − γ)q q 1 q Similarly = (1 − δ − γ) 1 + (1 − δ − γ)q 1 + (1 − δ − γ)q

−→ 1 − δ − γ as q → ∞.

q q q Finally, we have d2 −→ [1 − δ − (1 − δ − γ)] = γ as q → ∞. 46

So d2(T x, T y) = γ.

d (T x, T y) γ So 2 ≈ = 1. d1(x, y) γ

−1 Finally lim ||T ||loc x||T ||loc T x = 1. q→∞

This proves the theorem for dimension 2.

Next, we will prove the theorem in dimension n.

∞ q Let T : S(ln ) −→ S(ln) be the scaling down map.

Take x = (1, 1 − δ1,..., 1 − δn−1), and y = (1, 1 − δ1 − γ1,..., 1 − δn−1 − γn−1), where 0 < δi < 1, 1 ≤ i ≤ n − 1 are ﬁxed and γi → 0 as q → ∞ for 1 ≤ i ≤ n − 1. For the points x, y, we have, d1(x, y) = max {γi}. 1≤i≤n−1

To ﬁnd d2, we need to ﬁnd T x and T y.

T x = (1 − t)(1, 1 − δ1,..., 1 − δn−1), 0 < t < 1 such that

q q q q q (1 − t) + (1 − t) (1 − δ1) + ··· + (1 − t) (1 − δn−1) = 1. µ ¶ 1 1 q This gives 1 − t = q q . 1 + (1 − δ1) + ··· + (1 − δn−1)

q q Put K = 1 + (1 − δ1) + ··· + (1 − δn−1) Ãµ ¶ 1 µ ¶ 1 µ ¶ 1 ! 1 q (1 − δ )q q (1 − δ )q q Thus T x = , 1 ,..., n−1 . K K K

Similarly T y = (1 − s)(1, 1 − δ1 − γ1,..., 1 − δn−1 − γn−1), 0 < s < 1 such that q q q q q (1 − s) + (1 − s) (1 − δ1 − γ1) + ··· + (1 − s) (1 − δn−1 − γn−1) = 1. µ ¶ 1 1 q This gives 1 − s = q q . 1 + (1 − δ1 − γ1) + ··· + (1 − δn−1 − γn−1)

q q Put M = 1 + (1 − δ1 − γ1) + ··· + (1 − δn−1 − γn−1) . 47

Ã ! µ ¶q µ ¶ 1 µ ¶ 1 1 (1 − δ − γ )q q (1 − δ − γ )q q Thus T y = , 1 1 ,..., n−1 1 . M M M

d2 We need to ﬁnd an estimate for as q → ∞. We have d1 = max {γi}, we need to d1 1≤i≤n ﬁnd d2.

"µ ¶ 1 µ ¶ 1 #q "µ ¶ 1 µ ¶ 1 #q 1 q 1 q (1 − δ )q q 1 − δ − γ q dq(T x, T y) = − + 1 − 1 1 + ··· 2 M K K M

"µ ¶ 1 µ ¶ 1 #q (1 − δ )q q (1 − δ − γ )q q + n−1 − n−1 n−1 . K M

d To see what 2 approaches as q → ∞, we make some observations: d1 q q 1 · lim (1 + (1 − δ1) + ··· + (1 − δn−1) ) q = 1 since 0 < δi < 1 for all 1 ≤ i ≤ n − 1. q→∞ q q 1 · lim (1 + (1 − δ1 − γ1) + ··· + (1 − δn−1 − γn−1) ) q = 1 since q→∞

0 < δi, γi < 1 for 1 ≤ i ≤ n − 1. µ ¶ 1 µ ¶ 1 q q q (1 − δi) 1 · lim = lim (1 − δi) lim = 1 − δi, 1 ≤ i ≤ n − 1. q→∞ K q→∞ q→∞ K µ ¶ 1 µ ¶ 1 q q q (1 − δi − γi) 1 · lim = lim (1 − δi − γi) lim = 1 − δi − γi, 1 ≤ i ≤ n − 1. q→∞ M q→∞ q→∞ M From these observations, we get that

q q q d2 ≈ γ1 + ··· + γn−1 as q → ∞. £ ¤ 1 q q q So d2 ≈ γ1 + ··· + γn−1 . Which implies d2 −→ max {γi} as q → ∞. 1≤i≤n−1

d (T x, T y) So 2 ≈ 1. d1(x, y)

−1 Finally lim ||T ||loc x||T ||loc T x = 1. ¥ q→∞ CHAPTER 7

CONCLUSION AND FUTURE PROJECTS

The investigations and results in this thesis provide an opening for further research, which we will brieﬂy discuss here.

For the projections onto the unit ball investigated in Chapter three, it is natural to ask whether the assumption that outside points be projected onto the unit sphere can be removed. The following example in the Euclidean plane shows that projections which do not map outside points to the boundary can have as small Lipschitz norm as those which do.

Example: For each α, 0 ≤ α ≤ 1, consider the following projection onto the lower

2 half-plane in R . For y ≥ 0 , Pα(x, y) = (x, −αy). For y < 0, Pα(x, y) = (x, y). It is easy to see that ||Pα|| = 1, but only for α = 0 does Pα map outside points to the boundary.

Theorem 3.6 estimates the norm of the scaling down projection onto the unit ball for 1 < p < 2. It is natural to conjecture that for 1 < p ≤ 2, the scaling down projection is the projection onto the unit ball of Lp with minimal Lipschitz norm. However, the proof of Theorem 3.4 which gives the result for p = 1 does not generalize in any simple way. For 2 < p < ∞ we do not have a conjecture about what is the projection onto the unit ball in Lp with minimal Lipschitz norm. For p = ∞, it is not the scaling down projection. Since L1(0, 1) is a subspace of L∞ ( and l∞ ), theorem 3.4 gives that the scaling down projection in L∞ ( and l∞ ) has norm 2, but the truncation projection has norm 1.

In Chapters four, ﬁve, and six, the Banach-Mazur norm of the scaling down map

48 49

∞ 1 between the unit spheres of ln and ln is estimated to an exact order of magnitude. The problem about the order of magnitude of the Banach-Mazur distance between these unit spheres remains open. For the linear case it is easy to see that the Banach-

1 ∞ Mazur norm of the identity map between ln and ln is exactly n. But it is well known

1 ∞ √ that the Banach-Mazur distance between ln and ln is of the order of magnitude n. In this dissertation we have limited the investigations to the most classical Banach spaces. It should be an interesting task to study the corresponding problems for other

Banach spaces as well as for other topological linear spaces like the Lp-spaces where

0 < p < 1. BIBLIOGRAPHY

[1] Beauzamy, B., Maurey, B. Points minimaux et ensembles optimaux dans les espaces de Banach. J. Functional Analysis 24 (1977), no. 2, 107-139. [2] Lindenstrauss, J., Szankowski, A. On the Banach-Mazur distance between spaces having an unconditional basis. Aspects of positivity in functional analysis. J. (Tubingen, 1985), 119-136, North-Holland Math. Stud., 122, North-Holland, Am- sterdam, 1986. [3] Tomczak-Jaegermann, N. The Banach-Mazur distance between symmetric spaces. Israel Journal of Mathematics, v. 46 issue 1-2, 1983, p. 40-66. [4] Sanchez Perez, E. A. An estimate of the Banach-Mazur distances between Hilbert spaces and Banach spaces. Rendiconti del Circolo Matematico di Palermo Serie II, v. 46 issue 3, 1997, p. 465-476. [5] Giannopoulos, A. A. A note on the Banach-Mazur distance to the cube.(English summary) Geometric aspects of functional analysis. (Israel, 1992–1994), 67–73, Oper. Theory Adv. Appl., 77, Birkhuser, Basel, 1995. [6] Lovblom, G. Uniform homeomorphisms between the unit balls in Lp and lp. Proceedings of the American Mathematical Society, v. 123 issue 2, 1995, p. 405- 409. [7] Lovblom, G. Uniform homeomorphisms between unit balls in Lp-spaces. Math- ematica Scandinavica, v. 62 issue 2, 1988, p. 294-302. [8] Larsson, J. Sets of minimal points in Lp(0, 1). Mathematica Scandinavica, v. 63 issue 1, 1988, p. 151-168. [9] Benyamini, Y., Lindenstrauss, J. Geometric nonlinear functional analysis. Vol. 1. (English summary) American Mathematical Society Colloquium Publications, 48. American Mathematical Society, Providence, RI, 2000. xii+488 pp. ISBN: 0-8218-0835-4. [10] Benyamini, Y. Spheres in inﬁnite-dimensional normed spaces are Lipschitz con- tractible. Proceedings of the American Mathematical Society, v. 88 issue 3, 1983, p. 439-445. [11] Benyamini, Y. Nonexpansive selections of metric projections in spaces of continuous functions. Journal of Approximation Theory, v. 137 issue 2, 2005, p. 187-200.

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