Extreme points of function spaces
Jolien Janneke Kerssens
summer 2015
Bachelor thesis
Supervisor: prof. Jan Wiegerinck
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Korteweg-de Vries Institute for Mathematics Faculty of Science
Abstract
In this thesis we study extreme points, exposed points and strongly exposed points of the unit ball Σ = {x ∈ X | kxk ≤ 1} of several Banach spaces. First we calculate the extreme points of the unit ball in Lp(X). We introduce inner and outer functions in
order to describe the set of extreme points of the unit ball in Hardy spaces Hp. We will
prove the Rudin–de Leeuw theorem, which states that a function f ∈ H1 is an extreme point of Σ if and only if f is an outer function of norm 1. Finally we will examine general Hilbert spaces and the space C(X) of continuous functions on a compact Hausdorff space X.
Title: Extreme points of function spaces
Author: Jolien Janneke Kerssens, student number 10350462 Supervisor: prof. Jan Wiegerinck
Second grader: prof. Jan van Mill Date: summer 2015
Korteweg-de Vries Institute for Mathematics University of Amsterdam
Science Park 904, 1098 XH Amsterdam http://www.science.uva.nl/math
Contents
1 Introduction 4
2 Functions, Poisson kernels and factorizations 5
2.1 Notation . . . 5
2.2 Introduction to Lp and Hp . . . 5
2.3 Poisson kernels . . . 6
2.4 Extending functions to the closed unit disc . . . 10
2.5 Inner and outer functions . . . 11
3 Extreme and exposed points 14 3.1 The unit ball in Lp(X) . . . . 16
3.2 The unit ball in Hardy spaces . . . 19
3.3 The unit ball in Hilbert spaces . . . 22
3.4 The unit ball in C(X) . . . 23
4 Discussion 27
5 Populaire samenvatting 28
1 Introduction
In this thesis we will take a look at the so-called extreme points of several Banach spaces. The extreme points of a subset A ⊆ X are the points which do not lie on a line segment between two points in A. We may formally define this as follows:
Definition 1.1. Let X be a Banach space and let A ⊆ X. The extreme points of A are the elements in x ∈ A for which there do not exist distinct y1, y2 ∈ A and λ ∈ (0, 1)
such that
x = λy1+ (1 − λ)y2.
Let’s first determine the extreme points of the running track A depicted in Figure 1.1.
Figure 1.1: A running track.
Every point in the interior of A cannot be an extreme point, since for such x there exists > 0 such that B(x, ) ⊂ A and therefore x clearly lies on a line segment between two points in B(x, ) ⊂ A. Hence we only have to take a look at the boundary points of our running track. We see that the points on the two arcs are extreme points and the points on the interval are not, except for the points at the ends of the intervals.
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Figure 1.2: The extreme points of a running track.
It is relatively easy to determine the extreme points of a convex set A ⊆ Rn. However,
less is known about extreme points of more complicated spaces. In this thesis we will explore the properties of extreme points of several spaces, including Lp, Hp and Hilbert spaces.
2 Functions, Poisson kernels and
factorizations
In this chapter we will familiarize the reader with the concepts that are needed to determine the extreme points of some function spaces.
2.1 Notation
Throughout this thesis we will use the following notation: • Let D = {z ∈ C | kzk < 1}.
• Let C be the unit circle ∂D.
• Given a normed space X, we will use Σ to denote the closed unit ball in X.
2.2 Introduction to L
pand H
pLet (X, A, µ) be a measure space. For functions f : X → R we define the norms
kf kp = p s Z X |f |pdµ and kf k ∞= sup x∈X |f (x)|
to define Banach spaces of functions.
Definition 2.1. For 0 ≤ p ≤ ∞ we define Lp(X) as
Lp(X) = {f : X → R : f is measurable and kf kp < ∞} / ∼
where two functions f and g are equivalent if µ ({f 6= g}) = 0.
It turns out that L∞(X) and for p ≥ 1 the space Lp(X) are Banach spaces. The
following theorem is a well-known result in functional analysis.
Theorem 2.2. Let (X, A, µ) be a measure space and let 1 < p < ∞. The dual of Lp(X)
is isometrically isomorphic to Lq(X) where q is the unique solution to 1 p +
1 q = 1.
The norms on Lp(X) and L∞(X) can be used to create norms on analytic functions from the open unit disc D to C.
Definition 2.3. For 0 ≤ p ≤ ∞ we define the Hardy space Hp as the class of analytic functions on D such that for every 0 < r < 1 the p-norm in Lp([0, 2π]) of f
r(θ) := f (reiθ)
is bounded as r → 1. Together with the norm
kf k = sup
r∈(0,1)
kfrkp
the space Hp forms a Banach space. For f ∈ H∞ we define
kf k = sup
r∈(0,1)
kfrk∞= sup z∈D
|f (z)|.
This norm turns H∞ into a Banach space.
2.3 Poisson kernels
We will examine analytic functions on the open disc D. In this paragraph we will examine how to recapture such a function from its boundary values.
It is well-known that if f is analytic on a neighbourhood of D then the function is completely determined by its boundary values. Namely, for z ∈ D we have
f (z) = 1 2πi Z C f (ξ) ξ − zdξ.
However, this requires that f is analytic on a neighbourhood of D, which is not always the case. We will describe a way to reconstruct a harmonic function u from its boundary values. Suppose u is a real-valued harmonic function on an open neighbourhood E of D. Then there exists an analytic function f (z) =P anzn such that for all z ∈ E we have
u(z) = f (z) + f (z). Using the power series of f we obtain
u(z) = ∞ X n=0 anzn+ ∞ X n=0 an zn = 2 Re a0 + ∞ X n=1 anzn+ ∞ X n=1 an zn
so on the the circle with radius r this means
ur(θ) := u(reiθ) = ∞ X n=−∞ cnr|n|einθ with cn= a−n if n < 0 2 Re a0 if n = 0 an if n > 0 .
Note u1(θ) =
P∞
−∞cneinθ so if we know the boundary values of u we can calculate its
Fourier coefficients cnand therefore recapture u from it using ur(θ) =
P∞
n=−∞cnr |n|einθ.
Definition 2.4. The family of functions
Pr(θ) = ∞
X
n=−∞
r|n|einθ
is called Poisson’s kernel.
Since we will use this family of functions extensively, it is useful to have a few equiv-alent formulas.
Lemma 2.5. We can write the elements of Poisson’s kernel as
Pr(θ) = 1 − r2 1 − 2r cos θ + r2 = Re 1 + reiθ 1 − reiθ .
Proof. Using the geometric series we see that
Pr(θ) = ∞ X n=−∞ r|n|einθ = ∞ X n=0 reiθn+ ∞ X n=0 re−iθn− 1 = 1 1 − reiθ + 1 1 − re−iθ − 1 = 1 − r2 1 + r2− r (eiθ+ e−iθ) = 1 − r 2 1 − 2r cos(θ) + r2.
For the second equality note that
Re 1 + re iθ 1 − reiθ = Re " 1 + reiθ 1 − re−iθ (1 − reiθ) (1 − re−iθ) # = Re1 − re −iθ + reiθ− r2 1 + r2− r (eiθ+ e−iθ) = 1 − r2 1 − 2r cos(θ) + r2.
We need some more results from Fourier theory.
Definition 2.6. We define the convolution f ∗ g of two functions f, g ∈ L2(C) as
(f ∗ g)(x) = 1 2π
Z π
−π
f (x − t)g(t) dt.
Lemma 2.7. Let f, g ∈ L2(C). The nth Fourier coefficient of f ∗ g is equal to the product of the nth coefficients of f and g.
Proof. First we use a2+ b2 ≥ 2ab to see that Z π −π Z π −π e−inx |f (x − t)| |g(t)| dt dx ≤ 1 2 Z π −π Z π −π |f (x − t)|2+ |g(t)|2 dt dx = 1 2 Z π −π Z π −π kf k2 2+ kgk 2 2 dt dx < ∞.
This implies we can use Fubini to interchange the order of integration. Doing so, we obtain 1 2π Z π −π e−inx(f ∗ g)(x) dx = 1 2π Z π −π e−inx· 1 2π Z π −π f (x − t)g(t) dt dx = 1 2π Z π −π g(t) · 1 2π Z π −π e−inxf (x − t) dx dt ∗ = 1 2π Z π −π g(t) · 1 2π Z π −π e−in(t+y)f (y) dy dt = 1 2π Z π −π g(t)e−intdt · 1 2π Z π −π
e−inyf (y) dy,
where in (∗) we used a substitution y = x − t. This shows that indeed the nth Fourier coefficient is the product of the coefficients of f and g.
An important consequence is the following.
Corollary 2.8. For a real harmonic function in the closed disc we have ur(θ) = 1 2π Z π −π u1(t)Pr(θ − t) dt. Proof. We have u1(θ) = ∞ X n=−∞ cneinθ and Pr(θ) = ∞ X n=−∞ r|n|einθ
so the convolution has Fourier coefficients cnr|n| which are exactly those of ur(θ).
We found a way to recover a real harmonic function u from its boundary values. By combining the real and imaginary parts of a harmonic function this also holds for complex-valued harmonic function in the closed disc. Therefore Corollary 2.8 also holds for any analytic function in the closed disc. We can use this Corollary to prove a statement about continuous functions on the boundary of the disc.
Suppose we have a continuous function f : C → C and we want to construct a function u : D → C such that u is harmonic on D and equals f on C. The following theorem states that u(reiθ) = 2π1 R02πPr(θ − ϕ)f (eiϕ) dϕ fulfills our conditions. Note that u is
Theorem 2.9. Let f : C → C be continuous. Then for any θ ∈ [0, 2π] we have lim r→1 1 2π Z 2π 0 Pr(θ − ϕ)f (eiϕ) dϕ = f (eiθ).
Proof. First note that
1 2π
Z 2π
0
Pr(θ − ϕ) dϕ = 1
for all r ∈ (0, 1) since 1 is a real harmonic function in the closed disc. Using this we can write 1 2π Z 2π 0 Pr(θ − ϕ)f (eiϕ) dϕ − f (eiθ) = 1 2π Z 2π 0 Pr(θ − ϕ)f (eiϕ) − f (eiθ) dϕ.
Let > 0 be given. Since f is continuous there exists δ > 0 such that |f (eiϕ)−f (eiθ)| < 2
for |ϕ − θ| < δ, where we identify an angle of 0 with an angle of 2π. First note that for r ∈ [0, 1] both 1 − r2 ≥ 0 and 1 − 2r cos(θ − ϕ) + r2 ≥ 0. Consequently P
r(θ − ϕ) ≥ 0
for all r. We now see that 1 2π Z (θ−δ,θ+δ) Pr(θ − ϕ) f (eiϕ) − f (eiθ) dϕ < 1 2π · 1 · 2 < 2. Let us consider 1 2π Z [0,2π]\(θ−δ,θ+δ) Pr(θ − ϕ) f (eiϕ) − f (eiθ) dϕ,
the second part of the integral. The function f is continuous on [0, 2π] and therefore |f | is bounded by some M ∈ R. We have
lim r→1Pr(θ − ϕ) = limr→1 1 − r2 1 − 2r cos(θ − ϕ) + r2 ≤ lim r→1 1 − r2 1 − 2r cos(δ) + r2 = 0
since cos(δ) < 1, so there exists η2 > 0 such that |1 − r| < η2 implies Pr(θ − ϕ) < M .
This gives 1 2π Z [0,2π]\(θ−δ,θ+δ) Pr(θ − ϕ) f (eiϕ) − f (eiθ) dϕ < 1 2π · M · 2M < 2. Now for |1 − r| < min{η1, η2} we have
1 2π Z 2π 0 Pr(θ − ϕ) f (eiϕ) − f (eiθ) dϕ < as wanted.
2.4 Extending functions to the closed unit disc
However, we will also study functions f : D → C that only exist on the open disc. We will take a look at the boundary behaviour of these functions. That is, we want to know if limr→1f (reiθ) converges. The following theorem of Fatou makes sure that functions
that can be expressed as Poisson kernels of Borel measures can be extended onto the boundary of the unit disc. To state this theorem we need the notion of “differentiability of measures”.
Definition 2.10. Let µ be a measure on Rn. Let λ denote the Lebesgue measure. The
symmetric derivative of µ at x ∈ Rn is defined as
dµ
dλ(x) = limr→0
µ(B(x, r)) λ(B(x, r)), provided this limit exists.
Theorem 2.11 (Fatou). Let µ be a finite complex Borel measure on C and let g be the harmonic function on D defined by
g(r, θ) = Z
C
Pr(θ − t) dµ(t).
Let dθ denote Lebesgue measure on C. Then
lim g(r, θ) = 2π dµ dθ
(θ0)
as the point z = reiθ approaches eiθ0 along any path in the open disc which is not
tangential to the unit circle.
Proof. For a proof we direct the reader to Hoffman [1].
The following theorem is a direct consequence of Fatou’s theorem. Corollary 2.12. The radial limits
lim r→1g(r, θ0) = limr→1 Z C Pr(θ − t) dµ(t) = 2π dµ dθ (θ0) exist.
Recall that we wanted to examine if limr→1f (reiθ) exists for functions f : D → C.
The following theorem is the key to the answer for H1.
Theorem 2.13. For every f ∈ H1 we have
f (reiθ) = 1 2π
Z 2π
0
Proof. For a proof we direct the reader to Theorem 11.5.2 from Wiegerinck [2]. Corollary 2.14. Let f ∈ H1. Then the radial limits
lim
r→1f (re iθ)
exist almost everywhere.
Hence we can extend elements f : D → C in H1 to the boundary C of D. Note that for every p ∈ [1, ∞] we have Hp ⊆ H1 so we can extend these functions too. From now
on we will do so implicitly.
2.5 Inner and outer functions
To examine the elements of the Hardy space H1 we will factorize the elements into two types of functions: the so-called inner and outer functions.
Definition 2.15. An inner function is an analytic function I : D → C such that |I(z)| ≤ 1 on D and |I(eiθ)| = 1 almost everywhere on C.
Definition 2.16. Let D be a domain. An upper semi-continuous function f : D → R ∪ {−∞} is a function such that for every z ∈ D we have
lim sup
w→z
f (w) ≤ f (z).
A function f : D → R ∪ {−∞} is called subharmonic if it is upper semi-continuous and satisfies the Mean Value Inequality: for every a ∈ D there exists R > 0 such that
f (a) ≤ 1 2π Z 2π 0 f (a + reiθ) dθ for all 0 < r < R.
Lemma 2.17. Let F ∈ H1. We have log |F (reiϕ)| ≤ 1 2π
Z 2π 0
Pr(θ − ϕ) log |F (eiθ)| dθ
for all r ∈ (0, 1) and ϕ ∈ [0, 2π].
Proof. In Hoffman [1] the following variant of Jensen’s inequality for Borel measures µ on C and functions in g ∈ H1 is proven:
log R g dµ ≤R log |g| dµ.
Applying this to F and the measure µ = 2π1 Pr(θ − ϕ) dθ we obtain
log 1 2π Z 2π 0 Pr(θ − ϕ)F (eiθ) dθ ≤ 1 2π Z 2π 0 Pr(θ − ϕ) log |F (eiθ)| dθ.
Definition 2.18. Let F : D → C be holomorphic. If we have
log |F (reiϕ)| = Z 2π
0
Pr(θ − ϕ) log |F (eiθ)| dθ
we call F an outer function.
We now give two different equivalent ways to describe outer functions. Lemma 2.19. Every function F of the form
F (z) = λ exp 1 2π Z 2π 0 eiθ + z eiθ− zk(θ) dθ ,
where k is a real-valued integrable function on C and λ is a constant of modulus 1 is an outer function.
Proof. We have
log |F (reiϕ)| = log |λ| + log exp 1 2π Z 2π 0 eiθ+ z eiθ− zk(θ) dθ = 1 2π Z 2π 0 Re e iθ + z eiθ− z k(θ) dθ = 1 2π Z 2π 0 Pr(θ − ϕ)k(θ) dθ.
Now Theorem 2.9 implies that k(ϕ) = log(F (eiϕ)) almost everywhere on C. Hence F is an outer function.
Lemma 2.20. A function F ∈ H1 is an outer function if and only if for any f ∈ H1 such that |F | = |f | a.e. on C the inequality |F | ≥ |f | holds on D.
Proof. Suppose F is an outer function and let f be a function such that |F | = |f | almost everywhere on the unit circle. Recall Jensen’s inequality for Borel measures µ on C and functions in g ∈ H1:
R log |g| dµ ≥ log
R g dµ .
Applying this to f and the measure µ = 2π1 Pr(θ − t) dt we obtain
log |F (reiθ)| = Z 2π 0 log f (eit) dµ(t) ≥ log Z 2π 0 f (eit) dµ(t)
= log |f (reiθ)|
and therefore |F | ≥ |f | on D.
Now let F ∈ H1 be a function such that the second condition holds. Define G by
G(z) = exp 1 2π Z 2π 0 eiθ+ z eiθ− z log |F (e iθ)| dθ .
This function is an outer function by Lemma 2.19. Theorem 2.9 gives us |F | = |G| on the unit circle. Hence |F | ≥ |G| on D by our assumption. On the other hand we know that log |F | is subharmonic by Lemma 2.17 so we also have |F | ≤ |G| on D. Hence F/G is analytic and everywhere of norm 1. From the Open Mapping Theorem we deduce that F/G is constant. Since G is outer we now see that F also is an outer function.
We now arrive at our main result.
Theorem 2.21. Every f ∈ H1 has a factorisation f = IF with I an inner function
and F an outer function. This factorisation is unique up to a constant of modulus 1. Proof. Let f ∈ H1 be given. Extend f to the closed unit disc as described in §2.4. Define the outer function F by
F (z) = exp 1 2π Z 2π 0 eiθ+ z eiθ − zlog |f (e iθ )| dθ .
We want to show g = f /F is an inner function. Recall that for f we have
f reiθ = 1 2π
Z 2π
0
Pr(θ − t)f (eit) dt
by Theorem 2.13. Therefore on the unit circle we have |F | = |f | because we can apply Theorem 2.9 to log |f | so |g(z)| = 1 on C. Now Lemma 2.20 implies |g(z)| ≤ 1 for all z ∈ D since 1 is an outer function. This gives a decomposition f = gF . For uniqueness suppose we also have f = g0F0. Then |F | = |F0| = |f | on the unit circle. Hence F/F0 is
analytic and everywhere of norm 1. From the Open Mapping Theorem we deduce that F/F0 is a constant. This implies F = λF0. Now g0 = λg so indeed the factorisation is unique up to a constant of modulus 1.
Theorem 2.22. If f ∈ H1 is not the zero function then
Z π
−π
log |g(eiθ)| dθ > −∞.
Proof. For a proof we direct the reader to Theorem 11.5.2 from Wiegerinck [2]. We obtain the following result.
Corollary 2.23. If f ∈ H1 is not the zero function then f cannot vanish at a set of
3 Extreme and exposed points
Recall the following definition from the introduction.
Definition 3.1. Let X be a Banach space and let A ⊆ X. The extreme points of A are the elements in x ∈ A for which there do not exist distinct y1, y2 ∈ A and λ ∈ (0, 1)
such that x = λy1+ (1 − λ)y2. ] ] [ [
Figure 3.1: An illustration of the extreme points of a running track.
We give an equivalent condition, which will turn out to be very useful.
Lemma 3.2. A point x ∈ Σ is an extreme point of the unit ball if and only if for all t ∈ X the implication
kx + tk ≤ 1 and kx − tk ≤ 1 =⇒ t = 0 holds.
Proof. Suppose x is an extreme point and t ∈ X such that kx ± tk ≤ 1. Define y = x − t and z = x + t. By assumption kyk ≤ 1 and kzk ≤ 1 so y, z ∈ Σ. We see that x = 12(y + z) so y = z because x is an extreme point. Therefore t = 0.
Now suppose the implication holds and suppose x = λa + (1 − λ)b. Note that x lies on the line between a and b and since λ is not 0 or 1 there exist y and z on the line such that x = 12(y + z). Define t = x − y. Then kx + tk = kzk ≤ 1 and kx − tk = kyk ≤ 1 so t = 0 and x = y. Therefore our decomposition is trivial so x is extreme.
Returning to our example of the running track, we already noted that four extreme points seem to have a special property. Namely, they lie on the “boundary” of the set of extreme points. To make this more formal, we take a look at tangent lines at these points (see Figure 3.2). We see that these tangent lines intersect the running track at an interval, whereas for the other extreme points the tangent line only intersect the track at one point. The following definition generalises this property to arbitrary Banach spaces.
] ] [ [ P Q
Figure 3.2: The tangent line at Q intersects the running track in infinitely many points.
Definition 3.3. Let X be a Banach space and let A ⊆ X. A point a ∈ X is called an exposed point of A if there exists a bounded linear functional L ∈ X0 such that L(y) < L(a) for all y ∈ A \ {a}. We call L an exposing functional.
In our running track in Figure 3.2 the point P is exposed by its tangent line, whereas Q isn’t.
Definition 3.4. Let X be a Banach space and let A ⊆ X. A point a ∈ X is called a strongly exposed point if it is an exposed point and for all sequences (ai) ⊂ A we have
ai → a ⇐⇒ L(ai) → L(a)
where L is an exposing functional.
Note that the implication from the left to the right simply states L is continuous. Lemma 3.5. Every strongly exposed point is exposed and every exposed point is extreme. Proof. Clearly by construction every strongly exposed point is exposed. Now let X be a Banach space and let x be an exposed point of A ⊆ X with exposing functional L. Suppose there exist y1, y2 ∈ A and λ ∈ (0, 1) such that
x = λy1+ (1 − λ)y2.
This implies
L(x) = λL(y1) + (1 − λ)L(y2).
However, this means we cannot have both L(y1) < L(x) and L(y2) < L(x). This is a
contradiction. Therefore x is an extreme point.
It is not always the case that there exists a unique exposing functional. In case of the square in R2 we have infinitely many exposing functionals for the points on the edges
(see Figure 3.3). This is because the edges of the square are not ‘smooth’ enough. In R2 it is clear what smoothness means, but in an arbitrary Banach space this is not the case anymore. Therefore it is hard to determine when exposing functionals are unique. The following lemma shows that under some strict conditions this actually is the case. Lemma 3.6. Let X be a real Hilbert space. Let x be an exposed point of the unit ball Σ. Then the exposing functional L is unique up to multiplication by a constant.
x
Figure 3.3: A few exposing functionals of x.
Proof. Let x be an exposed point of Σ. Then kxk = 1 since all extreme points lie on the boundary of Σ. For any Hilbert space we know that every L ∈ X0 can we written as L = Lb for some b ∈ X where Lb(x) = hx, bi. Suppose L = Lb is an exposing functional
for x. Let B = {ei | i ∈ I} be an orthonormal basis for X with b/kbk ∈ B. Write
x = αb + ˜x where b ⊥ ˜x and |α| ≤ kbk1 . We have
L(x) = L(αb + ˜x) = hαb + ˜x, bi = |α|kbk2. We also have L b kbk = kbk ≤ |α|kbk2.
Since α ≤ kbk1 this in fact is an equality so kbkb = x. This proves that the exposing functional is up to multiplication by a constant equal to
Lx(y) = hy, xi.
3.1 The unit ball in L
p(X)
In this section we will discuss the extreme points of the unit ball in Lp(X).
Theorem 3.7. Let µ be the normalised Lebesgue measure on [a, b]. The unit ball Σ in L1(µ) has no extreme points.
Proof. Let f : [a, b] → R such that Z [a,b] |f | dµ ≤ 1 Define g : [0, b] → R by g(t) = Z [a,a+t] |f | dµ.
This is a continuous function and g(0) = 0, so there exists t0 ∈ (0, b) such that g(t0) = 1
2
R
[a,b]|f | dµ by the intermediate value theorem. This gives us
and
k2f · 1[a,a+t0]k = kf k ≤ 1
so both of these functions are elements of Σ. Clearly f = f · 1[a,a+t0]+f · 1(a+t0,b] so f is
not an extreme point.
Theorem 3.8. For 1 < p < ∞ the exposed points of the unit ball in Lp(X) are the
functions of norm 1.
Proof. Take q such that p−1+ q−1 = 1. Remember that (Lp(X))0 ∼
= Lq(X). Take f ∈ Lp
with kf k = 1. Define g by g(x) = |f (x)|f (x)p if f (x) 6= 0 and 0 otherwise. We have Z X |g|q dµ = Z X |f |pq |f |q dµ = Z X |f |p+q |f |q dµ = Z X |f |p dµ = 1
so g ∈ Lq(X). This function induces a functional Lg ∈ (Lp(X))0 ∼= Lq by
Lg(h) =
Z
X
gh dµ
for h ∈ Lp(X). Note that Lg(f ) = 1 since kf k = 1. By H¨older’s inequality we have
|Lg(h)| ≤ Z X |g||h| dµ ≤ Z X |g|qdµ 1q · Z X |h|pdµ 1p
with equality in the last step iff there exist constants α, β not both zero such that α|h|p = β|g|q almost everywhere. It follows that Lg(h) = 1 implies khkp = 1 and
α|h|p = β|g|q = β|f | pq |f |q = β |f |p+q |f |q = β|f | p
almost everywhere. Since kf kp = khkp = 1 we see that |h| = |f | almost everywhere. If
h would not have the same sign as f on a set of positive measure then L(h) < L(f ). Therefore Lg(h) = 1 if and only if h = f so f is an exposed point.
This shows that every extreme point of Σ ⊂ Lp(X) is even an exposed point. We
will show later on that for Hilbert spaces all extreme points of the unit ball are in fact strongly exposed. Hence, for L2(R) it follows that these points actually are strongly exposed. The only case left to consider is L∞(X). A function is of norm 1 if and only if ess supx∈X|f (x)| = 1.
Theorem 3.9. Let (X, A, µ) be a finite measure space. Then a function f ∈ Σ ⊆ L∞(X) is extreme if and only if
|f (x)| = 1
for almost all x ∈ X. Furthermore, these points also are exposed.
The theorem basically tells us that every function looks like the function plotted in Figure 3.4.
1
−1
Figure 3.4: A typical extreme point in L∞(X)
Proof. The implication from the left to the right is trivial. Suppose |f (x)| = 1
for almost all x ∈ X. Define A1 = {x ∈ X : f (x) = 1} and A−1 = {x ∈ X : f (x) = −1}.
Now define L ∈ (L∞(X))0 by
L(g) = Z
X
1A1− 1A−1 g dµ.
Clearly L(f ) = µ(X) and L(g) ≤ µ(X) for all g ∈ Σ.
Suppose g ∈ Σ and g 6= f . Then there exists B ∈ A with µ(B) 6= 0 and f (b) 6= g(b) for all b ∈ B. Since
0 < µ(B) = µ(B ∩ (A−1∪ A1)) = µ(B ∩ A−1) + µ(B ∩ A1)
we see that µ(B ∩ A−1) > 0 or µ(B ∩ A1) > 0. Without loss of generality assume
µ(B ∩ A−1) > 0. Since f (b) = −1 for all b ∈ B ∩ A−1 we have 1 ≥ g(b) > −1 for all
b ∈ B ∩ A−1 so −1 ≤ −g(b) < 1. Hence Z X − 1A−1f dµ = Z X 1A−1dµ > Z X − 1A−1g dµ Therefore also L(f ) > L(g).
Unfortunately this proof does not work for spaces with µ(X) = ∞ like R since the integral in the definition of L(f ) doesn’t converge. However, for R we can define
L(g) = Z R 1A1(x) − 1A−1(x) g(x)e −x2 dx
which does the trick in the same way. This can can be generalized to Rn by multiplying with the density of a multivariate normal distribution. Hence, the following theorem holds.
Theorem 3.10. A function f ∈ Σ ⊆ L∞(Rn) is extreme if and only if |f (x)| = 1
3.2 The unit ball in Hardy spaces
In this section we will discuss the extreme points of the unit ball in the Hardy spaces Hp. A cherished result on the extreme points of H1 was published for the first time by Rudin and de Leeuw in 1958 [3]. The proofs presented here are based on proofs from Hoffman [1].
Theorem 3.11 (Rudin- de Leeuw). A function f ∈ H1 is an extreme point of the unit
ball if and only if f is an outer function of norm 1.
Proof. Suppose f ∈ H1 is an outer function of norm 1. Let g ∈ H1 such that kf + gk1 ≤ 1 and kf − gk1 ≤ 1.
We want to show that g = 0 in order to use Lemma 3.2. We know that on the boundary of the disc f can only vanish on a set of Lebesgue measure 0 by Theorem 2.23. Therefore, on C we have g = φf almost everywhere for some function φ. Define the measure µ by dµ = 2π1 |f | dθ. Then φ is integrable with respect to µ and we have
Z
(|1 + φ| + |1 − φ|) dµ ≤ 2
since kf ± gk1 ≤ 1. However, |1 + φ| + |1 − φ| ≥ 2 and since µ has mass 1 we obtain
|1 + φ| + |1 − φ| = 2
almost everywhere. This implies φ is real-valued and −1 ≤ φ ≤ 1. Hence, |g| ≤ |f | on C. Just like in the proof of Lemma 2.20 we can show that |g| ≤ |f | on C implies |g| ≤ |f | on D. We see that φ := g/f defined on the unit disc is a bounded function with real boundary values. Because it is bounded, it equals its Poisson integral and hence will be real-valued on the unit disc. By the Open Mapping Theorem, it is constant. We have
|1 + φ| = |1 + φ| · kf k1 = k(1 + φ)f k1 = kf + gk1 ≤ 1
so φ ≤ 0. The same way we can show that φ ≥ 0 so φ = 0. This implies g = 0 so f is an extreme point of the unit ball by Lemma 3.2.
Now suppose f ∈ H1 is not an outer function. Then we write f = IF with I an inner and F an outer function with I non-constant. Since the factorisation is unique up to a constant of modulus 1 we can adjust λ in the definition of the outer function F . Note that this corresponds to multiplying the inner function with a constant of modulus 1.
We adjust λ in such a way that we have Z π
−π
|f (eiθ)| Re I(eiθ) dθ = 0. (3.1)
This is possible since for a given inner function I the continuous function
α 7→ Z π
−π
changes sign on [0, π] since ei·0 = 1 and eiπ = −1. Now define g = 12(1 + I2)F . This is non-zero and an element of H1. For z ∈ C we have
2 Re(z) = ¯z + z = 1 + z
2
z so we obtain
f (eiθ) Re I(eiθ) = f (eiθ)1 + I(e
iθ)2
2I(eiθ) = 1 2
1 + I eiθ2F (eiθ) = g(eiθ).
This implies
|f (eiθ) ± g(eiθ)| = |f (eiθ)| 1 ± Re I(eiθ) so
Z π
−π
|f (eiθ) ± g(eiθ)| dθ = Z π
−π
|f (eiθ)| 1 ± Re I(eiθ) dθ
(3.1)
= Z π
−π
|f (eiθ)|dθ.
In other words, we showed that kf ± gk1 = kf k1 but g 6= 0. Lemma 3.2 now implies f
cannot be an extreme point. This finishes the proof.
We will now prove a very surprising corollary. This corollary gives us not only in-formation about how many extreme points there are, but also about the spread of the extreme points.
Corollary 3.12. Let f ∈ H1 with kf k = 1. If f is not an extreme point of the unit ball
then we can write
f = 12f1+12f2
where f1, f2 ∈ H1 are extreme points of Σ.
Proof. Let f ∈ H1with kf k = 1 and suppose f is not an extreme point of Σ. Let f = IF be the decomposition of f into a non-constant inner function I and an outer function F . Define g = 12(1 + I2)F . Remember this is non-zero and an element of H1. We now
define f1 = f + g and f2 = f − g. We see that f = 12f1+12f2 and kf1k = kf2k = kf k = 1
since inner functions have norm 1 on the boundary. We will now show that f1 and f2
are both outer functions because then they are extreme points. We have
f1 = f + g = f + 12(1 + I2)F = 21F (I2+ 2I + 1) = 12F (1 + I)2.
Since the product of two outer functions is again outer, we only have to show that 1 + I is an outer function. For > 0 define I = 1 + + I. This is bounded by 2 + and is
bounded away from zero because of the . Hence also 1/I is bounded. Write
I = I1G1 and
1 I
where I1, I2 are inner functions and G1and G2 are outer functions. But now I1I2G1G2 =
id and since |G1G2| ≤ 1 by the Maximum Modulus Principle we obtain I1(x)| = |I2(x)| =
1 for all x ∈ D. By the open mapping theorem this implies I1 and I2 are constant. Hence
I is an outer function and we have
I(z) = exp 1 2π Z 2π 0 eiθ+ z eiθ− z log |I(e iθ)| dθ + i argI (0)
As → 0 we see that log |I| → log |I + 1| monotonically. Applying the Lebesgue
dominated convergence theorem we obtain
1 + I = lim →0I(z) = exp 1 2π Z 2π 0 eiθ+ z
eiθ− z log |1 + I(e
iθ)| dθ + i arg (1 + I(0))
Hence 1 + I is outer by Lemma 2.19, as wanted.
Theorem 3.13. For 1 < p < ∞ the extreme points of the unit ball in Hp are the functions of norm 1.
Proof. Let f ∈ Hp be a function of norm 1. First note that Hp ⊂ H1 so we can extend
functions to C as described in §2.4. We will first show that |f |p is subharmonic. We have
f (reiθ)
p
= exp p log f (reiθ) ≤ exp p 2π Z 2π 0 Pr(θ − ϕ) log f (eiθ) dθ (J) ≤ 1 2π Z 2π 0
exp log |f (eiθ)|pdθ = 1 2π Z 2π 0 f (eiθ) p dθ,
where we used Jensen’s inequality at (J). Since |f |p is subharmonic we see that kfrkp
increases as r increases. Hence,
kf kHp = lim
r→1kfrkp
Therefore we can identify any function of norm 1 in Hp with a function in Lp of norm
1. We already proved that the set of extreme points of Σ ⊆ Lp is exactly the boundary
in Theorem 3.8. If we cannot write a function as a convex combination of two functions in Lp we certainly cannot write f a convex combination of two functions from Hp.
Therefore the extreme points of the unit ball in Hp are the functions of norm 1 too.
Theorem 3.14. A function f ∈ H∞ is an extreme point of the unit ball if and only if |f (z)| ≤ 1 for all z ∈ D and
Z π
−π
Proof. Suppose |f (z)| ≤ 1 and Z π
−π
log(1 − |f (eiθ)|) dθ = −∞
and let g ∈ H∞ such that kf ± gk∞ ≤ 1. This means for all z ∈ D we have
|f (z)|2+ |g(z)|2 = 1 2(|f (z)| + |g(z)|) 2 +12(|f (z)| − |g(z)|)2 ≤ 1. We deduce that 2 Z π −π
log |g(eiθ)| dθ ≤ Z π −π log(1 − |f (eiθ)|2) dθ = Z π −π log(1 + |f (eiθ)|) dθ + Z π −π log(1 − |f (eiθ)|) dθ ≤ 2π log 2 + Z π −π log(1 − |f (eiθ)|) dθ = −∞.
Therefore g has to be the zero function by Theorem 2.22. From Theorem 3.2 we conclude f is an extreme point.
Now suppose the integral condition does not hold. This means log(1−|f |) is integrable over the unit circle. Define
g(z) = exp 1 2π Z π −π eiθ + z eiθ− z log(1 − |f (e iθ)|) dθ . We have |g(eiθ )| = 1 − |f (eiθ)|
by Lemma 2.19 so g is a bounded function. We have kf ±gk∞≤ 1 so f is not extreme.
This concludes our investigation of the extreme points of Hardy spaces.
3.3 The unit ball in Hilbert spaces
An exercise in Hoffmann [1] asks to determine the extreme points of the unit ball in an arbitrary Hilbert space H. In this paragraph we will do this and more. It turns out that every extreme point is even a strongly exposed point.
Theorem 3.15. Let H be a real Hilbert space. The exposed points of Σ ⊆ H are exactly the points of norm 1. Furthermore, these points are even strongly exposed.
Proof. Throughout this proof, let x ∈ Σ be such that kxk = 1. Define L ∈ H0 by L(y) = hx, yi. Clearly L(x) = 1 and −1 ≤ L(y) ≤ 1 for all y ∈ Σ. Now suppose L(y) = 1. This means hx, yi = 1 so
hx − y, x − yi = hx, xi − 2hx, yi + hy, yi = −1 + kyk2 ≤ 0
and therefore it equal zero so y = x. This shows that L is an exposing functional. Hence, x is an exposed point.
We will now show the functional L also strongly exposes x. Choose a basis B = {ei |
i ∈ I} of H (the Axiom of Choice is needed here). Without loss of generality we assume x = e1. As we saw, the unique (up to multiplication by constants) exposing functional
is given by L(y) = he1, yi. Clearly if xi → x we have L(xi) → L(x). Now suppose
L(xi) → L(x) = 1. Since kxik ≤ 1 we can write
xi = yie1 + ˜yi
with −1 ≤ 0 ≤ yi ≤ 1 and ˜yi ⊥ e1. We have L(xi) = yi → 1 and since kxik ≤ 1 and
0 ≤ k˜yik = kxik − |yi|ke1k
we see that k˜yik → 0. Hence ˜yi → 0 and indeed xi → x.
Corollary 3.16. All extreme points of Σ in a Hilbert space are strongly exposed
This concludes our analysis. Examples of Hilbert spaces are Rn, `2 and the previously
mentioned space L2(R).
3.4 The unit ball in C(X)
In this section we will take a look at the spaces CR(X) and CC(X). These are the spaces of continuous functions from X to R and C respectively. As we saw for Hilbert spaces, it is extremely useful to have a characterization of the bounded linear functionals of a space when determining exposed and strongly exposed points. The following theorem by Riesz states that under the right circumstances the bounded linear functionals on C(X) can be identified with so-called regular measures on X.
Definition 3.17. Let X be a locally compact space and let Ω be the Borel sigma algebra on X. A positive measure on (X, Ω) is called regular if and only if the following three conditions hold:
(i) For every compact subset K ⊆ X we have µ(K) < ∞ (ii) For any E ∈ Ω we have
µ(E) = sup{µ(K) : K ⊆ E and K is compact}
(iii) For any E ∈ Ω we have
µ(E) = inf{µ(U ) : U ⊇ E and U is open}.
Theorem 3.18 (Riesz representation theorem). If X is a locally compact space and µ is a regular measure on X, define Fµ: CC(X) → C by
Fµ(f ) =
Z
X
f dµ.
Then Fµ ∈ CC(X)
0 and the map µ 7→ F
µ is an isometric isomorphism of the set of all
complex-valued regular Borel measures on X endowed with the total variation norm onto CC(X)∗.
Theorem 3.19. The extreme points of Σ ⊆ CR[a, b] are exactly f = 1 and f = −1. These both are exposed points.
Proof. There are many exposing functionals, for example normalised Lebesgue integra-tion. Every measure that gives open sets in [a, b] a strictly positive measure in fact is an exposing functional.
We will now use the Riesz representation theorem to prove that these functions are not strongly exposed points.
Theorem 3.20. The unit ball Σ ⊆ CR[a, b] does not have any strongly exposed points. Proof. We will show 1 is not a strongly exposed point. As the Riesz representation theorem states, every bounded linear functional on CC(X) can be obtained by integrating against a complex-valued regular Borel measure on X. Let µ be such a measure on X. We have µ([a, b]) < ∞ by regularity. Hence there exists x0 ∈ [a, b] such that µ({x0}) = 0.
Now take a look at the functions depicted in Figure 3.5.
x0 x0 x0
Figure 3.5: Functions f1, f2 and f3 with a spike at x0.
We see that fi(x) → 1 µ-almost everywhere so
Z
fidµ →
Z 1 dµ
but clearly supx∈[a,b]|fi(x) − 1| = 1 for all i. Hence, fi 6→ 1. Therefore 1 is not a strongly
exposed point. The same way −1 is not a strongly exposed point.
Note that CC[a, b] is a much more interesting space, since it contains more elements of norm 1. We will now generalize this theorem. To do this, we need Urysohn’s lemma.
Lemma 3.21 (Urysohn’s lemma). Let X be a compact Hausdorff space. Suppose A, B ⊆ X are closed sets such that A ∩ B = ∅. Then there exists a continuous function f : X → [0, 1] such that f (a) = 0 for all a ∈ A and f (b) = 1 for all b ∈ B.
The following theorem (but not the proof) comes from Douglas [4].
Theorem 3.22. Let X be a compact Hausdorff space. The set of extreme points of Σ in CC(X) is the set of functions f for which we have
|f (x)| = 1 for all x ∈ X.
Proof. Suppose |f (x)| = 1 for all x ∈ X. We want to show that f is an extreme point. Suppose g ∈ CC(X) such that kf ± gk ≤ 1 and g 6= 0. Then there exists x0 ∈ X such
that g(x0) 6= 0. Write f (x0) = a + bi and g(x0) = c + di. We obtain
|f (x0) + g(x0)| = 1 + c2+ d2+ 2ac + 2bd
and
|f (x0) − g(x0)| = 1 + c2 + d2− 2ac − 2bd.
This implies |f (x0) + g(x0)| > 1 or |f (x0) − g(x0)| > 1 and this is a contradiction. By
Lemma 3.2 we conclude that f is an extreme point of CC(X).
For the other implication of the statement we will provide two different proofs. Before we start our first proof, we will give a proof for CR(X). Let f : X → R such that f ∈ Σ but not |f (x)| = 1 for all x ∈ X. Trivially f = id ◦ f where id : R → R is the identity function. Define h1, h2: R → R by
h1(x) = x + 12(x − 1)(x + 1)
h2(x) = x − 12(x − 1)(x + 1).
Both functions are drawn in Figure 3.6.
We have −1 ≤ hi(f (x)) ≤ 1 for all x ∈ X. Consequently, khi ◦ f k ≤ 1. Since
there exists x0 ∈ X such that f (x0) 6= ±1 we see that h1 ◦ f 6= h2 ◦ f . Since f = 1
2(h1 ◦ f ) + 1
2(h2 ◦ f ) we conclude f is not extreme.
Now, let’s generalize this proof to continuous functions f : X → C. We will use the h1 and h2 as defined in the first part. Define
g1 reiθ = h1(r)eiθ
g2 reiθ = h2(r)eiθ.
Again, 1 2g1 + 1 2g2 = 1 2h1(r)e iθ
+12h2(r)eiθ = reiθ
so 12(g1 ◦ f ) + 12(g2 ◦ f ) = f . Since there exists x0 ∈ X such that |f (x0)| 6= 1 so
x y h1 h2 1 −1 1 −1
Figure 3.6: The functions h1 and h2 with intersections (−1, −1) and (1, 1).
We will now give a different proof, which relies heavily on Urysohn’s lemma. Suppose f ∈ Σ with kf k = 1 is chosen such that |f (x0)| = 1 − 2 < 1 for some x0 ∈ X. Define
A = |f |−1[1 − , 1] B = |f |−1[1 − 3, 1 − 2]
Both sets are closed and non-empty since kf k = 1 and x0 ∈ B. Furthermore A ∩ B = ∅.
We now apply Urysohn’s lemma to obtain a continuous g : X → [0, 1] with g(a) = 0 for all a ∈ A and g(b) = 1 for all b ∈ B. Now define g1, g2: X → C by
h1(x) = f (x) + g(x)
h2(x) = f (x) − g(x).
Clearly 12h1+ 12h2 = f and kh1k ≤ 1 and kh2k ≤ 1 by the triangle inequality. Hence f
can be written as a non-trivial convex combination of two elements from Σ.
The proof that relies on Urysohn’s lemma provides us with crucial information. Namely, it shows the complexity of CC(X). It is not clear on first sight that for a general com-pact Hausdorff this space does not consists only of constant functions, whereas Urysohn’s lemma actually tells us there are many functions f : X → C that are continuous.
It follows from Theorem 3.22 that the extreme points of CC[a, b] are paths that lie entirely on the surface of the sphere {z ∈ C | kzk = 1}. As already noted in the proof of Theorem 3.22 we have the following result for CR(X).
Corollary 3.23. Let X be a compact connected Hausdorff space. The only extreme points of Σ in CR(X) are f = 1 and f = −1.
4 Discussion
We examined the extreme, exposed and strongly exposed points of the unit ball in several Banach spaces of holomorphic functions. Our finding that every point on the boundary of Σ in a Hilbert space is strongly exposed agrees with our intuition that the unit ball in a Hilbert space behaves nicely. However, it was surprising to see that Σ ⊂ L1(R) does
not have any extreme point. It was even more surprising to observe that every f ∈ H1
in the boundary of the unit ball can be written as f = 12f1+ 12f2,
where the functions f1 and f2 are extreme points of Σ ⊂ H1. This gives us not only
information about how many extreme points there are, but also about the spread of the extreme points.
In Theorem 3.9 we determined the extreme and exposed points of the unit ball in L∞(X), where X is a finite measure space. After that we generalized this result to Rn, however this argument is not valid in general measure spaces with infinite mass. It seems likely that a similar result does hold in a general setting. The same goes for L1(X). Furthermore, we proved that every point on the boundary of the unit ball in Lp with 1 < p < ∞ is exposed by constructing an exposing functional. We suspect that these points actually are strongly exposed, but we did not manage to show this yet.
5 Populaire samenvatting
In deze scriptie bekijken we de zogeheten extreme punten van bepaalde ruimten. De extreme punten van een verzameling A zijn de punten die niet het gemiddelde zijn van twee andere punten uit A. We zouden bijvoorbeeld de extreme punten van de opgevulde atletiekbaan in Figuur 5.1 kunnen bepalen.
Figuur 5.1: Een atletiekbaan.
Elk punt aan de binnenkant van de atletiekbaan kan geen extreem punt zijn, want we kunnen een punt een beetje links van ons punt en een ander punt beetje rechts van ons punt kiezen zodat ons punt precies ertussenin ligt. Om extreme punten te bepalen is het dus alleen nodig om naar de rand te kijken van ons figuur. In Figuur 5.2 zijn de extreme punten van de atletiekbaan rood gekleurd.
] ]
[ [
Figuur 5.2: De extreme punten van een atletiekbaan.
Het is interessant om verschillende soorten extreme punten te onderscheiden. Zoals in de figuur te zien is liggen er vier punten op de rand van de rode stukken. Om deze punten te karakteriseren bekijken we raaklijnen aan de atletiekbaan. We zien nu dat de lijnen aan deze vier punten de baan in nog veel meer punten raakt, terwijl de raaklijn aan de andere punten de atletiekbaan maar in ´e´en punt raakt. In Figuur 5.3 is dit verschil goed te zien. Wanneer de raaklijn alleen het extreme punt raakt noemen we het een ge¨exponeerd punt.
Nu bekijken we in plaats van figuren in het platte vlak de bol in ingewikkeldere ruimten. Zo bekijken we bijvoorbeeld de ruimte CR[0, 1]. Punten in deze ruimten zijn functies f : [0, 1] → R die continu zijn. Voor functies in onze ruimte houdt dat in dat de grafiek van f te tekenen is zonder het potlood van het papier af te halen. Voordat
] ] [ [ P Q
Figuur 5.3: De raaklijn aan het punt Q snijdt de atletiekbaan in nog veel andere punten.
we bollen van functies kunnen bekijken moeten we eerst deze functies een lengte geven. We defini¨eren de lengte van een functie f als
Lengte(f ) = max
x∈[0,1]|f (x)|.
De lengte van een functie is dus eigenlijk de maximale uitwijking van de functie. We kunnen nu spreken over de eenheidsbol van de ruimte CR[0, 1]. Dit is de verzameling van functies die hoogstens lengte 1 hebben.
Vervolgens willen we onderzoeken welke functies te schrijven zijn als gemiddelde van twee andere functies. In Figuur 5.4 is een voorbeeld gegeven van twee functies met hun gemiddelde. Dit laat dus zien dat de functie die met rood is aangegeven geen extreem punt is. Het blijkt dat alleen de constante functie 1 en de constante functie −1 extreme punten van de eenheidsbol in CR[0, 1] zijn.
x y
Figuur 5.4: Het gemiddelde van twee functies.
In deze scriptie wordt een soortgelijk resultaat bewezen voor de ruimte CR(X) waarbij X aan bepaalde eisen voldoet. Dit is de ruimte van continue functies van een ruimte X naar R. Het voorbeeld dat we zojuist hebben gezien is dus eigenlijk een speciaal geval van deze stelling. Verder worden er nog andere ruimten van functies bekeken. We bekijken bijvoorbeeld ook de collectie functies van een schijf naar de complexe getallen die voldaan aan strenge eisen en de ruimte van functies f : [0, 1] → R waarvoorR01f (t) dt
Bibliography
[1] Hoffman, K. (1962) Banach spaces of analytic functions. New York: Dover Publica-tions, Inc.
[2] Wiegerinck, J. J. O. O. (2011) Advanced Function Theory. Faculty of Mathematics, University of Amsterdam.
[3] de Leeuw, K. & Rudin, W. (1958) Extreme points and extremium problems in H1.
Pacific J. Math 8, 467-485.
[4] Douglas, R. G. (1976) Banach Algebra Techniques in Operator Theory. New York: Springer. DOI: 10.1007/978-1-4612-1656-8
[5] Beneker, P. (2002) Strongly exposed points in the unit balls of banach spaces of holo-morphic functions. Thesis at University of Amsterdam. ISBN: 90-5776-085-1.
