Math 516
Professor Lieberman
February 21, 2005
HOMEWORK #5 SOLUTIONS
Chapter 9
4. Let F1 and F2 be disjoint closed sets. Then for all x ∈ F1 and all y ∈ F2 , there are disjoint
open sets Ux,y and Vx,y such that x ∈ Ux,y and y ∈ Vx,y .
For a fixed x ∈ F1 , the family of sets hVx,y iy∈F2 is an open cover of F2 and Proposition
94 from 515 implies that F2 is compact, so there are finitely many y1 , . . . yn such that
n
[
F2 ⊂
Vx,yi .
i=1
(Of course, the choice and number of yi ’s will depend on the point x.) If we set
n
n
[
\
Vx =
Vx,yi , Ux =
Ux,yi ,
i=1
i=1
then Vx and Ux are disjoint open sets with x ∈ Ux and F2 ⊂ Vx . Now F1 is also compact,
so there are finitely may x1 , . . . , xm with
m
[
F1 ⊂
Uxj .
i=1
It follows that
U=
m
[
Uxj ,
V =
j=1
m
\
V xj
j=1
are disjoint open sets with F1 ⊂ U and F2 ⊂ V , so the space is normal.
10. Here’s a quick version.
(i) =⇒ (iii) This is just Corollary 35.
(iii) =⇒ (ii) Let f be a continuous real-valued function. Then h = |f |/(1+|f |) is a bounded
continuous real-valued function, so it must attain its maximum M . Since h can’t take on
the value 1, it follows that M < 1 and hence |f | ≤ M/(1 − M ), so f is bounded.
(ii) =⇒ (i) We prove the contrapositive. So suppose that X is not countably compact.
Then Proposition 32 gives an infinite sequence hxn i in X with no cluster point. It follows
that A = {xn } is closed and it has the discrete topology. Hence the function f , defined
on A by f (xn ), is a continuous real-valued function. The Tietze extension theorem then
gives a continuous real-valued function g on X with g = f on A. It follows that g is an
unbounded, continuous real-valued function.
Chapter 10
48. (a) Suppose that there are functions y and z along with a number λ ∈ (0, 1) such that
x = λy + (1 − λ)z. Minkowski’s inequality implies that
1 = kxk ≤ λkyk + (1 − λ)kzk ≤ 1,
so kyk = kzk = 1 and kλy + (1 − λ)zk = kλyk + k(1 − λ)zk. The condition for equality
in Minkowski’s inequality tells us that there are nonnegative constants α and β such that
αλy = β(1 − λ)z. It follows that y = z and hence x = y = z, so x is an extreme point.
1
2
(The problem doesn’t ask if there are other extreme points, but it’s easy to see that these
are the only ones.)
(b) First, suppose that x doesn’t satisfy |x(t)| = 1 a.e. Then there is a positive number δ
such that |x(t)| ≤ 1 − δ on some set E with positive measure. We now set
(
(
x(t)
if x ∈
/ E,
x(t)
if x ∈
/ E,
y(t) =
, quadz(t) =
x + δ if x ∈ E,
x − δ if x ∈ E,
Then kyk ≤ 1, kzk ≤ 1, and x = 12 y + 21 z. On the other hand, x 6= y on E, which has
positive measure, so x and y are different elements of L∞ , so x is not an extreme point.
If x does satisfy |x(t)| = 1 a.e., then suppose there are y and z in S and a number λ ∈
(0, 1) such that x = λy + (1 − λ)z. Then, for almost all t, we have x(t) = λy(t) + (1 − λ)z(t)
and |x(t)| = 1, |y(t)| ≤ 1, |z(t)| ≤ 1. Suppose x(t) = 1. Since y(t) ≤ 1, it follows that
z(t) ≥ 1 and hence z(t) = 1. Similarly y(t) = 1. The case x(t) = −1 is handled similarly,
so x = y = z a.e.
(c) Let x ∈ S. If x = 0, then we take y ≡ 1, z ≡ −1 to see that x = 12 y + 12 z, so this isn’t
an extreme point. If x 6= 0 on some set of positive measure, then kxk > 0, so there is a
number α ∈ (0, 1) such that
Z
Z α
1 1
|x|,
|x| =
2 0
0
and the integrals are not zero. We now define
y = 2xχ[0,α] ,
z = 2xχ[α,1]
to see that x = 12 y + 21 z. It’s easy to check that x 6= y and x 6= z, so x is not an extreme
point.
(d) If L1 were the dual of a normed linear space, then Algaoglu’s Theorem would imply that
S is weak∗ compact, and then, by the Krein-Milman theorem, S would be the closed convex
hull of its extreme point. But S doesn’t have any extreme points while it is nonempty since
it includes the element 0 (along with lots of others).
(e) For 1 < p < ∞, the arguments above show that the extreme points are all those
sequences x ∈ lp with kxk = 1. For p = ∞, it’s all those sequences x = hxm i with |xm | = 1
for all m.
For p = 1, the preceding argument doesn’t quite work. Instead, every sequence consisting
of ±1 in one position and 0 everywhere else is an extreme point. To see this, let x be such
a sequence and let y and z be in l1 with kyk, kzk ≤ 1 and x = λy + (1 − λ)z for some
λ ∈ (0, 1). For simplicity, we suppose that xn = 1 (the case xn = −1 is similar). Then
1 = λyn + (1 − λ)zn . Since kyk ≤ 1, it follows that yn ≤ 1, so (as before) zn = 1 and then
yn = 1. But kyk ≤ 1 implies that y = x and z = x, so x is an extreme point. Conversely,
if 0 < xn < 1 for some n, then we define
(
(
xm if m 6= n,
xm if m 6= n,
, quadzm =
ym =
1
if m = n,
0
if m = n.
Choosing λ = xn shows that x = λy + (1 − λ)z and x 6= y. A similar argument works if
−1 < xn < 0 for some n.
(f) To find the extreme points of S, the unit ball in C(X), we imitate the solution to part
(b). In particular, that argument shows that every x ∈ S satisfying |x(t)| = 1 for all t ∈ X
is an extreme point. Conversely, if x ∈ S and there is a point t0 ∈ X with |x(t0 )| < 1,
3
set δ = (1 − |x(t0 )|)/2. Since x is continuous, there is an open neighborhood N of t0 on
which |x(t) − x(t0 )| < δ and hence |x| < 1 − δ in N . From Urysohn’s Lemma, there is a
continuous real-valued function f such that f (t0 ) = 1, f = 0 outside N and 0 ≤ f ≤ 1.
Then y = x − δf and z = x + δf are in S, but not equal to x, and x = 12 y + 21 z.
When X = [0, 1], it follows that the only extreme points are the constant functions. (A
complete proof of this fact requires some version of the Intermediate Value Theorem. For
other spaces, this statement need not be true.) As in part (d), if C[0, 1] were the dual of
some normed linear space, then S would have to be the closed convex hull of its extreme
points and hence it would have to consist only of constant functions. Therefore C[0, 1] is
not the dual of a normed linear space.