Math 516
Professor Lieberman
March 27, 2009
HOMEWORK #5 SOLUTIONS
Chapter 10
43. Let x and y be points of closure of K and let λ ∈ [0, 1]. Set z = λx + (−1λ)y. We
want to show that z is also a point of closure of K, so let U be an open set containing
z. Then −z + U is an open set containing 0, and Proposition 10.14 gives a base B
satisfying properties (i)-(ii) of that Proposition, so there is V ∈ B with V ⊂ U .
Hence there is V1 ∈ V such that V1 + V1 ⊂ V .
Since V1 ∈ B, x + V1 and y + V1 are open sets containing x and y, respectively.
Hence there are points x1 ∈ K ∩ (x + V1 ) and y1 ∈ K ∩ (y + V1 ). It follows that
z1 = λx1 + (1 − λ)y1 is in K because K is convex. Moreover, x − x1 and y − y1 are in
V1 , so property (vi) of Proposition 10.14 implies that λ(x − x1 ) and (1 − λ)(y − y1 )
are in V1 . Therefore λ(x − x1 ) + (1 − λ)(y − y1 ) ∈ V because V1 + V1 ⊂ V . We now
write
z1 = z + λ(x − x1 ) + (1 − λ)(y − y1 )
to conclude that z1 ∈ z + V , so z1 ∈ K ∩ (z + V ). It follows that z1 ∈ K ∩ U , so z is
a point of closure of K. Hence the closure of K is convex.
48. (a) Suppose that kxk = 1 and there are elements y and z of S and a number
λ ∈ (0, 1) such that x = λy + (1 − λ)z. From Minkowski’s inequality, we have
1 = kxk = kλy + (1 − λ)zk ≤ kλyk + k(1 − λ)zk.
The properties of the norms imply that
kλyk + k(1 − λ)z = λkyk + (1 − λ)kzk.
Since kyk ≤ 1 and kzk ≤ 1, it follows that
kλyk + k(1 − λ)zk ≤ λ + (1 − λ) = 1,
so we must have
kλyk + k(1 − λ)zk = 1
which implies (by the second part of Minkowski’s inequality) that there are
nonnegative constants α and β, not both zero, such that αλy = β(1 − λ)z a.e.
From the inequalities kyk ≤ 1, kzk ≤ 1, and 0 < λ < 1, along with the equality
λkyk + (1 − λ)kzk = 1, we conclude that kyk = kzk = 1, so α and β must be
equal and nonzero. In other words y = z a.e., so x = λy + (1 − λ)y = y, which
means that x is an extreme point of S.
1
2
(b) First, Minkowski’s inequality shows that, if x is an extreme point, then kxk = 1.
Suppose first that |x(t)| = 1 a.e. and that there are elements y and z of S and a
number λ ∈ (0, 1) such that x = λy + (1 − λ)z. Then let t be a point such that
x(t) = 1 and such that |y(t)| ≤ 1 and |z(t)| ≤ 1. We then have
1 = λy(t) + (1 − λ)z(t).
Because z(t) ≤ 1, we have
1 ≤ λy(t) + (1 − λ)
so we must have y(t) = 1 and, similarly, z(t) = 1. Analogous reasoning shows
that, at any point t with x(t) = −1 and |y(t)| ≤ 1 and |z(t)| ≤ 1, we must have
y(t) = z(t) = −1. Since x = 1 or −1 a.e. and |z|, |y| ≤ 1 a.e., it follows that
x = y = z a.e., so x is an extreme point.
On the other hand, if x is a function which does not satisfy |x(t)| = 1 a.e, then
there are ε > 0 and a set A with positive measure such that |x(t)| < 1 − ε for
all t ∈ A. We now define y and z by
(
(
x(t)
if t ∈
/ A,
x(t)
if t ∈
/ A,
y(t) =
z(t) =
x(t) + ε if t ∈ A,
x(t) − ε if t ∈ A.
Simple algebra shows that x = 12 y + 12 y. In addition |y(t)| ≤ 1 a.e, and |z(t)| ≤ 1
a.e, so we can write x as a convex combination of functions in S which are
different from x. (Alternatively, we can take y = x + 21 (1 − |x|) and z =
x − 12 (1 − |x|).)
(c) Let x be a nonzero element of L1 , and define
Z t
F (t) =
|x|.
0
It follows from Lemma 5.7 that F is a continuous function of t with F (0) = 0
and F (1) = kxk, so there is T ∈ (0, 1) such that F (T ) = 21 kxk. If kxk ≤ 1, we
define y and z by
(
(
2x(t) if t ≤ T,
0
if t ≤ T,
y(t) =
z(t) =
0
if t > T,
2x(t) if t > T.
Then kxk = kyk = kzk and x = 12 y + 21 z. Because x isn’t zero, it follows that
kxk 6= 0, so x must be different from y and z. If x = 0, we can write x = 21 y + z1
with y(t) = 1 and z(t) = −1. Hence the unit sphere in L1 has no extreme points.
(d) Suppose L1 were the dual of some normed linear space X. Simple algebra shows
that the unit sphere in L1 is convex. In addition, Alaouglu’s theorem implies
that the unit sphere is compact in the weak∗ topology. Furthermore, the remark
at the top of page 241 implies that the weak∗ topology is locally convex. It
then follows from the Krein-Milman theorem that S is the closed convex hull of
its extreme points. However, part (c) shows that there are no extreme points.
Since the convex hull of the empty set is empty, we have a contradiction. Hence
L1 is not the dual of any normed linear space.
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(e) From the arguments in part (a), we conclude that the extreme points of the unit
sphere in lp are all sequences x with kxkp = 1 for 1 < p < ∞. The argument in
part (b) shows that the extreme points of the unit sphere in l∞ are all sequences
(xn ) with |xn | = 1 for all n. Finally, if x ∈ l1 has the form xn = 1 for some n and
xk = 0 for k 6= n, then x must be an extreme point because if x = λy + (1 − λ)z,
then 1 = λyn + (1 − λ)zn , so yn = zn = 1 and y = z = x, which means that
x is an extreme point. A similar argument shows that xn = 1 and xk = 0 for
k 6= n gives an extreme point. to see that these are all the extreme points, we
first observe that kxk < 1 implies that x is not an extreme point. (If kxk 6= 0,
then we can write x = λy + (1 − λ)z with y = x/kxk, λ = kxk and z = 0. If
kxk = 0, we can write x = λy + (1 − λ)z with λ = 12 and y = −z any nonzero
element with norm less than one.) If x ∈ l1 with kxk = 1 and |xn | < 1 for all
n, then there are two positive integers m and k such that |xm | ≥ |xk | > 0. We
define




xm + xk if n = m,
xm − xk if n = m,
yn = 0
zn = 2xk
if n = k,
if n = k,


x

otherwise,
xn
otherwise.
n
Then x = 12 x + 12 y and kyk = kzk = 1. Moreover, x 6= y and x 6= z, so x is not
an extreme point.
(f) The argument from part (b) shows that if x is an extreme point,of the unit
sphere in C(X), then |x(t)| = 1 everywhere. On the other hand, if there is a
t ∈ X such that |x(t)| < 1, then there is ε > 0 such that |x(t)| < 1 − ε. Since
x is continuous, there is an open set U with x ∈ U such that |x| < 1 − ε in
U . Since X is compact and Hausdorff, it is normal by Problem 9.4, and then
Urysohn’s Lemma gives a continuous function g such that g(t) = 1, g = 0 on
X \ U and 0 ≤ g ≤ 1 on X. (Note that {t} is closed by Proposition 8.6 because
X is Hausdorff.) We now set y = x + εg and z = x − εg. It’s clear that y and z
are continuous and that x = 21 y + 12 z. For w ∈ U , we have
|y(w)| ≤ |x(w)| + ε|g(w)| ≤ 1 − ε + ε = 1
and for w ∈
/ U , we have |y(w)| = |x(w)| ≤ 1, so y ∈ S. A similar argument
shows that z ∈ S, so x can’t be an extreme point of C(X).
X = [0, 1], the only functions with |x(t)| = 1 everywhere are x1 = 1 and
x2 = −1. Since there are nonconstant continuous functions in the unit sphere of
C[0, 1], the Krein-Milman theorem shows (just as in part (d)) that C[0, 1] can’t
be the dual of any normed linear space.
52. (a) Let (ϕα )α∈I be an orthonormal system. From Bessel’s inequality, we know that
X
(x, ϕα )2 ≤ kxk2
α∈E
for any finite subset E of I, so problem 2.21a implies that only countably many
coefficients (x, ϕα ) can be nonzero.
4
(b) If (ϕn )∞
n=1 is a (countable) complete orthonormal system, then the set of all finite
linear combinations of the form
k
X
αn ϕn
n=1
with k ∈ N and each αn ∈ Q is a countable dense subset of X. Hence, if a
Hilbert space has a countable complete orthonormal system, it’s separable. By
contraposition, in a nonseparable Hilbert space, every complete orthonormal
system is uncountable. The proof of Proposition 55 shows that each x ∈ X is
the sum
∞
X
(x, ϕi )ϕi
2
P∞
i=1
2
∞
and that kxk =
i=1 (x, ϕ) , where (ϕi )i=1 is the collection of all ϕα with
(x, ϕα ) 6= 0.
(c) This was done in class.
√
(d) Let (ϕα )α∈I be a complete orthonormal system. Then kϕα − ϕβ k = 2 if
α 6= β. Since a dense set must contain at least one point in each ball B(ϕα , 1),
it follows that a dense set must have at least as many elements as the complete
orthonormal system. On the other hand, the set of all finite linear combinations
of the ϕα ’s with rational coefficients is a dense subset with the same cardinality
as the complete orthonormal system.
(e) Let f : X → Y be an isometry and let (xα ) be a complete orthonormal system
for X. Then (f (xα )) is an orthonormal system for Y . Moreover, if (y, f (xα )) = 0
for all α, then (f −1 (y), xα ) = 0 for all α. Hence f −1 (y) = 0 and therefore y = 0,
so (f (xα )) is a complete orthonormal system for Y . It follows that X and Y
have the same dimension.
Conversely, if X and Y have the same dimension, let (xα )α∈I be a complete
orthonormal system for X and let (yα )α∈I be a complete orthonormal system
for Y . (Note that we can use the same index set I for both complete orthonormal
systems because they have the same cardinality.) We then define the function
f : X → Y by
³X
´ X
f
aα xα =
aα yα .
From Proposition 10.27, we conclude that the sum in Y converges if and only
if the sum is X converges, so this function is onto. A quick calculation shows
that f is also an isometry.
(f) Fix a cardinal number n, and let I be a set with cardinality n. We take H to be
the subset of all (xα )α∈I of RI such that only countably many xα ’s are nonzero
and such that
X
(xα )2 < ∞.
We define addition and scalar multiplication on H in the usual way:
(xα ) + (yα ) = (xα + yα ),
c(xα ) = (cxα )
5
for all x and y in H and all real c. In addition, we define the inner product by
X
((xα ), (yα )) =
xα y α .
(Since only countably many terms in this sum are nonzero, we conclude from
Cauchy’s inequality for l2 that the sum converges for all x and y in H.) Hence
H is an inner product space.
In addition, if (ξn ) is a Cauchy sequence in H with ξn = (xn,α ), then Bessel’s
inequality implies that (xn,α ) is a Cauchy sequence for each α. Setting xα =
lim xn,α , we conclude that ξn → (xα ), so H is compete, and hence it’s a Hilbert
space.
55. First, let hf1 , . . . , fn i be a finite orthonormal system in S. Fix x ∈ [0, 1] and define
the function f by
n
X
f (t) =
fi (x)fi (t).
i=1
Then f ∈ S, and Parseval’s identity implies that
n
X
2
kf k =
fi (x)2 .
i=1
But this sum is just f (x), so
kf k2 = f (x) = |f (x)| ≤ Kkf k.
It follows that kf k2 ≤ K 2 , so |f (x)| ≤ K 2 as well. Therefore
n
X
fi (x)2 ≤ K 2
i=1
for all x ∈ [0, 1].
We now integrate this inequality to infer that
Z 1
Z 1X
n
n Z
X
2
2
2
K =
K ≥
fi (x) dx =
0
R1
0
i=1
i=1
1
fi (x)2 dx.
0
Pn
But 0 fi (x)2 dx = 1 because the fi ’s are orthonormal, so K 2 ≥ i=1
1 = n. In other
words, any finite orthonormal system has at most K 2 elements,so the dimension of
S is at most K 2 .
Chapter 11
3. (a) First, we have
E1 = (E1 ∩ E2 ) ∪ (E1 \ E2 )
and
(E1 ∩ E2 ) ∩ (E1 \ E2 ) = ∅,
so
µE1 = µ(E1 ∩ E2 ) + µ(E1 \ E2 ).
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Next, E1 ∩ E2 ⊂ E2 and E1 \ E2 ⊂ E1 ∆ E2 , so
µE1 ≤ µE2 + µ(E1 ∆ E2 ) = µE2 .
Similarly µE2 ≤ µE1 , so µE1 = µE2 .
(b) Now we have
E2 = (E1 ∩ E2 ) ∪ (E2 \ E1 ).
Since E2 \ E1 ⊂ E2 ∆ E1 and E2 ∆ E1 has measure zero, it follows from the
completeness of µ that E2 \ E1 is also measurable. In addition,
E1 ∩ E2 = E1 \ (E1 \ E2 )
and E1 \E2 is measurable, so E1 ∩E2 is measurable. Hence E2 is also measurable.
11. Fix α ∈ R and set
Fα = {x ∈ X : f (x) < α},
Gα = {x ∈ X : g(x) < α},
N = {x : f (x) 6= g(x)}.
Then Gα \ Fα ⊂ N , so Gα \ Fα is measurable because µ is complete. In addition,
Gα ∩ Fα = Fα \ (Fα \ Gα ), so Gα ∩ Fα is measurable. Hence
Gα = (Gα \ Fα ) ∪ (Gα ∩ Fα )
is also measurable.
If µ is not complete, then there is a set N with µN = 0 which has a nonmeasurable
subset A. If we take f ≡ 0 and g = χA , then f is measurable because we have
(
X if α < 0,
{x : f (x) > α} =
∅
if α ≥ 0.
But g is not measurable because
1
{x : g(x) } = A,
2
which is not measurable. Since f = g on X \ N and µN = 0, it follows that f = g
a.e., so Proposition 11.8 is not true when µ is not complete.