Math 515
Professor Lieberman
October 27, 2004
HOMEWORK #9 SOLUTIONS
Chapter 6
15. By problem 14, k · k∞ is a norm on l∞ , so we only have to show that c and c0 are linear
spaces. But, if hξn i and hηn i are elements of c and if α and β are real numbers, then
lim αξn + βηn = α lim ξn + β lim ηn .
It follows that hαξn + βηn i is also a convergent sequence, so it’s in c. Moreover, if both
sequences have zero as a limit, then we conclude that hαξn + βηn i also has limit zero.
To see that they are complete, suppose hξn i is a Cauchy sequence of elements of c (so
ξn = hξn,m i). Again problem 14 shows that this sequence has a limit η = hηn i in l∞ . For
simplicity of notation, we write Ln for the limit of the sequence ξn . We claim that hLn i is
a Cauchy sequence, so it has a limit L and that L = lim ηn .
To prove this claim, let ε > 0 be given. Then there is a constant N so that
ε
sup |ξn,m − ξj,m | <
3
m
for all m, provided n, j ≥ N (because hξn i is Cauchy). Moreover, for n ≥ N , there is a
constant M (n) such that
ε
|ξn,m − Ln | <
3
provided m ≥ M (n). Fixing n, j ≥ N , we take m ≥ max{M (n), M (j)}. Then
|Ln − Lj | ≤ |Ln − ξn,m | + |ξn,m − ξj,m | + |ξj,m − Lj | < ε.
It follows that hLn i is a Cauchy sequence, and we write L for its limit. Given ε > 0, there
is an N so that
ε
sup |ξn,m − ηm | <
2
m
if n ≥ N . There is also a constant J such that |Ln − L| < 4ε if n ≥ J and a constant M
such that |ξn,m − Ln | < 4ε if m ≥ M and n = max{N, J}. For m ≥ M and n = max{N, J},
it follows that
|ηm − L| ≤ |ηm − ξn,m | + |ξn,m − Ln | + |Ln − L| < ε,
so L = lim ηm . It follows that c is complete. Moreover, if each ξn is in c0 , then lim ηm = 0,
so c0 is also complete.
17. First, we assume that p > 1. Here is the proof for finite measure spaces. Without loss of
generality kgkq > 0. Then, given ε, there is δ > 0 such that
Z
εq
|g|q <
(4M )q
E
for any set E with µE < δ. In addition, by problem 11.13b, the sequence converges in
measure to f , so there is a positive integer N such that µEn < δ for n ≥ N , where
ε
En = x : |fn − f |(x) >
.
2(µX)1/p kgkq
1
2
Writing Fn = X ∼ En , we have
Z
Z
Z
|fn − f ||g| +
|fn − f ||g| =
En
Z
q
|fn − f ||g|
Fn
1/q
ε
+
2(µX)1/p kgkq
Z
|g|
≤ kf − fn kp
|g|
Fn
ε ε
(µX)1/p kgkq = ε
≤ 2M
+
2M
2(µX)p kgkq
if n ≥ N . It follows that
Z
Z
fn g →
f g.
If X has infinite measure, then problem 11.21a implies that Y = {x ∈ X : g(x) 6= 0} is
σ-finite. Then
Z
Z
Z
Z
fn g =
fn g and
=
f g,
X
Y
X
Y
so we may assume that X is σ-finite. Writing X = ∪Xj , with Xj ⊂ Xj+1 and µXj < ∞
for all j, we conclude that
Z
Z
Z
(fn − f )g =
(fn − f )g +
(fn − f )g.
Xj
X∼Xj
Given ε > 0, there is an integer J such that
Z
ε q
|g|q <
4M
X∼Xj
if j ≥ J and hence
Z
ε
2
X∼XJ
for all n. The finite measure situation just proved shows that there is an integer N such
that
Z
ε
|fn − f ||g| <
2
Xj
if n ≥ N . It follows that
Z
(fn − f )g < ε
R
R
for n ≥ N and hence fn g → f g.
The result is not true for p = 1, even on a finite measure space. Take µ to be Lebesgue
measure on [0, 1], define fn by
(
n if 0 ≤ x < n1 ,
fn =
0 if n1 ≤ x ≤ 1,
R
R
and let g ≡ 1. Then fn → 0 a.e. but fn g = 1 and f g = 0.
(fn − f )g ≤
21. (a) If kgk = 0, take f = 1. Otherwise, take f = sgn g.
(b) Set E(ε) = {x : |g(x)| ≥ kgk∞ − ε}, and take f = sgn gχE(ε) . Then
Z
Z
fg =
|g| ≥ (kgk∞ − ε)mE(ε) = (kgk∞ − ε)kf k1 .
E(ε)
3
23. We first note that lp is just Lp (N, n), where N is the set of positive integers and n is counting
measure. The proof of Proposition 80 shows that, if η ∈ l1 , then the equation
∞
X
(1)
G(ξ) =
ξν ζn
ν=1
∞
l and
defines a bounded linear functional on
kGk = kηk1 . Hence this gives a bounded
linear functional on c and on c0 .
We first claim that every bounded linear functional F on c0 has this form. To prove this
claim, write eν = hen,ν i for the vector in c0 defined by
(
1 if n = ν,
eν,n =
0 if n 6= ν,
and set ην = F (eν ). We now define G by (1), and we show that G = F .
To show that G = F , we fix ξ ∈ l∞ , and we define a sequence hζn i of vectors in c0 by
(
ξn if ν ≤ n,
ζn,ν =
0
if ν > n.
Since ξ ∈ c0 , it follows that ξν → 0 as ν → ∞ and hence ζn → ξ in l∞ . By linearity of F
and G, we have G(ζn ) = F (ζn ) for all n and hence G(ξ) = F (ξ), so every bounded linear
functional on c0 has the form described above.
We now represent a bounded linear functional F on c as a sum. We define Ξ to be the
vector with Ξν = 1 for all ν and set α = F (Ξ). We also write F0 for the functional F
restricted to c0 . The claim now is that
F (ξ) = F0 (ξ − LΞ) + αL
with L = lim ξn . To prove this claim, we use linearity to infer that
F (ξ) = F (ξ − LΞ) + F (LΞ).
But ξ − LΞ ∈ c0 , so F (ξ − LΞ) = F0 (ξ − LΞ), and F (LΞ) = LF (Ξ) = αL. It follows from
(i) that there is a vector η ∈ l1 such that
X
F (ξ) =
ην [ξν − L] + αL.
Conversely, given a vector η ∈ l1 and any α ∈ R, this equation gives a bounded linear
functional on c, so this is our representation.