Chapter 5
Lp-spaces
5.1
Reduction to the Case `p and Lp
The main (and only) result of this section is the following Theorem.
Theorem 5.1.1. Let 1  p < 1 and let (⌦, ⌃, µ) be a separable measure
space, i.e. ⌃ is generated by a countable set of subsets of ⌦.
Then there is a countable set I so that Lp (⌦, ⌃, µ) is isometrically isomorphic to Lp [0, 1] p `p (I) or `p (I).
Moreover, if (⌦, ⌃, µ) has no atoms we can choose I to be empty and,
thus, Lp (⌦, ⌃, µ) is isometrically isomorphic to Lp [0, 1].
Proof. First note that the assumption that ⌃ is generated by a countable set
say D ⇢ P(⌦)Simplies that Lp (µ) is separable. Indeed, the algebra generated
by D is A = 1
n=1 An , where An is defined by recursively for every n 2 N as
follows: A1 = D, and, assuming, An is defined we let first
A0n+1 =
k
n[
j=1
Bj : k 2 N, Bj 2 An or Bjc 2 An
o
and then
An+1 =
k
n\
j=1
o
Bj : k 2 N, Bj 2 An or Bjc 2 An .
This proves that A is countable. Then we observe that span(1A : A 2 A) is
dense in Lp (µ)
We first reduce to the -finite case.
117
118
CHAPTER 5. LP -SPACES
Step 1: Lp (⌦, ⌃, µ) is isometrically isomorphic to a space Lp (⌦0 , ⌃0 , µ0 ) where
(⌦0 , ⌃0 , µ0 ) is a -finite measure space.
Let (fn ) ⇢ Lp (⌦, ⌃, µ) be a dense sequence in Lp (⌦, ⌃, µ) and define
[
⌦0 =
|fn | > 0 .
n2N
Since {|fn | > 0} is a countable union of sets of finite measure, namely
[
{|fn | > 0} =
{|fn | > 1/m}
m2N
⌦0 is also -finite. Moreover, for any f 2 Lp (⌦, ⌃, µ) it follows that {|f | >
0} ⇢ ⌦0 µ a.e. Therefore we can choose ⌃0 = ⌃|⌦0 = {A 2 ⌃ : A ⇢ ⌦0 } and
µ0 = µ|⌃0 .
Step 2: Assume (⌦, ⌃, µ) is a -finite measure space. Let I be the set of all
atoms of (⌦, ⌃, µ). SinceSµ is -finite, I is countable, and µ(A) < 1 for all
A 2 I. We put ⌦0 = ⌦ \ A2I A, ⌃0 = ⌃|⌦0 and µ0 = µ|⌃0 . Then
!
⇣Z
⌘
T : Lp (⌦, ⌃, µ) ! `p (I) p Lp (⌦0 , ⌃0 , µ0 ), f 7!
f dµ : A 2 I , f |⌦0 ,
A
is an isometry onto `p (I) p Lp (⌦0 , ⌃0 , µ0 ).
Now either µ0 = 0 or it is an atomless -finite measure.
In the next step we reduce to the case of µ being an atomless probability
measure.
Step 3: Assume that (⌦, ⌃, µ) is -finite, atomless and not 0. Then there
is an atomless probability µ0 on (⌦, ⌃) so that Lp (⌦, ⌃, µ) is isometrically
isomorphic to the space Lp (⌦, ⌃, µ0 ).
Since (⌦, ⌃, µ) is -finite there is an f 2 L1 (⌦, ⌃, µ), with f (!) > 0 for
all ! 2 ⌦ and kf k1 = 1. Let µ0 be the measure whose Radon Nikodym
derivative with respect to µ is f (thus µ0 is a probability measure) and
consider the operator
T : Lp (⌦, ⌃, µ) ! Lp (⌦, ⌃, µ0 ),
g 7! g · f
1/p
,
which is an isometry onto Lp (⌦, ⌃, µ0 ).
Step 4: Reduction to [0, 1]. Assume (⌦, ⌃, µ) is an atomless countably
generated probability space. Let (Bn ) ⇢ ⌃ be a sequence which generates
⌃. By induction we choose for each n 2 N0 a finite ⌃-partition Pn =
(n)
(n)
(n)
(P1 , P2 , . . . Pkn ) of ⌦ with the following properties:
(5.1)
{B1 , B2 , . . . Bn } ⇢ (Pn ) (the -algebra generated by Pn ),
5.1. REDUCTION TO THE CASE `P AND LP
(5.2)
(5.3)
(n)
µ(Pi )  2
n
119
, for i = 1, 2 . . . kn ,
Pn is a subpartition of Pn
each i 2 {1, . . . kn
1}
1
if n > 1, i.e. for
there are sn (i)  tn (i) in {1, . . . kn }, so that
tn (i)
(n 1)
Pi
[
=
(n)
Pj .
j=sn (i)
Put for n 2 N and 1  i  kn
h X
⌘
X
(n)
(n)
(n)
P̃i =
µ(Pj ),
µ(Pj ) , if j < kn and
ji 1
(n)
P̃kn =
h X
ji
(n)
µ(Pj ),
jkn 1
(n)
(n)
X
(n)
µ(Pj )
jkn
i
(n)
and P̃ (n) = (P̃1 , P̃2 , . . . , P̃kn ). Then P̃ (n) is a Borel partition of [0, 1],
S
(n)
(n)
with (P̃i ) = µ(Pi ), for each i  kn , and n2N P̃ (n) generate the Borel
-algebra on [0, 1].
For n 2 N put
Vn =
kn
nX
ai
i=1
(n)
Pi
Then Vn is a vector space and V =
Similarly Ṽ , with
Ṽn =
kn
nX
ai
i=1
o
: ai scalars ,
S
n Vn
is a dense subspace of LP (µ).
o
: ai scalars ,
(n)
P̃i
is a dense subspace of Lp [0, 1], and
T : V ! Ṽ ,
kn
X
i=1
ai
(n)
Pi
7!
kn
X
i=1
ai
(n)
P̃i
,
is an isometry whose image is dense in Lp [0, 1]. Thus T extends to an
isometry from Lp (µ) onto Lp [0, 1].
Exercises
120
1.
CHAPTER 5. LP -SPACES
Prove that for 1  p  1,
[
q>p
Lq [0, 1] ( Lp [0, 1].