18.3 Representation Theorems For The Dual Space Of L^{p}
Recall the concept of the dual space of a Banach space in the Chapter on Banach space
starting on Page 1235. The next topic deals with the dual space of L^{p} for p ≥ 1 in the
case where the measure space is σ finite or finite. In what follows q = ∞ if p = 1 and
otherwise,
-1
p
+
1
q
= 1.
Theorem 18.3.1(Riesz representation theorem) Let p > 1 and let (Ω,S,μ) be a finitemeasure space. If Λ ∈ (L^{p}(Ω))^{′}, then there exists a unique h ∈ L^{q}(Ω)(
1p
+
1q
= 1) suchthat
∫
Λf = hfdμ.
Ω
This function satisfies ||h||_{q} = ||Λ|| where
||Λ||
is the operator norm of Λ.
Proof: (Uniqueness) If h_{1} and h_{2} both represent Λ, consider
-- --
f = |h1 − h2|q−2(h1 − h2),
where h denotes complex conjugation. By Holder’s inequality, it is easy to see that
f ∈ L^{p}(Ω). Thus
This is because if x ∈ Ω, x is contained in exactly one of the A_{i} and so the absolute value
of the sum in the first integral above is equal to 1. Therefore |λ|(Ω) < ∞ because this was
an arbitrary partition. Also, if {E_{i}}_{i=1}^{∞} is a sequence of disjoint sets of S,
let
It is also clear from the definition of λ that λ ≪ μ. Therefore, by the Radon Nikodym
theorem, there exists h ∈ L^{1}(Ω) with
∫
λ(E) = hdμ = Λ(XE).
E
Actually h ∈ L^{q} and satisfies the other conditions above. Let s = ∑_{i=1}^{m}c_{i}X_{Ei} be a
simple function. Then since Λ is linear,
∑m ∑m ∫ ∫
Λ (s) = ciΛ(XEi) = ci hdμ = hsdμ. (18.3.9)
i=1 i=1 Ei
(18.3.9)
Claim: If f is uniformly bounded and measurable, then
∫
Λ (f) = hfdμ.
Proof of claim:Since f is bounded and measurable, there exists a sequence of
simple functions,
{sn}
which converges to f pointwise and in L^{p}
(Ω )
. This follows
from Theorem 9.3.9 on Page 644 upon breaking f up into positive and negative
parts of real and complex parts. In fact this theorem gives uniform convergence.
Then
by 18.3.10, and Holder’s inequality. This proves the theorem.
To represent elements of the dual space of L^{1}(Ω), another Banach space is
needed.
Definition 18.3.2Let (Ω,S,μ) be a measure space. L^{∞}(Ω) is the vector spaceof measurable functions such that for some M > 0,|f(x)|≤ M for all x outside ofsome set of measure zero (|f(x)|≤ Ma.e.). Define f = g when f(x) = g(x)a.e. and||f||_{∞}≡ inf{M : |f(x)|≤ Ma.e.}.
Theorem 18.3.3L^{∞}(Ω) is a Banach space.
Proof: It is clear that L^{∞}(Ω) is a vector space. Is ||||_{∞} a norm?
Claim:If f ∈ L^{∞}
(Ω)
, then
|f (x)|
≤
||f||
_{∞} a.e.
Proof of the claim:
{x : |f (x)| ≥ ||f|| + n− 1}
∞
≡ E_{n} is a set of measure zero
according to the definition of
||f||
_{∞}. Furthermore,
{x : |f (x)| > ||f|| }
∞
= ∪_{n}E_{n} and so
it is also a set of measure zero. This verifies the claim.
Now if
||f||
_{∞} = 0 it follows that f
(x)
= 0 a.e. Also if f,g ∈ L^{∞}
(Ω)
,
|f (x)+ g(x)| ≤ |f (x)|+ |g(x)| ≤ ||f||∞ + ||g||∞
a.e. and so
||f||
_{∞} +
||g||
_{∞} serves as one of the constants, M in the definition of
||f + g||
_{∞}. Therefore,
||f + g||∞ ≤ ||f||∞ + ||g||∞ .
Next let c be a number. Then
|cf (x)|
=
|c|
|f (x)|
≤
|c|
||f||
_{∞} and so
||cf ||
_{∞}≤
|c|
||f||
_{∞}.
Therefore since c is arbitrary,
||f||
_{∞} =
||c(1∕c) f||
_{∞}≤
||1||
c
||cf||
_{∞} which implies
|c|
||f||
_{∞}≤
||cf||
_{∞}. Thus ||||_{∞} is a norm as claimed.
To verify completeness, let {f_{n}} be a Cauchy sequence in L^{∞}(Ω) and use the
above claim to get the existence of a set of measure zero, E_{nm} such that for all
x
∕∈
E_{nm},
|fn(x)− fm(x)| ≤ ||fn − fm||∞
Let E = ∪_{n,m}E_{nm}. Thus μ(E) = 0 and for each x
∕∈
E,{f_{n}(x)}_{n=1}^{∞} is a Cauchy
sequence in ℂ. Let
{
0 if x ∈ E
f(x) = limn→ ∞ fn(x ) if x ∕∈ E = nl→im∞ XEC(x)fn(x).
Then f is clearly measurable because it is the limit of measurable functions.
If
Fn = {x : |fn (x)| > ||fn||∞ }
and F = ∪_{n=1}^{∞}F_{n}, it follows μ(F) = 0 and that for x
Hence ||f − f_{n}||_{∞}< ε for all n large enough. This proves the theorem.
The next theorem is the Riesz representation theorem for
( 1 )
L (Ω)
^{′}.
Theorem 18.3.4(Riesz representation theorem) Let (Ω,S,μ) be a finite measurespace. If Λ ∈ (L^{1}(Ω))^{′}, then there exists a unique h ∈ L^{∞}(Ω) such that
∫
Λ(f) = hfdμ
Ω
for all f ∈ L^{1}(Ω). If h is the function in L^{∞}(Ω) representing Λ ∈ (L^{1}(Ω))^{′}, then||h||_{∞} = ||Λ||.
Proof: Just as in the proof of Theorem 18.3.1, there exists a unique h ∈ L^{1}(Ω) such
that for all simple functions, s,
∫
Λ(s) = hsdμ. (18.3.11)
(18.3.11)
To show h ∈ L^{∞}(Ω), let ε > 0 be given and let
E = {x : |h(x)| ≥ ||Λ||+ ε}.
Let |k| = 1 and hk = |h|. Since the measure space is finite, k ∈ L^{1}(Ω). As in Theorem
18.3.1 let {s_{n}} be a sequence of simple functions converging to k in L^{1}(Ω), and pointwise.
It follows from the construction in Theorem 9.3.9 on Page 644 that it can be assumed
|s_{n}|≤ 1. Therefore
∫ ∫
Λ(kXE ) = lim Λ (snXE ) = lim hsndμ = hkdμ
n→ ∞ n→ ∞ E E
where the last equality holds by the Dominated Convergence theorem. Therefore,
This proves the existence part of the theorem. To verify uniqueness, suppose h_{1} and h_{2}
both represent Λ and let f ∈ L^{1}(Ω) be such that |f|≤ 1 and f(h_{1}− h_{2}) = |h_{1}− h_{2}|.
Then
and so η is one to one and in fact preserves norms. I claim that also η is onto. To see this,
let g ∈ L^{p}(Ω,
^μ
) and consider the function, r^{1p
}g. Then
∫ ∫ ∫
||r1pg||pdμ = |g|prdμ = |g|pd^μ < ∞
| |
Thus r^{}
1p
g ∈ L^{p}
(Ω,μ)
and η
( )
r1pg
= g showing that η is onto as claimed. Thus η is one
to one, onto, and preserves norms. Consider the diagram below which is descriptive of the
situation in which η^{∗} must be one to one and onto.
∗
p′ p ′ ^ η p ′
h,L (^μ) L (^μ),Λ → L (μ) ,Λ
p η p
L (^μ) ← L (μ)
Then for Λ ∈ L^{p}
(μ)
^{′}, there exists a unique
^Λ
∈ L^{p}
(μ^)
^{′} such that η^{∗}
^Λ
= Λ,
|| ||
||||^Λ||||
=
||Λ ||
.
By the Riesz representation theorem for finite measure spaces, there exists a unique
h ∈ L^{p′
}
(^μ )
which represents
^Λ
in the manner described in the Riesz representation
theorem. Thus
and represents Λ in the appropriate way.
If p = 1, then 1∕p^{′}≡ 0. This proves the Lemma.
A situation in which the conditions of the lemma are satisfied is the case where the
measure space is σ finite. In fact, you should show this is the only case in which the
conditions of the above lemma hold.
Theorem 18.3.6(Riesz representation theorem)Let (Ω,S,μ) be σ finite andlet
Λ ∈ (Lp (Ω, μ))′,p ≥ 1.
Then there exists a unique h ∈ L^{q}(Ω,μ),L^{∞}(Ω,μ) if p = 1 such that
∫
Λf = hf dμ.
Also ||h|| = ||Λ||.(||h|| = ||h||_{q}if p > 1,||h||_{∞}if p = 1). Here