23.5 Representation Theorems For The Dual Space Of L^{p}
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. Recall that the dual space of X is ℒ
(X,ℂ)
.
In fact, this theorem holds without the assumption that the measure space is σ finite in case p > 1 but this
is more technical to establish.
Theorem 23.5.1(Riesz representation theorem) Let p > 1 and let (Ω,S,μ) be a finite measure space. If
Λ ∈ (L^{p}(Ω))^{′}, then there exists a unique h ∈ L^{q}(Ω)(
1
p
+
1
q
= 1) such that
∫
Λf = Ωhf dμ.
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
0 = Λf − Λf =
∫
h |h − h |q−2(h-− h-)− h |h − h |q−2(h-− h-)dμ
1 1 2 1 2 2 1 2 1 2
∫
= |h1 − h2|qdμ.
Therefore h_{1} = h_{2} and this proves uniqueness.
Now let λ(E) = Λ(X_{E}). Since this is a finite measure space X_{E} is an element of L^{p}
(Ω)
and so it
makes sense to write Λ
(XE)
. In fact λ is a complex measure having finite total variation. This
follows from an easlier result but I will show it directly here. Let A_{1},
∫ n ∫
≤ ||Λ||( |∑ w X |pdμ)1p = ||Λ||( dμ)1p = ||Λ||μ (Ω )1p.
i=1 i Ai Ω
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 ) = E hdμ = Λ(XE).
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 Ei hdμ = hsdμ. (23.18)
i=1 i=1
(23.18)
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 19.1.6 on Page 1422 upon
breaking f up into positive and negative parts of real and complex parts. In fact this theorem gives uniform
convergence. Then
∫ ∫
Λ (f) = lnim→∞ Λ (sn) = lnim→∞ hsndμ = hfdμ,
the first equality holding because of continuity of Λ, the second following from 23.18 and the third holding
by the dominated convergence theorem.
This is a very nice formula but it still has not been shown that h ∈ L^{q}
(Ω)
.
Let E_{n} = {x : |h(x)|≤ n}. Thus |hX_{En}|≤ n. Then
To represent elements of the dual space of L^{1}(Ω), another Banach space is needed.
Definition 23.5.2Let (Ω,S,μ) be a measure space. L^{∞}(Ω) is the vector space of measurablefunctions such that for some M > 0,|f(x)| ≤ M for all x outside of some 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 23.5.3L^{∞}(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:
{ − 1}
x : |f (x)| ≥ ||f||∞ + n
≡ 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.
_{∞} 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||
_{∞} a.e. and so
||cf||
_{∞}≤
|c|
||f||
_{∞}. Therefore
since c is arbitrary,
||f||
_{∞} =
||c(1∕c)f||
_{∞}≤
| |
|1c|
||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
{
f(x ) = 0 if x ∈ E = lim XEC (x)fn(x).
limn →∞ fn(x) if x ∕∈ E n→∞
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
_{∞} by the triangle inequality.)
Thus f ∈ L^{∞}(Ω). Let n be large enough that whenever m > n,
||fm − fn||∞ < ε.
Then, if x
∕∈
E,
|f(x )− fn(x)| = lim |fm (x)− fn(x )| ≤ lim inf||fm − fn||∞ < ε.
m →∞ m →∞
Hence ||f − f_{n}||_{∞}< ε for all n large enough. ■
The next theorem is the Riesz representation theorem for
( 1 )
L (Ω)
^{′}.
Theorem 23.5.4(Riesz representation theorem) Let (Ω,S,μ) be a finite measure space. If Λ ∈ (L^{1}(Ω))^{′},then there exists a unique h ∈ L^{∞}(Ω) such that
∫
Λ (f ) = hf dμ
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 23.5.1, there exists a unique h ∈ L^{1}(Ω) such that for all simple
functions s,
∫
Λ(s) = hsdμ. (23.20)
(23.20)
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 23.5.1 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 19.1.6 on Page 1422 that it can be assumed |s_{n}|≤ 1. 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
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 23.5.6(Riesz representation theorem)Let (Ω,S,μ) be σ finite and let