25.3 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 1585. 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,
1p
+
1q
= 1.
Theorem 25.3.1(Riesz representation theorem) Let p > 1 and let (Ω,S,μ) be afinite measure space. If Λ ∈ (L^{p}(Ω))^{′}, then there exists a unique h ∈ L^{q}(Ω)(
1
p
+
1
q
= 1) suchthat
∫
Λf = hf dμ.
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 (X ).
E 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. (25.9)
i=1 i=1 Ei
(25.9)
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 7.1.6 on Page 445 upon breaking f up into positive and negative
parts of real and complex parts. In fact this theorem gives uniform convergence.
Then
∫ ∫
Λ(f) = lim Λ(s ) = lim hs dμ = hfdμ,
n→ ∞ n n→ ∞ n
the first equality holding because of continuity of Λ, the second following from 25.9 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
--
|hXEn |q−2(hXEn) ∈ Lp(Ω ).
By the claim, it follows that
∫
||hX ||q= h|hX |q−2(hX )dμ = Λ(|hX |q−2(hX ))
En q En En En En
To represent elements of the dual space of L^{1}(Ω), another Banach space is needed.
Definition 25.3.2Let (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 25.3.3L^{∞}(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.
_{∞} 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||
_{∞}≤
| |
|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 X C(x )f (x).
limn →∞ fn(x) if x ∕∈ E n→∞ 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 25.3.4(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) = Ωhf dμ
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 25.3.1, there exists a unique h ∈ L^{1}(Ω) such that
for all simple functions s,
∫
Λ (s) = hsdμ. (25.11)
(25.11)
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 25.3.1
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 7.1.6 on Page 445 that it can be assumed |s_{n}|≤ 1.
Therefore
∫ ∫
Λ (kX ) = lim Λ(s X ) = lim hs dμ = hkdμ
E n→ ∞ n E n→∞ E n 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
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 25.3.6(Riesz representation theorem)Let (Ω,S,μ) be σ finite andlet