19.8 The Riesz Representation Theorem
The Riesz representation theorem for the spaces L ^{p}
holds under certain conditions.
The proof follows the proofs given earlier for scalar valued functions.
Definition 19.8.1 If X and Y are two Banach spaces, X is isometric to Y if there
exists θ ∈ℒ
such that
This will be written as X
Y . The map θ is called an isometry.
The next theorem says that L ^{p′
}
is always isometric to a subspace of
^{′} for any Banach space,
X .
Theorem 19.8.2 Let X be any Banach space and let
be a finite measure
space. Let p ≥ 1
and let 1
∕p + 1
∕p ^{′} = 1
.(If p = 1
, p ^{′} ≡∞ . )
Then L ^{p′
} is isometric
to a subspace of ^{′} . Also, for g ∈ L ^{p′
} ,
|∫ |
|| ||
||sf|u|p≤1| Ω g(s)(f (s))dμ| = ||g||p′ .
p
Proof: First observe that for f ∈ L ^{p}
and
g ∈ L ^{p′
} ,
is a function in L ^{1}
. (To obtain measurability, write
f as a limit of simple functions.
Holder’s inequality then yields the function is in
L ^{1} .) Define
θ : Lp′ (Ω;X ′) → (Lp (Ω;X ))′
by
∫
θg(f) ≡ g(s)(f (s))dμ.
Ω
Holder’s inequality implies
||θg|| ≤ ||g||p′ (19.8.47)
(19.8.47)
and it is also clear that θ is linear. Next it is required to show
This will first be verified for simple functions. Let
where c _{i} ∈ X ^{′} , the E _{i} are disjoint and
Then
∈ L ^{p′
} . Let
ε > 0 be given. By the scalar Riesz representation theorem,
there exists
h ∈ L ^{p} such that
_{p} = 1 and
∫
||g(s)|| h (s)dμ ≥ ||g|| ′ − ε.
Ω X′ Lp(Ω;X′)
Now let d _{i} be chosen such that
ci(di) ≥ ||ci||X′ − ε∕||h||L1(Ω)
and
_{X} ≤ 1. Let
m∑
f (s) ≡ dih (s)XEi (s).
i=1
Thus f ∈ L ^{p}
and
_{Lp(Ω;X)
} ≤ 1. This follows from
∫ m∑
||f||pp = ||di||pX |h(s)|pXEi (s)dμ
Ω i=1
∑m ( ∫ p ) p ∫ p
= |h (s) |dμ ||di||X ≤ |h|dμ = 1.
i=1 Ei Ω
Also
||∫ ||
||θg|| ≥ |θg(f)| = || g(s)(f (s))dμ || ≥
Ω
||∫ m ( ) ||
|| ∑ ||ci|| ′ − ε∕||h||1 h(s)XE (s)dμ||
|Ω i=1 X L (Ω) i |
|∫ | |∫ |
≥ || ||g (s)|| h (s)dμ ||− ε|| h(s)∕||h|| dμ||
| Ω X′ | |Ω L1(Ω) |
Since ε was arbitrary,
and from 19.8.47 this shows equality holds in 19.8.48 whenever g is a simple
function.
In general, let g ∈ L ^{p′
}
and let
g _{n} be a sequence of simple functions converging
to
g in
L ^{p′
} . Then
||θg|| = lim ||θgn|| = lim ||gn|| = ||g||.
n→ ∞ n→∞
This proves the theorem and shows θ is the desired isometry.
Theorem 19.8.3 If X is a Banach space and X ^{′} has the Radon Nikodym property,
then if
is a finite measure space,
′
(Lp (Ω; X))′ ∼= Lp (Ω;X ′)
and in fact the mapping θ of Theorem 19.8.2 is onto.
Proof: Let l ∈
^{′} and define
F ∈ X ^{′} by
Lemma 19.8.4 F defined above is a vector measure with values in X ^{′} and
< ∞ .
Proof of the lemma: Clearly F
is linear. Also
||F (E)|| = sup ||F (E) (x)||
||x||≤1
1∕p
≤ ||l|| sup ||XE (⋅)x||Lp(Ω;X) ≤ ||l||μ (E) .
||x||≤1
Let { E _{i} } _{i=1} ^{∞} be a sequence of disjoint elements of S and let E = ∪ _{n<∞} E _{n} .
|| ∑n || || ∑n ||
||F (E )(x) − F (Ek)(x)|| = ||l(XE (⋅)x)− l(XEi (⋅)x)|| (19.8.49)
| k=1 | | i=1 |
|||| ∑n ||||
≤ ||l||||||XE (⋅)x− XEi (⋅)x||||
|| i=1 ||Lp(Ω;X )
( )1∕p
≤ ||l||μ ⋃ E ||x||.
k>n k
Since
μ < ∞ ,
( ⋃ )1∕p
nli→m∞ μ Ek = 0
k>n
and so inequality 19.8.49 shows that
|| n ||
lim ||||F (E) − ∑ F (E )|||| = 0.
n→∞ |||| k=1 k|||| ′
X
To show
< ∞ , let
ε > 0 be given, let
be a partition of Ω
, and
let
≤ 1 be chosen in such a way that
F (H ) (x ) > ||F (H )||− ε∕n.
i i i
Thus
|| ||
∑n ∑n ||||∑n ||||
− ε+ ||F (Hi)|| < l(XHi (⋅)xi) ≤ ||l|||||| XHi (⋅)xi||||
i=1 i=1 i=1 Lp(Ω;X )
( )1∕p
∫ ∑n 1∕p
≤ ||l|| Ω XHi (s)dμ = ||l||μ (Ω) .
i=1
Since ε > 0 was arbitrary,
∑n
||F (Hi)|| < ||l||μ(Ω)1∕p .
i=1
Since the partition was arbitrary, this shows
≤ μ ^{1∕p} and this proves the
lemma.
Continuing with the proof of Theorem 19.8.3 , note that
Since X ^{′} has the Radon Nikodym property, there exists g ∈ L ^{1}
such
that
Also, from the definition of F
,
( n∑ ) ∑n
l xiXEi (⋅) = l(XEi (⋅)xi)
i=1 i=1
∑n ∑n ∫
= F (Ei)(xi) = g (s)(xi)dμ. (19.8.50)
i=1 i=1 Ei
(19.8.50)
It follows from 19.8.50 that whenever h is a simple function,
∫
l(h ) = Ωg (s)(h (s))dμ. (19.8.51)
(19.8.51)
Let
Gn ≡ {s : ||g (s)||X′ ≤ n}
and let
j : Lp (G ;X ) → Lp (Ω;X )
n
be given by
{
jh (s) = h(s) if s ∈ Gn,
0 if s ∕∈ Gn.
Letting h be a simple function in L ^{p}
,
∫
j∗l(h) = l(jh) = g (s)(h(s))dμ. (19.8.52)
Gn
(19.8.52)
Since the simple functions are dense in L ^{p}
, and
g ∈ L ^{p′
} , it follows
19.8.52 holds for all
h ∈ L ^{p} . By Theorem
19.8.2 ,
∗
||g||Lp′(Gn;X′) = ||j l||(Lp(Gn;X))′ ≤ ||l||(Lp(Ω;X))′ .
By the monotone convergence theorem,
||g||Lp′(Ω;X′) = lim ||g||Lp′(Gn;X′) ≤ ||l||(Lp(Ω;X))′ .
n→ ∞
Therefore g ∈ L ^{p′
}
and since simple functions are dense in
L ^{p} ,
19.8.51
holds for all
h ∈ L ^{p} . Thus
l =
θg and the theorem is proved because, by
Theorem
19.8.2 ,
=
and the mapping
θ is onto because
l was arbitrary.
■
As in the scalar case, everything generalizes to the case of σ finite measure spaces.
The proof is almost identical.
Lemma 19.8.5 Let
be a σ finite measure space and let X be a Banach space
such that X ^{′} has the Radon Nikodym property. Then there exists a measurable function, r
such that r > 0
for all x , such that < M for all x , and ∫
rdμ < ∞ .
For
there exists a unique h ∈ L ^{p′
} (Ω;X ^{′} ), L ^{∞} (Ω;X ^{′} ) if p = 1 such that
Also || h || = || Λ|| . (|| h || = || h || _{p′} if p > 1, || h || _{∞} if p = 1). Here
Proof: First suppose r exists as described. Also, to save on notation and to
emphasize the similarity with the scalar case, denote the norm in the various spaces by
. Define a new measure
, according to the rule
∫
^μ(E) ≡ rdμ. (19.8.53)
E
(19.8.53)
Thus
is a finite measure on
S . Now define a mapping,
η :
L ^{p} (Ω;
X,μ )
→ L ^{p} (Ω;
X, )
by