For p > 1, you can represent the dual space of the Banach space L^{p}
(Ω )
. It turns out that it is just L^{q}
(Ω)
where 1∕p + 1∕q = 1. The proof involves the notion of uniform convexity. First recall Clarkson’s
inequalities. These fundamental inequalities were used to verify that L^{p}
(Ω)
is uniformly convex. More
precisely, the unit ball in L^{p}
∥ ∥ ∥ ∥ ( )
∥∥f+g∥∥q ∥∥ f −g∥∥q 1 p 1 p q∕p
∥ 2 ∥Lp + ∥ 2 ∥Lp ≤ 2 fLp + 2 gLp
Recall the following definition of uniform convexity.
Definition 14.3.2A Banach space, X, is said to be uniformly convex if whenever
∥xn ∥
≤ 1 and
∥∥xn+xm∥∥
2
→ 1 as n,m →∞, then {x_{n}} is a Cauchy sequence and x_{n}→ x where
∥x ∥
= 1.
Observe that Clarkson’s inequalities imply L^{p} is uniformly convex for all p > 1. Uniformly convex
spaces have a very nice property which is described in the following lemma. Roughly, this
property is that any element of the dual space achieves its norm at some point of the closed unit
ball.
Lemma 14.3.3Let X be uniformly convex and let ϕ ∈ X^{′}. Then there exists x ∈ X suchthat
x = 1,ϕ(x) = ϕ.
Proof: Let 
∥^xn ∥
≤ 1 and ϕ
(^xn)
→ϕ. Let x_{n} = w_{n}
^x
_{n} where w_{n} = 1 and
w ϕ^x = ϕ^x .
n n n
Thus ϕ
(xn)
= ϕ
(xn )
 = ϕ
(^xn)
→ϕ.
ϕ(x ) → ϕ, ∥x ∥ ≤ 1.
n n
We can assume, without loss of generality, that
ϕ (x ) = ϕ (x ) ≥ ϕ
n n 2
and ϕ≠0.
Claim 
xn+2xm
→ 1 as n,m →∞.
Proof of Claim: Let n,m be large enough that ϕ
(xn)
,ϕ
(xm )
≥ϕ−
ε2
where 0 < ε. Then
∥xn + xm ∥
≠0 because if it equals 0, then x_{n} = −x_{m} so −ϕ
By uniform convexity, {x_{n}} is Cauchy and x_{n}→ x,
∥x ∥
= 1. Thus ϕ
(x)
= lim_{n→∞}ϕ
(xn)
=
∥ϕ ∥
.
■
The proof of the Riesz representation theorem will be based on the following lemma which says that if
you can show a directional derivative exists, then it can be used to represent a functional in terms of this
directional derivative. It is very interesting for its own sake. It says that if ϕ ∈ X^{′} and
∥ϕ ∥
= ϕ
(x)
for some
∥x∥
= 1 and if
d
dt
∥x + ty∥
_{t=0} exists for each y, then you can represent ϕ in terms of this directional
derivative.
Lemma 14.3.4(McShane)Let X be a complex normed linear space and let ϕ ∈ X^{′}. Suppose there existsx ∈ X,
x
= 1 with ϕ
(x)
=
ϕ
≠0. Let y ∈ X and let ψ_{y}(t) = x + ty for t ∈ ℝ. Suppose ψ_{y}^{′}(0) existsfor each y ∈ X. Then for all y ∈ X,
′ ′ −1
ψy(0)+ iψ −iy(0) = ϕ ϕ(y).
Proof: Suppose first that ϕ = 1. Then by assumption, there is x such that
∥x∥
= 1 and
ϕ
(x )
= 1 =
∥ϕ∥
. Then ϕ
(y − ϕ(y)x)
= 0 and so
ϕ(x+ t(y − ϕ(y)x )) = ϕ(x) = 1 = ϕ.
Therefore,
x+ t(y− ϕ(y)x)
≥ 1 since otherwise
x + t(y− ϕ(y)x)
= r < 1 and so
( )
1 1 1
ϕ (x + t(y− ϕ(y)x))r = r ϕ(x) = r
which would imply that
ϕ
> 1.
Also for small t,
ϕ(y)t
< 1, and so
1 ≤ x + t(y− ϕ(y)x) = (1− ϕ(y)t)x + ty
∥∥ t ∥∥
≤ 1− ϕ (y)t∥∥x+ y∥∥.
1− ϕ(y)t
Divide both sides by
1− ϕ (y)t
. Using the standard formula for the sum of a geometric series,
. Hence the above is
dominated by an expression of the form
p p
Cp(x+ y)t (14.11)
(14.11)
The above lemma and uniform convexity of L^{p} can be used to prove a general version of the Riesz
representation theorem next. Let p > 1 and let η : L^{q}→ (L^{p})^{′} be defined by
∫
η(g)(f) = Ωgfdμ. (14.12)
(14.12)
Theorem 14.3.5(Riesz representation theorem p > 1) The map ηis 11, onto, continuous,and
ηg = g,η = 1.
Proof: Obviously ηis linear. Suppose ηg = 0.Then 0 = ∫gfdμfor all f ∈ L^{p}.Let f = g^{q−2}g. Then
f ∈ L^{p}and so 0 = ∫g^{q}dμ.Hence g = 0and ηis one to one. That ηg ∈ (L^{p})^{′}is obvious from the Holder
inequality. In fact,
η(g)(f) ≤ gqf p,
and so η(g)≤g_{q.} To see that equality holds, let

f = gq−2gg1−q q.
Then f_{p} = 1 and
∫
η(g)(f) = gqdμ∥g∥1−q= g.
Ω q q
Thus η = 1.
It remains to show ηis onto. Let ϕ ∈ (L^{p})^{′}.Is ϕ = ηgfor some g ∈ L^{q}?Without loss of generality, assume
ϕ≠0.By uniform convexity of L^{p}, Lemma 14.3.3, there exists g such that
ϕ (g) = ϕ,g ∈ Lp,g = 1.
For f ∈ L^{p}, define ϕ_{f}
(t)
≡∫_{Ω}
g+ tf
^{p}dμ. Thus
1
ψf(t) ≡ g + tfp ≡ ϕf(t)p.
Does ϕ_{f}^{′}(0)exist? Let
[g = 0]
denote the set
{x : g (x) = 0}
.
ϕf(t) − ϕf(0) ∫ (g+ tfp − gp)
= dμ
t t
is a finite
measure space. The argument will be a little different than the one given above for L^{p} with
p > 1.
Theorem 14.3.6(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: For measurable E, it follows that X_{E}∈ L^{1}
(Ω,μ)
. Define a measure
λ(E) ≡ Λ(XE )
This is a vector measure. If you have E = ∪_{i=1}^{∞}F_{i} where the F_{i} are measurable and disjoint, then if
E_{n}≡∪_{i=1}^{n}F_{i}, it follows from the dominated convergence theorem that X_{En}→X_{E} in L^{1}
(Ω,μ)
.
Therefore, by continuity of Λ,
∑n
Λ (XEn ) = Λ (XFi ) → Λ (XE) ≡ λ(E)
i=1
That is, ∑_{i=1}^{∞}λ
(Ei)
= λ
(∪∞i=1Ei)
. Then it follows from Theorem 14.2.5 that there exists a unique
h ∈ L^{1}