23.7. THE DUAL SPACE OF C0 (X) 639

Theorem 23.6.5 (Riesz representation theorem p> 1) The map η is 1-1, onto, con-tinuous, and

∥ηg∥= ∥g∥, ∥η∥= 1.

Proof: Obviously η is linear. Suppose ηg = 0. Then 0 =∫

g f dµ for all f ∈ Lp. Letf = |g|q−2g. Then f ∈ Lpand so 0 =

∫|g|qdµ. Hence g = 0 and η is one to one. That

ηg ∈ (Lp)′ is obvious from the Holder inequality. In fact, |η(g)( f )| ≤ ∥g∥q∥ f∥p, and so∥η(g)∥ ≤ ∥g∥q. To see that equality holds, let f = |g|q−2g ∥g∥1−q

q . Then ∥ f∥p = 1 and

η(g)( f ) =∫

Ω

|g|qdµ ∥g∥1−qq = ∥g∥q.

Thus ∥η∥= 1.It remains to show η is onto. Let φ ∈ (Lp)′. Is φ = ηg for some g ∈ Lq? Without loss

of generality, assume φ ̸= 0. By uniform convexity of Lp, Lemma 23.6.3, there exists gsuch that φg = ∥φ∥, g ∈ Lp, ∥g∥ = 1. For f ∈ Lp, define φ f (t) ≡

∫Ω|g+ t f |p dµ. Thus

ψ f (t)≡ ∥g+ t f∥p ≡ φ f (t)1p . Does φ

′f (0) exist? Let [g = 0] denote the set {x : g(x) = 0}.

φ f (t)−φ f (0)t

=∫

(|g+ t f |p−|g|p)t

dµ

From 23.10, the integrand is bounded by Cp (| f |p + |g|p) . Therefore, using 23.9, the dom-inated convergence theorem applies and it follows φ

′f (0) =

limt→0

φ f (t)−φ f (0)t

= limt→0

[∫[g=0]|t|p−1 | f |pdµ +

∫[g̸=0]

(|g+ t f |p−|g|p)t

dµ

]= p

∫[g̸=0]|g|p−2 Re(ḡ f )dµ = p

∫|g|p−2 Re(ḡ f )dµ

Hence ψ ′f (0) = ∥g∥−pq∫|g(x)|p−2 Re(g(x) f̄ (x))dµ . Note 1

p −1 =− 1q Therefore,

ψ′−i f (0) = ∥g∥

−pq

∫|g(x)|p−2 Re(ig(x) f̄ (x))dµ.

But Re(ig f̄ ) = Im(−g f̄ ) and so by the McShane lemma,

φ ( f ) = ∥φ∥ ∥g∥−pq

∫|g(x)|p−2[Re(g(x) f̄ (x))+ i Re(ig(x) f̄ (x))]dµ

= ∥φ∥ ∥g∥−pq

∫|g(x)|p−2[Re(g(x) f̄ (x))+ i Im(−g(x) f̄ (x))]dµ

= ∥φ∥ ∥g∥−pq

∫|g(x)|p−2g(x) f (x)dµ .

This shows that φ = η(∥φ∥ ∥g∥−pq |g|p−2g) and verifies η is onto. ■

23.7 The Dual Space of C0 (X)

Consider the dual space of C0(X) where X is a locally compact Hausdorff space. It willturn out to be a space of measures. To show this, the following lemma will be convenient.Recall this space is defined as follows.