626 CHAPTER 20. REPRESENTATION THEOREMS

Theorem 20.5.5 (Riesz representation theorem p > 1) The map η is 1-1, onto, continuous,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−qq .

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 20.5.3, there exists g suchthat

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 20.5.17, the integrand is bounded by Cp (| f |p + |g|p) . Therefore, using 20.5.16, thedominated 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

Henceψ′f (0) = ||g||

−pq

∫|g(x)|p−2 Re(g(x) f̄ (x))dµ .

Note 1p −1 =− 1

q . Therefore,

ψ′−i f (0) = ||g||

−pq

∫|g(x)|p−2 Re(ig(x) f̄ (x))dµ.