13.3. THE DUAL SPACE OF Lp (Ω) 325

Proof: (Uniqueness) If h1 and h2 in Lq 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 ∈Lp(Ω). Thus

0 = Λ f −Λ f =∫

h1|h1−h2|q−2(h1−h2)−h2|h1−h2|q−2(h1−h2)dµ =∫|h1−h2|qdµ.

Therefore h1 = h2 and this proves uniqueness.Now let λ (E) = Λ(XE). Since this is a finite measure space, XE is an element of

Lp (Ω) and so it makes sense to write Λ(XE). Is λ a complex measure?If {Ei}∞

i=1 is a sequence of disjoint sets of S , let Fn = ∪ni=1Ei, F = ∪∞

i=1Ei. Then bythe Dominated Convergence theorem, ∥XFn −XF∥p→ 0. Therefore, by continuity of Λ,

λ (F)≡ Λ(XF) = limn→∞

Λ(XFn) = limn→∞

n

∑k=1

Λ(XEk) =∞

∑k=1

λ (Ek).

This shows λ is a complex measure.It is also clear from the definition of λ that λ ≪ µ . Therefore, by the Radon Nikodym

theorem, there exists h ∈ L1(Ω) with λ (E) =∫

E hdµ = Λ(XE). Actually h ∈ Lq and satis-fies the other conditions above. This is shown next.

Let s = ∑mi=1 ciXEi be a simple function. Then since Λ is linear,

Λ(s) =m

∑i=1

ciΛ(XEi) =m

∑i=1

ci

∫Ei

hdµ =∫

hsdµ . (13.2)

Claim: If f is uniformly bounded and measurable, then Λ( f ) =∫

h f dµ.Proof of claim: Since f is bounded and measurable, there exists a sequence of simple

functions, {sn}which converges to f pointwise and in Lp (Ω) , |sn| ≤ | f |. This follows fromTheorem 6.1.10 on Page 140 upon breaking f up into positive and negative parts of realand complex parts. In fact this theorem gives uniform convergence. Then

Λ( f ) = limn→∞

Λ(sn) = limn→∞

∫hsndµ =

∫h f dµ,

the first equality holding because of continuity of Λ, the second following from 13.2 andthe third holding by the dominated convergence theorem.

This is a very nice formula but it still has not been shown that h ∈ Lq (Ω).Let En = {x : |h(x)| ≤ n}. Thus |hXEn | ≤ n. Then |hXEn |q−2(hXEn) ∈ Lp(Ω). By the

claim, it follows that

∥hXEn∥qq =

∫h|hXEn |q−2(hXEn)dµ = Λ(|hXEn |q−2(hXEn))

≤ ∥Λ∥∥∥|hXEn |q−2(hXEn)

∥∥p =

(∫|hXEn |qdµ

)1/p

= ∥Λ∥ ∥hXEn∥qpq ,

because q−1 = q/p and so it follows that ∥hXEn∥q ≤ ∥Λ∥. Letting n→ ∞, the monotoneconvergence theorem implies ∥h∥q ≤ ∥Λ∥.