20.3. REPRESENTATION THEOREMS FOR THE DUAL SPACE OF Lp 613

the last equality holding because q−1 = q/p and so(∫ ∣∣|hXEn |q−2(hXEn)∣∣p dµ

)1/p

=

(∫ (|hXEn |q/p

)pdµ

)1/p

= ||hXEn ||qpq

Therefore, since q− qp = 1, it follows that

||hXEn ||q ≤ ||Λ||.

Letting n→ ∞, the Monotone Convergence theorem implies

||h||q ≤ ||Λ||. (20.3.10)

Now that h has been shown to be in Lq(Ω), it follows from 20.3.9 and the density ofthe simple functions, Theorem 15.2.1 on Page 406, that

Λ f =∫

h f dµ

for all f ∈ Lp(Ω).It only remains to verify the last claim.

||Λ||= sup{∫

h f : || f ||p ≤ 1} ≤ ||h||q ≤ ||Λ||

by 20.3.10, and Holder’s inequality. This proves the theorem.To represent elements of the dual space of L1(Ω), another Banach space is needed.

Definition 20.3.2 Let (Ω,S ,µ) be a measure space. L∞(Ω) is the vector space of mea-surable functions such that for some M > 0, | f (x)| ≤ M for all x outside of some set ofmeasure zero (| f (x)| ≤M a.e.). Define f = g when f (x) = g(x) a.e. and || f ||∞ ≡ inf{M :| f (x)| ≤M a.e.}.

Theorem 20.3.3 L∞(Ω) is a Banach space.

Proof: It is clear that L∞(Ω) is a vector space. Is || ||∞ a norm?Claim: If f ∈ L∞ (Ω), then | f (x)| ≤ || f ||

∞a.e.

Proof of the claim:{

x : | f (x)| ≥ || f ||∞+n−1

}≡ En is a set of measure zero according

to the definition of || f ||∞

. Furthermore, {x : | f (x)|> || f ||∞}= ∪nEn and so it is also a set

of measure zero. This verifies the claim.Now if || f ||

∞= 0 it follows that f (x) = 0 a.e. Also if f ,g ∈ L∞ (Ω),

| f (x)+g(x)| ≤ | f (x)|+ |g(x)| ≤ || f ||∞+ ||g||

∞

a.e. and so || f ||∞+ ||g||

∞serves as one of the constants, M in the definition of || f +g||

∞.

Therefore,|| f +g||

∞≤ || f ||

∞+ ||g||

∞.