Measure, Integration & Real Analysis, 2021a

Section 7B L p (μ) 207
7.26 dual space of l p can be identified with l p′
Suppose 1 ≤ p < ∞. Forb =(b 1 , b 2 ,...) ∈ l p′ , define ϕ b : l p → F by
ϕ b (a) =
∞
∑ a k b k ,
k=1
where a =(a 1 , a 2 ,...). Then b ↦→ ϕ b is a one-to-one linear map from l p′ onto
( l
p ) ′ . Furthermore, ‖ϕb ‖ = ‖b‖ p ′ for all b ∈ l p′ .
Proof For k ∈ Z + , let e k ∈ l p be the sequence in which each term is 0 except that
the k th term is 1; thus e k =(0, . . . , 0, 1, 0, . . .).
Suppose ϕ ∈ ( l p) ′ . Define a sequence b =(b1 , b 2 ,...) of numbers in F by
Suppose a =(a 1 , a 2 ,...) ∈ l p . Then
b k = ϕ(e k ).
∞
a = ∑ a k e k ,
k=1
where the infinite sum converges in the norm of l p (the proof would fail here if we
allowed p to be ∞). Because ϕ is a bounded linear functional on l p , applying ϕ to
both sides of the equation above shows that
∞
ϕ(a) = ∑ a k b k .
k=1
We still need to prove that b ∈ l p′ . To do this, for n ∈ Z + let μ n be counting
measure on {1, 2, . . . , n}. We can think of L p (μ n ) as a subspace of l p by identifying
each (a 1 ,...,a n ) ∈ L p (μ n ) with (a 1 ,...,a n ,0,0,...) ∈ l p . Restricting the
linear functional ϕ to L p (μ n ) gives the linear functional on L p (μ n ) that satisfies the
following equation:
n
ϕ| L p (μ n )(a 1 ,...,a n )= ∑ a k b k .
k=1
Now 7.25 (also see Exercise 14 for the case where p = 1)gives
‖(b 1 ,...,b n )‖ p ′ = ‖ϕ| L p (μ n )‖
≤‖ϕ‖.
Because lim n→∞ ‖(b 1 ,...,b n )‖ p ′ = ‖b‖ p ′, the inequality above implies the inequality
‖b‖ p ′ ≤‖ϕ‖. Thus b ∈ l p′ , which implies that ϕ = ϕ b , completing the
proof.
The previous result does not hold when p = ∞. In other words, the dual space
of l ∞ cannot be identified with l 1 . However, see Exercise 15, which shows that the
dual space of a natural subspace of l ∞ can be identified with l 1 .
Measure, Integration & Real Analysis, by Sheldon Axler