Basic Analysis – Gently Done Topological Vector Spaces

9: The Dual Space of a Normed Space 101
Now suppose that X ∗ = X ∗∗∗ and suppose that X �= X ∗∗ . Then, by Corollary
7.30, there is λ ∈ X ∗∗∗ such that λ �= 0 but λ vanishes on X in X ∗∗ ,
i.e., λ(ϕ x ) = 0 for all x ∈ X. But then λ can be written as λ = ψ ℓ for some
ℓ ∈ X ∗ , since X ∗ = X ∗∗∗ , and so
λ(ϕ x ) = ψ ℓ (ϕ x )
= ϕ x (ℓ) = ℓ(x)
which gives 0 = λ(ϕ x ) = ℓ(x) for all x ∈ X, i.e., ℓ = 0 in X ∗ .
It follows that ψ ℓ = 0 in X ∗∗∗ and so λ = 0. This is a contradiction and we
conclude that X = X ∗∗ .
Corollary 9.5 Suppose that the Banach space X is not reflexive. Then the
natural inclusions X ⊆ X ∗∗ ⊆ X ∗∗∗∗ ⊆ ... and X ∗ ⊆ X ∗∗∗ ⊆ ... are all strict.
Proof Let X0 = X and, for n ∈ N, set Xn = X∗ n−1 . If Xn = Xn+2 , then it follows
that Xn−1 = Xn+1 . Repeating this argument, we deduce that X0 = X2 , which is
to say that X is reflexive. Note that the equalities here are to be understood as
the linear isometric isomorphisms as introduced above.
We shall now compute the duals of some of the classical Banach spaces. For
any p ∈ Rwith1 ≤ p < ∞, thespace ℓ p isthespaceofcomplex sequences x = (x n )
such that
�x� p = ��∞
n=1 |x n |p�1
p < ∞.
We shall consider the cases 1 < p < ∞ and show that for such p, �·� p is a norm
and that ℓp is a Banach space with respect to this norm. We will also show that
= 1. It therefore follows
the dual of ℓ p is ℓ q , where q is given by the formula 1
p
that these spaces are reflexive. At this stage, it is not even clear that ℓ p is a
linear space, never mind whether or not �·� p is a norm. We need some classical
inequalities.
+ 1
q
Proposition 9.6 Let a, b ≥ 0 and α, β > 0 with α+β = 1. Then
with equality if and only if a = b.
a α b β ≤ αa+βb
Proof We note that the function t ↦→ e t is strictly convex; for any x,y ∈ R,
e (αx+βy) < αe x +βe y . Putting a = e x , b = e y gives the required result.