Basic Analysis – Gently Done Topological Vector Spaces

100 Basic Analysis
and so �ϕ x � ≤ �x�. However, by the Hahn-Banach theorem, Corollary 5.24, for
any given x ∈ X, x �= 0, there is ℓ ′ ∈ X ∗ such that �ℓ ′ � = 1 and ℓ ′ (x) = �x�. For
this particular ℓ ′ , we then have
|ϕ x (ℓ ′ )| = |ℓ ′ (x)| = �x� = �x��ℓ ′ �.
We conclude that �ϕ x � = �x�, and the proof is complete.
Thus, we may consider X as a subspace of X ∗∗ via the linear isometric embedding
x ↦→ ϕ x . This leads to the following observation that any normed space
has a completion.
Proposition 9.2 Any normed space is linearly isometrically isomorphic to a dense
subspace of a Banach space.
Proof The space X ∗∗ is a Banach space and, as above, X is linearly isometrically
isomorphic to the subspace {ϕ x : x ∈ X}. The closure of this subspace is the
required Banach space.
Definition 9.3 A Banach space X is said to be reflexive if X = X ∗∗ via the above
embedding.
Note that X ∗∗ is a Banach space, so X must be a Banach space if it is to be
reflexive.
Theorem 9.4 A Banach space X is reflexive if and only if X ∗ is reflexive.
Proof If X = X ∗∗ , then X ∗ = X ∗∗∗ via the appropriate embedding. We see this
as follows. To say that X = X ∗∗ means that each element of X ∗∗ has the form ϕ x
for some x ∈ X. Now let ψ ℓ ∈ X ∗∗∗ be the corresponding association of X ∗ into
X ∗∗∗ ,
ψ ℓ (z) = z(ℓ) for ℓ ∈ X ∗ and z ∈ X ∗∗ .
We have X ∗ ⊆ X ∗∗∗ via ℓ ↦→ ψ ℓ . Let λ ∈ X ∗∗∗ . Any z ∈ X ∗∗ has the form ϕ x , for
suitable x ∈ X, and so λ(z) = λ(ϕ x ). Define ℓ : X → K by ℓ(x) = λ(ϕ x ). Then
It follows that ℓ ∈ X ∗ . Moreover,
|ℓ(x)| = |λ(ϕ x )|
≤ �λ��ϕ x �
= ��x�.
ψ ℓ (ϕ x ) = ϕ x (ℓ)
and so ψ ℓ = λ, i.e., X ∗ = X ∗∗∗ via ψ.
= ℓ(x) = λ(ϕ x )
Department of Mathematics King’s College, London