Measure, Integration & Real Analysis, 2021a

Section 6D Linear Functionals 181
6.72 ‖ f ‖ = max{|ϕ( f )| : ϕ ∈ V ′ and ‖ϕ‖ = 1}
Suppose V is a normed vector space and f ∈ V \{0}. Then there exists ϕ ∈ V ′
such that ‖ϕ‖ = 1 and ‖ f ‖ = ϕ( f ).
Proof
Let U be the 1-dimensional subspace of V defined by
Define ψ : U → F by
U = {α f : α ∈ F}.
ψ(α f )=α‖ f ‖
for α ∈ F. Then ψ is a linear functional on U with ‖ψ‖ = 1 and ψ( f )=‖ f ‖. The
Hahn–Banach Theorem (6.69) implies that there exists an extension of ψ to a linear
functional ϕ on V with ‖ϕ‖ = 1, completing the proof.
The next result gives another beautiful application of the Hahn–Banach Theorem,
with a useful necessary and sufficient condition for an element of a normed vector
space to be in the closure of a subspace.
6.73 condition to be in the closure of a subspace
Suppose U is a subspace of a normed vector space V and h ∈ V. Then h ∈ U if
and only if ϕ(h) =0 for every ϕ ∈ V ′ such that ϕ| U = 0.
Proof First suppose h ∈ U. If ϕ ∈ V ′ and ϕ| U = 0, then ϕ(h) =0 by the
continuity of ϕ, completing the proof in one direction.
To prove the other direction, suppose now that h /∈ U. Define ψ : U + Fh → F by
ψ( f + αh) =α
for f ∈ U and α ∈ F. Then ψ is a linear functional on U + Fh with null ψ = U and
ψ(h) =1.
Because h /∈ U, the closure of the null space of ψ does not equal U + Fh. Thus
6.52 implies that ψ is a bounded linear functional on U + Fh.
The Hahn–Banach Theorem (6.69) implies that ψ can be extended to a bounded
linear functional ϕ on V. Thus we have found ϕ ∈ V ′ such that ϕ| U = 0 but
ϕ(h) ̸= 0, completing the proof in the other direction.
EXERCISES 6D
1 Suppose V is a normed vector space and ϕ is a linear functional on V. Suppose
α ∈ F \{0}. Prove that the following are equivalent:
(a) ϕ is a bounded linear functional.
(b) ϕ −1 (α) is a closed subset of V.
(c) ϕ −1 (α) ̸= V.
Measure, Integration & Real Analysis, by Sheldon Axler