Measure, Integration & Real Analysis, 2021a

178 Chapter 6 Banach Spaces
If V is a real vector space, U is a subspace of V, and h ∈ V, then U + Rh is the
subspace of V defined by
U + Rh = { f + αh : f ∈ U and α ∈ R}.
6.63 Extension Lemma
Suppose V is a real normed vector space, U is a subspace of V, and ψ : U → R
is a bounded linear functional. Suppose h ∈ V \ U. Then ψ can be extended to a
bounded linear functional ϕ : U + Rh → R such that ‖ϕ‖ = ‖ψ‖.
Proof Suppose c ∈ R. Define ϕ(h) to be c, and then extend ϕ linearly to U + Rh.
Specifically, define ϕ : U + Rh → R by
ϕ( f + αh) =ψ( f )+αc
for f ∈ U and α ∈ R. Then ϕ is a linear functional on U + Rh.
Clearly ϕ| U = ψ. Thus ‖ϕ‖ ≥‖ψ‖. We need to show that for some choice of
c ∈ R, the linear functional ϕ defined above satisfies the equation ‖ϕ‖ = ‖ψ‖. In
other words, we want
6.64 |ψ( f )+αc| ≤‖ψ‖‖f + αh‖ for all f ∈ U and all α ∈ R.
It would be enough to have
6.65 |ψ( f )+c| ≤‖ψ‖‖f + h‖ for all f ∈ U,
because replacing f by f α
in the last inequality and then multiplying both sides by |α|
would give 6.64.
Rewriting 6.65, we want to show that there exists c ∈ R such that
−‖ψ‖‖f + h‖ ≤ψ( f )+c ≤‖ψ‖‖f + h‖ for all f ∈ U.
Equivalently, we want to show that there exists c ∈ R such that
−‖ψ‖‖f + h‖−ψ( f ) ≤ c ≤‖ψ‖‖f + h‖−ψ( f ) for all f ∈ U.
The existence of c ∈ R satisfying the line above follows from the inequality
( ) ( )
6.66 sup −‖ψ‖‖f + h‖−ψ( f ) ≤ inf ‖ψ‖‖g + h‖−ψ(g) .
f ∈U
g∈U
To prove the inequality above, suppose f , g ∈ U. Then
−‖ψ‖‖f + h‖−ψ( f ) ≤‖ψ‖(‖g + h‖−‖g − f ‖) − ψ( f )
= ‖ψ‖(‖g + h‖−‖g − f ‖)+ψ(g − f ) − ψ(g)
≤‖ψ‖‖g + h‖−ψ(g).
The inequality above proves 6.66, which completes the proof.
Measure, Integration & Real Analysis, by Sheldon Axler