Measure, Integration & Real Analysis, 2021a

Section 6D Linear Functionals 173
6.52 bounded linear functionals
Suppose V is a normed vector space and ϕ : V → F is a linear functional that is
not identically 0. Then the following are equivalent:
(a)
ϕ is a bounded linear functional.
(b) ϕ is a continuous linear functional.
(c) null ϕ is a closed subspace of V.
(d) null ϕ ̸= V.
Proof The equivalence of (a) and (b) is just a special case of 6.48.
To prove that (b) implies (c), suppose ϕ is a continuous linear functional. Then
null ϕ, which is the inverse image of the closed set {0}, is a closed subset of V by
6.11(d). Thus (b) implies (c).
To prove that (c) implies (a), we will show that the negation of (a) implies the
negation of (c). Thus suppose ϕ is not bounded. Thus there is a sequence f 1 , f 2 ,...
in V such that ‖ f k ‖≤1 and |ϕ( f k )|≥k for each k ∈ Z + .Now
f 1
ϕ( f 1 ) − f k
ϕ( f k ) ∈ null ϕ
for each k ∈ Z + and
lim
k→∞
Clearly
( f1
ϕ( f 1 ) − f k
ϕ( f k )
)
= f 1
ϕ( f 1 ) .
( f1
)
ϕ = 1 and thus
ϕ( f 1 )
This proof makes major use of
dividing by expressions of the form
ϕ( f ), which would not make sense
for a linear mapping into a vector
space other than F.
f 1
ϕ( f 1 )
/∈ null ϕ.
The last three displayed items imply that null ϕ is not closed, completing the proof
that the negation of (a) implies the negation of (c). Thus (c) implies (a).
We now know that (a), (b), and (c) are equivalent to each other.
Using the hypothesis that ϕ is not identically 0, we see that (c) implies (d). To
complete the proof, we need only show that (d) implies (c), which we will do by
showing that the negation of (c) implies the negation of (d). Thus suppose null ϕ is
not a closed subspace of V. Because null ϕ is a subspace of V, we know that null ϕ
is also a subspace of V (see Exercise 12 in Section 6C). Let f ∈ null ϕ \ null ϕ.
Suppose g ∈ V. Then
(
g = g − ϕ(g) )
ϕ( f ) f + ϕ(g)
ϕ( f ) f .
The term in large parentheses above is in null ϕ and hence is in null ϕ. The term
above following the plus sign is a scalar multiple of f and thus is in null ϕ. Because
the equation above writes g as the sum of two elements of null ϕ, we conclude that
g ∈ null ϕ. Hence we have shown that V = null ϕ, completing the proof that the
negation of (c) implies the negation of (d).
Measure, Integration & Real Analysis, by Sheldon Axler