Measure, Integration & Real Analysis, 2021a

190 Chapter 6 Banach Spaces
6.86 Principle of Uniform Boundedness
Suppose V is a Banach space, W is a normed vector space, and A is a family of
bounded linear maps from V to W such that
sup{‖Tf‖ : T ∈A}< ∞ for every f ∈ V.
Then
Proof
Our hypothesis implies that
sup{‖T‖ : T ∈A}< ∞.
V =
∞⋃
n=1
{ f ∈ V : ‖Tf‖≤n for all T ∈A} ,
} {{ }
V n
where V n is defined by the expression above. Because each T ∈Ais continuous, V n
is a closed subset of V for each n ∈ Z + . Thus Baire’s Theorem [6.76(a)] and the
equation above imply that there exist n ∈ Z + and h ∈ Vand r > 0 such that
6.87 B(h, r) ⊂ V n .
Now suppose g ∈ V and ‖g‖ < 1. Thus rg + h ∈ B(h, r). Hence if T ∈A, then
6.87 implies ‖T(rg + h)‖ ≤n, which implies that
T(rg + h)
‖Tg‖ = ∥
r
Thus
completing the proof.
− Th
r
sup{‖T‖ : T ∈A}≤
∥ ≤
‖T(rg + h)‖
r
+ ‖Th‖
r
n + sup{‖Th‖ : T ∈A}
r
≤ n + ‖Th‖ .
r
< ∞,
EXERCISES 6E
1 Suppose U is a subset of a metric space V. Show that U is dense in V if and
only if every nonempty open subset of V contains at least one element of U.
2 Suppose U is a subset of a metric space V. Show that U has an empty interior if
and only if V \ U is dense in V.
3 Prove or give a counterexample: If V is a metric space and U, W are subsets
of V, then (int U) ∪ (int W) =int(U ∪ W).
4 Prove or give a counterexample: If V is a metric space and U, W are subsets
of V, then (int U) ∩ (int W) =int(U ∩ W).
Measure, Integration & Real Analysis, by Sheldon Axler