Measure, Integration & Real Analysis, 2021a

Section 6C Normed Vector Spaces 169
We have now defined a function T : V → W. The reader should verify that T is a
linear map. Clearly
‖Tf‖≤sup{‖T k f ‖ : k ∈ Z + }
≤ ( sup{‖T k ‖ : k ∈ Z + } ) ‖ f ‖
for each f ∈ V. The last supremum above is finite because every Cauchy sequence is
bounded (see Exercise 4). Thus T ∈B(V, W).
We still need to show that lim k→∞ ‖T k − T‖ = 0. To do this, suppose ε > 0. Let
n ∈ Z + be such that ‖T j − T k ‖ < ε for all j ≥ n and k ≥ n. Suppose j ≥ n and
suppose f ∈ V. Then
‖(T j − T) f ‖ = lim
k→∞
‖T j f − T k f ‖
Thus ‖T j − T‖ ≤ε, completing the proof.
≤ ε‖ f ‖.
The next result shows that the phrase bounded linear map means the same as the
phrase continuous linear map.
6.48 continuity is equivalent to boundedness for linear maps
A linear map from one normed vector space to another normed vector space is
continuous if and only if it is bounded.
Proof Suppose V and W are normed vector spaces and T : V → W is linear.
First suppose T is not bounded. Thus there exists a sequence f 1 , f 2 ,...in V such
that ‖ f k ‖≤1 for each k ∈ Z + and ‖Tf k ‖→∞ as k → ∞. Hence
lim
k→∞
f
(
k
‖Tf k ‖ = 0 and T fk
)
= Tf k
‖Tf k ‖ ‖Tf k ‖
̸→ 0,
where the nonconvergence to 0 holds because Tf k /‖Tf k ‖ has norm 1 for every
k ∈ Z + . The displayed line above implies that T is not continuous, completing the
proof in one direction.
To prove the other direction, now suppose T is bounded. Suppose f ∈ V and
f 1 , f 2 ,...is a sequence in V such that lim k→∞ f k = f . Then
‖Tf k − Tf‖ = ‖T( f k − f )‖
≤‖T‖‖f k − f ‖.
Thus lim k→∞ Tf k = Tf. Hence T is continuous, completing the proof in the other
direction.
Exercise 18 gives several additional equivalent conditions for a linear map to be
continuous.
Measure, Integration & Real Analysis, by Sheldon Axler