Measure, Integration & Real Analysis, 2021a

168 Chapter 6 Banach Spaces
6.45 Example linear map that is not bounded
Let V be the normed vector space of sequences (a 1 , a 2 ,...) of elements of F such
that a k = 0 for all but finitely many k ∈ Z + , with ‖(a 1 , a 2 ,...)‖ ∞ = max k∈Z +|a k |.
Define T : V → V by
T(a 1 , a 2 , a 3 ,...)=(a 1 ,2a 2 ,3a 3 ,...).
Then T is a linear map that is not bounded, as you should verify.
The next result shows that if V and W are normed vector spaces, then B(V, W) is
a normed vector space with the norm defined above.
6.46 ‖·‖ is a norm on B(V, W)
Suppose V and W are normed vector spaces. Then ‖S + T‖ ≤‖S‖ + ‖T‖
and ‖αT‖ = |α|‖T‖ for all S, T ∈B(V, W) and all α ∈ F. Furthermore, the
function ‖·‖ is a norm on B(V, W).
Proof
Suppose S, T ∈B(V, W). Then
‖S + T‖ = sup{‖(S + T) f ‖ : f ∈ V and ‖ f ‖≤1}
≤ sup{‖Sf‖ + ‖Tf‖ : f ∈ V and ‖ f ‖≤1}
≤ sup{‖Sf‖ : f ∈ V and ‖ f ‖≤1}
+ sup{‖Tf‖ : f ∈ V and ‖ f ‖≤1}
= ‖S‖ + ‖T‖.
The inequality above shows that ‖·‖ satisfies the triangle inequality on B(V, W).
The verification of the other properties required for a normed vector space is left to
the reader.
Be sure that you are comfortable using all four equivalent formulas for ‖T‖ shown
in Exercise 16. For example, you should often think of ‖T‖ as the smallest number
such that ‖Tf‖≤‖T‖‖f ‖ for all f in the domain of T.
Note that in the next result, the hypothesis requires W to be a Banach space but
there is no requirement for V to be a Banach space.
6.47 B(V, W) is a Banach space if W is a Banach space
Suppose V is a normed vector space and W is a Banach space. Then B(V, W) is
a Banach space.
Proof
Suppose T 1 , T 2 ,...is a Cauchy sequence in B(V, W). Iff ∈ V, then
‖T j f − T k f ‖≤‖T j − T k ‖‖f ‖,
which implies that T 1 f , T 2 f ,... is a Cauchy sequence in W. Because W is a Banach
space, this implies that T 1 f , T 2 f ,... has a limit in W, which we call Tf.
Measure, Integration & Real Analysis, by Sheldon Axler