Basic Analysis – Gently Done Topological Vector Spaces

8: Banach Spaces 97
� � Tx� ≤ �T��x�. Therefore � � T� ≤ �T�. But since � T is an extension of T we have
that
� � T� = sup{� � Tx� : x ∈ X, �x� ≤ 1}
≤ sup{� � Tx� : x ∈ D(T), �x� ≤ 1}
= sup{�Tx� : x ∈ X, �x� ≤ 1}
= �T�.
The equality � � T� = �T� follows.
The uniqueness is immediate; if S is also a bounded linear extension of T to
the whole of X, then S − � T is a bounded (equivalently, continuous) map on X
which vanishes on the dense subset D(T). Thus S − � T must vanish on the whole
of X, i.e., S = � T.
Remark 8.17 This process of extending a densely-defined bounded linear operator
to one on the whole of X is often referred to as ‘extension by continuity’. If T is
densely-defined, as above, but is not bounded on D(T), there is no ‘obvious’ way
of extending T to the whole of X. Indeed, such a goal may not even be desirable,
as we will see later—for example, the Hellinger-Toeplitz theorem.
We can use the concept of Hamel basis to give an example of a space which is
a Banach space with respect to two inequivalent norms. It is not difficult to give
examples of linear spaces with inequivalent norms. For example, C[0,1] equipped
with the �·� ∞ and � ·� 1 norms is such an example. It is a little harder to find
examples where the space is complete with respect to each of the two inequivalent
norms.
Example 8.18 Let X = ℓ1 and Y = ℓ2 and, for k ∈ N, let ek be the element
ek = (δkm ) m∈N in ℓ1 and let fk denote the corresponding element in ℓ2 . For any t
with 0 < t < 1, let bt = (t,t2 ,t3 ,...). Then {ek : k ∈ N}∪{b t : 0 < t < 1} form
an independent set in ℓ1 , and {fk : k ∈ N}∪{b t : 0 < t < 1} form an independent
set in ℓ2 . These sets can be extended to Hamel bases B1 and B2 of ℓ1 and ℓ2 ,
respectively. Both B1 and B2 contain a subset of cardinality 2ℵ0 . On the other
hand, X ⊆ CN and Y ⊆ CN so we deduce that B1 and B2 both have cardinality
equal to 2ℵ0 . In particular, there is an isomorphism ϕ from B1 onto B2 which
sends ek into fk , for each k ∈ N. By linearity, ϕ defines a linear isomorphism from
ℓ1 onto ℓ2 .
For n ∈ N, let an = �n k=1 1
nek ∈ ℓ1 and let bn = �n k=1 1
nfk ∈ ℓ2 . Then
�an�1 = 1, for all n ∈ N, whereas �bn�2 = 1/ √ n.
We define a new norm |· | on ℓ1 by setting
|x| = �ϕ(x)� 2