Basic Analysis – Gently Done Topological Vector Spaces

96 Basic Analysis
Theorem 8.16 Suppose that X is a normed space and Y is a Banach space, and
suppose that T : X → Y is a linear operator defined on some dense linear subset
D(T) of X. Then if T is bounded (as a linear operator from the normed space
D(T) to Y) it has a unique extension to a bounded linear operator from all of X
into Y. Moreover, this extension has the same norm as T.
Proof By hypothesis, �Tx� ≤ �T��x�, for all x ∈ D(T), where �T� is the norm
of T as a map D(T) → Y. Let x ∈ X. Since D(T) is dense in X, there is a
sequence (ξ n ) in D(T) such that ξ n → x, in X, as n → ∞. In particular, (ξ n ) is a
Cauchy sequence in X. But
�Tξ n −Tξ m � = �T(ξ n −ξ m )� ≤ �T��ξ n −ξ m �,
and so we see that (Tξn ) is a Cauchy sequence in Y. Since Y is complete, there
exists y ∈ Y such that Tξn → y in Y. We would like to construct an extension
�T of T by defining � Tx to be this limit, y. However, to be able to do this, we
must show that the element y does not depend on the particular sequence (ξn ) in
D(T) converging to x. To see this, suppose that (ηn ) is any sequence in D(T) such
that ηn → x in X. Then, as before, we deduce that there is y ′ , say, in Y, such
that Tηn → y ′ . Now consider the combined sequence ξ1 ,η1 ,ξ2 ,η2 ,... in D(T).
Clearly, this sequence also converges to x and so once again, as above, we deduce
that the sequence (Tξ1 ,Tη1 ,Tξ2 ,Tη2 ,...) converges to some z, say, in Y. But
this sequence has the two convergent subsequences (Tξk ) and (Tηm ), with limits y
and y ′ , respectively. It follows that z = y = y ′ . Therefore we may unambiguously
define the map � T : X → Y by the prescription � Tx = y, where y is given as above.
Note that if x ∈ D(T), then we can take ξn ∈ D(T) above to be ξn = x for
every n ∈ N. This shows that � Tx = Tx, and hence that � T is an extension of T.
We show that � T is a bounded linear operator from X to Y.
Let x,x ′ ∈ X and let s ∈ C be given. Then there are sequences (ξn ) and (ξ ′ n )
′ ′ ′ ′ in D(T) such that ξn → x and ξn → x in X. Hence sξn +ξn → sx+x , and by
the construction of � T, we see that
�T(sx+x ′ ′
) = limT(sξ n
n +ξn ), using the linearity of D(T),
′
= limsTξ n
n +Tξ n
= s � Tx+ � Tx ′ .
It follows that � T is a linear map. To show that � T is bounded and has the same
norm as T, we first observe that if x ∈ X and if (ξ n ) is a sequence in D(T) such
that ξ n → x, then, by construction, Tξ n → � Tx, and so �Tξ n � → � � Tx�. Hence,
the inequalities �Tξ n � ≤ �T��ξ n �, for n ∈ N, imply (—by taking the limit) that
Department of Mathematics King’s College, London