Basic Analysis – Gently Done Topological Vector Spaces

92 Basic Analysis
Proposition 8.10 Suppose that T : X → Y is a bounded linear operator. Then
we have
�T� = sup{�Tx� : �x� ≤ 1}
= sup{�Tx� : �x� = 1}
= sup ��Tx�
�x� : x �= 0� .
Proof One uses the facts that if �x� ≤ 1 (and x �= 0), then �Tx� ≤ �Tx�/�x� =
�Tx/�x��.
Note that if T is bounded, then, by the very definition of �T�, we have �Tx� ≤
�T��x�, for any x ∈ X. Thus, a bounded linear operator maps any bounded set
in X into a bounded set in Y. In particular, the unit ball in X is mapped into the
ball of radius �T� in Y. We have seen in Corollary 6.20 that if T is a continuous
linear operator then it is bounded. Hence
�Tx ′ −Tx� = �T(x ′ −x)� ≤ �T��x ′ −x�
for any x,x ′ ∈ X. This estimate shows that T is uniformly continuous on X, and
that therefore, continuity and uniform continuity are equivalent in this context.
In other words, the notion of uniform continuity can play no special rôle in the
theory of linear operators, as it does, for example, in classical real analysis.
Definition 8.11 The set of bounded linear operators from a normed space X into
a normed space Y is denoted B(X,Y). If X = Y, one simply writes B(X) for
B(X,X).
Proposition 8.12 The space B(X,Y) is a normed space when equipped with its
natural linear structure and the norm �·�.
Proof For S, T ∈ B(X,Y) and any s, t ∈ K, the linear operator sS+tT is defined
by (sS +tT)x = sSx+tTx for x ∈ X. Furthermore, for any x ∈ X,
�Sx+Tx� ≤ �Sx�+�Tx� ≤ (�S�+�T�)�x�
and we see that B(X,Y) is a linear space. To see that �·� is a norm on B(X,Y),
note first that �T� ≥ 0 and �T� = 0 if T = 0. On the other hand, if �T� = 0,
then
0 = �T� = sup ��Tx�
: x �= 0�
�x�
which implies that �Tx� = 0 for every x ∈ X (including, trivially, x = 0). That
is, T = 0.
Department of Mathematics King’s College, London