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Measure Theory and Functional Analysis

Notes (c) T (N*) N* T is continuous, where T is a linear transformation of N into itself which is
not necessarily continuous.
Proof:

(a) x, y N and , be any scalars. Then
[T (f)] ( x + y) = f (T ( x + y))
Since T and f are linear, we get
f (T ( x + y)) = f (T (x) + f (T (y))

= [T (f)] (x) + [T (f)]y
part (a).
(b) Let f, g N and , be any scalars. Then
+
[T ( f + g) (x)] = ( f + g) (T (x)) = [T (f)] (x) + (T (g)] (x)

T is linear on N +
part (b)
(c) Let S be a closed unit sphere in N. Then we know that T is continuous T (S) is bounded
f (T (S)) is bounded for each f N*.

By definition of T , f (T (S)) is bounded if and only if [T (f)] (S) is bounded for each f in
N* = T (f) is in N* for each f in N*.
T (N) N*
part (c)

This completes the proof of the theorem.
Note: Part (c) of the above theorem enables us to restrict T to N* iff T is continuous. Hence
by making T continuous we define an operation called the conjugate of T by restricting T
to N*. We see it below.

21.1.2 The Conjugate of T

Definition: Let N be normed linear space and let T be a continuous linear transformation of N into
itself (i.e. T is an operator). Define a linear transformation T* of N* into itself as follows:

If f N*, then, T* (f) is given by
[T* (f)] (x) = f (T (x))
We call T* the conjugate of T.

Theorem 2: If T is a continuous linear transformation on a normed linear space N, then its
conjugate T* defined by
T* : N* N* such that

T* (f) = f.T where
[T* (f)] (x) = f (T (x)) f N* and all x N
is a continuous linear transformation on N* and the mapping T T* given by
: (N) (N*) such that

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