Page 240 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Unit 21: The Conjugate of an Operator

(T) = T* for every (N) Notes
is an isometric isomorphism of b (N) into b (N*) reverses products and preserves the identify
transformation.

Proof: We first show that T* is linear
Let f, g N* and , be any scalars
then [T* ( j + g)] (x) = ( j + g) (T (x))
= ( j) T (x) + ( g) T (x)

= [j (T (x))] + [g (T (x))]
= [T* (j)] (x) + [T* (g)] (x)
= [ T* (j) + T* (g)] (x) x N

Hence T* ( f + g) = T* (f) + T* (g)
T* is linear on N*.
To show that T* is continuous, we have to show that it is bounded on the assumption that T is
bounded.
T* = sup { T* (f) : f 1}
= sup { [T* (f)] (x) : f 1 and x }
= sup { f (T (x))|: f 1 and x }

= sup { f T x : x 1 and x } … (1)
T
T* is a bounded linear transformation on N* into N*. Hence by application of Hahn-
Banach theorem, for each non-zero x in N, a functional f N* such that
f = 1 and f (T (x)) = T (x) … (2)

T(x)
Hence T = sup : x 0
x

f T(x)
= sup : f 1, x 0 (by (2))
x

T *(f)(x)
= sup : f 1, x 0 (by (1))
x

T *(f) x
sup : f 1, x 0
x

= sup T *(f) : f 1 T * … (3)
From (1) and (3) it follows that

T = T* . … (4)

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