Page 241 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Measure Theory and Functional Analysis

Notes Now we show that
: (N) (N*) given by
(T) T * … (5)

for every T (N) is an isometric isomorphism which reverses the product and preserve the
identity transformation.
The isometric character of follows by using (5) as seen below:

(T) = T* = T .
Next we show that is linear and one-to-one. Let T, T (N) and , be any scalars. Then
1
( T + T ) = ( T + T)* by (3)
1
But [( T + T )* (f)] (x) = f ( T + T ) (x)
1 1
= f ( T (x) + T (x) )
1
Since f is linear, we get
[( T + T )* (f)] = f (T (x)) + f (T (x))
1 1
= [T* (f) (x) + T * (f) (x)]
1
= [T *(f)] [T * (f)](x)
1
x N. Hence we get
[( T + T )* (f)] = [T* (f)] + [T * (f) ]
1 1
= T * T * (f)
1
Hence ( T + T )* = T* + T * … (6)
1 1
Therefore ( T + T ) = ( T + T )* = T* + T * = (T) + (T )
1 1 1 1
is linear.
To show is one-to-one, let (T) = (T )
1
Then T* = T *
1
T * T * 1 = 0
Using (6) by choosing = 1, = – 1 we get

(T – T )* = 0 T – T = 0 or T = T .
1 1 1
is one-to-one.
Hence is an isometric isomorphism on (N) onto (N*).
Finally we show that reverses the product and preserves the identity transformation.

Now [(T T )* (f)] (x) = f ((T T ) (x))
1 1
= f (T (T (x))
1
= [T* (f)] [T (x)], since T (x) N and T* (f) N*.
1 1

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