Page 200 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Unit 16: Continuous Linear Transformations

Consider the vector x defined by Notes
y
x = i when y 0 and x = 0 otherwise. … (12)
i y i i i

y
Hence x max i 1 .
y i

n n
and |f (x)| = x y y i .
i i
i 1 i 1
n
Therefore y i f(x) f x f .
i 1

n
y f … (13)
i
i 1
n
It follows now from (11) and (13) that f y i so that y f is an isometric
i 1
isomorphism.

Hence,  n *  .
n
1
This completes the proof of the theorem.
Theorem 5: The conjugate space of  is  , where
p
q
1 1
1 and 1 < p < .
p q

or  * p  q .

p
Proof: Let x = (x )  so that x . … (1)
n p n
n 1
th
Let  = (0, 0, 0, …, 1, 0, 0, …) where 1 is in the m place.
n
e  for n = 1, 2, 3, …
p
n
*
We shall first determine the form of f and then establish the isometric isomorphism of  onto
p
 q .
n
By using (e ), we can write any sequence (x , x , …, x , 0, 0, 0, …) in the form x e and
n 1 2 n k k
k 1
n
x x e k = (0, 0, 0, …, x , x , …).
k
n+1 n+2
k 1

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