Page 94 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Unit 8: Bounded Linear Functional on the L -spaces
p

Conversely, let f is bounded. Then for any sequence (x ), we have Notes
n
|f (x )| k x n = 1, 2, … and k 0.
n n
Let x 0 as n then
n
f (x ) 0
n
f is continuous at the origin and consequently it is continuous everywhere.
This completes the proof of the theorem.
Theorem 3: If L is a linear space of all n-tuples, then
n
(i)  *  n q
p
n
(ii)  *  n
1
(iii)  n *  n 1

Proof: Let (e , e , …, e ) be a standard basis for L so that any x = (x , x , …, x ) L can be written
1 2 n 1 2 n
as x = x e + x e + … + x e .
1 1 2 2 n n
If f is a scalar valued linear function defined on L, then we get
f (x) = x f (e ) + x f (e ) + … + x f (e ) … (1)
1 1 2 2 n n
f determines and is determined by n scalars y = f (e ).
i i
Then the mapping
n
y = (y , y , … y ) f where f (x) = x y i is an isomorphism of L onto the linear space L of all
i
1 2 n
i 1
function f. We shall establish (i) – (iii) by using above given facts.
th
n
(i) If we consider the space L =  (1 p < ) with the p norm, then f is continuous and L
p
*
n
represents the set of all continuous linear functionals on  so that L =  n .
p p
Now for y f as an isometric isomorphism we try to find the norm of y’s.
For 1 < p < , we show that
 n p * =  .
n
p
For x  , we have defined,
n
p
1
n p
x = |x | p
i
i 1
n n
Now |f (x)| = x y i |x ||y |
i
i
i
i 1 i 1
By using Holder’s inequality, we get
1 1
n n p n q
|x y |  |x | p |y | q so that
i
i
i
i
i 1 i 1 i 1
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