Page 247 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Measure Theory and Functional Analysis

Notes 22.1.2 Theorems and Solved Examples

Theorem 1: If B is a Banach space and (f (x)) is sequence of continuous linear functionals on B such
i
that (|f (x)|) is bounded for every x B, then the sequence ( f ) is bounded.
i i
Proof: Since the proof of the theorem is similar to the theorem (1), however we briefly give its
proof for the sake of convenience to the readers.
For every m, let F B be the set of all x such that |f (x)| m n
m n
|f (x)| m n
n
Now F is the intersection of closed sets and hence it is closed.
m
As in previous theorem, we have
B =  F m . Since B is complete. It is of second category. Hence by Baire’s theorem, there is a

m 1
x F and a closed sphere S [x , r ] such that |f (x)| m n.
o m o o n
Let x be a vector with x r .
o
Now f (x) = f (x + x – x )
n n o o
= f (x + x ) – f (x )
n o n o
|f (x)| |f (x + x )| + |f (x )| … (1)
n n o n o
Since x + x – x = x < r , we have (x + r ) S [x , r ]
o o o o o o
|f (x + x ) | m … (2)
n o
Also we have |f (x )| k n … (3)
n o
From (1), (2) and (3), we have for x S [x , r].
o
|f (x)| < (m + k) n.
n
r x
Now for x B, consider the vector o .
x

x r x x f (x) m k
n
Then |f (x)| f o (m k) so that .
n n
r x r x r
o o o
In other words,
m k
f .
r
o
This completes the proof of the theorem.

Example 1: Show that the completeness assumption in the domain of (T ) in the uniform
i
boundedness theorem cannot be dropped.
Solution: Consider N= space of all polynomial x

n n , a 0
= x (t) = a t n
n 0
for finitely many n’s.

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