Page 248 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Unit 22: The Uniform Boundedness Theorem

It we define the norm on N as Notes
x = max {|a |, n = 0, 1, 2, …}
n
then N is an incomplete normed linear space.

n 1
Now define f (x) = a , n = 1, 2, …
n k
k 0
The functions {f } are continuous linear functional on N.
n
m
If we take x = a + a t + …… + a t then
0 1 m
|f (x)| (m + 1) max {|a |} = (m + 1) x ,
n k
so that (|f (x)|) is point-wise bounded.
n
n–1
2
Now consider x = 1 + t + t + … + t . Then x = 1 and from the definition of |f (x)| = n.
n
f (x)
Hence f n = n.
n x
( f ) is unbounded.
n
Thus if we drop the condition of completeness in the domain of (T ), the uniform boundedness
i
theorem is not true anymore.
Theorem 2: Let N be a normed linear space and B be a Banach space. If a sequence (T ) (B, N)
n
such that T (x) = lim T (x) exists for each x in B, then T is a continuous linear transformation.
n
Proof: T is linear.
T ( x + y) = lim T ( x + y)
n
= lim {T ( x) + T ( y)|}
n n
= lim T (x) + lim T (y)
n n
= T (x) + T (y) for x, y B and for any scalars and .
since lim T (x) exists, (T (x)) is a convergent sequence in N. Since convergent sequences are
n n
bounded, (T (x)) is point-wise bounded.
n
Hence by uniform bounded theorem, ( T ) is bounded so that a positive constant such that
n
T n.
n
Now T (x) T x x .
n n
Since T (x) T (x), we have
n
T(x) x

T is bounded (continuous) linear transformation. This completes the proof of the theorem.

Corollary 1: If f is a sequence in B* such that f (x) = lim f (x) exists for each x B, then f is
n n
continuous linear functional on B.

Example 2: Let (a ) be a sequence of real or complex numbers such that for each x = (x )
n n
c , a x converges. Prove that |a | .
o n n n
n 1 n 1

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