Page 249 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Measure Theory and Functional Analysis

Notes
n n
Solution: For every x c , let f a x . Since each a x i is a finite sum of scalars, (f ) is
i
o n i i n
i 1 i 1
n
sequence of continuous linear functional on c . Let f (x) = lim f (x) lim a x i . By cor. 1, f (x)
o n n n i
i 1
exists and bounded. f = |a | |a |
n < .
n . Since f is bounded,
n 1 n 1
Theorem 3: A non-empty subset X of a normed linear space N is bounded f (X) is a bounded set
of numbers for each f in N*.
Proof: Let X be a bounded subset of N so that a positive constant such that
1
x x X … (1)
1
To show that f (X) is bounded for each f N*. Now f N* f is bounded.
> 0 such that |f (x)| x x N … (2)
2 2
It follows from (1) & (2) that
|f (x)| x X.
1 2
f (X) is a bounded set of real numbers for each f N*.

Conversely, let us assume that f (X) is a bounded set of real numbers for each f N*.
To show that X is bounded. For convenience, we exhibit the vectors in X by writing X = {x }. We
i
now consider the natural imbedding J from N to N** given by
J : x F
i x i
From the definition of this natural imbedding, we have
F (f) = f (x) for each x N.
x
Hence our assumption f (X) = {f (x )} is bounded for each f N* is equivalent to the assumption
i
that F (f) is bounded set for each f N*.
x i

Since N* is complete F is bounded subset of N** by uniform boundedness theorem.
x i
That is, F is a bounded set of numbers. Since the norms are preserved in natural imbedding,
x i
we have F = x for every x X.
x i i i
Therefore ( x ) is a bounded set of numbers. Hence is bounded subset of N .
i i
This completes the proof of the theorem.
Theorem 4: Let N and N be normed linear space A linear transformation.
T : N N is continuous for each f N*, f o T N*.

Proof: We first note that f o T is linear. Also f o T is well defined, since T (x) N for every x N
and f is a functional on N so that f (T (x)) is well defined and f o T N*. Since T is continuous and
f is continuous, f o T is continuous on N.

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