Page 93 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Measure Theory and Functional Analysis

Notes
for every measurable set E. Thus we have ( E)  Eg d . The representation (f) fg d
then extends immediately to simple function f, and by a passage to the limit, to all f L since the
p
p
simple functions are dense in L , 1 p < . Also by lemma, we see that g q  .
L
To pass from the situation where the measure of X is finite to the general case, we use an
increasing sequence {E } of sets of finite measure that exhaust X, that is, X =  n 1 E n . According
n
to what we have just proved, for each n there is an integrable function g on E (which we can set
n n
c
to be zero in E ) so that
n
 (f) = fg d … (2)
n

p
whenever f is supported in E and f L . Moreover by conclusion (ii) of the lemma g p  .
n n L
Now it is easy to see because of (2) that g = g a.e. on E , whenever n m. Thus lim g (x)
n m m n n
= g (x) exists for almost every x, and by Fatou’s lemma, g L  . As a result we have that
q
 (f) f g du for each f L supported in E , and then by a simple limiting argument, for all f
p
n
q
p
L supported in E . The fact that  g L is already contained in Hölder’s inequality and
n
therefore the proof of the theorem is complete.
Theorem 2: Let f be a linear functional defined on a normed linear space N, then f is bounded
f is continuous.
Proof: Let us first show that continuity of f boundedness of f.
If possible let f is continuous but not bounded. Therefore, for any natural number n, however
large, there is some point x such that
n
|f (x )| n x … (1)
n n
x
Consider the vector y n so that
n
n x
n
1
y = .
n n
y 0 as n .
n
y 0 in the norm.
n
Since any continuous functional maps zero vector into zero, and f is continuous f (y ) f (0) = 0.
n
1
But |f (y )| = f(x ) … (2)
n n x n
n
It now follows from (1) and (2) that

|f (y )|> 1, a contradiction to the fact that
n
f (y ) 0 as n .
n
Thus if f is bounded then f is continuous.

86 LOVELY PROFESSIONAL UNIVERSITY