Page 91 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Measure Theory and Functional Analysis

Notes in the following sense: For every bounded linear functional  on L there is a unique g L so
p
q
that
 (f) = f(x)g(x)d (x), for all f L p
x
Moreover,  = g q .
 * L
This theorem justifies the terminology where by q is usually called the dual exponent of p.
The proof of the theorem is based on two ideas. The first, as already seen, is Hölder’s inequality;
p
to which a converse is also needed. The second is the fact that a linear functional  on L , 1 p
< , leads naturally to a (signed) measure . Because of the continuity of  the measure is
absolutely continuous with respect to the underlying measure , and our desired function g is
then the density function of in terms of .
We begin with:
Lemma: Suppose 1 p, q , are conjugate exponents.

q
(i) If g L , then g q sup fg .
L
f p
L 1
(ii) Suppose g is integrable on all sets of finite measure and
sup fg = M <
f L p 1
f simple

Then g L , and g q = M.
q
L
For the proof of the lemma, we recall the signum of a real number defined by

1 if x 0
sign (x) = 1 if x 0
0 if x 0

Proof: We start with (i). If g = 0, there is nothing to prove, so we may assume that g is not 0 a.e.,
and hence g 0 . By Hölder’s inequality, we have that
L q

g q sup fg .
L
f p
L 1
To prove the reverse inequality we consider several cases.

 First, if q = 1 and p = , we may take f (x) = sign g (x). Then, we have f 1 and
L
clearly fg g 1 .
L

q 1
 If 1 < p, q < , then we set f (x) = |g (x)| q–1 sign g(x) g q . We observe that
L
p p(q 1) p(q 1)
f g(x) d g 1 since p (q – 1) = q, and that fg g q .
L p L q L

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