Some Continuous Duals
I really wish I could just say in post titles.
Anyway, I want to investigate the continuous dual of for
. That is, we’re excluding the case where either
(but not its Hölder conjugate
) is infinite. And I say that when
is
-finite, the space
of bounded linear functionals on
is isomorphic to
.
First, I’m going to define a linear map . Given a function
, let
be the linear functional defined for any
by
It’s clear from the linearity of multiplication and of the integral itself, that this is a linear functional on . Hölder’s inequality itself shows us that not only does the integral on the right exist, but
That is, is a bounded linear functional, and the operator norm
is at most the
norm of
. The extremal case of Hölder’s inequality shows that there is some
for which this is an equality, and thus we conclude that
. That is,
is an isometry of normed vector spaces. Such a mapping has to be an injection, because if
then
, which implies that
.
Now I say that is also a surjection. That is, any bounded linear functional
is of the form
for some
. Indeed, if
then we can just pick
as a preimage. Thus we may assume that
is a nonzero bounded linear functional on
, and
. We first deal with the case of a totally finite measure space.
In this case, we define a set function on measurable sets by . It’s straightforward to see that
is additive. To prove countable additivity, suppose that
is the countable disjoint union of a sequence
. If we write
for the union of
through
, we find that
Since is continuous, we conclude that
, and thus that
is a (signed) measure. It should also be clear that
implies
, and so
. The Radon-Nikodym theorem now tells us that there exists an integrable function
so that
Linearity tells us that
for simple functions , and also for every
, since each such function is the uniform limit of simple functions. We want to show that
.
If , then we must show that
is essentially bounded. In this case, we find
for every measurable , from which we conclude that
a.e., or else we could find some set on which this inequality was violated. Thus
.
For other , we can find a measurable
with
so that
. Setting
and defining
, we find that
on
,
, and so
We thus find
and thus
Applying the monotone convergence theorem as we find that
.
Thus in either case we’ve found a so that
.
In the -finite case, we can write
as the countable disjoint union of sets
with
. We let
be the union of the first
of these sets. We note that
for every measurable set
, so
is a linear functional on
of norm at most
. The finite case above shows us that there are functions
on
so that
.
We can define if
, and let
be the sum of all these
. We see that
for every , and since
we find that
. Then Fatou’s lemma shows us that
. Thus the
-finite case is true as well.
One case in particular is especially worthy of note: since is Hölder-coonjugate to itself, we find that
is isomorphic to its own continuous dual space in the same way that a finite-dimensional inner-product space is isomorphic to its own dual space.
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