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Cooper
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 Posted: Mon May 28, 2007 11:01 am Post subject: Another question on dual space. |
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Hi!
In general, the dual of L_inf is not L_1.
To show this, we consider Lebesgue meassure m defined on I=[0,1].
Denote the collection of all bounded linear functional on L_inf as S
in which each element has the
form f -> integral_I {fg} dm for some g in L_1.
I showed first that S is isomorphic to L_1 and in the dual space of
L_inf, there is some bounded linear functional which is not element S.
Then, how can we prove that the dual of L_inf is not equal to L_1?
If we assume L_1 is isomorphic to the dual of L_inf, then we get only
S is contained in dual of L_inf and S is isomorphic to the dual of
L_inf. But it is insufficient to say S=the dual of L_inf our desired
goal to derive contradiction. There is counterexample it does not
establish.
(ex)Z is isomorphic to 2Z but 2Z is strictly contained in Z.
Help me please. |
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| Posted: Mon May 28, 2007 11:01 am Post subject: Re: Another question on dual space. |
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On May 28, 12:07 am, Cooper <cooper0...@gmail.com> wrote:
| Quote: |
Hi!
In general, the dual of L_inf is not L_1.
To show this, we consider Lebesgue meassure m defined on I=[0,1].
Denote the collection of all bounded linear functional on L_inf as S
in which each element has the
form f -> integral_I {fg} dm for some g in L_1.
I showed first that S is isomorphic to L_1 and in the dual space of
L_inf, there is some bounded linear functional which is not element S.
Then, how can we prove that the dual of L_inf is not equal to L_1?
If we assume L_1 is isomorphic to the dual of L_inf, then we get only
S is contained in dual of L_inf and S is isomorphic to the dual of
L_inf. But it is insufficient to say S=the dual of L_inf our desired
goal to derive contradiction. There is counterexample it does not
establish.
(ex)Z is isomorphic to 2Z but 2Z is strictly contained in Z.
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Let's clear up some terminology. Let F : L_p -> L_q^* be the map you
defined above: (F(f))(g) = integral_I fg dm. When people say things
like "the dual of L_q is equal to L_p", they mean that L_p is
isomorphic to L_q^* under the isomorphism F. The word "equal" is
really abused. L_p is not *equal* to L_q^*; the elements of L_p are
functions on I, but the elements of L_q are functions on L_q, so they
can't be identical. So "equal" here means "isomorphic under the
usual, natural isomorphism".
In that sense, you are done, because you have shown that the natural
map F : L_1 -> L_inf^* is not surjective and hence cannot be an
isometry.
However, as you say, you have not ruled out the possibility that there
exists some other isomorphism G. In fact there does not. Suppose
there did exist such G. L_1 is separable (1), so L_inf^* would be
separable as well. In general, if X is a Banach space and X^* is
separable, so is X (2). But L_inf is not separable (3), so this is a
contradiction. Each of these three statements is a good exercise to
try. |
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