Re: do only locally convex spaces have dual spaces?
- From: "Zig" <ziggurism@xxxxxxxxx>
- Date: Wed, 25 Jan 2006 03:00:07 +0000 (UTC)
Robert Israel wrote:
> In article <dr5o71$kvm$1@xxxxxxxxxxxxxxxx>, Zig <ziggurism@xxxxxxxxx> wrote:
> >
> >I conjecture therefore this theorem: if a space has a nontrivial dual
> >space, then the space is locally convex.
>
> False. Consider the direct sum X = A + B where A is a locally convex tvs
> and B is a tvs with trivial dual space
Well that puts the nail in the coffin of that idea. Thanks for the
counterexample.
I suppose there is no simple condition which implies a space will not
have a dual space?
Also, I found these notes online at
http://www.math.uconn.edu/~kconrad/blurbs/lpspace.pdf
about Lp spaces for 0<p<1. The notes say that the dual space of Lp is
nontrivial if the measure has an atomic set of finite measure. I guess
this means that while Lp(R) or Lp([0,1]) have trivial dual spaces,
Little lp(N), the space of sequences, will not, since it's Lp with the
counting measure, which is atomic. So that provides another
counterexample to my purported theorem.
I had a problem with that result though: the proof of the theorem in
those notes (that Lp has nontrivial dual space if the measure has an
atom) troubled me though. It rests on the fact that a measurable
function must be constant on an atomic set. This statement seems to be
untrue to me. For example, the Dirac measure m(E)=1 if a \in E and 0
otherwise seems to be atomic; every set is an atom. And yet there are
lots of nonconstant functions there. Can you straighten me out some
more?
.
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