Re: Lost on rigged Hilbert space
- From: Martin Wanvik <martinw@xxxxxxxxxxxx>
- Date: Mon, 02 Jun 2008 04:10:42 EDT
On Jun 2, 12:22 am, Martin Wanvik
<mart...@xxxxxxxxxxxx> wrote:
ownV' ~ (cl(V))' = H' ~ H (here ~ means "linearlyisometric"); H is
assumed to be real.
If V is dense and Hilbert then V = H.
The problem is when V is equipped with a finer
topology than H.
In this case, (V,<>_V) is a Hilbert space in its
the topology on V comes from an inner product, orright: it is
complete with respect to its own norm, but the
closure of V with
respect to the norm of H is H.
You are assuming a bit too much here, namely that
even a single norm. Typically, it does not.
of
In this setting V could well be a proper subset
H...
Consider the case of Sobolev spaces H^m(R^n) (=:H^m),
with duals H^(-
m), (m integer for instance). They are Hilbert spaces
with the
following *continuous* injections :
H^m <= H^0 <= H^(-m)
I am concerned with the fact that we have these
inclusions on the one
hand, but H^(m) isomorphic to H^(-m) on the other
hand.
The spaces H^m are all separable (at least that is what Wikipedia tells me), and hence isomorphic to H^0 (or any other separable, infinite dimensional Hilbert space for that matter).
.
- References:
- Re: Lost on rigged Hilbert space
- From: kalcifire
- Re: Lost on rigged Hilbert space
- Prev by Date: Re: You should know this man
- Next by Date: New mathematics / physical sciences positions at http://jobs.phds.org, Jun 02, 2008
- Previous by thread: Re: Lost on rigged Hilbert space
- Next by thread: Re: Lost on rigged Hilbert space
- Index(es):
Relevant Pages
|
