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Stephen MontgomerySmith1 science forum Guru
Joined: 01 May 2005
Posts: 487

Posted: Tue Jul 11, 2006 6:02 am Post subject:
Re: characterization of W^{1,p} norms



don wrote:
Quote:  Hi,
I have a question concerning W^{1,p} norms:
is it true that the W^{1,p} norm of a function $v$
can be
characterized as
\int u v + \nabla u \nabla v
with $u$ ranging in W^{1,q}?
(q is the dual exponent of p)
A possible proof is the followinig: a standard
representation theorem
for the dual of W^{1,p} in N dimensions states that
any dual F can be
represented through a N+1 ple of L^q functions f_i,
so that
F,v>=\int f_0 u + (f_1,...f_N) \nabla u
So, if I want to compute the norm of u, I must test
its gradient
against all vector fields. On the other hand the
vector field
(f_1,...,f_N) can be decomposed into a sum of a
solenoidal field and a
gradient. When testing with such decomposition, the
solenoidal field is
calceled out, and we remain with the scalar product
of two gradients.
Any opinions?
Thanks in advance.
joe
Thanks.
Joe
I was flipping through a book and saw something along the lines of
sup( \int_\Omega grad u grad v ) >= c  u 
where the norm is the W_0^{1,p}(\Omega) one and we are supping over v in W_0^{1,q} with norm of v <=1.
Here q is conjugate of p. They said result followed
from some Zygmund  Calderon theorem but I never figure out which one.
If you have this figure out I would really like to know some of the details or where to look.
I am in some open , bounded subset Omega of R^n .
thanks
craig

Here are ideas.
What you need to show is that if u is a vector field on Omega, and you
decompose it using the Hodge decomposition as nabla p + v, where div v
= 0, then v_p <= C u. You work out what the operator is  on
R^n it would be some kind of composition of Riesz projections, but on
Omega it would be something different, but still you can probably
calculate that it satisfies the hypothesis of the CalderonZygmund
Theorem. (Incidently I have a feeling that the smoothness of the
boundary of Omega will play a role, as I recall seeing somewhere that
it isn't necessarily true for all p in (1,infty) if the boundary of
Omega only satisfies Lipschitz boundary conditions.) 

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don11154 science forum beginner
Joined: 05 Jul 2005
Posts: 39

Posted: Fri Jul 07, 2006 9:13 pm Post subject:
Re: characterization of W^{1,p} norms



Quote:  Hi,
I have a question concerning W^{1,p} norms:
is it true that the W^{1,p} norm of a function $v$
can be
characterized as
\int u v + \nabla u \nabla v
with $u$ ranging in W^{1,q}?
(q is the dual exponent of p)
A possible proof is the followinig: a standard
representation theorem
for the dual of W^{1,p} in N dimensions states that
any dual F can be
represented through a N+1 ple of L^q functions f_i,
so that
F,v>=\int f_0 u + (f_1,...f_N) \nabla u
So, if I want to compute the norm of u, I must test
its gradient
against all vector fields. On the other hand the
vector field
(f_1,...,f_N) can be decomposed into a sum of a
solenoidal field and a
gradient. When testing with such decomposition, the
solenoidal field is
calceled out, and we remain with the scalar product
of two gradients.
Any opinions?
Thanks in advance.
joe
Thanks.
Joe

I was flipping through a book and saw something along the lines of
sup( \int_\Omega grad u grad v ) >= c  u 
where the norm is the W_0^{1,p}(\Omega) one and we are supping over v in W_0^{1,q} with norm of v <=1.
Here q is conjugate of p. They said result followed
from some Zygmund  Calderon theorem but I never figure out which one.
If you have this figure out I would really like to know some of the details or where to look.
I am in some open , bounded subset Omega of R^n .
thanks
craig
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To: scimathresearch@moderators.isc.org
Path: abergman
From: Aaron Bergman <abergman@physics.utexas.edu>
Newsgroups: sci.math.research
Subject: Re: Computing the Gerstenhaber Bracket
Date: Sat, 08 Jul 2006 02:28:45 0500
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In article <1152257239.068384.47280@k73g2000cwa.googlegroups.com>,
"Agusti Roig" <soquiso@hotmail.com> wrote:
Quote:  Aaron Bergman ha escrit:
I've been working on doing some computations in Hochschild cohomology,
and all the definitions I've seen of the Gerstenhaber bracket define it
either hopelessly abstractly or in terms of the bar resolution. I have a
different resolution I've been using to compute HH^*, and I was hoping
to be able to compute the Gerstenhaber bracket (or HH^2 to a specific
formal deformation of the algebra) without having to compute an explicit
quasiisomorphism to the bar resolution. Any ideas or references would
be appreciated.
I don't know if I understand the problem: do you mean you have a
resolution that allows you the computation of the Gerstenhaber bracket
and you want an explicit quasiisomorphism with the bar resolution
because your are not sure that your resolution actually computes
Hochschild cohomology?

No. I have an explicit projective resolution that isn't the bar
resolution. I was hoping to be able to compute the Gerstenhaber bracket
from it, but judging from the other response, I guess that's not so
likely.
The parenthetical comment is somewhat unclear, so, to elaborate, I was
also hoping to describe the explicit deformation for a given element of
HH^2 using my resolution, but perhaps that is also hard.
Thanx,
Aaron 

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don11154 science forum beginner
Joined: 05 Jul 2005
Posts: 39

Posted: Fri Jul 07, 2006 9:13 pm Post subject:
Re: characterization of W^{1,p} norms



Quote:  Hi,
I have a question concerning W^{1,p} norms:
is it true that the W^{1,p} norm of a function $v$
can be
characterized as
\int u v + \nabla u \nabla v
with $u$ ranging in W^{1,q}?
(q is the dual exponent of p)
A possible proof is the followinig: a standard
representation theorem
for the dual of W^{1,p} in N dimensions states that
any dual F can be
represented through a N+1 ple of L^q functions f_i,
so that
F,v>=\int f_0 u + (f_1,...f_N) \nabla u
So, if I want to compute the norm of u, I must test
its gradient
against all vector fields. On the other hand the
vector field
(f_1,...,f_N) can be decomposed into a sum of a
solenoidal field and a
gradient. When testing with such decomposition, the
solenoidal field is
calceled out, and we remain with the scalar product
of two gradients.
Any opinions?
Thanks in advance.
joe
Thanks.
Joe

I was flipping through a book and saw something along the lines of
sup( \int_\Omega grad u grad v ) >= c  u 
where the norm is the W_0^{1,p}(\Omega) one and we are supping over v in W_0^{1,q} with norm of v <=1.
Here q is conjugate of p. They said result followed
from some Zygmund  Calderon theorem but I never figure out which one.
If you have this figure out I would really like to know some of the details or where to look.
I am in some open , bounded subset Omega of R^n .
thanks
craig 

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Stephen MontgomerySmith1 science forum Guru
Joined: 01 May 2005
Posts: 487

Posted: Sat May 27, 2006 4:00 am Post subject:
Re: characterization of W^{1,p} norms



GT wrote:
Quote:  Hi,
I have a question concerning W^{1,p} norms:
is it true that the W^{1,p} norm of a function $v$ can be
characterized as
\int u v + \nabla u \nabla v
with $u$ ranging in W^{1,q}?
(q is the dual exponent of p)
A possible proof is the followinig: a standard representation theorem
for the dual of W^{1,p} in N dimensions states that any dual F can be
represented through a N+1 ple of L^q functions f_i, so that
F,v>=\int f_0 u + (f_1,...f_N) \nabla u
So, if I want to compute the norm of u, I must test its gradient
against all vector fields. On the other hand the vector field
(f_1,...,f_N) can be decomposed into a sum of a solenoidal field and a
gradient. When testing with such decomposition, the solenoidal field is
calceled out, and we remain with the scalar product of two gradients.
Any opinions?

I haven't worked out the solution, but I have a feeling that the
following fact might help you.
If a vector f with components in L_p(R^N) is decomposed into its
gradient part nabla u and its solenoidal part v (i.e. the Hodge
decomposition), then for 1<p<infinity, f_p is equivalent (i.e. the
ratio is bounded above and below by constants depending only upon p and
possibly N) to nabla u_p + v_p.
This follows basically because the decomposition is a singular integral
 indeed it can be made up from compositions of so called Riesz
tranforms. The result may in effect be found in
http://www.ams.org/mathscinetgetitem?mr=1442167
but this is somewhat using a sledgehammer to crack a nut.
My guess is that the result you are looking for is false if p=1 or
p=infinity.
I hope this helps.
Stephen 

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GT science forum beginner
Joined: 06 Oct 2005
Posts: 4

Posted: Thu May 25, 2006 3:00 pm Post subject:
characterization of W^{1,p} norms



Hi,
I have a question concerning W^{1,p} norms:
is it true that the W^{1,p} norm of a function $v$ can be
characterized as
\int u v + \nabla u \nabla v
with $u$ ranging in W^{1,q}?
(q is the dual exponent of p)
A possible proof is the followinig: a standard representation theorem
for the dual of W^{1,p} in N dimensions states that any dual F can be
represented through a N+1 ple of L^q functions f_i, so that
<F,v>=\int f_0 u + (f_1,...f_N) \nabla u
So, if I want to compute the norm of u, I must test its gradient
against all vector fields. On the other hand the vector field
(f_1,...,f_N) can be decomposed into a sum of a solenoidal field and a
gradient. When testing with such decomposition, the solenoidal field is
calceled out, and we remain with the scalar product of two gradients.
Any opinions?
Thanks in advance.
joe
Thanks.
Joe 

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