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## Re: [harmonic] question

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• Can you please state the theorem clearly!... ... From: raziye1361 Subject: [harmonic] question To: harmonicanalysis@yahoogroups.com
Message 1 of 15 , Oct 13, 2008
 Can you please state the theorem clearly!... --- On Mon, 13/10/08, raziye1361 wrote:From: raziye1361 Subject: [harmonic] questionTo: harmonicanalysis@yahoogroups.comDate: Monday, 13 October, 2008, 4:55 PMDear members please help me with the proof of this theorem Let S be a semigroup such that $l^(s)$ is amenable. then $|E(S)|^(1/2) \leq AM(l^(s))$ if the idempotents of S commute, then $|E(S)|\leq AM(l^(s))$ definition: let A be amenable Banach algebra we set AM(A)=inf{C> 0 :A is C-amenable}

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• Dear members Thanks for your reply to my questions Here there are some questions Let G is a locally compact abelian group and G^character group of G if G^
Message 2 of 15 , Oct 14, 2008
Dear members Thanks for your reply to my questions Here there are some
questions
Let G is a locally compact abelian group and G^character group of G if
G^ countable then G is compact why?
• I would think of the following argument: G is compact if and only if G^ is discrete (with its natural topology). If a group is countable it cannot be
Message 3 of 15 , Oct 14, 2008

I would think of the following argument:

G is compact if and only if  G^  is discrete (with its natural topology).

If a group is countable it cannot be anything but discrete, I guess,

because any “continouos” contribution (like R^d or T^s) would

induce infinitely many elements.

HGFEi

Von: harmonicanalysis@yahoogroups.com [mailto:harmonicanalysis@yahoogroups.com] Im Auftrag von ghorbani_aylin
Gesendet: Dienstag, 14. Oktober 2008 10:11
An: harmonicanalysis@yahoogroups.com
Betreff: [harmonic] question

Dear members Thanks for your reply to my questions Here there are some
questions
Let G is a locally compact abelian group and G^character group of G if
G^ countable then G is compact why?

• If I remember right, the contents of my previous mail regarding the relation between L1(G) and the character group G^ tells this for a locally compact abelian
Message 4 of 15 , Oct 14, 2008
 If I remember right, the contents of my previous mail regarding the relation between L1(G) and the character group G^ tells this for a locally compact abelian group. More precisely, that the spectrum of L1(G) as a Banach algebra is G^. The important fact that is to be noted here is that G^ is also locally compact since the existence of the multiplicative identity in L1(G) is not gaurenteed. When G is discrete, both L1(G) and M(G) coincide and as M(G) contains identity, the dirac measure or the point measure at the identity, L1(G) will contain such an identity in this case. Therefore G^ is compact. Infact more generally, if A is any commutative Banach algebra without identity, then the spectrum i.e., the set of all  nonzero continuous multiplicative linear functinals on A, is locally compact with respect to the weak star topology. If the Banach algebra contains the identity then the spectrum is compact. The reason why the spectrum of the algebra without identity is locally compact is the following. One can adjoin an identity to the algebra so that the algebra is a maximal ideal in the newly formed algebra. Therefore one of the multiplicative functionals on this new algebra is the one which is zero on all of our algebra and identity on the remailning part as the remaining part is just isomorphic to C.Best regardsShravan --- On Tue, 14/10/08, ghorbani_aylin wrote:From: ghorbani_aylin Subject: [harmonic] questionTo: harmonicanalysis@yahoogroups.comDate: Tuesday, 14 October, 2008, 1:40 PMDear members Thanks for your reply to my questions Here there are some questions Let G is a locally compact abelian group and G^character group of G if G^ countable then G is compact why?

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• Dear members Thanks for your reply to my questions Here there are some questions. let A be a Banach algebra with bounded approximate identity and E Banach
Message 5 of 15 , Oct 15, 2008
Dear members Thanks for your reply to my questions Here there are some
questions.
let A be a Banach algebra with bounded approximate identity and E
Banach A-bimodule if M={a.x.b: a,b in A, x n E}then A.(E/M)=0
• I suppose that it is clear from the definition of the quotient space E/M and the module action of A on E/M. Notice that A acts on E and this action is
Message 6 of 15 , Oct 16, 2008
 I suppose that it is clear from the definition of the quotient space E/M and the module action of A on E/M. Notice that A acts on E and this action is extended to E/M through the qoutient map. But by the definition of M A has to annihilate E/M. I suppose that explanation given is satisfactory. If not please mail me. Best regardsShravan --- On Thu, 16/10/08, ghorbani_aylin wrote:From: ghorbani_aylin Subject: [harmonic] questionTo: harmonicanalysis@yahoogroups.comDate: Thursday, 16 October, 2008, 11:32 AMDear members Thanks for your reply to my questions Here there are some questions.let A be a Banach algebra with bounded approximate identity and E Banach A-bimodule if M={a.x.b: a,b in A, x n E}then A.(E/M)=0

• Dear All please help me with the proof of this point. the group $G$ is discrete if the Haar measure $mu$ is discrete. In the proof of this point the writer
Message 7 of 15 , Apr 30 6:53 AM
Dear All
the group $G$ is discrete if the Haar measure $\mu$ is discrete.
In the proof of this point the writer has said if $G$ is not
discrete then it contains a compact set that is infinite. Is
there any one who can help me with the above line.
$G$ is also a locally compact Haussdorff space.
Thanks
Fatima
• Dear All please help me with the proof of this point. the group $G$ is discrete if the Haar measure $mu$ is discrete. In the proof of this point the writer
Message 8 of 15 , Apr 30 6:53 AM
Dear All
the group $G$ is discrete if the Haar measure $\mu$ is discrete.
In the proof of this point the writer has said if $G$ is not
discrete then it contains a compact set that is infinite. Is
there any one who can help me with the above line.
$G$ is also a locally compact Haussdorff space.
Thanks
Fatima
• If you have a locally compact Hausdorff space, then every point has a local basis of compact neighbourhoods. I.e. for any point there is a compact set which
Message 9 of 15 , Apr 30 12:28 PM
If you have a locally compact Hausdorff space, then every point has a
local basis of compact neighbourhoods. I.e. for any point there is a
compact set which contains an open set which also contains the point. Now,
if every compact is finite, than there are finite open sets. Take such a
set E, which contains p. As the space is Hausdorff for each x in E which
is not p one can find an open set U_x which contains p and does not
contain x. Then the intersection of E and all U_x is just {p}, and as an
intersection of finitely many open sets it should be open. I.e. the
topology is discrete.

I want to remark that even for a Hausdorff topological group, unless the
unity (and thus all other points) have a countable neighbourhood basis,
the statement does not hold. I.e. one can construct a Hausdorff
topological group in which only finite sets are compact, but it is not a
discrete.

> Dear All
> please help me with the proof of this point.
> the group $G$ is discrete if the Haar measure $\mu$ is discrete.
> In the proof of this point the writer has said if $G$ is not
> discrete then it contains a compact set that is infinite. Is
> there any one who can help me with the above line.
> $G$ is also a locally compact Haussdorff space.
> Thanks
> Fatima
>
>
>
• Dear All Thanks a lot Maria Roginskala. please help me with the following questions : 1-As I know when $G$ is a compact group then the Haar measure $mu$
Message 10 of 15 , Apr 30 10:57 PM
Dear All
Thanks a lot Maria Roginskala. please help me with the following
questions :
1-As I know when $G$ is a compact group then the Haar measure $\mu$
does belong to $M(G)$. Is it true when $G$ is a locally comapct abelian
group?
2-Let $G$ is a locally compact abelian group. if we prove that Haar
measure $\mu$ that is restircted to a relatively compact open subset
of $G$ , say $U$ , is not discrete , can we conclude that $\mu$ is not
discrete for the group $G$?
I wish to hear from whom can give his or her comments very soon.
Best Regards
Fatima
• Dear Fatima   1. Unless the haar measure is bounded it will not belong to M(G). 2. Ofcourse, one can conclude that the haar measure is not discrete. This
Message 11 of 15 , May 3, 2009
 Dear Fatima 1. Unless the haar measure is bounded it will not belong to M(G).2. Ofcourse, one can conclude that the haar measure is not discrete. This follows because of the fact that Haar measure is translation invariant. ByeShravan --- On Fri, 1/5/09, fatima22_m wrote:From: fatima22_m Subject: [harmonic] questionTo: harmonicanalysis@yahoogroups.comDate: Friday, 1 May, 2009, 11:27 AMDear AllThanks a lot Maria Roginskala. please help me with the followingquestions :1-As I know when $G$ is a compact group then the Haar measure $\mu$does belong to $M(G)$. Is it true when $G$ is a locally comapct abeliangroup?2-Let $G$ is a locally compact abelian group.. if we prove that Haarmeasure $\mu$ that is restircted to a relatively compact open subsetof $G$ , say $U$ , is not discrete , can we conclude that $\mu$ is notdiscrete for the group $G$?I wish to hear from whom can give his or her comments very soon.Best RegardsFatima

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• Dear Members can any one help me about some qusetions from these lemma. we fisrt have some assumptions: Asuumptions: Let $mu$ be a finite positive regular
Message 12 of 15 , Aug 16, 2009
Dear Members
can any one help me about some qusetions from these lemma.
we fisrt have some assumptions:
Asuumptions:
Let $\mu$ be a finite positive regular Borel Measure on the locally compact Hausdorff space $X$. we let $S^*(X. \mu)$ denote the elements of $L^\infty (\mu)$ that have absolute value one $\mu$-almost every where. Let $\rho: \Gamma\rightarrow S^*(X,\mu)$ be a group homomorphism from alocally compact abelian group $\Gamma$ such that $\rho(\gamma)\in CB(X)$ for all $\gamma$ and such that $\rho(\Gamma)$ seprates the points of $X$.
lemma: Under above assumptions , let $F$ be any weak limit of a net in $O(\rho)=\{ \rho(\gamma): \gamma\in \Gamma \}$. then $F$ has absolute value one every where as an element of $L^\infty (\mu)^{**}$.
proof: Fix a point $\omega$ in the support of $\mu$, considered as a measure on $\Delta(L^\infty)(\mu)$. then $|\rho(\gamma_{\alpha})|=1$
for all $\alpha$, since evaluation of $\rho(\gamma_{\alpha})$ at $\omega$ is given by the bounded linear functional $$\rho(\gamma_alpha)\rightarrow \int \rho(\gamma_\alpha)d\delta_omega=\prec \rho(\gamma_{\alpha}), \delta_\omega \succ$$
which maps $L^\infty(\mu)\rightarrow C$. in above equation the integration is aganist the unit point mass at $\omega\in \Delta(L^\infty)(\mu)$. and we are identifying elements of $L^\infty(\mu)$ with thier Gelfand transforms, hence the weak limit $F$ of elements
of $\rho(\Gamma)$ has $|F|=1$ as an element of $L^\infty(\mu)^**$.
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