## Re: [harmonic] question

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• 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 1 of 15 , Apr 30, 2009
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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
> 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 2 of 15 , Apr 30, 2009
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Dear All
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 3 of 15 , May 3, 2009
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 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 4 of 15 , Aug 16, 2009
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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 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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