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  • raziye1361
    Dear 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
    Message 1 of 15 , Oct 13, 2008
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      Dear 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}
    • shravan kumar
      Can you please state the theorem clearly!... ... From: raziye1361 Subject: [harmonic] question To: harmonicanalysis@yahoogroups.com
      Message 2 of 15 , Oct 13, 2008
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        Can you please state the theorem clearly!...

        --- On Mon, 13/10/08, raziye1361 <raziye1361@...> wrote:
        From: raziye1361 <raziye1361@...>
        Subject: [harmonic] question
        To: harmonicanalysis@yahoogroups.com
        Date: Monday, 13 October, 2008, 4:55 PM

        Dear 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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      • ghorbani_aylin
        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 3 of 15 , Oct 14, 2008
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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^ countable then G is compact why?
        • Hans G. Feichtinger
          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 4 of 15 , Oct 14, 2008
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            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

             

            www.nuhag.eu  

             


            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?

          • shravan kumar
            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 5 of 15 , Oct 14, 2008
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              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 regards
              Shravan

              --- On Tue, 14/10/08, ghorbani_aylin <ghorbani_aylin@...> wrote:
              From: ghorbani_aylin <ghorbani_aylin@...>
              Subject: [harmonic] question
              To: harmonicanalysis@yahoogroups.com
              Date: Tuesday, 14 October, 2008, 1:40 PM

              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?



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            • ghorbani_aylin
              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 6 of 15 , Oct 15, 2008
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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 A-bimodule if M={a.x.b: a,b in A, x n E}then A.(E/M)=0
              • shravan kumar
                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 7 of 15 , Oct 16, 2008
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                  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 regards
                  Shravan

                  --- On Thu, 16/10/08, ghorbani_aylin <ghorbani_aylin@...> wrote:
                  From: ghorbani_aylin <ghorbani_aylin@...>
                  Subject: [harmonic] question
                  To: harmonicanalysis@yahoogroups.com
                  Date: Thursday, 16 October, 2008, 11:32 AM

                  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


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                • fatima22_m
                  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, 2009
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                    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
                  • fatima22_m
                    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 9 of 15 , Apr 30, 2009
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                      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
                    • Maria Roginskaya
                      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 10 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
                        > 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
                        >
                        >
                        >
                      • fatima22_m
                        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 11 of 15 , Apr 30, 2009
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                          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
                        • shravan kumar
                          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 12 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.
                             
                            Bye
                            Shravan
                            --- On Fri, 1/5/09, fatima22_m <fatima22_m@...> wrote:

                            From: fatima22_m <fatima22_m@...>
                            Subject: [harmonic] question
                            To: harmonicanalysis@yahoogroups.com
                            Date: Friday, 1 May, 2009, 11:27 AM

                            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



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                          • fatima22_m
                            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 13 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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