Can you please state the theorem clearly!... --- On

**Mon, 13/10/08, raziye1361**wrote:*<raziye1361@...>*From: raziye1361 <raziye1361@...>

Subject: [harmonic] question

To: harmonicanalysis@yahoogroups.com

Date: 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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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 $L^\infty(\mu)^**$.