Amenability of Abelian Groups

# Amenability of Abelian Groups

 Theorem 1: Let $G$ be an abelian group. Then $G$ is an amenable group.

Recall that a group is said to be abelian (or commutative) if for all $g, h \in G$ we have that $gh = hg$.

• Proof: Let $A = \ell^1(G)$ and let $\sigma : A \to \mathbb{C}$ be defined for all $a \in A$ by:
(1)
\begin{align} \quad \sigma(a) = \sum_{g \in G} a(g) \end{align}
• Observe that $\sigma$ is a linear functional on $A$ since for all $a, b \in A$ and all $\alpha \in \mathbb{C}$ we have that:
(2)
\begin{align} \quad \sigma(a + b) = \sum_{g \in G} [a + b](g) = \sum_{g \in G} [a(g) + b(g)] = \sum_{g \in G} a(g) + \sum_{g \in G} b(g) = \sigma(a) + \sigma(b) \end{align}
(3)
\begin{align} \quad \sigma(\alpha a) = \sum_{g \in G} [\alpha a](g) = \alpha \sum_{g \in G} a(g) = \alpha \sigma(a) \end{align}
• Furthermore, $\sigma$ is multiplicative, since for all $a, b \in A$ and all $h \in G$ we have that:
(4)
\begin{align} \quad \sigma (ab) = \sum_{g \in G} (ab)(g) = \sum_{g \in G} \sum_{h \in G} a(h)b(h^{-1}g) = \sum_{h \in G} \sum_{g \in G} a(h)b(h^{-1}g) = \sum_{h \in G} a(h) \sum_{g \in G} b(h^{-1}g) = \sum_{h \in G} a(g) \sum_{g \in G} b(g) = \sigma(a) \sigma(b) \end{align}
• So $\sigma : A \to \mathbb{C}$ is bounded linear functional. Furthermore, for all $a \in A$ and all $g \in G$ we have that:
(5)
\begin{align} \quad \sigma (a \mathbf{g}) = \sum_{h \in G} (a \mathbf{g})(h) = \sum_{h \in H} \sum_{i \in G} a(i) \underbrace{\mathbf{g}(ih^{-1})}_{= 1 \: \mathrm{iff} i = gh} = \sum_{h \in H} a(gh) = \sum_{h \in H} a(h) = \sigma(a) \end{align}
(6)
\begin{align} \quad \sigma(\mathbf{g}a) = \sum_{h \in G} (\mathbf{g}a)(h) = \sum_{h \in G} \sum_{i \in G} \underbrace{\mathbf{g}(i)}_{=1 \: \mathrm{iff} \: i = g} a(ih^{-1}) = \sum_{h \in G} a(gh^{-1}) = \sum_{h \in G} a(h) = \sigma(a) \end{align}
• And by the multiplicativity of $\sigma$ we have also that:
(7)
\begin{align} \quad \sigma(a \mathbf{g}) = \sigma(a) \sigma \mathbf{g}) \end{align}
• From the three equalities above we see that for all $a \in A$ and for all $g \in G$ that:
(8)
\begin{align} \quad\sigma(a) \sigma(\mathbf{g}) \sigma(a \mathbf{g}) = \sigma(\mathbf{g}a) = \sigma(a) \end{align}
• Or equivalently:
(9)
\begin{align} \quad (\mathbf{g} \sigma )(a) = (\sigma \mathbf{g})(a) = \sigma (a) \quad \forall a \in A \end{align}
• Thus $\mathbf{g} \sigma = \sigma \mathbf{g} = \sigma$ on $A$. We can identity $A^*$ with $\ell^{\infty}(G)$ and $A^*_+$ with $\ell^{\infty}_+(G)$. Define:
(10)
\begin{align} \quad A^{**}_+ = \{ F \in A^{**} : F(f) \geq 0, f \in A^*_+ \} \end{align}
• And let:
(11)
\begin{align} \quad K = \{ F \in A^{**}_+ : \| F \| = F(\sigma) = 1 \} \end{align}
• Claim 1: $K$ is nonempty. Let $1 : A^{*}_+ \to \mathbb{C}$ be defined for all $f \in A^{*}_+$ by $1(f) = 1$. Then [[\$