ℓ^1(G) is Amenable if and only if G is an Amenable Group 2

Theorem 1: Let $G$ be a group. Then the group algebra $\ell^1(G)$ is amenable if and only if the group $G$ is an amenable group.

• $\Leftarrow$ Let $G$ be an amenable group. Then there exists a positive invariant mean $\mu$ on $G$, i.e., there exists a positive linear functional $\mu$ on $\ell^{\infty}(G)$ with the properties that $\mu(1) = 1$ and $\mu([T_h(m)]) = \mu(m)$ for all $m \in \ell^{\infty} (G)$.

• Let $X$ be a Banach $\mathfrak{A}$-module ($\ell^1(G)$-module). Let $D$ be a bounded $X^*$-derivation. Then for all $a, b \in \mathfrak{A}$ we have that $D(ab) = D(a)b + aD(b)$.

• Since $X^*$ is a normed (Banach) $\mathfrak{A}$-module, it satisfies axiom $BL1$ regarding the norm, that is, there exists a number $\kappa > 0$ such that for all $a \in \mathfrak{A}$ and for all $f \in X^*$ we have that:

(1)

\begin{align} \quad \| af \| \leq \kappa \| a \| \| f \| \quad \mathrm{and} \quad \| fa \| \leq \kappa \| a \| \| f \| \quad (\star) \end{align}

• For each $x \in X$ let $\mathbf{x} : G \to \mathbb{C}$ be defined for all $g \in G$ by:

(2)

\begin{align} \quad \mathbf{x}(g) = (\mathbf{g}D(\mathbf{g^{-1}}))(x) \end{align}

• Since $\mathbf{g^{-1}} \in \ell^1(G) = \mathfrak{A}$ and since $D : \mathfrak{A} \to X^*$, we have that $D(\mathbf{g^{-1}}) \in X^*$, and since $X^*$ is a Banach $\mathfrak{A}$-module, the multiplication $\mathbf{g} D(\mathbf{g^{-1}})$ is defined, and so by axiom $BM1$ at $(\star)$ we have that

(3)

\begin{align} \quad \| \mathbf{g} D(\mathbf{g^{-1}}) \| \leq \kappa \| \mathbf{g} \| \| D(\mathbf{g^{-1}}) \| = \kappa \| \mathbf{g} \| \| D \| \| \mathbf{g^{-1}} \| = \kappa \| D \| < \infty \end{align}

• So for each $x \in X$, the complex-valued function $\mathbf{x}$ defined above is such that for all $g \in G$:

(4)

\begin{align} \quad |\mathbf{x}(g)| =|(\mathbf{g} D(\mathbf{g^{-1}}))(x)| \leq \| \mathbf{g} D(\mathbf{g^{-1}}) \| \| x \| \leq \underbrace{\kappa \| D \| \| x \|}_{\mathrm{fixed}} \end{align}

• So each $\mathbf{x}$ is bounded on $G$, i.e., each $\mathbf{x} \in \ell^{\infty}(G)$ and moreover from above, $\| \mathbf{x} \|_{\infty} \leq \kappa \| D \| \| x \|$.

• Let $f : X \to [0, \infty)$ be defined for all $x \in X$ by:

(5)

\begin{align} \quad f(x) = \mu (\mathbf{x}) \end{align}

• Claim 1 $D = \delta_f$: For each $a \in \mathfrak{A} = \ell^1(G)$ observe that for every $h \in G$ we have that:

(6)

\begin{align} \quad \left ( \sum_{g \in G} a(g) \mathbf{g} \right )(h) = \sum_{g \in G} a(g) \mathbf{g}(h) = a(h) \end{align}

• So each $a \in \mathfrak{A}$ can be written in the form:

(7)

\begin{align} \quad a = \sum_{g \in G} a(g) \mathbf{g} \end{align}

• We want to show that $D(a) = \delta_f(a)$ for all $a \in \mathfrak{A}$, i.e., $D(a) = fa - af$, that is, $D(a)(x) = (fa - af)(x)$ for all $x \in X$, or equivalently show that for all $x \in X$:

(8)

\begin{align} \quad (D(\mathbf{g}))(x) &= (f\mathbf{g} - \mathbf{g}f)(x) \\ &=[f\mathbf{g}](x) - [\mathbf{g}f](x) \\ &= f(\mathbf{g}x) - \mathbf{g}(fx) \\ &= f(x\mathbf{g} - \mathbf{g}x) \quad (\star \star) \end{align}

• Let $x \in X$ and let $g \in G$. Let $z = x \mathbf{g} - \mathbf{g}x$. Then for each $h \in G$ we have that:

(9)

\begin{align} \quad \mathbf{z}(h) &= [\mathbf{h}D(\mathbf{h^{-1}})](z) \\ &= [\mathbf{h}D(\mathbf{h^{-1}})](x \mathbf{g} - \mathbf{g}x) \\ &= [\mathbf{h}D(\mathbf{h^{-1}})](x \mathbf{g}) - [\mathbf{h}D(\mathbf{h^{-1}})](\mathbf{g}x) \quad (\star \star \star) \end{align}

• Now observe that for each $h \in G$, since $D$ is an $X^*$-derivative we have that:

(10)

\begin{align} \quad D(\mathbf{h^{-1}}\mathbf{g}) &= D(\mathbf{h^{-1}}) \mathbf{g} + \mathbf{h^{-1}}D(\mathbf{g}) \\ \quad \mathbf{h}D (\mathbf{h}^{-1}\mathbf{g}) &= \mathbf{h} D(\mathbf{h^{-1}}) \mathbf{g} + \mathbf{h} \mathbf{h^{-1}} D(\mathbf{g}) \quad (\mathrm{Multiplication \: on \: the \: left \: by \:} \mathbf{h}) \end{align}

• Rearranging the above equation yields:

(11)

\begin{align} \quad -\mathbf{h}D(\mathbf{h^{-1}})\mathbf{g} &= \mathbf{h} \mathbf{h^{-1}} D(\mathbf{g}) - \mathbf{h} D(\mathbf{h^{-1}} \mathbf{g}) \\ &= D(\mathbf{g}) - \mathbf{h} D(\mathbf{h^{-1}} \mathbf{g}) \\ &= D(\mathbf{g}) - \mathbf{g}( \mathbf{g^{-1}}\mathbf{h}) D(\mathbf{h^{-1}} \mathbf{g}) \\ &= D(\mathbf{g}) - \mathbf{g}( \mathbf{g^{-1}}\mathbf{h}) D (\mathbf{g^{-1}}\mathbf{h})^{-1} \\ \end{align}

• Therefore:

(12)

\begin{align} \quad -\mathbf{h}D(\mathbf{h^{-1}})(\mathbf{g}x) &= (D\mathbf{g})(x) - \mathbf{g}[( \mathbf{g^{-1}}\mathbf{h}) D (\mathbf{g^{-1}}\mathbf{h})^{-1}](x) \\ &= (D\mathbf{g})(x) - [( \mathbf{g^{-1}}\mathbf{h}) D (\mathbf{g^{-1}}\mathbf{h})^{-1}](x \mathbf{g}) \quad (\star \star \star \star) \end{align}

• Set $y = x \mathbf{g}$. Then by $(\star \star \star)$ and $(\star \star \star \star)$ we have that:

(13)

\begin{align} \quad \mathbf{z}(h) & \overset{(\star \star \star)}= [\mathbf{h}D(\mathbf{h^{-1}})](x \mathbf{g}) - [\mathbf{h}D(\mathbf{h^{-1}})](\mathbf{g}x) \\ &= [\mathbf{h}D(\mathbf{h^{-1}})](y) - [\mathbf{h}D(\mathbf{h^{-1}})](\mathbf{g}x) \\ & \overset{(\star \star \star \star)} = [\mathbf{h}D(\mathbf{h^{-1}})](y) - (D\mathbf{g})(x) - [( \mathbf{g^{-1}}\mathbf{h}) D (\mathbf{g^{-1}}\mathbf{h})^{-1}](x \mathbf{g}) \\ &= [\mathbf{h}D(\mathbf{h^{-1}})](y) - (D\mathbf{g})(x) - [( \mathbf{g^{-1}}\mathbf{h}) D (\mathbf{g^{-1}}\mathbf{h})^{-1}](y) \\ &= \mathbf{y}(h) - (D\mathbf{g})(x) - \mathbf{y}(g^{-1}h) \end{align}

• Since this holds for every $h \in G$ we have that:

(14)

\begin{align} \quad \mathbf{z} = \mathbf{y} - T_g(\mathbf{y}) + (D\mathbf{g})(x)1 \end{align}

• Since $\mu$ is a positive invariant mean on $G$, we have that $\mu(\mathbf{y}) = \mu(T_g(\mathbf{y}))$ and $\mu(1) = 1$. Therefore:

(15)

\begin{align} \quad \mu(\mathbf{z}) &= \mu(\mathbf{y}) - \mu (T_g(\mathbf{y})) + \mu((D\mathbf{g})(x)1) \\ &= (D \mathbf{g})(x) \mu(1) \\ &= (D \mathbf{g})(x) \end{align}

• From the above equation and resubstituting $z = x\mathbf{g} - \mathbf{g}x$ we get:

(16)

\begin{align} \quad f(x\mathbf{g} - \mathbf{g}x) = f(z) = \mu(\mathbf{z}) = (D\mathbf{g})(x) \end{align}

• Thus $D = \delta_f$. So if $X$ is a Banach $\mathfrak{A}$-module then every bounded $X^*$-derivation is an inner bounded $X^*$-derivation. So $\ell^1(G)$ is amenable. $\blacksquare$