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

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.

Recall that if $G$ is a group then $\ell^1(G)$ is the Banach algebra of all complex-valued functions $f : G \to \mathbb{C}$ with the property that $\sum_{x \in G} |f(x)| <\infty$, with pointwise function addition, scalar multiplication of functions, and with the convolution product defined for all $f, g \in \ell^1(G)$ to be the function $f * g : G \to \mathbb{C}$ defined for all $x \in G$ by:

(1)

\begin{align} \quad (fg)(x) = \sum_{y \in G} f(y)g(y^{-1}x) \end{align}

And the norm on $\ell^1(G)$ is defined for all $f \in \ell^1(G)$ by:

(2)

\begin{align} \quad \| f \| = \sum_{x \in G} |f(x)| \end{align}

• Proof: Let $\mathfrak{A} = \ell^1(G)$. In the context of this notation, $\mathfrak{A}$ is a Banach algebra over $\mathbb{C}$ ith the operations of pointwise function addition, scalar multiplication of functions, and with convolution product defined for all $a, b \in \mathfrak{A}$ to be the function $ab$ defined for all $g \in G$ by:

(3)

\begin{align} \quad (ab)(g) = \sum_{h \in G} a(h)b(h^{-1}g) \end{align}

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

(4)

\begin{align} \quad \mathbf{g}(h) = \left\{\begin{matrix} 1 & \mathrm{h = g}\\ 0 & \mathrm{h \neq g} \end{matrix}\right. \end{align}

• Each $\mathbf{g}$ is simply the function which is equal to $1$ at $g \in G$ and $0$ for all other elements of $G$. Note that $\mathbf{g} \in \ell^1(G)$ since:

(5)

\begin{align} \quad \sum_{h \in G} |\mathbf{g}(h)| = |\mathbf{g}(g)| = 1 < \infty \end{align}

• Claim 1: If $e \in G$ denotes the identity element in $G$ then $\mathbf{e}$ is a unit for $\mathfrak{A}$: For each $a \in \mathfrak{A}$ and each $g \in G$ we have that:

(6)

\begin{align} \quad (a\mathbf{g})(h) = \sum_{i \in G} a(i) \mathbf{g}(i^{-1}h) \quad \forall h \in G \end{align}

• Since $\mathbf{g}(i^{-1}h) \neq 0$ if and only if $g = i^{-1}h$ $\Leftrightarrow$ $gh^{-1} = i^{-1}$ $\Leftrightarrow$ $i = hg^{-1}$, we see that:

(7)

\begin{align} \quad (a\mathbf{g})(h) = a(hg^{-1}) \quad (*) \end{align}

• Similarly, for each $a \in \mathfrak{A}$ and each $g \in G$ we have that:

(8)

\begin{align} \quad (\mathbf{g}a)(h) = \sum_{i \in G} \mathbf{g}(i) a(i^{-1}h) \quad \forall h \in G \end{align}

• Since $\mathbf{g}(i) \neq 0$ if and only if $g = i$ we see that:

(9)

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

• Replacing $g$ with $e$ at the equalities at $(*)$ and $(**)$ we see that for each $a \in \mathfrak{A}$ and all $h \in G$ both $(a\mathbf{e})(h) = a(he^{-1}) = a(h)$ and $(\mathbf{e}a)(h) = a(e^{-1}h) = a(h)$. So for all $a \in \mathfrak{A}$, $a\mathbf{e} = a = \mathbf{e}a$, so $e$ is a unit for $\mathfrak{A}$.

• Claim 2: The mapping $g \to \mathbf{g}$ is a group monomorphism of $G$ into $\mathrm{Inv}(\mathfrak{A})$: We have already proven that $\mathrm{Inv}(\mathfrak{A})$ forms a group. Let $g \in G$. Then $\mathbf{g}$ has inverse $\mathbf{g^{-1}}$ since for all $h \in H$:

(10)

\begin{align} \quad (\mathbf{g}\mathbf{g^{-1}})(h) = \sum_{i \in G} \mathbf{g}(i) \mathbf{g^{-1}}(i^{-1}h) \end{align}

• But $\mathbf{g}(i) \neq 0$ if and only if $i = g$ in which case $\mathbf{g^{-1}}(i^{-1}h) = \mathbf{g^{-1}}(g^{-1}h) \neq 0$ if and only if $g^{-1}h = g^{-1}$, implying that $h = e$. So $(\mathbf{g} \mathbf{g^{-1}}) = \mathbf{e}$. A similar argument shows that $(\mathbf{g^{-1}}\mathbf{g}) = \mathbf{e}$ as well.

• So indeed, the map $g \to \mathbf{g}$ maps elements of $G$ to elements of $\mathrm{Inv}(\mathfrak{A})$.

• Secondly, for all $g_1, g_2 \in G$ we have that for all $h \in G$, $\mathbf{g_1g_2}(h) = 1$ whenever $h = g_1g_2$ and $\mathbf{g_1g_2}(h) = 0$ otherwise. But similarly, we have that:

(11)

\begin{align} \quad (\mathbf{g_1} \mathbf{g_2})(h) = \sum_{i \in G} \mathbf{g_1}(i) \mathbf{g_2}(i^{-1}h) \end{align}

• The above function is equal to $1$ whenever both $\mathbf{g_1}(i) = 1$ and $\mathbf{g_2}(i^{-1}h) = 1$, and is equal to $0$ everywhere else. But $\mathbf{g_1}(i) = 1$ if and only if $i = g_1$ in which case $\mathbf{g_2}(g_1^{-1}h) = 1$ if and only if $g_1^{-1}h = g_2$, i.e., if and only if $h = g_1g_2$. So we see that:

(12)

\begin{align} \quad \mathbf{g_1g_2} = \mathbf{g_1} \mathbf{g_2} \end{align}

• So the map $g \to \mathbf{g}$ is a homomorphism of $G$ to $\mathrm{Inv}(\mathfrak{A})$. Lastly, it is injective since if $g_1, g_2 \in G$ are such that $\mathbf{g_1} = \mathbf{g_2}$ then for all $h \in H$ we have that $\mathbf{g_1}(h) = \mathbf{g_2}(h)$. But $\mathbf{g_1}(h) = 1$ if and only if $h = g_1$, in which case $\mathbf{g_2}(g_1) = 1$ implying that $g_1 = g_2$. So $g \to \mathbf{g}$ is a monomorphism of $G$ to $\mathrm{Inv}(\mathfrak{A})$.

• Creating a Multiplicative Linear Functional $\sigma$ on $\mathfrak{A}$: Define $\sigma : \mathfrak{A} \to \mathbb{C}$ for all $a \in \mathfrak{A}$ by:

(13)

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

• Let $a, b \in \mathfrak{A}$, $\alpha \in \mathbb{C}$. Then $\sigma$ is a multiplicative linear functional on $\mathfrak{A}$ since:

(14)

\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}

(15)

\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}

(16)

\begin{align} \quad \sigma (ab) = \sum_{g \in G} (ab)(g) = \sum_{g \in G} \left [ \sum_{h \in G} a(h)b(h^{-1}g)) \right ] = \sum_{h \in G} a(h) \sum_{g \in G} b(h^{-1}g) = \sigma(a) \sigma (b) \end{align}

• $\Rightarrow$ Suppose that $\mathfrak{A} = \ell^1(G)$ is amenable. By the proposition mentioned above, for the multiplicative linear functional $\sigma$ there exists an $F \in \mathfrak{A}^{**} \setminus \{ 0 \}$ with the properties that $F(\sigma) = 1$ and $F(fa) = \sigma (a) F(f)$ for all $a \in \mathfrak{A}$ and for all $f \in \mathfrak{A}^*$.

• For each $m \in \ell^{\infty}(G)$ let $m' : \mathfrak{A} \to \mathbb{C}$ be the linear functional on $\mathfrak{A}$ defined for all $a \in \mathfrak{A}$ by:

(17)

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

• Note that since $m \in \ell^{\infty}(G)$ there exists an $M > 0$ such that $|m(g)| \leq M$ for all $g \in G$, and since $a \in \mathfrak{A} = \ell^{1}(G)$, we have that $\sum_{g \in G} |a(g)| < \infty$, so for each $a \in \mathfrak{A}$, $\sum_{g \in G} |a(g)m(g)| \leq M \sum_{g \in G} |a(g)| < \infty$ so that $m(a)$ is a well-defined complex number for each $a \in \mathfrak{A}$.

• Claim 3: The map $m \to m'$ from $\ell^{\infty}(G)$ to $\mathfrak{A}^*$ is a isometry onto $\mathfrak{A}^*$: For all $m \in \ell^{\infty}(G)$ we have that:

(18)

\begin{align} \quad \| m' \|_{\mathrm{op}} = \sup_{\| a \|_1 =1} |m'(a)| = \sup_{\| a \|_1 = 1} \left | \sum_{g \in G} a(g)m(g) \right | \leq \sup_{\| a \|_1 = 1} \sum_{g \in G} |a(g)||m(g)| \leq \sup_{\| a \|_1=1} \sum_{g \in G} |a(g)| \| m \|_{\infty} = \| m \|_{\infty} \end{align}

• But actually, $\mathbf{e} \in \ell^{1}(G)$ is such that $\| \mathbf{e} \|_1 = \sum_{g \in G} |\mathbf{e}(g)| = 1$, so:

(19)

\begin{align} \quad \| m \|_{\infty} = \inf \{ M : | m(g) | \leq M, \: \forall g \in G \} &= \inf \{ M : |m(g) \mathbf{e}(g^{-1}g)| \leq M, \: \forall g \in G \} \\ &= \inf \left \{ M : \left | \sum_{h \in G} m(h) \mathbf{e}(h^{-1}g) \right | \leq M \right \} \\ &\leq \sum_{h \in G} \left | m(h)\mathbf{e}(h^{-1}g) \right | = \| m\mathbf{e} \|_1 \leq \| \mathbf{e} \|_1 \| m \|_1 = \| m \|_1 \end{align}

• Thus $\| m' \|_1 = \| m \|_{\infty}$ for all $m \in \ell^{\infty}(G)$, so the map $m \to m'$ is an isometry.

• Now for each $m \in \ell^{\infty}(G)$ and all $g \in G$ we define the left translation operator on $\ell^{\infty}(G)$ by $T_{g^{-1}}(m)$ to be the function in $\ell^{\infty}(G)$ defined for all $h \in G$ by $[T_{g^{-1}}(m)](h) = m((g^{-1})^{-1}h) = m(gh)$.

• Then for all $g \in G$, $m \in \ell^{\infty}(G)$, and $a \in \mathfrak{A}$ we have that:

(20)

\begin{align} \quad [T_{g^{-1}}(m)]'(a) &= \sum_{u \in G} a(g) [T_{g^{-1}}(m)](u) \\ &= \sum_{u \in G} a(g) m(gu) \\ &= \sum_{h \in G} a(g^{-1}h) m(h) \quad \left ( \mathrm{By \: the \: change \: of \: variables \:} h = gu \right ) \\ &= \sum_{h \in G} (\mathbf{g}a)(h)m(h) \quad \left ( \mathrm{By \: (**)} \right ) \\ &= m' (\mathbf{g}a) \end{align}

• Therefore $[T_{g^{-1}}(m)]' = m' \mathbf{g}$.

• Let $\phi : \ell^{\infty} (G) \to \mathbf{C}$ be defined for all $m \in \ell^{\infty}(G)$ by:

(21)

\begin{align} \quad \phi(m) = F(m') \end{align}

• Claim 4: $\phi$ is a bounded linear functional on $\ell^{\infty}(G)$: For all $m_1, m_2 \in \ell^{\infty}(G)$ and all $\alpha \in \mathbb{C}$ we have that:

(22)

\begin{align} \quad \phi(m_1 + m_2) = F((m_1 + m_2)') \overset{(\dagger)} = F(m_1' + m_2') \overset{(\dagger \dagger)} = F(m_1') + F(m_2') = \phi(m_1) + \phi(m_2) \end{align}

(23)

\begin{align} \quad \phi(\alpha m_1) = F((\alpha m_1)') \overset{(\dagger)} = F(\alpha m_1') \overset{(\dagger \dagger)} = \alpha F(m_1') = \alpha \phi(m_1) \end{align}

• (Where the equalities at $(\dagger)$ comes from the fact that $m \to m'$ is a linear, and the equalities at $(\dagger \dagger)$ comes from the fact that $F \in \mathfrak{A}^{**} \setminus \{ 0 \}$). Furthermore, $\phi$ is bounded since for all $m \in \ell^{\infty}(G)$ we have that:

(24)

\begin{align} \quad |\phi(m)| = |F(m')| \overset{(\dagger \dagger \dagger)} \leq \| F \| \| m' \|_{\mathrm{op}} \overset{(\dagger \dagger \dagger \dagger)} = \| F \| \| m \|_{\infty} \end{align}

• (Where the equality at $(\dagger \dagger \dagger)$ comes from $F \in \mathfrak{A}^{**} \setminus \{ 0 \}$ and where the equality at $(\dagger \dagger \dagger \dagger)$ comes from claim 3 saying that $m \to m'$ is an isometry.

• Claim 5: $\phi$ is an invariant mean (not necessary positive) on $\ell^{\infty}(G)$, that is $\phi(1) = F(\sigma) = 1$ and $\phi([T_g(m)]) = \phi(m)$ for all $g \in G$ and for all $m \in \ell^{\infty}(G)$: We have that $1' = \sigma$ by definition, and so:

(25)

\begin{align} \quad \phi(1) = F(1') = F(\sigma) = 1 \end{align}

• Now for all $g \in G$ and for all $m \in \ell^{\infty}(G)$ we have that:

(26)

\begin{align} \quad \phi([T_g(m)]) = F([T_g(m)]') = F(m'\mathbf{g^{-1}}) = \sigma(\mathbf{g^{-1}}) F(m') = \underbrace{\sum_{h \in G} \mathbf{g^{-1}}(h)}_{=1} F(m') = F(m') = \phi(m) \end{align}

• Establishing a Positive Invariant Mean on $\ell^{\infty}(G)$: Lastly, let:

(27)

\begin{align} \quad \ell^{\infty}_+(G) = \left \{ m \in \ell^{\infty}(G) : m \geq 0 \: \mathrm{on \:} G \right \} \end{align}

• For each [$x \in \ell^{\infty}_+(G)$ let:

(28)

\begin{align} \quad \theta(x) = \sup_{y \in \ell^{\infty}(G), \: |y| \leq x} \mathrm{Re}(\phi(y)) \end{align}

• Claim 6: $\theta$ is positively homogeneous and subadditive on $\ell^{\infty}_+(G)$: For all $\alpha > 0$ and all $x \in \ell^{\infty}_+(G)$ we have that:

(29)

\begin{align} \quad \mathrm{Positive \: Homogeneity:} \quad \theta(\alpha x) = \sup_{y \in \ell^{\infty}(G), \: |y| \leq x} \mathrm{Re}(\phi(\alpha y)) \overset{\mathrm{since} \: \alpha \in \mathbb{R}}= \alpha \sup_{y \in \ell^{\infty}(G), \: |y| \leq x} \mathrm{Re}(\phi(y)) = \alpha \theta(x) \end{align}

• And for all $x_1, x_2 \in \ell^{\infty}_+(G)$ we have that:

(30)

\begin{align} \quad \mathrm{Subadditivity:} \quad \theta(x_1 + x_2) = \sup_{y \in \ell^{\infty}(G), \: |y| \leq x_1 + x_2} \mathrm{Re} \phi(y) \leq \sup_{y \in \ell^{\infty}(G), \: |y| \leq x_1} \mathrm{Re}(\phi(y)) + \sup_{z \in \ell^{\infty}(G), \: |z| \leq x_2} \mathrm{Re}(\phi(z)) = \theta(x_1) + \theta(x_2) \end{align}

• Also note that $\theta(1) \neq 0$ since if otherwise, $\theta(1) = 0$ implies that $\mathrm{Re}(\phi(y)) = 0$ for all $y \in \ell^{\infty}(G)$ with $|y| \leq 1$, but $y =1$ is such that $\mathrm{Re}(\phi(1)) = 1$.

• Thus we can extend $(\theta(1))^{-1} \theta$ from $\ell^{\infty}_+(G)$ to the larger space, $\ell^{\infty}(G)$, which is the required positive invariant mean on $\ell^{\infty}(G)$. So $G$ is an amenable group. $\blacksquare$