A** (with First Arens Product) Has a Right I. IFF. A Has a B.R.A.I

# A** (with First Arens Product) Has a Right Identity IFF A Has a Bounded Right Approximate Identity

Recall from The First and Second Arens Products on A** that if $\mathfrak{A}$ is a Banach algebra then the second dual $\mathfrak{A}^{**}$ can be made into a Banach algebra by defining a multiplication (called the First Arens Product) on $\mathfrak{A}$.

The following theorem tells us exactly when a Banach algebra $\mathfrak{A}$ has a bounded right approximate identity.

 Theorem 1: Let $\mathfrak{A}$ be a Banach algebra. Then $\mathfrak{A}$ has a bounded right approximate identity if and only if the Banach algebra $\mathfrak{A}^{**}$ equipped with the first Arens product has a right identity.
• Proof: $\Rightarrow$ Suppose that $\mathfrak{A}$ has a bounded right approximate identity $(e_{\alpha}) \subset \mathfrak{A}$. Let $J : \mathfrak{A} \to \mathfrak{A}^{**}$ be the natural embedding of $\mathfrak{A}$ into $\mathfrak{A}^{**}$. Recall that $J$ is an isometry. Since $(e_{\alpha})$ is bounded, there exists an $M > 0$ such that $\| e_{\alpha} \| \leq M$ for all $\alpha$, and since $J$ is an isometry, $\| J(e_{\alpha}) \| = \| e_{\alpha} \| \leq M$ for all $\alpha$. So $(J(e_{\alpha}))$ is a bounded net in $\mathfrak{A}^{**}$.
• Recall by Alaoglu's Theorem that for a Banach space $X$, the closed unit ball of $X^*$ is weak* compact. In particular, the closed unit ball of $\mathfrak{A}^{**}$ is weak* compact and further, since $\mathfrak{A}^{**}$ equipped with the weak* topology is a locally convex topological vector space we have that any scaled closed ball of $\mathfrak{A}^{**}$ is weak* compact. Since $(J(e_{\alpha}))$ is contained in the scaled closed ball $MB_{\mathfrak{A}^{**}} = \{ T \in \mathfrak{A}^{**} : \| T \| \leq M \}$ and since $MB_{\mathfrak{A}^{**}}$ is weak* compact, we have that $(J(e_{\alpha}))$ has a weak* accumulation point. Let:
(1)
\begin{align} \quad E := \mathrm{wk}^* \lim_{\alpha} J(e_{\alpha}) \end{align}
• Then by definition of weak* convergence, this means that for all $f \in \mathfrak{A}^*$ we have that:
(2)
\begin{align} \quad E(f) = \lim_{\alpha} J(e_{\alpha})(f) \end{align}
• In particular, for all $f \in \mathfrak{A}^*$ and for all $a \in \mathfrak{A}$ we have that $fa \in \mathfrak{A}^*$ and so:
(3)
\begin{align} \quad E(fa) &= \lim_{\alpha} J(e_{\alpha})(fa) \\ &= \lim_{\alpha} (fa)(e_{\alpha}) \\ &= \lim_{\alpha} f(ae_{\alpha}) \\ &= f \left ( \lim_{\alpha} ae_{\alpha} \right ) \\ &= f(a) \\ \end{align}
• But by definition, $E(fa) = (Ef)(a)$. So the above equality shows that $(Ef)(a) = f(a)$ for all $a \in \mathfrak{A}$, and thus $Ef = f$ for all $f \in \mathfrak{A}^*$. So, for all $F \in \mathfrak{A}^{**}$ we have that:
(4)
\begin{align} \quad (FE)(f) = F(Ef) = F(f) \quad (\forall f \in \mathfrak{A}^*) \end{align}
• So $FE = F$, which shows that $E$ is a right approximate identity for Banach algebra $\mathfrak{A}^{**}$ equipped with the first Arens product.
• $\Leftarrow$ Suppose that the Banach algebra $\mathfrak{A}^{**}$ equipped with the first Arens product has a right approximate identity $E \in \mathfrak{A}^{**}$. Again, let $J : \mathfrak{A} \to \mathfrak{A}^{**}$ be the natural embedding of $\mathfrak{A}$ into $\mathfrak{A}^{**}$. By Goldstine's Theorem, $J(B_{\mathfrak{A}})$ is weak* dense in $B_{\mathfrak{A}^{**}}$. Since $J$ is an isometry, $J(kB_{\mathfrak{A}})$ is weak* dense in $kB_{\mathfrak{A}^{**}}$ for all $k > 0$. In particular, $J(\| E \|B_{\mathfrak{A}})$ is weak* dense in $\| E \| B_{\mathfrak{A}^{**}}$. Since $E \in \| E \| B_{\mathfrak{A}^{**}}$, there exists a net $(e_{\alpha}) \subset \| E \| B_{\mathfrak{A}}$ (and hence $(e_{\alpha})$ is bounded) such that:
(5)
\begin{align} \quad E = \mathrm{wk}^* \lim_{\alpha} J(e_{\alpha}) \end{align}
• By the definition of weak* convergence, this means that for all $f \in \mathfrak{A}^*$ we have that:
(6)
\begin{align} \quad E(f) &=\lim_{\alpha} J(e_{\alpha})(f) \end{align}
• In particular, for all $a \in \mathfrak{A}$ and for all $f \in \mathfrak{A}^*$ we have that $fa \in \mathfrak{A}^*$ and so:
(7)
\begin{align} \quad E(fa) &= \lim_{\alpha} J(e_{\alpha})(fa) \\ &= \lim_{\alpha} (fa)(e_{\alpha}) \\ &= \lim_{\alpha} f(ae_{\alpha}) \end{align}
• But we also have that for all $a \in \mathfrak{A}$ and for all $f \in \mathfrak{A}^*$ that:
(8)
\begin{align} \quad E(fa) = (Ef)(a) =J(a)(Ef) = (J(a)E)(f) = [J(a)](f) = f(a) \end{align}
• (Where the equality $(J(a)E)(f) = [J(a)](f)$ comes from the fact that $E$ is a right identity in $\mathfrak{A}^{**}$). Thus, for each $a \in \mathfrak{A}$:
(9)
\begin{align} f(a) = \lim_{\alpha} f(ae_{\alpha}) \quad (\forall f \in \mathfrak{A}^*) \end{align}
• So by definition:
(10)
\begin{align} a = \mathrm{wk} \lim_{\alpha} ae_{\alpha} \quad (\forall a \in \mathfrak{A}) \end{align}
• So $(e_{\alpha})$ is a bounded weak right approximate identity in $\mathfrak{A}$. By by the proposition on the Weak Approximate Identities in a Normed Algebra page, the existence of a bounded weak right approximate identity implies the existence of a bounded right approximate identity for $\mathfrak{A}$. $\blacksquare$