Let $\mathcal{A}$ be a non-reflexive Banach algebra. For the definition of Arens product, please refer to this link. Here we let $\square$ denote the first Arens product and $\diamond$ denote the second Arens product. Let $\pi: \mathcal{A} \rightarrow \mathcal{A}^{\ast\ast}$ be the canonical embedding. For each $a\in \mathcal{A}$, define:
$$ L_a: \mathcal{A}^{\ast\ast} \rightarrow \mathcal{A}^{\ast\ast}, \hspace{0.3cm} A\mapsto a\square A $$
and:
$$ R_a: \mathcal{A}^{\ast\ast} \rightarrow \mathcal{A}^{\ast\ast}, \hspace{0.3cm} A\mapsto A\diamond a $$
Then the mapping $L: a\mapsto L_a$ defines a representation of $\mathcal{A}$. My question is:
When $\mathcal{A} \square \mathcal{A}^{\ast\ast}$ (resp. $\mathcal{A}^{\ast\ast} \diamond \mathcal{A}$) is dense with respect to the norm topology, or equivalently, $\big[ L(\mathcal{A})(\mathcal{A}^{\ast\ast}) \big]^{\perp}$ (resp. $\big[ R(\mathcal{A})(\mathcal{A}^{\ast\ast}) \big]^{\perp}$) is $0$?
In the case when $\mathcal{A}$ has a bounded approximate unit (with respect to the norm topology) $(a_i)_{i\in I}$, let $E$ be a weak-$\ast$ cluster point of $\big\{ \pi(a_i) \big\}_{i\in I}$. Then when $E$ is THE unit of the Banach algebra $\big( \mathcal{A}^{\ast\ast}, \square \big)$, or $\big( \mathcal{A}^{\ast\ast}, \diamond \big)$?
Update: In the case when $\mathcal{A}$ has an approximate unit, one can show that $A\square E = E\diamond A = A$. Then define:
$$ Z\big( \mathcal{A}^{\ast\ast}, \square \big) = \big\{ A\in\mathcal{A}^{\ast\ast}: \forall\, B\in \mathcal{A}^{\ast\ast},\, A\square B = B\square A \big\} $$ In the case when $\mathcal{A}$ has an approximate unit, I proved that $\mathcal{A} \subseteq Z\big( \mathcal{A}^{\ast\ast}, \square \big)$ if and only if any cluster points of $\big\{ \pi(a_i) \big\}_{i\in I}$ commutes with all elements in $\mathcal{A}^{\ast\ast}$ with respect to the both Arens products. However, I wonder if there is a necessary condition on $\mathcal{A}$ that implies such $E$ is a unit and unique.
