Is there a Banach algebra which $A^2$ is not dense in $A$ but $(A^{**})^2$ is dense in $A^{**}$?

Question

Is there a Banach algebra which $A^2=\langle a_1a_2 ; a_1,a_2\in A\rangle$ is not dense in $A$ but $(A^{**})^2$ is dense in $A^{**}$?

Answer 1

Assuming that Matthew Daws's interpretation of your notation is correct, the answer is NO.

Let us suppose that the linear span of the set $\{ab \colon a,b\in A\}$ is not dense in $A$. By Hahn-Banach there is a non-zero functional $\psi\in A^*$ such that $\psi(ab)=0$ for all $a,b\in A$.

Recall the construction of the first Arens product on $A^{**}$:

• For $\phi\in A^*$ and $a\in A$ define $\phi\cdot a \in A^*$ by $(\phi\cdot a)(b) = \phi(ab) \qquad (b\in A)$.

• For $F\in A^{**}$ and $\phi\in A^*$ define $F_\phi \in A^*$ by $F_\phi(a)= F(\phi_a)\qquad (a\in A)$.$\newcommand{\sqprod}{\mathbin{\square}}$

• For $E\in A^{**}$ and $F\in A^{**}$ define $E\sqprod F \in A^{**}$ by $(E\sqprod F)(\phi) = E(F_\phi) \qquad (\phi\in A^*)$.

We see from this that $\psi\cdot a =0$ for all $a\in A$, so $F_\psi=0$ for all $F\in A^{**}$. Hence $$ (E\sqprod F)(\psi)=0 \qquad\hbox{for all $E,F\in A^{**}$.} $$

Choose any $x_0\in A$ such that $\psi(x_0)\neq 0$, and let $\widehat{x_0} \in A^{**}$ be the image of $x_0$ under the canonical embedding. Then by the previous paragraph, $\widehat{x_0}$ canot be in the weak-star closed linear span of the set $S=\{E\sqprod F \colon E,F\in A^{**}\}$, and in particular it is not in the norm-closed linear span.

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(A shorter version of the same proof uses the description of $\sqprod$ in terms of iterated weak-star limits. But I tend to prefer the purely algebraic description just to make sure I haven't accidentally mixed up my topologies.)