For a Banach algebra $A$ the bidual $A^{\*\*}$ may be given two natural products called the Arens products. By local reflexivity, there is an ultrafilter $U$ so that $A^{\*\*}$ embeds into the ultrapower $A^U$ isometrically via some map $h$. This map has a one-sided inverse $\sigma\colon A^U\to A^{\*\*}$ given by $\sigma([x\_i])=w^\*-\lim x\_i$.


Is this map an algebra homomorphism for the Arens products in $A^{\*\*}$?



Biduals of Banach algebras
In general, no: for certain $A$ one can get non-zero $c\in A$ and sequences $(a\_n)$ and $(b\_n)$ in $A$ that converge weakly to zero, such that $a\_n b\_n=c$ for all $n$; these will show that $\ker\sigma$ is not even closed under multiplication, let alone an ideal.


We can arrange for $A$ to be Arens regular, so that $A^{\*\*}$ is even a dual Banach algebra.


(As a loose analogy: one should suspect that the question has a negative answer just because $\lim\_i \lim\_j a\_{ij}$ can be very different from $\lim\_n a\_{nn}$ for a doubly-indexed sequence of scalars.)


Here is one example, which I probably learned from conversations with Matt Daws; there should be many others with similar behaviour. I use $\ell\_2$ to denote $\ell\_2({\bf N})$.
 Take $A= K(\ell\_2)$, let $a\_n$ be the operator that sends $\delta\_n$ to $\delta\_1$ and all other basis vectors to $0$; and let $b\_n$ be the operator that sends $\delta\_1$ to $\delta\_n$ and all other basis vectors to $0$.
 Clearly $a\_nb\_n$ is the projection onto the span of $\delta\_1$, for all $n$.


It only remains to justify the claim that $a\_n \stackrel{w}{\longrightarrow} 0$ and $b\_n \stackrel{w}{\longrightarrow} 0$. By trace duality every functional on $A$ is of the form $x\mapsto {\rm Tr}(yx)$ for some trace class operator $y$. A direct calculation shows that ${\rm Tr}(ya\_n) = \langle y\delta\_1, \delta\_n\rangle\to 0$ as $n\to\infty$, and likewise ${\rm Tr}(yb\_n)=\langle y\delta\_n,\delta\_1 \rangle = \langle \delta\_n, y^\*\delta\_1\rangle \to 0$.


