Is an ultrapower of an Arens regular (non-superreflexive) Banach algebra, Arens regular? 



Arens regularity of an ultrapower of an Arens regular Banach Algebra?
I am not 100% sure what the question is asking.



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> Is the ultrapower of an Arens regular Banach algebra also Arens regular? 
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As my comment said, if $A$ is a $C^\*$-algebra, then "yes".


$\newcommand{\mc}{\mathcal}\newcommand{\ip}[2]{\langle #1,#2\rangle}$To show it's not always true, here's a counter-example. Let $E$ be a reflexive Banach space, and let $A = \mc A(E)$ be the algebra of approximable (norm-closure of finite-rank) operators. Then $A$ is Arens regular. Let $\mc U$ be a non-principle ultrafilter on $\mathbb N$ (for example) and consider the ultrapower $(A)\_{\mc U}$ which, in the obvious way, can be considered as a closed subalgebra of $\mc B((E)\_{\mc U})$. [The idea behind the following proof is that if $(E)\_{\mc U}$ is not reflexive, then $\mc A( (E)\_{\mc U} )$ is not Arens regular, and "morally" $\mc A( (E)\_{\mc U} )$ is a subalgebra of $(A)\_{\mc U}$, so $(A)\_{\mc U}$ is not Arens regular. This is not true, so we work a bit harder.]


I'll use Heinrich's wonderful paper [Ultraproducts in Banach space theory](https://doi.org/10.1515/crll.1980.313.72) Sadly not open access (WHY?) Namely, Corollary 7.5 which says:



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> Let $E\_0\subseteq (E)\_{\mc U}$ be a separable subspace. For $\mu\in (E)\_{\mc U}'$ there is $(\mu\_i)\in (E')\_{\mc U}$ with $\|\mu\| = \|(\mu\_i)\|$ such that $\ip{\mu}{x} = \ip{(\mu\_i)}{x} = \lim\_{i\rightarrow\mc U} \ip{\mu\_i}{x\_i}$ for each $x=(x\_i)\in E\_0$.
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We use this to cope with the fact that we'd like that $(E)'\_{\mc U} = (E')\_{\mc U}$ but this is only true if $E$ is super-reflexive.


Suppose that $E$ is reflexive but not super-reflexive, so that $(E)\_{\mc U}$ is not reflexive. Hence we can find sequences $(x^{(n)})$ in $(E)\_{\mc U}$ and $(\mu^{(n)})$ in $(E)\_{\mc U}'$ such that
$$ \lim\_n \lim\_m \ip{\mu^{(n)}}{x^{(m)}} \not= \lim\_m \lim\_n \ip{\mu^{(n)}}{x^{(m)}}. $$
That is, the iterated limits exist, but are not equal. By the above result, we may suppose that each $\mu^{(n)} \in (E')\_{\mc U}$. For each $n$ let $x^{(n)} = (x^{(n)}\_i) \in (E)\_{\mc U}$ and $\mu^{(n)} = (\mu^{(n)}\_i) \in (E')\_{\mc U}$. Fix $y\in E, \phi\in E'$ with $\ip{\phi}{y}=1$. For each $n,i$ let
$$ a^{(n)}\_i = \mu^{(n)}\_i \otimes y, \qquad
b^{(n)}\_i = \phi \otimes x^{(n)}\_i. $$
Let $a^{(n)} = (a^{(n)}\_i) \in (A)\_{\mc U}$ and $b^{(n)} = (b^{(n)}\_i) \in (A)\_{\mc U}$. Define
$$ \psi : (A)\_{\mc U}\rightarrow\mathbb C; \quad
\psi((c\_i)) = \lim\_{i\rightarrow\mc U} \ip{\phi}{c\_i(y)}
\qquad ((c\_i)\in (A)\_{\mc U}). $$
Then
$$ \ip{\psi}{a^{(n)}b^{(m)}}
= \lim\_{i\rightarrow\mc U} \ip{\phi}{a^{(n)}\_i b^{(m)}\_i(y)}
= \lim\_{i\rightarrow\mc U} \ip{\mu^{(n)}\_i}{x^{(m)}\_i}
= \ip{\mu^{(n)}}{x^{(m)}}. $$
Hence the iterated limits of $\ip{\psi}{a^{(n)}b^{(m)}}$ are not equal, and so $(A)\_{\mc U}$ is not Arens regular.


