It is obvious that for a Banach space $E$, $E$ is reflexive iff $\ell^2(E)$ is reflexive. Let $\mathcal U$ be an ultrafilter. Is the reflexivity of $(E)\_\mathcal U$ equivalent to refelxivity of $(\ell^2(E))\_\mathcal U$?



About reflexivity of ultrapower
$\newcommand{\mc}{\mathcal}$The confusion seems to be over the following claim:



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> **Claim:** Let $\mc U$ be a countably incomplete ultrafilter, and let $E$ be a Banach space. If $(E)\_{\mc U}$ is reflexive, then $E$ is super-reflexive.
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To start with, we use the Eberlein-Smulian Theorem to observe that a Banach space $F$ is reflexive if and only if every *separable* subspace of $F$ is reflexive. [Question: Do we need Eberlein-Smulian to show this?]


Recall also that $\mc U$ being *countably incomplete* means that there is a nested sequence of sets $A\_1 \supseteq A\_2 \supseteq \cdots$ in $\mc U$ with $\cap\_i A\_i = \emptyset$. I think of this property as allowing us to embed sequential convergence into convergence along $\mc U$.


Finally, let us recall Theorem 6.3 in Heinrich's paper:



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> If $F$ is a separable Banach space finitely representable in $E$ then $F$ embeds isometrically into $(E)\_{\mc U}$ for any countably incomplete $\mc U$.
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Suppose towards a contradiction that $E$ is not super-reflexive, so there is a non-reflexive $F$ finitely representable in $E$. There is hence a separable subspace $F\_0$ of $F$ which is not reflexive. Clearly $F\_0$ is still finitely representably in $E$, and so isometric to a subspace of $(E)\_{\mc U}$. Hence $(E)\_{\mc U}$ is not reflexive, contrary to assumption.


We then proceed exactly as Jochen Glueck's comment.


**Edit:** As Tomek points out, if $E$ is separable (with no other condition) and $\mc U$ is countably *complete*, then $(E)\_{\mc U} = E$ canonically. Here's a proof (which I hadn't realised before). That $\mc U$ is countably complete is equivalent to the property that if $(A\_n)\_{n=1}^\infty$ is a sequence in $\mc U$ then also $\cap\_n A\_n \in \mc U$. Let $\mc U$ be on a set $I$, and let $A\_n\subseteq I$ be any sequence of subsets which cover $I$, so $\cup\_n A\_n = I$. We claim that then some $A\_n\in\mc U$. For if not, $I\setminus A\_n\in\mc U$ for all $n$ (as $\mc U$ is an ultrafilter) and so $\cap\_n (I\setminus A\_n) = \emptyset\in\mc U$, contradiction.


Let $(x\_n)$ be a dense sequence in $E$, and let $(y\_i)\in (E)\_{\mc U}$. Consider the sets
$$ A\_{n,m} = \{ i : \|y\_i - x\_n\| < 1/m \}. $$
As $(x\_n)$ is dense, for any fixed $i$ and $m$ there is some $n$ with $\|y\_i-x\_n\|<1/m$. So $(A\_{n,m})$ covers $I$, and so there is some $A\_{n,m}\in\mc U$. Fix this $n$ and consider $A\_{n,k}\subseteq A\_{n,m}$ for $k\geq m$. Repeating the argument finds that there is an increasing sequence $m \leq k\_1 < k\_2 < k\_3 < \cdots$ with $A\_{n,k\_i}\in\mc U$ for each $i$. Hence
$$ \bigcap\_i A\_{n,k\_i} = \{ i : y\_i=x\_n \} \in\mc U $$
and so $(y\_i) = x\_n\in E$ in $(E)\_{\mc U}$.


