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Hello I have the following construction: Let $(E,\|\cdot\|_E)$ be a Banach space, $E_n:=E^n$ and $\|x_n\|_n:=\frac{1}{n}\sum_{k=1}^n \|x_n(k)\|_E$ for all $x_n=(x_n(1),...,x_n(n))\in E^n$ and $n\in\mathbb{N}$. My question is whether the ultraproduct $(E_n)_{\mathcal{U}}$ along any (free) ultrafilter $\mathcal{U}$ is reflexive if $E$ is superreflexive?

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    $\begingroup$ If $E=\mathbb{R}$, then your ultraproduct is an infinite-dimensional AL-space (i.e. a Banach lattice whose norm is additive on the positive cone) and hence not reflexive. $\endgroup$
    – Jochen Glueck
    Commented Feb 19 at 14:14

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