Currently, I'm reading the paper "Towards the fixed point property for superreflexive spaces" by Andrzej Wiśnicki. In this article, given $X\_1,\dots,X\_n$ Banach spaces, he defines $(X\_1\oplus\dots\oplus X\_n)\_{\infty}$ as the product $X\_1\times\dots\times X\_n$ with the norm


\begin{equation\*}
\lVert (x\_1,\dots,x\_n)\rVert\_{\infty}=\max(\lVert x\_1\rVert\,\dots,\lVert x\_n\rVert)
\end{equation\*}


With this, in the next proposition, he uses the fact that, given a free ultrafilter on $\mathbb{N}$, $\mathcal{U}$, the following relation is satisfied:


$((X\_1\oplus\dots\oplus X\_n)\_{\infty}) \_{\mathcal{U}}=((X\_1) \_{\mathcal{U}}\oplus\dots\oplus (X\_n) \_{\mathcal{U}}) \_{\infty}$,


where $(X)\_{\mathcal{U}}$ denotes the ultrapower of $X$ respect to $\mathcal{U}$. Why this equality holds?


In addition, if you have access to the paper, I would like to ask one more question. In this same proposition, after using the first fact I asked about, they say that it is sufficient to prove that $(X\_1\oplus\dots\oplus X\_n)\_{\infty}$ has property $(S)$. Why? They have to prove it for $((X\_1) \_{\mathcal{U}}\oplus\dots\oplus (X\_n) \_{\mathcal{U}}) \_{\infty}$ in case they want to use the relation above, don't they?


Thank you very much!!



The ultrapower of the direct sum is the direct sum of ultrapowers
I'll attempt to record a careful proof here for your first question. My apologies for any errors or inefficiencies, as I am a bit rusty. I will write $\|\,\|\_j$ for the norm on $X\_j$ and $\|\,\|$ for the max-norm on $(X\_1\oplus\cdots\oplus X\_n)\_\infty$.


Writing $\Pi\_0$ for the space of bounded sequences in $(X\_1\oplus\cdots\oplus X\_n)\_\infty$, we have two seminorms on $\Pi\_0$: for $x\in\Pi\_0$, define
$$\|x\|\_{1,s}=\lim\_{i\to\mathcal{U}}\|x\_i\|$$
and
$$\|x\|\_{2,s}=\max\_{1\leq j\leq n}(\lim\_{i\to\mathcal{U}}\|x\_i^j\|\_j)$$


We have used subscripts to denote entries along the sequence direction, superscripts to denote the summand direction; e.g. $x\_i^j$ is the $i$th element of the sequence in $X\_j$. Observe that $\|\,\|\_{1,s}$ and $\|\,\|\_{2,s}$ essentially differ by commuting a $\lim\_{i\to\mathcal{U}}$ with a $\max\_{1\leq j\leq n}$. Thus we may write $\|\,\|\_{1,s}$ as


$$\|x\|\_{1,s}=\lim\_{i\to\mathcal{U}}(\max\_{1\leq j\leq n}\|x\_i^j\|\_j)$$


But this is just $\|x\|\_{2,s}$; indeed, if we write $c^j=\lim\_{i\to\mathcal{U}}\|x\_i^j\|\_j$, then for each $\epsilon>0$ and each $j$ we have


$$\{i:|\|x\_i^j\|\_j-c^j|<\epsilon\}\in\mathcal{U}$$


so by taking finite intersections


$$\{i:|\|x\_i^j\|\_j-c^j|<\epsilon\,\forall j\}\in\mathcal{U}$$


which in particular implies


$$\{i:\max\_{1\leq j\leq n}c^j-\epsilon\leq\max\_{1\leq j\leq n}\|x\_i^j\|\_j\leq\max\_{1\leq j\leq n}c^j+\epsilon\}\in\mathcal{U}$$


so (by arbitrariness of $\epsilon>0$) we have $$\lim\_{i\to\mathcal{U}}\max\_{1\leq j\leq n}\|x\_i^j\|=\max\_{1\leq j\leq n}c^j=\max\_{1\leq j\leq n}\lim\_{i\to\mathcal{U}}\|x\_i^j\|\_j$$


and $\|\,\|\_{1,s}=\|\,\|\_{2,s}$ on $\Pi\_0$. Finally, observe that $((X\_1\oplus\cdots\oplus X\_n)\_\infty)\_{\mathcal{U}}$ is the quotient of $((X\_1\oplus\cdots\oplus X\_n)\_\infty,\|\,\|\_{1,s})$ by the nullset $\|\,\|\_{1,s}=0$, and $((X\_1)\_{\mathcal{U}}\oplus\cdots\oplus(X\_n)\_{\mathcal{U}})\_\infty$ is the quotient of $((X\_1\oplus\cdots\oplus X\_n)\_\infty,\|\,\|\_{2,s})$ by the nullset $\|\,\|\_{2,s}=0$; the result follows.


For your second question: I'm not sure, but I suspect that the author intends to say that it suffices to show that it suffices to show that $(Y\_1\oplus\cdots\oplus Y\_n)\_\infty$ has property $(S)$, where $Y\_j=(X\_j)\_{\mathcal{U}}$ and the author is swapping out notation; I suppose this works as long as the condition $\varepsilon\_0(X\_i)<1$ is stable under ultrapowers. Indeed, it appears that the author makes exactly that claim midway through page 439: "On the other hand, $\varepsilon\_0(X)=\varepsilon\_0((X)\_{\mathcal{U}})$..."


