The ultrapower of the direct sum is the direct sum of ultrapowers

Question

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!!

Answer 1

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, $\varepsilon_0(X)=\varepsilon_0((X)_{\mathcal{U}})$..."