Ultra product of Banach spaces

Question

Let $\mathcal X_n$ be a sequence of Banach spaces and let $\omega \in\beta \mathbb N\setminus \mathbb N$ be a free ultrafilter. Define $\Pi_0=\{(\xi_n)_{n\in\mathbb N}:\xi_n\in\mathcal X_n\} $ and $N_\omega =\{(\xi_n)_{n\in\mathbb N}\in \Pi_0: \lim_{n\to\omega}\|\xi_n\|=0\} $.

I want to prove that for any $(\xi_n)\in \Pi_0 $, $\lim_{n\to\omega}\|\xi_n\|=\inf\{\sup_n\|\xi_n-\eta_n\|:{(\eta_n)_{n\in\mathbb N} \in N_\omega } \}$.

It is easy to see that $\lim_{n\to\omega}\|\xi_n\|\leq \inf\{\sup_n\|\xi_n-\eta_n\|:{(\eta_n)_{n\in\mathbb N} \in N_\omega } \}$. But I can't figure out how to show the other direction. Any help would be much appreciated.

Answer 1

For convenience let $L=\lim_{n\to\omega}\|\xi_n\|$, and suppose that

$$L<\inf\left\{\sup_n\|\xi_n-\eta_n\|:(\eta_n)_{n\in\Bbb N}\in N_\omega\right\}\;.$$

Let

$$\epsilon=\inf\left\{\sup_n\|\xi_n-\eta_n\|:(\eta_n)_{n\in\Bbb N}\in N_\omega\right\}-L>0\;;$$

then $\sup_n\|\xi_n-\eta_n\|-L\ge\epsilon$ for each $(\eta_n)_{n\in\Bbb N}\in N_\omega$.

Let

$$U=\left\{n\in\Bbb N:\big|\|\xi_n\|-L\big|<\frac{\epsilon}2\right\}\;;$$

$U\in\omega$ by the definition of $L$. Define $(\eta_n)_{n\in\Bbb N}\in\Pi_0$ by letting

$$\eta_n=\begin{cases} \xi_n,&\text{if }n\notin U\\ 0_{\mathcal{X}_n},&\text{if }n\in U\;; \end{cases}$$

clearly $\left(\eta_n\right)_{n\in\Bbb N}\in N_\omega$, so $\sup_n\left\|\xi_n-\eta_n\right\|-L\ge\epsilon$, and there is an $n(U)\in\Bbb N$ such that $\left\|\xi_{n(U)}-\eta_{n(U)}\right\|-L>\frac{\epsilon}2$.

• Show that $n(U)\in U$.
• Then get a contradiction by considering $\|\xi_{n(U)}\|-L$.

If you get stuck, mouse over the block below to see the argument.

But $\left\|\xi_n-\eta_n\right\|=0$ for $n\notin U$, and $L\ge 0$, so we must have $n(U)\in U$ and hence $\eta_{n(U)}=0_{\mathcal{X}_n(U)}$. But then $$\|\xi_{n(U)}\|-L=\|\xi_{n(U)}-\eta_{n(U)}\|-L>\frac{\epsilon}2\;,$$ contradicting the choice of $U$. It follows that $$L\ge\inf\left\{\sup_n\|\xi_n-\eta_n\|:(\eta_n)_{n\in\Bbb N}\in N_\omega\right\}\;,$$ as desired.