# Dual of the Ultraproduct of a Banach Space

Suppose we have a Banach space ultraproduct $(E_i)_U$. I can think of three natural objects which are "dual-like":

1. $(E_i^*)_U$, the ultraproduct of the duals of the ground spaces.
2. The space made up of objects $(f_i)_U$ such that:

a) Each $f_i:E_i\rightarrow\mathbb{R}$

b) There is a $B$ such that for every $i$ and every $x$, $|f_i(x)|\leq B||x||$

c) For any $(x_i)_U, (y_i)_U\in (E_i)_U$, $\lim_U||f_i(x_i+y_i)-f_i(x_i)-f_i(y_i)||=0$

d) For any $(x_i)_U\in (E_i)_U$ and any real $\alpha$, $\lim_U ||f_i(\alpha x_i)-\alpha f_i(x_i)||=0$

3. The ordinary dual $(E_i)_U^*$.

It's easy to see that the space in (1) is contained in the space in (2) which is contained in the space in (3). Furthermore, I think it's well known that (1) is, in general, strictly smaller than (3).

My question is what is known about (2). Does it have a name? I expect that it can be strictly smaller than (3), but are any examples known?

Edit: A little bit of motivation for why I'm thinking about (2). It has some nice properties that (3) doesn't, since its elements can be described: given $f$ in (2), for each $i$ we have a function $f_i$ on $E_i$, and by choosing $i$ large enough, we can ensure that $f_i$ "resembles" an element of the dual of $E_i$.

But (2) should still act a great deal like the true dual. (Formally, suppose $M$ is some model---say, of $ZFC$---where for every Banach space $X$, $\phi(X,X^*)$ holds, and which contains every element of our sequence $(E_i)$. Then in $(M)_U$, $(E_i)_U$ is a Banach space, and $M$ believes that (2) is the dual of $(E_i)_U$, so $M$ will satisfy $\phi((E_i)_U,(2))$. This may be enough for some purposes, or it may even imply that $\phi((E_i)_U,(2))$ holds in "the real world".)

-
Assuming that you mean the "Banach space ultraproduct" and not "ultraproduct of vector spaces" ... in 2. are you assuming that the $f_i$ are uniformly bounded on the unit balls of their domains? – Yemon Choi Jan 25 '11 at 20:31
If so, I think there might be spaces $E$ such that taking $E_i=E$ for all $i$, the space in (2) coincides with the space in (1), in which case it seems very likely that it will indeed be strictly smaller than the space in (3) – Yemon Choi Jan 25 '11 at 20:33
The answer to both your questions is yes (and I've rewritten the question to reflect that). – Henry Towsner Jan 25 '11 at 20:41
You need to be more careful in 2, I think: Why do we have, if $(x_i)$ and $(y_i)$ are bounded families with $\lim_U \|x_i-y_i\|=0$, then $\lim_U |f_i(x_i) - f_i(y_i)| = 0$? This is needed to ensure $(f_i)$ is well-defined as a functional. – Matthew Daws Jan 25 '11 at 21:04
For example, it would be enough to strengthen (b) to $|f_i(x_i)| \leq B \|x_i\|$ for each $i$ and $x_i\in E_i$. – Matthew Daws Jan 25 '11 at 21:06