Norms on an Ultraproduct Suppose $X$ is a Banach space and $\mathcal{U}$ is a non-principal ultrafilter on $\mathbb{N}$. I am interested in the Banach space $(X)_\mathcal{U}$, where we consider sequences $(x_i)_{i \in \mathbb{N}}$ with norm $\|(x_i)\| = \lim_\mathcal{U} \|x_i\|$ (quotiented out by the null sequences.)
My question is related to how this norm behaves on certain sequences. My main notion is that the limit behaves (roughly) like a Banach Limit, and thus assigns some notion of average to the sequence. I then think of $\|x\| = LIM(\|x_i\|)$, where $LIM$ is a Banach limit.
The first part of my question, is this true? Do ultraproducts of Banach spaces and Banach Limits mean roughly the same thing? (I have a gut feeling that this is either true, or can be made precise.)
If this is true, then I can say the following. If I have a sequence in the ultrapower given by $x = (x_1,\dots,x_m, x_1,\dots, x_m, x_1,\dots)$ then the norm of such an element is $\frac{1}{m} \sum_{i=1}^m \|x_i\|$. Indeed, apply shift invariance $m$ times, add the results, and apply shift invariance again. So, I can think of $\|x\| = \lim_n \frac{1}{n} \sum_{i=1}^n \|x_i\|$
So my real question is the following: suppose $Y = \{(x_i) :$ those sequences for which  $ \lim_n \frac{1}{n} \sum_{i=1}^n \|x_i\| $ converges $\}$. Does $\|x\|_{\mathcal{U}} = \lim_n \frac{1}{n} \sum_{i=1}^n \|x_i\|$  on $Y$?
I get the feeling that this is either correct or so completely wrong that my understanding of ultrapowers is unrescuable.
 A: 
So my real question is the following: suppose $Y = \{(x_i) :$ those sequences for which  $ \lim_n \frac{1}{n} \sum_{i=1}^n \|x_i\| $ converges $\}$. Does $\|x\|_{\mathcal{U}} = \lim_n \frac{1}{n} \sum_{i=1}^n \|x_i\|$  on $Y$? 

This is in fact a question about real sequences. You are asking whether $$\newcommand{\limti}[1]{\lim\limits_{#1\to\infty}}\limti n \frac{x_1+\dots+x_n}n=L$$ implies $$\newcommand{\Ulim}{\operatorname{{\mathcal U}-lim}}\Ulim x_n = L.$$ 
The above is not true. As a simple counterexample take $x_n=(0,1,0,1,0,1,\dots)$. It is easy to see that 
$$\limti n \frac{x_1+\dots+x_n}n=\frac12.$$
But the only possible values of $\Ulim x_n$ are $0$ and $1$. (This can be seen from the fact that every $\mathcal U$-limit is a cluster point of the given sequence.) 
Specifically, if you choose an ultrafilter such that $2\mathbb N\in\mathcal U$, then $\Ulim x_n=1$ and if you choose an ultrafilter such that $2\mathbb N\in\mathcal U$, then $\Ulim x_n=0$.
The above example also shows that $\mathcal U$-limit is not shift-invariant. (While Banach limit is shift-invariant. In fact, the above example returns $\frac12$ for every Banach limit.) This means that $\mathcal U$-limit and Banach limit are not the same thing.
However, $\mathcal U$-limit has some properties which Banach limit does not have. For example, you have $\Ulim (x_ny_n) = \Ulim x_n \cdot \Ulim y_n$. This is not true for Banach limit.
