Let $E$ be a Banach space and $(E)_{\mathcal U}$ be an ultrapower for some ultrafilter $\mathcal U$ on an index set $I$.
It is remarked in a paper that $(E')_{\mathcal U}$ can be naturally embedded into $(E)_{\mathcal U}'$ (where I use $'$ to indicate the normed space dual).
I just want to clarify that this natural embedding is given by the following action.
$$(\varphi_{i})_{\mathcal U}:(x_{i})_{\mathcal U}\mapsto \lim_{i\to\mathcal U}\varphi_{i}(x_{i})$$
Is this correct? (sorry if I chose poor tags)
$\bf{\text{Definition}}$:
$\lim_{i\to \mathcal U}x_i = x$ in a topological space $X$ if for every neighbourhood $U$ of $x, \{i\in I : x_i\in U\}\in \mathcal U$.
Yes, that’s correct. It works because $E_{\mathscr{U}}$ isn’t the full model-theoretic ultrapower: we consider only bounded sequences $x=\langle x_k:k\in\Bbb N\rangle$, so that $\mathscr{U}\text{-}\lim_k\|x_k\|$ exists, and we set $\|x\|_\infty=\mathscr{U}\text{-}\lim_k\|x_k\|$. That already cuts out the infinite elements of the full ultrapower, and in addition we take the quotient by the ideal $\{x:\|x\|_\infty=0\}$, identifying infinitesimally different elements.
Similarly, if $\langle\varphi_k:k\in\Bbb N\rangle_{\mathscr{U}}\in(E')_{\mathscr{U}}$, then by definition there is a $U\in\mathscr{U}$ such that $\{\|\varphi_k\|:k\in U\}$ is bounded. It follows that there is a $U\in\mathscr{U}$ such that $\{\varphi_k(x_k):k\in U\}$ is bounded and hence that $\mathscr{U}\text{-}\lim\varphi_k(x_k)$ exists, and it’s straightforward to verify that the map is a bounded linear functional on $E_{\mathscr{U}}$.