Let $(X_i)_{i\in I}$ be a family of Banach spaces and consider its ultraproduct $(X_i)_{\mathcal{U}}$ with respect to a non-trivial ultrafilter $\mathcal{U}$ over $I$. There is a canonical embedding $T:(X^*_i)_\mathcal{U}\longmapsto ((X_i)_\mathcal{U})^*$ given as $$ \langle Tx^*,x\rangle=\lim_{\mathcal{U}}\langle x^*_i,x_i\rangle$$
for $x^*=(x_i^*)_\mathcal{U}\in (X^*_i)_\mathcal{U}$ and $x=(x_i)_\mathcal{U}\in(X_i)_\mathcal{U}$
My questions is: How to prove this map $T$ is an isometry?, i.e. $||Tx^*||=||x^*||$
I tried the following: Let $x^*=(x_i^*)_\mathcal{U}\in (X^*_i)_\mathcal{U}$. By Hahn-Banach we can find $y=(y_i)_\mathcal{U}\in (X)_\mathcal{U}\subseteq ((X)_\mathcal{U})^{**}$ such that norms $Tx^*$, i.e. $$||Tx^*||=\langle Tx^*, y\rangle=\lim_\mathcal{U}\langle x_i^*,y_i\rangle$$ where last equality is by the definition of $T$. But I'm stuck at this point. Any help would be highly appreciated.
We don't need Hahn-Banach: the definition of the norm of a linear functional is enough. First, recall that $\|x^*\| = \lim_{\mathcal U} \|x_i^*\|$. For any $x\in (X_i)_{\mathcal U}$ we have $$ |\langle Tx^*,x\rangle| = \lim_{\mathcal{U}}|\langle x^*_i,x_i\rangle| \le \lim_{\mathcal{U}} \|x^*_i\| \|x_i\| = \|x^*\|\|x\| $$ which shows $\|Tx^*\|\le \|x^*\|$.
For the reverse inequality, pick some positive numbers $\epsilon_i$ such that $\lim_{\mathcal U} \epsilon_i = 0$. For each $i$ there is $x_i\in X_i$ such that $\|x_i\|=1$ and $\langle x^*_i,x_i\rangle\ge \|x_i^*\| - \epsilon_n$. Hence, for $x=(x_i)_{\mathcal U}$ we have $$ |\langle Tx^*,x\rangle| = \lim_{\mathcal{U}}|\langle x^*_i,x_i\rangle| \ge \lim_{\mathcal{U}} \|x^*_i\|-\epsilon_n = \|x^*\| $$ proving $\|Tx^*\|\ge\|x^*\|$.