We consider $(x _{n})$ a sequence of almost fixed points for $T$ in $C _{0}$. Since $C _{0}$ is weakly compact, we can assume that $(x _{n})$ is weak compact. Also, since the problem of the fixed point is invariant by translations, we can assume that $(x _{n})$ is weakly convergent to 0. let's consider $\mathscr{U}$ a free ultrafilter on the natural numbers and consider its corresponding ultrapower of Banach space \begin{align*} \widetilde{X} = (\ell ^{\infty}(X _{i})/\operatorname{ker}\mathcal{N},\Vert \cdot \Vert), \end{align*} where the norm $\Vert \cdot \Vert$ is define as \begin{align*} \Vert \widetilde{w} \Vert = \lim _{n, \mathscr{U}} \Vert w _{n} \Vert. \end{align*} for all $\widetilde{w} \in \widetilde{W}$. Let's call $\widetilde{x}$ is the equivalence class formed by the sequence $(x _{n})$. We define \begin{align*} D(\widetilde{w}) = \lim _{m, \mathscr{U}} \left( \lim _{n, \mathscr{U}} \Vert w _{m} - w _{n} \Vert \right). \end{align*}

this coefficient depends on each $\widetilde{w}$, but not on the representatives that form $\widetilde{w}$.

Let $T: C \to C $ a non-expansive function and $C$ a set of $X$. For the minimal set of \begin{align*} \mathscr{A} = \{ K \subseteq C: \ K \ \mbox{is not empty, weak compact, convex and $T$-invariant}\} \end{align*} called $C _{0}$. We know that $C _{0}$ is closed, convex and $T$-invariant. We define the set \begin{align*} \widetilde{C _{0}} = \{ \widetilde{w}: w _{n} \in C _{0}, \ \mbox{for all} \ n \in \mathbb{N} \} \in \widetilde{X} \mbox{.} \end{align*} Let \begin{align*} \widetilde{W} = \left\{ \widetilde{w} \in \widetilde{C _{0}}: \ \Vert \widetilde{w} - \widetilde{x} \Vert \leqslant \frac{1}{2} \ \mbox{and} \ D(\widetilde{w}) \leqslant \frac{1}{2} \right\} \mbox{.} \end{align*} Why $\widetilde{W}$ is closed?.