Let $E= \mathbb{F}_p(\!(u)\!)$, $E^s$ a separable closure of $E$ and write $G_E= \mathrm{Gal}(E^s/E)$ for the absolute Galois group of $E$. Take a lift of the $u$-adic valuation on $E$ to $E^s$ and write $\nu \colon E^s \rightarrow \mathbb{Q}$ for the resulting valuation. Let $\hat{E^s}$ be a completion of $E^s$ with regard to this valuation. We obtain a $G_E$ action on $\hat{E^s}$ by continuity. Is it true that $(\hat{E^s})^{G_E}=E$?
I would argue that yes, namely let $x \in \hat{E^s}$, then $x= \sum_{i=0}^{\infty} x_i$ with $x_i \in E^s$ and $\nu(x_i) < \nu(x_{i+1})$. Assume that that $x=g(x) = \sum_{i=0}^{\infty} g(x_i)$ for all $g \in G_E$. But $\nu(g(x_i))= \nu(x_i)$ as $g(u)=u$. Therefore $g(x_i)=x_i$ for all $i$ and we see $x_i \in E$ and thus $x \in E$.
Is my reasoning correct? My intuition would tell me that as $\hat{E^s}$ is 'a lot bigger' than $E^s$, also its fixed field under the $G_E$ action should be bigger, which makes me doubt my own argument.