Let $G$ be a polish group, $H$ an open subgroup of $G$. Now assume that $H$ acts by isometries (For all $h\in H$, the map $X\ni x\longmapsto X$ is an isometry) and continously on a metric space $(X,\delta)$. We define $$F=\{f:G\longrightarrow X|\,\forall h\in H,\,\,\forall g\in G,\,\,f(gh)=h^{-1}f(g)\}$$
$G$ act on $F$ by left translation $$G\times F\ni (g,f)\longmapsto \tau_{g}f \in F$$ where $\tau_{g}f(x)=f(g^{-1}x)$ for all $x\in G$.
My goal is to show that the previous action of $G$ on $F$ is continuous
We assume that we can equipped $F$ with the following metric: $$d:F\times F\ni(f,h)\longmapsto d(f,h)=\underset{g\in G}{\overset{}{\sup}}\,\delta(f(g),h(g))\in \mathbb{R}_{+}$$
If $F$ is equiped with the previuous action, then the action of $G$ on $F$ is by isometries
Since $H$ is open in $G$, we only need to show that the action of $H$ on $F$ is continuous. How to show that this fact? where $F$ is equipped by the toplogy induced by the previous metric.
Thank for any help
This question is my own research. It is not homework.
Thank for your previous answer. For this question, i have an idea. But i am not sure that i am right:
In fact, since the action is by isometries, the map $F\in f\longmapsto \tau_{g}f\in F$ is continuous for all $g\in G$. Now fix $f\in F$ and suppose that $g_{n}\longrightarrow g$ in $G$, then $g^{-1}_{n}a\longrightarrow g^{-1}a$ for all $a\in G$. Therefore, $f(g^{-1}_{n}a)\longrightarrow f(g^{-1}a)$ since $f$ is continuous. So we have $\tau_{g_{n}}f\longrightarrow \tau_{g}f$ pointwise. Now since $G$ is separable and $H$ is open, then the quotient space is countable(separable+discrete). Therefore the following $\underset{a\in G}{\overset{}{\sup}}\,\delta(f(g_{n}^{-1}a),f(g^{-1}a))$ is in fact a max. Therefore, We have: $$d(\tau_{g_{n}}f,\tau_{g}f)=\underset{a\in G}{\overset{}{\max}}\,\delta(f(g_{n}^{-1}a),f(g^{-1}a))\longrightarrow 0$$
Thank for givin your opinion about this idea.