Thus we have that for $f\in\mathcal{F}$ there exists $A_f\in {\operatorname{GL}}_d(\mathcal{O})$ depending on $f$ such that $$\label{der_nonzero}
              \min_{i=1,\cdots,d} |\partial_i^k f\circ A_f (A_f{^{\text{-}1}}({\mathbf{x}}_0) )|>0$$ in fact there exists a uniform $C>0$ such that $$\min_{i=1,\cdots,d} |\partial_i^k f\circ A_f (A_f{^{\text{-}1}}({\mathbf{x}}_0) )|>C.$$ This is because we can take $$C=\inf_{f\in\mathcal{F}}\sup_{A\in {\operatorname{GL}}_d(\mathcal{O})}\min_{i=1,\cdots,d} |\partial_i^k f\circ A (A{^{\text{-}1}}({\mathbf{x}}_0) )|,$$ which is nonzero. For if not, then there exists $\{f_n\} \in\mathcal{F}$ such that $$\sup_{A\in {\operatorname{GL}}_d(\mathcal{O})}\min_{i=1,\cdots,d} |\partial_i^k f_n\circ A (A{^{\text{-}1}}({\mathbf{x}}_0) )|<\frac{1}{n}.$$ Since $\mathcal{F}$ is compact, $\{f_n\}$ has a convergent subsequence $\{f_{n_k}\}\to f\in\mathcal{F}$. Taking limits, we get that $$\min_{i=1,\cdots,d} |\partial_i^k f\circ A (A{^{\text{-}1}}({\mathbf{x}}_0) )|=0 \ \forall \ A \in {\operatorname{GL}}_d(\mathcal{O}),$$ which is a contradiction to (\[der\_nonzero\]). Consider the following map $$\Phi_1: {\operatorname{GL}}_d({\mathbb{Q}}_\nu)\times C^l(U_\nu)\times U_\nu \longmapsto {\mathbb{Q}}_\nu$$ $$(A,f,{\mathbf{x}})\mapsto \min_{i=1,\cdots,d} |\partial_i^k f\circ A (A{^{\text{-}1}}({\mathbf{x}})|.$$ It can be easily verified that $\Phi_1$ is continuous. For every $f\in\mathcal{F}$ there exists $A_f \in {\operatorname{GL}}_d(\mathcal{O}) $ such that $\Phi_1(A_f,f,{\mathbf{x}}_0)\geq C>\frac{C}{2},$ so by continuity we have an open neighbourhood $U_{A_f}\times U_f\times U_{({\mathbf{x}}_0,f)}$ of $ (A_f,f,{\mathbf{x}}_0)$ such that $$\Phi_1(A,g,{\mathbf{x}}) >\frac{C}{2} \  
              \forall \  (A,g,{\mathbf{x}}) \in U_{A_f}\times U_f\times U_{({\mathbf{x}}_0,f)}.$$ In particular, $$\label{unicondition}
              \Phi_1(A_f,g,{\mathbf{x}})>\frac{C}{2} \ \forall g\in U_f \text{ and } \forall \ {\mathbf{x}}\in U_{({\mathbf{x}}_0,f)}.$$ Now $\mathcal{F}\subset \bigcup_{f} U_f,$ must have a finite subcovering $\{U_{f_i}\}_{i=1}^{r}$. So by (\[unicondition\]) we have that for 