Given a continuous map $G_K\xrightarrow{f} GL_n(\mathbb{C}_p)$, where $K$ is a finite extension of $\mathbb{Q}_p$,$G_K:=G_{\bar{K}/K}$ and $\mathbb{C}_p$ is the $p$ adic complex number field.
Fixed a constant $c>0$, and let $v_p$ be the p adic valuation of $\mathbb{C}_p$ such that $v_p(p)=1$, and for a matrix $A={(a_{i,j})}_{1\leq i,j\leq n}\in GL_n(\mathbb{C}_p)$, we define $v_p(A)=\text{min}\{v_p(a_{i,j})\}$.
Then can we choose a finite extension $L$ of $K$, such that all matrix $A\in f(G_L)$ satisfy $v_p(I-A)\geq c$($I$ is the identity matrix)?
Remark:I have edited my old question which is wrong. Sorry! This is a question used to prove all $p$-adic Galois representation of $G_K$ is overconvergent.
Thanks!
Not sure if this is correct, can you check :
With the profinite topology any sequence in $G_K=Gal(\overline{K}/K)$ has a convergent subsequence.
Given $r\in \Bbb{Q}_{>0}$ Let $U=GL_n(\Bbb{C}_p)/(I+p^r M_n(O_{\Bbb{C}_p}))$, it is almost a metric space with $$d(A,B) = \inf_{C\in I+p^r M_n(O_{\Bbb{C}_p})} \sup_{ij} |(AB^{-1} C-I)_{ij}|_p$$ $d$ is continuous on $GL_n(\Bbb{C}_p)$. Thus $f$ is continuous $G_K\to U$, ie. any subquence of $f(G_K)$ has a subsequence which is Cauchy for $d$.
If $f(G_K)\subset U$ isn't finite then this won't happen, because $d(A,B)\ne 0 \implies d(A,B) \ge p^{-r}$.
Thus $f(G_K)\subset U$ is finite and $H=f(G_K)\cap I+p^r M_n(O_{\Bbb{C}_p})$ has finite index and $ \overline{K}^{f^{-1}(H)}$ is your finite extension of $K$.