Are rings of power series over a local field complete? Let $K$ be a finite extension of $\mathbb{Q}_p$, and let $D$ be some disk $D = \{ x\in \overline{K} \mid |x| < c < 1\}$. Is the set of power series in $K[[T]]$ which converge on $D$ and are bounded on $D$ complete with respect to the supremum norm $\|f\|_\infty=\sup_{x\in \overline{K}, |x|<c} |f(x)|$?

Ok here is my shot at an answer, but I am not certain it is correct.
Suppose $|f_n(x) - f_m(x)| < \epsilon$ for all $|x|<c$. I show all terms of this difference must have absolute value less than $\epsilon$ implying an upper bound on the coefficients. Suppose $f_n(x)-f_m(x)$ has terms of absolute value greater than $\epsilon$ when evaluated at some $z$. Let $S = \{c_{i_1}z^{i_1},\ldots, c_{i_k}z^{i_k}\}$ be the list of all terms which have lowest valuation $u$, implying the valuation of $\sum_{j=1}^k c_{i_j}z^{i_j}$ is greater than $u$. There is some finite extension $K'$ of $K$ in which all of the elements of $S$ live. Now take $L$ to be some unramified extension of $K'$ such that the size of the residue field of $L$ is greater than $i_k$. Now suppose $\lambda$ is some uniformizer of $L$, so that there exists a power $\lambda^{e_1}$ such that the valuation of $\lambda^{e_1}$ is $u$. Then $c_{i_j}z^{i_j}\lambda^{-e_1}$ are all units, and the polynomial $g(T) = \sum_{j=1}^kc_{i_j}z^{i_j}\lambda^{-e_1}T^{i_j}$ must vanish for all $t$ in $L/\lambda$ else we can find $x = zt \in L$ with $|f_n(x)-f_m(x)| > \epsilon$ and $|x| < c$.  However $g$ cannot possibly vanish on all of $L/\lambda$ because $|L/\lambda| > i_k$ and $\deg g = i_k$. Thus all terms of $f_n(z)-f_m(z)$ have absolute value at most $\epsilon$, and this proves the result.
