Let $K$ be a local field with non-archimedean absolute value $|\cdot|$ and ring of integers $\mathcal{O}_K := \{x \in K : |x| \leq 1\}$. Moreover, let $n$ be a positive integer. It can be proved that the $K$-linear space $K[x_1, \ldots, x_n]$ of polynomials in $n$ variables is a Banach space with norm $\|\cdot\|$ defined as: $\|f\|$ is the maximum of the absolute values of the coefficients of $f$, for each $f \in K[x_1, \ldots, x_n]$.
I'm trying to prove the following statements:
If $(f_j)_{j \geq 0} \subseteq K[x_1, \ldots, x_n]$ is such that: $f_j$ has a root $\alpha_j \in \mathcal{O}_K^n$, for each $j \geq 0$; $f_j \to f$ for some $f \in K[x_1, \ldots, x_n]$; then $f$ has a root $\alpha \in \mathcal{O}_K^n$.
Same as 1 above but: each $f_j$ is homogeneous of fixed degree $m$; each $\alpha_j$ is not equal to $\mathbf{0}_{K^n}$ (the null vector of $K^n$); conclusion $\alpha \in \mathcal{O}_K \setminus \{\mathbf{0}_{K^n}\}$.
My idea is the following: $(f_j)_{j \geq 0}$ is convergent, hence is a Cauchy sequence this implies that $(\alpha_j)_{j \geq 0}$ is a Cauchy sequence (?), hence $\alpha_j \to \alpha$ for some $\alpha \in \mathcal{O}_K$, since $\mathcal{O}_K$ is closed. Now one can prove $f(\alpha) = 0$. Maybe an induction on $n$ can help.
However, I was unable to prove the statement in bold (and perharps it is false).
Thank you in advance for any suggestion.
Your ${\mathcal O}_K$ is compact, hence also $({\mathcal O}_K)^n$. Hence you can find a subsequence of $\alpha_j$, say $\alpha_{j_k}$, that is convergent to some $\omega\in ({\mathcal O}_K)^n$. For all $x\in ({\mathcal O}_K)^n$, you have $|f_{j_k}(x)-f(x)|\leq \|f_{j_k}-f\|$. Hence we get that $|f(\alpha_{j_k})|\to 0$. But of course $f(\alpha_{j_k})\to f(\omega)$, hence $f(\omega)=0$. Perhaps the second point can be proved in the same manner.