Let $\mathbb K$ be an algebraically-closed complete non-archimedean field whose absolute value is non-trivial. Consider the Tate algebra $T_n=\mathbb K\langle X_1,\dots, X_n \rangle$ and fix $f\in T_n$. We use $|\cdot|$ to denote the non-archimedean norm.

Then $f(x_1,\dots,x_n)$ converges for every $x=(x_1,\dots,x_n)\in B(\mathbb K)$ where $B(\mathbb K)$ denotes the unit ball consisting of $x$ with $|x_i|\le 1$.

Question: Suppose $f(x)=0$ for any $x$ with $|x_i|=1$, then can we conclude $f\equiv 0$ as an element of $T_n$?

For a polynomial this is obviously true due to the fundamental theorem of algebra. I am just curious about whether there is something similar in the non-archimedean world. More generally, let's assume $f=0$ on some subset $E$ of $T_n$. What conditions for $E$ can ensure $f\equiv 0$ in $T_n$?

**EDIT:** When $\mathbb K = \mathbb C((q))$ with $|q^n|=e^{-n}$, it seems that we can argue as follows: First note that $T_n \subset \mathbb C((q))[[X_1,\dots, X_n]]$. Suppose $f=\sum_{\nu\in \mathbb N^n} c_\nu X^\nu$ and also write $f=\sum_{n\ge -n_0} q^n f_n$ with $f_n\in \mathbb C[[X_1,\dots, X_n]]$. Actually we must have $f_n \in \mathbb C[X_1,\dots, X_n]$ as $|c_\nu|\to 0$ and there is at most finite terms $c_\nu X^\nu$ contributing to $q^nf_n$. Finally, $|x_i|=1$ is equivalent to $x_i\in \mathbb C$ and hence the assumption of the question means each $f_n\equiv 0$; so $f\equiv 0$ (maybe I had some mistakes) If my argument here was right, can we apply it for more general $\mathbb K$?