$p$-adic series bounded if and only if it has finitely many zeros Let $L\subseteq\mathbb{C}_p$ be a finite extension of $\mathbb{Q}_p$, $r$ be a positive real number, and $f$ be a series $\sum_{n\in \mathbb{Z}} a_nz^n$ convergent in $D= \{x\in \mathbb{C}_p|0<v(x)\leq r \}$ where $a_n$ are elements in $L$. Then I want to know if the following are equivalent.
(1) $f$ is a bounded function in the metric of $\mathbb{C}_p$
(2) $f$ only has finitely many zeros in $D$
(3) the set $\{\lvert a_n\rvert\}$ is bounded as a subset of $\mathbb{R}$(in the Euclid metric)
Symbols: $\lvert a_n\rvert\mathrel{:=}p^{-v(a_n)}$ and $v$ is the valuation of $\mathbb{C}_p$ extended by the valuation on $\mathbb{Q}_p$.
Motivations: I want to use this to prove some properties of the Robba ring over $L$, e.g., $\varepsilon^\dagger$ is a field.
Thanks!
 A: (2) and (3) are equivalent. This is corollary 3.3 in Laurent Berger's IHP course notes Galois representations and $(\varphi, \Gamma)$-modules in 2010. In the same way, we can prove (1) and (3) are equivalent (in one direction, convergence is used).
A: Example
Take $\mathbb Q_3$ itself.  Consider
$$
f(z) = \sum_{k=0}^\infty 3^{-k} z^{2k}.
$$
($a_n = 0$ when $n < 0$ or $n$ is odd.)  Now if $|z|<1$, then $|z| \le 3^{-1}$ and
$$
\big|3^{-k} z^{2k}\big| \le 3^k 3^{-2k} \to 0
$$
so $f$ converges on $\{z\;:\; |z|<1\}$.  But
$$
\big|a_{2k}\big| = \big|3^{-k}\big| = 3^k,
$$
and $\{|a_n|\}$ is not bounded.  Also
$$
f(z) = \frac{1}{1-3z^2} .
$$
Thus $f$ has no zeros in $D$.  And when $z \in D$ we have $|z|\le 3^{-1}$ so
$|z^2| \le 3^{-2}$ so $|3z^2| \le 3^{-3}$, so $|1-3z^3| = 1$ and therefore $|f(z)| = 1$ on $D$. Thus $f$ is bounded on $D$.

OK, but (note Dror's comment) what about boundedness on $\{z \in \mathbb C_3\;:\; 0<|z| < 1\}$?  There are $z \in \mathbb C_3$ with $3^{-1/2}<|z|<1$, and for such $z$ we have
$$
\big|3^{-k} z^{2k}\big| = 3^k |z|^{2k} > 3^{-k} (3^{-1/2})^{2k} > 1
$$
so the series does not converge.  

Let's try another one.
$$
f(z) = \sum_{k=0}^\infty 3^{-k} z^{3^k}
$$
Let $a_n \in \mathbb Q_3$ so that $a_{3^k} = 3^{-k}$ for subscripts of the form $3^k$ and zero otherwise.  Let $|z|<1$. Then is it true that $|z| = 3^r$ for some $r \in\mathbb Q$, $r < 0$ ??  If so,
$$
\big| a_{3^k} z^{3^k}\big| = \big|3^{-k} z^{3^k}\big| = 3^k (3^r)^{3^k}
= 3^{k+3^k r} \to 0
$$
and we get convergence.
But maybe that $|z| = 3^r$ is only for the algebraic closure of $\mathbb Q_3$, not the metric closure $\mathbb C_3$ if it??
