# $p$-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 $$D= \{x\in \mathbb{C}_p|0 where $$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!

• "only finite zeros" meaning only finitely many zeros? Nov 20 '19 at 12:12
• And $L$ is a subset, not an element, of ${\bf C}_p$, right? and the $a_n$ are in $L$? Nov 20 '19 at 12:15
• @GerryMyerson Sorry, I will edit my question right now!
– user141691
Nov 20 '19 at 12:16
• $L$ is still an element of ${\bf C}_p$? Nov 20 '19 at 12:23
• @GeraldEdgar $D=\{x: p^{-r}\leq|x|<1\}$, and $f$ is a bounded function when we regard $f$ as a function over $D$.
– user141691
Nov 20 '19 at 13:50

(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).

• How does your cited proof of equivalence square with @GeraldEdgar's counterexample? Nov 22 '19 at 17:16
• I don’t think it is a counterexample, since the domain of convergence of his series is clearly $\{z\in\Bbb C_p:v(z)>1/2\}$, as @DrorSpeiser has, in essence, pointed out. Nov 22 '19 at 18:13

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??

• Your answer wrote that if $|z|<1$, then $|z|<3^{-1}$?
– user141691
Nov 22 '19 at 14:10
• @Sssss, it says that if $\lvert z\rvert < 1$ then $\lvert z\rvert \le 3^{-1}$, which is true (in $\mathbb Q_3$). Nov 22 '19 at 17:14
• For this function there is no positive $r$ such that the function converges on $\{x\in\mathbb{C}_p :\ 0<v(x) \le r\}$. Nov 22 '19 at 17:51
• So you claim it does not converge on $\{x\;:\; 0 < v(x) \le 1\}=\{x\;:\; 3^{-1}\le |x| < 1\}$ ?? Nov 22 '19 at 17:53
• Yeah, @GeraldEdgar. Your last example is interesting. It is the logarithm of a formal group of height one defined over $\Bbb Z_3$, even $\Bbb Z_3$-isomorphic to the multiplicative formal group. it is convergent on the (open) unit disk in $\Bbb C_3$, has infinitely many zeros there, and is unbounded as a function on the open disk. Nov 22 '19 at 19:28