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.


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

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    $\begingroup$ How does your cited proof of equivalence square with @GeraldEdgar's counterexample? $\endgroup$ – LSpice Nov 22 '19 at 17:16
  • $\begingroup$ 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. $\endgroup$ – Lubin Nov 22 '19 at 18:13

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??

  • $\begingroup$ Your answer wrote that if $|z|<1$, then $|z|<3^{-1}$? $\endgroup$ – user141691 Nov 22 '19 at 14:10
  • $\begingroup$ @Sssss, it says that if $\lvert z\rvert < 1$ then $\lvert z\rvert \le 3^{-1}$, which is true (in $\mathbb Q_3$). $\endgroup$ – LSpice Nov 22 '19 at 17:14
  • $\begingroup$ For this function there is no positive $r$ such that the function converges on $\{x\in\mathbb{C}_p :\ 0<v(x) \le r\}$. $\endgroup$ – Dror Speiser Nov 22 '19 at 17:51
  • $\begingroup$ So you claim it does not converge on $\{x\;:\; 0 < v(x) \le 1\}=\{x\;:\; 3^{-1}\le |x| < 1\}$ ?? $\endgroup$ – Gerald Edgar Nov 22 '19 at 17:53
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    $\begingroup$ 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. $\endgroup$ – Lubin Nov 22 '19 at 19:28

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