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Let $K$ be a finite extension of $\mathbb{Q}_p$, and let $D$ be some disk $D = \{ x\in \overline{K} \mid |x| < c < 1\}$. Is the set of power series in $K[[T]]$ which converge on $D$ and are bounded on $D$ complete with respect to the supremum norm $\|f\|_\infty=\sup_{x\in \overline{K}, |x|<c} |f(x)|$?

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  • $\begingroup$ I used your answer to interpret your question, which is not good, though the question is interesting $\endgroup$
    – reuns
    Nov 28, 2021 at 5:12

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Your disc $D$ depends on $c$ and you never said anything about what kind of number $c$ is, other than it is in $(0,1)$. In particular, whether or not $c$ is in $|K^\times|$ could be an important distinction.

If $c$ is in $|K^\times|$, say $c=|\alpha|_p$, then by rescaling (replacing $f(T)$ with $f(\alpha T)$) we can put ourselves in the case $c=1$, so we can then suppose $D$ is the open unit disc in $\overline{K} = \overline{\mathbf Q_p}$. Or if we start with a series $f(T)$ converging on the open unit disc in $\overline{K}$, then $f(T/\alpha)$ converges on all $x$ with $|x|_p < c$. So when $c$ is in $|K^\times|$, there is no real difference between $c \not= 1$ and $c=1$. In any case, I have a surprise for you: the sup-norm $||f||_{\infty}$ need not be finite. So you need to be explicit about looking at series with a finite sup-norm on $D$. (Edit: the original version of the OP's question did not include finiteness of $||f||_\infty$ as a hypothesis, but the OP now fixed that, which makes this answer no longer directly relevant except to point out why the finiteness of $||f||_\infty$ has to be an assumption, just like in the definition of $L^p$-spaces.) I think your mistake is possibly due to overlooking a contrast between discs in finite extensions of $\mathbf Q_p$ and discs in the algebraic closure: your disc $D$ is not compact, so continuous functions on $D$ need not be bounded.

Specifically, the $p$-adic logarithm series $\log(1+T) = T - T^2/2 + \cdots$ in $\mathbf Q[[T]]$ converges on the open unit disc $\Delta$ in $\overline{\mathbf Q_p}$ since it converges on the open unit disc in $L$ for each finite extension $L$ of $\mathbf Q_p$ and its image on that disc is all of $\overline{\mathbf Q_p}$. Let’s prove that. Pick $z$ in $\overline{\mathbf Q_p}$. It lies in a finite extension $L$ of $\mathbf Q_p$. For big enough $n$, $|p^nz|_p < (1/p)^{1/(p-1)}$, so $p^nz = \log(1+y)$ for some $y$ in $L$ where $|y|_p < (1/p)^{1/(p-1)} < 1$ since the $p$-adic logarithm is a bijection (even an isometry) between the open discs in $L$ of radius $(1/p)^{1/(p-1)}$ around $1$ and $0$. Write $1+y$ as a $p^n$-th power in $\overline{\mathbf Q_p}$, say $1+y = w^{p^n}$. Then $|w|_p \leq 1$, and looking at both sides of the equation $1+y = w^{p^n}$ in the residue field of a finite extension of $\mathbf Q_p$ containing $y$ and $w$ shows $|w-1|_p < 1$. Then
$$ p^nz = \log(1+y) = \log(w^{p^n}) = p^n\log(w), $$ so $z = \log(w) = \log(1 + (w-1))$ where $w-1 \in \Delta$. Thus the surjectivity of $\log(1+T)$ from $\Delta$ to $\overline{\mathbf Q_p}$ is proved.

Instead of reducing to the case $c=1$, when $c \in (0,1)$ is in $|K^\times|$, say $c=|\alpha|_p$, let $f(T) = \log(1+T/\alpha)$, which has coefficients in $K$ and converges at all $x$ in $\overline{K}$ with $|x|_p < c$ since $|x/\alpha|_p < 1$. Then $f(T)$ on your disc $D$ has image $\overline{K}$ and thus its sup-norm on $D$ is infinite.

Similar results could be obtained for many other power series besides the logarithm by using Newton polygon arguments, but for the logarithm we can “see” its surjectivity onto $\overline{K}$ in the more direct way above.

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  • $\begingroup$ Yes obviously. You should have pinged me, only the last sentence of my answer is unclear/wrong $\endgroup$
    – reuns
    Nov 28, 2021 at 13:57
  • $\begingroup$ @reuns I think for the OP, the possible unboundedness of the sup-norm of a power series on $D$ may not have been obvious. I had not seen the end of your initial answer before I wrote mine. I will edit the part where I say the “whole premise” of the question is flawed. $\endgroup$
    – KCd
    Nov 28, 2021 at 14:08
  • $\begingroup$ I just edited thanks to you! I missed that the OP had in mind the ring of convergent power series, not that with bounded sup norm, and indeed I didn't clarify in my reasonning that they were different. You are right that this is the main answer to the question. $\endgroup$
    – reuns
    Nov 28, 2021 at 14:11
  • $\begingroup$ Hello @KCd I apologize for not including the boundedness condition in the premise. Thank you for pointing out this is necessary in order for the space to be a metric space. $\endgroup$ Nov 28, 2021 at 17:02
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Please take a look at kCd's answer before

I think the main theorem is that on the ring $R_K$ of power series $\in K[[T]]$ that converge on the closed unit disk $x\in \overline{K},|x|\le 1$ the sup norm of the coefficients and the sup norm of the values taken on the closed unit disk give the same norm.

This is because $R_K = \{f= \sum_{n\ge0} a_n T^n\in K[[T]], v(a_n)\to \infty\}$. If $f\ne 0$, with $v(f)=\inf_n v(a_n)$, then $p^{-v(f)} f\in O_{K(p^{v(f)})}/(\pi_{K(p^{v(f)})})[T]$ is a non-zero polynomial, it will be non-zero for some root of unity $\zeta\bmod (\pi_{K(p^{v(f)})})$ so that $v(p^{-v(f)} f(\zeta))=0$ ie. $$ \inf_{|x|\le 1} v(f(x))\le v(f(\zeta))=v(f)\le \inf_{|x|\le 1} v(f(x))$$

In particular $R_K$ is complete for both norm.

Then your ring is just $\{f\in K[[T]],\forall d\in \Bbb{Q}\cap (0,c), f(p^d T) \in R_{K(p^d)}\}$ which will be complete for the topology induced by the collection of $\sup_{|x|\le d} |f(x)|$ norms, but as kCd said $\sup_{|x|<c}|f(x)|$ doesn't need being finite so it is not a norm on your ring. The subring with $\sup_{|x|<c}|f(x)|$ finite is complete.

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  • $\begingroup$ So it is complete if we restrict the space to being bounded power series? $\endgroup$ Nov 28, 2021 at 15:51
  • $\begingroup$ Yes ${}{}{}{}{}$ $\endgroup$
    – reuns
    Nov 28, 2021 at 15:58
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Ok here is my shot at an answer, but I am not certain it is correct.

Suppose $|f_n(x) - f_m(x)| < \epsilon$ for all $|x|<c$. I show all terms of this difference must have absolute value less than $\epsilon$ implying an upper bound on the coefficients. Suppose $f_n(x)-f_m(x)$ has terms of absolute value greater than $\epsilon$ when evaluated at some $z$. Let $S = \{c_{i_1}z^{i_1},\ldots, c_{i_k}z^{i_k}\}$ be the list of all terms which have lowest valuation $u$, implying the valuation of $\sum_{j=1}^k c_{i_j}z^{i_j}$ is greater than $u$. There is some finite extension $K'$ of $K$ in which all of the elements of $S$ live. Now take $L$ to be some unramified extension of $K'$ such that the size of the residue field of $L$ is greater than $i_k$. Now suppose $\lambda$ is some uniformizer of $L$, so that there exists a power $\lambda^{e_1}$ such that the valuation of $\lambda^{e_1}$ is $u$. Then $c_{i_j}z^{i_j}\lambda^{-e_1}$ are all units, and the polynomial $g(T) = \sum_{j=1}^kc_{i_j}z^{i_j}\lambda^{-e_1}T^{i_j}$ must vanish for all $t$ in $L/\lambda$ else we can find $x = zt \in L$ with $|f_n(x)-f_m(x)| > \epsilon$ and $|x| < c$. However $g$ cannot possibly vanish on all of $L/\lambda$ because $|L/\lambda| > i_k$ and $\deg g = i_k$. Thus all terms of $f_n(z)-f_m(z)$ have absolute value at most $\epsilon$, and this proves the result.

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