It is known that a complex analytic function defined on an annulus, say, takes its maximum on the boundary. Does an analogue hold for $p$-adic analytic functions? 


More precisely suppose we have a doubly infinite power series $f(z) = \sum\_{n\in \mathbb{Z}}a\_n z^n $ with coefficients $a\_n \in K$ where $K$ is a finite extension of $\mathbb{Q}\_p$ (the rationals completed by a $p$-adic absolute value). Suppose further that $f$ converges for all $z \in K$ with $r\_1 \leq |z| < r\_2$. Does it hold that $|f(z)| \leq \max\{|f|\_{r\_1}, |f|\_{r\_2}\}$ for $r\_1\leq |z| < r\_2$? $|f|\_r $ denotes the maximum of $|f(z)|$ as $z$ varies over $z \in K$ with $|z| = r$.
If it isn't true I would be very grateful for a counterexample. Thanks a lot! 


As an aside: If true one should be able to take any complete non-archimedean field but I am unsure about whether compactness helps. 



P-adic functions on annuli
$\def\bQ{\mathbb{Q}}\def\bF{\mathbb{F}}\def\bZ{\mathbb{Z}}$This is false as stated because of the following important difference between $K$ and $\mathbb{C}$: the former is not algebraically closed. For example. consider $K=\bQ\_p(p^{1/k})$ with $k>1$ and the polynomial $$f(z)=z\prod\limits\_{a\in\bF\_p^{\times}}(z-[a])^2$$ with $r\_1=1/p,r\_2=1$(here $[x]$ denotes the Teichmuller representative of an element $x$ of the residue field of a complete non-archimedean field). Then for any $z\in K$ with $|z|$ equal to $1$ we have $|z-[a]|\leq p^{-1/k}$ for exactly one $a$ because the residue field of $\mathcal{O}\_K$ is equal to $\bF\_p$, so $|f(z)|\leq p^{-2/k}$. For $z$ with $|z|=p^{-1}$ we just get $|f(z)|=p^{-1}$. However, taking $z=p^{1/k}$ gives $|f(z)|=p^{-1/k}$ which is larger than any value of $f$ on the boundary.




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The statement becomes true if we assume that the residue field of $\mathcal{O}\_K$ is infinite(the counterexample above depends crucially on the finiteness of the residue field). The formation of the Newton polygon is a very convenient way to visualize the behavior of roots of a $p$-adic analytic function and the desired inequality in the case of an algebraically closed field $K$ follows quickly from the fact that the slopes of the Newton polygon of a Laurent series correspond to the valuations of the roots.


However, It might be instructive to unpack the proof of this theorem to get a direct argument for our boundary inequality: let $f(z)=\sum\limits\_{n\in \bZ}a\_nz^n$ be a Laurent series converging for $z$ satisfying $r\_1\leq |z|\leq r\_2$. Assume that $r\_1,r\_2$ are in the image of the norm map on $K$. We will prove that for every such $z$ there is $u$ with $|u|$ equal to $r\_1$ or $r\_2$ such that $|f(z)|\leq |f(u)|$.



> 
> **Lemma**. If the residue field of $\mathcal{O}\_K$ is infinite, then for every $z\in K$ there exists an element $z'\in K$ with $|z|=|z'|$
> and a number $n\in\bZ$ such that $|a\_n(z')^n|$ is larger or equal to
> any $|a\_m(z')^m|$ with $m\neq n$ and $|f(z)|=|a\_n(z')^n|$.
> 
> 
> Proof.
> Since the values $|a\_nz^n|$ tend to zero as $n$ tends to $\pm\infty$,
> there exists a finite set of indices $i\_1<\dots< i\_k$ such that
> $|a\_{i\_1}z^{i\_1}|=\dots=|a\_{i\_k}z^{i\_k}|$ and $|a\_mz^m|$ is less that
> this common value for any $m\notin\{i\_1,\dots, i\_k\}$. We want to find
> $z'$ such that the norm of the sum of these $k$ summands is precisely
> equal to the norm of each separate summand. To arrange that, pick
> $\lambda\in \mathcal{O}\_K/\mathfrak{m}\_K$ such that
> $1+\lambda^{i\_2-i\_1}\rho\left(\frac{a\_{i\_2}z^{i\_2}}{a\_{i\_1}z^{i\_1}}\right)+\dots+\lambda^{i\_k-i\_1}\rho\left(\frac{a\_{i\_k}z^{i\_k}}{a\_{i\_1}z^{i\_1}}\right)\neq 0$ where $\rho:\mathcal{O}\_K\to \mathcal{O}\_K/\mathfrak{m}\_K$ is the
> reduction map. Then $z'=[\lambda]z$ will do the job.
> 
> 
> 


We can now prove the statement: let $z\in K$ be any element in the annulus $r\_1\leq |z|\leq r\_2$. There exists $k\in\bZ$ such that $|f(z)|\leq |a\_kz^k|$. Assume that $k\geq 0$. Then $|a\_kz^k|\leq |a\_k|r\_2^k$. Using the lemma, find $u$ with $|u|=r\_2$ such that $|f(u)|=|a\_nu^n|$ and $|a\_nu^n|\geq |a\_mu^m|=|a\_m|r\_2^m$ for every $m$. It follows that $|f(u)|\geq |a\_k|r\_2^k\geq |f(z)|$. If $k<0$, then arguing in the same way with $r\_2$ replaced by $r\_1$ gives the result.


