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.

• Did you mean to write $r_1 \leq |z| \leq r_2$ instead of $r_1 \leq |z| < r_2$? – user1728 Jul 22 at 3:25

$$\def\bQ{\mathbb{Q}}\def\bF{\mathbb{F}}\def\bZ{\mathbb{Z}}$$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.

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.

• Instead of "are in the image of the norm map on $K$" it would be better to say "are in the image of the value group of $K^\times$" or "are in $|K^\times|$". The "norm map" on $K$ sounds like the multiplicative mapping $K \rightarrow \mathbb Q_p$ in field theory. – user1728 Jul 22 at 3:23
• Thanks a lot! I have entirely ignored the issue of the residue field being finite. – Gari Jul 28 at 13:24