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.

  • $\begingroup$ Did you mean to write $r_1 \leq |z| \leq r_2$ instead of $r_1 \leq |z| < r_2$? $\endgroup$
    – user1728
    Jul 22 '19 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.

  • $\begingroup$ 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. $\endgroup$
    – user1728
    Jul 22 '19 at 3:23
  • $\begingroup$ Thanks a lot! I have entirely ignored the issue of the residue field being finite. $\endgroup$
    – Gari
    Jul 28 '19 at 13:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.