\input{preamble.tex}
\unittitle{Convexity and monotonicity for subsidiary radii}

In this unit, we prove some theorems
governing the variation of the subsidiary
radii of a differential module on a disc or annulus.

\section{Setup}

Let $M$ be a finite free differential module of rank $n$
over $K \langle \alpha/t, t/\beta \rangle$, where $0 \leq \alpha \leq \beta$.
We are interested in the variation of the subsidiary radii of
$M \otimes F_\rho$ as $\rho$ ranges over $[\alpha,\beta]$.

The properties we are interested in are more convenient to describe in
logarithmic terms, so we set notation as follows.
For $\rho \in [\alpha, \beta]$, let $R_1(\rho), \dots, R_n(\rho)$ be the
extrinsic subsidiary radii of $M \otimes F_\rho$ in increasing order,
so that $R_1(\rho) = R(M \otimes F_\rho)$ 
is the generic radius of convergence of
$M \otimes F_\rho$.
For $r \in [-\log \beta, -\log \alpha]$, define 
\[
f_i(r) = -\log R_i(e^{-r}),
\]
so that $f_i(r) \geq r$ for all $r$. We will write $f_i(M,r)$ instead
of $f_i(r)$ in case there is ambiguity about which $M$ we are considering.

\section{Main results}

We now state the main results of this unit.
\begin{theorem} \label{T:subsidiary}
Let $M$ be a finite free differential module of rank $n$
over $K \langle \alpha/t, t/\beta \rangle$.
\begin{enumerate}
\item[(a)] (Linearity)
For $i=1, \dots, n$, 
the functions $f_i$ are continuous and piecewise affine.
\item[(b)] (Integrality)
If $i=n$ or $f_i(r_0) > f_{i+1}(r_0)$, then 
the slopes of
$f_1 + \cdots +f_i$ in some neighborhood of $r_0$ belong to $\ZZ$.
Consequently, the slopes of each $f_i$ belong to
$\frac{1}{1} \ZZ \cup \cdots \cup \frac{1}{n} \ZZ$.
\item[(c)] (Convexity)
For $i=1, \dots, n$,
the function $f_1 + \cdots + f_i$ is convex.
\item[(d)] (Monotonicity)
Suppose that  $\alpha = 0$. For $i=1, \dots, n$, 
for any point $r_0$ where $f_i(r_0) > r_0$,
the slopes of $f_1 + \cdots + f_i$ are nonpositive in some neighborhood of $r_0$.
(Remember that $f_i(r) = r$ for $r$ sufficiently large.)
\end{enumerate}
\end{theorem}

Note that we have integrality for the slopes of the $f_i$
but not for the values. Using the integrality result for intrinsic 
generic radii proved in a previous unit, we can at least deduce the following.
(One can prove something similar for subsidiary radii, but the statement
is a bit more complicated.)

\begin{theorem} \label{T:integrality}
Set notation as in Theorem~\ref{T:subsidiary}. 
Let $h$ be a nonnegative integer
and pick $m \in \{1,\dots,n\}$.
Suppose that the following hold for some $r_0$:
\begin{enumerate}
\item[(a)]
$f_m(r_0) = f_1(r_0)$;
\item[(b)]
either $m=n$, or $f_m(r_0)> f_{m+1}(r_0)$;
\item[(c)]
$f_1(r_0) - r_0 > \frac{p^{-h}}{p-1} \log p$.
\end{enumerate}
Then for $r$ in some neighborhood of $r_0$,
\[
\quad f_1(r) - r \in \frac{1}{m p^h} v(K^\times) + \frac{1}{m} r \ZZ.
\]
\end{theorem}
In fact, one cannot do better than this; see \S \ref{sec:key}.

\section{Variation of Newton polygons}

The formulation of Theorem~\ref{T:subsidiary} 
is motivated by the following result, which will also be used in the proof.
(Note that because of a sign discrepancy, convexity is traded for concavity
in (c), and nonpositive slopes are traded for nonnegative slopes in
(d).)
\begin{theorem} \label{T:newton}
Let $P \in K \langle \alpha/t,t/\beta \rangle[T]$ be a polynomial
of degree $n$. For $r \in [-\log \beta, -\log \alpha]$, let
$f_1(r), \dots, f_n(r)$ be the slopes of the Newton polygon of
$P$ under $|\cdot|_{e^{-r}}$, in increasing order.
\begin{enumerate}
\item[(a)] (Linearity)
For $i=1, \dots, n$, 
the functions $f_i$ are continuous and piecewise affine.
\item[(b)] (Integrality)
If $i=n$ or $f_i(r_0) < f_{i+1}(r_0)$, then the slopes of
$f_1 + \cdots +f_i$ in some neighborhood of $r$ belong to $\ZZ$.
Consequently, the slopes of each $f_i$ belong to
$\frac{1}{1} \ZZ \cup \cdots \cup \frac{1}{n} \ZZ$.
\item[(c)] (Concavity)
Suppose that $P$ is monic.
For $i=1, \dots, n$,
the function $f_1 + \cdots + f_i$ is concave.
\item[(d)] (Monotonicity)
Suppose that $P$ is monic and that $\alpha = 0$.
For $i=1, \dots, n$, the slope of
$f_1 + \cdots + f_i$ is nonnegative.
\end{enumerate}
\end{theorem}
\begin{proof}
Write $P = \sum_{i=0}^{n} P_i T^i$ with $P_i \in K \langle
\alpha/t,t/\beta \rangle$. Write $\NP_r(P)$ for the Newton polygon of
$P$ measured with respect to $v_r(\cdot) = -\log |\cdot|_{e^{-r}}$.

For $s \in \RR$ and $r \in [-\log \beta, -\log \alpha]$, put
\[
v_{s,r}(P) = \min_i \{v_r(P_i) + is\};
\]
that is, $v_{s,r}(P)$ is the $y$-intercept of the supporting line of
$\NP_r(P)$ of slope $s$.

The function $v_r(P_i)$ is continuous in $r$ and piecewise affine with
slopes in $\ZZ$; by the Hadamard three circles lemma, it is also concave.
Since $v_{s,r}(P)$ is the minimum of finitely many continuous,
piecewise affine, concave functions of $r$ with slopes in $\ZZ$, so then is
$v_{s,r}(P)$. 
Note also that $v_{s,r}(P)$ is concave as a function of the pair $(r,s)$,
since each function $(r,s) \mapsto v_r(P_i) + is$ has that property.

Note that 
$f_1(r) + \cdots + f_i(r)$ is the difference between the $y$-coordinates
of the points of $\NP_r(P)$ of $x$-coordinates $i-n$ and $-n$. That is,
\begin{equation} \label{eq:sum from vertices}
f_1(r) + \cdots + f_i(r) = \sup_s \{v_{s,r}(P) - (n-i)s\} - v_r(P_n).
\end{equation}
Moreover, the supremum in \eqref{eq:sum from vertices} is achieved
by some $s$ whose denominator is bounded by $n$.
Consequently, $f_1(r) + \cdots + f_i(r)$ is continuous and piecewise affine,
proving (a).

If $i=n$ or $f_i(r_0) > f_{i+1}(r_0)$, then the point of
$\NP_{r_0}(P)$ of $x$-coordinate $i-n$ is a vertex, and likewise for
$r$ in some neighborhood of $r_0$. In that case, for $r$ near $r_0$,
\[
f_1(r) + \cdots + f_i(r) = v_r(P_{n-i}) - v_r(P_n),
\]
proving (b).

Assume hereafter that $P$ is monic, so that $P_n = 1$ and \eqref{eq:sum
from vertices} reduces to
\[
f_1(r) + \cdots + f_i(r) = \sup_s \{v_{s,r}(P) - (n-i)s\}.
\]
It is not immediately clear from this that $f_1 + \cdots + f_r$ is concave, 
since we are taking the supremum rather
than the infimum of a collection of concave functions. To get around this,
pick $r_1, r_2 \in [-\log \beta, -\log \alpha]$ 
and put $r_3 = ur_1 + (1-u)r_2$ for some $u \in [0,1]$.
For $j \in \{1,2\}$, choose $s_j$ achieving the supremum
in \eqref{eq:sum from vertices} for $r = r_j$.
Put $s_3 = us_1 + (1-u)s_2$; then
using the convexity of $v_{s,r}(P)$ in both $s$ and $r$,
we have
\begin{align*}
f_1(r_3) + \cdots + f_i(r_3) &\geq v_{s_3,r_3}(P) - (n-i)s_3 \\
&\geq u(v_{s_1,r_1}(P) - (n-i)s_1)
+ (1-u)(v_{s_2,r_2}(P) - (n-i)s_2) \\
&= u (f_1(r_1) + \cdots + f_i(r_1))
+ (1-u) (f_1(r_2) + \cdots + f_i(r_2)).
\end{align*}
This yields concavity for $f_1 + \cdots + f_i$, proving (c).

Assume finally that $\alpha = 0$ (and $P$ is still monic). Then
each $v_r(P_i)$ is a nondecreasing function of $r$, as then is
each $v_{s,r}(P)$.
Since $v_r(P_n) = 0$, $f_1 + \cdots + f_r$ is nondecreasing
by \eqref{eq:sum from vertices},  proving (d).
\end{proof}

\section{Key differences}
\label{sec:key}

Having drawn an analogy between our original theorem and 
Theorem~\ref{T:newton}, we must now indicate some respects in which the
analogy falls short.

Besides having everything negated (flipping concavity to convexity),
our target theorem also has a boundary case that does not occur 
in the Newton polygon case. That is because subsidiary radii can ``max out''
by achieving equality in the bound $f_i(r) \geq r$,
at which one has an abrupt change of behavior which undermines
the nonpositivity property for slopes.

Another important difference must be noted between Theorem~\ref{T:newton}
and our desired results. In Theorem~\ref{T:newton}, the
function $f_1 + \cdots + f_n$ is piecewise of the form $mr + b$
where $m \in \ZZ$, but $b$ is also constrained: it must belong to the
additive value group of $K$. By contrast, this need not be the case
in Theorem~\ref{T:subsidiary} or~\ref{T:subsidiary},
as demonstrated by the following example. (See Theorem~\ref{T:integrality}
for the best possible affirmative result.)

Pick $\lambda \in K^\times$ and $0 < \alpha \leq \beta$ such that
for $\rho \in [\alpha, \beta]$,
\[
p^{1/(p-1)} < |\lambda| \rho^{-p} < p^{p/(p-1)}.
\]
Let $M$ be the differential module over $K \langle \alpha/t,t/\beta\rangle$
generated by $v$ satisfying $D(v) = -p \pi \lambda t^{-p-1}$.
For $\rho \in [\alpha, \beta]$, using the supremum norm on 
$M \otimes F_\rho$ given by $w$, we compute
\[
|D|_{M \otimes F_\rho} = p^{-p/(p-1)} |\lambda| \rho^{-p-1} < 
\rho^{-1};
\]
this tells us that
$|D|_{\tspect, M \otimes F_\rho} \leq \rho^{-1}$ but nothing stronger.

To compute $R(M \otimes F_\rho)$, we 
construct the module $M'$ over $K \langle \alpha^p/t^p, t^p/\beta^p \rangle$
with generator $w$ and $D'(w) = -\pi \lambda (t^p)^{-2}$.
In this case, we read off
\[
|D'|_{M' \otimes F'_\rho} = p^{-1/(p-1)} |\lambda| \rho^{-2p} >
\rho^{-p}
\]
so this also computes $|D'|_{\tspect,M' \otimes F'_\rho}$.
We thus have
\begin{align*}
R(M' \otimes F'_\rho) &= |\lambda|^{-1} \rho^{2p} \\
R(M \otimes F_\rho) &= |\lambda|^{-1/p} \rho^{2},
\end{align*}
where the latter holds by the Frobenius antecedent theorem
because $M = \phi^* (M')$. (More precisely, we first find
that $R(M \otimes F_\rho) \geq |\lambda|^{-1/p} \rho^{2} > p^{-1/(p-1)} \rho$,
so $M$ has a Frobenius antecedent; we then note that
$R(M' \otimes F'_\rho) > p^{-p/(p-1)} \rho^p$, so the Frobenius antecedent
of $M$ is forced to equal $M'$. We then get
$R(M \otimes F_\rho) = |\lambda|^{-1/p} \rho^{2}$.)

In particular, 
\[
f_1(r) = 2r + \frac{1}{p} \log |\lambda|
\]
has constant term which need not belong to the value group
of $K$.

\section{Convexity of the generic radius}

As a prelude to tackling Theorem~\ref{T:subsidiary},
we give a quick proof of convexity of the function $f_1$,
corresponding to the generic radius of convergence. This argument applies to
both discs and annuli, and can be (and historically was)
used in place of the full strength
of Theorem~\ref{T:subsidiary}
for many purposes.

Choose a basis of $M$,
and let $D_s$ be the basis via which $D^s$ acts on $M$. Then recall that
\[
R_1(\rho) = \min\{\rho, p^{-1/(p-1)} \liminf_{s \to \infty} |D_s|_\rho^{-1/s}
\}.
\]
For each $s$, the 
function $r \mapsto -\log |D_s|_{e^{-r}}^{-1/s}$ is convex in $r$ by
the Hadamard three circles lemma. This implies the convexity of
\[
f_1(r) = \max\{r, \frac{1}{p-1} \log p 
+ \limsup_{s \to \infty} (-\log |D_s|_{e^{-r}}^{-1/s}) \}.
\]

To improve upon this result, one might like to try to read off the
generic radius of convergence, and maybe even the other subsidiary radii,
from the Newton polygon of a cyclic vector. In order to do this,
we have to overcome two obstructions.
\begin{enumerate}
\item[(a)]
Some of the subsidiary radii may be greater than $p^{-1/(p-1)} \rho$,
in which case Newton polygons will not detect them.
\item[(b)]
One can only construct cyclic vectors in general for differential modules
over differential fields, not over differential rings.
\end{enumerate}
The first problem will be addressed using Frobenius descendants.
The second problem will be addressed by first using a cyclic vector over a fraction
field to establish linearity and integrality. We will then compare to a carefully
chosen lattice to deduce convexity and monotonicity.

\section{Twisted polynomials and Newton polygons}

Throughout this section,
let $F$ be a complete nonarchimedean differential field. 
We need a way to detect truncated spectral norms of differential modules over
$F$ without writing down cyclic vectors.
\begin{lemma}[Decomposition lemma] \label{L:decomp lemma}
Let $R$ be a complete subring of $F$.
Let $V$ be a finite differential module over $F$. Let $e_1,\dots,e_n$
be a basis of $V$ via which $D$ acts via a matrix $N$ which has the
same Newton and Hodge slopes less than $-\log |d|_F$, namely
$r_1,\dots,r_i$. Then in the decomposition of $V$ by spectral norm,
the component of spectral norm $s$ has dimension equal to the number of
$i$ such that $s = e^{-r_i}$.
\end{lemma}
\begin{proof}
If $N$ has no Hodge slopes less than $-\log |d|_F$, then $|D|_{\tspect,V}
\leq |D|_V \leq |d|_F$ and so we have nothing to check. We thus assume instead
that the least Hodge slope of $N$ is $r < -\log |d|_F$; it suffices to check
that we can separate off a component of $V$ accounting for that slope,
by making a change of basis over $\gotho_F$.

By the Hodge-Newton decomposition, we can find a matrix $U \in \GL_n(\gotho_F)$
such that $U^{-1} N U$ splits as a block diagonal matrix,
with the top left block accounting for the slope $r$. Put
$N_1 = U^{-1} N U + d(U)U^{-1}$; since $|d(U)U^{-1}| \leq 
|d|_F$, using the perturbation theorem for characteristic
polynomials, we see that
the matrix $N_1 = U^{-1} N U + d(U)U^{-1}$ again has the same Newton
and Hodge slopes less than $-\log |d|_F$, and moreover they agree with the
corresponding slopes of $N$. 

It is possible to repeat this process so as to
obtain a convergent sequence of change-of-basis matrices; we will make this
calculation in detail in the next unit.
\end{proof}

In order to apply Lemma~\ref{L:decomp lemma}, we need to produce good bases of
$V$. We can do this using twisted polynomials as follows.
(We made a similar calculation in a previous unit, in the case where $P$ had
only one slope.)
\begin{prop}\label{P:poly to norm}
Let $P = T^n + \sum_{i=0}^{n-1} P_i T^i \in F\{T\}$
be a monic twisted polynomial, and put $V = F\{T\}/F\{T\}P$.
Let $r_1 \leq \dots \leq r_n$ be the slopes of the Newton polygon of $P$,
and suppose that there exist $\lambda_i \in F$ of norm $e^{-r_i}$.
Let $N$ be the matrix via which $D$ acts on the basis 
\[
\lambda_{n}^{-1}\cdots \lambda_{n-i+1}^{-1} T^i \qquad (i=0,\dots,n-1).
\]
Then the Hodge slopes of $N$ are also $r_1,\dots, r_n$;
in particular, the Hodge slopes less than $-\log |d|_F$ compute
truncated spectral norms of constituents of $V$.
\end{prop}
\begin{proof}
Put $\mu_i = \lambda_{1} \cdots \lambda_{n-i-1}$
for $i=0,\dots, n-1$; then
\[
N = \begin{pmatrix}
0 & \cdots & 0 & \mu_0^{-1} P_0 \\
\lambda_{n-1} & \cdots & 0 & \mu_1^{-1} P_1 \\
& \ddots & & \vdots \\
0 & \cdots & \lambda_1 & \mu_{n-1}^{-1} P_{n-1} \end{pmatrix}.
\]
For $i=1,\dots, n-1$, we have
\begin{align*}
|\mu_i^{-1} P_i| &= e^{r_{i} + \cdots +r_{n-i-1}} |P_i| \\
&\leq e^{r_1 + \cdots + r_{n-1-1}} e^{-r_1 - \cdots - r_{n-i}} \\
&\leq e^{-r_{n-i}} = |\lambda_{n-i}|.
\end{align*}
Thus by using column operations over $\gotho_F$ (so as not to change the Hodge
polygon), we can clear everything in the
right column except $\mu_0^{-1} P_0$. By permuting rows and columns,
we end up with a diagonal matrix with entries of norm $e^{-r_1}, \dots, e^{-r_n}$.
This proves the claim.
\end{proof}

\section{Measuring small subsidiary radii}

We are now almost ready to prove the components of Theorem~\ref{T:subsidiary}
concerning large subsidiary radii. We postpone the proof of one more key
lemma until we can illustrate how it will be needed.
\begin{lemma} \label{L:convex1}
Fix $c_0 > 1/(p-1) \log p$, and define 
\[
f'_i(r) = \max\{f_i(r), c_0\}.
\]
\begin{enumerate}
\item[(a)] (Linearity)
For $i=1, \dots, n$, 
the functions $f'_i$ are continuous and piecewise affine.
\item[(b)] (Integrality)
If $i=n$ or $f'_i(r_0) > \max\{f'_{i+1}(r_0), r_0 + c_0\}$, then in a neigborhood of
$r_0$, 
\[
f'_1(r) + \cdots +f'_i(r) \in v(K^\times) + r \ZZ.
\]
Consequently, the slopes of each $f'_i$ belong to
$\frac{1}{1} \ZZ \cup \cdots \cup \frac{1}{n} \ZZ$.
\item[(c)] (Convexity)
For $i=1, \dots, n$,
the function $f'_1 + \cdots + f'_i$ is convex.
\item[(d)] (Monotonicity)
Suppose that $\alpha = 0$. For $i=1, \dots, n$, 
the slopes of $f'_1 + \cdots + f'_i$ are nonpositive at any point $r$ where
$f'_i(r) > r + c_0$.
\end{enumerate}
\end{lemma}
\begin{proof}
Put $F = \Frac K \langle \alpha/t,t/\beta \rangle$.
Choose a cyclic vector for $M \otimes F$ to obtain an isomorphism
$M \otimes F \cong F\{T\}/F\{T\}P$ for some monic twisted polynomial $P$
over $F$. We may then apply Theorem~\ref{T:newton} to deduce (a) and (b).

To deduce (c) and (d), we may work in a neighborhood of a single
value $r_0$ of $r$. 
There is no harm in enlarging $K$, so we may assume $v(K^\times) = \RR$.
Then we may reduce to the case $r_0 = 0$ by
replacing $t$ by $\lambda t$ for some $\lambda \in K^\times$.

Apply Proposition~\ref{P:poly to norm} to construct
$\lambda_{r,1},\dots, \lambda_{r,n} \in K$ such that
the basis of $M \otimes F_{e^{-r}}$ given by 
\[
\lambda_{r,n}^{-1}\cdots \lambda_{r,n-i+1}^{-1} T^i \qquad (i=0,\dots,n-1)
\]
satisfies the conclusion of the proposition. By Lemma~\ref{L:good lattice}
below, for any particular $c>1$, 
we may construct a basis $m_1, \dots, m_n$ of $M$ such that
the supremum norm defined by this basis differs from the 
supremum norm defined by the chosen basis of $M \otimes F_1$
by a multiplicative factor of at most $c$.
By continuity, for $r$ sufficiently close to 0,
the supremum norm defined by $m_1,\dots,m_n$ differs from the
supremum norm defined by the chosen basis of $M \otimes F_{e^{-r}}$
by a multiplicative factor of at most $c^2$.

Let $N$ be the matrix via which $D$ acts on $m_1,\dots, m_n$.
For $r$ close to 0, by the previous paragraph,
we can construct a change of basis matrix $U_r$
between $m_1, \dots, m_n$ and the chosen basis of 
$M \otimes F_{e^{-r}}$, such that $|U_r|, |U_r^{-1}| \leq c^{2}$.
For $c$ sufficiently close to 1 (and $r$ correspondingly close to 0),
we may conclude that the Newton polygons of $N$ and
$N + d(U_r) U_r^{-1}$ coincide in slopes less than $-r$.
The latter has the same Newton polygon as its conjugate
$U_r^{-1} N U_r + U_r^{-1} d(U_r)$;
by Lemma~\ref{L:decomp lemma}, we may conclude that for $r$ near 0,
the Newton polygon of $N$ under $|\cdot|_r$ computes subsidiary radii
less than $p^{-1/(p-1)} e^{-r}$. We may thus deduce (c) and (d)
from Theorem~\ref{T:newton}.
\end{proof}

In the previous proof, we needed to approximate a basis of $M \otimes F_1$
with a basis of $M$; the following lemma allows us to do this.
\begin{lemma}[Lattice lemma] \label{L:good lattice} 
Let $R$ be a complete $K$-subalgebra of $F_1$ (e.g.,
$K \langle \alpha/t, t/\beta\rangle$ with $1 \in [\alpha, \beta]$),
and put $R' = R \cap \gotho_{F_1}$.
Let $M$ be a finite free $R$-module of rank $n$,
and let $|\cdot|_M$ be a norm on $M \otimes F_1$ compatible with $|\cdot|_1$.
Assume that either:
\begin{enumerate}
\item[(a)]
$c>1$ and the value group of $K$ is not discrete; or
\item[(b)]
$c\geq 1$, the value group of $K$ is discrete, and the value group
of $M$ is the same as that of $K$.
\end{enumerate}
Then there exists a norm $|\cdot|'_M$ on $M \otimes F_1$
such that 
$\{m \in M: |m|'_M \leq 1\}$ is a finite free
$R'$-module of rank $n$,
and $c^{-1} |m|_M \leq |m|'_M \leq c |m|_M$ for all $m \in M$.
\end{lemma}
Although we will only need to apply this when $|\cdot|$ is the supremum
norm associated to some basis, we need the extra generality in order to
carry out the induction in the proof (at least in case (a)).
\begin{proof}
We induct on $n$. Pick any $m_1 \in M$ belonging to a basis of $M$,
so that $M_1 = M/R m_1$ is also free.
Using (a) or (b), we can rescale $m_1$ by an element of $K$ to force
$1 \leq |m_1|_M \leq c^{2/3}$.

Equip $M_1$ with the quotient norm
\[
|x_1|_{M_1} = \inf_{x \in M: x + M_1 = x_1} \{|x|_M\};
\]
this is a norm because $M_1$ is a closed subspace of $M$.
Moreover, the infimum is always achieved in case (b).
Apply the induction hypothesis to choose a 
basis $m_{2,1},\dots,m_{n,1}$ of $M_1$
such that the supremum norm $|\cdot|'_{M_1}$ defined by $m_{2,1},\dots,m_{n,1}$
satisfies $c^{-1/3}|x_1|_{M_1} \leq |x_1|'_{M_1} \leq c^{1/3}|x_1|_{M_1}$
for all $x_1 \in M_1$.
For $i=2,\dots,n$, choose $m_i \in M$ lifting $m_{i,1}$ such that 
$|m_i|_M \leq c^{1/3} |m_{i,1}|_{M_1} \leq c^{2/3}$.

Let $|\cdot|'_M$ be the supremum norm defined by $m_1, \dots, m_n$.
For $a_1, \dots, a_n \in R'$, we have
\[
|a_1 m_1 + \cdots + a_n m_n|_M \leq \max_{1\leq i \leq n}
\{|a_i| |m_i|_M\} \leq c^{2/3} \leq c.
\]
On the other hand, if $m \in M$ satisfies $|m|_M \leq 1$, 
we can uniquely write
$m = a_1 m_1 + \cdots + a_n m_n$ with $a_i \in R$.
By definition of the quotient norm, $|m|_{M_1} \leq 1$, so
$|m|'_{M_1} \leq c^{1/3}$. In other words, $|a_2|, \dots, |a_n| \leq c^{1/3}$,
so 
\[
|a_2 m_2 + \cdots + a_n m_n|_M \leq \max_{2 \leq i \leq n}
\{|a_i| |m_i|_M\} \leq c^{1/3} c^{2/3} = c.
\]
Since $|m|_M \leq 1 \leq c$, we have $|a_1m_1|_M \leq c$. Since $|m_1|_M 
\geq 1$,
we have $|a_1| \leq c$. This proves the desired inequalities.
\end{proof}

\section{Application of Frobenius}

We now prove parts (a), (b), (c) of Theorem~\ref{T:subsidiary}
without any lower bound restriction on the values of $f_i$.
Again, it suffices to work in a neighborhood of $r=0$.

We first prove an analogue of Lemma~\ref{L:convex1}
in which $c_0$ can be replaced by any positive value. We will accomplish this
using Frobenius descendants; if we tried to use
Frobenius
antecedents instead, we would encounter trouble in the boundary case
$f_i(r) = 1/(p-1) \log p$ and in the case where
$f_1(r) > 1/(p-1) \log p$ but $f_i(r) < 1/(p-1) \log p$.

\begin{lemma} \label{L:convex2}
Fix anonnegative integer $j$, and 
fix $c_j > p^{-j}/(p-1) \log p$. Define
\[
f'_i(r) = \max\{f_i(r), c_j\}.
\]
\begin{enumerate}
\item[(a)] (Linearity)
For $i=1, \dots, n$, 
the functions $f'_i$ are continuous and piecewise affine.
\item[(b)] (Integrality)
If $i=n$ or $f'_i(r_0) > \max\{f'_{i+1}(r_0), r_0 + c_j\}$, then in a neigborhood of
$r_0$, 
\[
f'_1(r) + \cdots +f'_i(r) \in v(K^\times) + r \ZZ.
\]
Consequently, the slopes of each $f'_i$ belong to
$\frac{1}{1} \ZZ \cup \cdots \cup \frac{1}{n} \ZZ$.
\item[(c)] (Convexity)
For $i=1, \dots, n$,
the function $f'_1 + \cdots + f'_i$ is convex.
\end{enumerate}
\end{lemma}
\begin{proof}
We proceed by induction on $j$, the case $j=0$ being
Lemma~\ref{L:convex1}. 
Let $R'_1(\rho^p), \dots, R'_n(\rho^p)$ be the subsidiary radii of
$\varphi_* M \otimes F'_{\rho}$ in increasing order. 
(The normalization is chosen this way because the series variable
in $F'_\rho$ is $t^p$, which has norm $\rho^p$.)
Put
$g_i(r) = -\log R'_i(e^{-r})$. 
By the Frobenius descendant theorem,
the list $g_1(pr), \dots, g_{pn}(pr)$ consists of
\[
\bigcup_{i=1}^n \begin{cases}
\{p f_i(r), pr + \frac{p}{p-1} \log p \mbox{ ($p-1$ times)}
\} & f_i(r) \leq r + 1/(p-1) \log p \\
\{\log p + (p-1)r + f_i(r) \mbox{ ($p$ times)}
\} & f_i(r) \geq r + 1/(p-1) \log p.
\end{cases}
\]
Thus we may deduce (a) from the induction hypothesis.

To check (b) and (c), it suffices to handle cases where 
$i = n$ or $f_i(0) > \max\{f_{i+1}(0), c_j\}$.
(We may linearly interpolate to establish convexity in
the other cases.)
In these cases, we have either
$f_i(0) > 1/(p-1) \log p$, in which case
in some neighborhood of $r=0$ we have
\begin{equation} \label{eq:frob1}
g_1(pr) + \cdots + g_{pi}(pr)
= p(f_1(r) + \cdots + f_i(r)) + pi \log p + (p-1)ipr,
\end{equation}
or $f_{i+1}(0) < 1/(p-1) \log p$ or $i=n$, in which case
in some neighborhood of $r=0$ we have
\begin{equation} \label{eq:frob2}
g_1(pr) + \cdots + g_{pi+(p-1)(n-i)}(pr)
= p(f_1(r) + \cdots + f_i(r)) + pn \log p + (p-1)npr.
\end{equation}
Moreover, $f_i(0) > c_j$ if and only if $g_{pi}(0) 
> c_{j-1}$ for $c_{j-1} = p c_j$.

If $f_i(0) > 1/(p-1) \log p$, apply \eqref{eq:frob1} and the induction
hypothesis to write piecewise
\begin{align*}
f_1(r) + \cdots + f_i(r) &= p^{-1} (g_1(pr) + \cdots + g_{pi}(pr) +
pi \log p + (p-1) ipr) \\
&= p^{-1} (m(pr) + *) \\
&= mr + p^{-1} *
\end{align*}
for some $m \in \ZZ$.
(Note that $*$ is not guaranteed to be in $p \cdot v(K^\times)$;
this explains the example of \S \ref{sec:key}.)
If $f_i(0) \leq 1/(p-1) \log p$, then $f_{i+1}(0) < 1/(p-1) \log p$,
so we may apply \eqref{eq:frob2} to write piecewise
\begin{align*}
f_1(r) + \cdots + f_i(r) &= p^{-1} (g_1(pr) + \cdots + g_{pi+(p-1)(n-i)}(pr) 
+ pn \log p + (p-1)npr) \\
&= p^{-1} (m(pr) + *) \\
&= mr + p^{-1} *
\end{align*}
for some $m \in \ZZ$.
(Here it was important that the domains of applicability of
\eqref{eq:frob1} and \eqref{eq:frob2} overlap: if
$f_i(0) = 1/(p-1) \log p$, then \eqref{eq:frob1} may not remain applicable
when we move from $r=0$ to a nearby value.)
\end{proof}

To finish proving (a), (b), (c) of Theorem~\ref{T:subsidiary},
we must check them in a neighborhood of $0$ under the hypothesis that
$f_i(0) = 0$; note that (b) will follow immediately from (a) given that (a),(b),
(c) are now known in a neighborhood of any $r$ for which $f_i(r) > r$.

 We first check continuity. 
By Lemma~\ref{L:convex2}, 
for any $\epsilon > 0$, 
we can find $0 <\delta 
< \epsilon$
such that 
\[
|\max\{f_i(r), \epsilon/4\}| < \epsilon/2 \qquad (|r| < \delta).
\]
For such $r$, $-\epsilon < -\delta < f_i(r) < \epsilon$; this yields 
continuity.

We next check piecewise affinity by induction on $i$. Given that
$f_1, \dots, f_{i-1}$ are linear in a one-sided neighborhood of $r=0$,
say $[-\delta,0]$, and given $f_i(0) = 0$, it suffices to check linearity of 
$f_i(r) - r$ in some $[-\delta',0]$. From what we know already, in a neighborhood
of each $r \in [-\delta,0]$ where $f_i(r) - r > 0$, $f_i(r) - r$ is
convex and piecewise affine with slopes in $\frac{1}{n!} \ZZ$.
Note that none of these slopes can be nonnegative, otherwise 
$f_i(r) - r$ would thereafter be nondecreasing and could not have
limit 0 at $r=0$. By the same argument, if $f_i(r_0) - r_0 = 0$ 
for some $r_0 \in [-\delta,0)$, then the slope of $f_i(r) - r$
at any point $r \in (r_0,0)$ with $f_i(r) - r > 0$ must simultaneously
be positive and negative; since this cannot occur, we must have
$f_i(r) -r =0$ for all $r \in [r_0,0]$.

If $f_i(r)-r=0$ for some $r<0$, we are then done, as $f_i(r) - r$
is constant in a one-sided neighborhood of 0. Otherwise,
the slopes of $f_i(r) - r$ in $[-\delta,0)$ form a sequence
of discrete values which are negative and nondecreasing. This
sequence must then stabilize, so $f_i(r) - r$ is linear in a one-sided
neighborhood of 0.

We finally check convexity by induction on $i$. 
Given that $f_1 + \cdots + f_{i-1}$ is convex and that $f_i(0) = 0$,
it suffices to check that $f_i(r)-r$ is convex in a neighborhood of 0.
But we already know that $f_i(r) - r$ is continuous and piecewise affine near 
0; it must then have nonpositive left slope and nonnegative right slope,
and so must be convex near 0.

\section{Monotonicity}

We still must prove (d) of Theorem~\ref{T:subsidiary}. 
Note that we have the desired
statement as part of Lemma~\ref{L:convex1} but not
Lemma~\ref{L:convex2}; that is because the Frobenius descendant has
a pole at $t=0$, throwing off the bound on slopes. 
To fix this, we must use the off-centered Frobenius
descendant theorem.

\begin{lemma} \label{L:convex3}
If $\alpha = 0$ and 
$f_i(0) > 0$, then the slope in a right neighborhood of $r=0$ is nonpositive.
\end{lemma}
\begin{proof}
We proceed as in the proof of Lemma~\ref{L:convex2}, but using 
the off-centered Frobenius $\psi$ instead of $\varphi$.
Let $R''_1(\rho^p), \dots, R''_n(\rho^p)$ be the subsidiary radii of
$\psi_* M \otimes F''_{\rho}$ in increasing order. 
Put
$g_i(r) = -\log R''_i(e^{-r})$. 
By the Frobenius descendant theorem,
if $f_i(0) > 1/(p-1) \log p$, then
\[
g_1(pr) + \cdots + g_{pi}(pr)
= p(f_1(r) + \cdots + f_i(r)) + pi \log p,
\]
whereas if $f_{i+1}(0) \leq 1/(p-1) \log p$ or $i=n$, then
\[
g_1(pr) + \cdots + g_{pi+(p-1)(n-i)}(pr)
= p(f_1(r) + \cdots + f_i(r)) + pn \log p.
\]
Moreover, $f_i(0) > c_j$ if and only if $g_{pi}(0) 
> c_{j-1}$ for $c_{j-1} = p c_j$.
Again, we deduce the claim from the corresponding claim
about $\psi_* M$; this sets up an induction with base case treated
by Lemma~\ref{L:convex1}.
\end{proof}

\section{Subharmonicity}

It is also worth noting the following harmonicity result.
For $\overline{\mu} \in \kappa^{\alg}_K$,
let $\mu$ be a lift of $\overline{\mu}$
in some complete extension $L$ of $K$.
If $\alpha \leq 1 \leq \beta$,
let $T_\mu: K \langle \alpha/t,t/\beta \rangle \to
L \langle \alpha/t,t/\beta \rangle$ be the map
$t \mapsto t + \mu$.

Let $s_{\infty,i}$ 
be the left slope of $f_1(M,r) + \cdots + f_i(M, r)$ at $r=0$.
Let $s_{\mu,i}$ be the right slope of 
$f_1(T_\mu^* M, r) + \cdots + f_i(T_\mu^* M, r)$ at $r=0$.
Define the $i$-th \emph{discrepancy} of $M$ at $r=0$ to be the sum
\[
\disc_i(M, 0) = -\sum_{\overline{\mu} \neq 0} s_{\mu,i};
\]
it is always nonnegative by Theorem~\ref{T:subsidiary}(d).
(Note that if we extend $K$ to some larger field and consider $\overline{\mu}$
transcendental over the original residue field, then $s_{\mu,i} =0$;
that is, the definition of discrepancy is impervious to extending $K$.)
We may extend the definition to other values of $r$ by rescaling $t$.

\begin{theorem}[Subharmonicity] \label{T:harmonicity}
Assume that $\kappa_K$ is algebraically closed
and that $1 \in (\alpha, \beta)$.
Fix $i \in \{1,\dots,n\}$ such that $f_i(0) > 0$.
Then
\[
s_{0,i} - s_{\infty,i} \geq \disc_i(M,0),
\]
with equality if $i=n$.
\end{theorem}
\begin{proof}
By applying $\varphi_*$ as needed, we can force $f_i(r) > 1/(p-1) \log p$.
Then this follows from an argument analogous to
Lemma~\ref{L:convex1}, but with Lemma~\ref{L:harmonicity} below used in place
of Theorem~\ref{T:newton}.
\end{proof}
\begin{lemma} \label{L:harmonicity}
Assume that $\kappa_K$ is algebraically closed
and that $1 \in (\alpha, \beta)$.
For $f \in K \langle \alpha/t, t/\beta \rangle$ nonzero,
put $v_r(f) = -\log |f|_{e^{-r}}$ for $r \in [-\log \beta, -\log \alpha]$.
Let $s_\infty$ be the left slope of $v_r(f)$ at $r=0$.
Let $s_\mu$ be the right slope of $v_r(T_\mu(f))$ at $r=0$. Then
\[
s_\infty = \sum_\mu s_\mu.
\]
\end{lemma}
\begin{proof}
Without loss of generality, we may assume that $|f|_1 = 1$.
The quotient of
$\gotho_{F_1} \cap K \langle \alpha/t, t/\beta\rangle$ by the ideal
generated by $\gothm_F$
is isomorphic to $\kappa_K[t,t^{-1}]$; let 
$\overline{f}$ be the image of $f$ in this quotient.
Then $s_\mu$ is the order of vanishing of $f$ at $\mu$,
whereas $s_\infty$ is the negative of the order of vanishing 
of $f$ at $\infty$. This gives the desired equality.
\end{proof}

Since discrepancy is nonnegative, Theorem~\ref{T:harmonicity}
includes the convexity inequality $s_{\infty,i} \leq s_{0,i}$
from Theorem~\ref{T:subsidiary}(c). One can turn things around
to get the following corollary.

\begin{cor}
With notation as in Theorem~\ref{T:harmonicity}, if
$s_{\infty,i} = s_{0,i}$, then $s_{\mu,i} = 0$ for all $\overline{\mu} \neq 0$.
\end{cor}

We also have a corollary that says that
for a finite differential module over $K \langle \alpha/t,t/\beta \rangle$,
the generic radius of convergence can be computed at any point in all but
finitely many residue discs, not just in a generic residue disc.

\begin{cor} \label{C:almost all zero}
With notation as in Theorem~\ref{T:harmonicity}, 
$s_{\mu,i} = 0$ for all but finitely many $\overline{\mu}$.
\end{cor}
\begin{proof}
The $s_{\mu,i}$ lie in the discrete subgroup $\frac{1}{n!} \ZZ$ of $\RR$
and are nonpositive, so their sum can only be bounded below if all but
finitely many of them are zero.
\end{proof}

\section{Radius and generic radius}

We can now interpret the radius of convergence of a differential
module on a disc in terms of the function $f_1$.

\begin{prop} \label{P:true radius}
Let $M$ be a differential module over $K \langle t/\beta \rangle$ for
some $\beta > 0$. Then the radius of convergence of $M$ equals $e^{-r}$,
for $r$ the smallest value such that $f_1(r) = r$.
Consequently, $f(r') = r'$ for all $r' \geq r$.
\end{prop}
\begin{proof}
By a result from a previous unit, the radius of convergence of $M$
is at least the generic radius of convergence of $M \otimes F_{e^{-r}}$,
which by hypothesis equals $e^{-r}$.
On the other hand, if $\lambda > e^{-r}$, then by hypothesis 
$f_1(-\log \lambda) > -\log \lambda$, or in other words
$R(M \otimes F_{\lambda}) < \lambda$. This means that 
$M \otimes K \langle t /\lambda \rangle$ cannot be trivial, so
the radius of convergence cannot exceed $\lambda$. This proves the
desired result.
\end{proof}

\begin{cor}
Let $M$ be a differential module over $K \langle t/\beta \rangle$ for
some $\beta > 0$. Then the radius of convergence of $M$ belongs to the
divisible closure of the multiplicative value group of $K$.
\end{cor}
\begin{proof}
By Theorem~\ref{T:subsidiary} and Theorem~\ref{T:integrality},
the function $f_1(r)$ is piecewise of the form $ar+b$ with $a \in \QQ$
and $b$ in the divisible closure of the additive value group of $K$.
By Proposition~\ref{P:true radius}, the radius of convergence of $M$
equals $e^{-r}$ for $r$ the smallest value such that $f_1(r) = r$.
To the left of this $r$, $f_1$ must be piecewise affine with slope $\neq 1$;
by comparing the left and right limits at $r$, we deduce that $r = ar+b$
for some $a \neq 1$ rational and some $b$ in the divisible closure of the
additive value group of $K$. Since this gives $r = b/(a-1)$, we deduce
the claim.
\end{proof}

One should be able to better control the denominators, as in the following
question.
\begin{question}
Let $M$ be a differential module over $K \langle t/\beta \rangle$ for
some $\beta > 0$. Does there necessarily exist
$j \in \{1,\dots,\rank(M)\}$ such that 
the $j$-th power of the radius of convergence of $M$
belongs to the $p$-divisible closure of the multiplicative value group of $K$?
\end{question}

We also have a criterion for when the radius of convergence equals the
generic radius.
\begin{cor}
Let $M$ be a differential module over $K \langle t/\beta \rangle$ for
some $\beta > 0$, such that for some $\alpha \in (0, \beta)$,
$R(M \otimes F_\rho)$ is constant for $\rho \in [\alpha,\beta]$. Then
$R(M) = R(M \otimes F_\rho)$. (A similar statement holds for the product
of the first $i$ subsidiary radii, for $i=1,\dots,\rank(M)$.)
\end{cor}
\section{Subsidiary radii as radii of convergence}

The generic radii of subsidiary convergence can be interpreted
as the radii of convergence of a well-chosen basis of local horizontal
sections at a generic point.

\begin{theorem}[after Young] \label{T:Young}
Let $(V,D)$ be a differential module over $F_\rho$ of dimension $n$
with subsidiary radii $s_1 \leq \cdots \leq s_n$.
Choose a basis $e_1,\dots, e_n$
of local horizontal sections of $V$ at a generic point
$\eta$. For $i=1,\dots,n$, let $\rho_i$ be the radius of convergence of 
$e_i$, and suppose that $\rho_1 \leq \dots \leq \rho_n$. Then
$\rho_i \leq s_i$ for $i=1,\dots,n$;
moreover, there exists a choice of basis for which $\rho_i = s_i$ for 
$i=1,\dots,n$.
\end{theorem}
\begin{proof}
We first produce a basis for which $\rho_i = s_i$ for $i=1, \dots, n$.
For this, we may apply the strong decomposition theorem to decompose
$V$ into components each with a single subsidiary radius, and thus
reduce to the case $s_1 = \cdots = s_n = s$.
By the geometric interpretation of generic radius, each Jordan-H\"older
constituent of $V$ admits a basis of local horizontal sections on
a generic disc of radius $s$. By a prior lemma, the same is true
for $V$ itself.

For the remaining inequality, we induct on $n$.
Let $m$ be the largest
integer such that $s_1 = s_m$. 
Let $V_1$ be the component of $V$
of subsidiary radius $s_1$, so that $\dim V_1 = m$. We will check that 
no local horizontal section of $V_1$ at a generic
point $\eta$ can have radius of convergence
strictly greater than $s_1$.

Suppose the contrary; then there would exist 
a local horizontal section of $V_1$
at $\eta$ which converges on a closed disc of radius $\lambda$ for
some $\lambda \in (s_1, \rho)$.
This would mean that $V_1 \otimes L \langle (t-\eta)/\lambda \rangle$
would have a trivial submodule, and so would have $\lambda$ as 
one of its subsidiary radii.
However, 
by arguing as in Theorem~\ref{T:harmonicity},
we see that the product of the subsidiary  radii of
$V_1 \otimes L \langle (t-\eta)/\lambda \rangle$ is equal to $s_1^m$
for $\lambda$ slightly smaller than $\rho$;
by Theorem~\ref{T:subsidiary}(c) and (d),
the equality holds for all $\lambda \in (s_1, \rho)$. This yields
a contradiction.

We conclude that any local horizontal section of $V$ that projects
nontrivially onto $V_1$ has radius strictly greater than $s_1$.
We can divide the given basis into $m$ sections that project
onto a basis of $V_1$, and $n-m$ sections that project onto a basis
of the complementary component. The first $m$ sections have radius
of convergence at most $s_1$ by above; the others have radii of 
convergence bounded by $s_{m+1},\dots,s_n$ by the induction hypothesis.
This yields the desired result.
\end{proof}
A basis of local horizontal sections for which $\rho_i = s_i$ for 
$i=1,\dots,n$ is sometimes called an \emph{optimal basis}.

\notes

For $f_1$, Christol and Dwork established convexity
[CD04, Proposition~2.4] (using essentially the same short proof
given here) and continuity at endpoints
[CD04, Th\'eor\`eme~2.5]. All other results in this unit are original.

We again remind the reader that subharmonicity, as a property of
suitable functions on Berkovich spaces, is addressed in the
work of Thuillier [Thu05].

When restricted to intrinsic subsidiary radii less than $p^{-1/(p-1)}$,
Theorem~\ref{T:Young} is a result of Young [You92, Theorem~3.1].
Young's proof is an explicit calculation using twisted polynomials;
it was limited to small radii because the Frobenius antecedent theorem
was not available at the time.

\exercises

\begin{enumerate}
\item
Given an example to show that in Theorem~\ref{T:newton}, $f_2$
need not be concave (even though $f_1$ and $f_2$ are concave).
\item
Here is a result of Dwork related to the example  in \S~\ref{sec:key}.
Suppose $\pi \in K$ satisfies $\pi^{p-1} = -p$. Prove that the power series
$E(t) = \exp(\pi t - \pi t^p)$ has radius of convergence strictly
greater than 1.
(By contrast, the series $\exp (\pi t)$ has radius of convergence $1$.)
Optional: prove that the radius of convergence is equal to $p^{(p-1)/p^2}$.
\item
Prove that if $K$ is discretely valued, then $\gotho_K \langle t \rangle$
is noetherian. It isn't otherwise, because then $\gotho_K$ itself is not
noetherian.
\item
Prove that each maximal ideal of $\gotho_K \langle t \rangle$ is generated by
$\gothm_K$ together with some $P \in \gotho_K[t]$
whose reduction modulo $\gothm_K$ is irreducible in
$\kappa_K[t]$.
\end{enumerate}

\end{document}

For this calculation, it suffices
to check everything in aonsider a finite free differential module $M$ of rank $n$ over
$R = K \langle \alpha/t, t/\beta \rangle$, and to check everything
in a neighborhood of some $\rho \in [\alpha, \beta]$. (Note that the
neighborhood is only one-sided if $\rho =\alpha$ or $\rho = \beta$.)

We first note that all quantities involved are invariant under enlarging $K$.
We may thus assume that the value group of $K$ is all of $\RR^+$.
By rescaling $t$, we may reduce to the case $\rho = 1$.

Let us first see what we get from characteristic polynomials. First, we
need a lemma left over from the numerical analysis unit. 

\begin{lemma} \label{L:change HN}
Let $F$ be a complete nonarchimedean field. Let $A$ be an $n \times n$
matrix over $F$ with Hodge slopes $s_{H,1}, \dots, s_{H,n}$ and Newton slopes
$s_{N,1},\dots, s_{N,n}$. Assume that $s_{N,1},\dots, s_{N,n}$ belong to
the additive value group of $F$.
Then there exists $U \in \GL_n(F)$ such that 
the Hodge and Newton slopes of $U^{-1} A U$ coincide,
$|U| \leq 1$,
and $|U^{-1}| \leq f$ for some continuous function
$f$ of the $s_{H,i}$ and $s_{N,i}$, which is equal to $1$ in case
$s_{H,i} = s_{N,i}$ for all $i$.
\end{lemma}
It would be an interesting problem to optimize the bound $f$; we have
not attempted to do so.
\begin{proof}
We argue by induction on $n$.
Let $m$ be the largest integer such that $s_{N,1} = s_{N,m}$.
By a result from the numerical analysis unit, we can choose $U_1 \in \GL_n(F)$
such that the first $m$ Hodge and Newton slopes of 
$U_1^{-1} A U_1$ coincide,
$|U| \leq 1$, and 
\[
|U^{-1}| \leq e^{(n-1) (s_{N,1} - s_{H,1})}.
\]
By the Hodge-Newton decomposition, we can choose $U_2 \in \GL_n(\gotho_F)$
such that $(U_1U_2)^{-1} A (U_1U_2)$ is block diagonal with the first
$m$ Hodge and Newton slopes occurring in the first block.
We now invoke the induction hypothesis to conclude.
\end{proof}