see that $h$ has no pole in $d(0, R^{-})$. So, in both cases, we have made $f$ in the form $\frac{h}{l}$ with $h, l \in \mathcal{A}(\mathbb{K})$, where $l$ has no zero in $d(0, r^{-})$. The proof is then similar to the case $f \in \mathcal{A}(\mathbb{K})$. $\square$

**Corollary 36.16.** Let $f \in \mathcal{M}(\mathbb{K})\setminus\mathcal{A}(\mathbb{K})$ (respectively $f \in \mathcal{M}(d(0, R^{-}))\setminus\mathcal{A}(d(0, R^{-}))$). If 0 is not a pole, $f(x)$ has a development in a power series whose radius of convergence is the minimal distance of poles of $f$ to 0. If 0 is a pole of order $q$ of $f$, then $f(x)$ has a development in a Laurent series $\sum_{k=-q}^{+\infty} a_k x^k$ with $a_{-q} \neq 0$ and the radius of convergence of the series $\sum_{k=0}^{\infty} a_k x^k$ is equal to the minimal distance of non-zero poles of $f$ to 0.

**Theorem 36.17.** Let $f \in \mathcal{M}(d(0, R^{-}))$ have no pole in an annulus $\Gamma(0, r, s)$ with $s < R$. Then $f(x)$ is equal to a Laurent series $\sum_{-\infty}^{+\infty} a_n x^n$ converging in all $\Gamma(0, r, s)$. For each $\mu \in [\log r, \log s]$, if $f$ has $q$ zeros and $t$ poles in $d(0, \theta^\mu)$ taking multiplicity into account, one has $\nu^+(f, \mu) = q - t$, and if $f$ has $q'$ zeros and $t'$ poles in $d(0, (\theta^\mu)^{-})$, one has $\nu^-(f, \mu) = q' - t'$. Then the functions in $\rho: \nu^+(f, \log \rho)$, $\nu^-(f, \log \rho)$ and $|f|(\rho)$ are increasing. Let $k = \nu^+(f, \log r)$. Then $|f|(\rho) \ge |a_k|\rho^k \forall \rho \in [r, s]$.

**Proof.** Since $f$ belongs to $\mathcal{M}(d(0, R^{-}))$ and since $s < R$, $f$ has finitely many zeros and poles in $d(0, s)$, hence we can write it $\frac{h}{l}$ with $h, l \in \mathcal{A}(d(0, R^{-}))$ having no common zero in $d(0, s)$. Since $f$ has no pole in $\Gamma(0, r, s)$, $l$ has no zero in $\Gamma(0, r, s)$. Let $\rho = \theta^\mu$. By Corollary 19.3, we have $\nu^+(f, \mu) = \nu^+(h, \mu) - \nu^+(l, \mu) = q - t$. Similarly, in $d(0, \rho^{-})$, we have $\nu^-(f, \mu) = \nu^-(h, \mu) - \nu^-(l, \mu) = q' - t'$.

We can write $f(x)$ in the form $\frac{h(x)}{Q(x)}$ with $h \in \mathcal{A}(d(0, s^{-}))$ and $Q \in \mathbb{K}[x]$. Then $Q$ has no zero in $\Gamma(0, r, s)$ and hence $\nu^+(Q, \mu)$ is constant in $[\log r, \log s]$. On the other hand, $\nu^+(h, \mu)$ is increasing hence so is $\nu^+(f, \mu)$. Consequently, the function $|f|(\rho)$ is increasing. Therefore, $|f|(\rho) \ge |a_k|\rho^k \forall \rho \in [r, s]$. $\square$

**Corollary 36.18.** Let $f \in \mathcal{M}(d(0, R^{-}))$ have no pole in $\Gamma(0, r, s)$, with $0 < r < s < R$ and let $q = \nu^-(f, \log s)$, $k = \nu^+(f, \log r)$. Then,

$$ \left(\frac{s}{r}\right)^k \le \frac{|f|(s)}{|f|(r)} \le \left(\frac{s}{r}\right)^q. $$

**Corollary 36.19.** Let $f \in \mathcal{M}(\mathbb{K})\setminus\mathbb{K}(x)$ have finitely many poles. For every $q \in \mathbb{N}$, $f$ satisfies $\lim_{r \to \infty} \frac{|f|(r)}{r^q} = +\infty$.
**Proof.** Let $f = \frac{h}{Q}$ with $Q$ a monic polynomial and $h \in \mathcal{A}(\mathbb{K})$. Since $f \notin \mathbb{K}(x)$, $h$ does not lie in $\mathbb{K}[x]$ hence it has infinitely many zeros and therefore infinitely many terms $a_n \neq 0$ when $n > 0$. $\square$

We also need the following lemma in the future:

**Lemma 36.20.** Let $f \in \mathcal{M}(\mathbb{K})$ be transcendental and have finitely many poles and let $P$ be a polynomial. There exists $s > 0$ such that $|f+P|(r) = |f|(r) \quad \forall r \ge s$ and then $f$ has the same number of zeros as $f+P$ in $d(0, r)$.

**Proof.** Let $R \in \mathbb{R}_+^*$ be such that all poles of $f$ and all zeros of $P$ lie in $d(0, R)$, let $q$ be the number of poles of $f$ and let $t = \deg(P)$. Then $f(x)$ is of the form $\frac{g(x)}{\prod_{j=1}^q (x-b_j)}$ with $g(x)$ of the form $\sum_{n=0}^\infty a_n x^n$. Now, when $r > R$, we have $|f|(r) = \frac{|g|(r)}{r^q}$ and $|P|(r) = \frac{|P|(R)}{R^t}r^t$. Consequently, by Theorem 22.22, $|f|(r)$ gets bigger than $|P|(r)$ when $r$ is big enough and hence there exists $s > R$ such that $|P|(r) < |f|(r) \quad \forall r > s$. But then, by Corollary 22.23, we have $\nu(f, \log r) = \nu(f+P, \log r) \quad \forall r > s$ and hence $f$ and $f+P$ have the same number of zeros in $d(0, r)$. $\square$

**Definitions.** Let $f \in \mathcal{M}(\mathbb{K})$ (respectively $f \in \mathcal{M}_u(d(a, R^{-}))$) and let $b \in \mathbb{K}$. Then $b$ is said to be an exceptional value for $f$ or a Picard value if $f-b$ has no zero in $\mathbb{K}$ (respectively in $d(a, R^{-})$) and $b$ is said to be a pseudo-exceptional value for $f$ if $\lim_{r \to \infty} |f-b|(r) = 0$ (respectively $\lim_{r \to R^{-}} |f-b|(r) = 0$). Moreover, if $f \in \mathcal{M}(\mathbb{K}) \setminus \mathbb{K}(x)$ (respectively if $f \in \mathcal{M}_u(d(a, R^{-}))$), $b$ is said to be a quasi-exceptional value for $f$ if $f-b$ has finitely many zeros in $\mathbb{K}$ (respectively in $d(a, R^{-})$) [49], [52], [81].

**Theorem 36.21.** Let $f \in \mathcal{M}(\mathbb{K}) \setminus \mathbb{K}$ (respectively $f \in \mathcal{M}_u(d(a, R^{-}))$). If $b$ is an exceptional value for $f$, then it is a pseudo-exceptional value for $f$. Let $f \in \mathcal{M}(\mathbb{K}) \setminus \mathbb{K}(x)$ (respectively $f \in \mathcal{M}_u(d(a, R^{-}))$. If $b$ is a quasi-exceptional value for $f$, then it is a pseudo-exceptional value for $f$.

**Proof.** Without loss of generality, we may assume that $a = b = 0$. Suppose first that $f \in \mathcal{M}(\mathbb{K}) \setminus \mathbb{K}$ and that $0$ is an exceptional value for $f$. So, $\frac{1}{f}$ has no pole in $\mathbb{K}$ (respectively in $d(0, R^{-})$), hence it is a function $h \in \mathcal{A}(\mathbb{K}) \setminus \mathbb{K}$ (respectively $h \in \mathcal{A}_u(d(0, R^{-}))$) so that $f = \frac{1}{h}$.
Then, by Corollary 14.18 (respectively by Theorem 14.20), we have $\lim_{r \to +\infty} |h|(r) = +\infty$ (respectively $\lim_{r \to R^{-}} |h|(r) = +\infty$).

Suppose now that $f \in \mathcal{M}(K)\K(x)$ and that $0$ is a quasi-exceptional value for $f$. Then $f$ is of the form $\frac{P(x)}{h(x)}$ with $P \in \K[x]$ and $h \in \mathcal{A}(K)\K(x)$. By Corollary 14.7, we have $\lim_{r \to +\infty} \frac{|P(r)|}{|h(r)|} = 0$, which proves the claim when $f \in \mathcal{M}(K)\K(x)$.

Next, suppose that $f \in \mathcal{M}_u(d(0, R^{-}))$ admits $0$ as a quasi-exceptional value. Then $f$ is of the form $\frac{P(x)}{h(x)}$ with $P \in \K[x]$ and $h \in \mathcal{A}_u(d(0, R^{-}))$. So $P$ is bounded in $d(0, R^{-})$, hence of course $\lim_{r \to +\infty} \frac{|P(r)|}{|h(r)|} = 0$, which ends the proof. $\square$

**Theorem 36.22.** Let $f \in \mathcal{M}(K)\K$ (respectively $f \in \mathcal{M}_u(d(a, R^{-}))$). Then $f$ admits at most one pseudo-exceptional value. Moreover, if $f \in \mathcal{A}(K)\K$ (respectively $f \in \mathcal{A}(d(a, R^{-}))$), then $f$ has no pseudo-exceptional value.

**Proof.** Suppose that $b$ is a pseudo-exceptional value for $f$. Without loss of generality, we may assume that $a = b = 0$. Let $t \in K^*$. Since $\lim_{r \to +\infty} |f|(r) = 0$ (respectively $\lim_{r \to R^{-}} |f|(r) = 0$), it is obvious that $\lim_{r \to +\infty} |f - t|(r) = |t|$ (respectively $\lim_{r \to R^{-}} |f - t|(r) = |t|$), so $t$ is not a pseudo-exceptional value for $f$.

Now, suppose $f \in \mathcal{A}(K)\K$. Since $\lim_{r \to +\infty} |f|(r) = +\infty$, of course, $0$ is not a pseudo-exceptional value of $f$. Finally, suppose $\mathcal{A}(d(0, R^{-}))$. Then if $f \in \mathcal{A}_u(d(0, R^{-}))$, we have $\lim_{r \to R} |f|(r) = +\infty$, hence $0$ is not a pseudo-exceptional value of $f$. And if $f \in \mathcal{A}_b(d(0, R^{-}))$, we have $\lim_{r \to R} |f|(r) = \|f\|_{d(0, R^{-})}$ which is not $0$, hence $0$ is not a pseudo-exceptional value of $f$ either. $\square$

**Corollary 36.23.** Let $f \in \mathcal{M}(K)\K$ (respectively $f \in \mathcal{M}_u(d(a, R^{-}))$). Then $f$ admits at most one exceptional value. Moreover, if $f \in \mathcal{M}(K)\K(x)$ (respectively $f \in \mathcal{M}_u(d(a, R^{-}))$), then $f$ admits at most one quasi-exceptional value. Further, if $f \in \mathcal{A}(K)\K$ (respectively if $f \in \mathcal{A}_u(d(a, R^{-}))$, then $f$ admits no exceptional value. And if $f \in \mathcal{A}(K)\K[x]$ (respectively if $f \in \mathcal{A}_u(d(a, R^{-}))$, then $f$ admits no quasi-exceptional value.
Chapter 37

Residues of Meromorphic Functions

Throughout this chapter, D is infraconnected, T is a hole of D, and
V is a disk of the form d(a, r) or d(a, r⁻), included in D̃, such that
Ṽ ∩ D ≠ ∅.

**Definition and notation.** Let $f \in \mathcal{M}(\mathbb{K})$ (respectively $f \in \mathcal{M}(d(0, R^{-}))$) have a pole $\alpha$ of order $q$ and let $f(x) = \sum_{k=-q}^{-1} a_k(x-\alpha)^k + h(x)$ with $a_{-q} \neq 0$ and $h \in \mathcal{M}(\mathbb{K})$ (respectively $f \in \mathcal{M}(d(0, R^{-}))$) and $h$ holomorphic at $\alpha$. Accordingly to previous notations for analytic elements in Chapters 11 and 15, the coefficient $a_{-1}$ is called residue of $f$ at $\alpha$ and denoted by $\text{res}(f, \alpha)$.

We can now compare residues on a hole defined for analytic ele-
ments and residues at a point, we just defined for a meromorphic
function:

**Theorem 37.1.** Let $a \in \mathbb{K}$, let $R \in \mathbb{R}^*_+$, let $f \in \mathcal{M}(d(a, R^{-}))$, and let $r \in ]0, R[$. Let $\alpha_j$, $1 \le j \le q$ be the poles of $f$ in $d(a, r)$, let $\rho \in ]0, \min_{i \ne j} |\alpha_i - \alpha_j|[ \text{ and for each } j = 1, \dots, q$, let $\rho_j \in ]0, \rho[$, and let $T_j = d(\alpha_j, \rho_j^{-})$. Let $D = d(\alpha, r) \setminus (\bigcup_{j=1}^q T_j)$. Then $f$ belongs to $H(D)$ and $\text{res}(f, \alpha_j) = \text{res}(f, T_j)$, $j = 1, \dots, q$.

**Proof.** By Corollary 36.2, $f$ belongs to $H(D)$. On the other hand, assuming that $\alpha_j$ is a pole of order $s_j$, by Corollary 36.16, $f(x)$ has a development at $\alpha_j$ in a Laurent series $\sum_{m=-s_j}^{\infty} b_{m,j}(x-\alpha_j)^m$. Consequently, by Theorem 15.1, the Mittag-Leffler term of $f$ on $T_j$ with respect to the infraconnected set $D$ is $\sum_{m=-s_j}^{-1} b_{m,j}(x-\alpha_j)^m$. Then $\text{res}(f, T_j) = b_{-1,j} = \text{res}(f, \alpha_j)$, which ends the proof. $\square$
**Corollary 37.2.** Let $f \in H_b(D)$ be meromorphic in $T = d(b, r^-)$ and admit only one pole $b$ inside $T$. Let $q$ be the multiplicity order of $b$. Then the Mittag-Leffler term of $f$ associated with $T$ is of the form $\sum_{j=1}^q \frac{a_j}{(x-b)^j}$, with $a_q \neq 0$ and also is of the form $\frac{P}{(x-a_j)^q}$, where $P$ is a polynomial of degree $s < q$. Moreover, it does not depend on $r$ when $r$ tends to 0.

**Definition.** Let $f \in \mathcal{M}(d(a, R^{-}))$, let $b$ be a pole of order $t$ of $f$, let $r > 0$ be such that $d(b, r^{-})$ contains no pole of $f$ other than $b$, and let $\frac{P(x)}{(x-b)^t}$ be the Mittag-Leffler term of $f$ associated with $d(b, r^{-})$. Then $\frac{P(x)}{(x-b)^t}$ is called the singular part of $f$ at $b$.

Thus, when we consider the Laurent development of $f$ around $b$, it is of the form $\sum_{-t}^{\infty} \frac{a_n}{(x-b)^n}$ and the singular part of $f$ at the pole $b$ is $\sum_{-t}^{-1} \frac{a_n}{(x-b)^n}$.

An element $f \in H(D)$ is said to be meromorphic in $V$ if there exist finitely many points $(a_i)_{(1 \le i \le n)}$ in $V$ such that $f$ has continuation to an element of $H((D \cup V)\setminus\{a_i| 1 \le i \le n\})$.

Let $f$ be meromorphic in $V$ and belong to $H((D \cup V)\setminus\{a_i| 1 \le i \le n\})$. For each $i = 1, \dots, n$, if $f \notin H((D \cup V)\setminus\{a_h| h \neq i\})$, then by Corollary 11.9 and Theorem 11.10, $a_i$ is a pole of $f$ as an element of $H((D \cup V)\setminus\{a_j| 1 \le j \le n\})$. Let $q_i$ be its order. Then $a_i$ is called a pole of $f$ of order $q_i$ in $V$. The polynomial $P(x) = \prod_{i=1}^n (x - a_i)^{q_i}$ is called the polynomial of the poles of $f$ in $V$.

**Lemma 37.3.** Let $D$ be bounded or belong to Alg and let $f \in H(D)$. If $f$ is meromorphic in $T$, the polynomial of its poles $P$ in $T$ satisfies $Pf \in H(D \cup T)$.

**Proof.** Indeed, let $D' = (D \cup T)\setminus\{a_1, \dots, a_n\}$. If $D$ is bounded, then so is $D'$, and therefore by Theorem 11.4, $Pf$ belongs to $H(D')$. But by construction $Pf$ is bounded at each point $a_i$, and therefore by Corollary 11.6, $Pf$ belongs to $H(D \cup T)$.

Now, suppose $D \in \text{Alg}$. Then by Theorem 17.9, $D'$ belongs to Alg and therefore $Pf$ belongs to $H(D')$, so we have the same conclusion. $\square$

**Lemma 37.4.** Let $D$ be bounded (respectively let $D \in \text{Alg}$) and let $f$ be invertible in $H(D)$. Then $f$ is meromorphic in $T$ if and only if so is $\frac{1}{f}$.
**Proof.** First, we suppose $f$ meromorphic in $T$. Let $P$ be the polynomial of its poles in $T$ and let $g(x) = f(x)P(x)$. Since $D$ is bounded (respectively belongs to Alg) by Lemma 37.3, $g$ belongs to $H(D \cup T)$. Let $Q$ be the polynomial of the zeros of $g$ in $T$. Since $f$ has no zero in $D$, $Q$ actually is the polynomial of the zeros of $g$ in $D \cup T$ and then $g$ is of the form $Q(x)h(x)$ with $h$ an element of $H(D \cup T)$ that has no zero in $T$. Hence, we have $\frac{1}{h} = \frac{1}{f} \frac{Q}{P}$. If $D \in \text{Alg}$, $\frac{1}{h}$ obviously belongs to $H(D)$. If $D$ is bounded, we have $\frac{Q}{P} \in R_b(D)$, and then by Theorem 11.4, $\frac{1}{h}$ belongs to $H(D)$. Thus, $\frac{1}{h}$ belongs to $H(D)$ anyway. Now, by Lemma 24.1, $h$ is invertible in $H(D \cup T)$. Hence, $f$ factorizes in the form $\frac{P}{Q}h$ and then $\frac{1}{f} = \frac{1}{Q} \frac{P}{h}$. But $\frac{P}{h}$ belongs to $H(D \cup T)$ and therefore $\frac{1}{f}$ is meromorphic in $T$ and admits $Q$ as the polynomial of its poles in $T$. We may obviously apply the same reasoning to $\frac{1}{f}$ and this shows the converse. $\square$

**Theorem 37.5.** Let $F \in H(D)$ be meromorphic in $T$ and satisfy $\|F - 1\|_D < 1$. Then $F$ has as many poles as many zeros in $T$.

**Proof.** Without loss of generality, we may obviously assume that $D$ is bounded because the hypothesis remains true in any set $D \cap d(0, R)$. We may also assume that $T = d(0, r^{-})$. Let $P$ (respectively $Q$) be the polynomial of the zeros (respectively the poles) of $F$ in $T$. Then by Lemma 37.3, $F$ factorizes in the form $\frac{f}{Q}$ with $f \in H(D \cup T)$. Now, the zeros of $f$ inside $T$ are just those of $F$, hence $f$ factorizes in the form $Pg$ with $g \in H(D \cup T)$, $g$ having no zero in $T$. Let $h = \frac{P}{Q}$. Since $g$ has no zero in $T$, by Theorem 22.14, $|g(x)|$ is equal to a constant inside $T$. Let $s \in ]0, r[$ be such that all zeros of $P$ and of $Q$ lie in $d(0, s)$. Then obviously $F$ belongs to $H(\Gamma(0, s, r))$. Now, by hypothesis, there exists $\lambda > 0$ such that $\Psi(F, \mu) \le \lambda$ for all $\mu \ge \log r$. Hence by continuity, there exists $\lambda' > 0$ and $s'$ in $]s, r[$ such that $\Psi(F, \mu) \le \lambda'$ for all $\mu \ge \log s'$. Thus, there exists $b \in ]0, 1[$ and $t$ in $]s', r[$ such that $|F(x) - 1| \le b$ for all $x \in \Gamma(0, t, r)$ and then, $|h(x)|$ is constant in $\Gamma(0, t, r)$. Hence, we have $\frac{d}{d\mu}\Psi(h, \mu) = 0$ for all $\mu \in [\log t, \log r]$. Since $h$ has neither any zero nor any pole in $C(0, t)$, by Corollary 4.17, $h$ has as many zeros as many poles in $d(0, t)$ and therefore in $d(0, r)$ and this ends the proof. $\square$

It is useful to consider again elements meromorphic at a point.
**Lemma 37.6.** Let $a \in \bar{D}$ and let $f \in H(D)$ be meromorphic but not holomorphic at $a$. Then $a$ is a pole of $f$.

**Proof.** By hypothesis, there exists $r > 0$ such that $f$ belongs to $H(D \cup (d(a,r)\setminus\{a\}))$. Suppose that $a$ is not a pole of $f$. Then by Theorem 36.4, $f$ belongs to $H(D \cup d(a,r))$ and then $f$ is holomorphic at $a$. $\square$

**Corollary 37.7.** Let $a \in \bar{D}$ and let $f \in H(D)$. Then $f$ is meromorphic at $a$ and admits $a$ as a pole of order $q$ if and only if there exists a disk $d(a,r)$ included in $\bar{D}$ and an element $h \in H(d(a,r))$ such that $f(x)(x-a)^q = h(x)$ whenever $x \in d(a,r)\setminus\{a\}$ and $h(a) \neq 0$.

**Corollary 37.8.** Let $D$ satisfy Condition (B) and let $f \in H(D)$. For every $a \in \bar{D}$, $f$ is meromorphic at $a$. For every $a \in D$, $f$ is holomorphic at $a$.

**Remark.** Let $a \in \bar{D} \setminus \hat{D}$ and let $f$ admit $a$ as a pole of order $q$. This does not imply that $f$ is meromorphic at $a$. Indeed, by [44], we know that there exist infraconnected sets $E$ with a point $a \in E \setminus \hat{E}$ and elements $h \in H(E)$ such that $\lim_{|x-a|\to 0} h(x) = 0$ and such that $\limsup_{|x-a|\to 0} \frac{|h(x)|}{|x-a|} = +\infty$. Let $E' = E \setminus \{a\}$ and let $g = \frac{1+h}{x-a}$. It is easily seen that $a$ is a pole of order 1 for $g$. But $h$ does not belong to any space $H(d(a,r))$, whenever $r > 0$ (because if it did, it should factorize in $H(d(a,r))$ in the form $(x-a)\ell(x)$, with $\ell \in H(d(a,r))$). Thus, it is seen that $(x-a)g$ does not belong to $H(E)$ and therefore $g$ is not meromorphic at $a$.

Concerning the derivation, Theorem 37.9 is easy and follows the classical rules:

**Theorem 37.9.** Let $f \in \mathcal{M}(\mathbb{K})$ (respectively $f \in \mathcal{M}(d(a, R^-)$). For each $\alpha \in \mathbb{K}$ (respectively $\alpha \in d(a, R^-)$) such that $f$ is holomorphic at $\alpha$, $f$ has a derivative $f'(\alpha)$ at $\alpha$. Further, given a point $\beta \in \mathbb{K}$ (respectively $\beta \in d(a, R^-)$) and the Laurent development of $f$ at $\beta$: $\sum_{k=-q}^{\infty} a_k (x-\beta)^k$ with $a_{-q} \neq 0$, the development of $f'$ at $\beta$ is $\sum_{k=-q}^{0} ka_k(x-\beta)^{k-1} + \sum_{k=1}^{\infty} ka_k(x-\beta)^{k-1}$.

**Proof.** Suppose first $f$ is holomorphic at $\alpha$. By Theorem 36.4, $f(x)$ is equal to a power series $\sum_{k=0}^{\infty} a_k (x-\alpha)^k$ converging inside a disk $d(\alpha, r^{-})$, where $r$ is the minimal distance from $\alpha$ to the various poles.
Then by Theorem 18.1, we know that $f$ has a derivative whose development is obtained by deriving term by term.

Suppose now that $\beta$ is a pole of order $q$ and let $r \in ]0, R[$ be strictly inferior to the minimal distance from $\beta$ to the other poles (with just $r < R$ if $\beta$ is the unique pole of $f$). By Lemma 36.1, for every $\rho \in ]0, r[$, $f$ belongs to $H(d(\beta, r)\setminus d(\beta, \rho^{-}))$ and the Laurent development of $f$ at $\beta$ is its Mittag-Leffler development as an element of $H(d(\beta, r)\setminus d(\beta, \rho^{-}))$: the Mittag-Leffler term associated with the hole $d(\beta, \rho^{-})$ is $\sum_{k=-q}^{-1} a_k(x - \beta)^k$ with $a_{-q} \neq 0$ and the term associated with $d(\beta, r)$ is $\sum_{k=0}^{\infty} a_k(x - \beta)^k$ with $a_{-q} \neq 0$. Consequently, by Theorem 18.19, the derivative has a Mittag-Leffler development at $\beta$: $\sum_{k=-q}^{-1} ka_k(x - \beta)^{k-1} + \sum_{k=1}^{\infty} ka_k(x - \beta)^{k-1}$. This is true for all $r \in ]0, R[$ strictly inferior to the minimal distance from $\beta$ to the other poles and for every $\rho \in ]0, r[$, which ends the proof. $\square$

Theorem 37.10 is an improvement of the classical upper bound $f'$ in function of $f$. That is due to J.P. Bézivin [10].

**Theorem 37.10.** For each $n \in \mathbb{N}$ and for all $r \in ]0, R[$, we have

$$|f^{(n)}|(r) \le |n!| \frac{|f|(r)}{r^n}.$$

Moreover, given $r \in ]0, R[$ such that $\nu^+(f, \log r) = \nu^-(f, \log r)$, if the residue characteristic $p$ does not divide $\nu(f, \log r)$, then $\nu(f', \log r) = \nu(f, \log r) - 1$ and

$$|f'|(r) = \frac{|f|(r)}{r}.$$

**Proof.** When $f \in \mathcal{A}(d(0, R^{-}))$, this is shown at Corollary 18.16. Now, consider the general case and set $f = \frac{U}{V}$ with $U, V \in \mathcal{A}(d(0, R^{-}))$. The stated inequality is trivial when $q = 1$. So, we assume it holds for $q \le n - 1$ and consider $f^{(n)}$. Writing $U = V(\frac{U}{V})$, by Leibnitz theorem, we have

$$U^{(n)} = \sum_{q=0}^{n} \binom{n}{q} V^{(n-q)} \left(\frac{U}{V}\right)^{(q)},$$

and hence

$$V\left(\frac{U}{V}\right)^{(n)} = U^{(n)} - \sum_{q=0}^{n-1} \binom{n}{q} V^{(n-q)} \left(\frac{U}{V}\right)^{(q)}.$$
Now, by Corollary 18.16, we have

$$ (1), \qquad |U^{(n)}|(R) \le |n!| \frac{|U|(R)}{R^n} $$

and for each $q \le n - 1$, we have $|V^{(n-q)}|(R) \le |(n-q)!| \frac{|V|(R)}{R^{n-q}}$

and $|(\frac{U}{V})^{(q)}|(R) \le |q!| \frac{|U|(R)}{|V|(R)R^q}$. Consequently,

$$ |\left(\frac{U}{V}\right)^{(q)}|(R)|\left(\frac{U}{V}\right)^{(n-q)}|(R) \le |((n-q)!)q!| \frac{|U|(R)}{|V|(R)R^n} $$

and then we can derive

$$ (2) \qquad \left| \binom{n}{q} \right| \left| \left(\frac{U}{V}\right)^{(q)} \right| (R) \left| \left(\frac{U}{V}\right)^{(n-q)} \right| (R) \le |n!| \frac{|U|(R)}{|V|(R)R^n}. $$

So, by (1) and (2), the first conclusion holds for $q = n$.

Suppose now that $\nu^+(f, \log r) = \nu^-(f, \log r)$ and that the residue characteristic of $K$ does not divide $\nu(f, \log r)$. Without loss of generality, we may assume that $f$ has no pole in $C(0, r)$ because all conclusions hold by continuity. In $C(0, r)$, $f(x)$ is equal to a power series $\sum_{-\infty}^{+\infty} a_n x^n$. Set $q = \nu(f, -\log(r))$. Then, $|f|(r) = |a_q|r^q$ and $|f'(r) = |q||a_q|r^{q-1} = |a_q|r^{q-1}$, which ends the proof. $\square$

It seems obvious that the condition for a meromorphic function to admit primitives is that all residues are null. This is stated by Theorem 37.11 but the proof is not this immediate.

Let us remark that the topology of uniform convergence in all disks of $K$ (respectively of all disk included inside an open disk $d(a, R^{-})$) is obviously defined on the algebra $M(K)$ (respectively $M(d(a, R^{-}))$) and that $M(K)$ (respectively $M(d(a, R^{-}))$) is complete for that topology.

Theorem 37.11 is a Mittag-Leffler theorem similar to this known in complex analysis.

**Theorem 37.11.** Let $(a_m, q_m)_{m \in \mathbb{N}}$ be a divisor of $K$ and for every $n \in \mathbb{N}$, and let $Q_m \in K[x]$ be of degree $< q_m$, prime with $(x - a_m)$ such that $\|Q_m\| \le 1$ $\forall m$. There exists $f \in M(K)$ admitting for poles each $a_m$ of order $q_m$ and no other pole and such that its singular part is $\frac{Q_m}{(x - a_m)^{q_m}}$.
**Proof.** Since $\deg(Q_m) < q_m$, we can check that the series $\sum_{n=0}^{\infty} \frac{Q_m(x)}{(x-a_m)^{q_m}}$ converges for all $x \notin \{a_0, \dots, a_m\}$ because when $|x-a_m| \ge |a_m| \ge 1$, then $\left|\frac{Q_m(x)}{(x-a_m)^{q_m}}\right| \le \|Q_m\|$. The series defines a meromorphic function $f$ in each disk $d(0, r^{-})$, and hence, by Theorem 36.9, it is a meromorphic function in $\mathbb{K}$. Then for each $m \in \mathbb{N}$, the function $f(x) - \frac{Q_m}{(x-a_m)^{q_m}}$ is holomorphic at $a_m$, hence $\frac{Q_m}{(x-a_m)^{q_m}}$ is its singular part at $a_m$. Moreover, since $Q_m$ is prime with $x-a_m$, then $a_m$ is a pole of order $q_m$. $\square$

Now, we can give an easy proof Theorem 37.12:

**Theorem 37.12.** $\mathbb{K}$ is supposed to have characteristic 0. A function $f \in \mathcal{M}(\mathbb{K})$ (respectively $f \in \mathcal{M}(d(a, R^{-}))$, $a \in \mathbb{K}$, $R > 0$) admits primitives in $\mathcal{M}(\mathbb{K})$ (respectively in $\mathcal{M}(d(a, R^{-}))$) if and only if all residues of $f$ are null.

**Proof.** Let $a$ be a pole of $f$. According to the Laurent series of $f$ at $a$, if $f$ admits primitives, then $f$ has no residue different from zero at $a$ because the function $\frac{1}{x-a}$ has no primitive in $\mathcal{M}(d(a, r))$ (whenever $r > 0$).

Now let $(a_m)_{m \in \mathbb{N}}$ be the sequence of poles of $f$, each of respective order $q_m$ and suppose that $\text{res}(f, a_m) = 0 \quad \forall m \in \mathbb{N}$. Then the singular part of $f$ at $a_m$ is of the form $\frac{Q_m(x-a_m))}{(x-a_m)^{q_m}}$ with $q_m \ge 2$ and $Q_m(x)$ is a polynomial of degree $\le q_m - 2$. Consequently, the singular part $l_m$ of $f$ at $a_m$ admits a primitive $L_m$ of the form $\frac{P_m(x-a_m)}{(x-a_m)^{q_m-1}}$ with $\deg(P_m) \le q_m - 2$. Suppose first that $f$ has finitely many poles $a_1, \dots, a_q$. Then $f(x)$ is of the form $g(x) + \sum_{m=1}^q l_m(x)$, where $g$ is a power series converging in $\mathbb{K}$ and hence $g$ admits a primitive $G$ vanishing at 0, which is also a power series converging in $\mathbb{K}$. Then $f$ admits a primitive $F(x) = G(x) + \sum_{m=1}^q L_m(x)$.

Consider now the general case and take a disk $d(0, R)$. Let $a_1, \dots, a_q(R)$ be the poles of $f$ in $d(0, R)$. Then in the same way, $i \, d(0, R)$ we can write $f(x) = g_R(x) + \sum_{m=1}^{q(R)} l_m(x)$, where $g_R$ is a power series converging in $d(0, R)$. Then, $g_R$ has a primitive $G_R$ converging in $d(0, R)$, and hence in $d(0, R)$, $f(x)$ admits primitive $S_{G_R + \sum_{m=1}^{q(R)} L_m(x) + c}$, where $C$ is a constant. Let $F_R$ be the primitive that vanishes at 0.
This holds for all $R > 0$, hence when $R < R'$, the decomposition in $d(0, R)$ is the restriction of that in $d(0, R')$. Thus, $f$ admits in each disk $d(0, R)$ a primitive $F_R$ vanishing at 0 such that $F_R(x) = F_{R'}(x) \forall x \in d(0, R)$. Consequently, we can define a function $F$ which is a primitive of $f$ in $\mathbb{K}$, and by Theorem 36.9, it is meromorphic in all $\mathbb{K}$. $\square$

**Corollary 37.13.** The field $\mathbb{K}$ is supposed to have characteristic 0. Let $f \in \mathcal{M}((\mathbb{K})$ (respectively $f \in \mathcal{M}(d(a, R^{-}))$, $a \in \mathbb{K}, R > 0$). Then $f'$ belongs to $\mathbb{K}(x)$ if and only if so does $f$.

**Proof.** If $f$ belongs to $\mathbb{K}(x)$, of course, so does $f'$. Now, suppose $f'$ belongs to $\mathbb{K}(x)$. We can write it in the form $\sum_{j=1}^{q} \frac{b_j}{(x-a_j)^{q_j}}$. And by Theorem 37.11, we have $q_j \ge 2 \forall j = 1, \dots, q$. Consequently, since $\mathbb{K}$ has characteristic 0, $f(x)$ is of the form $-\sum_{j=1}^{q} \frac{b_j}{q_j(x-a_j)^{q_j-1}} + c$ with $c \in \mathbb{K}$ and hence $f$ belongs to $\mathbb{K}(x)$. $\square$
# Chapter 38

## Bezout Algebras of Analytic Functions

Let us recall that the ring of analytic functions on a region of the complex number field is well known to be a Bezout ring. The two fundamental theorems necessary for a proof are the Weierstrass factorization theorem and the Mittag-Leffler theorem.

According to results of [72], it appears that in several hypotheses, rings of analytic functions on complete ultrametric algebraically closed fields are Bezout rings. However, that interesting property is not stated. Moreover, it derives from a Mittag-Leffler theorem referred in general topology whose justification is not relevant. Here we plan to give proofs of all these properties, using results on quasi-invertible analytic elements and on a Mittag-Leffler theorem for meromorphic functions similar to this of complex analysis but is quite different from Krasner's Mittag-Leffler theorem for analytic elements on an infraconnected subset of $\mathbb{K}$.

In Chapter 26, we showed that in $\mathcal{A}(\mathbb{K})$ every closed ideal is principal. Here, following the same methods, provided that $\mathbb{K}$ is spherically complete, we can now prove similar results with algebras $\mathcal{A}(d(a, R^{-}))$:

**Theorem 38.1.** Suppose $\mathbb{K}$ is spherically complete. Let $B$ be a divisor on $d(a, R^{-})$. There exists $f \in \mathcal{A}(d(a, R^{-}))$ such that $D(f) = B$.
**Proof.** Without loss of generality, we may obviously assume $a = 0$. Take $Q_m = 0 \forall m \in \mathbb{N}$. By Theorem 28.4, there exists $f \in \mathcal{A}(d(0, R^{-}))$ such that

(i) $f(0) = 1$,

(ii) $|f|(r) \le |B|(r) \forall r < R$,

(iii) $P_m$ divides $f$ in $\mathcal{A}(d(0, R^{-}))$.

By (iii), clearly $\mathcal{D}(f) \ge B$. Thus, we only have to check that $\mathcal{D}(f) \le B$. Indeed, for all $s \in \mathbb{N}$, we have

$$|B|(\rho_s) = \prod_{j=1}^{s} \prod_{i=1}^{u_m} \left|\left(1 - \frac{x}{a_{i,m}}\right)^{q_{i,m}}\right|(\rho_s) = \prod_{j=1}^{s} \prod_{i=1}^{u_m} \left(\frac{\rho_s}{\rho_m}\right)^{q_{i,m}}.$$

Now, suppose that $B \ne \mathcal{D}(f)$. Then there exists $\alpha \in d(0, R^{-})$ such that $\omega_{\alpha}(f) > B(\alpha)$. Let $s$ be such that $\rho_s > |\alpha|$. Since $f(0) = 1$, we have

$$|f|(\rho_s) \ge \frac{\rho_s}{|\alpha|} \prod_{j=1}^{s} \prod_{i=1}^{u_m} \left|\left(1 - \frac{x}{a_{i,m}}\right)^{q_{i,m}}\right|(\rho_s) > \prod_{j=1}^{s} \prod_{i=1}^{u_m} \left(\frac{\rho_s}{\rho_m}\right)^{q_{i,m}} = |T|(\rho_s),$$

a contradiction to (iii). □

By Theorems 27.10 and 38.1, we have Corollary 38.2:

**Corollary 38.2.** Suppose $\mathbb{K}$ is spherically complete. Let $B$ be a divisor on $d(a, R^{-})$. Then $\mathcal{T}_{a,R}(B)$ is principal [34].

**Corollary 38.3.** Suppose $\mathbb{K}$ is spherically complete. Then all closed ideals of $\mathcal{A}(d(a, R^{-}))$ are principal [34].

**Proof.** Let $I$ be a closed ideal of $\mathcal{A}(d(a, R^{-}))$ and let $E = \mathcal{D}(I)$. By Theorem 27.10, we have $I = \mathcal{T}_{a,R}(E)$. Now, by Corollary 38.2, $\mathcal{T}_{a,R}(E)$ is principal. □

**Lemma 38.4.** Let $E$ be a divisor of $\mathbb{K}$ (respectively a divisor of $d(a, R^{-})$, $a \in \mathbb{K}$, $R > 0$) and for each $r > 0$ (respectively $r \in ]0, R[$, let $g_r \in H(C(0, r))$. There exists $g \in \mathcal{A}(\mathbb{K})$ (respectively $g \in \mathcal{A}(d(a, R^{-}))$, not depending on $r$, such that $\mathcal{D}(g - g_r) \ge E_r$.

**Proof.** Let $f \in \mathcal{A}(\mathbb{K})$ (respectively $f \in \mathcal{A}(d(a, R^{-}))$) be such that $\mathcal{D}(f) \ge E$. By Theorem 37.11, there exists $F \in \mathcal{M}(\mathbb{K})$ whose principal parts at the poles located in $C(0, r)$ are respectively the same as
those of $g_r f^{-1}$ for each $r > 0$. Then $fF$ belongs to $\mathcal{A}(\mathbb{K})$ (respectively to $\mathcal{A}(d(a, R^{-}))$). Putting $g = fF$, we can see that $\mathcal{D}(g - g_r) \ge E_r$, which ends the proof. □

**Theorem 38.5.** Every ideal of finite type of $\mathcal{A}(\mathbb{K})$ is closed and is of the form $\mathcal{T}(E)$ with $E$ a divisor of $\mathbb{K}$.

**Proof.** Let $I$ be an ideal of finite type of $\mathcal{A}(\mathbb{K})$ generated by $f_1, \dots, f_q$ and let $E = \mathcal{D}(I)$. Then, $I = \mathcal{T}(E)$, and by Theorem 27.9, $\mathcal{T}(E)$ is closed, hence the closure of $I$ is $\mathcal{T}(E)$. Consequently, we can see that $E = \min(\mathcal{D}(f_1), \dots, \mathcal{D}(f_q))$. Let us fix $r > 0$. In $H(C(0, r))$, there exist $g_{1,r}, \dots, g_{q,r} \in H(C(0, r))$ such that $g = \sum_{j=1}^q g_{j,r}f_j$. For each $j = 2, \dots, q$, let $f_{j,r}$ be the polynomial of the zeros of $f_j$ in $C(0, r)$. By Lemma 38.4, there exists $g_j \in \mathcal{A}(\mathbb{K})$, not depending on $r$, such that $g_{j,r} - g_j$ be divisible in $H(C(0, r))$ by $\mathcal{D}(f_{1,r})$. Now, set $h = g - \sum_{j=2}^q g_j f_j$. We have $\mathcal{D}(h) \ge \mathcal{D}(f_1)$, hence $h$ factorizes in the form $g_1 f_1$ with $g_1 \in \mathcal{A}(\mathbb{K})$ (respectively $g_1 \in \mathcal{A}(d(a, R^{-}))$) and then

$$g = h + \sum_{j=2}^{q} g_j f_j = \sum_{j=1}^{q} g_j f_j. \quad \square$$

Similarly, when $\mathbb{K}$ is spherically complete, we have Theorem 38.6:

**Theorem 38.6.** Suppose $\mathbb{K}$ is spherically complete. Given $a \in \mathbb{K}$ and $R > 0$, every ideal of finite type of $\mathcal{A}(d(a, R^{-}))$ is closed and is of the form $\mathcal{T}_{a,R}(E)$ with $E$ a divisor of $d(a, R^{-})$ [34].

**Proof.** Let $I$ be an ideal of finite type of $\mathcal{A}(d(a, R^{-}))$ generated by $f_1, \dots, f_q$ and let $E = \mathcal{D}(I)$. Then, $I = \mathcal{T}_{a,R}(E)$, and by Theorem 27.10, $\mathcal{T}_{a,R}(E)$ is closed, hence the closure of $I$ is $\mathcal{T}_{a,R}(E)$. Consequently, we can see that $E = \min(\mathcal{D}(f_1), \dots, \mathcal{D}(f_q))$. Let us fix $r > 0$. In $H(C(0, r))$, there exist $g_{1,r}, \dots, g_{q,r} \in H(C(0, r))$ such that $g = \sum_{j=1}^q g_{j,r}f_j$. For each $j = 2, \dots, q$, let $f_{j,r}$ be the polynomial of the zeros of $f_j$ in $C(0, r)$. By Lemma 38.4, there exists $g_j \in \mathcal{A}(d(a, R^{-}))$, not depending on $r$, such that $g_{j,r} - g_j$ is divisible in $H(C(0, r))$ by $\mathcal{D}(f_{1,r})$. Now, set $h = g - \sum_{j=2}^q g_j f_j$. We have $\mathcal{D}(h) \ge \mathcal{D}(f_1)$, hence $h$ factorizes in the form $g_1 f_1$ with $g_1 \in \mathcal{A}(d(a, R^{-}))$ and then

$$g = h + \sum_{j=2}^{q} g_j f_j = \sum_{j=1}^{q} g_j f_j. \quad \square$$
By Theorem 38.5, we have Corollary 38.7:

**Corollary 38.7.** *A(ℝ) is a Bezout ring* [34].

**Proof.** Indeed, consider an ideal of finite type *I*. By Theorem 38.5, it is closed, and hence, by Theorem 27.11, it is principal. □

And by Theorem 38.6, we have Corollary 38.8:

**Corollary 38.8.** Let $a \in \mathbb{K}$ and let $R > 0$. If $\mathbb{K}$ is spherically complete, $\mathcal{A}(d(a, R^{-}))$ is a Bezout ring.

**Proof.** Indeed, consider an ideal of finite type *I*. By Theorem 38.7, it is of the form $T_{a,R}(E)$ with *E* a divisor of $d(a, R^{-})$ and it is closed. But then, by Corollary 38.3, it is principal. □
# Chapter 39

## Meromorphic Functions Out of a Hole

**Notation.** We fix $R > 0$ and denote by $I$ the interval $[R, +\infty[$. Throughout this chapter, we denote by $S$ the disk $d(0, R^{-})$ and put $D = \mathbb{K}\backslash S$.

We denote by $H_0(D)$ the $\mathbb{K}$-subvector space of the $f \in H(D)$ such that $\lim_{|x| \to +\infty} f(x) = 0$.

By classical properties of analytic elements, we know that given a circle $C(a, R)$ and an element $f$ of $H(C(a, R))$ i.e. a Laurent series $f(x) = \sum_{n=-\infty}^{+\infty} c_n (x-a)^n$ converging whenever $|x|=r$, then $|f(x)|$ is equal to $\sup_{n \in \mathbb{Z}} |c_n|r^n$ in all classes of the circle $C(a, r)$ except maybe in finitely many. When $a=0$, we put $|f|(r) = \sup_{n \in \mathbb{Z}} |c_n|r^n$. Then, $|f|(r)$ is a multiplicative norm on $H(C(0, r))$.

We denote by $\mathcal{A}(D)$ the $\mathbb{K}$-algebra of Laurent series converging in $D$ and by $\mathcal{A}^c(D)$ the set of $f \in \mathcal{A}(D)$ having infinitely many zeros in $D$. Similarly, we denote by $\mathcal{M}(D)$ the field of fractions of $\mathcal{A}(D)$ that we call field of meromorphic functions in $D$ and we denote by $\mathcal{M}^c(D)$ the set of functions $f \in \mathcal{M}(D)$ which have infinitely many zeros or poles in $D$. We denote by $\mathcal{M}_0(D)$ the set of functions $f \in \mathcal{M}(D)$ which have finitely many zeros and poles in $D$.

Similarly, as we did in $\mathbb{K}$ and inside a disk, here we define a pseudo-exceptional value and a quasi-exceptional value in $D$. Given a meromorphic function $f \in \mathcal{M}(D)$, a value $b \in \mathbb{K}$ is called a pseudo-exceptional value for $f$ if $\lim_{|x| \to +\infty} f(x) = b$ it is called a quasi-exceptional value for $f$ if $f-b$ has finitely many zeros in $D$, and it is called an exceptional value for $f$ if has no zero in in $D$.
**Proposition 39.1.** Let $f \in \mathcal{M}^c(D)$ and let $b$ be a quasi-exceptional value. Then $b$ is a pseudo-exceptional value.

**Proof.** Without loss of generality, we may assume that $b = 0$. Therefore, we can write $f$ in the form $\frac{P}{h}$ with $P$ a polynomial whose zeros lie in $D$ and $h \in \mathcal{A}^c(D)$. On the other hand, $h(x)$ is a Laurent series $\sum_{-\infty}^{+\infty} a_n x^n$ converging in all $D$, having infinitely many zeros, hence infinitely many coefficients $a_n$ with $n > 0$ are different from zero, therefore one sees that $\lim_{|x|\to+\infty} \frac{|P(r)|}{|h(r)|} = 0$, and hence $\lim_{|x|\to+\infty} \frac{|P(x)|}{|h(x)|} = 0$. $\square$

**Proposition 39.2.** Let $f \in \mathcal{M}(D)$. If $f$ has infinitely many zeros in $D$ (respectively infinitely many poles in $D$), the set of zeros (respectively the set of poles) is a sequence $(\alpha_n)_{n \in \mathbb{N}}$ such that $\lim_{n \to +\infty} |\alpha_n| = +\infty$.

**Proof.** Suppose first $f \in \mathcal{A}(D)$. For each $L > R$, $f$ belongs to $H(\Delta(0, R, L))$, and by Corollary 21.11, it is quasi-invertible in $H(\Delta(0, R, L))$, hence it has finitely many zeros in $\Delta(0, R, L)$, for every $L > R$. Consequently, if $f$ has infinitely many zeros in $D$, the zeros form a sequence $(\alpha_n)_{n \in \mathbb{N}}$ such that $\lim_{n \to +\infty} |\alpha_n| = +\infty$. Suppose now $f \in \mathcal{M}(D)$. Then $f$ is of the form $\frac{g}{h}$ with $g, h \in f \in \mathcal{A}(D)$. If $f$ has infinitely many zeros, so does $g$, and each zero of $f$ is a zero of $g$, hence the zeros of $f$ form a sequence $(\alpha_n)_{n \in \mathbb{N}}$ such that $\lim_{n \to +\infty} |\alpha_n| = +\infty$. Similarly, if $f$ has infinitely many poles, the $h$ has infinitely many zeros, and each pole of $f$ is a zero of $h$, hence the poles of $f$ form a sequence $(\beta_n)_{n \in \mathbb{N}}$ such that $\lim_{n \to +\infty} |\beta_n| = +\infty$. $\square$

**Theorem 39.3.** Let $f \in \mathcal{M}(D)$ have no zero and no pole in $D$. Then $f(x)$ is of the form $\sum_{-\infty}^{q} a_n x^n$ with $|a_q|r^q > |a_n|r^n \forall n < q, \forall r \ge R$, and $|f(x)| = |a_q|r^q \forall x \in D$.

**Proof.** For every $r \ge R$, $f$ belong to $H(C(0, r))$, and by Theorem 22.1, we have $\nu^+(f, \log(r)) = \nu^-(f, \log(r))$. Consequently, by continuity, $\nu(f, \log(r))$ is a constant $q$ in $\log(R)$, $+\infty$. It is then clear that $|f(x)| = |a_q|r^q \forall x \in D$. $\square$
**Theorem 39.4.** Let $f \in \mathcal{M}(D)$ have at least infinitely many zeros or infinitely many poles in $D$. Then $f$ admits at most one pseudo-exceptional value.

**Proof.** Suppose that $f \in \mathcal{M}^c(D)$ has two distinct pseudo-exceptional values $a$ and $b$. Without loss of generality, we can assume that $a=0$, and hence $f$ is of the form $\frac{\phi}{\psi}$ with $\phi$ and $\psi \in \mathcal{A}(D)$, $\psi$ admitting infinitely many zeros and satisfying $\lim_{r \to +\infty} \frac{\phi(r)}{|\psi(r)|} = 0$. Then, $f-b = \frac{\phi-b\psi}{\psi}$. But when $r$ is big enough, we have $|\phi-b\psi(r)| = |b\psi|(r)$, therefore $\phi - b\psi$ does not admit 0 as a pseudo-exceptional value, a contradiction. $\square$

**Corollary 39.5.** Let $f \in \mathcal{M}(D)$ have at least infinitely many zeros or infinitely many poles in $D$. Then $f$ admits at most one exceptional value.

**Definition.** Let $f \in H(D)$ have no zero in $D$, $f(x) = \sum_{n=-\infty}^{q} a_n x^n$ with $|a_q|R^q > |a_n|R^n \forall n < q$ and $a_q = 1$. Then $f$ is a Motzkin factor associated with $S$ and the integer $q$ is called the *Motzkin index of $f$* and is denoted by $mo(f, S)$.

**Theorem 39.6.** Let $f \in \mathcal{M}(D)$. We can write $f$ in a unique way in the form $f^S f^0$ with $f^S \in H(\mathbb{K}\setminus D)$ a Motzkin factor associated with $S$ and $f^0 \in \mathcal{M}(\mathbb{K})$, having no zero and no pole in $S$.

**Proof.** Suppose first $f \in \mathcal{A}(D)$ and take $V > R$. Then as a quasi-invertible element of $H(\Delta(0, R, V))$, by Theorem 29.16, $f$ admits a factorization in the form $f^S f^0$, where $f^S$ is a Motzkin factor and $f^0$ belongs to $H(d(0, V))$ and has no zero in $S$. Moreover, by Lemma 29.6, $f^S$ does not depend on $V$. Consequently, since $f^S$ is obviously invertible in $\mathcal{A}(D)$, we can factorize $f \in \mathcal{A}(D)$ in the form $f^S f^0$, where $f^0$ belongs to $\mathcal{A}(\mathbb{K})$ and has no zero in $S$.

Consider now the general case: $f = \frac{g}{h}$ with $g, h \in \mathcal{A}(D)$. Then we can write $g = g^S g^0$, $h = h^S h^0$, hence $f = \left(\frac{g^S}{h^S}\right)\left(\frac{g^0}{h^0}\right)$. Then we can check that this is the factorization announced in the statement: $f^S = \frac{g^S}{h^S}$ and $f^0 = \frac{g^0}{h^0}$.
Lemma 39.7 is immediate:

**Lemma 39.7.** The set of Motzkin factors associated with $S$ makes a multiplicative group. Let $f, g \in \mathcal{M}(D)$. Then, $(fg)^S = (f^S)(g^S)$, $(\frac{1}{f})^S = \frac{1}{f^S}$, $(fg)^0 = (f^0)(g^0)$, $(\frac{1}{f})^0 = \frac{1}{f^0}$, and $\mathrm{mo}(fg, S) = \mathrm{mo}(f, S) + \mathrm{mo}(g, S)$, $\mathrm{mo}(\frac{1}{f}, S) = -\mathrm{mo}(f, S)$.
Chapter 40

Shilov Boundary for Algebras $H(D)$

Given a holomorphic function $f$ in an open bounded connected subset $D$ of $\mathbb{C}$, $|f(z)|$ reaches its maximum on the boundary of $D$. Consider now a closed bounded infraconnected subset $D$ of $\mathbb{K}$ and $f \in H(D)$. We show that the supremum of the $\phi(f)$ when $\phi$ runs in $\text{Mult}(H(D), \| \cdot \|_D)$ is reached on the Shilov boundary that we characterize. We show that the set of circular filters is provided with a tree structure and that the diameter is an increasing function with values in $\mathbb{R}$, defining distances associated with this structure. The first remarks on that tree structure are due to E. Motzkin [74], and it was thoroughly examined in [14] and in [49]. Here the structure is helpful to determine the Shilov boundary for algebras $H(D)$.

Throughout this chapter and in the following, according to Theorem 14.1, we identify $\Phi(D)$ with $\text{Mult}(H(D), \| \cdot \|_D)$.

**Definition and notation.** Given $a \in \mathbb{K}$ and $r > 0$, we denote by $\mathcal{F}_{a,r}$ the circular filter of center $a$ and diameter $r$. Given a large circular filter $\mathcal{F}$, the set of its centers is denoted by $Q(\mathcal{F})$. Given two circular filters $\mathcal{F}$ and $\mathcal{G}$, $\mathcal{F}$ is said to surround $\mathcal{G}$ if either $\mathcal{G}$ is secant with $Q(\mathcal{F})$ or if $\mathcal{F} = \mathcal{G}$. Then, $\mathcal{F}$ is said to strictly surround $\mathcal{G}$ if $\mathcal{F}$ surrounds $\mathcal{G}$ and is different from $\mathcal{G}$.

Similarly, a circular filter $\mathcal{F}$ is said to surround a monotonous filter $\mathcal{G}$ if it surrounds the circular filter associated with $\mathcal{G}$. A monotonous filter $\mathcal{F}$ is said to surround a circular filter $\mathcal{G}$ if its associated circular filter surrounds $\mathcal{G}$ and $\mathcal{F}$ is said to surround a monotonous filter $\mathcal{G}$ if the circular filter associated with $\mathcal{F}$ surrounds the circular filter associated with $\mathcal{G}$.
We denote by $\leq$ the relation on the set of circular filters defined as $\mathcal{F} \leq \mathcal{G}$ if $\mathcal{G}$ surrounds $\mathcal{F}$ and by $\prec$ the relation defined as $\mathcal{F} \prec \mathcal{G}$ if $\mathcal{F} \preceq \mathcal{G}$ and $\mathcal{F} \neq \mathcal{G}$.

Given an order relation $\mathcal{R}$ defined on a set $E$, we call strict-order associated with $\mathcal{R}$ the relation $\overline{\mathcal{R}}$ defined as $a\overline{\mathcal{R}}b$ if $a\mathcal{R}b$ and $a \neq b$.

Let $D$ be an infraconnected subset of $\mathbb{K}$. If $D$ is bounded, of diameter $R$, the circular filter of center $a \in D$ and diameter $R$ is called the *peripheral* of $D$.

A circular filter $\mathcal{F} \in \Phi(\mathbb{K})$ is said to be *D-bordering* if it is secant with both $D$ and $\mathbb{K} \setminus D$. Moreover, it is said to be *strictly D-bordering* if it is the peripheral of $D$ or the peripheral of a hole of $D$.

**Remark.** If a circular filter $\mathcal{F} \in \Phi(\mathbb{K})$ is strictly *D-bordering* it has a center.

**Lemma 40.1.** Let $D$ be a closed infraconnected subset of $\mathbb{K}$. Every circular filter which is strictly *D-bordering* is *D-bordering*. Further, if $D$ is infraconnected affinoid, then a circular filter is *D-bordering* if and only if it is strictly *D-bordering*.

**Proof.** Let $\mathcal{F}$ be the peripheral of a hole $T = d(a, r^-)$ of $D$. All annuli $\Gamma(b, r', r'')$ with $b \in d(a, r)$ and $r' < r < r''$ have a non-empty intersection with $T$ and $D$, which shows that $\mathcal{F}$ is *D-bordering*. And we have a similar situation with the peripheral of $D$ if $D$ is bounded.

Now, if $D$ is infraconnected affinoid, the set of holes is finite. If a circular filter $\mathcal{F} \in \Phi(D)$ is not strictly *D-bordering*, then it admits annuli having an empty intersection with any hole and with $\mathbb{K} \setminus \tilde{D}$, hence it is not *D-bordering*.

Lemmas 40.2 and 40.3 are immediate:

**Lemma 40.2.** Let $\mathcal{F}, \mathcal{G}$ be two circular filters such that $\mathcal{Q}(\mathcal{F}) \neq \emptyset$ and $\mathcal{Q}(\mathcal{G}) \neq \emptyset$. Then, $\mathcal{F} \leq \mathcal{G}$ if and only if $\mathcal{Q}(\mathcal{F}) \subset \mathcal{Q}(\mathcal{G})$.

**Lemma 40.3.** Let $\mathcal{F}, \mathcal{G}$ be two circular filters such that $\mathcal{F} \preceq \mathcal{G}$. Then, $\mathrm{diam}(\mathcal{F}) \leq \mathrm{diam}(\mathcal{G})$. Moreover, $\mathcal{F} = \mathcal{G}$ if and only if $\mathrm{diam}(\mathcal{F}) = \mathrm{diam}(\mathcal{G})$.

**Theorem 40.4.** The relation $\leq$ is an order relation on $\Phi(\mathbb{K})$ and $\prec$ is the strict-order associated with this order relation.
**Proof.** Indeed, the relation is reflexive by definition. Let $\mathcal{F}, \mathcal{G}, \mathcal{H}$ be three circular filters. Suppose that $\mathcal{F} \preceq \mathcal{G}$ and $\mathcal{G} \preceq \mathcal{F}$. If $\mathcal{F} \neq \mathcal{G}$, then by definition $Q(\mathcal{F}) \neq \emptyset$, $Q(\mathcal{G}) \neq \emptyset$, $\mathcal{F}$ is secant with $Q(\mathcal{G})$, and $\mathcal{G}$ is secant with $Q(\mathcal{F})$. Consequently, $Q(\mathcal{F}) = Q(\mathcal{G})$, and therefore, by Lemma 40.3, we have $\mathcal{F} = \mathcal{G}$. Finally, suppose that $\mathcal{F} \preceq \mathcal{G}$ and $\mathcal{G} \preceq \mathcal{H}$, with $\mathcal{F} \neq \mathcal{G} \neq \mathcal{H}$. Then, $\mathcal{F}$ is secant with $Q(\mathcal{G})$ and with $Q(\mathcal{H})$, and therefore, $\mathcal{F} \preceq \mathcal{H}$. $\square$

**Corollary 40.5.** Let $D$ be infraconnected. The mapping $\mathrm{diam}$ is strictly increasing from $\Phi(D)$, ordered by $\preceq$, to $\mathbb{R}_+$.

**Theorem 40.6.** Let $D$ be infraconnected and let $\mathcal{F} \in \Phi(D)$. Then $\mathcal{F}$ is a minimal element in $\Phi(D)$ if and only if

(i) either it is punctual,

(ii) or it has no center,

(iii) or $Q(\mathcal{F}) \cap D = \emptyset$.

**Proof.** On one hand, it is easily seen that if $\mathcal{F}$ is punctual, or has no center, then it is minimal in $\Phi(\mathbb{K})$ and therefore in $\Phi(D)$. Suppose now that $Q(\mathcal{F}) \cap D = \emptyset$. Suppose that $\mathcal{G} \in \Phi(D)$ satisfies $\mathcal{G} \prec \mathcal{F}$. Then $\mathcal{G}$ is secant with $Q(\mathcal{F})$ and has a diameter strictly inferior to the diameter $r$ of $\mathcal{F}$, hence there exists a disk $d(b, s) \in \mathcal{G}$, with $s < r$. Consequently, $d(b, s) \subset Q(\mathcal{F})$. Since $d(b, s) \cap D = \emptyset$, we see that $\mathcal{G}$ is not secant with $D$.

Now, let $\mathcal{F} \in \Phi(D)$ be minimal in $\Phi(D)$. Suppose that $\mathcal{F}$ is not punctual and has a center and let $r = \mathrm{diam}(\mathcal{F})$. If $Q(\mathcal{F}) \cap D \neq \emptyset$, then for any $a \in Q(\mathcal{F}) \cap D$, of course, $\mathcal{F}$ surrounds the filter of neighborhoods of $a$, a contradiction to the hypothesis "$\mathcal{F}$ minimal in $\Phi(D)$". Thus, $Q(\mathcal{F}) \cap D = \emptyset$. If $Q(\mathcal{F}) = \emptyset$, then $\mathcal{F}$ has no center. Else, $Q(\mathcal{F}) \cap D = \emptyset$, which ends the proof. $\square$

**Lemma 40.7.** Let $\mathcal{F}, \mathcal{G}$ be two circular filters such that $Q(\mathcal{F}) \cap Q(\mathcal{G}) \neq \emptyset$. Then $\mathcal{F}$ and $\mathcal{G}$ are comparable with respect to the relation $\preceq$.

**Proof.** Since $Q(\mathcal{F}) \cap Q(\mathcal{G}) \neq \emptyset$ and since both $Q(\mathcal{F})$, $Q(\mathcal{G})$ are disks, we can suppose, for instance, $Q(\mathcal{F}) \subset Q(\mathcal{G})$. Then $\mathcal{F}$ is secant with $Q(\mathcal{G})$, hence $\mathcal{F} \preceq \mathcal{G}$. $\square$
Theorem 40.8 is a very typical theorem concerning circular filters:

**Theorem 40.8.** Let $\mathcal{F}$, $\mathcal{G}$ be two circular filters surrounding a certain circular filter $\mathcal{H}$. Then $\mathcal{F}$ and $\mathcal{G}$ are comparable with respect to the relation $\leq$.

**Proof.** If $\mathcal{F} = \mathcal{H}$, then $\mathcal{G}$ just surrounds $\mathcal{F}$. Now, suppose that both $\mathcal{F}$, $\mathcal{G}$ strictly surround $\mathcal{H}$. Then $\mathcal{H}$ is secant with both $\mathcal{Q}(\mathcal{F})$ and $\mathcal{Q}(\mathcal{G})$, hence $\mathcal{Q}(\mathcal{F}) \cap \mathcal{Q}(\mathcal{G}) \neq \emptyset$. Therefore, either $\mathcal{Q}(\mathcal{F}) \subset \mathcal{Q}(\mathcal{G})$ or $\mathcal{Q}(\mathcal{G}) \subset \mathcal{Q}(\mathcal{F})$, hence by Lemma 40.7, $\mathcal{F}$ and $\mathcal{G}$ are comparable. $\square$

Theorem 40.9 is also very important:

**Theorem 40.9.** Let $\mathcal{F} \in \Phi(\mathbb{K})$ and let $s > \text{diam}(\mathcal{F})$. There exists a unique disk $d(a, s)$ such that $\mathcal{F}$ is secant with it and there exists a unique circular filter on $\mathbb{K}$, of diameter $s$, surrounding $\mathcal{F}$.

**Proof.** Let $r = \text{diam}(\mathcal{F})$. We can check that there exists a unique disk $d(b, s)$ such that $\mathcal{F}$ is secant with this disk. Indeed, if $\mathcal{F}$ has a center $a$, this disk is just $d(a, s)$. Then $\mathcal{F}$ is surrounded by the filter $\mathcal{G}$ of center $a$ and diameter $s$.

Now, if $\mathcal{F}$ has no center, it admits as one of its elements a disk of $d(b, s')$ with $r < s' < s$ and hence $\mathcal{F}$ is secant with $d(b, s)$. Consider now another disk $d(c, s)$ such that $\mathcal{F}$ is secant with it. Then we have $d(b, s') \cap d(c, s) \neq \emptyset$, hence $d(b, s') \subset d(c, s)$, and hence $d(b, s) = d(c, s)$. So, we have shown again that $\mathcal{F}$ is surrounded by the filter $\mathcal{G}$ of center $a$ and diameter $s$.

Conversely, let $\mathcal{H}$ be another circular filter of diameter $s$, surrounding $\mathcal{F}$. Then, by Theorem 40.8, $\mathcal{G}$ and $\mathcal{H}$ are comparable. But they have the same diameter, hence by Lemma 40.3, they are equal. $\square$

<figure><img src="image_15.png" /></figure>

By Proposition 3.14, we obtain Theorem 40.10.

**Theorem 40.10.** Let $D$ be infraconnected, let $\mathcal{F}$ be a circular filter on $\mathbb{K}$ secant with $D$, and let $r \in]\text{diam}(\mathcal{F}), \text{diam}(D)[$. The unique circular filter of diameter $r$ surrounding $\mathcal{F}$ is secant with $D$.

**Proof.** By Theorem 40.9, there exists a unique disk $d(a, r)$ such that $\mathcal{F}$ is secant with it. By Proposition 3.14, the circular filter $\mathcal{G}$ of center $a$ and diameter $r$ is secant with $D$ and obviously surrounds $\mathcal{F}$.
This filter is unique by Theorem 40.9. Now, there exists a $\mathcal{G}$-affinoid $G$ containing annuli $\Gamma(a, s, r)$ and $\Gamma(a, r, t)$ such that $G \cap D = \emptyset$. On the other hand, by definition, $D$ being infraconnected admits points in $\Gamma(a, r, t)$ and also in $\Gamma(a, s, r)$, a contradiction. This shows that $G$ is secant with $D$. $\square$

**Proposition 40.11.** Let $\mathcal{F}$, $\mathcal{G}$ be two circular filters on $\mathbb{K}$ which are not comparable for the relation $\preceq$. There exist disks $F \in \mathcal{F}$ and $G \in \mathcal{G}$ such that $F \cap G = \emptyset$. Moreover, given disks $F' = d(a, r') \in \mathcal{F}$, $G' = d(b, s') \in \mathcal{G}$ such that $F' \cap G' = \emptyset$, we have $\delta(F, G) = \delta(F', G') > \max(\text{diam}(\mathcal{F}), \text{diam}(\mathcal{G}))$.

**Proof.** Suppose first that both $\mathcal{F}$, $\mathcal{G}$ have no center. Then $\mathcal{F}$ (respectively $\mathcal{G}$) admits a canonical basis $\mathcal{D}$ (respectively $\mathcal{E}$) consisting of a decreasing sequence of disks. But since the two filters are not secant, we can obviously find $F \in \mathcal{D}$ and $G \in \mathcal{E}$ such that $F \cap G = \emptyset$.

Suppose now that $\mathcal{F}$ has centers and let $d(a, r) = \mathcal{Q}(\mathcal{F})$. If all disks $G \in \mathcal{G}$ contain $d(a, r)$, then $\mathcal{G}$ surrounds $\mathcal{F}$, a contradiction to the hypothesis. Hence, there exists a disk $G \in \mathcal{G}$ which does not contain $d(a, r)$. Let $d(b, s) \in \mathcal{G}$ be a disk such that $d(b, s) \cap d(a, r) = \emptyset$. Then we have $|a-b| > r$, therefore we can take $r' \in ]r, s[$ and we have $d(a, r) \cap d(b, r') = \emptyset$.

Consider now disks $F' = d(a, r') \in \mathcal{F}$, $G' = d(b, s') \in \mathcal{G}$ such that $F' \cap G' = \emptyset$. Let $R = |a-b|$. Since $F' \cap G' = \emptyset$, we have $\delta(F', G') = R > \max(r', s')$ and $\mathcal{F}$ is secant with $d(a, R^{-})$, $\mathcal{G}$ is secant with $d(b, R^{-})$. Therefore, $F$ is included in $d(a, R^{-})$ and $G$ is included in $d(b, R^{-})$. Consequently, $\delta(F, G) = R$ and $\text{diam}(\mathcal{F}) < \text{diam}(F) < R$, $\text{diam}(\mathcal{G}) < \text{diam}(G) < R$. That ends the proof. $\square$

**Notation.** Let $\mathcal{F}$, $\mathcal{G}$ be two circular filters which are not comparable for $\preceq$. By Proposition 40.11, we note that given disks $F \in \mathcal{F}$ and $G \in \mathcal{G}$ such that $F \cap G = \emptyset$, $\delta(F, G)$ does not depend on the choice of the disks $F \in \mathcal{F}$ and $G \in \mathcal{G}$, so we can put $\lambda(\mathcal{F}, \mathcal{G}) = \delta(F, G)$ with $F, G$ disks such that $F \in \mathcal{F}, G \in \mathcal{G}, F \cap G = \emptyset$.

**Theorem 40.12.** Let $D$ be infraconnected and let $\mathcal{F}, \mathcal{G} \in \Phi(D)$. There exists $\sup(\mathcal{F}, \mathcal{G}) \in \Phi(D)$ and it is the unique circular filter of diameter $\lambda(\mathcal{F}, \mathcal{G})$ which surrounds both $\mathcal{F}$ and $\mathcal{G}$.
**Proof.** The claim is trivial when the two filters are comparable for $\leq$. So we assume they are not. Let $l = \lambda(\mathcal{F}, \mathcal{G})$. By Theorem 40.9, there exists a unique circular filter $\mathcal{S} \in \Phi(\mathbb{K})$ of diameter $l$ surrounding $\mathcal{F}$. Then $\mathcal{S}$ has centers and $\mathcal{Q}(\mathcal{S})$ is a disk $d(a, l)$. By Proposition 40.11, there are disks included in $d(a, l)$ which belong to $\mathcal{G}$ and therefore are secant with $d(a, l)$, hence $\mathcal{S}$ surrounds $\mathcal{G}$. We check that $\mathcal{S}$ is the smallest element of the set of filters on $\mathbb{K}$ surrounding both $\mathcal{F}$ and $\mathcal{G}$. Indeed, let $\mathcal{H}$ be a circular filter surrounding $\mathcal{F}$ and $\mathcal{G}$. Then both $\mathcal{F}$ and $\mathcal{G}$ are secant with $\mathcal{Q}(\mathcal{H})$. Let $d(b, s) = \mathcal{Q}(\mathcal{H})$. Consider disks $F = d(\alpha, \rho) \in \mathcal{F}$ and $G = d(\beta, \sigma) \in \mathcal{G}$ such that $F \cap G = \emptyset$. Since $F \cap d(b, s) \neq \emptyset$ and $G \cap d(b, s) \neq \emptyset$ and since $F \cap G = \emptyset$, it is easily seen that both $F$, $G$ are included in $d(b, s)$, hence $s \ge l$. But since both $\mathcal{S}$, $\mathcal{H}$ surround $\mathcal{F}$, by Theorem 40.8, they are comparable for $\leq$. Then, since $l \le s$, $\mathcal{H}$ surrounds $\mathcal{S}$ and therefore $\mathcal{S}$ is the smallest element of the set of filters on $\mathbb{K}$ surrounding both $\mathcal{F}$ and $\mathcal{G}$.

Now, suppose that both $\mathcal{F}$, $\mathcal{G}$ are secant with $D$. Then, $d(\alpha, \rho) \cap D \neq \emptyset$ and $d(\beta, \sigma) \cap D \neq \emptyset$. Since $\mathcal{S}$ is the circular filter of center $\alpha$ and diameter $l = |\alpha - \beta|$, it is secant with $D$ because $D$ is infraconnected. $\square$

**Notation.** Let $\mathcal{F}$, $\mathcal{G}$ be two circular filters and let $\mathcal{S} = \sup(\mathcal{F}, \mathcal{G})$. We set $\delta(\mathcal{F}, \mathcal{G}) = \text{diam}(\mathcal{S}) - \frac{\text{diam}(\mathcal{F})+\text{diam}(\mathcal{G})}{2}$.

**Remark.** Particularly, if $\mathcal{F} \preceq \mathcal{G}$, we have $\delta(\mathcal{F}, \mathcal{G}) = \text{diam}(\mathcal{G}) - \text{diam}(\mathcal{F})$. Moreover, given $a, b \in K$, the filter of neighborhoods $\mathcal{F}$ of $a$ and the filter of neighborhoods $\mathcal{G}$ of $b$ satisfy $\delta(\mathcal{F}, \mathcal{G}) = |a - b|$.

Lemma 40.13 is immediate:

**Lemma 40.13.** Let $\mathcal{F}, \mathcal{G}$ be circular filters. Suppose there are infraconnected subsets $F \in \mathcal{F}, G \in \mathcal{G}$ such that $F \cap G = \emptyset$. Then, $\delta(F, G) = \text{diam}(\sup(\mathcal{F}, \mathcal{G}))$, $2\delta(\mathcal{F}, \mathcal{G}) = 2\delta(F, G) - \text{diam}(\mathcal{F}) - \text{diam}(\mathcal{G})$, and $|\text{diam}(\mathcal{F}) - \text{diam}(\mathcal{G})|_{\infty} \le \delta(\mathcal{F}, \mathcal{G})$. If $\mathcal{F} \preceq \mathcal{G}$, we have $\delta(\mathcal{F}, \mathcal{G}) = \text{diam}(\mathcal{G}) - \text{diam}(\mathcal{F})$.

**Proof.** Let $\mathcal{S} = \sup(\mathcal{F}, \mathcal{G})$. Since both $F$ and $G$ are infraconnected, we have $\delta(F, G) = \text{diam}(\mathcal{S})$ hence $2\delta(F, G) = \delta(\mathcal{F}, \mathcal{S}) + \delta(\mathcal{G}, \mathcal{S}) = 2\text{diam}(\mathcal{S}) - \text{diam}(\mathcal{F}) - \text{diam}(\mathcal{G})$. The other claims are immediate. $\square$

<figure><img src="image_16.png" /></figure>

**Theorem 40.14.** $\delta$ is a distance on $\Phi(\mathbb{K})$.

**Proof.** We only have to show the triangular inequality. Let $\mathcal{F}, \mathcal{G}, \mathcal{H} \in \Phi(\mathbb{K})$, with $\text{diam}(\mathcal{F}) = r$, $\text{diam}(\mathcal{G}) = s$, $\text{diam}(\mathcal{H}) = t$. Let
$\mathcal{X} = \sup(\mathcal{F}, \mathcal{G})$, $\mathcal{Y} = \sup(\mathcal{G}, \mathcal{H})$, $\text{diam}(\mathcal{X}) = M$, $\text{diam}(\mathcal{Y}) = N$. Suppose, for instance, $M \le N$. We note that both $\mathcal{X}, \mathcal{Y}$ surround $\mathcal{G}$. Consequently, by Theorem 40.8, they are comparable. And since $M \le N$, we have $\mathcal{X} \subseteq \mathcal{Y}$. Consequently, $\mathcal{F} \subseteq \mathcal{Y}$, and hence, $\sup(\mathcal{F}, \mathcal{H})$ has a diameter $V \le N$. Then, $2\delta(\mathcal{F}, \mathcal{H}) = 2V - r - t \le N - r - t$.

On the other hand, $2\delta(\mathcal{F}, \mathcal{G}) + 2\delta(\mathcal{G}, \mathcal{H}) = 2M - r - s + 2N - s - t$. But since $s \le M$, we have $2\delta(\mathcal{F}, \mathcal{G}) + 2\delta(\mathcal{G}, \mathcal{H}) \ge 2N - r - t \ge 2V - r - t = 2\delta(\mathcal{F}, \mathcal{H})$. $\square$

**Remark.** $\delta$ is not a non-Archimedean distance on $\Phi(K)$. Indeed, consider three circular filters $\mathcal{F}_{a,r}$, $\mathcal{F}_{a,s}$, $\mathcal{F}_{a,t}$ with $r < s < t$. Then, $\delta(\mathcal{F}_{a,r}, \mathcal{F}_{a,t}) = t - r = \delta(\mathcal{F}_{a,r}, \mathcal{F}_{a,s}) + \delta(\mathcal{F}_{a,s}, \mathcal{F}_{a,t})$.

**Definition.** Henceforth, we call $\delta$-topology the topology defined by the distance $\delta$.

**Theorem 40.15.** Every bounded monotonous sequence of $\Phi(\mathbb{K})$ with respect to the order $\le$ has a limit with respect to the $\delta$-topology.

**Proof.** Let $(\mathcal{F}_n)_{n \in \mathbb{N}}$ be a bounded monotonous sequence of circular filters and for each $n \in \mathbb{N}$, let $r_n = \text{diam}(\mathcal{F}_n)$. Without loss of generality, we can obviously assume that the sequence is strictly monotonous. Clearly, for each $n \in \mathbb{N}^*$, each filter $\mathcal{F}_n$ has a center. Let $l = \lim_{n \to \infty} r_n$. Suppose first that it is a decreasing sequence. For each $n \in \mathbb{N}$, we can find a center $a_n \notin d(a_{n+1}, r_{n+1})$. Thus, the sequence $(a_n)_{n \in \mathbb{N}}$ is such that the sequence $(|a_{n+1} - a_n|)_{n \in \mathbb{N}}$ is strictly decreasing. Hence, by Proposition 3.18, there exists a unique circular filter $\mathcal{F}$ less thin than the sequence $(a_n)_{n \in \mathbb{N}}$ of radius $l$. And then, we check that the sequence $(\mathcal{F}_n)_{n \in \mathbb{N}}$ converges to $\mathcal{F}$ with respect to the $\delta$-topology. If the sequence is increasing, it is easily seen that the sequence converges to the circular filter of center $a_1$ and diameter $l$. $\square$

**Theorem 40.16.** $\Phi(\mathbb{K})$ is complete with respect to the $\delta$-topology.

**Proof.** Let $(\mathcal{F}_n)_{n \in \mathbb{N}}$ be a Cauchy sequence with respect to the $\delta$-topology. Without loss of generality, we can suppose that each $\mathcal{F}_n$ has a center. Indeed, for each $n \in \mathbb{N}$, we can take a circular filter $\mathcal{F}'_n$ with a center such that $\delta(\mathcal{F}'_n, \mathcal{F}_n) < \frac{1}{n}$ and then we check that the sequence $(\mathcal{F}'_n)_{n \in \mathbb{N}}$ is a Cauchy sequence again. Then, if it converges to a limit $\mathcal{G}$, so does the sequence $(\mathcal{F}_n)_{n \in \mathbb{N}}$.

So, we suppose that for each $n \in \mathbb{N}$, $\mathcal{F}_n$ has a center $a_n$. Let $r_n = \text{diam}(\mathcal{F}_n)$. For every $m, n \in \mathbb{N}$, let $S_{m,n}$ denote $\sup(\mathcal{F}_m, \mathcal{F}_n)$ and let
$s_{m,n} = \text{diam}(S_{m,n})$. Next, for each $n \in \mathbb{N}$, let $s_n = \sup_{m \ge n, t \ge n, (s_{m,t})}$. So, by definition, the sequence $(s_n)$ is decreasing. We show that the sequence $d(a_n, s_n)_{n \in \mathbb{N}}$ is decreasing with respect to the inclusion. Indeed, by definition, for every $m \ge n$, we have $Q(S_{n,m}) \subset d(a_n, s_n)$, therefore $Q(F_{n+1}) \subset d(a_n, s_n)$, hence $a_{n+1} \in d(a_n, s_n)$, and hence $d(a_{n+1}, s_{n+1}) \subset d(a_n, s_n)$.

Now, for every $n \in \mathbb{N}$, let $\mathcal{G}_n$ be the circular filter of center $a_n$ and diameter $s_n$. The sequence $(\mathcal{G}_n)_{n \in \mathbb{N}}$ is then decreasing with respect to the order $\leq$, and therefore, by Theorem 40.15, has a limit $\mathcal{G}$ with respect to the $\delta$-topologies. Let $s = \lim_{n \to \infty} s_n$. We prove that $\lim_{n \to +\infty} \delta(F_n, \mathcal{G}_n) = 0$.

Let $\epsilon > 0$. By definition of the sequence $s_n$, we can find a rank $t(\epsilon)$ such that

$$ (1) \quad s_n - s \le \frac{\epsilon}{4} \quad \forall n \ge t(\epsilon). $$

And since the sequence $(F_n)$ is a Cauchy sequence, there exists $k(\epsilon) \in \mathbb{N}$ such that $\delta(F_m, F_q) \le \frac{\epsilon}{4} \forall m, q \ge k(\epsilon)$. We can choose $k(\epsilon) \ge t(\epsilon)$. By Lemma 40.13, we can derive

$$ (2) \quad |r_n - r_m|_{\infty} \le \frac{\epsilon}{4} \quad \forall m, n \ge k(\epsilon). $$

On the other hand, since $\mathcal{G}_n$ surrounds $F_n$, we have

$$ (3) \quad \delta(\mathcal{G}_n, F_n) = s_n - r_n. $$

Now, let $q > m \ge n \ge t(\epsilon)$. By definition of the distance $\delta$, we have $\delta(F_m, S_{m,q}) \le \delta(F_m, F_q) \le \frac{\epsilon}{4}$, hence

$$ (4) \quad s_{m,q} - r_m \le \frac{\epsilon}{4}. $$

Now, we can take $m, q$ bigger than $n$ and such that $s_n - s_{m,q} \le \frac{\epsilon}{4}$, therefore $s_m - s_{m,q} \le \frac{\epsilon}{4}$ because $s_m \le s_n$. Consequently,

$$ \begin{aligned} \delta(\mathcal{G}_n, F_n) &= s_n - r_n = (s_n - s_m) + (s_m - s_{m,q}) \\ & \quad + (s_{m,q} - r_m) + (r_m - r_n) \\ & \le (s_n - s_m) + (s_m - s_{m,q}) + (s_{m,q} - r_m) + |r_m - r_n|_{\infty}, \end{aligned} $$

and hence, by (1), (2), (3), and (4), we obtain $\delta(\mathcal{G}_n, F_n) \le \epsilon \forall n \ge k(\epsilon)$. Thus, $\lim_{n \to \infty} \delta(F_n, \mathcal{G}) = 0$, and therefore, the Cauchy sequence $F_n$ has limit $\mathcal{G}$. $\square$
**Remark.** In [85], another distance $\tilde{\delta}$ is defined on the set of large circular filters, as $\tilde{\delta}(\mathcal{F}, \mathcal{G}) = 2 \log(\mathrm{diam}(\mathrm{sup}(\mathcal{F}, \mathcal{G}))) - \log(\mathrm{diam}(\mathcal{F})) - \log(\mathrm{diam}(\mathcal{G})).$ This last distance $\tilde{\delta}$ defines the same topology as $\delta$ on the set of large circular filters but does not admit the same Cauchy sequences and does not expand the usual distance on the field $\mathbb{K}$ defined by the absolute value. As a consequence, the set of large circular filters appears to be complete with respect to $\tilde{\delta}$ and therefore a sequence converging to a point $a \in \mathbb{K}$ in the usual sense, actually has no limit for $\tilde{\delta}$. We use $\tilde{\delta}$ in Chapter 40.

**Theorem 40.17.** Let $B$ be a totally ordered subset of $\Phi(\mathbb{K})$. Then $B$ admits an infimum $\mathcal{T}$ with respect to the order $\preceq$. Further, a subset $A$ of $\Phi(\mathbb{K})$ admits a supremum with respect to the order $\succeq$ if and only if it is bounded with respect to the $\delta$-topology.

**Proof.** Let $B$ be a totally ordered subset of $\Phi(\mathbb{K})$ and set $r = \inf\{\mathrm{diam}(\mathcal{G}) | \mathcal{G} \in B\}$. By Lemma 40.13, we can see that all sequences $(G_n)_{n \in \mathbb{N}}$ of $B$ satisfying $\lim_{n \to \infty} \mathrm{diam}(G_n) = r$ are Cauchy sequences and admit the same limit $\mathcal{T}$. Therefore, $\mathcal{T}$ is clearly the infimum of $B$ with respect to $\preceq$. Now, let $A$ be a subset of $\Phi(\mathbb{K})$. If it admits a supremum with respect to the order $\preceq$, it is obviously bounded. Conversely, assume that $A$ is bounded. We can clearly find a disk $d(b, s)$ such that all elements of $A$ are secant with $d(b, s)$. Consequently, the circular filter $\mathcal{F}$ of center $b$ and diameter $s$ is such that $Q(\mathcal{F})$ contains all elements of $A$. Now, by Theorem 40.8, the set $A^*$ of circular filters surrounding all elements of $A$ is totally ordered and therefore admits an infimum $S$ and we have $\mathcal{G} \preceq S$ for every $\mathcal{G} \in A$. Consequently, $S$ is the supremum of $A$. $\square$

**Remark.** By Lemma 3.15, it is easily seen that there exist circular filters without countable bases if and only if the residue class field of $\mathbb{K}$ is not countable.

**Theorem 40.18.** Let $D$ be a closed bounded subset of $\mathbb{K}$ and let $\varphi_{\mathcal{F}} \in \mathrm{Mult}(H(D), \| \cdot \|_D)$. There exists a basis of neighborhoods of $\varphi_{\mathcal{F}}$ in $\mathrm{Mult}(H(D), \| \cdot \|_D)$, with respect to the topology of pointwise convergence, consisting of the family of sets of the form $\mathrm{Mult}(H(E \cap D), \| \cdot \|_{E \cap D})$, where $E$ is a $\mathcal{F}$-affinoid.

**Proof.** Let $f_1, \dots, f_q \in H(D)$ and consider a neighborhood $W$ of $\varphi_{\mathcal{F}}$ of the form $\{\psi \in \mathrm{Mult}(H(D), \| \cdot \|_D) \mid |\varphi_{\mathcal{F}}(f_j) - \psi(f_j)|_{\infty} \le \epsilon\}$: we show that it contains a neighborhood of the
form Mult($H(E \cap D)$, $\| \cdot \|_{E \cap D}$), where $E$ is a $\mathcal{F}$-affinoid. Indeed, by Lemma 3.15 and Theorem 14.1, there exist infraconnected affinoid sets $E_j \in \mathcal{F}$, ($1 \le j \le q$) such that $|f_j(x)| - \varphi_{\mathcal{F}}(f_j)|_\infty \le \epsilon \quad \forall x \in E_j$ ($1 \le j \le q$). Then, by Proposition 2.16, $\bigcap_{j=1}^q E_j$ is an infraconnected affinoid set $E$ which belongs to $\mathcal{F}$. So, $E$ is a $\mathcal{F}$-affinoid such that $|f_j(x)| - \varphi_{\mathcal{F}}(f_j)|_\infty \le \epsilon \quad \forall x \in E_j$ ($1 \le j \le q$), therefore $|\psi(f_j)| - \varphi_{\mathcal{F}}(f_j)|_\infty \le \epsilon \quad \forall x \in E_j$ ($1 \le j \le q$), $\forall \psi \in \text{Mult}(H(E \cap D), \| \cdot \|_{E \cap D})$. Then, Mult($H(E)$, $\| \cdot \|_E$) is a neighborhood of $\varphi_{\mathcal{F}}$ in Mult($\mathbb{K}(x)$) and Mult($H(E \cap D)$, $\| \cdot \|_{E \cap D}$) is a neighborhood of $\varphi_{\mathcal{F}}$ in Mult($H(D)$, $\| \cdot \|_D$). $\square$

**Corollary 40.19.** Let $\varphi_{\mathcal{F}} \in \text{Mult}(\mathbb{K}[x])$. There exists a basis of neighborhoods of $\varphi_{\mathcal{F}}$ in $\text{Mult}(\mathbb{K}[x])$, with respect to the topology of pointwise convergence, consisting of the family of sets of the form $\text{Mult}(R(E), \| \cdot \|_E)$, where $E$ is a $\mathcal{F}$-affinoid.

**Theorem 40.20.** Let $D$ be a closed bounded set. The topology of pointwise convergence is weaker than the $\delta$-topology on $\text{Mult}(H(D), \| \cdot \|_D)$.

**Proof.** Let $\varphi_{\mathcal{F}} \in \text{Mult}(H(D), \| \cdot \|_D)$ and let $r = \text{diam}(\mathcal{F})$. Since the $\mathcal{F}$-affinoids form a basis of the filter $\mathcal{F}$, it is sufficient to show that given an $\mathcal{F}$-affinoid $E$ there exists $\epsilon > 0$ such that if $\delta(\mathcal{F}, \mathcal{G}) \le \epsilon$, then $\mathcal{F}$ is secant with $E$.

Suppose first that $\mathcal{F}$ has no center. There exists a disk $F$ included in $E$, of diameter $s > r$ with $s - r \le \frac{\epsilon}{2}$, and hence for every circular filter $\mathcal{G}$ secant with $F$, of diameter $t \in ]r, s]$, we have $\delta(\mathcal{F}, \mathcal{G}) \le 2(s-r)$. Consequently, fixing $\epsilon \le 2(s-r)$, we can see that all $\mathcal{G}$ secant with $F$ satisfy $\delta(\mathcal{F}, \mathcal{G}) \le \epsilon$, hence $F$ contains a neighborhood of $\mathcal{F}$.

Suppose now that $\mathcal{F}$ has a center $a$. Let $l$ be the minimum of the diameters of holes of $E$ inside $Q(\mathcal{F})$ and let $t$ be an annulus $\Gamma(a, r, t)$ containing no hole of $E$. Let us now fix $\epsilon < \frac{\min(t-r, r-l)}{2}$. Take $\mathcal{G}$ secant with $E$ such that $\delta(\mathcal{F}, \mathcal{G}) \le \epsilon$.

Suppose first that $\mathcal{G}$ is secant with $Q(\mathcal{F})$ and then, for instance, is secant with $d(a, r^{-})$. Then we can see that $\mathcal{G}$ is secant with $\Gamma(a, l, r)$ and hence is secant with $E$. Suppose now that $\mathcal{G}$ is not secant with $Q(\mathcal{F})$ and hence is secant with $\mathbb{K}\setminus d(a, r)$. Since $\epsilon < \frac{\min(t-r, r-l)}{2}$, we can see that $\mathcal{G}$ is secant with $\Gamma(a, r, t) \subset E$.

Consequently, we have found a neighborhood $V$ of $\varphi_{\mathcal{F}}$ with respect to the $\delta$-topology such that if $\delta(\mathcal{F}, \mathcal{G}) \le \epsilon$, then $\mathcal{G}$ is secant with $E$.
That finishes proving that every neighborhood of $\mathcal{F}$ with respect to
the pointwise topology contains a neighborhood with respect to the
$\delta$-topology. $\square$

**Remark.** The two topologies do not coincide in the general case. Suppose that a set $D$ has an increasing filter $\mathcal{F}$ of center $a$ and diameter $R$. Let $(a_n)_{n \in \mathbb{N}}$ be a sequence of $D$ such that $|a_n - a| < |a_{n+1} - a|$ and $\lim_{n \to +\infty} |a_n - a| = R$.

The sequence $(\varphi_{a_n})_{n \in \mathbb{N}}$ converges to $\varphi_{\mathcal{F}}$ with respect to the point-
wise topology but does not converge to any limit with respect to the
$\delta$-topology.

**Notation.** We denote by $\Sigma(D)$ the set of $D$-bordering circular filters and by $\Sigma_0(D)$ the set of strictly $D$-bordering circular filters.

**Remark.** Since the topology of pointwise convergence is weaker than the $\delta$-topology, the boundary of $\text{Mult}(R(D), \| \cdot \|_D)$ inside $\text{Mult}(\mathbb{K}[x])$, with respect to the $\delta$-topology, is obviously included in the boundary of $\text{Mult}(R(D), \| \cdot \|_D)$ inside $\text{Mult}(\mathbb{K}[x])$, with respect to the topology of pointwise convergence. The equality does not hold in the general case.

**Lemma 40.21.** Let $D$ be a closed bounded infraconnected subset of $\mathbb{K}$ and let $\mathcal{F}$ be a $D$-bordering but not strictly $D$-bordering circular filter. For every disk $M \in \mathcal{F}$, $\mathcal{F}$ is secant with $M \setminus D$ but not secant with any hole of $D$. Moreover, $\Sigma_0(D)$ is dense in $\Sigma(D)$ with respect to the topology of pointwise convergence.

**Proof.** Let $\varphi_{\mathcal{F}} \in \Sigma(D) \setminus \Sigma_0(D)$. Then every element of $\mathcal{F}$ contains holes of $D$ and hence is secant with $M \setminus D$. Suppose that $\mathcal{F}$ is secant with a hole $T = d(b, \rho^-)$ of $D$. Since $\mathcal{F}$ is not the peripheral of $T$, it is secant with a disk $d(c, s) \subset d(b, \rho^-)$ with $s < \rho$, and of course, it has elements $B$ of diameter $l \in ]s, \rho[$ that contain no hole of $D$, a contradiction. So, $\mathcal{F}$ is not secant with any hole of $D$. Now, take $h_1, \dots, h_n \in H(D)$ and $\epsilon > 0$. Since $\mathcal{F}$ admits a basis of infraconnected affinoid sets, we can find an infraconnected affinoid set $E \in \mathcal{F}$ such that $|h_j(x) - \varphi_{\mathcal{F}}(h_j)|_{\infty} < \epsilon \ \forall x \in E \cap D, \forall j = 1, \dots, n$. Then there exists a hole $T$ of $D$ included in $E$. Consequently, $|\varphi_T(h_j) - \varphi_{\mathcal{F}}(h_j)|_{\infty} < \epsilon \ \forall j = 1, \dots, n$. This shows that $\Sigma_0(D)$ is dense in $\Sigma(D)$. $\square$
**Corollary 40.22.** Let $D$ be a closed infraconnected subset of $\mathbb{K}$ such that $\Sigma_0(D)$ is finite. Then, $\Sigma(D) = \Sigma_0(D)$.

**Remark.** When $\Sigma_0(D)$ is not finite, it is not necessarily dense in $\Sigma(D)$ with respect to the $\delta$-topology. Consider a sequence $(a_n)_{n \in \mathbb{N}}$ with $|a_n| < |a_{n+1}|$ and $\lim_{n \to +\infty} |a_n| = 1$, let $r \in ]0, |a_1|$, and let

$$D = d(0, 2) \setminus (\bigcup_{n=1}^{\infty} d(a_n, r^{-})).$$

Then $\Sigma_0(D)$ consists of the set of circular filters $\mathcal{F}_{a_n, r}$ and $\mathcal{F}_{0,2}$. But $\Sigma(D)$ also contains $\mathcal{F}_{0,1}$. Then the sequence $(\mathcal{F}_{a_n, r})_{n \in \mathbb{N}}$ does converge to $\mathcal{F}_{0,1}$ with respect to the pointwise convergence but is not a Cauchy sequence with respect to the $\delta$-topology, and more precisely, $\delta(\mathcal{F}_{0,1}, \mathcal{F}_{a_n, r}) = 1 - r$.

**Theorem 40.23 ([15], [49], [50]).** Let $D$ be a closed bounded sub-set of $\mathbb{K}$. Then $\Sigma(D)$ is equal to the boundary of $\text{Mult}(H(D, \| \cdot \|_D)$ inside $\text{Mult}(\mathbb{K}[x])$ with respect to the topology of pointwise convergence.

**Proof.** First, we show that $\Sigma(D)$ is included in the boundary of $\text{Mult}(H(D, \| \cdot \|_D)$ inside $\text{Mult}(\mathbb{K}[x])$. Since $\Sigma_0(D)$ is dense inside $\Sigma(D)$ and since the boundary of $\text{Mult}(H(D, \| \cdot \|_D)$ inside $\text{Mult}(\mathbb{K}[x])$ is obviously closed, it is sufficient to show that $\Sigma_0(D)$ is included in that boundary. Let $\tilde{D} = d(a, r)$ and let $\varphi_{\mathcal{F}} \in \Sigma_0(D)$. Let $F$ be the subset of $\text{Mult}(\mathbb{K}[x]) \setminus \text{Mult}(H(D), \| \cdot \|_D)$ consisting of all $\varphi_{\mathcal{G}}$ such that $\mathcal{G}$ is secant with $\mathbb{K} \setminus d(a, s)$ for some $s > r$. If $\mathcal{F}$ is the peripheral of $d(a, r)$, it is obvious that $\varphi_{\mathcal{F}}$ belongs to the closure of $F$ and hence belongs to the boundary of $\text{Mult}(H(D), \| \cdot \|_D)$.

Consequently, we are led to assume that $\mathcal{F}$ is the peripheral of certain hole $d(b_j, r_j^{-})$ of $D$ and then symmetrically, it is also obvious that $\varphi_{\mathcal{F}}$ belongs to the closure of $\text{Mult}(\mathbb{K}[x]) \setminus \text{Mult}(H(D), \| \cdot \|_D)$ inside $\text{Mult}(\mathbb{K}[x])$.

Now, let us show that the boundary of $\text{Mult}(H(D, \| \cdot \|_D)$ inside $\text{Mult}(\mathbb{K}[x])$ is included in $\Sigma(D)$. Indeed, let $\varphi_{\mathcal{F}} \in \text{Mult}(H(D), \| \cdot \|_D)$ belong to the closure of $\text{Mult}(\mathbb{K}[x]) \setminus \text{Mult}(H(D), \| \cdot \|_D)$ in $\text{Mult}(\mathbb{K}[x])$. Since $\mathcal{F}$ is secant with $D$, we only need to show that it is also secant with $\mathbb{K} \setminus D$. Let $t = \text{diam}(\mathcal{F})$. Suppose it is not secant with $\mathbb{K} \setminus D$. Then we can find an $\mathcal{F}$-affinoid $F$ of the form
$d(b, u) \setminus (\bigcup_{j=1}^q d(c_j, \rho))$ that is included in $D$, with $t < u < r$, $\rho < t$ and $|c_i - c_j| = t \ \forall i \neq j$. Now, let us take $s' \in] \rho, t[$ and $s'' \in]t, u[$ and set

$$E = d(b, s'') \setminus (\bigcup_{j=1}^q d(c_j, s'^-)).$$

Let $h(x) = \prod_{j=1}^q (x - c_j)$. On one hand, we can check that when $x \in \mathbb{K} \setminus D$, either $|x - b| \ge u$, and then $|h(x)| \ge u^q$, or $|x - c_k| \le \rho$ for some $k$ and then $|h(x)| \le \rho(\prod_{j \neq k} |c_j - c_k|) = \rho t^{q-1}$.

On the other hand, when $x$ belongs to $E$, then we can check that $\rho t^{q-1} < s't^{q-1} \le |h(x)| \le s''^q < u^q$, which contradicts the hypothesis that $\varphi_F$ belongs to the closure of $\text{Mult}(\mathbb{K}[x]) \setminus \text{Mult}(H(D), \| \cdot \|_D)$ in $\text{Mult}(\mathbb{K}[x])$. Consequently, $\mathcal{F}$ is secant with $\mathbb{K} \setminus D$ and hence $\varphi_F$ belongs to $\Sigma(D)$. This finishes showing that the boundary of $\text{Mult}(H(D), \| \cdot \|_D)$ is included in $\Sigma(D)$ and hence is equal to $\Sigma(D)$. $\square$

**Definition.** A subset $S$ of $\Phi(D)$ is called a Shilov set if for every $f \in H(D)$, there exists $\mathcal{F} \in S$ such that $\varphi_{\mathcal{F}}(f) = \|f\|_D$. A Shilov set that is closed with respect to the topology of pointwise convergence is called Shilov boundary for $H(D)$ if it is the smallest closed Shilov set.

**Theorem 40.24.** Let $D$ be a closed bounded infraconnected subset of $\mathbb{K}$. Then $\Sigma(D)$ is equal to the Shilov boundary for $H(D)$.

**Proof.** First, we note that $\Sigma(D)$ is closed inside $\text{Mult}(\mathbb{K}[x])$ because, by Theorem 40.23, it is the boundary of $\text{Mult}(H(D), \| \cdot \|_D)$. By the Mittag-Leffler Theorem 16.1, it is easily seen that $\Sigma_0(D)$ is a Shilov set for $H(D)$.

Now, we show that $\Sigma(D)$ is the smallest closed Shilov set. Suppose it is not the smallest. Then, there exists another closed Shilov set $S$ which does not contain $\Sigma(D)$. Since $S$ is closed and since $\Sigma_0(D)$ is dense in $\Sigma(D)$, there exists $\psi \in \Sigma_0(D) \setminus S$. Let $\mathcal{G}$ be the strictly $D$-bordering filter such that $\psi = \varphi_{\mathcal{G}}$. Thus, $\psi$ is of the form $\varphi_{b,s}$.

Since $S$ is closed, there exists an infraconnected affinoid set $E$ of the form $d(b, s'') \setminus \bigcup_{j=1}^q d(c_j, s')$, with $s' < s < s''$ such that $\text{Mult}(H(E), \| \cdot \|_E) \cap S = \emptyset$. This means that for every $\varphi_F \in S$, $\mathcal{F}$ is not secant with $E$. Let $r = \text{diam}(D)$ and suppose first that $s = r$. Consider $g(x) = \prod_{j=1}^q (x - c_j)$. Then, $\varphi_{\mathcal{G}}(h) = r^q$, and for
every $\varphi_F \in S$, we can check that $\varphi_F(g) \le s'^q$, a contradiction to the hypothesis "S is a Shilov set".

Consequently, we have $s < r$, and therefore, since $\varphi_G \in \Sigma_0(D)$, the disk $d(b, s^-)$ is a hole of $D$. Let $h(x) = \frac{\prod_{j=1}^q (x-c_j)}{(x-b)^{q+1}}$. Then, $\varphi_G(h) = \frac{1}{s}$, and for every $\varphi_F \in S$, we can check that $\varphi_F(h) = \max(\frac{s'}{s''}, \frac{1}{s''}) < \frac{1}{s}$, a contradiction to the hypothesis "S is a Shilov set". Thus, we have proven that any closed Shilov set must contain $\Sigma(D)$ and this finishes proving that $\Sigma(D)$ is the Shilov boundary for $H(D)$. $\square$

**Corollary 40.25.** Let $D$ be an infraconnected affinoid subset of $\mathbb{K}$. The Shilov boundary for $H(D)$ is equal to $\Sigma_0(D)$.

**Corollary 40.26.** Let $D$ be a closed bounded infraconnected subset of $\mathbb{K}$. The following three sets are equal:

(i) $\Sigma(D)$,

(ii) the boundary of $\text{Mult}(H(D), \| \cdot \|_D)$ inside $\text{Mult}(\mathbb{K}[x])$ with respect to the topology of pointwise convergence,

(iii) the Shilov boundary for $(H(D), \| \cdot \|_D)$ with respect to the topology of pointwise convergence.

More precisely, we can state Corollary 40.27:

**Corollary 40.27.** Let $D$ be a closed bounded infraconnected subset of $\mathbb{K}$ having finitely many $D$-bordering filters. The following four sets are equal:

(i) $\Sigma(D)$,

(ii) the boundary of $\text{Mult}(H(D), \| \cdot \|_D)$ inside $\text{Mult}(\mathbb{K}[x])$ with respect to the topology of pointwise convergence,

(iii) the boundary of $\text{Mult}(H(D), \| \cdot \|_D)$ inside $\text{Mult}(\mathbb{K}[x])$ with respect to the $\delta$-topology,

(iv) the Shilov boundary for $(H(D), \| \cdot \|_D)$ with respect to the topology of pointwise convergence.

**Proof.** If $D$ has finitely many $D$-bordering filters, the boundary of $\text{Mult}(H(D), \| \cdot \|_D)$ inside $\text{Mult}(\mathbb{K}[x])$ is finite for both topologies and hence is reduced to $\Sigma(D) = \Sigma_0(D)$. $\square$

**Remark.** In particular, Corollary 40.27 applies to $H(D)$ when $D$ is an infraconnected affinoid set and then $H(D)$ is also called Krasner–Tate algebra [37].
Chapter 41

Mappings from $\Phi(D)$ to the
Tree $\Phi(\mathbb{K})$

We show that every element $f \in H(D)$ has continuation to a map-
ping $f_*$ from $\Phi(D)$ to $\Phi(\mathbb{K})$. Given a circular filter $\mathcal{F} \in \Phi(D)$, the
mapping that associates with each $f \in H(D)$ the circular filter $f_*(\mathcal{F})$
is uniformly continuous with respect to the norm of $H(D)$ and the
metric $\delta$ on $\Phi(\mathbb{K})$ [492].

Every analytic element on a closed bounded set $D$ has expansion to a function from $\text{Mult}(H(D), \|\cdot\|_D)$ to $\text{Mult}(\mathbb{K}[x])$ which is continuous with respect to both topologies on $\text{Mult}(\mathbb{K}[x])$. Results apply to meromorphic functions in $\mathbb{K}$ because, in particular, they define open continuous functions in $\text{Mult}(\mathbb{K}[x])$ with respect to the topology of pointwise convergence. Such a mapping is increasing with respect to the order of the tree $\text{Mult}(\mathbb{K}[x])$ if and only if it is an entire function.

**Notation.** Throughout this chapter, we denote by $D$ an infracon-
nected affinoid subset of $\mathbb{K}$.

Lemma 41.1 is easy:

**Lemma 41.1.** Let $\mathcal{F}, \mathcal{G} \in \Phi(\mathbb{K})$ be secant with a disk $d(a, r)$. Then, $\delta(\mathcal{F}, \mathcal{G}) \le r$.

**Proof.** Since $\mathcal{F}, \mathcal{G} \in \Phi(\mathbb{K})$ are secant with $d(a, r)$, their diameters are $\le r$. On the other hand, $\sup(\mathcal{F}, \mathcal{G})$ is also secant with $d(a, r)$ and hence has a diameter $\le r$. Consequently, $\delta(\mathcal{F}, \mathcal{G}) \le r$. $\square$
**Theorem 41.2.** Let the annuli $\Gamma(a, r', r)$ and $\Gamma(a, r, r'')$ be included in $D$ and let $h \in H(D)$. Then $h(\mathcal{F}_{a,r})$ is a basis of a filter thinner than a circular filter $\mathcal{H}$ of center $h(a)$ and diameter $\varphi_{a,r}(h-h(a))$.

**Proof.** Since $D$ is affinoid, up to a change of variable, we can assume that $d(a, r^{-}) \subset D$, hence $h \in \mathcal{A}(d(a, r^{-}))$. Then, putting $s = \varphi_{a,r}(h-h(a))$, we can see that, given $\rho < r$, $h(\Gamma(a, \rho, r))$ is an annulus of the form $\Gamma(h(a), \psi(\rho), s)$ with $\psi(\rho) < s$ and that $\lim_{\rho \to r} \psi(\rho) = s$. Consequently, $h(\mathcal{F}_{a,r})$ is a basis of a filter $\mathcal{C}$ secant with the increasing filter of center $h(a)$ and diameter $s = \varphi_{a,r}(h-h(a))$ and hence is secant with the circular filter $\mathcal{H}$ of center $h(a)$ and diameter $s$.

On the other hand, we have $\varphi_{a,r}(g \circ f) = \lim_{\mathcal{C}} |g(u)| \quad \forall g \in \mathbb{K}(u)$. Hence, for every $g \in \mathbb{K}(u)$, $|g(u)|$ admits a limit along $\mathcal{C}$. Therefore, by Theorem 4.22, $\mathcal{C}$ is thinner than a circular filter $\mathcal{H}'$ and hence $\mathcal{H}'$ is a circular filter secant with $\mathcal{H}$. But then it is equal to $\mathcal{H}$, which ends the proof. $\square$

**Theorem 41.3.** Let $h \in H(D)$ and let $\mathcal{F}$ be a circular filter with no center secant with $D$. Then $h(\mathcal{F})$ is a basis of a filter $\mathcal{F}'$ thinner than a circular filter $\mathcal{H}$ with no center.

**Proof.** Let $(D_n)_{n \in \mathbb{N}}$ be a canonical basis of $\mathcal{F}$. Since $\bigcap_{n \in \mathbb{N}} D_n = \emptyset$, without loss of generality, we can assume that there exists $q \in \mathbb{N}^*$ such that $D_q \subset D$. Consequently, by Theorem 22.9, each $h(D_n)$ is a disk and hence the sequence $h(D_n)$ is a decreasing sequence of disks. Suppose now the intersection of the sequence $(h(D_n))_{n \in \mathbb{N}}$ is not empty and let $b \in \bigcap_{n \in \mathbb{N}} h(D_n)$. For each $n \in \mathbb{N}$, there exists $a_n \in D_n$ such that $h(a_n) = b$. Since the equation $h(x) = b$ has finitely many solutions inside a disk, there exists $\alpha \in \bigcap_{n \in \mathbb{N}} h(D_n)$ such that $h(\alpha) = b$, a contradiction. This finishes showing that $h(\mathcal{F})$ is a basis of a circular filter with no center. $\square$

**Notation.** Let $h \in H(D)$ and let $\mathcal{F}$ be a circular filter secant with $D$. We denote by $h_*(\mathcal{F})$ the circular filter less thin than the filter admitting $h(\mathcal{F})$ for basis.

**Corollary 41.4.** Let $\mathcal{F}$ be a large circular filter on $D$ and let $h \in H(D)$. Then $h_*(\mathcal{F})$ has a center if and only if so does $\mathcal{F}$. Moreover, given $f \in H(h(D))$, then $\varphi_{h_*(\mathcal{F})}(f) = \varphi_{\mathcal{F}}(f \circ h)$.
**Proof.** We only have to check that $\varphi_{h_*(\mathcal{F})}(f) = \varphi_{\mathcal{F}}(f \circ h) \ \forall f \in H(h(D))$. Indeed,

$$ \varphi_{h_*(\mathcal{F})}(f) = \lim_{h_*(\mathcal{F})} (f(x)) = \lim_{\mathcal{F}} (h(f(x))) = \varphi_{\mathcal{F}}(h \circ f) \ \forall f \in H(h(D)). \quad \square $$

**Corollary 41.5.** Suppose $\Gamma(a,r,s) \subset D$. Let $h(x) = \sum_{-\infty}^{+\infty} a_n (x-a)^n \in H(D)$. Assume that $\nu(h, \log u) = m \neq 0 \ \forall u \in]r, s[$. Let $\mathcal{F}$ (respectively $\mathcal{G}$) be the circular filter of center $a$ and diameter $r$ (respectively $s$). Then, $h_*(\mathcal{F})$ (respectively $h_*(\mathcal{G})$) is the circular filter of center $0$ and diameter $|h|(r) = |a_m|r^m$ (respectively $|h|(s) = |a_m|s^m$). Moreover, $h_*(\mathcal{F})$ and $h_*(\mathcal{G})$ are circular filters comparable with respect to the order $\leq$.

**Proof.** By Lemma 41.2, $h(\mathcal{F})$ (respectively $h(\mathcal{G})$) is a basis of a filter $\mathcal{F}'$ (respectively $\mathcal{G}'$) thinner than the circular filter of center $a$ and diameter $\varphi_{a,r}(h-h(a))$ (respectively $\varphi_{a,s}(h-h(a)))$ i.e. $|a_m|r^m$ (respectively $|a_m|s^m$). The last statement is obvious. $\square$

**Definition.** Let $F$ be a field. Given $h = \frac{P(x)}{Q(x)} \in F(x)$ with $P, Q \in F[x]$ relatively prime, we call degree of $h$ the number $\deg(h) = \max(\deg(P), \deg(Q))$ and we call absolute degree of $h$ the number $|\deg(P) - \deg(Q)|_\infty$.

**Theorem 41.6.** Let $\mathcal{F} \in \Phi(D)$ and let $f, h \in H(D)$ be such that $\varphi_{\mathcal{F}}(f-h) < \text{diam}(h_*(\mathcal{F}))$. Then, $h_*(\mathcal{F}) = f_*(\mathcal{F})$. Further, if $E$ is a closed bounded set containing $f(D)$, then $\varphi_{f_*(\mathcal{F})} \in \text{Mult}(H(E), \| \cdot \|_E)$, and for all $g \in H(E)$, it satisfies $\varphi_{f_*(\mathcal{F})}(g) = \varphi_{\mathcal{F}}(g \circ f) \ \forall g \in H(E)$.

**Proof.** Suppose first that $\mathcal{F}$ has no center. By Theorem 41.3, we know that $h(\mathcal{F})$ is a basis of $h_*(\mathcal{F})$. Then, given a decreasing sequence of disks $(D_n)_{n \in \mathbb{N}}$ making a basis of $\mathcal{F}$, we have $D_n \subset D$ when $n$ is big enough and then we can check that $h(D_n) = f(D_n)$, which shows the conclusion.

Suppose now that $\mathcal{F}$ has a center $a$ and let $r$ be its diameter. Since $D$ is infraconnected affinoid, either it contains an annulus $\Gamma(a, r', r)$ or it contains an annulus $\Gamma(a, r, r'')$ where $f$ has no zero. In such an annulus, $f(x)$ is equal to a Laurent series $\sum_{-\infty}^{+\infty} a_n (x-a)^n$. Now, we have $\varphi_{\mathcal{F}}(f-h) < \text{diam}(h_*(\mathcal{F}))$. Then, $\varphi_{\mathcal{F}}(f-a_0) = \varphi_{a,r}(f-a_0) = \varphi_{a,r}(h-a_0) = \text{diam}(h_*(\mathcal{F}))$.
When $r'$ or $r''$ are close enough to $r$, if $D$ contains the annulus $\Gamma(a, r', r)$, we can check that the image of $\Gamma(a, r', r)$ by $f-a_0$ is equal to this by $h-a_0$, and similarly, if $D$ contains the annulus $\Gamma(a, r, r'')$, the image of $\Gamma(a, r, r'')$ by $f-a_0$ is equal to this by $h-a_0$. This holds for every such annuli in $D$ secant with $\mathcal{F}$ and shows that $h(\mathcal{F})$ and $f(\mathcal{F})$ are bases of a same filter. Consequently, $f(\mathcal{F} \cap D)$ generates the same filter as $h(\mathcal{F} \cap D)$. Then, by Lemma 41.2, there exists a unique circular filter $\mathcal{H} = h_*(\mathcal{F})$, less thin than the filter admitting for basis $f(\mathcal{F})$. Moreover, by Lemma 41.2, $\mathcal{H}$ admits $a_0$ as a center and its diameter, by construction, is $\varphi_{a,r}(h-a_0)$. Then, by Corollary 41.5, we check that $\text{diam}(f_*(\mathcal{F}))$ belongs to $|\mathbb{K}|$ if and only if so does $\text{diam}(\mathcal{F})$.

Now, for every $P \in \mathbb{K}[x]$, we set $\psi(P) = \varphi_{\mathcal{F}}(P \circ f)$. Then $\psi$ belongs to $\text{Mult}(\mathbb{K}[x])$ and therefore is of the form $\varphi_G$. Hence, by Theorem 4.10, $\mathcal{G} = \mathcal{H}$ and the last statement is immediate. $\square$

**Theorem 41.7.** Let $\mathcal{F} \in \Phi(D)$ and let $f, h \in H(D)$. Then, $\delta(f_*(\mathcal{F}), h_*(\mathcal{F})) \le \|f-h\|_D$.

**Proof.** Let $\mathcal{F}' = f_*(\mathcal{F})$ and let $\mathcal{F}'' = h_*(\mathcal{F})$. Let $r = \|f-h\|_D$. Suppose first $\text{diam}(\mathcal{F}') > r$. Then, by Theorem 41.6, we have $\mathcal{F}' = \mathcal{F}''$. Similarly, if $\text{diam}(\mathcal{F}'') > r$, we have $\mathcal{F}' = \mathcal{F}''$.

Suppose now $\text{diam}(\mathcal{F}') \le r$ and $\text{diam}(\mathcal{F}'') \le r$. Suppose first that $\mathcal{F}$ has no center. Then, by Corollary 41.4, both $\mathcal{F}'$ and $\mathcal{F}''$ have no center. Set $s' = \text{diam}(\mathcal{F}')$ and $s'' = \text{diam}(\mathcal{F}'')$. Let $(D_n)_{n \in \mathbb{N}}$ be a decreasing sequence of disks making a basis of $\mathcal{F}$ and let $E'_n = f(D_n)$, $E''_n = h(D_n)$, $n \in \mathbb{N}$. Then the sequence $(E'_n)_{n \in \mathbb{N}}$ is a basis of $\mathcal{F}'$ and the sequence $(E''_n)_{n \in \mathbb{N}}$ is a basis of $\mathcal{F}''$. But then, $\delta(E'_n, E''_n) \le r \quad \forall n \in \mathbb{N}$ and hence $E'_n = E''_n \quad \forall n \in \mathbb{N}$ because $\text{diam}(E'_n) \ge \text{diam}(\mathcal{F}')$ and $\text{diam}(E''_n) \ge \text{diam}(\mathcal{F}'')$. Consequently, $\mathcal{F}' = \mathcal{F}''$.

Suppose now that $\mathcal{F}$ has a center. Since $D$ is affinoid, without loss of generality, we can assume that $\mathcal{F}$ is the filter of center 0 and diameter 1 and that $d(0, 1) \subset D$. Then, $f(d(0, 1))$ is a disk $d(a, s)$ and $h(d(0, 1))$ is a disk $d(b, t)$ and we have $|a-b| \le r$, with $a = f(0)$ and $b = h(0)$. Now, by Theorem 41.2, $\mathcal{F}'$ is secant with $d(a, s)$ and $\mathcal{F}''$ is secant with $d(b, t)$ and we have $s = \text{diam}(\mathcal{F}')$, $t = \text{diam}(\mathcal{F}'')$. Thus, both $\mathcal{F}', \mathcal{F}''$ are secant with the disk $d(a, r) = d(b, r)$, and hence $\delta(\mathcal{F}', \mathcal{F}'') \le \|f-h\|_D$. $\square$
**Corollary 41.8.** For every $f \in H(D)$, $f_*$ is the uniform limit of a sequence $(h_n)_*$ with $h_n \in R(D)$, with respect to the distance $\delta$ on $\Phi(D)$.

**Definition and notation.** Let $D$ be a subset of $\mathbb{K}$ and let $u$ be $> 0$. We denote by $\Phi(D, u)$ the set of $\mathcal{F} \in \Phi(D)$ such that $\mathrm{diam}(\mathcal{F}) < u$. We call segment of $\Phi(D)$ a connected totally ordered subset of $\Phi(D)$ with respect to the $\delta$-topology.

**Lemma 41.9.** Let $\mathcal{F}$, $\mathcal{G}$ be circular filters such that $\mathrm{diam}(\mathcal{F}) < r$, $\delta(\mathcal{F}, \mathcal{G}) \le r$. Then $\mathrm{diam}(\mathcal{G}) < 2r$ and both $\mathcal{F}$, $\mathcal{G}$ are secant with a same disk $d(a, 2r)$.

**Proof.** If $\mathcal{G} \le \mathcal{F}$, then $\mathrm{diam}(\mathcal{G}) < r$ and there is nothing to show. Suppose now that $\mathcal{F} \le \mathcal{G}$. Then $\mathcal{G}$ has a center $a$. Since $\delta(\mathcal{F}, \mathcal{G}) \le r$, we have $\mathrm{diam}(\mathcal{G}) < s + r \le 2r$, hence both $\mathcal{F}$, $\mathcal{G}$ are secant with $d(a, 2r)$.

Suppose now that $\mathcal{F}$ and $\mathcal{G}$ are not comparable with respect to $\leq$. Let $\mathcal{S} = \sup(\mathcal{F}, \mathcal{G})$, and let $s = \mathrm{diam}(\mathcal{S})$. We have $\delta(\mathcal{F}, \mathcal{S}) < \delta(\mathcal{F}, \mathcal{G}) \le r$, hence $\mathrm{diam}(\mathcal{S}) \le \mathrm{diam}(\mathcal{F}) + \delta(\mathcal{F}, \mathcal{G}) < 2r$. Consequently, both $\mathcal{F}$ and $\mathcal{G}$ are secant with the disk $\mathcal{Q}(\mathcal{S})$ of diameter $s < 2r$, and therefore, $\mathrm{diam}(\mathcal{G}) < 2r$. $\square$

**Lemma 41.10.** Let $\sigma = \mathrm{codiam}(D)$, let $f \in H(D)$, and let $\epsilon$ be $> 0$. There exists $u \in ]0, \sigma[$ such that the restriction of $f_*$ to the set $\Phi(D, u)$ satisfies the following property: $\delta(f_*(\mathcal{F}), f_*(\mathcal{G})) \le \epsilon$ $\forall \mathcal{F}, \mathcal{G} \in \Phi(D, u)$ such that $\delta(\mathcal{F}, \mathcal{G}) < u$.

**Proof.** Since $f$ is uniformly continuous, there exists $t \in ]0, \sigma[$ such that $|f(x) - f(y)| \le \epsilon$ $\forall x, y \in D$ such that $|x - y| \le t$. Let $u = \frac{t}{2}$ and let $\mathcal{F}, \mathcal{G} \in \Phi(D, u)$ be such that $\delta(\mathcal{F}, \mathcal{G}) < u$. Then $\mathcal{F}$ is secant with a disk $d(a, u)$, and since $\delta(\mathcal{F}, \mathcal{G}) < u$, by Lemma 41.9, $\mathcal{G}$ is secant with $d(a, t)$. Consequently, $\delta(f_*(\mathcal{F})$ and $f_*(\mathcal{G}))$ are secant with a same disk of diameter $\epsilon$, and therefore, by Lemma 41.1 again, $\delta(f_*(\mathcal{F}), f_*(\mathcal{G})) \le \epsilon$. $\square$

**Lemma 41.11.** Let $\sigma = \mathrm{codiam}(D)$, let $h \in R(D)$, and let $u \in ]0, \sigma[$ be fixed. Let $w$ be the absolute degree of $h$. Let $\mathcal{F}_{a,r}, \mathcal{F}_{a,s}$ belong to $\Phi(D)$ with $u \le r < s$. Then,

$$ \delta(h_*(\mathcal{F}_{a,r}), h_*(\mathcal{F}_{a,s})) \le \|h\|_D \left( \frac{s^w - r^w}{u^w} \right). $$
**Proof.** Without loss of generality, we can assume $a = 0$. Suppose first that $h$ has no pole in $\Gamma(0, r, s)$. Then $h(x)$ is equal to a Laurent series $\sum_{-\infty}^{+\infty} a_n x^n$ converging in that annulus. Without loss of generality, we can suppose that $a_0 = 0$. Let $k = \nu^+(h, \log r)$ and $q = \nu^-(h, \log s)$. Suppose first $|h|(s) > |h|(r)$. Then, by Corollary 36.18, we have $1 < \frac{|h|(s)}{|h|(r)} \le \left(\frac{s}{r}\right)^q$. Now, suppose $|h|(s) < |h|(r)$. Then we have $1 < \frac{|h|(r)}{|h|(s)} \le \left(\frac{s}{r}\right)^{-q}$. On the other hand, by Theorem 36.17, we have $|q| \le w$. Consequently, when $|h|(s) > |h|(r)$, we obtain

$$0 < |h|(s) - |h|(r) \le |h|(r) \left( \left( \frac{s}{r} \right)^w - 1 \right) \le \|f\|_D \left( \left( \frac{s}{r} \right)^w - 1 \right),$$

and when $|h|(s) < |h|(r)$, we obtain

$$0 < |h|(r) - |h|(s) \le |h|(s) \left( \left( \frac{s}{r} \right)^w - 1 \right) \le \|f\|_D \left( \left( \frac{s}{r} \right)^w - 1 \right).$$

So, in both cases, we have

$$0 < \left| |h|(s) - |h|(r) \right|_{\infty} \le \|f\|_D \left( \left( \frac{s}{r} \right)^w - 1 \right) \le \|f\|_D \left( \frac{s^w - r^w}{u^w} \right),$$

therefore

$$\delta(h_*(\mathcal{F}_{0,r}), h_*(\mathcal{F}_{0,s})) \le \|h\|_D \left( \frac{s^w - r^w}{u^w} \right).$$

Consider now the general case when $h$ may admit poles in $\Gamma(0, r, s)$. Let $C(0, r_1), \dots, C(0, r_k)$ be the circles included in $\Gamma(0, r, s)$, containing at least one pole of $h$, with $r < r_1 < \dots < r_k < s$. Set $r_0 = r$ and $r_{k+1} = s$. Thus, we have $\mathcal{F}_{0,r} \subset \mathcal{F}_{0,r_1} \subset \dots \subset \mathcal{F}_{0,r_k} \subset \mathcal{F}_{0,s}$ and hence $\delta(\mathcal{F}_{0,r}, \mathcal{F}_{0,s}) = \sum_{j=0}^k \delta(\mathcal{F}_{0,r_j}, \mathcal{F}_{0,r_{j+1}})$. As we just showed, $\delta(h_*(\mathcal{F}_{0,r_j}), h_*(\mathcal{F}_{0,r_{j+1}})) \le \|h\|_D \left( \frac{r_{j+1}^w - r_j^w}{u^w} \right)$, $j = 0, \dots, k$. Therefore,

$$\delta(h_*(\mathcal{F}_{0,r}), h_*(\mathcal{F}_{0,s})) \le \|h\|_D \sum_{j=0}^k \left( \frac{r_{j+1}^w - r_j^w}{u^w} \right) = \|h\|_D \left( \frac{s^w - r^w}{u^w} \right). \quad \square$$

**Theorem 41.12.** Let $\sigma = \text{codiam}(D)$, let $h \in R(D)$, let $u \in ]0, \sigma[$ be fixed, and let $w$ be the absolute degree of $h$. Let $\mathcal{F}, \mathcal{G} \in \Phi(D)$ with $u \le \text{diam}(\mathcal{F}) = r < \text{diam}(\mathcal{G}) = s$ and $\mathcal{F} \prec \mathcal{G}$. Then,

$$\delta(h_*(\mathcal{F}), h_*(\mathcal{G})) \le \|h\|_D \left( \frac{s^w - r^w}{u^w} \right).$$
**Proof.** By Lemma 41.10, we only have to check the conclusion when $\mathcal{F}$ has no center. Then $\mathcal{F}$ has a basis consisting of a decreasing sequence of disks $(D_n)_{n \in \mathbb{N}}$ and these disks are included in $D$ when $n$ is big enough. The sequence $(h(D_n))_{n \in \mathbb{N}}$ is then a basis of $h_*(\mathcal{F})$. For each $n \in \mathbb{N}$, there is a unique circular filter $\mathcal{F}_n \in \Phi(D)$ such that $Q(\mathcal{F}_n) = D_n$. Setting $r_n = \text{diam}(D_n) = \text{diam}(\mathcal{F}_n)$, we have $\lim_{n \to \infty} (r_n) = r$, and we can check that each filter $h_*(\mathcal{F}_n)$ has a center and, by Lemma 41.11, satisfies

$$ \delta(h_*(\mathcal{F}_n), h_*(\mathcal{G})) \le \|h\|_D \left( \frac{s^w - r_n^w}{u^w} \right). $$

Now, when $n$ tends to $+\infty$, $\delta(h_*(\mathcal{F}_n), h_*(\mathcal{F}))$ tends to 0 and $\|h\|_D \left( \frac{s^w - r_n^w}{u^w} \right)$, which ends the proof. $\square$

**Theorem 41.13.** Let $\text{diam}(D) = R$ and let $f \in H(D)$. The mapping $f_*$ from $\Phi(D)$ to $\Phi(\mathbb{K})$ is uniformly continuous with respect to the $\delta$-topology on both sets.

**Proof.** By Corollary 41.8, $f_*$ is a uniform limit of a sequence $((h_*)_n)_{n \in \mathbb{N}}$ with $h_n \in R(D)$, so it is sufficient to prove the statement when $h \in R(D)$, what we assume. For convenience, we can also suppose that $D \subset d(0, R)$ with $R < 1$. Let $\sigma = \text{codiam}(D)$, let $h \in R(D)$, and let $\epsilon$ be $> 0$. By Lemma 41.10, there exists $u \in ]0, \sigma[$ such that the restriction of $h_*$ to the set $\Phi(D, u)$ satisfies the following property: $\delta(h_*(\mathcal{F}), h_*(\mathcal{G})) \le \epsilon \ \forall \mathcal{F}, \mathcal{G} \in \Phi(D, u)$ such that $\delta(\mathcal{F}, \mathcal{G}) \le u$. So, we fix $u$ in that way. Let $\mathcal{F}, \mathcal{G} \in \Phi(D)$ be such that

$$ \delta(\mathcal{F}, \mathcal{G}) \le \min \left( \frac{\epsilon u^w}{2^{w+1} \|f\|_D w R^2}, \frac{u}{2} \right). $$

Suppose first that $\text{diam}(\mathcal{F}) < \frac{u}{2}$. Since $\delta(\mathcal{F}, \mathcal{G}) \le \frac{u}{2}$, by Lemma 41.9, we can check that $\mathcal{F}, \mathcal{G}$ are secant with a same disk $d(a, u)$, and hence by Lemma 41.10, $\delta(h_*(\mathcal{F}), h_*(\mathcal{G})) \le \epsilon$.

Now, suppose that $\text{diam}(\mathcal{F}) \ge \frac{u}{2}$, $\text{diam}(\mathcal{G}) \ge \frac{u}{2}$. Let $\mathcal{S} = \text{sup}(\mathcal{F}, \mathcal{G})$, let $r = \text{diam}(\mathcal{F})$, and let $t = \text{diam}(\mathcal{S})$. Then, $\delta(f_*(\mathcal{F}), f_*(\mathcal{G})) \le \delta(f_*(\mathcal{F}), f_*(\mathcal{S})) + \delta(f_*(\mathcal{S}), f_*(\mathcal{G}))$. Since $\mathcal{F} \prec \mathcal{S}$,
by Theorem 41.12, we have

$$ \delta(f_*(\mathcal{F}), f_*(\mathcal{S})) \le \|f\|_D \left( \frac{t^w - r^w}{\left(\frac{u}{2}\right)^w} \right). $$

But now, since $r < t < 1$, we note that $t^w - r^w \le (t - r)wR^2 = \delta(\mathcal{S}, \mathcal{F})wR^2$, hence

$$ \delta(f_*(\mathcal{F}), f_*(\mathcal{S})) \le \|f\|_D \left( \frac{(t-r)wR^2}{\left(\frac{u}{2}\right)^w} \right) = \|f\|_D \left( \frac{\delta(\mathcal{F}, \mathcal{S})wR^2}{\left(\frac{u}{2}\right)^w} \right) \le \frac{\epsilon}{2}. $$

Similarly, we have $\delta(f_*(\mathcal{G}), f_*(\mathcal{S})) \le \frac{\epsilon}{2}$. Therefore, $\delta(f_*(\mathcal{F}), f_*(\mathcal{G})) \le \epsilon \forall \mathcal{F}, \mathcal{G} \in \Phi(D)$ such that $\delta(\mathcal{F}, \mathcal{G}) \le \min(\frac{\epsilon u^w}{2^{w+1}\|f\|_D}, \frac{u}{2})$. Consequently, $f_*$ is uniformly continuous. □

**Notation.** Given $f \in \mathcal{M}(K)$, we denote by $\Pi(f)$ the set of poles of $f$.

**Lemma 41.14.** Let $f \in \mathcal{M}(K)$ and suppose that $\Pi(f) \cap D = \emptyset$. Then, $f \in H(D)$.

**Proof.** We can write $f$ in the form $\frac{q}{h}$, where $g, h \in \mathcal{A}(K)$ have no common zeros and $h(x) \ne 0 \forall x \in D$, hence $h$ is invertible in $H(D)$. □

<figure><img src="image_17.png" /></figure>

**Theorem 41.15.** Let $f \in H(D)$ and let $B = f(D)$. Then, $f_*(\Phi(D)) = \Phi(B)$.

**Proof.** First, let $\mathcal{F} \in \Phi(D)$. The circular filter $f_*(\mathcal{F})$ is clearly secant with $B$ and hence belongs to $\Phi(B)$.

Now, let $\mathcal{G} \in \Phi(B)$. Let $(b_n)_{n \in \mathbb{N}}$ be a sequence of $B$ thinner than $\mathcal{G}$ and let $(a_n)_{n \in \mathbb{N}}$ be a sequence of $D$ such that $f(a_n) = b_n \forall n \in \mathbb{N}$. Since the sequence $(a_n)_{n \in \mathbb{N}}$ is bounded, by Corollary 3.20, we can extract a subsequence $(a_{\zeta(m)})_{m \in \mathbb{N}}$ which is thinner than a circular filter $\mathcal{H}$. Then, $\mathcal{H} \in \Phi(D)$. And then the sequence $(b_{\zeta(m)})_{m \in \mathbb{N}}$ is thinner than $f_*(\mathcal{H})$, hence $f_*(\mathcal{H}) = \mathcal{G}$. □

<figure><img src="image_18.png" /></figure>

**Lemma 41.16.** Let $f \in H(D)$ and $\mathcal{F}_{a,r} \in \Phi(D)$. Then $f_*(\mathcal{F}_{a,r})$ is of the form $\mathcal{F}_{b,s}$. Moreover, $s$ belongs to $|K|$ if and only if so does $r$. Let $\mathcal{F}_{e,u} \in \Phi^\circ(f(D))$ with $u \in |K|$. There exists $\mathcal{F}_{c,t} \in \Phi^\circ(D)$ such that $f_*(\mathcal{F}_{c,t}) = \mathcal{F}_{e,u}$, with $t \in |K|$. □
**Proof.** By Corollary 41.5, it is obvious that $f_*(\mathcal{F}_{a,r})$ is a circular filter with a center $b$ and a diameter $s$. Moreover, by Corollary 41.5 again, $s$ belongs to $|\mathbb{K}|$ if and only if so does $r$. Now, consider $\mathcal{F}_{e,u}$. By Lemma 41.15, there exists $\mathcal{F}_{c,t} \in \Phi(D)$ such that $f_*(\mathcal{F}_{c,t}) = \mathcal{F}_{e,u}$ and then, as we just saw, $t$ belongs to $|\mathbb{K}|$ if and only if so does $u$. $\square$

**Theorem 41.17.** Let $f \in \mathcal{M}(\mathbb{K})$ (respectively $f \in \mathcal{M}(d(a, R^-))$). Then $f_*$ is increasing from $\Phi(\mathbb{K})\setminus\Pi(f)$ to $\Phi(\mathbb{K})$ if and only if $f \in \mathcal{A}(\mathbb{K})$ (respectively $f \in \mathcal{A}(d(a, R^-))$).

**Proof.** Let $f \in \mathcal{A}(\mathbb{K})$ (respectively $f \in \mathcal{A}(d(a, R^-))$). Let $\mathcal{F}, \mathcal{G} \in \Phi(\mathbb{K})$ be such that $\mathcal{F} \prec \mathcal{G}$. Let $s = \text{diam}(\mathcal{G})$ and $r = \text{diam}(\mathcal{F})$, hence $r < s$. Suppose first that $\mathcal{F}$ has a center $a$. Without loss of generality, we can assume $a = 0$. Let $f(x) = \sum_{n=0}^{\infty} a_n x^n$, and let $k = \nu(f, \log r)$, $q = \nu(f, \log s)$. Then, by Theorem 41.6, we have $f_*(\mathcal{F}) = |a_k|r^k$, $f_*(\mathcal{G}) = |a_q|r^q$, and of course, $|a_k|r^k = |f|(r)$, $|a_q|r^q = |f|(s)$, hence $|f|(r) < |f|(s)$. Let $\rho = |f|(r)$, $\sigma = |f|(s)$. So we have $f_*(\mathcal{F}) = \mathcal{F}_{0,\rho}$, $f_*(\mathcal{G}) = \mathcal{F}_{0,\sigma}$, hence $f_*(\mathcal{F}) \prec f_*(\mathcal{G})$. Now, suppose that $\mathcal{F}$ has no center. Then $\mathcal{F}$ admits a center in a spherically complete extension $\tilde{\mathbb{K}}$ of $\mathbb{K}$ and $f$ has an expansion $\tilde{f}$ in $\tilde{\mathbb{K}}$, hence the relation $f_*(\mathcal{F}) \prec f_*(\mathcal{G})$ holds in $\tilde{\mathbb{K}}$, so it holds in $\mathbb{K}$.

Now, consider $f \in \mathcal{M}(\mathbb{K})\setminus\mathcal{A}(\mathbb{K})$ (respectively $f \in \mathcal{M}(d(a, R^-))\setminus\mathcal{A}(d(a, R^-))$) and let $a$ be a pole of $f$. Let $d(a, l)$ be a disk such that $f$ admits neither zeros nor poles different from $a$ inside $d(a, l)$. Without loss of generality, we can assume $a = 0$. Let $r, s \in ]0, l[$ be such that $r < s$. Set $\omega_0(f) = -q$. Inside $d(0, l)$, $|f|(r)$ is of the form $|a|r^{-q}$. Let $\rho = |f|(r)$, $\sigma = |f|(s)$. Then $f_*(\mathcal{F}_{0,r}) = \mathcal{F}_{0,\rho}$, $f_*(\mathcal{F}_{0,t}) = \mathcal{F}_{0,\sigma}$, hence of course $\mathcal{F}_{0,\sigma} \prec \mathcal{F}_{0,\rho}$, which shows that $f_*$ is not increasing. $\square$
Chapter 42

Injective Analytic Elements

In this chapter, D is an infraconnected affinoid subset of $K$.

This chapter is aimed at characterizing the injective meromorphic functions in a subset that is a chained union of infraconnected affinoid subsets. The relation satisfied by such injective functions recalls the one obtained in [58] by Yvette Perrin. The equality

$$|f(x) - f(y)|^2 = |x - y|^2 |f'(x)f'(y)| \quad \forall x, y \in D$$

was shown for various kinds of injective analytic functions. In Chapter 49 of Ref. [44], it was suggested that all injective analytic elements should satisfy that by making the conjecture that an injective analytic element should be the product of a major Moebius function by a train of minor Moebius functions.

In Ref. [80], J. Rivera Letelier stated the above equality for all injective meromorphic functions. Here we give a complete proof of this equality by using tools of circular filters and the Shilov boundary.

**Notation.** Recall that we denote by $U$ the unit disk $d(0, 1)$ and by $\mathcal{K}$ the residue class field of $K$. We use the hyperbolic distance $\tilde{\delta}$ and defined on $\Phi^{\circ}(K)$ as follows: consider two circular filters $\mathcal{F}, \mathcal{G} \in \Phi^{\circ}(K)$, and let $\mathcal{S} = \sup(\mathcal{F}, \mathcal{G})$. We set $\tilde{\delta}(\mathcal{F}, \mathcal{G}) = 2\log(\text{diam}(\mathcal{S})) - \log(\text{diam}(\mathcal{F})) - \log(\text{diam}(\mathcal{G}))$.

Thus, Lemma 42.1 is immediate:

**Lemma 42.1.** Let $a, b \in K$, $r, s \in \mathbb{R}^*$. Then,

$$\tilde{\delta}(\mathcal{F}_{a,r}, \mathcal{F}_{b,s}) = 2 \max(\log r, \log s, \log |a - b|) - \log r - \log s.$$
**Lemma 42.2.** $\tilde{\delta}$ is a distance on $\Phi^\circ(\mathbb{K})$ that defines the same topology as $\delta$ on $\Phi^\circ(\mathbb{K})$.

**Proof.** We only have to show that $\tilde{\delta}$ satisfies the triangular inequality. The proof is similar to this of Theorem 40.14, by replacing each diameter by its logarithm, but it also takes into account the concavity of the logarithm. Next, as the logarithm function is locally bicontinuous, it is easily seen that each neighborhood with respect to $\delta$ is a neighborhood with respect to $\tilde{\delta}$ and vice versa. $\square$

Next, we need several lemmas.

**Lemma 42.3.** Let $\mathcal{F}_{a,r} \in \Phi^\circ(\mathbb{K})$ and let $f(x) = \frac{1}{x}$. Then,
$$ \text{if } |a| \le r, \text{ then } f_*(\mathcal{F}_{a,r}) = \mathcal{F}_{0,\frac{1}{r}}, $$
$$ \text{if } |a| > r, \text{ then } f_*(\mathcal{F}_{a,r}) = \mathcal{F}_{\frac{1}{a},\frac{r}{|a|^2}}. $$

**Proof.** Suppose first $|a| \le r$. The annuli $\Gamma(0, r - \epsilon, r + \epsilon)$ belong to $\mathcal{F}_{a,r}$ and their images by $f$ are annuli of the form $\Gamma(0, \frac{1}{r+\epsilon}, \frac{1}{r-\epsilon})$. But the unique circular filter admitting all these annuli for elements is clearly $\mathcal{F}_{0,\frac{1}{r}}$.

Suppose now that $|a| > r$. The annuli $\Gamma(a, r - \epsilon, r + \epsilon)$ belong to $\mathcal{F}_{a,r}$ and their images are annuli of the form $\Gamma(\frac{1}{a}, \frac{r-\epsilon}{|a|^2}, \frac{r+\epsilon}{|a|^2})$. But the unique circular filter admitting these annuli for elements is clearly $\mathcal{F}_{\frac{1}{a},\frac{r}{|a|^2}}$, which ends the proof. $\square$

**Lemma 42.4.** Let $f \in H(D)$ be a Moebius function. Then $f_*$ is isometric in $\Phi^\circ(D)$ with respect to the distance $\tilde{\delta}$.

**Proof.** $f$ is equal to a composition of particular Moebius functions of three types:

(i) $h_1(x) = x + c, c \in \mathbb{K}$,

(ii) $h_2(x) = \lambda x, \lambda \in \mathbb{K}$,

(iii) $h_3(x) = \frac{1}{x}$.

Consequently, it is sufficient to prove that each $h_j$ is isometric. $h_1$ is obviously isometric with respect to $\tilde{\delta}$. Let us check that so is $h_2$. For simplifying notation, put $g = f_2$ and $l = h_3$. Consider two circular filters $\mathcal{F}_{a,r}$, $\mathcal{F}_{b,s}$ with $r \le s$. It is immediately seen that
$g_*(\mathcal{F}_{a,r}) = \mathcal{F}_{\lambda a, |\lambda|r}$ and $g_*(\mathcal{F}_{b,s}) = \mathcal{F}_{\lambda b, |\lambda|s}$. Then,

$$
\begin{align*}
\tilde{\delta}(g_*(\mathcal{F}_{a,r}), g_*(\mathcal{F}_{b,s}))
&= 2 \max(\log(|\lambda|r), \log(|\lambda|s), \log(|\lambda a - \lambda b|)) - \log(|\lambda a|) - \log|\lambda s| \\
&= 2 \max(\log r, \log s, \log(|a - b|)) - \log r - \log s = \tilde{\delta}(\mathcal{F}_{a,r}, \mathcal{F}_{b,s}).
\end{align*}
$$

We now consider $l = h_3$ and check that $l$ preserves the distance in
each of the following six cases:

(1) Suppose first $a \neq 0, b \neq 0, |a| \le |b|, |a-b| = |b|, r < |a|, s < |b|$.
By Lemma 42.3, we have $l_*(\mathcal{F}_{a,r}) = \mathcal{F}_{\frac{r}{a}, \frac{|a|}{2}}$, $l_*(\mathcal{F}_{b,s}) = \mathcal{F}_{\frac{s}{b}, \frac{|b|}{2}}$, hence

$$
\begin{align*}
\tilde{\delta}(l_*(\mathcal{F}_{a,r}), l_*(\mathcal{F}_{b,s})) = 2 \max & \left( \log \frac{r}{|a|^2}, \log \frac{s}{|b|^2}, \log \left( \left| \frac{1}{a} - \frac{1}{b} \right| \right) \right) \\
& - \log \frac{r}{|a|^2} - \log \frac{s}{|b|^2}.
\end{align*}
$$

Since $\frac{1}{|a|} \ge \frac{1}{|b|}$ and $\frac{r}{|a|^2} < \frac{1}{|a|}$, $\frac{s}{|b|^2} < \frac{1}{|b|}$, we get

$$
\begin{align*}
\tilde{\delta}(l_*(\mathcal{F}_{a,r}), l_*(\mathcal{F}_{b,s})) &= 2 \log \frac{1}{|a|} - \log \frac{r}{|a|^2} - \log \frac{s}{|b|^2} \\
&= \log |a-b| - \log r - \log s = \tilde{\delta}(\mathcal{F}_{a,r}, \mathcal{F}_{b,s}).
\end{align*}
$$

(2) Suppose $|a| = |b|$, $|a-b| < |b|$, $r < |a-b|$, $s < |a-b|$. By Lemma 42.3, we have again $l_*(\mathcal{F}_{a,r}) = \mathcal{F}_{\frac{r}{a}, \frac{|a|}{2}}$, $l_*(\mathcal{F}_{b,s}) = \mathcal{F}_{\frac{s}{b}, \frac{|b|}{2}}$, hence

$$
\begin{align*}
\tilde{\delta}(l_*(\mathcal{F}_{a,r}), l_*(\mathcal{F}_{b,s})) ={}& 2 \max \left( \log \frac{r}{|a|^2}, \log \frac{s}{|b|^2}, \log \left( \left| \frac{1}{a} - \frac{1}{b} \right| \right) \right) \\
& - \log \frac{r}{|a|^2} - \log \frac{s}{|b|^2}.
\end{align*}
$$

Now, $\log(\frac{1}{a} - \frac{1}{b}) = \log \frac{|b-a|}{|a|^2} > \max(\log \frac{r}{|a|^2}, \log \frac{s}{|b|^2})$, so

$$
\begin{align*}
\tilde{\delta}(l_*(\mathcal{F}_{a,r}), l_*(\mathcal{F}_{b,s})) &= 2 \log \left( \frac{|b-a|}{|a|^2} \right) - \log \frac{r}{|a|^2} - \log \frac{s}{|b|^2} \\
&= 2 \log |b-a| - \log r - \log s = \tilde{\delta}(\mathcal{F}_{a,r}, \mathcal{F}_{b,s}).
\end{align*}
$$
(3) Suppose $a = b \neq 0$, $r < s < |b|$. Then, $l_*(\mathcal{F}_{a,r}) = \mathcal{F}_{\frac{1}{a}, \frac{r}{|a|^2}}$, $l_*(\mathcal{F}_{a,s}) = \mathcal{F}_{\frac{s}{a}, \frac{|a|^2}{2}}$, hence

$$ \begin{aligned} \tilde{\delta}(l_*(\mathcal{F}_{a,r}), l_*(\mathcal{F}_{a,s})) &= 2 \max \left( \log \frac{r}{|a|^2}, \log \frac{s}{|a|^2} \right) - \log \frac{r}{|a|^2} - \log \frac{s}{|a|^2} \\ &= \log r - \log s = \tilde{\delta}(\mathcal{F}_{a,r}, \mathcal{F}_{b,s}). \end{aligned} $$

(4) Suppose that $a = 0, b \neq 0$, $r < |b|, s < |b|$. Then, $l_*(\mathcal{F}_{0,r}) = \mathcal{F}_{0,\frac{1}{r}}$, $l_*(\mathcal{F}_{b,s}) = \mathcal{F}_{\frac{s}{b}, \frac{|b|^2}{2}}$, hence

$$ \begin{aligned} \tilde{\delta}(l_*(\mathcal{F}_{a,r}), l_*(\mathcal{F}_{b,s})) &= 2 \max \left( \log \frac{1}{r}, \log \frac{s}{|b|^2}, \log \left( \left| \frac{1}{b} \right| \right) \right) - \log \frac{1}{r} - \log \frac{s}{|b|^2} \\ &= 2 \log \frac{1}{r} - \log \frac{1}{r} - \log \frac{s}{|b|^2} = 2 \log |b| - \log r - \log s \\ &= \tilde{\delta}(\mathcal{F}_{0,r}, \mathcal{F}_{b,s}). \end{aligned} $$

(5) Suppose finally $a = b = 0$, $r < s$. Then, $l_*(\mathcal{F}_{0,r}) = \mathcal{F}_{0,\frac{1}{r}}$, $l_*(\mathcal{F}_{0,s}) = \mathcal{F}_{0,\frac{1}{s}}$, so

$$ \begin{aligned} \tilde{\delta}(l_*(\mathcal{F}_{0,r}), l_*(\mathcal{F}_{0,s})) &= \tilde{\delta}(\mathcal{F}_{0,\frac{1}{r}}, \mathcal{F}_{0,\frac{1}{s}}) \\ &= 2 \max \left( \log \frac{1}{r}, \log \frac{1}{s} \right) - \log \frac{1}{r} - \log \frac{1}{s} \\ &= \log s - \log r = \tilde{\delta}(\mathcal{F}_{0,r}, \mathcal{F}_{0,s}). \end{aligned} $$

(6) Suppose now that $b = 0$, $r < |a|$, $s \ge |a|$. As in the previous case, we have $l_*(\mathcal{F}_{0,s}) = \mathcal{F}_{0,\frac{1}{s}}$, $l_*(\mathcal{F}_{a,r}) = \mathcal{F}_{\frac{r}{|a|}, \frac{|a|^2}{2}}$, hence

$$ \begin{aligned} \tilde{\delta}(l_*(\mathcal{F}_{a,r}), l_*(\mathcal{F}_{b,s})) &= 2 \max \left( \log \frac{r}{|a|^2}, \log \left( \left| \frac{1}{|a|} \right| \right) \right) - \log \frac{1}{s} - \log \frac{r}{|a|^2} \\ &= -2 \log |a| + \log s - \log r + 2 \log |a| = \log s - \log r \\ &= \tilde{\delta}(\mathcal{F}_{0,r}, \mathcal{F}_{b,s}). \end{aligned} $$

Now, we check that one of those six cases occurs:
(A) suppose that $\mathcal{F}_{a,r} \prec \mathcal{F}_{b,s}$. Then we can assume $a = b$. If $a = b = 0$, we are in case (5). If 0 is not a center of both filters, we are in case (3). If $a = 0$ but $b$ is not a center of $\mathcal{F}_{b,s}$, we are in case (6). There is no other possibility when $\mathcal{F}_{a,r} \prec \mathcal{G}_{b,s}$.

(B) Suppose now that $\mathcal{F}_{a,r}$ and $\mathcal{G}_{b,s}$ are not comparable for the relation $\preceq$. If $a = 0$ and $b \neq 0$, we have $r < |b|$ and $s < |b|$, so we are in case (4). Suppose now $a \neq 0$, $b \neq 0$. Then we are in case (1) or in case (2). That finishes the proof of the lemma. $\square$

In order to prove Lemma 42.7, we have to recall a classical lemma in algebra:

**Lemma 42.5.** Let $F$ be an algebraically closed field and let $f, g$ be rational functions such that $g \circ f$ is the identity on $F$. Then both $f$ and $g$ are Moebius functions.

We also need Lemma 42.6:

**Lemma 42.6.** Let $h(x) = \frac{ax+b}{cx+d}$ with $a, b, c, d \in U$ and $|ad-bc| = 1$ and let $\mathcal{G} = \mathcal{F}_{0,1}$. Then, $h_*(\mathcal{G}) = \mathcal{G}$.

**Proof.** Let $\bar{h}$ be the residue class of $h$ in $\mathcal{K}(u)$. The Moebius function $\bar{h}$ is injective and maps all elements of $\mathcal{K}$ to $\mathcal{K}$ except at most one. Consider now a $\mathcal{G}$-affinoid $D$. Since $\mathcal{G} = \mathcal{F}_{0,1}$, all classes of $U$ are included in $D$ except at most finitely many, and therefore, $D$ is sent by $h$ onto a set containing all classes of $U$ except a finite number of them. Therefore, $h_*(\mathcal{G})$ is less thin than the filter $\mathcal{T}$ generated by the sets consisting of all classes of $U$ except finitely many. And clearly, $\mathcal{G}$ is the unique circular filter thinner than $\mathcal{T}$. Hence, $h_*(\mathcal{G}) = \mathcal{G}$. $\square$

**Lemma 42.7.** Suppose $\tilde{D} = U$. Let $f \in H(D)$ be such that $f_*(\mathcal{F}_{0,1}) = \mathcal{F}_{0,1}$. Then $|f(x)| = 1$ holds in all classes of $U$ except maybe finitely many and there exists $h \in R(D)$ such that

(i) $\|f - h\|_D < 1$,

(ii) $h = \frac{P}{Q}$ with $P, Q \in \mathbb{K}[x]$, $\|P\| = 1$, $\|Q\| = 1$,

(iii) $h_*(\mathcal{F}_{0,1}) = \mathcal{F}_{0,1}$.

Moreover, given any function $h \in R(D)$ such that $\|f - h\|_D < 1$, then $h$ satisfies (ii) and (iii).
**Proof.** We know that $|f(x)| = |f|(1)$ holds in all classes of $U$
except maybe a finite number of them. But since $f_*(\mathcal{F}_{0,1}) = \mathcal{F}_{0,1}$,
we have $|f|(1) = 1$. Consider now $h \in K(x)$ satisfying $\|f-h\|_D < 1$.
By Theorem 41.6, we have $h_*(\mathcal{F}_{0,1}) = f_*(\mathcal{F}_{0,1}) = \mathcal{F}_{0,1}$. Then we
can write $h$ in the form $\frac{P}{Q}$ with $P, Q \in K[x]$, $\|P\| = 1$,
$\|Q\| = 1$. $\square$

**Notation.** Let $\mathcal{F}_{0,1} \in \Phi(D)$ and let $f \in H(D)$ be such that $f_*(\mathcal{F}_{0,1}) = \mathcal{F}_{0,1}$. Let $h \in R(D)$ satisfy $\|f-h\|_D < 1$.

**Lemma 42.8.** Suppose $\tilde{D} = U$ and let $\mathcal{F}_{0,1} \in \Phi(D)$, let $f \in H(D)$ be bianalytic and assume that $f_*(\mathcal{F}_{0,1}) = \mathcal{F}_{0,1}$. There exists a neighborhood $V$ of $\mathcal{F}_{0,1}$ with respect to the metric $\tilde{\delta}$ such that the restriction of $f_*$ to $V$ is isometric. Moreover, the image $\bar{f}$ of $f$ in the residue class field $\mathcal{K}(t)$ is a Moebius function.

**Proof.** Let us set again $\mathcal{G} = \mathcal{F}_{0,1}$. Since $f$ is bianalytic, the inverse function $g = f^{-1}$ belongs to $H(f(D))$. We define a neighborhood $V$ of $\mathcal{F}_{0,1}$ such that $f_*$ is isometric in $V$ with respect to the distance $\tilde{\delta}$. Consider the image $\bar{f}$ of $f$ in the residue class field $\mathcal{K}(u)$. Since $\varphi_{0,1}(f) = 1$, $|f(x)|$ is equal to 1 in all classes of $U$ except maybe finitely many. Let $B$ be the union of the classes of $U$ included in $D$. Then $B$ is of the form $U \setminus (\cup_{j=1}^q d(a_i, 1^-))$ with $|a_j| \le 1$, $|a_i - a_j| = 1 \ \forall i, j \le q$. By definition, $f$ is injective in $B$. Since $f_*(\mathcal{G}) = \mathcal{G}$, for each class $\Lambda$ of $U$ included in $D$, by Corollary 36.11, $f(\Lambda)$ is a class of $U$, and hence, $\bar{f}$ takes different values in different classes of $U$. Now, similarly, the function $g$ maps all classes of $U$ but finitely many onto classes of $U$. Moreover, when $f$ takes values in a class $\Lambda$ of $C(0, 1)$, then $g$ is defined in $\Lambda$. So, similarly, $\bar{g}$ is not a constant and satisfies $\bar{g} \circ \bar{f}(u) = u$ for all elements $u \in \mathcal{K}$ but finitely many. Consequently, $\bar{g} \circ \bar{f}$ is the identity in $\mathcal{K}(u)$, and hence by Lemma 42.5, $\bar{f}$ is a Moebius function. Then we can find a Moebius function $h(x) = \frac{ax+b}{cx+d}$ such that $\bar{h} = \bar{f}$, which implies $a, b, c, d \in U$ and $|ad - bc| = 1$.

Therefore, $f(x)$ is of the form $h(x) + \tau(x)$ with $\tau \in H(D)$ satisfying $\varphi_{0,1}(\tau) < 1$. Then, by Lemma 42.6, we have $h_*(\mathcal{G}) = \mathcal{G}$. Now, since $\varphi_{0,1}(\tau) < 1$, by Theorem 41.6, we have $h_*(\mathcal{G}) = f_*(\mathcal{G}) = \mathcal{G}$. Moreover, by Theorem 41.12, both functions $h_*$ and $f_*$ are continuous with respect to the $\delta$-topology, so there exists a neighborhood
V of $\mathcal{G}$ in $\Phi^\circ(D))$ such that $\mathrm{diam}(\tau_*(\mathcal{H})) < \mathrm{diam}(h_*(\mathcal{H})) \forall \mathcal{H} \in V$, therefore $(h+\tau)_*$ and $h_*$ give the same image of $\mathcal{H}$ for every $\mathcal{H} \in V$. But by Lemma 42.4, $h_*$ is isometric, hence $f_*$ is isometric in $V$. $\square$

**Corollary 42.9.** Let $f \in H(D)$ be bianalytic and let $\mathcal{F}_{a,r} \in \Phi^\circ(D)$ with $r \in |\mathbb{K}|$. There exists a neighborhood $V$ of $\mathcal{F}_{a,r}$ with respect to the metric $\tilde{\delta}$ such that the restriction of $f_*$ to $V$ is isometric.

**Proof.** Since $r \in |\mathbb{K}|$, without loss of generality, we can suppose that $\mathcal{F}_{a,r} = \mathcal{F}_{0,1}$ and consider the restriction of $f$ to $D \cap U$. Then we can apply Lemma 42.8 to this restriction. $\square$

**Lemma 42.10.** Let $I$ and $J$ be closed bounded intervals in $\mathbb{R}_+^*$ with a homeomorphism $\phi$ from $I$ onto $J$ and assume that for each $a \in I$, there exists a neighborhood $V$ of a such that $|\log a - \log b|_\infty = |\log(\phi(a)) - \log(\phi(b))|_\infty \forall a, b \in V$. Then, $|\log a - \log b|_\infty = |\log(\phi(a)) - \log(\phi(b))|_\infty \forall a, b \in I$.

**Proof.** By elementary properties of real functions in an interval, since $\phi$ is a homeomorphism from $I$ onto $J$, it is strictly monotonous. Without loss of generality, we can assume that it is increasing. Now, let $a, b \in I$ with $a < b$. By compacity, we can find finitely many open intervals $V_k$, $k = 1, \dots, q$ such that $[a, b] = \bigcup_{k=1}^q V_k$ and $|\log x - \log y|_\infty = |\log(\phi(x) - \log(\phi(y))|_\infty \forall x, y \in V_k, \forall k = 1, \dots, q - 1$. Set $V_k =]a_k, a_{k+1}[$. Without loss of generality, we can assume that $a < a_k < a_{k+1} < a_q < b$. Then,

$$ \begin{align*} |\log(\phi(a)) - \log(\phi(b))|_{\infty} &= \log(\phi(b)) - \log(\phi(a)) \\ &= \log(\phi(a_1)) - \log(\phi(a)) + \sum_{k=1}^{q-1} \log(\phi(a_{k+1})) - \log(\phi(a_k)) \\ &\quad + \log(\phi(b)) - \log(\phi(a_q)) \\ &= \log(a_1) - \log(a) + \sum_{k=1}^{q-1} \log(a_{k+1}) - \log(a_k) + \log(b) - \log(a_q) \\ &= \log(b) - \log(a), \end{align*} $$

which ends the proof. $\square$
**Definition.** Let $\mathcal{F}_{a,r}$, $\mathcal{F}_{b,s}$ be two circular filters on $\mathbb{K}$. If $\mathcal{F}_{a,r} \prec \mathcal{F}_{b,s}$, we call shortest path from $\mathcal{F}_{a,r}$ to $\mathcal{F}_{b,s}$ the path $\gamma$ defined by the following function $\gamma$ from $[r, s]$ to $\Phi^{\circ}(\mathbb{K})$: $\gamma(u) = \mathcal{F}_{a,u}$.

Now, if $\mathcal{F}_{a,r}$ and $\mathcal{F}_{b,s}$ are not comparable, we set $S = \sup(\mathcal{F}_{a,r}, \mathcal{F}_{b,s})$, with $R = \text{diam}(S)$, and we call shortest path from $\mathcal{F}_{a,r}$ to $\mathcal{F}_{b,s}$ the path $\gamma$ defined by the following injective function from $[r, 2R - s]$ to $\Phi^{\circ}(\mathbb{K})$: $\gamma(u) = \mathcal{F}_{a,u}$, $\forall u \in [r, R]$ and $\gamma(u) = \mathcal{F}_{b,2R-u}$, $\forall u \in [R, 2R - s]$.

We prove Proposition 42.12 in two steps. Lemma 42.11 is immediate and just comes from the definition of $\tilde{\delta}$ and lets us define the distance $\tilde{d}$ on intervals in $\mathbb{R}$:

**Lemma 42.11.** Let $\mathcal{F}_{a,r}$, $\mathcal{F}_{a,s}$ be two circular filters in $\mathbb{K}$ such that $\mathcal{F}_{a,r} \prec \mathcal{F}_{a,s}$. Let $\gamma$ be the shortest path defined in the interval $I = [r, s]$ with values in $\Phi^{\circ}(\mathbb{K})$, as $\gamma(u) = \mathcal{F}_{a,u}$ $\forall u \in [r, s]$. On $I$, let $\tilde{d}$ be defined by $\tilde{d}(t, u) = |\log u - \log t|_{\infty}$. Then $\tilde{d}$ is a distance on $I$ and $\gamma$ is an isometric homeomorphism from $I$ to $\gamma(I)$ when $I$ is equipped with the distance $\tilde{d}$ and $\gamma(I)$ is equipped with the distance $\tilde{\delta}$.

Lemma 42.11 is generalized in Proposition 42.12:

**Proposition 42.12.** Let $\mathcal{F}_{a,r}$, $\mathcal{F}_{b,s}$ be two circular filters in $\mathbb{K}$ and let $\mathcal{F}_{a,R} = \mathcal{F}_{b,R} = \sup(\mathcal{F}_{a,r}, \mathcal{F}_{b,s})$. Let $\gamma$ be the shortest path from $\mathcal{F}_{a,r}$ to $\mathcal{F}_{b,s}$ defined in the interval $I = [r, 2R - s]$. Then $\gamma$ is an isometric homeomorphism from $I$ to $\gamma(I)$ when $I$ is equipped with the distance $\tilde{d}$ and $\gamma(I)$ is equipped with the distance $\tilde{\delta}$.

**Proof.** If $\mathcal{F}_{a,r} \prec \mathcal{F}_{b,s}$ or if $\mathcal{F}_{b,s} \prec \mathcal{F}_{a,r}$, then the claim is given by Lemma 42.11. Now, assume that $\mathcal{F}_{a,r}$ and $\mathcal{F}_{b,s}$ are not comparable. Then the restriction of $\gamma$ to both sections $[r, R]$, $[R, 2R - s]$ is isometric. So, we only have to check that given $u \in [r, R]$ and $w \in [R, 2R - s]$, we have $\tilde{\delta}(\mathcal{F}_{a,u}, \mathcal{F}_{b,w}) = \tilde{d}(u, w)$. But actually we check that

$$ \begin{aligned} \tilde{d}(u, w) &= |\log(u) - \log(w)|_{\infty} \\ &= |(\log(R) - \log(u)) + (\log(w) - \log(R))|_{\infty} \\ &= |\log(R) - \log(u)|_{\infty} + |\log(w) - \log(R)|_{\infty} \\ &= \tilde{d}(u, R) + \tilde{d}(R, w), \end{aligned} $$
**Theorem 42.14.** Let $f \in H(D)$ be bianalytic. Then the function $f_*$ is isometric with respect to the distance $\tilde{\delta}$.

**Proof.** Since the set of circular filters with a center is dense in the whole set $\Phi^\circ(D)$ with respect to the $\delta$-topology and therefore with respect to the topology defined by $\tilde{\delta}$, it is sufficient to prove that for any two circular filters $\mathcal{F}_{a,r}, \mathcal{F}_{b,s} \in \Phi^\circ(D)$, we have $\tilde{\delta}(f_*(\mathcal{F}_{a,r}), f_*(\mathcal{F}_{b,s})) = \tilde{\delta}(\mathcal{F}_{a,r}, \mathcal{F}_{b,s})$.

Let $\mathcal{F}_{a,r}, \mathcal{F}_{b,s} \in \Phi^\circ(D)$. Let $\mathcal{F}_{a,R} = \mathcal{F}_{b,R} = \sup(\mathcal{F}_{a,r}, \mathcal{F}_{b,s})$. Let $\mathcal{F}_{\alpha,\rho} = f_*(\mathcal{F}_{a,r})$ and $\mathcal{F}_{\beta,\nu} = f_*(\mathcal{F}_{b,s})$ and let $\mathcal{F}_{\alpha,\nu} = \sup(\mathcal{F}_{\alpha,\rho}, \mathcal{F}_{\beta,\tau})$.

Consider the shortest path $\gamma$ defined in the interval $I = [r, 2R-s]$, with values in $\Phi^\circ(D)$. Similarly, consider the shortest path $\sigma$ defined in the interval $J = [\rho, 2\nu - \tau]$, with values in $\Phi^\circ(f(D))$ such that $\sigma(\rho) = \mathcal{F}_{\alpha,\rho}$, $\sigma(2\nu - \tau) = \mathcal{F}_{\beta,\tau}$.

By Theorem 42.13, $f_*$ is injective, hence the mapping $h = \sigma^{-1} \circ f_* \circ \gamma$ is injective and hence is a bijection from $I$ onto $J$. Moreover, since both $\sigma$, $\phi$ are homeomorphism, $h$ is continuous. Consequently, it is a continuous bijection from $I$ onto $J$ with respect to the metric $\tilde{d}$. But the topology defined by the metric $\tilde{d}$ on a bounded interval of $\mathbb{R}$ is clearly the same as the usual topology. Consequently, $h$ is a continuous bijection from $I$ onto $J$ with respect to the usual topology of $\mathbb{R}$ on both $I$ and $J$. Therefore, $h$, as an injective continuous function, is a strictly monotonous function. Without loss of generality, we can assume that $h$ is strictly increasing.

Now, by Corollary 42.9, for every $u \in I \cap |\mathbb{K}|$, there exists a neighborhood $\mathcal{V}(u)$ such that $h$ is isometric in $\mathcal{V}(u)$ with respect to $\tilde{d}$. This means that in such an interval $\mathcal{V}(u)$, we have $|\log(h(y)) - \log(h(x))|_{\infty} = \log y - \log x \ \forall x, y \in \mathcal{V}(u), x < y$.

But since $|\mathbb{K}|$ is dense in $\mathbb{R}_+$, by compacity, we can find finitely many intervals $[a_j, a_{j+1}]$, with $a_j \in |\mathbb{K}|$, $j = 0, \dots, q-1$, such that $I = [a_0, a_q]$ and that $h$ is isometric in each interval $]a_j, a_{j+1}[$. Actually, by continuity, $h$ is then isometric in each interval $[a_j, a_{j+1}]$. So, in each interval $[a_j, a_{j+1}]$, we have $\frac{h(y)}{h(x)} = \frac{y}{x}$, and then, by continuity, we can check that the equality $\frac{h(y)}{h(x)} = \frac{y}{x}$ holds in all $I$. For instance, let $a \in [a_k, a_{k+1}]$ and let $b \in [a_{k+1}, a_{k+2}]$. Then,

$$ \frac{h(b)}{h(a)} = \left( \frac{h(b)}{h(a_{k+1})} \right) \left( \frac{h(a_{k+1})}{h(a)} \right) = \left( \frac{b}{a_k + 1} \right) \left( \frac{a_{k+1}}{a} \right) = \frac{b}{a}. $$
So, by induction, we can extend the relation to the whole interval I. And similarly, if h is decreasing, then we have

$$ \frac{h(b)}{h(a)} = \frac{a}{b} \quad \forall a, b \in I, a < b. $$

Consequently, h is isometric.

Now, by Proposition 42.12, both $\gamma$, $\sigma$ are isometric homeomorphisms from I (respectively J) to their respective images, when I (respectively J) is provided with $\tilde{d}$ and $\Phi^\circ(D)$ is provided with $\tilde{\delta}$. Consequently, $f_*$ is isometric with respect to $\tilde{\delta}$. $\square$

**Notation.** We denote by $\mathcal{P}$ the peripheral of D.

**Theorem 42.15.** Suppose $\tilde{D} = U$ and let $f \in H(D)$ be bianalytic and such that $f(\tilde{D}) = U$ and $f_*(\mathcal{P}) = \mathcal{P}$. Then $f_*$ preserves the order $\leq$ and the distance $\delta$ on circular filters of $\Phi(D)$. Moreover, given $\mathcal{G}, \mathcal{G}' \in \Phi(D)$, $f_*$ satisfies $f_*(\sup(\mathcal{G}, \mathcal{G}')) = \sup(f_*(\mathcal{G}), f_*(\mathcal{G}'))$.

**Proof.** Since the set of centered circular filters is dense inside $\Phi(D)$, it is sufficient to prove the statement when both $\mathcal{G}, \mathcal{G}'$ are centered and large. So we assume $\mathcal{G} = \mathcal{F}_{a,r}, \mathcal{G}' = \mathcal{F}_{a',r'}$ with $rr' > 0$.

Let us first show that $f_*$ preserves the order and hence take $a' = a, r < r'$. Set $f_*(\mathcal{F}_{a,r}) = \mathcal{F}_{b,s}, f_*(\mathcal{F}_{a,r'}) = \mathcal{F}_{b',s'}$. Since $f_*(\mathcal{P}) = \mathcal{P}$ and since $\mathcal{P}$ also is the peripheral of $f(D)$, by Theorem 42.14, we have $\tilde{\delta}(\mathcal{F}_{a,r}, \mathcal{P}) = -\log r, \tilde{\delta}(\mathcal{F}_{b,s}, \mathcal{P}) = -\log s$, hence $r = s$, and similarly, $r' = s'$. Consequently, $\tilde{\delta}(\mathcal{F}_{b,s}, \mathcal{F}_{b',r'}) = 2 \max(r, r', \log|b' - b|) - r - r'$. But by Theorem 42.14, $\tilde{\delta}(\mathcal{F}_{b,r}, \mathcal{F}_{b',r'}) = \tilde{\delta}(\mathcal{F}_{a,r}, \mathcal{F}_{a,r'}) = \log r' - \log r$. Thus, $\log(|b' - b|) \le r'$, hence $\mathcal{F}_{b',r'} = \mathcal{F}_{b,r'}$, and therefore, $f_*(\mathcal{F}_{a,r}) < f_*(\mathcal{F}_{a,r'})$, i.e. $f_*$ preserves the order.

Consequently, we can derive that $f_*$ preserves the distance $\delta$ on the set $\{\mathcal{F}_{a,\rho} | \rho \ge r\}$. Indeed, we have $\delta(\mathcal{F}_{a,\rho}, \mathcal{F}_{a,r}) = \rho - r$, but we have seen that $f_*(\mathcal{F}_{a,\rho}), f_*(\mathcal{F}_{a,r})$ are of the form $\mathcal{F}_{b,\rho}, \mathcal{F}_{b,r}$ respectively, hence $\delta(f(\mathcal{F}_{a,\rho}), f(\mathcal{F}_{a,r})) = \rho - r = \delta(\mathcal{F}_{a,\rho}, \mathcal{F}_{a,r})$. Particularly, since $f_*$ preserves $\mathcal{P}$, it is obvious that $f_*$ also preserves the diameters of circular filters.

Now, we prove that $f_*$ preserves the supremum. We consider $\mathcal{F}_{a,r}, \mathcal{F}_{a',r'}$ and set $\mathcal{M} = \sup(\mathcal{F}_{a,r}, \mathcal{F}_{a',r'})$ and $\mathcal{N} = \sup(f_*(\mathcal{F}_{a',r}), f_*(\mathcal{F}_{a,r}))$. The statement is trivial when $a = a'$. So, we can assume
that the two filters are not comparable. Therefore, $\text{diam}(\mathcal{M}) = |a - a'| > \max(r, r')$ and $\tilde{\delta}(\mathcal{F}_{a,r}, \mathcal{F}_{a,r'}) = 2 \log |a - a'| - \log r - \log r'$. Moreover, since $f_*$ preserves the diameter, we have $\text{diam}(f_*(\mathcal{M})) = \text{diam}(\mathcal{M}) = |a - a'|$. On the other hand, we have $f_*(\mathcal{F}_{a,r}) = \mathcal{F}_{b,r}$, $f_*(\mathcal{F}_{a',r'}) = \mathcal{F}_{b',r'}$. Since $f_*$ preserves the order, we have $\mathcal{F}_{b,r} \prec f_*(\mathcal{M})$, $\mathcal{F}_{b,r'} \prec f_*(\mathcal{M})$, hence $\mathcal{N} \preceq f_*(\mathcal{M})$. On the other hand, since $f_*$ preserves the distance $\tilde{\delta}$, we have

$$2 \max(\log r, \log r', \log |a - a'|) - \log r - \log r' \\ = 2 \max(\log r, \log r', \log |b - b'|) - \log r - \log r'$$

hence $\log |a - a'| = \max(\log r, \log r', \log |b - b'|)$, and hence $|b - b'| = |a - a'|$. Consequently, $\text{diam}(\mathcal{N}) = |a - a'| = \text{diam}(f_*(\mathcal{M}))$, which proves the equality, and therefore, $f_*(\mathcal{M}) = \mathcal{N}$. Consequently, $f_*(\sup(\mathcal{G}, \mathcal{G}')) = \sup(f_*(\mathcal{G}), f_*(\mathcal{G}'))$. Moreover, since $f_*$ preserves the distance $\delta$ on a totally ordered subset of $\Phi(D)$ and since $f_*(\mathcal{P}) = \mathcal{P}$, we can check that $\delta(\mathcal{M}, \mathcal{P}) = \delta(\mathcal{N}, \mathcal{P})$, and hence $\text{diam}(\mathcal{N}) = \text{diam}(\mathcal{M})$.

Now, let us show that $f_*$ preserves the distance $\delta$ in general. Let $R = \text{diam}(\mathcal{M}) = \text{diam}(\mathcal{N})$. We have $\sup(f_*(\mathcal{F}_{a,r}), f_*(\mathcal{F}_{a',r'})) = \mathcal{F}_{b,R} = \mathcal{F}_{b',R}$. Then, $f_*(\mathcal{F}_{a,r}) = \mathcal{F}_{b,r}$, $f_*(\mathcal{F}_{a',r'}) = \mathcal{F}_{b',r'}$ and $f_*(\mathcal{F}_{a,r}) = \mathcal{F}_{b,R} = \mathcal{F}_{b',R}$. Consequently, $\delta(\mathcal{F}_{a,r}, \mathcal{F}_{a',r'}) = 2R - r - r'$, and hence $\delta(f_*(\mathcal{F}_{a,r}), f_*(\mathcal{F}_{a',r'})) = \delta(\mathcal{F}_{b,r}, \mathcal{F}_{b',r'}) = 2R - r - r'$, which ends the proof. $\square$

**Theorem 42.16.** Let $f \in H(D)$ be bianalytic. Then, $f_*(\Sigma(D)) \subset \Sigma_0(f(D))$.

**Proof.** Let $\mathcal{P}$ be the peripheral of $D$, let $R = \text{diam}(\mathcal{P})$, and let $E = f(D)$. Then $f_*(\mathcal{P})$ is a circular filter $\mathcal{F}_{a,r}$. Suppose $\mathcal{F}_{a,r}$ does not belong to $\Sigma_0(E)$. By definition of $\Sigma_0(E)$, there exists no annulus $\Gamma(a, r', r)$ with empty intersection with $E$, and similarly, there is no annulus $\Gamma(a, r, r'')$ with empty intersection with $E$. Now, since $D$ is infraconnected, by Theorem 21.12, so is $E$. Therefore, by Proposition 3.17, we can find circular filters $\mathcal{F}_{a,r'}$, $\mathcal{F}_{a,r''}$ secant with $E$, with $r' < r < r''$ and $r', r'' \in |\mathbb{K}|$. Consequently, by Theorem 41.6 and Lemma 41.16, there exist circular filters $\mathcal{F}_{\alpha',\rho'}$, $\mathcal{F}_{\alpha'',\rho''}$ secant with $D$ such that $f_*(\mathcal{F}_{\alpha',\rho'}) = \mathcal{F}_{a,r'}$ and $f_*(\mathcal{F}_{\alpha'',\rho''}) = \mathcal{F}_{a,r''}$ (with $\rho', \rho'' \in |\mathbb{K}|$). Next, since $f$ is injective, by Theorem 42.13, $f_*$ is injective on the set of circular filters $\mathcal{F}_{\alpha,\rho} \in \Phi(D)$, hence both $\mathcal{F}_{\alpha',\rho'}$, $\mathcal{F}_{\alpha'',\rho''}$ are
different from $\mathcal{P}$, therefore, $\rho' < R$, $\rho'' < R$. Let $R' = \max(\rho', \rho'')$. Since $D$ is infraconnected, there exists a path $\gamma'$ from $[0, 1]$ to $\Phi(D)$ such that $\gamma(0) = \mathcal{F}_{\alpha', \rho'}$, $\gamma(1) = \mathcal{P}$, and similarly, there exists a path $\gamma''$ from $[1, 2]$ to $\Phi(D)$ such that $\gamma(1) = \mathcal{P}$, $\gamma(1) = \mathcal{F}_{\alpha'', \rho''}$. Thus, we can derive a path $\gamma$ from $[0, 2]$ to $\Phi(D)$ such that $\gamma(0) = \mathcal{F}_{\alpha', \rho'}$, $\gamma(2) = \mathcal{F}_{\alpha'', \rho''}$. Then $f_* \circ \gamma$ defines a path from $[0, 2]$ to $\Phi(E)$ such that $f_* \circ \gamma(0) = \mathcal{F}_{a, r'}$ and $f_* \circ \gamma(2) = \mathcal{F}_{a, r''}$. Let $s \in ]R', R]$.

The function in $t$: $\text{diam}(f_* \circ \gamma(t))$ is obviously continuous in $t$, and therefore, by intermediate values theorem, there exists $\tau' \in ]0, 1[$ and $\tau'' \in ]1, 2[$ such that $\text{diam}(f_* \circ \gamma(\tau')) = \text{diam}(f_* \circ \gamma(\tau''))$, a contradiction to the injectivity of $f_*$. This proves that one of the two annuli $\Gamma(a, r', r)$ and $\Gamma(a, r, r'')$ has an empty intersection with $E$, and therefore, $\mathcal{F}_{a, r}$ belongs to $\Sigma_0(E)$.

By inversion, we can generalize this proof to any other circular filter of $\Sigma(D)$. Consider a circular filter $\mathcal{F} \in \Sigma(D)$ that is not $\mathcal{P}$. Since $D$ is infraconnected affinoid, by Corollary 40.25, it is the peripheral of a hole $T = d(b, s^-)$ of $D$. Without loss of generality, we can assume that $b = 0$. Let $u(x) = \frac{1}{x}$ and let $B = u(D)$. Obviously, $u_*(\mathcal{F})$ is the peripheral $\mathcal{P}'$ of $B$ and $\Sigma(f(D)) = \Sigma(f \circ u(B))$. So, $(f \circ u)_*(\mathcal{P}')$ belongs to $\Sigma_0(f \circ u(B))$. $\square$

**Remark.** Actually, we know that given an infraconnected affinoid set $D$, $H(D)$ is a Krasner–Tate algebra [37], so for every $f \in H(D)$, $f(D)$ also is infraconnected affinoid, hence $\Sigma(f(D)) = \Sigma_0(f(D))$. On the other hand, if the field $\mathbb{K}$ has characteristic zero, an injective analytic element of the Krasner–Tate algebra $H(D)$ is a universal generator [39], which proves that the inverse function belongs to $H(f(D))$ and then $f_*$ is a bijection from $\Sigma(D)$ onto $\Sigma(f(D))$. And if $\mathbb{K}$ has characteristic $p \neq 0$, an injective function is a $p$-th root of a universal generator [39], which leads to the same result.

**Theorem 42.17.** Let $f \in H(D)$ be bianalytic. Then,

$$|f(x) - f(y)|^2 = |x - y|^2 |f'(x)f'(y)| \quad \forall x, y \in D.$$

**Proof.** Since $D$ is infraconnected affinoid, we can assume that $\tilde{D} = U$. Let $E = f(D)$.

Suppose first that $f_*(\mathcal{P})$ is the peripheral of $E$. Without loss of generality, we can assume that $0 \in E \cap D$. Let $R = \text{diam}(E)$. By Lemma 41.16, $R$ belongs to $|\mathbb{K}|$, so we can take $\lambda \in \mathbb{K}$ such that
$|\lambda| = \frac{1}{R}$. Let $g = \lambda f$. Then, $g_*(P) = P$, hence we can apply Theorem 42.15 to $g: g_*$ preserves the distance $\delta$ on $D$, so we have $|g(x)-g(y)| = |x-y| \quad \forall x, y \in D$.

We can deduce that $|g'(x)|$ is identically equal to 1 in all $D$. Indeed, let us fix $y$ in $D$. Put $a = y$ and $b = g(y)$. Then of course, $|\lim_{x \to a} \frac{g(x)-g(a)}{x-a}| = 1$, hence $|g'(x)| = 1 \quad \forall x \in D$. Therefore, we can write

$$|g(x) - g(y)|^2 = |x - y|^2 |g'(x)g'(y)| \quad \forall x, y \in D.$$

Suppose now that $f_*(P)$ is not the peripheral of $f(D)$. By Theorem 42.16, $f_*(P)$ is the peripheral of a hole $T = d(a, r^-)$ of $E$. Without loss of generality, we can suppose that $0 \in T$. By Lemma 41.16, $\text{diam}(T)$ belongs to $|\mathbb{K}|$, so we can find $\lambda \in \mathbb{K}$ such that $|\lambda| = \text{diam}(T)$. Consider now the function $h(x) = \frac{\lambda}{f(x)}$. Clearly, $h_*(P) = P$. Consequently, we have $|h(x)-h(y)|^2 = |x-y|^2|h'(x)h'(y)| \quad \forall x, y \in D$. This means

$$\left| \frac{\lambda(f(x) - f(y))}{f(x)f(y)} \right|^2 = |x - y|^2 \left| \left( \frac{\lambda f'(x)}{f(x)^2} \right) \left( \frac{\lambda f'(y)}{f(y)^2} \right) \right| \quad \forall x, y \in D,$$

and hence $|f(x) - f(y)|^2 = |x - y|^2 |f'(x)f'(y)| \quad \forall x, y \in D$ again. $\square$

**Remark.** In Ref. [58], Yvette Perrin showed that in a Hederic field, an injective meromorphic function is a Moebius function, i.e. it satisfies

$$(f(x) - f(y))^2 = (x - y)^2 f'(x) f'(y) \quad \forall x, y \in D.$$

Here we obtain a formula that looks like that one.

In Corollary 23.13, we saw that a strictly injective analytic element on a disk $d(a, r)$ is bianalytic. This is much more general and concerns particularly analytic elements on affinoid sets, but the theory of Krasner-Tate algebras is then necessary to give a clear explanation of this [37], [39]. In order to generalize Theorem 42.17, we must recall those statements.

**Theorem 42.18.** Let $D$ be an infraconnected affinoid subset of $\mathbb{K}$. Then, $H(D)$ is a Krasner-Tate algebra.

**Theorem 42.19.** Let $D$ be an infraconnected affinoid subset of $\mathbb{K}$ and let $f \in H(D)$. Then, $f(D)$ is an affinoid subset of $\mathbb{K}$.
Now, when $\mathbb{K}$ is of characteristic 0, we can add a few precisions.

**Theorem 42.20.** Let $\mathbb{K}$ have characteristic 0, let $D$ be an infraconnected affinoid subset of $\mathbb{K}$, and let $f \in H(D)$ be injective. For every $b \in sp(f)$, either $f-b$ is invertible in $H(D)$, of $(f-b)H(D)$ is a maximal ideal of $H(D)$. Moreover, $f^{-1}$ belongs to $H(f(D)$.

By Corollary 36.2, we can also derive Corollary 42.21:

In a consequence of Theorems 42.17 and 42.20, we obtain the following theorem:

**Theorem 42.21.** Let $\mathbb{K}$ have characteristic 0, let $D$ be an infraconnected affinoid subset of $\mathbb{K}$, and let $f \in H(D)$ be injective. Then,

$$|f(x) - f(y)|^2 = |x - y|^2 |f'(x)f'(y)| \quad \forall x, y \in D.$$

By Corollary 36.2, we can also derive Corollary 42.20:

**Corollary 42.22.** Let $\mathbb{K}$ have characteristic 0. Let $a \in \mathbb{K}$, $r \in \mathbb{R}_+^*$ and let $f \in \mathcal{M}(d(a, R^-))$ and let $D$ be an affinoid infraconnected subset of $d(a, R)$. Then $f$ is injective in $D$ if and only if $f'$ has no zero in $D$ and $f$ satisfies

$$|f(x) - f(y)|^2 = |x - y|^2 |f'(x)f'(y)| \quad \forall x, y \in D.$$

**Corollary 42.23.** Let $\mathbb{K}$ have characteristic 0. Let $E$ be a chained union of infraconnected affinoid subsets $(D_j)_{j \in J}$ of $\mathbb{K}$ and let $f$ be a function from $E$ to $\mathbb{K}$ whose restriction to each $D_j$ belongs to $H(D_j)$ and is injective in $E$. Then for every $x, y \in E$, we have

$$|f(x) - f(y)|^2 = |x - y|^2 |f'(x)f'(y)|.$$

**Proof.** Indeed, given $x, y \in E$, there exists a finite chained family of infraconnected subsets $D_{j(1)}, \dots, D_{j(q)}$ whose union $E'$ contains $x, y$. By Proposition 2.13, $E'$ is infraconnected, and by Proposition 2.16, it is affinoid. Moreover, by Theorem 15.12, $f$ belongs to $H(E')$, so the conclusion follows from Theorem 42.17. $\square$

**Examples.**

(1) Let $(a_n)_{n \in \mathbb{N}}$ be a sequence of $\mathbb{K}$ such that $\lim_{n \to \infty} |a_n| = 1$, let $(r_n)_{n \in \mathbb{N}}$ be a sequence of $|\mathbb{K}|$ such that
0 < $r_n < |a_n|$, and let $D = \{x \in d(0, 1^-) \setminus \bigcup_{n=1}^{\infty} d(a_n, r_n^-)$. Let $f \in H(D)$ be injective.

For every $r \in |K| \cap ]0, 1[$, the set $D_r = D \cap d(0, r)$ is clearly affinoid, hence the restriction of $f$ to $D_r$ is bianalytic, and hence, we have

$$|f(x) - f(y)|^2 = |x - y|^2 |f'(x)f'(y)| \quad \forall x, y \in D_r.$$

This holds for every $r < 1$, and hence, this holds for all $x, y \in D$.

(2) Let $f$ be a meromorphic function in $\mathbb{K}$ which is injective in the annulus $\Gamma(0, r, s)$. Given $s', r' | K$ such that $r|eqR' < s' | eqs$, then $\Gamma(0, r', s')$ is an affinoid, and hence,

$$|f(x) - f(y)|^2 = |x - y|^2 |f'(x)f'(y)| \quad \forall x, y \in \Gamma(0, r', s').$$

This holds for every $r'$, $s'$, and hence,

$$|f(x) - f(y)|^2 = |x - y|^2 |f'(x)f'(y)| \quad \forall x, y \in \Gamma(0, r, s).$$

**Proof.** This is given by Theorem 41.5 for Example 1 and Theorem 41.7 for Example 2. □
# Chapter 43

## Counting Functions and
Nevanlinna Theory

The Nevanlinna theory was made by Rolf Nevanlinna on complex functions [79] and widely used by many specialists of complex functions, particularly Walter Hyman [65]. It consists of defining counting functions of zeros and poles of a meromorphic function $f$ and giving an upper bound for multiple zeros and poles of various functions $f-b, b \in \mathbb{C}$.

A similar theory for functions in a $p$-adic field was constructed and correctly proved by A. Boutabaa [21] in the field $\mathbb{K}$, after some previous work by Ha Huy Khoai [63]. In [23], the theory was extended to functions in $M(d(0, R^{-}))$ by taking into account Lazard's problem [72].

**Notations.** Recall that given three functions $\phi, \psi, \zeta$ defined in an interval $J =]a, +\infty[$ (respectively $J =]a, R[$), with values in $[0, +\infty[$, we write $\phi(r) \le \psi(r) + O(\zeta(r))$ if there exists a constant $b \in \mathbb{R}$ such that $\phi(r) \le \psi(r) + b\zeta(r)$. We write $\phi(r) = \psi(r) + O(\zeta(r))$ if $|\psi(r) - \phi(r)|$ is bounded by a function of the form $b\zeta(r)$.

Similarly, we write $\phi(r) \le \psi(r) + o(\zeta(r))$ if there exists a function $h$ from $J =]a, +\infty[$ (respectively from $J =]a, R[$) to $\mathbb{R}$ such that $\lim_{r \to +\infty} \frac{h(r)}{\zeta(r)} = 0$ (respectively $\lim_{r \to R} \frac{h(r)}{\zeta(r)} = 0$) and such that $\phi(r) \le \psi(r) + h(r)$. And we write $\phi(r) = \psi(r) + o(\zeta(r))$ if there exists a function $h$ from $J =]a, +\infty[$ (respectively from $J =]a, R[$) to $\mathbb{R}$ such that $\lim_{r \to +\infty} \frac{h(r)}{\zeta(r)} = 0$ (respectively $\lim_{r \to R} \frac{h(r)}{\zeta(r)} = 0$) and such that $\phi(r) = \psi(r) + h(r)$.
Throughout this chapter, we denote by I the interval $[t, +\infty[$ and
by J an interval of the form $[t, R[$ with $t > 0$.

We have to introduce the counting function of zeros and poles of
a meromorphic function $f \in \mathcal{M}(\mathbb{K})$ or $f \in \mathcal{M}(d(0, R^{-}))$ counting or
not multiplicity. Here we choose a presentation that avoids assuming
that all functions we consider admit no zero and no pole at the origin.

**Definitions.** We denote by $Z(r, f)$ the counting function of zeros of $f$ in $d(0, r)$ in the following way.

Let $(a_n)$, $1 \le n \le \upsilon(r)$, be the finite sequence of zeros of $f$ such that $0 < |a_n| \le r$ of respective order $s_n$.

We set $Z(r, f) = \max(\omega_0(f), 0) \log r + \sum_{n=1}^{\upsilon(r)} s_n(\log r - \log |a_n|)$,
and so, $Z(r, f)$ is called the counting function of zeros of $f$ in $d(0, r)$,
counting multiplicity.

In order to define the counting function of zeros of $f$ without multiplicity, we put $\overline{\omega}_0(f) = 0$ if $\omega_0(f) \le 0$ and $\overline{\omega}_0(f) = 1$ if $\omega_0(f) \ge 1$.

Now, we denote by $\bar{Z}(r, f)$ the counting function of zeros of $f$
without multiplicity:

$\bar{Z}(r, f) = \overline{\omega}_0(f) \log r + \sum_{n=1}^{\upsilon(r)} (\log r - \log |a_n|)$, and so, $\bar{Z}(r, f)$ is called the counting function of zeros of $f$ in $d(0, r)$ ignoring multiplicity.

In the same way, considering the finite sequence $(b_n)$, $1 \le n \le \tau(r)$, of poles of $f$ such that $0 < |b_n| \le r$, with respective multiplicity order $t_n$, we put $N(r, f) = \max(-\omega_0(f), 0) \log r + \sum_{n=1}^{\tau(r)} t_n(\log r - \log |b_n|)$ and then $N(r, f)$ is called the counting function of the poles of $f$, counting multiplicity.

Next, in order to define the counting function of poles of $f$ without multiplicity, we put $\overline{\omega}_0(f) = 0$ if $\omega_0(f) \ge 0$ and $\overline{\omega}_0(f) = 1$ if $\omega_0(f) \le -1$ and we set $\bar{N}(r, f) = \overline{\omega}_0(f) \log r + \sum_{n=1}^{\tau(r)} (\log r - \log |b_n|)$ and then $\bar{N}(r, f)$ is called the counting function of the poles of $f$, ignoring multiplicity.

Now we can define the Nevanlinna function $T(r, f)$ in $I$ or $J$ as
$T(r, f) = \max(Z(r, f), N(r, f))$ and the function $T(r, f)$ is called
characteristic function of $f$ or Nevanlinna function of $f$.

Finally, if $S$ is a subset of $\mathbb{K}$, we denote by $Z_0^S(r, f')$ the counting function of zeros of $f'$, excluding those which are zeros of $f-a$ for any $a \in S$.
**Remark.** If we change the origin, the functions $Z, N, T$ are not changed, up to an additive constant.

By Corollary 21.2, Lemma 43.1 is easy:

**Lemma 43.1.** Let $\hat{\mathbb{K}}$ be a complete algebraically closed extension of $\mathbb{K}$ whose absolute value extends that of $\mathbb{K}$ and let $f \in \mathcal{M}(\mathbb{K})$ (respectively let $f \in \mathcal{M}(d(0, R^{-}))$). Let $\hat{d}(0, R) = \{x \in \hat{\mathbb{K}} \mid |x| < R\}$. The meromorphic function $\hat{f}$ defined by $f$ in $\hat{d}(0, R)$ has the same Nevanlinna functions as $f$.

In a $p$-adic field such as $\mathbb{K}$, the first main theorem is almost immediate and is an immediate consequence of Corollary 22.27.

**Theorem 43.2.** Let $f \in \mathcal{M}(\mathbb{K})$ (respectively $f \in \mathcal{M}(d(0, R^{-}))$) have no zero and no pole at 0. Then $\log(|f|(r)) = \Psi(f, \log r) = \log(|f(0)|) + Z(r, f) - N(r, f)$ and $\log^+(|f|(r)) \le \log(|f(0)|) + T(r, f) - N(r, f)$.

**Proof.** We can write $f(x) = \frac{h}{l}$ with $h, l \in \mathcal{A}(\mathbb{K})$ (respectively $h, l \in \mathcal{A}(d(0, R^{-}))$) such that $l(0)h(0) \ne 0$. By Corollary 22.27, we have $\log(|h|(r)) = \log(|f(0)|) + Z(r, h)$, $\log(|l|(r)) = \log(|l(0)|) + N(r, l)$, so the conclusion is obvious. $\square$

Theorem 43.3 is now immediate:

**Theorem 43.3.** Let $f, g \in \mathcal{M}(\mathbb{K})$ (respectively $f, g \in \mathcal{M}(d(0, R^{-}))$). Then, $Z(r, fg) \le Z(r, f) + Z(r, g)$, $N(r, fg) \le N(r, f) + N(r, g)$, $T(r, fg) \le T(r, f) + T(r, g)$, $T(r, f + g) \le T(r, f) + T(r, g) + O(1)$, $T(r, cf) = T(r, f) \quad \forall c \in \mathbb{K}^*$, $T(r, \frac{1}{f}) = T(r, f)$, $T(r, \frac{l}{g}) \le T(r, f) + T(r, g)$.

Suppose now $f, g \in \mathcal{A}(\mathbb{K})$ (respectively $f, g \in \mathcal{A}(d(0, R^{-}))$). Then, $Z(r, fg) = Z(r, f) + Z(r, g)$, $T(r, f) = Z(r, f)$, $T(r, fg) = T(r, f) + T(r, g) + O(1)$, and $T(r, f + g) \le \max(T(r, f), T(r, g))$. Moreover, if $\lim_{r \to +\infty} T(r, f) - T(r, g) = +\infty$, then $T(r, f + g) = T(r, f)$ when $r$ is big enough.

**Lemma 43.4.** Let $\alpha_1, \dots, \alpha_n \in \mathbb{K}$ be pairwise distinct, let $P(u) = \prod_{i=1}^n (u - \alpha_i)$, and let $f \in \mathcal{M}(d(0, R^{-}))$. Then, $Z(r, P(f)) = \sum_{i=1}^n Z(r, f - \alpha_i)$ and $\bar{Z}(r, P(f)) = \sum_{i=1}^n \bar{Z}(r, f - \alpha_i)$.

**Lemma 43.5.** Let $f \in \mathcal{M}(\mathbb{K})$. Then $f$ belongs to $\mathbb{K}(x)$ if and only if $T(r, f) = O(\log r)$.
**Proof.** If $f$ belongs to $\mathbb{K}(x)$, one can write it $\frac{P(x)}{Q(x)}$ with $P, Q \in \mathbb{K}[x]$ having no common zeros, hence $Z(r, f) = Z(r, P)$ and $N(r, f) = Z(r, Q)$, and hence $T(r, f) = O(\log r)$.

Now suppose that $f \notin \mathbb{K}(x)$. Suppose, for instance, that $f$ has infinitely many zeros $(a_n)$ of respective order $q_n$. Then let us fix $s \in \mathbb{N}$ and let $t \in \mathbb{N}$ be $> s + 1$. For $r$ big enough, we have $Z(r, f) > \sum_{n=1}^t (\log r - \Psi(a_n)) > s \log r$, hence $Z(r, f)$ is not $O(\log r)$. Similarly, if $f$ has infinitely many poles, we get to the same conclusion. □

Applying Lemma 43.1 and Theorem 37.10 to $\frac{f'}{f}$, up to a change of origin, we can derive Corollary 43.6:

**Corollary 43.6.** Let $f \in \mathcal{M}(\mathbb{K})$ (respectively $f \in \mathcal{M}(d(0, R^-))$). Then,

$$Z\left(r, \frac{f'}{f}\right) - N\left(r, \frac{f'}{f}\right) \le -\log r + O(1).$$

**Theorem 43.7.** Let $f \in \mathcal{A}(\mathbb{K})$ (respectively $f \in \mathcal{A}(d(0, R^-))$) and let $b \in \mathbb{K}$. Then, $Z(r, f) = Z(r, f - b) + O(1)$, $r \in I$ (respectively $r \in J$).

**Proof.** Let $f(x) = \sum_{n=0}^{\infty} a_n x^n$ and let $\rho \in \mathbb{R}_+^*$ (respectively $\rho \in ]0, R[$) be such that $\nu^+(f, \log \rho) > 0$ and $\nu^+(f - b, \log \rho) > 0$. Then we have $\nu^+(f, \mu) = \nu^+(f - b, \mu) \forall \mu > \log \rho$ (respectively $\forall \mu \in ]\log \rho, \log R[$). Consequently, on each circle $C(0, r)$ such that $r > \rho$ (respectively $r \in ]\rho, R[$), $f$ and $f - b$ have the same number of zeros, taking multiplicity into account. Let $(a_n)$ be the sequence of zeros of $f$, with respective multiplicity $q_n$, with $|a_n| \le |a_{n+1}|$, $n \in \mathbb{N}^*$ and $|a_n| > \rho$ if and only if $n \ge t$.

Similarly, let $(b_n)$ be the sequence of zeros of $f - b$, with respective multiplicity $s_n$, with $|b_n| \le |b_{n+1}|$, $n \in \mathbb{N}^*$ and $|b_n| > \rho$ if and only if $n \ge u$. Since $f$ and $f - b$ have the same number of zeros in $d(0, \rho)$, we also have

$$(1) \quad \sum_{n=1}^{t} q_n = \sum_{n=1}^{u} s_n.$$

Consequently, for all $r > \rho$ (respectively $r \in ]\rho, R[$), we have

$$\sum_{\substack{n \ge t, \\ |a_n| \le r}} q_n(\log r - \Psi(a_n)) = \sum_{\substack{n \ge u \\ |b_n| \le r}} s_n(\log r - \Psi(b_n)).$$
Now, suppose that both $f(0)$, $f(0) - b$ are not 0. Then,

$$Z(r, f) = \sum_{|a_n| \le r} q_n(\log r - \Psi(a_n)),$$

$$Z(r, f - b) = \sum_{|b_n| \le r} s_n(\log r - \Psi(b_n)).$$

Therefore, $Z(r, f) - Z(r, f - b)$ is reduced to

$$\sum_{\substack{|a_n| \le r, \\ |a_n| \le \rho}} (\log r - \Psi(a_n)) - \sum_{\substack{|b_n| \le r, \\ |b_n| \le \rho}} (\log r - \Psi(b_n)) \\ = \sum_{n=1}^{t} q_n(\log r - \Psi(a_n)) - \sum_{n=1}^{u} s_n(\log r - \Psi(b_n))$$

that this is a constant by (1), for $r > \rho$ (respectively for $r \in ]0, R[$).

And now, suppose that 0 is a zero of order $q_1$ of $f$. Then,

$$Z(r, f) = q_1 \log r + \sum_{n \ge 2, |a_n| \le r} q_n(\log r - \Psi(a_n)),$$

and therefore, $Z(r, f) - Z(r, f - b)$ is reduced to

$$q_1 \log r + \sum_{\substack{n \ge 2, \\ |a_n| \le \rho}} q_n(\log r - \Psi(a_n)) - \sum_{|b_n| \le \rho} s_n(\log r - \Psi(b_n)) \\ = q_1 \log r + \sum_{n=2}^{t} q_n(\log r - \Psi(a_n)) - \sum_{n=1}^{u} s_n(\log r - \Psi(b_n)),$$

and we check that this is a constant again thanks to (1).

Similarly, if $f(0) = b$, then $f$ and $f - b$ playing the same role, we have the same conclusion. □

**Theorem 43.8 (First Main Theorem).** Let $f, g \in \mathcal{M}(\mathbb{K})$ (respectively let $f, g \in \mathcal{M}(d(0, R^{-}))$). Then $T(r, f + b) = T(r, f) + O(1)$ and $\log^+|f|(r) \le T(r, f) - N(r, f) + O(1)$.

Let $h$ be a Moebius function. Then, $T(r, f) = T(r, h \circ f) + O(1)$. Let $P(X) \in \mathbb{K}[X]$. Then, $T(r, P(f)) = \deg(P)T(r, f) + O(1)$ and $T(r, f'P(f)) \ge T(r, P(f))$.
Suppose now $f$ and $g$ have finitely many poles. Then, $Z(r, fg) = Z(r, f) + Z(r, g)$, $T(r, f) = Z(r, f) + O(1)$, $T(r, fg) = T(r, f) + T(r, g) + O(1)$, and $T(r, f + g) \le \max(T(r, f), T(r, g)) + O(\log(r))$. Moreover, if $\lim_{r \to +\infty} T(r, f) - T(r, g) = +\infty$, then $T(r, f + g) = T(r, f)$ when $r$ is big enough.

**Proof.** $T(r, f + b) \le T(r, f) + O(1) \le T(r, f + b) + O(1)$, hence $T(r, f + b) = T(r, f) + O(1) \forall b \in \mathbb{K}$. Now, consider $T(r, f + g)$ when $f, g$ have finitely many poles. We have $T(r, f + g) = Z(r, f + g) + O(\log r) \le \max(Z(r, f), Z(r, g)) + O(\log(r))$.

Let $h(X) = \frac{aX+b}{cX+d}$ be a Moebius function and let $g(x) = h \circ f(x)$. We can write $h(X) = \frac{a}{c} + \frac{\lambda}{cX+d}$ with $\lambda = d(1 - \frac{a}{c})$. Then,

$$ 
\begin{aligned}
T(r, g) &= T\left(r, \frac{\lambda}{cf(x) + d}\right) + O(1) = T(r, cf(x) + d) + O(1) \\
&= T(r, f(x)) + O(1).
\end{aligned}
$$

Now, let $P(X) = \prod_{k=1}^{q}(X - a_k) \in \mathbb{K}[x]$ be a polynomial of degree $q$ and let $F(x) = P(f(x))$. Then, $T(r, f - a_k) = T(r, f) + O(1) \forall k = 1, \dots, q$, and hence, $T(r, F) = qT(r, f) + O(1)$. Moreover, zeros of $P(f)$ are not poles of $f'$ and poles of $f'$ are poles of $f$ and hence are not zeros of $P(f)$. Consequently, $N(r, f'P(f)) = N(r, P(f)) + N(r, f') = N(r, P(f)) + N(r, f) + \overline{N}(r, f)$, $Z(r, f'P(f)) = Z(r, P(f)) + Z(r, f')$. Therefore, $T(r, f'P(f)) \ge T(r, P(f))$.

It now only remains to prove that $T(r, P(f)) = qT(r, f) + O(1)$. Let $P(X) = \prod_{j=1}^{q}(X - a_j)$. It is immediate to check that $Z(r, P(f)) = \sum_{j=1}^{q} Z(r, f - a_j) = qZ(r, f) + O(1)$ and that $N(r, P(f)) = qN(r, f)$, therefore $T(r, P(f)) = qT(r, f) + O(1)$. $\square$

**Theorem 43.9.** Let $f \in \mathcal{M}(\mathbb{K})$ (respectively $f \in \mathcal{M}(d(0, R^{-}))$). There exists $\phi, \psi \in \mathcal{A}(\mathbb{K})$ (respectively $\phi, \psi \in \mathcal{A}(d(0, R^{-}))$) such that $f = \frac{\phi}{\psi}$ and $\max(T(r, \phi), T(r, \psi)) \le T(r, f) + O(1)$, $r \in I$ (respectively ($r \in J$)).

**Proof.** Let $V_1 = \mathcal{D}(f)$ and let $V_2 = \mathcal{D}(\frac{1}{f})$. Suppose first $f \in \mathcal{M}(\mathbb{K})$. By Theorem 27.4, there exists $\phi, \psi \in \mathcal{A}(\mathbb{K})$ such that $\mathcal{D}(f) = \mathcal{D}(\phi), \mathcal{D}(\frac{1}{f}) = \mathcal{D}(\psi)$. Consequently, $Z(r, f) = Z(r, \phi), N(r, f) = Z(r, \psi)$ and the claim is immediate. Now, suppose $f \in \mathcal{M}(d(0, R^{-}))$. By Theorem 27.14, there exists $\phi \in \mathcal{A}(\mathbb{K})$ such that $\mathcal{D}(f) \le \mathcal{D}(\phi)$ and
such that $|\mathcal{D}(\phi)|(r) \le |V_1|(r)+1$, $r \in J$, hence $Z(r, \phi) \le Z(r, f)+1$, $r \in J$. Let $\psi = \frac{\phi}{f}$. Then $\psi$ lies in $\mathcal{A}(d(0, R^{-}))$ because $\mathcal{D}(f) \le \mathcal{D}(\phi)$. And $\mathcal{D}(\psi) = \mathcal{D}(\frac{1}{f}) + \mathcal{D}(\phi) - \mathcal{D}(f)$. Consequently, $|\mathcal{D}(\psi)|(r) \le |\mathcal{D}(\frac{1}{f})|(r) + 1$. But $T(r, \phi) = Z(r, \phi) + O(1) = \log(|\mathcal{D}(\phi)|(r)) + o(1)$ and $T(r, \psi) = Z(r, \psi) + O(1) = \log(|\mathcal{D}(\psi)|(r)) + O(1)$. Therefore, $\max(T(r, \phi), T(r, \psi)) \le \max(Z(r, f), N(r, f)) + O(1)$, $r \in I$ (respectively $r \in J$). $\square$

**Theorem 43.10.** Let $f \in \mathcal{M}(d(0, R^{-}))$. Then $f$ belongs to $\mathcal{M}_b(d(0, R^{-}))$ if and only if $T(r, f)$ is bounded in $[0, R[$.

**Proof.** Suppose first $f \in \mathcal{A}(d(0, R^{-}))$. Without loss of generality, we can obviously suppose that $f(0) \ne 0$. By Theorem 43.2, we have $\log|f|(r) = \log(|f(0)|) + Z(r, f)$. And $|f|(r) = \sup\{|f(x)| \mid x \in d(0, r^{-})\}$, so the claim is clear. Now, consider the general case. Suppose $T(r, f)$ is not bounded, so either $Z(r, f)$ or $N(r, f)$ is not bounded. Let $f = \frac{\phi}{\psi}$ with $\phi, \psi \in \mathcal{A}(d(0, R^{-}))$. If $Z(r, f)$ is not bounded, then $\phi \notin \mathcal{A}_b(d(0, R^{-}))$. If $N(r, f)$ is not bounded, then $\psi \notin \mathcal{A}_b(d(0, R^{-}))$. Thus, $f$ cannot be put in the form $\frac{\phi}{\psi}$ with $\phi, \psi \in \mathcal{A}_b(d(0, R^{-}))$, and therefore, $f \notin \mathcal{A}_b(d(0, R^{-}))$.

Conversely, if $f \in \mathcal{A}_b(d(0, R^{-}))$, then it is of the form $\frac{\phi}{\psi}$ with $\phi, \psi \in \mathcal{A}_b(d(0, R^{-}))$, hence both $Z(r, \phi)$, $Z(r, \psi)$ are bounded. But since $Z(r, f) \le Z(r, \phi)$ and $N(r, f) \le Z(r, \psi)$, $T(r, f)$ is clearly bounded in $[0, R[$. $\square$

**Corollary 43.11.** Let $f \in \mathcal{M}_u(d(a, R^{-}))$, and let $h \in \mathcal{M}_b(d(a, R^{-}))$, $h \ne 0$. Then, $fh$ belongs to $\mathcal{M}_u(d(a, R^{-}))$.

By Theorems 43.8 and 43.10, we can also derive Corollary 43.12.

**Corollary 43.12.** Let $f \in \mathcal{M}(d(a, R^{-}))$ and let $P \in \mathbb{K}[x]$. Then $P(f)$ belongs to $\mathcal{M}_b(d(a, R^{-}))$ if and only if so does $f$.

**Theorem 43.13.** Let $\mathbb{K}$ be of characteristic 0. Let $f \in \mathcal{M}(\mathbb{K})$ (respectively $f \in \mathcal{M}(d(0, R^{-}))$). Then $Z(r, f') - N(r, f') \le Z(r, f) - N(r, f) - \log r + O(1)$, $r \in I$ (respectively $r \in J$). Moreover, $N(r, f^{(k)}) = N(r, f) + k\bar{N}(r, f) + O(1)$, $r \in I$ and $Z(r, f^{(k)}) \le Z(r, f) + k\bar{N}(r, f) - k \log r + O(1)$, $r \in I$ (respectively $r \in J$).

**Proof.** Without loss of generality, we can assume that $f, f', \dots, f^{(k)}$ have no zero and no pole at 0.
The first statement is immediate and just comes from this basic property: if $\alpha$ is a pole of $f$ of order $q$, then it is a pole of $f^{(k)}$ of order $q + k$. Next, by Theorem 43.2, we have $Z(r, f) - N(r, f) = \Psi(f, \log r) - \log(|f(0)|)$ and $Z(r, f') - N(r, f') = \Psi(f', \log r) - \log(|f'(0)|)$. But $\Psi(f', \log r) \le \Psi(f, \log r) - \log r$, hence we obtain $Z(r, f') \le N(r, f') - N(r, f) + Z(r, f) - \log(r) + O(1)$. Actually, $N(r, f') - N(r, f) = \overline{N}(r, f)$, hence $N(r, f^{(k)}) = N(r, f) + k\overline{N}(r, f)$. Next, $Z(r, f') \le Z(r, f) + \overline{N}(r, f) - \log(r) + O(1)$. Now, suppose that the second statement has been proved for $k \le t$. Thus, we have $Z(r, f^{(t+1)}) \le Z(r, f^{(t)}) + \overline{N}(r, f^{(t)}) - \log(r) = O(1)$. But as we just noted, $\overline{N}(r, f^{(t)}) = \overline{N}(r, f)$, hence $Z(r, f^{(t+1)}) \le Z(r, f) + (t + 1)\overline{N}(r, f) - (t + 1)\log(r) + O(1)$. $\square$

Lemma 43.14 is now classical and easily checked:

**Lemma 43.14.** Let $\alpha_1, \dots, \alpha_q \in \mathbb{K}$ be pairwise distinct, let $S = \{\alpha_1, \dots, \alpha_q\}$, and let $P(x) = \prod_{j=1}^q (x - \alpha_j)$. Let $f \in \mathcal{M}(\mathbb{K})$ (respectively $f \in \mathcal{M}(d(0, \mathbb{R}^-))$). Then,

$$ \sum_{j=1}^{n} Z(r, f - \alpha_j) = Z(r, P(f)), \quad \sum_{j=1}^{n} \overline{Z}(r, f - \alpha_j) = \overline{Z}(r, P(f)) \quad \forall r \in I $$

(respectively $\forall r \in J$). Moreover, assuming that $\mathbb{K}$ is of characteristic 0, we have

$$ Z(r, P(f)) - \overline{Z}(r, P(f)) \le Z(r, f') \le Z(r, f) + \overline{N}(r, f) - \log(r) + O(1) \quad \forall r \in I \quad (\text{respectively } \forall r \in J). $$

**Corollary 43.15.** We assume that $\mathbb{K}$ is of characteristic 0. Let $f \in \mathcal{M}(\mathbb{K})$ (respectively $f \in \mathcal{M}(d(0, \mathbb{R}^-))$). Then, $T(r, f^{(k)}) \le (k+1)T(r, f) + O(1)$ ($r \in I$) (respectively $r \in J$).

**Theorem 43.16.** We assume that $\mathbb{K}$ is of characteristic 0. Let $f \in \mathcal{M}(\mathbb{K})$ (respectively $f \in \mathcal{M}(d(0, \mathbb{R}^-))$). Then, $T(r, f) - Z(r, f) \le T(r, f') - Z(r, f') + O(1)$. Further, given $\alpha \in \mathcal{M}(d(0, \mathbb{R}^-))$, we have $T(r, \alpha f) - Z(r, \alpha f) \le T(r, f) - Z(r, f) + T(r, \alpha)$.

**Proof.** By Theorem 43.13, the first statement is immediate. Let us check the last one. On one hand, $T(r, f) - Z(r, f) =$
$\max(Z(r, f), N(r, f)) - Z(r, f) = \max(0, N(r, f) - Z(r, f))$ $r < R$.
On the other hand,

$$ 
\begin{aligned}
T(r, f') - Z(r, f') &= \max(Z(r, f'), N(r, f')) - Z(r, f') \\
&= \max(0, N(r, f') - Z(r, f')) \\
&= \max(0, N(r, f) + \overline{N}(r, f) - Z(r, f')) \quad r < R.
\end{aligned}
$$

But by Theorem 43.13, $-Z(r, f') \ge -Z(r, f) - \overline{Z}(r, f) + \log(r) + O(1)$ $r < R$, hence $T(r, f') - Z(r, f') \ge \max(0, N(r, f) - Z(r, f) + \log(r)) + O(1) \ge T(r, f) - Z(r, f) + O(1)$, $r < R$.

Now, let $\alpha \in \mathcal{M}(d(0, R^-))$. Suppose $N(r, \alpha f) \ge Z(r, \alpha f)$, $r < R$. Then, $T(r, \alpha f) - Z(r, \alpha f) = N(r, \alpha f) - Z(r, \alpha f)$ $r < R$. We can write $\alpha$ in the form $\frac{\beta(x)}{\lambda(x)}$ with $\beta, \lambda \in H(d(0, r))$, $\beta, \lambda$ having no common zero. Next, we can write $\lambda$ in the form $\lambda_1\lambda_2$ were each zero of $\lambda_1$ is not a zero of $f$ and each zero of $\lambda_2$ is a zero of $f$. Then we can check that $N(r, \alpha f) = N(r, f) + Z(r, \lambda_1)$ and $Z(r, \alpha f \ge Z(r, f) - Z(r, \lambda_2)$. Consequently, $N(r, \alpha f) - Z(r, \alpha f) \le N(r, f) + Z(r, \lambda_1) - (Z(r, f) - Z(r, \lambda_2)) = N(r, f) - Z(r, f) + Z(r, \lambda_1) + Z(r, \lambda_2) = N(r, f) - Z(r, f) + Z(r, \lambda) \le N(r, f) - Z(r, f) + T(r, \lambda)$ $r < R$.

Suppose now that $N(r, \alpha f) \le Z(r, \alpha f)$. We can do a symmetric reasoning with the zeros of $\beta$. $\square$

Lemma 43.17 is an immediate consequence of Corollary 22.27 and Theorem 37.10:

**Lemma 43.17.** Let $f \in \mathcal{M}(\mathbb{K})$ (respectively $f \in \mathcal{M}(d(0, R^-))$) and let $G = \frac{f'}{f}$. Then, $G$ satisfies $Z(r, G) \le N(r, G) - \log r + O(1)$ $r \in I$ (respectively ($r \in J$)).

**Proof.** Without loss of generality, we can assume that 0 is neither a pole of $f$ nor a zero for $ff'$. By Theorem 37.10, $G$ satisfies $\Psi(G, \log r) \le -\log r$. On the other hand, by Theorem 43.2, we have $\Psi(G, \log r) = \log |G(0)| + Z(r, G) - N(r, G)$. Consequently, we obtain $\log |G(0)| + Z(r, G) - N(r, G) \le -\log r + O(1)$, which proves the claim. $\square$

**Corollary 43.18.** Let $f \in \mathcal{M}(\mathbb{K})$ (respectively $f \in \mathcal{M}(d(0, R^-))$) and let $G = \frac{f'}{f}$. Then, $T(r, G) = N(r, G) + O(1)$.

**Proof.** By Lemma 43.17, we have $T(r, G) \le N(r, G) + O(1)$. But as all meromorphic functions, $G$ satisfies $T(r, G) \ge N(r, G)$. $\square$
# Chapter 44

## A Non-Clean Entire Function

In Chapter 31, we noted the following question:

Let $f(x) = \sum_{n=0}^{\infty} b_n x^n \in \mathcal{A}(\mathbb{K})$ be such that $0 < \rho(f) < +\infty$. Do we have

$$ \rho(f)\sigma(f) = \psi(f) = \frac{\limsup_{n \to +\infty} \left( n \sqrt[n]{|b_n|^{\rho(f)}} \right)}{e}? $$

Actually, the answer is no in the general case. Using the counting function of an entire function, we construct a counter-example. In order to prove that, we need two basic lemmas.

**Lemma 44.1.** Let $f_1, f_2$ be two functions from $\mathbb{R}_+$ to $\mathbb{R}_+$ such that $\lim_{x \to +\infty} f_1(x) = \lim_{x \to +\infty} f_2(x) = +\infty$ and

$$ \limsup_{x \to +\infty} \frac{f_1(x)}{f_2(x)} = b \in \mathbb{R}_+, \quad \liminf_{x \to +\infty} \frac{f_1(x)}{f_2(x)} = a > 0. $$

Then,

$$ \lim_{x \to +\infty} \frac{\mathrm{Log}(f_1(x))}{\mathrm{Log}(f_2(x))} = 1. $$

**Lemma 44.2.** Let $\alpha, \beta \in \mathbb{R}_+$ and let $g(x) = e^{-x}(\alpha x - \beta)$. Then $g'$ has a unique zero at $1 + \frac{\beta}{\alpha}$ and $g(1 + \frac{\beta}{\alpha}) = \alpha e^{-(1+\frac{\beta}{\alpha})}$. Moreover, $g$ is increasing in $[0, 1 + \frac{\beta}{\alpha}]$ and is decreasing in $[1 + \frac{\beta}{\alpha}, +\infty[$ and tends to 0 when $x$ tends to $+\infty$.
**Theorem 44.3.** Suppose that $\mathbb{K}$ is of characteristic 0. There exist regular non-clean functions $f \in \mathcal{A}(\mathbb{K})$ such that $\psi(f) > \rho(f)\sigma(f)$.

**Proof.** We begin the definition of positive increasing sequences $(r_m)_{m \in \mathbb{N}}$, $(\alpha_m)_{m \in \mathbb{N}}$ in $\mathbb{N}^*$, $(\beta_m)_{m \in \mathbb{N}}$, where $r_0 = 1$, $r_{2m} \in |\mathbb{K}|$, $2r_{2m} \le \alpha_m < 2r_{2m}+1$, We put $q_m = \alpha_m - \alpha_{m-1}$, $\nu_m = \text{Log}(r_m)$, $\beta_0 = 0$, and $\beta_m = \beta_{m-1} + q_m(\text{Log}(r_{2m}))$.

In $[0, +\infty[$, we define $g_k(\nu) = e^{-\nu}(\alpha_k\nu - \beta_k)$ up to the rank $m$ and suppose that the function $g_k$ satisfies $1 \le g_k(\nu_{2k}) \le 1 + \frac{1}{4k^2}$ and $1 \le g_k(\nu_{2k+2}) \le 1 + \frac{1}{4(k+1)^2}$ $\forall k = 1, \dots, m-1$ and $g_{k-1}(\nu_{2k}) = g_k(\nu_{2k})$.

By Lemma 44.2, $g_k$ is increasing in $[\frac{\beta_k}{\alpha_k}, 1 + \frac{\beta_k}{\alpha_k}]$ from 0 to a maximum equal to $\alpha_k e^{1+\frac{\beta_k}{\alpha_k}}$ and is decreasing to 0 when $\nu$ tends to $+\infty$. Hence, $g_k$ takes the value 1 at a unique point $\lambda_{2k}$ in $[\frac{\beta_k}{\alpha_k}, 1 + \frac{\beta_k}{\alpha_k}]$ and at a unique point $\lambda_{2k+2} \in [1 + \frac{\beta_k}{\alpha_k}, +\infty[$. We then have $g_k(\lambda_{2k}) = e^{-\lambda_{2k}}(\alpha_k\lambda_{2k} - \beta_k)$ and $g_k(\lambda_{2k+2}) = e^{-\lambda_{2k+2}}(\alpha_k\lambda_{2k+2} - \beta_m)$, hence $\lambda_{2k} = \frac{e^{\lambda_{2k}+\beta_k}}{\alpha_k}$ and $\lambda_{2k+2} = \frac{e^{\lambda_{2k+2}+\beta_k}}{\alpha_k} > \nu_{2k+1}$ and we can take the value $r_{2k+2} \in |\mathbb{K}|$ close enough to $e^{\lambda_{2k+2}}$ such that, putting $r_{2k+2} = e^{\nu_{2k+2}}$, we then have

$$1 \le g_k(\nu_{2k+2}) = 1 + x_k \le 1 + \frac{1}{4(k+1)^2}$$

and

$$(1) \qquad 1 \le g_k(\nu_{2k}) = 1 + y_k \le 1 + \frac{1}{4(k)^2}.$$

We note that $r_{2k+2} > r_{2k+1}$, hence $\nu_{2k+2} > \nu_{2k+1}$. Next, the function $g_{k+1}$ is defined in the same way in $[\nu_{2k+2}, \nu_{2k+4}]$ as $g_{k+1}(\nu) = e^{-\nu}(\alpha_{k+1}\nu - \beta_{k+1})$. And we can check that $g_{k+1}(\nu_{2k+2}) = g_k(\nu_{2k+2})$.

Then by Lemma 44.2, $g_m$ has a maximum at

$$(2) \qquad \nu_{2m+1} = 1 + \frac{\beta_m}{\alpha_m},$$

and $g_{m+1}$ has a maximum at $\nu_{2m+3} = 1 + \frac{\beta_{m+1}}{\alpha_{m+1}}$ and $g_{m+1}(\nu_{2m+3}) = \alpha_{m+1}e^{\frac{\beta_{m+1}}{\alpha_{m+1}}} > 1$, hence $\nu_{2m+3} > \nu_{2m+2}$. Consequently, the sequence $(r_n)_{n \in \mathbb{N}}$ and is strictly increasing. This way, the sequences are now
defined for all $m \in \mathbb{N}$. Recall that $q_m = \alpha_m - \alpha_{m-1}$. We put $\Theta_m = \nu_{2m+1} - \nu_{2m}$. Then, $\nu_{2m} = \frac{\beta_m + e^{\nu_{2m}(1+x_m)}}{\alpha_m}$, and hence by (1) and (2), we obtain

$$ (3) \quad \Theta_m = 1 - \frac{e^{\nu_{2m}}(1 + y_m)}{\alpha_m} = 1 - \frac{r_{2m}(1 + y_m)}{\alpha_m} = 1 - \frac{r_{2m}(1 + y_m)}{2r_{2m} + \eta_m}, $$

where $(\eta_m)_{m \in \mathbb{N}}$ is a positive sequences bounded by 1 and the sequence $(y_m)$, by (1), satisfies $0 \le y_m \le \frac{1}{4(m)^2}$. Then,

$$ (4) \qquad \Theta_m \ge \frac{1}{2} - \frac{1}{8(m)^2} > \frac{15}{32}. $$

We can now define by induction the sequences $(r_m)$, $(\nu_m)$, $(g_m)$, $(y_m)$, $(\Theta_m)$ and then $\lim_{n \to +\infty} r_n = +\infty$. Consequently, by (3) and (4),

$$ (5) \qquad \lim_{m \to +\infty} \Theta_m = \frac{1}{2}. $$

We now obtain

$$ g_m(\nu_{2m+1}) = e^{-\nu_{2m+1}}(\alpha_m \nu_{2m+1} - \beta_m) = \alpha_m e^{-(1+\frac{\beta_m}{\alpha_m})}, $$

and hence, by (2),

$$ (6) \qquad g_m(\nu_{2m+1}) = \frac{(2r_{2m} + \eta_m)}{r_{2m+1}} = 2e^{-\Theta_m} + \zeta_m, $$

where $(\zeta_m)_{m \in \mathbb{N}}$ is a positive sequence of limit 0, since $\lim_{m \to +\infty} r_{2m+1} = +\infty$.

We can now define a function $g$ in $[0, +\infty[$ as $g(\nu) = g_m(\nu)$ when $\nu \in [\text{Log}(r_{2m}), \text{Log}(r_{2m+2})]$.

So, by (5), we have

$$ (7) \qquad \lim_{m \to +\infty} g(\nu_{2m+1}) = \frac{2}{\sqrt{e}}. $$

Thus, we can check that

$$ (8) \qquad \limsup_{\nu \to +\infty} g(\nu) = \limsup_{m \to +\infty} g(\nu_{2m+1}) < 2. $$

Now, by Theorem 27.4, we can consider the entire function $f$ admitting $q_m$ zeros on each circle $C(0, r_{2m})$ and no other zero. Let $(a_{j,m})_{(1 \le j \le q_m)}$ be the zeros of $f$ on the circle $C(0, r_{2m})$.
Then, when $2m \le r < r_{2m+2}$, the counting functions of zeros of $f$ (counting multiplicity) is of the form

$$Z(f, r) = \sum_{k=1}^{m} \sum_{j=1}^{q_k} (a_{k,j}(\text{Log}(r) - \text{Log}(r_{2k}))) \\ = \sum_{k=1}^{m} q_k(\text{Log}(r) - \text{Log}(r_{2k}))),$$

and hence, putting $\alpha_m = \sum_{k=1}^{m} q_k$ and $\beta_m = \sum_{k=1}^{m} q_k r_{2k}$, the function $g$ appears as the quotient of the counting function of zeros of $f$ (counting multiplicity) by $e^\nu$ when we put $\nu = \text{Log}(r)$. So, we have $\frac{\text{Log}(|f|(r))}{r} = g(\nu)$ whenever $\text{Log}(r) = \nu \in [\nu_{2m}, \nu_{2m+2}]$, and therefore by $(1)$, we can see that

$$\liminf_{r \to +\infty} \frac{\text{Log}(|f|(r))}{r} = 1$$

and

$$(9) \quad \limsup_{r \to +\infty} \frac{\text{Log}(|f|(r))}{r} = \lim_{m \to +\infty} |g(\nu_{2m+1})| = \frac{2}{\sqrt{e}}.$$

Moreover, by (9) and Lemma 44.1, we have

$$\lim_{r \to +\infty} \frac{\text{Log}(\text{Log}(|f|(r)))}{\text{Log}(r)} = 1,$$

hence $\rho(f) = 1$.

Further, since $\limsup_{r \to +\infty} \frac{\text{Log}(|f|(r))}{r} = \frac{2}{\sqrt{e}}$ and $\liminf_{r \to +\infty} \frac{\text{Log}(|f|(r))}{r} = 1$, we can see that $\sigma(f) = \frac{2}{\sqrt{e}}$ and $\tilde{\sigma}(f) = 1$. Thus, $f$ is not clean though it is regular.

More precisely, by construction, for every $r \in [r_{2m}, r_{2m+2}]$, we have $\psi(f, r) = \frac{2r_{2m}+\eta_m}{r}$, and hence $\psi(f, r_{2m})$ is of the form $2 + y_m$, where $(y_m)_{m \in \mathbb{N}}$ is a sequence of limit 0. Therefore, $\psi(f) \ge 2$, while $\rho(f) = 1$. This shows that $f$ does not satisfy the relation $\psi(f) = \rho(f)\sigma(f)$, and hence, this is not always satisfied when a function $f$ is not clean. $\square$
**Remark 1.** Of course, by Theorem 31.5, we know that the function $f$ built in the proof of Theorem 44.3 satisfies $\psi(f) > \tilde{\psi}(f)$. But we can directly verify this: on one hand, $\psi(f) = 2$, and on the other hand, we can see that $\psi(f, r_{2m+1}) = \frac{\alpha_m}{r_{2m+1}}$, and hence by (5), $\tilde{\psi}(f) \le \frac{2}{\sqrt{e}}$.

Next, $f$ must satisfy Theorem 31.5: $\rho(f)\sigma(f) \le \psi(f) \le \rho(f)$ ($e\sigma(f) - \tilde{\sigma}(f)$). Let us check. We have seen that $\psi(f) = 2$, $\rho(f) = 1$, $\sigma(f) = \frac{2}{\sqrt{e}}$, $\tilde{\sigma}(f) = 1$. Then, $\rho(f)(e\sigma(f) - \tilde{\sigma}(f)) = 2\sqrt{e}-1 > 2$. That is okay.

**Remark 2.** By Corollary 31.7, a clean entire function such that $\sigma(f) > 0$ is regular. The converse is not true, as shown in Theorem 44.3.

In complex analysis, given an entire function $f$, we put $M(f, r) = \sup\{|f(z)|_\infty, |z|_\infty = r\}$, where $|\cdot|_\infty$ is the archimedean modulus on $\mathbb{C}$. In [4], the authors claimed that if a complex entire function $f$ satisfies

$$ \limsup_{r \to +\infty} \frac{\mathrm{Log}(\mathrm{Log}(M(f, r))))}{\mathrm{Log}(r)} = \liminf_{r \to +\infty} \frac{\mathrm{Log}(\mathrm{Log}(M(f, r))))}{\mathrm{Log}(r)}, $$

then

$$ \limsup_{r \to +\infty} \frac{\mathrm{Log}(M(f, r))}{r^\rho} = \liminf_{r \to +\infty} \frac{\mathrm{Log}(M(f, r))}{r^\rho}, $$

where $\rho = \lim_{r \to +\infty} \frac{\mathrm{Log}(\mathrm{Log}(M(f,r))))}{\mathrm{Log}(r)}$. In the field $\mathbb{K}$, we just checked that such a theorem does not hold. Actually, the proof of [4] is put in doubt by the following argument held in Lemma 2 of [4]:

since

$$ \int_{r_0}^{+\infty} \frac{\exp(\mathrm{Log}(M(r, f)))}{(\exp(r^\lambda))^{t-\varepsilon+1}} dr = +\infty, $$

“then”

$$ \liminf_{r \to +\infty} \frac{\exp(\mathrm{Log}(M(r, f)))}{(\exp(r^\lambda))^{t-\varepsilon}} = +\infty. $$

Suppose, for example, that in $[r_0, +\infty[$, $M(r, f)$ is equivalent to $\frac{\exp(r\lambda(t-\varepsilon+1))}{r}$.
Then, $\frac{\exp(r\lambda(t-\varepsilon+1))}{\exp(r\lambda(t-\varepsilon))}$ is equivalent to $\frac{1}{r}$, and hence

$$\int_{r_0}^{+\infty} \frac{\exp(\log(M(r, f)))}{(\exp(r^\lambda))^{t-\varepsilon+1}} dr = +\infty,$$

but

$$\liminf_{r \to +\infty} \frac{\exp(\log(M(r, f)))}{(\exp(r^\lambda))^{t-\varepsilon}} = 0.$$
Chapter 45

Nevanlinna Theory in $K$ and
Inside a Disk

We can now prove the second main theorem under different forms. Lemma 45.1 is essential and directly leads to the theorems.

**Lemma 45.1.** Let $f \in \mathcal{M}(K)$ (respectively $f \in \mathcal{M}_u(d(0, R^{-}))$). Suppose that there exists $\xi \in K$ (respectively $\xi \in \mathcal{M}_b(d(0, R^{-}))$) and a sequence of intervals $I_n = [u_n, v_n]$ such that $u_n < v_n < u_{n+1}$, $\lim_{n \to +\infty} u_n = +\infty$ (respectively $\lim_{n \to +\infty} u_n = R$) and

$$ \lim_{n \to +\infty} \left( \inf_{r \in I_n} T(r, f) - Z(r, f - \xi) \right) = +\infty $$

(respectively $\lim_{n \to +\infty} (\inf_{r \in I_n} T(r, f) - Z(r, f - \xi)) = +\infty$).

Let $\tau \in K$ (respectively let $\tau \in \mathcal{M}_b(d(0, R^{-}))$, $\tau \neq \xi$. Then, $Z(r, f - \tau) = T(r, f) + O(1) \forall r \in I_n$, when $n$ is big enough.

**Proof.** We know that the Nevanlinna functions of a meromorphic function $f$ are the same in $K$ and in an algebraically closed complete extension of $K$ whose absolute value extends that of $K$. Consequently, without loss of generality, we can suppose that $K$ is spherically complete because we know that such a field does admit a spherically complete algebraically closed extension whose absolute value expands that of $K$. If $f$ belongs to $\mathcal{M}(K)$, we can obviously set it in the form $\frac{g}{h}$, where $g, h$ belong to $\mathcal{A}(K)$ and have no common zero. Next, since $K$ is supposed to be spherically complete, if $f$ belongs to $\mathcal{M}(d(0, R^{-}))$, we can also set it in the form $\frac{g}{h}$, where $g, h$
belong to $\mathcal{A}(d(0, R^{-}))$ and have no common zero. Consequently, we have $T(r, f) = \max(Z(r, g), Z(r, h))$.

When $\xi$ is a constant, we can obviously suppose that $\xi = 0$. Suppose now $\xi \in \mathcal{M}_u(d(0, R^{-}))$. Then, $f - \xi$ also belongs to $\mathcal{M}_u(d(0, R^{-}))$ and $\tau - \xi$ belongs to $\mathcal{M}_b(d(0, R^{-}))$. Consequently, in both cases, we can assume $\xi = 0$ to prove the claim. Next, up to a change of origin, we can also assume that none of the functions we consider have a pole or zero at the origin.

Now, we have $\lim_{n \to +\infty} (\inf_{r \in I_n} T(r, f) - Z(r, f)) = +\infty$ i.e.

$$ (1) \quad \lim_{n \to +\infty} \left( \inf_{r \in I_n} (Z(r, h) - Z(r, g)) \right) = +\infty. $$

Particularly, we note that $T(r, f) = Z(r, h) + O(1)$ whenever $r \in I_n$ when $n$ is big enough.

Consider now $Z(r, f - \tau) = Z(r, g - \tau h)$. But by (1), we can see that $|g|(r) < |\tau|h|(r) \forall r \in I_n$ when $n$ is big enough, hence $Z(r, g - \tau h) = Z(r, \tau h) \forall r \in I_n$ when $n$ is big enough, hence $Z(r, \tau h) = Z(r, h) \forall r \in I_n$ when $n$ is big enough. Therefore, $Z(r, f - \tau) = Z(r, h) + O(1) = T(r, f) + O(1), \forall r \in I_n$ when $n$ is big enough. So the claim is proven when $\tau$ is a constant.

Suppose now that $f \in \mathcal{M}(d(0, R^{-}))$ and $\tau \in \mathcal{M}_b(d(0, R^{-}))$. By Theorem 36.11, we can write $\tau$ in the form $\frac{\phi}{\psi}$, where $\phi, \psi \in \mathcal{A}_b((d(0, R^{-}))$ have no common zero. Consider $Z(r, f - \tau h) = Z(r, \frac{\psi g - \phi h}{\psi h})$. Since $g$ and $h$ have no common zero and since both $\phi, \psi$ are bounded, we have $Z(r, \frac{\psi g - \phi h}{\psi h}) = Z(r, \psi g - \phi h) + O(1)$. By (1), in $I_n$, we have $|\psi g|(r) < |\phi h|(r)$ when $n$ is big enough, and since $|\cdot|(r)$ is an absolute value, $|\psi g - \phi h|(r) = |\phi h|(r)$ in $I_n$ when $n$ is big enough. Therefore, we have $Z(r, \psi g - \phi h) = Z(r, \phi h) = Z(r, h) + O(1)$ in $I_n$ when $n$ is big enough. Consequently, $Z(r, f - \tau) = Z(r, h) + O(1) = T(r, h) + O(1) = T(r, f) + O(1) \forall r \in I_n$ when $n$ is big enough. That finishes proving Lemma 45.1. $\square$

**Theorem 45.2.** Let $f \in \mathcal{M}(\mathbb{K})$ and let $a_1, \dots, a_q \in \mathbb{K}$ be distinct. Then,

$$ (q-1)T(r, f) \le \max_{1 \le k \le q} \left( \sum_{j=1, j \ne k}^{q} Z(r, f - a_j) \right) + O(1). $$
**Corollary 45.3.** Let $f \in \mathcal{M}(\mathbb{K})$ and let $a_1, \dots, a_q \in \mathbb{K}$ be distinct. Then, $(q-1)T(r, f) \le \sum_{j=1}^q Z(r, f - a_j) + O(1)$.

**Theorem 45.4.** Let $f \in \mathcal{M}(d(0, R^{-}))$ and let $\tau_1, \dots, \tau_q \in \mathcal{M}_b(d(0, R^{-}))$ be distinct. Then,

$$ (q-1)T(r, f) \le \max_{1 \le k \le q} \left( \sum_{j=1, j \neq k}^{q} Z(r, f - \tau_j) \right) + O(1). $$

**Corollary 45.5.** Let $f \in \mathcal{M}(d(0, R^{-}))$ and let $\tau_1, \dots, \tau_q \in \mathcal{M}_b(d(0, R^{-}))$ be distinct. Then, $(q-1)T(r, f) \le \sum_{j=1}^q Z(r, f - \tau_j) + O(1)$.

**Proof.** (Proof of Theorems 45.2 and 45.4) Suppose Theorems 45.2 (respectively Theorem 45.4) is wrong. In order to make a unique proof for the two theorems, in Theorem 45.2, we set $\tau_j = a_j$. Thus, there exists $f \in \mathcal{M}(\mathbb{K})$ (respectively $f \in \mathcal{M}(d(0, R^{-}))$) and $\tau_1, \dots, \tau_q \in \mathbb{K}$ (respectively $\tau_1, \dots, \tau_q \in \mathcal{M}_b(d(0, R^{-}))$) such that $(q-1)T(r, f) - \max_{1 \le k \le q} \left( \sum_{j=1, j \neq k}^{q} Z(r, f - \tau_j) \right)$ admits no superior bound in $]0, +\infty[$. So, there exists a sequence of intervals $J_s = [w_s, y_s]$ such that $w_s < y_s < w_{s+1}$, $\lim_{s \to +\infty} w_s = +\infty$ (respectively $\lim_{s \to +\infty} w_s = R$) and two distinct indices $m \le q$ and $t \le q$ such that

$$ \lim_{s \to +\infty} \inf_{r \in J_s} (T(r, f) - Z(r, f - \tau_m)) = +\infty $$

and

$$ \lim_{s \to +\infty} \inf_{r \in J_s} (T(r, f) - Z(r, f - \tau_t)) = +\infty. $$

But by Lemma 45.1, this is impossible. This ends the proof of Theorems 45.2 and 45.4. $\square$

**Remark.** Theorem 45.2 does not hold in complex analysis. Indeed, let $f$ be a meromorphic function in $\mathbb{C}$ omitting two values $a$ and $b$, such as $f(x) = \frac{e^x}{e^x - 1}$. Then, $Z(r, f - a) + Z(r, f - b) = 0$.

**Theorem 45.6.** Let $\mathbb{K}$ be of characteristic 0. Let $\alpha_1, \dots, \alpha_q \in \mathbb{K}$, with $q \ge 2$, let $S = \{\alpha_1, \dots, \alpha_q\}$, and let $f \in \mathcal{M}(\mathbb{K})$ (respectively
$f \in \mathcal{M}(d(0, R^{-})))$. Then, $(q-1)T(r, f) \le \sum_{j=1}^{q} \bar{Z}(r, f - \alpha_j) + Z(r, f') - Z_0^S(r, f') + O(1) \forall r \in I$ (respectively $\forall r \in J$).

Moreover, if $f$ belongs to $f \in \mathcal{A}(\mathbb{K})$ (respectively $\mathcal{A}(d(0, R^{-})))$, then $qT(r, f) \le \sum_{j=1}^{q} \bar{Z}(r, f - \alpha_j) + Z(r, f') - Z_0^S(r, f') + O(1) \forall r \in I$ (respectively $\forall r \in J$).

**Theorem 45.7 (Second Main Theorem) A. Boutabaa.** Let $\mathbb{K}$ be of characteristic 0. Let $\alpha_1, \dots, \alpha_q \in \mathbb{K}$, with $q \ge 2$, let $S = \{\alpha_1, \dots, \alpha_q\}$, and let $f \in \mathcal{M}(\mathbb{K})$ (respectively $f \in \mathcal{M}_u(d(0, R^{-})))$. Then,

$$ (q-1)T(r, f) \le \sum_{j=1}^{q} \bar{Z}(r, f - \alpha_j) + \bar{N}(r, f) - Z_0^S(r, f') - \log r + O(1) \\ \forall r \in I \text{ (respectively } \forall r \in J). $$

**Proof.** (Proof of Theorems 45.6 and 45.7) By Theorem 45.2 (respectively 45.4), there exists a constant $B > 0$, and for each $r > 0$ (respectively for each $r \in ]0, R[$), there exists $k(r) \in \mathbb{N}$, $k(r) \le q$, such that $(q-1)T(r, f) \le \sum_{j=1, j \ne k(r)}^{q} Z(r, f - a_j) + B$ i.e. $(q-1)T(r, f) \le \sum_{j=1}^{q} Z(r, f - a_j) - Z(r, f - a_{k(r)}) + B$. Now, $\sum_{j=1}^{q} Z(r, f - a_j) = \sum_{j=1}^{q} \bar{Z}(r, f - a_j) + Z(r, f') - Z_0^S(r, f') - \log r$. Consequently, $(q-1)T(r, f) \le \sum_{j=1}^{q} \bar{Z}(r, f - a_j) + Z(r, f') - Z_0^S(r, f') - Z(r, f - a_{k(r)}) + B$, and this proves the first claim of Theorem 45.6. Particularly, if $f \in \mathcal{A}(\mathbb{K})$ (respectively if $f \in \mathcal{A}(d(0, R^{-})))$, then we have $Z(r, f - a_j) = T(r, f - a_j) = T(r, f) + O(1) \forall j = 1, \dots, q$, hence $Z(r, f - a_{k(r)}) = T(r, f) + O(1)$, and therefore, $qT(r, f) \le \sum_{j=1}^{q} \bar{Z}(r, f - a_j) + Z(r, f') - Z_0^S(r, f') + O(1)$, which ends the proof of Theorem 45.6.

Henceforth, by Theorem 43.13, there exists a constant $c_j > 0$ such that $Z(r, f') \le Z(r, f - a_j) = N(r, f - a_j) - \log r + c_j$. Let $c = \max(c_1, \dots, c_q)$. Then, $Z(r, f') - Z_0^S(r, f') - Z(r, f - a_{k(r)}) \le \bar{N}(r, f - a_{k(r)}) + c - \log r = \bar{N}(r, f) + c - \log r$. Consequently,

$$ \sum_{j=1}^{q} Z(r, f - a_j) = \sum_{j=1}^{q} \bar{Z}(r, f - a_j) + \bar{N}(r, f) - \log r + O(1). $$

That finishes the proof of Theorem 45.7. □
**Remark.** In Theorem 45.7, in the hypothesis $f \in \mathcal{M}(d(0, R^{-}))$, the term $-\log r$ has no veritable meaning since $r$ is bounded.

**Corollary 45.8.** Let $\mathbb{K}$ be of characteristic 0. Let $\alpha_1, \dots, \alpha_q \in \mathbb{K}$, with $q \ge 2$, let $S = \{\alpha_1, \dots, \alpha_q\}$, and let $f \in \mathcal{M}(\mathbb{K})$ (respectively $f \in \mathcal{M}(d(0, R^{-}))$). Then,

$$ \sum_{j=1}^{q} (Z(r, f - \alpha_j) - \bar{Z}(r, f - \alpha_j)) \le T(r, f) + \bar{N}(r, f) - Z_0^S(r, f') \\ - \log r + O(1) \quad \forall r \in I \text{ (respectively } \forall r \in J). $$

We can check that given $f \in \mathcal{A}(\mathbb{K})$ such that $0 < \rho(f) < +\infty$ and $0 < \sigma(f)$, we have $\sigma(f, r) = \frac{T(r, f)}{r^{\rho(f)}}$. Consequently, Corollary 45.9 is just a corollary of Corollary 45.8.

Using $\rho(f)$, we can derive a new form of the main Nevanlinna theorem for entire functions:

**Corollary 45.9.** Let $\mathbb{K}$ be of characteristic 0. Let $f \in \mathcal{A}(\mathbb{K})$ be such that $0 < \rho(f) < +\infty$ and let $a_1, \dots, a_q \in \mathbb{K}$. Then,

$$ (q-1)\sigma(f) \le \limsup_{r \to +\infty} \left( \frac{1}{r^{\rho(f)}} \sum_{i=1}^{q} \bar{Z}(r, f - a_i) \right). $$

**Proof.** We have $\sigma(f, r) = \frac{\text{Log}(|f|(r))}{r^{\rho(f)}}$, hence by Theorem 45.7,

$$ r^{\rho(f)}(q-1)\sigma(f, r) \le \sum_{i=1}^{q} \bar{Z}(r, f - a_i) - \text{Log}(r) + O(1). $$

The conclusion is then obvious. $\square$

<figure><img src="image_19.png" /></figure>

**Corollary 45.10.** Let $\mathbb{K}$ be of characteristic 0. Let $f \in \mathcal{A}(\mathbb{K})$ be clean. Let $a_1, \dots, a_q \in \mathbb{K}$. Then,

$$ (q-1)\sigma(f) \le \liminf_{r \to +\infty} \left( \frac{1}{r^{\rho(f)}} \sum_{i=1}^{q} \bar{Z}(r, f - a_i) \right) $$

and

$$ (q-1)\psi(f) \le \liminf_{r \to +\infty} \left( \frac{\rho(f)}{r^{\rho(f)}} \sum_{i=1}^{q} \bar{Z}(r, f - a_i) \right). $$
# Chapter 46

## Nevanlinna Theory Out of a Hole

Throughout Chapters 46-57, the field $\mathbb{K}$ is supposed to be of characteristic 0

Now, we mean to construct a Nevanlinna theory for meromorphic functions in the complement of an open disk thanks to the use of specific properties of the analytic elements on infraconnected subsets of $\mathbb{K}$ already examined.

Given $f \in \mathcal{M}(D)$, for $r > R$, here we denote by $Z_R(r, f)$ the counting function of zeros of $f$ between $R$ and $r$, i.e. if $\alpha_1, \dots, \alpha_m$ are the distinct zeros of $f$ in $\Delta(0, R, r)$, with respective multiplicity $u_j$, $1 \le j \le m$, then $Z_R(r, f) = \sum_{j=1}^m u_j(\log(r) - \log(|\alpha_j|))$. Similarly, we denote by $N_R(r, f)$ the counting function of poles of $f$ between $R$ and $r$, i.e. if $\beta_1, \dots, \beta_n$ are the distinct poles of $f$ in $\Delta(0, R, r)$, with respective multiplicity $v_j$, $1 \le j \le m$, then $N_R(r, f) = \sum_{j=1}^n v_j(\log(r) - \log(|\beta_j|))$. Finally, we put $T_R(r, f) = \max(Z_R(r, f), N_R(r, f))$.

Next, we denote by $\bar{Z}_R(r, f)$ the counting function of zeros without counting multiplicity: if $\alpha_1, \dots, \alpha_m$ are the distinct zeros of $f$ in $\Delta(0, R, r)$, then we put $\bar{Z}_R(r, f) = \sum_{j=1}^m \log(r) - \log(|\alpha_j|)$.

Similarly, we denote by $\bar{N}_R(r, f)$ the counting function of poles without counting multiplicity: if $\beta_1, \dots, \beta_n$ are the distinct poles of $f$ in $\Delta(0, R, r)$, then we put $\bar{N}_R(r, f) = \sum_{j=1}^n \log(r) - \log(|\beta_j|)$.

Finally, putting $W = \{a_1, \dots, a_q\}$, we denote by $Z_R^W(r, f')$ the counting function of zeros of $f'$ on points $x$ where $f(x) \notin W$.

Given two functions defined in an interval $I = [b, +\infty[$, we write $\phi(r) = \psi(r) + O(\log(r))$ (respectively $\phi(r) \le \psi(r) + O(\log(r))$)
if there exists a constant $B > 0$ such that $|\phi(r) - \psi(r)|_\infty \le B \log(r)$, $r \in I$ (respectively $\phi(r) - \psi(r) \le B \log(r)$, $r \in I$).

We write $\phi(r) = o(\psi(r))$, $r \in I$, if $\lim_{r \to +\infty} \frac{\phi(r)}{\psi(r)} = 0$.

**Theorem 46.1.** Let $f \in \mathcal{M}(D)$. Then, $\log(|f|(r)) - \log(|f|(R)) = Z_R(r, f) - N_R(r, f) + \text{mo}(f, S)(\log r - \log R)$ ($r \in I$).

**Proof.** By Theorem 39.6, we have $f = f^S f^0$. Since $f^S$ has no zero and no pole in $D$, by Theorem 39.3, it satisfies $|f^S|(r) = r^{\text{mo}(f,S)} \forall r \in I$, hence $\log(|f^S|(r)) - \log(|f^S|(R)) = \text{mo}(f,S)(\log r - \log R)$ ($r \in I$). Next, since $f^0$ has no zero and no pole in $S$, we have $\log(|f^0|(r)) - \log(|f^0|(R)) = Z_R(r, f^0) - N_R(r, f^0)$ ($r \in I$), therefore the statement is clear. $\square$

**Corollary 46.2.** Let $f \in \mathcal{M}(D)$. Then $T_R(r, f)$ is identically zero if and only if $f$ is a Motzkin factor.

**Corollary 46.3.** Let $f, g \in \mathcal{A}(D)$ satisfy $\log(|f|(r)) \le \log(|g|(r)) \forall r \ge R$ ($r \in I$). Then, $Z_R(r, f) \le Z_R(r, g) + (\text{mo}(g, S) - \text{mo}(f, S))(\log(r) - \log(R))$, ($r \in I$).

**Theorem 46.4.** Let $f \in \mathcal{A}(D)$. Then, $Z_R(r, f') \le Z_R(r, f) + O(\log(r))$ ($r \in I$).

**Proof.** Indeed, by Theorem 18.2, we have $|f'|(r) \le \frac{|f|(r)}{r}$. Therefore, the conclusion comes from Theorem 46.1. $\square$

We can now characterize the set $\mathcal{M}^c(D)$:

**Theorem 46.5.** Let $f \in \mathcal{M}(D)$. The three following statements are equivalent:

(i) $\lim_{r \to +\infty} \frac{T_R(r, f)}{\log(r)} = +\infty$ ($r \in I$),

(ii) $\frac{T_R(r, f)}{\log(r)}$ is unbounded,

(iii) $f$ belongs to $\mathcal{M}^c(D)$.

**Proof.** Consider an increasing sequence $(u_n)_{n \in \mathbb{N}}$ in $\mathbb{R}_+$ such that $\lim_{n \to +\infty} u_n = +\infty$ and let $(k_n)_{n \in \mathbb{N}}$ be a sequence of $\mathbb{N}^*$.
Clearly, we have

$$ \lim_{r \to +\infty} \frac{\sum_{u_n \le r} k_n(\log(r) - \log(u_n))}{\log(r)} = +\infty. $$

Consequently, if a function $f \in \mathcal{M}^c(D)$ has infinitely many zeros (respectively infinitely many poles in $D$), then $\lim_{n \to +\infty} \frac{Z_R(r,f)}{\log(r)} = +\infty$ (respectively $\lim_{n \to +\infty} \frac{N_R(r,f)}{\log(r)} = +\infty$), hence in both cases, $\lim_{n \to +\infty} \frac{T_R(r,f)}{\log(r)} = +\infty$. Conversely, if $f$ has finitely many zeros and finitely many poles in $D$, then we check that $\lim_{n \to +\infty} \frac{T_R(r,f)}{\log(r)} < +\infty$. Thus, the equivalence of the three statements is clear. $\square$

**Theorem 46.6.** The set $M_0(D) = \mathcal{M}(D)\backslash\mathcal{M}^c(D)$ is a subfield of $\mathcal{M}(D)$. Every $f \in \mathcal{M}^c(D)$ is transcendental over $M_0(D)$.

**Proof.** By Theorem 46.5, $M_0(D)$ is a subfield of $\mathcal{M}(D)$. Now, consider a polynomial $P(Y) = \sum_{j=0}^n a_j Y^j \in M_0(D)[Y]$ with $a_n = 1$. Let $f \in \mathcal{M}^c(D)$ and suppose that $P(f) = 0$. Then, $f^n = -\sum_{j=0}^{n-1} a_j f^j$. Set $\Xi = \sum_{j=0}^{n-1} a_j f^j$ and $f = f^0 g$ with $g, h \in \mathcal{A}(D)$ having no zero in $S$. Then, $\Xi = \frac{\sum_{j=0}^{n-1} a_j g^j h^{n-1-j}}{h^{n-1}}$.

Since $\sum_{j=0}^{n-1} a_j g^j h^{n-1-j}$ belongs to $\mathcal{A}(D)$, by Theorem 46.7, we have

$$ T_R \left( r, \sum_{j=0}^{n-1} a_j g^j h^{n-1-j} \right) \le (n-1)T_R(r, f) + O(\log(r)), \quad (r \in I), $$

and of course, $T_R(r, h^{n-1}) \le (n-1)T_R(r, f)$, $(r \in I)$. Consequently

$$ T_R(r, \Xi) \le (n-1)T_R(r, f) + O(\log(r)), \quad (r \in I). $$

But on the other hand, by Theorem 46.7, $T_R(r, f^n) = nT_R(r, f)$. Therefore, we should have $nT_R(r, f) \le (n-1)T_R(r, f) + O(\log(r), (r \in I)$, which is impossible by Theorem 46.6 because $f$ belongs to $\mathcal{M}^c(D)$. Consequently, the equality $P(f) = 0$ is impossible, which proves that $f$ is transcendental over $M_0(D)$. $\square$

Operations on $\mathcal{M}(D)$ work almost like for meromorphic functions in the whole field.
**Theorem (First Main Theorem) 46.7.** Let $f, g \in \mathcal{M}(D)$. Then for every $b \in \mathbb{K}$, we have $T_R(r, f+b) = T_R(r, f) + O(\log(r))$ ($r \in I$), $T_R(r, f,g) \le T_R(r, f) + T_R(r, g) + O(\log(r))$ ($r \in I$), $T_R(r, \frac{1}{f}) = T_R(r, f)$, $T_R(r, f+g) \le T_R(r, f) + T_R(r, g) + O(\log(r))$ ($r \in I$), and $T_R(r, f^n) = nT_R(r, f)$.

Let $h$ be a Moebius function. Then, $T_R(r, h \circ f) = T_R(r, f) + O(\log(r))$ ($r \in I$).

Moreover, if both $f$ and $g$ belong to $\mathcal{A}(D)$, then

$$T_R(r, f+g) \le \max(T_R(r, f), T_R(r, g)) + O(\log(r)) \quad (r \in I)$$

and $T_R(r, fg) = T_R(r, f) + T_R(r, g)$, ($r \in I$). Particularly, if $f \in \mathcal{A}^c(D)$, then $T_R(r, f+b) = T_R(r, f) + O(1)$ ($r \in I$). Given a polynomial $P(X) \in \mathbb{K}[X]$ of degree $q$, then $T_R(r, P \circ f) = qT_R(r, f) + O(\log(r))$.

**Proof.** Suppose first $f, g \in \mathcal{A}(D)$. It is immediate to check that $T_R(r, fg) = Z_R(fg) = Z_R(f) + Z_R(r, g) = T_R(r, f) + T_R(r, g)$, that $T_R(r, f^n) = nT_R(r, f)$, and that $T_R(r, \frac{1}{f}) = T_R(r, f)$.

Then, $T_R(r, f+g) = Z_R(r, f+g) = \log(|f+g|(r)) - mo(f+g,S)(\log(r) - \log(R))$, ($r \in I$). But $\log(|f+g|(r)) \le \max(\log(|f|(r)), \log(|g|(r)))$, hence

$$Z_R(r, f+g) \le \max(Z_R(r, f) + mo(f, S)(\log(r) - \log(R)), \\ Z_R(r, g) + mo(g, S)(\log(r) - \log(R)))$$

and hence $T_R(r, f+g) \le \max(Z_R(r, f), T_R(r, g)) + O(\log(r))$.

Particularly, given $b \in \mathbb{K}$, we have $T_R(r, f+b) \le T_R(r, f) + O(\log(r)) \le T_R(r, f) + O(\log(r))$, hence $T_R(r, f+b) = T_R(r, f) + O(\log(r))$.

Now, given a polynomial of degree $q$, we have $Z_R(r, P \circ f) = qZ_R(r, f) + O(\log(r))$ and $N_R(r, P \circ f) = qN_R(r, f)$, hence $T_R(r, P \circ f) = qT_R(r, f)$.

Now, suppose $f \in \mathcal{A}^c(D)$. Then $f(x)$ is a Laurent series $\sum_{k=-\infty}^{+\infty} a_k x^n$ convergent in all $D$ such that $\lim_{r \to +\infty} |f|(r) = +\infty$. Let $b \in \mathbb{K}$ and take $V$ be such that $|f|(r) > |b| \forall r \ge V$. Then for every $r > V$, $|f|(r)$ is of the form $|a_k|r^k$ with $k > 0$, $|a_n|r^n < |a_k|r^k \forall n > k$ and the number of zeros of $f$ in $\Delta(0, R, r)$ is $k - mo(f, S)$. Next, $f-b$ is of the form $\sum_{k=-\infty}^{+\infty} c_n x^n$ with $c_n = a_n \forall n \ne 0$ and $c_0 = a_0 - b$. Consequently, $f-b$ has the same number of zeros in $\Delta(0, R, r)$ and
in each circle $C(0, r)$ for $r > V$. Therefore, $T_R(r, f) = T_R(r, f - b)$ when $r$ is big enough.

Next, consider the general case: $f, g \in \mathcal{M}(D)$. First, it is immediate to check that $T_R(r, fg) \le T_R(r, f) + T_R(r, g)$. Similarly, for $T_R(r, \frac{1}{f})$. By definition, we have $Z_R(r, \frac{1}{f}) = N_R(r, f)$ and $N_R(r, \frac{1}{f}) = Z_R(r, f)$, hence $T_R(r, \frac{1}{f}) = T_R(r, f)$.

Now, consider $T_R(r, f + g)$ in the general case: $f, g \in \mathcal{M}(D)$. By Theorem 39.6, we can write

$$f + g = f^S \left( \frac{f_1^0}{f_2^0} \right) + g^S(g_1^0, g_2^0),$$

hence

$$T_R(r, f + g) = T_R \left( r, \frac{f^S f_1^0 g_2^0 + g^S g_1^0 f_2^0}{f_2 g_2} \right)$$

with $f_1^0, f_2^0, g_1^0, g_2^0 \in \mathcal{A}(\mathbb{K})$, having no zero in $T$ and $f^S, g^S$ Motzkin factors associated with $S$. Then, $Z_R(r, f^S f_1^0 g_2^0) = Z_R(r, f_1^0) + Z_R(r, g_2^0), Z_R(r, g^S g_1^0 f_2^0) = Z_R(r, g_1^0) + Z_R(r, f_2^0)$, hence, by what we just saw, $Z_R(r, f^S f_1^0 g_2^0 + g^S g_1^0 f_2^0) \le \max(T_R(r, f), T_R(r, g)) + O(\log(r))$. And obviously, $Z_R(r, f_2 g_2) \le T_R(r, f) + T_R(r, g)$. So we obtain in the general case $T_R(r, f + g) \le T_R(r) + T_R(r, g) + O(\log(r))$.

Finally, consider a Moebius function $h$. Then, $h \circ f(x)$ is of the form $C + \frac{\epsilon}{\alpha f(x)+\beta}$, and thereby, $T_R(r, h \circ f) = T_R(r, f) + O(\log(r))$. □

**Theorem 46.8.** Every $f \in \mathcal{M}^c(D)$ is transcendental over $\mathcal{M}_0(D)$.

**Proof.** Consider a polynomial $P(Y) = \sum_{j=0}^n a_j Y^j \in \mathcal{M}_0(D)[Y]$ with $a_n = 1$. Let $f \in \mathcal{M}^c(D)$ and suppose that $P(f) = 0$. Then, $f^n = -\sum_{j=0}^{n-1} a_j f^j$. Set $\Xi = \sum_{j=0}^{n-1} a_j f^j$ and $f = f^0 g^h$ with $g, h \in \mathcal{A}(D)$ having no zero in $S$. Then, $\Xi = \frac{\sum_{j=0}^{n-1} a_j g^j h^{n-1-j}}{h^{n-1}}$.

Since $\sum_{j=0}^{n-1} a_j g^j h^{n-1-j}$ belongs to $\mathcal{A}(D)$, by Theorem 46.7, we have

$$T_R \left( r, \sum_{j=0}^{n-1} a_j g^j h^{n-1-j} \right) \le (n-1)T_R(r, f) + O(\log(r)), (r \in I),$$

and of course, $T_R(r, h^{n-1}) \le (n-1)T_R(r, f)$, $(r \in I)$. Consequently,

$$T_R(r, \Xi) \le (n-1)T_R(r, f) + O(\log(r)), (r \in I).$$
But on the other hand, by Theorem 46.7, $T_R(r, f^n) = nT_R(r, f)$. Therefore, we should have $nT_R(r, f) \le (n-1)T_R(r, f) + O(\log(r))$, $(r \in I)$, which is impossible by Theorem 46.6 because $f$ belongs to $M^c(D)$. Consequently, the equality $P(f) = 0$ is impossible, which proves that $f$ is transcendental over $M_0(D)$. $\square$

**Theorem 46.9.** Let $f \in \mathcal{M}(D)$. Then, $N_R(r, f^{(k)}) = N_R(r, f) + k\overline{N}_R(r, f)$, $(r \in I)$ and $Z_R(r, f^{(k)}) \le Z_R(r, f) + k\overline{N}_R(r, f) + O(\log(r))$, $(r \in I)$.

**Proof.** The inequality $N_R(r, f^{(k)}) = N_R(r, f) + k\overline{N}_R(r, f) + O(1)$, $r \in I$, is obvious. Next, consider $f$ in the form $\frac{g}{h}$ with $g, h \in \mathcal{A}(\mathbb{K})$. Recall that we can write $h$ in the form $\overline{hh}$ with $\overline{h}$ and $\tilde{h}$ in $\mathcal{A}(\mathbb{K})$, each zero of $\overline{h}$ being of order one and all zeros of $h$ being a zero of $\overline{h}$. So, $h'$ is of the form $\overline{hh}$, where $\overline{h}$ belong to $\mathcal{A}(\mathbb{K})$ and none of the zeros of $\overline{h}$ is a zero of $h$. Then, $f'$ is of the form $\frac{g'\overline{h}-g\overline{h}}{h\overline{h}}$. So, $Z_R(r, f') \le Z_R(r, g'\overline{h}-g\overline{h})$, and hence, by Theorem 46.7,

$$ (1) \qquad Z_R(r, f') \le \max(Z_R(r, g'\overline{h}), Z_R(r, g\overline{h})). $$

On one hand, by Theorem 46.5, $Z_R(r, g') \le Z_R(r, g) + O(\log r)$, and by Corollary 46.4, we have $Z_R(r, g') \le Z_R(r, f) + O(\log(r))$. Obviously, $Z_R(r, \overline{h}) \le \overline{Z}_R(r, h) = \overline{N}_R(r, \ell) = \overline{N}_R(r, f)$, hence $Z_R(r, g'h) \le Z_R(r, f) + \overline{N}_R(r, f)$.

Now, let us estimate $Z_R(r, \overline{h})$. Since $\log(|h'|(r)) \le \log(|h|(r)) - \log r$, we have $Z_R(r, h') \le Z_R(r, h) + O(\log(r))$. But since $h' = \overline{hh}$, we have $Z_R(r, \overline{h}) = Z_R(r, h') - Z_R(r, \overline{h}) \le Z_R(r, h) - Z_R(r, \overline{h}) + O(\log(r)) = Z_R(r, \overline{h}) + O(\log(r)) = \overline{N}_R(r, f) + O(\log(r))$. Consequently,

$$ \begin{aligned} Z_R(r, g\overline{h}) &\le Z_R(r, g) + \overline{N}_R(r, f) + O(\log(r)) \\ &= Z_R(r, f) + \overline{N}_R(r, f) + O(\log(r)). \end{aligned} $$

Thus, by (1), we have proven the claim when $k = 1$ and then it is immediately derived by induction on $k$. $\square$

Similar to Lemmas 43.17 and 46.9, now we have Lemma 46.10 and Corollary 46.11:

**Lemma 46.10.** Let $f \in \mathcal{M}(D)$ and let $G = \frac{f'}{f}$. Then, $G$ satisfies $Z_R(r, G) \le N_R(r, G) - \log r + O(1)$ $(r \ge R)$. $\square$
**Corollary 46.11.** Let $f \in \mathcal{M}(D)$ and let $G = \frac{f'}{f}$. Then, $T_R(r, G) - N_R(r, G) = 0 + O(1)$.

Lemma 46.12 is necessary in the proof of Theorem 46.13.

**Lemma 46.12.** Let $f \in \mathcal{M}(D)$. Suppose that there exists $\xi \in \mathbb{K}$ and a sequence of intervals $J_n = [u_n, v_n]$ such that $u_n < v_n < u_{n+1}$, $\lim_{n \to +\infty} u_n = +\infty$, and $\lim_{n \to +\infty} \left[ \inf_{r \in J_n} \frac{T_R(r, f) - Z_R(r, f - \xi)}{\log(r)} \right] = +\infty$.

Let $\tau \in \mathbb{K}$, $\tau \neq \xi$. Then, $Z_R(r, f - \tau) = T_R(r, f) + O(\log(r))$ $\forall r \in J_n$ when $n$ is big enough.

**Proof.** Without loss of generality, we can obviously suppose that $\xi = 0$. By Theorem 39.6, $f$ is of the form $f^S f^0$ and $f^0$ is of the form $\frac{g}{h}$ with $g, h \in \mathcal{A}(D)$, having no zero in $S$. Set $w = f^S$. Thus, we have

$$ \lim_{n \to +\infty} \left[ \inf_{r \in J_n} \frac{Z_R(r, h) - Z_R(r, g)}{\log(r)} \right] = +\infty. $$

Consequently, by Theorem 46.1,

$$ (1) \quad \lim_{n \to +\infty} \left[ \inf_{r \in J_n} \frac{\log(|h|(r) - \log(|g|(r)))}{\log(r)} \right] = +\infty. $$

Consider now $f - \tau$. We have $f - \tau = \frac{wg - \tau h}{h}$, hence

$$ \log(|f|(r)) = \log(|wg - \tau h|(r) - \log(|h|(r)). $$

But by (1), we have $\log(|\tau h|(r)) > \log(|wg|(r))$ because $\log(|w|(r) = O(\log(r))$, therefore $\log(|wg - \tau h|(r)) = \log(|\tau h|(r))$ $\forall r \in J_n$ when $n$ is big enough, and hence

$$ (2) \quad \lim_{n \to +\infty} \left[ \sup_{r \in J_n} \frac{\log(|\tau h - wg|(r) - \log(|h|(r))}{\log(r)} \right] = 0. $$

Consequently, by (2) and by Theorem 46.1,

$$ \lim_{n \to +\infty} \left[ \sup_{r \in J_n} \frac{Z_R(r, \tau h - wg) - Z_R(r, h)}{\log(r)} \right] = 0 $$

i.e.

$$ \lim_{n \to +\infty} \left[ \sup_{r \in J_n} \frac{Z_R(r, f - \tau) - T_R(r, f)}{\log(r)} \right] = 0, $$

which proves the claim.

□